Airflow In Ducts 9780880690188

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AIRFLOW IN DUCTS LEO A. MEYER

TECHNICIAN'S LIBRARY

I N D O O R

E N V I R O N M E N T

-

1 3 1

LAMABOOKS

H YAC B O O K S

T H AT W O R K

AIRFLOW IN

DUCTS by

LEO A. MEYER H. LYNN WRAY, P.E., Technical Advisor

INDOOR ENVIRONMENT IET TECHNICIAN'S LIBRARY

LAMA BOOKS

H VAC B O O KS

THAT WORK

FOREWORD Yo u a r e p r o b a b l y w o r k i n g a s a t e c h n i c i a n i n o n e o f t h e i n d o o r e n v i r o n m e n t f i e l d s . This means that you have at least some understanding of air flow in ducts. H o w e v e r, d o n ' t f a l l i n t o t h e t r a p o f t h i n k i n g . " I k n o w a l l t h i s s t u f f." R e a d e a c h c h a p t e r. T h e n d o t h e R e v i e w. I n m y e x p e r i e n c e , e v e r y t i m e I s t u d i e d

material | "knew all about,'

learned new ideas and corrected misunderstandings.

I f y o u s t u d y e a c h c h a p t e r c a r e f u l l y, y o u w i l l g a i n n e w i d e a s . M o r e i m p o r t a n t , y o u w i l l g i v e y o u r s e l f a s o l i d u n d e r s t a n d i n g o f b a s i c p r i n c i p l e s t h a t y o u will be able to

apply in the field. You will also be able to apply your knowledge to more advanced technical principles covered in later books in this series.

Indoor Environment Technician's Librar T h i s b o o k i s p a r t o f t h e I n d o o r E n v i r o n m e n t Te c h n i c i a n ' s L i b r a r y. T h e s e a r e practical books that you can use as training or as reference. These books apply to all areas of the indoor environment industry:

Heating, ventilating, and air conditioning Energy management Indoor air quality Service work Te s t i n g , a d j u s t i n g , a n d b a l a n c i n g

If You Are Training Others If you are a supervisor training others, you will find that the Indoor Environment Te c h n i c i a n ' s L i b r a r y c a n m a k e i t e a s i e r. A S u p e r v i s o r ' s G u i d e i s a v a i l a b l e f o r s o m e books. It includes teaching suggestions and key questions you can ask to make sure the student understands the material.

Le o A . M e y e r LAMA Books 2 3 8 1 S l e e p y H o l l o w Av e Hayward CA 94545-3429 888-452-6244

510-785-1099 fax

w w w. l a m a b o o k s . c o m ISBN 978-0-88069-018-8

© Copyright 1996 by LAMA Books 2nd printing 2004, 3rd printing 2007 All rights reserved. No part of this publication may be reproduced, stored in an electronic retrieval

system or transmitted in any form or by any meanselectronic, mechanical, or otherwise-without the p r i o r p e r m i s s i o n o f t h e p u b l i s h e r.

LAMA Books specifically disclaims any and all liability for damages of any type whatsoever that may result directly or indirectly from a person's reliance upon or utilization of the information contained in this book.

TA B L E O F C O N T E N T S 1. Basics of Air flow 2. Calculating Duct Sizes

11 --

3. Air Quantity and Velocity

26

4. Pressures in a Duct

38

5. Air flow in a Duct and Dynamic Losses

49

6. Sizing Duc twork

58

7. Calculating Pressure Losses in Ductwork

71

8. Duct Fittings

83

9. Measuring Air flow

97

Review Answers Appendix-Equations Used in This Book

Index

108

116 119

1

BASICS OF AIRFLOW

If you work in the HVAC industry, you are concerned with t h e p ro c e s s o f c o n d i t i o n i n g a i r a n d m ov i n g a i r f ro m o n e a re a t o a n o t h e r. I t m a ke s s e n s e t h a t yo u n e e d t o k n ow h ow a i r f l ows i n d u c ts a n d w h a t fa c to rs a f fe c t t h e f l ow.

When you complete this book, you will understand airflow i n a d u c t b e t t e r t h a n m o s t p e o p l e i n t h e H VAC i n d u s t r y d o .

This will help you fabricate and install ductwork more

e f fe c t i ve l y a n d h e l p yo u m a ke c h a n g e s i n d u c t wo r k s i ze w i t h c o n f i d e n c e . A g o o d u n d e r s t a n d i n g o f a i r f l ow i s n e e d e d fo r d i f fe re n t s p e c i a l t i e s i n t h e H VAC i n d u s t r y, s u c h a s :

.

O

Service work Te s t i n g , a d j u s t i n g , a n d b a l a n c i n g

0

Energy management

O

Indoor air quality work

WHY MOVE AIR? Think of what HVAC means-heating, ventilating, and air conditioning:

.

o Ventilating is bringing outside air into a building. Heating and air conditioning means heating, cooling, and cleaning air, and regulating its moisture content.

2

C o n d i t i o n e d a i r m u st b e d e l i ve re d to s e l e c te d a re a s o f a b u i l d i n g a n d t h e n re m ove d f ro m t h o s e a re a s a n d re t u r n e d fo r re - c o n d i t i o n i n g . C o n d i t i o n e d a i r i s u s u a l l y t r a n s p o r te d

through ductwork. The p r o c e s s o f m ov i n g a i r i n d u c ts d i s t r i b u te s e n e r g y :

. .

I n c o l d we a t h e r, h e a t t a ke n f ro m a n e n e rg y s o u rc e s u c h a s gas or elec tricity is added to air. This heat is d e l i ve re d to t h e c o n d i t i o n e d s p a c e . I n h o t we a t h e r, h e a t i s re m ove d f ro m a i r by t h e u s e of elec trical energy. In this case, heat energy is b e i n g r e m ove d f r o m t h e c o n d i t i o n e d s p a c e .

THE HVAC SYSTEM An HVAC system has different components: O

.

Central air handling system-Generally contained in a mechanical room. It includes the fan to move t h e a i r a n d e q u i p m e n t to c o n d i t i o n t h e a i r b e fo re i t i s d e l i ve re d to t h e c o n d i t i o n e d s p a c e .

Boiler-Provides hot water to heat the air.

0 Refrigeration unit-Provides a means to cool the air.

O

Duct system-Distributes the conditioned air where needed.

Central Air Handling System F i g u re 1 s h ows a t y p i c a l c e n t r a l a i r h a n d l i n g syste m . T h i s i s a schematic drawing that shows the par ts of the system and h ow t h ey a re re l a te d to e a c h o t h e r. I t d o e s n o t s h ow t h e l o c a t i o n o f t h e c o m p o n e n ts i n a n a c t u a l i n s t a l l a t i o n . T h e re a re m a ny d i f fe re n t sys te m va r i a t i o n s . Yo u n e e d to u n d e rs t a n d t h e b a s i c c o m p o n e n ts o f a sys t e m a n d t h e i r

relationship to each other. Then you can identify the components on any job.

3

EA DAMPER

EA

RA

(EXHAUST AIR)

(RETURN AIR) RA DAMPER

OA •OUTSIDE AIR)

OA DAMPER

MIXED AIR -

SA ( S U P P LY A I R )

F I LT E R S U P P LY A I R FA N

Fig. 1: Central air handling system

S u p p l y a i r ( S A ) i s t h e c o n d i t i o n e d a i r d e l i ve re d to t h e

building. In Fig. 1 the supply air is in the lower right hand corner. When supply air is delivered into a room, an equal amount o f a i r m u s t b e re m ove d f ro m t h e ro o m . T h i s re t u r n a i r ( R A ) is transported back to the central air system for reprocessing. In Fig. 1 the retur n air is in the upper right

corner.

O n l y a p e rc e n t a g e o f t h e re t u r n a i r c a n b e re u s e d . T h e a i r wo u l d b e c o m e s t a l e i f t h e s a m e a i r we re u s e d ove r a n d ove r a g a i n . To avo i d t h i s , f re s h a i r i s b ro u g h t i n to t h e syste m

through the outside air (OA) intake. In Fig. 1 the outside

air inlet is in the lower left.

When outside air enters the building, the same amount of a i r m u s t b e re m ove d f ro m t h e b u i l d i n g t h ro u g h t h e ex h a u s t air (EA) outlet or other exhaust air systems. T h e ex h a u st a i r ( E A ) , o u ts i d e a i r ( OA ) , a n d re t u r n a i r ( R A ) d u c ts a l l o p e r a te to g e t h e r. A n a u to m a t i c c o n t ro l syste m o p e r a t e s t h e d a m p e r m o t o rs t o m a i n t a i n t h e p ro p e r m i x o f a i r. T h e OA a n d t h e E A d a m p e r s o p e n t o g e t h e r a n d c l o s e together to balan ce the air en terin g an d leavin g the

4

building. As the OA damper opens, the RA damper closes s o t h a t t h e s a m e a m o u n t o f a i r re m a i n s i n t h e syste m . Af t e r t h e a i r i s m i xe d i n t h e p ro p e r p ro p o r t i o n s , t h e m i xe d a i r i s d r aw n t h ro u g h a f i l te r b e fo re i t e n te rs t h e fa n a n d retur ns to the conditioned spaces. T h e sys t e m s h ow n i n F i g . 1 d o e s n o t p rov i d e fo r h e a t i n g o r

cooling the air.

Heating System F i g u re 2 s h ows h ow h e a t i s a d d e d . A h e a t i n g c o i l ( F i g . 3 ) i s added to the system before the air enters the fan. It is located after the filter so that there is less chance of dirt

clogging the coil. The coil is similar to a car radiator. Hot water flows through tubes in the coil. Metal fins are attached to the tubes. The EA

RA

FA N H O T WAT E R C O I L HOT WATER SUPPLY,

P U M P.

HWR H OT WAT E R R E T U R N Fig. 2: Hot water system added to air handling system

5

h e a t f ro m t h e h o t wa te r i s con d u c te d to t h e ou ts i d e of t h e tubes and to the fins. The air passing across the coil is heated by absorbing heat from the fins and the tubes. T h e h o t wa te r fo r t h e c o i l i s

heated by a boiler. A pump

TE

c i rc u l a te s t h e h o t wa te r t h ro u g h

the coil.

Fig. 3: A heating coil

Cooling System F i g u re 4 s h ows a c h i l l e d wa t e r c o i l a d d e d t o t h e sys t e m t o

cool the supply air. The chilled water coil is similar to the

h o t wa t e r c o i l . C h i l l e d wa t e r f l ows t h ro u g h t h e t u b e s t o c o o l

the air.

CHILLER-

RA

EA

CWR

cWS

SA

FAN C H I L L E D WAT E R C O I L

- HWR• Frg. 4: Chilled water system added

6

The water is cooled by a chiller, which is a refrigeration u n i t . A p u m p c i rc u l a te s wa te r t h ro u g h t h e c h i l l e r w h e re i t i s

cooled and then returned to the chilled water coil.

The system shown in Fig. 4 is a complete air handling system. It can: 0

M i x t h e re t u r n a i r a n d o u ts i d e a i r to t h e c o r re c t

.

proportions.

0

Heat the air.

Filter the air.

0 Cool the air.

WHY AIR FLOWS A wa te r sys te m i n a h o u s e i s u s u a l l y u n d e r a p re ss u re o f a b o u t 3 0 p s i ( p o u n d s p e r s q u a re i n c h ) . W h e n a fa u c e t o n t h e wa te r l i n e i s o p e n e d , t h e l i n e p re ss u re a t t h e o p e n fa u c e t d e c re a s e s t o ze ro. T h e wa t e r f l ows t o t h e l ow p re ss u re a re a .

Air is a fluid just like water. It also flows from one area to a n o t h e r b e c a u s e o f a d i f fe re n c e i n p re s s u re :

0

.

I n t h e o p e n a i r, a i r f l ows f ro m a h i g h e r p re ss u re to a l owe r p re ss u re . W i n d i s a i r t h a t i s m ov i n g f ro m a h i g h e r p re ss u re a re a to a l owe r p re ss u re . In a duc t, air also flows from a higher pressure to a lower pressure. A fan (Fig. 5) creates the higher p re ss u re . T h e o p e n e n d o f t h e d u c t h a s a l owe r pressure, so the air flows out.

All air in an o p e n sys t e m i s u n d e r a t m o s p h e r i c p re s s u re , which is normally 14.7 psi at sea level. Pressure in a duct refers to the pressure that is higher or lower than the atmospheric pressure of 14.7 psi. A positive pressure (+) is above atmospheric pressure. A negative pressure (-) is

below.

7

HIGH PRESSURE,

AIRFLOW

FA N LOWER PRESSURE

Fig. 5: The fan creates pressure that causes air to flow

T h e a m o u n t o f a i r f l ow i n g t h ro u g h a d u c t i s re g u l a te d by t h e a m o u n t o f p re s s u re d i f fe re n c e a n d by t h e sys t e m resistance. The higher the pressure difference, the greater the air velocity and the greater the quantity of air that will

flow from the duct.

Friction is a resistance which slows down airflow. The flow of air creates friction as it rubs against the side of the duct, and the fric tion creates resistance to the air flow. Think of b l ow i n g t h ro u g h a p i e c e o f g a rd e n h o s e 6 i n c h e s l o n g . Yo u c a n fe e l a g o o d s t re a m o f a i r c o m i n g o u t o f t h e t u b e . N ow t r y to b l ow t h ro u g h a h o s e 5 0 fe e t l o n g . L i t t l e o r n o a i r comes out the other end. This is because fric tion created by the sides of the long hose reduces the pressure at the open end of the hose.

MEASURING DUCT PRESSURES Since the air pressure in a duc t direc tly affec ts the flow of air, measuring the air pressure is impor tant. Air pressure in a

tire is measured in psi (pounds per square inch). Air pressure in a duct is measured in inches water g a g e ( i n c h e s wg o r " wg ) . B o t h p s i a n d i n c h e s wg m e a s u re t h e s a m e

thing-the amount of pressure on a given amount of area. H oweve r, i n c h e s wg i s u s e d fo r d u c t b e c a u s e i t i s s u i t a b l e fo r m e a s u r i n g s m a l l va l u e s . C o m p a r e t h e m e a s u r e m e n t s t a ke n by a c a r p e n te r a n d a m a c h i n i s t . T h e c a r p e n te r u s e s a r u l e d i v i d e d i n to e i g h t h s a n d s i x te e n t h s o f a n i n c h . T h e

8

machinist uses instruments that measure in thousandths of

an inch. C o m p a re t h e m e a s u re m e n t s o f p s i a n d wg . T h e p re s s u re i n a d u c t m i g h t b e 1" wg . T h i s s a m e d u c t p re ss u re m e a s u re d i n

psi would be just 0.04 psi. You can see why psi would not b e a g o o d s c a l e fo r m e a s u r i n g s m a l l c h a n g e s i n a i r p re ss u re

in ducts.

THE WATER GAGE T h e t e r m i n c h e s wg (o r " wg ) m e a n s

inches of water differential in a water

g a g e . A wa te r g a g e ( F i g . 6 ) i s t h e b a s i c d ev i ce fo r m e as u r i n g a i r p re ss u re i n d u c t . In ac tual prac tice, other instruments that a re m o re c o nve n i e n t to u s e , s u c h a s a n

electronic manometer (Fig. 7), are used t o m e a s u r e " wg . T h e s e w i l l b e d e s c r i b e d in another book in this series. But all i n s t r u m e n ts m e a s u r i n g i n c h e s wg a re b a s e d o n t h e p r i n c i p l e o f t h e wa te r g a g e ,

which is explained in Figs. 8 and 9. Fig. 6: A water gage

F i g u re 8 s h ows a s m a l l , f l ex i b l e , p l a st i c h o s e c o n n e c te d to a U - s h a p e d g l a ss t u b e at B. The pressure at the open ends of the tubes (A and E) is equal (because it is only atmospheric pressure). T h e re fo re t h e l eve l o f t h e wa te r a t C

and D is the same. F i g u re 9 s h ows t h e o p e n e n d o f t h e

hose placed in a duct in which the air is flowing. Therefore the pressure at A is greater than the pressure at E. This p re s s u re p u s h e s t h e wa t e r l e ve l d ow n a t C . T h i s m e a n s t h a t t h e wa te r l eve l a t D m u s t r i s e by t h e s a m e a m o u n t .

F i g . 7: A n e l e c t ro n i c m a n o m e te r

9

FLEXIBLE HOSE.

A I R F LOW

GLASS TUBE.

AIR

PRESSURE

DUCT

LEVEL

LINE

Fig. 8: U tube with equal pressure on both ends

Fig. 9: U tube with " wg pressure

re a d i n g

If the pressure pushes the water down 0.4" at point C, the level must rise 0.4" at point D. Therefore there is a difference of 0.8" between the two levels (Fig. 9). The p re ss u re re a d i n g o n t h i s U -t u b e i s 0.8" wg .

REVIEW I f yo u c a n a n swe r t h e fo l l ow i n g q u e s t i o n s w i t h o u t re fe r r i n g to the text, you have lear ned the contents of this chapter. Tr y t o a n swe r e ve r y q u e s t i o n b e fo re yo u c h e c k t h e a n swe r s

in the back of the book.

1 . What do the letters HVAC stand for? 2. List the four basic components of an HVAC system. 3. How is the conditioned air usually transported from the central HVAC to the conditioned space?

10

4. What do the following letters stand for?

A. RA B. EA C. SA

D. OA 5. What is the purpose of a coil in a central air handling syste m ?

6. How is heat added to the mixed air in a central air h a n d l i n g sys te m ?

7. How does the chilled water coil cool the air? 8. W h a t i s a t m o s p h e r i c a i r p re ss u re a t s e a l eve l i n p s i ? 9. W hy i s d u c t p re s s u re m e a s u re d i n " wg i n s t e a d o f p s i ?

10. What does wg stand for? 11. What would a pressure of 1"wg do to the water level of a U tube?

11

2

CALCULATING DUCT SIZES USING MATH To work with airflow in ducts you need to know how to wo r k s i m p l e e q u a t i o n s a n d h ow t o c h a n g e e q u a t i o n s t o a d i f fe re n t fo r m . T h e s e t wo s k i l l s a re e ss e n t i a l fo r a l l H VAC te c h n i c i a n s . I f yo u n e e d to rev i e w t h i s b a s i c m a t h , u s e a n o t h e r b o o k i n t h i s s e r i e s e n t i t l e d M a t h fo r t h e I n d o o r E nv i ro n m e n t Te c h n i c i a n . T h i s b o o k i s d i re c t a n d e a sy t o u n d e r s t a n d . I t re v i e ws o n l y t h e k i n d o f m a t h n e e d e d o n t h e

job. It will give you skills that will make the rest of your s t u d i e s m u c h e a s i e r. This chapter assumes that you know how to use simple

equations and know how to use a calculator. You will use equations to calculate duc t sizes (covered in this chapter) a n d a i r q u a n t i t y a n d ve l o c i t y (c ove re d i n t h e n ex t c h a p te r ) . T h e s e a n d m a ny o t h e r c a l c u l a t i o n s a re i m p o r t a n t fo r a ny

HVAC technician.

You should have a calculator that has a key for pi (7t) and a key for square root:

It is also useful to have keys for parentheses, marked:

For some industry areas, such as service work, energy mana g e m e n t , a n d TA B ( t e s t i n g , a d j u s t i n g , a n d b a l a n c i n g ) , i t i s a l s o u s e f u l t o h ave a c a l c u l a t o r w i t h a ke y t h a t g i ve s c u b e r o o t s :

C u b e ro o ts a re n o t n e e d e d fo r t h i s b o o k .

12

WORKING WITH DUCT SIZES Calculating duc t sizes is ver y impor tant because the q u a n t i t y a n d ve l o c i t y o f a i r d e l i ve re d by a sys te m a re re l a te d t o t h e d u c t s i ze . Yo u m ay n e e d t o c a l c u l a t e w h a t s i ze d u c t

is needed to deliver a given amount of air. Or you may have to c a l c u l a te w h a t s i ze re c t a n g u l a r d u c t h a s t h e e q u i va l e n t a re a o f a g i ve n ro u n d d u c t . To wo r k w i t h d u c t s i ze s , yo u

must be able to do the following: . Calculate the area of a rectangular duct when you know the duct dimensions.

.

. .

.

C o nve r t s q u a re i n c h e s to s q u a re fe e t a n d c o nve r t square feet back to inches. F i n d t h e d i m e n s i o n o f o n e s i d e o f a re c t a n g u l a r d u c t

when the duct area and one side are known.

