Solutions manual to accompany Principles of extractive metallurgy 9780070538481, 0070538484

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Technical hlformation Center & Library National Mine Health & Safety Academy 1301 Airport Road Beaver, WV 25813-9426 (304) 256-3266 -.,,,..-~------r-----4

TN 665 .R698 1974

Terkel Rosenqvist Professor of Extractive Metallurgy University of Trondheim, Norway

McGraw-Hill Book Company New York

St. Louis San Francisco Dusseldorf Johannesburg Kuala Lumpur London Mexico Montreal New Delhi Panama Paris Sao Paulo Singapore Sydney Tokyo Toronto

"

,i

Solutions Manual to accompany PRINCl,PLES ·OF EXTRACTIVE METALLURGY

Copyright @ 1974 by McGraw-Hill, Inc. All rights reserved. Printed in the United States of America. The contents, or parts thereof, may be reproduced for use with PRINCIPLES OF EXTRACTIVE METALLURGY, by Terkel Rosenqvist provided such reproductions bear copyright notice, but may not be reproduced in any form for any other purpose without permission of the publisher. 0-07-053848-4 1234567890WHWH7987654

P RE F A C E.

.1

;•.,.,,.,,1

This manual gives the answers to problems ~nsted at the end of each chapter in the text "Principles of Extractive Metallurgy".

In

order to save space complete derivation of the answers have not been given.

In cases where the procedure is obvious, the manual gives

only the final answer .

In other cases a few steps in the derivation

are indicated . Before assigning a problem, the instructor should familiarise himself with it , and make sure that he himself can solve it .

Some

problems are a bit tricky, and involve some intricate thinking .

On

the other hand, the problems may serve as good starting points for further discussions on the topics involved . Owing to difficulties in accurate reading of thermodynamic graphs and inaccuracy in arithemetic calculations various students may get slightly different numerical answers, without this being regarded as errors .

In a few cases the problems as given contain

too much or too little information .

These cases are pointed out in

the manual , and in the latter cases the missing information is estimated .

A few misprints that occur in the first printing of the

text are also pointed out .

Although the author has used great care

in solving the problems he can not guarantee that mistakes do not occur, and he should be grateful for information from instructors either about errors or misprints in the problems or in the solution manual.

Solutions to Problems. 1-1

19.5 at.% Cr,

1-2

33.2 wt.% 02,

1-3

Principles of Extractive Metallurgy.

7.52 at.% Ni,

73.33 at.% Fe .

75.4 wt.% N2' 1.4 wt.% Ar. 3 Volume of furnace gas= 2.45 Nm with composition (wet) = 18.8 % CO , 3.2 % 02' 10.2 % H o, 2 2 CO , 3.5 % 02' 75.5 % N2. 2

3 Theoretical air requirement = 2447 Nm. b) Actual air 3 requirement= 6070 Nm. c) 6.3 % so , 80.95 % N2, 12.85 % 02. 2 3 d) Volume of roast gas at 500°C = 16 800 m

1-4

a)

1-5

a)

35.3 g sulfur.

that n 1-6

a)

1. 22 mole% s2, increases on heating).

tot

b)

1-8

Pg

= 8.36 mm Hg (Notice 2

117 .6 kg hot metal,

127.6 kg slag with 40.35 wt% Sio , 2 3 43.95 wt.% CaO, b) 369 . 5 Nm furnace gas

16.70 wt% Al o , 2 3 with 25.2 mole% co, 1-7

67.8 % N , (dry) = 21.0 % 2

a)

X

c)

n

a)

0

17.9 mole% CO , 2

%, a~ 0.122 %, b) 4 2 (0.122 X 2.6/0.05) = 40

= 1.00 ~

s

0.039

~

~

0 . 04 %

s 0.173 %. The difference between the two means is 0.33 % with a preeision ' 2 27 s = /(0.125) + (0.173) = 0.21 %. Thus the second ore is ~

0.25 %,

s

56.9 mole% N2.

~

0.125 %,

b)

likely to be richer than the first one. values for

0 1S

0

~

c)

0.35,

~

The estimated

less reliable because the number of samples

less .than in Problem 1-7.

1

1S

2-1

a)

6H

3750 cal/mole independent of path.

b)

0 to 500°c at 1 atm:

q

1 to 10 atm at 500°:

q = -3536.5 cal, w = -3536.5 cal

Total, path 1

q =

=

0

= 3750

cal, w =

992

cal

213.5 cal, w= -254:4, 5 cal

1 to 10 atm at o c :

q = -1250

cal, w = -1250

cal

0 to 500°c at 10 atm:

q =

3750

cal, w =

992

cal

Total, path 2

q =

2500

cal, w =

-258

cal

In both cases 6E = q-w = 2758 cal and 6H

=

2758 + 992 =

3750 cal/mole. c)

w = 388 cal,

q = 388 - 2758 = -3270 cal/mole.

