Rotary Reactor Engineering [1st ed] 9780444530264, 0444530266

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Rotary Reactor Engineering

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Rotary Reactor Engineering Daizo Kunii and Tatsu Chisaki 3-8-8 Minami Ikebukuro, Toshima-ku, Tokyo, Japan 171-0022

AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD PARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO

Elsevier Radarweg 29, PO Box 211, 1000 AE Amsterdam, The Netherlands Linacre House, Jordan Hill, Oxford OX2 8DP, UK

First edition 2008 Copyright © 2008 Elsevier B.V. All rights reserved No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (+44) (0) 1865 843830; fax (+44) (0) 1865 853333; email: [email protected]. Alternatively you can submit your request online by visiting the Elsevier web site at http://elsevier.com/locate/permissions, and selecting Obtaining permission to use Elsevier material Notice No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. Because of rapid advances in the medical sciences, in particular, independent verification of diagnoses and drug dosages should be made Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN: 978-0-444-53026-4

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Printed in Hungary 07 08 09 10 11

10 9 8 7 6 5 4 3 2 1

Preface

The rotary reactor is a processing unit which rotates on its axis and in which a chemical transformation takes place between gas and solids flowing through the unit, usually at high temperature. These units have been used in the past and are used today in a variety of basic industries and are often called rotary kilns. They are inclined slightly and solids pass downward through the unit by rotation of the unit and also by the influence of gravity, and sometimes with the help of baffles. Their operations are fairly well established; however, today’s social and economic circumstances and environmental pressures require products of higher quality and at lower cost. To satisfy these needs we look to a different design of unit, and this leads us to a different type of rotary reactor, which lies horizontal and rotates, as desired, by suitable partition plates attached with guide plates. These units have a great flexibility in that one part of the unit can have one type of flow, while the other section can have a different type of flow. In addition, streams of solids can be made to flow counter-currently or co-currently in different parts of the reactor, as desired. All of this can be achieved with the suitable arrangement of partition plates within the unit. On the basis of physical and chemical fundamentals, design equations are derived theoretically to predict the performance characteristics of these reactors. These equations can be extended to all types of rotary reactors and kilns. With these extra degrees of freedom we have found that the rotary reactor is able to create new contacting patterns between gas and solids, which give it a definite advantage in certain situations over the other gas/solid contacting patterns being used today. We have developed a variety of practical experiences with this technology which has led to the development and commercialization of a variety of new processes. In this book we plan to systematize these experiences quantitatively, so that the reader can easily understand their behavior. The first part of this book presents the fundamentals: solid movement, conversion of solids, and heat transfer. The middle part of the book applies these equations to a variety of processes which have been developed so far and shows how they are used. In the last part, conceptual designs of novel rotary reactors are proposed, the performance characteristics of which are predicted on the basis of the above equations, particularly in the gasification of solid wastes. Some of this book comes originally from our Japanese book, entitled “Development Engineering of Rotary Kilns”, which was distributed to a limited circle of engineers in the lime industry. However, the present book is not merely a translation of the original but is a new book in which significant modifications and additions to the original have been made to meet a broader coverage of subjects and readers.

vi

Preface

We would like to express our appreciation to Dr. Octave Levenspiel and Dr. Edmund Immergut for their valuable advice and help towards realizing the publication of this book, and our thanks to Mrs. Ruriko Nobe, Mrs. Kumi Komatsu and Miss Ikuko Kasai for their typing and drawing efforts. Daizo Kunii Tatsu Chisaki 2007

Notation Symbols and constants used locally are not included here. Equations are indicated for reference to the location where the symbol appears first. A B Bf b C CA CAb CAe CAi CA∗ Ca CC Cg Cs Cs1 , Cs2 DA DAe df dp dt dti dt0 dts FB Ff Fp Fs Fs∗ Ga GC Gg HB Hc Hc∗ HG

gaseous reactant; Eq. (1.5.1) solid reactant; Eq. (1.5.1) amount of fuel per unit mass of solids; Eq. (7.2.2) constant; Eq. (1.5.1) fraction of carbon; Eq. (7.1.4) concentration of A ; Eq. (3.2.1) CA in bulk flow; Eq. (5.1.2) CA in exit gas flow; Eq. (5.2.5) CA in inlet gas flow; Eq. (5.2.3) CA in equilibrium with B; Eq. (3.2.1) specific heat of air; Eq. (7.2.2) specific heat of CO2 ; Eq. (7.2.2) specific heat of gas; Eq. (6.2.3) specific heat of solids; Eq. (6.3.6) that of feed stock; Eqs. (7.2.2) (7.2.3) diffusivity of gaseous reactant; Eq. (3.2.16) effective diffusivity; Eq. (3.2.14) diameter of flame; Eq. (6.3.1) diameter of solids; Eq. (3.2.14) diameter of cylinder; Eq. (2.2.7) inner diameter; Eq. (6.3.1) outer diameter; Eq. (6.3.11) diameter of screw; Eq. (2.3.1) flow rate of solid reactant B; Eq. (5.2.8) feed rate of fuel; Eq. (6.2.2) rate of solid product; Eq. (6.3.16) flow rate of solids; Eq. (2.2.6) circulating rate of solids; Eq. (2.3.1) air for combustion of fuel; Eq. (7.2.2) CO2 issued from feed stock; Eq. (7.2.2) volume of wet combustion gas; Eq. (6.2.3) heat of reaction; Eq. (3.4.2) necessary heat for endo-thermic reaction; Eq. (7.1.7) evolved heat from exo-thermic reaction; Eq. (9.1.1) gross calorific value of dry gas; Eq. (7.1.2)

[–] [–] [kg, Nm3 /kg]  gm-mol B  gm-mol A [–] [gm-molA/cm3 ] [gm-mol A/cm3 ] [gm-mol A/cm3 ] [gm-mol A/cm3 ] [gm-mol A/cm3 ] [kcal/Nm3 ◦ C] [kcal/Nm3 ◦ C] [kcal/Nm3 ◦ C] [kcal/kg ◦ C] [kcal/kg ◦ C] [cm2 /sec] [cm2 /se] [cm, m] [cm, m] [cm, m] [cm, m] [cm, m] [cm, m] [gm-mol B/sec] [kg, Nm3 /hr] [kg/hr] [kg/hr] [kg/hr] [Nm3 /kg, Nm3 ] [Nm3 /kg] [Nm3 /kg, Nm3 ] [kcal/gm-mol B] [kcal/kg] [kcal/kg] [kcal/Nm3 ]

viii

Notation

Kd Kp Kr k¯ kc

necessary heat to supply; Eq. (7.1.8) convectional heat transfer coefficient; Eq. (6.3.11) heat transfer coefficient, inside of retort; Eq. (8.3.5) heat transfer coefficient, outside of retort; Eq. (8.4.4) convectional heat transfer coefficient of single solid; Eq. (3.4.2) heat transfer coefficient of wall surface, contacting with rotating solids; Eq. (6.3.6) radiant heat transfer coefficient; Eq. (3.4.2) radiant heat transfer coefficient from hot gas, on the basis of inner surface area of reactor; Eq. (6.2.3) radiant heat transfer coefficient from inner surface of retort to rotating solids layer, based on surface area of inner wall, to which rotating solids contact; Eq. (6.3.5) radiant heat transfer coefficient, from inner surface of insulation cover to outer surface of retort; Eq. (8.3.1) mass transfer coefficient; Eq. (3.2.16) equilibrium constant; Eq. (3.5.4) rate constant for a first order reaction; Eq. (3.3.4) overall rate constant; Eq. (3.2.14) rate constant of chemical reaction; Eq. (3.2.4)

kr

rate constant of chemical reaction; Eq. (3.2.2)

kr∗ ke

rate constant; Eq. (4.1.1) effective thermal conductivity of solids packet; Eq. (6.3.6) thermal conductivity of gas thermal conductivity of sticked solids layer, and retort; Eq. (8.2.1) thermal conductivity of solids; Eq. (4.2.2) heat loss per solids; Eq. (7.2.2) width of solids layer; Eq. (5.2.6) height of solids layer; Eq. (5.1.2) pitch length of screw; Eq. (2.3.1) length of reactor; Eq. (2.2.11) thickness of refractory; Eq. (8.3.2) rate of rotation; Eq. (2.2.2) excess air ratio; Eq. (7.2.2) partial pressure of gaseous component; Eq. (3.5.4) rate of heat transfer from electric heater; Eq. (9.1.4) gross calorific value of feed stock; Eq. (7.1.2) rate of heat transfer to rotary retort; Eq. (9.2.3) rate of heat transfer between two flows of solids; Eqs. (9.1.2) (9.1.5)

Hn hc hi h0 hp (hp )HC hr hrg (hrs )HC

(hrs )SH

kg kHi , kH0 ks L le lp lps lr lW N n pgas Q∗ QG QH QH1 , QH2

[kcal/kg] [kcal/m2 hr ◦ C] [kcal/m2 hr ◦ C] [kcal/m2 hr ◦ C] [kcal/m2 hr ◦ C] [kcal/m2 hr ◦ C] [kcal/m2 hr ◦ C] [kcal/m2 hr ◦ C] [kcal/m2 hr ◦ C]

[kcal/m2 hr ◦ C] [cm/sec] [–] [1/sec] [cm/sec] [cm/sec]  1

 gm-mol A  cm3

(sec)



[1/sec, hr] [kcal/m hr ◦ C]

[kcal/m hr ◦ C] [kcal/m hr ◦ C] [kcal/m hr ◦ C] [kcal/kg] [cm, m] [cm, m] [cm, m] [cm, m] [cm, m] [1/hr] [–] [kPa, atm] [kcal/kg] [kcal/kg] [kcal/kg] [kcal/kg]

Notation

Qlr QN QR qH qW R Re Rv r rc Ta Tb TC Tce Tf Tg TH Ti T0 TS Ts TW T∗ ∆Tav t tr U Ua ug u0 Va Vg Vw VO2 Vr W XB x ∆x YB = 1 − XB y

ix

rate of heat loss; Eq. (6.3.15) net calorific value of fuel; Eq. (6.2.2) rate of heat transfer; Eq. (6.3.13) heat flux; Eq. (8.4.4) flux of heat loss; Eq. (6.3.11) radius; Eqs. (2.2.1) Reynolds number; Eq. (3.2.16) ratio of surface area; Eq. (6.4.3) radius; Eqs. (2.2.2) radius of reacting interface; Eq. (3.2.4) temperature of air; Eq. (7.2.4) temperature of surroundings; Eq. (3.4.2) temperature of heated surface; Eq. (6.3.4) temperature of discharged solids; Eq. (7.2.2) temperature of flame; Eq. (6.3.1) temperature of gas; Eq. (6.2.3) temperature of retort; Eq. (8.3.2) temperature of interface; Eq. (4.2.3) ambient temperature; Eq. (7.2.1) temperature of surface S; Eq. (8.3.1) temperature of solid; Eq. (3.4.2) temperature of outer surface; Eq. (6.3.12) average temperature of wall and solids surfaces; Eq. (6.3.14) average temperature difference; Eq. (6.4.1) time; Eq. (3.2.3) nominal residence time of solids, Eq. (1.3.1) overall heat transfer coefficient; Eq. (8.4.6) volumetric heat transfer coefficient; Eq. (6.4.1) velocity of gas; Eq. (3.2.16) superficial velocity of gas; Eq. (5.2.7) volume of air; Eq. (7.2.4) volume of dry gas; Eq. (7.1.5) volume of water vapor volume of necessary oxygen for partial oxidation; Eq. (7.1.6) volume of reactor; Eq. (6.4.1) electric input; Eq. (8.3.2) conversion of solid reactant; Eq. (3.2.2) distance; Eq. (5.1.1) length of element; Eq. (2.2.2) Eq. (4.1.1) distance; Eq. (5.2.3)

[kcal/hr] [kcal/kg] [kcal/hr] [kcal/m2 hr] [kcal/m2 hr] [cm, m] [–] [–] [cm, m] [cm, m] [◦ C] [◦ C] [◦ C] [◦ C] [◦ C] [◦ C] [◦ C] [◦ C] [◦ C] [◦ C] [◦ C] [◦ C] [◦ C] [◦ C] [sec, hr] [sec, hr] [kcal/m2 hr ◦ C] [kcal/m3 hr ◦ C] [cm, m/sec] [cm, m/sec] [Nm3 /kg] [Nm3 /kg] [Nm3 /kg] [Nm3 /kg] [cm3 , m3 ] [kW] [–] [cm, m] [cm, m] [–] [cm, m]

x

Notation

Greek Symbols α∗ β β∗ γ γ∗ γs δ δHi δH0 δ∗ ω ω0 ω∗ εH , εC εf εg εi εm εv ε∗ η1 η2 ηc ηs ηG ηr ϑr κf χ µ ρB ρ¯ ρg ρs τ ϕ ζ

ratio of gasified carbon; Eq. (7.1.4) (product)/(feedstock); Eq. (7.2.6) ratio of preheat; Eq. (7.1.4) volumetric fraction of bulk solids; Eq. (2.2.1) ratio of heat in exit flow; Eq. (7.1.4) volumetric fraction of bulk solids in screw cylinder; Eq. (2.3.1) width of gas passage thickness of sticked solids layer; Eq. (8.2.1) thickness of retort; Eq. (8.2.1) ratio of heat loss; Eq. (7.1.4) angle of surface; angle of rotation axis; Eq. (2.2.5) constant; Eq. (2.2.5) ratio of heat added from outside; Eq. (7.1.4) emissivity of surface H and C; Eq. (6.3.3) emissivity of flame; Eq. (6.3.1) emissivity of gas; Eq. (6.3.2) intrinsic void ratio; Eq. (3.4.5) average emissivity of solid surfaces; Eq. (6.3.1) void fraction of bulk solids; Eq. (5.1.1) ratio of tar fraction; Eq. (7.1.4) transport efficiency; Eq. (2.3.3) transport efficiency; Eq. (2.3.6) carbon efficiency; Eq. (7.1.4) transport efficiency; Eq. (2.3.1) gasification efficiency; Eq. (7.1.2) reactor efficiency; Eq. (5.2.1) repose angle; Eq. (2.2.5) rate constant of flame; Eq. (6.2.3) fraction of wall surface, which contacts rotating solids; Eq. (6.3.3) viscosity of gas; Eq. (3.2.16) molar density of solid; Eq. (3.3.1) bulk density of solids; Eq. (2.2.9) density of gas; Eq. (3.2.16) density of solid time for complete conversion of solids; Eq. (3.2.7) inclination angle of guide plates; Eq. (2.3.3) (calorific value of tar)/(value of feed stock); Eq. (7.1.4)

[–] [–] [–] [–] [–] [–] [cm, m] [cm, m] [cm, m] [–] [deg.] [deg.] [–] [–] [–] [–] [–] [–] [–] [–] [–] [–] [–] [–] [–] [–] [deg.] [1/m] [–] [gm/cm·sec] [gm-mol B/cm3 ] [gm/cm3 , kg/m3 ] [gm/cm3 , kg/m3 ] [gm/cm3 , kg/m3 ] [sec, hr] [deg.] [–]

Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

v

Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

vii

1

2

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Contacting methods between gas and solids . . . . . . . . . . . . 1.2 Contact operation between gas and solids . . . . . . . . . . . . . 1.3 Residence time characteristics of solids . . . . . . . . . . . . . . 1.3.1 Plug flow (rod-like flow) . . . . . . . . . . . . . . . . . . 1.3.2 Complete back-mix flow of solids . . . . . . . . . . . . . 1.3.3 Improvement of residence time characteristics in a rotary reactor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Enhancement of gas-solid contacting in rotary reactors . . . . . . 1.5 Examples of industrial application . . . . . . . . . . . . . . . . . 1.6 Cooperation with mechanical engineers . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1 1 2 4 4 4

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. . . . .

5 5 9 9 10

Movement of Solids in Rotary Cylinder . . . . . . . . . . . . . . . . . . . . 2.1 Experimental studies on solids flow in a horizontally rotating cylinder 2.1.1 Movement of solids within a sectional area, perpendicular to the rotation axis . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Both sides of cylinder closed . . . . . . . . . . . . . . . . . . 2.1.3 Circular weir at one side of cylinder . . . . . . . . . . . . . . . 2.1.4 Circular weir to replace solids quickly . . . . . . . . . . . . . 2.2 Theoretical studies on movement of solids . . . . . . . . . . . . . . . 2.2.1 Simple model . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Transportation rate of solids . . . . . . . . . . . . . . . . . . . 2.2.3 Equations to predict the performance of a rotating cylinder . . 2.2.4 Discussions on obtained equations . . . . . . . . . . . . . . . 2.3 Improvement of residence time characteristics for rotating solids . . . 2.3.1 Application of screw cylinders . . . . . . . . . . . . . . . . . . 2.3.2 Research and development of U-Turn system . . . . . . . . . 2.3.3 Theoretical equations on transfer rate of solids . . . . . . . . . 2.3.4 Transfer rates of solids in annular space . . . . . . . . . . . . 2.3.5 Comparison of partition plates with screw cylinders . . . . . . Example 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11 11 11 12 13 14 14 14 15 16 17 18 18 19 20 21 22 23 23 24 24 25

xii

3

Contents

Conversion of Solids with Gaseous Reactant . . . . . . . . . . . . . 3.1 Reaction rate of solid conversion . . . . . . . . . . . . . . . . . 3.2 Kinetic models of gas–solid reactions . . . . . . . . . . . . . . 3.3 Relation between rate constants of chemical reaction, based on different models . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Application of kinetic models to oxidation of carbon . . . . . . 3.4.1 Graphite . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Petroleum coke . . . . . . . . . . . . . . . . . . . . . . 3.4.3 Char from coal . . . . . . . . . . . . . . . . . . . . . . 3.4.4 Stable temperature of an isolated carbon particle . . . . 3.4.5 Gaseous reactant around particle . . . . . . . . . . . . 3.4.6 Carbon dispersed in inorganic solids . . . . . . . . . . 3.5 Gasification of carbon . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Boudouard’s reaction . . . . . . . . . . . . . . . . . . . 3.5.2 Gasification of carbon by steam . . . . . . . . . . . . . 3.6 Activation of carbonaceous pellet . . . . . . . . . . . . . . . . 3.7 Roasting of zinc sulfide . . . . . . . . . . . . . . . . . . . . . . 3.8 Reduction of iron ore . . . . . . . . . . . . . . . . . . . . . . . Example 3.1 . . . . . . . . . . . . . . . . . . . . . . . . Example 3.2 . . . . . . . . . . . . . . . . . . . . . . . . Example 3.3 . . . . . . . . . . . . . . . . . . . . . . . . Example 3.4 . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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27 27 30

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32 33 33 33 34 35 35 36 37 37 38 40 41 41 42 43 44 44 46

4

Thermal Decomposition and Conversion of Composite Pellets . . 4.1 Elimination of trace species in solids . . . . . . . . . . . . . . 4.2 Calcination of limestone . . . . . . . . . . . . . . . . . . . . 4.3 Decomposition of manganese sulfate . . . . . . . . . . . . . 4.4 Thermal cracking of organic solids . . . . . . . . . . . . . . . 4.5 Composite made of iron ore and oil . . . . . . . . . . . . . . 4.6 Reduction of composite pellet, ferro-chromium ore and coke Example 4.1 . . . . . . . . . . . . . . . . . . . . . . . Example 4.2 . . . . . . . . . . . . . . . . . . . . . . . Example 4.3 . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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47 47 47 49 50 50 52 52 53 53 55

5

Conversion of Solids in Rotary Reactors . . . . . . . . . . . . . . . . . . . 5.1 Conversion of gas and solids within solids layer . . . . . . . . . . . 5.1.1 Simplified model . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Effect of layer thickness on time necessary for conversion of solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Enhancement of contact by sending gaseous reactant into a rotating layer of solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Simplified model . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Conversions of solids, calculated from rate constant Kr for gaseous reactant . . . . . . . . . . . . . . . . . . . . . . . . 5.2.3 Conversion of solids, calculated from rate constant kr . . . .

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57 57 57

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59

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61 61

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62 63

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Contents

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64 64 64 64 64 65 65 66 67 68

Heat Transfer in a Rotary Reactor, Direct Heating . . . . . . . . . . . . 6.1 Combustion of fuels . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Combustion model of a gas burner . . . . . . . . . . . . . . 6.1.2 Liquid fuel . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.3 Pulverized coal and coke . . . . . . . . . . . . . . . . . . . . 6.1.4 Inside combustion and reverse flame . . . . . . . . . . . . . 6.1.5 Volume of combustion region . . . . . . . . . . . . . . . . . 6.2 Temperature profile in turbulent flame . . . . . . . . . . . . . . . . . 6.3 Heat transfer in a rotary reactor at high temperature . . . . . . . . . 6.3.1 Radiant heat transfer from flame and combustion gas . . . . 6.3.2 Radiant heat transfer from inner wall surface to surface of rotating solids layer . . . . . . . . . . . . . . . . . . . . . . . 6.3.3 Heat transfer coefficient by direct contacting of solids from the hot wall surface . . . . . . . . . . . . . . . . . . . . . . . 6.3.4 Temperature of the inner wall surface . . . . . . . . . . . . . 6.3.5 Heating capacity of a rotary reactor . . . . . . . . . . . . . . 6.4 Enhancement of heat transfer . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Lifters in a rotary dryer . . . . . . . . . . . . . . . . . . . . . 6.4.2 Discussions on volumetric heat transfer coefficient . . . . . 6.4.3 Partition plates . . . . . . . . . . . . . . . . . . . . . . . . . Example 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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69 69 69 71 72 72 74 74 76 76

.

78

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80 81 82 84 84 85 86 86 87 87 88 89 90

Performance of Rotary Reactors, Direct Heating . . . . . . . . . . 7.1 Prediction of performance . . . . . . . . . . . . . . . . . . . 7.1.1 Mass and enthalpy balances . . . . . . . . . . . . . . 7.1.2 Enthalpy balance, complete combustion . . . . . . . 7.1.3 Enthalpy balance, partial combustion and gasification 7.1.4 Special cases . . . . . . . . . . . . . . . . . . . . . . 7.1.5 Applicability of equations . . . . . . . . . . . . . . . 7.2 Calcination of limestone . . . . . . . . . . . . . . . . . . . . 7.2.1 Procedure for design calculation . . . . . . . . . . . . 7.2.2 Estimation of heat loss . . . . . . . . . . . . . . . . .

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93 93 93 93 94 96 97 97 97 98

5.3

6

7

xiii

5.2.4 Different devices . . . . . . . . . . . . . . . . . . . . . . . 5.2.5 Rotary sealing of distribution manifold . . . . . . . . . . . High temperature stability of isolated solids in exothermic reaction 5.3.1 Volumetric fraction of falling solids . . . . . . . . . . . . . 5.3.2 High temperature stability of falling solids . . . . . . . . . 5.3.3 High temperature near nozzles of injection gas . . . . . . . Example 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . Example 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . Example 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

xiv

Contents

7.3

7.4

7.5

7.2.3 Prediction of overall performance . . . . . . . . . . . . . . . 7.2.4 Prediction of solids conversion and gas temperature . . . . . Pre-reduction of composite pellets, made of ferro-chromium ore and coke . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 Conversion of solids . . . . . . . . . . . . . . . . . . . . . . 7.3.2 Rotary kiln . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.3 Direction of improvement . . . . . . . . . . . . . . . . . . . Activation of char . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Model of a rotary reactor . . . . . . . . . . . . . . . . . . . . 7.4.2 Application of equations . . . . . . . . . . . . . . . . . . . . 7.4.3 Direction of improvement . . . . . . . . . . . . . . . . . . . Gasification of combustible feed stock . . . . . . . . . . . . . . . . . Example 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 7.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. .

99 102

. . . . . . . . . . . . . . .

105 105 106 108 109 109 109 111 111 113 113 115 119 121 125

8

Heat Transfer in Rotary Reactors, Indirect Heating . . . . . . . . . . 8.1 Necessary information for satisfactory design . . . . . . . . . . . 8.1.1 Material of the retort . . . . . . . . . . . . . . . . . . . . 8.1.2 Thickness of the rotary retort . . . . . . . . . . . . . . . 8.1.3 Emissivity of the retort surface . . . . . . . . . . . . . . 8.1.4 Sticking of solids and formation of thick layer . . . . . . 8.2 Heat transfer within the rotary retort . . . . . . . . . . . . . . . . 8.3 Heat transfer from an electric heater . . . . . . . . . . . . . . . . 8.4 Heat transfer from gas flow . . . . . . . . . . . . . . . . . . . . . 8.4.1 Examples of practical design . . . . . . . . . . . . . . . 8.4.2 Proposed design for gas flow . . . . . . . . . . . . . . . 8.4.3 Simplified model . . . . . . . . . . . . . . . . . . . . . . 8.4.4 Overall heat transfer coefficient and temperature of retort Example 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . Example 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . Example 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . Example 8.4 . . . . . . . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

127 127 127 127 128 128 128 129 132 132 132 132 136 136 138 139 140 142

9

Performance of Rotary Reactors, Indirect Heating . . . . . . . . . . . . 9.1 Electric heating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.1 Enthalpy balance . . . . . . . . . . . . . . . . . . . . . . . . 9.1.2 Improvement of electric heating . . . . . . . . . . . . . . . . 9.1.3 Working equations for design calculation of the new heating system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.4 Prediction of performance . . . . . . . . . . . . . . . . . . . 9.1.5 Direction of improvement . . . . . . . . . . . . . . . . . . .

. . . .

143 143 143 144

. . .

146 147 147

. . . . . . . . . . . . . . . . . .

Contents

9.2

xv

. . . . . . . .

. . . . . . . .

148 148 150 152 153 155 157 159

Application of a Rotary Reactor for the Re-utilization of Solid Wastes . 10.1 Material and energy recovery from solid wastes . . . . . . . . . . . . 10.2 Proposed rotary reactors for re-utilization of secondary resources . . 10.2.1 De-lacquering of spent cans . . . . . . . . . . . . . . . . . . 10.2.2 Activation of char . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Gasification of solid wastes . . . . . . . . . . . . . . . . . . . . . . . 10.3.1 Matrix presentation of gasification processes . . . . . . . . . 10.3.2 Gasification processes of MSW developed by the authors . . 10.3.3 Proposal for a novel rotary reactor to produce rich gas from MSW . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Gasification of sewage sludge . . . . . . . . . . . . . . . . . . . . . 10.4.1 Conventional incineration . . . . . . . . . . . . . . . . . . . 10.4.2 Proposal for a rotary reactor to gasify sewage sludge . . . . 10.5 Possibility for application to gasification of low grade coal . . . . . Example 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . Example 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . Example 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . Example 10.4 . . . . . . . . . . . . . . . . . . . . . . . . . . Example 10.5 . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . .

163 163 164 164 165 167 167 167

. . . . . . . . . . .

169 173 173 174 175 176 179 181 186 188 195

Brief Careers of the Authors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

197

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

199

10

Heating by combustion gas . . . . . . . . . . . . . . . . . . . . . 9.2.1 Oxidation of residual carbon in spent catalyst . . . . . . 9.2.2 Direction of improvement . . . . . . . . . . . . . . . . . 9.2.3 Application to thermal cracking of solid waste materials Example 9.1 . . . . . . . . . . . . . . . . . . . . . . . . . Example 9.2 . . . . . . . . . . . . . . . . . . . . . . . . . Example 9.3 . . . . . . . . . . . . . . . . . . . . . . . . . Example 9.4 . . . . . . . . . . . . . . . . . . . . . . . . .

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–1– Introduction In a rotary reactor, transformation of solids takes place within a rotating layer of solids. In this chapter, the advantages and disadvantages of rotary reactors are discussed in comparison with other contacting methods, for example, moving bed, fluidized bed and entrained flow of solids. Its disadvantages can be remedied by injecting gaseous reactant directly into the rotating layer of solids. Furthermore, using a suitable arrangement of a partition with guide plates or screw plates, different types of solid flow can be achieved in a single rotary reactor. This possibility is another advantage of this contacting method.

1.1 CONTACTING METHODS BETWEEN GAS AND SOLIDS Fundamental methods of contacting solids with flowing gas are illustrated in Fig. 1.1. For physical and chemical transformation of solids at elevated temperature at steady state, Table 1.1 summarizes the main characteristics of each contacting method. In moving and fluidized beds, gas flows through the beds of solids, whereas in entrained solids flow the solids are dispersed in the gas stream. Therefore, these contacting methods have been applied for many industrial processes, in which chemical reactions take place between gas and solids. In a rotary reactor on the other hand, gas flows along the surface of the solids layer. Gaseous reactant should diffuse across the boundary layer on the surface of the rotating solids, and then through the void space in the layer. This diffusional process drastically reduces the overall rate of reaction, sometimes one order of magnitude less than the other contacting methods mentioned above. Taking into account the above diffusional resistance, the conversion of solids in a rotary reactor will be discussed quantitatively in Chapter 5. Such diffusional resistance as above is a definite disadvantage of the rotary reactor, in the case where the process involves chemical reaction between gas and solids. This is the reason why the rotary reactor has been mainly applied for special cases of gas-solid process systems, in which heat transfer and/or thermal decomposition of solids are the rate controlling step. To overcome the above disadvantage, a variety of devices has been applied for better contact between gas and solids. For example, steam is injected into the rotating layer of char to produce active carbon commercially. Suppose it is necessary to develop a novel gas-solid system and none of the conventional contacting methods could satisfy its special requirements. Let us try to test the rotary reactor, emphasizing its advantages and remedying its disadvantages. This may be a chance for

2

1. Introduction

Figure 1.1

Fundamental contacting methods between gas and solids. flat hearth

us to create an innovative gas-solid process system. Thus, we need sufficient understanding of the exact phenomena in a rotating reactor, on the basis of physical and chemical fundamentals.

1.2

CONTACT OPERATION BETWEEN GAS AND SOLIDS

At steady state operation, the contact between gas and solids is simplified as shown in Fig. 1.2. In Fig. 1.1, the flow pattern in the vertical moving bed is counter-current, i.e. Fig. 1.2 (a). In the horizontal and/or inclined moving bed, it is (c), i.e. cross flow. In the entrained solids flow, it is always co-current flow, Fig. 1.2 (b). A fluidized bed is a typical case of complete back mix flow, i.e. Fig. 1.2 (d). Both counter current (a) and co-current (b) are feasible in rotary cylinder and flat hearth.

Comparison of contacting methods in gas-solids reactor process Size of solids (mm)

Industrial application

Advantage

Disadvantage

Vertical moving bed (Shaft Kiln)

10 ∼ 80

Reduction of iron ore. Gasification of coal/coke. Gasification of MSW. Calcination of limestone.

Counter current flow. High conversions. High thermal efficiency. Large capacity.

Coarse and hard solids are needed. High pressure drops for small size.

Horizontal and/or inclined moving bed (grate)

5 ∼ 50

Combustion of coal/MSW. Quenching of cement clinker. Pre-heating of limestone.

Size distribution allowed. Moderate pressure drop.

Cross flow for gas and solids. Low thermal efficiency. Frequent maintenance is necessary for metal ware.

Fluidized bed

0.05 ∼ 10

Combustion/gasification of Coal/MSW. Roasting of metal sulfide or sulfate. Production of TiCl4 , SiHCl3 , active carbon.

Steady state processing of solid particles. Easy automatic control. Uniform temperature. Large capacity possible.

Complete back mix flow of solids. Exponential residence time for solids. Large bed volume needed for high solid conversion. High pressure drop. High carry over dust.

Entrained flow of solids

0.01 ∼ 0.2

Combustion and gasification of coal. Roasting of sulfide ore.

Pulverized solids react with gas shortly. Large capacity.

Only fine solids. Only co-current flow for gas and solids. Low thermal efficiency.

Rotary cylinder

0.01 ∼ 30

Clinkering of cement. Calcination of limestone. Reduction of ore. Regeneration of spent catalyst. Activation of char.

Wide range of solids possible. Solids plug flow. Suitable for solids which may sinter or agglomerate. Large capacity.

Temperature not uniform. Slow rate of reaction between gas and solids. Low thermal efficiency. High construction cost.

Multi-stage rotary hearth

0.1 ∼ 10

Activation of char. Incineration of sewage sludge. Regeneration of adsorbent in sugar plant.

Suitable for solids liable to sinter or melt.

Temperature gradients are severe and difficult to control. 3

Contacting method

1.2. Contact operation between gas and solids

Table 1.1.

4

1. Introduction

Figure 1.2 Steady state flow in gas-solid system.

1.3 1.3.1

RESIDENCE TIME CHARACTERISTICS OF SOLIDS

Plug flow (rod-like flow)

Suppose solids are fed into the vertical moving bed in Fig. 1.1 at steady state. Solids move down slowly, and exit at its bottom. The residence time tr for all solids is approximately the same, and this flow pattern is called a plug flow or rod-like flow. tr =

Volume of solids layer [m3 ] Volumetric rate of bulk solids [m3 /hr]

(1.3.1)

In an ordinary rotary reactor, the flow pattern of solids at steady state can be approximately assumed to be a plug blow (rod-like blow). The following are the results of a test by stimulus response in a commercial rotary reactor, the aspect ratio of which is 10 m/0.9 m = 11.1. Slightly coarser solids were used as tracer, and the weight fraction of the tracer in the discharge solids was measured. Time after impulse Weight fraction of tracer

[min] [%]

60 0.3

68 3.2

71 15.2

74 2.1

77 0.5

80 0.3

We can determine the average residence time of solids to be 71 ± 3 min in this case. Where the aspect ratio of a rotary reactor is smaller than 10, for instance 4, the residence time distribution of tracer solids deviates considerably from plug flow. 1.3.2

Complete back-mix flow of solids

In a fluidized bed into which solids are fed at steady state, as seen in Fig. 1.2 (d), the residence time characteristics are quite different from plug flow. Let us mark the solids which are fed into the bed at an instance. They mix into the bed instantaneously, reside there for a while then exit from the bed. At the nominal value of residence time tr calculated with Eq. (1.3.1), nearly 63% of the fed solids have been already discharged, while 37% still remain in the bed. A very long time is needed to discharge all the marked solids. The above characteristics indicate that the residence time of fed solids distributes from zero to infinity. Therefore, the fluidized bed is not suitable for a gas-solid reaction system,

1.4. Enhancement of gas-solid contacting in rotary reactors

5

Figure 1.3 Improvement of the flow pattern of solids in a rotary reactor.

in which the residence time of all solids should be nearly the same. In spite of the above disadvantages, the fluidized bed has been applied in a variety of gas-solid reaction systems because of its specific advantages (see Kunii and Levenspiel [1]). 1.3.3

Improvement of residence time characteristics in a rotary reactor

When developing a novel gas-solid reaction process, it sometimes happens that neither plug flow nor complete back-mix flow of solids can be satisfactorily applied. For instance, suppose thermal decomposition of MnSO4 to MnO2 at 1000 ◦ C. Solids are liquefied at first, and then solidified after releasing SO2 . The residence time of the final solid should be nearly uniform. Because sintering of solids and clogging of the path are serious, this kind of reaction cannot be processed in either moving beds and/or fluidized beds. The only possibility is the rotary reactor. Suppose the inner space of a rotary reactor is separated into two parts, for instance by a partition plate; solids can circulate within the reactor as seen in Fig. 1.3 (a), or just return back as (b). This kind of solid flow cannot be realized in other contacting modes, as presented in Fig. 1.1. The flow pattern of solids in Fig. 1.3 (a) can solve the preceding problem. Feed solids are mixed into the “dry” solids which re-circulate from the hot region in the reactor. They are liquefied within packets of “dry” solids, and then transformed to coarse “dry” grain, which can circulate easily. The flow of solids in Fig. 1.3 (c) is a combination of (a) and (b). This is possible by positioning a co-axial cylinder within the rotary reactor. Coupled with direct injection of gaseous reactant into rotating solids, the feasibility of combined flow pattern for solids is an apparent advantage of the rotary reactor.

1.4

ENHANCEMENT OF GAS-SOLID CONTACTING IN ROTARY REACTORS

To remedy the disadvantage of rotary reactors, namely poor contacting between gas and solids, a variety of devices has been developed, as summarized in Table 1.2. IHI Co. in

6

1. Introduction Table 1.2. Devices to enhance contacting between gas and solids in rotary reactors

Contacting device

Remarks Incineration of municipal solid waste. Problems: Maintenance of seal plates. Low pressure of steam. Falling down of fines.

Activation of char. Problems: Sealing of rotary manifold. Many tubes are needed for even distribution of steam. Reactor efficiency is low.

Japan developed a rotary incinerator for municipal solid waste as seen in Table 1.2 [A], applying the same principle as the well known Roto-louvre dryer. For production of active carbon a number of tubes are positioned from outside the rotary reactor (see Table 1.2 [B]). Residual carbon in spent catalyst from a petroleum refinery is regenerated by the air, injected from the rotating horizontal tubes, as seen in Table 1.2 [C]. Rotary sealing of the gas distribution manifold should be well designed to realize high reactor efficiency. The authors have tested the other devices outlined in Table 1.3: [A] uses a rotary manifold to prevent leakage; [B] proposes an intermittent rotation of a reactor, in which the

1.4. Enhancement of gas-solid contacting in rotary reactors

7

Table 1.2. (Continued) Contacting device

Remarks Regeneration of spent catalyst from refinery. Decomposition of dioxins in fly ash. Problems: Sealing of gas distribution manifold. Reactor efficiency is intermediate.

Activation of char by steam. Problems: Reactor efficiency is not high. Torsion force due to friction is high. Strength of tubes should be large enough.

gaseous reactant is sent from flat distributor; and in [C], the gaseous reactant can contact solids similarly to an ordinary moving bed. When lifters are positioned in a rotary reactor, solids flow across their edges, and fall down to its bottom. Because solids are dispersed in its inner space, the conversion of solids appears to be enhanced. Quantitative prediction is carried out by the authors, suggesting that enhancement by lifters is limited, as long as the solid temperature is kept at its planned level.

8

1. Introduction

Table 1.3. Devices to enhance gas solid contacting, proposed by the authors [2] Contacting device

Remarks Example: Combustion of char. Advantage: Contacting is good. Porous ceramic plate can be used as distributor. Problem: Perfect rotary sealing is needed.

Example: Activation of very fine carbon. Advantage: Good contacting. Even very fine particles are well fluidized. Disadvantage: Periodic variation of gaseous components.

Example: Gasification of char. Advantage: Simple and counter current contacting is possible. Problem: With fine solids, pressure drop is high.

1.5. Examples of industrial application

9 Table 1.4.

Examples of practical rotary reactors Equation

Reaction

Temperature of solids [◦ C]

• Oxidation of carbon in spent catalyst • Elimination of carbon in pure silica • De-lacquering of spent cans • Activation/gasification of char • Oxidation and reduction of ferrite powder • Gasification of MSW

500 ∼ 600 1,000 450 ∼ 500 800 ∼ 900 400 ∼ 500 800 ∼ 900

• Calcination of magnesite, limestone • Agglomeration of composite solid for fertilizer • Clinkering of calcia, dolomite • Clinkering of cement • Reduction of ferro-chromium ore • Spinel transformation of ferrite • Decomposition of metal sulfate • Thermal cracking of solid waste

600 ∼ 900 600 ∼ 900 1,600 ∼ 1,800 1,400 ∼ 1,450 1,200 ∼ 1,300 500 1,000 500 ∼ 800

(1.5.1)

(1.5.2)

1.5

EXAMPLES OF INDUSTRIAL APPLICATION

Let us classify gas-solid reactions as follows:

A (gas) + b B (solid) →

B (solid) →



 gaseous product

solid product gaseous and solid products

solid product gaseous and solid product

(1.5.1)

(1.5.2)

Table 1.4 summarizes examples of conventional rotary reactors in industries. To keep the reactors at high temperature, considerable amount of thermal energy should be supplied. The consumption of electricity or fuel in rotary reactors should be reduced as much as possible. Thermal insulation and heat recovery from hot solids and flue gas are most crucial for drastic reduction of energy consumption.

1.6 COOPERATION WITH MECHANICAL ENGINEERS When it is planned to develop a novel rotary reactor, chemical engineers should consult with material and mechanical engineers regarding the following aspects: • Selection of material • Confirmation of strength • Appropriate design of rotary sealing

10

1. Introduction

• Feeding device • Discharging device In conventional rotary reactors, rotary sealing is usually insufficient as it allows appreciable leakage. When complete sealing is required for any novel processes, the rotary sealing and feeding devices should be carefully designed. Intimate cooperation with mechanical engineers is definitely needed for this purpose.

REFERENCES [1] D. Kunii, O. Levenspiel, Fluidization Engineering, 2nd edition (Butterworth-Heinemann, Stoneham, 1991). [2] D. Kunii, T. Chisaki, Development Engineering of Rotary Kilns (CHISAKI Co., 2005), in Japanese.

–2– Movement of Solids in Rotary Cylinder Rotating solids in cylinders flow easily, just like a liquid. The results of experimental studies by the authors are briefly presented. The movement of solids is theoretically analysed, giving useful equations to predict the flow rate of solids as well as the volumetric fraction of solids. In a normal rotary reactor, movement of solids is only in one direction. To improve residence time characteristics of solids in a single rotary reactor, the authors propose to position a re-circulation region inside the reactor. Theoretical equations to predict the flow rate of solids are presented.

2.1 2.1.1

EXPERIMENTAL STUDIES ON SOLIDS FLOW IN A HORIZONTALLY ROTATING CYLINDER

Movement of solids within a sectional area, perpendicular to the rotation axis

Research associates of the author have carried out visual observations on the movement of solids in a sectional area of a rotating cylinder. Black tracer solids were put on the surface of rotating white solids. Their findings are represented by a simple model, illustrated in Fig. 2.1. Solids rotate just like a rigid body, and then collapse to slide down along a flat surface, the inclination angle of which is nearly the same as their repose angle. In Fig. 2.1, the solids that slide down mix into the main body of rotating solids. Due to this movement, the solids are well mixed within the cross section, which is perpendicular to the rotation axis. This results in uniform mixing of solids and then uniform temperature, in the case of a rotary reactor. This movement of solids is crucial for any process in a rotary reactor, and the slipping of the main body of solids should be seriously prevented. Once slipping of solids occurs, solids do not mix at the cross section, deteriorating the performance of the rotary reactor. Slipping of solids occurs under the following conditions: • Inner surface is smooth and slippery, and then solids are fine. • Volumetric fraction of bulk solids is small. • Gas issues within the solids layer. When the above possibility occurs in a new gas-solid system, the authors advise that the volumetric fraction of the bulk layer be increased, for instance, up to 20–30%. This has been proved in a satisfactory commercialization of a rotary reactor made of quartz cylinder and operated at high temperature. Coarse solids usually rotate in a reactor with a rough surface as illustrated in Fig. 2.1, even in the case where the fraction of bulk solids is less than 10%.

12

2. Movement of Solids in Rotary Cylinder

Figure 2.1 Experimental findings on movement of solids in a rotating cylinder.

Figure 2.2 Movement of solids in a horizontal cylinder.

2.1.2

Both sides of cylinder closed

Particles are filled within a cylinder, as shown in Fig. 2.2, in which the volumetric fraction of bulk solids is γ = 0.5. When the cylinder is turned to be horizontal, the surface of the solids layer collapses to settle at their repose angle, ϑr . The cylinder is rotated once, and the solids move to form a flat surface at an inclination angle of ω, referring to the axis.

2.1. Experimental studies on solids flow in a horizontally rotating cylinder

13

Further rotation makes the angle ω decrease, as seen in Table 2.1. For normal solids, cumulative rotation 10 is enough to achieve an almost horizontal surface of the bulk solids. This verifies that rotating solids flow easily, just like a liquid. 2.1.3

Circular weir at one side of cylinder

When one side of the rotary cylinder has a circular weir as shown in Fig. 2.3, the volumetric fraction of the bulk solids γe at the end is determined by a graph in the figure. In accordance Table 2.1. Decrease of inclination angle ω with cumulative rotation of the cylinder Solids

Size [mm]

Bulk density [gm/cm3 ]

Repose angle [deg.]

Styrene sphere Rice grain Urethane pellet NiLi compounds

2.5 3×5 3×3 0.01

0.022 0.93 0.72 0.80

24 33 34 65

Solids

Angle of inclined surface, ω [deg.] Cumulative rotation 0 1

Styrene sphere Rice grain Urethane pellet NiLi compounds

24 33 34 65

8 10 17 21

2

4

4 5 8 14

2 2 4 9

10 0 1 1.5 4

Figure 2.3 Volumetric fraction of rotating solids layer inside a circular weir.

15 0 0 0.5 2

14

2. Movement of Solids in Rotary Cylinder

with similar experiments to Fig. 2.2, we found that the angle of the inclined surface ω inside the circular weir is 0.5 ∼ 2◦ , depending upon the flow rate of the solids across the weir. Thus we can estimate an average value of γ¯ inside the weir, by means of geometrical manipulation. Usually, γ¯ > γe 2.1.4

Circular weir to replace solids quickly

If it is required to replace rotating solids quickly, the circular weir interferes to remove the solids completely. It is plausible, then, to make an opening in the weir. The flow rate of the solids through the opening should be less than the feed rate of the solids, so that surplus solids overflow across the weir, and keep the volumetric fraction of the solids γ¯ to be nearly constant.

2.2 2.2.1

THEORETICAL STUDIES ON MOVEMENT OF SOLIDS

Simple model

Fig. 2.4 is a simple model proposed by Kunii on the basis of the previous experimental findings. Let us use the following notation. Notation Fs : N: R = dt /2: r: vs : ∆x: ρ: ¯ κdt : ϑr : ψ:

feed rate of solids rotation radius of the cylinder radius, variable linear velocity of sliding solids element length mean bulk density of solids average thickness of solids layer, sliding down on the surface of rotating solids repose angle angle against sliding layer

[kg/hr] [r. p. hr] [m] [m] [m/hr] [m] [kg/m3 ] [m] [deg.] [deg.]

The cylinder is assumed to be very long, and solids are fed into its upper side continuously at steady state. In Fig. 2.4 [A], the main body of the solids rotates just like a rigid body, and collapses into a layer of sliding solids, the inclination of which coincides with the repose angle of solids ϑr . Assuming κdt ≪ R, we can approximately calculate the volumetric fraction of rotating solids, γ as, γ=

πR 2



 ψ  ψ  ψ 360 − R sin 2 R cos 2 πR 2

=

sin ψ2 cos ψ2 ψ − 360 π

(2.2.1)

Fig. 2.4 [B] is a side view of the rotating cylinder, where ω is the angle of the rotation axis to the horizontal surface.

2.2. Theoretical studies on movement of solids

15

Figure 2.4 Kunii’s model of solids movement in an inclined cylinder.