Calculate the area of round duct when the diameter is known. C a l c u l a t e t h e d i a m e t e r o f ro u n d d u c t w h e n t h e a re a

is known. A l l o f t h e s e o p e r a t i o n s a re ex p l a i n e d i n t h i s c h a p te r. E a c h o n e i s a s i m p l e o p e r a t i o n , e s p e c i a l l y i f yo u u s e a c a l c u l a to r. B e s u re t h a t yo u u n d e rst a n d e a c h o f t h e m b e fo re l e av i n g t h e

chapter. You will need all of these in order to work the problems with air quantity and velocity in the next chapter and in your work.

FIND THE AREA OF DUCT The cross sec tion of rec tangular duc t is a re c t a n g l e ( F i g . 1 ) .

You often have to f i n d t h e a re a o f t h e

Fig. 1: Cross section of a duct

13

c ro ss s e c t i o n . To f i n d t h e a re a o f a ny re c t a n g l e ,

multiply the dimensions o f o n e s i d e by t h e o t h e r.

For example, to find the a re a o f t h e re c t a n g l e i n

Fig. 2, multiply the h eig h t (2 inche s) by th e

width (4 inches): AREA - 8 IN.?

Fig. 2: The area of a rectangle 2" x 4" is 8 square inches

Area = Width x Height Area = 4" x 2" A r e a = 8 s q u a re i n c h e s

As yo u c a n s e e i n F i g . 2, t h e re re a l l y a re 8 b oxe s o n e i n c h square in the rec tangle that is 4" x 2". U s e t h i s p ro c e d u re to f i n d t h e a re a o f a d u c t . S i n c e d u c t s i ze s a re g i ve n i n i n c h e s , i f yo u m u l t i p l y o n e s i d e by t h e o t h e r s i d e yo u h ave t h e a re a i n s q u a re i n c h e s . T h e a b b rev i a t i o n i s s q . i n . o r i n ? .

NOTE: When identifying rectangular duct, the width (horizontal) is given first. The height (ver tical) is given

second. For duct drawn in an HVAC system, the dimension o f t h e s i d e s h ow n i n t h e d r aw i n g i s g i ve n f i rst . T h u s a d u c t c o u l d b e d e s i g n a t e d a s 1 8" x 1 2 " o r 1 2 " x 1 8" d e p e n d i n g o n t h e v i e w o f t h e d r aw i n g . D i m e n s i o n s fo r re c t a n g u l a r d u c t a r e u s u a l l y e ve n n u m b e r s . E ve n n u m b e r s c a n b e d i v i d e d by

2 (for example, 8, 10, 12, 14, etc.).

EXAMP LE Find the area of a duc t that is 10" x 8":

Area = Width x Height

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Area = 10" x 8" Area = 80 sq. in. PROBLEMS D o t h e s e p ro b l e m s b e fo re yo u g o o n . Af t e r yo u wo r k t h e proble ms , ch eck t he an swers at t h e en d of t h is ch apter.

1. A duct is 30" x 24". What is its area in square inches? 2. What is the area of an 18" x 18" duct in square inches?

CHANGE SQUARE INCHES TO SQUARE FEET

For most problems in HVAC, you n e e d t o k n ow d u c t a re a s i n s q u a re fe e t r a t h e r t h a n s q u a re i n c h e s . A i r ve l o c i t y i s m e a s u re d i n fe e t p e r m i n u t e a n d a i r q u a n t i t y i s m e a s u re d i n

cubic feet per minute. Therefore, you need to know duct a re a i n s q u a re fe e t i n o rd e r to use it with these other

values. A s q u a re fo o t m e a s u re s o n e fo o t o n e a c h s i d e o f a s q u a re . T h a t m e a n s i t i s 1 2 inches on each side (Fig. 3). T h e re fo re t h e a re a o f o n e s q u a re fo o t i s 1 4 4 s q u a re

inches:

Area = Width x Height Area = 12" x 12" 12°

Area = 144 sq. in. AREA - 144 IN.2

Fig. 3: One square f o o t e q u a l s 1 4 4 s q u a r e i n c h e s

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S i n c e 1 4 4 s q . i n . i s t h e s a m e a s o n e s q u a re fo o t , to c h a n g e s q u are i n ch e s to s q u are fe e t (s q . ft . or ft ?), d i v i d e by 144: sq. ft. =

sq. in.

144

EXAMPLE A re c t a n g u l a r d u c t m e a s u re s 24 " x 1 8". W h a t i s i ts cross-sectional area in square feet? S te p 1 : F i n d t h e a re a i n s q u a re i n c h e s .

Area = Width x Height A re a = 24 " x 1 8"

Area = 432 sq. in. Step 2: Find the area in square feet.

Area in sq. ft.

sq. in. 144

Area in sq. ft.=

432 sq. in. 144

Area = 3 sq. ft. T h e s e t w o m a t h steps can be combined into one equation:

Area =

Area =

Width x Height 144

24" x 18" 144

Area = 3 sq. ft.

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N OT E : T h i s p r o b l e m c a n a l s o b e w o r ke d by c h a n g i n g 2 4 "

and 18" to feet before multiplying them. Do this by dividing the inches by 12. You can probably do this in your head for these numbers:

24" = 2 ft. 18" = 1.5 ft.

2' x 1.5' = 3 sq. ft. E i t h e r m e t h o d p ro d u c e s t h e s a m e a n swe r, a n d b o t h a re e a sy

to do on the calculator.

Change Square Feet to Square Inches Yo u m ay n e e d to c h a n g e s q u a re fe e t to s q u a re i n c h e s ,

especially in order to find the dimension in inches for a duct if the area is given in square feet. To change square feet to

square inches, multiply by 144: Area in sq. in. = Area in sq. ft. x 144 F o r ex a m p l e , f i n d t h e a re a i n i n c h e s o f a d u c t w i t h a n a re a of 3.8 sq. ft.:

Area in sq. in. = Area in sq. ft. x 144 Area in sq. in. = 3.8 sq. ft. x 144 Area = 547.2 sq. in.

PROBLEMS 3. A d u c t m e a s u re s 3 0" x 1 6 ". W h a t i s i t s a re a i n s q u a re fe e t ( to t wo d e c i m a l p l a c e s ) ?

17

4. W h a t i s t h e a re a i n s q u a re fe e t ( t o t wo d e c i m a l p l a c e s ) of a duct that measures 42" x 34"? 5. W h a t i s t h e a re a i n s q u a re i n c h e s o f a d u c t t h a t h a s a n area of 1.75 sq. ft.? 6. If a duc t has an area of 5 sq. ft., what is the area in square inches? C h e c k yo u r a n swe rs a t t h e e n d o f t h e c h a p t e r.

FIND A DUCT SIDE DIMENSION S o m e t i m e s yo u k n ow w h a t t h e a r e a o f a d u c t s h o u l d b e a n d m u s t f i n d t h e d i m e n s i o n s o f a d u c t t h a t wo u l d h ave t h a t s a m e a re a . F o r ex a m p l e , s u p p o s e yo u a re i n s t a l l i n g a 24 " x 24 " d u c t . T h e d u c t d i m e n s i o n s m u s t c h a n g e to 1 4 " h i g h i n o rd e r t o f i t b e t we e n a b e a m a n d a p i p e . T h e p ro b l e m i s t o d e te r m i n e t h e w i d t h t h a t t h e n e w d u c t s h o u l d b e s o t h a t t h e

area is the same as a 24" 24" duct. C h a n g e t h e e q u a t i o n t o s o l ve fo r w i d t h :

Area = Width x Height

Width =

Area Height

D u c t s i z e s a r e u s u a l l y e ve n n u m b e r s . A l w ay s r o u n d t o t h e n ex t l a rg e r eve n n u m b e r, n o t a s m a l l e r o n e , b e c a u s e i t i s u s u a l l y a c c e p t a b l e to h ave a s l i g h t l y l a rg e r a re a t h a n n e c e s s a r y, b u t n o t a s m a l l e r o n e .

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EXAMPLE A duc t 32" x 32" must be changed to a height of 28" so that

it can fit between a beam and a pipe. How wide should the d u c t b e i n o rd e r to ke e p t h e s a m e a re a ?

Step 1: Find the area. Area = Width x Height A re a = 3 2 " x 3 2 "

Area = 1024 sq. in. S te p 2 : F i n d t h e w i d t h o f t h e n e w d u c t .

Width =

Width :

A re a Height 1024 sq. in. 28"

Width = 36.57" Round up to 38") S i n c e d u c t s i ze s a re eve n n u m b e rs , t h e a n swe r o f 3 6.5 7 i s ro u n d e d u p t o t h e n ex t h i g h e s t e ve n n u m b e r, w h i c h i s 3 8".

PROBLEMS 7. A duct run is 20" x 16". A transition must be made so that the duc t keeps the same area but is only 10" high. ( Re m e m b e r t h a t h e i g h t i s t h e s e c o n d d i m e n s i o n . ) W h a t s h o u l d t h e n e w d u c t s i ze b e ? 8. A d u c t i s 2 0" x 8". I t m u s t ke e p t h e s a m e a re a , b u t o n l y b e 1 8" w i d e . ( Re m e m b e r t h a t w i d t h i s t h e f i rst d i m e n s i o n . ) W h a t i s t h e n e w d u c t s i ze ? C h e c k yo u r a n swe rs a t t h e e n d o f t h e c h a p te r.

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FIND THE AREA OF ROUND DUCT T h e s i ze o f ro u n d d u c t i s g i ve n by t h e d i a m e te r ( F i g . 4 ) i n i n c h e s . T h e r a d i u s i s h a l f t h e d i a m e te r ( F i g . 4 ) . D i v i d e t h e d i a m e t e r by 2 t o f i n d t h e r a d i u s . Yo u c a n p ro b a b l y d o t h i s i n yo u r h e a d . F o r ex a m p l e , 1 2 " ro u n d d u c t h a s a r a d i u s o f

6".

The equation to find the area of a circle uses the symbol r, called pi, and the radius:

Area of circle = T x Radiusí T h e sy m b o l u re p re s e n t s a n u m b e r a l i t t l e ove r 3. M a ny c a l c u l a to rs h ave a key fo r 7 t c a r r i e d to m a ny d e c i m a l p l a c e s . I f yo u d o n o t h ave a T key o n yo u r c a l c u l a to r, u s e t h e

n u m b e r 3.1416.

The term radius? means

radius squared. To square a number is to multiply it DIAMETER - 12"

by i ts e l f. F o r ex a m p l e , i f

the radius is 6", multiply t h a t n u m b e r by i ts e l f :

Radius' = 6 x 6 12 DUCT

Radius-: 36

EXAMPLE RADIUS - 6'

A round duct has a 16" diameter. What is the a re a ? ( Re m e m b e r t h a t t h e radius is half the diameter, sO the radius is

12" DUCT Fig. 4: The radius is half the diameter

8".) If the radius is given

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i n i n c h e s , t h e a re a i s g i ve n i n s q u a re i n c h e s .

Area = T x RadiusArea = 3.1416 x 8" x 8" Area = 201.06 sq. in. J u st a s w i t h re c t a n g u l a r d u c t , yo u w i l l o f te n h ave to c h a n g e t h e s q u a re i n c h e s to s q u a re fe e t . D i v i d e t h e a re a by 1 4 4 to change square inches to square feet:

Area (sq. ft.) =

Area (sq. ft.) =

T x Radius?

144 3.1 41 6 x 8" x 8" 144

Area = 1.4 sq. ft. (Rounded off)

PROBLEMS 9. How many square inches (to two decimal places) are t h e re i n a 9" d i a m e te r c i rc l e ?

10. How many square feet are there in the 9" diameter c i rc l e ( to t wo d e c i m a l p l a c e s ) ?

11. How many square feet (to two decimal places) are t h e re i n a 2 2 " d i a m e te r d u c t ?

Check your answers a t t h e e n d o f t h e c h a p te r.

FIND THE DIAMETER OF

ROUND DUCT

Yo u m ay k n ow t h e a re a o f a re c t a n g u l a r d u c t i n s q u a re fe e t a n d n e e d t o d e t e r m i n e w h a t d i a m e t e r o f ro u n d d u c t w i l l h ave a p p r ox i m a t e l y t h e s a m e a r e a .

21

C h a n g e t h e e q u a t i o n fo r t h e a re a o f a c i rc l e to s o l ve fo r t h e

radius:

Area = T ×Radius?

Radius2

=

Area Tt

Radius =

Area T

The symbol means the square root of a number. The s q u a re ro o t i s t h e o p p o s i te o f a s q u a re d n u m b e r. F o r ex a m p l e , t h e s q u a re ro o t o f 1 6 i s 4 b e c a u s e 4 s q u a re d i s 1 6

(4 x 4 = 16). The square root of 16 is written 16: VT6 = 4 U s e a c a l c u l a to r to f i n d s q u a re ro o ts .

EXAMPLE A re c t a n g u l a r d u c t h a s a n a re a o f 1 9 2 s q . i n . W h a t d i a m e t e r ro u n d d u c t h a s t h e e q u i va l e n t a re a ?

Radius =

Radius =

/Area

T 192 sq. in. TC

Radius = 7.82" D i a m e t e r = 7.8 2 " x 2

Diameter = 15.64" (Round to 16") Th e diame te r is rou n de d off to th e n ex t h ig h e st eve n number (16").

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NOTE: Round duct over 10" diameter is available in even whole numbers (12", 14", 16", etc.). Round duct under 10" diameter is available with a diameter of every whole number (9", 8", 7", etc.).

PROBLEMS 12. A round duct must have an area of 240 sq. in. What s h o u l d t h e d u c t d i a m e te r b e ?

13. A round duct should have an area of 2 sq. ft. What should the diameter be?

14. A rectangular duct measures 18" x 12". It must be c h a n g e d t o a ro u n d d u c t o f t h e s a m e a re a . W h a t s h o u l d t h e d i a m e te r o f t h e ro u n d d u c t b e ?

C h e c k your a n swe rs a t t h e e n d o f t h e c h a p te r.

REVIEW I f yo u c a n a n swe r t h e fo l l ow i n g q u e s t i o n s w i t h o u t re fe r r i n g to the text, you have lear ned the contents of this chapter. Tr y t o a n swe r e ve r y q u e s t i o n b e fo re yo u c h e c k t h e a n swe r s in the back of the book. For these problems, give duc t diameters to the nearest w h o l e n u m b e r. G i ve d u c t d i m e n s i o n s t o t h e n e a r e s t e ve n

number. 1. What is the area in square inches of 18" x 12" duct? 2. W h a t i s t h e a re a i n s q u a re fe e t ( to 2 d e c i m a l p l a c e s ) o f 1 2 " x 1 0" d u c t ? 3. W h a t i s t h e a re a i n s q u a re fe e t ( to 2 d e c i m a l p l a c e s ) o f 44" x 26" duct?

23

4. U s e m e n t a l a r i t h m e t i c to f i n d t h e a re a i n s q u a re fe e t o f 48" x 36" duc t.

5. A duct must have an area of 240 square inches. One s i d e m u s t m e a s u re 1 2 ". W h a t s h o u l d t h e d i m e n s i o n o f the other side be (to the nearest even number)? 6. A 2 0" x 1 8" d u c t m u s t b e t r a n s i t i o n e d t o a d u c t w i t h t h e s a m e a re a t h a t i s 1 4 " h i g h . W h a t w i l l t h e s i ze o f t h e duct be?

7. A 12" diameter round duct must be changed to a re c t a n g u l a r d u c t o f t h e s a m e a re a . T h e re c t a n g u l a r d u c t m u st b e 8" h i g h . W h a t a re t h e d i m e n s i o n s o f t h e re c t a n g u l a r d u c t ? 8. A 2 0 " x 1 8 " r e c t a n g u l a r d u c t m u s t b e c h a n g e d t o r o u n d duc t of the same area. What is the diameter of the ro u n d d u c t ( to t h e n e a re st eve n n u m b e r ) ?

9. Figure 5 shows a run of duct. Give the dimensions of the duct at points A, B, C, and D.

SAME AREA

12' HIGH

AIRFLOW DIRECTION . 8 1

X . O

1 . 7 5 S Q . F T. 14' HIGH

7 2 S Q. I N . 6° HIGH

Fig. 5: Problem 9

SAME AREA AS C

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SUMMARY OF EQUATIONS USED TO FIND DUCT SIZES F i n d t h e a re a o f re c t a n g u l a r d u c t i n s q u a re i n c h e s :

Area = Width (in.) x Height (in.)

Find the area of rec tangular duc t in square feet: Area (sq. ft.) =

Width (in.) x Height (in.)

144

C h a n g e s q u a re i n c h e s to s q u a re fe e t : sq. ft.=

sq. in.

144

C h a n g e s q u a re fe e t to s q u a re i n c h e s :

sq. in. = sq. ft. x 144

Find one side of a duc t if the area and another side is

known:

Width =

Are a Height

25

Find the area of round duct: Area = 7 X Radius-

F i n d t h e r a d i u s o f ro u n d d u c t i f t h e a re a i s k n ow n :

Radius =

Area TL

ANSWERS

TO PROBLEMS 1. 720 sq. in. 2. 324 sq. in. 3. 3.3 3 s q . f t . 4. 9.9 2 s q . f t 5. 25 2 s q . i n . 6. 7 2 0 s q . i n . 7. 3 2 " x 1 0 " 8. 18" x 10" 9. 6 3.6 2 s q . i n . 1 0. 0.4 4 s q . f t . 1 1. 2 .6 4 s q . f t . 12. 18" (Remember to round u p t o h i g h e r e v e n n u m b e r. ) 13. 20 (Remember to change the sq. ft. to sq. in.)

1 4. 1 8 " ( R e m e m b e r t o r o u n d u p . )

26

AIR QUANTITY AND VELOCITY

If you work with a heating, ventilating, and air conditioning sys t e m , yo u n e e d t o b e a b l e t o c a l c u l a t e t h e q u a n t i t y a n d

velocity of the air. O

Air quantity is the volume of air delivered by a duct in a given period of time. It is also called rate of flow. Air quantity is measured in cubic feet per minute (CFM).

0 A i r ve l o c i t y i s t h e s p e e d o f a i r t h ro u g h t h e d u c t . A i r velocity is measured in feet per minute (FPM) All HVAC technicians must be able to calculate how much air a duc t is delivering to a space and how fast the air is

moving. They need to measure air delivery conditions and calculate what changes are needed. A i r f l ow a f fe c ts c o m fo r t a n d i n d o o r a i r q u a l i t y b e c a u s e i t

delivers heated, cooled, or outside air to the conditioned s p a c e s . Even if the air is at the proper temperature:

0

If the velocity (speed) of the air in the duc t is too fa st , t h e syste m w i l l b e n o i sy a n d t h e c o n d i t i o n e d

space will be drafty. 0

If the quantity of air delivered to the space is too

much or too little, the space will be too hot or too

cold.

When the airflow is not correct, the room is uncomfortable.

27

AIR QUANTITY DEPENDS ON AIR VELOCITY AND DUCT SIZE T h e q u a n t i t y o f a i r d e l i ve re d by a d u c t i s a l s o c a l l e d t h e

volume. It is measured in cubic feet per minute. A cubic fo o t o f a i r wo u l d f i t i n a c u b e t h a t i s o n e fo o t o n a l l s i d e s

(12" x 12" x 12") (Fig. 1). It is easy to see that the quantity of air that a duct

can deliver in one

minute depends on two things:

O Velocity (speed) of the air O

S i ze o f t h e d u c t Fig. 1 A cubic foot is one foot long on

all sides

F i g u re 2 s h ows t h a t

.

velocity affects how much air is delivered:

O

I f d u c t o n e fo o t s q u a re h a s a i r f l ow i n g a t 1 0 0 0 F P M (feet per minute), it will it deliver 1000 CFM (cubic feet of air per minute) (Fig. 2A). I f d u c t o n e fo o t s q u a re h a s a i r f l ow i n g a t 4 0 0 0 F P M ( fe e t p e r m i n u te ) , i t w i l l d e l i ve r 4 0 0 0 C F M (c u b i c fe e t per minute) (Fig. 2B).

4000 CFM

1000 CFM 1000 FPM

4000 FPM

(FEET PER MINUTE)

(FEET PER MINUTE)

'TE

(CUBIO FEET PER MINUTE

(CUBIO FEET PER MINUTE)

:

Fig. 2: The amount of air delivered depends partly on the air velocity

28

AREA

FT'

1000 FPM 1000 CFM AREA = 4 FT2

1000 FPM

00 CFM

Fig. 3: The amount of air delivered depends partly on the area of the duct

Figure 3 shows that the size of the duc t also affec ts how

much air is delivered:

.