2-2

H - H 298 473

= 16 381 cal/mole 1 . 2

2-3

H - H = 19 236 cal/mole Fe. 1873 298

2-4

a)

=

+ 2 o = CO + 2H 0(g) : tH = -191.6 kcal 2 2 2 298 4 Stoichiometric air: adiabatic temp.~ 2040°c, 20 % air

CH b)

excess: adiabatic temp.~ 1780°c. c)

Available heat above 1600°C:

Stoichiometric air= 46 kcal,

20 % air excess= 23 kcal. d)

Stoichiometric air:

= 118

heat above 1600°C temp.~ 2470°c, e)

adiabatic temp.~ 2700°C, available kcal.

20 % air excess:

available heat above 1600°C

Stoichiometric amount of oxygen enriched air: 0

•

0

temp.~ 2600 C, avairable heat above 1600 C = 20 % excess of oxygen enriched air: 0

2300 C, 2-5

a)

tE

b)

6H

298

0

kcal,

=

109 kcal.

adiabatic 84 kcal.

adiabatic temp.~

available heat above 1600 C =

= -400

adiabatic

67 kcal.

= -401 kcal/mole Al o 298 2 3 = [tV-T(o6V/oT)p]49 = 0.007 kcal

tH

- 6H 50 atm. 1 atm which is insignificant compared to the precision given for tH50 atm• 2

2-6

0

a)

Heat capacity of calorimeter: 16 cal/ C.

b)

6H of alloy formation

= -9.0 cal, refers to o

0

c =

273°K

and 0.01 mole of alloy.

= Au(diss): 6H523

Au(s)

c)

=

-5970 cal/mole

Ag(diss): 6 N523 = +3800 cal/mole (Au, Ag)(s) = Au, Ag(diss): 6H 523 = -1330 cal/mole

Ag(s)

=

6 H298 of alloy formation 2-7

a) b)

=

-904.6 cal/mole.

ZnS + 1,5 02 = ZnO + S0 2 6 0 6 0 -105.95 kcal, = Hll73 H298 ZnS + 2.25 02 + 8.46 N (25°C) 2 Heat surplus

2-8

=

=

-106.55 kcal

=

ZnO + so

+ 0.75 2 8.46 N (900°C) 2

0

2

+

23.80 kcal, -130 kcal

=

As the curves in Appendix B do not go beyond the melting points of Al o and Cr, the adiabatic temperature can not be calculated, 2 3 but can be estimated to be above the melting point of Al o 2 3 (2035°C). 2-9

Al o + Si0 = Al Si0 (andJ: 6H = -1.99 kcal 2 2 969 2 3 5 3 Al o + 2 Sio = Al si o (mull.): 6H = +5,44 kcal 2 3 2 4 2 13 968 2 Al Si0 (and,) + Al o = Al Si o (mull.):6H = +9.42 kcal 2 5 2 3 6 2 13 968

3-1

8

3-2

6S(Zn)

3-3

3-4

0 473

0

8

= =

29s

=

38. 5 eu,

0 S 473

-2.52 eu, 6S (cont.)

=

=

66.4 eu

+2.59 eu,6S(overall) = +0.07eu

3 = 10.4•10-3 cal/atm, Sn(.ll,): 6Vf = 0.43 cm 30 Tf(2 atm) - Tf (1 atm) = 5.2·lo- c 0 a) 68 -37.9 eu. b) Reversible: q = -11.3 kcal, 298 = w' = +57 kcal. Irreversible: q = -68.3 kcal. For reversible: 0 q/T = -37.9 eu = Mo. For irreversibel q/T = -229 eu < 6S Sn(s)

=

in agreement with 2. law.

For both cases 6H

-68,3 kcal in agreement with 1. law. 3

=

q - w'

=

-3-5

Fe(a.)

=

b.C

(lnT-ln 1183) = -3 -1.45•10 cal/atm. b.G 215 - 1.45•10- 3 (P-1) + 3-6

C(graph) = C(diam): b.V a)

At 1000°K:

At 2000°K: b) 3-7

P

0

0 = -1.8 cal/ C. b.S = b.S1183 + b.C p 3 p 12.92 - 1.8 ln T eu. b.V = -0.06 cm =

Fe(y):

P =

0

b.V(P-1) - Tb.S + 1183 b.S1183 1.8 T ln T - 12.92 T. =

=

-1.91 ml

=

=

-46.3•10- 3 cal/atm.