2.2.2

Transportation rate of solids

In Fig. 2.4, P and Q are the gravity centers of the two parts of the thin layer of sliding solids respectively. The transfer rate of the thin layer is represented by the movement of P to Q. In accordance with the inclination angle of the rotation axis ω, the gravity center P moves vertically to Q′ , during one sliding. The surface of the sliding solids layer inclines ω degrees to the horizontal. Thus the sliding route of solids deviates only a little from the vertical, just at a small angle ω0 in Fig. 2.4 [B], and then reaches position Q′′ . In Fig. 2.4 [A], the volumetric rate of the rotating solids can be calculated by

R r=R cos

ψ 2

(∆xdr)(2πrN) = 2πN∆x



R

r=R cos

=

π ψ N∆xdt2 sin2 4 2

ψ 2

rdr = πN∆xR 2 sin2

ψ 2 (2.2.2)

The volumetric rate of sliding solids, κdt in thickness, passing section OA in Fig. 2.4 [A] is given by ∆x(κdt )vs

(2.2.3)

Since Eq. (2.2.2) equals Eq. (2.2.3), we obtain, κvs =

π ψ Ndt sin2 4 2

(2.2.4)

On the other hand, the velocity component of the sliding solids in an axial direction is calculated by the following equation. vs sin ϑr sin(ω + ω0 )

(2.2.5)

16

2. Movement of Solids in Rotary Cylinder Table 2.2. Numerical values calculated from Eq. (2.2.1)

γ [–]

0

sin3 ψ2 γ / sin3 ψ2

0.05

0.1

0.2

0.3

0.4

0.5

0

0.19

0.37

0.66

0.85

0.96

1.0

0.25

0.26

0.27

0.30

0.35

0.41

0.50

Therefore, the volumetric transfer rate of the sliding solids in an axial direction is calculated by  Fs ψ = (κdt ) 2R sin vs sin ϑr sin(ω + ω0 ) ρ¯ 2

(2.2.6)

Substitution of Eq. (2.2.4) into Eq. (2.2.6) gives Fs π ψ = N dt3 sin3 sin ϑr sin(ω + ω0 ) ρ¯ 4 2 On the basis of Eq. (2.2.1), the relation between γ and sin3 give the values shown in Table 2.2. In the case of γ  0.20, we can approximately use, sin3

(2.2.7) ψ 2

is numerically calculated to

ψ ∼ = 3.75γ 2

(2.2.8)

Substitution of Eq. (2.2.8) into Eq. (2.2.7) results in the following equation Fs = 2.945dt3 N γ sin ϑr sin(ω + ω0 ) ρ¯

(2.2.9)

Eq. (2.2.9) is valid in the case where the ratio of length lr to diameter dt is large enough, and the end effect at the exit side can be neglected. 2.2.3

Equations to predict the performance of a rotating cylinder

We can derive various quantities from Eq. (2.2.7), which are useful to predict the performance characteristics of the rotary reactor. Volume of discharged solids per one rotation, Fs ρ¯

N = 2.945dt3 γ sin ϑr sin(ω + ω0 )

(2.2.10)

Residence time, tr mass of solids tr = = feed rate

π

2 4 d t lr

Fs

 γ ρ¯

 1 γ = sin ϑr sin(ω + ω0 ) sin3

ψ 2



lr /dt N

2.2. Theoretical studies on movement of solids

The average volume of γ / sin3 tr =

ψ 2

17

is found in Table 2.2 to be 0.267 at γ  0.2. Thus,

0.267 lr /dt sin ϑr sin(ω + ω0 ) N

(2.2.11)

The volumetric fraction of bulk solids, γ is calculated from Eq. (2.2.9) as, γ=

Fs /ρ¯ 2.945dt3 N

sin ϑr sin(ω + ω0 )

(2.2.12)

The apparent linear velocity of solids in an axial direction, us 

Fs π 2 dt γ us = 4 ρ¯

us = 3.750dt N sin ϑr sin(ω + ω0 ) 2.2.4

(2.2.13)

Discussions on obtained equations

Fig. 2.5 compares Eq. (2.2.9) with practical data on the inner diameter of a rotary kiln vs its nominal production capacity. It is suggested to use ω0 = 2◦ for adequate curve fitting. Eq. (2.2.11) indicates that residence time tr of solids is independent of feed rate Fs . When the feed rate is doubled, the average volumetric fraction of the bulk solids is also doubled, as shown by Eq. (2.2.12), resulting in a constant residence time.

Figure 2.5 Comparison of theoretical equation Eq. (2.2.9) with practical data. [B] is plotted from Ref. [3].

18

2. Movement of Solids in Rotary Cylinder

When the volumetric fraction of the bulk solids is increased at the position of the solids inlet, the solids are apt to roll out from the inlet side of the rotating cylinder. To prevent this, the rotation rate is usually increased, sacrificing decreasing residence time. From Eq. (2.2.11), we find the residence time is proportional to the aspect ratio, lr /dt whereas it is inversely proportional to the rotation rate, N . The first thing to do in the above case is to improve the inlet device in order to prevent solids loss, instead of rotation increase. When the exit side of the rotary cylinder is open, the volumetric fraction of the bulk solids decreases forward the exit. Taking into account the above end effect, it is required to use an average value of γ throughout the length in the preceding equations.

2.3

2.3.1

IMPROVEMENT OF RESIDENCE TIME CHARACTERISTICS FOR ROTATING SOLIDS Application of screw cylinders

The movement of solids in a rotary cylinder is characterized as a plug flow (rod-like flow), only toward one direction. This is an advantage of a rotary reactor for ordinary feed stock. Nonetheless, depending upon the operating conditions planned for some novel processes, such a flow of solids should be improved considerably to meet their new requirements. The flow patterns of solids such as shown in Fig. 1.3 can be realized by the application of screw cylinders. Examples of conceptual design are found in Fig. 2.6. A screw plate is fixed to its outer cylinder, and rotates together. Solids are transferred into two screw cylinders but in opposite directions, establishing sound re-circulation of solids. In the case of Fig. 2.6 [B], the re-circulating flow rate of the solids in a cylinder Fs∗ is approximately calculated by the following equation. Fs∗

=



π 2 ¯ d lps γs ηs ρN 4 ts

(2.3.1)

where, dts is the diameter of the screw, lps is the length of its pitch, γs is the average volumetric fraction of solids in the screw cylinder, and ηs is the efficiency of transportation.

Figure 2.6 Application of screw cylinders to improve residence characteristics of rotating solids.

2.3. Improvement of residence time characteristics for rotating solids

19

By applying a transparent model simulating a screw cylinder (Fig. 2.6 [B]), the authors measured the volume of the solids that were transported due to rotation of the main cylinder. On the basis of Eq. (2.3.1), the numerical value of γs ηs was determined to be 0.22, under the following conditions. The volumetric fraction of the solids in the main cylinder was γ = 0.2 for both sides of the screw cylinder. The authors put an intake device at the inlet of the screw cylinder. It was found that the transportation rate of the solids was just the same for two cases, i.e. with the intake device and without it, under the above operating conditions. In case such screw cylinders as Fig. 2.6 [B] are designed, we can ignore any intake device. For practical design calculation, it is plausible to use the safer side of the numerical value, for instance, γs ηs = 0.20 In Example [10.3] and [10.5], the above value is applied to the design calculations of conceptual reactors. 2.3.2

Research and development of U-Turn system

The screw cylinders shown in Fig. 2.6 can be applied for improving the residence characteristics, and for better contact between gas and rotating solids (see Table 1.3 [C]). Gas should flow through the rotating solids and it is necessary to design a complete rotary sealing to prevent any leakage of gas. In reaction processes such as thermal cracking of organic materials, decomposition of metal sulfate and dehydration of inorganic compounds, intimate contact between gas and solids is not necessary. To meet the above requirement and to add a disintegrating action to agglomerated lumps of solids, the authors started research and development, on a so-called “U-Turn” system. Fig. 2.7 illustrates its working principle. A partition plate is put into a horizontally rotating

Figure 2.7 Models for U-Turn system. [A]: solids, one direction. [B]: solids, circulating around the partition plate.

20

2. Movement of Solids in Rotary Cylinder

Figure 2.8 Movement of solids on the partition plate.

cylinder, separating the inner space into two parts. A number of inclined guide plates are attached to both surfaces of the partition plate. Depending upon the angle of inclination of the guide plates, solids move toward one direction [A], or circulate around the partition plate [B]. Fig. 2.8 shows how solids move in the rotating cylinder during one rotation. Solids A at (1) in Fig. 2.8 are flattened on the partition plate after a rotation of 90◦ (2). At 135◦ (3), the solids slide down along the inclined partition plate. Because the guide plates are inclined, the solids are transferred in an axial direction. When the guide plates are attached as shown in Fig. 2.7 [B], the moving direction of B in Fig. 2.8 is opposite that of A. By continuous rotation, the rotating solids move to circulate around both edges of the partition plate. 2.3.3

Theoretical equations on transfer rate of solids

In Fig. 2.7 [B], we use the following notations. Fs∗ : ∆x: ϕ: η1 :

flow rate of solids distance of solids, transferred by one rotation angle of inclined guide plate efficiency

[kg/hr] [m] [deg.] [–]

Let us focus our attention on the solids within the section between the two guide plates. After one rotation of the horizontal cylinder, the solids are transferred into the next section. According to Kunii et al. [Ref. [2]], ∆x is given by ∆x = dt cot ϕη1

(2.3.2)

Efficiency η1 takes account of solids which fall out at the edge of the guide plates. After one rotation, the volume of the bulk solids Vb , transferred in an axial direction, in one side of the inner space, can be calculated as, Vb =

 1 π 2 dt (γ1 )(∆x) 2 4

2.3. Improvement of residence time characteristics for rotating solids

21

Substitution of Eq. (2.3.2) into the above one results in, Vb =

π 3 d cot ϕγ1 η1 8 t

When the cylinder rotates N [1/hr], the rate of solids flow in one side of the inner space Fs∗ is calculated as follows: ¯ = Fs∗ = Vb ρN

π 3 ¯ d cot ϕγ1 η1 ρN 8 t

(2.3.3)

For guide plates attached as shown in Fig. 2.7 [A], the flow rate of solids toward one direction Fs is given by Fs = 2Fs∗

(2.3.4)

In the paper by the authors [Ref. [2]], experiments were carried out using two visible models. 8 cm ID, 19.7 cm long 19.4 cm ID, 50.0 cm long Solids

dp [cm]

ρ¯ [gm/cm3 ]

ϑr [deg.]

Urethane pellet Rice grain Fluffy styrene sphere

0.3 × 0.3 0.3 × 0.5 0.25

0.72 0.93 0.022

34 33 24

In the case of urethane pellet, efficiency η1 is obtained as, Volumetric fraction γ [–]

0.1

0.2

0.3

η1 [–]

0.63

0.57

0.35

Additional experiments indicate that adequate conditions to obtain good circulation are as follows: ϕ = 60 ∼ 70◦ 2.3.4

breadth of guide plate ≈ 0.2dt

Transfer rates of solids in annular space

In order to replace the device in Fig. 2.6 [A], the authors designed partition plates to which inclined guide plates are attached, in the annular space between two co-axial cylinders. Let us focus our attention on the portion of solids P on the partition plate in Fig. 2.9 [A]. Due to 90◦ rotation, the solids flow down from the lower edge of the partition plate, as seen in Fig. 2.9 [B]. Thus, the displacement distance ∆x from the initial portion P to the solids

22

2. Movement of Solids in Rotary Cylinder

Figure 2.9 Partition plates within annular space between two co-axial cylinders.

settled on the bottom of outer cylinder Q is given by the following equation.  1 cot ϕη2 ∆x = (d1 − d2 ) 2

(2.3.5)

where η2 is the efficiency of the guide plates. In one side of the annular space, the volume of bulk solids Vb is transferred in the axial direction, due to one rotation.   1 π 2 Vb = d − d22 (γ2 )(∆x) 2 4 1 Therefore the re-circulating rate of solids Fs∗ is given by the next equation, in the case where the rotation rate is N [1/hr]. Fs∗ =

 π 2 ¯ d1 − d22 (d1 − d2 ) cot ϕγ2 η2 ρN 16

(2.3.6)

Then the transport rates of solids in the axial direction is given by Fs = 2Fs∗ 2.3.5

(2.3.7)

Comparison of partition plates with screw cylinders

In the case where the gaseous reactant in the main gas flow does not need to contact the solids in a rotary reactor, it is better to apply the partition plate for axial circulation of solids. If the gaseous reactant is required to contact solids intimately, we had better apply screw cylinders. Perfect rotary sealing is inevitable to apply the screw contactor. When it is planned to re-circulate solids between two regions in a single rotary reactor, and leakage of gas is seriously prevented between them, the screw cylinder design shown in Fig. 2.6 looks very attractive. The feasibility of their application will be discussed in Chapter 10.

Example 2.1

23

Example 2.1 Estimate the capacity for the following two rotary kilns: (1) Calcination of limestone, dt = 3 m ID N = (1 r.p.m.)(60 min/hr) = 60 1/hr ϑr = 35◦ , ω = 1.85◦ , ω0 = 2◦ , γ = 0.1, ρ¯ = 1,000 kg/m3 (2) Clinkering of cement, dt = 5 m ID N = (1.5 r.p.m.)(60 min/hr) = 90 1/hr ϑr = 35◦ , ω = 3◦ , ω0 = 2◦ , γ = 0.08, ρ¯ = 1,250 kg/m3 Solution For calcination of limestone, Eq. (2.2.9) gives:     hrs 1 kg kg 24 Fs = 24 × 2.945(3 m)3 60 (0.1) sin 35◦ sin(1.85◦ + 2◦ ) 1,000 3 day hr hr m = 441,000 kg/day = 441 tons/day For clinkering of cement:     kg kg hrs 1 24 Fs = 24 × 2.945(5 m)3 90 (0.08) sin 35◦ sin(3◦ + 2◦ ) 1,250 3 day hr hr m = 3,975 × 103 kg/day = 3,975 tons/day A similar calculation gives the lines shown in Fig. 2.5.

Example 2.2 A rotary kiln calcinates limestone under the following conditions: Estimate the average value of volumetric fraction γ and residence time tr to produce 500 and 600 tons of quicklime. dt = 2.8 m ID, l = 54 m, ω = 1.72◦ , ω0 = 2◦ bulk density of limestone 1,350 kg/m3 repose angle ϑr = 35◦ rotation rate 2 r.p.m. Solution Feed rate of limestone for 500 tons pr/day,

 500,000 kg pr kg pr kg 0.56 = 37,200 24 hrs kg hr With Eq. (2.2.12),

γ=

 kg  kg  1,350 3 37,200 hr m = 0.0955 = 9.55%  1  sin 35◦ sin(1.72◦ + 2◦ ) 2.945(2.8 m)3 2 × 60 hr

For 600 tons/day, similarly

γ = 0.1146 = 11.46% Residence time tr is calculated with Eq. (2.2.11) as tr =

0.267 54 m/2.8 m = 1.153 hr sin 35◦ sin(1.72◦ + 2◦ ) 2 × 60 1 hr

for both cases.

24

2. Movement of Solids in Rotary Cylinder

Example 2.3 A horizontal cylinder rotates at N = 4 r.p.m., in which a partition plate such as that shown in Fig. 2.7 [B] is positioned. Calculate the circulation rate of solids Fs∗ in the cylinder. Data dt = 2 m ID, ρ¯ = 1,000 kg/m3 , γ1 = 0.20, η1 = 0.57 To get the same circulation of solids, it is planned to design such cylinders as shown in Fig. 2.6 [B]. Determine the pitch length of the screw. Data diameter of 4 inner cylinders dts = 0.76 m γs ηs = 0.20 Solution For a cylinder with a partition plate, Eq. (2.3.3) gives,   kg π 1 Fs∗ = (2 m)3 cot 65◦ (0.20)(0.57) 1,000 3 = 40,080 kg/hr = 40.08 tons/hr 4 × 60 8 hr m In the case of four screw cylinders, the flow rate of solids in one cylinder is given by Eq. (2.3.1) as,   40,080 kg/hr kg π 1 = (0.76 m)2 (lps )(0.2) 1,000 3 4 × 60 2 4 hr m In the case of lps = dt , lps = 0.92 m For circulation of solids in a rotary reactor, the former looks simpler than the latter.

Example 2.4 The authors developed a rotary reactor to eliminate dioxins in fly ash from an incinerator, which was composed of co-axial cylinders as shown in Fig. 2.9. Under the following operating conditions, estimate the transport efficiency η2 . Data dt0 = 0.90 m, dti = 0.62 m ρ¯ = 450 kg/m3 , N ∗ = 2 r.p.m. ϕ = 65◦ , Fs = 100 kg/hr, γ2 = 0.185 Solution With Eq. (2.3.6),  kg π  (0.9 m)2 − (0.62 m)2 (0.9 m − 0.62 m) cot 65◦ = hr 8   kg 1 × (0.185)η2 450 3 2 × 60 hr m

Fs = 100

Thus η2 = 0.459 For an ordinary partition plate such as that shown in Fig. 2.7, we have η1 = 0.57 at γ = 0.20. This difference would be attributed to the distance between the guide plates in Fig. 2.9, where l1 ≪ l2 . Instead, in Fig. 2.7, l1 ≈ l2 . In some novel rotary reactors, the surface of the inner cylinder needs to contact the rotating solids in the annular space. We may use a screw plate as shown in Fig. 2.6 [A], but the solids rotate only on the lower surface of the outer cylinder, and hardly contact the inner cylinder. This is the reason why the authors used partition plates in the commercial reactor.

References

25

REFERENCES [1] D. Kunii, T. Chisaki, Development Engineering of Rotary Kilns (CHISAKI Co., Tokyo, 2005), in Japanese. [2] D. Kunii, T. Chisaki, O. Levenspiel, Powder Technology 96 (1998) 1–5. [3] W.H. Duda, Cement Data Book, vol. 1, 3rd edition (International Process Engineering in Cement Industry, Bauverlag GmbH, Wiesbaden and Berlin, 1985).

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–3– Conversion of Solids with Gaseous Reactant This chapter deals with transformation of a solid by chemical reaction with a gaseous reactant. Several models on conversion of a single solid are summarized in Table 3.2. Some of the familiar reactions are analyzed by equations from the models. Examples of how to apply the above models to the conversion of a single solid are: Experimental data of oxidation, gasification and activation of carbonaceous substances, roasting of sulfide ore and reduction of iron ore. Quantitative understanding of the conversion of a single solid is the most important requisite to predict the conversion of solids in any rotary reactor.

3.1

REACTION RATE OF SOLID CONVERSION

Reactions between gases and solids are simply represented by Eq. (3.1.1) (see Kunii and Levenspiel [2])

A (gas) + bB (solid) →

 gaseous product

solid product gaseous and solid products

(3.1.1)

When it is planned to develop a gas–solid reaction process, we have to collect fundamental kinetic data for it. However, it is rather difficult to find them under exactly the same conditions as needed. The conversion rate of solids depends enormously upon their physical characteristics. Rigid or porous, crystalline or amorphous, pure or catalytic impurities, such features of solids used to give different experimental data. For example, in the familiar Boudouard’s reaction, C + CO2 → 2CO, the conversion rate of carbon varies within one order of magnitude for chars from different coals (see Table 3.6). The fundamental kinetic data of a gas–solid reaction can be acquired from experimental studies in a thermo-balance such as that seen in Fig. 3.1. A monolithic or pelletized sample of a solid, either spherical or cylindrical, reacts with the flowing gaseous reactant, and the decreasing rate of its weight is continuously and precisely measured by means of a quartz spring. When solids of small size are tested, we can use a basket made of quartz wire such as that illustrated in Fig. 3.2. The sample solids are sandwiched between two thin layers of

28

3. Conversion of Solids with Gaseous Reactant

Figure 3.1

Thermo-balance (from Takamura et al., Ref. [13]).

Figure 3.2 Example of sample holder (from Adshiri, Ref. [9]).

loose quartz wool. This device makes the gaseous reactant flow through the mono-layer of the sample solids, ensuring good contact between gas and solids. In place of the thermo-balance, a couple of other devices have been utilized to measure the reaction rate of a gas–solid system, as illustrated in Table 3.1. The product gas issuing from the horizontal tube should be precisely measured to give the correct conversion of solids. Furthermore, the thickness of the solids layer in the sample boat should be as small as possible to prevent diffusional resistance in it. A fluidized bed reactor is sometimes applied to measure the conversion of solids.

3.1. Reaction rate of solid conversion

29

Table 3.1. Experimental device to measure the reaction rate of a solid sample Device

Remarks Uniform temperature and intimate contact between gas and solids are needed. The smaller the thickness of the solids layer the better. Effect of diffusional resistance should be evaluated, and deducted from the overall rate of reaction.

Bed temperature can be kept uniform. Temperature of solid in (a) can be measured. In (b), fluidizing particles can contact with the solid reactant, passing across the wire gauze.

Applied for a process in which it is necessary to measure the composition of product gases. Flow of solids is completely back mixing. Average residence time of solids is adjusted by feeding of inert solids.

30

3. Conversion of Solids with Gaseous Reactant

3.2

KINETIC MODELS OF GAS–SOLID REACTIONS

For quantitative understanding of gas–solid reaction processes, we need appropriate kinetic models to represent them. Let us focus our attention on the transformation of a single solid containing reactant B by a chemical reaction with gaseous reactant A. When the rate of chemical reaction is high enough, diffusional resistance usually intrudes to its transformation. For instance, mass transfers through boundary layer controls when a carbon particle is oxidized in the air at high temperature. We should take into account such diffusional resistance to analyze the transformation of a single solid. Table 3.2 summarizes kinetic models for the conversion of a single solid, and definition of chemical rate constants, based upon several simple models (see Kunii and Levenspiel [2]). When the experimental data are ready for a novel gas–solid reaction process, we should choose an appropriate model, and obtain fundamental rate constants for the reaction. In Table 3.2 we assume CA∗ , the equilibrium concentration of gaseous reactant A, is nearly zero on solid B. If CA∗ should be taken into account, we have to replace CA with ∆CA ∆CA = CA − CA∗

(3.2.1)

When the rate of chemical reaction is comparable to the rate of diffusion in the shrinking core model, we can approximately use Eqs. (3.2.6) and (3.2.7), by application of the overall rate constant k¯ in place of kc (see Ref. [2]) 1 1 1 ≈ + k¯ kc 12DAe /dp

(3.2.14)

In case mass transfer through the boundary layer around a solid should be taken into consideration, the overall rate constant k¯ can be used in Eqs. (3.2.4)–(3.2.7) 1 1 1 = + ¯k kc Kd

(3.2.15)

where mass transfer coefficient Kd is calculated with the following equation for Sherwood Number Kd dp 1/2 = 2 + 0.6Sc1/3 Rep DA

(3.2.16)

Here Sc is the Schmidt number and then, Rep = (ρg dp ug )/µ = Reynolds number dp = 2R, ρg is gas density, ug the relative velocity between gas and solid, and µ is the viscosity of the gas. In the decreasing size model, k¯ from Eq. (3.2.15) can be applied in place of kc in Eqs. (3.2.10) and (3.2.11).

Kinetic models for conversion of a single solid, gas-solid reaction, Refs. [1,2] Model

Rate of reaction

Uniform reaction model for porous solid, size unchanging.

dXB = kr CA (1 − XB ) dt

Shrinking core model. Blanket of product remains.

−1 dNA −1 dNB = 4π rc2 dt 4π rc2 b dt

Decreasing size model.

= kc CA

(3.2.2)

(3.2.4)

Integration

Time for complete conversion, τ

Typical example

1 − XB = exp(−kr CA t) (3.2.3)

At 1 − XB = 0.01 4.61 kr CA

Activation of char

Chemical reaction controls. t rc =1− (3.2.6) τ R

ρB R bkc CA

(3.2.7)

ρB R 2 6bDAe CA

(3.2.9)

Fick’s law for diffusion of reactant A through product blanket  3 rc = 1 − XB (3.2.5) R

Diffusion through the product layer controls.  3  2 rc rc t +2 =1−3 τ R R (3.2.8)

Eqs. (3.2.4), (3.2.5)

Chemical reaction controls. t rc =1− (3.2.10) τ R Diffusion through boundary layer controls  2 rc t (3.2.12) =1− τ R

Notation and unit

ρB R bkc CA

(3.2.11)

ρB R 2 2bDA CA

(3.2.13)

stoichiometric value

[gm-mol B/gm-mol A]

kr

rate constant defined by Eqs. (3.2.2) (3.2.3)

CA DA DAe dp Kd kc

concentration of A diffusivity of A that through product blanket size of solid mass transfer coefficient rate constant of chemical reaction

[gm-mol A/cm3 gas] [cm2 /sec] [cm2 /sec] [cm] [cm3 gas/cm2 solid sec] [cm3 gas/cm2 solid sec]

dNA /dt R, rc t XB ρB τ

rate of reaction radius of solid time conversion of B molar density of B time for complete conversion of B

Oxidation of metal sulfide Oxidation of carbon in ash rich pellet

Oxidation of carbon at high temperature

   gm-mol A 1 (sec) 3 cm gas

[gm-mol A/sec] [cm] [sec] [–] [gm-mol B/cm3 solid] [sec]

31

b

3.2. Kinetic models of gas–solid reactions

Table 3.2.

32

3. Conversion of Solids with Gaseous Reactant

3.3

RELATION BETWEEN RATE CONSTANTS OF CHEMICAL REACTION, BASED ON DIFFERENT MODELS

In Table 3.2, we have two different constants for the reaction of solid reactant, kr and kc . On the basis of the unit volume of the solid, the decreasing rate of B is given by the following equation  gm-mol B dXB = ρ = ρB kr CA (1 − XB ) B dt (cm3 solid)(sec)

(3.3.1)

Suppose solids are spherical with diameter dp . The surface area of solids per unit volume is,   6 cm2 dp cm3 The decreasing rate of reactant gas A is represented by Eq. (3.2.4). Combining it with Eq. (3.3.1) gives, 1 6 ρB kr CA (1 − XB ) = kc CA b dp

(3.3.2)

We get,

kc =



ρB dp (1 − XB ) kr 6b

cm3 gas (cm2 solid)(sec)



(3.3.3)

The rate constant of chemical reaction kc is based on the surface area of the single solid. In addition, let us introduce another constant Kr [1/sec] based on the unit volume of the solid. The decreasing rate of gaseous reactant A is defined as follows:  dNA  − = 4πrc2 kc CA = dt



4 3 πr Kr CA 3 c

(3.3.4)

Thus, Kr =

3 6 kc = kc rc dp



cm3 gas 3 (cm solid)(sec)



(3.3.5)

Combining with Eq. (3.3.3),

Kr =

ρB (1 − XB ) kr b



cm3 gas (cm3 solid)(sec)



(3.3.6)

3.4. Application of kinetic models to oxidation of carbon

33

3.4

APPLICATION OF KINETIC MODELS TO OXIDATION OF CARBON

3.4.1

Graphite

Parker and Hottel [3] carried out experimental studies on the oxidation of a graphite surface in the air and presented the rate of the chemical reaction based on the external surface area of the sample. Assuming the rate is proportional to the concentration of O2 in the air, the rate constant of the chemical reaction is calculated by application of Eq. (3.2.4). C + O2 = CO2 + 94.0

kcal gm-mol

(3.4.1)

(exothermic)

Combined with data from similar experiments, the approximate values of the rate constant kc are given in Table 3.3. 3.4.2

Petroleum coke

LaNauze and Jung [4] applied a similar setup to Table 3.1 [B] b. Their experimental conditions are summarized as follows: dt = 10.5 cm ID, size of fluidizing sand 0.65 mm, wire gauze basket sampler 6.5 cm ID, wire aperture 2.5 mm × 2.5 mm. A single spherical sample of petroleum coke, size ranging from 8 mm to 14 mm, was added to the basket in the bed, fluidized by the air at about 900 ◦ C. Unconstrained oxidation of the sample continued for a specified time, at 30 or 60 sec intervals. At the end of each period the air was switched to nitrogen, and the sample was transferred to a flask of liquid nitrogen for a short time. This technique enabled the size and weight of the reacting coke to be measured. An example of their results is illustrated in Fig. 3.3. The decrease in size is explained very well by Eqs. (3.2.5) and (3.2.10) until XB ≈ 0.93. The average density of the shrinking core was found to decrease a little. They pointed out that this was due to internal oxidation taking place inside the sample. Eq. (3.2.7) is applied to obtain the rate constant from Fig. 3.3. kc = 14.8 cm/sec at 900 ◦ C for dpi = 12 mm It is consistent with Table 3.3 for graphite. Table 3.3. Rate constant of chemical reaction kc for oxidation of a graphite surface [3], kc Temperature [◦ C] [cm/sec] kc Temperature [◦ C] [cm/sec] kc

520 0.05 800 5

550 0.075 900 17

570 0.12 1,000 42

600 0.2 1,100 85



cm3 gas cm = sec (cm2 solid)(sec)

620 0.28 1,200 200

650 0.60 1,300 400



700 1.4 1,400 750

34

3. Conversion of Solids with Gaseous Reactant

Figure 3.3 Examples of data from LaNauze and Jung [4].

3.4.3

Char from coal

Jung and Stanmore [5] applied another technique similar to Table 3.1 [B] b, and carried out experiments to dry, devolatize and burn very wet Victorian brown coal. Fluidized bed 7.6 cm, river sand 360 µm, dp = 2 ∼ 7.4 mm. Drying and devolatization was completed within about 60 sec, as shown in Fig. 3.4. Char particles of a size ranging from 2 ∼ 5.5 mm were oxidized in the sampler. The char particles were removed after a set reaction time by withdrawing fine sands from the sam-

Figure 3.4 Data on thermal cracking of Victorian brown coal, adopted from Junk and Stanmore [5].

3.4. Application of kinetic models to oxidation of carbon

35

pler. From the measured loss of mass, the average oxidation rate of the char particles was obtained to be, 4.8 × 10−4 gm B/cm2 sec Average temperature of the bed, 750 ◦ C. By application of Eq. (3.2.4), rate constant kc is calculated as, kc = 15.9 cm/sec The above value corresponds to kc = 17 cm/sec at 900 ◦ C for graphite in Table 3.3. The temperature of the char was found to be higher than the bed. 3.4.4

Stable temperature of an isolated carbon particle

Suppose a single sphere of carbon is oxidized in a reactor, kept at constant temperature Tb . The stable temperature of the sphere Ts is determined by the following equation. ¯ A bHB = (hp + hr )(Ts − Tb ) kC

(3.4.2)

where HB means heat of reaction [kcal/gm-mol B], and hp and hr are convectional and radiant heat transfer coefficients, respectively [kcal/m2 hr ◦ C]. In Example 3.3, we estimate the stable temperature of carbon particles under similar experimental conditions to the previous reports. Eq. (3.4.2) is solved graphically, as illustrated in Fig. 3.5, where we find that the temperature of the carbon particle is higher than its surroundings. The temperature difference depends on its size, as follows: Results of calculation in Example 3.3. dp Ts − Ta

[mm] [◦ C]

2 180

4 110

12 70

Taking into account the above temperature rise, rate constant kc for char is nearly consistent with Table 3.3 for graphite. 3.4.5

Gaseous reactant around particle

LaNauze [4] pointed out that carbon particle coarser than 3 ∼ 5 mm was surrounded by a CO flame. This justifies the hypothesis proposed by previous investigators that CO forms at the external surface of carbon. CO diffuses through the boundary layer around the particle and reacts with O2 , which diffuses counter-wise. At high temperature, therefore, Eq. (3.4.1) should be represented as a combination of two chemical equations. 1 kcal C + O2 = CO + 26.4 2 gm-mol

(exothermic)

1 kcal CO + O2 = CO2 + 67.6 2 gm-mol

(exothermic)

(3.4.3) (3.4.4)

36

3. Conversion of Solids with Gaseous Reactant

Figure 3.5 Stable temperature of carbon particle, oxidized in the air at 800 ◦ C.

3.4.6

Carbon dispersed in inorganic solids

Hellinkx [6] burnt dense coal shale in the air; from the results the diffusion of oxygen through the blanket of ash was found to be the controlling step. A value of effective diffusivity of oxygen through the ash blanket was estimated. DAe = 0.042 cm2 /sec

at 900 ◦ C

For design calculation of a similar case, we could assume Eq. (3.4.5). DAe ≈ εi DA

(3.4.5)

where DA is the diffusivity of gaseous reactant, and εi is the effective porosity of the solid. In the case of dense coal shale, we estimate εi = 0.029. Another example is spent catalyst from a petroleum refinery. It contains residual carbon, which should be oxidized in order to be used again. (Size 0.85 mm OD and ca. 4 mm in length, residual carbon 5 wt.%) Experiments of its oxidation in a micro-reactor at 550 ◦ C gave the data shown in Table 3.4. At a low temperature such as 550 ◦ C, oxidation of dilutely dispersed carbon within the catalyst could be analyzed by application of the uniform reaction model in Table 3.2. Concentration of O2 in the air CA is, CA =

gm-mol A 273 1 gm-mol A (0.21) = 3.11 × 10−6 273 + 550 22.4 × 103 cm3 cm3

3.5. Gasification of carbon

37 Table 3.4.

Rate of chemical reaction. oxidation of residual carbon in spent catalyst, 550 ◦ C Reaction time t [min]

15

20

30

Residual carbon [wt.%]

0.11

0.12

0.03

2.2

2.4

0.6 × 10−2

1,364

1,000

914

1 − XB  kr

1

gm-mol A (sec) cm3



Average

1,093

Substituting 1 − XB , CA and t into Eq. (3.2.3), we get kr in Table 3.4. Similar calculation gives, 1 kr = 1,770  gm-mol A  (sec) cm3

3.5 3.5.1

at 570 ◦ C

GASIFICATION OF CARBON

Boudouard’s reaction

By application of a similar technique to Table 3.1 [A], Mayers [7] measured the gasification rate of graphite by carbon dioxide. C + CO2 = 2CO − 41.22

kcal gm-mol

(3.5.1)

(endothermic)

Similar experiments were carried out by Walker [8]. Eq. (3.2.4) is used to analyze their data, giving the rate constant of the chemical reaction in Table 3.5. The rate constant should be applied only for a monolithic and/or a thick layer of graphite. Adschiri [9] carried out experimental studies on the gasification of particulate char, 0.5 ∼ 0.6 mm, by sending the O2 –N2 mixture into a thermo-balance, similar to Fig. 3.1. He used a quartz basket to hold the particulate char (see Fig. 3.2). Fig. 3.6 gives an example of his results. Eq. (3.2.3) for the uniform reaction model is applied to obtain the rate constant of gasification, summarized in Table 3.6. We find that the rate constant varies in char samples from different sources. Table 3.5. Rate constants for gasification of graphite Ref. [7] kc [(cm3 gas)/(cm2 solid)(sec)] Temperature

900

950

1,000

1,100

1,200

1,300

C + O2 = CO2 C + H2 O = CO + H2

0.005 0.020

0.026 0.050

0.06 0.14

0.20 0.44

0.55 1.40

1.1 –

38

3. Conversion of Solids with Gaseous Reactant

Figure 3.6 Gasification rate of char with CO2 , mean size 0.55 mm (taken from Adschiri [9]).

Table 3.6. Approximate rate constant of Boudouard’s reaction, calculated from data given by Adschiri [9] Coal from which char sample was prepared

Yallourn I Yallourn II Taiheiyo Hokutaku Baiduri

kr



1 (gm-mol CO2 /cm3 gas)(sec)

800 ◦ C

900 ◦ C

1,000 ◦ C

1,100 ◦ C

17.8

156 117 70.2 31.2 23.4

– 339 339 296 233

–

5.5

22.8

10.9

Electrode carbon by Nakao, 1974

3.5.2



Gasification of carbon by steam

Gasification of carbon by steam has been extensively carried out by a number of investigators (refer to Johnson [10]). Mayers [7] measured the gasification rate of a thick graphite sample by steam with a similar experimental reactor to Table 3.1 [A]. From his data, rate constant kc is calculated with Eq. (3.2.4) and compared with those by CO2 , as shown in Table 3.5. Little appreciable differences are found between the two gaseous reactants in case of graphite. At the surface of carbon, a chemical reaction is supposed to occur. C + H2 O = CO + H2 − 31.38

kcal gm-mol

(endothermic)

(3.5.2)

3.5. Gasification of carbon

39 Table 3.7. Equilibrium constant Kp

Temperature

[◦ C]

400

500

600

700

800

900

1,000

Kp

[–]

0.1

0.2

0.35

0.58

0.88

1.26

1.7

Mean values of data from Hahn, Haber-Richardt, Engels and Neuman-Köhler

Figure 3.7 Conversion of char (1.1 mm) from Soya coal, gasified by steam (adopted from Hashimoto [11]).

At high temperature, the product species must react in gas phase, towards attaining reaction equilibrium. CO + H2 O ⇄ CO2 + H2 + 9.838

kcal gm-mol

(exothermic)

(3.5.3)

Equilibrium constant Kp is given by Table 3.7, where, Kp =

pCO pH2 O pCO2 pH2

(3.5.4)

Hashimoto performed extensive experiments on the gasification of char from different coals by steam at ambient and high pressure in a thermo-balance reactor [11]. Conversion XB increases with elapsing time, as shown in Fig. 3.7. By application of Eq. (3.2.3) in Table 3.2, the rate constant of the chemical reaction is calculated to give Table 3.8. Comparison of Fig. 3.7 with Fig. 3.6 clearly indicates gasification of char by steam is remarkably faster than that by CO2 . For instance, to attain XB = 0.5, CO2 needs 40 min, whereas H2 O needs 4 min at 854 ◦ C. In the gasification of char, the chemical reaction takes place on the internal surface of the pore, and then the volumetric fraction of the pore, εi increases with increasing conversion XB . Consequently, the internal surface area increases from its initial value and attains its maximum at approximately XB = 0.5. By further gasification, XB approaches 0.8 ∼ 0.9 until the particle starts to collapse.

40

3. Conversion of Solids with Gaseous Reactant Table 3.8. Rate constant of chemical reaction, for gasification of char by steam, calculated from Fig. 3.7

Char from Soya coal

kr



1 (gm-mol H2 O/cm3 gas)(sec)

[◦ C]

3.6



787

887

985

1,085

1,185

105

758

2,168

5,149

9,212

ACTIVATION OF CARBONACEOUS PELLET

By applying a fluidized bed reactor, as shown in Table 3.1 [B] c, pellets composed of powdery char were activated by sending a mixture of steam and N2 . The experimental data of XB vs ◦ C are shown in Fig. 3.8. For active carbon, the reaction is usually stopped at XB = 0.5. A uniform reaction model is the most appropriate in this case. Applying Eq. (3.2.3) in Table 3.2, the rate constant is calculated to give Table 3.9. Compared with the gasification of char from Soya coal shown in Table 3.8, the rate constant of dense pellet is nearly one third at 800 ◦ C.

Figure 3.8 Conversion vs time, activation of cylindrical carbonaceous pellets, 4.2 mm OD, 6.4 mm long. Table 3.9. Rate constants for activation of carbonaceous pellets, calculated from Fig. 3.8 Dense pellets 4.2 mm OD, 6.4 mm long kr 1

 gm-mol A  cm3

(sec)

Temp. [◦ C] Base, coconut Base, coal

700 – 20

750 9 40

800 30 54

850 46 65

3.7. Roasting of zinc sulfide

41

3.7

ROASTING OF ZINC SULFIDE

Ogawa [12] carried out experimental roasting of zinc blend of a size ranging from 3.38 ∼ 17.3 mm, and pointed out that diffusion of oxygen through the product blanket was the controlling step. He utilized Eq. (3.2.8) to analyze his data, from which the effective diffusivity was calculated as, DAe = 0.0506 cm2 /sec (900 ◦ C) From Eq. (3.4.5), we estimate εi = 0.034. Takamura et al. [13] carried out experimental oxidation of a single pellet composed of zinc sulfide powder in a thermo balance, shown in Fig. 3.1. Spherical pellets, 9.9 and 11.5 mm in size. O2 /N2 = 21/79 The experimental data of solid conversion vs time were analyzed to give a relation of dXB /dt vs time. At 580 ◦ C, for instance, 1 − XB [–] dXB  1  dt

0.2 0.005

sec

0.4 0.01

0.6 0.015

0.8 0.020

Since dXB /dt is proportional to 1 − XB , they found the uniform reaction model is applicable in this case. With Eq. (3.2.2), kr = 8, 167

1 (gm-mol O2 /cm3 gas)(sec)

at 580 ◦ C.

Similar analysis results in the following conclusion. 560 ∼ 600 ◦ C: continuous reaction model is applicable, Eq. (3.2.3). 600 ∼ 690 ◦ C: widespread reaction zone. above 690 ◦ C: approach to shrinking core model, Eq. (3.2.8).

3.8

REDUCTION OF IRON ORE

For the purpose of developing novel processes for the direct reduction of iron ore, an enormous number of experimental investigations have been reported (see Refs. [14]–[16]). Let us choose Hamada’s work [16] as an example. He applied a fluidized bed reactor of 28 mm ID, in which a batch of iron ore was reduced. His experimental conditions are summarized as follows: Iron ore Brazil, 0.25 ∼ 4 mm Static height of fluidized bed 15 cm Gas H2 , CO, CH4 Temperature 800 ∼ 1,000 ◦ C

42

3. Conversion of Solids with Gaseous Reactant

Figure 3.9

Reduction of iron ore (Hamada [16]).

Fig. 3.9 represents some of his results. Even though sintering of ore within a fluidized bed is very troublesome in any direct reduction process, he was able to perform his experiments to a 95% conversion. He analyzed his data and found autocatalytic reaction of CH4 took place at conversion XB higher than 50%. He interpreted that sintering was prevented by carbon formation at the particle surface. Looking at Fig. 3.9, the existence of CH4 in the fluidizing gas seems to have little influence on XB vs t curve. The average value of XB in Fig. 3.9 is represented well by the continuous reaction model until XB = 0.90. From Eq. (3.2.3), we obtain, kr = 216

1 at 900 ◦ C. (gm-mol A/cm3 )(sec)

Example 3.1 In the case where diffusion through the product blanket is controlling in the shrinking core model, derive Eqs. (3.2.8) and (3.2.9), on the basis of Fick’s law. Solution Suppose a quasi-steady state diffusion of gaseous reactant A through the product blanket, in the shrinking core model in Table 3.2.  dC ′ (4π r 2 ) DAe A = constant (3.8.1) dr ′ is the concentration of A within the product blanket. Eq. (3.8.1) is solved with the following boundary where CA conditions, giving Eq. (3.8.2).

r = ri , ′ CA

CA

=

′ = 0, CA

1 − rrc

1 − rRc

r = R,

′ =C CA A

(3.8.2)

Example 3.2

43

The mass balance at the interface, r = rc , gives  ′   dC  (4π r 2 )(−drc )(ρB ) = (4π r 2 )DAe  A  dt dr rc

The concentration gradient is calculated from Eq. (3.8.2), and substituted to the above equation to give, ρB (−drc ) =

CA DAe · dt rc 1 − rRc

(3.8.3)

Integration of Eq. (3.8.3) results in Eqs. (3.2.8) and (3.2.9).

Example 3.2 A graphite sphere is oxidized by the air in a fluidized bed of inert particles, the temperature of which is kept at 800 ◦ C. Predict the stable temperature of the sphere for dp = 2, 4 and 12 mm. Assume HB = 94 kcal/gm-mol B. Data The relative velocity of the air is 0.5 m/sec. The convectional heat transfer coefficient between the sphere and fluidized bed is 344 kcal/m2 hr ◦ C. Temp.

[◦ C]

800

1,000

1,200

1,400

DA kc

[cm2 /sec]

1.35 5

1.75 42

2.17 200

2.63 750

[cm/sec]

kc from Table 3.3. Solution Radiant heat transfer coefficient hr is calculated with the next equation. hr =

 +273 4  800+273 4  (0.9)(4.88) Ts100 − 100 Ts − 800

For example, at Ts = 1,000 ◦ C, hr = 285.6 kcal/m2 hr ◦ C. Thus the right-hand term of Eq. (3.4.2) is given by  kcal (344 + 285.6) (1,000 ◦ C − 800 ◦ C) = 125,950 kcal/m2 hr = 3.498 × 10−3 kcal/cm2 sec 2 ◦ m hr C Similar calculation at other temperatures gives a line in Fig. 3.5, designated by the right-hand term of Eq. (3.4.2). On the other hand, for dp = 4 mm = 0.4 cm, Re =

(0.4 cm)(3.282 × 10−4 gm/cm3 )(50 cm/sec) = 14.27 4.6 × 10−4 gm/cm sec

From Eq. (3.2.16), Kd dp = 2 + 0.6(0.75)1/3 (14.27)1/2 = 4.06 DA Kd = (4.06)(1.75 cm2 /sec)/(0.4 cm) = 17.76 cm/sec With Eq. (3.2.15) cm 1 = 12.48 k¯ = 1 1 sec + 42 17.76

44

3. Conversion of Solids with Gaseous Reactant

With b = 1(gm-mol B)/(gm-mol A), HB = 94 kcal/gm-mol B, the left-hand term of Eq. (3.4.2) gives    gm-mol B gm-mol A kcal (12.48) 2.011 × 10−6 94 = 2.36 × 10−3 kcal/cm2 sec 3 gm-mol A gm-mol B cm Similar calculations are carried out at the other temperatures and sizes. The results of the calculations are plotted to give the left term lines in Fig. 3.5. Eq. (3.4.2) is graphically solved in Fig. 3.5, giving a stable temperature as follows: dp [mm] 2 4 12

Ts [◦ C] 980 910 870

Ts − Tb [◦ C] 180 110 70

With the smaller size, the larger the temperature rise.

Example 3.3 A single particle of char is gasified in stationary gas which contains 30% of steam. Calculate the overall rate constant k¯ for different size dp , and discuss the controlling step in this reaction. Data Particle temperature 985 ◦ C  2 2  985+273 1.5 = 0.872 cm DA = 0.277 cm sec 312.6+273 sec ρ = 1.5 gm , ρ = 0.125 gm-mol C , X = 0.5 s

cm3

B

B

cm3

Solution From Table 3.8 we take kr = 2, 168  gm-mol1A  cm3 gas

(sec)

With Eq. (3.3.3),

 gm-mol B  (dp cm solid)(1 − 0.5) 0.125 2,168 (cm3 gas) cm3 solid ×  gm-mol A  = 22.58dp kc =  gm-mol B  (cm2 solid)(sec) (sec) 6 1 gm-mol A cm3 gas

According to Eq. (3.2.15),

1 1 1 + = 22.58dp Kd k¯

(3.8.4)

In the stationary gas, Eq. (3.2.16) gives, Kd =

2DA 2(0.872) 1.744 = = dp dp dp

Overall rate constant k¯ is calculated with Eq. (3.8.4), giving Table 3.10. Looking at Table 3.10, we find diffusion through the boundary layer controls in gasification of char by steam, which diameter is larger than 1.0 cm.

Example 3.4 Steady state azotation of calcium carbide was carried out by Yagi et al. [17] in a fluidized bed reactor of 57 mm ID, similar to Table 3.1 [C]. N2 + CaC2 = CaCN2 + C + 72

kcal gm-mol

(exothermic)

(3.8.5)

From the experimental data of solids conversion, the necessary time for complete conversion of solids was calculated on the basis of their theoretical equations (see Table 3.11).