I f d u c t o n e s q u a re fo o t h a s a i r m ov i n g a t 1 0 0 0 F P M

(feet per minute), it will deliver 1000 CFM (cubic feet of air per minute) (Fig. 3A). O

4 If duct 4 square feet has air moving at 1000 FPM ( fe e t p e r m i n u te ) , i t w i l l d e l i ve r 4 0 0 0 C F M (c u b i c

feet of air per minute) (Fig. 3B).

FINDING AIR QUANTITY As you can see, a small duct with the air flowing rapidly m ay d e l i ve r i n o n e m i n u t e t h e s a m e a m o u n t a s a l a r g e d u c t w i t h t h e a i r f l ow i n g m o re s l ow l y. T h i s fa c t g i ve s u s a ve r y

useful equation: Quantity = Area x Velocity

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EXAMPLE Air is flowing at 1200 FPM through 28" x 18" duct. How m u c h a i r i s b e i n g d e l i ve re d ? Step 1: Find the cross-sec tional area of the duc t in s q u a re fe e t . ( T h i s e q u a t i o n wa s c ove re d i n C h a p te r 2. )

Area =

Area =

Width x Height 144 2 8" x 1 8"

144

Area = 3.5 sq. ft. S te p 2 : F i n d t h e a i r q u a n t i t y.

Quantity = Area x Velocity Quantity = 3.5 sq. ft. x 1200 FPM Quantity = 4200 CFM

FINDING AIR VELOCITY T h e e q u a t i o n c a n b e w r i t te n i n a d i f fe re n t fo r m to f i n d t h e

velocity if the quantity of air and the duct area are known: Velocity =

Quantity Area

EXAMPLE A sys t e m i s t o d e l i ve r 2 1 1 0 C F M . T h e d u c t s i ze i s 2 0" x 1 6 ". W h a t a i r ve l o c i t y i s n e e d e d t o h a n d l e t h i s a i r q u a n t i t y ?

30

Step 1: Find the area of the duc t in square feet.

Area = Width x Height 144

Area =

2 0" x 1 6 "

144

A re a = 2.2 2 s q . f t . S te p 2 : F i n d t h e a i r ve l o c i t y.

Velocity =

Velocity =

Quanity Area

2110 CFM 2.22

Ve l o c i t y = 9 5 0 F P M

FINDING DUCT AREA T h e e q u a t i o n c a n a l s o b e c h a n g e d t o f i n d t h e d u c t a re a n e e d e d i f t h e re q u i re d a i r q u a n t i t y a n d a i r ve l o c i t y a re

known:

Area = Quanity

Velocity

EXAMPLE A sys t e m n e e d s t o d e l i ve r 8 5 0 0 C F M . T h e ve l o c i t y i n t h e d u c t i s t o b e 1 6 0 0 F P M . W h a t s h o u l d t h e a re a o f t h e d u c t be?

Area =

Quanity Velocity

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Area =

8500 CFM 1600 FPM

Area = 5.31sq. ft. PROBLEMS D o t h e s e p ro b l e m s b e fo re you go on. After you work the p ro b l e m s , c h e c k t h e a n swe rs a t t h e e n d o f t h i s c h a p te r. Standard trade prac tice is to:

O Round off CFM or FPM to the nearest 5. O

Ro u n d o f f d i m e n s i o n s fo r re c t a n g u l a r d u c t to t h e n e a re s t e ve n n u m b e r.

1. Air is moving at 1500 FPM through 32" X 18" duct. How much air is being delivered? 2. Duct that is 30" x 16' is delivering 4035 CFM. What is the air velocity? 3. If the air velocity is 1400 FPM and the duct is 24" x 24", how much air is being delivered? 4. How wide should a duct be if 5200 CFM must be d e l i ve r e d a t 1 5 0 0 F P M ? T h e d u c t m u s t b e 1 8 " h i g h .

AIR VELOCITY AND AIR QUANTITY OF ROUND DUCT T h e e q u a t i o n s fo r a i r ve l o c i t y a n d a i r q u a n t i t y a re u s e d fo r ro u n d a s we l l a s re c t a n g u l a r d u c t .

32

As yo u l e a r n e d i n C h a p te r 2, t h e e q u a t i o n fo r f i n d i n g t h e cross-sec tional area for round duc t is this:

Area = TX Radius' O n c e yo u h ave c a l c u l a te d t h e c ro ss - s e c t i o n a l a re a o f ro u n d d u c t , yo u c a n u s e t h e t h re e ve rs i o n s o f t h e a i r f l ow e q u a t i o n :

Quantity = Area x Velocity

Velocity =

Quantity A re a

Area = Quantity

Velocity

EXAMPLE-FINDING AIR QUANTITY A i r i s f l ow i n g a t 1 2 0 0 F P M t h ro u g h a 2 2 " d i a m e t e r d u c t .

How much air is being delivered? Step 1 : Find the area of the round duc t in s q u a r e feet.

Area =

Area =

T x Radius?

144

3.1416x11x11 144

A re a = 2.6 4 s q . f t . S te p 2 : F i n d t h e a i r q u a n t i t y.

Quantity = Area x Velocity Q u a n t i t y = 2.6 4 s q . f t . X 1 2 0 0 F P M

Quantity = 3168 CFM (Round to 3170 CFM)

33

EXAMPLE-FINDING AIR VELOCITY A sys t e m i s t o d e l i ve r 2 1 1 0 C F M . T h e ro u n d d u c t h a s a d i a m e t e r o f 1 8". W h a t a i r ve l o c i t y i s n e e d e d t o d e l i ve r t h i s

air quantity? Step 1: Find the area of the duc t in square feet.

Area =

Area

T x Radius* 144

3.1416x9x9 144

Area = 1.767 sq. ft. Step 2: Find the air velocity. Velocity =

Velocity =

Quantity Area

2110 CFM 1.767 sq. ft.

Velocity = 1194FPM (Round to 1195)

EXAMPLE-FINDING THE AREA AND DIAMETER A sys t e m n e e d s t o d e l i ve r 8 5 0 0 C F M . T h e ve l o c i t y i n t h e duct is to be 1600 FPM. What should the diameter of the round duct be? S te p 1 : F i n d t h e a re a i n s q u a re i n c h e s .

Area =

Quantity

Velocity

34

Area :

8500 CFM 1600 FPM

Area = 5.31sq. ft. Area (in sq. in.) = 5.31sq. ft.x144 Area = 765 sq. ft.

NOTE: You can do this in one operation on your calculator: 8500 CFM

Area = 144x

1600 FPM

Area = 765

S t e p 2 : F i n d w h a t d i a m e t e r h a s a p p rox i m a t e l y t h a t

area.

Radius =

Area

Radius

765

Radius

243.5

Radius = 15.6 D i a m e t e r = 1 5.6 x 2 D i a m e t e r = 3 1.2 " ( R o u n d u p t o 3 2 " )

N OT E : Yo u c a n d o t h e t wo p a r t s o f t h i s p ro c e s s i n o n e

operation on your calculator: Diameter

5.31 sq. ft. x 144 x2 T

Diameter = 31.2" (Round up to 32")

35

PROBLEMS F o r t h e s e p ro b l e m s :

O

Ro u n d o f f C F M o r F P M t o t h e n e a re s t 5.

O

Ro u n d u p d u c t d i a m e t e r t o t h e n ex t e ve n n u m b e r fo r

duct over 10". Round up to the next whole number fo r d u c t u n d e r 1 0".

5. If the air velocity is 1255 FPM and the round duct has a diameter of 26", how much air is being delivered? 6. I f 3 0" d i a m e t e r d u c t d e l i ve r s 74 8 0 C F M , w h a t i s t h e a i r

velocity? 7. What size round duct is needed to handle 1200 CFM at 800 FPM? C h e c k yo u r a n swe rs a t t h e e n d o f t h e c h a p te r.

REVIEW I f yo u c a n a n swe r t h e fo l l ow i n g q u e s t i o n s w i t h o u t re fe r r i n g to the text, you have lear ned the contents of this chapter. Tr y t o a n swe r e ve r y q u e s t i o n b e fo re yo u c h e c k t h e a n s we r s

in the back of the book.

Follow trade practice in rounding off. 1. What does CFM mean? 2. What does FPM mean? 3. A 12" x 12" duct has an air velocity of 2000 FPM. What is the CFM?

4. A 24" x 12" duct delivers 4000 CFM. What is the velocity of the air in FPM?

36

5. A duct measures 24" x 16". The air velocity in the duct i s 1 5 0 0 F P M . W h a t i s t h e C F M ( to t h e n e a re st 5 ) ?

6. A duct is 32" x 18". It is delivering 4800 CFM. What is the velocity in FPM (to the nearest 5)?

7. A duct must deliver 6000 CFM at 2000 FPM. The re c t a n g u l a r d u c t m u s t b e 1 8" h i g h . W h a t a re t h e dimensions of the duct?

8. If the duct in item 7 must be round, what is the duct d i a m e te r ( to t h e n e a re s t w h o l e n u m b e r ) ?

9. The air in 22" diameter round duct has a velocity of 1550 FPM. What is the CFM? 10. A 20" round duct delivers 6000 CFM. What is the air velocity? 11. What diameter round duct is needed to deliver 2500 CFM at 1300 FPM? 12. If the duct in item 11 must be rectangular with one side 1 4 ", w h a t i s t h e d u c t s i ze ?

37

SUMMARY OF EQUATIONS USED WITH AIR QUANTITY AND VELOCITY Find air quantity: Quantity = Area x Velocity Find air velocity: Velocity =

Quantity Area

Find duct area if air quantity and velocity are known: Area =

Quantity Velocity

ANSWERS TO PROBLEMS NOTE: Do all your calculations on the calculator and do not r o u n d o f f u n t i l t h e f i n a l a n s we r. If you round off numbers at

va r i o u s p o i n t s i n t h e calculation, your answers may be a little different from the

a n s w e r s g i v e n b e l o w. 1. 6,000 CFM

2. 1,2 1 0 F P M 3. 5600 CFM

4. 2 8" 5. 4625 CFM

6. 1 5 2 5 F P M 7. 1 8 " d i a m e t e r ( R e m e m b e r t o r o u n d u p t o t h e n ex t e v e n n u m b e r. )

38

4

PRESSURES IN A DUCT

You now understand how air velocity (FPM) and the duct area affec t the quantity of air (CFM) that flows in a duc t. Yo u h ave a l s o l e a r n e d t h a t t h e re m u st b e a d i f fe re n c e i n pressure (measured in "wg) for air to flow in duc ts. But there are still more fac tors that affec t air flow in a duc t.

DUCT PRESSURES W h e n e ve r a i r f l ows i n a d u c t , t h re e a i r p re s s u re s in the duct

are related to the airflow:

. .

0 Static pressure Ve l o c i t y p re s s u re To t a l p re s s u re

Two fac tors cause loss of pressure (resistance) in a duc t

.

system: F r i c t i o n l o ss

O Dynamic loss T h i s c h a p te r ex p l a i n s t h e s e t h re e k i n d s o f p re ss u re w h e n a i r f l ows i n d u c ts . I t a l s o ex p l a i n s h ow p re ss u re l o ss e s o c c u r

and how they affect the airflow.

Atmospheric Pressure Con s ide r a p ie ce of straig h t du c t th at is op e n on th e e n ds ( F i g . 1 ) . N o a i r f l ows t h ro u g h t h e d u c t b e c a u s e t h e re i s n o pressure difference to cause the air to flow. There is

39 AT M O S P H E R I C

PRESSURE

AT M O S P H E R I C

PRESSURE

AT M O S P H E R I C

PRESSURE

N O AIRF LOW

AT M O S P H E R I C

PRESSURE Fig. 1: Atmospheric pressure is the ordinar y pressure of the air

pressure on the duc t, but it is the same pressure on both ends, on the outside, and on the inside of the duc t. This is t h e o rd i n a r y p re ss u re o f t h e a i r c a l l e d a t m o s p h e r i c p re ss u re . At m o s p h e r i c p re ss u re i s 1 4.7 p s i ( p o u n d s p e r s q u a re i n c h ) a t s e a l eve l . As yo u r i s e a b ove s e a l eve l , t h e re i s l e ss a i r a b ove yo u a n d t h e re fo re a l owe r a t m o s p h e r i c p re ss u re . At 5,0 0 0 fe e t t h e a t m o s p h e r i c p re ss u re i s n o r m a l l y 1 2.2 5 p s i . Atmospheric pressure is always present. It is regarded as the zero point for duc t pressures: O

A positive duct pressure (+) is greater

than

a t m o s p h e r i c p re ss u re . A d u c t p re ss u re o f 0.1 0" wg m e a n s t h a t t h e p re ss u re i s 0.1 0" wg a b ove atmospheric pressure. O

A negative duct pressure (-) is less than atmospheric p re ss u re. A du c t p re ss u re of -0.10" wg means that the p re ss u re i s 0.1 0" wg b e l ow a t m o s p h e r i c p re ss u re .

Since atmospheric pressure is the zero point, it does not a f fe c t t h e c a l c u l a t i o n s i n t h i s b o o k . H oweve r, t h e d i f fe re n c e in atmospheric pressure at different elevations above sea l eve l i s a fa c to r i n m o re a d va n c e d c a l c u l a t i o n s t h a t w i l l b e

covered in a later book in this Indoor Environment

40

Te ch n i ci a n ' s L i b r a r y. T h e s e

calculations require knowing t h e we i g h t o f a c u b i c fo o t o f

G R E AT E R T H A N

AT M O S P H E R I C

AT M O S P H E R I C

PRESSURE

a i r. A i r we i g h s l e ss a t h i g h e r

PRESSURE

elevations.

Velocity Pressure Now add a fan to one end of the duc t (Fig. 2). When the fan

Fig. 2: Higher pressure created by the fan causes the air to flow

op e rate s , th e fan blade s c o m p re ss t h e a i r s l i g h t l y by putting pressure on it. Air has weight. Just as it requires pressure to push a car, it requires pressure to push air. Since the air at the fan outlet (the duc t inlet) is under pressure which is greater than atmospheric pressure, the air m ove s f ro m t h e h i g h e r p re ss u re t o t h e l owe r p re ss u re a t t h e open end of the duc t (Fig. 2).

The pressure that is a result of the air in motion in the duc t i s c a l l e d ve l o c i t y p re ss u re . Ve l o c i t y m e a n s s p e e d . T h e greater t h e ve l o c i t y o f t h e a i r, t h e h i g h e r t h e ve l o c i t y pressure

will be.

In the open air, when you feel wind in your face, it is the velocity pressure of the air that you feel. Velocity pressure c re a t e d by t h e w i n d m a ke s t re e b r a n c h e s sway a n d m a ke s a

kite fly.

Static Pressure

S TAT I C PRESSURE

T h e a i r p u s h e d by t h e fa n a l s o exer ts pressure on the sides of the duc t. This is called the static p re ss u re . S t a t i c m e a n s n o t moving. Compare this to a balloon Fig. 3: Static pressure keeps a balloon inflated ( F i g . 3 ) . A b a l l o o n st ays i n f l a te d

41

because the air inside it presses against it on all sides. In the same way, s o m e o f t h e p re ss u re f ro m the fan is static pressure against the sides of the duct (Fig. 4).

STATIC PRESSURÉ

If

Fig. 4: Static pressure against the sides of the duct does not move the air

you release an inflated

balloon, it flies through the air until the balloon is out of air (Fig. 5). This is because the static pressure

in the balloon has been

released. The static pressure turns to velocity pressure as it l e ave s t h e b a l l o o n . T h e re a c t i o n to ve l o c i t y p re ss u re p ro p e l s t h e b a l l o o n t h ro u g h t h e a i r u n t i l a l l t h e st a t i c pressure is

released.

Static pressure is an impor tant consideration in the con st ru c t i on of a d u c t . Th e re h ave b e e n cas e s w h e re a d u c t wa s m a d e o f l i g h t g a g e material that was n o t st ro n g e n o u g h

BALLOON REACTS TO THE V E LO C I T Y P R E S S U R E .

S

4S $

F i g . 5 : Ve l o c i t y p r e s s u r e m ove s a b a l l o o n

R E S AT I C -

42

to resist the static p r e s s u r e

produced by the fan. Too much st a t i c p re ss u re b a l l o o n e d t h e wa l l s of the duc t (Fig. 6) and caused

damage.

F i g . 6 : S t a t i c p r e s s u r e w i l l m a ke t h e sides of the duct balloon out if the

Total Pressure

duct is not strong enough

Total pressure in the duc t is the sum of the velocity pressure a n d t h e st a t i c p re ss u re . T h i s m e a n s t h a t we d e a l w i t h t h re e k i n d s o f a i r p re ss u re i n t h e d u c t : O

Static pressure (SP)-The pressure on the sides of the duct.

0 Velocity pressure (VP)-The pressure that is the result of air in motion in the duct.

O

Total pressure (TP)-The total pressure on the air. It is the sum of static pressure and velocity pressure at

any one location. These three kinds of pressure are closely related. Total pressure (TP) equals velocity pressure (VP) plus static pressure (SP):

TP = VP + SP I f to t a l p re ss u re re m a i n s c o n st a n t , w h e n o n e o f t h e t wo changes, the other changes in the opposite direc tion. For ex a m p l e , w i t h T P c o n s t a n t , w h e n V P i n c re a s e s , S P decreases. If SP increases, VP decreases.

43

PRESSURE CHANGES IN A DUCT Resistance Think of a fan with a five foot length of air tight straight duc t at t ach e d to its d is ch arg e . Th e air q u an t it y in CF M (cu b ic fe e t p e r m in u te ) at t he fan ou tl et an d at t h e end of th e duc t wil l b e t h e s a m e . H oweve r, i f yo u a d d a h u n d re d fe e t o f s t r a i g h t duc t, there is less air volume at the end of the duc t even t h o u g h t h e fa n d i s c h a rg e p re ss u re re m a i n s t h e s a m e . I f yo u c o n t i n u e to a d d st r a i g h t d u c t , a t s o m e p o i n t ve r y l i t t l e a i r wo u l d c o m e o u t o f t h e e n d o f t h e d u c t . T h i s i s b e c a u s e there is resistance to the air movement in the duc t. Re s i st a n c e to a i r m ove m e n t i n a d u c t syste m i s c a u s e d by b o t h f r i c t i o n l o ss e s a n d d y n a m i c l o ss e s d u e to d i st u r b an ce o f f l ow. T h e t e r m d i s t u r b a n c e o f f l ow re fe rs t o a ny c h a n g e

in the direction of airflow or change in air velocity.

Friction Losses W h e n wa t e r f l ows d ow n a s t re a m , f r i c t i o n i s c re a t e d a s t h e water rubs along the banks. As a result of the fric tion, the wa t e r a l o n g t h e b a n k f l ows s l owe r t h a n t h e wa t e r i n t h e ce n te r of t h e st re am . In much the same way, static pressure pushes air against the s i d e s o f t h e d u c t a s t h e a i r f l ows a l o n g . T h e a i r r u b b i n g against the sides of the duc t causes fric tion. This fric tion causes some loss of the static pressure. This is called

friction loss.

As a result of friction loss, the total pressure is greatest at the fa n o u t l e t a n d g r a d u a l l y d e c re a s e s a s t h e a i r m ove s a l o n g the duc t.

44

AT M O S P H E R I C P R E S S U R E

9022399

Islaiz9 T O TA L

PRESSURET

FRICTION LOSS S TAT I C P R E S S U R E

T O TA L

PRESSURE

VELOCITY PRESSURE

F i g . 7: F r i c t i o n c a u s e s a l o s s o f s t a t i c p r e s s u r e

In a straight duc t with no size changes, the velocity of the a i r a n d t h e re s u l t i n g ve l o c i t y p re ss u re r e m a i n t h e s a m e . However, fric tion causes a loss of static pressure (SP). Figure 7 shows straight duc t that is open on the end. At the fan outlet (point A) the total pressure is at its highest, and

consists of both static pressure and velocity pressure. The V P re m a i n s t h e s a m e t h ro u g h o u t t h e d u c t , b u t f r i c t i o n causes a steady loss of static pressure. At the duct outlet (point B), static pressure has disappeared. At this point, total pressure is the same as the velocity pressure.