= (1600/46.3)10 3 ~ 35 000 atm. 3 (2700/46.3)10 ~ 60 000 atm.

High temperature is used for kinetic reasons. 0

= 29330 cal/mole, T = 1153°K, Pzn(693 K) = a) b.H V V 4 2.o· 10- atm. b) b.Go = · 30915 - 27.8 T cal, log Pzn(s)= sub. - 6755/T + 6.08. c) The heat capacity of solid, liquid and

gas are assumed equal. 3-8

0 = -2.83 cal/ C, b.Go = 34512 + p 2.83 T ln T - 45.53 T, log pMg = -7543/T - 1. 425 log T + 9.95. By approximations the boiling point is found to be 0 1377°K = 1104 °c with b.H = 30615 cal and b.So = 22.23 eu.

Mg(£)

=

Mg(g):

b.C

V

V

3-9a)

00

C

p

0

~

co N

r-r-1

0

500

1000

4

1500

3-9

b)

0

c)

500

1000

1500

s

-

--==-

Lrn°

·-

-

T

/

!:iC

0

2000°K

p

is assumed~ 0

500

1000

5

1500

M !::,Go

--3-10

b)

23 15 9 Po atm = 3•lo- (300°Cl, 2·10~ (500°C), 3•10- (800°C). 7 21 13 2 P = 1: % 0 2 3•10- (300°C) 2·10- (500°C) 3·10tot ·' •' = 10: % 0 2 = 3·10- 22 (300°C) ' 2•10- 14 ( 800°) P ' tot 8 (500°C), 3·10- (800°).

c)

500°c is used because it combines a low

a)

o2

pressure with

a high reaction rate. 3-11

a)

b)

3-12

o + 3 H2 = 2 Cr+ 3 H20: K 2 3 2.7°10- 9 % H o = 0.14 for both 2 0 6H = + 90 kcal, 6S 0 = +22 eu. Cr

2 CO: 6G

0

973 By approximations are found:

o,

K

a)

=

F

l(pressure or temperature),

gas composition), c) F

=

3-14

a)

=

=

b) F

2(temp. and pressure)

(pressure or temperature), f) F

85 % co, tot 62 % CO, 38 % CO 2 , for ptot =

for p

15 % co , for p = 1 atm: tot 2 10 atm: 27 % CO, 73 % CO . 2 3-13

1

=

e) F

=

2(temp. and d) F

=

1

l(pressure or temperature)

O(non-variant).

All phases may coexist, giving F

1.

=

b)

For fixed

4: F 2 which 4. c) For P max may be any two of temp., pressure, or alloy composition. temp. and pressure:

3-15

P

0

= A1 si o 3 Al Sio + Si0 : 6G = +11 4410 - 7,3,Tcal 2 2 5 6 2 13 0 = A1 Si o 2 Al Si0 + A1 o : 6G = - 9 420 + 7.4•Tcal 2 6 2 13 5 2 3 0 For reaction (1) 6G = 0 at 1564°K and andalusite is unstable

a)

above this temperature.

For reaction (2) 6G

0

= 0 at 1272°K

and mullite is unstable below this temperature.

6

3-15

b) Si0

-

1600

2

+ Mull.

-

1564°K Mull. +

-

1400

And. + Mull. Si0

2 And.

And, + A1

c)

-

50

-

o

2 3

a

45

o

2 3

1272°K

....

1200

A1

+

55 Mole% Al 203

I

60

65

The temp. for coexistence between andalusite, mullite and

silica increases with increasing pressure. 3-16

B + ½ N2 = BN: b.G 10- 3 atm at 2185°K

0

=

-60 000 + 20.6 T cal pN becomes 1 2

= B4c + 2 N2 : 6G 0 = +226 000 - 82,0 T cal)pN -3 0 2 becomes 10 atm at 2065 K. 4 BN + C

4-1

4-2

NZn

0

0.231

0.484

0.495

0.748

1.000

azn

0

0.312

0.581

0.596

o. 787

1.000

'Yzn

::::1.5

1.35

1.20

1.20

1.05

1.00

0.1

0.3

0.5

0.7

0.9

6SA 1 (eu)

4.26

3.47

2.80

1.30

0.28

aAl (900) 0 yA 1 (900)

0,0061

0.104

0.460

0.722

0.906

0,061

0.346

0.920

1.030

1.007

0

aAl (1100 )

0.0093

0.112

0.420

0.688

0.901

0

0,093

0.373

0.841

0.983

1.001

NAl 0

yA 1 (1100)

Aluminium obeys approximately Raoult's Law at NAl > 0.7. 7

4-2

4-3

(cont.,) Silver obeys Henry's Law in the same range. NPb l'iG(cal)