Example 3.4

45 Table 3.10. Calculated results in Example 3.3

Size dp [cm] [cm/sec] [cm/sec] [cm/sec]

Kd kc k¯

0.1

0.2

0.4

0.6

1.0

2.0

4.0

17.4 2.26 2.00

8.7 4.52 2.98

4.4 9.03 2.94

2.9 13.6 2.48

1.7 22.4 1.62

0.87 46.2 0.856

0.44 90.3 0.433

Table 3.11. Necessary time τ [hr] for complete conversion of solids, azotation of calcium carbide [17] Temperature [◦ C]

Mean size of CaC2 [mm]

0.17 0.21 0.36 Mean value of kc [(cm3 gas)/(cm2 solid)(sec)]

1,060

1,080

1,100

1,200

8.6 hr 10.0 17.0

4.4 5.0 9.0

2.5 3.4 5.0

0.9 1.2 2.0

0.94

1.9

2.4

9.0 × 10−3

Estimate the controlling step in this reaction, and calculate the rate constant at each temperature. Data ρs = 2.0

gm cm3

= 0.0312

gm-mol cm3

Solution The size ratio is 0.36 mm/0.17 mm = 2.12. The ratio of τ is 17/8.6 = 1.98 at 1,060 ◦ C. Similarly we get, Temperature [◦ C] τ0.36 mm /τ0.17 mm

1,060 1.98

1,080 2.05

1,100 2.0

1,120 2.22

We find τ is proportional to size dp in this case, and choose the shrinking core model with chemical reaction controlling, i.e. Eq. (3.2.7). CA = kc =

273 1 gm-mol A 0.01219 gm-mol A · = 273 + T 22.4 × 103 cm3 273 + T cm3

ρB dp 2bCA τ

For dp = 0.17 mm = 0.017 cm, at T = 1,060 ◦ C CA = 9.145 × 10−6 gm-mol A/cm3 , kc =

τ = 8.6 hr = 3.096 × 104 sec

cm (0.0312)(0.017) = 9.37 × 10−4 sec 2(9.145 × 10−6 )(3.096 × 104 )

The mean value of kc is calculated for three sizes and added to Table 3.11.

46

3. Conversion of Solids with Gaseous Reactant

REFERENCES [1] O. Levenspiel, Chemical Reaction Engineering (John Wiley and Sons, New York, 1962); Chemical Reactor Omnibook (OSU Book Stores, Corvallis, OR, 1989). [2] D. Kunii, O. Levenspiel, Fluidization Engineering, 2nd edition (Butterworth-Heinemann, Stoneham, MA, 1991). [3] A.L. Parker, H.C. Hottel, Ind. Eng. Chem. 28 (1936) 1334. [4] R.D. LaNauze, K. Junk, Nineteenth Symposium (International) on Combustion, the Combustion Institute, 1,057 (1982); Proceedings, 7th International Conference on Fluidized Bed Combustion, 1040 (1983); R.D. LaNauze, Chem. Eng. Res. Dev. 63 (Jan. 1985). [5] K. Junk, B.R. Stanmore, Fuel 59 (Feb. 1980) 74. [6] L.J. Hellinkx, Chem. Eng. Sci. 3 (1954) 201. [7] M.A. Mayers, J. Am. Chem. Soc. 56 (9) (1934) 1879; 61 (Aug. 1939) 2053. [8] P.L. Walker, et al., Ind. Eng. Chem. 45 (1953) 1703. [9] M. Adschiri, PhD Thesis (University of Tokyo, 1986). M. Adschiri, et al., Proc. 1987 1st. Conf. on Coal Science (Elsevier, Amsterdam, 1987), p. 515. [10] J.L. Johnson, Kinetics of Coal Gasification (John Wiley & Sons, New York, 1979). [11] K. Hashimoto, et al., IEC Proc. Des. Dev. 72 (1979) 18; J. Fuel Soc. Japan 62 (674) (1983) 421; 65 (10) (1986) 798; 66 (6) (1987) 418; Fuel 65 (Nov. 1986) 1516. [12] Y. Ogawa, J. Mine & Met. Inst. Japan (45) (1929) 554. [13] T. Takamura, et al., J. Chem. Eng. Japan 1 (4) (1974) 276. [14] E.G. Hamilton, B.O. Holland, The Direct Reduction of Iron Ores (1902–1967). [15] W. Wenzel, et al., Arch. Eisenhütten W. 43 (1972) 805. [16] T. Hamada, D. Kunii, Proc. 84th Annual Meeting of Iron and Steel Inst. of Japan, Vol. 58, Oct 1972, S 328; T. Hamada, et al., loc. cit., Vol. 61, Apr. 1975, S 407. [17] S. Yagi, D. Kunii, Chem. Eng. Sci. 16 (1961) 364.

–4– Thermal Decomposition and Conversion of Composite Pellets In this chapter, conversion of a single solid, which does not need a gaseous reactant, is presented and explained on the basis of proposed models. Thermal decomposition of inorganic and/or carbonaceous substances, and reduction of inorganic compounds are sometimes carried out in large-scale rotary reactors. The quantitative analysis of the performance data from the above processes indicates that heat transfer is a crucial step to perform these processes.

4.1

ELIMINATION OF TRACE SPECIES IN SOLIDS

In some cases, a small amount of substance should be eliminated from the product solids by heating them to an adequate temperature. Elimination of dioxins in fly ash and of silanol from ultra pure silica are two examples of this case. Since the initial concentration of the substance is usually small, let us use the ratio of residual substance B as follows: YB =

CB = 1 − XB CBi

Suppose the above reactions are independent of the outside atmosphere, then use kr∗ in place of kr CA in Eq. (3.2.2). Taking into account the equilibrium value of CB at a given temperature, YBe , integration gives the following equation.   YB − YBe = exp −kr∗ t YBi − YBe

(4.1.1)

See Example 4.1.

4.2

CALCINATION OF LIMESTONE

For decomposition of limestone, dolomite and/or magnesite, it is appropriate to apply the shrinking core model. In this case, however, the core shrinks due to thermal decomposition of the solid reactant at the reacting interface. Heat conduction through the product blanket controls the shrinking rate of the core, instead of reaction with gaseous reactant.

48

4. Thermal Decomposition and Conversion of Composite Pellets

Figure 4.1 Partial pressure of CO2 at the interface of decomposition for CaCO3 and MgCO3 , 760 mmHg = 1 atm.

Decomposition of limestone needs considerable thermal energy. CaCO3 = CaO + CO2 − 42.5

kcal gm-mol

(20 ◦ C, endothermic)

(4.2.1)

Assume a constant temperature Ts at the surface of solid B and then Ti at the decomposing interface. When resistance to CO2 diffusion across the product blanket is negligible, Ti is estimated from Fig. 4.1, which gives the relation between the equilibrium pressure of CO2 and the interface temperature. Quasi-steady state conduction of heat through the product blanket is represented by the following equation, in the case where Ts > Ti and rc  r  R.  dT (4πr 2 ) ks = constant dr

(4.2.2)

Eq. (4.2.2) is solved with the following boundary conditions, giving Eq. (4.2.3). r = rc

T = Ti ;

1− T − Ti = T s − Ti 1 −

r =R

T = Ts

rc r rc R

(4.2.3)

Letting HB the molar heat of reaction and ks be thermal conductivity of the product blanket, the heat balance at the interface, r = rc gives      dT  2 2 4πrc (−drc )(ρB HB ) = 4πrc ks  dr

   dt  rc

(4.2.4)

4.3. Decomposition of manganese sulfate

49

Figure 4.2 Necessary time τ for complete conversion of limestone, ∆T = Ts − Ti .

The temperature gradient is calculated from Eq. (4.2.3), and is substituted to Eq. (4.2.4). (ρB HB )(−drc ) =

ks Ts − Ti dt · rc 1 − rRc

Integration results in Eq. (4.2.5), where rc /R is given by Eq. (3.2.5). The time necessary for complete conversion τ is calculated by τ=

ρB HB dp2 ρB HB R 2 = 6ks (Ts − Ti ) 24ks (Ts − Ti )

(4.2.5)

In Example 4.2, τ is calculated for granular limestone under reasonable operating conditions of calcination, giving Fig. 4.2, where ∆T = Ts − Ti . In limestone of considerable size, the diffusion rate of CO2 through the product blanket should be taken into account. In this case, we have to take a slightly higher temperature than that from Fig. 4.1 (see Example 4.3).

4.3

DECOMPOSITION OF MANGANESE SULFATE

Rotary reactors are sometimes utilized to decompose metal sulfate for further processing. For example, the following reaction is important to produce the necessary manganese for an electric cell. 1 1 MnSO4 = Mn3 O4 + SO2 + O2 3 3

(4.3.1)

50

4. Thermal Decomposition and Conversion of Composite Pellets Table 4.1. Experimental data on thermal decomposition of manganese sulfate Temperature 1,030 ◦ C Time [min] 1 − XB [–]

12.5 0.13

15.5 0.085

20 0.042

The above reaction was carried out in a small rotary reactor of 15 cm ID, the results of which are shown in Table 4.1. From Eq. (4.1.1) we get, kr∗ = 2.68 × 10−3

1 sec

A similar calculation can be applied for other inorganic substances, such as CaSO4 , Ca(OH)2 etc.

4.4

THERMAL CRACKING OF ORGANIC SOLIDS

When a particle of organic substance is suddenly fed into a region at high temperature, it is decomposed very quickly, issuing a variety of gaseous and vapour species. Char remains, in the case where the flowing gas does not contain oxygen or steam. Look at Fig. 3.4, observed by Junk and Stanmore [1]. Wet particles of Victorian brown coal are dried and de-volatized quickly within 50 seconds. Hashimoto [2] fed the platinum basket containing Taiheiyo coal particles of 1.1 mm suddenly into a thermo-balance kept at 787 ◦ C. N2 was used as its carrier gas and the composition of the effluent gas was measured. Just after heating the sample, the composition of cracked gas increases rapidly, and at nearly 1 min, peak concentrations were observed for CH4 , H2 , CO and CO2 . Approximately three minutes later, such gases were no longer detected. From an engineering point of view, we can assume that thermal cracking of hydrocarbon solids finishes at high temperature within a minute. For coarse solids, however, a longer time is needed to raise the temperature of its centre to the planned value. In contrast with the above discussions, thermal cracking of coal and/or biomass at intermediate temperature, i.e. 400–500 ◦ C, is of practical importance to produce char of suitable quality for further utilization. The slow rate of temperature rise is necessary to obtain suitable char for production of active carbon. Thus the optimum schedule of temperature vs time in a micro-reactor must be established beforehand.

4.5

COMPOSITE MADE OF IRON ORE AND OIL

Hasegawa [3] used a similar system to Fig. 3.1, in which a spherical composite made of iron ore and heavy oil was heated at high temperature. The composition of the gas produced

4.5. Composite made of iron ore and oil

51

Figure 4.3 Issue of gases from a spherical composite pellet, made of iron ore and straight asphalt, at 1,050 ◦ C (from Hasegawa [3]).

in the effluent flow was measured until the reaction was completed. Size of sphere 3.5 and 7 mm Weight ratio 0.18 ∼ 0.2 gm oil/gm ore Iron ore from Mt. Newman Total Fe 58.37%, 8.9 ∼ 74 µm Fig. 4.3 shows some of his results, with straight asphalt, 1,050 ◦ C. Under these conditions, the conversion of the iron ore XB was 0.445. We find that straight asphalt in the composite was cracked within ca. 60 sec at high temperature, which is consistent with data given by previous investigators, [1,2]. Steady state experiments of simultaneous reaction for a composite of powdery iron ore and straight asphalt were carried out by Hasegawa [3], using a setup similar to Table 3.1 [C]. The mean size of the pellets was 0.77 mm, and the composite ratio was 0.2 gm oil/gm ore. To adjust the mean residence time tr of the pellets, 20 min for example, sands of 0.36 mm were fed into the fluidized bed together with the pellets continuously. After elapsed time, conversion of pellet (reduction) XB and components in effluent gas attained constant values at steady state. Temperature [◦ C] Conversion XB [–]

850 0.23

950 0.52

Composition of gas, at 950 ◦ C (carrier gas N2 deducted) Components [%]

H2 32

CO 29

H2 O 17

CO2 11

CH4 6

C2 H4 5

Total 100

Simultaneous reactions, namely thermal and auto-catalytic cracking of hydrocarbon and then reduction of iron ore take place inside the pellets.

52

4. Thermal Decomposition and Conversion of Composite Pellets

Figure 4.4 Experimental results (adopted from Katayama [4]). pCO = 0.35 atm.

4.6

REDUCTION OF COMPOSITE PELLET, FERRO-CHROMIUM ORE AND COKE

The reduction of composite pellets, which are composed of ferro-chromium ore and coke, is important as a preliminary process for the production of ferro-chromium. Katayama [4] used a thermo-balance similar to Fig. 3.1, and measured the reduction rate of a spherical pellet, the diameter of which was 8.6 mm and porosity 0.3. Chemical analysis of the pellet was as follows: Components mass [%]

Cr2 O3 37.4

Fe∗ 10.6

Si O2 7.6

Al2 O3 8.5

CaO 0

MgO 11.8

C 14.7

Ig.loss 3.7

∗ As FeO and Fe O 2 3

Three size ranges were chosen for the ore and coke, namely −150, −250, −325 mesh. Fig. 4.4 presents examples of his results. At 1,300 ◦ C, it was found that the grain size of both the ore and coke did not affect the reduction rate very much. Instead, concentrations of CO, CO2 and H2 have a considerable effect on the total conversion XB vs time curves. The existence of CO2 around the pellet deteriorates its conversion considerably. In practical operation, carbon is added to convert CO2 to CO, in accordance with the Boudourd’s reaction, i.e. Eq. (3.5.1). Example 4.1 Ultra pure particles of silica, 150 µm, are heated to eliminate a residual substance at high temperature. Calculate the rate constant kr∗ and predict the necessary time for heating to reduce its content down to 60 ppm.

Example 4.2

53

Data of silanol concentration t = 0, 300 ppm t = 10 hrs, 120 ppm Equilibrium with water vapour in flowing nitrogen 20 ppm

Solution In Eq. (4.1.1), (120 − 20)/300 = 0.357 (300 − 20)/300 We get, kr∗ = 0.1031/hr Thus, necessary time t is calculated by Eq. (4.1.1), (600 − 20)/300 = exp(−0.103t) (300 − 20)/300 t = 18.9 hrs

Example 4.2 Limestone is calcined in combustion gas at high temperature. When the temperature difference ∆T = Ts − Ti is kept constant, calculate the time necessary for its complete conversion τ . Data ρB =

2.9 gm/cm3 gm-mol B 100.1 gm/gm-mol = 0.029 cm3 cal kcal ks = 1.8 m hr ◦ C = 0.0050 cm sec ◦ C HB = 42.5 × 1,000 gm-cal mol B

Solution From Eq. (4.2.5), for dp [cm] and ∆T [◦ C], τ=

(0.029)(42.5 × 1,000)dp2 24(0.005)(∆T )

= 10,270

dp2 ∆T

The above equation gives the lines in Fig. 4.2. Under ordinary operating conditions, for example, dp = 3 cm and τ = 2 hrs, the temperature difference ∆T = Ts −Ti is found to be only 2 ◦ C. The above analysis suggests to us that the calcination of the limestone is definitely controlled by the rate of heat transfer. In practical calcination of limestone, the operating conditions are controlled so that its residence time is nearly the same as the time necessary for complete conversion τ . Suppose limestone of 2 cm size is calcined in a rotary reactor (kiln), where its residence time τ is 1 hr. From Fig. 4.2 we have ∆T = 12 ◦ C. This means it is necessary to supply thermal energy to the limestone particles so as to keep its surface temperature Ts 12 ◦ C higher than Ti . When some fraction of coarser size, 3 cm for example, is mixed in main solids of 2 cm, their surface temperature should be just the same under a given condition of heat supply. The necessary time for the coarser solid τ is estimated to be 2.3 hr from Fig. 4.2, longer than the residence time of solids, 1 hr. It is obvious that existence of coarse solids deteriorates the overall conversion of limestone in a rotary kiln.

Example 4.3 Limestone 2 cm in size is calcinated in a hot gas stream, containing 30% of CO2 . Estimate the temperature of the interface where decomposition takes place. Its residence time in the calcination zone is 1 hr. As the first approximation, assume 900 ◦ C.

54

4. Thermal Decomposition and Conversion of Composite Pellets

Solution The pressure of CO2 is (760 mmHg)(0.3) = 228 mmHg From Fig. 4.1, the equilibrium temperature is found to be 800 ◦ C. In the case of coarse limestone, the time necessary for complete conversion τ can be calculated from Eq. (3.2.9), taking into account CO2 diffusion through the product blanket as: τ=

ρB dp2

(4.6.1)

24bDAe ∆CA

Where ρB = 0.029 gm-mol B/cm3 , b = 1 gm-mol B/gm-mol A  cm2 273 + 900 1.5 DA = (0.142 cm2 /sec) = 1.26 273 sec Lime is porous so then let us assume εi = 0.1 in Eq. (3.4.5), DAe ∼ = (0.1)DA = 0.126

cm2 sec

Substituting to Eq. (4.6.1),  gm-mol B  (2.0 cm)2 0.029 cm3 3,600 sec =   2 gm-mol B 24 1 gm-mol A 0.126 cm sec (∆CA )

∆CA is calculated to be,

gm-mol A cm3 gm-mol 273 1 gm-mol (0.3) CA = = 3.12 × 10−6 273 + 900 22.4 × 103 cm3 cm3 ∆CA = 1.066 × 10−5

Thus the concentration at the interface should be, CA = 3.12 × 10−6 + 1.066 × 10−5 = 1.38 × 10−5

gm-mol A cm3

The concentration of gas at 1 atm = 760 mmHg and 900 ◦ C, gm-mol gas 1 gm-mol gas 273 · = 1.039 × 10−5 22.4 × 103 cm3 273 + 900 cm3 The pressure of CO2 at the reacting interface is given by (760 mmHg)

1.38 × 10−5 = 1,009 mmHg 1.039 × 10−5

From Fig. 4.1, we can estimate Ti = 910 ◦ C. It is confirmed that the first approximation is appropriate. Similarly, for dp = 3 cm and τ = 1.5 hrs, Ti = 930 ◦ C From Fig. 4.2, we get ∆T = 12 ◦ C for ∆T = 18 ◦ C for

dp = 2 cm, τ = 1 hr dp = 3 cm, τ = 1.5 hrs

The results are summarized in Table 4.2.

References

55 Table 4.2. Results of Example 4.3 dp [cm]

τ [hr]

Ti [◦ C]

∆T [◦ C]

Ts [◦ C]

2 3

1.0 1.5

910 930

12 18

922 948

REFERENCES [1] K. Junk, B.R. Stanmore, Fuel 59 (Feb. 1980) 74. [2] K. Hashimoto, et al., J. Soc. Fuel, Japan, 62 (1983) 421, 674. [3] Y. Hasegawa, et al., Chem. Eng. J. 18 (1979) 24; Kagaku Kogaku Ronbunshu, 7 (1981) 278, 285; 8 (1982) 604. [4] H. Katayama, Doctor Thesis (Tohoku University, Faculty of Engineering, 1984), No 702.

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–5– Conversion of Solids in Rotary Reactors On the basis of the kinetic models presented in the preceding chapters, equations are obtained to predict the overall conversion of solids and gaseous reactant, while solids are processed in a rotary reactor. When the gaseous reactant diffuses within the rotating layer of solids, the overall rate of solids conversion is found to be extremely slow on a practical scale. Injecting the gaseous reactant directly into the rotating layer of solids enhances the rate of conversion appreciably. Practical data on oxidation of residual carbon in spent catalyst, and activation of char are analysed quantitatively, and numerical values of reactor efficiency for both cases are acquired for further application.

5.1 5.1.1

CONVERSION OF GAS AND SOLIDS WITHIN SOLIDS LAYER Simplified model

In the light of solids movement in a rotary cylinder, let us assume that the solids are well mixed in its sectional area, and their temperature is uniform there. Fig. 5.1 shows a simplified model for conversion of solids by gaseous reactant, which diffuses from outside the bulk solids layer. In this chapter, it is convenient to use rate constant Kr , which is easily calculated from kc or kr (refer to Chapter 3, Eqs. (3.3.5) and (3.3.6)). Letting DAe be the effective diffusivity of gaseous reactant A in a layer of mean thickness lp , the mass balance of A within an infinitesimal volume (1 cm2 )(dx cm) is given by the following equation.     dCA d 2 CA dCA DAe − dx dt = dx(1 − εv )Kr CA + dt − DAe − dx dx dx 2 This gives DAe

d 2 CA = (1 − εv )Kr CA dx 2

(5.1.1)

where εv is the void fraction of the bulk solids layer, and Kr is the rate constant of solids, based on the unit volume of solid.

58

5. Conversion of Solids in Rotary Reactors

Figure 5.1 A model for conversion of solids within a bulk solids layer in a rotary reactor.

Eq. (5.1.1) is solved under the following boundary conditions, where Kd is the mass transfer coefficient across the boundary layer. x = 0 Kd (CAb − CAs ) = −DAe x=0

dCA dx

dCA =0 dx

The solutions are as follows: β e−αx + e−αlp · e−α(lp −x) CA = · CAb β + 1 1 + β−1 e−2αlp

(5.1.2)

CAs β 1 + e−2αlp = · CAb β + 1 1 + β−1 e−2αlp β+1

(5.1.3)

β+1

where α=



(1 − εv )Kr DAe

1/2

,

β=

Kd [(1 − εv )DAe Kr ]1/2

(5.1.4)

The average concentration is given by, CA 1 = CAb lp



lp

x=0

  β CA 1 1 − e−2αlp dx = · CAb (β + 1)α lp 1 + β−1 e−2αlp β+1

(5.1.5)

In case the value of αlp is large, (5.1.5) is simplified to, CA β 1 = · CAb (β + 1)α lp

(5.1.6)

5.1. Conversion of gas and solids within solids layer

59

Figure 5.2 Concentration profile and average concentration of reactant gas A in rotating layer of solids B, from Example 5.1.

In Example 5.1 on a rotary reactor to convert solids B, the concentration profile of gaseous reactant A within the rotating layer of solids is calculated with Eqs. (5.1.2), (5.1.3) and (5.1.4), giving Fig. 5.2 [A]. To calculate Kr , we take   kr = 8.10 1/(gm-mol A/cm3 )(sec)

At the intermediate stage of reaction, i.e. XB = 0.5, the average concentration of gaseous reactant is calculated, the results of which are shown in Fig. 5.2 [B]. Similar profiles are obtained for such reaction processes, as elimination of residual carbon in spent catalyst at 520 ◦ C, and reduction of powdery ferrite by hydrogen at 530 ◦ C (see Kunii [1]). Fig. 5.2 [B] clearly shows that the mass transfer coefficient of the boundary layer does not appreciably affect the average concentration of gaseous reactant, in the case where lp is larger than 8 cm. The conversion rate of solids per unit volume of bulk solids is given by,   gm-mol B b(1 − εv )Kr C A (5.1.7) (cm3 solid)(sec) 5.1.2

Effect of layer thickness on time necessary for conversion of solids

To test the conversion of solids, a small rotary reactor is sometimes applied. Suppose a batch of char is activated with pure steam in it. The experimental data are given in Fig. 5.3 and Table 5.1. In the case where the rate constant for single solids is taken to be 8.1 [1/(gm-mol A/ cm3 gas)(sec)], let us estimate the average concentration of gaseous reactant A in the layer of solids, the average thickness of which is 2.5 cm.

60

5. Conversion of Solids in Rotary Reactors

Figure 5.3 Activation of char from coconut shell in a small rotary reactor. Table 5.1. Quality of active carbon, char from coconut shell, at conversion XB = 0.5 Mean size 1.2 mm

Residence time [hr] Inner surface area [m2 /gm] Acetone adsorbed [%]

Temperature [◦ C] 900

950

2.3 2,080 44

2.0 2,080 –

From Fig. 5.2, we obtain C A /CAb = 0.55 for Kd = 4 cm/sec We find such data from a small rotary reactor are still affected by the diffusional transport of gaseous reactant through the thin layer of solids. In a commercial rotary reactor of 2 m ID, suppose the average thickness of the rotating solids layer to be 25 cm, and the average concentration is calculated by Eq. (5.1.6). The numerical values of α and β are given in Example 5.1 to be 0.5379 cm−1 and 5.312. With Eq. (5.1.6), CA 5.312 1 = · = 0.0626 CAb (5.312 + 1) (0.5379)(25)

5.2. Enhancement of contact by sending gaseous reactant into a rotating layer of solids

61

The above calculation suggests that a large reactor needs a far longer residence time than a small one. Suppose solids of 1.2 mm at 900 ◦ C, the necessary residence time is 2.3 hr. For a large reactor, (2.3 hrs)

0.550 = 20.2 hrs 0.0626

A long residence time needs a large volume of reactor. To remedy this disadvantage, gaseous reactant should be sent into the rotating layer of solids to enhance gas–solid contact. The above discussion is based on the intermediate value of the rate constant Kr [1/sec]. In case Kr is large enough, the lines in Fig. 5.2 fall down sharply in the vicinity of the layer surface, and then conversion is restricted within a thin layer. On the contrary, for a small value of Kr , the lines in Fig. 5.2 penetrate deeper from the surface of the solid layer, so that the effect of layer thickness on the average concentration of gaseous reactant is considerably reduced.

5.2

5.2.1

ENHANCEMENT OF CONTACT BY SENDING GASEOUS REACTANT INTO A ROTATING LAYER OF SOLIDS

Simplified model

As mentioned previously, the major disadvantage of rotary reactors for gas solid processes is due to the diffusional resistance of the gaseous reactant within the rotating layer of bulk solids. In order to improve the performance characteristics of the rotary reactor, the gaseous reactant is sometimes injected into the rotating layer of solids. Let us focus our attention on a rotary reactor, illustrated in Fig. 5.4, in which rotating tube tuyeres are positioned. Gaseous reactant percolates through the rotating layer of bulk solids, and contacts with solids. When solids are coarse, the flow pattern of the gaseous reactant must be similar to an ordinary fixed bed reactor. For quantitative discussion, let us introduce a reactor efficiency ηr by the following definition. ηr =

Conversion rate in reactor, unit volume Conversion rate of single solids, unit volume

(5.2.1)

The numerical value of ηr is dependent upon the shape of the solids, mean size, range of size distribution, design of tuyere or distributor and then the rate constant of chemical reaction, Kr . In case of very slow reaction, ηr will be near unity. Under ordinary operating conditions of gas solid reactors, ηr is less than unity. Calculations in Examples 5.2 and 5.3 give practical data for it. When fine solids are charged, they are apt to partially fluidize and the surface of the rotating bulk solids is flattened. The reactor efficiency ηr can be predicted numerically if the operating conditions are known (refer to Kunii and Levenspiel [2]).

62

5. Conversion of Solids in Rotary Reactors

Figure 5.4 Flow pattern of gas from rotating tube tuyeres.

5.2.2

Conversions of solids, calculated from rate constant Kr for gaseous reactant

Let lp and u0 be the average depth of the solids layer and the average superficial gas velocity, respectively. By application of the rate constant Kr [(cm3 gas/cm3 solid)/sec], and distance from the inlet of gas y [cm], we have −u0 dCA = (1 − εv )Kr ηr CA dy

(5.2.2)

Integration gives,   CA (1 − εv )Kr ηr = exp − y CAi u0

(5.2.3)

The average value of CA throughout the depth lp is,   (1 − εv )Kr ηr lp CA u0 = 1 − exp − CAi (1 − εv )Kr ηr lp u0  CAe CAi = 1− ln CAi CAe

(5.2.4)

where   (1 − εv )Kr ηr lp CAe = exp − CAi u0 The volume of bulk solids in a rotary reactor is given by, 

π 2 dt lr γ 4

(5.2.5)

5.2. Enhancement of contact by sending gaseous reactant into a rotating layer of solids

63

where lr is the length of the solids layer, and γ is the volumetric fraction of the bulk solids. Approximately, let us define an effective width of the solids layer by le . 

π 2 d γ = le l p 4 t

(5.2.6)

The conversion rate of the gaseous reactant is given by   (1 − εv )Kr ηr lp (le lr )u0 (CAi − CAe ) = (le lr )u0 CAi 1 − exp − u0

(5.2.7)

Using the mass flow rate of solid reactant FB , the conversion of solids XB is calculated as   (1 − εv )Kr ηr lp FB XB = b(le lr )u0 CAi 1 − exp − u0

(5.2.8)

Combination of Eqs. (5.2.4) and (5.2.8) results in Eq. (5.2.9). (5.2.9)

FB XB = b(le lr lp )(1 − εv )Kr ηr C A 5.2.3

Conversion of solids, calculated from rate constant kr

The fundamental equation for conversion of solids, Eq. (3.2.3) is modified to apply in a rotary reactor as follows: 1 − XB = exp[−kr ηr C A tr ]

(5.2.10)

where C A is calculated by Eq. (5.2.4), and tr is the residence time of solids in the reaction region. tr =

(le lp lr )(1 − εv )ρB FB

(5.2.11)

Eq. (3.3.6) gives the relation between kr and Kr . Kr =

ρB (1 − XB ) kr b

(3.3.6)

Let us use a time-averaged value of Kr . By application of Eq. (5.2.10), we obtain, Kr =

ρ B kr btr



tr

t=0

(1 − XB )dt =

ρB bC A ηr tr



 1 − exp(−kr C A ηr tr )

(5.2.12)

Substitution of Eq. (5.2.11) and (5.2.12) into Eq. (5.2.10) gives an identical equation to Eq. (5.2.9), in the case where K r = Kr .

64

5. Conversion of Solids in Rotary Reactors

5.2.4

Different devices

A high value of the reactor efficiency ηr can be attained by other devices shown in Table 1.3 [A], namely rotating distributor. In practice, this is a rotary moving bed for coarse solids, or a rotary fluidized bed for fines. On the other hand, such tube tuyeres of limited number as shown in Table 1.2, [B] must give a smaller value of ηr than the others. [C] in Table 1.2 gives good contact between gas and solids. Gaseous reactant percolates through the rotating layer of solids, achieving cross flow contact. The value of ηr in this case is supposed to be close to that of a fixed bed or a fluidized bed. 5.2.5

Rotary sealing of distribution manifold

Looking at Fig. 5.4, it is needed to send the gaseous reactant selectively to tubes which are just immersed in the rotating layer of solids. A rotary distribution manifold is usually applied in industrial rotary reactors. Because considerable pressure drop is required to send the gaseous reactant uniformly into the rotating layer of solids, appropriate rotary sealing is important to prevent possible leakage of the gaseous reactant. The design and development of a satisfactory rotary manifold, which can completely prevent leakage of reactant gas, is of the utmost importance.

5.3

5.3.1

HIGH TEMPERATURE STABILITY OF ISOLATED SOLIDS IN EXOTHERMIC REACTION

Volumetric fraction of falling solids

In rotary driers, lifters are usually positioned to enhance the rate of heat transfer. At first glance, lifters appear to enhance contacting. Kunii analysed the movement of solids, falling down across the edges of the lifters under ordinary operating conditions. It is concluded that the volumetric fraction of the falling solids is nearly 1% of the total solids in the reactor. In the case of endothermic reaction, the small fraction of solids does not appreciably enhance the total rate of conversion. 5.3.2

High temperature stability of falling solids

When small solids fall in a reactor, where an exothermic reaction takes place between gas and solids, the solids temperature rises rapidly to its stable value, which can be predicted with Eq. (3.4.2). Once the above phenomenon occurs in the rotary reactor, the overall rate of conversion increases enormously, even the volumetric fraction of solids is as small as 1%. If only the rate of conversion is required to increase, and we do not care about the temperature rise of the particles, lifters could be designed in practice. However, the above phenomenon should be avoided in such a case as spent catalyst, so as not to deteriorate its quality.

5.3. High temperature stability of isolated solids in exothermic reaction

65

A similar calculation to Example 3.2 is carried out, in the case where a single graphite particle of 1.14 mm is falling in the air of 800 ◦ C. The stable temperature is found as follows: O2 in gas Stable temperature

[%] [◦ C]

10 950

21 (air) 1,460

The deterioration of solids due to high temperature can be prevented when the concentration of O2 is reduced to 10%. When we are dealing with carbonaceous solids in hot air, it should be kept in mind that particles dispersed in the air flow are all ready to ignite and raise their temperature to white hot (see Ref. [3]). 5.3.3

High temperature near nozzles of injection gas

High temperature stability of carbonaceous solids occurs not only in their dispersion state, but also in the vicinity of the gas nozzles. If solids stay at the inlet for a while during a rotation of the reactor, they must be in a position to ignite similarly. High temperature due to such a local ignition is not too easy to detect. However, its occurrence may result in an appreciable deterioration of quality for the solids. Example 5.1 Dense grains of solids B are converted by gaseous reactant A in an experimental reactor. Suppose the rate constant is obtained from the data to be 8.1 [1/(gm-mol A/ cm3 gas)(sec)] at 900 ◦ C. A rotary reactor is designed for the production of half converted solids. Predict the concentration profile of the gaseous reactant A within the rotating layer of bulk solids, the average thickness lp of which is given to be 20 cm. Next, calculate the relation between the average concentration of the gaseous reactant and the average thickness of the layer. Data ρB = 1.6

gm B -mol B = 0.1333 gmcm 3 cm3 2 cm DAe = 1.40 sec εv = 0.5

The mass transfer coefficient is estimated as, Kd = 4

cm3 gas (cm2 solid)(sec)

=4

cm sec

and

∞

The average conversion of the solids. XB = (0 + 0.5)/2 = 0.25 Solution From Eq. (3.3.6),

Kr =



0.1333

  gm-mol B  (1 − 0.25) cm3 gas 1 1 cm3 solid = 0.810 = 0.810 × 8.1 gm-mol A 3 solid)(sec) gm-mol B sec (cm ·sec 1 gm-mol A cm3 gas

With Eqs. (5.1.4) and (5.1.5) we get, α=



 1 (1 − 0.5)(0.810) 1/2 = 0.5379 1.40 cm

66

5. Conversion of Solids in Rotary Reactors

β=

4 = 5.312 [–] [(1 − 0.5)(1.40)(0.810)]1/2

e−αlp = e−0.5312×20 = 2.44 × 10−5 ≪ 1 In the case Kd = 4 cm/sec, Eqs. (5.1.2) and (5.1.3) give CA = 0.8416e−0.5379x CAb   1 − e−1.076lp CA 1.565 = −1.076l p CAb lp 1 + 0.6828e In the case Kd = ∞, similarly we get CA 1.859 1 − e−1.076lp = · CAb lp 1 + e−1.076lp The above equations give Fig. 5.2.

Remarks The numerical value of Kd is estimated on the basis of the following assumption. The laminar boundary layer of length lb [cm] is renewed by turbulence from the main gas flow, the average velocity of which is u0 [cm/sec].  ρg u0 lb 1/2 Kd lb = 0.664Sc1/3 DA µ gm Letting DA = 2.8 cm2 /sec, Sc = 0.7, ρg = 3.01×10−4 cm 3 , lb = 25 cm, u0 = 2.4 m/sec = −4 240 cm/sec, µ = 5 × 10 gm/cm·sec, we calculate as Kd = 4.0 cm/sec.

Example 5.2 Spent catalyst from a petroleum refinery contains 5% carbon, which should be eliminated for re-utilization of the carrier solids. A rotary retort is heated by combustion gas from kerosene burners. The surplus amount of air is injected into the rotating layer of solids, through tube tuyeres as seen in Table 1.2 [C]. A rotary manifold is positioned in the retort. Performance data are given as follows: Inner diameter of retort 0.9 m Its total length 10 m Product 250 kg/hr 20 min 520 ◦ C Product 150 kg/hr 20 min 570 ◦ C

residual carbon 0.88% residual carbon 0.12%

Estimate the numerical values of reactor efficiency ηr in this system. Predict the residual carbon when the temperature is raised to 620 ◦ C and the residence time is reduced to 15 min. Stoichiometric air to oxidize the residual carbon is given as 0.505 Nm3 /kg pr, whereas surplus air is sent into the rotating solids layer. Thus the average fraction of oxygen in the percolating air is given to be 18.8%.

Example 5.2

67

Solution The average concentration of oxygen A at 520 ◦ C is CA =

gm-mol A 1 gm-mol A 273 (0.188) = 2.89 × 10−6 273 + 520 22.4 × 103 cm3 cm3

1 − XB = (0.88%)/(5%) = 0.176 In Eq. (5.2.10), tr = 20 × 60 = 1,200 sec. ln 0.176 = −(kr )(2.89 × 10−6 )(ηr )(20 × 60)    gm-mol A kr ηr = 501 1 (sec) cm3 The numerical values of kr can be found in Chapter 3, Table 3.4 as 550 ◦ C

kr = 1,093

570 ◦ C

kr = 1,770

To estimate kr at different temperatures, let us utilize rate constant kc from Table 3.3 for graphite cylinder. Dependency on temperature is assumed to be the same in both constants, kc and kr ; then we get the following numerical values.

Temperature kc ratio of kc

[◦ C] [cm/sec] [–] 

1  gm-mol A  (sec) 3

kr

cm



520 0.05 0.667

550 0.075 1

570 0.12 1.60

600 0.20 2.67

620 0.28 3.73

729

1,093

1,749

2,918

4,081

Reactor efficiency is ηr = 501/729 = 0.687 = 69% A similar calculation gives ηr at 150 kg pr/hr and 570 ◦ C, ηr = 0.645 = 65% If the reaction temperature can be raised to 620 ◦ C, the capacity of the reactor increases to 333 kg pr/hr, and the residual carbon is reduced to 0.011% in accordance with the following calculation. tr = (20 min)(250 kg/hr)/(333 kg/hr) = 15 min   1 − XB = exp −(4,081)(2.566 × 10−6 )(0.65)(15 × 60) = 0.00219

Residual carbon is (0.00219)(5%) = 0.011%.

Example 5.3 In a commercial reactor to produce active carbon, the volume of rotating solids is approximately represented by le = 1.2 m, lp = 0.25 m and lr = 12 m. A number of tube tuyeres are positioned to send steam into the rotating layer of solids. Estimate the numerical value of reactor efficiency ηr in this case. Data Feed rate of carbon FB = 400 kg/hr = 9.259 gm-mol B/sec Active carbon 235 kg/hr XB = 1 − 235/400 = 0.413 Temperature 900 ◦ C

68

5. Conversion of Solids in Rotary Reactors

Steam 720 kg/hr Void ratio 0.50 Rate constant of activation kr = 16.1

1

 gm-mol A 

gm B/cm3

cm3

(sec)

Density of solids ρB = 1.6 = 0.1333 gm-mol B/cm3 Mean value of XB = 0.413/2 = 0.207 Solution Eq. (3.3.6) gives, ρB (1 − XB ) (0.1333)(1 − 0.207) 1 kr = (16.1) = 1.702 b 1 sec 720 kg 22.4 Nm3 900 + 273  m cm · · (1.2 m × 12 m) = 0.0743 = 7.43 u0 = 3,600 sec 18 kg 273 sec sec

Kr =

with Eq. (5.2.5),   CAe (1 − 0.5)(1.702)ηr (25) = exp − = exp(−2.863ηr ) CAi 7.43 CAi =

gm-mol A 273 1 gm-mol A · = 1.039 × 10−5 22.4 × 103 cm3 273 + 900 cm3

Conversion of the gaseous reactant is calculated by the following equation.    gm-mol A CAe cm (120 cm)(1,200 cm) 1.039 × 10−5 1 − 7.43 sec CAi cm3  C gm-mol A = 11.12 1 − Ae CAi sec On the other hand, conversion of solids is  gm  (400 − 235) 103 kg gm-mol B gm-mol B · = 3.819 3,600 sec 12gm B sec Since b = 1 gm-mol B/gm-mol A,  C 11.12 1 − Ae = 3.819 CAi Thus, CAe /CAi = 0.6566 = exp(−2.863ηr ) ηr = 0.1469

REFERENCES [1] D. Kunii, Chemical Engineering, Japan, No. 10, (1993) p. 72. [2] D. Kunii, O. Levenspiel, Fluidization Engineering, 2nd edition (Butterworth-Heinemann, Stoneham, USA, 1991). [3] M. Aoyaki, D. Kunii, Chem. Eng. Comm. 1 (1974) 191.

–6– Heat Transfer in a Rotary Reactor, Direct Heating The rate of heat transfer sometimes controls the overall rate of conversion in a rotary reactor, in which solids are heated directly by flowing gas. Heat evolution by combustion of fuel is briefly discussed from the hydrodynamic point of view. In a rotary reactor operating at high temperature, radiant heat transfer from burning flame and/or combustion gas prevails. Heat transfer mechanisms, including thermal conduction from the inner surface of the reactor to the rotating solids, are analysed quantitatively, giving several equations and graphs for application to practical design calculation.

6.1 COMBUSTION OF FUELS 6.1.1

Combustion model of a gas burner

The heat necessary for thermal processing at high temperature in a rotary reactor is usually supplied by combustion of fuel, applying a suitable burner. The optimum fuel is selected from gas, liquid, and pulverized solid fuel mainly based on economic optimization, in accordance with its location and then environmental requirements. The combustion of fuel from the burner is substantially controlled by turbulent (eddy) mixing and diffusion of combustible gas with the air stream. Fig. 6.1 [A] is a simplified model for combustion of fuel gas in a rotary reactor. Pure fuel gas or gas premixed with primary air issues from a nozzle, the inner diameter of which is dtn . A jet stream of combustible gas has a far higher velocity than the secondary air, and a large discontinuity of linear velocities takes place at the boundary of the two gas flows. According to hydrodynamic fundamentals, a sheet of intense vortexes are generated along the discontinuous interface. While they flow downstream, they vary their size and intensity, and exchange between gas and air flow regions. Since the core part of each vortex is conserved during its movement, an intense exchange of gas and air takes place between two flows. This type of transport phenomenon is called turbulent diffusion. At the interface between two kinds of vortexes the molecular diffusion of the gaseous reactant is very rapid, leading to an instantaneous reaction. This means that the overall rate of combustion is controlled by the above hydrodynamic phenomenon, i.e. turbulent diffusion. In the case of Fig. 6.1 [A], the length of the flame lf can be estimated as shown in Table 6.1, summarized from experimental and theoretical studies by Hottel [1] and Yagi [2].

70

6. Heat Transfer in a Rotary Reactor, Direct Heating

Figure 6.1 Models of combustion with burners.

When the fuel gas contains higher hydrocarbons like C3 H8 , and is burnt in a flame without premixed air, the molecules of such species are cracked thermally, forming soots and combustible gas of lower molecular weight. We call this a luminous flame. Premixing of the air into the fuel gas makes the flame front attach to the edge of the burner nozzle, forming a so-called blue cone. Here propagation of the flame front is balanced with the issuing flow of gas and air mixture. Combustible gas, passed through the flame front attached to the burner nozzle, is chemically transformed to combustible gas of lower molecules, and is burnt with the secondary air, by the similar turbulent diffusion mechanism to pure combustible gas.

6.1. Combustion of fuels

71 Table 6.1. Ratio of turbulent flame length to nozzle diameter

Fuel

Premix air [%]

Gas

H2 CO C2 H2 C3 H8

Fuel gas 4,150 kcal/Nm3

Pulverized coal 6,860 kcal/kg

lf /dtn

Exp. data

0 0 0 0

125 41 168 280

Hottel [1] dtn = 3.2–6.4mm

0 10 20 30 40 50

68 62 57 54 56 47

Yagi [2] dtn = 1 ∼ 10 mm

26 21

70 77

Yagi [2]

Thus, premixing of air makes the gas flame shorter and more intense (see Table 6.1). However, too much premixing of primary air makes the flame rather unstable and sometimes leads to a dangerous phenomenon such as blow off. The rate of the turbulent diffusion is moderate in the set-up illustrated in Fig. 6.1 [A], which gives a long flame. When a shorter and more intense flame is required for a specific process, high speed whirling of the secondary air around the burner nozzle is effective, in addition to its premixing. The above discussion is restricted to the complete combustion of fuel gas. A flame such as in Fig. 6.1 [A] can be established even when the total air is less than the stoichiometrical necessary amount. In this case, oxygen in the air flow is completely consumed and the gas flow after the flame contains combustible gas components such as H2 , CO, CH4 and soots. We call this partial combustion or oxidation, and apply it for the production of lean fuel gas. If we use a mixture of steam and oxygen in place of the air, we can produce H2 and CO to use in succeeding processes. 6.1.2

Liquid fuel

In the case where kerosene is atomized in Fig. 6.1 [B], droplets are vaporized almost instantaneously and the mechanism of its combustion is then the same as the fuel gas. When heavy oil is atomized to form droplets ranging from 10 ∼ 100 µm in Fig. 6.1 [B], they are heated by thermal radiation and by mixing with hot gas, and their vaporization and carbonization then take place in succession. Combustible gas from fine droplets mixes with primary air and the ignition of the mixture makes the starting region of the flame. Inside the flame, coarser droplets issue combustible gas, so that the overall rate of combustion is controlled by the turbulent diffusion, similarly to the gas flame. In Fig. 6.1 [B], each droplet remains a one order of magnitude smaller particle of char. A cloud of such fine char behaves just like a combustible gas and the controlling step in such a flame is still turbulent diffusion.

72

6.1.3

6. Heat Transfer in a Rotary Reactor, Direct Heating

Pulverized coal and coke

Fig. 6.1 [C] shows a model for a pulverized coal burner. The usual size of the coal ranges from 20 ∼ 100 µm. When fine coal particles are injected from the nozzle into a hot region, 5 ∼ 10 milliseconds is needed to heat up and carbonize them, from which combustible gas issues. Since the linear velocity of the air–solids mixture is high, considerable length is needed until the combustible gas is ignited with the primary air. This portion is called a black cone. For stable combustion of pulverized coal, ignition of the combustible gas is crucial. After the region of ignition, the turbulent diffusion of the mixture, composed of combustible gas and small char particles, to the secondary air flow controls the overall rate of combustion. Experimental results of flame length are compared with fuel gas in Table 6.1, suggesting that the controlling step is nearly the same for two cases, namely ordinary fuel gas and pulverized coal. Coarse char particles drop onto the rotating solids to be processed and some of intermediate size must be entrained by the combustion gas, flowing out of the flame region. Sherman [3] carried out combustion of pulverized Pocahontas coal, transported by all the necessary air for combustion (primary air only). At 1,320 ◦ C, nearly 180 milliseconds was needed to achieve an unburnt ratio of coal of 10%. To predict the time necessary for complete conversion of coarse char, Eqs. (3.2.11) and (3.2.15) can be applied (see Example 6.1). In the case where solids contain little volatile matter, we have to add some amount of ordinary fuel in order to ensure stable ignition of the mixture flow. Combustion of fuel is one of the established fields in industrial technologies and there are a number of specialists working on their specialised subjects. For selecting the best burner to position in a novel reactor, we should consult such specialists and learn to use the necessary know-how. In this case, however, we should determine the specifications of the burners as precisely as possible to meet our requirements for the planned process. In addition, we should require them to supply burners with a reliable ultra violet or photo sensor, so that we are confident of the stable and safe operation of the burners. 6.1.4

Inside combustion and reverse flame

In some rotary reactors, combustible gas issues from the rotating layer of solids and is burnt by the air, which is sent from outside the reactor. Let us call this “inside combustion”. Usually, the amount of air is less than stoichiometrically needed for combustion, so that the effluent gas contains H2 , CO and other combustible components. In other words, partial combustion (oxidation) of issued gas takes place inside the reactor. Fig. 6.2 illustrates simplified models for such a reactor: [A] is for a conventional one to produce active carbon; [B] is a reactor developed by the authors for activation of larch chips; [C] shows the combustion of CO from the rotating layer of composite, made of ferro-chromium ore and coke. The authors have developed a rotary reactor to gasify solid wastes, which is shown in [D]. In Fig. 6.2 [A], [B] and [D], the air is injected into combustible gas at high temperature. The concentration profile for the gaseous reactants is just the opposite to ordinary flame, which is formed by injection of combustible gas into surplus air flow. From a hydrodynamic point of view, the mechanism of turbulent diffusion between two flows of gases is similar in the above two cases. Therefore, a similar turbulent diffusional flame is formed,

6.1. Combustion of fuels

73

Figure 6.2 Examples of inside combustion.

the rate of combustion of which is controlled by mixing of vortexes (eddies) between two flows of gases. Let us call this a reverse flame. For stable operation of the above reverse flame, the following conditions are required. • The combustion chamber should be heated to 800 ◦ C, before starting the reverse flame. • Reliable devices for ignition and detection of flame should be prepared.