Dynamic Losses Wa t e r f l ow i n g i n a s t re a m a l ways h a s s o m e d i s t u r b a n c e t o

its flow. A large rock or tree root in a stream will make the wa t e r sw i r l a n d e d d y a ro u n d i t . A s h a r p b e n d i n t h e s t re a m w i l l c a u s e c o m p l ex c u r re n ts . A l l t h e s e va r i a t i o n s d i st u r b t h e

smooth flow of water.

Airflow in duct moves in much the same way. Dampers, c o i l s , o r o t h e r o b s t a c l e s m a ke t h e a i r sw i r l a n d e d d y. A n

elbow breaks up the smooth flow of air. All these variations disturb the air flow (Fig. 8). Any losses due to disturbance in o t h e r w i s e s t r a i g h t f l ow o f a i r a re c a l l e d d y n a m i c l o ss e s .

45

Dynamic losses can be caused by any change

in direction or velocity DA M P E R

in a run of duc t, as a result of coils, elbows, offsets, and transitions. T h ey a l l c a u s e t h e a i r to

change direction or velocity in some way. Any change in direction or velocity results in dynamic loss. Chapter 5

explains how particular Fig. 8: Air disturbance causes dynamic loss

fittings affect air flow. Designers use tables to

e s t i m a t e h ow m u c h d y n a m i c l o s s a f i t t i n g w i l l c a u s e .

Dynamic Losses Due to Duct Size Change A ny c h a n g e i n d u c t s i ze c a u s e s a c h a n g e i n ve l o c i t y, s o a c h a n g e i n s i ze c a u s e s a d y n a m i c l o ss .

Compare the airflow to a river. When a river channel n a r rows , t h e wa te r f l ows fa s te r. W h e n t h e s a m e r i ve r

channel widens, the water slows down. In the same way, if d u c t s i ze b e c o m e s s m a l l e r, t h e ve l o c i t y ( F P M ) a n d t h e resulting velocity pressure (VP) increase. If duct size b e c o m e s l a rg e r, ve l o c i t y a n d t h e re fo re ve l o c i t y p re ss u re decrease. Remember the equation: Quantity = Area x Ve l o c i t y. I f t h e q u a n t i t y re m a i n s t h e s a m e , w h e n t h e a re a i n cre a s e s , t h e ve l o ci t y m u st d e cre a s e . W h e n t h e a re a decreases, the velocity must increase.

46

COIL

S TAT I C P R E S S U R E

DEg

V E LO C I T Y P R E S S U R E

T O TA L PRESSURE

-3UNSS3dd TV1O1-

Fig. 9: Change in duct size causes change in air velocity

F i g u re 9 d e m o n s t r a te s w h a t h a p p e n s w h e n a d u c t c h a n g e s

size:

0 At p o i n t A , n o f r i c t i o n o r d y n a m i c l o s s e s h ave ye t occurred, so the total pressure is at its greatest. The a m o u n t o f t h e V P d e p e n d s o n t h e s i ze o f t h e d u c t a n d t h e C F M b e i n g d e l i ve re d . T h e re st o f t h e to t a l pressure is static pressure.

0 F ro m p o i n t A to p o i n t B , t h e d u c t i s st r a i g h t a n d d o e s n o t c h a n g e s i ze . T h e V P re m a i n s t h e s a m e f ro m A to B b e c a u s e t h e a i r ve l o c i t y d o e s n o t

.

change. The static pressure decreases from point A to point B because of fric tion losses. All the pressure loss is in static pressure. F ro m p o i n t B t o p o i n t C , t h e d u c t b e c o m e s s m a l l e r, sO the VP increases (in order to deliver the same C F M . ) T h e st a t i c p re ss u re d e c re a s e s m o re r a p i d l y because of fric tion and dynamic losses. The result is a l owe r t o t a l p re ss u re .

47

O

From point C to point D the duct is straight. The VP re m a i n s t h e s a m e b e c a u s e t h e a i r ve l o c i t y re m a i n s the same. The static pressure decreases, but the loss is less than it was from B to C, where there was a size change. The total pressure decreases only a

little.

PROBLEMS T h e fo l l ow i n g p ro b l e m s a re b a s e d o n F i g . 9.

Problems to 3 refer to the duct from points D to E: 1. W h a t c a u s e s t h e l o s s i n t o t a l p re s s u re f ro m p o i n t D t o E?

2. D o e s t h e ve l o c i t y p re ss u re i n c re a s e o r d e c re a s e ? 3. W h a t c a u s e s t h e c h a n g e i n ve l o c i t y p re s s u re ? P ro b l e m s 4 a n d 5 re fe r to t h e d u c t f ro m p o i n ts E to F. N o te that this is a coil. 4. W hy d o e s t h e to t a l p re ss u re d e c re a s e i n t h i s s e c t i o n ? 5. What par t of the loss is in static pressu re an d what par t is in velocity pressure?

Problems 6 and 7 refer to points G to H: 6. W hy i s t h e re n o s t a t i c p re s s u re a t p o i n t H ? 7. W h a t h a p p e n s t o t h e ve l o c i t y p re s s u re a f t e r p o i n t G ? C h e c k yo u r a n swe rs a t t h e e n d o f t h e c h a p t e r.

48

REVIEW I f yo u c a n a n swe r t h e fo l l ow i n g q u e st i o n s w i t h o u t re fe r r i n g to the text, you have lear ned the contents of this chapter.

Tr y t o a n swe r eve r y q u e s t i o n b e fo re you check the answers

in the back of the book.

Matching Some items may have more than one answer. 1. Atmospheric

A. Pressure on the outside walls of the duc t

pressure

2. Total pressure 3. Friction loss 4. Static pressure 5. Dynamic loss 6. Velocity pressure

B.

Pressure losses due to c h a n g i n g t h e d i re c t i o n o f t h e air

C. The result of air in motion D. Caused by air rubbing against the sides of the duc t

E.

Presses against the inside

walls of the duct

F. SP + VP G. TP - VP

7. ls the resistance to air flow greater at the fan outlet or at the end of the duct run? ANSWERS

TO PROBLEMS

8. A duct changes size from a small c ro s s - s e c t i o n a l a re a t o a m u c h l a rg e r a re a . T h e C F M re m a i n s t h e s am e . Doe s t h e tot al p re ss u re i n c re a s e o r d e c re a s e ?

Yo u r a n s w e r s n e e d n o t b e t h e e x a c t

wording of those below but should contain the same general ideas.

1. Dynamic loss due to change in velocity (On this short piece, friction loss is not significant.)

2. Decrease 3. The duct size becomes larger so

9. At a certain point in the duct run, the static pressure is measured at

1.75" wg. The total pressure is m e a s u r e d a t 2 .0 0 " wg . W h a t i s t h e ve l o c i t y p re ss u re ?

the air velocity decreases. 4. Because of dynamic loss due to

disturbance created by the coil. 5 . M o s t o f t h e c h a n g e i s i n S P. 6. As the air leaves the duct there is nothing to contain it and it spreads out into the atmosphere. 7. I t g r a d u a l l y d e c r e a s e s a s t h e a i r

moves into the room.

49

5

AIRFLOW IN A DUCT AND DYNAMIC

LOSSES

This chapter and Chapter 6 deal with different aspec ts of designing duc ts and duc t fittings. A fitting is any sec tion of a duc t run that is not straight duc t. Typical fittings are elbows, offsets, and transitions. E ve r y t e c h n i c i a n i n t h e H VAC i n d u s t r y s h o u l d u n d e r s t a n d t h e b a s i c s c ove re d i n t h e s e c h a p te rs . T h i s k n ow l e d g e w i l l

help you deal with problems in the shop and in the field. It will also let you recognize potential problems. T h e a i r f l ow i n a d u c t i s o f t e n m i s u n d e rs t o o d . I f yo u u n d e rst a n d t h e s i m p l e p r i n c i p l e s o f C h a p te rs 4, 5, a n d 6,

you will have a much better knowledge of air handling than m o s t t e c h n i c i a n s i n t h e i n d u s t r y. H oweve r i t i s i m p o r t a n t t o u n d e rs t a n d t h a t t h e m a t e r i a l i n t h e s e t h re e c h a p te rs i s b a s i c . I t w i l l a l l ow yo u to m a ke d u c t f i t t i n g s t h a t ke e p p re ss u re l o ss e s to a m i n i m u m ; to m a ke m i n o r c h a n g e s i n a d u c t e q u i p m e n t a n d d u c t sys t e m s ; a n d to s o l ve p ro b l e m s i n d u c t fa b r i c a t i o n , f i e l d i n st a l l a t i o n , a i r

balancing, service work, and indoor air quality. However, c o m p l e t e d u c t sys t e m d e s i g n a n d e q u i p m e n t s p e c i f i c a t i o n s

for a building are much more complex than the material p re s e n t e d h e re . D e s i g n i n g a d u c t e d sys t e m s h o u l d b e d o n e

by a mechanical engineer who is experienced in air conditioning design.

50

AIRFLOW PATTERNS The two patterns of airflow in straight duct are laminar and turbulent:

.

Laminar flow means air traveling in layers in a straight line (Fig. 1). Because of fric tion, the

AIRFLOW

layer of air against the s i d e s o f t h e d u c t m ove s

.

more slowly than the air in the middle of the duct. Turbulent flow means that the air tumbles and swirls as it moves down

F i g . 1 : L a m i n a r a i r f l ow

t h e d u c t ( F i g . 2 ) . C o m p a re t u r b u l e n t f l ow t o wa t e r a s it flows over rapids.

Laminar airflow is stratified. This means that the air temperatures

tend to remain in layers of air and do not mix. Since the air is the

medium for moving heat, this re s u l ts i n p o o r h e a t t r a n s p o r t a t i o n .

Fig. 2: Turbulent air flow

However, completely laminar a i r f l ow s e l d o m o c c u r s i n d u c t .

Turbulent airflow is the most common condition of airflow i n d u c t . Tu r b u l e n c e h e l p s ke e p t h e a i r t e m p e r a t u re m o re eve n l y s p re a d . H oweve r, i t a l s o re s u l ts i n m o re f r i c t i o n l o ss .

Too much turbulence may create too much air noise. T h e a m o u n t o f t u r b u l e n c e i n t h e a i r f l ow d e p e n d s o n a number of different fac tors such viscosity, fric tion, and air

velocity. A complex calculation determines what is known a s a Re y n o l d s n u m b e r. I f t h e Re y n o l d s n u m b e r i s a b ove a c e r t a i n va l u e , t h e a i r f l ow i s t u r b u l e n t . F o r p r a c t i c a l

51

p u r p o s e s , a l m o st a l l a i r f l ow c o n d i t i o n s i n a d u c t a re t u r b u l e n t . Tr u e l a m i n a r f l ow o c c u r s o n l y i n d u c t s t h a t h ave a n a i r ve l o c i t y t h a t i s to o l ow fo r p r a c t i c a l u s e .

AIRFLOW AND DUCT FITTINGS It is a law of physics that a moving objec t tends to travel in a straight line. It resists any change in direc tion. The greater the velocity, the more resistance there is to a change of

direction. You have all experienced this when trying to change direc tion in a car that is traveling too fast. Air that is t r ave l i n g a l s o wa n ts to t r ave l i n a st r a i g h t l i n e . T h e g re a t e r the velocity, the greater the resistance to changing direc tion.

Consider a s q u a r e throat elbow without turning vanes (Fig. 3). The airflow tends to

• HEEL

be straight until it is forced to t u r n . W h e n t h e a i r h i ts t h e f l a t LOW PRESSURE HERE

h e e l o f t h e e l b ow, p re ss u re build-up forces the air to tur n. Th e re s u lt is exce ss ive

T H R O AT

turbulence with a low pressure Fig. 3: Turbulence in a square throat elbow without vanes

a re a a t t h e t h ro a t o f t h e e l b ow. This turbulence causes

dynamic loss. A poorly designed fitting can develop a

dynamic loss equal to the

HEEL

f r i c t i o n l o ss d eve l o p e d by 5 0 feet or more of straight duc t.

A radius elbow with a throat radius that is too small (Fig. 4) has almost the same loss as a T H R O AT R A D I U S TOO SMALL

s q u a re throat elbow. The air at

the heel of the elbow

Fig. 4: Turbulence in a small radius elbow gradually turns, but the air

52

near the throat tends to travel in a straight line. The result is a build-up of pressure and resulting turbulence at the heel near the end of the turn. W h e n a f i t t i n g c re a te s t u r b u l e n c e , t h e t u r b u l e n c e c o n t i n u e s for several feet in the straight duc t that is downstream of the fitting (Figs. 3 and 4). Du c t fi t t i n g s ch an g e t h e d i re c t i on of ai r flow (e lb ows ), or change the size of the duct (transitions). Duct accessories (dampers, coils) inter rupt the air flow. Therefore all fittings an d acce ss orie s in t h e d u c t cre ate d yn amic loss . Th e designer tries to keep that loss as small as prac tical. The general rule for fittings is to make the tur ns as gradual as poss ib l e and to m ake t ran s ition s as l on g as prac t ical .

"Persuade" the air to make gradual turns-you can't make it turn a square corner even if you build the duct that way.

Radius Throat Elbows To ke e p a i r m ov i n g s m o o t h l y, a r a d i u s e l b ow ( F i g . 5 ) m u s t h ave a t h ro a t r a d i u s t h a t i s l a rg e e n o u g h . D e s i g n m a n u a l s

recommend that a radius elbow have a centerline radius at least 1 ½ times the width of

HEEL

the cheek (Fig. 5). This is the

.

s a m e a s s ay i n g t h a t : /CHEEK

The throat radius should equal the duct

width.

T H R O AT R A D I U S

CENTERLINE RADIUS:

F o r ex a m p l e , a d u c t w i t h a

12 x 15 = 18°

cheek 12" wide should have a c e n te r l i n e r a d i u s o f 1 8" ( 1 2 "

T H R O AT R A D I U S - Ç R A D I U S - M O T H

x 1.5 = 18"). The throat radius

T H R OAT R A D I U S = 1 8 ° - #

should be 12" (the same as

T H R OAT R A D I U S = 1 8 ° - 1 2 '

the duct width).

T H R OAT R A D I U S = C H E E K W I DT H Fig. 5: Recommended throat radius

53

Except on a drawing, when an elbow is identified, the first dimension is the width of the cheek.

Square Throat Elbows To b e e f fe c t i ve , s q u a re t h ro a t e l b ows m u s t h ave t u r n i n g va n e s ( F i g . 6 ) . T h e s e h e l p t h e a i r to c h a n g e d i re c t i o n m o re s m o o t h l y ( F i g . 7 ) . Re s e a rc h h a s s h ow n t h a t s i n g l e wa l l va n e s ( F i g . 8 ) p ro d u c e t h e s m a l l e st a m o u n t o f d y n a m i c l o ss .

&

ES

1

Fig. 6: Square throat elbow with turning vanes

Fig. 7: Vanes allow air flow to change direction more smoothly

Fig. 8: Single wall turning vane

54

Double wall vanes (Fig. 9) were commonly used in the past. However research has shown that these are much less effec tive than single wall va n e s .

Properly installed, a trailing edge on t h e va n e ( F i g . 1 0 ) w i l l f u r t h e r re d u c e d y n a m i c l o s s e s . H owe ve r, t e s t s h ave l e d re s e a rc h e rs t o re c o m m e n d eliminating the trailing edge. This is because ver y often the vanes are not

installed properly. The result is that they cause more loss instead of less. Th e van e s are e ffe c tive on ly if th ey are installed carefully so that the trailing edge is parallel to the duc t

Fig. 9: Double wall vanes

sides (Fig. 10). If they are not, they i n c re a s e d y n a m i c l o s s e s .

Transitions

MUST POINT

PA R A L L E L TO DUCT SIDE

Tr a n s i t i o n s a re d e s i g n e d to c h a n g e t h e s i ze o f t h e d u c t

(Fig. 11). To make the change

TRAILING EDGE

gradual, the sides of transitions s h o u l d b e 3 0° o r l e ss f ro m t h e

VA N E

straight (Fig. 11).

F i g . 1 0 : Va n e s w i t h t r a i l i n g e d g e

Offsets D u c t o f fs e ts ( F i g . 1 2 ) c a n b e m a d e w i t h a n g l e s o r w i t h smooth curves. The best design is the S offset (Fig. 12). W h e n t h e p a t t e r n s a re d e ve l o p e d by a p p rove d m e t h o d s o f layou t, the chan ge is gradu al, the air is gu ided throu gh a s m o o t h c u r ve , a n d t h e c ro ss - s e c t i o n a l a re a i s m a i n t a i n e d . This means the least possible dynamic loss.

55

F i g . 1 1 : Tr a n s i t i o n s i d e s s h o u l d n o t b e o v e r 3 0 ° f r o m s t r a i g h t

CHEEK

S OFFSET

S OFFSET

Fig. 12: Offsets

56

It is important to make S offsets as long as practical. This m a ke s t h e c u r ve s g r a d u a l . C o m m o n p r a c t i c e i s to c u t t h e cheek out of a 36" sheet. This makes the offset about 34" a f te r a l l owa n c e s a re m a d e fo r t r a n sve rs e c o n n e c t i o n s . F o r extreme offsets, the cheeks are sometimes made longer. Remember that the shorter the offset, the greater the

dynamic loss will be. Ve r y l a rg e o f fs e ts a re o f te n m a d e u s i n g 4 5 ° o r 3 0° e l b ows w i t h s t r a i g h t d u c t i n b e t we e n ( F i g . 1 3 ) . T h e r a d i u s e l b ows s h o u l d fo l l ow g o o d l ayo u t p r a c t i c e : t h e t h ro a t r a d i u s s h o u l d equal the width of the cheek.

45° ELBOW

STRAIGHT

DUCT

45° ELBOW

Fig. 13: Offset with 45° elbows

REVIEW 1. The throat radius of an elbow with an 18" cheek should b e a t l e a st

inches.

57

Matching 2. Turbulent flow

A.

3. Trailing edge 4. Reynolds number

B.

5. Heel

6. Best type of turning vane 7. Transitions 8. Laminar air flow

9. Cheek 10. Dynamic loss

11. Throat

Üo

E.

The flat side of an elbow A i r t u m b l e s a n d sw i r l s

The back of an elbow The small side of an elbow A straight length of metal on a

turning vane A i r f l ows i n s e p a r a te l aye rs

u'O

Double wall vanes

H. Single wall vanes I. Re l a te d to t h e a m o u n t o f turbulence in the duct J.

K.

Changes duc t size Re s u l t s f ro m c h a n g i n g a i r f l ow

direction

58

6

SIZING DUCTWORK

C h a p t e r 3 s h owe d yo u h ow t o c a l c u l a t e d u c t s i ze s u s i n g t h e

equation Quantity = Area x Velocity. By calculating duct sizes with this equation it is possible to change duc t sizes a n d m a i n t a i n t h e s a m e q u a n t i t y ( C F M ) a n d ve l o c i t y ( F P M ) i n the duct. H owe ve r, a s yo u l e a r n e d i n C h a p t e r s 4 a n d 5, t h e p re s s u re l o ss c h a n g e s e a c h t i m e t h e d u c t s i ze c h a n g e s . T h e e q u a t i o n

Q =AxV does not take this into account. The duct designer needs to know what the total pressure loss is for a duc t run in order to selec t the proper size fan. The static pressure at the fan outlet must be equal to the resistance of the duc t system.

Using Q = A x V to calculate duct size changes with each c h a n g e i n C F M m a i n t a i n s t h e s a m e ve l o c i t y, b u t t h e f r i c t i o n l o ss fo r e a c h s i ze w i l l n o t re m a i n t h e s a m e . T h i s i s n o t to

say that the equation Q = A x V is not an important e q u a t i o n . U n d e rst a n d i n g t h e re l a t i o n s h i p s o f t h i s e q u a t i o n i s e ss e n t i a l to u n d e rst a n d i n g a i r f l ow i n d u c t .

EQUAL FRICTION

LOSS METHOD OF SIZING DUCT

T h e i n d u st r y h a s g e n e r a l l y a d o p te d t h e e q u a l f r i c t i o n l o ss m e t h o d o f s i z i n g d u c t . T h i s g i ve s t h e e q u i va l e n t d u c t s i ze b a s e d u p o n m a i n t a i n i n g t h e s a m e f r i c t i o n l o ss . T h e e q u a l f r i c t i o n l o ss m e t h o d va r i e s t h e ve l o c i t y b u t m a i n t a i n s t h e same fric tion loss per 100 feet of duct run. By maintaining the same fric tion loss per 100 feet, it is only necessary to d e te r m i n e t h e to t a l l e n g t h o f t h e d u c t r u n to d e te r m i n e t h e to t a l f r i c t i o n l o ss . T h e m e t h o d i s ex p l a i n e d i n t h i s c h a p te r.

