SOLUTIONS MANUAL
Basic Principles and
Calculations in
Chemical Engineering
Eighth Edition
David M. Himmelblau
James B. Riggs
Upper Saddle River, NJ • Boston • Indianapolis • San Francisco
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Solutions Chapter 2
2.1.1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
2.2.1
N/mm or nm (nanometer)
°C/M/s
100 kPa
273.15 K
1.50m, 45 kg
250°C
J/s
250 N
a. Basis: 60.0 mile/hr
60.0 mile 5280 ft 1 hr
ft
= 88
hr
1 mile 3600 sec
sec
b. Basis: 50.0 lbm/(in)2
50.0/lbm 454 g 1 kg
1(in)2 (100 cm)2
kg
= 3.52 × 104 2
2
2
2
(in)
1 lb 1000 g (2.54 cm)
(1 m)
m
c. Basis: 6.20 cm/(hr)2
6.20 cm 1 m 109 nm 1(hr) 2
nm
= 4.79 2
2
2
(hr) 100 cm 1 m (3600 sec)
sec
a.
0.04 g 60 min (12 in )3 1 lb m
lb m
= 9.14
3
3
(min )( m ) 1 hr 1 ft 454g
(hr ) (ft 3 )
b.
2L 3600 s 24 hr 1 ft
ft 3
= 6.1× 10 3
s 1 hr 1 day 28.32L
day
2.2.2
3
c.
1 ft 2
(1 in)2
1 yr 1 day 1 hr 2.2 lbm
6 (in.)(cm 2 )
2
2
2
(yr)(s)(lbm )(ft ) (12 in) (2.54 cm) 365 days 24 hr 3600s 1 kg
m
1 ft 0.3048 m
–11
= 1.14 × 10
12 in
1 ft
( kg) (s2 )
2–1
Solutions Chapter 2
2.2.3
a.
Basis: 1 mi3
1mi3 ! 5280 ft # 3 ! 12 in # 3 ! 2. 54 cm # 3 ! 1 m # 3
" 1 mi $ " 1 ft $ " 1 in $ " 100 cm $
9
3
= 4.17 × 10 m
b.
Basis: 1 ft3/s
1 ft 3 60 s 7.48 gal
= 449 gal / min
1 s 1 min 1 ft 3
2.2.4
1 hr 2200 gal 1000 mile
= 4190.5 gal
525 mile 1 hr
1 hr 2000 gal 1000 mile 4210 gal
=
475 mile 1 hr
(20 gal)
None: 20 gal more are needed.
2.2.5
Let tA be the time for A to paint one house; tB for B
A does a house in 5 hours, or 1 house/5 hr. B does one house
in 3 hours, or 1 house/3 hr.
1 house tA hr 1 house tB hr
+
= 1 house
5 hr
3 hr
Also tA = tB so that
tA =
3
5
8
tA +
tA = 1 or
tA =1
15
15
15
15
hr = tB = 1.875 hr or 112.5 min
8
2.2.6
(a) mass, because masses are balanced
(b) weight, because the force exerted on the mass pushes a spring
2.2.7
1.0 Btu
24 hrs 1 ft 2
1 in 2
(100 cm)2 1.8o F 2.54 cm
⎛ o F ⎞ 1 day (12 in)2 (2.54 cm 2 ) 1 m 2
in
1oC
2
(hr)(ft ) ⎜ ⎟
⎝ ft ⎠
kJ
252 cal 12 in 4.184 J 1 kJ
= 1.49 × 104
2
1 Btu 1 ft 1 cal 1000 J
(day)(m )(°C / cm)
2–2
Solutions Chapter 2
2.2.8
Basis: 1 lb H2O
3
a.
1
1 1 lb m ! 3 ft "
(s 2 )(lb f )
KE= mv 2 =
= 0.14(ft)(lbf )
#
$
2
2
% s & 32.174(ft)(lb m )
b.
Let A = area of the pipe and v = water velocity. The flow rate is
⎛ πD 2 ⎞
(v)
q = Av = ⎜
⎝ 4 ⎟⎠
π (2 in)2 (1 ft)2 3 ft 60 s 7.48 gal
= 29.37 gal/min
4
(12 in)2 S 1 min 1 ft 3
2.2.9
PE =
75 gal 8.33 lbm 32.2 ft 100 60 s s 2 -lbf
Btu
= 4818 Btu/hr
2
min
ft hr 32.2 lbm -ft 778 ft-lbf
gal
sec
2 hp 2545 Btu
= 5090 Btu/hr
hp-hr
Rate of energy input for heating = PW - PE =5090 - 4818 = 272 Btu/hr
Pump Work =
2.2.10
The object has a mass of 21.3 kg (within a precision of ± .1 kg). The weight is the force
used to support the mass.
2.2.11
Basis: 10 tons at 6 ft/s
2
1
1 20,000 lbm ! 6 ft " 1(s 2 )(lbf )
KE =
m v2 =
= 11,200(ft)(lbf )
#
$
2
2
% s & 32.2(ft)(lbm )
2.2.12
Basis: 1 mRNA
1 nRNA 3x ribonucleotides 1 active ribosome
nRNA
1200 amino acids
protein
264 ribonucleotides min-active ribosome x amino acids
13.6 protein molecules formed per min per nRNA
2–3
=
Solutions Chapter 2
2.3.1
2.3.2
A
is
in
g/cm3
B
is
in
g/(cm3)(°C)
Since the exponent of e must be dimensionless
C
is
in
atm-1
a 1.
a 2.
a 3.
b1
A=
lb
1.096 g (30.48)3cm3 1 lb m
= 68.4 m3
cm3
ft 3
454g
ft
b2
B=
lb
0.00086 g (30.48)3cm3 1 lbm 1oC
= 0.0298 3 mo
3 o
3
o
(cm )( C)
ft
454 g 1.8 R
(ft )( R)
b3
C=
0.000953 1 atm
1
= 0.0000648
2
atm 14.7 lbf /in
lb f /in 2
Introduce the units. The net units are the same on both sides of the equation.
⎛ ft ⎞
(ft)(ft) ⎜ 2 ⎟
⎝s ⎠
1.5
1/2
=
ft 3
s
1/ 2
2.3.3
No.
!
"
#
$
-3
3
1 $
2 # (2) 9.8m 10 m 50×10 (kg)(m)
q = 0.6(2m ) #
2 $
s2
kg
(s 2 )(m 2 )
% 2&
#
1- ' ( $
#
) 5 * $,
+
The net units on the right hand side of the equation are
1/2
! m3 "
m $ 4 %
&s '
2
≠
m3
s
Consequently, the formula will not yield 80.8 m3/s, presumably in the formula the g
should be gc for use in the AE system.
2–4
Solutions Chapter 2
1.19
2.3.4
(2Δp)/ρ
Q = 0.61S
assume hole is open to atmosphere
⎡
lbf ⎤
14.7
⎢
⎥
73 in gas. 0.703 H 2O 1 ft
144 in ⎢ 23 lbf
in 2 ⎥
+
Δp =
1 gas 12 in 33.91 ft H 2O ⎥
1 ft 2 ⎢ in 2
⎢
⎥
⎣
⎦
2
= 3,579 lbf /ft
2
62.4
lb
ρ=
0.703ft3
lbH 2O
ft 3H 2O
3
1 lb/ft H 2O
=43.87
lbm
ft 3
2
⎛ 1 ⎞
S = π⎜
= 3.41× 10-4 ft 2
⎟
⎝ (4)12 ⎠
Q = (3600)(0.61)(3.41× 10−4 ) (2)(3579)g c / 43.87 = 54
2.3.5
ft 3
hr
⎡⎛ τ ⎞ 1/2 ⎛ N ⎞ 1/2 ⎛ m 3 ⎞ 1/2 ⎤
⎡m ⎤
⎥
u ⎢ ⎥ = k[1] ⎢⎜ ⎟ ⎜ 2 ⎟ ⎜
⎢⎝ ρ ⎠ ⎝ m ⎠ ⎝ kg ⎟⎠ ⎥
⎣s⎦
⎣
⎦
To get u in ft/s, substitute for τ and for ρ , and multiply both sides of the equation by
3.281 ft/1 m (k is dimensionless).
2
τ′lbf ⎛ 3.281 ft ⎞
N
1N
τ 2 =
2 ⎜
⎟
1
m
0.2248
lbf
⎝
⎠
m
ft
3
ρ′ lbm ⎛ 3.281 ft ⎞
kg
1 kg
ρ 3 =
m
ft 3 ⎜⎝ 1 m ⎟⎠ 0.454 lbm
⎛ τ′ ⎞
u ′ = 2.57 k ⎜ ⎟
⎝ ρ′ ⎠
2–5
Solutions Chapter 2
2.3.6
Place units for the symbols in the given equation, and equate the units on the left and
right hand sides of the equation by assigning appropriate units to the coefficient 0.943.
LHS
RHS
1
3
2
("
+ 4
(
)
(
)
%
Btu
Btu
lb
ft
Btu
hr
ft
"
m%
'
?$
(hr ) ft 2 (Δ °F) *)# (hr )(ft )(Δ °F)& # ft 3 & (hr)2 lb m ft lb m Δ °F -,
( )
The units are the same on the right and left so that 0.943 has no units associated with
it.
2.3.7
The equation is
Δp = 4fρ ⎡⎣(v 2 / 2g) (L/D) ⎤⎦
The units on the right hand side (with f dimensionless) in SI are
ρ
⎞
kg ⎛ m 2
⎜
m ⎟⎟
1
m 3 ⎜ s2
→ 2
⎜ (kg)(m) m ⎟
m
⎟
⎜ s2
⎠
⎝
hence the equation is not dimensionally consistent because Δp has the units of N/m2.
If g is replaced with g, the units would be correct.
2.3.8
For dimensional consistency:
B – absolute temperature in either ºR or K
C – absolute temperature in either ºR or K
A – dimensionless
In most cases the argument of a logarithm function should be dimensionless, but in
this case it will not be. Therefore, the numerical values of A, B, and C will depend
upon the units used for temperature and the units used for p*.
2.4.1
Two because any numbers added to the right hand side of the decimal point in 10 are
irrelevant.
2.4.2
The sum is 1287.1430. Because 1234 has only 4 significant figures to the left of an
implied decimal point, the answer should be 1287 (no decimal point).
2–6
Solutions Chapter 2
2.4.3
Two significant figures (based on 6.3). Use 4.8 × 103.
2.4.4
The number of significant figures to the right of the decimal point is 1 (from 210.0m),
hence the sum of 215.110 m should be rounded to 215.1 m.
2.4.5
Step 1: The product 1.3824 is rounded off to 1.4
Step 2: Calculate errors.
For absolute error, the product 1.4 means 1.4 + 0.05
Thus
0.05
× 100% = 3.6% error
1.4
Similarly 3.84 has
0.005
× 100% = 0.13% error
3.84
and 0.36 has
0.005
× 100% = 1.4% error
0.36
Total 2.7% error
2.6.1
(a)
16.1 lb mol HCl 36.5 lb HCl
= 588 lb HCl
1 lb mol HCl
(b)
19.4 lb mol KCl 74.55 lb KCl
= 1466 lb KCl
1 lb mol KCl
(c)
(d)
11.9 g mol NaNO3
85 g NaNO3 2.20 ×10-3lb
= 2.23 lb NaNO3
1 g mol NaNO3
1g
164 g mol SiO2 60.1 g SiO2 2.20 ×10-3lb
= 21.7 lb SiO2
1 g mol SiO2
1g
2–7
Solutions Chapter 2
2.6.2
a)
4 g mol mg Cl 2 (95.23)g MgCl 2
= 380.9 g
gmol MgCl 2
b)
2 lb mol C3H8 (44.09 )lb C 3H8 454g C3 H8
4
= 4 × 10 g C3 H8
lb mol C3 H8 1 lb C3 H8
c)
16 g N2
(d)
2.6.3
gmol N 2 1 lb mol N 2
= 1.26 × 10−3 lb mol N 2
(28.02)g N2 454 gmol N 2
3 lb C2 H6O 1 lb mol C2 H6O 454 g mol
= 29.56 g mol C2 H6O
(46.07) lb C2 H6O 1 lb mol
Basis: 100 g of the compound
comp.
C
O
H
g
42.11
51.46
6.43
MW
12
16
1.008
g mol
3.51
3.22
6.38
13.11
Ratio of Atoms
1.09
1
2
Multiply by 11 to convert the ratios into integers
The formula becomes C12O11H22
Checking MW: 12(12) + 11(16) + 22(1.008) = 342 (close enough)
2–8
Solutions Chapter 2
2.6.4
Vitamin A, C20O H30, Mol Wt.: 286
Vitamin C, C6 H8O6, mol. wt: 176
Vitamin A:
a.
Vitamin A =
2.00 g mol 286 g 1 lb
= 1.26 lb
1 g mol 454 g
16 g 1 lb
= 0.0352 lb
454 g
Vitamin C =
2.00 g mol 176 g 1 lb
= 0.775 lb
g mol 454 g
16 g 1 lb
= 0.0352 lb
454 g
b.
Vitamin A =
1.00 lb mol 286 lb 454 g
= 130,000 g
1 lb mol 1 lb
Vitamin C =
1.00 lb mol 176 lb 454 g
= 79,900 g
1 lb mol 1 lb
For both
2.6.5
12 lb 454 g
= 5450 g
1 lb
Mass fraction to mole fraction:
ω1
MW1
x1 =
ω1
(1− ω1 )
+
MW2
MW1
Mole fraction to mass fraction
ω1 =
x1MW1
(x1 )(MW1 ) + (1− x1 )(MW2 )
2–9
Solutions Chapter 2
2.6.6
Basis: 100 g mol gas
Comp.
CO2
N2
O2
H 2O
mol = %
19.3
72.1
6.5
2.1
100.0
Avg. mol. wt =
MW
44
28
32
18
g
849.2
2018.8
208.0
37.8
3113.8
3113.8
= 31.138
100.0
2.6.7
Basis: 100 lb mol
Comp.
CO2
CO
N2
% = mol
16
10
30
MW
44
28
28
lb
2640
280
840
3760
Avg. MW = 37.6 lb/lb mol
2.7.1
(a) A gas requires a convenient basis of 1 or 100 g moles or kg moles (if use SI
units).
(b) A gas requires a convenient basis of 1 or 100 lb moles (if use AE units).
(c) Use 1 or 100 kg of coal, or 1 or 100 lb of coal because the coal is a solid and mass
is a convenient basis.
(d) Use 1 or 100 moles (SI or AE) as a convenient basis as you have a gas.
(e) Same answer as (e).
2.7.2
Since the mixture is a gas, use 1 or 100 moles (SI or AE) as the basis.
2–10
Solutions Chapter 2
2.8.1
Basis: 1000 lb oil
lb oil
lb H2 O
62.4
3
ft
ft 3 = 57.78 lb oil
lb H2 O
ft 3 oil
1.00
ft 3
0.926
1000 lb oil
2.8.2
1 ft 3 oil 7.48 gal
57.78 lb oil 1 ft 3 = 129.5 gal
Basis: 10,010 lb
10,010 lb 1 gal 0.134 ft3
=
3
8.80 lb
gal
152 ft
Basis: 1 g mol each compound
2.8.3
g mol
mw
g
ρ (g/cm3)
V (cm3)
Pb
1
207.21
207.21
11.33
18.3
Zn
1
65.38
65.38
7.14
9.16
C
1
12.01
12.01
2.26
5.31
V̂ =
2.9.1
Component
Na
C1
O
g mol
1
1
3
5
mass (g)
density (g/cm 3 )
mol fraction
0.20
0.20
0.60
2–11
Set 2 is the correct one
MW
23
35.45
16
g
23
35.45
48
106.45
mass fraction
0.22
0.33
0.45
1.00
Solutions Chapter 2
2.9.2
Basis: 1 gal of solution.
Mass of solution:
1.0824 lb soln
lb H O
62.4 3 2
3
ft H 2O
ft H2 O 1 ft 3 1 gal
= 9.03 lb soln .
lb H2 O
7.
481
gal
1.00 3
ft H 2O
Mass fraction KOH =
0.813 lb
9.03 lb
Mass fraction H2O = 1 – 0.09
2.9.3
a)
1000
b)
No
=
= 0.09
0.91
1.00 Total
c)
Yes, because for solids and liquids the ratio in ppb is mass, whereas for gases
the ratio is in moles.
2.9.4
Basis: 100 g of the sample
The biomass sample is
g = % dry weight of cells
C
O
N
H
P
other
Total
50.2
20.1
14.0
8.2
3.0
95.5
4.5
100.0
10.5 g cells 50.2 g C 1 g mol C
= 0.439 g mol C/g mol ATP
g mol ATP 100 g cells 12 g C
2–12
Solutions Chapter 2
2.9.5
Basis: 190,000 ppm
190,000 g PCB
× 100 = 19%
106
2.9.6
MMM(s) → NN(s) + 3CO2 (g)
a.
b.
2.9.7
2×107 disintegrations 1 min
min
60
1 curie
106µ curie
3×1010 disintegrations/s
1 curie
= 11µ curie
2 ×107 disin tegrations 0.80
= 1.6 ×107 cpm
min
The relation to use is
t1/ 2 = l n(2) /(k)(OH − )
with (OH − ) = 1.5 ×106
k
Methanol
Ethanol
MTBE
t1/ 2 (sec onds)
0.15 ×10−12
1×10−12
0.60 ×10−12
30.8 ×105
4.6 ×105
7.7 ×105
The order is in increasing persistence ethanol, MTBE, and methanol.
2.9.8
Yes.
Bases are first entries.
4800 g mol CC14 103 g mol air kg mol air 154 g CC14 103 mg CC14
=
109 g mol air
kg mol air 22.8 m3 g mol CC14
g CC14
32.4 mg CC14 /m3 which exceeds the NIOSH standards
2–13
Solutions Chapter 2
2.9.9
a.
25,600 ton P 2.6 yr
1
1 gal 2000 lb 454 g 106 µ g
yr
1.2 × 1014 gal 3.785 L 1 ton 1 lb
1 g
= 133.1µ g / L
b.
19,090 lb(P )municipal
= 63. 4%
30,100 lb(P)total
c.
19,090 lb(P )municipal 0.70 lb(P)det. 100 lb(P )tota l
= 44.4%
30,100 lb(P)total 1 lb(P)municipal 30,100 lb(P)tota l
For d. and e. assume that (P)outflow remains unchanged.
d.
P retained/yr = 2,240 + 6,740 + 0.7 × 19,090 – 4,500 = 17,843 ton/yr.
P conc. in ppb =
17,843 ton 2.6 yr
1
2000 lb 1 g / cm 3
109 g
= 92.6 ppb
yr
1.2 × 1014 gal ton 8.345 lb / gal 1billion g
This is greater than 10 ppb. Eutrophication will not be reduced.
e.
P retained/yr = 2,240 + 6,740 + 0.3 × 19,090 – 4,500 = 10,207
ton
yr
P conc. in ppb =
10,207 ton 2.6 yr
2000 lb 1 g / cc
109 g
= 53.2 ppb
1.2 × 1014 gal ton 8.345 lb / gal 1 billion g
yr
This will not help.
Basis: 106 g mol gas
2.9.10
350 g mol H 2S 1 g mol gas 34g H 2S 1 g mol CO2
106g mol gas 1 g mol CO2 1 g mol H 2O 44 g CO2
=
270 g H 2S
270
g H 2S
= 6
6
10 g CO2
10 g total liquid
mass fraction H2S = 2.70 ×10−4
2–14
Solutions Chapter 2
2.9.11
(a)
1 mol O2
104 mol O2
⇒
or 104 ppm
6
100 mol gas
10 mol gas
(b)
Basis: 100 mol gas
answer
Comp.
SO3
SO2
O2
Total
% = mol
55
3
1
59
mol fr.
0.932
0.051
0.017
1.000
or mol %
93.2
5.1
1.7
100.0
2.9.12
On a mol basis, the carbon dioxide concentration in air is about 350 parts per million
(ppm), while that of oxygen is about 209,500 ppm. If the atmospheric concentration
of carbon dioxide is increasing at about 1% per year (i.e., from 350 ppm this year to
353.5 ppm next year), and not to 1%, the 3.5-ppm change in dioxide concentration
causes the oxygen concentration to fall from 209,500 to about 209,497 ppm, which is
less than a 0.002% decrease. So, there is no need to worry about an oxygen deficit at
present.
2.10.1
TK = –10 + 273 = 263K
T°F = –10 (1.8) + 32 = 14°F
T°R = 14 + 460 = 474°R
a)
10°C 1.8°F
1.0°C + 32 = 50°F
b)
10°C 1.8°F
1.0°C + 32 + 460°R = 510°R
2.10.2
c)
d)
−25o F – 32o F 1.0oC
+ 273K = 241.3K
1.8
1.8o F
150K 1.8°R
1.0K = 270°R
2–15
Solutions Chapter 2
2.10.3
2.4346 × 10-5TK

C p = 8.41 +
⎛
⎞
J
⎜ 2.4346×10-5
⎟ (T )
⎜⎝
(gmol)(K)2 ⎟⎠ K
Substitute
1.8 TK = T°R
8.41 + 2.4346 × 10œ5
Cp
=
8.41 + 1.353 ×10−5 To R
=
2.10.4
J
" T°R $
2#
(gmol)(K) 1.8 %
The instrument does not contain mercury, but has to contain a fluid that responds
at –76°C and can be calibrated to measure temperature.
Basis: 15cm3 water
2.11.1
a.
b.
m = ρV=
1000 kg 10 m 10 m 0.15 m
= 15,000 kg
m3
F
m g 15,000 kg 9.80 m
1 N 1 Pa
=
=
= 1470 Pa
2
A
A
s
10 m 10 m 1 (kg)(m) N
s2
m2
= 1.47 kPa
1.470 kPa
or
1 atm 14.7 psi
= 0.21 psi
101.3 kPa atm
15 cm H2O
1 in. 1 ft 14.696 psi
= 0.21
2.54 cm 12 in. 33.91 ft H2 O
2–16
Solutions Chapter 2
2.11.2
ρconcrete = 2080 kg/m 3
ρwater = 1000 kg/m 3
Volume of concrete:
Sides
Vs =
5 m 30.2 m 200 mm
1m
2 sides
3
= 60.4 m
1000 mm
5 m 27.2 m 200 mm
1m
2 ends
= 54.4 m3
1000 mm
Ends
Ve =
Floor
Vf =
27.4 m 30.4 m 200 mm
1m
1000 mm
= 166.592 m
Total volume = 281.392 m3
Mass of concrete =
281.392 m3 2080 kg
= 585,295
m3
Volume of displaced water required to float:
586,543.36 kg 1 m 3 H2 O
= 586.543 m 3
1000 kg H2 O
V = LWh → h =
h=
V
LW
586.54 m 3
27.4m 30.4 m
2–17
= 0.703 m
3
Solutions Chapter 2
2.11.3
lbf in.2
= lbf → lbm in the AE system
in.2
so the procedure is ok, although the unit conversion is ignored.
2.11.4
Neither John is necessarily right. The pressure at the top of Pikes Peak is
continually changing.
2.11.5
(a) p ABS = pGauge + patm
! 22.4 lbf 144 in.2 " ! 28.6 in. Hg 14.7 psia 144 in.2 "
2
$ = 5250 lbf /ft
#
$+#
2
2
2
1 ft & %
1
29.92 in. Hg 1 ft &
% 1 in.
⎛ 22.4 psig 29.92 in. Hg ⎞
+ 28.6 in. Hg = 74.2 in. Hg
14.7 psia ⎟⎠
⎝
(b) ⎜
⎛ 22.4 psig 1.013 × 105 N/m 2 ⎞ ⎛ 28.6 in. Hg 1.013 × 105 N/m 2 ⎞
= 2.51× 105 N/m 2
⎟⎠ + ⎜⎝
14.7 psia
27.92 in. Hg ⎟⎠
1
1
⎝
(c) ⎜
! 22.4 psig 33.91 ft H2 O " ! 28.6 in. Hg 33.91 ft H 2O "
$ +#
$ = 84.1 ft H 2 O
1
14.7 psia & %
1
29.92 in. Hg &
%
(d) #
2–18
Solutions Chapter 2
2.11.6
Δp = ρgh
a.
=
1 g 9.8 m 13.1 m 1 kg (100 cm )3 Pa
1 kPa
1 kg 1000 Pa
cm 3 s2
1000 g
m3
m s2
= 128. 4 kPa
b.
mg
ρVg
ρ(S.A.)(t )g
F
=
=
=
A A Bottom A Bottom
ABottom
2
π 2 π (30.5 m )
2
A Bottom = D =
= 730.62 m
4
4
2
ATop = 730.62 m
Aside = πDh = π(30.5 m)(13.1m) = 1255.2 m
2
S.A.= ATop + A Bottom + ASide
2
= (730.62 + 730.62 +1255.22 ) m = 2716.46 m
2
ρ(S.A.)(t )g
=
A
(
7.86 g 100 cm
cm 3
m3
3
)2,716.46 m2 9.35 × 10 œ3 m
1 kg
1 Pa
1 kPa
=
2
1000 g 1 kg / ms 1000 Pa
2–19
2.68 kPa
9.8 m
s2
730.6 m
2
Solutions Chapter 2
2.11.7
p A + ρ1gh1 + ρ 2gΔh = PD
p B + ρ1gh1 + ρ 2gΔh = PD
pA -pB = ρ1gh1 +ρ2gΔh −
ρ 1gh1ρ− 1gΔ h
= ρ2gΔh1 − ρ1gΔh = gΔh(ρ2 − ρ1 )
=
9.8 m 0.78 in 1 m 13.546 œ0.91g 1 kg
s2
3.28 ft
cm 2
1000 g
1 N
1 ft ! 100 cm # 3 m 2 1 Pa 760 mm H2
12 in " 1 m $ 1 kg 1 N 101.3 × 10 3 Pa
(m )(s)2 m 2
=18.4 mm Hg
The pressure at A is higher than the pressure at B.
Alternate solution:
Δp =
(0.78 in.) (13.546 − 0.91) 760 mm Hg
= 18.5 mm Hg
13.546
29.92 in. Hg
2–20
Solutions Chapter 3
3.1.1
Boundary:
Around both pumps and include the soil at the end of the pipes.
It’s an open system.
It’s at steady-state except at startup (note system boundary limits fluid), or
if fluid enters, unsteady state.
3.1.2
Either open (flow) or closed (batch) is acceptable if explanation is given
(1)
flow – material comes in and out continuously over a suitably long
period of time
(2)
batch – material is injected into the system, and then in a short period
of time, a reaction occurs with the system valves closed.
(a)
open if you have to replace water, and water evaporates; otherwise closed
(b)
open
3.1.3
3.1.4
a) closed
b)open
c)open
d)closed
3–1
Solutions Chapter 3
3.1.5
system
contains
ice + H2O
Ice
boundary
boundary
(c)
Unsteady state for any assumptions
b!
"open
# assume water (melted ice) leaves the system $
a%
&flow
b!
"closed
# assume water stays in sytem because the ice and water $
a%
&batch
remain on melting
3.1.6
P
F1
F
2
F
unsteady state
open
P
unsteady state
open
steady state
open
(a)
(b)
(c)
3–2
Solutions Chapter 3
3.1.7
(a)
(b)
(c)
(d)
x
(e)
(f)
x
(g)
1.
2.
x
x
____________________________________________________
3.
x
x
4.
Depends on the time period considered
____________________________________________________
5.
x
x
6.
x
x
x
____________________________________________________
7.
x
x
3.1.8
If the overall flows in and out over a time period for several batches are
considered, and the local batches ignore, the process can be treated as continuous.
3.1.9
Basis: 1 minute
Accumulation (kg)
In (kg)
mfinal − 0
= 300 + 100
n final = 20 kg
20 kg 60 min
= 1200 kg
min 1 hr
3–3
Out (kg)
−
380
Solutions Chapter 3
3.1.10
(c)
27.0 g
3.1.11
Basis: Data shown on flowsheet units are MTA
In
2.5 × 106
Out
404 ×103
228 ×103
152 ×103
101 ×103
67 ×103
36 ×103
100 ×103
1190 ×103
2,278 × 103
mass in does not equal to mass out
Reasons:
(1) Some of the material was burned as fuel
(2) Some of the material formed gases that were exhausted to atmosphere
(such as H2O, CO2).
(3) errors in measurement.
(4) Some process streams are not shown.
3.1.12
Based on the process measurements, there are 5,000 lb/h more flow for the
process leaving the heat exchanger than the feed rate to the heat exchanger; therefore, the
material balance for the process fluid does not close. The reason for this discrepancy
could be faulty flow sensor readings or possibly a leak of the condensate or steam into the
process stream.
3–4
Solutions Chapter 3
3.1.13
Basis: 1 week
in (tons) = 920 + 0.6 = 920.6
out (tons) = 3.8 + 620 + 0.01 + 0.01 + 0.08 + 1.1 + 275 + 20 = 920
Yes
3.1.14
The density of the crystalline silicon in the cylinder is 2.4 g/cm3.
Basis: 62 kg silicon
The system is the melt, and there is no generation or consumption. Let Δmt be the
accumulation.
Accumulation
Δmt
=
Input
0
-
Output
0.5(62 kg)
Δm t = − 31 kg
Let t be the time in minutes to remove one-half of the silicon
2.4 g π(17.5cm)2 0.3 cm t min 1
= (62,000 g)
cm3
4
min
2
t = 179min
3.1.15
Basis: 1 hr
Overall material balance (kg): 13,500 + 26,300 ? 39,800
Yes. The balance is satisfied
NaC1 balance: 0.25 (13,500) + 0.05 (26,300) ? 0.118 (39,800)
4690 ≅ 4696
The balance is closely satisfied but not exactly.
The closure is good for industrial practice.
3–5
Solutions Chapter 3
3.1.16
Basis: 1 hour
The overall material balance is
In (lb)
106,000
?
Out (lb)
74,000 + 34,000 = 108,000
The error in the overall material balance is 2000 lb/h or 1.9%; therefore, the overall
material balance is within the expected error for industrial flow sensors.
Propylene balance:
0.7 × 106,000 ? (0.997) (74,000) + (0.1) (34,000)
74,200 ? 73,778 + 3,400 = 77,178 (3.8% error)
Propane balance:
0.4 × 106,000 ? (0.003) (74,000) + (0.9) (34,000)
31,800 ? 222 + 30,600 = 30,622 (3.8% error)
Note that the relative error for the component balances are twice as large as the relative
error for the overall material balance, indicating that there is additional error in the
composition measurements used for the component balances.
3.2.1
(a) water and air.
(b) Insulation, air and what is in the atmosphere.
(c) Yes (cold water in, hot water out)
(d) Yes
3–6
Solutions Chapter 3
3.2.2
Four balances are possible, 3 components plus 1 total.
0.10F1 + 0.50F 2 + 0.20F 3 = 0.35P
0.20F1 + 0F2
+ 0.30F 3 = 0.10P
0.70F1 + 0.50F 2 + 0.50F 3 = 0.55P
Total balance F1 + F2 +F3 = P
Only 3 of the equations are independent.
3.2.3
If you specify F, P, W, you can calculate all of the stream variables.
a)
Unknown: three stream values F, P, W (plus two compositions if you take into
account all of the variables).
b)
The two known compositions are not given but may be calculated from Σxi = 1,
c)
Two components exist; hence two independent material balances can be written.
The problem cannot be solved unless one stream value is specified.
3.2.4
(a)
No. The equations have no solution – they are parallel lines.
The rank of the coefficient matrix is only 1 because the
⎡1 2⎤
det ⎢
=0
⎣1 2⎥⎦
The rank of the augmented matrix is 2
⎡1 2 1⎤
Largest non zero det. of ⎢
is of order 2
⎣1 2 3⎥⎦
Thus, although the 2 equations are independent and the number of variables is 2
(the necessary conditions), the sufficient conditions are not met.
3–7
Solutions Chapter 3
x2
x + 2x =1
1
2
x + 2x = 3
1
2
x1
(b)
x2
•
(x1 − 1)2 + (x2 − 1)2 = 0
No; 2 solutions
•
The equations are independent
x1
x1 + x 2 = 1
Two solutions exist as can be seen from the plot hence no unique solution exists.
3.2.5
a)
No for both
⎡1 1 1 ⎤
⎢1 2 3 ⎥ rank of coefficient matrix is2.
⎢⎣3 5 7 ⎥⎦ rank of augmented matrix is2.
r = 2, n = 3
b)
multiple solutions exist
⎡1 0 1⎤
⎢5 4 9⎥ 3d column issum of1st two columns,so rank is2.
⎢⎣2 4 6⎥⎦ rank of augmented matrix is2, also.
r = 2, n = 3
multiple solutions exist
3–8
Solutions Chapter 3
3.2.6
(a) F; (b) F; (c) F (the maximum can be more than the number of independent
equations).
3.2.7
P = 16 kg
F = 10 kg
ωF1= 0.10
ωF2= ?
ωF3= 0 (assumed)
ωP1= 0.175
ωP2= ?
ωP3= ?
A = 6 kg
ωA1= 0.30
ωA2= ?
ωA3= 0.20
Unknowns (4): ω2F ,ω2A ,ω2P ,ω3P
Equations:
Mass balances:
(1):
F(0.10) + A(0.30) = P(0.175)
or
10(0.10) + 6(0.30) = 16(0.175)
(2):
F (ω 2 F ) + A (ω 2 A ) = P (ω 2 P )
or
10 (ω 2 F ) +6 (ω 2 A ) = 16 (ω 2 P )
(3):
F(0)
+A(0.20) = P (ω 2 P )
or
10(0)
+ 6(0.20) = 16 (ω 2 P )
redundant
ω 2 F ,ω 2 A ,ω 2 P ,ω3P
10
6 -16 0
0
0 -0 16
coeff. matrix
Two independent material balances exist (the rank of the coefficient matrix is 2).
In addition three sum of mole fraction equations exist. The total number of equations is
5, hence the degrees of freedom = -1. The problem is overspecified, and will require at
least squares solution.
3–9
Solutions Chapter 3
3.2.8
A
x1
Basis: D =100 lb
B
x2
C
x3
D
0 M 1.4 "
! 5.0 0
#90.0 10.0 0 M 31.2 $
#
$
Coefficient matrix is # 5.0 85.0 8.0 M 53.4 $
#
$
# 0 5.0 80.0 M 12.6 $
#% 0
0 12.0 M 14 $&
No unique solution exists with 5 equations and 4 variables. A least squares solution
could be determined, or the 4 equations with the most accurate data solved.
The rank of the coefficient matrix is 3
The rank of the augmented matrix is 4
Hence no unique solution exists.
3.2.9
Number of unknowns:
F, W, P and 9 compositions:
Equations
W
Specifications: x CF
=0
Sum of mole fractions = 1
Material balances (3 species)
Degrees of freedom =
12
1
3
3
7
5
You can make any set of measurements that results in independent equations (assuming
equal accuracy). For example, you cannot use all the flows, or all the compositions in
one stream, as the resulting set of material balances will not consist of 5 independent
balances, but a lesser number.
3–10
Solutions Chapter 3
3.2.10
You can see by inspection that no combination of tanks 1,2 and 3 will give a mole
fraction of 0.52 for the mixture.
Basis: 2.50 mol of tank 4
Let xi = be the total moles of tank i
The balances are
0.23x1 + 0.20x2 + 0.54x3 = 0.25(2.5) = 0.625
0.36x1 + 0.33x2 + 0.27x3 = 0.23(2.5) = 0.575
0.41x1 + 0.47x2 + 0.19x3 = 0.52(2.5) = 1.300
The coefficient matrix has a rank of 3 as does the augmented matrix so the set of
equations has a solution. However, the solution is
x1 = −4.00
x 2 = 6.36
x 3 = − 0.327
The values of x1and x3 are not physically realizable!
3.2.11
The inerts are treated as a compound.
Unknowns: 5 compounds × 5 streams plus 5 streams =
30
Equations: Specifications:
Concentrations not shown in diagram assumed
to be zero
8
Concentrations with % given
7
E = 11 kg
1
Material balances: 5
5
Implicit equations (∑ ωi = 1)
5
26
a.
Degrees of freedom (additional specifications) =
4
You must make 4 measurements that result in independent equations
b.
No.
3–11
Solutions Chapter 3
3.2.12a
a.
b.
The remaining gas is 100% minus the N2, and was put on the figure as R.
c.
Basis: 100 mol A
d.
Unknowns: A, B, C
Equations: N2 and R material balances
Basis = A = 100 mol
Note: You could treat the values of R as unknowns in each stream,
and then there would be 3 more unknowns and 3 more independent
equations ( ∑ x i = 1 or ∑ mi = total mass flow i )
e. & f. N2:
100(0.90) + B(0.30) = C(0.65)
R:
100(0.10) + B(0.70) = C(0.35)
Total:
100 + B
=C
Two of the above equations are independent
g.
Solution:
B = 71.4 mol
C = 171.4 mol
A
100
=
= 1.40
B 71.4
3–12
Solutions Chapter 3
3.2.12b
a.
DT = dry timber
b.
The DT is the balance of each stream.
c.
Basis: F = 100 kg
d.