74

6. Heat Transfer in a Rotary Reactor, Direct Heating Table 6.2. Examples of loading capacity in the combustion region

Combustion

Purpose

kcal m3 hr

Burner Burner Burner in Combustor Burner in Combustor Burner & inside combustion Inside combustion Inside combustion

Rotary kiln, limestone RK for sintering of CaO for de-lacquering of cans for fluid bed reactor Reduction of pellets Activation of char gasification of solid waste

200,000 ∼ 300,000 300,000 800,000 1,000,000 50,000 15,000 100, 000 ∼ 200,000

6.1.5

Volume of combustion region

Suppose a process requires us to determine the volume of the combustion region (chamber) as its first approximation. In this case, we had better refer to the following practical criteria. [Loading capacity in combustion region] =

(Heat evolved by combustion) (Volume of combustion region)

(6.1.1)

Given the aspect ratio of the region (chamber), we can determine its dimension.

6.2 TEMPERATURE PROFILE IN TURBULENT FLAME Fig. 6.3 gives a model to estimate the temperature profile within a burning flame. The ratio of unburnt fuel is defined as Yf and κf is the rate constant of combustion. Yf = exp(−κf x)

(6.2.1)

The heat evolved in an infinitesimal section, the thickness of which is dx, is given by Ff QN



dYf dx = Ff QN κf exp(−κf x) − dx

where: Cg : Ff : Gg :

specific heat of gas feed rate of fuel volume of gas

[kcal/Nm3 ◦ C] [kg/hr] [Nm3 /kg]

(6.2.2)

6.2. Temperature profile in turbulent flame

Figure 6.3

hrg : QN : Tg : T ∗:

75

Model to estimate temperature profile in flame.

radiant heat transfer coefficient, per unit surface area of inner surface net calorific value temperature of gas average temperature of wall and solids

[kcal/m2 hr ◦ C] [kcal/kg] [ ◦ C] [ ◦ C]

The heat balance leads to Eq. (6.2.3).   Ff QN κf exp(−κf x)dx = Ff Gg Cg dTg + (πdt dx)(hrg ) Tg − T ∗

(6.2.3)

The boundary condition is, x = 0,

Tg = Tgi

Integration results in, Tg = T ∗ +



   a2 a2 − T ∗ − Tgi exp(−a1 x) − exp(−κf x) κf − a1 κf − a1

(6.2.4)

76

6. Heat Transfer in a Rotary Reactor, Direct Heating

Figure 6.4 Temperature profile of flame, from Example 6.2.

where a1 =

πdt hrg , Ff Gg Cg

a2 =

QN κf Gg Cg

(6.2.5)

In Example 6.2, practical operation data in a rotary reactor (kiln) to calcinate limestone 9,600 kg pr/hr = 230 tons pr/day (dti = 2.4 m, 40 m in length) are analysed on the basis of above equations. Fig. 6.4 shows the results of the calculation, where we estimate the rate constant of flame combustion κf is ca. 0.3 m−1 , and the flame length is approximately 15 m in the kiln.

6.3 6.3.1

HEAT TRANSFER IN A ROTARY REACTOR AT HIGH TEMPERATURE

Radiant heat transfer from flame and combustion gas

In the region where combustion takes place, radiant heat transfer from the burning flame is controlling, whereas convectional heat transfer is one order of magnitude less. For conve-

6.3. Heat transfer in a rotary reactor at high temperature

77

Figure 6.5 Model of heat transfer mechanism in the combustion and heating region.

nient application to practical design calculation, the authors propose a simplified model to represent the heat transfer mechanism in this region. In Fig. 6.5, radiant heat (mainly infra-red ray) is emitted from the burning flame at high temperature, some fraction of which arrives at the surface of the rotating solids layer, and the other arrives at the inner refractory surface. The rest is absorbed in the combustion gas around the flame. For predicting radiant heat transfer, sophisticated equations – so called “Enclosure theory” – have so far been presented (see Refs. [4–7]). In this book, however, radiant heat transfer is simplified to apply to the complex mechanism of heat transfer in a rotary reactor, without losing significant fundamentals. Radiant heat absorbed in combustion gas around the flame is converted to thermal energy, which should be emitted as infra-red ray from the gas to the rotating solids and inner wall surface. Thus we can assume that all the radiant heat emitted from the flame arrives at the surface of the rotating solids and the inner wall surface. The radiant heat transfer coefficient, hrg , to two solid surfaces is calculated by,

hrg =

 Tf +273 4 df dti (εf εm )(4.88) 100 Tf − T ∗

−

 T ∗ +273 4  100

[kcal/m2 hr ◦ C]

(6.3.1)

where df is the outer diameter of the flame, εf is the emissivity of the flame, εm is the average emissivity of the solids layer surface and the inner wall surface, and T ∗ is the average temperature of the above two surfaces. The emissivity of the luminous flame was measured very precisely by members of an international committee. They employed a large experimental furnace with a square section of 1 m × 1 m and 6.5 m in length, located in

78

6. Heat Transfer in a Rotary Reactor, Direct Heating Table 6.3. Emissivity of luminous flame εf , fuel oil (Ref. [8])

Condition

Distance from burner [m]

Mean size of droplet [µm]

60 28

1

2

3

4

5

0.85 0.80

0.9 0.88

0.72 0.68

0.50 0.45

0.42 0.38

Imijden (Ref. [8]). Table 6.3 gives examples of their results indicating scarce effects from mean size of oil droplets. Other results show that the measured value of εf decreases by nearly 0.1 when atomizing air is increased so much. Atomization by steam in place of the air reduces εf by 0.05. On the other hand, increasing the feed rate of fuel has practically no effect on εf . Provided the flame length is 6.5 m, we could take the average value of εf in Table 6.3 to be approximately 0.7. The average diameter of flame df depends strongly on the hydrodynamic feature of turbulent diffusion. By visual observation of the flame in a practical rotary reactor, in which a burner is operated under similar flow conditions to the one planned, we can estimate the approximate value of df /dt . From the end of the flame, combustion gas flows in the reactor. In this region, Eq. (6.3.2) should be applied.  T +273 4  T ∗ +273 4  − 100 (εg εm )(4.88) g100 hrg = ∗ Tg − T

[kcal/m2 hr ◦ C]

(6.3.2)

The emissivity of combustion gas εg is obtained by a procedure established by Hottel [5,6]. The calculation for combustion gas in an ordinary rotary kiln gives 0.1–0.3, depending on its scale (refer to the above text book on radiant heat transfer). In Example 6.2, the radiant heat transfer coefficient is calculated at normal conditions for a rotary reactor to calcinate limestone (rotary kiln), dt = 2.5 m. The results are presented in Fig. 6.6. 6.3.2 Radiant heat transfer from inner wall surface to surface of rotating solids layer

In Fig. 6.5, the fraction of inner surface area χ , to which rotating solids contact, is needed for further calculation. By simple calculation, χ versus the volumetric fraction of solids γ is given in Fig. 6.7. Letting TH and TC be the temperatures of the inner wall surface and the rotating solids, respectively, we calculate the radiant heat transfer coefficient from the inner surface to the rotating solids layer. Radiant heat emitted from the wall surface per unit length of the reactor is given by,  TH + 273 4 (πdti )(1 − χ)εH (4.88) 100

6.3. Heat transfer in a rotary reactor at high temperature

79

Figure 6.6 Radiant heat transfer coefficient in a rotary reactor, dt = 2.5 m.

The geometrical view (angle) factor from the inner surface πdti (1 − χ) to the rotating layer of solids is represented by FHC . πdti (1 − χ)FHC = πdti χFCH χ . Since FCH = 1, we have FHC = 1−χ Radiant heat, emitted from the inner surface, is mainly infrared ray. When it passes through the flame and combustion gas, some part of the infrared ray is absorbed by them. Let us take εg∗ to be the average value of emissivity for the flame and combustion gas. Thus, the rate of radiant heat transfer from the inner wall surface to the rotating solids layer is calculated approximately with Eq. (6.3.3), in the case where εH and εC are close to unity.

   TH + 273 4 TC + 273 4 − πdti (1 − χ)(εH εC )(4.88) FHC (1 − εg∗ ) 100 100

(6.3.3)

80

6. Heat Transfer in a Rotary Reactor, Direct Heating

Figure 6.7

Fraction of wall surface area which contacts rotating solids.

We define the radiant heat transfer coefficient from the hot inner wall to the layer of solids by (hrs )HC , on the basis of the surface area of the inner wall, to which the rotating solids contact. (6.3.4)

πdti χ(hrs )HC (TH − TC ) Combination of Eqs. (6.3.3) and (6.3.4) leads to the following equation,

(hrs )HC =

 +273 4  TC +273 4  εH (1 − εg∗ )εC (4.88) TH100 − 100 T H − TC

[kcal/m2 hr ◦ C] (6.3.5)

6.3.3 Heat transfer coefficient by direct contacting of solids from the hot wall surface

In a rotary reactor, solids are heated by direct contact with the hot wall surface. The inner wall surface functions as a kind of regenerator, changing the surface temperature periodically during the rotation. Theoretical calculation reveals that the amplitude in the periodical change of surface temperature is not too much, as long as the rotation is larger than 2 r.p.m. In this section, let us take the time averaged temperature of the wall. For a packet of solids, which suddenly contact the hot surface and then leave it after residing there for a time t , Kunii and Levenspiel [9] give the following equation to calculate

6.3. Heat transfer in a rotary reactor at high temperature

81

the time averaged value of the heat transfer coefficient due to the above contact, on the basis of contacting surface area. 

¯ s ke ρC (hp )HC = hpacket = 1.13 t

1/2

(6.3.6)

where ke is the effective thermal conductivity of a packet of solids, ρ¯ is the bulk density, and Cs is the specific heat of solids. (6.3.7)

t = χ/N Substitution of Eq. (6.3.7) to (6.3.6) gives (hp )HC = 1.13



ke ρC ¯ sN χ

1/2

(6.3.8)

Effective thermal conductivity has been studied by many research groups. For design calculation, it is convenient to apply the theoretical equation by Kunii and Smith [10]. 6.3.4

Temperature of the inner wall surface

In Fig. 6.5, the temperature of the inner wall surface TH is determined by a given value of the flame temperature Tf . On the basis of unit length, we have the heat that the inner wall surface receives, (πdti )(1 − χ)hrg (Tf − TH )

(6.3.9)

Heat transfer from the inner wall to the rotating solids by thermal radiation is given by Eq. (6.3.4). Heat transfer from the inner wall to the rotating solids by direct contact (πdti )χ(hp )HC (TH − TC )

(6.3.10)

Heat loss to outside, π(d¯t )

ks (TH − TW ) = (πdt0 )qW = π(dt0 )(hc + hr )W (TW − T0 ) lW

(6.3.11)

Heat balance at the inner surface of the reactor gives the following equation.   d¯t ks (1 − χ)hrg (Tf − TH ) = χ (hrs )HC + (hp )HC (TH − TC ) + (TH − TW ) dti lW (6.3.12) From Eq. (6.3.11), TW is determined as a function of TH . Thus we can determine TH at a given value of the flame temperature Tf by Eq. (6.3.12). In Eq. (6.3.11) (hc + hr )W is a

82

6. Heat Transfer in a Rotary Reactor, Direct Heating

Figure 6.8 Heat transfer coefficient from the outer surface of a reactor, natural convection and thermal radiation.

summation of two coefficients from the outer surface of the reactor, by natural convection and by thermal radiation. For design calculation, it is convenient to use Fig. 6.8. The temperature of the inner wall surface TH is predicted in an ordinary rotary reactor for calcinating limestone in Example 6.3, the results of which are illustrated in Fig. 6.9. We find the wall temperature is fairly high in the region where combustion takes place. 6.3.5

Heating capacity of a rotary reactor

If the radiant heat transfer coefficient hrg and the temperature of the inner wall surface TH are predicted in a rotary reactor, we can calculate its heating capacity. For example, let us look at a rotary reactor for calcinating limestone, usually called a rotary kiln and illustrated in Fig. 6.10. For practical calculation, the long reactor is divided into three regions in this case. The rate of heat transfer in the reactor QR is then calculated by,     QR = (πdti lr1 )hrg1 Tf − T1∗ + (πdti lr2 )hrg2 Tg2 − T2∗   + (πdti lr3 )hrg3 Tg3 − T3∗

(6.3.13)

T ∗ = χTC + (1 − χ)TH

(6.3.14)

The average temperature T ∗ is calculated using the following equation.

where TC and TH are the temperature of the rotating solids and of the inner surface, respectively. Fraction χ is obtained from Fig. 6.7.

6.3. Heat transfer in a rotary reactor at high temperature

83

Figure 6.9 Prediction of temperatures: inner wall surface and outer shell surface.

Heat loss from the outer surface is obtained as follows: Qlr = (πdt0 lr1 )(hc + hr )W1 (TW1 − T0 ) + (πdt0 lr2 )(hc + hr )W2 (TW2 − T0 ) + (πdt0 lr3 )(hc + hr)W3 (TW3 − T0 )

(6.3.15)

In Fig. 6.10, radiant heat transfers from the combustion region to the inner surface of the hood, the amount of which is designated as Qlh . The heating capacity of the reactor, Fp [kg/hr], can be predicted with Eq. (6.3.16). Fp Hn = QR − Qlr − Qlh

(6.3.16)

84

6. Heat Transfer in a Rotary Reactor, Direct Heating

Figure 6.10 Simplified model of a rotary kiln.

where Hn is the heat necessary for transformation of the solids in the reactor. In Example 6.5, the capacity is predicted in two rotary reactors, i.e. 2.5 m ID, 40 m and 2.8 m ID, 54 m. The above calculation procedure is not restricted to a rotary reactor such as that shown in Fig. 6.10, but is applicable to any reactor in which the necessary heat is generated inside. In Chapter 7, the performance of several different rotary reactors is predicted on the basis of the above calculation procedure.

6.4 6.4.1

ENHANCEMENT OF HEAT TRANSFER

Lifters in a rotary dryer

In a conventional rotary dryer, lifters are usually positioned to enhance heat transfer between the solids and the flowing gas. Unlike previous reactors, the temperature of the flowing gas is low and the radiant heat transfer coefficient is not appreciable compared with convectional heat transfer. For design calculation, the volumetric heat transfer coefficient Ua is practically employed, on the basis of the following equation.    kcal (6.4.1) (Rate of heat transfer) = Vr m3 Ua 3 ◦ (∆Tav ◦ C) m hr C

6.4. Enhancement of heat transfer

85

where Vr is the volume of the unit and ∆Tav is the averaged temperature difference. From commercial performance data, we find in Ref. [11],

6.4.2

Lifter   Ua 3kcal◦ m hr C

Simple 100 ∼ 200

Complex 200 ∼ 300

Discussions on volumetric heat transfer coefficient

In an attempt to use Ua for the design of rotary reactors, let us discuss the heat transfer mechanism in a rotary cylinder in which lifter and/or partition plates are positioned. Ua is supposed to be composed of three terms as follows: (6.4.2)

Ua = Ual + Uac + Uar

where Ual is by lifter, Uac is by convectional heat transfer and Uar is by radiant heat transfer from gas. In the case of a rotary dryer, Uar ≈ 0 because of low temperature. At high temperature in a rotary reactor, however, Uar becomes appreciable. Letting hrg be the radiant heat transfer coefficient defined by Eq. (6.3.2), it can be predicted as: 

π 2 dt Uar ∆Tav = (πdt )Rv hrg ∆Tav 4

Then, Uar =

4Rv hrg dt

(6.4.3)

where Rv is the ratio of the total surface area, referring to the inner surface of the reactor. Let us estimate the value under the following conditions. εm = 0.8 εg = 0.19 Tg = 600 ◦ C T ∗ = 300 ◦ C dt = 2 m Rv = 2.0 From Eq. (6.3.2) we calculate hrg = 11.7 kcal/m2 hr ◦ C. Eq. (6.4.3) gives Uar = (4)(2)(11.7)/(2) = 46.8 kcal/m3 hr ◦ C For a simple lifter, Ual + Uac = 100 ∼ 200 kcal/m3 hr ◦ C, At 600 ◦ C we have Ua ∼ = 150 ∼ 250 kcal/m3 hr ◦ C

at drying

86

6. Heat Transfer in a Rotary Reactor, Direct Heating

6.4.3

Partition plates

When partition plates and baffle plates are positioned in a reactor, and lifters are removed to avoid carry-over of fines, we can predict Ua by the following equation. Ua = Uac + Uar =

4Rv (hc + hrg ) dt

(6.4.4)

In the case of two perpendicular partition plates, Rv =

πdt + 4dt = 2.27 πdt

With hc = 10 kcal/m2 hr ◦ C, Eq. (6.4.4) gives Ua =

4(2.27)(10 + 11.7) kcal = 98.5 ∼ 100 3 ◦ 2 m hr C

Example 6.1 Pulverized coal is burnt in a rotary reactor for calcinating limestone. Calculate the time for complete conversion of char in the flame. If the length of the flame is found to be 15 m, estimate the possible state of their conversion within and outside of the flame. Data Diameter of kiln 2.5 m, Length 40 m Average velocity of flame stream 20 m/sec Average gas velocity after combustion 8 m/sec Mean temperature of burning flame 1,400 ◦ C Size of char 10 ∼ 100 µm Density of porous char 0.8 gm/cm3 = 0.0667 gm-mol B/cm3 Diffusivity of oxygen 2.7 cm2 /sec Average fraction of O2 inside the flame 3% Solution Assuming Rep = 0 in Eq. (3.2.16), we have Kd = 2(2.7)/dp = 5.4/dp From Table 3.3, kc = 750 cm/sec CA =

gm-mol A 1 gm-mol A 273 (0.03) = 2.185 × 10−7 273 + 1,400 22.4 × 103 cm3 cm3

Eqs. (3.2.11), (3.2.15) give, τ=

  (0.0667)(dp /2) dp 1 + −7 (1)(2.185 × 10 ) 750 5.4

With Eqs. (3.2.10) and (3.2.5), XB = 1 −



 rc 3 t 3 =1− 1− f R τ

Here, tf = 15 m/(20 m/sec) = 0.75 sec.

Example 6.2

87

The results of calculation are as follows: dp [µm]

[cm]

10 30 60 100 150

0.001 0.002 0.006 0.010 0.015

τ [sec]

1 − XB

XB

0.232 0.865 2.24 4.86 9.41

0 0.0024 0.294 0.606 0.779

1 0.998 0.706 0.396 0.221

Coarser char particles are entrained by combustion gas from the flame region, and oxidized with oxygen from the excess air. The unburnt char stays in the kiln for a while. (40 m − 15 m)/(8 m/sec) = 3.13 sec The conversion of char coarser than 100 µm is fairly low.

Example 6.2 In a practical rotary kiln, estimate the rate constant of combustion κf [m−1 ], by comparison with the measured temperature of its shell. Data Ff = 1,475 kg oil/hr 9,600 kg pr/hr = 230.4 tons pr/day dt = 2.5 m, lr = 40 m Gg = 15.2 Nm3 /kg oil, Cg = 0.34 kcal/Nm3 ◦ C QN = 10,000 kcal/kg oil hrg = 191.4 kcal/m2 hr ◦ C T ∗ = 1,100 ◦ C (average value of inner surface and product) Tgi = 334 ◦ C Solution With Eq. (6.2.5), a1 =

π(2.5)(191.4) = 0.1972 m−1 (1,475)(15.2)(0.34)

a2 =

(10,000)(κf ) = 1,935κf m−1 (15.2)(0.34)

For κf = 0.3, a2 = 580.6 Eq. (6.2.4) gives, Tg = 1,100 +



 580.6 − (1,100 − 334) exp(−0.1972x) − 5,647.9 exp(−0.30x) 0.3 − 0.1972

= 1,100 + 4,881.9 exp(−0.1972x) − 5,647.9 exp(−0.30x) By similar calculations to the above, Fig. 6.4 is obtained. By comparison with the measured temperature of the shell, we can take κf = 0.3 m−1 , and estimate that the flame length is ca. 15 m.

Example 6.3 Calculate the radiant heat transfer coefficient in a rotary reactor under the following conditions.

88

6. Heat Transfer in a Rotary Reactor, Direct Heating

Combustion and heating region dt = 2.5 m, df = 1.0 m, εf = 0.7 εm = (0.8 + 0.84)/2 = 0.82 T ∗ = (1,200 ◦ C + 950 ◦ C)/2 = 1,075 ◦ C Combustion gas flowing region εg = 0.19 T ∗ = (1,000 ◦ C + 900 ◦ C)/2 = 950 ◦ C Solution Combustion and heating region With Eq. (6.3.1) and Tf = 1,400 ◦ C, 1.0 (0.7)(0.82)(4.88) 1,400+273 4 −  1,075+273 4  100 100 hrg = 2.5 = 156.2 kcal/m2 hr ◦ C 1,400 − 1,075

Combustion gas flowing region With Eq. (6.3.2) and Tg = 1,200 ◦ C, hrg =

 4  950+273 4  (0.19)(0.82)(4.88) 1,200+273 − 100 100 1,200 − 950

= 75.1 kcal/m2 hr ◦ C

A similar calculation to the above gives Fig. 6.6.

Example 6.4 In a rotary kiln of 2.5 m ID, predict the temperatures of the inner wall surface and the outer shell surface. Apply the radiant heat transfer coefficient given in Fig. 6.6. Data Combustion and heating region εH = 0.8, εC = 0.84, εg∗ = 0.4, TC = 950 ◦ C ρ¯ = 950 kg/m3 , Cs = 0.24 kcal/kg ◦ C, γ = 0.08 ke = 1.5 kcal/mhr ◦ C ks = 1.3 kcal/mhr ◦ C, lW = 0.15 m Combustion gas region I εg = 0.19, TC = 900 ◦ C ks = 1.3 kcal/mhr ◦ C, lW = 0.15 m Combustion gas region II εg = 0.19, TC = 850 ◦ C ks = 1.2 kcal/mhr ◦ C, lW = 0.125 m ks = 0.4 kcal/mhr ◦ C, lW = 0.025 m εg = 0.19 is the result of calculation by application of Hottel’s procedure [5]. Solution Combustion and heating region With Eq. (6.3.5), we calculate (hrs )HC =

 +273 4  950+273 4  (0.8)(1 − 0.4)(0.84)(4.88) TH100 − 100 TH − 950

TH

[◦ C]

1,000

1,100

1,200

1,300

(hrs )HC

[kcal/m2 hr ◦ C]

153.0

172.7

194.4

218.4

Example 6.5

89

From Fig. 6.7, χ is given to be 0.23 for γ = 0.08. With Eq. (6.3.8), then   kcal (1.5)(950)(0.24)(2 × 60) 1/2 = 477.3 2 ◦ (hp )HC = 1.13 0.23 m hr C We find (hrs )HC is comparable to hrg , but (hp )HC is much higher. In the case of dt0 /dt ≈ 1.0, Eq. (6.3.12) gives,   1.3 (1 − 0.23)hrg (Tf − TH ) = 0.23 (hrs )HC + 477.3 (TH − 950) + (TH − TW ) 0.15

(1)

On the other hand, from Eq. (6.3.11) we use, (hc + hr )W =

(1.3/0.15)(TH − TW ) TW − T0

(2)

At a given value of Tf , Eqs. (1) and (2) are two equations for two unknown variables, namely TH and TW . Applying Fig. 6.8, TH and TW are solved graphically, giving the lines in Fig. 6.9. Combustion gas region I In Eq. (6.3.5), we use εg = 0.19 in place of εg∗ . Thus, (hrs )HC =

 +273 4  900+273 4  (0.8)(1 − 0.19)(0.84)(4.88) TH100 − 100 TH − 900

TH

[◦ C]

950

1,000

1,100

(hrs )HC

[kcal/m2 hr ◦ C]

182.3

196.7

220.5

Two equations corresponding to Eqs. (1) and (2) are graphically solved to give the lines shown in Fig. 6.9. In combustion gas region II, a similar calculation is carried out, the results of which are added to Fig. 6.9.

Example 6.5 Predict the heating capacity of a rotary reactor for calcinating limestone (usually called a rotary kiln), the dimensions of which are given as follows: dti = 2.5 m, dt0 = 2.8 m lr1 = 10 m, lr2 = 14 m, lr3 = 16 m (total 40 m) Qlh = 200,000 kcal/hr (heat loss from hood, by preliminary calculation) Data γ = 0.08, χ = 0.23 (from Fig. 6.7) Heat of decomposition 753 kcal/kg pr conversion of solids inlet XB5 = 0.10 outlet XB3 = 0.96 temperature of solids inlet 800 ◦ C outlet 950 ◦ C Solution From Fig. 6.10, we take the average temperatures in three regions as shown in Table 6.4. TH and TW are evaluated, and then T ∗ is calculated with Eq. (6.3.14). The heat transfer coefficient (hc + hr )W is obtained from Fig. 6.8, and the heat flux qW is calculated by qW = (hc + hr )W (TW − T0 )

(3)

90

6. Heat Transfer in a Rotary Reactor, Direct Heating Table 6.4. Calculation results in Example 6.5

Region

Comb. & heating

Comb. gas I

Comb. gas II

Length

[m]

lr1 = 10

lr2 = 14

lr3 = 16

Tf , Tg TH TC TW

[◦ C] [◦ C] [◦ C] [◦ C]

1,460 1,150 950 310

1,280 980 900 280

1,080 890 850 230

T∗

[◦ C]

1,104

962

881

(hc + hr )W qW

[kcal/m2 hr ◦ C] [kcal/m2 hr]

25.0 7,250

23.0 5,980

20.0 4,200

hrg

[kcal/m2 hr ◦ C]

165

84

60

The results of the above calculation are summarized in Table 6.4. Thus Eq. (6.3.13) gives QR = (π)(2.5)(10)(165)(1,460 − 1,150) + (π)(2.5)(14)(84)(1,280 − 962) + (π)(2.5)(16)(60)(1,080 − 881) = 8,454,890 kcal/hr Heat loss from the shell is calculated as (π)(2.8)(10)(7,250) + (π)(2.8)(14)(5,980) + (π)(2.8)(16)(4,200) = 1,965,310 kcal/hr Heat for conversion of solids is then, 8,454,890 − 1,965,310 − 200,000 = 6,289,600 kcal/hr The necessary heat for conversion of solids per unit mass of product is given by     kcal 1 − 0.96 kcal kcal (0.96 − 0.10) + 0.28 (0.96) (950 ◦ C − 20 ◦ C) + 0.24 753 kg pr kg ◦ C 0.56 kg ◦ C    kcal kcal kcal 1 − 0.10 0.24 (0.10) (800 ◦ C − 20 ◦ C) = 510.7 − 0.28 + kg ◦ C 0.56 kg ◦ C kg pr Fp =

6,289,600 kcal/hr kg pr tons pr = 12,316 = 296 510.7 kcal/kg pr hr day

The heating capacity of the rotary kiln is estimated to be 296 tons per day.

REFERENCES [1] H.C. Hottel, et al., Third Symposium on Combustion, Flame and Explosion Phenomena, 254, 266 (The Williams & Wilkins Co, 1949). [2] S. Yagi, K. Saji, Fourth Intern. Symposium on Combustion (Sept. 1952). [3] R.A. Sherman, Trans. ASME 56 (1934) 401. [4] S. Yagi, D. Kunii, Japan Science Review 2 (4) (1952) 397.

References

[5] [6] [7] [8]

91

H.C. Hottel, W.H. McAdams, Heat Transmission, 3rd. edition (McGraw Hill, New York, 1954) Chapter 4. H.C. Hottel, A.F. Sarofim, Radiative Transfer (McGraw Hill, New York, 1957). A.K. Oppenheim, Trans. ASME 78 (1956) 725. Flame Radiation Research Joint Committee, J. Inst. Fuel 32, No. 222,338,344 (1959-7), No. 180, 23 (19561). [9] D. Kunii, O. Levenspiel, Fluidization Engineering, 2nd. edition (Butterworth-Heinemann, Stoneham, 1991). [10] D. Kunii, J.M. Smith, AIChE Journal 3 (3) (1957) 273. [11] D. Kunii, ed., Handbook of Drying Technology (Sogo Gijutsu Center, 1991).

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–7– Performance of Rotary Reactors, Direct Heating To carry out a gas–solid reaction at high temperature, thermal energy should be supplied. In this chapter, procedures are presented for predicting the performance of rotary reactors, which are directly heated by the internal combustion of fuel and/or other combustible substances. Examples of design calculation are presented for four cases, namely: calcination of limestone, pre-reduction of ferro-chromium pellets, activation of char and gasification of solid wastes. The direction of improvement for lower fuel consumption is discussed for conventional rotary reactor systems.

7.1 7.1.1

PREDICTION OF PERFORMANCE

Mass and enthalpy balances

When we plan to develop a novel reactor process, mass and enthalpy balances are most important in preparing working equations for design calculation. On the basis of sound information from conventional reactors for a similar purpose, let us take a simple flow sheet as the first approximation, and make contours around each unit, as seen in Fig. 7.1. Equations of mass and enthalpy balances should be made with respect to contour 1 at first, to predict the overall performance of the process. If further information is required, balance equations with respect to contours 2–6 should be formulated (refer to Example 7.1). For a rotary reactor operated at high temperature, reduction of fuel consumption is crucial for its economic advantage. In some industrial sectors, net calorific value is preferably employed. When the fuel is burnt completely within the process, this is all right. However, in the case of gasification, we have to use gross calorific value. For enthalpy balance, we need the amount of heat loss from component units and pipelines. It is calculated by application of Eq. (6.3.11) and Fig. 6.8 for any units, tentatively designed as the first approximation. 7.1.2

Enthalpy balance, complete combustion

Enthalpy balance at steady state is given by,       Net heat of Heat in Heat in + + combustion inlet gas solid feed

94

7. Performance of Rotary Reactors, Direct Heating

Figure 7.1 Example of a simplified flow sheet.

⎡

⎤     Heat for Latent heat of water vapour Heat in ⎣ ⎦ = endothermic + + from moisture in the feed stock exit gas reaction ⎡ ⎤   Heat in Heat ⎣ ⎦ + product + (7.1.1) loss solids Unit of Eq. (7.1.1) is usually [kcal/hr], but sometimes [kcal/kg or Nm3 fuel] or [kcal/kg feed or product], depending on the purpose of its calculation. 7.1.3

Enthalpy balance, partial combustion and gasification

In the case where partial combustion (oxidation) and gasification of hydro-carbon is carried out, it is necessary to know the amount of combustible gas. Kunii [1] used the simple model shown in Fig. 7.2, and presented Eqs. (7.1.3) and (7.1.4) to predict the gasification efficiency and the amount of oxygen necessary to operate the process at steady state. Letting QG [kcal/kg] and HG [kcal/Nm3 ] be the gross calorific values of the feed stock and product dry gas, respectively, and Vg [Nm3 /kg] be the volume of product dry gas, the gasification efficiency ηG is given by, Vg HG = ηG QG

(7.1.2)

The gasification efficiency ηG is obtained with the following equation. ηG = (α ∗ + β ∗ + ω∗ ) − (γ ∗ + δ ∗ + ζ · ε ∗ )

(7.1.3)

7.1. Prediction of performance

95

Figure 7.2 Model for partial combustion and gasification.

⎧ 7,838C(1 − ηc ) ⎪ ⎪ C : fraction of carbon α∗ = 1 − ⎪ ⎪ ⎪ QG ⎪ ⎪ ⎪ ⎪ ⎪ ∗ (heat of inlet gas, including latent heat) ⎪ ⎪ β = ⎪ ⎪ QG ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (heat added from outside) ⎪ ⎪ ω∗ = ⎪ ⎪ QG ⎪ ⎨ (heat carried out by products) ⎪ γ∗ = ⎪ ⎪ QG ⎪ ⎪ ⎪ ⎪ ⎪ (heat loss) ⎪ ⎪ ⎪ δ∗ = ⎪ ⎪ QG ⎪ ⎪ ⎪ ⎪ ∗ ⎪ ε = mass fraction of tar ⎪ ⎪ ⎪ ⎪ ⎪ (gross calorific value of tar) ⎪ ⎪ ⎩ζ = QG

(7.1.4)

From the enthalpy balance we obtain,

  Vg (H2 ) + (CO) + Σ(Cm Hn )f1 (Cm Hn ) = 3.279 × 10−4 QG ηG

(7.1.5)

where (H2 ) or (Cm Hn ) are the molar fraction of gases in dry product gas. The necessary amount of oxygen is given by, Vo2 = 0.5818ηc C + 1.639 × 10−4 QG (α ∗ − ηG ) − Vg Σ(Cm Hn )f2 (Cm Hn ) − 2.0ε ∗

(7.1.6)

96

7. Performance of Rotary Reactors, Direct Heating Table 7.1. Gross calorific value of combustible gas, and factors f1 and f2 Gas H2 CO

[kcal/Nm3 ]

f1 (Cm Hn )

f2 (Cm Hn )

3,050 3,020

CH4 C2 H6 C3 H8 n-C4 H10

9,500 16,644 23,688 30,469

3.1149 5.4572 7.7667 9.9903

0.4426 0.7714 1.1167 1.5049

C2 H4 C3 H6 C4 H8 C2 H2 C6 H6

15,055 21,964 28,993 13,867 35,227

4.9362 7.2015 9.5064 4.5467 11.5502

0.5319 0.8993 1.2468 0.2267 1.7249

where 2.0ε∗ is an approximate term, which depends on the mass ratio of hydrogen to carbon in tar (see Kunii [1]). The gross calorific value of combustible gas and factors f1 and f2 in Eqs. (7.1.5) and (7.1.6) are given in Table 7.1. In Examples 7.4 and 7.5, the above equations are applied to predict the performance of given reactors for the activation of char and gasification of solid waste as well. 7.1.4

Special cases

In the case where we take β ∗ = 0, γ ∗ = 0 and δ ∗ = 0, the heat necessary for chemical reaction Hc is given by ω∗ QG . Combination of Eqs. (7.1.2) and (7.1.3) leads to, Vg HG + QG (ζ · ε ∗ ) = QG (α ∗ + ω∗ ) = QG − 7,838C(1 − ηc ) + QG ω∗ Thus,   Hc = ω∗ QG = Vg HG + (ζ · ε ∗ )QG + 7,838C(1 − ηc ) − QG     gross calorific value gross calorific value = − of products of feed stock

(7.1.7)

Eq. (7.1.7) is the so-called Hess’s law. Equations (7.1.2)–(7.1.4) are not restricted to partial combustion, but are also applicable to the gasification of combustible feed stock. When such feed stock is externally heated in a reactor, a similar calculation to the above results in Eq. (7.1.8), which is useful for estimating the necessary heat Hn to operate it.   Hn = Hc + QG (γ ∗ + δ ∗ ) − β ∗

(7.1.8)

7.2. Calcination of limestone

7.1.5

97

Applicability of equations

Equations (7.1.2)–(7.1.8) are obtained from mass and enthalpy balances and then applicable for all processes on the gasification of combustible feed stock. For gasification of coal by partial combustion (oxidation), refer to Kunii [1]. Eqs. (7.1.5) and (7.1.6) can predict the gasification efficiency and the necessary amount of oxygen very well. The heat of decomposition depends on the chemical composition of the feed stock. For example, Material Hc [kcal/kg]

poly-ethylene 455

p.-propylene 619

p.-styrene 616

When we fail to find numerical data for Hc in the literature, the next step is to utilize Eq. (7.1.7). Since Hc is calculated as the difference between two values, its accuracy is not too reliable. Experimental data on thermal cracking of bio-mass reported by Sadakata [3] were analysed to calculate Hc . It is found that Hc at a temperature higher than 700 ◦ C has a minus sign and that thermal cracking of the bio-mass is exothermic at the temperature range.

7.2 7.2.1

CALCINATION OF LIMESTONE

Procedure for design calculation

Fig. 7.3 is a simplified model of a rotary reactor for calcinating limestone. This type of rotary reactor is usually called a rotary kiln and is applied to decompose and sinter a variety of inorganic feed stocks. Thermal energy in the exit gas from the rotary kiln should be recovered by application of a preheater as seen in the figure. For feed stock with an average size coarser than 5 mm, counter-current exchange of heat is the best contacting method for heat recovery. The preheater in Fig. 7.3 is designed to

Figure 7.3 Rotary kiln to calcinate limestone.

98

7. Performance of Rotary Reactors, Direct Heating

achieve uniform contacting of the solids with the gas flow, by applying a rotating hearth and a number of pusher rods. The product solids exit from the rotary kiln at high temperature, containing much heat. This should be recovered by sending the secondary air into a so-called cooler. Regardless of its name, it is an important unit for heat recovery. Depending on the size of the feed stock, other types of preheater and cooler have so far been employed, for instance, horizontal moving beds, fluidized beds or entrained flow. When the size of the feed stock is 5 ∼ 20 mm, however, the system in Fig. 7.3 is found to be superior to the other combinations for calcination of limestone. For predicting the thermal performance of a rotary kiln, it is suggested that the following procedure of calculation be applied. (1) (2) (3) (4) (5) (6)

Determine dimensions of units as the first approximation. Calculate heat loss from each unit. Apply equations of enthalpy balance, Eq. (7.1.1). Determine process variables. Compare the results of calculation with requirements. Modify the first dimensions to be the second approximation. Proceed the calculation, similarly.

7.2.2

Estimation of heat loss

Suppose it is required to design a rotary kiln to produce 600 tons of quick lime per day. As the first approximation take dti = 2.8 m, dt0 = 3.1 m and lr = 54 mm. Assume that the refractory lining is the same as Example 6.4. In Example 6.5, we have calculated the values of heat flux qw from the outer surface of a kiln (dti = 2.5 m). Region   qw mkcal 2 hr

Combustion & heating

Comb. gas I

Comb. gas II

7,250

5,980

4,200

Letting the above three regions all be 18 m in length, the heat loss from the outer surface of the kiln is calculated as (3.1)(π)(18)(7,250) + (3.1)(π)(18)(5,980) + (3.1)(π)(18)(4,200) = 3,055,500 kcal/hr Rate of product is, Fp =

600,000 kg pr = 25,000 24 hr

Heat loss per unit mass of product is, ⎫ Combustion & heating region L3 = 50.8 ⎪ ⎬ kcal Combustion gas I L4 = 41.9 Lr = 122.2 ⎪ kg pr Combustion gas II L′4 = 29.5 ⎭

7.2. Calcination of limestone

99

Considerable heat is lost from the outer surface of the hood at the burner side. According to a detailed calculation on radiant heat transfer from the combustion and heating region to the inner surface of the hood, the average temperature of the surface, TH is calculated to be TH = 1,104 ◦ C Eq. (6.3.11) is simplified to (hc + hr )W =

(ks / lW )(TH − TW ) T W − T0

(7.2.1)

Eq. (7.2.1) is graphically solved as shown in Fig. 6.8. Substituting the above numerical values and T0 = 20 ◦ C, we find TW = 250 ◦ C qW = (1.2/0.2)(1,104 − 250) = 5,124

kcal m2 hr

Heat loss per unit mass of product from the hood is given by Lh1 =

(57 m2 )(5,124 kcal/m2 hr) = 11.7 kcal/kg pr 600,000 kg/24hr

For other units in Fig. 7.3, a similar calculation is carried out, the results of which are summarized as follows.

7.2.3

Rotary kiln Hood, burner side Hood, chute side Cooler Preheater

Lr = 122.2 kcal/kg pr Lh1 = 11.7 Lh2 = 4.9 L3 = 14.5 L5 = 28.0

Total

Lt = 181.3 kcal/kg pr

Prediction of overall performance

In Fig. 7.3, let us use two unknown variables, namely the amount of fuel, Bf [kg/kg pr] and the temperature of the exit gas Tge [ ◦ C]. Other notations are as follows: Ca , Cg , CC :

specific heat of air, wet combustion gas and CO2

Cs1 , Cs2 :

specific heat of feed and product

Ga , Gg , GC :

volume of necessary air, wet combustion gas in the case of stoichiometric combustion and CO2

Hc :

heat to decompose



kcal 3◦

C  Nm kcal ◦

 kgNmC3  kg fuel

 kcal  kg pr



100

7. Performance of Rotary Reactors, Direct Heating

n:

excess air ratio

QN :

net calorific value

XBr :

conversion of solids

[–]  kcal  kg fuel

[–]

β = (product)/(feed stock)

[–]

To represent the process system of Fig. 7.3, let us use the simplified model given in Fig. 7.1. For contour 1 we obtain, [Input] Heat of combustion

[kcal/kg pr] Bf QN

[Output] Heat of decomposition Wet combustion gas Excess air CO2 from decomposition Solid product Heat loss, total

Hc XBr Bf Gg Cg (Tge − T0 ) Bf Ga nCa (Tge − T0 ) GC XBr CC (Tge − T0 ) Cs2 (Tce − T0 ) Lt

[Input] = [Output] leads to the next equation.   Bf QN = Bf (Gg Cg + Ga nCa ) + XBr GC CC (Tge − T0 ) + Hc XBr + Cs2 (Tce − T0 ) + Lt

(7.2.2)

In the preheater, a countercurrent exchange of heat takes place within a vertical moving bed. Hot combustion gas from the rotary kiln flows into the bottom of the bed and calcinates limestone to some extent, usually XB5 = 0.1 ∼ 0.2. Thus the temperature of the combustion goes down to that of decomposition, Tc ≈ 900 ◦ C (refer to Example 4.2). Let us focus our attention on the upper half of the moving bed, where only heat exchange occurs without decomposition reaction, and set contour 6 around this region (see Fig. 7.1). Letting Tg5 and Tc6 be the decomposition temperature of limestone, the enthalpy balance across contour 6 gives the following working equation. 

 1 Bf (Gg Cg + Ga nCa ) + XBr GC CC (Tg5 − Tge ) = Cs1 (Tc6 − T0 ) β

(7.2.3)

Two unknown values, namely Bf and Tge are easily determined by simultaneous solution of Eqs. (7.2.2) and (7.2.3). In Example 7.1, the above calculation is carried out to predict the performance of a rotary kiln to produce 600 tons of quick lime per day. The results are summarized in Table 7.2. The fuel heat consumption, Bf QN [kcal/kg pr] depends on the efficiency of heat recovery in the preheater. Chisaki has retro-fitted his novel preheaters to several rotary kilns, which were constructed many years ago, as shown in Fig. 7.3. Nearly 30% increase of capacity and 25% decrease of fuel heat consumption have been achieved, as predicted before their installation. Fig. 7.4 shows practical data, which almost coincide with the predicted values.

7.2. Calcination of limestone

101 Table 7.2.

Heat balance predicted for calcination of limestone, 600 tons pr/day, 2.8 m ID, 54 m long, Bf = 0.1062 kg/kg pr, Tge = 235 ◦ C, n = 0.20 Input Fuel

(0.1062)(10, 000)

Output Decomposition Comb. gas Excess air CO2 Solid product Heat loss

(0.96)(753) (0.1062)(11.8)(0.34)(235 − 20) (0.1062)(11.0)(0.1)(0.31)(235 − 20) (0.96)(0.3994)(0.5)(235 − 20) (0.24)(60 − 20) rotary kiln hood, burner side hood, chute side cooler preheater

Total

Figure 7.4

[kcal/kg pr]

[%]

1,062

100

[kcal/kg pr]

[%]

722.9 91.6 15.6 41.2 9.6 181.3

68.1 8.6 1.4 3.9 0.9 17.1

122.2 11.7 4.9 14.5 28.0 1,062

100

Practical performance data on rotary kilns to which Chisaki’s preheaters were retro-fitted.

102

7. Performance of Rotary Reactors, Direct Heating

Figure 7.5 Effect of heat loss in the process shown in Fig. 7.3.