59

Aspect Ratio U n d e rs t a n d i n g a s p e c t r a t i o i s i m p o r t a n t fo r t h e e q u a l fric tion loss method. Choosing the best aspec t ratio of a duc t can reduce fric tion loss. Fric tion loss in a duc t is the result of the air molecules rubbing against the inside of the duc t. For the same air q u an t it y an d ve l ocit y, a du c t wit h a g reater sur face for th e a i r to r u b a g a i n st w i l l d eve l o p m o re f r i c t i o n l o ss . T h i s means that if the quantity of air and the area of the duc t

remain the same: O

The greater the perimeter (distance around) of a duc t, t h e m o re f r i c t i o n l o ss t h e re w i l l b e .

As p e c t r a t i o i s a way to d e te r m i n e t h e b e st p r a c t i c a l p e ri m e te r for a d u c t . As pec t rat io is th e ratio bet ween th e

width and height of a duct. Divide the width by the height to find the first number of the aspec t ratio: As p e c t R a t i o =

Width Height

A s q u a re d u c t h a s a n a s p e c t r a t i o o f 1 t o 1 : As p e c t R a t i o :

12"

12"

Aspect Ratio = 1 1 A 24" x 8" duc t has an aspec t ratio of 3 to 1:

Aspect Ratio =

24" g"

:

:

Aspect Ratio = 3 1

60

A 3 0" x 1 2 " d u c t h a s a n a s p e c t r a t i o o f 2.5 to 1 :

As p e c t R a t i o =

30" 2.5 12"

Aspect Ratio = 2.5 1 Figure 1 compares the aspec t ratios of different size duc ts that have nearly the same area (about 200 square inches) b u t i n c re a s i n g p e r i m e te rs . ( Ac t u a l d u c t i s s i ze d i n eve n

numbers. Odd numbers are used in Fig. 1 in order to m a i n t a i n a b o u t t h e s a m e a re a . ) N o te t h a t t h e a s p e c t r a t i o increases as the length of the perimeter increases. Notice also that the round pipe has the shor test perimeter and t h e re fo re t h e l e a st f r i c t i o n l o ss fo r a g i ve n C F M a n d ve l o c i t y.

Size

Area (sq. in.) Perimeter Aspect Ratio

16" Dia.

201

50"

NA

14" x 14"

196

56"

1 :1

17" X 12"

204

58"

1.42 : 1

2 0" x 1 0"

200

60"

2 :1

25" X 8"

200

66"

3.12: 1

29" X 7"

203

72"

4.14 :1

33" x 6"

198

78"

5.5 :1

Fig. 1: Perimeters and aspect ratios of ducts with almost the same area

W h e n u s i n g a s p e c t r a t i o to c h o o s e d u c t s i ze s , fo l l ow t h e s e

general principles: O Round duct has the least friction loss.

:

O

Next to round duc t, square duc t is best.

. .

61

As the aspect ratio increases, the friction loss increases. Avo i d u s i n g d u c t w i t h a n a s p e c t r a t i o g re a te r t h a n 3

to 1, if possible. In addition to increasing friction loss, it costs more to fabricate and install-more l a b o r, m o re m a t e r i a l , a n d h e av i e r g a g e s o f m e t a l .

Using the Equal Friction Loss Method T h e e q u a l f r i c t i o n l o ss m e t h o d o f s i z i n g d u c t i s b a s e d u p o n maintaining the same fric tion loss for ever y 100 feet of duc t. B y m a i n t a i n i n g t h e s a m e f r i c t i o n l o ss p e r 1 0 0 fe e t , i t i s o n l y necessary to deter mine the total length of the duc t run to deter mine the total fric tion loss. For example, if the duc t is s i ze d to m a i n t a i n a f r i c t i o n l o ss o f 0.1" wg p e r 1 0 0 fe e t a n d the total length of the duc t run is 225 feet, calculate the

friction loss:

225

0.1" x - - = 0.225" wg 100 The duc t designer can use the fric tion loss of 0.225" wg for the duc t run to deter mine the size of fan needed for the syste m. In order to size duc t for the fric tion loss chosen, an equal fric tion char t (Fig. 2) or a duc t calculator (described on page 67) is used. To illustrate the use of the fric tion char t, Fig. 3 shows a run of duct supplying 8000 CFM. The CFM of each branch run is given. Assume that a fric tion loss of 0.1" wg per hundred feet has been selec ted for sizing the duc twork for the run.

62

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63 1700 CFM 1250 CFM

1400 CFM

3000 CFM

600 CFM

1000 CFM 1300 CFM

750 CFM

F R I C T I O N L O S S : 0. 1 " W G P E R 1 0 0 F T.

Fig. 3: Duct run

Data for Point A The first step is to deter mine the duc t size and the velocity

at point A. O Lo c a t e 8 0 0 0 C F M o n t h e s c a l e a t t h e b o t t o m o f t h e

. . .

friction chart (Fig. 2). Locate 0.1" wg fric tion loss on the scale at the left side of the char t.

Follow the vertical 8000 CFM line up and follow the horizontal 0.1" wg line across to where they intersec t. This is marked on the char t. Th e marke d p oin t in dicate s th at th e ve locity (th e diagonal lines slanting down to the right) is

approximately 1640 FPM. O

T h e m a r ke d p o i n t i n d i c a te s t h a t t h e d u c t s h o u l d b e the equivalent of 30" diameter (the diagonal lines slanting up to the right).

T h e m a r ke d p o i n t re q u i re s a 3 0" d i a m e t e r d u c t , d e l i ve r i n g 8 0 0 0 C F M a t 1 6 4 0 F P M . I f t h e d u c t r u n i s t o b e re c t a n g u l a r, the 30" diameter must be conver ted to a rec tangular size t h a t w i l l g i ve t h e s a m e f r i c t i o n l o ss . ( I t w i l l n o t b e a n e q u a l are a.)

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Converting to Rectangular Duct To c o nve r t t h e 3 0" d i a m e te r to a n e q u i va l e n t re c t a n g u l a r duc t size, use a char t that gives rec tangular equivalents for

.

round duc t (Fig. 4):

.

The numbers on the left side of the char t are the d i a m e te rs o f ro u n d d u c t . T h e n u m b e r s i n t h e c h a r t i n groups o f t wo a re t h e w i d t h a n d h e i g h t fo r re c t a n g u l a r d u c t .

F o r ex am p l e , t h e f i rst g ro u p o f t wo n u m b e rs to t h e r i g h t o f 3 0" d i a m e t e r a re 2 7 a n d 2 7. T h i s m e a n s t h a t t h e e q u i va l e n t o f 3 0" d i a m e te r i s a s q u a re d u c t w h i c h m e a s u re s 2 7 " x 2 7 ". F o r p r a c t i c a l p u r p o s e s , 2 7 x 2 7 " i s n o t a g o o d c h o i c e . F i rst , re c t a n g u l a r d u c t i s u s u a l l y g i ve n i n eve n n u m b e rs . S e c o n d ,

a duct 27" deep will generally not fit into the available ceiling space. Moving to the right across the char t, 40" x 20" and 48" x 16" are also given as equivalents of a 30" diameter round duc t. Either size is acceptable. Assume that 40" x 20" duct is the size that best fits the space available. The aspec t ratio of a 40" x 20" duc t is 2 to 1:

Aspect Ratio

40" 2 20"

Aspect Ratio = 2 :1 This is an acceptable aspec t ratio. Therefore, the duc t c h o s e n fo r p o i n t A to p o i n t A' i n F i g . 3 i s 4 0" x 2 0".

Data for Point B N ow d e t e r m i n e t h e d a t a fo r p o i n t B o n t h e d u c t r u n i n

Fig. 3. The CFM required for this point of the duct will be

66

l e ss b e c a u s e t wo b r a n c h l i n e s h ave b e e n t a ke n o f f. T h e C F M fo r t h e s e t wo b r a n c h l i n e s t o t a l s 3 0 0 0 C F M :

1700 CFM + 1300 CFM = 3000 CFM T h e re fo re t h e q u a n t i t y n e e d e d a t p o i n t B i s 5 0 0 0 C F M :

8000 CFM - 3000 CFM = 5000 CFM Locate the intersec tion on the equal fric tion char t (Fig. 2) for

the 5000 CFM line and the 0.1" line. This point indicates a 2 5 " d i a m e t e r d u c t w i t h a ve l o c i t y o f a p p rox i m a t e l y 1 4 5 0 F P M . S i n c e t h e p o i n t i s i n b e t we e n l i n e s , t h e n u m b e rs m u s t be estimated. Estimates could var y slightly, but there will be little difference in the final selec tion of the rec tangular duc t

size. Duct dimensions are rounded up to even numbers. F i g u re 5 s h ows t h e d a t a fo r p o i n ts A a n d B .

Point

CFM

FPM

Dia. (Round)

Rect. Duct

Aspect

A

8000

1640

30"

40" x 20"

2 1

B

5000

1450

25"

30" x 18"

1.7 1

Ratio

Fig. 5: Data for points A and B

PROBLEMS 1. C o m p l e t e t h e t a b l e fo r p o i n ts C t o J. Ro u n d o f f d u c t d i a m e t e r t o t h e n e a re s t w h o l e n u m b e r :

Point

:

:

:

:

B

CFM

FPM

Dia. (Round)

Rec t.

8000

1640

30"

40" x 20"

2 1

5000

1450

25"

30" x 18"

1.7 1

Duct

C

X 1 8"

D

x 10"

Aspect Ratio

67

Point

CFM

FPM

Dia.

Rect.

(Round)

Duct

Aspect Ratio

x 10' x 8" x 8"

H

x 8°

x 8' x 8'

C h e c k yo u r a n swe rs a t t h e e n d o f t h e c h a p t e r.

Duct Calculators D u c t c a l c u l a to rs ( F i g . 6 ) a re a l s o ava i l a b l e to g i ve t h e s a m e results as the equal friction loss chart (Fig. 2) and the equivalent duc t sizes char t (Fig. 4). There are different for ms o f d u c t c a l c u l a t o rs

available. All these calculators include the U CT U L ATO R

following information:

.

O

.

O

Fig. 6: Duc t calculator

O

CFM F r i c t i o n l o ss p e r 1 0 0 feet of duc t

FPM Round duct diameter A selection of

equivalent rectangular duct sizes

If a ny t wo o f t h e s e va l u e s a re k n ow n , t h e o t h e rs c a n b e deter mined.

68

40" x 20"

F R I C T I O N L O S S : 0 . 1 ' W G P E R 1 0 0 F T.

F i g . 7: E x te n d e d p l e n u m va r i a t i o n o f F i g . 3. I t e l i m i n a te s o n e c o st l y t r a n s i t i o n .

OTHER METHODS OF SIZING DUCT T h e e q u a l f r i c t i o n l o ss m e t h o d i s t h e m o s t c o m m o n l y u s e d m e t h o d o f s i z i n g d u c t . H owe ve r o t h e r m e t h o d s a re u s e d .

The extended plenum method is a modification of the equal f r i c t i o n m e t h o d . I t e l i m i n a te s s o m e o f t h e s i ze c h a n g e s o f the equal fric tion method. This reduces the cost of fa b r i c a t i o n a n d i n s t a l l a t i o n . F i g u re 7 s h ows t h e ex t e n d e d p l e n u m d e s i g n a p p l i e d to t h e d u c t sys te m i n F i g . 3. T h e s t a t i c re g a i n m e t h o d re d u c e s t h e a i r ve l o c i t y i n t h e duc t. This reduces the velocity pressure. The result is that st a t i c p re ss u re i s l a rg e r t h a n i t wo u l d b e u n d e r t h e e q u a l f r i c t i o n m e t h o d . ( Re m e m b e r t h a t T P = S P + V P ) . I n o t h e r words, static pressure "regains" some of its pressure.

69

REVIEW 1. The most common method of sizing duct is the loss method. 2. T h i s m e t h o d i s b a s e d o n t h e l o s s per

feet of

duct run.

3. T h e e q u a t i o n fo r a s p e c t r a t i o i s

divided by

4. A 12" x 12" duct has an aspect ratio of 5. A 36" x 12" duct has an aspect ratio of 6. W h i c h h a s t h e m o s t d e s i r a b l e a s p e c t r a t i o, a 24 " x 6 " duct or a 12" x 12" duct? C o m p l e t e t h e c h a r t b e l ow, b a s e d o n t h e e q u a l f r i c t i o n c h a r t

in Fig. 2.

Friction 0.1" w g

800 1200

9.

11.

CFM

FPM

Loss

12.

7.

Duct Dia. 8.

5000 5000

10. 24 "

1 3. F o r i t e m 8, i f o n e s i d e o f t h e re c t a n g u l a r d u c t m u s t b e 6", what is the other side?

14. For item 10, if one side of the rectangular duct must be 1 8", w h a t i s t h e o t h e r s i d e ?

15. For item 10, what size rectangular duct would have the best aspect ratio?

70

ANSWERS TO PROBLEMS N OT E : Yo u r a n s w e r s f o r F P M m a y v a r y s l i g h t l y. H o w e v e r, t h i s

:

:

:

:

:

:

:

:

should affect the other figures little if at all.

Dis. (Round)

Rect. Duct

1250

20"

18" x 18"

1 1

600

850

11"

10" x 10"

1 1

E

1700

1100

16"

20" x 10°

2 1

F

1300

1040

15"

24" x 8"

3 1

C

1250

1040

15"

24" x 8"

3 1

H

1000

1000

14"

20" x 8"

2.5 1 3 1

1 .7 5 1

CFM

FPM

C

2750

D

1400

1050

15*

24 x 8"

750

900

12"

14" x 8"

Aspect

Ratio

71

CALCULATING PRESSURE LOSSES IN DUCTWORK

7

C h a p te r 6 t a u g h t yo u h ow to s i ze d u c ts u s i n g t h e e q u a l f r i c t i o n l o s s m e t h o d . T h i s c h a p t e r ex p l a i n s h ow t o c a l c u l a t e f r i c t i o n a n d d y n a m i c l o ss e s , w h i c h o c c u r i n a l l sys t e m s . The duct designer (usually a mechanical engineer) calculates the pressure loss in a duc t system. This calculation is needed as par t of the process of selec ting a fan for the system. If the system pressure loss is kept low, less energy is needed to operate the fan. I t i s a l s o u s e f u l fo r t h e i n d o o r e nv i ro n m e n t te c h n i c i a n to u n d e r s t a n d h ow t o c a l c u l a t e d u c t p re s s u re . T h i s c a l c u l a t i o n is direc tly related to designing effec tive duc t fittings. On i n st a l l e d syste m s , a p ro b l e m ca n o f te n b e t r a ce d to a p o o r l y d e s i g n e d d u c t f i t t i n g w h i c h h a s a n exc e ss i ve d y n a m i c l o ss .

TOTAL PRESSURE LOSSES The total pressure loss for a supply air duc t system is the loss for the longest path from the outside air intake to the far thest outlet. It is NOT the sum of all the paths. In Fig. 1, the total system loss equals the total resistance from fric tion

and dynamic loss of the following: O

Outside air intake

O

Filters

.

Coils

72 COILS

F I LT E R

A

OA I N TA K E

F i g . 1 : S yste m p re ss u re l o ss i s f ro m OA i n t a ke to fa r t h e st o u t l e t ( G )

O

Fa n sys te m e f fe c t

.

Transitions B, C, and D

0

Elbow E

O

Take-off F

0 S t r ai g h t d u c t

O Outlet G Fan system effec t is the pressure loss created by the duc t c o n n e c t i o n a t t h e fa n i n l e t a n d d i s c h a rg e . Fa n s a re r a te d w i t h n o i n l e t d u c t a n d a st r a i g h t o u t l e t d u c t o f t h e s a m e d i m e n s i o n as t h e fa n o u t l e t . A ny o t h e r c o n d i t i o n a f fe c ts fan per for mance and is par t of the total pressure loss. T h e to t a l p re ss u re l o ss , c a l c u l a te d f ro m t h e o u ts i d e a i r intake to the end of outlet G, is the amount of pressure that t h e fa n m u st p rov i d e to ove rc o m e t h e syste m re s i st a n c e .

73

PRESSURE LOSSES

.

Remember that: F r i c t i o n l o ss i s t h e re s u l t o f a i r m o l e c u l e s r u b b i n g o n the inside sur face of the duc t. Fric tion is related to the roughness of the inside duc t sur face, the length of the duc t, and the velocity of air in the duc t. Fric tion loss increases greatly as air velocity increases. Sometimes it is incor rec tly assumed that fric tion loss

includes dynamic loss. O Dynamic loss is the result of a disturbance to airflow i n t h e d u c t . D y n a m i c l o s s e s o c c u r w h e n e ve r a d u c t c h a n g e s d i re c t i o n o r s h a p e . T h e d e s i g n a n d fabrication of a duc t fitting affec ts the amount of

dynamic loss. Changes in airflow direction or ve l o c i t y s h o u l d b e m a d e a s s m o o t h l y a n d g r a d u a l l y as possible. LENGTH -

Dynamic losses and friction losses make up the pressure l o ss e s o f a i r f l ow i n a d u c t .

LOSSES IN

STRAIGHT DUCT

ENGTH -

If the duct system has been Fig. 2: Lengths of ducts are calculated from centerlines

d e s i g n e d by t h e e q u a l f r i c t i o n method, calculating the fric tion loss for straight duc t is

simple. At elbows, the length of straight duc t is measured from centerline to centerline (Fig. 2). Use this equation to calculate pressure loss for straight duct: Lo s s = " wg p e r 100 feet X

Length of duc t (in feet)

100

74

For example, suppose a straight run of duc t is 322 feet long. It is designed by the equal fric tion method for a pressure l o ss o f 0.1 5 " wg p e r 1 0 0 fe e t . W h a t i s t h e t o t a l p re ss u re loss? Lo s s = " wg p e r 100 feet X

Loss = 0.15' Wg X

Length of duc t (in feet)

100

322' 100

Loss = 0.15' wg X 3.22 Loss = 0.48" wg

LOSSES IN FITTINGS Fittings add additional losses. This dynamic loss is calculated separately for each fitting. Losses for fittings used to be calculated as equivalent duc t lengths. Tables indicated that a type of fitting had the pressure loss equal to a par ticular length of straight duc t. For example, a table might show that the pressure loss for a c e r t a i n e l b ow wa s t h e e q u i va l e n t o f t h e p re ss u re l o ss o f 70 feet of straight duct. This was added to the total length of

totaled

the straight duct. If the straight duct in the system 1 0 0 fe e t , t h e syste m l o ss wo u l d b e c a l c u l a te d fo r 1 70 fe e t o f straight duct. A newer and better method of calculating losses for fittings i s n ow re c o m m e n d e d . I t i s b a s e d o n f i n d i n g t h e l o s s coefficient for each fitting from tables. A coefficient is s i m p l y a n u m b e r d e te r m i n e d by l a b o r a to r y te sts fo r a par ticular type of fitting. Once this coefficient is deter mined, the loss for the fitting is deter mined by the

following equation: Loss = Coefficient x Velocity pressure

75

For example, suppose you had to find the pressure loss for

the elbow in Fig. 3. The air is flowing at a velocity of 2000 F P M . U s e t h e c h a r t i n F i g . 4, w h i c h i s t a ke n f ro m t h e

SMACNA manual HVAC Systems Duct Design, and follow

.

these steps.

Identify key dimensions: The drawing in Fig. 4 has letters to identify necessary dimensions needed to c a l c u l a te t h e l o ss c o e f f i c i e n t :

R= Radius of the vane

S 3 6

x

Vane spacing

1 8

Identify these dimensions on

the Fig. 3 drawing: 36 x 18

Radius = 4.5" Spacing = 3.25" There are two standard sets of these radius and spacing d i m e n s i o n s t o c h o o s e f ro m .

R ( R A D I U S ) = 4.5 S ( S PAC I N G ) = 3.2 5 Fig. 3: 36" x 18" elbow

L O S S C O E F F I C I E N T S ( C ) F O R S I N G L E T H I C K N E S S VA N E S V E L O C I T Y, F P M

DIMENSIONS, INCHES

S

2.0 4.5

1000

1500

2000

2500

1.5

0.24

0.2 3

0.2 2

0.2 0

3.25

0.2 6

0.24

0.23

0.22

SMACNA, HVAC Systems Duct Design

Fig. 4: Loss coefficient chart for a square elbow with vanes

76

.