0.1

0.3

0.5

0.7

0.9

-540

-945

-1010

-840

-520

-600

-1150

-1750

-2500

0.74

0.56

0.41

0.28

-1750

-900

-450

-300

0.41

0.64

0.80

0.86

l'iGAg(cal) -220 0.90 aA_g l'iGPb(cal) -3400 0.18 aPb

The value for l'iG and a are approximate. 4-4

[oln a/o(l/T)]

4-5

[oln a/oP)T

4-6

By means of Eq.(4-8a) is derived: l'iHA

4-7

azn

4-8

a)

=

l'iH/R

=

p

0

=

l'iV/RT < 0

=

75/750

2 aNB and l'iHB

=

=

2 aNA

0.1

=

1.0 800°C

t l000°C

0

0.01

0.5

1.0

1.5

2.0

wt.% C 2

b)

o.5 % c:

c)

For Pco + Pco = 0.2 atm: 95 % Co, 5 % CO2 2 o C(graph) = C(l %) : tG = +340 cal, l'ic 01273= ... 1020 cal. 1073 At graphite saturation: aC(800°C) = 0.85, a~(l000°C) = 1.5

ac

=

o.~9, (Pco) /pco

8

=

3.5.

4-8

d)

6G~

will increase, 6G~ will become more negative, 073 273 a~(800°C) will decrease, a~(l000°C) will increase.

e)

Silicon will decrease the solubility of graphite in austenite

f)

Solubility of diamond in austenite will be about 3.0 % at l000°C Solution:

4-9

a

a)

NaCl(0.9)

aNCl(0.5) b) 4-10

NaCl-KBr

NaCl-KCl

NaCl-CaBr 2

0.9

0.81

0.736

0.5

0.25

0.167

6T(0.9) NSi

aAg

0.0000

1.000

0.530

0.484

0.0238

0.987

0.633

0.410

0.0546

0.933

0.736

0.343

0.111

0.835

0.843

0.251

0.292

0.663

0.928

0.113

0.358

0.590

The Ag and Si activites are shown below, curves A and B, together with those of Probl. 4-11. 4-11

% Si

aAg(T)

aAg(1773)

1.0

1.000

1.0

% Si

aSi(T)

asi(l773)

1.000

5.9

0.328

0.285

0.962

0.962

6.1

0.362

0.311

2.1

0.922

0.926

8.1

0.455

0.401

3.0

0.881 aSi(T)

0.886 aSi(l773)

11.8

0.604

0.549

23.1

0.750

0.721

3.0

0.153

0.133

43.3

0.839

0.830

3.5

0.176

0.154

64.4

0.906

0.905

4.0

0.203

0.179

100.0

1.000

1.000

4.8

0.240

0.212

9

4-11 (cont.) 1.0

0.8

0.6

0.4

0.2

Ag

A and C:

0.6

0.4

0.2

Si activities,

A and B from Problem 4-10,

5-1

a)

b)

c)

Band D:

0.8

Ag activities

C and D from Problem 4-11.

K = k 1 /k 2 = 790 app exp(-6G /RT) ~ 500 CH = c1 = 15.25•10-6 mole/cm 3 initially. 6 2• 2 At equi·1·b i rium: CHI = 28 ·l 0- , 3 mole/cm . 3 17 dCH /d0 = l.465•10mole/cm sec, initially. 1 -17 3 half has reacted: 0.364•10 mole/cm sec.

H + 1 2 2 Kcalc =

=

2 HI:

Si

o

10

1.25,10

After

-6

5-1

d)

CHI(equil) 30

20

'°....,0

.

u

10

e+ e)

30

'°....,0

20

.

CH

u

2

10

e+ 5-2

300°K kl 500°K kl

5-3

2.06·10

-24

3

mole/cm sec. -13 2 1.37·10 mole/cm sec.

= =

3 a) H = 2 H(diss): CH = 165 ~ cm /100 g Pd. 2 3 At 0.1 atm: C = 51.8 cm 3 /100 g Pct2 = 6.16 cm3 /cm, at H 3 3 3 1 atm: CH = 164 cm /100 g Pd = 19.5 cm /cm Pd. 5 3 2 b) Flux = 3·8•10- (19.5-6.16)•3600/0.01 = 182.5 cm /cm h (In some copies of the text the diffusivity is erroneously

11

5-3 (cont.) given to 3.8·10-

3

2 cm /sec, which will make the answer 100 times

larger), 5-4

Flux

=

3 597.2 cm /h

5-5

e e b) e

=

1 h:

=

100 h: s/M =

=

1 h: C

a)