The effect of heat loss per unit mass of product is predicted in Example 7.1, the results of which are shown in Fig. 7.5. Suppose the total heat loss Lt in the system of Fig. 7.3 is reduced to (181.3 kcal/kg pr)(2/3) = 121 kcal/kg pr, we estimate Bf = 0.095 kg oil/kg pr and Tge = 190 ◦ C from Fig. 7.5. It is not difficult to achieve a fuel heat consumption of (0.095)(10,000) = 950 kcal/kg pr in a brand new plant of 600 tons pr/day. In Fig. 7.4, we find that the fuel heat consumption appears to increase sharply with decreasing capacity. This is because the total heat loss Lt is quite large in a small-scale rotary kiln system. Even for a very small-scale rotary kiln, however, Lt can be reduced considerably by utilization of the appropriate refractory and then by shortening the length of the kiln. The smallest, but brand new, rotary kiln system, coupled with CHISAKI’s preheater, calcinates dolomite 2–5 mm in size, the capacity of which is 18 tons pr/day. In spite of its small size, practical data of fuel heat consumption is 1,300 kcal/kg pr, which corresponds to the 300 tons pr/day kiln in Fig. 7.4. 7.2.4

Prediction of solids conversion and gas temperature

Fig. 7.1 is a simplified flow sheet of the rotary kiln process shown in Fig. 7.3. Suppose we send the secondary air into the cooler, the amount of which is just the same as that stoichiometrically needed. From the enthalpy balance across contour 2 we take, Cs2 (Tc3 − Tce ) = Va Ca (Ta − T0 ) + L2

(7.2.4)

7.2. Calcination of limestone

103

In case, Bf = 0.1058 kcal oil/kg pr, Va = (0.1058)(11.0) = 1.164 Nm3 /kg pr, Tc3 = 950 ◦ C, Tce = 60 ◦ C, L2 = 14.5 kcal/kg pr, Ta = 20 +

(0.24)(950 − 60) − 14.5 = 572 ◦ C (1.164)(0.31)

To find the unknown variables in Fig. 7.1, equations of enthalpy balance are obtained for contours 3 and 4, where β = Ca O/Ca CO3 = 0.56. Contour 3 Combustion and heating region [Input] Heat of combustion Secondary air Solids [Output] Heat of decomposition Product Combustion gas CO2 Heat loss

[kcal/kg pr] Bf QN Bf Ga Ca (Ta − T0 )   B4 + Cs2 XB4 (Tc4 − T0 ) Cs1 1−X β (XBr − XB4 )Hc Cs2 (Tc3 − T0 ) Bf (Gg Cg + Ga nCa )(Tg3 − T0 ) (XBr − XB4 )GC CC (Tg3 − T0 ) L3

[Input] = [Output] leads to 

 1 − XB4 Bf QN + Bf Ga Ca (Ta − T0 ) + Cs1 + Cs2 XB4 (Tc4 − T0 ) β = (XBr − XB4 )Hc + Cs2 (Tc3 − T0 ) + Bf (Gg Cg + Ga nCa )(Tg3 − T0 ) (7.2.5) + (XBr − XB4 )GC CC (Tg3 − T0 ) + L3 Contour 4 Heating region by combustion gas [Input] Combustion gas CO2 Solids [Output] Heat of decomposition Solids Combustion gas CO2 Heat loss

Bf (Gg Cg + Ga nCa )(Tg3 − T0 ) (XBr − XB4 )GC CC (Tg3 − T0 )   B5 + Cs2 XB5 (Tc5 − T0 ) Cs1 1−X β (XB4 − XB5 )Hc   B4 Cs1 1−X + Cs2 XB4 (Tc4 − T0 ) β Bf (Gg Cg + Ga nCa )(Tg4 − T0 ) (XBr − XB5 )GC CC (Tg4 − T0 ) L4

[Input] = [Output] gives, Bf (Gg Cg + Ga nCa )(Tg3 − T0 ) + (XBr − XB4 )GC CC (Tg3 − T0 )   1 − XB5 + Cs2 XB5 (Tc5 − T0 ) + Cs1 β

104

7. Performance of Rotary Reactors, Direct Heating

= Bf (Gg Cg + Ga nCa )(Tg4 − T0 ) + (XBr − XB5 )GC CC (Tg4 − T0 )   1 − XB4 + Cs1 + Cs2 XB4 (Tc4 − T0 ) + (XB4 − XB5 )Hc + L4 β

(7.2.6)

Contour 5 Pre-calcination region in the preheater [Input] Combustion gas CO2 Solids [Output] Heat of decomposition Solids Combustion gas CO2 Heat loss

Bf (Gg Cg + Ga nCa )(Tg4 − T0 ) (XBr − XB5 )GC CC (Tg4 − T0 ) Cs1 β1 (Tc6 − T0 ) (XB5 )Hc   B5 Cs1 1−X + Cs2 XB5 (Tc5 − T0 ) β Bf (Gg Cg + Ga nCa )(Tg5 − T0 ) (XBr − 0)GC CC (Tg5 − T0 ) L5

[Input] = [Output], Bf (Gg Cg + Ga nCa )(Tg4 − T0 ) + (XBr − XB5 )GC CC (Tg4 − T0 ) 1 + Cs1 (Tc6 − T0 ) β = XB5 Hc + Bf (Gg Cg + Ga nCa )(Tg5 − T0 ) + XBr GC CC (Tg5 − T0 )   1 − XB5 + Cs1 + Cs2 XB5 (Tc5 − T0 ) + L5 β

(7.2.7)

In the above equations, we can assume the following: Tc3 = 950 ◦ C

Tc4 = 900 ◦ C

Tc5 = Tc6 = 800 ◦ C

Tg5 = 800 ◦ C By applying the information on heat transfer given in Chapter 6, we use the following equations within contours 3 and 4, respectively. Contour 3 (πdti l3 )hrg3 (Tg3 − T3∗ )/Fp  Cs1 = (XBr − XB4 )Hc + (1 − XBr ) + Cs2 XBr (Tc3 − T0 ) β  Cs1 − (1 − XB4 ) + Cs2 XB4 (Tc4 − T0 ) + L3 β

(7.2.8)

7.3. Pre-reduction of composite pellets, made of ferro-chromium ore and coke

105

Contour 4 (πdti l4 )hrg4 (∆Tav )/Fp

 Cs1 (1 − XB4 ) + Cs2 XB4 (Tc4 − T0 ) β  Cs1 − (1 − XB5 ) + Cs2 XB5 (Tc5 − T0 ) + L4 β

= (XB4 − XB5 )Hc +

(7.2.9)

where ∆Tav =

(Tg3 − T3∗ ) − (Tg4 − T4∗ ) T −T ∗

ln Tg3 −T3∗ g4

(7.2.10)

4

T ∗ is given by Eq. (6.3.14). From four independent equations, namely Eqs. (7.2.5), (7.2.6), (7.2.8) and (7.2.9), four unknown variables, i.e., XB4 , XB5 , Tg3 and Tg4 can be determined numerically. In Example 7.2, numerical calculation is carried out on a similar basis to Example 7.1, the results of which are as follows: Tg3 = 1,515 ◦ C XB4 = 0.42

Tg4 = 1,045 ◦ C XB5 = 0.19

The above values, as well as Bf = 0.1058 kg oil/kg pr and Tge = 230 ◦ C, almost coincide with the performance data in a practical plant.

7.3

7.3.1

PRE-REDUCTION OF COMPOSITE PELLETS, MADE OF FERRO-CHROMIUM ORE AND COKE

Conversion of solids

Composite pellets made of ferro-chromium ore and coke are pre-reduced in a rotary kiln, an example of which is illustrated in Fig. 7.6. An example of their composition can be found in Chapter 4, Section 4.6. As explained there, the pellets are reduced at a temperature higher than 1,250 ◦ C. The conversion process is as follows: 

Cr2 O3 + 3C = 2Cr + 3CO (−1,823 kcal/kg Cr) FeO + C = Fe + CO (−683 kcal/kg Fe)

From the fundamental data shown in Fig. 4.4, we find at 1,250 ◦ C and 2 hrs, (XB )Cr = 0.55 (XB )Fe = 0.90

(7.3.1)

106

7. Performance of Rotary Reactors, Direct Heating

Figure 7.6

7.3.2

Model of a rotary kiln for pre-reduction of composite pellets of ferro-chromium ore and coke.

Rotary kiln

In the model rotary kiln shown in Fig. 7.6, the feed rate of the pellets is 18,000 kg/hr (14,720 kg pr/hr = 353 tons pr/day). The operating variables are assumed as follows: Feed pellet 1.223 kg/kg pr Fuel oil 0.085 kg/kg pr Primary and secondary air 1.678 Nm3 /kg pr Tertiary air 0.077 Nm3 /kg pr Coke in pellets 0.205 kg/kg pr Coke fed to kiln 0.065 kg/kg pr Mean size of pellet 8 mm Inclination angle of center axis ω = 2 deg. Fraction of bulk solids γ = 0.05 Bulk density of product pellets 1,700 kg/m3 Conversion Cr 55% Fe 90% Reduced pellet Total Cr 32% Total Fe 17% Within the reduction region, ca. 30 m, CO is issued from the ore and fills the void space of the rotating solids layer. CO percolates through the layer of rotating solids and then flows out from the surface of the layer. Therefore, turbulent diffusion flame takes place above the surface of the rotating pellets, as seen in Fig. 6.2 [C]. The concentration profile within the kiln is also shown in Fig. 7.7. The volume of CO issued from the unit mass of the feed is calculated as  kg Nm3 3 × 22.4 Nm3 0.32 = 0.1137 (0.55) kg feed 2 × 52 kg kg feed  kg Nm3 22.4 Nm3 0.17 = 0.0613 (0.90) kg feed 55.9 kg kg feed Total

0.1750

Nm3 kg feed

7.3. Pre-reduction of composite pellets, made of ferro-chromium ore and coke

107

Figure 7.7 Concentration profile in the model rotary kiln shown in Fig. 7.6.

The residence time in the reduction region, tr is found

tr =

 π kg kg (4.2 m)2 (30 m)(0.05) 1,700 3 = 2.4 hr 14,720 4 hr m

Applying the above data as the first approximation, fundamental equations of enthalpy balance as well as those of heat transfer are utilized to predict the performance characteristics of the rotary kiln in Example 7.3. Table 7.3 summarizes its results.

108

7. Performance of Rotary Reactors, Direct Heating Table 7.3. Results of Example 7.2. Pre-reduction of pellets, ferro-chromium ore and coke Rotary kiln 4.2 m ID 100 m in length Capacity 18,000 kg feed/hr, 14,720 kg pr/hr Tf = 1462 ◦ C Tge = 727 ◦ C Bf = 0.08772 kg oil/kg pr

Input

kcal/kg pr

Fuel CO from reduction Comb. of coke

(0.08772)(10,000) = (0.1750)(3,045) = (0.205 × 0.25 + 0.065 × 0.5)(6,800) =

Total Output Unburnt CO Heat of reaction Heat of product Heat of exit gas Heat loss

(1.854)(0.06)(3,045) = (0.20)(1,250 − 20) = (1.987)(0.34)(699 − 20) =

Total

7.3.3

%

877.2 532.9 569.5

44.3 26.9 28.8

1,979.6

100.0

kcal/kg pr

%

338.7 425.3 246.0 477.6 492.0

17.1 21.5 12.4 24.1 24.9

1, 979.6

100.0

Direction of improvement

The operating conditions of the model kiln shown in Fig. 7.6 are not very economic in terms of high fuel costs. When drastic improvement of fuel consumption is required, the operating conditions should be improved as follows: (1) (2) (3) (4)

Increase of γ in the reduction region, leading to capacity increase. Increase of temperature Preheating of secondary air Enhancement of heat exchange in the preheating region

The volumetric fraction of the bulk solids γ could be increased by an appropriate design of the exit configuration in the kiln. The heating capacity of this kiln is far smaller than that for calcination of limestone. Pre-reduction Calcination

kg pr/hr 14,720 kg pr/hr = 10.6 2 m) (100 m) m3

π 4 (4.2

kg pr/hr 550,000 kg pr/24 hr = 68.9 π 2 m3 4 (2.8 m) (54 m)

As far as heating capacity is concerned, a considerable increase can be achieved if we raise the flame temperature from 1,462 ◦ C to 1,515 ◦ C; the latter is predicted in a rotary kiln for calcinating limestone in Example 7.2. The effects of the above improvement could be evaluated by applying a similar calculation procedure to Example 7.3. In Table 7.3 we find that the heat of unburnt CO in the exit gas is considerably high. It can be burnt well before exiting from its stack, and the combustion heat could be recovered by

7.4. Activation of char

109

preheating the secondary air. The above improvement might be necessary in future because of more stringent environmental regulations.

7.4 7.4.1

ACTIVATION OF CHAR

Model of a rotary reactor

Activation of char is discussed in Chapters 3 and 5 (see also Figs. 3.8 and 5.3). Assuming the conversion of char is ca. XB = 0.5 for activated carbon, the time necessary for activation appears to be ca. 1 hr at 800 ◦ C on a small scale. However in a commercial reactor, ca. 7 hrs are needed in practice. This difference is attributed to the low average concentration of water vapour in the rotating layer of solids. Fig. 7.8 shows a simplified model of a rotary reactor in which char is activated at 900 ◦ C, by injecting steam. At steady state, the necessary thermal energy is supplied from the combustion of H2 and CO by secondary air, which is sent from outside. The dimensions and operating conditions of the model are assumed as follows: dti = 2.1 m, drying and preheating region 12 m activation region 12 m thickness of refractory 0.3 m Inclination angle of center axis 0.3 deg. Feed rate of dry char 400 kg/hr Water contained 0.2 kg/kg d.f. Average size of char 5 mm Volumetric fraction of bulk solids γ = 0.10 Average bulk density of solids 420 kg/m3 Residence time in activation region 7.2 hr Table 7.4 shows the mass balance estimated in the model reactor. 7.4.2

Application of equations

In the model reactor, char particles are gasified by water vapour as seen in Fig. 6.2 [A]. H2 and CO issue out of the rotating layer of char and are burnt by the air which is sent from outside, giving the necessary heat for the endothermic reaction, Eq. (3.5.2).

Figure 7.8

A model of a rotary reactor for activation of wet char.

110

7. Performance of Rotary Reactors, Direct Heating Table 7.4. Mass balance in the model Basis, 1 kg of dry char Input

Output

Item

kg/kg

Dry char Moisture Air Steam

1.0 0.2

Nm3 /kg

Item

kg/kg

2.7

Act. carbon Gasified carbon Carbon loss

0.52 0.45 0.03

1.40

Table 7.5. Results of Example 7.4. Activation of char on the basis of 1 kg dry feed Operating conditions:

Activation temperature Air Steam Moisture

900 ◦ C 2.436 Nm3 1.4 kg 0.2 kg

Product

Active carbon Dry gas H2 [%] 16.7 HG = 754 kcal/Nm3 Unreacted H2 O

0.52 kg 3.215 Nm3 N2 Total 59.8 100

CO 8.1

CO2 15.4

1.235 Nm3

Enthalpy balance QG = 8,000 kcal/kg d.f. Input Carbon gasified Steam

kcal/QG α∗ β∗

0.515 0.1183

Total

0.6333

Output Combustible gas Dry exit gas Water vapour Solids Heat loss Total

kcal/QG ηG γ∗ γ∗ γ∗ δ∗

kcal/d.f.

%

4,120 946

81.3 18.7

5,066

100.0

kcal/d.f.

%

0.3046 0.1129 0.1242 0.0128 0.0788

2,437 903 994 102 630

48.2 17.8 19.6 2.0 12.4

0.6333

5,066

100

As explained in Chapter 6, the above flame is the reverse situation, compared with an ordinary flame. Since turbulent diffusion is the controlling phenomenon in this case, the mechanism of combustion is just the same as the ordinary one. For good quality active carbon, we have to predict the percentage of H2 and CO in the reactor. In Example 7.4, performance characteristics including H2 and CO percentage are predicted on the basis of Eqs. (7.1.2)–(7.1.6). The results are summarized in Table 7.5.

7.5. Gasification of combustible feed stock

7.4.3

111

Direction of improvement

In the light of quantitative prediction, such a rotary reactor as that illustrated in Fig. 7.8 could be improved enormously. Its procedure is proposed as follows: (1) Injection of steam is changed in Table 1.2 from [B] to [C]. Reactor efficiency ηr could be doubled, reducing residence time tr to half. (2) Volumetric fraction of solids γ is increased from 0.1 to 0.2. (3) Inlet tubes of injection air should be improved for better combustion of issued combustible gas. The above improvements might result in a much more efficient reactor, in the case where it is planned to install a brand new one.

7.5

GASIFICATION OF COMBUSTIBLE FEED STOCK

The authors have developed a novel rotary reactor to gasify combustible materials that are difficult to use in conventional gasification reactors. Fig. 7.9 presents a simplified model to explain how it works. Any type of combustible solid stock, granules or slurry, can be fed into its re-circulation region, where a partition plate with inclined guide plates are positioned (refer to Fig. 2.7 where the function of the partition plate is fully explained). In accordance with the rotation of the reactor, hot char particles re-circulate to the location of feeding. The feed stock falls onto the massive hot solids, mixes into the flow of hot particles, and is heated up to the temperature necessary for thermal decomposition. During its re-circulation in this region, the combustible gas issues from the feed stock, and the char particles are re-circulated within the region. Reactors of 0.9 m ID have been tested to gasify two kinds of pellets, derived from solid waste material. It was confirmed that thermal cracking of solids was carried out as designed. From the opposite end of the reactor, the gasification medium, i.e. O2 and H2 O, is injected into the issuing flow of combustible gas and vapour. A turbulent diffusion flame is

Figure 7.9

A simplified model of a U-Turn rotary reactor, proposed for the gasification of solid waste materials.

112

7. Performance of Rotary Reactors, Direct Heating Table 7.6. Results of prediction, gasification of plastic waste by oxygen and steam, in Example 7.5

Feed stock Dry basis

poly- propylene 90% QG = 9,400 kcal/kg moisture 0.1 kg/kg feed rate 500 kg/hr

waste pulp 10%

Gasification conditions, on the basis of dry feed 1 kg Steam Exit gas

1.0 kg/kg 1,000 ◦ C

0.6974 Nm3 /kg

Oxygen

Produced gas Dry gas

2.731 Nm3 /kg

= 2,700 kcal/Nm3

HG Unreacted H2 O

H2

CO

CO2

CH4

Total

46.9

30.1

19.2

3.8

100

1.12 kg

Enthalpy balance Input Feed stock, gasified Steam

kcal/QG α∗

0.9659 0.07191

β∗

Total

1.0378

Output Combustible gas Heat in dry gas Heat in water vap. Heat loss Total

kcal/kg d.f.

kcal/QG ηG γ∗ γ∗ δ∗

%

9, 079.3 676.0

93.1 6.9

9,755.3

100.0

kcal/kg d.f.

%

0.7887 0.1042 0.1259 0.019

7,414.0 979.5 1,183.5 178.3

76.1 10.0 12.1 1.8

1.0378

9,755.3

100.0

formed in the combustion region, and O2 is consumed along the flame. Combustion gas at high temperature is mixed into the surplus flow of combustible gas and vapour, and reforming of higher hydrocarbons then takes place in the mixture, reducing its temperature somewhat. Reforming of the higher hydrocarbons continues within the hot hood, the volume of the hood being of practical importance for the after reforming. Some fraction of carbon in char particles is gasified in the combustion region, but the remainder flows out of the rotary reactor. To attain high gasification efficiency, it is necessary to gasify the residual carbon accompanied with inorganic solids. Thus some amount of gasification medium is sent into their discharge region. This type of rotary gasification reactor is suitable for utilizing combustible waste materials in which the ash, S and Cl contents are not too high. In this case, hot combustible gas can be directly sent to another furnace or kiln. In Example 7.5, equations of enthalpy balance, i.e. Eqs. (7.1.2)–(7.1.6) are applied to predict its performance, in the case where waste plastics are gasified by an oxygen and steam mixture at 1,000 ◦ C. The results are summarized in Table 7.6.

Example 7.1

113

Example 7.1 Limestone is calcinated in the rotary kiln illustrated in Fig. 7.3. Predict its performance under the following conditions. Production 600 tons/day Kiln 2.8 m ID 54 m in length Fuel oil QN = 10,000 kcal/kg Air Ga = 11 Nm3 /kg, Ca = 0.31 kcal/Nm3 ◦ C Combustion gas Gg = 11.8 Nm3 /kg, Cg = 0.34 kcal/Nm3 ◦ C CO2 GC = 0.3994 Nm3 /kg, CC = 0.50 kcal/Nm3d ◦ C Excess air n = 0.2 Conversion XBr = 0.96 Solids Cs1 = 0.28 kcal/kg ◦ C, Cs2 = 0.24 kcal/kg ◦ C, Tc5 = Tc6 = 800 ◦ C, Tce = 60 ◦ C CaO = 0.56, Hc = 753 kcal/kg pr, Lr = 181.3 kcal/kg pr β = CaCO 3 Solution Substitution of the above numerical values to Eq. (7.2.2) gives,   Bf (10,000) = Bf (11.8 × 0.34 + 11.0 × 0.2 × 0.31) + (0.96)(0.3994)(0.5) (Tge − 20) + (753)(0.96) + (0.24)(60 − 20) + 181.3

Rearranging,

Bf =

913.78 + 0.1917(Tge − 20) 10,000 − 4.694(Tge − 20)

(1)

From Eq. (7.2.3),   1 (0.28)(800 − 20) Bf (11.8 × 0.34 + 11.0 × 0.2 × 0.31) + (0.96)(0.3994)(0.5) (800 − Tge ) = 0.56 Then,

Bf =

83.08 − 0.04084 800 − Tge

(2)

The above two equations are solved graphically to give, Bf = 0.1062 kcal oil/kg pr,

Tge = 235 ◦ C

Using the above values, the heat balance of the process is given in Table 7.2. For a different value of heat loss, Lt , a similar calculation is carried out, the results of which are shown in Fig. 7.5.

Example 7.2 In the same rotary kiln as Example 7.1 (model illustrated in Fig. 7.1), predict the conversion of solids and gas temperature in the rotary kiln. Numerical data are given as follows: l3 = 15 m, l4 = 18 + 21 = 39 m

114

7. Performance of Rotary Reactors, Direct Heating

Heat loss per unit mass of product cooler hood, burner side rotary kiln, comb. & heating rotary kiln, heating region hood, chute side preheater

[kcal/kg pr] 14.5  11.7 62.5 50.8 71.4  4.9 32.9 28.0

= L2 = L3 = L4 = L5

Temperature of solids Tce = 60 ◦ C, Tc3 = 950 ◦ C, Tc4 = 900 ◦ C, Tc5 = 800 ◦ C, Tc6 = 800 ◦ C, T0 = 20 ◦ C Suppose we have the following data from preliminary calculation in the case where n = 0.2. Bf = 0.1058 kcal oil/kg pr, Tge = 230 ◦ C, XBr = 0.96 From Table 6.4 T ∗ [◦ C] comb. & heating heating region

hrg

1,104 mean 922



kcal m2 hr ◦ C

165 mean 72



Solution Numerical values are put into Eq. (7.2.5), (0.1058)(10,000) + (0.1058)(11)(0.31)(572 − 20) +



 0.28 (1 − XB4 ) + 0.24XB4 (900 − 20) 0.56

= (0.96 − XB4 )(753) + (0.24)(950 − 20) + (0.1058)(11.8 × 0.34 + 11.0 × 0.2 × 0.31)(Tg3 − 20) + (0.96 − XB4 )(0.3994)(0.5)(Tg3 − 20) + 62.5 Rearranging, Tg3 = 20 +

688.57 + 524.2XB4 0.6864 − 0.1997XB4

(1)

From Eq. (7.2.8), π(2.8)(15)(165)(Tg3 − 1,104) 600,000/24

 0.28 (1 − 0.96) + 0.24 × 0.96 (950 − 20) 0.56  0.28 (1 − XB4 ) + 0.24XB4 (900 − 20) + 62.5 − 0.56

= (0.96 − XB4 )(753) +

Rearrangement results in, Tg3 = 1,768 − 601.9XB4 Equations (1) and (2) are solved graphically to give XB4 = 0.42,

Tg3 = 1,515 ◦ C

(2)

Example 7.3

115

Similarly, another working equation is obtained from Eq. (7.2.6), Tg4 = 20 +

561.4 + 550.2XB5 0.4966 + 0.1997(0.96 − XB5 )

(3)

Eq. (7.2.7) gives the same equation as the above. For XB5 = 0.1 ∼ 0.3, Tg4 is calculated to give 943 ∼ 1,176 ◦ C. In Eq. (7.2.10), we use Tg3 = 1,515 ◦ C, T3∗ = 1,104 ◦ C and T4∗ = 922 ◦ C to calculate ∆Tav . Thus a relation between ∆Tav and XB5 can be plotted in a graph. On the other hand Eq. (7.2.9) gives,   0.28 π(2.8)(39)(72) ∆Tav = (0.42 − XB5 )(753) + (1 − 0.42) + 0.24 × 0.42 (900 − 20) 25,000 0.56   0.28 (1 − XB5 ) + 0.24XB5 (800 − 20) + 71.4 − 0.56 Rearrangement leads to 0.9880∆Tav = 341.56 − 550.2XB5

(4)

Two relations between ∆Tav vs XB5 are crossed in a graph to determine both values simultaneously. Tg4 is determined from Eq. (3). Thus we obtain, XB5 = 0.19,

∆Tav = 240 ◦ C,

Tg4 = 1,045 ◦ C

Heat transferred in the rotary kiln is calculated as ⎫ π(2.8)(15)(165)(1,515 − 1,104) ⎪ = 357.9 ⎪ ⎪ ⎬ 25,000 ⎪ π(2.8)(39)(72)(240) ⎪ ⎭ = 237.1 ⎪ 25,000

595

kcal kg pr

Example 7.3 In a rotary kiln to pre-reduce composite pellets of ferro-chromium ore and coke (model illustrated in Figs. 7.6 and 7.7), determine the following variables. Temperature of exit gas Average temperature of flame Necessary fuel oil Table of heat balance is required to compose. Postulates Rotary kiln dti = 4.2 m, dt0 = 4.6 m, lr = 100 m, ks = 1.2 kcal/m hr ◦ C, lW = 0.2 m Hood Surface area 194 m2 , ks = 1.2 kcal/m hr ◦ C, lW = 0.15 m Capacity 18,000 kg dry pellets/hr, 1.223 kgkgpr , 14,720 kg product/hr, N = 2.4 r.p.m. Fuel oil 0.0850 kg oil/kg pr, first approximation QN = 10,000 kcal/kg, C = 86%, H = 14% Coke carbon content 85%, 6,800 kcal/kg Coke in pellets 0.205 kg/kg dry pellet Coke fed to the kiln 0.065 kg/kg dry pellet Primary plus secondary air 1.678 Nm3 /kg pr Tertiary air (74 m) 0.077 Nm3 /kg pr Dry gas 1.854 Nm3 /kg pr, Wet exit gas 1.987 Nm3 /kg pr

116

7. Performance of Rotary Reactors, Direct Heating

Percentage of coke burnt by the air in pellets 25%, outside of pellets 50%     0.205 kgkgpr (0.85)(0.25) + 0.065 kgkgpr (0.85)(0.50) = 0.07119

kg Carbon kg pr

γ = 0.05, χ = 0.19

Component of dry gas

70 m

100 m

CO2 CO O2 N2

16 8 1 75

17 6 3 74

Total

100

100

Solution Volume of CO issued from ore is calculated as, (0.32)(0.55)

3 × 22.4 Nm3 22.4 Nm3 Nm3 + (0.17)(0.90) = 0.1750 2 × 52 kg 55.9 kg kg pr

The heat necessary for the endothermic reactions is given by, (0.32)(0.55)(1,823) + (0.17)(0.90)(683) = 425.3

kcal kg pr

Heat transfer coefficients in the combustion region are calculated using the equations in Chapter 6. Putting df = 1.7 m, dti = 4.2 m, εf = 0.7 m, εm = 0.82, Tf = 1,500 ◦ C, T ∗ = 1,250 ◦ C into Eq. (6.3.1), we obtain, hrg = 232.2

kcal m2 hr ◦ C

In Eq. (6.3.5) use εH = 0.8, εC = 0.85, TH = 1,400 ◦ C and TC = 1,250 ◦ C, εg∗ = 0.4, (hrs )HC = 325.7

kcal m2 hr ◦ C

In Eq. (6.3.6), let us put ke = 1.5 kcal/m hr ◦ C, ρ¯ = 900 kg/m3 , Cs = 0.20 kcal/kg ◦ C, N = 2.4 r.p.m. (hp )HC = 511.2

kcal m2 hr ◦ C

From Fig. 6.7, we get χ = 0.19 for γ = 0.05 in this case. Eq. (6.3.12) leads to

(1 − 0.19)(232.2)(1,500 − TH ) = (0.19)(325.7 + 511.2)(TH − 1,250) +

4.4 1.2 · (TH − TW ) 4.2 0.2

(1)

From Eq. (6.3.11), (hc + hr )W =

5.739(TH − TW ) TW − 20

(2)

By application of Fig. 6.8, TH and TW are graphically determined. TW = 286 ◦ C,

TH = 1,370 ◦ C

qW = (hc + hr )W (TW − 20) = 6,251

kcal m2 hr

Example 7.3

117

In the preheating region, let us assume values of temperatures as the first approximation. Tg = 900 ◦ C,

TC = 700 ◦ C,

TH = 800 ◦ C

Similar calculation results in, hrg = 37.9

kcal , m2 hr ◦ C

TW = 195 ◦ C,

(hrs )HC = 114.0

TH = 720 ◦ C,

kcal , m2 hr ◦ C

qW = 3,063

(hp )HC = 511.2

kcal m2 hr ◦ C

kcal m2 hr ◦ C

When the temperature of the inner surface is 1,000 ◦ C in the burner side hood, combination of Eq. (6.3.11) and Fig. 6.8 determines TW and qW in the case where lW = 0.15 m. TW = 280 ◦ C,

qW = 5,850

kcal m2 hr

Suppose thermal insulation of the hood at the gas exit side is poor, and the temperature of its inner surface is 300 ◦ C lower than the exit gas, the heat loss by thermal radiation from the end of kiln is estimated as, π (4.2m)2 (0.8)(0.8)(4.88) 700+273 4 −  400+273 4  kcal  4 100 100 m2 hr

14,720 kg pr/hr

= 20.3

kcal kg pr

The estimated heat loss per unit mass of product is summarized as follows: Location Surface area [m2 ] [ ◦ C] TH TW [ ◦ C] qW [kcal/m2 hr] L [kcal/kg pr]

Rotary kiln

Hood burner side

30 m

70 m

433.5 1,370 286 6,251 184.1

1,011.6 720 195 3,063 210.5

194 1,000 280 5,850 77.1

The total heat loss is given by, 184.1 + 210.5 + 77.1 + 20.3 = 492

kcal kg pr

In this case, enthalpy balance, Eq. (7.1.1) leads to the following equations. Bf [kg fuel/kg pr]. [Input] Fuel, combustion Comb. CO from ore Comb. of coke [Output] Unburnt CO Heat of reaction Heat of pellets Exit gas Heat loss

 kcal  kg pr

(10,000)Bf (0.1750)(3,045) = (0.205 × 0.25 + 0.065 × 0.5) × (6,800) = (1.854)(0.06)(3, 045) = (0.20)(1,250 − 20) = (1.987)(0.34)(Tge − 20)

532.9 569.5 338.7 425.3 246.0

492.0

[Input] = [Output] gives the next equation, Bf = 0.03996 + 6.756 × 10−5 (Tge − 20)

(3)

118

7. Performance of Rotary Reactors, Direct Heating

The enthalpy balance in the preheating region gives, [Input] Heat of wet gas Comb. heat of CO

(1.910)(0.34)(1,400 − 20) = (1.854)(0.08)(3,045) =

896.3 451.6

[Output] Heat of exit wet gas Comb. heat of CO Heat of solids Heat loss

(1.987)(0.34)(Tge − 20) (1.854)(0.06)(3,045) = (0.20/0.8177)(1,250 − 20) = 210.5 + 20.3 =

338.7 300.8 230.8

[Input] = [Output], we obtain Tge = 727 ◦ C The necessary heat in the combustion-reduction region is given by,  kcal kcal kcal kcal kcal (1,250 ◦ C − 1,100 ◦ C) + 184.1 + 0.20 + 77.1 = 716.5 Hn = 425.3 kg pr kg kg pr kg pr kg pr In this region, TH is estimated to be 1370 ◦ C previously. The mean temperature of the heated surface T ∗ is calculated by Eq. (6.3.14). T ∗ = (0.19)(1,250) + (1 − 0.19)(1,370) = 1,347 ◦ C Letting Fp be the rate of product [kg pr/hr], the heat balance gives,   kcal kcal = π(4.2 m)(30 m) 232.2 2 ◦ (Fp ) 716.5 (Tf − 1,347 ◦ C) kg pr m hr C In the case where Fp = 14,720 kg pr/hr, we get Tf = 1,462 ◦ C. Heat transfer is calculated in the preheating region. The necessary heat here is estimated as,   kg kcal kcal kcal kcal 1.223 0.20 (1,250 ◦ C − 100 ◦ C) + 210.5 + 20.3 = 512.1 kg pr kg ◦ C kg pr kg pr kg pr With Eqs. (7.2.9), we use    kg pr kcal kcal ∆Tav 14,720 512.1 = π(4.2 m)(70 m) 37.90 2 ◦ hr kg pr m hr C Thus, ∆Tav = 215 ◦ C. In the preheating region, temperature profiles are taken as follows: Tg T∗

1,462 ◦ C −→ 727 ◦ C 1,347 ◦ C ←− T ∗

∆T1

115 ◦ C ←− ∆T2

∆Tav = 215 ◦ C =

∆T2 − 115 ∆T

ln 1152

We calculate ∆T2 = 360 ◦ C, T ∗ = 367 ◦ C. With Tge = 727 ◦ C, Eq. (3) gives Bf = 0.08772 kg/kg pr The above value almost coincides with its first approximation. With the above results, the enthalpy balance of the kiln is given as seen in Table 7.3.

Example 7.4

119

Example 7.4 Active carbon is produced in the rotary reactor illustrated in Fig. 7.8. Predict the necessary amounts of the air and exit gas. The composition of the exit gas as well as its gross calorific value are also required. Operating conditions Dry char 400 kg/hr Wet char 480 kg/hr Water 0.2 kg/kg d.f. Dry char C = 90%, QG = 8,000 kcal/kg Steam 1.40 kg/kg d.f., 200 ◦ C Reaction temperature 900 ◦ C, Temperature of exit gas 850 ◦ C Product Active carbon 0.52 kg/kg d.f., Dust loss 0.03 kg/kg d.f. Refractory 1.0 kcal/m hr ◦ C

0.1 m

0.3 kcal/m hr ◦ C

0.2 m

Solution Preliminary calculation gives approximate temperatures as follows:  kcal 

Region

Tg [◦ C]

Tc [◦ C]

T ∗ [◦ C]

TW [◦ C]

qW

Activation Preheating

1,100 900

900 600

1,000 700

105 80

1,063 660

m2 hr

Heat loss from the reactor and hoods in both its sides is calculated by application of the previous procedure. Lt = 630 kcal/kg d.f. Later, kg dry feed is simplified to kg. The carbon efficiency ηc is calculated as, ηc =

0.9 − (0.9)(0.52 + 0.03) = 0.45 0.9

The amount of gasified carbon is given as, 0.9 − (0.52 + 0.03)(0.9) = 0.405 kg carbon/kg The consumed water vapour is (0.405)(18/12) = 0.6075 kg H2 O/kg The unreacted water vapour is 0.2 + 1.4 − 0.6075 = 0.9925 kg H2 O/kg = 1.235 Nm3 /kg As the first approximation, let us assume the amount of dry gas to be Vg = 3.2 Nm3 /kg. With Eq. (7.1.4) we calculate as follows: α ∗ = 1 − (7,838)(0.9)(1 − 0.45)/8,000 = 0.5150   β ∗ = (1.4) 586 + (0.5)(200 − 20) /8,000 = 0.1183

γ ∗ (dry gas) = (3.2)(0.34)(850 − 20)/8,000 = 0.1129   γ ∗ (water vapour) = (0.9925) 586 + (0.5)(850 − 20) /8,000 = 0.1242

120

7. Performance of Rotary Reactors, Direct Heating

γ ∗ (solids) = (0.52 + 0.03)(0.2)(950 − 20)/8,000 = 0.0128 δ ∗ = (630 kcal/kg)/(8,000 kcal/kg) = 0.0788 ω∗ = 0,

ε∗ = 0

Gasification efficiency is obtained from Eq. (7.1.3). ηG = α ∗ + β ∗ − (γ ∗ + δ ∗ ) = 0.5150 + 0.1183 − (0.1129 + 0.1242 + 0.0128 + 0.0788) = 0.3046 From Eqs. (7.1.5) and (7.1.6),   Vg (H2 ) + (CO) = 3.279 × 10−4 (8,000)(0.3046) = 0.7990 Nm3 /kg

VO2 = (0.5818)(0.9)(0.45) + 1.639 × 10−4 (8,000)(0.515 − 0.3046) = 0.5115 Nm3 /kg Vair = 2.436 Nm3 /kg,

VN2 = 1.924 Nm3 /kg

Carbon balance leads to the next equation.   Vg (CO) + (CO2 ) = (0.405)(22.4/12) = 0.756 Nm3 /kg

Let us assume CO, CO2 , H2 and H2 O are in equilibrium at 900 ◦ C. With Eq. (3.5.4) and Table 3.7 we have, Kp = 1.26 =

(CO)(H2 O) (CO2 )(H2 )

Taking Z = Vg (CO), it is determined from the following equation. 1.26 =

Z(1.235) (0.756 − Z)(0.799 − Z)

Z = 0.260 Nm3 /kg The components of dry gas are calculated as follows: [gas] CO H2 CO2 N2

[Nm3 /kg d.f.] 0.260 0.539 0.496 1.924

[%] 8.1 16.7 15.4 59.8

Total

3.215

100.0

The gross calorific value of the above dry gas is, HG = (0.081)(3,020) + (0.167)(3,050) = 754.0 kcal/Nm3 The calculated value of Vg coincides with its first approximation. Thus the results are summarized in Table 7.5. Eq. (3.5.2) gives the necessary heat for gasification of carbon by water vapour.     kg-mol kg kg kcal 0.405 400 31,380 = 423,630 kcal/hr kg hr 12 kg kg-mol Heat loss in the combustion region is estimated to be,    2 kg kcal 400 = 168,000 kcal/hr 630 kg 3 hr Assuming combustion takes place with 12 m, the loading capacity is calculated as kcal 423,630 + 168,000 = 14,234 3 π (2.1)2 (12) m hr 4

Example 7.5

121

The radiant heat transfer coefficient hrg in the activation region is obtained from Fig. 6.6 to be 67 kcal/m2 hr ◦ C. (Flame is non-luminous.) Letting the average temperature difference be ∆Tav , the rate of heat transfer from gas flow to solid surface is given by π(2.1 m)(12 m)(67 kcal/m2 hr ◦ C)(∆Tav ◦ C) = 423,630 + 168,000 kcal/hr Thus we obtain, ∆Tav = 112 ◦ C In the case where T ∗ = 1,000 ◦ C, we estimate Tg = 1,000 + 112 = 1,112 ◦ C

Example 7.5 In a paper-pulp plant, a mixture of poly-propylene sheet and pulp is usually surplus. To produce hydrogen from the stock, a similar rotary gasification reactor to Fig. 7.9 is designed under the following conditions. Predict the necessary amount of oxygen as well as its possible performance. Conditions Feed stock moisture 0.100 kg/kg d.f. dry basis poly-propylene 90% waste pulp 10% carbon 0.8183 kg/kg d.f.; hydrogen 0.1354 kg/kg d.f. QG 9,400 kcal/kg d.f. Feed rate 500 kg d.f./hr Gasification medium Oxygen, steam 1.0 kg/kg d.f. Temp. of exit gas 1,000 ◦ C; Carbon efficiency ηc = 0.95 Heat needed for decomposition at 500 ◦ C, 893 kcal/kg d.f. Rotary reactor dti = 1.8 m, partial combustion region 3.6 m refractory 0.06 m ks = 0.40 kcal/m hr ◦ C 0.06 m ks = 0.17 kcal/m hr ◦ C γ = 0.15 χ = 0.30 N = 3 r.p.m. ρ¯ = 450 kg/m3 ke = 0.5 kcal/m hr ◦ C Cs = 0.24 kcal/kg ◦ C ϕ = 60◦ η1 = 0.57 dtf = 0.8 m εH = εC = εm = 0.8 εf = 0.7 εg∗ = 0.4 Hood 0.5 m × 1.6 m × 4.2 m Solution (1) Heat Transfer Coefficient and Temperature In the combustion region, let us assume, Tf = 1,100 ◦ C

TC = 900 ◦ C

From Eq. (6.3.1) we obtain hrg = 112.7 kcal/m2 hr ◦ C

T ∗ = 1,000 ◦ C

122

7. Performance of Rotary Reactors, Direct Heating

Eq. (6.3.5) gives, (hrs )HC = 137.3 kcal/m2 hr ◦ C With Eq. (6.3.8), (hp )HC = 166.1 kcal/m2 hr ◦ C Substituting the above numerical values to Eq. (6.3.12), (1 − 0.3)(137.3)(1,100 − TH ) = (0.3)(137.3 + 166.1)(TH − 900) +

1.92 1 (TH − TW ) · 1.8 0.06 + 0.06 0.40 0.170

(1)

From Eq. (6.3.11),  1.92  2.04

(hc + hr )W =

1 0.06 + 0.06 0.4 0.17

(TH − TW ) =

TW − 20

1.871(TH − TW ) TW − 20

Two variables, TH and TW are determined by solving Eqs. (1) and (2) with Fig. 6.8. TH = 992 ◦ C

TW = 130 ◦ C

(hc + hr )W = 14.0 kcal/m2 hr ◦ C qW = (14)(130 − 20) = 1,540 kcal/m2 hr In the thermal cracking region we take, Tg = 600 ◦ C εm = 0.8

TC = 500 ◦ C

T ∗ = 550 ◦ C

εg = εg∗ = 0.40

A similar calculation to the above results in the following values. hrg = 38.1

(hrs )HC = 38.1

TH = 520 ◦ C

(hp )HC = 166.1

kcal m2 hr ◦ C

TW = 90 ◦ C

TW = 805 kcal/m2 hr The outer surface areas are combustion and hood region 47.5 m2 thermal cracking region 19.9 m2 Heat loss per 1 kg of dry feed (47.5)(1,5400) + (19.9)(805) kcal = 146.3 + 32.0 = 178.3 500 kg d.f. On the basis of its calorific value, δ∗ =

1,783 = 0.0190 9,400

(2)

Example 7.5

123

(2) Heat for Thermal Decomposition Let us assume 50% of decomposition takes place in the re-circulation region. The necessary heat in the region is,      kcal kcal kcal kg H2 O kcal 586 (500 ◦ C − 20 ◦ C) + 893 (0.5) = 529.1 0.1 + 0.5 ◦ kg d.f. kg kg C kg d.f. kg d.f. 529.1 + 32.0 = 561.1

kcal kg d.f.

(561.1)(500) = 280,550 kcal/hr The re-circulation rate of the solids is calculated with Eq. (2.3.3),  1 π = 9,157 kg solid/hr Fs∗ = (1.8 m)3 cot 60◦ (0.15)(0.57)(450 kg/m3 ) 3 × 60 8 hr Letting Ts be the temperature of the solids, re-circulating from the combustion region,   kcal kcal kg 0.24 (Ts − 500 ◦ C) = 280,550 9,157 hr kg ◦ C hr Ts = 628 ◦ C The rate of radiant heat transfer from flame to the re-circulation region, and to inner surface of combustion region is calculated as follows:  π kcal (1.8 m)2 112.7 2 ◦ (1,100 ◦ C − 628 ◦ C) 4 m hr C  kcal (1,100 ◦ C − 800 ◦ C) + π(1.8 m)(0.8 m) 112.7 2 ◦ m hr C = 135,363 + 152,953 = 288,316 kcal/hr where the length of the inner surface is assumed to be 0.8 m, and the average temperature T ∗ is taken as 800 ◦ C. The above calculation tells us that the heat necessary in the re-circulation region is supplied by radiant heat transfer from the flame in the combustion region. In the combustion region, the necessary heat is calculated as (893)(0.5) + 146.3 = 592.8 kcal/kg d.f. (592.8)(500) = 296,400 kcal/hr The average temperature difference between the flame and the solid surface ∆Tav is calculated by the following equation.   π(1.8 m)(3.6 m − 0.8 m)(112.7 kcal/m2 hr ◦ C) ∆Tav ◦ C = 296,400 kcal/hr ∆Tav = 166 ◦ C

T ∗ = 1,000 ◦ C

Tg = 1,000 + 166 = 1,166 ◦ C (3) Product Gas and Oxygen Required In the case where the temperature of the exit gas is 1,000 ◦ C, it still contains some amount of CH4 . From experimental data of gasification, let us assume here as the first approximation. (CH4 ) = 0.05 (H2 ) + (CO) Volume of dry gas Vg = 2.70 Nm3 /kg d.f. Water vapour to send 1.0 kg/kg d.f.

(1)

124

7. Performance of Rotary Reactors, Direct Heating

From Eq. (7.1.4), α∗ = 1 −

(7,838)(0.8183)(1 − 0.95) = 0.9659 9,400

β∗ =

(1.0)[586 + (0.5)(200 − 20)] = 0.07191 9,400

γ1:∗ =

(2.70)(0.37)(1,000 − 20) = 0.1042 9,400

γ2:∗ =

(1.10)[586 + (0.5)(1,000 − 20)] = 0.1259 9,400

γ ∗ = γ1:∗ + γ2∗ = 0.2301 In Eq. (7.1.3), ω∗ = 0 and ε∗ = 0 ηG = α ∗ + β ∗ − (γ ∗ + δ ∗ ) = 0.9659 + 0.07191 − (0.2301 + 0.019) = 0.7887 From Eq. (7.1.5) and Table 7.1,   Nm3 Vg (H2 ) + (CO) + 3.1149(CH4 ) = 3.279 × 10−4 (9,400)(0.7887) = 2.431 kg d.f.

(2)

Combination of Eqs. (1) and (2) gives,

  Vg (H2 ) + (CO) = 2.103 Nm3 /kg d.f.

(3)

Vg (CH4 ) = 0.1052 Nm3 /kg d.f. From the carbon balance, (0.8183)(0.95)

  12 kg C = Vg (CO) + (CO2 ) + (CH4 ) × kg d.f. 22.4

  Nm3 Vg (CO) + (CO2 ) + Vg (CH4 ) = 1.451 kg d.f.

  Nm3 Vg (CO) + (CO2 ) = 1.451 − 0.1052 = 1.346 kg d.f.

(4)

Letting Vw be the amount of unreacted water vapour, the hydrogen balance gives 0.1354 + (1 + 0.1)

2 2 4 2 kg = 0.2576 = Vg (H2 ) + Vg (CH4 ) + Vw 18 kg d.f. 22.4 22.4 22.4

Vg (H2 ) + Vw = (0.2576)

22.4 − 2(0.1052) = 2.675 2

Combined with Eq. (3), we have Vw = 2.675 − (2.103 − Z) = 0.572 + Z where Z = Vg (CO). The equilibrium constant is Kp = 1.7 at 1,000 ◦ C from Table 3.7. Eq. (3.5.4) gives, Kp = 1.7 =

[CO][H2 O] Z(0.572 + Z) = [CO2 ][H2 ] (1.346 − Z)(2.103 − Z)

By graphical solution, Z = Vg (CO) = 0.823 Nm3 /kg

(5)

References

125

From Eqs. (3), (4) and (5) we obtain, Vg (H2 ) = 1.280 Vg (CO) = 0.823 Vg (CO2 ) = 0.523 Vg (CH4 ) = 0.1052 Vg = 2.731

46.9% 30.1 19.2 3.8

Vw = 1.395 Nm3 /kg = 1.12kg/kg d.f.

100.0%

Vg coincides with its first approximation, 2.70 Nm3 /kg d.f. The necessary amount of oxygen is calculated with Eq. (7.1.6) and Table 7.1. VO2 = (0.5818)(0.95)(0.8183) + 1.639 × 10−4 (9,400)(0.9659 − 0.7887) − (2.731)(0.038)(0.4426) = 0.6794 Nm3 /kg d.f. The gross calorific value of the dry gas is, Hg = (0.469)(3,050) + (0.301)(3,020) + (0.038)(9,500) = 2,700 kcal/Nm3 ηG =

(2.731)(2,700) = 0.784 9,400

The results of the calculations are summarized in Table 7.6.

REFERENCES [1] D. Kunii, Proc. Pan Pacific Synfuel Conference (Tokyo, 1982). [2] D. Kunii, T. Chisaki, Development Engineering of Rotary kilns (CHISAKI Co., Dec. 2005). [3] M. Sadakata, et al., Chemical Engineering Journal 22 (1981) 221.

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–8– Heat Transfer in Rotary Reactors, Indirect Heating This chapter deals with heat transfer mechanisms in a rotary retort, which is heated by electric heater or hot gas from a burner of fuel. Equations to predict the heat transfer coefficients are presented for numerical design calculation.

8.1 8.1.1

NECESSARY INFORMATION FOR SATISFACTORY DESIGN

Material of the retort

Suppose it is planned to develop a gas–solid reaction process and it is decided to apply a rotary retort. We need thorough information on experimental results in the laboratory. Moreover, for the satisfactory design of a commercial rotary reactor, practical information on material for the retort is definitely crucial. For the necessary temperature of the retort, the following materials should be selected: Stainless steel Special alloy for high temperature SiO2 , SiC

700 ◦ C 1,000 ◦ C 1,200 ◦ C

When corrosive gas issues during conversion of solids at a high temperature, it is usually necessary for the selection to be on the safer side, in order to prevent unexpected destruction of the retort after its commercialization. For the safer design of a large retort to operate under critical conditions, therefore, mathematical simulation is recommended at an early stage of the project to predict the distributions of stress and deformation in the retort. Close cooperation with material and mechanical engineers is absolutely necessary at the start of the project. 8.1.2

Thickness of the rotary retort

The thickness of the rotary retort is first determined to give it enough mechanical strength. In addition, its temperature should be as uniform as possible. Let us focus our attention on a portion of the rotary retort. It gets heat from an electric heater or hot gas and raises its temperature. The portion rotates and contacts the solids of low temperature, giving its heat to the solids and reducing its own temperature.