Unknowns: F, W, P
OR
Equations:
Basis: F = 100 kg
Material Balances: Water, DT
(total); 2 independent
Degrees of freedom = 0
F, W, P, DTF, DTW, DTP, H2OF,
H2 OP , H2 OW
Equations:
F = 100 kg
Material Balances: Water, DT,
(total); 2 independent
Specifications: 3 for water
Implicit equations: 3 of ∑ ω i = 1
Degrees of freedom = 0
e.&f. Introducing the specifications and basis into the material balances:
Water: (0.201)100 = (1)W + (0.086)P
DT:
(0.80) 100 = 0
+ (0.914)P a tie element
Total:
100 = W + P
g.
W = 12.5 kg
P = 87.5 kg
W 12.5 kg
=
= 0.125kg/kg
F
100 kg
3–13
Solutions Chapter 3
3.2.12c
a. & b.
A N2 1.00
mol. fr.
N2 0.70
CH4 0.30
1.00
d.
CH4 n CH 4
C 2H 6 n C 2 H 6
N2
n
P
F
N2
B
c.
or
N2
C 2H 6
0.90
0.10
1.00
∑=P
∑
x CH 4
x C2 H 6
x N2
=1.00
Basis: B = 100 mol
Unknowns:
Assume all of the variables except those that are specified as zero are
included in the analysis.
F
A
P
B
stream variable
1
1
1
1
4
+
component variables
2
1
3
2
8
=
12
Equations:
Basis:
Material balances:
N2, CH4, C2H6
Specified ratio: n CH /n C H = 1.3
Specified values of n: 2 + 1 + 0 + 2 =
Implicit equations (∑ xi = 1 in P)
(Note: the implicit equations of streams
F, A, and B are redundant with the
specifications)
Total
Degrees of freedom =
4
2
6
1
3
1
5
1
11
1
More information is needed to solve the problem uniquely.
3–14
Solutions Chapter 3
3.2.13
Here are some possibilities. Consult the tables at the end of the Chapter for more
suggestions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Rephrase the problem to make sure you understand it?
Draw a simple diagram of what was happening?
Think about what was going into the tank and what was coming out?
Imagine yourself inside the tank, and ask what was going on around you?
Ask whether there were any physical laws to consider (such as conservation
of matter or energy)?
Try to imagine the answer as a number, graph, table, or whatever?
Try to identify essential variables?
Choose a notation?
Look for a ready-made formula for the answer?
Look for simplifying assumptions?
Try to find an easier version of the problem?
Look for bounds (simple models that would definitely underestimate
or overestimate the answer)?
3–15
Solutions Chapter 4
4.1.1
Initial
Final
plant 5 ppm
soil 6 ppm
Plant 755 ppm
Soil 5 ppm
Final − Initial
Final − Initial
Arsenic balance: S (5) − S(6) =
P(755) − P(5)
750P = S or
S
= 750
P
4.1.2
Step 5: Basis is 1 ton (2000 lb) sludge
Steps 1, 2, 3, 4:
Sludge
F
0.70 H 2 O
0.30 Solids
1.00
Drier
P
0.25 H 2O
0.75 Solids
W
1.00
H2 O
1.00
Steps 6 and 7:
Balances: H2O
Solids
Dried Sludge
Basis: 2000 lb F
Unknowns: P, W
Steps 8 and 9:
Total: 2000 = P + W
Solids: 2000 (0.30) = P (0.75)
W = 2000 − 800 = 1200 lb
P = 800 lb
4–1
Solutions Chapter 4
4.1.3
Step 5:
Basis: 1 min
Steps 2, 3, 4:
D
220 mL
V 215 mL
B
Urea 2.30 mg/mL
H2O 220 mL
P
1.70 mg/mL
H2O 215 mL
Steps 6 and 7: Unknowns: mP urea and mP H2O with two equations: urea and H2O. Degrees
of freedom = 0.
Steps 8 and 9:
(a)
H2O balance: 220 mL − 215 mL = 5 mL
Urea balance:
(b)
2.30 mg 220 mL
1.70 mg 215 mL
−
= 141 mg
mL
mL
Dialisate: 1500 + 5 = 1505 mL
Urea: 141 mg
Concentration: 141/1505 = 0.0934 mg/mL
4–2
Solutions Chapter 4
4.1.4
Step 5:
Basis: 1 day
Steps 2, 3, 4:
W ton
1.00 H2O
F=2 ton
P (ton) NaOH
mass fr
NaOH 0.03
H2O 0.97
1.00
H2O
Step 6: Unknowns, 2: P, W
Step 7: Balances 2: NaOH, H2O, total (2 of the 3)
Steps 8 & 9:
Total: 2 = W + P
NaOH: 2 (.03) = P(.18)
P = 1/3 ton
(666 lb)
4.1.5
W = 1 2/3 ton
(3334 lb)
Steps 2, 3, 4:
500 lb 10% solution
A
20% solution
3000 lb
13% polymer
pure solvent
S
Step 5:
Basis: 3000 lb of final product
Steps 6 and 7: Unknowns: A and S; balances: total and polymer
4–3
Mass fr.
0.18
0.82
1.00
Solutions Chapter 4
Steps 8 and 9:
Total wt
A + S + 500 = 3000
A + S = 2500
Polymer
0.2A + 50 = 390
A = 5 (340) =
1700
S = 2500 – 1700 = 800
4.1.6
Steps 1, 2, 3 and 4:
N = nitrocellulose
S – solvent
Treat the problem as a steady state flow
problem, or as an unsteady state batch
system. As a flow system:
M (lb) N 100%
F (lb)
N
S
Plant
P 1000 lb
0.055
0.945
1.000
Step 5:
Mass fr.
N 0.08
S 0.92
1.00
Basis: 1000 lb P
Steps 6 and 7: Two unknowns, F and M. Two balances can be made, N and S.
Steps 8 and 9:
Solve to get
N:
F ( 0. 055) + M (1. 00) = 1000 (0. 008)
S:
F (9. 045) + M ( 0) = 1000 ( 0. 92)
! F = 973. 5 lb
#
"
M = 26. 5 lb
#
$
P = 1000.0 lb
4–4
Solutions Chapter 4
4.1.7
Steps 1, 2, 3 and 4:
F = 100 kg mol/min
CH4
He
Mol fr.
0.80
0.20
1.00
W (kg mol) CH
4
He
System
P (kg mol)
CH4
He
Mol.
n CH
4
n He
Mol fr.
0.50
0.50
1.00
Step 5:
Basis
Step 4:
To get P, calculate first the average mol. wt. of F and P, or transform the
mole fractions to mass fractions.
Step 5:
Basis: 1.00 mol F
Mol
CH4
He
1 min → 100 kg mol F
0.80
0.20
1.00
Basis : 1.00 mol P
Mol
CH4
He
0.50
0.50
1.00
100 kg mol F
MW
kg
16.03
4
12.8
0.8
13.6
MW
kg
16.03
4
8.01
2.0
10.01
13.6 kgF 0.20 kgP 1kgmol P
1.00kg mol F 1kgF 10.01kgP
P = 27.2 kg mol
Steps 6, 7, 8 and 9:
In W
100(0.80) = 27.2(0.50) + n CH 4 ⎫
⎬
100(0.20) = 27.2(0.50) + nHe ⎭
n CH 4
nHe
4–5
Mol
Mol.fr.
66.4
6.4
0.912
0.088
72.8
1.00
Solutions Chapter 4
4.1.8
a. Basis: 1 kg mole of mixture
Three components exist: (CH4)x, (C2H6)x, (C3H8)x
Let A, B, C respectively represent kg mol of each mixture; these are the unknowns.
Equations:
0.25A + 0.35B + 0.55C = 0.30
(CH4)x balance
0.35A + 0.20B + 0.40C = 0.30
(C2H6)x balance
0.40A + 0.45B + 0.05C = 0.40
(C3H8)x balance
There is a unique solution to the set of equations (in kg mol)
The solution is
A = 0.600
B = 0.350
C = 0.05
b. It is proposed to prepare the final mixture by blending four different compounds (A,
B, C, D); there will still be three equations, but now there will be four unknowns. Since
the rank is now less than n, there will be an infinite number of possible blends of the four
mixtures. Not required: (An optimization of a revenue function subject to the equations
is needed.)
4–6
Solutions Chapter 4
4.1.9
a.
A
100% CO2 100 lb = 2.27 lb mol
CH4 100%
(MW = 16)
P (lb) = ?
mol fr.
CH4 0.9796
CO2 0.0204 (average)
1.000
F=?
(lb)
100 lb 1 lb mol
= 2.27 lb mol
44 lb
Steps 2, 3, 4:
Step 5: Basis 1 min
Step 6: Unknowns: F, P
Step 7: Balances: CH4, CO2
Steps 8 and 9:
Balances in moles
CH4: F (1.00) + A(0) = P (0.9796)
CO2: F (0) + 2.27 = P (0.0204)
F=
b.
P = 111.41 lb mol
111.27 lb mol 0.9796 lb mol CH4 16 lb CH4
= 1746 lb/min
1 lb mol P
1 lb mol CH 4
(a)
Redo the problem with a new composition for F:
CH4
CO2
0.99
0.01
1.00
CH 4 : F (0.99) + A(0) = P (0.9796) !
" F = 3520 lb/min
CO2 : F (0.01) + 2.72 = P (0.0204) #
error
4–7
3520-1744
100 = 50% (b)
3520
Solutions Chapter 4
4.1.10
Steps 1, 2, 3, and 4:
10 lb/hr
Mass fr.
H2 O 1- 0.00012
0.00012
P Br
1.00
F
Mass fr.
H 2 O 1.00
Br
0
1.00
Step 5: Basis: 1 hr
Step 6: Unknowns:
F, P
Step 7: Balances:
Steps 8 and 9:
Total:
Br:
H2O:
H2O, Br
F + 10 =
F (0) + 10
F (1.00) + 0
P
=
=
P (0.00012) ; P = 8.33 × 104 lb/hr
P (1-.00012)
F = 8.33 × 104 − 10 = 8.33 ×10 4 lb / hr
4–8
Solutions Chapter 4
4.1.11
Step 5: Basis: 1 year
Steps 2, 3, 4:
P = 15 x106 lb
F = ? lb
PET
PVC
mass fr.
0.98
0.02
1.00
W
PVC 100%
Steps 6 and 7:
Two unknowns: F, W
Two balances: PET, PVC
Steps 8 and 9:
F (.98) = P (.999990) ≅ 15 × 106
F (.02) = W (1.00) + P (0.000010)
= W + (15 × 106 (10-5)
Total balance could be used in lieu of one of the above
PET balance:
PVC balance:
F = 15 × 106 + W
From PET: F ≅ 15.3 × 106 lb
W = (15.3 × 106) (0.02) – 15 × 106 (10-5) = 0.31 × 106 lb
4–9
mass fr.
PET 0.999990
PVC 0.000010
(10 ppm)
Solutions Chapter 4
4.1.12
Assume steady state flow problem (alternate is unsteady state batch problem).
Steps 1-4:
200 g H2O
1.00
100 g
Na2B4O7
1.00
No rxn.
No accum.
P
(g)
Na2B4O7 10H2O
F (g)
Final solution
mol fr.
Na2B4O7 0.124
H2O
0.876
1.000
Calculate composition of P: Basis: 100 mol Na2B4O7.10 H2O
Na2B407
H2 O
mol
1
10
Step 5:
Basis: 200 g H2O + 100 g Na2 B4O7
Step 6:
Unknowns: F, B
Step 7:
balances: Na2B4O7, H2O
Step 8:
Step 9:
Na 2 B4O7 :
g
201.27
180.0
381.27
mol fr.
0.528
0.472
1.00
H 2O:
100 = P (0.528) + F (0.124) !
" 2 indept.
200 = P (0.472) + F (0.876) #
Total:
100 + 200 = P + F
Use H2O and Total to get P = 155.5g and F = 144.5 g
ratio:
Step 10:
MW
201.27
18
155.5
(100) = 51.8 g Na 2 B4O7 /100 g H 2O
300
Check use Na2B4O7
100.0 = 100.0 ok
4–10
Solutions Chapter 4
4.1.13
Assume the process is a steady state one without reactions.
Steps 1, 2, 3, 4:
F1
.
FeC13 6H2O
1000 kg
F2
.
FeC13 H2O
MW
P
.
FeC13 2.5 H2O
162.22
270.32
180.24
207.26
18.02
Fe C13
Fe C13.6H2O
Fe C13.H2O
Fe C13.2.5 H2O
H 2O
Calculate the compositions for just one iron compound. FeC13 is the simplest to use.
For F1:
Basis: 1 kg mol
For F2:
FeC13 ⋅ 6H 2 O
FeC13
H 2O
Basis: 1 kg mol
FeC13 ⋅1H 2 O
kg mol
1
6
MW
162.22
18.03
kg
162.22
108.18
270.40
FeC13
H 2O
kg mol
1
1
MW
162.22
18.03
For P Basis: 1 kg mol
FeC13
H 2O
kg mol
1
2.5
MW
162.22
18.03
kg
162.22
45.08
207.30
Step 5: Basis: 1000 kg FeC13.6H2O
Step 6: Unknowns P, F
Step 7, 8, 9: Balances (kg)
IN
OUT
! 162.22 kg FeC13 "
! 162.22 "
! 162.22 "
$ + F2 #
$ = P#
$
kg tot &
% 270.40 &
% 180.25 &
% 207.30
FeC13: 1000 #
Total: 1000 + F2 = P
4–11
kg
162.22
18.03
180.25
Solutions Chapter 4
soln
Step 10:
F2 = 1555.7 kg
P = 2555.7 kg
Check: F2 + F1 = 1554.5 + 1000 = 2554.5
ok
4.1.14
M.W. Na2 S2 O2 = 142
M.W. Na 2 S2 O2⋅ 5H2 0 = 232
feed = F
60% Na2S2 O2
1% impurity
39% H 2O
H2 O
W
0
H2 O
Wi
S
I
C
10°C
Na 2 S2 O2⋅ 5H2 O
crystals
0.06 lb satd. soln/lb crystals
II
Na 2 S2 O2⋅ 5H2 O
0.1 % impurity
0% H 2O as H 2 O
D
Process II Compositions:
a)
Stream D
Comp. lb mol
Na2S2O2
H2O
Total
satd soln
1.4 lb Na 2 S2 O2⋅ 5H2 O
1 lb H 2O
? lb impurity
dried cyrstals
Basis: 1 lb mol Na2S2O2 ⋅ 5 H2O, impurity free
mol wt
1
5
6
lb
142
18
wt fr
142
90
232
0.612
0.388
1.000
Basis: 100 lb D (Na2S2O2⋅ 5 H2O = 99.9 lb)
Comp.
Na2S2O2
H2 O
impurity
Total
Stream C
99(0.612)
99(0.388)
lb
=
=
61.1
38.8
0.1
100.0
Let y = lb impurity/100 lb free water in saturated solution
4–12
Solutions Chapter 4
Basis: 100 lb dry crystals, impurity free
lb from dry
Comp.
crystals
+
lb from adhering soln
from calculation below
= Total
Na2S2O2
61.2
⎛ 85.7 ⎞
⎟
(6)⎜
⎝ 240 + y ⎠
⎛ 85.7 ⎞
⎟
61.2 + 6 ⎜
⎝ 240 + y ⎠
H2O
38.8
⎛ 154.3 ⎞
⎟
(6)⎜
⎝ 240 + y ⎠
⎛ 154.3 ⎞
⎟
38.8 + ⎜
⎝ 240 + y ⎠
impurity
-----
⎛ y ⎞
⎟
(6)⎜
⎝ 240 + y ⎠
⎛ y ⎞
⎜
⎟
⎝ 240 + y ⎠
Totals
100.0
6
106
Stream S
Comp.
Basis: 100 lb free water in satd. soln., impurity free
lb from salt lb from free water
Na2S2O2
1.4(61.2) = 85.68
H2O
1.4(38.8) = 54.32
Total
100
wt. fr.
wt.fr.total
85.7
0.357
85.7
(240 + y)
154.3
0.643
154.3
(240 + y)
__
140.00
100
240.0
1.000
y
240 + y
impurity =
Total
Basis: 100 lb F
(Rather than use the impurity balance, use the total balance instead if you want)
Water in, Wi, comes from balances on Unit I.
Total:
100 + Wi = S + C
4–13
Solutions Chapter 4
⎛ 85.7 ⎞ ⎫
⎟
61.2 + 6⎜
⎛⎜ 85.7 ⎞
⎝ 240 + y ⎠ ⎪ Unknowns: W i ,S,C, y
⎟S+
Na 2 S2 O2
60=
C⎪
⎝ 240 + y ⎠
106
⎬
Total:4
⎛⎜ 154.3 ⎞
⎟ ⎪
38.8+ 6
⎛ 154.3 ⎞
⎝ 240 + y ⎠ ⎪
⎟S +
39 + W i = ⎜
C
H 2O
⎝ 240 + y ⎠
106
⎭
Balances on Unit II needed as well because 4 unknowns exist.
Total:
C = W0 + D
Na 2 S2 O2
H 2O
⎛ 85.7 ⎞
⎟C
61.2 + 6⎜
⎝ 240 + y ⎠
⎫
⎪ Unknowns:W o ,D
= 0.611D
⎪
106
⎬
Total:2
⎛⎜ 154.3 ⎞
⎟C
⎪
38.8+ 6
⎝ 240 + y ⎠
= W o + 0.388D⎪
⎭
106
6 equations and 6 unknowns
Note that the complex term involving y can be eliminated to solve for D, W0, Wi, and S,
i.e., in effect making total Na2S2O2 and H2O balances overall the process.
The solution is
(a)Wi = 23.34 lb
(b)66.5%
4.1.15
Basis: 100 lb pulp as received
Comp
H2 O
lb = %
22
Pulp
78
Total
100
Freight cost =
$1.00 2000 lb
100 lb 1 ton
Assume air dried pulp means the 12% moisture pulp.
Allowed water =
78 lb pulp 12 lb H 2 O
88 lb pulp
= 10.65 lb H 2 O
4–14
= $20.00/ton
Solutions Chapter 4
Pulp on contract basis = 78 + 10.65 = 88.65 lb
Basis: 1 ton pulp as received
Cost =
88.65 lb 12% pulp rec'd 1 ton shipped $60.00
100 lb shipped
1 ton
$53.19/ton
4.1.16
You pay for soap plus transportation.
Basis: 100 kg soap with 30% water
Convert soap in wet soap to soap in dry soap. Data:
W (wet soap)
H2O 30 kg
soap
D(dry soap)
5%
70 kg
95%
100 kg
100
Soap balance: 0.70W = 0.95D
or 0.70(100) = 0.95D
D = 73.68 kg of which 70 kg is soap
$6.05
= $36.05
100
Cost of W at your site (containing 70 kg soap):
100 ($0.30 +
Cost of D at your site (containing 70 kg soap):
! x$ $6.05 "
73.68 #
+
$ = $36.05
% kg 100 &
x = $ 0.43/kg
4–15
Solutions Chapter 4
4.1.17
The problem here is to decide on the balances to make. Not all will be indept balances.
Steps 1, 2, 3, and 4:
F = 100 lb
H(lb)
Mixer
Mass fr.
H2 O 0.124
VM 0.166
0.575
C
0.135
Ash
1.000
(lb)
P
Mass fr.
H2 O
VM
C
Ash
0.10
ω VM
ωC
0.10
1.00
Mass fr.
VM 0.082
C
0.887
H2O 0.031
1.000
Mass
mH O
2
m VM
mC
m Ash
P
Step 5:
Basis: F = 100 lb
Step 6:
Unknowns:
m H 2 O , mVM, mC, mAsh, P, H (6)
Step 7:
Balances:
H2O, VM, C, Ash, Σmi = P, m H 2 O /P = 0.10
mAsh/P = 0.10: (7), and presumably 6 are independent
Steps 8 and 9:
H2O (lb):
In
100 (0.124) + H (0.082)
=
Out
m H 2O
VM :(lb)
100 (0.166) + H (0.082)
=
mVM
C (lb):
100 (0.575) + H (0.887)
=
mC
Ash (lb)
100 (0.135) + H (0)
=
mAsh
=
P
Total
100 + H
4–16
Solutions Chapter 4
Solution:
mAsh = 13.5 lb
m H 2 O = 13.51
Not all of the equations have to be written down as above. Note that ash is a tie element to
P, so that the ash balance gives
100(0.135) + H(0) = P(0.10)
P = 135 lb
From a total balance
H + 100 = P
Step 10:
H = 35 lb
Check via H2O balance
12.4 + 35(0.031) = 13.5 !
" ok
13.49
=13.5#
4–17
and mAsh = 13.5 lb
Solutions Chapter 4
4.1.18
Steps 1, 2, 3, and 4:
Assume the other components have the same density as water in all flows.
3.785 × 106 L/day
150 mg/L BOD
a.
B
5 mg/L BOD
0.3 m 3 /s
A
C
BOD = ? = x
Step 5: Basis: 1 day
0.3m 3 3600s 24hr 1000 L
7
A=
= 2.592 ×10 L / day
3
s
hr day m
Steps 6 and 7:
Unknowns: C, x
Balances:
H2O, BOD
Steps 8 and 9:
Total mass balance A + B = C
1day A L 1kg 1day B L 1kgB 1day C L 1kg C
+
=
day L
day L
day L
BOD mass balance
−3
−3
1day A L 5×10 gBOD 1day B L 150 ×10 gBOD 1day C L x g BOD
+
=
day
L
day
day
day
L
2.592 × 107 kg + 3.785 × 106 kg = C = 2.9705 × 107 kg
5 × 10-3 A + 150 × 10-3 B = xC
5× 10 −3A +150 ×10 −3 B
x=
C
4–18
Solutions Chapter 4
=
5 × 10−3(2.592 × 10 7 ) + 150 × 10 −3(3.785 × 106 )
2.9705 × 107
b.
g/L
N BOD = ?
B
530 x 106 L/day
3x10-3 g/L BOD
Total mass balance
−3
= 23.475 × 10 g / L
= 15.8x106 L/day
C
A
5x10-3 g/L BOD
A+B=C
1day A L 1kg 1day B L 1kg 1day C L 1kg
+
=
day L
day L
day L
530 × 106kg + 15.8 × 106 kg = C = 5.458 × 108 kg
BOD mass Balance
−3
−3
−3
1day A L 3× 10 gBOD 1day B L N ×10 g 1day C L 5×10 g
+
=
day
L
day
L
day
L
3 × 10-3A + N × 10-3 B = 5 × 10-3C
5 × 10 −3 C − 3 × 10 −3 A
N=
B
(5 × 10−3 )(5.458 × 10 8 ) − (3 × 10−3 )(5.30 × 108 )
−3
≅ 72.1x10 g / L
=
7
1.58 × 10
4–19
ok
Solutions Chapter 4
4.1.19
Step 5:
Basis: 1 day
Steps 2, 3, and 4:
%
C3
C3
i − C4
n − C4
C5 +
%
1.9
51.6
46.0
0.5
100.0
i − C4
n − C4
D
F=5804
kg mol/day
3.4
95.7
0.9
100.0
%
B
i−C
n- C 4
C 5+
1.1
97.6
1.3
100.0
Steps 6 and 7:
There are two unknowns, and we have 4 independent equations, but the precision of
measurement is not the same in each equation, hence different results are obtained
depending on the equations used. Choose the most accurate to use.
C3
(0.019)(5804) =
i-C4
n-C4
C5+
Total:
In
110
2992
2667
37
5804
=
=
=
=
=
Steps 8 and 9:
Use total + i-C4 balances:
kg mol/day
B = 2563
D = 3240
2992 = 0.975(5804-B) + 0.011 B
Use total + n-C4 balances:
B = 2704
D = 3100
2667 = 0.009 (5804 - B) + 0.976B
The other two balances give
4–20
Out
0.034D
0.057D + 0.011B
0.009D + 0.976B
0.013B
D+B
Solutions Chapter 4
C3 as a tie component:
kgmol D
5804 kgmol F 0.019kg mol C3
1kg mol D
= 3243
1.00 kgmol F 0.034 kg mol C3
day
1day
C5+ as a tie element:
5804 kgmol F 0.006kg mol C5 +
1kgmol B
= 2679 kgmol B / day
1.00 kg mol F
1day
0.013 kgmol C5 +
4.1.20
Assume the process is in the steady state, and no reaction occurs. Solution assumes no S is
in stream A and no N2 in stream B.
Steps 1, 2, 3 and 4:
mass fr.
A
ω NH
3
ω HA2
1.00
mass fr.
0.0633
0.9367
1.00
NH 3
N2
mol fr.
A
x NH
3
NH 3
x NA2
S
1.00
mass fr.
ω BNH 3
ω SB
1.00
Gas
1000 kg
mol fr.
NH 3
0.10
N2
0.90
1.00
Step 5: Basis 1.0 hr
A
, ω NA2 , ω BNH 3 , ω SB , a total of 6.
Step 6: The unknowns are A, B, ω NH
3
Steps 7 and 8: Three compound balances can be made, N2, NH3, and S. One relation is
given:
A
B
ωNH 3 = 2 ωNH 3
4–21
Solutions Chapter 4
Two summations exist: ∑ ω Ai = 1 and ∑ ω Bi = 1
hence, we have an adequate number of balances (unless some are not independent) to solve
the problem.
Material balance equations in grams
Total:
F+S= A+B
1000 + 1000 = A + B
NH3:
A
1000(0.0633) + 0 = Aω NH
+ Bω BNH 3
3
N2 :
A
1000(0.9367) + 0 = Aω NH
+0
2
S:
0 + 1000 = 0 + Bω SB
Constraints
A
ω NH
+ ω NA2 = 1
3
ω BNH 3 + ω SB = 1
A
ω NH
= 2ω BNH 3
3
Solve in Polymath
Note: The equations can be converted to linear equations by letting
A
A
m NH
= ω NH
A, m NA2 = ω NA2 A, etc. The set can be solved in
3
3
A
.
Polymath or reduced to a quadratic equation in m BNH 3 or m NH
3
4–22
Solutions Chapter 4
4.1.21
Basis: 12 hours
The process is illustrated in the figure
MTBE
H2O
mass fr. = 1.00
MTBE
H2O
H2O
MTBE
The MTBE entering the pond in 12 hours is
25 boats 0.5 L gasoline 1000 cm3 0.72 g 0.10 g MTBE
= 900 g MTBE
1 boat
IL
1 cm3 1 g gasoline
The pond holds (ignoring the MTBE in the pond which is negligible)
3000 m 1,000 m 3m 1000 kg H 2O
= 9 × 109 kg H 2O
1 m 3 H 2O
The increase in the concentration of MTBE is
900 g MTBE 1 kg
= 1× 10-10 g MTBE/g H 2O
9
1000
g
9 × 10 kg H 2O
4–23
Solutions Chapter 5
BaCl2 + Na2SO4 → BaSO4 + 2NaCl
5.1.1
Mol. wt.: 208.3
142.05
233.4
58.45
a. Basis: 5.0 g Na2SO4
Example:
5 g Na 2 SO 4 1 g mol Na 2 SO 4 1 g mol BaCl 2 208.3 g BaCl 2
=
142.05 g Na 2 SO 4 1 g mol Na 2 SO 4 1 g mol BaCl 2
7.33 g BaCl 2
Problem
b.
c.
d.
e.
f.
g.
h.
i.
Basis
5 g BaSo4
5 g NaCl
5 g BaCl2
5 g BaSO4
5 lb NaCl
5 lb BaCl2
5 lb Na2SO4
5 lb NaCl
Answer
4.47 g BaCl2
8.91 g BaCl2
3.41 g Na2SO4
3.04 g Na2SO4
6.08 lb Na2SO4
5.59 lb BaSO4
8.21 lb BaSO4
9.98 lb BaSO4
AgNO3 + NaCl → AgCl + NaNO3
5.1.2
Mol. wt.: 169.89
58.45
143.3
85.01
a. Basis: 5.0 g NaCl
Example: 5.0 NaCl
1 g mol NaCl 1 g mol AgNO3 169.89 g AgNO3
=
58.45 g NaCl 1 g mol NaCl 1 g mol AgNO3
14.53 g AgNO3
Problem
b.
c.
d.
e.
f.
g.
h.
i.
Basis
5 g AgCl
5 g NaNO3
5 g AgNO3
5 g AgCl
5 lb NaNO3
5 lb AgNO3
5 lb NaCl
5 lb AgNO3
5–1
Answer
5.92 g AgNO3
10.00 g AgNO3
1.721 g NaCl
2.04 g NaCl
3.44 lb NaCl
4.22 lb AgCl
12.27 lb AgCl
4.22 lb AgCl
Solutions Chapter 5
5.1.3
a.
The balanced equation is:
3 As2S3 + 4 H2O + 28 HNO3 = 28NO + 6 H3AsO4 + 9 H2SO4
and the reaction is unique within relative proportions.
b.
2 KC1O3 + 4 H C1 = 2 KC1 + 2 C1O2 + C12 + 2 H2O
or
KC1O3
or
3 KC1O3 + 10 HC1 = 3 KC1 + 2 C1O2 + 4 C12 + 5 H2O
+ 6 H C1 = KC1
+ 3 C12
+ 3 H2 O
We classify these reactions as non-unique since they are not simply proportional
equations, but are linearly independent, and infinitely many equations can be obtained by
linear combination.
5.1.4
Basis: Data given in problem statement
1.603g CO2 1 gmol CO2 1 gmol C 12.00g C
= 0.437 g C
1
44.01g CO2 1 gmol CO2 1 gmol C
0.2810gH2 O 1 gmol H2 O 2 gmol H 1.008g H
= 0.03144 g H
1
18.02g H2 O 1 gmol H2 O 1 gmol H
0.6349 g CxHyOz – 0.437g C – 0.03144g H = 0.1661g O
Based on O:
0.1661g O 1 gmol O 0.01038 gmol O
=
= 1 gmol O
1
16.00g O
0.01038
0.437 g C 1 gmol C
0.03642 gmol C
=
= 3.5 gmol C
1
12.00 g C
0.01038
0.03144 g H 1 gmol H
0.03119 gmol H
=
= 3 gmol H
1
1.008 g H
0.01038
The formula in integers is
C 7 H6 O2
5–2
Solutions Chapter 5
5.1.5
a)
C
H
Zn
O
LHS
RHS
4
10
1
14
4
10
1
14
Yes, the equation is balanced.
b) Basis: 1.5 kg ZnO
1.5 kg ZnO 1000 g ZnO 1 gmol ZnO 1 gmol DEZ 123.4 gDEZ
1.0 kg ZnO 81.40 g ZnO 1 gmol ZnO 1 gmol DEZ
×
c)
1.0 kg DEZ
= 2.27 kg DEZ
1000 g DEZ
20 cm 3 H2 O 1.0 g H2 O 1.0 gmol H2 O 1 gmol DEZ 123.4 g DEZ
1 cm 3 H2 O 18.0 g H2 O 5 gmol H 2O 1 gmol DEZ
= 27.4 g DEZ
5.1.6
Basis: 2 g mol
C: 6 × 12 = 72 H: 8 × 1 = 8
mol. wt. = 72 + 8 + 96 = 176
O: 6 × 16 = 96
2 g mol 176 g 1 lb
= 0.775 lb
1 g mol 454 g
5–3
Solutions Chapter 5
5.1.7
Basis: Data given in the problem
Mass of hydrate
Mass of dry sample
Mass of water
10.407 g
-9.520 g
0.887 g
9.520 g BaI2 1 g mol BaI2
= 0.0243 g mol BaI2
391 g BaI2
0.887 g H2 O 1 g mol H2 O
= 0.0493 g mol H2 O
18.02 g H2 O
0.0243
= 0.49, or 2 H 2 O for 1 BaI2 . Thus the hydrate is BaI2 ⋅ 2H 2 O.
0.0493
5.1.8
Basis: 2000 tons 93.2% H2SO4 (1 day)
a.
2000 tons soln 0.932 ton H 2SO4 1 ton mole H 2SO4
1 mol S
32 ton S
1 ton soln
98 ton H 2SO4 1 mol H 2SO4 1 ton mol S
= 609 ton S
1.5 mol O2 32 ton O2
b. 609 ton S 1 ton mol S
= 913.5 ton O2
32 ton S 1 mol S 1 ton mol O2
1 mole H 2 O 18 ton H 2 O
c. 609 ton S 1 ton mole S
= 342.6 ton H2O
1 mol S 1 ton mol H 2 O
32 ton S
5–4
Solutions Chapter 5
5.1.9
FeSO4 ⋅7H2O
MW
277.9
FeSO4 ⋅H 2O
FeSO4 ⋅4H2O
223.9
FeSO4
MW
169.9
151.9
It is best to evaluate the three types of ferrous sulfate in terms of the cost/ton of FeSO4
Basis: 475 ton material
a.
FeSO4 ⋅7H2O:
$83,766.25
= $176.35/ton FeSO4 ⋅ 7H2 O , which is equivalent to
475 ton FeSO4 ⋅ 7H 2 O
$176.35
277.9 ton FeSO4 ⋅ 7H2 O 1 ton mol FeSO4 ⋅7H2 O
1 ton FeSO4 ⋅ 7H2 O 1 ton mol FeSO4 ⋅ 7H2 O
1 ton mol FeSO4
1 ton mol FeSO4
= $323/ton FeSO4
151.9 ton FeSO4
b.
FeSO4 ⋅4H2O:
$323 151.9 ton FeSO4
1 ton mol FeSO4
1 ton mol FeSO4 ⋅ 4H2 O
ton FeSO 4 1 ton mol FeSO4 1 ton mol FeSO4 ⋅ 4H2 O 223.9 ton FeSO 4 ⋅ 4H 2O
= $219 / ton FeSO4 ⋅ 4H2O
c.
FeSO4 ⋅H 2O:
$323 151.9 ton FeSO4
1 ton mol FeSO4
1 ton mol FeSO4 ⋅H 2O
ton FeSO 4 1 ton mol FeSO4 1 ton mol FeSO4 ⋅ H2 O 169.9 ton FeSO4 ⋅ H2 O
= $289 / ton FeSO4 ⋅ H2O
5–5
Solutions Chapter 5
5.1.10
The reaction is C 4H10 + 6 12 O2 → 4CO2 + 5H2O
Basis: 1 mol C4H10
! mol O2 for complete combustion "
Minimum O2 ≅ (LFL) #
$
mol C4 H10
%
&
≅ (1.9%) (6.5/1) = 12.4%
5.1.11
Basis: 1 ton dolomite
Assume complete
reactions
CaCO3 + H2SO4 → CaSO4 + H2O + CO2
MgCO3 + H2SO4 → MgSO4 + H2O + CO2
a.
Comp.
%
lb
mol.wt.
CaCO3
MgCO3
SiO2
68
30
2
1360
600
40
100.0
84.3
60.0
Total
100
2000
lb mol
that react
13.60
7.13
20.73
20.73 lb mol react 1 mol CO2 produced
44 lb
1 mol react
ton dolomite
1 lb mol CO2 = 912 lb CO2
b. 20.73 lb mol react
ton dolomite
1 mol H 2 SO4 reqd 98 lb H 2 SO4 100 lb acid
= 2160 lb acid
1 mol react
1 mole H 2 SO4 94 lb H 2 SO4
5–6
Solutions Chapter 5
5.1.12
Basis: 1 kg dichlorobenzene (DCB)
The balanced stoichiometric reaction for dichlorobenzene is shown below:
C6 H 4 C12 + 6.5 O2 → 6 CO2 + H 2 O + 2 HC1
Therefore, for 1 kg of dichlorobenzene, the following moles of HC1 are produced:
1 kg DCB 1 kg mol DCB 2 kg mol HC1
= 0.0136 kg mol HC1
147 kg DCB 1 kg mol DCB
The balanced stoichiometric reaction for tetrachlorobiphenyl is as shown below:
C12 H6 C14 + 12.5 O 2 → 12 CO 2 + H 2O + 4 HC1
Therefore, for 1 lb of tetrachlorobiphenyl (TCB), the following moles of HC1 are
produced:
Basis: 1 kg TCB
1 kg TCB 1 kg mol TCB 4 kg mol HC1
= 0.0138 kg mol HC1
290 kg TCB 1 kg mol TCB
Thus, the amount of acid produced does not change significantly (≈ 1.5%) , and neither will
the amount of base required for neutralization.
5–7
Solutions Chapter 5
5.2.1
Basis: 1 L solution = 1000 g H2O
50 g H2 S 1 g mol H2 S 1 g mol HOCl 52.45g HOCl 100 g HOCl soln
10 6 g soln 34 g H2 S 1 g mol H2 S 1 g mol HOCl
5 g HOCl
1000 g soln 2
=
1 L soln
3.08 g HOCl solution
You can use the density of H2O for the density of the solution as the H2 S content has
negligible effect on the density.
5.2.2
Basis: 135 mol CH4
ξ=
0 − 135
= 22.5 reacting mols
−6
5.2.3
Basis: 10 lb moles phosgene
For phosgene:
ξ=
n f − n i 10 − 0
=
= 10 reacting moles
1
1
n CO,i = n CO,f -ν COξ = 7 − (−1)(10)
=
17 lb mol
n C12 ,i = n C12 ,f -νC12 ,iξ = 3 − (−1)10
=
13 lb mol
5–8
Solutions Chapter 5
5.2.4
Basis: 1080 lb bauxite
A12 O3 +3H 2SO4 → A12 (SO4 )3 + 3H 2 O
1080 lb bauxite 55.4 lb A12 O3 1 lb mol A12 O3
= 5.87 lb mol A12 O3
100 lb bauxite 101.9 lb A12 O3
ξ=
0 − 5.87
= 5.87 reacting moles
−1
Basis: 2510 lb H2SO4 (77%)
2510 lb H2SO4 (77%) 0.77 lb H 2SO4 1 lb mol H 2SO 4
= 19.70 lb mol H 2SO4
1 lb H2SO4 (77%) 98.1 lb H 2SO4
ξ=
0 − 19.70
= 6.57 reacting moles
−3
(a)
A12O3 is the limiting reactant and H2SO4 is the excess reactant.