Find the coefficient: Next find the loss coefficient on t h e t a b l e i n F i g . 4. S i n c e t h e

velocity is 2000 FPM, follow the column for 2000 FPM down to the row for 4.5 r a d i u s a n d 3.2 5 s p a c i n g . T h e i n te rs e c t i o n o f t h e

column and row has a

.

Ve l o c i t y

Velocity Pressure

FPM

in. wg.

2000

0.2 5

2050

0.2 6

2100

0. 2 7

2150

0.29

2200

0.3 0

2250

0.3 2

2300

0.3 3

2350

0.3 4

c o e f f i c i e n t o f 0.2 3. ( T h e

2400

0.3 6

coefficient is not a

2450

0.3 7

2500

0.3 9

2550

0.41

m e a s u r e m e n t o f a ny s o r t . ) F i n d ve l o c i t y p re s s u re : T h e equation for pressure loss in a fitting (Loss = Coefficient x VP) requires the figure for the velocity pressure. This c a n b e fo u n d f ro m a t a b l e

2600

0.4 2

2650

0.4 4

2700

0.4 5

2750

0. 4 7

2800

0. 4 9

2850

0.5 1

that conver ts velocity (FPM) to ve l o c i t y p re ss u re ( V P )

2900

0.5 2

2950

0.5 4

( F i g . 5 ) . T h e ve l o c i t y i n F P M

3000

0.5 6

for airflow in this elbow is

2 0 0 0 F P M . Lo o k i n g t o t h e right of 2000 FPM on the t a b l e i n d i c a te s a V P o f 0.2 5 "

Fig. 5: A portion of a velocity and velocity pressure chart

wg . ( T h e re a re m a ny s o u rc e s fo r ve l o c i t y/ ve l o c i t y p re ss u re tables.) O

C a l c u l a te t h e l o ss : T h e p re ss u re l o ss fo r t h e e l b ow

can now be calculated:

Lo ss = C o e f f i c i e n t x V P Lo s s = 0.2 3 x 0.2 5 " wg Lo s s = 0.0 5 7 5 " w g

The manual HVAC Systems Duct Design by SMACNA has loss coefficient tables for many types of fittings. Each table is different because they use different ratios and methods to d e te r m i n e t h e l o ss c o e f f i c i e n t o f t h e f i t t i n g . H oweve r, e a c h t a b l e h a s a d r aw i n g o f t h e f i t t i n g to s h ow t h e d i m e n s i o n s

77

WHEN A = 180°

USE THE VP OF THE DOWNSTREAM SECTION. 6

+YA

10° 15°-40° 0.0 5 0.0 5

0.05 0º

0.0 5

50°-60° 90° 120°

150° 180°

0.12 0.18 0.17 0.27

0.35 0.41

0.1 9 0.2 9

0.37 0.43

0.0 5 0.0 4 0.04 0.0 5

0.0 6 0.07

0.0 7 0.0 8

0.24

0.2 6

0.18 0.28 0.36 0.42

ußisag iong SWaISAS OVAH 'VNJVWS

Fig. 6: Loss coefficients for a transition

n e e d e d . F o r ex a m p l e , F i g . 6 i s a t a b l e fo r a t r a n s i t i o n w i t h the sides slanting at the same angle:

. The letters A and A, stand for the area at each end of

.

the transition. T h e sy m b o l 0 i n F i g . 6 st a n d s fo r t h e i n c l u d e d a n g l e of th e tap e rin g s ide s (Fig . 7 ). De te r min e th e in clu de d a n g l e by a d d i n g to g e t h e r t h e a m o u n t o f t a p e r o n each side.

30°

a +% 15° + 15° = 30°

Fig. 7: Add the angle of taper on each side to determine the included angle

78

For the transition in Fig. 8:

A, =18" x 18" A = 9" x 9" 0 = 30° (because 2 sides slant 15°)

e

Calculate A, /A:

A, _ 18" x 18" A 9" x 9"

N O T E : 1 . A L L S I D E S S L A N T AT 1 5 °

2. VELOCITY - 2000 FPM

Fig. 8: Transition

A-4

A

W i t h t h e s e f i g u re s , yo u c a n f i n d t h e c o e f f i c i e n t i n t h e t a b l e (Fig. 6). For the included angle (30°), choose the column for 1 5 ° t o 4 0°. Tr a c e t h a t c o l u m n t o i n t e rs e c t w i t h t h e row

under A,/A marked 4. The coefficient is 0.04. T h e a i r ve l o c i t y i n t h e l e av i n g d u c t i s 2 0 0 0 F P M . T h e t a b l e in Fig. 5 indicates that the velocity pressure for 2000 FPM is

0.25" wg. Now you have the information needed to calculate the total pressure loss for this fitting: Lo ss = Coefficient x Velocity pressure Lo s s = 0.0 4 x 0.2 5 " wg LO s s = 0.0 1" wg

79

SOURCES OF COEFFICIENTS T h e b e st s o u rc e o f c o e f f i c i e n t c h a r ts i s H VAC S yste m s D u c t

Design, a manual published by SMACNA (Sheet Metal and A i r C o n d i t i o n i n g C o n t r a c t o rs ' N a t i o n a l Ass o c i a t i o n ) . T h i s

SMACNA manual is probably the most complete and a u t h o r i t a t i ve re fe re n c e o n d u c t d e s i g n . Ta b l e s fo r p re ss u re l o ss fo r f i t t i n g s c a n a l s o b e fo u n d i n H a n d b o o k o f F u n d a m e n t a l s p u b l i s h e d by A S H R A E ( A m e r i c a n S o c i e t y o f H e a t i n g , Re f r i g e r a t i n g , a n d A i r- C o n d i t i o n i n g E n g i n e e rs ) . P re ss u re l o ss e s c a n a l s o b e c a l c u l a te d by co m p u te r programs. Data for each fitting and straight duc t in a run is e n t e re d i n t h e c o m p u t e r. T h e c o m p u t e r d o e s a l l t h e necessary calculations and gives the total pressure loss for

the complete run.

PROBLEMS 1. Calculate the pressure loss for the elbow shown in

Fig. 9.

3 2

x

2 0

VA N E R A D I U S = 4. 5 ° VA N E S PAC I N G = 3. 2 5 ° 32 x 20 A I R V E LO C I T Y: 2 5 0 0 F P M

Fig. 9: Calculate the pressure loss

80

2. Give the data for each of the items below for Fig. 10. A. Total feet of straight duc t for the fric tion loss

calculation. (Do not include the transition.) B. Pressure loss for straight duct C. R and S for the elbow D. Coefficient for the elbow E.

VP for the elbow

F. Pressure loss for the elbow G. A,/A for the transition H. e for the transition I.

J.

Coefficient for the transition Pressure loss for the transition (For a transition, use t h e V P fo r t h e d ow n st re a m s e c t i o n . )

K. Total pressure loss for the entire duct run

-3'-0-

60-0-

2 4

x

1 8

NOTES: FRICTION LOSS = 0.15° WG/100' VELOCITY IN 30° x 24° = 2000 FPM 5 4 ' - 0 °

VELOCITY IN 24' x 18' = 3000 FPM VA N E R A D I U S = 4. 5 ° VA N E S PAC I N G = 3. 2 5 ' 30 x 24

Fig. 10: Problem 2

81

REVIEW 1. A i r r u b b i n g a g a i n st t h e s i d e s o f t h e d u c t c re a te s loss. 2. A ny t h i n g t h a t c re a t e s a d i s t u r b a n c e o f t h e a i r f l ow i n a loss. duc t creates 3. T h e f r i c t i o n l o ss fo r a d u c t sys te m i s 0.1 5 " wg p e r 100 ft. The length of straight duc t in the system totals 160 feet. What is the total fric tion loss for the straight duct?

4. Calculate the friction loss for this elbow. VA N E R A D I U S - 4.5 °

5. For the duct below,

VA N E S PAC I N G = 3. 2 5 °

determine the following:

24 x 18 AIR VELOCITY: 2000 FPM

A . P re ss u re l o ss fo r t h e

B. C. D. E.

straight duct.

R and S for the elbow Coefficient for the elbow VP for the elbow Pressure loss for the elbow

F. A, /A for the transition G. 0 for the transition H. Pressure loss for the transition

I.

Total pressure loss for entire duc t run -45'-0°

• 5'-0°

X

32 x 22 -

Z

E

2 4

x

1 6

NOT TO SCALE

N OT E S : F R I CT I O N LO S S = 0.1 5 ° WG/ 1 0 0' V E LO C I T Y I N 3 2 ' x 2 2 ' = 2 0 0 0 F P M 4 8 ' - 0 °

VELOCITY IN 24' x 16° = 3000 FPM

-

VA N E R A D I U S = 4 . 5 ° 32 x 22

VA N E S PAC I N G - 3. 2 5 *

82

ANSWERS TO PROBLEMS 1. 0.0 8 5 8" wg [ C o e f f i c i e n t = 0. 2 2 V P = 0.3 9" wg Loss = Coefficient x VP Loss = 0.22 x 0.39" wg]

2A. 120 [54' + 3" + 1' - 3" (half of 30")

+ 1' - 3" + 3" + 3' + 60] 2B. 0.1815" wg loss for straight duct [Loss

121 x 015*wg]

2C. R = 4.5 S = 3. 2 5

2 D. 0. 2 3 2 E . 0.2 5 " wg [2000 FPM = 0.25" wg VP]

2 F. 0.0 5 7 5 " w g l o s s f o r e l b o w [Loss = Coefficient x VP LOss = 0.23 x 0.25° wg)

2G. 1.667

4.30 x24 24"x18° ^-16671 2H. 30° [15° + 15° - 30°] 21. 0.05 2 1. 0.0 2 8" wg [Loss = Coefficient x VP

L o s s = 0.0 5 x 0.5 6 |

2 K . 0.2 6 7 wg [ 0.1815" wg for straight duct

0.0 5 75 " wg fo r e l b ow + 0.0 2 8 " w g f o r t r a n s i t i o n

0.2 6 7 " wg

83

DUCT FITTINGS

8

You have already learned how air flows in ducts, what p re ss u re s a n d p re ss u re l o ss e s o c c u r i n d u c t , a n d h ow d u c t syste m s a re d e s i g n e d . T h i s c h a p te r a p p l i e s t h e k n ow l e d g e to duc t fittings. You will lear n to reduce resistance in fittings a n d t o re c o g n i ze d u c t sys t e m p ro b l e m s .

If an HVAC system is not producing the volume of air n e e d e d , t h e c a u s e c a n b e p o o r f i t t i n g s . Yo u s h o u l d b e a b l e to identify poor fitting applications. The chapter does not deal with the engineering aspec ts of duc t design. It explains design in ter ms of prac tical applications for duc t fittings for i n d o o r e nv i ro n m e n t t e c h n i c i a n s w h o s e r v i c e , a d j u s t , o r

operate HVAC systems.

PRESSURE DROP IN FITTINGS Any duc t fitting that changes the direc tion of air flow or that changes the size of the duc t is a potential source of large

dynamic losses. F o r ex a m p l e , i n P ro b l e m 2 F i n C h a p t e r 7, t h e p re ss u re l o ss fo r t h e e l b ow w a s d e t e r m i n e d t o b e 0.0 5 7 5 " wg . T h e fric tion loss for the straight duc t was 0.15" wg per 100 feet of duc t. From this, it can be calculated that the fric tion loss of the elbow is equal to 38 feet of straight duc t:

0.0 5 75 "

wg ×

0.15" wg

100 feet = 38 feet (38.333 rounded off)

84

Consider the duct elbow in Fig. 1 It is a poorly designed fitting because the throat radius is too s m a l l . T h e d u c t syste m h a s a fric tion loss of 0.1" wg per 100 feet

20 x 18

of straight duct. The pressure loss

1 0 ° T H R O AT RADIUS

for this elbow with a 10" throat

r a d i u s i s 0.0 6 7 2 " wg . T h i s e q u a l s

POOR FTTING

the loss for 67 feet of straight duct. S i n c e a ny d u c t f i t t i n g c a u s e s d y n a m i c l o ss e s , i t i s i m p o r t a n t to keep these losses to a prac tical

NOTES:

1 . F R I C T I O N L O S S - 0. 1 ' W G / 1 0 0 F T 2 . V E LO C I T Y - 2 2 5 0 F P M

Fig. 1: Elbow with a small throat radius

minimum. Even properly designed duc t fittings impose substantial pressure losses. Poorly designed fittings have much larger losses. On some jobs the standards for duc t fittings are carefully specified. On other jobs the shop or the individual sheet

or

m e t a l wo r ke r m a ke s d e c i s i o n s re g a rd i n g t h ro a t r a d i u s length. These decisions deter mine whether a fitting has a large pressure loss or a comparatively small one. I f a d u c t l i n e i s n o t d e l i ve r i n g e n o u g h a i r, l o o k a t t h e d e s i g n of the fittings to see if they might be the cause of unusual dynamic losses.

GOOD PRACTICES FOR FITTING DESIGN I f yo u u n d e rst a n d t h e p r i n c i p l e s o f g o o d d u c t d e s i g n , yo u can ofte n locate p rob le m are as i n t h e d u c t syste m . Th e re are i m p o r t a n t p r a c t i c e s t h a t s h o u l d b e o b s e r ve d w h e n fa b r i ca t i n g o r i n st a l l i n g d u c t .

85 PROPER HEEL

Avoid Choking a Fitting

WRONGI

C U RV E

AREA IS

C H O K E D A f i t t i n g i s c h o ke d i f t h e a re a i n

the middle of the fitting is less than the area of the ends (Fig. 2). This is the result of poor patter n drafting. Fig. 2: Choking means reducing the fitting area

Dog-leg offsets (Fig. 3) should be

avoided:

.

T h ey c re a te a h i g h

dynamic loss.

BISECT ANGLES

O

They can easily be choked if not laid out properly.

D o g - l e g s a re s e l d o m fa b r i c a te d i n

PROPER OFFSET

t h e s h o p. S o m e t i m e s t h ey a re c u t

in the field as an emergency m e a s u re , b u t i t i s n o t g o o d

WRONGI

A R E A I S C H O K E D p r a c t i c e . I f a d o g - l e g o f fs e t m u s t b e

fa b r i c a t e d i n a n e m e rg e n c y, t h e

angle must be bisected (Fig. 3A). If proper methods are not followed and the angle is not bisec ted, the Fig. 3: A dog-leg offset must be

carefully laid out to avoid choking

area of the straight duc t will be smaller than the rest of the duc t (Fig. 3B).

An S offset should be used instead of a dog-leg offset. If proper patter n drafting methods are followed for all

fittings, there will be no choking. Computerized layout p ro g r a m s n eve r c h o ke a f i t t i n g .

86

Roughness

Material

Rating

compared to

aluminum

Aluminum

Smooth

x1

Galvanized steel

Medium smooth

x3

Rigid fibrous glass

Medium rough

x10

Rigid fibrous liner

Medium rough

x10

Flexible

Rough

x100

Rough

x100

Concrete

Fig. 4: Comparing the roughness of duct lining

Keep Duct Linings Smooth Fric tion loss is the result of air rubbing against the duc t sur face. The smoother the inner sur face of the duc t, the less

friction will be developed. If an installed HVAC system does n o t d e l i ve r e n o u g h a i r, o n e o f t h e re a s o n s c o u l d b e excessive fric tion loss because of the roughness of the duc t. T h e t a b l e i n F i g . 4 s h ows t h e c o m p a r a t i ve ro u g h n e ss o f some of the commonly used duct materials. Aluminum is the smoothest of these, so it has a value of 1 . All the other materials are compared to

it. Galvanized steel is most

M A X I M U M 3' - 0"

commonly used for duct. For special applications, other metals such as a l u m i n u m , st a i n l e ss ste e l , a n d c o p p e r a re u s e d . Flexible duc t (Fig. 5) is

M E TA L D U C T

comparatively rough on the

FLEX DUCT

inside. This is one of the r e a s o n s w hy m o s t b u i l d i n g cod e s li m i t t h e le n g t h s of

Fig. 5: The length of flexible duct that should be used is limited because it causes so much friction loss

87

TRANSVERSE JOINT

F i g . 6 : Tr a n s v e r s e j o i n t s

runs for flexible duc t. The job specifications for the duc t

shown in Fig. 5 limit flexible duct to 3 foot lengths. Flex d u c t i s c o m m o n l y u s e d to c o n n e c t d i f f u s e rs to d u c t wo r k to c u t d ow n l a b o r c o s t s . O n m o s t j o b s t h e m a te r i a l a s we l l a s t h e c o n s t r u c t i o n s t a n d a r d s fo r t h e d u c t a r e s p e c i f i e d . U s u a l l y S M AC N A d u c t

standards are followed.

Avoid Air Leakage A i r l e a ka g e f ro m s e a m s a n d j o i n ts i n t h e d u c t syste m c a n b e a m a j o r c a u s e o f e n e rg y l o ss . E ve n w i t h c a re f u l workmanship, an unsealed system can leak as much as 30%

of the total CFM.

Leakage can be controlled by careful workmanship and by sealing all transverse joints (Fig. 6) with a s e a l a n t . Pa te n te d d u c t connec tors (Fig. 7) provide the tightest air seal for transverse joints. These require gaskets or caulking for the tightest

seal.

GASKET D U CT WA L L

Fig. 7: Typical patented duct connector

88

Unfortunately, duct air leakage is commonly ignored on s m a l l j o b s . O n l a rg e p ro j e c ts , t h e b u i l d i n g s p e c i f i c a t i o n s often state the acceptable leakage as a percentage of total

airflow. A commonly specified leakage rate is 5% of total C F M . A ny t h i n g l e ss t h a n t h i s p e rc e n t a g e re q u i re s exc e ss i ve fa b r i c a t i o n a n d i n st a l l a t i o n c o sts a n d i s o n l y re q u i re d fo r special situations such as nuclear energy installations.

Avoid Heat Loss H e a t c a n b e c o n d u c t e d t h ro u g h u n i n s u l a t e d m e t a l d u c t walls. For heating supply duc t, heat is lost to the s u r ro u n d i n g a i r. F o r c o o l i n g s u p p l y d u c t , h e a t i s g a i n e d f ro m t h e s u r ro u n d i n g a i r. I n e i t h e r c a s e , m o re e n e rg y i s n e e d e d to s u p p l y t h e c o n d i t i o n e d s p a c e w i t h t h e d e s i re d a i r t e m p e r a t u re . Supply duct in an unconditioned space, such as an attic s p a c e , s h o u l d b e i n s u l a te d to p reve n t t h i s u nwa n te d t r a n s fe r o f h e a t . Wr a p t h e o u ts i d e o f t h e d u c t w i t h i n s u l a t i o n to prevent heat loss or heat gain. Some duc t materials, such as r i g i d f i b ro u s d u c t b o a rd , a re i n s u l a t i n g m a te r i a l s t h e m s e l ve s

and do not need additional insulation.

Avoid Air Noise A i r n o i s e f ro m t h e H VAC sys t e m t h a t i s a p ro b l e m i n t h e conditioned space can come from three sources: O

Fan noise

O

Turbulence noise

o Excessive air velocity If there is too much pressure loss in a system, the fan speed m ay h ave to b e i n c re a s e d to p rov i d e t h e re q u i re d s t a t i c pressure. Generally, the greater the fan speed (RPM), the g r e a t e r the fan noise.

89

Turbulence noise is produced by the fan, by improperly reinforced duc twork, and by all the duc t fittings. Diffusers, registers, and grilles can also add to the noise.

.

A i r n o i s e c a n b e p re ve n t e d by g o o d d u c t d e s i g n : Ke e p t h e a i r ve l o c i t y a s l ow a s p r a c t i c a l . I n g e n e r a l , b r a n c h l i n e s to t h e o u t l e ts a re d e s i g n e d fo r a l owe r

velocity than the main duct. . Design the duct fittings for low dynamic loss. O Locate diffusers, registers, and grilles at least 3 feet

away from any duct fittings. . Apply acoustic lining inside the duct walls. Acoustic lining must be installed very carefully to prevent exc e ss i ve f r i c t i o n a n d e ro s i o n o f t h e m a te r i a l . A d eve l o p i n g t e c h n o l o g y i s e l e c t ro n i c n o i s e c o n t ro l . T h e b a s i c i d e a i s t h a t a m i c ro p h o n e p i c ks u p t h e d u c t n o i s e a n d t h e n o i s e i s a n a l y z e d by a c o m p u t e r. T h e c o m p u t e r generates a counter sound that is out of phase with the air noise. This sound is transmitted to a speaker in the duc t. It effec tively silences the air noise.