128

8. Heat Transfer in Rotary Reactors, Indirect Heating

The rotary retort functions similarly to a regenerative heat exchanger, and its temperature varies during one rotation. In the case of stainless steel and rotation of 2 r.p.m. the amplitude of its temperature is calculated as follows: Thickness Amplitude

[mm] [◦ C]

1 39

4 9.7

To reduce temperature amplitude, we should use a retort thicker than 4 mm. 8.1.3

Emissivity of the retort surface

When a retort is heated at a high temperature, thermal radiation (mainly infra-red ray) is absorbed on its outer surface. Therefore, the absorptivity of the surface should be high enough. According to fundamentals on radiant heat transfer, we know the following relation. Absorptivity is nearly equal to emissivity, and then unity minus reflectivity. The emissivity for the metal surface of a brand new retort is usually very small. It is recommended at first, to make its value large, for instance, by means of preliminary oxidation of the surface. Inside the retort, the radiant heat transfer from its inner surface to the rotating solids is also important. The emissivity of the inner surface should be large, similar to its outer surface. 8.1.4

Sticking of solids and formation of thick layer

When a small amount of impurities is contained, the fine solids are apt to stick to the heating surface and form a thick layer at a considerably lower temperature than the melting point of the main solid particles. In the case of a rotary retort, such phenomena drastically reduce the heat transfer and then lessen its capacity. The formation of coarse lumps sometimes takes place due to fragments being separated from the thick solids layer. At a fairly early stage of the project, the existence of the above phenomena and possible methods of how to get rid of them, should be studied. It is preferable to fabricate a test rotary reactor, specially to clarify the above problems. When the authors developed a rotary reactor to eliminate polychloro-dibenzo-dioxins in fly ash from an incinerator of MSW, they encountered the above problem which they solved by positioning scraping plates inside (see Example 9.2).

8.2

HEAT TRANSFER WITHIN THE ROTARY RETORT

Fig. 8.1 shows a simplified model of a rotary retort, the inner diameter of which is dti and the thickness δH0 . Let us take into consideration the solid layer sticking to its inner surface, the thickness of which is δHi and assume that the emissivity of gas filling the retort can be ignored at its operating conditions. In the case where δHi , δH0 ≪ dti , we can take dti = dt0 = dt

8.3. Heat transfer from an electric heater

129

Figure 8.1 Heat transfer model in rotary retort.

In Fig. 8.1, the heat transfer coefficient (hrs )HC is calculated by Eq. (6.3.5) with εg∗ = 0, and (hp )HC is given by Eq. (6.3.8). The heat transfer coefficient hi , based on the inner surface area of the retort, summarizes the transfer mechanisms within the retort as follows: hi =

1 δH0 kH0

+

δHi kHi

+

1 χ[(hrs )HC +(hp )HC ]

(8.2.1)

Where kH0 and kHi are the thermal conductivity of retort and of the sticking solid layer, respectively. When a partition plate is positioned as seen in Figs. 2.7 and 2.8, Eq. (6.3.8) is not rigorously applied. Instead, an approximate calculation gives, (hp )HC = 1.4(hp )HC

8.3

from Eq. (6.3.8)

(8.2.2)

HEAT TRANSFER FROM AN ELECTRIC HEATER

Fig. 8.2 illustrates a model of heat transfer from electric heating elements to the outer surface of a rotary retort. Radiant heat transfer from the inner surface of the insulation cover cylinder to the outer surface of the retort is represented by a heat transfer coefficient, given by Eq. (8.3.1).  +273 4  TH0 +273 4  εS εH (4.88) TS100 − 100 (hrs )SH = TS − TH0

(8.3.1)

In case we can neglect the heat losses at both end sections of the retort and the insulation cover, the heat transfer mechanism is designated by the following equations approximately,

130

8. Heat Transfer in Rotary Reactors, Indirect Heating

Figure 8.2 Model of heat transfer to rotary retort, electric heating.

on the basis of the unit length of the heating region. 

kcal 860 kW·hr



W kW lr m

 ks 1 = (πdt0 )(hrs )SH (TS − TH0 ) + (π d¯t ) (TS − TW ) 2 lW (8.3.2)

ks (π d¯t ) (TS − TW ) = (πdtW )(hc + hr )W (TW − T0 ) (8.3.3) lW    kcal W kW 1 860 + (πdt0 )(hrs )SH (TS − TH0 ) = (πdt )hi (TH0 − TC ) kW·hr lr m 2 (8.3.4) hi (TH0 − TC ) =

kHi (TH0 − THi ) δHi

(8.3.5)

From Eqs. (8.3.2) and (8.3.4), we obtain Eq. (8.3.6).

(2πdt0 )(hrs )SH (TS − TH0 ) + (π d¯t )

ks (TS − TW ) = (πdt )(hi )(TH0 − TC ) lW

(8.3.6)

8.3. Heat transfer from an electric heater

131

Figure 8.3 Results of Example 8.1.

With the given values of TC and TS , we can calculate TH0 , THi , TW and W/ lr from the above equations and Fig. 6.8. Example 8.1 explains the algorithm of the above calculation in a rotary retort of dt = 1 m, the results of which are shown in Fig. 8.3. We find considerable difference takes place between the temperatures of the retort and the solids, due to the layer of solids sticking to its inner surface. After numerical determination of TS and TH0 at a given value of W/ lr , we can calculate the approximate temperature of the elements with the following equation.

 ∗    TS + 273 4 TE + 273 4 W − = 860 (πdtE )(nE )(εH εE )(4.88) 100 100 lr

(8.3.7)

where dtE , nE and εE are the diameter, number and emissivity of heating elements, respectively. TS∗ =

TS + TH0 2

(8.3.8)

The calculated values of TE are added to Fig. 8.3. When electric heating elements are evenly distributed on the inner surface of the insulation cover or panel heaters, the temperature of the elements is approximately given by TS

132

8. Heat Transfer in Rotary Reactors, Indirect Heating

in the preceding equations. Eqs. (8.3.2) and (8.3.4) are unified to give,  ks W = (πdt ) (TS − TW ) + (πdt )hi (TH0 − TC ) (860) lr lW

(8.3.9)

This algorithm is simpler than the former case.

8.4 8.4.1

HEAT TRANSFER FROM GAS FLOW

Examples of practical design

In Table 8.1, we find examples of practical design, in the case where solids are heated indirectly by flowing hot gas. Example [A] is applied for processes of intermediate capacity at moderate temperature, such as elimination of residual carbon in spent catalyst at ca. 500 ∼ 600 ◦ C. Example [B] is designed for larger capacity and higher temperature. In the case where a rotating bundle of retorts is needed to be heated to 1,000 ◦ C, local heating of the retort should be definitely prevented and the gas flow at uniform temperature must then be sent into the space outside the bundle. At high temperature, heat is transferred from the combustion gas to the rotary retort, mainly by thermal radiation. When special baffle plates are positioned around the retort, the gas can contact the retort surface with considerably high linear velocity, giving an appreciable value of convectional heat transfer coefficient. Solids can be heated by contact with a number of heating tubes, as illustrated in Table 8.1 [C]. This design can be utilized in the case where solids do not clog the spaces between tubes. When hot air flows through the heating tubes, only convectional heat transfer takes place. Suppose the air is replaced by combustion gas at the same temperature. Thermal radiation from carbon dioxide and water vapour increases the total rate of heat transfer appreciably. 8.4.2

Proposed design for gas flow

For a rotary retort of considerable capacity, the surface area of heat transfer should be large, so that a design such as shown in Table 8.1 [B] and [C] is justified. In Chapter 3, the authors pointed out that a partition plate in the rotating cylinder can be applied extensively to improve the residence time characteristics of rotating solids. Such a partition plate can be utilized as an extended surface for heat transfer. Fig. 8.4 is a design for this purpose, proposed by the authors. 8.4.3

Simplified model

Heat transfer coefficient from the hot gas to the surface of the retort h0 can be calculated by application of a simple model, shown in Figs. 8.5 and 8.6. In Fig. 8.6, an infra-red ray is emitted from the flowing gas at high temperature, arrives at both surfaces of the retort and insulation cover cylinder. For design calculation, it is convenient to employ heat transfer coefficients by thermal radiation as follows:

8.4. Heat transfer from gas flow

133

Table 8.1. Indirect heating by hot gas flow Gas flow Outside rotary reactor

Example of design

Remarks Radiant heating from solid surface.

Large surface area. Prevention of hot spots at high temperature.

Inside rotary reactor

Large surface area. In case solids do not clog between tubes. Hot combustion gas is superior to the hot air.

134

8. Heat Transfer in Rotary Reactors, Indirect Heating

Figure 8.4 Extension of heat transfer surface area.

Figure 8.5 Heat transfer to retort surface.

From gas to retort, approximately  T +273 4  TH0 +273 4  εg εH (4.88) g100 − 100 hrg = Tg − TH0

(8.4.1)

From the inner surface of the insulation cover to the outer surface of the retort,

(hrs )SH =

 +273 4  TH0 +273 4  εS (1 − εg )εH (4.88) TS100 − 100 TS − TH0

(8.4.2)

8.4. Heat transfer from gas flow

135

Figure 8.6 Model of heat transfer.

It should be kept in mind that the above approximation is possible in the case where εH and εS are close to unity (i.e. black body). Otherwise we have to calculate them in accordance with a rigorous enclosure theory. In practical design, however, εH and εS are usually close to unity, justifying the above approximation. In the case where the linear velocity of the gas flow at the heating surfaces is large, the convectional heat transfer coefficient hC must be comparable to hrg in parallel. Suppose the insulation is thick enough and heat loss is negligible compared with the heat flux from flowing gas, the thermal energy received by the inner surface of the cover cylinder is emitted as infra-red ray, and then transferred to the surface of the retort. Based on the unit surface area of the retort, the mechanism of heat transfer within the space between the insulation cover cylinder and the retort can be represented by the following equations. At the inner surface of the insulation cover cylinder (hc + hrg )(Tg − TS ) = (hrs )SH (TS − TH0 )

(8.4.3)

The total heat transferred to the retort qH is given by (hc + hrg )(Tg − TH0 ) + (hrs )SH (TS − TH0 ) = qH = h0 (Tg − TH0 )

(8.4.4)

TS is obtained from Eq. (8.4.3), and submitted to Eqs. (8.4.4). Thus the summarized heat transfer coefficient h0 is calculated by the following equation. h0 = hc + hrg +

1 1 hc +hrg

+

1 (hrs )SH

(8.4.5)

136

8.4.4

8. Heat Transfer in Rotary Reactors, Indirect Heating

Overall heat transfer coefficient and temperature of retort

The overall heat transfer coefficient U is calculated by a combination of Eq. (8.2.1) and (8.4.5). U=

1 1 hi

+

(8.4.6)

1 h0

From the heat balance of the surface of the retort we get, TH0 =

h0 Tg + hi TC h0 + hi

(8.4.7)

We can predict the temperature of the inlet gas flow, which is allowed at a critical temperature of the retort. In the illustrative calculations of Examples 8.1 and 8.4, the main resistance to heat transfer is the layer of solids sticking to the inner surface of the retort. For deterministic design, its scraping is the most preferable to maintain a high rate of heat transfer. Example 8.1 Calculate the temperatures and heat transfer coefficient to the solids in the rotary retort shown in Fig. 8.2. Data Retort dt = 1 m, δHi = 8 mm, kHi = 0.4 kcal/m hr ◦ C, δH0 = 10 mm, kHi = 20 kcal/m hr ◦ C W/ lr = 20 kW/m TC = 500 ◦ C, εc = 0.7, εH = 0.8, γ = 0.20, χ = 0.34 From Fig. 6.7 kg kcal ◦ Cs = 0.24 kg ¯ = 450 m ◦C , ρ 3 , N = 3 r.p.m., ke = 0.5 kcal/m hr C Insulation cover cylinder dtS = 1.02 m + 0.06 m × 2 = 1.14 m dtW = 1.14 m + 0.15 m × 2 = 1.44 m d¯t = (1.14 + 1.44)/2 = 1.29 m ks = 0.14 kcal/m hr ◦ C, lW = 0.15 m, εS = 0.8 Heating elements dtE = 30 mm, nE = 12, εE = 0.8 Solution Let us calculate the heat transfer coefficients within the retort. From Eq. (6.3.5) with εg∗ = 0 we obtain (hrs )HC =

 4  500+273 4  (0.8)(0.7)(4.88) 600+273 − 100 100 600 − 500

= 61.17

From Eq. (6.3.8),   (0.5)(450)(0.24)(3 × 60) 1/2 kcal (hp )HC = 1.13 = 191.1 2 ◦ 0.34 m hr C Thus Eq. (8.2.1) gives, kcal 1 hi = 0.01 0.008 = 31.10 2 ◦ 1 m hr C 20 + 0.4 + (0.34)(61.17+191.1)

kcal m2 hr ◦ C

Example 8.1

137

On the other hand, the heat transfer coefficient from the inner surface of the insulation cover cylinder to the outer surface of the retort is calculated by Eq. (8.3.1) as

(hrs )SH =

 4  600+273 4  (0.8)(0.8)(4.88) 650+273 − 100 100 650 − 600

= 90.53

kcal m2 hr ◦ C

From Eq. (8.3.5) we get (31.10)(TH0 − 500) =

0.40 (TH0 − THi ) 0.008

Rearranging, THi = 0.3780TH0 + 311

(1)

Eq. (8.3.3) gives, π(1.29)

0.14 (TS − TW ) = π(1.44)(hc + hr )W (Tw − 20) 0.15

Thus, (hc + hr )W =

0.8361(TS − TW ) TW − 20

(2)

From Eq. (8.3.6), (2)(1.02)(90.53)(TS − TH0 ) + (1.29)

0.14 (TS − TW ) = (1)(31.10)(TH0 − 500) 0.15

Rearranging, TH0 = 72.1 + 0.8614TS − 0.00558TW

(3)

For a given value of TS , TW is calculated by application of Eq. (2) and Fig. 6.8. Thus TH0 and THi are calculated with Eqs. (3) and (1). On the other hand, Eq. (8.3.2) gives, 

860 2



W lr



= π(1.02)(90.53)(TS − TH0 ) + π(1.29)

0.14 (TS − TW ) 0.15

Thus we get, W = 0.6747(TS − TH0 ) + 0.00880(TS − TW ) lr

(4)

The temperature of the heating elements, TE , is calculated by application of Eq. (8.3.7) as follows: π(0.03)(12)(0.8)(0.8)(4.88)



   ∗ TS + 273 4 W TE + 273 4 = 860 − 100 100 lr

(5)

The results of the above calculation are given in Fig. 8.3, from which we obtain the following values for the temperature. TE = 790 ◦ C,

TS = 680 ◦ C,

TH0 = 650 ◦ C,

The temperature difference due to the sticking layer of solids is 650 − 550 = 100 ◦ C

THi = 550 ◦ C,

TW = 65 ◦ C

138

8. Heat Transfer in Rotary Reactors, Indirect Heating

Example 8.2 The residual carbon in spent catalyst from a petroleum refinery is oxidized in a rotary retort, which is heated by combustion gas of kerosene (see Example 5.2). In this case, the solids do not stick to the inner surface of the retort. Combustion gas is sent to the unit as illustrated in Table 8.1 [A], and induced from its bottom. Predict the overall heat transfer coefficient from gas to solids under the following conditions. The properties of solids and the notation are the same as for Example 5.2. Fuel oil C 86% H 12% Air needed stoichiometrically 10.84 Nm3 /kg oil Wet combustion gas 11.51 Nm3 /kg oil In case of 50% excess air Volume of wet gas 16.93 Nm3 /kg oil Gas

CO2

H2 O

N2

O2

Total

%

9.5

7.9

75.9

6.7

100

Effective length of emissive gas lG = 0.4 m Heat transfer coefficient by convection 8 kcal/m2 hr ◦ C Tg = 1,000 ◦ C, TC = 550 ◦ C εH = 0.8, εS = 0.8 As the first approximation let us assume TS = 900 ◦ C, TH0 = 750 ◦ C Solution The heat transfer coefficient hi within the retort is given by Eq. (8.2.1), applying the calculated values in Example 8.1. 1 kcal = 82.24 2 ◦ hi = 0.01 1 m hr C 20 + (0.34)(61.17+191.1) Graphs presented by Hottel [1] are very useful to estimate the emissivity of gas εg . PCO2 lG = (0.095)(0.4) = 0.038 m·atm PH2 O lG = (0.079)(0.4) = 0.032 m·atm At 1,000 ◦ C, we have εCO2 = 0.08,

εH2 O = 0.04,

εg ≈ 0.08 + 0.04 = 0.12

From Eq. (8.4.1),

hrg =

 4  750+273 4  (0.12)(0.8)(4.88) 1,000+273 − 100 100 1,000 − 750

= 28.69 kcal/m2 hr ◦ C

Eq. (8.4.2) gives

(hrs )SH =

 4  750+273 4  (0.8)(1 − 0.12)(0.8)(4.88) 900+273 − 100 100 900 − 750

= 146.2 kcal/m2 hr ◦ C

Example 8.3

139

With Eq. (8.4.5) we calculate h0 = 8 + 28.69 +

1 1 1 8+28.69 + 146.2

= 66.02 kcal/m2 hr ◦ C

From Eq. (8.4.6), U=

1 1 1 82.24 + 66.02

= 36.62

kcal m2 hr ◦ C

The temperature of the retort is calculated Eq. (8.4.7), TH0 =

(66.02)(1,000) + (82.24)(550) = 750 ◦ C 66.02 + 82.24

The calculated value is just the same as its first approximation. The maximum heat flux across the surface of the retort is predicted by,  q = U (Tg − TC ) = 36.62

kcal (1,000 ◦ C − 550 ◦ C) = 16,380 kcal/m2 hr ◦ m2 hr C

Example 8.3 In a rotary retort, Table 8.1 [B], thermal decomposition of metal sulfate particles is carried out, the maximum temperature of which is 1,000 ◦ C. In order to prevent its destruction, it is strictly required to keep the temperature of the retort lower than 1,080 ◦ C. Predict the allowable temperature of the inlet combustion gas. Data Retort δH0 = 10 mm, ks = 20 kcal/m hr ◦ C, N = 6 r.p.m. δHi = 15 mm, ks = 0.7 kcal/m hr ◦ C Rotating solids ρ¯ = 1,000 kg/m3 , Cs = 0.2 kcal/kg ◦ C ke = 0.80 kcal/m hr ◦ C γ = 0.10, χ = 0.25, εH = 0.80, εC = 0.85 Outside the retort εH = 0.8, εS = 0.8, εg = 0.12 As the first approximation, assume as follows: Tg = 1,150 ◦ C, TS = 1,100 ◦ C, TH0 = 1,080 ◦ C, hC = 8 kcal/m2 hr ◦ C Solution Let us use Eq. (6.3.5) with εg∗ = 0. (hrs )HC =

 4  1,000+273 4  (0.80)(0.85)(4.88) 1,080+273 − 100 100 1,080 − 1,000

= 300.7 kcal/m2 hr ◦ C

With Eq. (6.3.8),   (0.8)(1,000)(0.20)(6 × 60) 1/2 = 542.4 kcal/m2 hr ◦ C (hP )HC = 1.13 0.25 With Eq. (8.2.1), 1 hi = 0.01 0.015 = 37.74 kcal/m2 hr ◦ C 1 20 + 0.70 + (0.25)(300.7+542.4)

140

8. Heat Transfer in Rotary Reactors, Indirect Heating

From Eq. (8.4.1),  4  1,080+273 4  (0.12)(0.8)(4.88) 1,150+273 − 100 100

hrg =

1,150 − 1,080

= 50.14 kcal/m2 hr ◦ C

With Eq. (8.4.2),

(hrs )SH =

 4  1,080+273 4  (0.80)(1 − 0.12)(0.8)(4.88) 1,100+273 − 100 100 1,100 − 1,080

= 278.4 kcal/m2 hr ◦ C

With Eq. (8.4.5) we obtain, h0 = 8 + 50.14 +

1 1 1 8+50.14 + 278.4

= 106.2 kcal/m2 hr ◦ C

From Eq. (8.4.6), U=

1 1 1 37.74 + 106.2

= 27.71

kcal m2 hr ◦ C

Eq. (8.4.7) gives TH0 =

(106.2)Tg + (37.74)(1,000) = 0.7378Tg + 262 106.2 + 37.74

TH0 = 1,080 ◦ C at

Tg = 1,109 ◦ C

The allowable heat flux is calculated as q = h0 (Tg − TH0 ) = (106.2)(1,109 − 1,080) = 3,080 kcal/m2 hr Where the thickness of the solids layer sticking to the inner surface of the retort δHi can be reduced to 5 mm by proper design of a scraper, we calculated as follows: hi = 80.73 kcal/m2 hr ◦ C

U = 45.86 kcal/m2 hr ◦ C

We can achieve a 45.86/27.71 = 1.66 times increase of heating capacity.

Example 8.4 The rotary reactor illustrated in Fig. 8.5 is designed to gasify combustible waste materials issuing from a plant in the chemical industry. Calculate the heat transfer coefficients under the following conditions. Breadth of channels for gas flow 3 cm CO2 9.5%, H2 O 7.9% Average gas temperature 900 ◦ C Average solid temperature 700 ◦ C Average gas velocity 20 m/sec Assume hi in Example 8.1 can be used. Solution The emissivity of the gas is calculated in accordance with a procedure presented by Hottel [1]. The effective thickness of the gas flow is lg = (3 cm) × 2 = 6 cm

Example 8.4

141

Thus, PCO2 · lG = (0.095)(0.06) = 0.006 m·atm PH2 O · lG = (0.079)(0.06) = 0.005 m·atm From graphs we get, εCO2 = 0.035,

εH2 O = 0.010

Approximately, εg = 0.035 + 0.010 = 0.045. As the first approximation, let us take TH0 = 800 ◦ C, TS = 850 ◦ C. With Eq. (8.4.1),  4  800+273 4  (0.045)(0.8)(4.88) 900+273 − 100 100

hrg =

900 − 800

= 9.97 kcal/m2 hr ◦ C

The convectional heat transfer coefficient is calculated by the Sieder–Tate equation. hc dte = 0.023Pr0.4 Re0.8 kg  gm  cm  (3 cm × 2) 3 × 10−4 2,000 sec cm3 Re = = 8,000 gm 4.5 × 10−4 cm·sec

(1)

Pr = 0.65 Thus, we have hc = (0.023)(0.65)0.4 (8,000)0.8 ×



 0.065 mkcal hr ◦ C = 27.80 kcal 0.03m × 2 m2 hr ◦ C

From Eq. (8.4.2), (hrs )SH =

 4  800+273 4  (0.8)(1 − 0.045)(0.8)(4.88) 850+273 − 100 100 850 − 800

= 158.0 kcal/m2 hr ◦ C

Eq. (8.4.5) gives, h0 = 27.8 + 9.97 +

1 1 1 27.80+9.97 + 158.0

= 68.3

kcal m2 hr ◦ C

From Eq. (8.4.6), U=

1 1 1 82.24 + 68.3

= 37.3

kcal m2 hr ◦ C

With Eq. (8.4.7) we get, TH0 =

(68.3)(900) + (82.24)(700) = 791 ◦ C 68.3 + 82.24

almost the same as its first approximation. The above values of U and TH0 are obtained for the outer surface of the retort, which can get radiant heat from the inner surface of insulation cover. If a hot solid surface does not exist, the rate of heat transfer decreases. Thus, U = 25.88

kcal m2 hr ◦ C

and TH0 = 763 ◦ C

142

8. Heat Transfer in Rotary Reactors, Indirect Heating

A 30% lower value of U needs more surface area. However, this disadvantage can be eliminated if we put a black plate inside, parallel to the surface of the partition plate. For the heating surface shown in Fig. 8.4, it is plausible to do so to enhance heat transfer.

REFERENCE [1] H.C. Hottel, Chapter 3 in “Heat Transmission, 3rd edition” by McAdams (McGraw Hill, New York, 1954)

–9– Performance of Rotary Reactors, Indirect Heating Based on the equations acquired in Chapter 8, the procedure for designing a rotary retort heated by electric heater is presented. To reduce consumption of electricity, a novel rotary reactor is proposed, which recovers the apparent heat of solids by heat exchange between two solid flows. When a rotary retort is heated by hot gas flow, radiant heat from the inner surface of the insulation cover should be taken into account. For oxidation of residual carbon in spent catalyst, the performance of a novel rotary reactor is predicted and compared with conventional ones. Directions for improvement to drastically reduce energy consumption is proposed.

9.1 9.1.1

ELECTRIC HEATING

Enthalpy balance

Fig. 9.1 shows a simplified model of a rotary retort which is heated by electric heaters. The enthalpy balance in this special case is represented by the following equations, per unit mass of the feed stock. (860)

W = Cs (Tcm − Tc1 ) − Hc∗ + Va Ca (Tae − T0 ) + L1 + L2 + L3 Fs

(9.1.1)

where Va and Tae are the volume of the carrier gas per unit mass of the feed stock and its exit temperature. L1 , L2 and L3 are heat losses from the inlet part of the retort, from the insulation cover of the electric heater and from the outlet part of the retort, respectively [kcal/kg d.f.]. Furthermore, Hc∗ is the thermal energy, evolved from exothermic reaction, occurring in the retort. In the case of such an endothermic reaction as thermal decomposition of limestone, the minus sign of Hc∗ is changed to a plus sign. For satisfactory design, we should determine an appropriate and safe temperature of the retort, taking into account its strength, sticking of solids, corrosion and also its cost. Suppose a desirable diagram of solid temperature vs time is acquired from experimental data in a batch reactor. To reproduce the scheme in a practical rotary reactor under steady state operation, the heating elements should be divided into three or more regions, to each of which the input of electric power should be independently controlled (see Example 9.1). If we apply the rotary retort shown in Fig. 9.1, we cannot avoid a high consumption of electricity. Suppose fly ash containing toxic dioxins is heated to 500 ◦ C and cooled. In this

144

9. Performance of Rotary Reactors, Indirect Heating

Figure 9.1 Model of an electrically heated rotary retort.

case we obtain the following data by preliminary calculation, L1 = 10

L2 = 20 3

Va = 0.25 Nm /kg d.f. Tae = 500 ◦ C

L3 = 6 kcal/kg d.f. Ca = 0.31 kcal/Nm3 ◦ C

Cs = 0.24 kcal/kg ◦ C

Eq. (9.1.1) gives  1  W = (0.24)(500 − 20) + (0.25)(0.31)(500 − 20) + 10 + 20 + 6 Fs 860 =

1 kW·hr kW·hr [115.2 + 37.2 + 36] = 0.2191 = 219.1 860 kg d.f. ton d.f.

The cost of electric power is considerably high. 9.1.2

Improvement of electric heating

In Fig. 9.1, the solids are heated to a high temperature and physical and chemical transformations then take place within an appropriate residence time. However, their heat content is not needed and should be removed in a succeeding cooler. To reduce electric input drastically for the same purpose, the authors have developed a novel rotary reactor that can recover the heat content of the hot solids within itself. Fig. 9.2 illustrates its principle.

9.1. Electric heating

145

Figure 9.2 Simplified model of a novel retort for heat recovery from hot solids.

Hot solids, heated to their maximum temperature, move into a coaxial inner cylinder, and are transferred toward its outlet. Meanwhile, the hot solids give their heat to cold solids, moving in the annular space between two cylinders. Heat exchange takes place between the hot and cold solids. Partition plates, attached with inclined guide plates, are satisfactory to achieve the above flow pattern of solids. If the heat exchange region is designed adequately, the temperature of the exiting solids can be reduced to 120 ◦ C. Replacing Tcm by Tc3 in Eq. (9.1.1) and applying the same numerical values as the previous paragraph, we have,  1  W = (0.24)(120 − 20) + (0.25)(0.31)(120 − 20) + 10 + 20 + 6 Fs 860 = 0.0672

kW·hr kW·hr = 67.2 kg d.f. ton d.f.

The system shown in Fig. 9.2 can reduce the consumption of electricity to 67.2/219.1 = 0.307. Such a reduction of electricity input is more pronounced when the maximum temperature is higher than 500 ◦ C, for example, when heating ferrite powder at 800 ◦ C. In Fig. 9.2, the coaxial inner cylinder has another function, by rotating at a little slower speed than the main retort. Partition plates, positioned on the outer surface of the inner cylinder, scrape the solids sticking to the inner surface of the retort. The counter-current flow of air is effective in this system because of the following functions: (1) to recover heat from the exiting solids and utilize it to preheat the cold solids, thus reducing the surface

146

9. Performance of Rotary Reactors, Indirect Heating

area necessary for heat exchange; (2) to oxidize residual carbon in the ash and reduce input of electric energy. 9.1.3

Working equations for design calculation of the new heating system

The novel heating system shown in Fig. 9.2 reduces the cost of electric energy to 1/3 ∼ 1/4. It has been commercialized for eliminating toxic polychloro-dibenzo-dioxins in fly ash from a municipal solid-waste incineration plant by the authors and Takuma Co. (see Example 9.2). Fig. 9.2 appears to be complicated but it can be designed theoretically on the basis of fundamental equations for heat balance and rate of heat transfer. The amount of heat exchange between two solids flows in the heat recovery region, QH1 [kcal/kg d.f.], is given by, QH1 = Cs (Tc2 − Tc1 ) − Va Ca (Ta2 − Ta3 ) + L1  ∗  = Cs Tc2 − Tc3 − Va Ca (Ta2 − Ta1 )

(9.1.2)

From Eq. (9.1.2) we obtain,

 ∗   Cs Tc2 − Tc2 − (Tc3 − Tc1 ) = L1 + Va Ca (Ta3 − Ta1 )

(9.1.3)

Q∗ + QH2 + Hc∗ + Va Ca (Tam − Ta2 ) = Cs (Tcm − Tc2 ) + L3

(9.1.4)

From the heat balance in the annular space of the electrically heating region,

From the heat balance in the inner cylinder,   ∗ QH2 = Cs Tcm − Tc2 − Va Ca (Tam − Ta2 )

(9.1.5)

where Q∗ is the heat transferred from electric heaters, QH2 is the heat exchanged between two solids flows, and Hc∗ is the heat of exothermic reaction [kcal/kg d.f.]. Based on information from Chapter 8, we calculated the temperature difference between gas flows and wall surface in the above system and found the following equations can be approximately assumed. ∗ Tc2 + Tc2 2 Tam = Tcm

Ta2 =

(9.1.6)

Combination of Eqs. (9.1.4) and (9.1.5) gives  ∗  Q∗ = Cs Tc2 − Tc2 + L3 − Hc∗

(9.1.7)

Input to the heaters, W/Fs [kW hr/kg d.f.], is given by   kcal W 860 = Q∗ + L2 kW·hr Fs

(9.1.8)

9.1. Electric heating

147

Substitution of Eq. (9.1.7) into Eq. (9.1.8) and rearrangement by means of Eq. (9.1.3) results in Eq. (9.1.9).  W (860) = Cs (Tc3 − Tc1 ) + Va Ca (Ta3 − Ta1 ) + L1 + L2 + L3 − Hc∗ Fs

(9.1.9)

In the case of endothermic reaction, the minus sign of Hc∗ is changed to a plus sign. The values of heat transfer, i.e. Q∗ , QH1 and QH2 , are represented as follows: Fs Q∗ = (πdt l2 )hi (∆Tav )∗2 (∆Tav )∗2 =

(TH02 − Tc2 ) − (TH0m − Tcm ) H02 −Tc2 ln TTH0m −Tcm

  Fs QH1 = πdt∗ l1 U1 (∆Tav )1

(∆Tav )1 =

∗ (Tc2

− Tc2 ) − (Tc3 − Tc1 ) T ∗ −T

c2 ln Tc2 c3 −Tc1

  Fs QH2 = πdt∗ l2 U2 (∆Tav )2

(∆Tav )2 =

∗ (Tc2

− Tc2 ) − ∆Tcm ln

∗ −T Tc2 c2 ∆Tcm

(9.1.10) (9.1.11) (9.1.12) (9.1.13) (9.1.14) (9.1.15)

where dt∗ is the diameter of the inner cylinder, and ∆Tcm is the temperature difference between two flows of solids at the left end of the reactor in Fig. 9.2. 9.1.4

Prediction of performance

Example 9.2 explains the design procedure of this reactor for elimination of polychlorodibenzo-dioxins contained in fly ash from an incineration plant. Compared with the results obtained in Example 9.1 for a conventional rotary reactor, we find a two-thirds reduction in electric energy is possible in practice. Since the retort and inner cylinder are horizontal, appropriate partition plates, attached with inclined guide plates, are needed to keep the volumetric fraction of solids in its annular space and inner cylinder at γ = 0.1 ∼ 0.2, at a given feed rate. Eqs. (2.3.6) and (2.3.7) are utilized to design such a structure (see Example 2.4). When the retort rotates by 3 r.p.m. and the inner cylinder rotates by 2 ∼ 2.5 r.p.m., the partition plate positioned on the outer surface of the inner cylinder functions as an effective scraper. The construction cost of the model shown in Fig. 9.2 must be higher than that of Fig. 9.1. However, the model shown in Fig. 9.2 excludes a cooler, which is inevitable in the case of Fig. 9.1. 9.1.5

Direction of improvement

To apply a rotary reactor for larger capacity, we need a wider surface area to exchange the heat between two flows of solids. Thus the authors propose to design the heat exchange

148

9. Performance of Rotary Reactors, Indirect Heating

Figure 9.3 Proposed scheme of a heat exchanger between two flows of solid.

region by a unit shown in Fig. 9.3. Based on previous equations in this book, the performance of the unit can be predicted quantitatively. This improvement must broaden the applicability of the system in Fig. 9.2, further to any processes, electrically heated. Another improvement is to install tubes to inject the air into the rotating layer of ash. The scheme in Table 1.2 [C] can be easily applied (see Example 9.2). As shown in Eq. (9.1.9), the evolved heat from the oxidation of residual carbon in the fly ash reduces the necessary electrical energy considerably. By injecting air into the fly ash, we can estimate that 50% of the residual carbon is oxidized. On the same basis as Example 9.2, W/Fs is calculated and compared with the results of the other cases. Case W Fs

Fig. 9.1 Ex. 9.1 182.4

 kW·hr  ton

Fig. 9.2 Ex. 9.2 60.3

Air injection 0.921

By injection of air, it is possible to operate a reactor such as that shown in Fig. 9.2, without input of electrical energy at steady state.

9.2 9.2.1

HEATING BY COMBUSTION GAS

Oxidation of residual carbon in spent catalyst

Fig 9.4 shows a model of a rotary reactor for removing the residual carbon in spent catalyst from a petroleum refinery. Hot gas from the burning flame is sent into heating compartments composed of refractory wall, and then heats the rotary retort. In this case, flue gas goes out from its bottom. A number of horizontal tubes are positioned in the retort, from which the air is sent into the rotating layer of catalyst to oxidize the residual carbon. For adjustment of residence time characteristics in a practical reactor, four spiral plates are attached to the inner cylinder. The operating conditions for the reactor are supposed to be, Catalyst 250 kg/hr initial carbon 5%, which should be reduced to 0.05% at 600 ◦ C and tr = 20 min. Necessary air for complete conversion of carbon 0.51 Nm3 /kg feed.

9.2. Heating by combustion gas

149

Figure 9.4 Model of a rotary reactor for elimination of residual carbon in spent catalyst.

Retort 0.9 m ID, total 10 m, heating region 6 m. 6 compartments for combustion and heating region, angle of rotation axis 2◦ . Fundamental information on this process is found in Chapters 3 and 5. To predict the thermal performance of this reactor, let us apply the simplified model shown in Fig. 9.5 where Contour 1 covers the system as a whole, and Contours 2 and 4 are for the rotary retort. Contour 3 is for the refractory insulation cover, and Contour 5 is for its combustor. Enthalpy balances, on the basis of 1 kg feed stock, are formulated as follows: Contour 1 Bf QN + Vaf Ca (Taf − T0 ) + Vap Cg (Tap − T0 ) + Hc∗ = Cs (Tce − T0 ) + Vgp Cg (Tgp − T0 ) + Vge Cg (Tge − T0 ) + L2 + L3 + L4 + L5

(9.2.1)

Contour 2 QH + Vap Ca (Tap − T0 ) + Hc∗ = Cs (Tcm − T0 ) + Vgp Cg (Tgm − T0 ) + L2 (9.2.2) Contour 3 Vge Cg (Tgi − Tge ) = QH + L3

(9.2.3)

Contour 4 Cs (Tcm − Tce ) + Vgp Cg (Tgm − Tgp ) = L4

(9.2.4)

Contour 5 Bf QN + Vaf Ca (Taf − T0 ) = Vge Cg (Tgi − T0 ) + L5

(9.2.5)

150

9. Performance of Rotary Reactors, Indirect Heating

Figure 9.5

Model of heating system for rotary reactor.

Summation of Eqs. (9.2.2)–(9.2.5) results in Eq. (9.2.1), where QH is the heat transfer from the combustion gas to the outer surface of the retort [kcal/kg d.f.]. For notation, refer to Fig. 9.5. In Fig. 9.5, the temperature of the combustion gas Tgi should be controlled precisely, in order to prevent local heating of the retort. Letting the average temperature difference between the combustion gas and the rotating solids be ∆Tav , the heat transfer across the retort is given by, Fs QH = (πdt lr )U (∆Tav )

(9.2.6)

The overall heat transfer coefficient U can be calculated with equations in Chapter 8. 9.2.2

Direction of improvement

In Example 9.3, a design calculation is carried out for the oxidation of residual carbon in spent catalyst, by application of reasonable postulates. The results are summarized in Table 9.1 from which we can see the enormous effect of heat recovery from process gas and flue gas. Compared with Case I which excludes heat recovery, Case II with heat recovery can reduce fuel consumption from 0.2223 kg oil/kg feed to 0.08593 (i.e., to 39%). To meet further requirements to decrease fuel consumption, the authors propose the reactor shown in Fig. 9.6. Compared with Case I and II in Table 9.1, the following improvements could be made.

9.2. Heating by combustion gas

151 Table 9.1.

Prediction of performance for oxidation of residual carbon in spent catalyst, results of Example 9.3 Case

I (Fig. 9.5)

II (Fig. 9.5)

III (Fig. 9.6)

Feed

[kg/hr]

250

250

375

Heat loss  kcal 

L2 L3

29.7 281.6

29.7 281.6

19.8 57.3

L4 L5

263.5 40.5

263.5 40.5

19.8 27.0

Total

615.3

615.3

123.9

Temp. [◦ C]

Tap Taf Tge Tce

20 20 720 420

220 420 720 420

400 520 720 600

[kg oil/kg feed]

Bf

kg feed

Enthalpy Balance Input

!

kcal kg

0.2223 "#

%

$

Fuel Oxidation Process air Comb. air

2,223 388 0 0

85.1 14.9 0 0

Total

2,611

100.0

Output

kcal kg

Solids Process gas Flue gas Heat loss, total ⎞ ⎛ Feed end ⎟ ⎜ Insulation ⎟ ⎜ ⎝ Cooling region⎠

96.0 489.6 1,410.0 615.3 ⎛ 29.7⎞ ⎜281.6⎟ ⎟ ⎜ ⎝263.5⎠

Combustor

Total

40.5

2,611

% 3.7 18.8 53.9 23.6 ⎛ 1.1⎞ ⎜10.8⎟ ⎟ ⎜ ⎝10.1⎠ 1.6

100.0

!

kcal kg

0.08593 "#

859.3 388.0 223.2 275.4

%

$

!

kcal kg

0.00857 "# %

49.2 22.2 12.8 15.8

85.7 388.0 282.7 34.3

10.8 49.1 35.8 4.3

1,746

100.0

790.7

100.0

kcal kg

%

kcal kg

96.0 489.6 545.0 615.3 ⎛ 29.7⎞ ⎜281.6⎟ ⎟ ⎜ ⎝263.5⎠ 40.5

1,746

5.5 28.0 31.2 35.3 ⎛ 1.7 ⎞ ⎜16.1 ⎟ ⎟ ⎜ ⎝15.2 ⎠ 2.3

100.0

139.2 473.3 54.36 123.9 ⎛19.8 ⎞ ⎜57.3 ⎟ ⎟ ⎜ ⎝19.8 ⎠ 27.0

790.7

$

% 17.6 59.8 6.9 15.7 ⎛ 2.5⎞ ⎜ 7.3⎟ ⎜ ⎟ ⎝ 2.5⎠ 3.4

100.0

To increase the volumetric fraction of solids from γ = 0.15 to 0.15 × 1.5 = 0.225 To position the rotary manifold outside the retort for prevention of air leakage into the retort. To design a compact and efficient insulation cover cylinder, instead of a massive brick structure. In the proposed case III, only (0.00857)(375) = 3.2 kg oil/hr is needed to keep the reactor at steady state, because of the evolved heat from the oxidation of carbon.

152

9. Performance of Rotary Reactors, Indirect Heating

Figure 9.6 Proposed model of a rotary reactor for elimination of residual carbon in spent catalyst.

In the above discussion, we assume that the injection of air into the rotating solids does not deteriorate the quality of the catalysts, even in the case where high temperature stability could occur locally. If the feed stock is very sensitive to higher temperature, we had better reduce the O2 concentration to 10% by re-circulating the effluent gas, so that a stable temperature can be kept close to that of the rotating solids. 9.2.3

Application to thermal cracking of solid waste materials

A rotary retort has been applied for the thermal cracking of solid waste materials. Rather than the massive structure of the refractory insulation unit shown in Fig. 9.4, it is far less expensive to construct a simple parallel flow of combustion gas to the surface of the retort. The advantages are as follows: (1) The maximum temperature of the retort can be precisely controlled. (2) There is far less heat loss from the outer surface of the insulation cover, made of low density refractory. Fig. 9.7 illustrates the results of design calculations to carbonize solid waste materials containing a considerable amount of NaCl from a food industry plant. Because its melting temperature is as low as 800 ◦ C, conventional incineration and/or gasification reactors encountered great problems caused by the NaCl. Example 9.4 deals with the fundamental design calculations to carbonize this material at 500 ◦ C. Combustible gas, tar, liquor and char are its intermediate products. Feed rate 1,000 kg/hr Retort 1.6 m ID Heating region 4.8 m long Fuel (heavy tar) 72.5 kg/hr Products Dry gas 150 Nm3 /hr (2,040 kcal/Nm3 )

9.2. Heating by combustion gas

153

Figure 9.7 Design example of a rotary retort for thermal cracking of solid waste materials.

Liquor 140 kg/hr Tar, gained 400 − 72.5 = 327.5 kg/hr With the above calculation, we find that such a simple heating scheme is enough for carbonization of solid waste materials, as long as the feed rate is within the order of 1,000 kg/hr. Even in cases where the operating temperature is 500 ◦ C, far lower than the melting point of NaCl, sticking and/or agglomeration are supposed to take place in the retort. To prevent such problems, we should install a partition plate with inclined guide plates, the so-called U-Turn system. The circulation of hot char takes place as seen in Fig. 9.7 (refer to Fig. 2.7 and Example 9.4). The rotary retort shown in Fig. 9.7 is applicable to the gasification of any solid material; for instance, car shredder dust is gasified to produce rich fuel gas. The fundamental design calculation is just the same as Example 9.4, but the heat necessary due to its endothermic reaction should be estimated precisely (refer to Chapter 7, item 7.1.4) and Eqs. (7.1.7) and (7.1.8)). Example 9.1 Fly ash is separated by bag filters from flue gas in a municipal solid waste incineration plant. It contains polychloro-dibenzo-dioxins, 1.0 nanogram TEQ (toxic equivalent quantity), 95% of which should be eliminated to comply with strict environmental pollution regulations.

154

9. Performance of Rotary Reactors, Indirect Heating

PCDD in fly ash can be eliminated when it is kept at 450 ∼ 500 ◦ C for 20 min. Design a rotary retort such as that shown in Fig. 9.1, under the following conditions. Since fly ash contains a variety of salts and other substances, it is very sticky at the heating surface, resulting in serious resistance to heat transfer. By installing an adequate scraper, the thickness of the ash layer is kept at 8 mm. Data Feed rate of dry ash 500 kg/hr Heat transfer coefficient, inside of the retort hi = 31.1 kcal/m2 hr ◦ C γ = 0.20, ρ¯ = 450 kg/m3 , Cs = 0.24kcal/kg ◦ C Fly ash inlet 60 ◦ C Max 500 ◦ C Carrier air Va = 0.25 Nm3 /kg d.f. Maximum temperature of retort 530 ◦ C Heat loss, preliminarily calculated. L1 = 6, L2 = 20, L3 = 10 kcal/kg d.f. From performance data in a commercial plant, we found that 1.4% of carbon remained in the feed stock, and that 20% of the carbon was oxidized by the carrier air. Solution Let us design three region heaters as illustrated in Fig. 9.1, and assume the following temperature distribution. Temperature [◦ C] Av. Retort 225 450 400 490 475 505

Solids 60 → 350 350 → 450 450 → 500

Region (a) Region (b) Region (c)

∆Tav 225 90 30

Eq. (9.1.1) gives, 860

W = (0.24)(500 − 60) − (0.014)(0.2)(7,838) Fs + (0.25)(0.31)(500 − 20) + 6 + 20 + 10 = 156.9

kcal kg

W/Fs is obtained to be 0.1824 kW·hr/kg feed = 182.4 kW·hr/ton feed. For Fs = 500 kg/hr, we need 91.2 kW. Letting the outer diameter of the retort be dt and the length of each section la , lb and lc , we calculate as follows: In region (a), 

500

kg hr

  kcal ◦ C − 60 ◦ C) ∼ (πd l ) 31.1 kcal 0.24 (350 (225 ◦ C) = t a kg ◦ C m2 hr ◦ C

Thus, dt la = 1.583 m2 Similarly, dt lb = 1.365 m2 ,

dt lc = 2.047 m2

If we take dt = 1.2 m, la = 1.32 m,

lb = 1.14 m,

lc = 1.706 m

Example 9.2

155

Taking into consideration the distances between neighbouring regions, and the two end parts of the retort, the total length of the retort is estimated to be 4.17 m + (0.2 m) × 4 + (0.8 m) × 2 = 6.57 m The residence time of ash in Region c) is given by π (1.2 m)2 (1.706 m + 0.2 m × 2)(0.2)450 kg  4 m3 = 0.429 hr = 25.7 min

tr =

500 kg/hr

The above is the case where the thickness of the sticking solids layer is kept at 8 mm. If it is thicker due to the inadequate design of the scraping device, a much longer retort is needed.

Example 9.2 For drastic reduction of electrical energy consumption, it is planned to install the rotary reactor shown in Fig. 9.2, in place of a conventional rotary retort, for elimination of PCD Dioxins in fly ash. Use similar numerical data to Example 9.1. Data Carrier air Va = 0.25 Nm3 /kg d.f. γ1 = 0.2, γ2 = 0.2, χ = 0.34 L1 = 10, L2 = 20, L3 = 10 kcal/kg d.f. Tcm = 500 ◦ C, Tc1 = 60 ◦ C, Tc3 = 170 ◦ C, Ta1 = 20 ◦ C, Ta3 = 115 ◦ C TH0,max = 505 ◦ C, TH02 = 450 ◦ C, ∆Tm = 0.1 ◦ C Solution With Eq. (9.1.9) we calculate,   W 1 (0.24)(170 − 60) + (0.25)(0.31)(115 − 20) = + 10 + 20 + 10 − (0.014)(0.2)(7,838) Fs 860 = 0.06025 kW·hr/kg d.f. = 60.25 kW·hr/kg d.f. Compared with Example [9.1], the electricity input is reduced to 1/3 of Fig. 9.1. From Eq. (9.1.3),

Then,

   ∗ − Tc2 − (170 − 60) = 10 + (0.25)(115 − 20) (0.24) Tc2 ∗ − T = 182.3 ◦ C Tc2 c2

With Eq. (9.1.6), Ta2 =

∗ Tc2 + Tc2

2

= Tc2 + 91.15

Eq. (9.1.2) gives, QH1 = (0.24)(Tc2 − 60) − (0.25)(0.31)(Tc2 + 91.15 − 115) + 10 = 0.1625Tc2 − 2.552

[kcal/kg d.f.]