(b)
%xs H2SO4:
(19.70)(1/ 3) − 5.87
6.57 − 5.87
(100) =
(100) = 11.9%
5.87
5.87
or H2SO4 (77%) required:
%xs =
(5.87)(3)(98.1)
= 2244 lb
0.77
2510 − 2244
= 12%
2244
Basis: 2000 lb A12(SO4)3
2000 lb A12 (SO4 )3 1 lb mol A12 (SO4 )3 1 lb mol A12 O3
342.1 lb A12 (SO4 )3 1 lb mol A12 (SO4 )3
101.9 lb A12 O3 1 lb bauxite
=1075 lb bauxite
1 lb mol A12 O3 0.554 lb A12 O3
1075
× 100 = 99.5%
1080
5–9
Solutions Chapter 5
5.2.5
Basis: 100 kg of fusion mass
BaSO 4
+
4C → BaS
+
4CO
Comp.
Wt.% = kg
MW
kg mol
BaSO4
BaS
C
Ash
11.1
72.8
13.9
2.2
100.0
233.3
169.3
12.0
____
0.0476
0.430
1.16
____
BaSO4 orig. = 0.0476 + 0.430 = 0.4776 mol
Corig. = 1.16 + 4 (0.430) = 2.88 mol
Carbon is the limiting reactant.
Needed C for complete reaction = 4 (0.4776) = 1.9104 mol
% excess =
2.88 − 1.91
× 100 = 50.8% excess C
1.91
Degree of completion =
0.4776 − 0.0476
= 0.900
0.4776
You could calculate the extent of reaction from the BaSO4, and use it to get the original
amounts of reactants.
5–10
Solutions Chapter 5
5.2.6
CO + C12 → COC12
Basis: 20 kg products
Comp.
C12
COC12
CO
kg
3.00
10.00
7.00
20.00
kg
mol
0.0423
0.1011
0.250
0.393
mol.wt.
70.91
98.91
28.01
The extent of reaction is
CO:
0 − 0.351
= 0.351
−1
C12:
0 − .143
= 0.143 (smallest ξ )
−1
C12 is the limiting reactant; CO is the excess reactant
a.
0.351 − 0.143
(100) = 145% excess CO
0.143
b.
0.1011
(100) = 70.7% C12 converted
0.143
c.
0.1011 kg mol COC12
= 0.205
(0.143 + 0.351) kg mol reactants
5–11
kg mol
C12 in
0.0423
0.1011
0.143
kg mol
CO in
0.1011
0.250
0.351
Solutions Chapter 5
5.2.7
Basis: 100 kg mol A
100 kg mol C 6H 5NO2 86.9 kg mol C 6H 8NOCl 0.95 kg mol C 6H 7NO 3 kg mol C 8H 9NO2
100 kg mol C 6H 5NO2 kg mol C 6H 8NOCl 4 kg mol C 6H 7NO
→ 0.619 fraction overall conversion .
5–12
Solutions Chapter 5
5.2.8
C5 H7 O2 N
C5
H7
O2
N
60
7
32
14
113 kg/k mol
NH +4
N
14
H4
4
18
20,000 kg C 5 kg NH +4
= 1000 kg NH +4
100 kg C
1000 kg NH +4 kg mol NH 4+
= 55.556 k mol NH +4
+
18 kg NH 4
55.556 kg mol NH +4 1 kg mol C5H 2O2 N 95 kg mol NH +4 113 kg C5H 2O2 N
55 kg mol NH +4 100 kg mol fed kg mol C5H 7O 2 N
= 108.43 kg C5H7O2N
5.2.9
2C2 H4 + O2 → 2C2 H4O
For 100% conversion of C2H4 according to reaction (a):
2 mol C2 H 4O / time
=1
2 mol C2 H 4 / time
For reaction b3:
1 mol C2 H 4 O/time
= 0.75
1.33 mol C2 H 4 /time
5–13
Solutions Chapter 5
5.2.10
Reaction on which to base excess is
Fe2O3 + 3C → 2Fe + 3CO
Basis: 1 ton Fe2O3
a. Theoretical C =
1 ton Fe 2O3 2000 lb Fe 2O3 1 lb mol Fe 2O3
3 lb mol C 12 lb C
= 451 lb
1 ton Fe 2O3 159.7 lb Fe 2O3 1 lb mol Fe 2O3 lb mol C
600 œ451
× 100 = 33% excess C
451
b.
1 ton Fe 2O3 1200 lb Fe lb mol Fe 1 lb mol Fe 2O3 159.7 lb Fe 2O3
1 ton Fe 2O3 55.85 lb Fe 2 lb mol Fe 1 lb mol Fe 2O3
= 1720 lb Fe2O3 reacts
1720 × 100 = 86.0% Fe 2O3
2000
c. (i)
1 ton Fe 2O3 183 lb FeO lb mol FeO 1 lb mol CO 28 lb CO
= 35.62 lb CO
1 ton Fe 2O3 71.85 lb FeO 2 lb mol FeO 1 lb mol CO
1 ton Fe 2O3 1200 lb Fe 1 lb mol Fe 3 lb mol CO 28 lb CO
= 902.4 lb CO
1 ton Fe 2O3 55.85 lb Fe 2 lb mol Fe 1 lb mol CO
Total (35.62 lb CO + 902.4 lb CO) = 938 lb CO
(ii)
1 ton Fe 2O3 1200 lb Fe lb mol Fe 3 lb mol C 12 lb C
= 387.1 lb C
1 ton Fe 2O3 55.85 lb Fe 2 lb mol Fe lb mol C
1 ton Fe 2O3 183 lb FeO lb mol FeO 1 lb mol C 12 lb C
= 15.3 lb C
1 ton Fe 2O3 71.85 lb FeO 2 lb mol FeO lb mol C
387.1 + 15.3 = 402.4 lb C used / 1ton Fe2O3
5–14
Solutions Chapter 5
or use
938 lb CO 1 lb mol CO 1 lb mol C 12 lb C
28 lb CO 1 lb mol CO 1 lb mol C
Selectivity: Mass basis:
d.
1200 lb Fe formed
= 6.56
183 lb FeO formed
1200 lb Fe lb mol Fe
= 21.49 lb mol Fe
55.85 lb Fe
183 lb FeO ×
Mole Basis:
lb mol FeO
= 2.55 lb mol FeO
71.85 lb FeO
21.49 lb mol Fe
= 8.44
2.55 lb mol FeO
5.2.11
Cl2 + 2NaOH → NaCl + NaOCl + H2O
MW: 71.0 40.01
58.5
74.5
18.0
Basis: 1145 lbm NaOH + 851 lb Cl2 (→ 618 lbm NaOCl)
a. Determine limiting reactant
Assume all Cl2 reacts, calculate lb of NaOH required
NaOH required =
2 lb mol NaOH lb mol Cl 2 851 lb Cl 2 40.01 lb NaOH
71 lb Cl 2
lb mol NaOH
lb mol Cl 2
= 959.11 lb NaOH required
C12 is limiting reactant
or use extent of reaction
NaOH:
1145
= 28.63 lb mol
40
(0-28.63)/-2 = 14.3
C12:
851
= 11.99 lb mol
71.0
(0-11.99)/ − 1 = 12.0 (minimum)
5–15
Solutions Chapter 5
b. % excess NaOH =
=
lb mol NaOH in excess
(100)
lb mol NaOH for rxn
1145 lb m/NaOH
lb mol NaOH
1
œ 959.11
40.01 lb NaOH
40.01
(100)
1
959.11
40.01
= 19.4% excess NaOH
618
lb mol C12 that reacted
c. Degree of completion = =
= 74.5 = 0.692
lb mol C12 available to react 851
71.0
ξ
8.30
or
=
=0.69
ξ min
12.8
lb
d. Yield of NaOCl = 618 lb NaOCl = 0.726
851 lb Cl 2
lb
e.
618
= 8.30 lb mol NaOC1
745
ξ=
8.30 − 0
= 8.30 reacting moles
1
5–16
Solutions Chapter 5
5.2.12
Basis: 30 mol CO, 12 mol CO2, 35 mol H2O feed = 1 hr
CO
CO2
H2O
mol
30
12
35
mol
F
CO + H2O
MW:
a)
CO
CO2
H2
18
H2O
P
28
18
CO2 + H2
fed
in eqn
CO 1
CO 30
=
= 0.86 <
= =1
H2 O 1
H2 O 35
44
2
CO is limiting reactant
b)
H 2O is the excess reactant
c)
18
= 0.514
35
d)
18
= 0.60
30
e)
18 kg mol H2 2.016 kg H 2
1 kg mol H 2O
1 kg mol H2 35 kg mol H2 O 18 kg H2 O
(for each mol of H2 produced, 1 mol CO reacts)
= 0.0576 kg H2
f)
1 mol of CO2 is produced for each mol of H 2
mol CO 2 produced
18
= 0.60
30
mol CO fed
(g)
ξ=
18 − 0
= 18 reacting moles using H 2
1
5–17
Solutions Chapter 5
5.2.13
Basis: 100 lb mesitylene (C9H12)
The reactions are
C9H12
MW
120.1
C9 H12 + H 2 → C8 H+10 CH 4
C8H10 106.08
C8 H10 + H 2 → C7 H+8 CH 4
C7 H8
92.06
The initial selectivity is: C8 H10 /C7 H8 = 0.7 and 0.8
The number of moles of C9H12 in the basis is
100 lb C9 H12 1 lb mol C9 H12
= 0.833 lb mol
120.1 lb C9 H12
For the selectivity of 0.7 (the equations are in lb mol)
(1) C8 H10 + (1)C7 H8 = 0.833
C8 H10 / C7 H8 = 0.7
with the results:
(lb mol)
(in lb)
C7 H8 = 0.490
C8 H10 = 0.343
45.1
36.4
For the selectivity of 0.8 the results are
C7 H8 = 0.463
C8 H10 = 0.370
42.6
39.3
The catalyst used in the respective cases is
Selectivity of 0.7
Selectivity of 0.8:
0.490 lb mol C7 H8 92.06 lb C7 H8
1 lb catalyst
500 lb C7 H8
1 lb mol C7 H8
0.463 lb mol C7 H8 92.06 lb C7 H8
1 lb catalyst
500 lb C7 H8
The economic analysis of the change is
5–18
1 lb mol C7 H8
= 0.090 lb
= 0.085 lb
Solutions Chapter 5
Income change
Increase in C8H10:
Decrease in C7H8:
$
(39.3 − 36.4)($0.65) = +1.89
(45.1 – 42.6) ($22) =
− 0.55
Expense change (decrease)
Change in catalyst used: (0.090-0.085) ($25) = + 0.13
$1.47
Net change
5.3.1
Basis: 1 mol A
A → 3B
50 percent conversion gives as the product:
A
B
mol
0.5
1.5
2.0
(a)
Mole fraction A = 0.5/2.0 = 0.25
(b)
Extent of reaction ξ =
0.5 − 1.0
= 0.5 using A
−1
5–19
Solutions Chapter 5
5.3.2
Steps 1, 2, 3 and 4:
mole %
Product
Methanol F
CH3 OH 100%
Reacts
P
A
Air
O2
N2
mol frac
0.21
0.79
1.00
Step 5:
Basis: 100 mol P
Step 6: Unknown
F, A
Step 7: Balances
N, H, O, C
H2 O
5.9
O2
13.4
N2
62.6
CH2 O
4.6
CH3 OH
12.3
HCOOH
1.2
100.0
More balances than unknowns. Are they independent? Consistent?
Steps 8 and 9:
Balances in moles
Check by first solving C, N (tie elements)
C:
F (1) = 4.6 + 12.3 + 1.2 = 18.1
F = 18.1 mol
2N:
A (0.79) = 62.6
A = 79.24 mol
F
A
Check via O: (2) (79.24)(.21) + 18.1
51.38
?
=
13.4(2) + 5.9+4.6+12.3+1.2(2)
?
=
52.0
Close but not exactly the same, more oxygen out than in. Might be caused by round off.
Check via H:
4(18.1)
?
=
5.9(2) + 4.6(2) + 12.3(4) + 1.2(2)
?
72.40
= 72.6
Not exactly the same; more hydrogen out than in, but again may be caused by round off.
The conclusion is that probably no error exists in the measurements.
5–20
Solutions Chapter 5
5.3.3
a.
Additional information was the molecular weights of the compounds and the
chemical reaction equations.
MW CaCO3 = 100.0, MW MgCO3 = 84.3, MW CO2 = 44.
b.
Assume a batch process
Final
Initial
F
CaCO3
CaO
MgCO3
MgO
P
G
CO2
c.
Basis: 1 g sample
d.
F = 1.000 g; P is not explicitly given but can be calculated: P = 0.550 g;
G is not explicitly given and can be calculated, G = 0.450 g
e.
Mass fraction of CaO in P and F, mass fraction MgO in P and F
(or mass of each)
f.
Species balances
g.
It appears to be a carbon balance
h.
0 degrees of freedom
i.
Yes
5–21
Solutions Chapter 5
5.3.4
Steps 1, 2, 3, and 4:
G(kg mol)
Ore
Cus
Cus*
inert
SO3
P(kg mol)
Reactor
n CuS (kg mol)
Y (kg mol)
CuO
A(kg mol)
Air
O2
N2
*
100%
SO2
mol fr.
0.072
O2
0.081
N2
0.847
1.000
unreacted CuS
inert
mol fr.
0.21
0.79
1.00
This CuS doesn’t react.
This is a steady state open process with reaction.
Step 5: Basis: 100 kg mol P
Steps 6 and 7: Unknowns: n CuS ,A, G, Y
Element balances: S, O, N, Cu
Steps 8 and 9: Balances are in kg mol on the elements
S:
n CuS (1) = G(1) + 0.072 (100)
2O:
A(0.21) = G(1.5) + 100(0.072) + 100(0.081) + Y( )
2N:
Cu:
A(0.79) = 100(0.847)
n Cus (1) = Y(1)
1
2
5–22
Solutions Chapter 5
Solution yields (in kg mol): G = 1.81, nCuS = Y = 9.01
1.81 mol SO3 1 mol S
1 mol SO3
= 0.20 of CuS that reacts goes to SO3 20%
9.01 mol CuS that reacts 1 mol S
1 mol CuS
If you use the extent of reaction, you need to write down the balanced reactions
CuS + O2 → CuO + SO2
SO2 +
1
2
(1)
(2)
O2 → SO3
Steps 6 and 7:
The unknowns: 4 above plus ξ1 and ξ2
Balance: 6 species balances
Steps 8 and 9:
CuO:
CuS
Y
0
=
=
0 + (1) (ξ1 )
n CuS + (−1)(ξ1 )
O2 :
P(0.081)
=
A(0.21) + (−1)(
+ξ1−) (
N2
SO2
SO3:
P(0.847)
P(0.072)
G
=
=
=
A(0.79)
ξ1 = 9.01
ξ2 = 1.81
0 + (1)(ξ1 ) + ( 1)(ξ2−)
0 + (1)(ξ2 )
5–23
1
2
)(ξ 2 )
Solutions Chapter 5
5.3.5
Assume a steady state flow process with reaction.
Step 5: Basis: 1 hr (6.22 kg mol = F)
Steps 1, 2, 3 and 4:
H2 (g)+SiHC13 (g) → Si(s)+3HC1(g)
F = 6.22 kg mol
Mol fr.
0.580 H 2
Reactor
Mol fr. mol
W(kg mol) HCl x HCl nHCl
SiHCl3 x SiHCl 3 n SiHCl 3
H2
P
0.420 SiHCl 3
1.00
Si
0.223 nH 2
Mol fr.
1.00
MW Si = 28.086
n H 2 /W = 0.223
Step 6:
Unknowns:
W, P, 3 n i
Step 7:
Balances:
element balances are H, Si, Cl, ∑ ni = W, n H 2 /W = 0.223
Or, using the extent of reaction
Step 6:
Unknowns = 6:
W, P, 3ni ,ξ
Step 7:
Balances = 6
Species balances: H2, SiHC13, HC1, Si
∑ ni = W
n H2 /W = 0.223
Steps 8 and 9:
Balances are in kg mol.
5–24
Solutions Chapter 5
Cl (kg mol):
Out
In
=
6. 22(0. 420)(3) nHCl + n SiHCl3 (3)
Si (kg mol):
6.22(04.20)(1) = P(1) + n SiHCl3
H (kg mol):
6.22[( 0.580 )( 2) + 0.420(1)] = n HCl + nSiHCl3 + nH 2 (2 )
Balances using ξ :
H2 : "%n H2 − (0.580)(6.22) #& = ( 1)(ξ )
SiHC13 : "%nSiHC13 − (0.420)(6.22) #& = ( 1)(ξ )
−
−
HC1:
[n HC1 − 0] = (+3)(ξ )
Si : [nSi − 0] = (+1)(ξ )
ξ = 1.83
Solution:
W = 8.05
P=1.83
n H 2 = 1.78
n HCl = 5.49
n SiHCl3 = 0.7824
1.83kgmol Si 1hr 20 min 28.086 kgSi
= 17.13kgSi
1hr
60min
1kgmol Si
1.46 kg Si initial
18.59 kgfinal Si
5–25
Solutions Chapter 5
5.3.6
6HF(g) + SiO2 (s) → H2SiF6 (l ) + 2H2O(l )
(1)
H2SiF6 (g) → SiF4 (g) + 2HF(g)
(2)
The net reaction is
4HF + SiO2 → SiF4 + 2H2O
Steps 2, 3, and 4:
SYSTEM
mol fr.
0.50 HF
0.50 N2
1.00
F
SiO2
on wafers
P
The process is an unsteady state, open process
Step 5:
Basis: 1 mol F entering
Steps 6 and 7:
Unknowns:
4 ni , P1 ξ, loss SiO2 from wafer (L)
Balances:
HF, N2 , SiF4 , H2O, SiO2 (5 species)
Other equations: ∑ ni = P; 10% HF reacts
Steps 8 and 9:
SiO2:
HF:
n HF = 0.50 + (−4)ξ = 0.50(.90) = 0.45
ξ = 1.25 ×10−2 reacting moles
SiF4:
n SiF4 = 0 + (1)(1.25×10-2 ) = 1.25 × 10-2 mol
H2O:
N2 :
n H2O = 0 + (2)(1.25 × 10-2 ) = 2.50 × 10-2 mol
n N2 = 0.50 mol
n SiO2 ,f − n SiO2 ,i =(−1)(ξ ) = −1.25 × 10-2 mol
5–26
nHF
nN2
nSiFH
nH2O
Solutions Chapter 5
Composition
HF
SiF4
H2 O
N2
mol
0.45
1.25 10-2
2.50 10-2
0.50
0.9875
mol fr.
0.456
0.013
0.025
0.506
1.000
If you use element balances:
Step 6 and 7:
The unknowns: 4ni, P, L
The balances: H, F, Si, O, N (presumably 4 are independent)
Other equations: ∑ n i = P, 10% HF reacts
Steps 8 and 9:
H:
0.50(1) = n HF + 2n H O
F:
0.50(1) = n HF + 4nSiF
Si:
(nSiO ,f -nSiO ,i ) = nSiF
O:
2(nSiO ,f -nSiO ,i ) = n H O
N:
2(0.50) = 2n N
2
4
2
2
2
4
2
2
2
n HF = 0.90(0.50)=0.45
5–27
Solutions Chapter 5
5.3.7
Steps 1, 2, 3 and 4:
SiH4 +O2 → SiO2 +2H2
4PH3 + 5O2 → 2P2O+5 6H2
100% O2
100% SiH4
A
B
D
C
100% PH 3
SiO2
H2
100%
E
P2 O5
Step 5: Basis:
1 g mol PH3
Step 6: No. Unknowns:
A, B, D, E, n EP 2 O 5
5
Step 7: Element balances:
Si, H, P, O, ωEP2O5
5
in
Out
Note: n
E
P 2O 5
+n
E
SiO2
=E
Steps 8 and 9:
P Bal (in grams):
!
1 g mol PH3 2 g mol P2O5 2 g mol P
31gP
= 31g P "
4 g mol PH3 1 g mol P2O5 1 mol P
"
# in P2O5
1 g mol PH3 2 g mol P2O5 5 g mol O
16 g O
"
= 40 g O "
4 g mol PH3 1 g mol P2O5 1 g mol O
$
Total:
or since all the P is in stream E:
________
71 g P2O5
1 g mol PH3 2 g mol P2O5 142 g P2O5
= 71g P2O5
4 g mol PH3 1 g mol P2O5
Si Bal:
in
Out
5–28
Solutions Chapter 5
B g mol SiH4 1g mol SiO2 1g molSi 28.1g Si
= 28.1B g Si
1g mol SiH4 1g mol SiO2 1g mol Si
B g mol SiH4 1g mol SiO2 2 g mol O 16 g O
= 32 B g O
1g mol SiH4 1g mol SiO2 1g mol O
71
= 0. 05
71 + 60.1B
1 g mol PH 3
Total:
3.55 + 3.0058B = 71
34 g PH 3
= 34 g PH 3
1g mol PH 3
23.5
. g mol SiO2 32.1g SiH 4
= 7.54 g SiO2 O2
1g mol SiH4
754
34
= 22.2
gSiO 2
gPH
5–29
60.1 B g SiO2
B = 23.5 g mol
Solutions Chapter 5
5.3.8
Steps 1, 2, 3 and 4:
Cu
Cu(NH 3 ) 4 Cl 2
NH 4 OH
A
B
C
D
E
Cu(NH 3 ) 4 Cl
H2 O
F
Cu
Step 5:
Basis: 1 board
3
A:
4in 8in 0.03in (2.54cm) 8.96g l gmol Cu
= 2.219g mol Cu
Cu foil
in 3
cm 3 63.55g Cu
F:
Cu remaining
1 − 0.75 2.219 gmol Cu
Steps 6 and 7: Unknowns:
= 0.555g mol Cu
B, C, D, E
Balances: Cu, N, Cl, H, O should be enough if one is redundant
Steps 8 and 9:
(Balances to be solved:)
Cu Balance:
2.219g mol Cu+
Bg mol Cu(NH3 )4 Cl 2
1gmol Cu
1g mol Cu(NH3 )4 Cl 2
D gmol Cu (NH3 ) 4 Cl
1gmol Cu
1gmol Cu (NH3 )4 Cl
= 1.664 gmol Cu +
5–30
Solutions Chapter 5
Cl Balance:
Bg mol Cu(NH3 )4 Cl 2
D g mol Cu(NH3 )4 Cl
2 gmol Cl
1gmol Cl
=
gmol Cu(NH3 )4 Cl 2
gmol Cu(NH3 ) 4 Cl
N Balance:
Bg mol Cu(NH3 )4 Cl 2
=
D gmol Cu(NH3 ) 4 Cl
Solution:
Cu overall:
Cg mol NH4 OH 1gmol N
4 gmol N
+
gmol Cu(NH3 )4 Cl 2
gmol NH4 OH
4gmol N
g mol Cu(NH4 )Cl
A + B = F + D →A + B = F + (2B)
2.219 + B = 0.555 + (2B)
B = 2.219 - 0.555 = 1.664 g mol Cu(NH3)4Cl2
Cl:
D = 2B= 2(1.664) = 3.328 g mol Cu(NH3)4Cl
N:
4B + C = 4D
C = 4(D-B) = 4(3.328 - 1.664) = 6.656 g mol NH4OH
MW of NH4OH
N
14
H4
4
O
16
H
1
35
35gNH4 OH 6.656gmol
= 232.0 g NH4 OH
gmol
Cu(NH3)4CL2
Cu
N
H
Cl2
63.55
56.00
12.00
70.90
202.45
202.45 (g/gmol) (1.664 g mol) = 336.88 g Ca ( NH 3 ) 4 Cl 2
5–31
Solutions Chapter 5
5.3.9
F
Feed
Mixer
contaminated
sand
( 16,000 lb )
F'
Incinerator
60%
combustible
sand plus
combustion
products
H
a)
O2 enriched air
x O 2 = 0.4
hexane
100% excess
Basis: 8 tons of contaminated sand containing 30% PCB
4, 4' dichlorobiphenyl ⇒
Cl
Cl
C12H8Cl2, MW = 222
8tons sand + PCB 0.3ton PCB 2000lb PCB
= 4800 lbPCB
tonsand + PCB ton PCB
If we want the net feed to be 60% combustible, we see that the sand is a tie component:
A sand balance around the feed mixer: 16000(0.7) = F'(0.4) so F' =28,000 lb.
A total balance around the feed mixer.: F + H = F' ⇒ H = 12, 000lb m hexane
b)
C1 2H8Cl 2 + 13.5O2 → 12CO2 + 2HCl + 3H 2 O
C 6 H1 4 + 9.5O2 → 6CO2 + 7H2 O
Even with total combustion, we will produce undesirable HCl, which will adversely
affect the environment if it is simply vented. Therefore, it would be wise to condense
the HCl and H2O by cooling. Perhaps the acid solution might be sold to help pay for the
process.
(c)
21.62 moles PCB
P
16.54% CO2
12.20% O2
139.5 moles nC 6
H2 O = ?
HCl = ?
N2 = ?
A
enriched air
(40% O2, 60% N 2 )
5–32
Solutions Chapter 5
By a tie component for CO2 , we know:
12(21.62) + 139.5 (6) = 0.1654 P hence P = 6629 mol
By an O tie: 0.40(2)A = 0.1654(2)(6629) + 0.1220(2)(6629) + x H 2 O (1) (6629)
x H 2 O (6629) =
21.62 PCBReacted 3H2 O 139.6 C6 Reacted 7 H2 O
=
= 1042
1 PCB
1 C6
so x H 2 O = 0.1572
From O tie balance: A = 6065.4 mol of 40% O2 air
Thus: (N2) in = (N2) out = 0.6(6065.4)=3639.2 mol N2
xN2 =
3639.2
= 0.549
6629.0
We can get HCl by a Cl balance:
21.6PCBReacted 2 HCl
= 43.2mol H Clin product
1PCB
xHCl= 0.0065
Product
Composition
Component
CO2
O2
H 2O
HCl
xi
0.1654
0.1220
0.1572
0.0065
N2
0.5490
We know % excess O2 used is : % excess =
100 [O2 fed − O2 req' d ]
(O2 req' d)
O2 fed = 0.4 (air fed) = 0.4 (6065.4) = 2426.2 mol O2
Also, since we have total combustion,
(O2 req'd) = (O2 consumed) = (O2 fed - O2 out) = 2426.2 - 0.1220(6629)
= 1617.5 moles req'd
⎛ 2426.2 − 1617.5⎞
50%excessO 2 used
Thus: % O2 = 100 ⎝
⎠ =
1617.5
5–33
Solutions Chapter 5
5.4.1
Steps 1, 2, 3, and 4: To solve this problem you must recall that CaCO3 reacts with
H2SO4 to form CaSO4 and that CaCO3 when heated yields CO2 and CaO whereas when
CaSO4 is heated at the same time it does not decompose. Although the problem seems to
be underspecified, when you draw a diagram of the process and place the known data on
it, the situation becomes clearer. Assume the process is a steady state open one with
reaction. Then the element balances are just in = out. If you use the extent of reaction,
you make species balances.
Note that we have designated the composition of P and A in terms of the mass of
each component, m Pi , rather than the mass fraction ω Pi , because this choice makes the
element or species balances linear (avoids products such as ωP).
The CO2 liberated from the sludge is equivalent to
1lbCO2 1lb mol CO2 1lb mol CaCO 3 100.09lb CaCO3
10 lbP 44lb CO2 1lb mol CO2 1lb mol CaCO3
= 0.227 lb CaCO3/lb P
A (lb) waste acid
lb
A
H2 SO4
m H SO
H 2O
m
2
Y (lb mol)
4
A
H 2O
CO2 gas
∑ =A lb
Feed F (lb)
Reaction
Vessel
P (lb)
dry
sludge
mass fr.
CaCO 3
inert
0.95
0.05
1.00
1.000 = mol fr.
Water
W (lb)
H2 O mass fr. = 1.00
Step 5:
lb
CaCO 3
m PCaCO
CaSO 4
m CaSO
inert
m Pinert
3
P
4
∑ = P (lb)
also
m PCaCO
P
3
= 0.227
Basis: F = 100 lb
A
Step 6:
The variables whose values are unknown are A, m AH 2 SO 4 ,m H 2 O ,m PCaCO3
m PCaSO 4 , m Pinert , P, W, and Y, a total of 9.
5–34
Solutions Chapter 5
Step 7:
The element balances that can be made are Ca, C, O, S, H, inert (6 total)
and we know Σ m Ai = A , Σ m Pi = P, and m PCaCO3 / P = 0.227 for a total of 9. If the
equations are independent, we can find a unique solution.
(
)
Steps 8 and 9: The equations (in = out) are (except for the inert balance which is in lb)
in moles
1 lb mo lCa
95 lb CaCO3 1 lb mol CaCO3
100.09 lb CaCO3 1lb mol CaCO3
Ca:
m PCaCO3 lbCaCO3 1lb mol CaCO3
1lb mol Ca
100.09lb CaCO3 1lb mol CaCO3
=
m PCaCO3 lbCaSO4 1lb mol CaSO4
1lb mol Ca
136.15lb CaSO4 1lb mo l CaSO4
+
1 lb mol C
0.95 lb CaCO 3 1 lb mol CaCO3
Y lb mol CO2 1 lb mol C
=
100.09 lb CaCO3 1lb mol CaCO 3
1 lb mol CO2
C:
+
m PCaCO3 lbCaCO3 1 lb mol CaCO3
1 lb mol C
100.09 lb CaCO3 1 lb mol CaCO 3
A
m H 2 SO 4 lb H2 SO4 1 lb mol H2 SO4
S:
1 lb molS
98.08 lb H2 SO4 1 lb mol H2 SO4
m PCaSO4 lb CaSO4 1lb mol CaSO4
1lb molS
=
136.25lbCaSO4 1lb mol CaSO4
2H :
m AH 2 SO 4 lb H2 SO 4 1 lb mol H2 SO 4 1 lb mol H 2
98. 08 lb H 2 SO4 1 lb mol H2 SO4
m AH 2 O lb H2 O 1 lb mol H2 O 1 lb mol H 2
+
18. 016 lb H2 O 1 lb mol H 2 O
=
O:
+
W lb H2 O 1 lb mo l H2 O 1 lb mol H2
18. 016 lb H2 O 1 lb mol H2 O
95 lb mol CaCO 3 1 lb mo l CaCO3
3 lb mol O
100.09 lb CaCO3 1 lb mol CaCO 3
m AH 2 SO4 lb H2 SO4 1 lb mol H2 SO4
4 lb mol O
98.08 lb H2 SO4 1 lb mol H2 SO4
5–35
Solutions Chapter 5
A
m H 2 O lb H2 O 1lb mol H2 O
+
1lb mol O
18.016lb H2 O 1lb mol H 2 O
=
Y lb mol CO2 2lb mol O
1mol CO2
+
m PCaCO3 lbCaCO3 1lb mol CaCO3
3lb mol O
100.09lb CaCO3 1lb mol CaCO 3
+
+
m PCaSO4 lb CaSO4 1lb mol CaCO 3
4lb mol O
136.15CaCO 4 1lb mol CaSO4
W lb H2 O lb mol H 2 O 1lb mol O
18.016lb H2 O l lb mol H 2O
Inert: 5 lb inert = m Pinert lb inert
The solution of these equations using a computer code would give all of the values of the
unknown quantities. However, a review of the set of equations, after having gone into
great detail about each equation, indicates that only 4 of the equations have to be solved
to get the desired answer, namely the Ca balance, the Σm Pi = P, and the inert balance plus
m PCaCO3 = 0.227 :
Σm Pi = P:
P
m PCaCO3 + m PCaSO4 + 5 = P = m CaCO
/ 0.227
3
Ca:
0.949 = 0.010 m PCaCO3 + 0.00734 m PCaSO4
m PCaCO3 = 28.55 lb and 100
28.55
= 30%unreacted .
95
5–36
Solutions Chapter 5
5.5.1
Basis: 100 kg mol
synthesis
gas
Burner
flue gas
air
One composition is unknown, the flue gas. After selection of a basis, the air flow is
known. Hence, this problem can be solved by direct addition and subtraction
Unknowns:
(CO2, H2O, N2,O2) = 4
Balances:
(C, H, O, N) = 4
Comp.
%=mol
CO2
O2
CO
H2
N2
6.4
-------------0.2
----------(0.2)
40.0 CO+1/2 O2→CO2
20.0
50.8 H2+1/2 O2→H2O
25.4
2.6
-------------XS O2: (0.40)(45.2) =
18.08
Total O2 in
63.28
N2 in with O2 (63.28)(79/21) =
Totals
Comp.
Reaction
mol
%
CO2
46.4
13.0
H 2O
50.8
14.3
N2
240.65
67.6
18.1
5.1
356.0
100.0
O2
Totals
reqd.O2
5–37
CO2
H2 O
N2
O2
6.4
40.0
50.8
2.6
18.08
46.4
50.8
238.05
240.65
18.1
Solutions Chapter 5
5.5.2
Steps 1,2 3 and 4:
H2 O 100 %
mol %
W
P
mol frac F (mol) = ?
CH 4 x
N2
y
A
1.00
O2
N2
air
CO
1.0
C O2
8.7
O2
3.8
N2
86.5
100.0
mol frac
0.21
0.79
1.00
Step 5: Basis: 100 mol P
Step 6: Unknowns:
x, y, A, W, F Total 5
Step 7: Element Balances:
Equations:
C, H, N, O
a)
If x and y are moles, not mole fractions,
add: x + y = F
b)
If x, y are mole fr., add x + y = 1.00
Steps 8 and 9: Use x and y as mol:
C balance:
x = 8.7 + 1.0 = 9.7
2N balance:
y + .79A = 86.5
2H balance:
2x = W = 2(9.7) = 19.4
2O balance:
0.21A = (1/2) (W + 8.7) + 1.0 + 3.8
Use 2O bal:
A=
A and P will be mol
9.7 + 8.7 + 0.5 + 3.8
= 108
0.21
5–38
Total 5
Solutions Chapter 5
Use 2N bal:
y = 1.10
required O2: CH4 +2O2→ CO2+2H2O
(1)
(9.7)(2) = 19.4 mol
total O2 in: (0.21) (108) = 22.7
% xs air = % xs O2 =
3.3
(100)
19.4
= 17%
(a)
(b)
excess O2= 22.7 - 19.4 = 3.3 mol
Percentage of CH4, N2: CH4 =
9.7
(100) = 89.8%
9.7 + 1.1
N2 =
10.2%
Alternate solution:
Use extent of reaction and species balances.
Another equation needed for CO:
(2)
CH 4 + 1.5O2 → CO + 2H 2 O
Species balances to get ξ1 and ξ 2 :
F
CH 4 : O = n CH
+ ( −1)+ξ1 − ( 1)ξ 2
4
CO2 : 8.7 = 0 + (1)ξ1
CO: 1.0 = 0 + (1)ξ 2
ξ1 = 8.7 reacting moles
ξ 2 = 1.0 reacting moles
F
Thus n CH
= 8.7 + 1.0 = 9.7 mol and other compounds follow.