Change Duct Direction Smoothly L i ke a h i g h s p e e d c a r t a k i n g a c u r ve , a i r te n d s to m ove i n a straight line. It resists changing direction. The result is that e l b ows c re a te ex t r a f r i c t i o n a n d d y n a m i c l o ss f ro m t u r b u l e n c e . T h e re fo re w h e n eve r a n e l b ow o r o f fs e t i s u s e d to c h a n g e d i re c t i o n , t h e f i t t i n g s h o u l d b e d e s i g n e d to m a ke t h e a i r f l ow c h a n g e d i re c t i o n a s s m o o t h l y a s p o ss i b l e . A n offset should always be as long as prac tical to keep the

airflow smooth.

E l b ows c a n t u r n a ny a n g l e , t h ey c a n o f fs e t u p o r d ow n , a n d t h ey c a n c h a n g e s i ze f ro m o n e e n d to t h e o t h e r.

90

To keep air moving smoothly and to re d u c e t h e d y n a m i c l o ss e s , va n e s a r e a d d e d . Tu r n i n g va n e s ( F i g . 8 ) a re u s e d i n s q u a re t h ro a t e l b ows . A s yo u

learned in Chapter 5, laboratory tests have shown that single thickness vanes create less p re ss u re l o ss t h a n d o u b l e

thickness vanes. However, vanes longer than 36" should be

reinforced or should be double

F i g . 8 : Tu r n i n g v a n e s

t h i c k n e ss i n o rd e r to h o l d t h e i r shape. Tu r n i n g va n e s a re h e l d i n r a i l s ( F i g . 9 ) . Va n e s a n d r a i l s a re p u rc h a s e d f ro m m a n u fa c t u re rs . R a i l s c o m e i n st r i p s t h a t a re c u t to f i t t h e e l b ow. T h ey m u st ex te n d t h e f u l l d i st a n c e b e t we e n t h e t h ro a t a n d t h e h e e l . Do n o t i n st al l s h o r t r ai l s . Pre-punched slots on the rails fit the cur ve of the vane. The vanes are slipped into the slots in the rails and the metal is

crimped over to hold them in place. The standard spacing on these rails is 3¼" between vanes. Va n e s m u s t a l w a y s b e

placed in each slot. Using only every other pair of slots is a poor prac tice

which increases dynamic l o ss e s . Te sts h ave s h ow n t h a t t h i s p r a c t i c e m o re t h a n d o u b l e s t h e p re ss u re loss for the fitting. After the vanes are fastened in the rails, the r a i l s a n d va n e s a re

installed inside the elbow. Fig. 9: Rail for turning vanes

91

Th e rails mu st be p os ition e d s o th at th e van e s are in lin e

with the airflow: O

T h e l e a d i n g e d g e m u s t b e a l i g n e d w i t h t h e e n te r i n g

air (Fig. 10). O

T h e l e av i n g e d g e m u st d i re c t t h e a i r i n l i n e w i t h t h e duc t tur n (Fig. 10). If tur ning vanes are not

positioned properly, they can create more disturbance than an L E AV I N G E D G E

elbow without vanes

(Fig. 11).

LEADING EDGE

S p l i t te r va n e s ( F i g . 1 2 ) a re u s e d i n r a d i u s t h ro a t

elbows to reduce A

I

R

L

F

O

d y n a m i c l o ss . A s p l i t te r

W

Fig. 10: Vane is aligned with air flow

vane runs the full length

of the elbow. The number of splitter vanes used

d e p e n d s o n t h e t h ro a t r a d i u s o f t h e e l b ow. S p l i t te r va n e s fo r r a d i u s t h ro a t e l b ows

are not used as often as tur ning vanes in square throat

elbows. However, they may be required, especially if the throat radius is small.

WRONGI

F i g . 1 1 : Va n e s o u t o f a l i g n m e n t c a u s e

turbulence

Fig. 12: Splitter vane

92

T h e t h ro a t r a d i u s i s very

important in determining pressure loss. For

example, Fig. 13 shows t wo ve rs i o n s o f t h e s a m e

AT 2 0 0 0 F P M P R E S S U R E LO S S I S 0.1 3" WG

POOR FITTING

elbow. One has a 6" 6 ' T H R O AT RADIUS

t h ro a t r a d i u s a n d t h e o t h e r has a 24" throat radius:

. .

Loss for 6" radius

AT 2 0 0 0 F P M

P R E S S U R E LO S S I S 0.0 5 ° WG

elbow is 0.13" wg. Loss for 24" radius

elbow is 0.05" wg. 24 X 12

2 4 " T H R OAT

RADIUS

The elbow with the 6" throat has nearly three

Fig. 13: Comparing pressure loss

times the pressure loss as

the elbow with the 24" throat.

Splitter vanes are effec tive because increasing the throat r a d i u s o f a r a d i u s e l b ow re d u c e s t h e d y n a m i c l o ss . T h e e l b ow i n F i g . 1 3 w i t h a 6 " t h ro a t r a d i u s g e n e r a te s a l a rg e pressure loss (0.1 3" wg) because the throat radius is only ¼ o f t h e c h e e k w i d t h . B y a d d i n g t wo s p l i t te r va n e s ( F i g . 1 4 ) , t h e e l b ow i s , i n e f fe c t , t u r n e d i n to t h re e e l b ows , e a c h o n e w i t h a t h ro a t r a d i u s c l o s e to o r g re a te r t h a n t h e c h e e k w i d t h . T h e re i s a l a rg e c h a n g e i n p re ss u re l o ss : O

Lo ss fo r e l b ow w i t h o u t s p l i t te r va n e s i s 0.1 3" wg .

0 Lo ss fo r e l b ow w i t h 2 s p l i t te r va n e s i s 0.0 1 2 5 " wg . Because of the splitter vanes, the pressure loss has dropped

to about 10% of the original 0.13" wg.

93

24 X 12

8 X 12

+, p

8 X 12

v. p

8 X 12

F i g . 1 4 : Tw o s p l i t t e r v a n e s t u r n s t h e e l b o w i n t o t h r e e e l b o w s

Keep Transitions Gradual To re d u c e d y n a m i c l o ss , m a ke t r a n s i t i o n s a s l o n g a s

practical. Doing this keeps the included angle (Fig. 15) between the two sides of a transition as small as possible. A

small included angle means a small angle of change (Fig. 15). The included angle should not be more than 30° if possible.

15° 15°

T R A N S I T I O N TO O A B R U P T INCLUDED ANGLE

Fig. 15: Included angle and angle of change

ANGLE OF CHANGE

94

F i g u re 1 5 s h ows t wo ve rs i o n s o f a t r a n s i t i o n . O n e i s s h o r t a n d h a s a 6 0° i n c l u d e d a n g l e . T h e o t h e r i s l o n g e r a n d h a s a 3 0° i n c l u d e d a n g l e . T h e f i t t i n g w i t h t h e s m a l l i n c l u d e d angle (30°) has about 45% less pressure loss than the fitting w i t h t h e larg e i n clu d e d an g le (60°). O f cou rs e t h i s ass u m e s that both are flat on the sides and both have the same air

velocity.

L-qW

Take-offs

LIS 4" MINIMUM

Ta ke - o f fs re q u i re t h e a i r to

change direction. Figure 16 shows the best design for branch take-offs that are 90° to

LOnG NIVW

the main duct. The length of the take-off should be ¼ of the

width of the branch duct and

n o l e ss t h a n 4 ". I t s h o u l d h ave a Fig. 16: Branch take-off

45° angle.

SUMMARY OF DUCT FITTING APPLICATIONS Po o r l y d e s i g n e d a n d i n st a l l e d d u c t a n d f i t t i n g s c a n r u i n t h e p e r fo r m a n c e o f a n H VAC s y s t e m . At h i g h p r e s s u r e s a n d h i g h ve l o c i t i e s , d u c t d e s i g n i s eve n m o re c r i t i c a l . C o n s i d e r all the principles of good duct and fitting applications. O

O

Ke e p d u c t a s l a rg e a s p r a c t i c a l fo r l owe r a i r ve l o c i t y a n d t h u s l owe r l o ss e s . Keep duc t runs as straight as possible. Each change

in direction in a duct run creates a dynamic loss. 0

Keep air velocity low to reduce air noise. Use acoustic lining or electronic noise control if necessary.

95

0 U s e t h e fo l l ow i n g e q u a t i o n t o d e t e r m i n e p re s s u re loss in a fitting:

Loss = Coefficient x VP O

U s e g r a d u a l t u r n s a n d c h a n g e s . M a ke t r a n s i t i o n s a n d offsets as long as practical.

. Use 45° angle on take-offs. O

U s e t u r n i n g va n e s o r s p l i t te r va n e s i n e l b ows .

O

F o l l ow g o o d p a t t e r n d e ve l o p m e n t p r a c t i c e s t o avo i d

.

O

ch oki n g fi t t i n g s . Ke e p f l ex i b l e d u c t a s s h o r t a s p o ss i b l e to avo i d fric tion loss. Insulate supply duc t to reduce heat loss or heat gain a n d s ave e n e rg y.

O

U s e c a re f u l wo r k m a n s h i p o n d u c t a n d u s e s e a l a n t t o

avoid air leakage.

REVIEW 1. I f t h e a re a o f a f i t t i n g i s l e ss i n t h e m i d d l e t h a n o n t h e ends, the fitting is

2. Flexible duct and other rough duct linings increase loss.

3. A commonly specified leakage rate for a job is not to exceed % of total CFM. 4. L i s t t h re e c o m m o n c a u s e s o f a i r n o i s e i n a d u c t sys t e m .

5. List four ways to reduce turbulence noise.

96

6. T h e t h ro a t o f a t a ke - o f f s h o u l d b e a t a 7. T h e l e n g t h o f t h e t a ke - o f f s h o u l d b e

o

angle.

of the

width of the branch duct, and a minimum of inches. 8. F o r t h e s a m e C F M , d o e s l a rg e r d u c t i n c re a s e o r re d u c e

air velocity? 9. Does higher air velocity increase or reduce air noise?

97

MEASURING AIRFLOW

9

T h e c h a p te rs i n t h i s b o o k h ave d e a l t w i t h st a t i c p re ss u re (SP), velocity pressure (VP), air velocity (FPM), and air q u a n t i t y ( C F M ) . T h i s l a s t c h a p te r ex p l a i n s h ow i n s t r u m e n ts a re u s e d to m e a s u re t h e s e q u a n t i t i e s i n d u c t a n d a t o u t l e ts . This chapter does not cover detailed instruc tions for

measuring airflow. These will be covered in another book in t h i s s e r i e s . M e a s u r i n g a i r f l ow p re c i s e l y re q u i re s m o re k n ow l e d g e a n d p r a c t i c e t h a n o n e c h a p t e r c a n c ove r. D e te r m i n i n g t h e C F M i n d u c ts , b r a n c h d u c ts , a n d o u t l e ts i s one of the basic processes of the TAB (testing, adjusting, a n d b a l a n c i n g ) te c h n i c i a n . I t i s a l s o e ss e n t i a l fo r t h e I AQ

(indoor air quality) technician because IAQ problems are o f te n re l a te d to t h e a i r c i rc u l a t i o n i n a ro o m a s p rov i d e d by

the HVAC system. The HVAC installer must know how to t a ke a i r f l ow m e a s u re m e n ts i n o rd e r to d i s c u ss p ro b l e m s with engineers and technicians. T h e a i r f l ow q u a n t i t y ( C F M ) i n a d u c t i s n o t m e a s u re d d i re c t l y. I t i s d e te r m i n e d by g o i n g t h ro u g h o t h e r m e a s u re m e n t s a n d c a l c u l a t i o n s : O

O

.

M e a s u re V P ( ve l o c i t y p re ss u re ) . C o nve r t V P to ve l o c i t y ( F P M ) a n d ave r a g e t h e re a d i n g s . Calculate CFM using the equation CFM = Area X

Velocity. T h i s s o u n d s l i ke a n i nvo l ve d p ro c e s s . H owe ve r, m o s t d i g i t a l air pressure measuring instruments at least conver t the

98

ve l o c i t y p re ss u re ( V P ) t o ve l o c i t y ( V ) . S o m e e l e c t ro n i c i n s t r u m e n ts a l l ow yo u t o i n p u t t h e d u c t s i ze a n d

automatically compute the CFM.

DETERMINING AIRFLOW IN DUCT I n st r u m e n ts t h a t m e a s u re a i r f l ow i n d u c t c o n s i st o f t wo basic components:

.

O

A sensing device (such as a pitot tube) A re a d o u t d e v i c e ( s u c h a s a m a n o m e t e r )

The sensing device senses the air flow pressures (velocity pressure or static pressure). The readout device receives the s i g n a l f ro m t h e s e n s i n g d ev i c e a n d t u r n s i t i n to a n a n a l o g o r d i g i t a l re a d i n g . ( A n a n a l o g re a d i n g i s a s c a l e a n d i n d i c a to r,

such as a needle and dial. A digital reading is in numbers.)

Pitot Tube T h e p i to t ( p e a ' -to e ) t u b e ( F i g . 1 ) i s a c o m m o n l y u s e d s e n s i n g d ev i c e fo r m e a s u r i n g ve l o c i t y p re ss u re . I t i s a c t u a l l y two tubes, one inside the other (Fig. 2). The tube is inser ted i n t h e d u c t w i t h i ts t i p p o i n te d i n to t h e a i r s t re a m :

. .

To t a l p re ss u re i s s e n s e d t h ro u g h t h e h o l e i n t h e t i p

of the tube.

Static pressure is sensed

through the holes a ro u n d t h e o u ts i d e o f the tube. The pitot tube has two outlets,

called ports (Fig. 2). One transmits total pressure (TP) and the other transmits static pressure (SP). If the static

Fig. 1 P i t o t t u b e

99 AIRFLOW

S TAT I C PRESSURE ORIFICES

INNER TUBING ( I M PACT T U B E )

T O TA L PRESSURE

O R IF ICE

S TAT I C PRESSURE ORIFICES

OUTER TUBING ( S TAT I C T U B E ) S TAT I C

PRESSURE PORT

• T O TA L P R E S S U R E P O R T

SECTION A-A

Fig. 2: Pitot tube

p re ss u re i s s u b t r a c te d f ro m t h e to t a l p re ss u re , t h e re m a i n i n g p re ss u re i s ve l o c i t y p re ss u re ( V P ) . ( T h i s i s b a s e d o n t h e e q u a t i o n

TP = VP + SP, which can be rewritten

V P = T P - S P. )

Manometers A m a n o m e t e r i s u s u a l l y u s e d a s t h e re a d o u t Fig. 3: Inclined manometer

d ev i c e t h a t i s c o n n e c te d to t h e t wo p o r ts o n

the pitot tube-the total pressure port and the static pressure por t. The manometer s u b t r a c ts S P f ro m T P to g i ve a re a d i n g i n V P.

.003

The inclined manometer (Fig. 3) is generally used for ver y accurate readings and for c a l i b r a t i n g o t h e r d ev i c e s .

The electronic manometer (Fig. 4) provides a digital readout of velocity pressure or static pressu re . Th ere are different makes an d models, but they commonly use a pitot tube to s en s e air flow press u res . Fig. 4: Electronic manometer

100

An electronic manometer is a small computer and can be p ro g r a m m e d t o d o va r i o u s o p e r a t i o n s . U s u a l l y i t c o nve r t s a series of VP readings to velocities (FPM), and provides an average of these velocities.

Magnehelic Gages M a g n e h e l i c g a g e s ( F i g . 5 ) c a n a l s o b e u s e d to re a d ve l o c i t y pressures. They use special tips to sense static pressure.

Magnehelic is a trade name, but it is so widely used that it has become a generic

term.

Magnehelic gages are often used to indicate

MAGNEHEUS, wewhtoi vetien

differential pressure. They have sensors permanently mounted at selected places in a n a i r h a n d l i n g sys t e m t o s h ow t h e difference in static pressure across a

c o m p o n e n t . F o r ex a m p l e , t h e s t a t i c p re s s u re F i g . 5 : A M a g n e h e l i c g a g e s e n s o rs o f a M a g n e h e l i c g a g e c a n b e m o u n te d o n e a c h s i d e o f a b a n k o f f i l te rs .

and sensor

The gage indicates the difference in pressure b e t we e n t h e t wo l o c a t i o n s . S u c h a d i f fe re n t i a l p re ss u re re a d i n g s h ows w h e n t h e f i l te rs a re to o d i r t y. M a g n e h e l i c g a g e s a re g e n e r a l l y n o t u s e d fo r d e t e r m i n i n g ve l o c i t y p re ss u re i n a d u c t h av i n g a ve l o c i t y l e ss t h a n

4000 FPM.

Taking a Pitot Tube Traverse B e c a u s e t h e a i r f l ow i n a d u c t i s t u r b u l e n t a n d t h e ve l o c i t y i s n o t u n i fo r m , a s i n g l e re a d i n g o f ve l o c i t y p re ss u re a t a ny o n e p o i n t i s n o t a g o o d i n d i c a t i o n o f t h e ve l o c i t y i n t h e d u c t . E ve n a fe w r a n d o m s a m p l e s w i l l n o t g i ve a n a c c u r a t e p i c t u re o f t h e ve l o c i t y i n t h e d u c t . T h e re fo re s eve r a l readings are taken in a single cross sec tion and the readings a re c o nve r t e d t o ve l o c i t y i n F P M a n d ave r a g e d . T h e l o c a t i o n a n d t h e p ro c e ss fo r t h e re a d i n g s i s c a l l e d a p i to t

tube traverse.

101

There must be a systematic method of taking these readings. Th e du c t cross s e c tion ch os e n s h ou ld be at a s p ot w h e re th e a i r f l ow ve l o c i t y i s a s u n i fo r m a s p o ss i b l e . T h e re fo re i t should be as far downstream of fans and fittings as is p r a c t i c a l . T h i s s h o u l d b e a t l e a st 7.5 d u c t d i a m e te rs d ow n s t re a m a n d 2.5 d i a m e t e r s u p s t re a m o f a ny d i s t u r b a n c e c a u s e d a by a f i t t i n g .

Square Duct Traverse

PITOT TUBE READING IN

C E N T E R O F E AC H A R E A

F o r a square duct traverse (used for any rec tangular duc t), the cross section of the d u c t i s d i v i d e d i n to a n HOLES DRILLED IN DUCT FOR

P I TOT T U B E Fig. 6: Pitot tube traverse points

imaginary grid with s q u a re s o f

approximately 6" x 6" (Fig. 6). The number of s q u a re s d e p e n d s u p o n the size of the duc t.

T h e g e n e r a l r u l e i s t h a t t h e re s h o u l d b e n o fe we r t h a n 1 6 s q u a re s a n d n o m o re t h a n 6 4. D e t a i l s o f t h e l ayo u t o f t h e traverse squares is covered in another book in this series. A p i to t t u b e re a d i n g i s t a ke n i n the center of each square.

Holes are drilled in the duct at

t h e p ro p e r l o c a t i o n s to i n s e r t the pitot tube (Fig. 6). These h o l e s m ay b e i n t h e s i d e o r t h e bottom of the duct. The pitot t u b e i s m a r ke d ( F i g . 7 ) to b e sure it is inser ted the proper a m o u n t to h i t t h e c e n te r o f e a c h s q u a re . P i to t t u b e s c a n b e Fig. 7: Pitot tube marked for depth of

inser tion

obtained in different lengths to

reach into any size duct.

102

Each of these velocity pressure (VP) readings is conver ted to ve l o c i t y ( F P M ) . T h e n a l l o f t h e F P M re a d i n g s a re ave r a g e d to o b t a i n t h e F P M o f t h e a i r f l ow a t t h a t l o c a t i o n o f t h e d u c t . (Th e VP re adin g s can n ot be ave rag e d be cau s e th e VP is a s q u a re d f u n c t i o n o f ve l o c i t y a n d t h e re fo re c a n n o t b e

averaged.) I f o n l y a V P re a d i n g i s ava i l a b l e , ve l o c i t y ( F P M ) c a n st i l l b e

determined by one of the following two methods: O

U s e a t a b l e to c o nve r t V P to ve l o c i t y ( F P M ) . ( T h e re i s a p a r t i a l t a b l e o n p a g e 76 o f t h i s b o o k . )

O

Use the following equation to convert VP to velocity: Velocity = 4005x / Velocity Pressure

F o r ex a m p l e , t h e V P i s 0.2 5 " wg . W h a t i s t h e

velocity in FPM? Velocity = 4005x /VP Velocity = 4005x /0.25" wg Velocity = 4005x0.5 Ve l o c i t y = 2 0 0 2 .5 F P M

When the velocity is determined, the airflow quantity (CFM) is calculated with the equation CFM = Area x Velocity. If t h e re a d o u t d ev i c e d o e s n o t d o t h i s a u t o m a t i c a l l y, yo u c a n u s e yo u r own calculator to determine CFM.