(1)

From Eq. (9.1.5),     QH2 = (0.24) 500 − (Tc2 + 182.3) − (0.25)(0.31) 500 − (Tc2 + 91.15) = 44.56 − 0.1625Tc2

(2)

156

9. Performance of Rotary Reactors, Indirect Heating

From Eq. (9.1.7), Q∗ = (0.24)(182.3) + 10 − (0.014)(0.2)(7,838) = 31.81 kcal/kg d.f.

(3)

Heat transfer coefficients are calculated as follows: Heat exchange region, Eq. (6.3.5),

(hrs )HC =

 4  200+273 4  (0.8)(0.85)(4.88) 250+273 − 100 100 250 − 200

= 16.44 kcal/m2 hr ◦ C

With Eq. (6.3.8),   ((0.5)(450)(0.24)(3 × 60)) 1/2 kcal (hp )HC = 1.13 = 191.1 2 ◦ 0.34 m hr C The effect of the partition plate is estimated by Eq. (8.2.2), (hp )HC = (1.4)(191.1) = 267.5 kcal/m2 hr ◦ C From Eq. (8.2.1), kcal 1 = 32.4 2 ◦ hi = 0.01 0.008 1 m hr C + + 20 0.40 (0.34)(16.44+267.5) The overall heat transfer coefficient U1 between the two solid flows is given by U1 = hi /2 = 16.2 kcal/m2 hr ◦ C The heat transfer coefficients in the electric heating region is calculated with Eq. (6.3.5),

(hrs )HC =

 4  450+273 4  (0.8)(0.85)(4.88) 500+273 − 100 100 500 − 450

= 55.64 kcal/m2 hr ◦ C

(hp )HC = 267.5 From Eq. (8.2.1), kcal 1 = 33.78 2 ◦ hi = 0.01 0.008 1 m hr C + + 20 0.40 (0.34)(55.64+267.5) The overall heat transfer coefficient U2 between the two solids flows is given by h kcal U2 = i = 16.89 2 ◦ 2 m hr C Average temperature differences are calculated as follows: (∆Tav )∗2 =

(∆Tav )1 =

(∆Tav )2 =

(450 − Tc2 ) − (505 − 500) 450−T

c2 ln 505−500

182.3 − (170 − 60) 182.3 ln 170−60

182.3 − 0.1 ln 182.3 0.1

= 143.1

= 24.28

(4)

(5)

(6)

With Eqs. (1), (2) and (3), as well as the numerical values of hi , U2 and U1 , Eqs. (9.1.10), (9.1.12) and (9.1.14) are utilized to obtain l2 and l1 , if we take dt = 1.2 m and dt∗ = 0.8 m.

Example 9.3

157

From Eq. (9.1.10), l2 =

124.9 (500)(31.81) 1 = · π(1.2)(33.78) (∆Tav )∗2 (∆Tav )∗2

(7)

From Eq. (9.1.14), l2 =

(500)(44.56 − 0.1625Tc2 ) 1 · = 21.62 − 0.07883Tc2 π(0.8)(16.89) 24.28

(8)

From Eq. (9.1.12), l1 =

500(0.1625Tc2 − 2.552) = 0.01395Tc2 − 0.219 π(0.8)(16.2)(143.1)

(9)

The numerical values of Tc2 and l2 are determined by graphical solution of Eqs. (7) and (8) with Eq. (4). Tc2 = 245 ◦ C,

l2 = 2.30 m

From Eq. (9), we get l1 = 3.20 m and ∗ = 245 + 182.3 = 427.3 ◦ C Tc2

The total length of the retort is estimated as lr = 2.30 m + 0.2 m × 4 + 3.20 m + (0.8 m) × 2 = 7.90 m Compare it with the length of the conventional rotary reactor, lr = 6.57 m, calculated in Example 7.1. Suppose 70% of the solids in the electric heating region is at 450 ∼ 500 ◦ C, the residence time tr is estimated to be, π (1.2 m)2 (2.30 m + 0.2 m × 4)(0.2)(0.7)450 kg  4 m3 = 0.418 hr = 26.5 min

500 kg/hr

It should be noticed that the retort in Fig. 9.1 needs a cooler for the hot solids, thus increasing the construction cost.

Example 9.3 In a rotary retort, the inner diameter of which is 0.9 m and the total length is 10 m, spent catalyst is heated to oxidize residual carbon from 5% to 0.05% at 600 ◦ C within 20 min. Process air is injected from tubes such as those shown in Table 1.2 [C] or Fig. 9.4. Predict the performance of the reactor system in three cases. Case I Case II Case III

Without heat recovery from process air and flue gas, 250 kg/hr With heat recovery system, 250 kg/hr Ideal heating system proposed in Fig. 9.6, 375 kg/hr

Use the following data. γ = 0.2, χ = 0.34 (hrs )HC = 16.44, (hp )HC = 267.5 kcal/m2 hr ◦ C ksi = 0.40 kcal/m2 hr ◦ C, δHi = 2 mm = 0.002 m From Eq. (8.2.1), hi = 0.01 0.002 1 = 63.06 2kcal◦ 1 m hr C 20 + 0.4 + (0.34)(16.44+267.5) hc = 4.6 kcal/m2 hr ◦ C (preliminary calculation)

158

hrg (hrs )SH

9. Performance of Rotary Reactors, Indirect Heating

Upper region 17.15 86.71

Lower region 7.39 85.13

Heat loss [kcal/kg feed] L2 29.7 19.8

Case I, II Case III

L3 281.6 57.3

L4 263.5 19.8

Total 615.3 123.9

L5 40.5 27.0

Tgi = 1,100 ◦ C gas to heating chamber Tcm = 600 ◦ C, Tam = 600 ◦ C Vap = 3.6 Nm3 /kg for Case I and II Vap = 2.4 Nm3 /kg for Case III Solution With Eq. (8.4.5) we calculate for the upper part of the heating chamber. h0 = 4.6 + 17.45 +

1 1 1 4.6+17.45 + 86.71

= 39.63

kcal m2 hr ◦ C

For the lower part, similarly h0 = 22.50 Eq. (8.4.6) gives for the upper part, U=

1 1 1 63.06 + 39.63

= 24.33

kcal m2 hr ◦ C

For the lower part, similarly U = 16.58 Therefore, the average value of U is, 24.33 + 16.58 kcal = 20.46 2 ◦ 2 m hr C In Case I, we estimate Tce and Tgp , on the basis of Eq. (9.2.4) (refer to Fig. 9.5). (0.24)(600 − Tce ) + (3.6)(0.34)(600 − Tgp ) = 263.5

kcal kg

In the case where Tce = Tgp , we obtain, Tce = Tgp = 420 ◦ C From Eq. (9.2.2) we have for Case I, Tap = T0 , QH = (0.24)(600 − 20) + (3.6)(0.34)(600 − 20) + 29.7 − (0.05 − 0.0005)(7,838) = 490.8 kcal/kg With Eq. (9.2.6), ∆Tav is calculated.    kcal kcal kg 490.8 = π(0.9 m)(6 m − 0.1 m × 4) 20.46 2 ◦ (∆Tav ) 250 hr kg m hr C ∆Tav = 378.8 ◦ C

Example 9.4

159

As the first approximation, let us assume Tge = 700 ◦ C. Thus the average temperature of the gas within the heating chamber is approximately given by Tg = (1,100 ◦ C + 700 ◦ C)/2 = 900 ◦ C Along the retort, (∆T )av is given by ∆Tav =

(900 − 20) − (Tge − 600) ln T 880 −600

= 378.8 ◦ C

ge

From the above equation, Tge is calculated to be 720 ◦ C. Let us estimate the necessary amount of excess air n to make Tgi = 1,000 ◦ C in Case I, by application of Eq. (9.2.5). (10,000)Bf = Bf (11.8 + 11.0 × n)(0.34)(1,100 − 20) + 40.5 As the first approximation, assume Bf = 0.20 kg/kg. Then we obtain excess air n = 1.35 = 135%. The necessary fuel Bf is calculated with Eq. (9.2.1) for Case I. (10,000)Bf + (0.05 − 0.0005)(7,838) = (0.24)(420 − 20) + (3.6)(0.34)(420 − 20) + Bf (11.8 + 11.0 × 1.35)(0.34)(720 − 20) + 29.7 + 281.6 + 263.5 + 40.5 We obtain, Bf = 0.2223 kg/kg The enthalpy balance in Table 9.1 is then completed from the above value. Similar calculations are carried out for Case II, which recovers heat from the exit gases and preheats inlet air flows. The results are compared with Case I, indicating the enormous effect of preheat to reduce fuel consumption (see Table 9.1). Case III is an ideal one, proposed by the authors. It is possible to drastically reduce fuel consumption to 4% of Case I.

Example 9.4 Solid waste materials, including considerable sodium chloride are required to be converted to clean fuel for domestic utilization. Since the melting point of NaCl is 800 ◦ C, thermal cracking in a rotary retort is selected as the first step of the project. Design the rotary reactor under the following conditions. Composition of feed stock Component wt. [%]

FC 18.0

VM 76.0

NaCl 5.7

Ash 0.3

Total 100

Gross calorific value 4,500 kcal/kg d.f. Dry feed rate 1,000 kg/hr Product pattern, at 500 ◦ C, from laboratory test Dry gas Liquor Tar Char

0.21 kg/kg 0.14 kg/kg 0.40 kg/kg 0.25 kg/kg

Total

1.00 kg/kg

0.15 Nm3 /kg

2,040 kcal/Nm3 1.43 kg/Nm3

160

9. Performance of Rotary Reactors, Indirect Heating

Composition of dry gas, vol.% CO 52.5

H2 2.5

CO2 35.6

CH4 7.5

C2 H4 1.9

Total 100

Postulates (1) Thermal cracking of biomass at higher temperature is exothermic. Nevertheless, let us assume here heat evolution at 500 ◦ C is negligible for the safer side of the design. (2) To keep the maximum temperature of the retort lower than 800 ◦ C, the inlet temperature of the combustion gas Tgi is set to be 1,100 ◦ C. (3) For heat transfer characteristics inside rotary retort, let us use the numerical values obtained in Example 9.3. (4) According to preliminary calculation, we can assume heat losses in Fig. 9.7 as follows: L2 ≈ 0,

L3 = 22.6,

L4 ≈ 0,

L5 = 25.4

kcal kg d.f.

(5) Heavy tar to burn Net calorific value 7,500 kcal/kg Stoichiometric air 8.0 Nm3 /kg Stoichiometric combustion gas 8.5 Nm3 /kg Excess air to burn n = 0.2 Solution Necessary heat Hn for the retort is given by Dry gas   Nm3 kcal kcal 0.15 0.41 (500 ◦ C − 20 ◦ C) = 29.52 kg kg Nm3 ◦ C Liquer     kcal kcal kg ◦ C − 20 ◦ C) = 115.6 kcal 586 (500 0.14 + 0.5 kg kg kg ◦ C kg Tar     kcal kcal kcal kg 70 (500 ◦ C − 20 ◦ C) = 143.2 + 0.6 0.40 ◦ kg kg kg C kg Char   kcal kcal kg 0.20 (500 ◦ C − 20 ◦ C) = 24.0 0.25 kg kg ◦ C kg Total

Hn = 312.3

kcal kg

In Fig. 9.7, the enthalpy balance for the combustion gas flowing along the retort gives its volume Vg1 [Nm3 /kg] as follows:  kcal kcal (1,100 ◦ C − 700 ◦ C) = 312.3 + 22.6 = 334.9 Vg1 0.34 kg Nm3 ◦ C Vg1 = 2.463 Nm3 /kg

Example 9.4

161

Letting Bf2 be the necessary tar to burn in the combustor, the mass and enthalpy balances around it are represented by Eqs. (1) and (2), where Vg2 is the volume of combustion gas, which is re-circulated to the combustor. 2.463 = Vg2 + (8.5 + 8.0 × 0.2)Bf2

(1)

(7,500)Bf2 + Vg2 (0.34)(500 − 20) + Bf2 (8.0)(1 + 0.2)(0.31)(500 − 20) = (2.463)(0.34)(1,100 − 20) + 25.4

(2)

Bf2 and Vg2 are obtained from Eqs. (1) and (2). Vg2 = 1.727 Nm3 /kg

Bf2 = 0.07249 kg/kg,

Air (0.07249)(8)(1 + 0.2) = 0.6959 Nm3 /kg. The results of the above calculation are shown in Fig. 9.7. The temperature of the recycle gas, 500 ◦ C, is restricted by the safe running temperature of the re-circulation blower. In Fig. 9.7, we design a partition plate installed with inclined guide plates. This ensures the satisfactory dispersion of the feed stock into the circulating flow of hot solids. This scheme can prevent possible problems caused by NaCl contained in the product char. As the first approximation, the width of the annular space δ, where the combustion gas flows parallel to the axis of the retort, is taken to be 4 cm. The emissivity of the combustion gas εg = 0.06 is obtained for the following condition by applying H.C. Hottel’s Procedure (Ref. [5, Chapter 6]). lG = 2 × 0.04 = 0.08 m,

PCO2 = 0.103,

PH2 O = 0.101 atm,

Tg = 900 ◦ C

With Eq. (8.4.1),

hrg =

 4  700+273 4  (0.06)(0.8)(4.88) 900+273 − 100 100 900 − 700

= 11.68

kcal m2 hr ◦ C

From Eq. (8.4.2),

(hrs )SH =

 4  700+273 4  (0.8)(1 − 0.06)(0.8)(4.88) 800+273 − 100 100 800 − 700

= 126.0

kcal m2 hr ◦ C

The convectional heat transfer coefficient hc is calculated by Sieder and Tate’s equation as follows:

Re =

 gm  cm  (4 cm × 2) 3.0 × 10−4 3 2,000 sec cm gm

4.8 × 10−4 cm·sec

= 10,000

where the hydraulic equivalent diameter of the annular space is dte = (4 cm) × 2 = 8 cm hc dte = 0.023(Pr )0.4 (Re )0.8 = 0.023(0.65)0.4 (10,000)0.8 = 30.68 kg hc = (30.68)

0.065 kcal/m hr ◦ C kcal = 24.9 2 ◦ 0.08 m m hr C

With Eq. (8.4.5), h0 = 24.9 + 11.68 + In Example 9.3 we find, hi = 63.06

kcal m2 hr ◦ C

1 1 1 24.9+11.68 + 126.0

= 64.93

kcal m2 hr ◦ C

162

9. Performance of Rotary Reactors, Indirect Heating

Thus Eq. (8.4.6) gives, U=

1 1 1 63.06 + 64.93

= 32.0

kcal m2 hr ◦ C

The average temperature difference in Fig. 9.7 is calculated as ∆Tav =

(1,100 ◦ C − 500 ◦ C) − (700 ◦ C − 400 ◦ C) ln 1,100−500 700−400

= 433 ◦ C

The heat transfer surface area of the retort is given by (312.3 kcal/kg)(1,000 kcal/hr) = 22.53 m2 (32.0 kcal/m2 hr ◦ C)(433 ◦ C) Letting l = 3dt , we design, dt = 1.55 m ∼ 1.6 m l = 4.8 m The width of the annular space δ is calculated as follows: Flow rate of gas 

2.463

Nm3 kg



1,000 kg 3,600 sec



900 + 273 273

Then, δ = 0.0293 m ∼ 0.03 m = 3 cm



= 2.94

 m3 m = π (1.6 m)(δ) 20 sec sec

– 10 – Application of a Rotary Reactor for the Re-utilization of Solid Wastes Rotary reactors are suitable for re-utilization of solid wastes, which are hard to treat by other contacting methods. The flow pattern of the solids is improved by installing partition plates and/or screws inside, which broaden their applicability to processing solid waste materials. The authors’ practical experience in the research and development of new processes is summarized in this chapter. The possibilities for further application of their concept to the gasification of solid waste and low grade coal is confirmed by the presentation of design examples on a commercial scale.

10.1

MATERIAL AND ENERGY RECOVERY FROM SOLID WASTES

In parallel with the development of modern civilization, the amount of solid wastes increases enormously. The first step of their disposal has simply been to bury them underground, i.e., in landfill. In many countries, however, the available space for landfill is becoming scarce. Furthermore, landfill is apt to alter the natural environment considerably. The shortage of available space for landfill has urged communities to decrease the volume of solid wastes by incineration. For municipal solid wastes, grate incinerators (horizontal or inclined moving beds) and fluidized bed ones have mainly been installed and operated in the suburbs of many cities. The thermal energy in the flue gas from the incinerator is usually recovered and used to generate steam by application of a waste heat boiler. In the newer types of incineration plant, this steam is sent to a turbine, which drives an electric power generator. However, the conversion efficiency of thermal energy to electric power is usually rather low. Flue gas from a conventional incinerator contains a variety of corrosive impurities, which attack the surface of the super-heater in the boiler. Consequently, the working pressure of the boiler is limited to prevent the destruction of its super-heater, resulting in a rather low conversion efficiency to electric power. The recent trend for utilization of solid wastes is to separate the valuable substances in them by applying advanced separation technologies. Metals are selectively collected and sent to metallurgical plants as raw materials with high purity. Plastics and biomass are separated to make so-called refuse derived fuel (RDF) or refuse derived paper-plastic fuel (RPF). In this sense, municipal solid waste (MSW) is a secondary resource, produced by communities.

164

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

The remainder of the MSW still contains combustible substances, mixed with considerable amounts of inorganics. Rather than landfill or incineration, this secondary resource should be re-utilized by the application of suitable processes. The gasification of such stock must be one solution to this requirement, and can supply clean fuel gas to the local community. The possibility of applying rotary reactors for the gasification of MSW is discussed in this chapter.

10.2 10.2.1

PROPOSED ROTARY REACTORS FOR RE-UTILIZATION OF SECONDARY RESOURCES

De-lacquering of spent cans

Fig. 10.1 illustrates a rotary reactor for carrying out de-lacquering spent cans, developed and commercialized by the authors. Spent cans, collected from the vendors, are loosely pressed and sent to the reactor. Both surfaces of the thin metal plate (150 µm) are coated with thin layers of organic film (25 µm). Inside the spent cans, there are residual liquids and/or fragments of cigarettes which should be eliminated before sending to the smelting furnace. To prevent the oxidation of the aluminum or steel plate, the spent cans are heated to 450 ∼ 500 ◦ C by combustion gas in which the composition of O2 is precisely controlled. In Example 10.1 a fundamental design calculation is presented for de-lacquering of steel cans, 3 tons/hr. Since the operation temperature is not too high, light insulation material between the metal plates is applied in place of heavy refractory. Making suitable openings at the exit end of the cooling section, the feed stock can be easily switched, from steel to aluminum for example. The practical performance is almost the same as that predicted in Table 10.1. An improvement would be to increase capacity, say to 4,000 kg/hr, by increasing the volumetric fraction of the cans from γ = 0.2 to 0.27 for example. In addition, the preheating of the feed stock by recovered heat from hot cans is very effective for drastically

Figure 10.1

Rotary reactor for de-lacquering spent cans.

10.2. Proposed rotary reactors for re-utilization of secondary resources

165

Table 10.1. Results of calculation in Example 10.1 Rotary reactor 2 m ID, Heating region 9.4 m long 2 r.p.m., ω = 2 deg. De-lacquering of spent steel cans, 3,000 kg/hr at 500 ◦ C Temperature of gas from combustor 900 ◦ C Volume of recycle gas 0.512 Nm3 /kg Fuel oil 0.01655 kg/kg feed = 16.55 kg/ton feed Enthalpy Balance, on the basis of feed stock 1 kg Input

kcal kg

Heat of fuel oil Combustible substance Preheat air

165.5 24.0 27.9

Total

217.4

100.0

Output

kcal kg

%

Solid Water vapour Combustion gas Excess air Heat loss Total

% 76.2 11.0 12.8

75.0 20.5 58.5 14.9 48.5

34.5 9.4 26.9 6.9 22.3

217.4

100.0

Remarks 0.3%, 8,000 kcal/kg 400 ◦ C

500 ◦ C 900 ◦ C 900 ◦ C 900 ◦ C

reducing fuel consumption. If we use such a heat exchanger between two flows of cans, as suggested in Fig. 9.3, we can dry and preheat the feed cans up to 250 ◦ C. A similar calculation to Example 10.1 gives us, Bf = 0.00687 kg/kg feed = 6.87 kg oil/ton feed It is possible to achieve an enormous reduction in fuel consumption. 10.2.2

Activation of char

An organization connected with the forestry industry planned to produce active carbon from chips of raw larch, the water content of which was 46% wet basis. (0.852 water/ dry biomass). The authors developed a novel rotary reactor in which drying, thermal cracking of chips and activation of char were carried out consecutively in a single cylinder. Fig. 10.2 gives an outline of the pilot plant. Heat for drying is supplied by re-circulating the hot char coming from the carbonization region. The hot char, heated in the burningactivation region, transports thermal energy to the carbonization region. In Fig. 10.2, steam is injected into a rotating layer of char, in a similar way to Table 1.2 [D]. To burn combustible gas without contacting the active char, air is issued from orifices to impinge on the upper wall surface. Steady state operation with a feed rate of 50 kg/hr verified that the reactor worked out well, producing active carbon of a satisfactory quality.

166

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

Figure 10.2 chips.

Outline of a pilot plant developed by the authors to produce active carbon directly from wet wood

Figure 10.3

Conceptual design of a rotary reactor for active carbon, proposed by the authors, 100 kg feed/hr.

Biomass is an excellent feed stock for producing clean fuel gas, which can be sent to a gas engine or turbine for power generation (see Fig. 9.7 and Example 9.4). However, a considerable amount of char is formed which should be converted to valuable products such as active carbon. By improving the steam injection tubes in Table 1.2 from [B] to [C], and then increasing the volumetric fraction of the char from γ = 0.1 to 0.2, we could design a rotary reactor for activation of char, with a capacity 4 times larger than the conventional one shown in Fig. 7.8. Applying engineering know-how acquired from the above R&D work, the authors propose to design a rotary reactor to activate char as shown in Fig. 10.3. The procedure for

10.3. Gasification of solid wastes

167

its design calculation is found in Example 10.2, in which char from spent coffee powder is activated to produce 60 kg/hr of active carbon.

10.3 10.3.1

GASIFICATION OF SOLID WASTES

Matrix presentation of gasification processes

A variety of reactors have been developed for the gasification of municipal solid wastes (MSW), by applying almost all the contacting methods illustrated in Fig. 1.1. Some examples are represented in a matrix, given in Table 10.2 (see Kunii [1]). As one genre of gas–solid reaction systems, the gasification of MSW is designated by two components, namely the gas–solid contacting method and the way to supply thermal energy. Rotary reactors, either indirectly or directly heating, have been applied in the development of many processes, some of which have been commercialized in practice. 10.3.2

Gasification processes of MSW developed by the authors

In 1972–1981, a large scale R&D project was carried out to develop the Kunii–Kunugi Process for producing olefins from very heavy oil such as asphalt. It was adopted as a national project in Japan and verified to work out satisfactorily as planned, by the operation of a large demonstration plant processing 120 tons per day of very heavy oil. In the process, Table 10.2. Matrix representation of reactors for gasification of MSW Gas solid contacting method

To supply thermal energy Heating outside

Partial oxidation

Moving solids

Vertical retorts (Pressed block, horizontal.) First step of Thermo-Select

Vertical beds Torrax Shin Nihon Iron Purox, UCC Pyrox II, Kunii-TSK∗ Onoda Cement Co. (spent tires)

Fluidized solids

Entrained solids Rotary cylinder

Heat carrier

Pyrox, Kunii-TSK∗ ICFG, Ebara (Bio-mass) Battelle/FERCO (Bio-mass) Flash, Occidental

Ruhr Kohle AG PKA Conrad Siemens Hitachi Fig. 9.7, Fig. 10.4

Bubbling in molten bath ∗ (TSK: Tsukishima Kikai Co.)

Landguard, Monsanto Fig. 7.9

Second step of Thermo-Select

Fig. 10.5, Fig. 10.8

168

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes Table 10.3. Operation data of Pyrox process, gasification of MSW, Funabashi City, Igarashi et al. [3]

Basis 1 kg of dry MSW MSW gasified, less than 100 mm Component wt [%]

∗ From practical data in 1989

Chemical∗

wt [%]

Paper 46.9 C 48.43 Plastics 15.1 H 7.06 Wood 2.7 N 0.99 Cloths 3.3 S 0.15 Food waste 17.7 Cl 0.64 Metal etc 6.3 O 29.42 Others 8.0 Inorganics 13.31 Total 100.0 Total 100.00 Proximate wt [%] Gross calorific value of dry MSW 3,291 kcal/kg Water 45.0 Organics 41.5 Inorganics 13.5 Total 100.0 Gasification/combustion 720 ◦ C/820 ◦ C Steam 0.524 kg/kg d.M.S.W, Air 3.2 Nm3 /kg 52.3% of dry gas was sent to the combustors as fuel.

Dry gas produced Component

vol [%]

H2 15.32 O2 0.17 2.41 N2 CO2 16.72 CO 31.34 CH4 17.04 2.33 C2 H6 C2 H4 9.79 C3 H8 0.11 C3 H 6 3.09 Others 1.68 Total 100.00 Gross calorific value 5,596 kcal/Nm3 Vol. of dry gas 0.445 Nm3 /kg

the necessary heat for thermal cracking was supplied by re-circulating coke particles between two deep fluidized beds (see Shoji [2]). Kunii applied his re-circulation system to the gasification of very wet municipal solid wastes, but using sand in place of coke. Tsukishima Kikai Co. developed the system called the Pyrox Process and successfully constructed three units in Funabashi City, the capacity of each was 150 tons/day (refer to Igarashi [3] and Koike [4]). The Pyrox units were operated satisfactorily for 8 years from 1984. Examples of practical gasification of MSW are presented in Table 10.3. The Pyrox Process was accepted because it was the only process which was able to comply with the environmental requirements from the municipal government at that time. It was the first commercial large-scale plant for the gasification of MSW and produced plenty of rich gas. Nonetheless, the plant was replaced afterwards by conventional incinerators, because of the following reasons. (1) The out-of-date regulations there prohibited the sale of product rich gas to other organizations. In other words, the Funabashi plant was adopted only as an incineration plant. (2) Within 8 years, technologies for the abatement of pollution were developed for conventional incinerators which complied with the strict regulations. In 1999 a feasibility study was carried out by another engineering company to use Pyrox for producing hydrogen for fuel cell cars. It evaluated Pyrox to be promising, even in the west coast of the USA where they could find enough space for landfill. Kunii and Tsukishima also developed a single fluidized bed reactor, which gasified a wet mixture of plastics and spent pulp from a paper-pulp plant. The air was sent to its bottom;

10.3. Gasification of solid wastes

169 Table 10.4.

Gasification of plastic waste in commercial reactor, single fluidized, partial oxidation by the air, 750 ◦ C, 500 kg/hr, Hasegawa [5] Feed stock

Product Composition N2 O2 CO2 H2 CO

Mixture of polyethylene film and pulp 2 ∼ 3 mm diameter, ca. 100 mm in length Component, dry basis, wt% PE Pulp Total Component, wet basis, wt% Combustible Incombustible Water Total Dry basis Net cal. val. Air injected Dry gas prod.

90 10 100 48.7 1.3 50.0 100.0 9,400 kcal/kg 3.73 Nm3 /kg 3.36 Nm3 /kg

CH4 C2 H6 C2 H4 C2 H2 C3 H6 1–3 C4 H6 C6 H6 C6 H5 CH3 , etc. Total Net cal. value 2,109 kcal/Nm3

vol.% 63.11 0.08 12.28 4.28 6.00 6.14 0.27 6.15 0.39 0.43 0.24 0.55 0.08 100.00

Gasification efficiency, ηG = (3.36)(2,109) = 0.754 9,400

product gas was scrubbed and sent to a boiler, to replace a considerable amount of fuel oil. Examples of practical data are summarized in Table 10.4. To gasify car shredder dust for fuel gas, the authors and Taiheiyo Cement Co. developed a novel rotary reactor, which is illustrated in Fig. 10.4. This pilot reactor gasified 140 kg/hr of the stock at 800 ◦ C, producing rich dry gas, ca 7,000 kcal/Nm3 . Unlike a conventional rotary retort, it was positioned within a refractory cover cylinder, which rotated by an ordinary device. The cover cylinder was responsible for the mechanical strength, so that the retort could be designed for high temperature without concerns for its deformation and then destruction. Solids, mainly metal pieces covered with carbon, were discharged through an inner cylinder, after giving their heat to the cold feed stock. Residual carbon in the discharged solids was burnt in a rotary combustor, shown in Table 1.3 [A], to generate hot gas for heating the reactor. In the same category as the above, pilot reactors such as shown in Fig. 9.7 have been tested to satisfactorily gasify solid wastes composed of biomass and plastics. For waste from food plant, see Example 9.4. 10.3.3

Proposal for a novel rotary reactor to produce rich gas from MSW

In accordance with the current requirements for fuel cell cars, the authors propose a novel rotary reactor to gasify MSW for production of rich gas, from which hydrogen is obtained

170

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

Figure 10.4 Pilot plant for gasification of car shredder dust, developed by the authors and Taiheiyo Cement Co., retort 0.63 m OD 2.2 m long.

by conventional processes. Solids carrier processes like the Pyrox Process could be applied for this purpose on a large scale. However, its circulation system of solids is too sophisticated to operate in the local community where MSW is collected and re-utilized on a small scale. A conceptual design of the rotary reactor is illustrated in Fig. 10.5, which could correspond to a unit of the Pyrox Process (by Kunii and TSK, wet MSW 150 tons/day). Wet MSW is half dried to 0.20 kg water/kg dry MSW and fed into the gasification region of the reactor. A partition plate, with a number of guide plates attached, is positioned in it, and char is circulated in the region. The wet MSW is fed onto the re-circulating hot char, mixed into it, and heated quickly up to its decomposition temperature. The necessary heat for vaporization of water and then thermal cracking of MSW is supplied by the hot char, re-circulating from the oxidation and heating region of the reactor, across 4 screw cylinders (see Fig. 10.6). The char particles in the screws function as a good seal against leakage of gas between the two regions. The rate of solid circulation between the two regions is simply determined by the rate of rotation, and predicted with Eq. (2.3.1).

10.3. Gasification of solid wastes

171

Figure 10.5 Conceptual design of a novel rotary reactor to gasify wet MSW, proposed by the authors.

Figure 10.6 Model of screw cylinders in Fig. 10.5.

In the oxidation-heating region, air is injected into the rotating char at high temperature, from tube distributors such as shown in Table 1.2 [C]. An appropriate design for the rotary manifold is crucial in this process. Heavy fraction of tar, separated from tar scrubber is sent into the region as auxiliary fuel, if necessary. In Example 10.3, a conceptual design calculation of the reactor is carried out to process very wet MSW, the results of which are listed in Table 10.5. The calorific efficiency to fluid fuel is as follows: Dry gas 65.33% Tar

(0.38%)

⎫ ⎪ ⎬

Total 66.1% 7,000 = 0.81% ⎪ ⎭ 3,291

172

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes Table 10.5. Results of Example 10.3, design calculation of conceptual rotary reactor for gasification of very wet MSW, proposed in Fig. 10.5 Feed stock, wet municipal solid waste in Table 10.3 Initial water content 0.818 kg/kg dry MSW (45% wet basis) Dry MSW Organics 75.5% Inorganics 24.5% Gross calorific value 3,291 kg/kg dry MSW Fees rate 3,438 kg dry MSW/hr, corresponding to initial wet MSW 150 tons/day Temperature Rotation 5.0 r.p.m. Gasification 720 ◦ C, Oxidation 820 ◦ C Product Dry fuel gas: gross calorific value 5,000 kcal/Nm3 0.430 Nm3 /kg dry MSW 1,478 Nm3 /hr Gasification efficiency ηG = 0.6533 Tar from scrubber: 0.06 kg/kg, available tar 0.0038 kg/kg Calorific efficiency for fluid fuel ηT = ηG + (0.0038)(7,000/3,438) = 0.661 = 66.1% Dimension of reactor Inner diameter 3m Length, gasification region 4.5 m Screw cylinders 3m Oxidation region 4(1 + 0.2) ∼ =5m Total length 13 m Power generation estimated Very wet MSW, 150 tons/day 2,609 kW

If a gas engine is applied for electric power generation, 150 tons/day of very wet MSW could generate,   kg kW·hr kcal = 2,609 kW 3,438 (0.661)(0.3) 3,291 kg hr 860 kcal This appears attractive as a source of new energy. In the course of the above calculation, we find that the calorific efficiency to fluid fuel increases with decreasing water content, and then with higher preheating temperature of the air and water vapour. In the case of dry MSW, as discarded in western communities, the amount of tar to be burnt is minimized, resulting in higher calorific efficiency. A similar calculation to Example 10.3 gives the following values for MSW with 0.20 kg water/kg dry MSW. Necessary amount of tar to burn 0.0435 kg/kg dry MSW Available tar 0.06 − 0.0435 = 0.0165 kg/kg dry MSW Calorific efficiency to fluid fuel is, gas 65.33% tar

(1.65%)(7,000/3,291) = 3.51%

+

68.34%

10.4. Gasification of sewage sludge

173

In some communities, plastic wastes are selectively collected without mixing with ordinary MSW. They must be the most suitable feed stock to gasify in such reactor as shown in Fig. 10.5. Very rich gas 8,000 ∼ 10,000 kcal/Nm3 as well as tar are produced, and solid materials are then discharged after complete burning. The high price of crude oil must be a spur for the development of local gasification of plastic wastes in many places.

10.4 10.4.1

GASIFICATION OF SEWAGE SLUDGE

Conventional incineration

A huge amount of sewage sludge is usually incinerated in rotary, multi-stage or fluidized bed reactors, consuming an enormous amount of fuel. The content of ordinary sewage sludge is shown in Table 10.6. The gross calorific value of dry solids is less than the necessary heat for vaporization of the water content. Therefore, preheating of combustion air is important to reduce the consumption of auxiliary fuel (see Table 10.7), obtained in Example 10.4. Letting the preheating temperature be 400 ◦ C in the case of exit gas 900 ◦ C, we still have to use 0.25 kg oil/kg d.f., i.e., 0.05 kg oil/kg wet feed = 50 kg oil/ton wet feed. If the Table 10.6. Example of sewage sludge, de-watered by filter press Components, based on 1 kg of dry solid + Inorganics 0.2 kg 1 kg + Organics 0.8 kg 5 kg Water 4.0 kg (Water content, wet basis 80%) Calorific Gross Net

Value of dry stock 4,150 kcal/kg 3,783 kcal/kg

Stoichiometric volumes in combustion Air 4.8 Nm3 /kg Gas (wet) 5.4 Nm3 /kg

Table 10.7. Reduction of fuel oil by preheating of air Temp. of air [◦ C]

20

200

400

600

kg oil/kg d.f.

0.445

0.338

0.245

0.171

174

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

capacity of an incineration plant is 1,000 wet-tons/day, 50 tons of oil will be burnt every day. In order to reduce this consumption of fuel oil, we have to develop better processes. 10.4.2

Proposal for a rotary reactor to gasify sewage sludge

The ideal approach to dealing with very wet solids must be the so-called auto vapour compression process. Since feed substance is very hard to handle, however, the authors propose a gasification process in place of the above ideal one. Fig. 10.7 is a simplified flow sheet of the proposed model. In Example 10.4, a conceptual process system is designed to generate electric power from very wet sewage sludge, which could save fuel oil in a conventional incineration plant. Very wet sewage sludge is dried in a rotary dryer which is heated indirectly by hot exhaust gas from gas and tar engines, as seen in Fig. 10.7. Since the amount of water is as much as 4 kg/kg d.f., additional thermal energy should be supplied. In Example 10.4, the authors suggest installing a similar reactor to the rotary reactor of Fig. 10.5 in parallel, which gasifies dry RDF, RPF biomass and/or collected mass of plastic waste. Water vapour from the dryer has an unbearable odour. When it is introduced to the oxidation region of the rotary reactor the odour is completely eliminated. In the example, a design calculation is carried out to dry sewage sludge satisfactorily by indirect heating. Re-circulation of the dry granules within the dryer is essential to prevent the feed materials sticking to the heat transfer surface. The results of Example 10.4 are summarized as follows:

Figure 10.7 Hypothetical model for electric power generation from very wet sewage sludge.

10.5. Possibility for application to gasification of low grade coal

175

Very wet sewage sludge, 4 kg water/kg d.f., 100 tons/day RDF in parallel reactor of similar dimension 18 tons/day Electric power generation 2,008 kW No odour in exhaust water vapour.

10.5

POSSIBILITY FOR APPLICATION TO GASIFICATION OF LOW GRADE COAL

Gasification of coal has long been an established technology. Four types of contacting methods, namely vertical moving bed, fluidized bed, entrained flow and molten bath, have been used for conventional gasification reactors (see Kunii [1]). Reflecting the high price of crude oil, utilization of low grade coal appears to be attractive for local power generation or other purposes. Low grade coal usually contains much ash and impurities, which might cause serious environmental problems. Let us discuss whether the new reactor proposed in this chapter can be applied or not. Unlike municipal solid waste, much carbon remains in the ash when low grade coal is thermally decomposed (see Table 10.8). To utilize low grade coal or lignite, we have to gasify the carbon contained in the ash particles with reasonable gasification efficiency. Fig. 10.8 is a conceptual model of a novel gasification reactor proposed by the authors. In Example 10.5 calculations are made to design a rotary reactor to gasify 400 tons of wet lignite per day, on the basis of the assumed product pattern given in Table 10.9. The results of the above calculation give a design for a promising gasification reactor, the operation of which is expected to be much easier and more efficient than conventional ones (see Table 10.10). In the course of the above design calculation, the authors are confident that the novel rotary reactors as shown in Fig. 10.5 and/or Fig. 10.8 can be well designed for the gasification of any combustible feed stock, if the fundamental data of thermal cracking are supplied. Compared with fluidized bed reactors, the rotary reactor is more suitable to gasify such material, to be processed locally. A feasibility study for gasification of materials such as oil-shale, oil sand and/or peat is likely to contribute to current energy problems.

Table 10.8. A model of low grade lignite, dry basis Proximate analysis Volatile matter Fixed carbon Ash

[%] 26.3 23.7 50.0

Total 100.0 Gross calorific value 3,863 kcal/kg d.f.

Ultimate analysis

[%]

C H N S O Ash Total

38.5 3.1 0.6 0.2 7.6 50.0 100.0

176

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

Figure 10.8

Conceptual model for gasification of lignite for rich product gas.

Table 10.9. Product pattern, assumed for thermal cracking of model lignite in Table 10.8 Conditions, assumed Average size 3 mm Temperature 850 ◦ C Mean residence time of solids 28 min Carrier gas, N2 Experimental reactor, Table 3.1 [C] Water content of feed stock 0.281 kcal/kg d.f. Wet basis 21.94% Product, estimated

[%]

Composition of dry gas, estimated [%]

Char

20.1 50.0 (70.1)

CO CO2 H2 CH4 C2 H4 C2 H 6 C3 H 6 C3 H 8 C4 Total

Tar Water Dry gas Total

carbon ash

13.2 2.3 14.4 100.0

21.7 7.2 41.0 20.2 4.3 1.4 2.4 0.0 1.8 100.0

Dry gas 0.138 Nm3 /kg 1.04 kg/Nm3 Gross calorific value 5,778 kcal/Nm3 ηG = (5,778)(0.138) = 0.2064 3,863

Example 10.1 A rotary reactor, shown in Fig. 10.1, is designed to remove organic substances in spent steel cans. Predict its performance under the following conditions. Rotary reactor 2 m ID, 9.4 m in length Spent steel cans Feed rate 3,000 kg/hr Bulk density 200 ∼ 250 kg/m3 Water content 2%

Example 10.1

177 Table 10.10.

Predicted data for conceptual rotary reactor to gasify low grade lignite, shown in Fig. 10.8. Results of Example 10.5 Basis dry feed stock 1 kg Feed stock

Composition Gross calorific value Water content

Table 10.8 3,863 kcal 0.281 kg

Products by thermal cracking, Table 10.9 Feed rate

400 tons of wet feed/day = 16,670 kg wet feed/hr = 11,983 kg d.f./hr

Temperature

gasification 850 ◦ C oxidation-heating 950 ◦ C ≈0 1.072 Nm3 /kg d.f. 0.5025 kg/kg d.f. 0.5132 Nm3 3,773 kcal/Nm3 0.132 kg ηT = ηG + (0.132)(7,000/3,863) = 0.501 + 0.239 = 0.740 = 74% volumetric fraction of solid 20% inner diameter 4.0 m 4 screw cylinders 1.6 m rate of rotation 4 r.p.m.

Tar as fuel Injection air Injection steam Dry gas produced Gross cal. value Tar, available Calorific efficiency for fluid fuel Dimension of reactor

gasification region re-circulation region oxidation-heating region total length

6m 4m 7m 17 m

Estimation of power generation 11,950 kW

Accompanied organic substance (VOC) 0.3 wt% Net calorific value of VOC 8,000 kcal/kg Fuel oil Net calorific value 10,000 kcal/kg Stoichiometric air 11.0 Nm3 /kg Stoichiometric combustion gas 11.8 Nm3 /kg Excess air 30% Operating conditions Volumetric fraction of bulk solids 20% Residence time 0.44 hr = 26.6 min Heating temperature 500 ◦ C Temp. of exit gas from combustor 900 ◦ C for decomposition of PCDD Temp. of recycle gas 500 ◦ C restricted by recycle blower

178

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

From preliminary calculation heat losses are given by Rotary reactor Combustor Cyclone, recycle line

18.63 kcal/kg feed 16.20 kcal/kg feed 13.70 kcal/kg feed

Total

48.53 kcal/kg feed

Solution Letting Bf be the amount of fuel needed for 1 kg of feed stock, the enthalpy balance on contour 1 in Fig. 10.1 is formulated as follows: [Input] Heat of fuel Heat of VOC Preheat air

10,000Bf (0.003)(8,000) (11.0)(Bf )(1 + 0.3)(0.31)(400 − 20)

[Output] Product Water vapour Combustion gas Excess air Heat losses

(1 − 0.02 − 0.003)(0.160)(500 − 20) (0.02)[586 + (0.5)(900 − 20)] (11.8)(Bf )(0.34)(900 − 20) (11.0)(Bf )(0.3)(0.31)(900 − 20) 48.53

From Input equals to Output, we obtain Bf = 0.01655 kg oil/kg feed = 16.55 kg oil/ton feed. The enthalpy balance on Contour 1 is given in Table 10.1. The enthalpy balance on Contour 2 in Fig. 10.1 is given by the following equation, where Vg is the volume of gas from the combustor to the rotary reactor. [Input] Inlet VOC

Vg (0.34)(900 − 20) (0.003)(8,000)

[Output] Product Gas Heat loss

(1 − 0.02 − 0.003)(0.16)(500 − 20) Vg (0.34)(500 − 20) 18.63

From Input = Output we get, Vg = 0.512 Nm3 /kg (twice of exit gas) Stoichiometric air is (11.0)(0.01655) = 0.1821 Nm3 /kg, and exit gas is (11.8 + 11.0 × 0.3)(0.01655) = 0.2499 Nm3 /kg. The necessary heat in the rotary reactor is given by   (1 − 0.02 − 0.003)(0.16)(500 − 20) + (0.02) 586 + (0.5)(500 − 20) + 18.63 − (0.003)(8,000) = 86.18 kcal/kg feed

(86.18)(3,000) = 258,540 kcal/hr Letting the mean temperatures of gas and solid be 700 ◦ C and 300 ◦ C, respectively, the radiant heat transfer coefficient from the combustion gas hrg is calculated with Eq. (6.3.2), where the emissivity of the gas is taken

Example 10.2

179

εg = 0.19.

hrg =

 4  300+273 4  (0.19)(0.8)(4.88) 700+273 − 100 100 700 − 300

= 14.62

kcal m2 hr ◦ C

The volumetric heat transfer coefficient is measured in the rotary dryer with simple lifter plates as presented in Chapter 6. Ua ≈ 100 kcal/m3 hr ◦ C The contribution of the radiant heat transfer from the combustion gas is estimated with Eq. (6.4.3) where Rv ≈ 1. ∆Ua = 4 ×

hrg 14.62 kcal/m2 hr ◦ C =4× = 29.2 kcal/m3 hr ◦ C dt 2m

Let us assume here, Ua = 100 + 29.2 ∼ 130 kcal/m3 hr ◦ C The volume of the reactor is large enough that the temperatures of gas and solids approach each other very quickly. Letting ∆Tav be the average temperature difference, and ∆T be that at their exit, we can estimate ∆Tav and ∆T as follows: From Eq. (6.4.1) we have, 258,540

 kcal kcal π = (2 m)2 (9.4 m) 130 3 ◦ (∆Tav ) hr 4 m hr C

∆Tav = 67.3 ◦ C In the case where the exponential distribution of the two temperature is assumed, ∆Tav =

(900 − 20) − ∆T ln 900−20 ∆T

At ∆T = 0.2 ◦ C, ∆Tav is calculated to be 104 ◦ C > 67.3 ◦ C. This means both temperatures are nearly equal, in the middle part of the reactor. The residence time of cans in the reactor is π (2 m)2 (9.4 m)(0.20)(225 kg/m3 ) 4 = 0.443 hr = 26.6 min

3,000 kg/hr

Example 10.2 In a food company, spent coffee powder is carbonized to obtain fuel gas, from which dry char is discharged. Design a rotary reactor to activate it for domestic application. Dry char 100 kg feed/hr Temperature 900 ◦ C, γ = 0.20 Water content 0.1 kg/kg d.f. Mass balance and enthalpy balance are given in Tables 7.4 and 7.5. εg = 0.19, εH = εC = 0.8 Solution The necessary residence time tr in the conventional rotary reactor shown in Fig. 7.8, is given in Example 5.3, to be tr = 7.2 hr and reactor efficiency ηr = 0.1469, in the case of γ = 0.1. Let us design such tube distributions of steam as illustrated in Table 1.2 [C] and γ = 0.20. In this case, the reactor efficiency ηr is supposed to be doubled.