4
5–39
1.1
(100)=
9.7+1.1
Solutions Chapter 5
5.5.3
Steps 2, 3, and 4:
H2 O 100%
W
C
H
inert
P
F
mass frac
0.90
0.06
0.04
Furnace
A
1.00
O2
N2
mole %
R
mol frac
0.21
0.79
mass frac
0.10
C
inert 0.90
1.00
Step 5: Basis: 100 kg mol P
CO2
13.9
C
13.9
CO
0.8
0.8
O2
4.3
N2
81.0
100.0
14.7
1.00
Steps 6 and 7:
Unknowns: F, R, A, W
4
Probably ok but check for redundancy
Balances:
C, H, O, inert, N2
5
Steps 8 and 9:
Balances (in moles):
C:
F(0.90)
R(0.10)
= P(0.139+0.008) +
+ W(0)
12
12
H:
F (0.06)
= P (0) + R(0) + W (2)
1.005
2N:
A (0.79) = P(0.81) + R (0) + W (0)
2O:
A (0.21) = P (0.139 +
0.008
W
+ 0.043) +
2
2
Inert (mass): F (0.04) = R (0.90)
5–40
Solutions Chapter 5
From 2O,
100(.81)
= 102.53kgmol
0.79
W = 5.863 kg mol
From H,
F = 196.98 kg
From inert,
R = 8.7548 kg
A=
From 2N,
196.98(0.06)
= 5.94
2
5.94
O2 in:
= 2.96
2
196.98(0.90)
C in =
= 14.77
12
H in:
Step 10:
Check via C: 196.98 (0.90) /12 = 100 (0.147) + (8.7548 (0.10)/12)
ok
Required O2: 14.7 + (5.863/2) = 17.63 kg mol Total O2in = 102.53 (0.21) = 21.53
Excess (21.53 - 17.63)/17.63 = 0.221 or 22%
Alternate solution using extent of reaction:
C+O2 → CO2
1
C+ O2 → CO
(1)
(2)
2
1
2H+ O 2 → H 2 O
2
CO 2 :
(3)
P
F
n CO
=13.9 = n CO
+(1) ξ1
2
2
ξ1 = 13.9
CO:
n =0.8 = n +(1) ξ2
ξ 2 = 0.8
N2 :
n PN2 = 81.0 = n AN2
A = 102.5 mol
P
CO
F
CO
n OA2 = 0.21(102.5) = 21.5 mol
1
1
2
2
O2:
4.3 = 21.5 + (-1) ( ξ1 ) + (- )( ξ 2 ) + (- ) ( ξ 3 )
C:
0 = nCF + (-1)(13.9) + (-1)(0.8)
H:
0 = nFH + (-2) (2.90)
nCF = 14.7 mol
nHF = 5.80 mol (some round off)
5–41
ξ 3 = 2.90
Solutions Chapter 5
5.5.4
Steps 1, 2, 3 and 4:
H2 O
100%
W
CH4 100%
P
F
A
CO2 20.2
O2
4.1
N2 75.7
100.0
enriched O2 40
N2 60
air
100
Step 5:
Basis 100 mol P
Step 6:
Unknowns: F, A, W (all compositions known)
Step 7:
Element balances: C, H, O, N (1 extra balance)
Steps 8 and 9:
C balance: 20.2 = F
2N balance: 75.7 = 0.6A, A = 126.2
2H balance: 2F = W, W = 40.4
2O balance: 0.4A = 1/2 W + 20.2 + 4.1, 50.5 ≠ 20.2 + 20.2 + 4.1 = 44.5
Analysis isnot correct.
5–42
Solutions Chapter 5
5.5.5
Basis: 100 mol F
P
F = 100 mol
Moles = %
CH4 : 60
C 2 H6 : 20
CO:
5
O2 :
5
N 2:
10
100
50% xs air:
CH4 + 2O2
n CO2
A
0.21 O2
0.79 N
1.00 2
Moles
281.25
1058
or
x CO2
n H 2O
x H 2O
n O2
xO2
nN2
x n2
∑ =P
∑ =1.00
Calculate O2 and N2 in with the air in
reqd O2 (mols)
60 (2) = 120
→ CO2 + 2H2 O
C 2 H6 + 312 O2 → 2CO2 + 3H 2O
20(3.5) = 70
2.5
⎛1
5⎝ ⎞⎠ =
192.5
2
1
CO + O 2 → CO 2
2
Less O2 in gas
5.00
Net required O2
xs O2: 187.5 (.5) =
Total
N2 in with O2
187.5
93.75
281.25
281.25 0.79
= 1058
0.21
Material balances (elemental):
in (F)
+
in (A)
=
out (P)
C:
60 + 2 (20) +5 +
0
=
n CO2
H:
60 (4) + 20 (6) +
0
=
n H 2 O (2)
5
+5
2
+
281.25 =
n CO2 +
O as O2:
5–43
n H 2O
2
+ nO 2
Solutions Chapter 5
N as N2:
10
+
n O 2 = 288.75 - 105 -
1058
180
= 93.75
2
moles
composition of stack in % moles
n CO2 = 105
7.26
12.44
73.82
6.48
n H 2 O = 180
n N 2 = 1068
n O 2 = 93.75
Total moles =
out
nN2
=
1446.75
100.00
5.5.6
Steps 2, 3, and 4:
7.9%N2
21%O2
A
F = 100 mol
P
CS2 40.0
SO2 10.0
H2O 50.0
xCO
2
xSO
2
xO
2
xN
W
100% H2O
Step 5:
Basis: 100 mols feed; assume complete combustion
Step 4:
Equations:
CS2 + 3O2 → CO2 + 2SO2
O2 required:
% excess:
40 mol CS2 3 mol O2
= 120.0 mol O2
1
1 mol CS2
O 2 excess
× 100 =% excess
O 2 required
Steps 6, 7, 8 and 9:
5 unknowns and 5 equations → unique solution
5–44
2
Solutions Chapter 5
Element balances
Component
C
Mol in
40.0
Mol out
Px CO
2S
1
1
2O
10.0 + (50.0) + .21A
(x C O2 + 0.02 + x O2 )P+ W
2N
2H
.79A
50.0
(1-x O2 − 0.02 − x CO2 )P
W
2
2
(10.0) + 40.0
2
1
(0.02)P
1
2
2
2(45.0)
= 4500
.02
x CO2 = 0.0089
P=
% Excess =
0.1829(4500 molO2 )
(100) = 686%
120 mol
Alternate solution: Use extent of reaction and species balances.
5.5.7
Steps 2, 3 and 4:
W
Fuel
F
mass fr.
C 0.74
H 0.14
ash 0.12
1.00
Flue gas
A
air O2 0.21
N2 0.79
1.00
Step 5:
H2O 1.00
R
ash
1.00
Basis: P = 100 mol
Step 6: Unknowns:
F,A,R,W !" a discrepancy unless one
#
Step 7: Balances: C,H,O,N,ash "$ balance is not independent
Steps 8 and 9:
all balances in moles
5–45
P
CO2
CO
O2
N2
mol
12.4
1.2
5.7
80.7
100.
Solutions Chapter 5
C balance:
F(0.74)
= 12.4 + 1.2 = 13.6
12
F=
(13.6)(12)
= 220.5 lb
0.74
2N balance:
A(0.79) = 80.7
A = 102.2 mol
2H balance:
F(0.14)
= W;
2
W = 15.44 mol
2O balance:
A (.21) = W
(1.00)
1.2
+ 12.4 +
+ 5.7
2
2
21.46 = 26.42
not equal hence
Some discrepancy exists
A total balance can be used, but must be in mass, and the mass of P calculated.
5.5.8
Basis: 100 moles exhaust gas
Comp
CO2
O2
N2
Total
exit gas
%
exhaust gas mol
16.2
tie element
13.1
4.8 ⎫
⎬
79.0⎭ 83.8
100.0
16.2 mol CO2
100.0 mol exit gas
= 123.6mol exit gas
100.0 mol exhaust gas 13.1mol CO2
exhaust gas = 100.0 mol
air leaked in = 23.6 mol
23.6
= 0.236 mol air / mol exhaust gas
100.0
5–46
mol
16. 2
Solutions Chapter 5
5.5.9
Steps 2, 3, and 4:
H2 O 100%
W mol
Sludge
S mass
mass frac
C
S
H2
O2
0.40
0.32
0.04
0.24
mol
1.00
O2
N2
mole %
P mol
Furnace
A
F mass
mol frac
Fuel oil lb
0.21
0.79
mC
mH
1.00
∑ F (lb)
CO
2.02
CO2
10.14
O2
4.65
N2
81.67
SO 2
1.52
100.00
Step 5:
Basis: 100 mol P
Steps 6 and 7:
Unknowns:
S, W, F (or mH),A, mC
5 Element Balances: S, C, H, O, N
Element Balances (moles)
S:
0.32S
=
32
2N:
A(.79) = 81.67
C:
S(0.40) m C
+
=10.14+2.02
12
12
2O:
W
2.02
0.24(152)
+ A (0.21) =
+ 1.52 + 10.14 + 4.65 +
2
2
32
1.52
S = 152 lb
A= 103.38 lb mol (2998.01 lb)
mC = 85.12 lb
W = 11.04 lb mol (198.71 lb)
5–47
Solutions Chapter 5
0.04(152) m H
+
=W
2.03
2.03
2H:
mH = 16.32 lb
Check via a total mass balance
152 + 16.32 + 85.12 + 2998.01
3251.45
?
=
?
=
198.71 + 3035.56
3234.27
close enough
lb
85.12
16.32
101.44
C
H
152 lb
S
=
=1.50
F 101.44 lb
mass fr
0.84
0.16
1.00
B
5.5.10
Basis: 1 min
[CaO]
theor.
=
ft 3
1000 gal
min
1.05×62.4 lb soln. 0.02lb H 2SO4 mol H 2SO4
7.48 gal
56.1 lb CaO
mol CaO
ft
3
lb soln.
mol CaO
98 lb H 2SO4 mol H 2SO4
=100.3 lb CaO/min
Feed rate of CaO = 1.2 [CaO ]theor . = 120.4 lb CaO/min
CaSO4 production rate =
×
100.3 lb CaO
min
60 min 24 hr 365 d
hr
d
yr
mol CaO
mol CaSO4 136.1 lb CaSO4
56.1 lb CaO mol CaO
= 63,947 ton/yr
5–48
mol CaSO4
ton
2000 lb
Solutions Chapter 5
5.5.11
Basis: 100 mol of product gas (20.5%, C2H4O; 72.7 N2; 2.3 O2; and 4.5%CO2
C2 H 4 +
1
O2 → C2 H 4O
2
C2H4 + 3O2 → 2CO2 + 2H2O
Calculate ξ1 :
n iout − n iin
ξ1 =
υi
Use C2H4O:
Calculate ξ2 :
Use CO2:
ξ2 =
ξ1 =
20.5 − 0
= 20.5
1
4.5 − 0
= 2.25
2
Calculate the entering air using a N2 balance:
N2: 0.79A = 72.7; therefore, A = 92.03 mol
Calculate the moles of C2H4 entering (F)
in
n out
CH4 − n CH4 = υ1ξ1 + υ2 ξ2
0 − F = −ξ1 − ξ2 hence F = 22.75
O2 feed rate = 0.21A = 19.33 mol
(O2)theor = F/2 = 11.375 mol
% Excess O2 = [(19.33-11.375)/11.375](100) = 70.0%
To get the ethylene feed:
22.75 C2 H4 100,000 ton C2 H4O 2000 lb C2 H4O mol C2 H 4O
20.5 mol C2 H4O
yr
ton C2 H4O 44 lb C2 H 4O
×
yr
d 28 lb C2 H 4
= 16,123 lb C2 H 4 /h
365 d 24 h mol C2 H 4
5–49
Solutions Chapter 5
5.5.12
F = 100 kg mol
CO2
N2
C
1.00
O2 0.21
N2 0.79
C + O2 → CO2
Basis: 100 kg mol coke
Step 5:
a.
O2 in = 100 kg mol
100 kg mol C 1 mol O 2 79 mol N 2
= 376 kg mol N 2
1 mol C 21 mol O2
Steps 6 and 7:
Unknowns: moles (ni) CO2, O2, and N2 in P, and P
Balances: elements C, O, N
Equations:
P
i
∑n = P
or add ξ and have C, CO2, N2, O2 species balances
Steps 8 and 9:
Element balances
C:
2O:
2N:
in
100
100
376
=
=
=
out
n CO
n CO +n O
nN
2
2
2
n O2 = 0
2
5–50
in
Solutions Chapter 5
Component in P
N2
O2
CO2
kg mol
376
0
100
476
mol %
79
0
21
100
With 50% xs air, xsO2, is 50 mol
The total O2 in is 150 kg mol
The N2 in is
150 0.79
= 564
0.21
The N and C balance are the same but the 2O balance is
In
150
out
n O2
−
reacts
=
100
Component in P
N2
O2
CO2
n O2 = 50
kg mol
564
50
100
714
mol %
79
7
14
100
c. Same 150 mol O2 enter and 564 mol N2 exit but O2 is different
in
Oxygen balance:
Component
CO2
CO
O2
N2
Total
150
mol CO2 used
–
90
kg mol
90
mol%
12.51
10
1.39
55
564
719
7.65
78.45
100.00
5–51
–
Used CO
5 = 55 kg mol O2 lb fg
Solutions Chapter 5
5.5.13
Steps 1, 2, 3 and 4:
B
CH3CHO 1.00
P = 100 mol (Basis)
F
CH3CH2OH 1.00
A
mol. wt.
CH3CHO
44
CH3CH2OH 46
CO2
O2
CO
H2
CH4
N2
W H2O
1.00
Air .21 O2
.79 N2
1.00
%
0.7
2.1
2.3
7.1
2.6
85.2
100.0
C
0.7
2.3
O2
0.7
2.1
2.3/2
7.1
5.2
2.6
5.6
H2
3.95
12.3
Step 5: Basis: 100 mol P
Step 6: Unknowns: F, A, B, W
Step 7: Balances:
C, O, H, N
ok
Steps 8 and 9:
Via algebra
2N balance: A(.79) = 85.2
A = 107.85 mol
C balance:
F(2) = B(2) + 5.6
2H balance: F(3) = B(2) + W(1) + 12.3
2O balance:
1
1
1
2
2
2
F( ) + (.21)A = B ( ) + W ( ) + 3.95
Solution:
B = 44.1 mol
F = 46.9 mol
W = 40.2 mol
44.1 mol acet. 44 kg acet.
100 mol P 1 kg mol EtOH
= 0.899
100 mol P 1 kg mol acet. 46.9 mol EtOH
46 kg
kg acetaldehyde
kg ethanol
Alternate solution: Use extent of reaction and species balances, but it is more
complicated.
5–52
Solutions Chapter 5
5.5.14
Assume ammonia and glucose (MW = 180.1) are fed in stoichiometric proportions.
The reaction equation has to be set up and balanced to use the value given of 60% (mole
assumed) conversion of glucose.
a C6 H12O6 + b NH3 → c CH1.8O0.5 N0.2 + d C2 H6O + e CO2 + f H2O
Pick a basis of 4.6 kg of EtOH (C2H6O, MW = 46.07)
4600 g EtOH 1 g mol EtOH
= 100 g mol EtOH produced
46.07 g EtOH
To balance the chemical reaction equation use element balances. Take a basis of a = 1. Assume
60% conversion of the glucose means that 1 mole of glucose produces 0.6 mol of ethanol, not that
one mole of glucose produces 3 mol of ethanol.
C:
H:
O:
N:
6 = c + 2d + e
12 + 3b = 1.8c + 6d + 2f
6 = 0.5c + d + 2e + f
b = 0.2c
Specifications 0.60 = d
Results
a = 1 (the basis)
b = 0.8
c=4
d = 0.6
e = 0.8
f = 1.8
167 g mol C6 H12O .08 g mol NH 3
= 133.6 g mol NH 3
1 g mol C6 H12O
167 g mol C6 H12O6 180.1 g mol C6 H12O6
= 30,076 g or 30.1 kg C6 H12O6
1 g mol C6 H12O6
133.6 g mol NH 3 17.03 g NH 3
= 2,275 g or 2.28 kg NH 3
1 g mol NH 3
5–53
Solutions Chapter 6
6.1.1
4 (such as 2 component balances for each unit)
6.1.2
2 (The two overall balances on A and B sum up to the overall balance on
the system)
6.1.3
Unit I involves A, B, C
Unit II involves D, E
Unit III involves A, B, C, D
Components
3
2
4
Number of independent balances
9
If A and B are always combined in the same ratio, then you have to reduce the
independent balances by 1 for Unit I and 1 for Unit III, a total of 2.
6.1.4
Unknowns:
stream flows including F:
Mass fractions (7 plus 3 in F) 10
Independent material balances:
Species in 1 : 3
2:2
3:3
6
16
8
Other independent equations:
∑ ω i = 1 in each stream including F 8
6–1
16
Solutions Chapter 6
6.1.5
Basis: F = 100 kg
mass fr.
A 0.60
B 0.20
C 0.20
1.00
% = kg
A 15
B 30
C 55
100
F = 100 kg
m = mass fraction
P1
(1)
(2)
mass fr.
A 0.10
B 0.85
P2 C 0.05
1.00
W2=20 kg
W1
mass fr.
A m1A
B m1B
C m1C
1.00
(a) SEPARATE ANALYSIS
Column (1) balances
!
= W1 (m1A ) + P1 (0.60) !
"
any
3
"
(2) B: 0.30(100) = W1 (m1B ) + P1 (0.20) "
" 4 indept. eqns.
are indept."
"
1
(3) C: 0.55(100) = W1 (m C ) + P1 (0.20) "
eqns. "
"
"#
100 = W1
+ P1
"
(4)
m1A + m1B + m1C = 1
"
#
(1) A: 015(100)
Unknowns: W1 ,P1 ,m1A ,m1B ,m1C
5 – 4 = 1 degree of freedom
5 unknowns
6–2
A 0.002
B m2B
C m2C
1.00
Solutions Chapter 6
Column (2) balances
(5) A: m1A W1
!
= 0.10 P2 + 0.002 (20) !
" any 3 "
2
(6) B: m W1 = 0.85P2 + m B (20)
" are indept." 5 indept. eqns.
"
"
1
2
(7) C: m C W1 = 0.05P2 + m C (20)
" eqns. "
"
"#
W1 = P2 + 20
"
m1A + m1B + m1C = 1
"
#
2
2
0.002 + m B + mC = 1
1
B
Unknowns: W1 , P2 , m1A , m1B , m1C , m 2B , m C2 > unknowns
7 – 5 = 2 degrees of freedom
JOINT ANALYSIS
Adding the 2 units to make a joint system
No. of
Unknowns
• From the separate calculations start with:
• Delete the 4 common variables
W1 , m1A , m1B , m1C
that are treated as unknowns
in both subsystems leaving:
• Delete the equation
m1A + m1B + m1C
that will no longer be independent in
the joint system (it appears in both
subsystems) once leaving:
No. of
Degrees
indept. Eqns. of freedom
12
9
3
8
9
-1
8
8
0
ANALYSIS OF OVERALL SYSTEM
System overall balances
(1) A: 100 (0.15)
= P1 (0.60) + P2 (0.10) + 20 (0.002) !
"
(2) B: 100 (0.30) = P1 (0.20) + P2 (0.85) + 20 (m 2B ) "
4 indept.
(3) C: 100 (0.55) = P1 (0.20) + P2 (0.05) + 20 (mC2 ) "
" eqns.
"
"
0.002 + m2B + mC2 = 1
"#
2
2
Unknowns:
4 unknowns
P1 , P2 , m B , m C
4 - 4 = 0 degrees of freedom as expected
6–3
Solutions Chapter 6
6.2.1
The results show that the system of equations is very sensitive to small perturbations in
the coefficients (measurements).
6–4
Solutions Chapter 6
6.2.2
Basis: 1 hr (1000 kg)
a. Overall balances:
Total: 1000 = W + D
Salt: 1000 (0.0345) = 0 + 0.069D
W= 500 kg
D = 500 kg
b. Salt balance on the freezer:
1000 (0.0345) = 0 + 0.048B
B = 718.75 kg
Total balance on the freezer:
1000 = A + B
A = 281.25 kg
Total balance around filter:
B = 718.75 = C + D
C =218.75 kg
6–5
Solutions Chapter 6
6.2.3
Two subsystems exist, hence 4 component balances can be written. No reaction
occurs and the process is assumed to be in the steady state. Steps are omitted here. The
balances are
System : Splitter
System : Stock Chest
Total: R = E + P
Fiber : 2.34 = xE+xP
Water: 7452 = 4161+3291
Total: P +N = L
Fiber: xP+xN =103.26
Water: 3291 + 18 = 3309
(a)
Also N(0.15) = 18 N = 120
0.85 (N) =xN = 0.85 (120)
(b)
Overall balances
Total: R+ N = E + L
Fiber: 2.34 + xN = xE + 103.26
Water: 7452 + 18 = 4161 + 3309
(c)
Substitute (b) into (c), solve (c) for xE , and then solve (a) for xP.
x N =102
x P =1.26
x E =1.08
all in kg
6.2.4
Step 5:
Basis: 100 kg mol = F
Steps 1, 2, 3, and 4:
H2 O 100%
W
F (kg mol)
E (kg mol)
P (kg mol)
Furnace
CH4
H2
C2 H 6
mol frac
0.70
0.20
0.10
1.00
A (kg mol)
O2
N2
mol frac
0.21
0.79
Duct
mole frac.
mol
CO 2
-
n ECO
O2
0.02
n
H2 O
N2
-
n EH O
mole frac.
Air ?
2
E
O2
B (kg mol)
2
E
-
nN
1.00
E
1.00
2
O2
N2
mol frac
0.21
0.79
mol
CO 2
-
n PCO
O2
0.06
n
N2
-
nPN
1.00
P
2
P
O2
2
1.00
Assume an air leak occurs first and use material balances to calculate the amount. This is
a steady state flow process without reaction.
6–6
Solutions Chapter 6
Overall system
Step 6:
Unknowns are:
A, B, W, P, x N 2 (or n N2 ), x CO2 (or n CO2 )
Step 7:
Balances are:
C,H, O, N, Σ x Pi = 1 (o r ∑ n Pi = P)
Steps 8 and 9:
In
Out
=
n CO2
=
W (2)
2O (kg mol): 0.21A + 0.21B
=
190(1.00)
+ 90 + 0.06P
2
2N (kg mol): 0.79A + 0.79B
=
n PN 2
=
P
C(kg mol):
0.70(100) + 0.10 (100) 2
n CO2 = 90 kg mol
H (kg mol):
0.70 (100)(4) + 0.20 (100)(2)
+ 0.10 (100)(6)
W
=
190 kg mol
90 + 0.06 P + n PN 2
A balance about the furnace or about the duct is needed; or these two alone would have
been sufficient, omitting the overall balance.
System: Furnace (or Duct)
E
Step 6: Unknowns:
A, n ECO2 ,n EO 2 , nH 2 O , n EN2
Step 7: Balances:
C, H, O, N2, ∑ nEi = E
Steps 8 and 9:
C (kg mol):
H (kg mol):
0.70 (100) + 0.10 (100)(2)
n ECO2 = 90 kg mol
=n ECO2
E
0.70 (100)(4) + 0.20 (100)(2) + 0.10 (100)(6) = n H 2 O
E
n H 2 O = 190
190
2O (kg mol): 0.21A
= n EO 2 + 90 +
2
6–7
Solutions Chapter 6
= n EN 2
2N (kg mol): 0.79A
90 + 190 + n EO 2 + n EN 2 = E
0.02
= n EO 2 / E
Solution:
kg mol in E
kg mol
n ECO2
90
A
982
n H 2O
E
190
E
1077
n EO 2
21.5
n EN 2
776
1077.5
From the overall N2 balance:
O2 balance:
0.79 (982) + 0.79B = n PN 2
(1)
0.21 (982) + 0.21B = 185 + 0.06P
(2)
(1)
22.2 + 0.21B = 0.06P
B = 207 kg mol
(3)
90 + (0.79)(982) + 0.79B = 0.94P
P = 1095 kg mol
It is an air leak.
207 kgmol / 100 kgmol F
6–8
Solutions Chapter 6
6.2.5
Steps: 1,2,3, and 4: This is a steady state process with no reaction taking place. All the
compositions are known except for stream C. We can pick the overall system first to get
D, and then make balances on the first (or second) units to get C and its composition.
Step 5:
Basis:
F = 290 kg (equivalent to 1 second)
Step 6:
Unknowns:
A, B, C, D and E and the composition of C
Step 7:
Balances:
NaCl, HCl, H2SO4, H2O, inert solid
Step 8:
For the overall system there are 4 unknowns (C is excluded) and 5 species
balances:
Overall balances (kg)
In
Out
NaCl:
HCl:
H2SO4:
H2O:
A(0.040)
A(0.050)
A(0.040)
A(0.870)+B(0.910)
=
=
=
=
290(0.0138)
290(0.0255)+D(0.020)+E(0.015)
290(0.0221)+D(0.020)+E(0.015)
290(0.9232)+D(0.960)+E(0.970)
Inerts:
Totals:
B(0.09)
A+B
=
=
290(0.0155)
D + E +290
Steps 6 and 7:
Two of the equations are not independent: HC1 and H2SO4.
HC1: D(0.020) + E (0.015) = 7.40 – 5.00 = 2.40
H2SO4: D(0.020) + E (0.015) = 6.41 – 4.00 = 2.41
Thus, overall the degrees of freedom are 0.
Steps 8 and 9:
The solution of the equations is (in kg/s)
A = 100 ,
B = 50 ,
D = 60 and E = 80
from which C = 150 via a total balance about the first unit.
6–9
Solutions Chapter 6
6.2.6
Notation: Subscripts Benzene = bz, Xylene = xy, Solids = s
W = mass fraction, Fi = liquid stream, Si = solids stream
0
ωS bz = 1.0
0
ωS xy = 0
2
2
ωS bz = (1-ωSxy)
1
1
ωS bz = (1-ωSxy)
2
ωS xy
1
ωS xy
2
ωS s = 0
1.0
ωS s = 0
1.0
ωS s = 0
1.0
S0
S1
S2
0
1
1
2
F1
F2
2
F0
1
0
ωF bz = (1-ωFxy)
ωF bz = (1-ωFxy)
ωF bz = 0
ωF xy
ωF xy
ωF xy=0.1
2
1
2
1
2
ωF s = 0.9
1.0
1
ωF s = 0.9
1.0
0
0
ωF s = 0.9
1.0
Step 5: Basis is 1 hr (F0 = 2000 kg and S0 = 1000 kg)
Steps 6 and 7:
2
1
1
Unknowns:
Unit 1: ωFxy , ωSxy , ωFxy , S1 , F1 , F2 !"
# Net = 8
S1
F1
S2
1
1
2
Unit 2: ωxy , xy
ω, xy
ω, S , F , S "$
Material Balances:
Unit 1: bz, xy, s
Unit 2: bz, xy, s
2
1
1
2
Additional Equations: ωFxy = ωSxy and ωFxy = Sxy
ω (the related equations for benzene are
redundant). Total equations = 8.
The solution is simplified if two balances are made first (not an essential step):
All material balances are in kg.
Solids balance on Unit 2 and also Unit 1:
2000 (0.9) = F1 (0.9)
2000 (0.9) = F2 (0.9)
F1 = 2000
F2 = 2000
6–10
Solutions Chapter 6
Total balance on Unit 1 and also Unit 2:
2000 + S1 = 2000 + S2
1000 + 2000 = 2000 + S1
S1 = S 2
S1 = 1000
S2 = 1000
The other balances are
Unit 2:
1
1
2
Benzene:
1000(1 − ωSxy ) = 2000(1 − ω Fxy ) + 1000(1 − ωSxy )
Xylene:
2000(0.1) + 1000ωSxy = 2000 Fxy
ω + 1000 Sxy
ω
1
1
1
2
2
10ωFxy = Sxy
ω
Unit 1:
1
2
1
Benzene:
1.0(1000) + 2000(1 − ωF ) = 2000(1 − ωFxy ) + 1000(1 −ωSxy )
Xylene:
0 + 2000ωFxy = 2000ωFxy + 1000(ωSxy
1
2
2
1
)
1
10ωFxy = Sxy
ω
Solve the equations to get the compositions of the streams:
b)
Stream
Component
wt fr
Stream
Component
wt fr
S0
Bz 1.0
Xy 0.0
F0
Bz 0.9
Xy 0.1
S1
Bz 0.97
Xy 0.03
F1
Bz 0.082
Xy 0.018
S2
Bz 0.82
Xy 0.18
F2
Bz 0.097
Xy 0.003
Xylene in Feed
= 2000 × 0.1 = 200 kg
Xylene in Product
= 1000 × 0.18 = 180 kg
% Recovery =
180
×100 = 90%
200
6–11
Solutions Chapter 6
6.2.7
Steps 1, 2, 3 and 4:
mol %
64.29H2
14.29SiCl 4
21.42 H2 SiCl 2
HCl (g)
A
100% Si
B
C
Reactor
D
Separator
H 2 ,SiCl 4 ,HSiCl 3
H 2 SiCl 2
Step 5: Basis:
100%
E
HSiCl 3
100 kg B
100 kg Si 1kg mol Si
= 3.560 kg mol Si
28.09 kgSi
Steps 6 and 7: Unknowns:
A, D, E
Balances:
H, Cl, Si
Steps 8 and 9: Balances to be solved:
System: overall process
Si overall mole balance
kg mol Si = 3.560
= D kg mol gas
+
! 0.1429 kg mol SiCl 4 1 kg mol Si
0. 2142 kg mol H 2 SiCl2 1 kg mol Si $
+
#"
kg mol gas
kg mol SiCl 4
kg mol gas
kg mol H 2Si Cl 2 &%
E kg mol HSiCl3
1 kg mol Si
= 0. 3571D + E
kg mol HSiCl3
Cl overall mol balance
6–12
Solutions Chapter 6
⎡ 0.1429kg mol SiCl 4 4 kg mol Cl
A kgmol HCl 1kg mol Cl
= D kg mol gas
⎢⎣
kgmol SiCl 4
kg mol gas
kg mol HCl
+
0.2142 kgmol H2 SiCl 2 2 kgmol Cl ⎤ E kg mol HSiCl3 3kg mol Cl
+
kg mol gas
kgmol H2 SiCl2 ⎥⎦
kg mol HSiCl 3
A=
D + 3E
H overall mol balance
⎡ 0.6429kg mol H2 2 kgmol H
A kgmol HCl 1kgmol H
= D kg mol gas
⎢⎣
kg mol H2
kg mol HCl
kgmol gas
+
0.2142 kgmol H2 SiCl 2 2 kgmol H ⎤ E kg mol HSiCl3 1kg mol H
+
kg mol gas
kgmol H2 SiCl2 ⎥⎦
kg mol HSiCl 3
A=
Solution:
2
1
H
Si
Cl3
3
1.71421D + E
3.560 = 0.3571 1D + E
1
A
= D + 3E
2
A
= 1.7142D + E
3
2E
= 0.7142 1D
E
= 0.3571 1D
4 → 3.56
4
= 0.3571D + 0.3571 D
D
= 4.98
E
= 1.78
A
= 10.32
MW
1.01
28.09
106.35
6–13
Solutions Chapter 6
135.45 kg/mol
135.45 kg HSiCl 3 1.78 kg mol
= 241 kg HSiC13
kg mol HSiCl 3
6.2.8
Assume steady state process
P = 100 mol
100% H2 O
W
gas
F2
F
1
mol fr.
C 0.357
H 0.643
1.00
oil
A
A1
air
A2
Basis: 100 mol natural gas
Comp.
mol=% atoms C
atoms O
atoms H
CH4
C2H2
CO2
96
2
2
96
4
2
--4
384
4
--
Total
100
102
4
388
Basis: 100 mol oil
Comp.
atoms C
atoms H
(CH1.8)n
100
180
Basis: 100 mol flue gas (dry)
Comp. mol%
CO2
CO
O2
N2
Total
mol C mol O
atoms H
10.0 10.00 20.00
-0.63 0.63 0.63
-4.55 -9.10
84.82 --100.00 10.63
29.73
6–14
----
mol N2
84.82
84.82
Solutions Chapter 6
Note:
If have separate air streams, we have 5 unknowns and can't solve.
Basis: 100 mol natural gas (or use 100 mol dry gas product.)
Let F2
= mol oil
F1 or P
= mol dry flue gas
A
= mol air to natural gas fed boiler plus oil fed boiler
W
= mol water associated with the dry flue gas
Four balances: In = Out
C balance
102 + F2
= 0.1063 P
(1)
= 0.8482 P
(2)
N2 balance
0.79A
H balance
388 + 1.80 F2 = 2 W
(3)
O balance
4 + 0.42A
= 0.2973P +W
(4)
P
= 1729 mol dry flue gas
W
= 268 mol H2O
F2
= 82 mol oil and 1 mol C = 1 mol oil so
C in oil = F2
= 82 mol
Total C = 102 + 82
= 184 mol
% C burned from oil = (0.82)(100) = 44.5%
184
6–15
Solutions Chapter 6
6.2.9
Step 5: Basis: 100 kg mol = F
Steps 1, 2, 3, 4:
F (kg mol)
mol fr.
CH4 0.70
H2 0.20
C2H6 0.10
1.00
W
E (kg mol)
Furnace
mol fr. mol
A (kg mol)
mol fr.
0.21 O2
0.79 N2
1.00
CO2
nECO2
O2 0.02 nEO2
H2O
nEH 2O
N2
nN2
1.00
E
100%
H2O
P (kg mol)
Duct
mol fr. mol
B (kg mol)
Air ?
mol fr.
0.21 O2
0.79 H2
1.00
nPCO2
CO2
O2 0.06 nPO 2
nPN2
N2
1.00
P
Assume first an air leak occurs and use material balances to calculate the amount. This is
a steady state flow process without reaction.
Overall system
Step 6: Unknowns are: A, B, W, P, x N2 (or n N2 ), x CO2 (or n CO2 )
Step 7: Balances are:
C, H, O, N, ∑ x iP =1 (or ∑ n iP = P)
Step 8:
C (kg mol):
H (kg mol):
2O (kg mol):
2N (kg mol):
In
0.70(100) + 0.10(100)2
Out
= n CO2
n CO2 = 90 kg mol
0.70(100)(4) + 0.20(100)(2) + 0.10(100)(6) = W(2)
W = 190 kg mol
190(1.00)
0.21A + 0.21 B
=
+ 90 + 0.06P
2
0.79A + 0.79 B
= n PN2
90 + 0.06P + n PN2
=P
A balance about the furnace or about the duct is needed, or those two alone would have
been sufficient, omitting the overall balance.
System: Duct
6–16
Solutions Chapter 6
E
Step 6: Unknowns: A, E, n CO
, n OE 2 , n EH2O
2
Step 7: Balances:
C, H, O, N2 , ∑ niE = E
Steps 8 and 9:
C (kg mol):
0.70(100) + 0.10(100) (2) =
E
n CO
2
E
= 90 kg mol
n CO
2
0.70(100)(4) + 0.20 (100) (2) + 0.10(100) (6) = n EH2O
H (kg mol):
n EH2O = 190
2O (kg mol):
0.21A = n OE 2 + 90 +
2N (kg mol):
0.79A = n EN2
190
2
90 + 190 + n OE 2 + n EN2 = E
0.02 =
Solution
n O2
E
kg mol in E
90
A
kg mol
982
190
E
1077
n
E
CO2
n
E
H2 O
n
E
O2
21.5
n
E
N2
776
1077.5
From the overall 2N balance
0.79(982) + 0.79B = n PN2
2O balance
(1') 22.2 + 0.21B = 0.06P
(1)
0.21(982) + 0.21B = 185 + 0.06P
(2)
! B = 207 kg mol
"
(3) 90 + (0.79)(982) + 0.79B = 0.94P # P = 1095 kg mol 207 kg mol / 100 kg mol F
6–17
Solutions Chapter 6
6.2.10
Step 5: Basis = 100 lb A
Steps 2, 3, and 4:
Fe added: 36(100) = 3600 lb
MW TiO2 = 79.9 lb/lb mol
MW Fe = 55.8 lb/lb mol
MW H2SO4 = 98.1 lb/lb mol
MW TiOSO4 = 159.7 lb/lb mol
For Part (a), the system boundary has been drawn on a light solid line.
Steps 6 and 7:
F
Unknowns (9): B, C, J, K, H, F, ξ1 , ξ 2 , m Fe
Species balances (7): TiO2 , Fe, H2SO4 , H2O, TiOSO4 , O2 , inert
Other equations (2): Reactions 1 and 2 are complete
Degrees of freedom = 0
6–18
Solutions Chapter 6
Steps 8 and 9
a. lb H2O removed in evaporator (J):
Mol TiO2 in slag:
Mol Fe in Slag:
(0.70)(100 lb) lb mol
= 0.876 lb mol
79.9 lb
(0.80)(100 lb) lb mol
= 0.143 lb mol
58.8 lb
Amount of H2SO4 based on theoretical requirements of Equations (1) and (2)
0.876 + 0.143 = 1.02 lb mol H2SO4
1.02 lb mol H 2SO 4 98.1 lb
= 100 lb H 2SO 4
lb mol
100 lb
B=
= 149.2 lb Water in stream B = 49.2 lb
0.67
O2 in:
0.143 lb mol Fe 0.5 lb mol O2 32 lb O2
= 2.29 lb O2 =C (0.0715 lb mol)
1 lb mol Fe 1 lb mol O2
Using TiO2:
ξ1 =
0 − 0.876
= 0.876 moles reacting
( − 1)
Using O2:
ξ2 =
0 − 0.0715
= 0.143 moles reacting
(0.5)
TiOSO4 in stream K
K
n TiOSO
= 0 + (1) (0.876) = 0.876 lb mol TiOSO4
4
K=
0.876 lb mol 159.7 lb 1
= 170.6 lb
1 lb mol 0.82
H2O in K = 170.6 (0.18) = 30.7 lb
Water formed in the reactions (use ξ1 and ξ2 ) :
n out,rxn
= (1)(0.876) + (1)(0.143) = 1.019 lb mol (or 8.36 lb H2O)
H2 O
Water exiting from the evaporator J (lb):
49.2 + 18.36 – 30.7 = 36.9 lb
6–19
Solutions Chapter 6
b.
Exit H2O from dryer:
inlet air =
18 lb mol 29 lb
= 522 lb dry air
lb mol
inlet water =
0.036 mol H 2 O 18 mol air 18 lb
= 11.7 lb H 2O
mol air
lb mol
water added to air = (0.18) 170.9 lb – (0.876 lb mol)
18 lb
= 15.0 lb
lb mol
Humidity = (15.0 lb H2O + 11.7 lb H2O)/522 lb air
= 0.051 lb H2O/lb air
c.