=

s/M =

1.8 %/h

= 3, (C-C 0 )/(C S -C) 0

=

0.3, (C-C )/(C -C ) 0

0.54 %,

0

S

0

0.04, C = 0.06 % =

0.75, C

= 1.12%

= 100 h: C = 1.25 %. The

density of iron is unnecessary. 5-6

e e b) e e a)

0.066, (C -C )/(C -C) = 0.15, Cm = 0.2% m o s o = 100 h: MIL = 0.66,(C -C )/(C -C) = 0.92, C = 1.4 % m o s o m = 1 h: MIL = 0.33, ( C -c ) / (C -C ) = 0.6, C = 0.9 % m o s o = 100 h: MIL = 3,3, ( C -c ) / (C -c ) 1.0, C 1.5 % m o s o (In some copies of the text the abscissa scale of Fig 5-5b =

1 h: M/L

=

~

~

is distorted.) 5-7

FeO + H2

=

Fe+ H o, K = p /pH (out) 2 H20 2 = K/(1 + K)•dnH (in)/d0

a) d~e /d0 b) "Rate" l.S independent of Ehe lump size or of the amount of wustite. c) I

I

'Fe

I

Fe I+ I FeO IFeO I

% Fe

% FeO

The reaction zone will move towards the right with time.

12

5-8

CO

+ C

2

-dn C

=

2

=

2 lip > T pa' lip e pe a

6-15

a)

e

7-1

a)

1

b)

"Theoretical" energy

b) c)

5 h. b) Cu/Fe m -4 1. 35 • 10 (after 10 hr) =

mm+

0.05 mm:

E

42.2 kWh/1000 kg

=

= 1 cm and ~ 1 4 3.8·10 erg/g ~ 10 10- 3 12 ~ 0.05 %

For d =

~ 7-2

3 mesh screen screen

=

10- 3 ( imrnedia tely) and

=

1.75 cr 6/p (1/d -l/d ). 1 2 = 0.1 cm: Theoretical energy

3 kWh/ton

2

"Theoretical" efficiency

0.067 m, 6 mesh screen

=

2

0.38 m, 20 mesh screen

=

2 0.38 m, 10 mesh

2 (In some = 0.023 m.

copies of the text the screen analysis is erroneously given as +3 -6 mesh, etc., instead of -3 +6 mesh, etc.).

16

7-3 ...,1""""""0.-----~-------.-----....------.------,------..--=----+

80

60 oO

i::::

•..-1

(I) (I)

40

Cd

p..

§

u

s..g

- 20

-1

-1.2

-0.8

-0.6

-0.4

-0.2

0

0.2

log diam (nnn}

7-4

a) b)

Stokes' range: dl/d2 = 1.47, Newton 1 s range: d/d 2 Stokes' range: d/d 1. 55, Newton's range; d /d = 2 1 2

7-5

2.9 kg ferrosilicon to 1 kg water

7-6

a)

= 2.40.

2 Permeability~ 0.075 (liter) /(min,atm) for all three time intervals. at 1 atm

7-7

= 2.16.

b) Time for 10 liter ~

~

222 min.

c} 10 liter

666 min.

= 6 m3 /sec. (at l00°C) = 4.4 Nm3 /sec.

a)

Volume flow

b)

d.

d)

On the moon the gravitational force is much less than on

min

= 49 µm.

c) Ford

=

20 µm:

t

=

120 m.

Earth, which would make the answers under (b) and (c} correspondingly larger.

17

7-8

a)

w = 0.138 m,

b)

From average fractional retentions for the -5 µm, +5 .,..10 um

= 2.63 m/sec.

0.276 m, y

=

H

fractions the percentage of retained dust is found and with approximate screen analyses:

5 %

- 5 um,

~

60,

5 %

+5 -10 µm, 23 % +10 -20 µm, and 67 % +20 µm.

8-1

a)

Calorific power

=

7750 kcal/kg (net}, 8030 kcal/kg (gross)

b)

Molar composition of combustion gas:

66.5 CO , 26.6 H o, 2 2 0.4 so , 22 o , 415 N + 78 gash (c ~ 0.2 kcal/kg°C). 2 2 2 By trial and error the adiabatic flame temperature is found ~ l 700°c.

8-2

3 3 Volume of air= 4.44 Nm, of gas= 5.37 Nm /kg C. power of gas= 5630 kcal/kg carbon, of carbon=

Calorific 8070 kcal/kg.

The difference represents heat losses from the reactor and physical heat in the producer gas. 8-3

Molar ratio of o ~

8-4

2

to CO

1/1.