180

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

Let us focus our attention on a single particle of char. The progress of its conversion is represented by Eq. (5.2.10).   1 − XB = exp −kr ηr C¯ A tr

When ηr is doubled, we estimate.

tr = (7.2 hr)(1/2) = 3.60 hr Letting the necessary volume of the activation region be Vr [m3 ], we calculate,   kg AC kg AC 60 (3.60 hr) = (Vr )(0.20) 450 hr m3  π 2 Vr = 2.40 m3 = d lr 4 t In the case of dt = 1.2 m, lr = 2.12 m. For safer design, we take dt = 1.2 m,

lr = 2.4 m

The temperature profile in the preheating region is assumed as follows: Tg T∗

1,100 ◦ C → 850 ◦ C 900 ◦ C ← 450 ◦ C

T ∗ is given by Eq. (6.3.14). The average temperature difference ∆Tav is then calculated as ∆Tav =

(850 − 450) − (1,100 − 900) 850−450 1,100−900

= 288 ◦ C

The radiant heat transfer coefficient from the combustion gas to the inner surface of the reactor is estimated by Eq. (6.3.2) as, hrg =

 4  700+273 4  (0.19)(0.8)(4.88) 1,000+273 − 100 100 1,000 − 700

= 42.77

kcal m2 hr ◦ C

From Table 7.5, let us assume the heat loss from the preheating region to be 630 kcal/kg d.f. The rate of heat transfer is given by the following equation.  kcal π(1.2 m)(lr ) 42.77 2 ◦ (288 ◦ C) m hr C   kcal kg 0.20 (900 ◦ C − 20 ◦ C) = 100 hr kg ◦ C      kcal kg water kcal kg + 0.5 0.1 586 (900 ◦ C − 20 ◦ C) + 100 ◦ hr kg kg kg C   kg kcal + 100 630 hr kg = 17,600 + 10,260 + 63,000 = 90,860

kcal hr

Thus we get lr = 1.957 m ≈ 2 m A conceptual design is proposed in Fig. 10.3, on the basis of the above calculation.

Example 10.3

181

Example 10.3 Very wet MSW (0.818 kg H2 O/kg dry MSW) was gasified in a Pyrox Process to produce 0.445 Nm3 /kg dry MSW of 5,596 kcal/Nm3 (gross) (see Table 10.3). However, 52.3% of the product gas was sent to its combustion/heating reactor as process fuel. The gasification efficiency of the Pyrox Process is, ηG =

(0.445)(5,596) = 0.757 = 75.7% 3,291

The efficiency of the available dry gas to send to any succeeding process, e.g., power generation, is reduced to, (0.757)(1 − 0.523) = 0.361 = 36.1% The main reason for the above low efficiency is the high content of liquid water in the feed MSW, which consumes more thermal energy than that for thermal cracking of dry feed. To achieve a high value for the available fluid fuel, possibly up to 66%, let us apply the rotary reactor illustrated in Fig. 10.5. It is planned to dry the MSW shown in Table 10.3 from 0.818 kg H2 O/kg dry MSW to 0.20 in a dryer, which is heated indirectly by hot exhaust gas from a gas engine or exit gas from the reactor. Since the water vapour from MSW is unbearably odorous, it is directly sent to a space in the oxidation-heating region of the reactor to eliminate odour. Tar is separated in a tar cooler scrubber, a heavy fraction of which is sent to the oxidationheating region as fuel. Predict the performance of the reactor, and determine its principal dimension under the following conditions. MSW Characteristics Feed rate of wet MSW Feed rate of dry MSW Gross calorific value Bulk density

Similar to Table 10.3 150 tons/day = 6,250 kg/hr 3,438 kg/hr 3,291 kcal/kg dry MSW 280 kg dry MSW/m3

Operating conditions Gasification region Oxidation/heating region Bulk density of carrier solids Preheating temperature of the air and water vapour

720 ◦ C 820 ◦ C 700 kg/m3 450 ◦ C

Product pattern and product gas modified from Igarashi [3] and Hasegawa [5] Dry gas Tar Condensed liquor Carbon Inorganics

0.49 kg/kg dry MSW 0.06 kg/kg dry MSW 0.08 kg/kg dry MSW 0.13 kg/kg dry MSW 0.24 kg/kg dry MSW 1.00 kg/kg dry MSW

182

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

0.430 Nm3 /kg dry MSW 5,000 kcal/Nm3 (0.430)(5,000) = 0.6533 3,291

Dry gas Gross calorific value Gasification efficiency Data Specific heat, dry gas Specific heat, tar vapour Specific heat of solids Heat of vaporization, tar Excess air for oxidation γs ηs in Eq. (2.3.1) is assumed 0.20.

0.413 kcal/Nm3 ◦ C 0.80 kcal/kg ◦ C 0.24 kcal/kg ◦ C 70 kcal/kg n = 0.20

Solution (1) Prediction of the necessary heat to the gasification region Let us make a contour around the gasification region in Fig. 10.5, and apply Eqs. (7.1.2)–(7.1.4) to predict the necessary heat. In the gasification region, carbon efficiency ηc is calculated as, ηc =

0.4843 − 0.130 = 0.7316 0.4843

With Eq. (7.1.2), (7,838)(0.4843)(1 − 0.7316) = 0.6904 3,291

α∗ = 1 −

No steam is sent into the region. Thus, β∗ = 0 The fraction of heat in exiting flows of products is calculated as follows: γ∗ =

(0.430)(0.413)(720 − 20) 3,291 +

(0.06)[70 + (0.8)(720 − 20)] 3,291

+

(0.08)[586 + (0.5)(720 − 20)] 3,291

condensed liquor

+

(0.20)[586 + (0.5)(720 − 20)] 3,291

water vapour

+

(0.24)(0.24)(720 − 20) 3,291

= 0.1411 Preliminary calculation on heat loss gives, δ ∗ = 0.01 From the given data we have, ηG = 0.6533 ζ=

dry gas

7,000 = 2.127, 3,291

ε∗ = 0.06

tar

inorganics

Example 10.3

183

The necessary heat to supply is calculated from Eq. (7.1.3), ηG = 0.6533 = α ∗ + 0 + ω∗ − (γ ∗ + δ ∗ + ζ · ε∗ ) = 0.6904 + 0 + ω∗ − (0.1411 + 0.01 + 2.127 × 0.06) Thus, ω∗ = 0.2416 Necessary heat = (0.2416)(3,291) = 795.1 kcal/kg dry MSW (2) Enthalpy balance in the oxidation-heating region In the region, 0.13 kg of carbon/kg dry MSW is transported from the gasification region, and burnt by sending air into the rotating layer of heat carrier solids. The amount of carbon is not sufficient to give the above heat to the gasification region. Let us send Bf kg of tar/kg dry MSW as auxiliary fuel to the oxidation-heating region. The enthalpy balance gives the following equations where fs [kg/kg dry MSW] is the circulating rate of solids between the oxidation-heating and gasification regions. 

[Input] Heat of combustion

Preheat, air Preheat, water vapour Heat in carrier solids [Output] Heat in heat carrier to gasification region

kcal kg dry MSW



(0.13)(7,838) = 1,018.9 (7,000)Bf     22.4 1 (0.13) + (7.6)Bf (1 + 0.2) 12 0.21 ×(0.31)(450 − 20) (0.818 − 0.2)[586 + (0.5)(450 − 20)] fs (0.24)(720 − 20)

fs (0.24)(820 − 20) 

 ⎤ 22.4 1 (1 + 0.2) ⎦ × (0.34)(820 − 20) 12 0.21 +Bf (8 + 7.6 × 0.2)

Combustion gas

⎡

Water vapour Inorganics Heat loss

(0.818)[586 + (0.5)(820 − 20)] (0.24)(0.24)(820 − 20) (0.01)(3,291)

⎣ (0.13)

From [Input] = [Output] and fs (0.24)(820 − 720) = 795.1 kcal/kg dry MSW, we calculate, Bf = 0.0562 kg/kg dry MSW A heavy fraction of product tar is sent to the oxidation-heating region, and then a light fraction is utilized as liquid fuel. Therefore, the calorific efficiency of the available fluid fuel from MSW is predicted to be ηT = ηG + ηL = 0.6533 + (0.06 − 0.0562)(7,000/3,291) = 0.6533 + 0.0081 = 0.6613 The amount of air to be sent to the oxidation-heating region is,     1 22.4 + (7.6)(0.0562) (1 + 0.2) = 1.899 Nm3 /kg dry MSW (0.13) 12 0.21 Odorous water vapour, 0.618 kg/kg dry MSW is sent to the oxidation-heating region and the odour is completely eliminated. (3) Dimension of the re-circulation region, four screw cylinders Preliminary calculation shows that the circulating solids are mainly inorganics, mixed with a small percentage of carbonaceous substance. The rate of circulation fs [kg/kg dry MSW] is obtained from the following

184

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

equation.  kcal kcal fs 0.24 (820 ◦ C − 720 ◦ C) = 795.1 ◦ kg C kg dry MSW Thus, fs = 33.1 kg/kg With Eq. (2.3.1) and γs ηs = 0.2, we have    kg π 2 kg =2 dts ls (0.2) 700 3 N (33.1) 3,438 hr 4 m In the case of ls = dts , the rate of rotation is given by, N = 517.5/dts3

[1/hr]

dts

[m]

1.2

1.3

1.4

N

[1/hr] [r.p.m.]

300 5.0

236 3.9

189 3.1

Geometrically, dts /dt = 0.4. If we take dts = 1.2 m, the inner diameter of the reactor dt is 3 m, and 5 r.p.m. Let us determine the length of the screw cylinder to be, 1.2 m × 2.5 = 3 m (4) Dimension of the gasification region In the gasification region, wet MSW is fed onto a hot flow of solids and dispersed in it. Let us take a look at Fig. 3.4, in which very wet brown coal of 7.4 mm is dried and cracked at 720 ◦ C within 60 sec. We can assume the necessary residence time for complete thermal cracking is 60 sec, until the solids reach the other end of the partition plate in Fig. 10.5. The circulation rate of the solids due to rotation of the partition plate is calculated with Eq. (2.3.3) in the case where dt = 3 m, γ1 = 0.2, η1 = 0.57, f = 65◦ , ρ¯ = 700 kg/m3 and N = 5.0 r.p.m. = 300 [1/hr].   kg π 1 Fs∗ = (3 m)3 (cot 65◦ )(0.2)(0.57) 700 3 = 118,360 kg/hr 300 8 hr m On the basis of 1 kg dry MSW, 118,360/3,438 = 34.43 kg/kg dry MSW The volumetric flow rate of the solids is given by   kg  kg m3 118,360 690 3 (3,600 sec) = 0.0476 hr sec m where 690 kg/m3 is the average bulk density of the mixture, composed of circulating solids and feed stock. The volume of solids in one side of partition plate is then,  m3 0.0476 (60 sec) = 2.86 m3 sec The volume of the gasification reactor is calculated as, (2.86 m3 )(2)/0.2 = 28.6 m3 Therefore, the length of the gasification region is, π (3 m)2 = 4.04 m For safer design 4.5 m (28.6 m3 ) 4

Example 10.3

185

The bulk density of the solids in the region is checked. 

280

  kg kg (1 + 34.4) = 688 ≈ 690 kg/m3 (1) + 700 (34.4) m3 m3

The flow rate of gas which issues from the rotating layer of solids is,     3  Nm3 3,438 kg 720 + 273 m 0.06 22.4 = 2.393 m3 /sec 0.43 + + (0.20) 6.0 18 kg 3,600 sec 273 Nm3 The surface area of the layer is (3 m)(0.87)(4.50 m) = 11.8 m2 The average velocity of the gas at the surface is given by, U0 = 2.393/11.8 = 0.203 m/sec (5) Dimension of the oxidation-heating region Carbon to be oxidized is,   kg 3,438 kg kg 0.13 = 0.1242 kg 3,600 sec sec The volumetric flow rate of air to oxidize it is,   22.4 820 + 273 m3 1 (0.1242) (1 + 0.2) = 5.304 12 0.21 273 sec To prevent fluidization of fine solids, let us design u0 = 0.50 m/sec. Then the surface area of the rotating solids layer is calculated as, (5.304)/(0.5) = 10.61 m2 The length of the region is (10.61 m2 )/(3 m)(0.87) = 4.06 m (6) Overall rate constant K¯ r for a layer of solids in the oxidation-heating region Let us postulate that char in this region is similar to that used by Jung and Stanmore [5, Chapter 3]. Initial size dp = 3.18 mm,

750 ◦ C,

kc = 15.9 cm/sec

Assuming the same dependency on temperature with graphite, given in Table 3.3, we estimate the rate constant at 820 ◦ C to be, kc = (15.9)(6.0/2.4) = 39.8 cm/sec The mass transfer coefficient Kd is calculated with Eq. (3.2.16), by application of the following numerical values. dp = 0.318 cm,

DA = 1.43 cm2 /sec,

µ = 4.2 × 10−4 gm/cm·sec,

Sc = 0.75,

From Eq. (3.2.15), k¯ = 13.8 cm/sec. With Eq. (3.3.5), Kr =



6 0.318 cm



13.8

cm sec



= 260.4

1 sec

ρ = 0.322 × 10−3 gm/cm3 ug = 100 cm/sec,

Kd = 21.1 cm/sec

186

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

If the excess air is 20%, 0.20 CAe = = 0.1667 CAi 1 + 0.2 Define the volumetric fraction of char referring to the total volume of solids in the region to be C1∗ . K¯ r = C1∗ Kr = 260.4C1∗ Let us estimate C1∗ which gives 20% excess for percolating air. From Eq. (5.2.5), (1 − εv )K¯ r ηr lp = − ln 0.1667 = 1.792 u0 In the case where εv = 0.5, ηr = 0.65, lp = (300)(0.2) = 60 cm, u0 = 50 cm/sec, we obtain, K¯ r = 4.595

1/sec

and

C1∗ = 0.0176 = 1.76%

(7) Carbon balance The carbon balance between the gasification and oxidation-heating regions gives the following equation. 0.130

  kg  ∗ kg = 34.43 C2 − C1∗ kg kg

C2∗ and C1∗ are the mass fraction of carbon to circulating solids. C2∗ is obtained by, C2∗ = 0.0176 + 0.00378 = 0.02138 In the system shown in Fig. 10.5, the majority of circulating solids is heat carrier, containing a little bit of carbon. (8) Device to eliminate residual carbon in discharge ash Residual carbon in discharge ash should be oxidized before leaving the oxidation-heating region. Plug flow of solids is needed at the exit end of the region. Design a 20% length of partition plate for this purpose. (9) Electric power generation From fluid fuel produced in 150 tons/day of very wet MSW, we estimate the amount of power generation as follows: (3,291 kcal/kg)(3,438 kg/hr)(0.661)(0.3) = 2,609 kW 860 kcal/kW·hr The results of the above calculation are summarized in Table 10.5.

Example 10.4 When sewage sludge in Table 10.6 is incinerated in a reactor, predict the necessary amount of fuel oil. Assume the temperature of the exit gas is 900 ◦ C. To delete additional fuel oil, a conceptual process system illustrated in Fig. 10.7 is planned. Compare its advantages over a conventional incinerator.

Example 10.4

187

Solution Letting Ta be the preheat temperature of the air, the necessary fuel oil Bf [kg/kg d.f.] is calculated from the enthalpy balance.   kcal [Input] kg d.f. Dry solid Fuel oil Preheat air

3,783 (10,000)Bf (4.8 + 11.0Bf )(1 + 0.3)(0.31)(Ta − 20)

[Output] Water vapour Combustion gas Excess air Heat loss

(4)[586 + (0.5)(900 − 20)] = 4,104 [5.4 + 11.8Bf ](0.34)(900 − 20) [4.8 + 10.0Bf ](0.3)(0.31)(900 − 20) 200

From [Input] = [Output] we obtain, Bf =

2,529.5 − 1.9344(Ta − 20) 5,651.0 + 4.433(Ta − 20)

(1)

The results of Eq. (1) are given in Table 10.7. At Ta = 400 ◦ C, we obtain Bf = 0.245 kg/kg d.f. For wet feed 100 tons/day = 4,166.7 kg/hr, the necessary feed rate of the additional oil is, (4,166.7)(0.2)(0.245) = 204.2 kg oil/hr = 4.9 ton soil/day In Example 10.3 and Table 10.5, we obtain ηT = 0.661. When the sewage sludge is dried and sent to the rotary reactor illustrated in Fig. 10.7, the following value of heat is supplied to the combustor of a gas/tar engine.  kcal kcal 4,150 (0.661) = 2,743 kg kg d.f. Suppose the generation efficiency is 30%, the generated electric power is,   kcal kW kcal 860 (0.3) = 0.957 2,743 kg d.f. kW·hr kg d.f. On the other hand, the thermal energy in exhaust gas from a gas engine or turbine is,  kcal kcal (1 − 0.30) = 1,920 2,743 kg d.f. kg d.f. We can utilize the above energy from 400 ◦ C to 200 ◦ C for indirect heating to the dryer as follows:  kcal 400 ◦ C − 200 ◦ C kcal 1,920 = 1,011 kg d.f. 400 ◦ C − 20 ◦ C kg d.f. The water to be vaporized in the dryer is, 1,011 kcal/kg d.f. kcal = 1.62 586 kcal/kg + (0.5 kcal/kg ◦ C)(100 ◦ C − 20 ◦ C) kg d.f. To vaporize 4.0 − 1.62 = 2.38 kg/kg d.f., we have to supply more hot gas to the dryer. Let us use dry RDF, the gross calorific value of which is 7,000 kcal/kg. When we gasify 0.90 kg/kg d.f. of the RDF in the parallel rotary reactor shown in Fig. 10.7, a similar calculation to the above leads to the following results. The power generation from the dry solids is calculated as, (0.90)(7,000)(0.661)(0.3) kW·hr = 1.453 860 kg d.f.

188

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

The thermal energy in the exit gas from the gas engine is, (0.9)(7,000)(0.661)(1 − 0.3) = 2,915

kcal kg d.f.

The water to be vaporized in the dryer is, (2,915) 400−200 400−20 586 + (0.5)(150 − 20)

= 2.36

kg H2 O kg d.f.

It is consistent with the required value. The above results are compared with a conventional incinerator, for 100 tons per day of wet sewage sludge. • Conventional incinerator, preheat temperature of the air 400 ◦ C Fuel oil 4.9 tons/day • System in Fig. 10.8 Fuel oil zero RDF or waste plastics 18 tons/day Power generation

(0.957 + 1.453)

 kW·hr kg 100,000 (0.2) = 2,008 kW kg d.f. 24 hrs

Example 10.5 It is planned to utilize low grade lignite, the contents of which are summarized in Table 10.8, for the production of rich fuel gas. Suppose thermal cracking of the lignite is carried out in an experimental reactor, giving the product pattern in Table 10.9. Applying the concept of Fig. 10.8 to the project, predict the performance of the novel rotary reactor, and determine its dimensions to gasify 400 tons of wet lignite per day. Assume the temperature of the gasification region is 850 ◦ C, and that of the oxidationheating region is 950 ◦ C. Solution (1) Estimation of dry gas produced On the basis of dry feed 1 kg, the gasification efficiency by thermal cracking is 0.2064, shown in Table 10.9. Product tar, 0.132 kg/kg d.f. is utilized as liquid fuel from lignite. On the other hand, 0.201 kg of carbon remains, accompanied by ash, 0.5 kg/kg d.f. Nearly half of the residual carbon can be gasified to form H2 + CO, by sending steam into the gasification region in Fig. 10.8. As the first approximation, let us assume 50% of the carbon is gasified, and 50% is oxidized in the oxidationheating region to supply the necessary heat for the gasification. The amount of H2 + CO formed in the region is,  0.201

kg 22.4 × 2 Nm3 (0.50) = 0.3752 Nm3 /kg d.f. kg d.f. 12 kg

The overall gasification efficiency is calculated as, ηG =

(0.138)(5,778) + (0.3752)(3,035) = 0.5012 3,863

(2) Steam to send, and calorific value of product gas Letting the conversion of steam be 50%, the amount of necessary steam to send is calculated as, (0.3752)

1 18 1 · · = 0.3015 kg/kg d.f. 2 22.4 0.5

Unreacted steam (0.3015)(1 − 0.5) = 0.1508 kg/kg d.f.

Example 10.5

189

Volume of dry gas 0.138 + 0.3752 = 0.5152 Nm3 /kg d.f. Gross calorific value of mixed gas is, kcal (0.138)(5,778) + (0.3752)(3,035) = 3,773 0.5152 Nm3 Product tar 0.132 kg/kg d.f. (7,000 kcal/kg) (3) Estimation of necessary heat to supply Let us apply Eqs. (7.1.2)–(7.1.4) to the gasification region in Fig. 10.8. From Table 10.8, C = 0.385 The carbon efficiency in the gasification region ηc is calculated as, ηc = 1 −

(0.201)(0.5) = 0.7390 0.385

Dimensionless values are calculated as follows: α∗ = 1 − β∗ =

(7,838)(0.385)(1 − 0.7390) = 0.7961 3,863

(0.3015)[586 + (0.5)(300 − 20)] = 0.0567 3,863

δ ∗ = 0.01

assumed

(0.138)(0.368)(850 − 20) γ∗ = 3,863

dry gas from cracking

+

(0.3752)(0.320)(850 − 20) 3,863

+

(0.132)[70 + (0.8)(850 − 20)] 3,863

+

(0.023 + 0.281 + 0.1508)[586 + (0.5)(850 − 20)] 3,863

+

(0.5)(0.24)(850 − 20) 3,863

H2 , CO from char tar water vapour

ash

= 0.2055  7,000 (0.132) = 0.2392 ζ · ε∗ = 3,863 From Eq. (7.1.3), 0.5012 = 0.7961 + 0.0567 + ω∗ − (0.2055 + 0.01 + 0.2392) We obtain ω∗ = 0.1031,

(3,863)(0.1031) = 398.3 kcal/kg d.f.

(4) Necessary fuel in oxidation-heating region It is planned to send some fraction of product tar to the oxidation-heating region, for heating up and temperature control. Combustion air is 20% in excess, preheated to 400 ◦ C. The necessary air to oxidize the carbon transported from the gasification region is given by   1 22.4 (1 + 0.2) = 1.072 Nm3 /kg d.f. (0.201)(0.5) 12 0.21

190

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

Let us check if residual carbon is enough to give heat for gasification. 

[Input]



Heat of combustion Preheat of air Heat carrier

(0.201)(0.5)(7,838) = 787.7 (1.072)(0.31)(400 − 20) = 126.3 fs (0.24)(850 − 20)

[Output] Heat carrier Combustion gas Ash Heat loss

fs (0.24)(950 − 20) (1.072)(0.34)(950 − 20) = 339.0 (0.5)(0.24)(950 − 20) = 116.0 (3,863)(0.01) = 38.6

kcal kg d.f.

where fs (0.24)(950 − 850) = 398.3 [Input] − [Output] = (787.7 + 126.3) − (339.0 + 116.0 + 38.6) − 398.3 = 22.1 > 0 We find no additional fuel is needed at steady state operation. Therefore, product tar can be evaluated as fluid fuel. The total efficiency for the available fluid fuel is then, ηT = ηG + ηL = 0.5012 + (0.132)(7,000/3,863) = 0.7404 (5) Fraction of carbon in solids The majority of solids in the system of Fig. 10.8 is ash from the lignite. Let us estimate the fraction of carbon in solids, C1∗ and C2∗ [kg/kg solids] in the oxidation-heating region and gasification region, respectively. In the oxidation-heating region, let us design that half of the carbon is oxidized in the circulating zone with the partition plate, and half is oxidized in the plug flow zone in order to eliminate residual carbon in the discharge ash. The carbon to be oxidized in the plug flow zone is,  0.201

 kg 1 kg (0.5) = 0.05025 kg d.f. 2 kg d.f.

The carbon balance gives, 0.05025

  kg kg ash  ∗ = 0.50 C1 − 0 kg d.f. kg d.f.

C1∗ = 0.1005 kg/kg ash

The rate of solids in the re-circulation region is obtained as, fs (0.24)(950 − 850) = 398.3,

fs = 16.60 kg/kg d.f.

The mass balance gives,   (0.201)(0.5) = (16.60) C2∗ − 0.1005

Then, C2∗ = 0.1066 kg/kg ash

(6) Dimension of re-circulation region, four screw cylinders The circulation rate of the solids is, (16.60)

(400,000 kg)(1 − 0.2194) = (16.60)(13,010 kg/hr) = 216,000 kg/hr 24 hr

Example 10.5

191

With Eq. (2.3.1) and γs ηs = 0.2, we have   kg π 2 216,000 = 2 dts ls (0.2) 800 3 N 4 m In the case where ls = dts N = 859.4/dts [1/hr] dts

[m]

1.6

1.7

N

[1/hr] [r.p.m.]

209.8 3.5

174.9 2.92

Geometrically dts /dt = 0.4. We take dts = 1.6 m, dt = 4.0 m, N = 3.5 r.p.m. The length of the cylinder should be, 1.6 m × 2.5 = 4.0 m (7) Dimension of the oxidation-heating region The rate constant of the chemical reaction is found to be 15.9 cm/sec at 750 ◦ C for char particle of dp = 3.18 mm (see Chapter 3, Section 3.4). Assuming the same dependency of temperature with graphite, we estimate it at 950 ◦ C to be, kc = (1.59)(29.0/2.4) = 192.1 cm/sec Taking DA = 1.69 cm2 /sec and εi = 0.1, Eq. (3.4.5) gives, DAe = 0.169 cm2 /sec With Eq. (3.2.14) and dp = 0.3 cm, 1 1 1 + , = 192.1 12(0.169)/0.3 k¯

cm k¯ = 6.530 sec

The diffusion of O2 within ash is found to be controlling. From Eq. (3.3.5), Kr =

6 6 ¯ 1 k= (6.530) = 130.6 dp 0.3 sec

In the plug flow zone in Fig. 10.8, the average value of the carbon fraction is given by, (0 + 0.1005)/2 = 0.05025 kg/kg solids If the density of the char is nearly the same as the ash, we can use the above value as the volumetric fraction. For the layer of solids then, K¯ r = (0.05025)(130.6) = 6.563 1/sec In the plug flow zone of solids, let us apply Eq. (5.2.5), and the following data. εv = 0.5,

ηr = 0.65, lp = 0.2, lp = 0.2(400 cm) = 80 cm,   CAe (1 − 0.5)(6.563)(0.65)(80) = 0.0582 = exp − CAi 60

The excess air ratio n is calculated by, n = 0.0582, 1+n

n = 0.0618

and u0 = 60 cm/sec,

192

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

The volumetric flow of the air, percolating in the zone is given by, (0.201)(0.5) (0.4743)

   Nm3 22.4 1 + 0.0618 1 = 0.4743 2 12 0.21 kg d.f.

(400,000)(1 − 0.281) 273 + 950 × = 7.073 m3 /sec (24)(3,600) 273

The surface area of the solids layer is,   m m3 0.600 = 11.79 m2 7.073 sec sec The length of the zone is, (11.79 m2 )/(4.0)(0.87) = 3.39 m In the re-circulating zone where the partition plate is positioned, K¯ r = (0.1066)(130.6) = 13.14 1/sec With εv = 0.5, K¯ r = 13.14, ηr = 0.65, lp = 80, u0 = 60 cm/sec we obtain (1 − εv )K¯ r ηr lp (1 − 0.5)(13.14)(0.65)(80) = = 5.694 u0 60 CAe n = 0.00337 = CAi 1+n n = 0.00338 = 0.338% The volumetric flow of the air percolating in the above zone is calculated as,    1 22.4 1 + 0.00337 (0.201)(0.5) = 0.4482 Nm3 /kg d.f. 2 12 0.21  950 + 273 (400,000)(1 − 0.281) = 6.684 m3 /sec × (0.4482) (24)(3,600) 273 The surface area of the solids layer is given by, (6.684 m3 /sec)/(0.60 m/sec) = 11.14 m2 The length of the zone is then, (11.14)/(4)(0.87) = 3.20 m The total length of the oxidation-heating region is 3.39 m + 3.20 m = 6.59 m The amount of air to be sent to the oxidation-heating region is given by 0.4743 + 0.4482 = 0.9225 Nm3 /kg d.f. The necessary air for complete oxidation is,   1 22.4 × (1 + 0.2) = 1.072 Nm3 /kg d.f. (0.201)(0.5) 12 0.21 Surplus air, 1.072 − 0.9225 = 0.1495 Nm3 /kg d.f. is sent into the space in the oxidation-heating region.

Example 10.5

193

(8) Dimension of the gasification region Unlike MSW in Example 10.3, the gasification rate of carbon should be taken into account. By interpolation of the data in Table 3.8 for the gasification of carbon with water vapour, Eq. (3.5.2), we estimate. kr = 420

1 (gm-mol A/cm3 gas)(sec)

at 850 ◦ C

From Eqs. (3.3.5) and (3.3.6), kc =

dp ρB (1 − XB ) kr 6b

(1)

Since XB = 0 ∼ 0.5 in the gasification region, we take the average value of 1 − XB to be 0.75. The carbon content in solids there is calculated by (0.385)(0.75) = 0.3660 (0.385)(0.75) + 0.50 where C = 0.385 is from Table 10.8. Therefore, the molar density of carbon contained in ash particle is given as   gm gm-mol B 1 = 0.03660 0.3660 ρB = 1.2 12 gm/gm-mol cm3 cm3 With b = 1 [gm-mol B/gm-mol A] and dp = 0.30 cm Eq. (1) gives, kc =

(0.3)(0.03660)(0.75) 1 (420) = 0.5765 6(1) sec

From Eq. (3.2.14) 1 1 1 = + , 0.5765 12(0.169)/0.3 k¯

1 k¯ = 0.5312 sec

It is found that the diffusion of the reactant gas in the ash can be neglected. With Eq. (3.3.5) Kr =

6 6 ¯ 1 k= (0.5312) = 10.62 dp 0.3 sec

In the previous calculation, the carbon content of the solids circulating in the gasification region is, C2∗ = 0.1066 kg C/kg solids The overall rate constant in the layer of solids within the gasification region is then given by, K¯ r = (10.62)(0.1066) = 1.132 1/sec In the case where u0 = 0.30 m/sec = 30 cm/sec (1 − εv )K¯ r ηr lp (1 − 0.5)(1.132)(0.65)(80) = = 0.9811 u0 30 From Eq. (5.2.5), CAe /CAi = 0.375 When conversion of the water vapour is 50%, the exit concentration of the water vapour is given by, CAe 0.5 = = 0.330 CAi 0.5 + 0.5 × 2 Two values of CAe /CAi are consistent with each other.

194

10. Application of a Rotary Reactor for the Re-utilization of Solid Wastes

The amount of steam to be sent to the gasification region is then, (0.201)(0.5)

(0.3752)

 18 1 kg Nm3 = 0.3015 = 0.3752 12 0.5 kg kg

(400,000)(0.719) 850 + 273 m3 · = 5.138 (24)(3,600) 273 sec

The surface area of the solids layer is (5.138 m3 /sec)/(0.30 m/sec) = 17.13 m2 The length needed for the gasification of carbon is, 17.13 m2 = 4.92 m ∼ 5 m (4.0 m)(0.87) The circulating rate of solids within the gasification region due to the partition plate is calculated with Eq. (2.3.3). dt = 4.0 m,

ϕ = 65◦ ,

γ1 = 0.2,

η1 = 0.57,

ρ¯ = 700 kg/m3 ,

N = 3.5 r.p.m. = 210 1/hr    1 kg π (4.0 m)3 (cot 65◦ )(0.2)(0.57) 700 3 = 196,400 kg/hr 210 Fs∗ = 8 hr m The volumetric flow rate is, 196,400  3,600 = 0.0779 m3 /sec 700 The necessary time for heating and thermal cracking of the feed stock is assumed to be 90 sec. Thus, the volume of solids in one side of the partition plate is given by, 

0.0779

m3 (90 sec) = 7.01 m3 sec

The total volume of the gasification region is, (7.01)(2)/0.2 = 70.1 m3 The length of the region is, π 2 70.1 (4) = 5.58 m 4

For safer design, we take this as 7 m. (9) Electric power generation Where the efficiency of power generation is 30% with an appropriate gas engine, power generation is predicted as follows: (400,000)(1 − 0.281)(3,863)(0.740)(0.30) = 11,950 kW (24)(860) The results of the above calculation are summarized in Table 10.10.

10.5. References

195

REFERENCES [1] D. Kunii, Chem. Eng. Sci. 35 (9) (1980) 1887. [2] Y. Shoji, et al., Fourth International Conference on Fluidization, D. Kunii and R. Toei (Eds.) (Engineering Foundation, Kashikojima, 1983). [3] M. Igarashi, et al., Journal of Energy Resources Technology 106/377 (Sept. 1984). [4] T. Koike, M. Nagao, 12th General Conference on Research for Disposal of MSW (Feb. 1991). [5] M. Hasegawa, T. Ishida, Research on Disposal of MSW (Japan, Oct. 1983).

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Brief Careers of the Authors Daizo Kunii 1963 Professor, University of Tokyo 1983 President, Society of Chemical Engineers, Japan 1984 Professor Emeritus, University of Tokyo Tatsu Chisaki 1955 Engineer, the Tanabe Kakoki Co. 1975 President, the CHISAKI Co.

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Index

A accompanied organic substance (VOC) 177 activation – of carbon – – in commercial rotary reactor, deriving reactor efficiency 67 – – mass balance 110 – of carbonaceous pellet 40 – of char 6, 7, 93, 109, 165 – – from coconut shell 60 activation/gasification of char 9 active carbon 67 – directly from wet wood chips 166 agglomeration of composite solid for fertilizer 9 angle of inclined guide plate 20 angle of inclined surface 13 application of kinetic models 33 appropriate design of rotary sealing 9 average concentration – of gaseous reactant 58 – of reactant gas A 59 azotation of calcium carbide 44 B black cone 72 blow off 71 blue cone 70 Boudouard’s reaction 27, 37 bubbling in molten bath 167 C calcination – of limestone 47, 93, 97 – – necessary time for complete conversion 53 – – temperature of decompositing interface 53 – of magnesite, limestone 9 calorific efficiency for fluid fuel 172, 177 capacity of rotary kilns 23 car shredder dust 153 carbon dispersed in inorganic solids 36 carbon efficiency 94 char from coal 34 CHISAKI’s preheater 101, 102

circular weir 13, 14 circulation rate of solids 24 clinkering of calcia, dolomite 9 clinkering of cement 9 clogging of the path 5 co-axial cylinder 5 co-current flow 2, 4 combustion – and heating region 98, 103 – gas region 83, 98 – model of a gas burner 69 – of char 8 – of fuel 69, 72 complete back-mix flow 4 composite made of iron ore – and heavy oil 50 – and oil 50 composite pellets made of ferro-chromium ore and coke 105 concentration – profile in the model rotary kiln 107 – profile of reactant gas A in rotating layer of solids B 59 conceptual rotary reactor for gasification of very wet MSW 172 confirmation of strength 9 consumption of electricity 143, 145 contact operation 2 contacting methods 1–3 – comparison 3 convectional heat transfer coefficient 35, 43, 135 conversion – of char gasified by steam 39 – of composite pellets 47 – of gas and solids within solids layer 57 – of iron ore 51 – of pellet and components in effluent gas 51 – of solids 1, 7, 31, 100, 105 – – by gaseous reactant, in rotating layer of solids 57 – – in rotary reactors 57 – – in rotating layer of solids 62, 63 – – with gaseous reactant 27 conversion efficiency of thermal energy to electric power 163

200

Index

conversion rate of the gaseous reactant in rotating layer of solids 63 cooler 98, 147 cost of electric power 144 counter-current exchange of heat 97, 100 counter-current flow 2, 4 critical temperature of the retort 136 cross flow 2, 4 D de-lacquering of spent cans 9, 164 decomposition – of dolomite 47 – of limestone 47, 48 – of magnesite 47 – of manganese sulfate 49 – of metal sulfate 9, 19 decreasing size model 31 dehydration of inorganic compounds 19 device – to enhance contacting between gas and solids 6 – to enhance gas solid contacting 8 – to measure the reaction rate of a solid sample 29 diffusional process 1 diffusional resistance 1 diffusivity 31 – of gaseous reactant 36 – through product blanket 31 dioxins 24 direct injection of gaseous reactant 5 direct reduction of iron ore 41 direction of improvement – activation of char 111 – for pre-reduction of composite pellets 108 – for rotary reactor, electrically heated 147 – for the oxidation of residual carbon in spent catalyst 150 discharging device 10 distributor 8 E eddies 73 effect of heat loss in calcination of limestone effective diffusivity of gaseous reactant 57 efficiency of transportation 20 electric heating elements 131 electrically heated rotary retort 144 elimination – of carbon in pure silica 9 – of dioxins in fly ash 47 – of residual carbon in spent catalyst 59 – – deriving reactor efficiency 66 – of silanol in ultra pure silica 47, 52

102

– of trace species 47 emissivity – of combustion gas 78 – of luminous flame 78 – of retort surface 128 enhancement – of contact by sending gaseous reactant 61 – of gas-solid contact simplified model, 5, 6, 8 – of heat transfer 84 enthalpy balance 95, 102, 103, 107, 143 – complete combustion 93 – partial combustion and gasification 94 entrained flow of solids 3 entrained solids 167 entrained solids flow 1, 2 equations for design calculation of the new heating system 146 equilibrium constant 39 excess air ratio 100 exothermic reaction 64 extension of heat transfer surface area 134 F feeding device 10 Fick’s law 31 fixed bed 2 flame front 70 flame region 83 flat distributor 7 flat hearth 2 flow pattern – of gas from rotating tube tuyeres 62 – of solids in a rotary reactor 5 fluidized bed 1–4, 34, 43 – incinerator 163 – of fine inert particles 29 – reactor 28, 40, 41, 44 – steady state operation 29 fluidized solids 167 fly ash from an incinerator 24 formation of thick layer 128 fraction – of carbon 95 – of inner surface area 78 – of wall surface area 80 G gas–solid reaction 9 – process 27 gas–solid reactor process 3 gaseous reactant around particle gasification – efficiency 94, 172 – of carbon 37

35

Index

– of carbon by steam 38 – of char 8 – of combustible feed stock 111 – of lignite for rich product gas 176 – of low grade coal 175 – of low grade lignite 188 – of mixture of poly-propylene sheet and pulp 121 – of MSW 9, 167 – of plastic waste in commercial reactor, single fluidized 169 – of sewage sludge 173, 186 – of solid wastes 93, 167 – of spent coffee powder 179 – rate of char with CO2 38 graphite 33 graphite sphere 43 grate incinerator 163 gross calorific value 94 – of combustible gas, factors f1 and f2 96 – of product dry gas 94 guide plates 20 H heat – of combustion 103 – of decomposition 97, 103 – of reaction 35 – to decompose 99 heat balance – of the surface of the retort 136 – predicted for calcination of limestone 101 heat exchanger between two flows of solid 148 heat flux from the outer surface of kiln 98 heat loss – from the outer surface 83 – per unit mass of product 99 – – in a rotary kiln 98 heat transfer – coefficient – – at outer surface of the reactor 99 – – by direct contacting of solids from the hot wall surface 80 – – from gas to retort 134 – – from the hot gas to the surface of the retort 132 – – from the inner surface of the insulation cover 134 – – from the outer surface of a reactor 82 – – inside the retort 129 – – of wall surface, contacting with rotating solids 81 – direct heating 69 – from gas flow 132 – in a rotary reactor 76

201

– indirect heating 127 – mechanism 69 – – in the combustion and heating region 77 – model in rotary retort 129 – to retort surface 134 – to rotary retort, electric heating 130 – within the rotary retort 128 heating capacity of a rotary reactor 82, 89 heating region by combustion gas 103 heating system for rotary reactor, heated by combustion gas 150 Hess’s law 96 high temperature – near nozzles of injection gas 65 – stability of falling solids 64 – stability of isolated solids 64 horizontal tube 29 I ignition of the combustible gas 72 improvement – for lower fuel consumption 93 – of electric heating 144 – of residence time characteristics 18 incineration – of municipal solid waste 6 – of sewage sludge 173 – plant 163 incinerator 6 inclination angle 12 indirect heating by hot gas flow 133 information for design of rotary reactors, indirect heating 127 inside combustion 72, 73 inside rotary reactor 133 insulation cover cylinder 132 issue of gases from a spherical composite pellet 51 K kinetic models – for conversion of a single solid 31 – of gas–solid reactions 30 Kunii–Kunugi Process 167 Kunii’s model of solids movement 15 L laminar boundary layer 66 landfill 163 lifters 7, 64 – in a rotary dryer 84 lignite 175

202

liquid fuel 71 loading capacity in the combustion region low grade lignite 175 luminous flame 70

Index

74

– of pulverized coal particles, time necessary for complete conversion 86 – of residual carbon in spent catalyst 37, 148 oxidation and reduction of ferrite powder 9

M

P

mass and enthalpy balances 93 mass transfer coefficient 30, 31 – of boundary layer 59 material of the retort 127 matrix presentation of gasification processes 167 mixing and diffusion of combustible gas 69 model – for electric power generation from very wet sewage sludge 174 – of a rotary kiln 84 – of heat transfer 135 molecular diffusion 69 movement of solids 11, 12 – on partition plate 20 moving bed 1–3, 7 moving solids 167 MSW, see municipal solid waste multi-stage rotary hearth 3 municipal solid waste (MSW) 163

partial pressure of CO2 at the interface of decomposition for CaCO3 and MgCO3 48 partition 19 – plate 5, 86, 129, 132 – – compared with screw cylinder 22 – – with a number of guide plates 170 – – with inclined guide plates 111 – plates – – within annular space 22 performance – for oxidation of residual carbon in spent catalyst 151 – of rotary kiln for pre-reduction of composite pellets 108 – of rotary reactor – – activation of char 110 – – for de-lacquering of spent cans 165 – predicted for gasification of plastic waste by oxygen and steam 112 petroleum coke 33 plug flow 4 Pocahontas coal 72 practical performance data on rotary kilns 101 practical rotary reactors 9 pre-calcination region in the preheater 104 pre-reduction – of composite pellets 105 – of ferro-chromium pellets 93 predicted temperaturs in retort, electrically heated 131 prediction – of overall performance 99 – of production capacity, rotary kilns 17 preheater 97, 100 premixed air 70 premixing of air 70, 71 prevention of air leakage 151 product blanket 42 product pattern, assumed for thermal cracking of model lignite 176 production capacity of rotary kilns 17 pulverized coal 86 pulverized coal and coke 72 Pyrox process 168, 170

N necessary amount of oxygen 95 necessary heat Hn to operate gasification 96 necessary time τ for complete conversion of limestone 49 net calorific value 75, 100 novel retort for heat recovery from hot solids 145 novel rotary reactor – to eliminate polychloro-dibenzo-dioxins in fly ash 155 – to gasify wet MSW 171 – to produce rich gas from MSW 169 O operation data of Pyrox process, gasification of MSW 168 outside rotary reactor 133 overall heat transfer coefficient 136, 150 overall rate – constant 30, 35, 44 – of conversion 69 – of reaction 1 oxidation – of carbon 33 – – in spent catalyst 9

Q quality of active carbon 60 quality of catalysts 152

Index

R radiant heat 77 – transfer 69 – – coefficient 35, 75, 77, 82, 87, 114 – – from flame and combustion gas 76 – – from inner wall surface to surface of rotating solids layer 78 rate – of heat transfer 69 – of product 98 – of solid circulation 170 – of solids flow 21 rate constant 32, 61, 62 – for activation of carbonaceous pellets 40 – for gasification of graphite 37 – of activation 68 – of Boudouard’s reaction 38 – of chemical reaction 30–33, 37, 61 – – for gasification of char by steam 40 – of combustion 74, 87 – of solids 57 ratio – of total surface area, referring to the inner surface of the reactor 85 – of turbulent flame length to nozzle diameter 71 re-circulating – flow rate of the solids 18 – rate of solids 22 re-circulation 11 – region 111 – – composed of 4 screw cylinders 171 reaction rate of solid conversion 27 reactor – for elimination of polychloro-dibenzo-dioxins contained in fly ash 147 – for gasification of MSW 167 – for re-utilization of secondary resources 164 reactor efficiency 6, 61, 64, 66, 67 reduction – of composite pellet, ferro-chromium ore and coke 52 – of composite pellet, ferro-chromium ore and coke, experimental data 52 – of ferro-chromium ore 9 – of fuel consumption 93 – of fuel oil by preheating of air 173 – of iron ore 41, 42 – of powdery ferrite by hydrogen 59 refuse derived fuel (RDF) 163 refuse derived paper-plastic fuel (RPF) 163 regeneration of spent catalyst 7 relative velocity 30 repose angle 12, 13 residence time 4, 16, 18, 177 – characteristics 5

203

– in reduction region 107 – of solids 23, 63 residual carbon 66 reverse flame 72, 73 Reynolds number 30 roasting of zinc sulfide 41 rod-like flow 4 rotary – cylinder 2, 3, 167 – dryer 174 – fluidized bed 64 – gasifier for combustible feed stock 111 – kiln 97 – – for limestone 113 – – for pre-reduction of composite pellets 106, 115 – – to calcinate limestone 97 – manifold 6, 66, 151 – moving bed 64 rotary reactor 1 – advantages 3 – direct heating 93 – disadvantages 3 – electric heating 143 – for activation of char 109 – for active carbon 119 – – proposed by the authors 166 – for de-lacquering of spent cans 164, 176 – for elimination of residual carbon in spent catalyst 149, 152 – for gasification of MSW 167 – for the re-utilization of solid wastes 163 – heated by combustion gas 148 – practically applied 9 – to eliminate polychloro-dibenzo-dioxins in fly ash 153 – to gasify car shredder dust 169, 170 – to gasify low grade lignite 177 – to gasify sewage sludge 174 – to gasify solid waste materials, containing much NaCl 159 – to gasify waste materials, design calculation 140 rotary retort 127, 143, 152, 153 – for thermal cracking of solid waste materials 153 – for thermal decomposition of metal sulfate, design calculation 139 – to oxidize residual carbon in spent catalyst 157 – – prediction of its performance 138 rotary sealing of distribution manifold 64 rotating – cylinder 16 – distributor 64 – hearth 98 Roto-Louvre dryer 6

204

S Schmidt number 30 screw cylinder 18, 19 seal plate 6 selection of material 9 sewage sludge, de-watered by filter press 173 shrinking core model 31, 42 simple model on movement of solids 14 simultaneous reaction for a composite of powdery iron ore and straight asphalt 51 single fluidized bed reactor, which gasified a wet mixture of plastics and spent pulp 168 sintering of ore 42 sintering of solids 5 slipping of solids 11 solid wastes 163 spent catalyst 64 spinel transformation of ferrite 9 stable temperature of isolated carbon particle 35, 36 steady state 1 sticking of solids 128 summarized heat transfer coefficient 135 T temperature – difference between gas and solid 35 – distribution in rotary kiln 90 – inner wall surface and outer shell surface 83 – inner wall surface and the outer shell surface – – in rotary kiln 88 – of inner wall surface 81, 82 – profile – – in flame 75 – – in turbulent flame 74 – – of flame 76 temperatures and heat transfer coefficients in rotary retort 136 theoretical equations on transfer rate of solids 20 thermal cracking – of MSW 170 – of organic materials 19 – of organic solids 50 – of solid waste 9 – of solid waste materials 152

Index

– of Victorian brown coal 34 thermal decomposition 47 – of manganese sulfate, experimental data 50 thermal radiation 71 thermo-balance 27, 28 – reactor 39 thickness of rotary retort 127 time for complete conversion 31 – of B 31 – of char 86 time necessary for complete conversion 54 time necessary for conversion of solids, affected by layer thickness 59 transfer rates of solids in annular space 21 transportation rate of solids in rotary cylinder 16 tube tuyeres 64 turbulent diffusion 69–72, 110 turbulent flame, longitudinal temperature distribution 87 U U-Turn rotary reactor 111 U-Turn system 19 uniform reaction model 31 V Victorian brown coal 34 void fraction of the bulk solids layer 57 volume – of combustion region 74 – of discharged solids per one rotation 16 – of product dry gas 94 volumetric fraction 13 – of bulk solids 12, 17, 18 – of falling solids 64 – of rotating solids 14 – of solids 23 volumetric heat transfer coefficient 84 volumetric transfer rate 16 vortexes 69, 73 X XB conversion of solid reactant 35