The pounds of TiO2 produced (P):
By reaction (3), 0.876 lb mol of TiOSO4 goes to TiO2.
(0.876) (79.9) = 70 lb TiO2
6–20
Solutions Chapter 6
6.3.1
a. 1; b. 3; c. 0; d. 7
a.
b.
6–21
Solutions Chapter 6
c.
d.
6–22
Solutions Chapter 6
6.3.2
Step 5: Basis: 60 kg W
Pick the overall system
Steps 6 and 7:
Unknowns: F, P
Balances: A, B
Steps 8 and 9:
Total F= 60 + P ! F = 380 kg
"
B: 0.80F = 0 + 0.95P # P = 320 kg
Pick mixing point as the system
Steps 6 and 7:
Unknowns: G, R
Balances: A, B (or total as alternate)
Steps 8 and 9:
Total: 380 + R = G
!
" R = 126.7
B: 0.80(380) + R(0) = 0.60G #
R 125.7
=
= 0.33 kg R/kg F
F
380
6–23
Solutions Chapter 6
6.3.3
Basis: 1 hour
An overall balance shows that Rout from the adsorber must equal Rin to the adsorber if a
steady state exists. Let Ri = R.
Adsorber balance of U (units are U):
600 mL 1.37U
R mL 5.2U
600 mL 0.08U
R mL 19.3U
+
=
+
1 mL
1 mL
1 mL
1 mL
600 (1.37-0.08) = R(19.3-5.2)
R = 55 mL/hr
6.3.4
Steps 2, 3, and 4:
B1 0.75
B2 0.25
S 0.00
1.00
B1 = butene
B2 = butadiene
S = solvent
A
5,000 lb/hr
I
C (lb) B1
B2
S
E (lb) = 10,000
II
B1 0.0
B2 0.01
S 0.99
Step 5: Basic: 1 hr (A = 5000 lb)
Select whole process as the system.
6–24
B (lb) B1 1.00
B2 0
S 0
D (lb) B1 0.05
B2 0.95
S 0
Solutions Chapter 6
Steps 6 and 7:
Unknowns: B, D
Balances: Total, B1, B2, S (not all independent)
Steps 8 and 9:
Total balance: 5000 = B + D
B1 balance:
5000 (0.75) = B(1) + D(0.05)
D = 1316 lb
B = 3684
Apparently the calculated values are not correct. (The value for C can be obtained from
balances on Unit I or II).
6.3.5
Steps 1, 2, 3 and 4:
H 2 O 100%
10,000 kg/hr 20%
W
KNO3 Solution
P
Evaporator
mass frac.
H2 O
0.50
KNO3
0.50
1.00
Boundary line
for overall balance
300°F
mH O
m KNO
F
mass frac.
H2 O
0.80
KNO3
0.20
Feed
1.00
2
M
50%
KNO3
3
∑P
Crystallizer
R
Recycle 100 °F
H2O
0.625
0.6 kg KNO 3
HNO3
0.375
kg H 2O
C
mass frac.
H2 O
0.04
KNO3
mass frac.
Saturated Solution
1.000
Boundary line for
balance around
crystallizer
0.96
1.00
Step 5: Basis: 1 hr = 10,000 kg KNO3 solution
Step 4 cont’d:
Compute the weight fraction composition of R. On the basis of 1 kg of water, the
saturated recycle steam contains (1.0 kg of H2O + 0.6 kg of KNO3) = 1.6 kg total. The
recycle steam composition is
6–25
Solutions Chapter 6
0.6 kg KNO3
1 kg H2 O
= 0.375 kg KNO3 / kg solution
1 kg H2 O 1.6 kg solution
or 37.5% KNO3 and 62.5% H2O (which has been added to the figure).
Analysis of complete process ( 6 streams):
Unknowns:
W, M, C, R, m PH 2 O m PKNO 3
=6
Balances:
(2 units + 1 mix point) × 2 components
=6
Other compositions are (in %)
M
H2O 50
KNO3 50
F
80
20
W
100
0
C
4
96
R
62.5
37.5
Start with overall balances as substitute for unit balances
Steps 6 and 7:
Unknowns:
W, C,
Balances:
2 components
Overall balance to calculate C:
Total:
10,000
=W+C
KNO3: 10,000 (0.20) = W (0) + C (0.96)
10, 000 kgF 0.20 kgKNO3 1kgcrystals
= 2083 kg / hr crystals = C
1kg F
l hr
0.96 kgKNO3
W = 10,000 - 2083 = 7917 kg
To determine the recycle stream R, we need to make a balance that involves the stream R.
Either (a) a balance around the evaporator or (b) a balance around the crystallizer will do.
The latter is easier since only three rather than four (unknown) streams are involved.
Total balances on crystallizer:
M=C+R
M = 2083 + R
a
Component (KNO3) balance on crystallizer:
M ω M = Cω C + Rω R
6–26
Solutions Chapter 6
0.5M = 0.96C + R (0.375)
Solving equations a and
b
b we obtain
0.5 (2083 + R) = 0.375R + 2000
R = 7670 kg/hr
6.3.6
Single pass conversion is:
a.
fsp =
−υLR ξ
feed
n reactor
LR
Single pass conversion based on H2 as the limiting reactant:
1.979 − 3.96
= 0.99
−2
0 − 3.96
ξ max =
= 1.98
−2
ξ=
SP conversion =
−(−2)(0.99)
= 0.50
3.96
Use Eq. (12.1) and Eq. 12.2 with H2 the limiting reactant
b.
0.99 =
fsp =
c.
− υLR ξ
−( − 2) ξ
=
ξ = 0.980
fresh feed
n LR
1.98
−(−2)(0.98)
= 0.986
1.989
Overall conversion of H2
f overall of H 2 =
d.
−(−2)(0.99)
= 1.0
1.98
Overall conversion of CO
f overall of CO =
−(−1)(.99)
= 0.99
1
6–27
Solutions Chapter 6
6.3.7
Steps 2, 3, and 4:
Recycle R(mol)
CO, H2O
F(mol)
Reactor
mol
%
52 CO nCO
48 H2O nH2O
L(mol)
Separator
H2: 3 mol %
Step 5:
P(mol)
CO2
H2
CO
Basis: P = 100 mol
Pick the overall process as the system.
Step 6:
F
Unknowns: n CO
, n HF O
Step 7:
Balances:
2
C, H, O
Steps 8 and 9: Element balances
C (mol): n CO = 48 + 4 = 52
Step 10:
!"
# total 100
H (mol): n H2O (2) = 48 (2)
n H2O = 48"$
O (mol): n H2O (1) + n CO (1) = 48(2) + 4 check is ok.
a.
Composition of fresh feed
40% H 2 O and 52% CO
Alternate solution: Use extent of reaction
Based on
Then
48 − 0
= 48
1
F
nCO
= 4 + ξ = 4 + 48 = 52 mol
F
n H2O = 0 + ξ = 48 mol
CO2:
ξ=
To get the recycle, make a balance about the separator (no reaction occurs)
6–28
mol % = mol
48
48
4
100
Solutions Chapter 6
H 2 balance (mol):
Total (mol):
L(0.03) = 48! L=1600 mol
"
L=100 + R # R=1500 mol
1500 mol R
= 31.3
48 mol H 2
b.
6.3.8
Steps 2, 3, and 4:
MW Ca (Ac)2 = 158.1
MW HAc = 60
100 kg Ca(Ac)2 1 kg mol Ca(Ac)2
= 0.633 kg mol Ca(Ac)2
158.1 kg Ca(Ac)2
Step 5:
Basis: 1000 kg of Ca(Ac)2 feed
Inspection of the diagram shows that the overall conversion of Ca(Ac)2 is 100%
(none leaves the process). Thus, fOA = 1. You are given fSP = 0.90. Then
R = 0.703 kg mol
a.
fSP
0.90
6.33
=
=
f OA
1
6.33 + R
b.
The single pass conversion for Ca(AC)2 is 0.90.
0.90 =
− υ LR ξ
−( − 1)ξ
=
feed
n LR
6.33
or 111 kg
ξ = 5.697
Overall HAc balance:
6.33 kg mol Ca (Ac) 2
2 kg mol HAc
= 12.66 kg mol HAc or 760kg
1 kg mol Ca(Ac) 2
6–29
Solutions Chapter 6
6.3.9
Step 5:
Basis: 100 kg F1 ≡ 1 hr
Overall balance
Step 6 and 7:
Unknowns: F2, P1, P2
Element balances: Ca, Na, C1, CO3 (enough, not all independent)
CO3 balance:
900 kg CaCO3 1 kg mol CaCO3 1 kg mol CO3
100 kg CaCO3 1 kg mol CaCO3
= 9.00 kg mol CO3
100 kg Na 2 CO3 1 kg mol Na 2 CO 3 1 kg mol CO 3
0.94 kg mol CO3
=
106 kg Na 2 CO3 1 kg mol Na 2 CO 3
9.94 kg mol CO3
a.
9.94 kg mol CO3 1 kg mol Na 2 CO3 106 kg Na 2 CO3
= 1054 kg Na 2 CO3
1 kg CO3
1 kg mol Na 2 CO3
6–30
Solutions Chapter 6
Species balance about the reactor plus the separator on CaCO3:
IN


OUT
[1000(0.90) + R] (.24) = R



UNREACTED
R = 284 kg
Alternate solution:
Use the extent of reaction to get the same results.
6.3.10
This is steady state process with reaction and recycle. Pick the overall process as the
system.
Step 5: Basis: 100 kg mol CH4
Steps 1, 2, 3, and 4:
The system is as shown in the diagram with recycle added.
0.21
0.79
Air
kgmol
O2 230
N2 865.24
CH4
CO2
O2
100 kg mol
H 2O
NO
N2
R
Calculate the moles of each component entering.
C H4 (g) + 2O2 (g)→C O2 (g)+ 2 H2 O (g)
Calculation of the required O2 and accompanying N2 in moles
req'd O2
100(2) = 200
xs
200 (0.15)
Total O2
= 30
230
6–31
Solutions Chapter 6
⎛ 230 .79 ⎞
⎟ =
N2 in ⎜
.21 ⎠
⎝
865.24
The exit concentrations (via stoichiometry) are
CO2
O2
H2 O
NO
N2
Mol
100
30
200
415 × 10-6
865.24
1195.24
Next, consider adding recycle as shown in the diagram. Recycle is not involved in the
overall balance, hence the concentration of NO will not be affected because the extent of
reaction with and without recycle remains the same. The recycle does reduce the
combustion temperature, which in turn will reduce the exit concentration of NO.
6.3.11
Basis: 1 hr ≡ 1L = F
all concentrations are g/L.
W
F=1L
D
Zn 100
Ni 10.0
Zn 0.10
Ni 1.00
H2O
P0
Zn 190.1
Ni 17.02
Figure P12.9
Pick the overall process as the system.
Steps 6 and 7:
Balances:
3
H2O, Zn, Ni
Unknowns:
3
W, D, P
6–32
(all in L)
Solutions Chapter 6
Steps 8 and 9:
D(L) 0.10g
+ P 0 (190.1g / L)
L
Zn:
100 (1) + 0
=
Ni:
10(1) +0
= D (1.00) + P0 (17.02)
Solve to get P0 = 0.525L
D = 1.056L
Pick Unit 3 as the system:
Steps 6 and 7:
Balances:
Unknowns:
2
2
Zn, Ni
R32, P2 (all in L)
Steps 8 and 9:
Zn:
P2 (3.50) + 0 = 0.10(1.056) + R32 (4.35)
Ni:
P2 (2.19) + 0 = 1.00(1.056) + R32 (2.36)
Solve to get R 3 2 =2.75L/hr
6.3.12
H
Mass frac.
P
oil + dirt ω o+d
P
H2 O
ω HO
D
mass fr.
0.229
ω
H
ω H O 0.771
mass fr.
H
o+d
oil + dirt 0.229
2
2
1.00
P
H2 O
1.000
2910 gal/day
R
G
G
ω o+d
ω HGO
mass fr.
0.9927
Mass frac.
oil + dirt 0.229
H2 O
0.771
1.000
2
H2 O
0.771
1.000
Filter
oil + dirt 0.0073
90 gal/day
F
3000 gal/day
1.000
6–33
Solutions Chapter 6
Pick the total process as the system
Step 5:
Basis: 1 day (equivalent to D = 90 gal, F = 3000 gal, P = 2910 gal)
Steps 6 and 7:
Unknowns:
P
ωo+d
,
P
H2 O
Balances:
oil + dirt, H2O
ω
Steps 8 and 9:
Overall oil + dirt balance:
In
Out
P
3,000(0.0073) = 2,910(ωo+d ) + 90(0.229)
P
ωo+d
=
21.9 − 20.61
= 4.43 10−4 ×
2,910
(a)
To solve for R, make balances on the mixing point, the filter, and the splitter. Not all of
the balances are independent:
Total
oil + dirt
Splitter:
H = 90 + R
0.229H = 0.229(90) + 0.229(R)
Filter:
G = 2910 + H
ωGo+d G = 2910 (4.43 × 10-4) + H(0.229)
0.0073 (3000) + 0.229R = ωGo+d G
Mixing point: 3000 + R = G
Unknowns: H, R, G, ω Go+d
Balances:
6 (4 independent)
Steps 8 and 9:
The solution is
R = 57.2 gal/day
6–34
ωGo+d = 0.01145
Solutions Chapter 6
6.3.13
F°
Organic Solvent
100%
I
Aqueous Phase
FA = 10 L/min
F°
FA = 10 L/min
A
C0 = 100 g/L
A
C1 = 0.1 g/L
O
C1 = 0.01g / L
Fermentation product balance:
100
g 10L / min
10 L / min 0.1g / L
= F°(0.01) +
L
F°=
1000 − 1 = 99, 900 L
min
0.01
B:
F2 °
F1 °
I
II
A
C1
FA = 10 L/min
CA = 100 g/L
o
10L/min
O
A
C1 = 0.1C1
Fermentation
balance:
F3 °
10L/min
III
A
C2
O
A
C 2 = 0.1C 2
A
C3 =0.1 g/L
O
A
C3 = 0.1C 3
Assuming F1° = F2° = F3° = F°/3
100
I:
1
A
II:
g
L
L 10 min = C A ⋅10 L
C1
g
L
L 10 min = C A ⋅10 L
2
6–35
F°
A
mn + 3 × 0.1C (1)
F°
A
min + 3 × 0.1C (2)
F =10L
A
Solutions Chapter 6
A
III:
C2
g
L
L 10 min = C A ⋅10 L
3
F°
A
min + 3 × 0.1C3 (3)
Solving (1), (2) and (3) Simultaneously
C 1 = 10.0 g L
C 2 = 1.0 g L
A
C 3 = 0.1 g L
A
A
F = 2702 L min
C:
Fermentation Product Balance:
I:
g
100 L 10 L min
A
A
A
+ 0.1 C ⋅ F°= 0.1 C ⋅ F°+ C ⋅ 10
II:
C1 ⋅10 + 0.1(0.1)F°= 0.1C 2 ⋅ F°+ C2 ⋅10
III:
C 2 ⋅10 + 0 = 0.1(0.1) ⋅ F°+ 0.1 (10)
2
A
1
A
A
A
Solving (1), (2) and (3) Simultaneously
A
C1 = 10.36g / L
F = 960 L min
A
C 2 = 1.06 g / L
A
C3 = 0.10 g / L
This one is the least
6.3.14
Steps 1, 2, 3, and 4:
Mol. Wt: Na2S, 78; CaCO3, 100; Na2 CO3, 106; CaS, 72
Step 5: Basis 1 hr
6–36
1
Solutions Chapter 6
Steps 6, 7, 8, and 9:
System: Overall (System A)
Na Balance:
=
b.
(1000)(0.45)lb Na 2S 1 lb mol Na 2S 2 lb mol Na
78 lb Na 2S 1 lb mol Na 2S
m Na 2CO3 lb 1 lb mol Na 2CO3
106 lb Na 2CO3
2 lb mol Na 100 lb Na 2CO3 soln
1 lb mol Na 2CO3
80 lb Na 2CO3
mNa2CO3 = 764 lb/hr
System: Reactor plus separator (System B)
in
out
gen. Consumption accum
Na 2S balance: !100(0.45)
R" R
!100(.45) R "
+
−
+0 −$
+ % 0.90=0
$
%
(lb mol) & 78
78 ' 78
78 '
& 78
[1000(0.45) + R] (0.10) = R
R = 50 lb/hr
a.
Alternate solution
reacts.
Na2S is the limiting reactant (LR). Mole Na2S = (0.45)(1000)/78 = 5.77. All of it
1.00 =
( − 1)( − 1)ξ
,
5.77
ξ = 5.77
Overall fraction conversion of Na2S is 100%; fOA = 1.
fSP
n freshfeed
0.90
5.77
LR
=
= freshfeed
=
recycle
f OA
1.00 n LR +n LR
5.77+R
5.19 + 0.9R = 5.77
R = 0.64 lb mol (or 50 lb)
6–37
Solutions Chapter 6
6.3.15
A basis of 1 hr requires the listing and solution of many simultaneous equations.
A basis of 100 mol of toluene in G is more convenient
Makeup
M
F
100% H2
H2
mol
1
CH4
1
total
2
Feed
Mixer
100% toluene
P Pure
RG
n wH
only
2
n
W
w
CH 4
One system
Gross
Reactor
Feed
G
4H 2
4CH 4
1toluene
3450 lb/hr
C7 H
8
Gas Recycle
B
Separator
Benzene
100%
<-- 4H2 to 1 toluene
Liquid Recycle
D
Diphenyl
100%
RL
100% toluene
Basis: 100 mol toluene in stream G (gross feed)
System: Reactor plus separator
Species balances
In
Out
Gen
Cons.
100
-RL
+0
-100 (.80 +.08)
100 (4)
− nHW2
+0
⎛1
-100 (4) (.80 +.08 ⎝ ⎞⎠ ) = 0
2
W
− nCH
4
+ 100(4)(.80 + .08)
(1) toluene:
(2) H2:
(3) CH4:
100 (4)
Solve above for
RL = 12 mol
n W
H 2 = 64 mol !
" Total = 816 mol
nW
=
752
mol
CH 4
#
6–38
-0
=0
=0
Solutions Chapter 6
System : Mixer plus makeup point
F
(Species balances but no reaction so In = Out)
RG
M
(1) Toluene:
F (1.0) +
0
+
(2) H2:
0
+
M(1.0)
+
(3)CH4:
0
+
0
+
0
R
+
! 64 "
RG #
$ +
% 816 &
! 752 "
RG #
$ +
% 816 &
G
L
12
=
100
0
=
400
0
=
400
Solve (3) for RG = 434 mol
Solve (1) for F = 88 mol
M = 366
Change basis to 1 hour
3450 lb 1 lb mol
= 37.5 lb mol of toluene in F
92 lb
100 lb mol tol inG 37.5 lb mol F 12 lb mol R L
= 5.11 lb mol R L /hr
88 lb mol F
1 hr
100 lb mol tol in G
100 37.5 434 lb mol R G
= 185 lb mol R G /hr
88 1 100 lb mol tol in G
If F = 37.5 lb mol of toluene is selected as the basis, you have to make the same
balances as above plus benzene and diphenyl balances on the reactor plus separator
because F is a known but G becomes an unknown.
The unknowns would be
G, RL, RG, D, B, M n PH 2 ,n PCH4
Alternate solution
Calculate the extent of reaction for each reaction, and use them as shown in the
book to get the outputs of the reactor plus separator system instead of using generation
and consumption terms.
6–39
Solutions Chapter 6
6.3.16
This is a steady state problem with reaction and recycle.
Step 5:
Basis: 100 mol F
Steps 1, 2, 3 and 4:
Make the overall balances first
1
SO2 + O2 ↔ SO3
2
SO2
O2
N2
kg mol
10.0
9.0
81.0
F (mol)
Overall
Process
P(mol)
Calcd.
Calcd.
Use stochiometry mol
mol fr
SO3 0.95 (10) = 9.50 0.100
SO2 0.05 (10) = 0.50 0.00525
O2 9-4.75
= 4.25 0.0446
N2
81.0 0.850
95.25 1.000
Step 6:
The unknowns are the 4 exit compositions plus the extent of reaction if it
is to be used.
Steps 7, 8, and 9: We have 3 element balances plus the fraction conversion of the SO2 to
SO3, or 4 species balances (SO3, SO2, N2,O2). We can use a mixture of element and
species balances.
6–40
Solutions Chapter 6
Element S:
10
= 0.95 (10) + 0.05 (10)
Compound
N2:
81.0 = 81.0
check is ok
Select unit 1 as the system
mol
SO 2
10.0
O2
9.0
N2
81.0
fract.
calcd. mol
SO 3 7.5
conv.
SO 2
2.5
0.75
O2
5.25
1
N2
81.00
0.75
SO 3 : 7.5
S : 10
SO 2 : 2.5
0.25
= 1.5 (7.5) + 2.5 + n O 2 = 11.25 + 2.5 + n O 2
Element 2O: 9.0 + 10.0 = 19
n O 2 = 5.25
Balance around converter 2 plus separator: Note we need H = P + R, and observe that the
composition of H, P and R is the same.
kg mol
mol fr.
SO 3
9.50
0.100
SO 2
0.50
0.00525
SO 3
mol
7.5
SO 2
2.5
fract.
conv.
O2
5.25
0.65
O2
4.25
0.0446
N2
81.00
2
N2
81.00
0.85
1.000
H
P
R
System
same composition as P
R
Make a species balance on SO2
In
Out
Consumption
[2.5 + R(0.00525)] - [0.50 + R(0.00525)] +0
R = 111 kg mol
6–41
Generation
-[2.5 + R(0.00525)]0.65 = 0
Solutions Chapter 6
6.3.17
Steps 2, 3, and 4:
(Non Bz is non-benzene)
Step 5: Basis: 1 hr
Process is steady state, no reaction
Steps 2, 3 and 4:
Compositions:
at P
SO2
Bz
0.15 kg SO2
= 0.130
1+ 0.15
1 − 0.130
=
mWBz
at W
W
NonBz
m
mWSO2
0.870
1.00
Pick the overall process as the system
Steps 6 and 7:
6–42
!
"
$ ∑ =W
"
%
Solutions Chapter 6
Unknowns: P, 3miW (4)
Balances:
Bz, SO2, non Bz,
Steps 8 and 9:
Balances are in kg
Total
3000 + 1000 = P + W
Bz:
1000 (0.70) = P !#
1.00 "
W
$ + m Bz
% 1.15 &
Non Bz: 1000 (0.30) = P (0) + mW
N-Bz
SO2:
W
W
mBz
/mnonBz
= 0.25 (4)
0.15 "
W
$ + n SO2
% 1.15 &
3000 = P !#
W
mnonBz
= 300 kg
W
mBz
= 300 (0.25) =75 kg
a. Solution: P = 719 kg
W
mSO
= 2906 kg
2
W = 3281 kg
System: Unit 1
Steps 6 and 7:
D
Unknowns: m DBz , m DnonBz , mSO
Balances:
2
Bz, nonBz, SO2
Steps 8 and 9: Balances are in kg
b.
Bz + nonBz: D! = A + 375
D! = 1000 + 375 = 1375 kg
System: Unit 2
6–43
Solutions Chapter 6
Steps 6 and 7:
Steps 8 and 9: Balances are in kg
Bz + nonBz: C! = 749 + G !
System: mixing point
D' = 1375 kg
G'
F = 1000 kg
G ! = 1375 – 1000 = 375 kg
c.
C! = 749 + 375 = 1124 kg
6.3.18
Steady state process with reaction
Step 1, 2, 3, and 4:
Get the amount of {reqd., excess} HNO3 in G.
Step 5:
Basis: 1 hr
Step 4:
1 ×103 kg Glycerine 1 kg mol Gly
= 10. 86 kg mol Gly
92.11 kg Gly
Glycerine
C3 H8 O3 + 3HNO3 →C3 H5 O3 (NO2 )3 + 3H2 O
Required:
HNO3 is 3 (10.86)
6–44
= 32.58 kg mol
Solutions Chapter 6
Excess:
Steps 6 and 7:
HNO3 is 32.56 (.20) = 6.516
Total:
39.10 kg mol in G
System is overall
W
Unknown are F, P,m W
H 2 SO 4 ,and m H 2 O
(Σmi=W)
Equations are: C, H, S, O, N
(or you can use stoichiometry and make species balances, or use the extent
of reaction)
C:
10.86 = 0.9650 P
P = 11.25 kg mol or 11.25 (227.09)
= 2556kg
(a)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - -- - - - - - - - - - - - - System is the mixing point
5
Unknowns:
F, m H 2 SO 4 , m H 2 O ,R, G
Balances:
H 2SO4 , H 2O , HNO3 and m H 2 O + m H 2 SO 4 + 39.10 =G
(a) Glycerine Feed
=
Nitroglycerine Produced =
1000
= 10.86 kg mol
92.11
1086 kg mol 227.98 kg
= 2466 kg
lb mol
0.9650 P = 2465
(a)
P = 2555 kg
(b) Mol HNO3 req'd
10.86 mols Glyc. 3mol HNO3
mol Glyc
= 32.58 mol
=
Actual HNO3 Req'd
= 1.200 (32.58) = 39.10 mol
HNO3 in R
= 39.10 - 32.58 = 6.52 mol
= 6.52 (63.01) = 411 kg
HNO3 is 70.00% R, 0.7000R = 411
6–45
4
Solutions Chapter 6
R = 587lb
(c)
(b)
HNO3 in G = 39.10 mol = 2464 kg
HNO3 in R = 411 kg
HNO3 Balance:
F+R=G
F + 411 = 2464
F = 2053 lb HNO3
F is 43.00% HNO3 so that 0.4300 F = 2053
F = 4774lb
(d)
H 2SO4 in F
(c)
= 0.5000 (4774) = 2387 kg
= H 2SO4 in W
H2O in F
H2O Generated by reaction
= 0.0700 (4774) = 334 kg
= 10.86 Mole Glyc × 3
= (10.86) (3) (18.02) = 587 kg
H2O in Stream P
= 0.0350 (2555) = 89 kg
H2O in W
= H2O in F + H2O generated - H2O in P
= 334 + 587 - 89 = 832 kg
Component
kg
wt%
H 2SO4
2387
74.15
H 2O
832
3219
25.85
100.00
Stream W = 3219 kg/ hr
H2 SO 4
= 74.15 %
H2 O
= 25. 85%
(d)
Step 10:
Do numbers check?
Glycerine + F
1000 + 4774
5774
=
=
=
P+W
2555 + 3219
5774
6–46
Solutions Chapter 6
6.4.1
By inspection you can see that the flow of C12 and H2 in the separator is going the
wrong way, hence any calculations make no sense.
6.4.2
No purge stream exists out of the separator so that CO will build up.
6.4.3
Step 1, 2, 3 and 4:
CO2
Makeup
3 : 1 H2 /CO 2
6
Recycle
7
O2
3 : 1 H 2 /CO 2
CO
Converter
99 kg mol CH 4
1 kg mol N 2
100%
9
Condenser
10
Methanol
CH 4 feed
11
Reactor
Reformer
217.8 kg mol
12
CO conv.
3
steam
13
8
5
4
Pure 5% N 2
Methanol
100% CH4
conversion
90% CO2
yield
55%
conversion
solution
Chemical Reactions
a)
CH4 + 2H2 O → CO2 + 4H2
(main reformer rxn)
b)
CH4 + H2 → CO + 3H2
(reformer side rxn)
c)
2CO + O2 → 2CO2
(CO converter rxn)
d)
CO2 + 3H 2 → CH 3 OH + H 2O
(methanol rxn)
CH4 feed is 1% N2 or 1 kg mol N2
steam feed is 10% excess based on reaction (a).
⎛ kgmol H 2O ⎞
⎟ = 198kgmol steam
99 kg mol CH4 ⎜ 2
⎝ kgmol CH4 ⎠
1.1 (198) = 217.8 kg mol steam
6–47
Solutions Chapter 6
Step 5:
Basis: 100 kg mol CH4 in feed
Steps 6 and 7:
Unknowns:
6-CO2 makeup
10-reactor product
3-reformer product
7-3:1 H2/CO2
11-Methanol solution
4-O2 feed, stoichiometric
8-recycle, H2/CO2=3
12-condenser tops
5-CO conv. products
9-reactor feed, H2/CO2=3
13-purge, 5% N2
Balances:
Reformer balance
Condenser balance
CO conv. balance
purge/recycle balance
CO2 makeup balance
Feed/recycle balance
Methanol reactor balance
Steps 8 and 9:
Solve balances serially.
Reformer balance gives stream 3
CO2 =
99 kgmol CH4 conv. 0.9conv by(a) 1kgmol CO2
= 89.1kgmol CO2
1kgmol CH 4
1conv.
CO = 99(0.1) = 9.9 kg mol CO
H 2O reacted =
+
2 kgmol H 2 O 89.1kg mol CO2
1kgmol CO2
1kg mol H2 O 9.9kg mol CO
= 188.1kgmol H2 O
1kgmol CO
H 2O remaining = 217.8 - 188.1 = 29.7 kg mol H 2O
H2 = 4(89.1) + 3(9.9) = 386.1 kg mol H2
N2 = 1 kg mol
CO conv. balance gives streams 4 & 5:
Stream 4:
6–48
Solutions Chapter 6
O2 =
9.9 kgmol CO (1 / 2) kgmol O2
= 4.95kgmol O 2
1kgmol CO
Stream 5:
CO2 = 89.1 + 9.9 = 99 kg mol CO2
H2O = 29.7 kg mol H2O
H2 = 386.1 kg mol H2
N2 = 1 kg mol N2
CO2 makeup gives streams 6 and 7:
stream 7 is 3:1 H2/CO2
CO2 = 386.1/3 = 128.7 kg mol CO2 needed
Stream 6:
CO2= 128.7 - 99 = 29.7kg mol CO2
b.
purge/recycle gives stream 13:
N2 is inert species:
stream 13 =
1.0 kgmol N2
= 20 kgmol instream 13
0.05kg mol N2 / kg mol stream13
Stream 13:
H2/CO2 = 3
Let x = mol frac. of CO2 in stream 13
1 = 0.05 + 3x + 1x
4x = 0.95, x= 0.2375
N2 = 1 kg mol N2
H2 = 20 (3) (0.2375) = 14.25kg mol H2
CO2 = 20 (0.2375) = 4.75 kg mol CO2
Special balance gives stream 11:
6–49
a.
Solutions Chapter 6
128.7 CO2
29.7 H2O
386.1 H2
1.0 N 2
1 rxn occurs
13
7
11
4.75 CO2
14.25 H 2
1.00 N2
CO2 reacted = 128.7 - 4.75 = 123.95 kg mol CO2 reacted
H2 reacted = 386.1 - 14.25 = 371.85 kg mol H2 reacted
CH3OH produced = 123.95 kg mol CH3OH
H2O produced = 123.95 kg mol H2O
Stream 11:
H2O = 29.7 + 123.95 = 153.65 kg mol H2O
CH3OH = 123.95 kg mol CH3OH
mass stream 11 = 153.65 (18) + 123.95 (32) = 6732.1kg
123.95 (32)
wt. % CH3OH =
= 58.9% CH3 OH wt. %
6732.1
d.
Methanol reactor balance for 55% conversion
From special balance, each pass uses 123.95 kg mol CO2
123.95 = 0.55 (CO2)in so (CO2)in = 225.36 kg mol CO2
Stream 8 = stream 9 - stream 7
Stream 8: CO2 = 225.36 - 128.7 = 95.66
So
recycle
95.66
=
= 20.35
purge
4.75
c.
6–50
Solutions Chapter 7
7.1.1
14.7 psia
100 ft 3
psia ft 3
(296)(1.8)o R
10.73
o
(lb mol) ( R)
n=
pV 15.5 mm Hg
=
RT
760 mm Hg
n=
(15.5)(14.7)(100)
= 0.00524 lb mol
(760)(10.73)(533)
lb H 2 O = (0.00524)(18) = 0.0944 lb H 2O
7.1.2
(pV)1 = (pV)2
1×1 = p×1.2
p=
7.1.3
1
= 0.83 atm
1.2
Specific volume:
V̂ =
MW = molecular weight
(R)(T) (R/MW)T (1545.3/28.97) (78 + 460)
=
=
= 13.56 ft 3 /lb m
p
p
(14.7)(144)
Molal specific volume:
V̂m =
(R)(T) (1545.3) (78 + 460)
=
= 392.8 ft 3 /lb mol
p
(14.7)(144)
Note that
V̂ =
V̂m
392.8
=
= 13.56 ft 3 /lbm
MW 28.97
7–1
Solutions Chapter 7
7.1.4
Assume ρH2O = 62.4 lb/ft3 , pbar = 14.696 psia
p=
500 f tH 2O 14.696 psia
+ 14.696 psia = 231.387 psia
33.91 ft H2 O
ft 3
359 ft3 14.696 45 + 460
Vˆ =
= 23.4
lb mol
lb mol 231.387 32 + 460
7.1.5
pV = nRT
3
T=
pV
121 kPa 25 L 1 m
(kg mol)(K)
=
3
nR 0.0011kg mol
1000 L 8.314 (kPa)(m )
= 330.8 → 331K
7.1.6
You must first convert the temperatures and pressures into absolute units:
460 + 70 = 530°R
460 + 75 = 535°R
atmospheric pressure = 29.99 in. Hg = 14.73 psia
final pressure = 29.99 in. Hg +
4 in. H 2 O 29.92 in. Hg
12 in H 2 O 33.91 ft H 2 O
ft H 2 O
= 29.99 + 0.29 = 30.28 in. Hg absolute
The simplest way to proceed, now that the data are in good order, is to apply the ideal gas
law. Take as a basis, 16.01 ft3 (do not forget to include the volume of the O2 tank in your
system) of O2 at 75°F and 30.28 in. Hg. Determine the initial pressure in the O2 tank
alone.
7–2
Solutions Chapter 7
! V "! n "! T "
p1 = p2 # 2 $# 1 $# 1 $
% V1 &% n2 &% T2 &
! 16.01 ft 3 " ! 530°R "
p1 = 30.28 in. Hg #
$#
$ = 480 in. Hg absolute
3
% 1 ft & % 535°R &
In gauge pressure,
p1 =
7.1.7
a)
(480 − 29.99) in. Hg 14.696 psia
= 221 psig
29.92 in. Hg
Basis: 1 hr
Q= ν A
3
11.3 ft 3600 s (18.0 in)2 1 ft 2
5 ft
Q=
2.875
×
10
π
=
hr
s
1 hr
(12 in)2
5 3
In 1.0 hr 2.875 × 10 ft
b)
m = ρν s
m = 0.0796
In 1 day:
lb m
lbm 11.3 ft π (18.0 )2 in 2 1 ft 2
3
2 = 6.358
s
ft
s
(12 in )
6.358 lb m 3600s 24 hr
5
= 5.49 × 10 lb m
s
1 hr
7–3
Solutions Chapter 7
7.1.8
Basis: air at 30 psi and 75°F in volume of tire
(1)
Initial State
(2)
Final State
p1 = 44.7 psia
p2 = ?
T1 = 535°R
T2 = 600°R
V1 = ?
V2 =
N1 = ?
n2 = ?
! n "R! T "
p1V1
=# 1$ # 1$
p2 V2 % n 2 & R % T2 &
Assume volume is the same
! T2 "
#
! 600 "
$ p2 = p1 % & = (44.7) %
& = 50.2 psia
The number of mols are the same )
' 535 (
' T1 (
50.2 – 14.7 = 35.5 psia
just over the limit
7.1.9
Apply pV = nRT twice or use it once with R
Basis: 1 pound mole fg
V (0.7302)(560)
=
n
1.54
(1) 100°F (560°R) and 1.54 atm
(2) 32°F (492°R) and 1 atm (SC)
! T "! p "
p1V1
T
= 1 so V1 = V2 # 1 $# 2 $
p2 V2
T2
% T2 &% p1 &
=
359 ft 3 560o R 1 atm
265 ft 3
=
1 lb mol 492o R 1.54 atm
lb mol
7–4
Solutions Chapter 7
7.1.10
(a) 1.987 cal/(g mol)(K);
(b) 1.987 Btu/(lb mol)(°R);
14.7 psia 359 ft 3
3
(c )
= 10.73 ( psia ) ft / (°R )(lb mol )
492°R 1 lb mol
( )
5
2
1.013 × 10 N/m 22.4 m 3
1J
1 kg mol
(d)
3
273 K
1 kg mol 1 (N) (m)
10 g mol
= 8.314 J/(g mol) (K)
(e)
1 atm 22,400 cm3
3
= 82.06 cm (atm) /( K)( g mol)
273 K 1 g mol
( )
1 atm 359 ft 3
3
(f)
= 0.7302 ft ( atm)/ (lb mol )(°R )
492°R 1 lb mol
( )
7.1.11
Basis: 1 lb mol gas at 760 mm Hg and 60°F
Sp. gr. =
(
p C3H8 MWC3H8
RTC3H8
)
p air (MW) air
RTair
=
800(44) 520 1
=1.483
560 760 29
7.1.12
Basis: 1 m3 H2 at 5°C and 110 kPa
a.
1 m 3 273 110 kPa 1 kg mol 2 kg
3
= 0.0952 kg/m
3
278 101.3 kPa 22.4 m 1 kg mol
b.