2

Effluent gas temperature

700°C. 1000

1000

I

I 1

oc

Caco

3

CaO

oc

1 CaC0 3 Temperature I

~

900 860

~

------

- -

- - -

-

·-·

1.0 0.5

0.5 atm

920 900

Pco

2

atm

I

l (a) (b) In case (b) the interface temperature (920°) and pressure

(1 atm) are chosen to show that chemical equilibrium does not prevail. At 920°C the equilibrium decomposition pressure is about 1.5 atm. 18

8-5

0

0 Fe (so ) : t ~ 720 C, CuS0 : t 840°c, 820 C, Znso : t 2 4 3 4 4 CoS0 : t ~ 930°c. 4 For Cuso and Znso the values refer to decomposition to form 4 4 basic sulphates. For decomposition of the basic sulphates

higher temperatures are needed.

The procedure used was to

calculate the equilibrium so

pressure at various temperatures. 3 Inserted in the equilibrium constant for the reaction so = 3 so + ! o , and remembering that p = 2 p 0 , the partial 80 2 2 pressures of so and o are calculat~d. The total pressure is 2 2 plottet as function of temperature, and the temperature for 1 atm is derived.

ptot 8-6

pressure for the reaction cu s + 2 cu o = 6 Cu+ so 2 2 2 2 0 reaches o.21 atm at about 570 C, corresponding to the reaction

The so

Cu s + o (air) = 2 2 that temperature. 8-7

Pso

=

3 8-8

4•10-11,

Pso

5·10

=

2

+ 3.76 N being possible above 2

-10

atm.

2

o2 (air) = ZnO + S0 2 + 0.75 % so , 7.35 % 02, 82.85 % N2 2

ZnS + 2.25 a)

9.8

b)

Heat surplus pr kg of ZnS

c)

Gross heat surplus Net heat surplus

8-9

2 Cu+ so

a)

= =

02

= 246.5 kcal

1233 Mcal/h,

Heat loss= 133 Mcal/h,

1 100 Mcal/h

From Eq.(6-29) the minimum gas velocity is: v. = 1.12 min 2 m/sec, G . = 0.39 kg/(m sec). From Eq.(6-32) the min

maximum gas velocity is: 2 kg/ (m sec). b)

Feeding rate of pyrite

c)

Height of bed

V

=

max

~

17 m/sec,

G

max

5.8

2

0.16 kg/(m sec)

d)

= 2.15 m 2 Pressure drop = 28 350 N/m

e)

By decreasing particle diameter the gas velocity, and

212 mm Hg

consequently the feeding rate, must be lowered.

Because

of increased chemical reaction rate a shorter retention

19

8-9 (cont} time will be needed, and the bed hight will be considerably lowered. 8-10

8-11

a)

Theoretical amount of air

b)

Fan energy

a)

kg oil/kg pellets

= ~p-vol 2.6 Wh/kg Fe o . 2 3

magnetite

~

~

360°c.

3

0.61 Nm /kg Fe o . 2 3 4 1.013·10 •0.93 = 9 400 N•m

0.01.

Temperature of preheated

Temperature of cooled hematite

~

670°c b)

With heat losses the oil consumption and the temperature of preheated magnetite will increase, the temperature of cooled hematite will decrease.

9-1

a}

b) c)

9-2

0

0

PbO: any temperature, Sn0 : above 600 C, FeO above 680 C, 2 Sio : above 1600°c. 2 Si may reduce Cr o at any temperature. 2 3 About 980°c. Total pressure 2 atm with 0.5 % CO , 2 49,5 % CO and 50 % Zn.

In order to solve this problem it is necessary to stipulate the effluent gas temperature.

0

If this is set at say 100 Ca

total gas requirement of 8.0 moles/kg ore is obtained, of which 2.9 moles are needed for the reduction of Fe o to Fe o , and 2 3 3 4 5.1 moles for the heating of the ore where it is burned with 4.26 moles of air.

This gives the following heat balances for

the three zones, basis 1 kg ore, from which also the stoichiometry is seen; heat losses being disregarded:

20

9-2 (cont} Zone I Preheating Innput

l

2.95 Fe 2o 3 8.8 Si0 J 2 0.89 02 3.37 N 2 1. 28 co

kcal ore 25°C

l

2.95 Fe 0

2.41 0

8.03

2

1. 28

8

0.45

3

1.28 0.51

650°c 2 H0 2 N 2 co -t.H 2 298 H 0(g) 2

Total input

0.96

6

0.51 H 2 1.13 CO 4.66

8.8

air 25°C

kcal

Output

o

l

2 3 Sio 2 CO 2 H0 2 N 2 co -Lrn

l

61 650°C

87 2

100°c

1 5 34

298

Total output

190

21 120 20 190

Zone II Reduction Input 2.95 Fe 2o 3 8.8 Si0 2 2.0 co

l

kcal 61 650°c

9 3

0.16

1

1.97 0.73 0 29

600°C 2 H0 2 N 2 Fe o -t.H298 3 4 11 CO 2 11 H 0(g) 2

Total input

kcal

1.97 Fe o 3 4 8.8 Sio 2 1.28 co

89

0.8 H 2 0.41 CO 4.66

Output

0.51 H2 1. 13 CO

3

l

65 675°c

6 2

2

650°C

69

0.45 H o 2 4.66 N 2 2.95 Fe o -t.H 2 3 298 11 o. 73 co

17

Total output

20 526

796

21

93

8 3 21 579 19 796

9-2 (cont) Zone III Cooling Input 1.97 Fe 3o 4 8.8 Si0 2 2.0 co

l

0.8 H2 0.41 CO

kcal 65 675

1.97 Fe o 3 4 8.8 Sio 2 2.0 co

93

gas 25°C

2 0.16 H o 2 4.66 N 2

l

0.8 H2 0.41 CO

0

Total input

kcal

Output

600°c

2 0.16 Hz° 4.66 N2

158

540°C

Total output

122

36

158

The net calorific power of the gas is 182 kcal/kg ore= 182 Meal/ton ore of which 122 kcal/kg is found in the cooled ore giving this a temperature of about 540°c.

For zone II the in-

put gas temperature 600°c and consequently the output ore 0

temperature, 675 C, are

estimated, but will have no effect on

the overall heat balance. 9-3

13.4 kg of carbon consumed by direct reduction.

31.5 kg of

oxygen removed by indirect reduction. 9-4

a)

Top gas temperature will decrease,

decrease, 9-5

c)

b)

Top gas volume will

co /CO ratio in top gas will increase. 2

Increased blast temperature will give higher hearth temperature and increased Si content in the iron.

9-6

a) b)

357 g slag with 37.8 % Sio , 13.2 % Al o , 49.0 % Cao 2 2 3 Necessary carbon = 335 g, gas weight = 845 g and volume

3

0.62 Nm /kg hot metal.

22

9-6 (contl Input

kcal

1560 g ore 335 g coke 313 Caco 23.1 co

1 kg metal

j 25°C

2 1.37 CaSi0

23.1 co 610

" "

0.36 Si

27

"

11

153 16

100°c

2353

Total output

819

Deficiency is 1534 kcal

= 1785 Wh/kg hot metal.

a) b)

l

II

300

4 2 8.5 Fe o ,-,f,H298 1668 2 3 0.36 Sio 78 " 2 3. 13 ( CaCO caO +.CO ) 134 3 2

30

Total input

10-1

4.6 CO

141

3 0.88 CaiiO/

1400°C

357 g slag

0

3 -Lm298

1.5 CO

kcal

Output

"\

0.81

0.19,

the total pressure and the Cd content of the vapor will increase.

10-2

a) b)

c)

0

0.96 atm(900°C), 2.5 atm(l000°C), 5.8 atm(ll00°C) Pzn NZn = 0.52 (900°C), 0.20 (1000°C), 0.086 (ll00°C) 0 0.1 would give NZn > 1 which is At 900 and 1000 C, Yzn impossible, implying that Yzn can not be 0.1 at high Zn contents. For Yzn

0

= 0.1 would give NZn = 0.86.

At 1100, Yzn

= 10 : NZn = 0.052 (900),

0.0096 (ll00°C)

23

0.020

(l000°C),

10-3

1.0 . 900° 1000° aZn

1100° (a}

0.4

0.17 __

At.% Zn 1600

I

Zn

+

'

1400 oc 1200 (b)

1000

PZn

=

1 atm

800

Fe 10-4

a)

Zn =

0

1 atm at about 885 C which gives Pco

0.03 atm 2

b)

and Pzn ~ 0.09 atm, ptot Zinc recovery = 91.5 %

c)

Vapor losses can be decreased by working under increased pressure (and temperature).

24

1. 12 atm

10-5

a)

1.4 % ot the zinc will reoxidize.

b)

Intial condensation at about 840 C.

0

about 590°c.

99 % complete at

Note that after 99 % condensation Pzn

=

0.01

atm. 10-6

a)

Input: 1 kg sinter, Output: 186 g Pb, 1.96 Nm

b)

3

324 g C,

temperature~

3

air.

300 g slag with 33 % Si0

top gas with 7 % Zn,

62,25 % N . 2 Heat deficiency

1.55 Nm

~

10.25 % CO , 2

2

and 67 % FeO, 20.5 % CO,

25 kcal/kg sinter requires a blast

70°C.