Round Duct Traverse For a 20-point round duc t traverse, the duc t cross sec tion is d i v i d e d i n to f i ve c o n c e n t r i c c i rc l e s w i t h t h e p a t te r n o f points shown in Fig. 8. These two sets of traverse points m u s t b e 9 0° t o e a c h o t h e r. F i g u re 9 s h ows h ow e a c h o f t h e s e p o i n ts i s n u m b e re d . Po i n ts 1 t o 1 0 a re t a ke n by

103

i n s e r t i n g t h e p i to t t u b e i n o n e h o l e in the du c t (Fig. 8). Poin ts 11 to 20

are taken from another hole in the duct. The traverse points are located by

multiplying the duct diameter by t h e c o n st a n ts i n F i g . 9. T h i s g i ve s t h e d i s t a n c e m e a s u re d i n f ro m t h e wa l l o f t h e d u c t fo r t h e p i to t t u b e

HOLES FOR

re a d i n g . F o r ex a m p l e , fo r a 1 0"

PITOT TUBE

diameter duct, readings 1 and 11

Fig. 8: Round duct traverse

a re t a ke n ¼ " f ro m t h e h o l e i n t h e

duct:

0.9 74

20

.

which is close to ¼"

from the hole in the

0.9 1 8 0.854

9 e

1 0" X 0.0 2 6 = 0.2 6 ",

duct.

0.7 74 • 0.6 5 8

Readings 10 and 20 are t a ke n 9 ¾ " f ro m t h e h o l e :

+

O

15

10" x 0.974 = 9.74", w h i c h i s c l o s e to 9 ¾ " f ro m t h e h o l e i n t h e

• 0.3 4 2 0. 2 2 6

duct.

0146 0.0 8 2 11

0.0 2 6

Fig. 9: Numbered points and multipliers for a round duct traverse

T h e re st o f t h e p ro c e d u re is the same as for square duct.

DETERMINING AIRFLOW AT OUTLETS

A pitot tube cannot be used to determine CFM at outlets (registers, grilles, and diffusers). This is because there is no duc twork, so there is no static pressure.

104

Direct Reading Instruments D i re c t re a d i n g i n s t r u m e n ts u s e d t o b e c o m m o n l y u s e d t o m e a s u re t h e ve l o c i t y a t o u t l e ts . H oweve r, i t i s d i f f i c u l t to m e a s u re a c c u r a te l y w i t h t h e s e i n st r u m e n ts . T h e a i r f l ow i s turbulent, has different spreading patter ns, and is par tially b l o c ke d by b a rs o r d i f f u s e r c o n e s . B o t h ro t a t i n g va n e

anemometers and velometers provide very inaccurate readings, since the air flow is so turbulent and ir regular. If d am p e rs b e h i n d t h e fa c e o f t h e o u t l e t a re p a r t i a l l y cl o s e d , t h e re a d i n g s a re e ve n m o re i n a c c u r a t e . Ve l o m e t e r s c a n n o t be used with registers and grilles. In a addition, a great deal of mathematics is required to d e te r m i n e C F M u s i n g t h e s e i n st r u m e n ts . T h i s t a ke s t i m e a n d t h e re s u l ts a re s u b j e c t to e r ro rs i n c a l c u l a t i o n .

Flow Hoods F l ow h o o d s a r e a n a c c u r a t e m e t h o d o f m e a s u r i n g a i r f l ow. A

flow hood (Fig. 10) is also called a cap t u re h ood . It con s ists of a hood, also called a skir t, that fits

tightly over an outlet or inlet and

Low rooo

a m e te r t h a t i n d i c a te s C F M . T h e h o o d i s p l a c e d o n a f l a t s u r fa c e a ro u n d t h e o u t l e t . A s h o r t p e r i o d

of time is required to obtain an a c c u r a t e re a d i n g .

E

When the flow hood is placed ove r a d i f f u s e r, i t c re a te s b a c k

pressure, an additional pressure on the airflow from the outlet. Back pressure can introduce a

large error in the CFM reading. Newer digital hoods can c o m p e n s a t e fo r b a c k p re s s u re .

On older hoods and analog

O) IN3WOMISNI JONTY

F i g . 1 0 : F l ow h o o d

105

h o o d s t h e b a c k p re ss u re c o m p e n s a t i o n m u s t b e e s t i m a te d a c c o rd i n g t o t h e m a n u fa c t u re r ' s i n fo r m a t i o n . S o m e e l e c t ro n i c m e t e rs n o t o n l y m e a s u re C F M , b u t a l s o c o r re c t fo r a l t i t u d e , i n d i c a te s u p p l y a i r te m p e r a t u re , p rov i d e b a c k p re ss u re c o m p e n s a t i o n fo r h i g h a i r f l ow r a te s , a n d record several readings for future recall.

Flow hoods can have analog or digital readings. A flow h o o d w i t h a n a n a l o g re a d o u t i s fa s t e r. H owe ve r, i t m u s t b e cor rec ted for air density other than standard air, and it c a n n o t p r ov i d e b a c k p r e s s u r e c o m p e n s a t i o n .

A flow hood with a digital readout is a little slower, but it h a s exc e l l e n t a c c u r a c y a n d i t c a n h ave b a c k p re ss u re

compensation. F l ow m e te rs a re ex p e n s i ve a n d s h o u l d b e t re a te d as t h e f i n e precision instruments that they are.

REVIEW 1. T h e o r i f i c e a t t h e e n d o f t h e p i to t t u b e s e n s e s p re s s u re . 2. T h e o r i f i c e a t t h e s i d e o f t h e p i to t t u b e s e n s e s p re ss u re . 3. F o r ve r y a c c u r a t e re a d i n g s t h e

manometer is

used.

4. A

g a g e is often used to indicate pressure on

two sides of an HVAC component. 5. To t a ke a p i to t t u b e t r ave rs e o n a re c t a n g u l a r d u c t , t h e cross sec tion of the duc t is divided into a grid of

approximately

s q u a re s .

106

6. A p i to t t u b e t r ave rs e u s u a l l y s h o u l d n o t h ave l e ss t h a n nor more than p o i n ts .

7. For

15" diameter duc t, give the distance the pitot tube

m u s t b e i n s e r te d i n t h e d u c t fo r e a c h o f t h e fo l l ow i n g grid numbers. (Refer to Fig. 8. Answers to nearest ⅛".)

#1. #7.

#10.

#11. #15. #20. 8. T h e b e s t i n s t r u m e n t fo r m e a s u r i n g a i r f l ow a t d u c t outlets is a

9. A pitot tube traverse on a 36" x 18 duct results in the following VP readings. Calculate the velocity for each o f t h e s e V P re a d i n g s u s i n g t h e e q u a t i o n

V = 4005x/VP. Round answers to the nearest whole n u m b e r.

#1. 0.32" wg #2. 0.33" wg # 3.0.3 2 " Wg #4.0.3 4 " wg # 5.0.3 2 " wg # 6. 0.2 9 " Wg

#7.0.32" wg #8. 0.34" wg #9.0.42" Wg #10. 0.47" W g

#11. 0.36" Wg

# 1 2.0.2 9" Wg #13. 0.30" wg #14. 0.30" wg # 1 5.0.2 9" Wg #16. 0.32" wg

#17.0.29" wg #18.0.27" wg

107

10. What is the velocity in FPM for this traverse (to the n e a re s t w h o l e n u m b e r ) ?

11. Based on the velocity in item 10, what is the CFM of the duct at this traverse point? (Round off to the nearest 5.)

108

REVIEW ANSWERS

Chapter 1

1. Heating, ventilating, and air conditioning 2. Central air handling system Boiler or furnace Refrigeration unit Duct system

3. Through ductwork

4. A. Return air B. Exhaust air C. Supply air D. Outside air 5. To heat or cool the air 6. By a heating coil 7. T h e c h i l l e d wa t e r c o i l c a u s e s t h e h e a t t o t r a n s fe r f ro m

the air to the coil.

8. 14.7 psi 9. B e c a u s e d u c t p re ss u re s a re ve r y low. If measured in psi

the number would be very small and inconvenient to use.

10. Water gage

109

11. It makes a difference of 1" between the low side and hi g h s id e of t he U t u b e . In ot her word s, it rais es t he wa t e r l e ve l ½ " o n o n e s i d e a n d l owe r s t h e wa t e r l e ve l

1" on the other.

Chapter 2

1. 216 sq. in. 2. 0.83 sq. ft. 3. 7.94 sq. ft.

4. 12 sq. ft. 5. 20" 6. 26" x 14"

7. 14" x 8" 8. 22" diameter

9. A. 30" x 12" B. 12" x 6" C. 18" x 14"

D. 18" diameter

110 Chapter 3

1. Cubic feet per minute 2. Feet per minute 3. 2 0 0 0 C F M 4. 2000 FPM 5. 4000 CFM

6. 1200 FPM 7. 24" x 18" 8. 24" diameter 9. 4090 CFM 1 0. 2 75 0 F P M 11. 20" diameter 12. 20" x 14" ( 2 5 0 0 ÷ 1 3 0 0 x 1 4 4 = 2 76.9 2 3 s q . i n .

276.923 + 14" = 19.78" If you took the answer of 20" from item 11 and c h a n g e d i t to a re c t a n g u l a r d u c t , t h e a n swe r wo u l d b e 2 2.4 4 ". H oweve r t h e 2 0" d i a m e t e r wa s t h e re s u l t o f rounding off so using that figure is not as accurate as

reworking the whole problem.)

Chapter 4

1. A 2. F

111

3. D 4. E,G 5. B 6. C 7. At the fan outlet 8. Decrease 9. 0.25" wg

Chapter 5

1. 18 2. B 3. E

4. 5. C

6. H 7. J 8. F

9. A 10. K

11. D

112 Chapter 6 1. equal friction

2. 100 3. width, height

4. 1 :1

5. 3 1 6. 12" x 12" 7. 470 CFM 8. 10" dia 9. 0.06" wg 10. 28" dia

11. 0.14" wg 12. 1600 FPM 13. 14" 14. 36" 15. 26" x 26"

Chapter 7

1. friction

:

2. dynamic

113

3. 0.24" 4. 0.058"

5. A. 0.1515"wg B. R = 4.5" S = 3.5"

C. 0.23 D. 0.25" wg

E. 0.0575"

F. 1.83 G. 30° H. 0.028" I. 0.237" wg

Chapter 8

1. choked 2. friction

3. 5% 4. Can be in any order: Fan noise Turbulence noise E xc e ss i ve a i r ve l o c i t y

114

5. Can be in any order: Keep air velocity low. Ke e p d y n a m i c l o ss i n f i t t i n g s l ow. Keep outlets at least 3 feet away from duc t fittings. Use acoustic lining.

6. 45°

7. ¼, 4" 8. Reduce 9. Increase

Chapter 9

1. total 2. static

3. inclined 4. M a g n e h e l i c 5. 6" x 6"

6. 16, 64 7. #1.⅝" #7. 3⅝" # 1 0. 1 4 ⅝ "

#11.⅝" #15.5⅛"

# 2 0. 1 4 ⅝ "

8. flow hood

115

9. #1. 2266 FPM

#2.2301 FPM #3.2266 FPM #4.2335 FPM #5.2266 FPM #6. 2157 FPM # 7.2 2 6 6 F P M #8. 2335 FPM # 9. 2 5 9 6 F P M

#10. 2746 FPM #11. 2403 FPM

#12.2157 FPM # 1 3. 2 1 9 4 F P M

#14.2194 FPM #15.2157 FPM #16. 2266 FPM

#17.2157 FPM #18.2081 FPM 10. 2286 FPM

11. 10,285 CFM [CFM = Area x Velocity CFM = 36" x 18". • × 144

2286

C F M = 10,287 (round to 10,285)l

116

APPENDIX EQUATIONS FOR DUCT SIZES F i n d t h e a re a o f re c t a n g u l a r d u c t i n s q u a re i n c h e s :

Area = Width (in.) x Height (in.) Find the area of rec tangular duc t in square feet:

Area (sq. ft.)

= Width (in.) x Height (in.) 144

C h a n g e s q u a re i n c h e s to s q u a re fe e t sq.

ft.

=

sq.

in.

144

C h a n g e s q u a re fe e t to s q u a re i n c h e s :

sq. in. = sq. ft. x 144 Find one side of a duc t if the area and another side is

known:

Width =

Area

Height

F i n d t h e a re a o f ro u n d d u c t :

Area = T x Radius2

117 F i n d t h e r a d i u s o f ro u n d d u c t i f t h e a re a i s k n ow n :

Radius

=

Area

Find aspect ratio: Width (long side) Aspect ratio =

Height (short side)

EQUATIONS FOR AIR QUANTITY AND VELOCITY Find air quantity:

Quantity = Area X Velocity

Find air velocity:

Quantity Velocity = Area

Find duct area if air quantity and velocity are known: A re a =

Quantity Velocity

C o nve r t ve l o c i t y p re ss u re t o ve l o c i t y :

Velocity = 4005 x /Velocity Pressure

118

EQUATIONS FOR PRESSURE Estimate pressure loss for straight duct:

Loss = "wg per 100 feet x Length of duc t (in feet) 100

C a l c u l a t e p re s s u re i n d u c t :

Total pressure = Static pressure + Velocity pressure Velocity pressure = To t a l p re s s u re Static pressure

119

INDEX A

Air conditioning, 1-2 A i r f l ow p a t t e r n s , 5 0 - 5 7 A i r l e a ka g e , 8 7 - 8 8 Air p re ss u re , 6 m e a s u re d i n p s i , 6 m e a s u re d i n " wg ( i n c h e s wa te r g a g e ) , 7 A i r q u a n t i t y, 2 6 - 3 7

Air velocity, 26-37 Air volume, See Air quantity A n a l o g a n d d i g i t a l re a d i n g s , 9 8 Angle of change, 93 Area of duc t (cross sec tion), 12-22, 30-31

rectangular duct, 12-17 round duct, 19-22 ASHRAE, Handbook of Fundamentals, 79, 64 As p e c t r a t i o, 5 9 At m o s p h e r i c p re ss u re , 6, 3 8 - 3 9

B

Boiler, 5

C

Calculating duct size, 11-25 c ro s s - s e c t i o n a l a re a , 1 2 - 1 7

duct side, 17-18 round duct, 19-22 C e n t r a l a i r h a n d l i n g sys te m , 2 - 6

Chilled water coil, 5 Chiller, 5-6 C h o ke d f i t t i n g , 8 5 Coefficients for fittings, 74-79, 95 C o o l i n g syste m , 5 - 6

D

D i f fe re n t i a l p re ss u re re a d i n g s , 1 0 0 D i g i t a l a n d a n a l o g re a d i n g s , 9 8 Dog-leg offset, 85

120

Duct calculator, 67 Duct connectors, 87 Duct design, 49 Duc t fitting design, 83-96 Duct fittings, 49-57 definition, 49 Duc t fittings, dynamic losses, 74-80 D u c t l i n i n g ro u g h n e ss , 8 6 - 8 7 D u c t s i ze , S e e C a l c u l a t i n g d u c t s i ze D y n a m i c l o s s , 3 8, 4 4 - 4 7, 4 9 - 5 7, 7 3, 8 4

LU

Elbows, 51-54 E q u a l f r i c t i o n l o ss m e t h o d , 5 8 - 6 7

SMACNA chart, 62 Equations, 116-117 Equivalent duc t lengths for calculating pressure loss, 74 E q u i va l e n t d u c t s i ze s c h a r t , A S H R A E , 6 4 Ex h au st air, 3 Extended plenum method, 68

1

Fan system effec t, 72 F l ex i b l e d u c t , 86 -8 7

Flow hood, 104-105 Friction, 7 Fric tion loss, 38, 43, 58-61, 73, 83-84

H

H e a t l o ss , 8 8 Heating coil, 4-5 Heating system, 4-5

IAQ (indoor air quality), 97 I n c l u d e d a n g l e , f i t t i n g , 7 7, 9 3

Laminar flow, 50

M

Magnehelic gages, 100

Manometer, 8, 99-100 Math, 11 Measuring airflow, 97-107 a t o u t l e ts , 1 0 3 - 1 0 5

121

Mixed air, 4

N

O P

N o i s e , H VAC s y s t e m , 8 8 - 8 9, 9 4 Offsets, 54-56

Outside air, 3 Pitot tube, 98-103 Pitot tube traverse

rectangular duct, 101-102 round duct, 102-103 Pressure difference, 7 Pressure in a duc t, 38-48 Pressure loss, calculating, 71-82 P re ss u re l o ss c o m p u te r p ro g r a m s , 79

R

Re s i s t a n c e i n d u c t , 7, 4 3

Return air, 3 Reynolds number, 50 Rotating vane anemometer, 104 Round duct, area and diameter, 19-22 air velocity and air quantity, 31-35

S

S offset, 85

S i z i n g d u c t wo r k , 5 8 - 7 0

SMACNA, HVAC System Duct Design, 62, 76-77, 79 S p l i t te r va n e , 9 1 - 9 3 S q u a re i n c h e s a n d s q u a re fe e t , 1 4 - 1 7 Static pressure, 38, 40-42, 44, 46, 58, 98-99 Static regain method, 68

Supply air, 3

T

TA B ( te st i n g , a d j u st i n g , a n d b a l a n c i n g ) , 97

Take-off, 94

Total pressure, 38, 42, 98-99 To t a l p re ss u re l o ss , 7 1 - 7 2

Transitions, 54, 93-94 Turbulent airflow, 50

122

Tu r n i n g va n e s , 5 3 - 5 4, 9 0 l e a d i n g e d g e a n d l e av i n g e d g e , 9 1 rails , 90

C V

U -t u b e , 8 - 9 Ve l o c i t y a n d ve l o c i t y p re ss u re , c o nve rs i o n c h a r t , 76 Velocity pressure, 38, 40-42, 44, 46

Velometer, 104 Ventilation, -

W

Water gage, 8

AIRFLOW IN DUCTS ANOTHER BOOK FROM THE INDOOR ENVIRONMENT TECHNICIANS'S LIBRARY

INDOOR ENVIRONMENT

TECHNICIAN'S LIBRARY

When you complete this book, you will understand air flow in duc t better than most p e o p l e i n t h e H VA C i n d u s t r y d o . Yo u w i l l b e a b l e t o f a b r i c a t e a n d i n s t a l l d u c t w o r k m o r e effectively and make changes in ductwork size with confidence. Learn how duct fittings a f f e c t a i r f l o w. U n d e r s t a n d y o u r a i r d e l i v e r y s y s t e m a n d w h a t h a p p e n s i n s i d e t h e d u c t .

U n d e r s t a n d s t a t i c p re s s u re , ve l o c i t y p re s s u re , t o t a l p re s s u re , d y n a m i c l o s s , a n d friction losses-in practical terms that you can use on the job.

"Airflow in Ducts is a very valuable book for our students at J & J Air Conditioning. found the book very concise and clear for technician level students and a great help teaching the subject. Although fluid dynamics (including airflow) is a complicated subject, this training book presents the material in such a way that persons who do not normally use such technology were interested and excited to apply it in their work. Ke n n e t h M a r t i n , P E

For over 30 years Leo A. Meyer has dedicated his career to writing and publishing quality training materials for the

H VA C i n d u s t r y. H i s b a c k g r o u n d a s s h e e t m e t a l j o u r n e y m a n , fo r e m a n , i n s t r u c t o r, t e a c h e r t r a i n e r, a n d w r i t e r m a ke h i m t h e fo re m o s t a u t h o r o f p r a c t i c a l , d ow n -t o - e a r t h H VAC t r a i n i n g

m a t e r i a l fo r a l l a s p e c t s o f t h e i n d u s t r y.

A practical manual for anyone who works with HVAC system, including those in installation, service work, energy management, indoor air quality o r TA B ( t e s t i n g , a d j u s t i n g a n d b a l a n c i n g ) . ISBN 9780880690188

G E T M O R E H VA C B O O K S BY PHONE OR ONLINE 1-888-452-6244 :

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