2 kg H 2 1 kg mol air
= 0.069 = specific gravity
1 kg mol H 2 29 kg air
7–5
Solutions Chapter 7
7.1.13
Basis: 2m3 at 200 kPa and 25°C
pCO2 = 200(0.8) = 160 kPa
7.1.14
Basis: 1 liter final volume
Assumptions:
(1)
Temperature is constant
(2)
Ideal gas law applies
pT = pO2 + p N2 = (760 + 760) = 1,520 mm Hg
7.1.15
a. F
b. F
c. T
7–6
Solutions Chapter 7
7.1.16
V=?
90°F
4 in. H 2O gauge
1 ft 3
70°F
215 psia
460 + 70 = 530°R
460 + 90 = 550°R
atmospheric pressure = 29.92 in. Hg = std atm = 14.7 psia
initial pressure =
200 psig + 14.7 psia 29.92 in. Hg
= 437 in. Hg
14.7 psia
final pressure = 29.92 in. Hg +
4 in. H2 O 29.92 in. Hg
12 in. H 2 O 33.91 ft H2 O
ft H2 O
= 29.92 + 0.29 = 30.21 in. Hg
Basis: 1 ft3 of oxygen at 70°F and 200 psig
3
final volume =
1.00 ft 550°R 437 in. Hg
530°R 30.21 in. Hg
3
= 15.0 ft at 90°F and 4in. H2 O gauge
Formally, the same calculation can be made using
p T
V2 = V1 1 2
since n 1 = n 2
p 2 T1
7–7
Solutions Chapter 7
7.1.17
?
20 ft 3
10 lb
CO2
30°C
10 lb CO2
known vol.
at S.C.
Solution
We can write (the subscript 1 stands for standard conditions, 2 for the conditions
in the tank)
V T
p2 = p1 1 2
V2 T 1
14.7 psia 10 lb CO 2 1 lb mol CO2 359 ft 3
303K
= p2 = 66.6
3
44 lb CO 2 1 lb mol 20 ft 
273K


psia


 
V2
V1
T2
T1
Hence the gauge on the tank will read (assuming that it reads gauge pressure and that the
barometer reads 14.7 psia) 66.6 – 14.7 = 51.9 psig
7.1.18
Basis: Data in the diagram
State 1 is before corking; state 2 is after equilibrium is reached after filling. Assume pA =
pZ1 = 29.92 in. Hg abs. pZ2 = pA + ρ Hg gh = 29.92 + h where pZ2 is in inches Hg.
Use the ideal gas law: pZ2 VZ2 = pZ1VZ1
pZ2 = 29.92 (8/6) = 39.89 in. Hg
39.89 = 29.92 + h
or
h = 9.97 in. Hg
9.97 + 14 = 24 in. Hg
7–8
Solutions Chapter 7
7.1.19
1.
Basis: Flue gas at 1800°F, constant pressure, and given volume
Calculate the inlet cross-sectional area A:
Ai = π[(Di )2 ]/4 = π(42 )/4 = 12.57 ft 2
2.
Calculate the inlet volumetric flow rate Q:
Q = (velocity)×(cross-sectional area) = 25(12.57) = 314.16 ft 3/s
3.
Calculate the outlet volumetric flow rate using the ideal gas law:
Qo = Qi (To /Ti ) = 314.16(460 + 550)/(460 + 1800) = 140.40 ft 3/s
4.
Calculate the outlet cross-sectional area:
Ao = Qo /νo = 140.40/20 = 7.02 ft 2
5.
Calculate the outlet duct diameter:
(Do )2 = 4(A0 )/π = 4(7.02)/π
Do = (4(7.02)/π)0.5 = 2.99 ft
7.1.20
Basis: 10 kg FeS
(MW = 87.9)
FeS + 2HC1 → H2S + FeC13
10 kg FeS 1 kg mol FeS
= 0.114 kg mol FeS
87.9 kg FeS
Assume pressure most likely is gage. Absolute pressure = 15.71 + 76.0 = 91.71 cm Hg =
122.2 kPa
nRT 0.114 kg mol FeS 8.314(kPa)(m3 ) (30 + 273)K
V=
=
p
(kg mol)(K) 122.2 kPa
= 2.35 m3 @ 30o C and 15.71 cm Hg gauge
7–9
Solutions Chapter 7
7.1.21
Basis: 1 hr = 500 lb waste (W)
Get μg/ft3 at SC
1 μg/mL = 0.001g/L; ignore HC in totaling W
500 lb W 454 g 0.01 g HC
= 5.32×10-3g HC/ft 3SC
3
1 lb
1 g W 427,000 ft SC
The minimum volume of flue gas is
1 ft 3SC
1 g 10 µg 25 mL
= 0.047 ft 3SC
-3
6
5.32 × 10 g HC 10 µg mL
a.
0.047 ft 3SC
b.
1 hr
= 1.1 × 10-7 hr
3
427,000 ft SC
7.1.22
Basis: 100 ppm of TCE (MW 131.5) in air
ρ=
p MW
where MW is the average MW. At 100 ppm, MW =
RT
(100 ×10−6 )(131.5) + (1 − 100 ×10−6 )(29) 29.013
=
The density is essentially the same as that of air.
7.1.23
Basis: 1 m3 gas passing at 10°F and 1 atm. vs 60°F and 1 atm.
(assume p1 = p2), T1 = 520°R; T2 = 470°R
1 m3 520o R
=1.106 m3
470o R
Since the moles, and mass, are directly proportional to volume at constant pressure, %
increase = 10.6%
7–10
Solutions Chapter 7
7.1.24
Basis: CO2 in cylinder
Volume of cylinder
2
" 0.75 #
3
∏r h = ∏$
% (4.33) = 1.91 ft
& 4 '
2
Assume 100% CO2 in cylinder at 0°C
1.91 ft 3 82.6 psia 530o R
= 11.55 ft 3 at 14.7 psia, 70o F
o
14.7 psia 492 R
If the machine has a 4 gal capacity
4 gal
1 ft 3
= 0.535 ft 3 at 14.7 psia and 70oF
7.48 gal
You have more than enough CO2 to fill the machine, if it operates under atmospheric
pressure.
7–11
Solutions Chapter 7
7.1.25
Basis: 100 kg mol gas.
a)
Total
b)
Mol
87
12
1
100
CH4
C2 H6
C3H8
MW
16
30
44
kg
1392
360
44
1796
composition in vol. percent = mol percent
CH4 87%
C2H6 12%
C3H8 1%
c)
MW of gas =
kg
1796 kg
= 17.96
kg mol
100 kg mol
R = 8.314 (kPa)(m3)/(kg mol)(K)
no. of moles of gas =
T = 9°C = 282.15K
80 kg
= 4.45 kg mol
17.96 kg kg mol
( 3 ) 282.15 K
nRT 4.45 kg mol 8.314(kPa) m
V=
=
p
(kg mol )(K)
d)
e)
600 kPa
= 17.4m3
17.96 kg
1 kg mol
density =
= 0.801 kg/m3
3
1 kg mol 22.415 m at SC
Specific gravity =
density of gas at 9°C and 600kPa
density of gas at SC
600
1 17.96
8.314 282
_______________
101.3
29
8.314 273
= 3.55
kg/m3gas at 9oC, 600 kPa
kg/m3 gas at SC
7–12
wt%
77.5
20.0
2.5
100.0
Solutions Chapter 7
7.1.26
Basis: 100 mol mixture
a.
Comp
Mol
Mol. Wt.
lb
Wt %
Air
99
29
2870
94.72
Br2
1
159.8
160
5.28
3030
100.00
100
3030
lb
= 30.3
100
lb mol
b.
Avg. Mol. Wt. =
c.
sp.gr. at SC =
d.
sp.gr. =
e.
(30.3) (492) (114.7)
(359) (560) (14.7)
sp.gr. =
= 7.54
(29) 492
30
(359) 520 29.92
30.30
= 1.045
29
30.30
= 0.1895
159.8
7–13
Solutions Chapter 7
7.1.27
Steps 1, 2, 3:
CH4 + 2 O2 → CO2 + 2 H2O
Step 4: Unknowns:
P and all the compositions of P (5 unknowns)
Step 5: Balances:
Step 6: Basis:
CH4 + 3/2 O2 → CO + 2 H2O
C, H, O, N + fact that 30% of C → CO
100 mol F
Steps 7, 8:
C balance: 100 mol in CO = 30 mol !
"
CO 2 = 70 mol # out
100 mol "$
2N balance:
Reqd. O2:
200 mol
XS O2: 0.20 (200) =
40
Total O2 in
240 mol
total N2 in =
240 O2 0.79 N2
= 903 mol = n N2 out
0.21 O2
2H balance: mol H 2O out = mol H 2 in =
400
= 200 mol
2
2O balance: O2 in – O2 in CO, CO2, H2O = O2 out
240 –
30
200
+ 70 +
= 55 mol
2
2
7–14
Solutions Chapter 7
alternate: XS O2 + O2 not used for CO = 40 +
30
= 55 mol
2
pCO = pTyCO
30
1258
= (740)
= 17.7 mm Hg
7.1.28
Steps 2, 3, 4:
Step 5: Basis: Gas listed in the figure
Steps 6 and 7:
Unknowns p (final), V (final); Equations: ideal gas laws
Vfinal = 1m3
n1 + n2 = nfinal
Steps 8 and 9:
n1 =
p1V1
pV
pV
n2 = 2 2 nf = f f
RT1
RT2
RTf
(600 kPa)(0.5m3 ) (150 kPa)(0.5m3 ) pf (kPa)(1.0m3 )
+
=
293K
303K
288K
pfinal = 366 kPa
7–15
Solutions Chapter 7
7.1.29
Steps 2, 3, and 4: p1 = 55 + 14.7 = 69.7 psia
Step 5: Basis: Gas with given data
Steps 6 and 7: Unknowns: T1 , T2 , Tf , pf (Vf = 400 + 50 = 450 ft 3 )
Equations:
p1V1 = n1 RT1
T1 = T2 = Tf
p2V2 = n 2 RT2
n1 + n 2 = nf
pf Vf = nf RTf
n1 =
p1V1
400(69.7)
=
RT
RT
n2 =
(50)(14.7)
RT
! 50(14.7) + 400(69.7) "
pf Vf = #
$ RT
RT
%
&
pf =
50(14.7) + 400(69.7)
28635
=
= 63.8 psia = 48.9 psig
450
450
7–16
Solutions Chapter 7
7.1.30
Step 5: Basis: 1 min
Steps 2, 3, and 4:
28.8 m3 at SC
1 kg mol
= 1.29 kg mol
22.4 m3 at SC
Step 6:
Unknowns: P, F
Step 7:
Indept. balances: air, SF6
Steps 8-9:
Balances are in moles
SF6 : F(0) + 1.29 = P(4.15×10-6 ) hence P = 3.098 × 105 kg mol
3.098 × 105 kg mol 22.4 m3 at SC 60 + 273
= 8.46×106 m3 /min
1 kg mol
273
Alternate solution:
28.8 m3 at SC 333K
1
= 8.46 ×106
6
273K 4.15 ×10
7–17
Solutions Chapter 7
7.1.31
Step 5: Basis: 15 lb CO2 added ≡ 30 min
Steps 2, 3, and 4:
Step 6:
Unknowns: F, P
Step 7:
Balances: Gas, CO2
Steps 8 and 9: Balance overall on process in moles
Gas:
0.988 (F) = P(0.966)
CO2:
0.012 (F) + 0.341 = P(0.034)
Total: F + 0.341 = P
F = 14.79 lb mol
F=
P = 15.13 lb mol
14.79 lb mol 21.9(in Hg)(ft 3 ) (60 + 460)o R
= 182 ft 3 /min
o
30 min 31.2 in Hg (lb mol)( R)
at 60oF and 31.2 in Hg
7–18
Solutions Chapter 7
7.1.32
Steps 2, 3, and 4:
Step 5: Basis: 2 min
Steps 6, 7, 8 and 9:
mol of gas out =
30,000ft 3 720 mm Hg 492o R 1 mole
= 75 mol
760 mm Hg 520o R 359 ft 3
60 mol gas 0.162 mol CO2
= 9.7 mol CO2 in
1 mol gas
75 mol gas 0.131 mol CO2
= 9.7 mol CO2 out
1 mol gas
mol of gas in =
47,800ft 3 740 mm Hg 492o R 1 mole
= 60 mol
760 mm Hg 1060o R 359 ft 3
Thus since there are more moles out than in, one may assume the system does contain a
leak. Also, assuming a leak would bring no additional CO2 into the system, the above
CO2 balance calculations indicate the given analysis is probably correct.
7–19
Solutions Chapter 7
7.1.33
Steps 1, 2, 3, 4: Steady state problem with a reaction.
Step 5:
Basis: 1 kg mol n-butane
Step 6:
Unknowns are the moles in P (4) plus P.
Step 7:
The element balances are C, H, O, N, and we can use Σni = P so we can
get a unique solution. You can use element balances or compound
balances. The former is easier.
Step 4:
Calculate A given the 40.0% excess air.
C 4 H10 + 6 1 2 O2 → 4 CO2 + 5 H2 O
In
1.00 kg mol C4 H10 6 1 2 kg mol O 2
1 kg mol C4 H10
7–20
= 6.5 kg mol O 2
Solutions Chapter 7
6.5(0.40)
= 2.60 kg mol O2 xs
= 9.10 kg mol O2 total
9.10 (0.79 0.21)
= 34.23 kg mol N2
Steps 7, 8, and 9:
Balances (kg mol)
In
C:
1(4)
H:
Out
=
n CO2
n CO2 = 4
1(10) =
n H 2 O (2)
n H2O = 5
O2:
9.10
n CO2 +n O2 +
N2
34.23 =
=
n H2O
n O2 = 2.6
2
n N2
n N2 = 34.23
P = 4 + 5 + 2.6 + 34.23 = 45.83 kg mol
( 3) 533.1K
nRT 45.83 kg mol 8.314(kPa) m
V=
=
p
100.0 kPa
(kg mol)(K )
= 2.03x103m3
a.
flue gas analysis kg mol
mol%
CO 2
4
9
H 2O
5
O2
2.6
34.23
11
6
N2
Total
45.83
7–21
75
100
Solutions Chapter 7
7.1.34
H2 100%
F1
F2
Reactor
1000°C
HSiC3 100%
2
3
644.95 mol
π
2
2
L (Do – DI ) =
4
2. 33 g π 100 cm (10 cm) – (1 cm)
cm
W
P
Si 100%
P = (D)(V) = D
HC 
100%
4
2
g mol
= 644. 95 g mol Si product
28. 086 g
Si balance
F 2 g mol HSiCl 3
1 g mol Si
= P g mol Si, hence F 2 = P
1 g mol HSiCl 3
H g mol balance
F1 g mol H 2 2 g mol H F2 g mol HSiCl 3 1 g mol H
W g mol HCl 1 g mol H
+
=
1 g mol H 2
1 g mol HSiCl3
1 g mol HCl
Cl mol balance
F 1(0) +
F 2 g mol HSiCl 3
3 g mol Cl
W g mol HCl 1 g mol Cl
=
1 g mol HSiCl 3
1 g mol HCl
Solution:
F2 = P = 1 g mol HSiCl3
3F2 = W = 3 g mol HCl
2F1 + F2 = W
F1 =
3 g mol – 1 g mol
= 1 g mol H 2
2
7–22
Solutions Chapter 7
1 g mol H 2 644.95 g mol Si
= 644.95 g mol H 2
g mol Si
pV = nRT
V=
nRT
p
( 3)
644.95 g mol 82.06 cm ( atm) 1273 K
1L
=
( g mol)( K) 1 atm 1000 cm3
= 67,373 L
7.1.35
Basis: 1 hr
a.
2.5×10-7 g mol O2 2.9×106 cells 104 mL 103 mg mol
725 mg mol
=
3
(10 cells)(hr)
mL
1 g mol
hr
b.
45 L gas 0.40 L O2 110 kPa 1 m3
(kg mol)(K)
3
h
1 L gas
10 L 298 K 8.314(kPa)(m3 )
= 7.99 × 10-4 kg mol/hr
c.
or
799 mg mol/hr
Increase
7–23
Solutions Chapter 7
7.1.36
Basis: 106m3 at SC of CH4 burned completely
Ignore the oxidation of S, N, and C going to CO in the material balance to simplify the
calculations. Including those products will have negligible effect on the N2 and CO2 produced.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H 2O(g)
Calculate the kg mol of CH4 burned:
106 m3at SC
kg mol
= 44,613 kg mol
22.415 m3
Assume 10% excess air is used. Base the CO2 on the emission factor value.
Required O2 entering: 44,613(2) =
Excess O2 (0.1)(89,226)
Total O2:
N2 entering: 98,149(79/21) =
Exit gas
CO2 produced:
N2 exiting:
H2O exiting: (2)(44,613) =
O2 exiting: (0.10)(89,226) =
SO2: (9.6)/64 =
kg mol
89,226
8,923
98,149
369,227
44,613
369,227
89,226
8,923
(kg mol) Emission factor
43,812
0.15
0.086
0.722
0.174
0.017
2.9 × 10-7
!3040 / 46 =
"
NO 2 : #1600 / 46 =
"1500 / 46 =
$
66
1.3 ×10−7
35
6.8 × 10−5
33
6.4 ×10−5
!1344 / 28 =
CO "
#640 / 28 =
48
23
9.3 ×10−5
511,200
1.00
!!
→
!!
→
!!
→
Total (about)
On a dry basis:
mol fr.
SO2
NO2
CO
3.5 ×10−7
1.6×10-4 , 8.2×10-5 , and 7.8×10-5 respectively
1.1×10-4 and 5.5×10-5 , respectively
7–24
4.5 ×10−5
Solutions Chapter 7
7.1.37
Basis: 103 L of No. 6 fuel oil
103L oil 0.86 g 1000 cm3
= 8.6×105 g or 8.6×102 kg of oil
1 cm3
1L
To get the moles of pollution components alone in the gas produced, the kg have to be converted
using MW (ignore the particulate matter)
kg
MW
g mol
m3 at SC on the basis
of 103/kg
SO2
SO3
NO2
19 (0.84) = 15.96
0.69 (0.84) = 0.580
8
64
80
46
249
7.3
174
6.5
0.18
4.5
CO
CO2
0.6
3025
28
44
21
6.875 × 104
0.55
1,790
Example of calculating the m3 at SC on the basis of 1 metric ton (103 kg) of No. 6 fuel oil burned:
SO2:
0.249 kg mol 22.415 m3 at SC 103 kg oil
= 6.5 m3 at SC
2
8.6×10 kg oil
1 kg mol
7–25
Solutions Chapter 7
7.1.38
7–26
Solutions Chapter 7
7.1.39
Process: Steady state with reaction and recycle
Step 5:
Basis: 1 min
Steps 2, 3, and 4:
260L C6H6
950L H2
C6 H 6 + 3 H 2 → C6 H12
Covert L → mol
pV
feed
n=
RT
p = 150 kPa
n H 2 = 45.96 g mol
T = 100°C + 273 = 373K
(kPa ) m3
R = 8.314
(kg mol)(K)
( )
n C 6 H 6 = 12.58 g mol
Steps 6, 7, 8 and 9:
Select the overall system for the first balances.
H2 :
In –
45.96 –
Out
n °H 2
+ generation
+
0
– Consumption = 0
– 0.75(45.96) = 0
C 6 H6 :
12.58 –
n °C 6 H 6
+
–
C6H12:
0
n °C 6 H12
+ 0.75(45.96) 13 –
–
0
0.75(45.96) 13 = 0
from 1: n °H 2 = 11.5 g mol
from 2: n °C 6 H 6 = 1.09 g mol
from 3: n °C 6 H12 = 11.5 g mol
b.
Volumetric flow rates:
T = 200°C = 473K
P = 100 kPa = 0.987 atm
( n ) (RT) = 452 liters/min
V =
o
H2
H2
p
VC6H6 = 42.8 liters/min
C6 H12 = 452 liters/min
7–27
0
= 0
Solutions Chapter 7
c.
H2 balance on Reactor + Separator
H2: (45.95 + 0.90R) – (0.90R +11.5 ) + 0 – 0.48 (45.95 + 0.90R) = 0
Substitute for n °H 2 and solve to get R = 28.7 kg mol
Volumetric flow rate (R) = 890 liters/min
7.1.40
Basis: 1 day = 104 kg C2H2O
If the overall system is chosen, the unknowns are F1 and ni, and three independent
equations can be written from C, H, and O balances plus nCO2 = nH2O.
Pick the system of the reactor plus separator, and use the data for the conversion.
Step 5: Basis
10, 000 kg 1.00 kg mol
= 227.3 kg mol C 2H 4O/day
day
44 kg
Steps 6, 7, 8 and 9:
Mixing point: F1 + R = F2 ⇒ 0.40F1 + R = n FC22 H 4 , 0.60F1 = nFO22 . Unknowns:
F1 ,F2 ,R,n FO 2 , nFC 2 H 4
Reactor plus separator:
7–28
Solutions Chapter 7
C2H4 balance:
consumed
! in
$ out 
#"0.4(F1 )+ R &% – R – [0.4(F1 ) + R]0.5 = 0
generation


C H O balance: 0–227.3+[F1 (0.4) + R]( 0.50)(0.70) = 0
2 4
F1 = 811.79 kg mol/day
R = 324.7 kg mol/day
F1 + R = F2 = 1136.49
1136.5 kg mol 22.40 m3 at SC 1 day
3
(a)
= 1059 m at SC/hr
day
1 kg mol
24 hr
adjust recycle


324.71 kg mol of C2 H4 in R
22.40 m at SC 283K 101.3 kPa
3
(
)( )
(b) 811.79 0.40 kg mol of C 2 H 4 in F1 22.40 m at SC 273K 100 kPa = 1.05
3
Overall balances (kg mol): unknowns n1 tr 3
F1
C:
811.79 (.40) (2) = 227.3 (2) + nCO2
Balances 3
nCO2 = 194.83
H:
811.79 (.40) (4) = 227.3 (4) + nW (2) nW
O:
811.79 (.60) (2) = 227.3 (1) + 194.83 (2)
+ 194.83 (1) + nO2 (2)
nO2 = 81.18
G = 194.83 + 194.83 + 81.18 = 470.84
470.84 kg mol P 2 22.4 m 3 at SC 353 101.3
day
273 100
kg mol
(c)
4
3
= 1.38 × 10 m gases/day at 80°C and 100 kPa
7–29
= 194.83
Solutions Chapter 7
7.1.41
Basis: 1 lb mol of 50% H2 and 50% C2H4O
p1V1 = n1RT1
p1 n1
=
p2 n2
p2V2 = n2RT2
T1 = T 2
760 mm Hg 1
hence n 2 = 0.922 lb mol
=
700 mm Hg n 2
Let y = mol of C2H6O formed
(0.5 – y) + (0.5 – y) + y = 0.922
from which y = 0.079 lb mol
Degree of completion =
C2H6O formed
0.079
×100 = 15.8%
0.50
7.1.42
C6 H12O6 is glucose and C3H8O3 is glycerol
The reaction equation is
a CH1.8O0.5N 0.5 + b NH 3 + c C6 H12O6
→ d C3H8O3 + e CO2 + f N2 + g H2 O
The amount of CO2 measured (52.4L at 95 kPa and 300 K) is
52.4 L CO2 1 m3 95 kPa
(kg mol)(K)
= 2 ×10−3 kg mol or 2.00 g mol
3
1000 L
300 K 8.314 (kPa)(m )
Use element balances and specifications to get the coefficients in the reaction equation. Let c = 1
(divide both sides by c) be the basis for obtaining the coefficients: then there are 6 unknowns, 4
equations, and two specifications:
C:
a + 6c = 3d + e
H:
1.8a + 3b + 12 = 8d + 2g
7–30
Solutions Chapter 7
O:
0.5a + 6 = 3d + 2.00(2) + g
N:
0.5a + b = 2b
Specifications:
f
=1
b
and
e
=2
c
The solution of the equations is:
a = 7.27,
b = 3.64,
d = 1.09,
f = 3.64,
g = 1.64
The glycerol produced was 1.09 g mol , and
The biomass reacted was 7.27 g mol
7.2.1
1.a
Immediate but approximate
2.e
Exact but requires access to a handbook
3.b
Good accuracy, needs calculation of z
4.c
Good accuracy, more complicated calculation than b even in a program
with a data base
5.e
May require a long search to filter out a good value
For other answers, look at the answer to P15.1.
7.2.2
Will get the most accurate value, and be quick, assuming you have the handbook that
contains the required data. For other answers, look at the answer to P15.1.
7–31
Solutions Chapter 7
7.2.3
Some valid answers are
(1)
continuous analytical functions, and these functions can be differentiated
and integrated
(2)
higher accuracy (with complex equations)
7.2.4
B
C
D
+ 2 + 3
V
V
V
ˆ
pV
The results for
= z are shown in the Excel table below. You can decide how much
RT
z=1+
accuracy is needed.
p(atm)
4 terms
2 terms
1 term
% difference
2 from 4
1 from 4
0
10
50
100
200
500
1000
5000
1.00
0.95
0.88
0.77
0.80
0.90
0.95
0.99
1.00
0.95
0.88
0.76
0.79
0.90
0.95
0.99
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
0
0
0
-0.8
1.6
0
0
0
7.2.5
⎛ L ⎞
a ⇒ (atm)⎜
⎝ g mol⎠
2
b ⇒ L/g mol
α ⇒ dimensionless
7–32
0
-4.9
-13.3
-29.8
-25.1
-10.7
-3.6
-1.1
Solutions Chapter 7
7.2.6
Basis: 460 kg CO2
a. RK Equation:
460 kg CO2 1 kg mol
= 10.455 kg mol CO2
44 kg
Tc = 304.2K, pc = 7385 kPa
a = 6458 (kPa) (m6) (K0.5)/(kg mol)2
b = 0.0297 m3/kg mol
(kPa )( m3 )
R = 8.314
(kg mol)(K)
V̂ =
V
10.4 m3
m3
=
= 0.995
n
10.455 kg mol
kg mol
p=
(8.314)(290)
6458
−
0.995 − 0.0297 0.995(0.995 + 0.0297)(290)1/ 2
= 2.126 ×103 kPa
b. SRK Equation:
Data as above:
T/Tc = 290/304.2 = 0.953
a! =
0.42748(8.314)2 (304.2)2
= 379.2(kPa)(m6 )/(kg mol)2
7358
b=
0.08664(8.314)(304.2)
= 0.0297 m3 /kg mol
3
7358 × 10
p=
(8.314)(290)
(379.2)(1.040)
−
0.995 − 0.0297 0.995(0.995 + 0.0297)
2
κ = 0.480 + (1.574)(0.225) − 0.176(0.225)
=
0.843
1
2
λ = [1 + 0.843(1 − 0.953
= )]2 1.040
No significant difference.
p = 2.111×103 kPa
7–33
Solutions Chapter 7
7.2.7
Basis: 1 min
(1)
CO2
(2)
CO2
2000 m3/min
20oC
500 kPa
110oC
4800 kPa
Calculate the moles entering the compressor:
500
atm
p r1 = 101.3
= 0.068
72.9 atm
4800
atm
p r2 = 101.3
= 0.650
72.9 atm
Tr1 =
(20 + 273)K
= 0.96
304.2 K
Tr2 =
(110 + 273)K
= 1.26
304.2 K
2
! cm3 "
! m3 "
a = 3.6 × 10 atm $
=
364.68
kPa
%
$
%
& g mol '
& kg mol '
2
6
b = 42.8
cm3
m3
(kPa)(m3 )
= 0.0428
, R = 8.314
g mol
kg mol
(kg mol)(K)
Use Van der Waals’ equation:
!
n 2a "
p
+
$
% (V − nb) = nRT
V2 '
&
As the initial guess for the number of moles in State 1 use the ideal gas law
n1 =
p1V1
(500)(2000)
=
= 410.5 kg mol
RT1
(8.314)(293)
The number of moles at condition (1) is obtained from
!
n12 (364.68) "
=
$ 500 +
% (2000 − 0.0428 n1 ) (8.314)(293)n
1
20002 '
&
From Polymath (on the CD)
n1 = 420 kg mol
7–34
Solutions Chapter 7
Calculate the m3 exiting the compressor:
Insert into Vander Waals’ equation new parameters:
p = 4800 kPa
T = 383 K
n = 420 kg mol
Solve for V
V = 246 m3 at 383K and 4800 kPa
7.2.8
Basis: CO2 at 81 psig and 25oC
Pressure in can (81.0 psig + 14.7 psia) (1 atm/14.7 psia) = 6.51 atm
Temperature 25°C ⇒ 298.15K
Tc = 304.2 K
V = π (4.05 cm )2 (17.0 cm) = 876 cm 3
pc = 72.9 atm
n = ? CO2
ω = 0.225
MW= 84.00
C6 H8 O7 + NaHCO 3 →
MW =44.01
CO2
+ H2 O + NaC 6 H7O 7
Using Peng-Robinson equation
n 3 baα – b 2 RT – b3 p – n2 (V) aα – 2bRT – 3b 2 p – n V2 (bp – RT) – V3 p = 0
(
Equations ⇒
)
(
) ( )
f ( n) = n3 q – n2 (V)(z ) – n V2 ( bp – RT ) – V3 p = 0
( )
q = baα – b 2 RT – b3 p
z = aα – 2bRT – 2b2 p
! R 2Tc 2 "
6
a = 0.45724 $
% = 3.908 10
& pc '
×
7–35
Solutions Chapter 7
b = 0.07780(RTc / pc ) = 26.64
1/ 2
α = !1 + κ (1 − (T
= / Tc ) "
&
'
2
Tr = 298.15/304.2 = 0.980
(Tr)1/2 = 0.990
1.007
2
κ = 0.37464 + 1.54226ϖ – 0.26992ϖ = 0.7080
The ideal gas law gives 0.233 mol CO2
R = 82.06 (cm3)(atm)(g mol)(K)
T = 298.18K
p = 6.51 atm
Introduce the values into the PR equation to get n = 0.2420 mol CO2
0.2420 mol CO2 1 mol NaHCO3 84.00 g NaHCO3
= 20.3 g NaHCO 3
1 mol CO2
1 mol Na HCO3
7.2.9
For CO2, ω = 0.225 and Tc = 304.K, pc = 72.9 atm or 7385 kPa
R = 8.314 (kPa) (m3)/(kg mol) (K)
a = 6458 (kPa) (m6) (K0.5)/(kg mol)2
b = 0.0297 m3/(kg mol)
1200 =
(8.314)(290)
6458
−
−
2
ˆ − 2.97 ×10 ) V(V
ˆ ˆ + 2.97 × 10−2 )(290)1/ 2
(V
V̂ = 1.876 m3 /kg mol from Polymath
n CO2 =
n CO2 =
V 10.4
=
= 5.54 kg mol
ˆ 1.876
V
5.54 kg mol
48 kg
= 266 kg CO2
1 kg mol
7–36
Solutions Chapter 7
7.2.10
nRT
n2 a
– 2
V − nb V
p=
T = 492.0°R
nRT ⎞ ⎛ V2 ⎞
⎛
⎜–
⎟
a = ⎝p –
V – nb ⎠ ⎝ n2 ⎠
ft 3 )(atm)
(
R = 0.7302
(lb mol )(°R )
Basis: 1.0 lb mol
ft 3 )(atm )
(
1.0 lb mol 0.7302
492.0°R
( lb mol)(°R)
200 atm =
200 =
(1.806 ft3 – (1.0 lb mol) b)
359.3 ft 3 ( atm)
( )
(1.860 ft 3 – b)
–
(1.0 1b mol)2 a
– 0.2891 a
similarly
1000 =
359.3 ft 3 (atm)
( )
( 0.741 ft 3 – b )
⎛
a = ⎝ 1000 –
–1.82a
359.3 ⎞ ⎛ 1 ⎞
–
0.741 – b ⎠ ⎝ 1.82 ⎠
200 =
359.3 ⎞
359.3
⎛ 1 ⎞⎛
1000 –
– (0.289)⎝ –
⎠
⎝
0.741 – b ⎠
(1.860 – b )
1.82
200 =
57.07
359.3
+ 158.8 –
(1.860 – b)
(0.741 – b )
41.20 =
57.07
359.3
–
(1.860 – b) (0.741 – b)
ft 3
b = 0.4808
lb mol
! ft 3 "
a = 209.3(atm) #
$
% lb mol &
2
7–37
(1.86ft 3 )
2
Solutions Chapter 7
7.2.11
a. Solution by compressibility factor method:
Basis: 80 lb water at 900°K
10 ft 3 18 lb
1cm3 /g mol
= 140.8 cm3 /g mol
80 lb 1 lb mol 0.016 ft 3 /lb mol
V̂ =
Tc = 647.4K;
pc = 218.3 atm from Appendix D1 (217.6 atm for the CD)
Vci =
RTc
82.06(cm3 )(atm) 647.4 K
=
= 243 cm3 /g mol
o
pc
( K)(g mol) 218.3 atm
Vr! =
V̂
140.8
T
900
=
= 0.579; Tr =
=
= 1.39
Vci
243
Tc
647.4
From the compressibility charts, pr = 1.9
p = pr pc = (1.9)(218.3) = 415 atm
b. solution using the Redlich-Kwong equation of state:
p=
R 2 Tc
RT
a
where a = 0.4278
− 1/2
ˆ
ˆ ˆ
pc
(V-b)
T V(V+b)
RTc
b = 0.0867
pc
2.5
2
!
(cm3 )(atm) "
0.4278 #82.06
(647.4 K)2.5
$
(K)(g mol) &
(cm3 ) 2 (atm)(K)1/2
%
a=
= 1.407 × 108
218.3 atm
(g mol)
b=
0.0867 82.06(cm3 )(atm) 647.4 K
cm3
= 21.1
(K)(g mol)
218.3 atm
g mol
p=
(82.06)(900)
1.407 × 108
−
= (617 206) atm
−
(140.8 − 21.1) (900)1/ 2 (140.8)(140.8 + 21.1)
= 411 atm
7–38
Solutions Chapter 7
7.2.12
For ethane: Tc = 516.3 and pc = 63.0 atm
a. Use Van der Waals’ equation.
Basis: Ethane at 100oF and 2000 psig (MW = 30)
p = 2014.7 psia
V = 1.0 ft3
T = 560ºR
2
2
!
! cm3 " "
! ft 3 "
6
-3
4
a = # 5.50 × 10 atm #
$ $ (3.776 × 10 ) = 2.07 ×10 psia #
$
#
g mol ' $
&
& lb mol '
&
'
!
cm3 "
ft 3
-2
b = $ 65.1
1.60
10
=1.04
×
(
)
%
g mol '
lb mol
&
!
an 2 " ! V
"
p
+
− b % = RT
$
2 %$
V
n
'
&
'&
! ba "
!a"
n 3 $ 2 % − n 2 $ % + n(pb + RT) − pV= 0
&V '
&V'
n3 − 0.962n 2 + 0.376n − 0.0936 = 0
Iterate to find value of n.
Use Excel to solve this cubic equation to obtain a positive, real answer.
n = 0.594 lb mol or 17.8 lb ethane
b. Use the compressibility factor method.
pc
Tc
pr
Tr
709 psia
549ºR
2.84
1.02
mass =
z from the compressibility charts
1(ft)3 2014.7 psia (oR)(lb mol) 30 lb
= 23.4 lb ethane
560oR
0.43
10.73(psia)(ft)3 mol
7–39
0.43
Solutions Chapter 7
7.2.13
a.
Van der Waals equation
2
! 27 " R Tc
a =# $
% 64 & pc
! 1 " RT
b=# $ c
% 8 & pc
Methane
Propane
Tc = 190.7K
Tc = 369.9K
pc = 45.8 atm
pc = 42.0 atm
Methane:
190.7
= 4.16
45.8
Propane:
369.9
= 8.81
42.0
Propane will be higher
for both coefficients
b.
SRK equation:
a! =
0.42748 R 2 Tc
pc
b=
0.08664 RTc
pc
Answer is same as for (a); propane
will be higher for both coefficients
7.2.14
pabs = pg + barometric pressure
T = 273 – 50 = 223 K
pabs = 39 + 1 = 40 atm
Basis: 1 g mol H2
H2 Coefficients
a
Van der Waals
0.246 ×106 (atm)(cm)6 /(g mol)2
26.6(cm)3 /g mol
Redlich-Kwong
1.439 ×106 (atm)(K1/2 )(cm)6 /(g mol)2
18.5(cm)3 /g mol
Computer results
7–40
b
Solutions Chapter 7
Van der Waals:
471.78 cm3 /g mol
Redlich-Kwong
471.26 cm3 /g mol
5L 1000 cm3 1 g mol
= 10.60 g mol(van der Waals)
L
471.78 cm
5 (1000)/471.26 = 10.61 g mol (Redlich − Kwong)
7.3.1
Basis: 7 lb N2 or 7/28 = 0.25 lb mol N2
a.
b.
nRT 0.25 lb mol (120 + 460)o R 0.732(atm)(ft 3 )
p=
=
V
0.75 ft 3
(lb mol)(o R)
= 141 atm
Tc = 126.3K
pc = 33.54 atm
580
= 2.55
(126.3)(1.8)
Tr =
0.75 ft 3
V̂
0.25 lb mol
Vr! =
=
= 0.607
RTc
(atm)(ft 3 )
(1.314)
(126.3K)
pc
(lb mol)(K)
33.54 atm
From the compressibility charts
pr = 4.3
so that
p = pr pc = (4.3) (33.54) = 144 atm
Note: You calculate z from Tr and Vr! , and then use p =
7–41
znRT
V
Solutions Chapter 7
7.3.2
Basis: 3 lb mol of gas at 252°C and 463 psia.