With heat losses the actual blast

temperature is much higher. 0.214 atm (1200°C), 1.34 atm (1400°C)

10-7 =

11-1

1 atm at about 1373°c

a}

1800 oc

1600

CaTiSi0

b)

1600°C:

homogeneous melt,

with 39 % cao,

37 % Sio , 2 + CaSi0 (s) + CaTiSi0 (s). 3 5

25

5

1400°C:

CaTio (s) + melt 3 24 % Tio , 1200°C: CaTi0 (s) 3 2

as.

=

i

1.7·10

,...6

2.4·10

... 2

4.5·10

-3

,

=

. 4·10- 4

11-3

,

wt% FeO

'

~

2.10 -2 .

50 g/(sec•cm), v = 5.1·10- 2 cm/sec (d = 1 mm), 4 5.1·10- cm/sec (d = 0.1 mm).

11-4

µ

=

11~5

First choice:

11-6

Iron crucible.

graphite, second choice mullite.

0.05 atm,

11-7

0.1 atm.

Under the con-

ditions given the tube will have a limited lifetime.

12-1

a)

b)

c)

12-2

a)

b)

c)

d)

12-3

a)

nFe =

6.24,

nSi0 = 0.86 2 ncu 8 = 1.57, ~es= 2.84, ~eO = 3.40, n 8 i = 1.70 02 Addia Sio = 1.70 - 0.86 = 0.84 moles= 50.4 g/kg cone. 2 Added o = 1.70 + 8.06 = 9.76 moles= 218.6 Nl/kg cone. 2 Heat surplus without N : 204 kcal corresponding to 20.4 2 moles N . Furnace gas = 8.06 so + 20.4 N . Blast = 2 2 2 9.76 o + 20.4 N i.e. with 32.4 % o . 2 2 2 26.0 at% Cu,

nCu =

3.14,

n8 =

29.6 at% Fe,

12.47,

35.8 at% S,

8.6 at% 0

The ratio (n 8 + n )/(nFe + ½ ncu) = 44.4/42.6 = 1.04 0 i.e., slightly more than stoichiometric. 2 a Cu s = (26.0/55.6) (35.8/44.4) = 0.177 2 = (29.6/55.6)(35.8/44.4) = 0.429 aFeS = 1 atm at 1220°c Pso 2 5 = l.1·10a Cu o 2 7 4 p0 = 4.5·10- atm. b) aFeS = 7.3·10- , c) a!e ~ 0.3,acu 20 304 2 ~ 0.04 d) (1) Higher FeO activity will give higher Fe o and cu o. (2) Higher so 2 3 4 2 pressure will give lower activity of FeS, and higher activities of FeS, activities of Fe

o

3 4

and cu 26

o.

2

(3) Lover temperature will

12-3 (contl.

o, and higher activity 2 (In order to solye this problem it was necessary

give lower activities of .FeS and Cu

o . 3 4 0 to extrapolate the 6G curve for Cu

of Fe

12-4

PbS a)

+

2 PbO

6G

0

=

= 0

at about 845°c.

13-1

a)

~

from 1200 to 1300°C}

so 2

3 Pb+

giving aPbO

o

2

= -2.5

b)

kcal

0.6.

Since Table 13-1 does not list e~n the effect of Mn must be disregarded. c) [% o]

=

=

0.0796.

0.0283.

b)

a

0= 0.0251

-4

13-2

a)

H S/H = 5.26,10 c) -dn /d~ = 2 2 8 3 2 5.26°10- [% s]. For 1 kg steel: -dln % S /dVH = 0.752. 2 Integrated from 0.1 to 0.02[% SJ this gives V = 2.14 3 H2 Nm /kg steel.

13-3

a'

=

a

C

Pcs

2

= 0.199.

This gives a~

8

5.6,

13-5

[% o] = 5.8·10- 2

S-.8·10

9.3·10

.,...7

,

-4 ,

=

=

atm. 3. 33 [% cu],

[%

cu]

O.ll

1 atm at about 1850°c 8.7,10 -3 atm at 1850 0 C

a)

Pco

b)

Pea = With increasing temperature the Cac

c)

=

content of the melt 2 as well as the Ca losses by evaportion will increase. A

pure Cac 14-2

= 0.10,

= 154.

(S)/[_S]

13-6

aJ

5 Psis = 2.4•10- atm.

=

13-4

14-1

b)

2

melt can not be obtained.

0.15 atm.

27

14-3

Or""":;:::;=:::-----~-----------.--..., 0 -0. -10.5 -18.5

-so -60 -80

I

-52.5

-so

t

-100

T ~ 1000°c

t

,....._

-125

,-f

co

LJ

~

'-'

0

-75

,.c.,