Volume = 50 ft3
R=
(1 atm)(359 ft 3 )
(atm)(ft 3 )
= 1.315
(1 lb mol) (273 K)
(lb mol)(K)
!
"
# 463 psia $
50 ft 3 )
#
psia $ (
# 14.7
$
atm
%
&
z=
= 0.76
!
(atm)(ft 3 ) "
o
(3 lb mol) #1.315
$ (273 + 252 C)
(lb mol)(K) &
%
The Pitzer ascentric factor is not as convenient to use as the compressibility charts.
From compressibility charts (see the CD for better precision)
z = 0.76
Tr =
525
= 1.05
500
pr = 0.7
Thus
463 psia
= 0.7
pc
pc ≅ 661 psia (45.0 atm)
7–42
Solutions Chapter 7
7.3.3
Basis: 1 lb n-octane
Tc = 1025°R
pr =
pc = 24.6 atm
MW = 114
27
= 1.10
24.6
To get the temperature of the gas
1 lb 1 lb mol 359 ft 3SC 1 atm T K z
114 lb
27 atm 273 K
Tz = 845
= 0.20 ft 3
zTR = 845/1025 = 0.825
From the compressibility Chart Fig. 14.4b (expanded in the CD)
read Tr = 1.16, so that T = (1.16)(1025) = 1189o R (729°F)
Alternate solution: Calculate z = 0.71 and use pV = znRT.
7.3.4
Basis: 50 lb CO2 (50/44) = 1.138 lb mol CO2
V = 5.0 ft3
p = 1600 + 14.7 = 1614.7 psia
Pc = 1070 psia
Tc = 547.5°R
pr =
1614.7
= 1.51
1070
V̂ =
5.0
= 4.40 ft 3 /lb mol
1.138
Vci = 5.50 ft 3 /lb mol so that Vri =
4.40
= 0.800
5.50
Tr from Figure 14.4b is (expanded in the CD) is 1.44
T = (1.44) (547.5) = 788o R (328°F)
The Pitzer ascentric factor is less convenient to use for this problem.
7–43
Solutions Chapter 7
7.3.5
Basis: 1ft3 CH4
pV = nRTz
R=
p = 214.7 psia
10.73(psia)(ft 3 )
(lb mol)(oR)
Tc = 191°K → 344°R
pc = 45.79 atm → 672.9 psia
Tr = 540/344 = 1.57
pr =
214.3
= 1.0
214.7
z = 0.975
n=
(214.3)(1)
= 0.038 lb mol
(10.73)(540)(0.975)
weight (mass) = (0.038) (16.03) = 0.608 lb
By use of the Pitzer ascentric factor (ω = 0.008 for CH4)
z = z0 + z1ω = 0.976 + 0.024 (0.008) = 0.976
7–44
Solutions Chapter 7
7.3.6
Basis: 25L of CO2 at 25°C and 200 kPa
For CO2:
pc = 7385 kPa
Tc = 304.2 K
pr =
p
200 kPa
=
= 0.0271
pc
7385 kPa
Tr =
T
(25 + 273) K
=
= 0.98
Tc
304.2o K
From the compressibility chart, at these coordinates, z ≅ 0.99
pV = nzRT
n=
200 kPa
1 atm 25L
(g mol)(K)
101.3 kPa
0.08206(L)(atm) 298 K 0.99
= 2.04 g mol
m = (2.04) (44) = 87.9 g
or 0.0879 kg
7.3.7
Basis: 106 ft3 at 60°F and 14.7 psia
pr Vr = zr n r R r Tr
at the reservoir
pV = znRT
at SC
Tr =
T
120 + 460 580
=
=
= 1.69
Tc
191(1.8)
344
pr =
p
1000
1000
=
=
= 1.48
pc
(45.79)(14.7)
674
zr = 0.95
! p " ! T "! z "
! 14.7 "! 120 + 460 "! 0.95 "
3
Vr = V # $# r $# r $ =106 #
$#
$#
$ = 15,600 ft
1000
60
+
460
1
%
&%
&%
&
% pr & % T &% z &
7–45
Solutions Chapter 7
7.3.8
Basis: 1 kg propane
For propane: Tc = 370 K
pc = 4255 kPa
and R = 0.18855 (kPa)(m3 )/(kg)(K)
Reduced temperature = Tr =
T
230 + 273.2
=
= 1.36
Tc
370
Reduced pressure
p
6000
=
= 1.408
pc
4255
= pr =
From a generalized compressibility chart
at pr = 1.408 and Tr = 1.36, z ≈ 0.80
Since pV = znRT,
V̂=
V̂ =
zRT
p
(0.80)(0.18855)(503.2)
= 0.0127 m3/kg
(6000)
7.3.9
a)
Yes
g)
No
b)
No
h)
No
c)
Yes
d)
No
e)
Yes
f)
No
7–46
Solutions Chapter 7
7.3.10
The answer is it depends on the gas temperature and the pressure. If pV = znRT applies,
the ideal gas law is represented by z = 1, and for a non-ideal gas, z can be greater or less
znRT
than 1. The pressure is p =
. So preal = pideal z, for fixed V and for a fixed number
V
of moles and temperature (the volume of cylinder is fixed). The pressure prediction by
the ideal gas law will be conservative (higher than the true pressure) for z < 1, and lower
than the true pressure for z > 1.
7.3.11
Basis: Data on gases cited in problem.
a.
The cubic feet associated with “165 ft3” and “240 ft3” probably means that the
values are the respective volumes of an ideal gas at standard conditions, because the
volumes calculated for the gases in part b. below are quite different than the stated
values.
b.
To find the amount of gas in the cylinder, first find the approximate volume in the
cylinder.
Vcyl =
π d2h
4
(3.14)(9)2 (52)
=
= 1.914 ft 3
(4)(1728)
Use the gas law as ratios:
p 2 V2
z RT
= 2 2
p1V1
z1RT1
! p "! z T "
V2 = V1 # 1 $# 2 2 $
% p2 &% z1T1 &
For C2H4:
Tc = 283 K, pc = 50.5 atm
Assume T1 = 80°F = 540°R = 300°K
300
!
= 1.06
""
283
# z1 = 0.35
1515
p r1 =
= 2.05"
(14.7)(50.5)
"$
Tr =
7–47
Solutions Chapter 7
(At standard conditions: z2 = 1.00)
! 1515 "! 1 "! 492 "
3
V2 = (1.914) #
$#
$#
$ = 514 ft
14.7
0.35
540
%
&%
&%
&
For CH4:
Tc = 191 K, pc = 45.8 atm
Tr =
300
= 1.57
191
pr2 =
2015
= 2.98
(14.7)(45.8)
z2 = 0.84
! 2015 "! 1 "! 492 "
3
V2 = (1914) #
$#
$#
$ = 285 ft
% 14.7 &% 0.84 &% 540 &
MW of C2H4 = 28
(514)
wt of C2H4 =
wt of CH4 =
MW of CH4 = 16
28
(359)
(285)
16
(359)
Therefore
c.
= 12.7
wt C2 H4 > wt CH 4
For C2H4:
pV = znRT
m=
= 40.1
= z
m
RT
MW
(1515)(1.91)(28)
pV(MW)
=
(0.35)(10.73)(590)
zRT
= 40 lb C2 H4
For CH4:
m=
(2015)(1.91)(16)
= 12.7 lb CH 4
(0.84)(10.73)(540)
Check with cylinder:
163 – 90 = 123 lb cylinder
105 – 9.5 = 135 lb cylinder
7–48
Solutions Chapter 7
7.3.12
a)
The density of the gas must be 2.0 g/ m 3 . MW is Si is 28.086 so
= (2/28.086) = 0.0712 g mol/cm3. Let T - 298K. Also z = 1 at Tr = 1.98
V̂
and Vri very large.
3
zRT 1 8.314(Pa)(m3 ) ! 100 cm " 298 K
=
= 2.48 ×109 Pa
p=
$
%
ˆ
(g mol)(K) & 1 m '
V
b)
For a lower pressure, use a less dense gas (lower MW).
c)
At too high a temperature, you cannot get adequate density.
7.3.13
a)
p1V1 = z,n,RT
R = constant
pa V1 = z 2 n 2 RT1
V1 = constant, the vol. of cylinder
zn
p1
= 1 1
p2 z 2 n 2
Tc = 133.0 K
Tr = T T =
24.44 + 273
= 2.24
133
c
pr1 =
pc = 34.5 atm
2000 psig + 14.7
= 3.97
14.7 psig/atm 34.5 atm
pr2 = 3.80
z1 = 1.0 ⎫
⎬ Ideal gas behavior
z 2 = 1.0 ⎭
p1 ! n1 " ! 1.0 " n1
= # $# $ =
p2 % n 2 & % 1.0 & n 2
7–49
Solutions Chapter 7
n CO =
pV 14.7 psia
=
RT
536°R
Vcylinder =
10.73
175 ft 3
(psia ) ft 3
( )
= 0.447 lb mol
( lb mol)(°R)
nCO RT1
p1
( 3)
0.447 lb mol 10.73(psia ) ft 536°R
Vc =
= 1.276 ft 3
2014.7psia (lb mol)(°R)
n final,CO =
p2 Vc
=
RT
1924.7 psia
1.276 ft 3
= 0.427 lb mol
( psia ) ft 3 536°R
10.73
(lb mol )(°R )
( )
n CO, lost = (0.447 – 0.427)lb mol ≅ 0.020 lb mol
rate of loss =
b)
0.020 lb mol
48 hr
pV 14.7 psia
n air =
=
RT 536.0°R
= 4.167 ×10−4
lb mol
hr
1600 ft 3
= 4.09 lb mol air
( psia ) ft 3
10.73
(lb mol)(°R)
( )
( –6 )(4.09 lb mol) = 4.09 × 10 –4 lb mol
100 ppm = 100 10
4.09 × 10-4
t=
= 0.982 hr
-4 lb mol
4.167 × 10
hr
c)
t = 67 hr
CCO =
d)
n CO =
4.167 × 10-4 67
= 0.0279 lb mol
0.0279 lb mol
lb mol
= 1.75×10-5
3
1600 ft
ft 3
The CO alarms would go on to alert people in the building.
7–50
Solutions Chapter 7
7.4.1
T = 180°F (640°R)P =
2400 + 14.7 = 2415 psia
Basis: 33.6 lb gas
pV = znRT
Basis: 100 mol gas
mol
mol.wt.
CO2
10
44
440
CH4
40
16
640
C 2 H4
100
50
28
2560
1400
n=
lb
avg. mol wt. =
2480
= 24.8
100
33.6
= 1.35 lb mol
24.8
Tc# = 0.10 (304.2) + 0.40 (190.7) + 0.50 (283.1) = 248.25 K
pc# = 0.10 (72.9) + 0.40 (45.8) + 0.50 (50.5) = 50.86 atm
p′r =
p
2415
= 3.23
′ =
p c (50.86 )(14.7)
Tr′ =
T
640
= 1.43
′ =
Tc (248.25)(1.8)
From Fig. in text, z ≅ .75
3
znRT 0.75 1.35 lb mol 10.73(psia ) ft 640°R
V=
=
(lb mol)(°R)
p
2415 psia
( )
3
= 2.88 ft at 180 °F, 2415 psia of gas is cylinder volume
7–51
Solutions Chapter 7
7.4.2
Basis: 540 kg gas @ 393 K, 3500 kPa absolute
Component
kg
kg mol
mole fraction
pc, atm
Tc , K
CH4
C2 H6
C3H8
N2
100
240
150
50
6.25
8.00
3.41
1.79
0.321
0.412
0.175
0.092
45.8
48.2
42.1
33.5
191
305
370
126
Total
540
19.45
1.000
Tc# = (0.32l) (191) + (0.412) (305) + (0.175) (370) + (0.092) (126) = 262.8 K
pc# = (0.321) (45.8)+(0.412) (48.2)+(0.175) (42.1)+(0.092) (33.5) = 44.96 atm
T r′ =
3.93
= 1.50
262.8
p r′ =
3500 1 atm
= 0.77
44.96 101.3
From Nelson and Obert Chart, z# = 0.935
MW (average) = 540/19.45 = 27.8 kg/kg mol
ρ=
=
3500 kPa
27.8 kg
393 K 0.935
(K) (kg mol)
3
kg mol 8.31 (kPa) (m )
31.9 kg
m at 393K and 3500 kPa
3
7–52
Solutions Chapter 7
7.4.3
0.20 EtOH
0.80 CO2
1.00
T = 500K
V = 180 cm3 /g mol
p=?
Basis: gas as above
pc# = (0.20) (63.0) + (0.80) (72.9) = 70.92 atm
Tc# = (0.20) (516.3) + (0.80) (304.2) = 346.6 K
p r′ = ? =
T r′ =
500
= 1.44
346.6
Vci′ =
′
Vri =
p
p
=
p c 70.92
RTc′
p′c
=
(1 atm )( 22.4 L) 1000 cm3 346.6 K
= 401 cm3 /g mol
(273 K )(1 g mol) 1 L 70.92 atm
180
= 0.45
401
From the compressibility chart, pr ≅ 2.5 , p ≅ 177 atm
7–53
Solutions Chapter 7
7.4.4
Basis: 30ft3 mixture of 100 atm and 300°F
Comp
Mol. frac
Tc
pc
Tc!
p!c
C2H4
0.60
283
50.5
170
30.3
A
0.40
150.7
48
60.2
19.2
230.2
49.5
Tr! = T/Tc! =
p!r =
760
= 1.835
(230.2)(1.8)
p
100
=
= 2.02
p!c
49.5
z = 0.95
Avg. Mol. wt. = (0.60)(28) + (0.40)(40) = 32.8 lb/lb mol
pV = znRT
n=
pv
(100)(14.7)(300)
=
= 57.0 lb mol
zRT (0.95)(10.71)(760)
Assume capacities are given at storage conditions. Then type IA is selected because it is
the cheapest.
Number of tanks required =
(57.0)(32.8)
= 30
62
7–54
Solutions Chapter 7
7.4.5
Steps 2, 3, and 4:
Assume a continuous process although a batch process would be acceptable and give the
same results.
F=?
mol fr.
0.20 C2H4
0.80 N2
1.00
Step 5: Basis:
P=?
Mixer
G 100% C2H4 @ 1450 psig and 70oF
G at given conditions
Steps 6 and 7:
Unknowns:
Balances:
F, P
!
"
C2 H 4 , N 2 , (or total) # degrees of freedom = 0
"
$
Steps 8 and 9:
Calculate G:
Tr =
mol fr.
C2H4 0.50
N2 0.50
1.00
70 + 460
= 1.044
508.3
7–55
MW
28
28
Solutions Chapter 7
1450 + 14.7
= 1.992
735
pr =
z = 0.34
V=
(359)(n)(14.7)(530)(0.34) (2640)
=
(1465)(492)
(1728)
n=
(2640)(1465)(492)
= 1.157 lb mol
(1728)(359)(14.7)(530)(0.34)
Balances:
C2H4
F(0.20) + 1.157 = P(0.50)
Total
F + 1.157 = P
F = 1.928 lb mol
P = 3.085 lb mol
7.4.6
Basis: Gas at 50 atm and 600°R
From the compressibility charts:
pr =
50
= 3.5
14.3
Tr =
z = 0.58
600
= 1.2
40.0 + 460
100,000 std ft 3 1 lb mol 359(0.58) act ft 3 ⎛ 1 ⎞ 600
3
Q=
= 1415 actual ft /hr
3
⎝
⎠
hr
1 lb mol
359 std ft
50 492
or use pV = znRT twice
p2 V2 z 2 n2 RT2
=
p1V1
z1n1 RT1
V2
⎛ T ⎞⎛ p ⎞⎛ z ⎞
= V1 ⎜ 2 ⎟ ⎜ 1 ⎟ ⎜ 2 ⎟
⎝ T1 ⎠ ⎝ p 2 ⎠ ⎝ z1 ⎠
⎛ 600 ⎞ ⎛ 1 ⎞ ⎛ 0.58⎞
= 100,000⎝
492 ⎠ ⎝ 50 ⎠ ⎝ 1.00 ⎠
= 1415 actual ft 3 /hr
7–56
Solutions Chapter 7
7.4.7
Step 5:
Basis: 1 hr.
Steps 2, 3, and 4: Steady state, open system
liquid
30,000 = F(kg)
mass fr.
Bz 0.50
Tol 0.30
Xy 0.20
1.00
mole fr.
Bz 0.912
Tol 0.072
Xy 0.016
1.000
Separator
vapor
P1(kg)
MW
78.11
92.13
106.16
kg
71.24
6.63
1.70
79.57
liquid
P3(kg)
mass fr.
Bz 0.06
Tol 0.09
Xy 0.85
1.00
liquid
P2 = 9,800 (kg)
mass fr.
0.895
0.084
0.021
1.00
mBz
mTol
mXy
Σ = 9,800
mass fr.
Convert F to moles and then to mass:
Basis: 100 kg = F
Component
kg
MW
kg mol
mol fr.
Tc(K)
pc(atm)
Bz
50
78.11
0.640
0.592
562.6
48.6
Tol
30
92.13
0.324
0.299
593.9
40.3
Xy
20
106.16
0.118
0.109
619
34.6
1.082
1.000
100
7–57
Solutions Chapter 7
Calculate the kg of F; get z:
Tc! = 562.6 (0.592) + 593.9 (0.299) + 619 (0.109) = 578.1
p!c = 48.6 (0.592) + 40.3 (0.299) + 34.6 (0.109) = 44.6
Tr =
607K
=1.05
578.1K
pr =
26.8 atm
= 0.60
44.6 atm
z = 0.80
Together mass of feed:
pV
26.8 atm 483 m3
101.3 kPa (kg mol) (K)
n=
=
zRT
0.80 1 atm 8.314(kPa)(m3 ) 607 K
= 324.7 kg mol
F = (324.7) (100)/(1.082) = 30,017 kg, say 30,000 kg
The material balances are in kg.
Step 6:
Unknowns: P1 , P3 , m Bz , m Tol , m Xy
Step 7:
Balances:
Bz
Tol
mBz 3
= = 1.5
mxy 2
Xy
∑ mi = 9,800
Step 8:
The balances in kg are (only 3 are independent mass balances)
(1)
Bz
30,000 (0.50) = P1 (0.895) + m Bz + P3 (0.06)
(2)
Tol:
30,000 (0.30) = P1 (0.084) + mTol + P3 (0.09)
(3)
Xy:
30,000 (0.20) = P1 (0.021) + mXy + P3 (0.85)
(4)
Total 30,000
= P1
+ 9,800 + P3
7–58
Solutions Chapter 7
(5)
m Bz + mTol + m Xy = 9800
(6)
m Bz = 1.5m Xy
Step 9:
Solution: Using (1), (2), (3), (5), and (6) via Polymath
kg
mass fr.
P3 = 5500 kg/hr
mBz = 1518
0.156
P1 = 14,700 kg/hr
mtol = 7270
0.745
0.103
mXy = 1012
9800
7–59
1.00
Solutions Chapter 8
8.2.1
a. (1); b. (2); c. (3); d. (4); e. (3); f. (1)
8.2.2
8.2.3
C=1
P=2
F=2–2+1=1
8.2.4
F=2–P+C
(a)
(b)
F=2–2+1= 1
F=2–3+2= 1
8.2.5
C = 3 (O2, N2, H2O)
P=2
F=2–2+3= 3
8–1
Solutions Chapter 8
8.2.6
(1) F = 2 – P + C
The rank of
Ca
C
O
CaCO3(s)
1
1
3
CaO(s)
1
0
1
!1 0 9 "
#1 1 0$ is only 2 so C = 2
#
$
#%3 2 0 $&
P = 3 hence F = 2 – 3 + 2 = 1
8.2.7
(a) 40ºC; (b) 190ºC; (c) 60ºC; (d) Compound B
8.2.8
(e)
8.2.9
a.
F=2–P+C
C=3
P = 2 (assume that p is not unreasonably high)
F=2–2+3= 3
b.
Possible variables are: pressure
mol or mass fraction of
H2O
!
"
acetic acid # only 2 are independent
ethyl alcohol "$
8.3.1
All true
8–2
CO2(g)
0
1
2
Solutions Chapter 8
8.3.2
(a) higher; (b) lower; (c) higher; (d) lower; (e) no change; (f) lower;
(g) higher; (h) lower; (i) higher; (j) lower; (k) lower; (l) higher
8.3.3
From the p-T diagram for water you can see that raising the pressure by a large amount
on ice at constant temperature causes the water to go from the solid phase to the liquid
phase.
8.3.4
The vapor pressure of methanol is less than that of gasoline so that at lower temperatures
the fuel-to-air mixture is insufficient for combustion.
Automotive parts can be made of alcohol tolerant materials such as stainless steel or other
corrosion resistant materials and additives to the methanol, such as 15% gasoline, can
help solve the problem of difficult cold-weather engine starts.
8.3.5
Fit the Antoine Equation using the three data points to get A, B, and C. In mm Hg
kPa
2.53
15.0
58.9
mm Hg
18.98
112.54
441.90
B
= 2.9434
C + (−40+ 273)
B
1n(112.54) = A −
= 4.7233
C + (−10+ 273)
B
1n(441.90) = A −
= 6.0911
C + (−20+ 273)
1n(18.98) = A −
The solution from Polymath is
A = 16.5386
B = 2707.43
C = -33.8541
2707.43
At 40ºC, ln (p*) = 16.538 −
−33.85 + (40 + 273)
p* = 939 mm Hg (125 kPa)
8–3
Solutions Chapter 8
8.3.6
At the triple point, all the phases (solid, liquid and vapor) exist in equilibrium. Therefore,
the vapor pressure of liquid ammonia = the vapor pressure of solid ammonia
15.16 – 3063/T = 18.7 – 3754/T
T = 195 K
8.3.7
No. It is a typo. Probably the number was 98.8ºC because if you equate the pressures
from the two equations, T = 371.75K, or 98.7oC.
8.3.8
Merge the two equations (A1 MW = 27)
W = 10-4 = 5.83 × 10-2
1/2
(p v )(27)
T1/2
where p v =
[10
1.594×104
)
T
(8.79 −
1/ 2
T
The solution from Polymath is
T = 1492K
8.3.9
(a) T; (b) F; (c) T; (d) T; (e) F; (f) T
8–4
]
Solutions Chapter 8
8.3.10
Basis: 2.10 lb water
Specific volume:
ˆ = (1 − x) V
ˆ +xV
ˆ
V
l
g
Specific volume of water = Vˆl =
0.35
= 0.175m3 /kg
2
From the steam tables:
Specific volume of saturated liquid,
Vˆl = 0.001 088 m3 /kg
specific volume of saturated vapor =
Vˆg = 0.414 m3 /kg
Then, 0.175 = (1 – x) 0.001 088 + x (0.414)
The quality = x = 0.42
8.3.11
Since the two phases coexist in equilibrium, we have a saturated mixture, and the
pressure must be the saturation pressure at the given temperature:
p = p* @ 90ºC = 70.14 kPa
ˆ and V
ˆ values are V = 0.001036 m3 /kg and V
ˆ = 2.361 m3 /kg
At 90ºC, V
l
l
g
g
Add the volume occupied by each phase:
ˆ +m V
ˆ
V= Vl + Vg = mV
l
g g
= (8 kg)(0.001 m3/kg) + (2 kg)(2.361 m3/kg) = 4.73 m3
8–5
Solutions Chapter 8
8.3.12
The hot oil vaporized the water, and the increase in volume in the system caused the
damage.
8.3.13
Basis: 10 lb of water
Specific volume:
V̂ =
10.0
= 4.975 ft 3 /lb
2.01
From the paper steam tables (in ft3/lb) at 80 psia:
V̂liq = 0.0176
V̂vapor = 5.476
Mass balance:
m L + m V = 2.01
Volume balance:
10.0 = m L (0.0176) + m V (5.476)
Mass of liquid is 0.184 lb
Mass of vapor is 1.826 lb
The volume of liquid is
Vliq = m liq V̂liq = (0.184)(0.0176) = 3.28 × 10−3 ft 3
The volume of vapor is
Vvapor = m vap V̂vapor = (10.0 − 3.28 × 10−3 ) = 9.997 ft 3
Quality is 1.826 / 2.01 = 0.908
8–6
Solutions Chapter 8
8.3.14
Use data from the steam tables at the initial condition of p = 101.33 kPa and saturated
water (i.e. two phases):
V̂liq = 1.044 cm3/g and V̂vap = 1673.0 cm3/g.
Then the mass of both phases in the container can be calculated:
m liq = (0.03 m3) (100 cm/m)3 / 1.044 cm3/g = 28,376g
m vap = 2.97 m3 (100 cm/m)3 / 1673.0 cm3/g = 1,775 g
= 30,511 g
mt otal
At the final conditions all of the liquid has been evaporated, and only saturated vapor will
exist having a specific volume of:
Vv ap = (3.00 m3) (100 cm/m)3 / 30,511 g = 98.3 cm3/g
Interpolating in the steam tables
(212-T) ºC/(212-214) ºC = (99.09-98.3) cm3/g/(99.09-95.28) cm3/g
T = 212oC
(1985.2-p) kPa/(1985.2-2065.1) kPa = (99.09-98.3) cm3/g/(99.09-95.28) cm3/g
p = 2002 kPa
8.3.15
Based on data from the steam tables:
State 1: liquid
State 3: solid
State 5: liquid (saturated)
State 2: superheated vapor
State 4: saturated vapor
State 6: vapor (saturated)
You can calculate the properties of a mixture of vapor and liquid in equilibrium (for a
single component) from the individual properties of the saturated vapor and saturated
liquid by averaging the properties of the two saturated phases. The weights are the
respective amounts of each phase.
8–7
Solutions Chapter 8
8.3.16
Based on data from the steam tables:
State 1: saturated vapor
State 3: two phase (saturated liquid and vapor)
State 2: saturated liquid
State 4: superheated vapor
8.3.17
Prepare a Cox chart as described in the text. Use semi-log paper with the horizontal axis
log10. Obtain vertical scale rules by use of steam properties at even integers for the
temperature. Or you can use Antoine Equation, Appendix G, to get 2 points and draw a
line between them.
(a) Acetone
(b) Heptane
(c) Ammonia
(d) Ethane
Tcritical (o F)
p critical
cox chart (psia)
pcritical (psia)
454
512
270
89.7
670
700
1500
750
691
397
1636
708
8–8
Solutions Chapter 8
8.3.18
The procedure is the same as outlined for P16.28. The result from the Cox chart is that
the vapor pressure for aniline at 350o is 22 atm .
8.3.19
From the CD that accompanies the text the respective vapor pressures at 300K are in
increasing value
Ethanol
Methanol
MTBE
p* in mm Hg
PEL (ppm)
65.8
139.7
294
1000
200
100
The relative values of p* are about the same as the inverse of the relative values of the PEL.
8–9
Solutions Chapter 8
8.4.1
(a) Volume increases
(b) Pressure increases
8.4.2
T= − 20oC
p* =14.1 mm Hg
O2
p t =760 mm Hg
N2
C 6 H1 4
pair =760 − 14.1= 745.9 mm Hg
pC6H14 = pC6H14 =14.1 mm Hg
Basis: 760 mol saturated gas
p ⇒ moles
O2
N2
C6H14
(or use 100 mol)
mol
= 156.6
= 589.3
= 14.1
760
.21 (745.9)
.79 (745.9)
For complete combustion
C6H14 + 9 1/2 O2 → 6 CO2 + 7 H2O
14.1 mol C6H14 required 14.1 (9.5) = 134 mol O2 required
156.6 – 134 = 22.60 mol O2 excess
22.60
× 100 = 17 %
134
8–10
Solutions Chapter 8
8.4.3
mol. wt. CCl3NO2 = 164.5
Basis: 1 mol saurated gas
mol fr.
0.02
0.98
1.00
CCl3NO2
air
100 kPa 0.02 mol CCl 3NO3 760 mm Hg
= 15.0 mm Hg
1 mol gas
101.3 kPa
(a)
Using linear interpolation, which may introduce a slight error, 15.0 mm Hg
corresponds to
" 15.0 − 13.8 #
o
$
% (5) = 1.3 + 15 = 16 C
& 18.3 − 13.8 '
(a)
(289.5K)
Basis: 100 m3 at 100 kPa and 16.5°C
3
100 m air 100 kPa 273 K
101.3 kPa 289.5 K
1 kg mol air
3
22.4 m air at SC
0.02 mol CCl3 NO2 164.5kgCCl 3 NO2
= 13.95 kg
0.98mol air
1kgmol CCl 3 NO2
8.4.4
Assume the tank with air fills with the hazardous liquid vapor at 80ºF. The pressure in
the tank with air will become after the transfer (10 + 14.7) + 13 = 37.7 psia hence the
seal will possibly rupture during the transfer
8–11
Solutions Chapter 8
8.4.5
At 60ºF, p*H O =0.52 in Hg; at 75ºF, p* = 0.87 in Hg
2
Basis: 12,000 ft3 air at 75ºF and 29.7 in Hg absolute
12,000 ft 3
0.52 in Hg H 2 O 492oR 18 lb H 2 O
1 lb mol
= 9.62 lb H 2 O
359 ft 3 at SC 29.92 in Hg total 535oR lb mol
8.4.6
Basis: 1 gal benzene (sp. gr. 0.879, MW 78.1) at 750 mm Hg, 70ºF
Moles of air =
(3600 ft 3 ) (750) (14.7)
= 9.18 lb mol
(10.73) (760) (530)
Moles of benzene =
(1) (1ft 3 )(0.879)(62.4)
= 0.094 mol
(7.48)(78)
nTotal = 9.18+0.094 = 9.274 mol
yBz =
0.094
= 0.0101
9.274
mol % = 1.01% therefore beneath explosive limit
8–12
Solutions Chapter 8
8.4.7
n1 =
Basis: 350 ft3 C2H2 – O2 mixture at 25ºC, 745 mm Hg
pV
(745)(14.7)(350)
=
= 0.874 lb mol
RT
(760)(10.73)(298)(1.8)
! 745 "! 14.7 "! 300 "! 1 "
n2 = #
$#
$#
$#
$ = 0.671 lb mol
% 760 &% 10.73 &% 333 &% 1.8 &
p*H2O at 60ºC = 149.4 mm Hg
149.4
y H2O at2 =
= 0.201
745
n H2O = (0.201)(0.671) = 0.135 lb mol
C2 H 2 +
5
O 2 → 2CO 2 + H 2 O
2
You get 1 mol H2O/mol C2H2 on reaction
Therefore moles C2H2 = 0.135
mol O2 = 0.874 – 0.135 = 0.739 lb mol
y O2 at 1 =
0.739
= 0.845
0.874
VO2 =(0.845)(350) = 296 ft 3 at 745 mm and 25ºC
VC2H2 = 350 − 276 = 54 ft 3 at 745 mm and 25ºC
8.4.8
Elephant seals have an average body temperature several degrees higher than ours. That
means their breath can hold a bit more moisture than ours, and on that day, that extra
moisture could have been just enough for condensation to occur.
But there are other reasons the elephant seals’ breath may have been easier to see. It
could be that it was a few degrees colder on the beach than where the observer was sited.
And, as you know if you have ever experimented with seeing your breath, the bigger the
volume of air, the more moisture available to condense out and so the condensed
moisture was more visible.
8–13
Solutions Chapter 8
8.4.9
(a) p*Bz exists if saturation occurs. Then, with ptot = 100 kPa
pBz = 100 (0.014) = 1.4 kPa, which
corresponds to
−11oC from Perry
or use Antoine Eqn. and solve for T ( p*Bz = 10.5 mm Hg)
ln (10.5) = 15.9008 b)
(-15oC from the CD)
2788.51
T − 52.36
Repeat for p*Bz = 8.0 kPa (60 mm Hg) corresponding to 15.4oC
8.4.10
Assume equilibrium.
At 75°F, from the Antoine equation, the vapor pressure of benzene is 90 mmHg.
Assume the barometer is 760 mm Hg abs.
p∗B
90
mol Bz
mol Bz
=
=
= 0.13
∗
mol air
mol air pt − pB 760 − 90
a) The OSHA limit is 1 ppm over 8 hours. The above greatly exceeds this limit.
b) No, because the garage is probably not well ventilated, and also if a water heater is in
the garage, the LEL (Lower Explosive Limit) may be exceeded.
8–14
Solutions Chapter 8
8.4.11
Basis: 1 lb air
If the air is saturated, p*C8 = 2.36 in Hg and pair = 29.66 - 2.36 = 27.30 in Hg.
p*C8
pair
=
2.36
mol C8
= 0.0864
27.30
mol air
(a)
lb air
lb air 27.30 mol Air 29 lb air 1 lb mol C 8
= 2.94
=
l bC 8
lb C8 2.36 mol C 8 1 lb mol air 114.2 lbC 8
(b)
nc
2.36 mol C 8
= 8 = 0.080 which is also the fraction of the volume.
(27.30 + 2.36) mol total n tot
2.94lb air 1 lb mol air 359 ft 3 at SC 580 R 29.92in Hg
lb C8
29 lb air 1lb mol air 492 R 29.66in Hg
(c)
3
= 43.3 ft at 120 °F and 29.6 in. Hg/lb octane
8.4.12
At equilibrium, the pressure of Hg is its vapor pressure
−4
p* H g 1.729× x10 g mol Hg
=
=
=
pt
nt
pair
99.5g mol air
pHg
nHg
1.729×10 −4 gmol Hg
Note: pt = pair + pHg = pair for all purposes so
99.5gmol total gas
-4
Basis: 1.729 × 10 g mol Hg
−4
1.729×10 gmol Hg 200.59g 1000mg
= 38.14 mgHg
1gmol Hg 1g
nRT
=
p
3
99.5gmol air 293.15K 8.314(kPa)(m ) 1kg
3

= 2.44m at 20 C and 99.5kPa
99.5 kPa (kg mol)(K) 1000g
V=
8–15
Solutions Chapter 8
15.65
mg
at 20°C and 99.5 kPa.
m3
Level is not acceptable.
A mercury spill can be cleaned up reasonably effectively by dusting with sulfur, then
vacuuming with a vacuum cleaner specifically designed for mercury pick-up, which
prevents the escape of mercury vapor.
The instructor should point out to the student that the assumption of the “no
ventilation” is not a very good approximation unless the surface of the mercury is rather
large or there is truly no ventilation in the storeroom.
8.4.13
Basis: 1 hr
The question really is: does the Hg condense at 150ºF, or is it still a vapor (and thus not
collected).
The moles of Hg are: 40,000 (0.023 × 10-2)
lb Hg 1 lb mol Hg
= 0.046 lb mol
200 lb Hg
The moles of gas are (ignoring the Hg):
40,000 lb gas 1 lb mol gas
= 1250 lb mol
32 lb gas
y Hg =
0.046
= 3.68 10−5
0.046 + 1250
×
The partial pressure of the Hg in the gas is yp total
(3.68 × 10-5)(14.7) = 5.4 × 10-4 psia
The Hg will remain a vapor and not condense.
8–16
Solutions Chapter 8
8.4.14
Basis: 20,000 ft3 moist air at entering conditions in 1 day
20,000 ft 3 273 800 1 lb mol
= 57.18 lb mol total
280 760 359 ft 3 at SC
(0.532 lb mol H2O)
System: Refrigerator is system
Steps 6 and 7:
Unknowns: W2 and P; balance: H2O and air
Steps 8 and 9:
!
" 0.99.12 #
" 0.13 #
H 2 O: 57.18 $
% = W2 (1) + P $
%
& lb mol
' 106.63 (
' 106.63 (
&
) P = 56.72
" 105.64 #
" 106.5 #
&
Air: 57.18 $
% = W2 (0) + P $
% not accurate & W2 = 0.4613
' 106.63 (
' 106.63 (
*
Total: 57.18
= W2
+P
Basis: 30 days
30 (0.4613) (18) = 249 lb H 2O/30 days
8–17
Solutions Chapter 8
8.4.15
Steps 1-4:
air
F(mol)
H2O
25oC
100 kPa
remove 50%
H2O
(saturated)
air
H2O
P
100 kPa
W (mol)
100% H2O
dew point =
16oC
mol fr.
100-p*P/100
p*P/100
1.00
at exit p* = ?
At 16ºC
0.57 in Hg
0.28 psia
*
p H2O ≅ 13.6 mm Hg =1.81 kPa From steam tables p* = 1.92 kPa
Thus pair = 100 – 1.81 = 98.18 kPa
pair = 100 – 1.92 = 98.08
Step 5: Basis is 100 Mol F
Step 6: Unknowns: W, P
Step 7: Balances: H2O, air
Steps 8 & 9: Balances around process in mol
Air
! 100 − p*H2O "
! 98.08 "
100 $
=
W(0)
+
P
$
% in P
%
100
& 100 '
&
'
H2O
! p*H2O in P "
! 1.92 "
100 #
=
W(1)
+
P
#
$
$
% 100 &
% 100 &
Total
100
Also, p* = f(T)
=W+P
vapor pressure relation
•
The exit gas has one-half the entering H2O, or ½ (1.92) = 0.96 mol thus W = ½
(1.92) = 0.96 mol H2O.
•
The exit air has 98.08 mol dry air.
•
At the exit p H O = p*H O = y H O (100) =
2
2
2
0.96
100 ≅ 0.97 kPa
0.96 + 98.08
8–18
Solutions Chapter 8
•
From the steam tables, this corresponds to about
280 K
(7oC)
, or use Antoine Eq.
Note: Taking ½ of (1.92) = 0.96 and dividing by 100 instead of (pH O +pair ) is wrong. The
answer is close to correct because of the very small amount of water.
2
8.4.16
Basis: 1000m3 sat. air at 99 kPa and 30ºC
p*H2O at 30ºC = 0.6155 psia = 31.8 mm Hg = 4.24 kPa
1000 m3
1 kg mol
99 kPa 273 K
= 39.3 kg mol
3
22.415 m a tSC 101.3 kPa 303 K
99 kPa 1000 m3 (kg mol)(K)
n=
=39.3 kg mol
303 K 8.314 (kPa)(m3 )
or
Initial:
4.24 "
$ = 1.68 kg mol H 2 O
% 99 &
Initial mol H2O = 39.3 !#
99 − 4.24 #
% = 39.3 − 1.68 = 37.62 kg mol air
99
&
'
Initial mol air = 39.3 "$
Final:
At 14ºC and 133 kPa, the air is still saturated, and nair = 37.62 kg mol
p*14oC H2O = 0.2302 psia = 11.9 mm Hg = 1.59 kPa = p H 2O @14oC
p H2 O
pair
=
n H2 O
n air
so
n H O ! There is the same as
1.59
= 2 "
(133 − 1.59) 37.62 $ an H2O material balance
n H2O = 0.46 kg mol
1.68 – 0.46 = 1.22 kg mol or 22 kg
8–19
Solutions Chapter 8
8.4.17
Steps 2, 3, and 4:
C2H6
CO2
H2O
O2
N2
Combustion
chamber
AIR
100 kPa
(20% excess)
Assume combustion is complete and the gases are ideal. The process is open, steady
state with reaction. The first objective is to make material balances.
C2 H6 +3.5O2 → 2CO2 + 3H2 O
Step 5:
Basis: 100 mol C2H6
Step 4:
The entering O2 is
The excess O2 is (0.20) (3.50) =
Total O2 is
350 mol
70 mol
420 mol
0.79 "
$=
% 0.21 &
The N2 is 420 !#
1580 mol
Steps 6 and 7:
Unknowns: 4 (The components of the flue gas)
Balances: C, H, O, N, total of 4
Degrees of freedom = 0
Steps 8 and 9 (balances in moles, in = out):
C:
2(100) = n CO
n CO 2 = 200 mol
2N:
H:
O:
1580 = n N
6 (100) = 2n H O
420(2) = 2n CO +n H O +2n O
Total
2
n N2 = 1580 mol
2
n H2O = 300 mol
2
2
2
n O2 = 700 mol
2150 mol
2
The dewpoint is the temperature at which the flue gas is saturated
p*H2O = y H2O pTotal =
300
(100 kPa) = 14.0 kPa
2150
This pressure of H2O corresponds (from the steam tables) to 325.7 K (52.6o C)
8–20
Solutions Chapter 8
8.4.18
Basis: 100 mol gases leaving absorber
CH4 + 2O2 →CO2 + 2 H2 O
0.8335mol N2 0.21mol O 2 1mol CO2
= 0.1108 mol CO2
0.79 mol N2 2 mol O2
1 - 0.8335 - 0.1108 = 0.0557 mol H2O; 0.00557 (20) = 1.114 psia

Dew point ≈ 105 F at 1.114 psia
b) At constant temperature of 130 °F, the gas must be compressed until the partial
pressure increases and becomes equal to the vapor pressure in order to reach the point of
condensation.
From Steam tables, the vapor pressure of water at 130 °F is 2.223 psia.
The total pressure at which the partial pressure of water (mol fraction = 0.0557) will
be 2.223 psia is
pH 2 O = p t y H 2 O
2.223
= 40.1psia
0.0557
8.4.19
Steps 2, 3, and 4:
The dewpoint of 140ºF gives p*H2O = 149.3 mm Hg
Step 5:
Basis: 100 lb wet solids in (BDS is bone dry solids)
8–21
Solutions Chapter 8
Steps 8 and 9:
BDS balance: 60 lb BDS = 0.90 (wet solids out)
wet solids out = 66.7 lb
From a total balance 100 – 66.7 = 33.3 lb H2O evaporated = 1.85 lb mol H2O evaporated
For the air,
Assume the air exits at 1.0 atm
Composition in:
H2O is
10
(100)
800
790
(100)
800
149.3
H2O is
(100)
800
Air is
Composition out:
Air
=
1.25%
=
98.75%
=
=
18.7%
81.3%
Overall balance (mol) on the gas phase
Ain + 1.85 = Aout
Air balance (mol) on the gas phase
0.9875 Ain = 0.813 Aout
Ain = 8.619 lb moles wet air in
8.619 lb mol 359 ft 3 660o R 760
= 3940 ft 3 of wet air @ 800 mm Hg and 200ºF
o
lb mol 492 R 800
8–22
Solutions Chapter 8
8.4.20
Basis: 1 lb mol of benzene free air
Barometric Pressure
=
mm Hg
742
p* benzene at 40ºC
=
181
Partial pressure of air
=
561
Partial pressure of benzene
=
181
At 25 psig total pressure or
p* benzene at 10ºC
Partial pressure of air
2052
=
45.4
2006.6
Final partial pressure of benzene in air
0.323 – 0.023
⇒ 0.323 lb mol benzene
=
45.4 ⇒ 0.023 lb mol benzene
= 0.3 lb mol C6H6 recovered
a.
(0.3)(100)
= 92.9% recovery
(0.323)
b.
2 psig is 103 + 742 = 845 mm Hg
(0.023)(845)
= 18.9 mm Hg partial pressure of benzene in the recycled air.
(1.023)
8–23
Solutions Chapter 8
8.4.21
Basis: 1 hr
Calculate the amounts of the various components in the gas phase:
Water
H 2O in = 0 (by specification)
H 2O out (saturated):
p* at 300 K = 3.51 kPa = p H 2O
y H2O =
3.51 kPa
= 0.0319
110 kPa
CO2
CO2 in:
y CO2 = 0.00055
CO2 out: y CO2 = 0.125
O2
O2 in: yO2 = 0.21
O2 out: yO2 = 0.0804
Assume the other gas in and out is N2.
The entering gas is
600 m3 120 kPa
(kg mol)(K)
= 28.87 kg mol
300 K 8.314(kPa)(m3 )
The exit gas can be obtained from a N2 balance
28.87(1 − 0 − 0.00055 − 0.21) = n out (1 − 0.0319 − 0.125 − 0.0804)
n out = 29.89 kg mol
In one hour in the steady state the moles of CO2 produced were
29.89(0.125) − 28.87(0.00055) = 3.72 kg mol
The mole of O2 consumed were
28.87(0.21) − 30.14(0.0804) = 3.64 kg mol
RQ =
3.72
= 1.02
3.64
8–24
Solutions Chapter 8
8.5.1
(a) F; (b) F;(c) F; (d) T; (e) T; (f) T
8.5.2
For multiple components, the vapor pressure varies with composition as well as with
temperature, and to be precise the composition should be stated.
8.5.3
Henry’s Law is p = Hx, a straight line with no intercept. A plot of the data indicates
Henry’s law fits the data well.
8–25
Solutions Chapter 8
8.5.4
p = Hx where p is in kPa and x is mol fraction
The mole fraction of O2 in the gas phase is y O
p O2
y O2 =
pTotal
=
2
Hx O2
pTotal
The moles of O2 and H2O are:
O2 :
6 mg
1g
103 L 1 g mol O 2
= 0.1875 g mol/m3 in water
3
L 1000 mg 1 m
32 g O 2
H 2 O:
3
3
10×10
= 55.6 × 103g mol/m 3
18
0.1875
= 3.37 10−6
×
55.6 × 103 + 0.1875
4.02×106 kPa
1
3.37×10-6 mol fr.
pTotal = 1 atm
y O2 =
= 0.134
mol fr.
101.3 kPa
x O2 =
Basis: 1 L gas
nO =
2
pO
2
RT
V=
(101.3 kPa)(0.134)
1 atm
101.3 kPa
1L
(g mol)(K)
18 g
0.08206(L)(atm) 290 K 1 g mol
1000 mg
1g
0.101 mg
=
per L
8.5.5
Use Raoult’s Law.
p Total = p*P (1 − x B ) +p*B x B
From Perry the vapor pressures are:
p*Propane = 16.8 atm
p*Butane = 4.8 atm
pTotal = (100/14.7) atm = 6.80 atm
6.80 = 16.8(1 − x B ) + 4.8x B
x Butane = 0.83 assuming the liquid phase is essentially all of the mixture.
8–26
Solutions Chapter 8
8.5.6
Mole fraction in the liquid phase is given by solving the following equations for x Bz :
and yTol =p*Tol xTol /pTotal to get xBz = (pTotal − p*Tol )/(p*Bz − p*Tol )
yBz = p*Bz x Bz /pTotal
The corresponding equilibrium mole fraction in vapor phase (y) is,
yi = p*i x i /pTotal
Using these equations for the various temperatures given, the following data for x and y
are obtained for the total pressure of 1 atm
T(ºC)
80
92
100
110.4
xBz
1.000
0.508
0.256
0.000
yBz
1.000
0.720
0.453
0.000
From the above-calculated data, the following T-x diagram can be drawn.
8–27
Solutions Chapter 8
8.5.7
8.5.8
The mole fractions of each component are needed to apply Raoult’s law. Assuming a
basis of 100 g of solution, we can construct the following:
Water
Methanol
g
Molecular
weight
25
75
18
32
Mol
Mol
fraction
1.39
2.34
3.73
0.37
0.63
1.00
Raoult’s law is used to compute the vapor pressure (p*) of pure methanol based on the
partial pressure required to flash:
p = xp*
p* = p/x = 62/0.63 = 98.4 mm Hg.
Use the Antoine equation to get T
ln(p* ) = 18.5875 −
3626.55
-34.29 + T
T = 293 K (20o C)
8–28
Solutions Chapter 8
8.5.9
Steps 2, 3, and 4:
LH O = ?
2
100%
H2O
SO2
Air
y1 = 0.03
G = 84.9 m3
293 K
pTotal = 1 atm
y2 = 0.003 SO
2
Air
290K
pTotal = 1 atm
LH O
2
nH O=?
2
nSO2= ?
saturated
Basis: 1 min (85 m3 entering gas)
Step 5:
Steps 6 and 7:
Unknowns:
L
n LH2O , n SO
, Air
2
Equations. Material balances:
H2O, SO2, Air
yi > Hxi
Steps 8 and 9:
The liquid and vapor are assumed to be in equilibrium at the exit, and the water is
saturated with SO2 so that
y1 = Hx1
0.03 = (43) x SO = so that x SO2 = 0.00070 mol fraction
2
Material balance SO2 (moles):
G (0.03 – 0.003) = L (0.00070 – 0)
b.
g mol H 2 O
L
= 38.6
G
g mol air
The respective flow rates are
G=
85 m3 1 g mol gas
= 3540 g mol gas
0.024 m3
8–29
Solutions Chapter 8
L = (3540) (38.4) = 137 ×103 g mol water
In terms of kg
a.
L = (136) (18) = 2450 kg/min
8.5.10
Use the Antoine equation or the physical property package on the CD to get the vapor
pressure of pentane (P) and heptane (H). Assume ideal liquid and vapor.
Basis: Data given in problem statement
ln p*P (mm) = 15.8333 −
2477.07
T-39.94
p*P = 283.6 mm Hg (5.48 psia)
ln p*H (mm) = 15.8737 −
2911.32
T − 56.51
p*H = 20.6 mm Hg (0.40 psia)
Convert mass fractions to mole fractions:
Basis: 100 g liquid
P
H
g
20
80
100
MW
72.15
86.17
g mol
0.28
0.93
1.21
mol. fr.
0.23
0.77
1.00
Equations to use, i = P and H
∑ yi = 1
a.
yi =
p*i x i
pTotal
p*P
p*H
xP +
xH = 1
pTotal
pTotal
pTotal = (283.6)(0.23) + 20.6(0.77) = 80.95 mm Hg
8–30
Solutions Chapter 8
3.19 in Hg
10.8 kPa
1.57 psia
16
(0.23) = 0.90
4.07
0.5
yH =
(0.77) = 0.10
4.07
1.00
yP =
b.
yi =
p*i x i
pTotal
8.5.11
Data:
MW
Benzene (B) 78.1
Toluene (T) 92.1
p* at 60ºC
51.3kPa
18.5 kPa
Assume ideal gas and liquid. Apply Raoult’s Law. Calculate mole fractions in the
liquid.
b.
xB =
1
78.1
= 0.541
1
1
+
78.1 92.1
xT =
a.
1
92.1
= 0.459
1
1
+
78.1 92.1
The equations to use are
x B p tot = x B p*B
yT p tot = x T p*T
yB + yT = 1
yB + x T = 1
(1 − y B )p tot = x T p*T
" x B p* B #
*
$1 −
% p tot = x T p T
p tot '
&
tot
p tot = x B p*B + x T p*T = (0.541)(51.3) + (0.459)(18.5) = 36.2 kPa
(b)
What will be the composition of this first bubble?
yB =
x B p* B
(0.541)(51.3)
=
= 0.766
p tot
36.2
y T = 1 − y=B
0.234
8–31
Solutions Chapter 8
8.5.12
Data:
p* at –31.2ºC (kPa)
Propane (P) 160.0
n-butane (B) 27.6
Assume ideal solution and vapor. Use Raoult’s law
(a)
yP ptotal = x P p*P
yP +yB = 1
yBptotal = x Bp*B
yP +yB = 1
x p p*p
yP =
(1 − y P )p total = (1 − x P )p*B
p total
More manipulations give
xP =
p total − p*B 101.3 − 26.7
=
= 0.56
p*P − p*B 160.0 − 26.7
x B = 1 − 0.56
=
0.44
(0.56)(160.0)
= 0.884
101.3
(b)
yP =
(c)
For propane
TºC
-0.5
-16.3
-31.2
-42.1
x
0.0
0.196
0.560
1.0
yB = 1 − 0.884
=
y
0.0
0.577
0.884
1.0
8–32
0.116
Solutions Chapter 8
8.5.13
Assume ideal solution
For the dewpoint
1
p Total
=
y1 y 2
+
p*1 p*2
(1)
p Total = 100 psia or 5, 171 mm Hg
yi (mol fr.)
0.20
n-pentane
!
2477.07 "
p*1 = exp $15.8333 −
(−39.94 + T) %'
&
(2)
0.80
n-hexane
!
2697.55 "
p*2 = exp $15.8366 −
(−48.78 + T) %'
&
(3)
p*i is mm Hg and T is in K
Solve Equation (1) to get from Polymath T = 413 K
8–33
Solutions Chapter 8
8.5.14
Assume ideal solution.
For the bubblepoint
(1)
pTotal = p*1 (T)x1 + p*2 (T)x 2
pTotal = 200 psia or 10,342 mm Hg
xi (mol fr.)
0.20
n-pentane
!
2477.07 "
p*1 = exp $15.8333 −
(−39.94 + T) %'
&
(2)
0.80
n-hexane
!
2697.55 "
p*2 = exp $15.8366 −
(−48.78 + T) %'
&
(3)
p*i is mm Hg and T is in K
Solve Equation (1) to get from Polymath T = 447 K
8.5.15
Calculate the bubble point and dew point temperatures at 39.36 in. Hg = 1000 mm Hg.
Use the Antoine equations to get p* based on the assumption the solution is ideal.
ln p* = A −
K1 =
p*1
pt
B
C+T
1 = Benzene
2 = Toluene
ln(K1 ) = 1n(p*1 ) − 1n(pt ) = 1n(p*1 )− 6.9078
Calculate 1nK1 and lnK2. Solve the bubble point equation and the dew point equation by
a computer code (or Newton’s method) starting with T = 365K (slightly above p* for
benzene).
a. Bubble point temperature
∑ yi = 1 =
If solution is ideal
K
∑i x i
Ki xi =
p*i x i
p*
p*
and 0.5 1 + 0.5 2 − 1 = 0
pt
pt
pt
8–34
(1)
Solutions Chapter 8
Use the Antoine equation to get p*i
Benzene p*1 = 15.9008 −
2788.51
−52.36 + T
Toluene p*2 = 16.0137 −
3096.52
−53.67 + T
Solution of Equation (1) using Polymath: Bubble point temperature is 375 K
b. Dew point temperature
∑ xi = 1 =
y
0.5 0.5
∑i =
+
Ki
Ki
K2
(2)
Solution of Equation (2): Dewpoint temperature is 381.5 K
8.5.16
Assume pentane is the major contributor to the gas-phase composition.
yi =
pi
p tot
For Raoult's law,
*
pi = p i x i
*
pi =
so
Using the Antoine equation,
ln (36) =15.8333−
2477.07
−39.94 + T
T = 242 K
8–35
yi p tot (0.018)(100)
=
= 36kPa
xi
0.05
Solutions Chapter 8
8.5.17
Assume that the pressure at the bottom of the lake = pCO2
p = pgh varies with height
Avg. depth = 225 m
pCO2 =
1000 kg 9.8 m 225 m
= 2.21 × 103 kPa + 1.01 × 102 kPa
m3
s2
= 23 Atm
pCO2 = HxCO 2 ⇒
x CO2
=
23atm
= 0.0134
1.7×10 3 atm mol fr
If the entire 200,000 tons were saturated at pCO but not supersaturated
200, 000 ton H2 O 1000 kg 1kgmol
= 1.11 × 107 kg mol
ton
18kg
0.0134 =
n CO2
nCO 2 is mol of CO2
1.11×107 + nCO 2
n CO2 = 1.51 × 105 kg mol
(1.5×10 5 )(8.314)(273)
nRT
6
3
= VCO2 =
= 3.39×10 m
p
101.3kPa
Above would be the worst case. The CO2 would be less; p CO2 average might be a better
choice to use in which case VCO 2 would be, say ½ the calculated amount.
8–36
Solutions Chapter 8
8.5.18
VI
y1I
FI
z1I
VII
yII
2
I
II
LI
x1I = z1II
LII
x1II
1 designates component A
2 designates component B
For each of the separation units we can write a mass balance and an equilibrium
relationship, e.g., Raoult’s Law.
Material balance on Comp. 1
V
y1
F
z1
Fz1 = Vy1 + Lx1
(1)
Raoult’s Law
Sep.
Unit
L
x1
y1pTotal = x1p*1
(2)
(1-y1 )pTotal = (1-x1 )p*2
(3)
Solve for y1 in (1):
! Fz " ! L "
y1 = $ 1 % − $ % x1
& V ' &V'
(4)
Introduce in (2):
!# Fz1 $ # L $ "
*
(& V ' − & V ' x1 ) pTotal = x1p 1
+ * + ,*
Basis: 1 minute
Introduce the given values
p*1 = 10 kPa
p*2 = 100 kPa
FI = 100 mol
LI = 50 mol
8–37
Solutions Chapter 8
V I = 50 mol
VII= 25 mol
LII = 25 mol
z1 = 0.50
The solution is
p*1 =
10kPa
p*1 =
10kPa
p*2 =
100kPa
p*2 =
100kPa
Unit I
FI =
LI =
VI =
z1I =
Unit II
100 mol/s
50 mol/s
50 mol/s
0.5
x1 =
I
0.75975 = z1
I
0.24025 = z 2
I
0.24025
0.75975
x2 =
y1 =
I
y2 =
FII =
LII =
VII =
z1II =
50 mol/s
25 mol/s
25 mol/s
0.75975
II
x1 =
II
0.93391
II
x2 =
II
0.06609
y1 =
II
0.58559
II
0.41441
y2 =
I
II
p Total = 31.62 kPa
pTotal = 15.95 kPa
8.6.1
Apply K values to solve the problem. Get the mole fraction of the compound in
the liquid phase.
Basis: 100 g water (5.555 g mol)
liquid
g
MW
g mol
g mol
x (mol fr.)
vapor
K
y (mol fr.)
H 2O
Glycerol
5.5
92.08
0.0597
5.555
0.0106
1.2 × 10-7
1.27 × 10-9
MEK
1.1
72.10
0.0153
5.555
0.00275
3.065
0.00843
Phenol
2.1
94.11
0.0223
5.555
0.00400
0.00485
1.94 × 10-5
The mole fractions in the gas phase are in the far right hand column. Glycerol and phenol
have the lowest concentrations. If volatization is proportional to the vapor phase
concentration, only MEK might be a problem.
8–38
Solutions Chapter 9
9.1.1
Basis: 1 lbm
252 cal 1 lb m
4
= 2.5 × 10 cal / kg
a. 45.0 Btu
1 Btu 0.454 kg
lb m
3
1.055 × 10 J 1 lb m
5
45.0
Btu
= 1.048 × 10 J / kg
b.
1 Btu
0.454 kg
lb m
–4
2.930 × 10 (kW)(hr) 1 lb m
– 2 ( kW)(hr )
45.0
Btu
= 2.91× 10
c.
1 Btu
0.454 kg
lb m
kg
2
7.7816 × 10 (ft) (lb f)
4
d. 45.0 Btu
= 3.5 × 10 ( ft )(lb f ) / lb m
lb m
1 Btu
9.1.2
2
cal
252 cal hr
ft
a. 6000 Btu
= 0.4521
2
2
2
(s) cm 2
(hr) (ft ) Btu 3600 s 30.48 cm
( )
cal
(Δ) 1.8°F
b. 2.3 Btu 252 cal lb
= 2.3
( g)(°C )
(lb) (°F) Btu 453.6 g (Δ) °C
c.
J
200 Btu 1, 055 J hr
(Δ )1.8°F
ft
= 3.46
(s)(cm )(°C)
(hr )( ft )(°F ) Btu 3600 30.45 cm ( Δ )°C
3
d.
10.73 (lb f) (ft ) 144 in. 2
(in. 2 ) (lb mol) (°R)
= 8.31
ft
2
3
Btu
lb mol 1.055×10 J (Δ) 1.8°R
Btu
778 (ft) (lb f) 453.6 g mol
(Δ) °K
J
(g mol )(°K)
9–1
Solutions Chapter 9
9.1.3
Basis: 1011 watts
108 W 8.6057 ×105 cal 1 hr
1 (min)(cm2 ) 1 m2
103 (W)(hr) 60 min 0.10 32.0 cal (100 cm) 2
= 4.5 × 104 m2 larg e
9.1.4
(a) T; (b) T; (c) F; (d) F; (e) T
9.1.5
(a) T; (b) T; (c) T; (d) T; (e) T; (f) F; (g) F
9.1.6
partial pressure
volume
specific gravity
potential energy
relative saturation
specific volume
surface tension
refractive index
intensive
extensive
intensive
extensive
intensive
intensive
intensive
intensive
9.1.7
To convert to J/(min) (cm2) (°C), we set up the dimensional equation as follows
2
2
0.026 G 0.6 Btu 1055 J 1 hr ! 1 ft " ! 1 in. " 1.8o F
h = 0.4
#
$ #
$
D (hr) (ft 2 ) (o F) 1 Btu 60 min % 12 in. & % 2.54 cm & 1o C
= 8.86 × 10-4
G 0.6 J
D0.4 (min)(cm2 )(o C)
9–2
Solutions Chapter 9
9.1.8
975 W A m 2 320 days 24 hr 5 0.21 3.6 × 106 J
m2
1 day 25
1( kW)( hr )
a.
20
= 3 × 10 J
11
A = 2.5 × 10 m
2
( 2.5×10 km )
5
2
The capital cost is probably too great and the maintenance too high relative to other
power sources. Also, the location would have to be in the far western united so power
would have to be transmitted over long distances.
3×1020 J =
b.
10,000 Btu 2,000 lb T tons 1055 J 0.70
lb
ton
1 Btu
T = 2.0 ×1010 tons
2.0 ×1010
= 0.012
1.7 ×1012
9.1.9
a.
1 lb fat 1 kg 7,700 k cal day = 7 days
2.2 lb
kg
500 k cal
b.
1 lb fat 1 kg 7,700 k cal 4.184 k J km 0.6214 mi = 22.75 mi
2.2 lb
kg
1 k cal 400 k J 1 km
c. Total energy consumed is the same. The higher power consumption is compensated
for in the shorter travel time.
9–3
Solutions Chapter 9
9.1.10
First substitute for T F
o
ToF = To C (1.8) + 32
Next change units
k=
{a + b !#T (1.8) + 32"$} Btu 1055 J 1 ft
o
C
(hr)(ft)(o F)
= 1.05 !#a + b(To C (1.8) + 32 "$
1 in
1 hr 1.80o F
1 Btu 12 in 2.54 cm 60 min 10 K
J
(min)(cm)(K)
or k = 1.05 a + 1.8 bT C + 33.2
o
9.1.11
900,000 J (s)(10-3 ) kW
= 105 kW or 100 MW
-3
9×10 s
1J
The answer is yes .
9.1.12
A main meal is equivalent to about 4000 kJ. A ¼ hour is 900 seconds, so that (700 J/s)
(900 s) / 1000 = 630 kJ, an insufficient amount of time.
9.1.13
A closed system because no mass exchange occurs with the surroundings.
9.1.14
Yes. Watts are power, and if the energy is transferred for a very short period to time, the
power can be quite large. For example, 1J (a very small amount of energy) transferred in
10–6s is 1 MW!
9–4
Solutions Chapter 9
9.2.1
The work done by the cylinder-piston system is W = -p(V2-V1)
=−
350 kPa 0.15 − 0.02 m3
1 kJ
= 45.5 kJ
(1 kPa)(1 m3 )
−
9.2.2
No work is done because the boundary of the system remains fixed.
9.2.3
Basis: 2 kg mass
PE = mgh =
12 kg 9.80 m 25 m 1(J)(s2 )
= 2940 J
s2
1(kg)(m2 )
9.2.4
Any correct examples will be satisfactory for (b), (c), and (d). But for (a), liquid
to solid, there are probably no examples. If any, they would be pathological cases.
9.2.5
Basis: 10 lb water (5 lb vaporized)
The work done by the piston is caused by the water vapor expanding at constant
temperature and pressure. Ignore the volume of the liquid water at the initial conditions,
and assume the initial amount of vapor is also negligible. From state to state 2
ΔV̂vapor = 4.897 ft 3 / lb from the steam tables. For the 5 lb
V2 = (5 lb)(4.897 ft 3 /lb) = 24.49 ft 3
W=−
89.65 lbf 144 in 2 (24.49 ft 3 )
= −3.16 × 105 (lbf )(ft)
2
2
in
1 ft
9–5
Solutions Chapter 9
9.2.6
(a) T; (b) F; (c) T; (d) F; (e) F; (f) F; (g) T
9.2.7
The comment is not a correct use of the term heat, which is a transfer of energy.
9.2.8
(a)
Wrong. Heat is energy, not material.
(b)
Wrong. Heat is energy by definition.
(c)
Wrong. Heat is energy; cold is related to temperature.
(d)
Wrong. See (c).
(e)
Wrong. Heat is a transfer of energy.
(f)
Heat can be measured, but indirectly via temperature, mass, etc.
(g)
Wrong. See (a).
(h)
OK except for “or can be stored,” which is wrong.
(i)
Burning produces a change in the state of the reactants and products, but the
change is not heat. Heat transfer can result from the change in state.
(j)
Heat cannot be stored and hence destroyed. Heat transfer can be terminated.
9.2.9
A better use of words than “absorbing heat” would be “to transfer heat.” Heat is really
not “concentrated.” The internal energy of the medium increases.
9–6
Solutions Chapter 9
9.2.10
(a) F; (b) F; (c) T; (d) F; (e) T; (f) T but debatable; (g) F; (h) F
9.2.11
Basis: 1 minute
KE = (1/2 ) mv2
v=
500,000 g 1 cm3
4
= 22,143 cm/min
min
1.15g π(5)2cm2
2
2
2
1 500 kg ! 22,143 cm " 1(J)(s 2 ) ! 1 min " ! 1 m "
KE =
#
$
#
$ #
$ = 6810 J
2 min % min & 1(kg)(m2 ) % 60s & % 100 cm &
9.2.12
Basis: 1 lbm H2O
KE = 1/2 mv2
2
1 1 lbm ! 10 ft " (lbf )(s2 )
KE =
= 1.55 (ft) (lbf )
#
$
2
% s & 32.2 (lbm )(ft)
9–7
Solutions Chapter 9
9.2.13
Basis: 1 kg steam
1
2
250.5°C
4000 kPa
650°C
10,000 kPa
ΔU = ΔH – ΔpV or use U values directly from SI tables or computer code.
Data here taken from the steam tables.
State 1 (assumed saturated) State 2 (superheated)
250.5°C (523.5 K)
650°C (923 K)
482°F
1202°F
4000 kPa abs.
10,000 kPa abs.
p
580 psia
1450 psia
Vˆ
0.498 m3/kg
0.04117 m3/kg
ΔU
2602 kJ/kg
3338 kJ/kg
ΔH (ref. satd liquid at 0°C)
2801 kJ/kg
3742 kJ/kg
T
9.2.14
Internal energy is the energy contained in the material of a system because of its
molecular arrangement. Heat is energy that flows between the system and the
surroundings because of a temperature gradient. Therefore they are not the same class of
energy.
9.2.15
ΔH = ΔU + Δ (pV)
When a liquid vaporizes Δ (pV) adds to ΔU.
9.2.16
∆H and ∆U are zero because ∆H and ∆U are point (state) functions (variables).
9–8
Solutions Chapter 9
9.2.17
None. The initial and final states for U and H are the same.
9.2.18
(a)
(b)
(c)
(d)
(e)
T
F
F
T
F
(f)
(g)
(h)
(i)
(j)
F
F
F
T
T
F
F
F
T
T
(f)
(g)
(h)
(i)
(j)
T
F
T
F
F
9.2.19
(a)
(b)
(c)
(d)
(e)
(k)
(l)
(m)
(n)
T
F
T
T
9.2.20
(a) 1; (b) 3; (c) 5; (d) 2; (e) 4
The boiling temperature is at 4 reading to the left scale (100ºC)
The freezing temperature is at 1 reading to the left scale (0ºC)
9.2.21
Tc = 540.2 K
pc = 27 atm
!T "
0.0331$ b % − 0.0327 + 0.0297 log (p c )
kJ
& Tc '
ΔH v = Tb
(g mol) (K)
!T "
1.07 − $ b %
& Tc '
ΔH v = 31.679
kJ
g mol
The percent error is negligible in view of the precision of the data.
9–9
Solutions Chapter 9
9.2.22
ΔH = 2∫
250 + 273
50 + 273
(27.32 + 0.6226 ×10−2 T − 0.0950 ×10−5 T 2 )dT
ΔH = 11,980 J (The CD gives 11,784 J)
9.2.23
Basis: 5 ft3 vessel
Vapor = 4 ft3, Liquid = 1 ft3
Steam tables: Sat'd steam at 1000 psia
3
Vv = 0.44596
ml =
mv =
1 ft
3
lb
0.02159 ft
4 ft
3
ft
lb
3
3
lb
0.44596 ft
3
Vl = 0.02159
ft
lb
lb
mass fr.
46.32
0.838
8.97
0.162
55.29
1.000
=
=
mt =
Quality = 0.162
9.2.24
⎛ πD 2 ⎞
(0.25)2
⎜
⎟
The volume of the wire is
5.5 = π
( 5.5) = 0.27 cm 3 . The density of Al is
⎝ 4 ⎠
4
19.35 g/cm3, so 19.35(0.27) = 5.22g or 0.194 g mol Al. From Perry, Cp = 20.0 +
0.0135T (T in K, Cp in J/(g mol)(°C) and ∆Hfusion = 10,670 J/g mol at 660°C.
⎡660 +273
⎤
ΔH = 0.194 ⎢ ∫ (20.0 + 0.0135T)dT + 10,670 ⎥ = 5560 J
⎣ 25+ 273
⎦
5560 2.773 × 10 –7 = 1.54 × 10 –3 kWh
(
)
9–10
Solutions Chapter 9
9.2.25
Integrate the heat capacity equation, use Table D3 in the Appendix, or use CD
From the CD ΔH = 2580 J/g mol
9.2.26
We will use the steam tables for this problem.
Basis: 1 lb of H2O at 60°F
From steam tables (ref. temp. = 32°F):
Ĥ = 28.07 Btu/lb
at 60°F
Ĥ = 1604.5 Btu/lb
at 1150°F and 240 psig (254.7 psia)
Δ Ĥ = (1604.5 – 28.07) = 1576.4 Btu/lb
ΔH = 1576(8.345) = 13.150 Btu/gal
Note: The enthalpy value that has been used for liquid was taken from the steam tables
for the saturated liquid under its own vapor pressure. Since the enthalpy of liquid water
changes negligibly with pressure, no loss of accuracy is encountered for engineering
purposes if the initial pressure on the water is not stated.
9–11
Solutions Chapter 9
9.2.27
Basis: 3 kg H2O
The initial conditions are obtained from the SI steam tables for saturated liquid. Enthalpy
is a function of the temperature and pressure, but the effect of pressure on liquid water
under these conditions is negligible. Therefore the enthalpy of water at 300K and 101.3
kPa can be approximated by the enthalpy of saturated water at the same temperature but
at the vapor pressure of 3.536 kPa.
The final conditions are presumably a state in which water is all vapor. A check of the
steam tables shows this assumption to be true.
At
T(K)
p(kPa)
300
800
3.536
1500
ˆ
ΔH(kJ/kg)
111.7
3384.3
The enthalpy change is
ΔH =
3 kg (3384.3-111.7) kJ
= 9817.8 kJ
kg
9.2.28
Yes. 40ºF and saturated liquid is an arbitrary reference state for enthalpy that can be
assigned a value of zero. Only enthalpy changes can be computed using the chart; the
enthalpy itself cannot be determined.
9–12
Solutions Chapter 9
9.2.29
Basis: 1.2 ft3 gas at 7.3 atm
p of 1 atm = 15,450 lbf/ft2
pV1.3 = const.
1.3
V1 = (1.2)1.3 = 1.27
1.3
p1V1 = (15450)(1.27) = 19,620
19,620
W = − ∫=pdV − ∫= − 1.3 dV
V
1.3
V2
19,620 ∫
dV
! 19,620 " V2
= −$
1.3
0.3 %
V
& −0.3V ' V1
1.3
=
(V1 )(p1 ) (1.27)(7.3)
=
= 9.27
p2
(1.0)
V2 = 5.54 ft 3
0.3
V1 = (1.2)0.3 = 1.06
V2
0.3
W=
= (5.54)0.3 = 1.67
19,620 ! 1
1 "
(19,620)(0.344)
−
=−
= −22,500(ft)(lbf )
$
%
0.3 &1.67 1.06 '
0.3
9–13
Solutions Chapter 9
9.2.30
The density of air at 27ºC and 1 atm is ρ =
ρ=
101.3 kPa
Power =
(p)(MW)
(RT)
29 kg air
(kg mol)(K)
= 1.18 kg / m3
1 kg mol air 8.314(kPa)(m3 ) 300 K
1
1
1
 2 = (ρAv) v 2 = ρAv 3
mv
2
2
2
Assume air flow through the windmill is equivalent to the average flow in a pipe of
diameter 15m.
20 mi 1.61×103m 1 hr
v=
= 8.94 m/s
hr
1 mi
3600 s
2
! 1 " 1.18 kg
Power = # $
3
%2& m
! 15 m "
3
$
2
% 2 & ! 8.94 m " 1(s )J (s)(W) 1 kW = 74.46 kW
#
$
2
% s & 1(kg)(m ) 1 J 1000W
π#
Electrical energy = (74.46)(0.30) = 22.3 kW
9–14
Solutions Chapter 9
9.2.31
Basis: 1 lbm of vehicle
a.
25, 000 mi 1 hr 5280 ft
= 36, 600 ft/s
hr
3600 s 1 mi
9 2
1 1.34 × 10 ft 1 lb m 1
7 (ft) (lb f)
K.E. =
= 2.08 × 10
lb m
2
2
gc
s
7
2.08 × 10 (ft) (lb f)
1 Btu
=
K.E. =
778 (ft) (lb f) 2.68 × 10 4 Btu / lb
lb m
b.
m
Heat capacity of, say, 1 Btu/(lb) (°F) gives
1 Btu 20°F
= 20 Btu/lb m .
(lb) (°F)
Essentially all the energy must be transferred to the surroundings
9–15
Solutions Chapter 9
9.2.32
The heat capacity of a monoatomic ideal gas is
volume is:
5
R , and the heat capacity at constant
2
5
3
R −R = R
2
2
You can show that C p = C Vˆ + R
ˆ " # ∂U
ˆ + ∂ (pV)
ˆ $ ! ∂U "
! ∂H
! ∂V "
Cp = &
' =(
) =&
' + p&
'
∂T
* ∂T +p
* ∂T +p ,
- p * ∂T +p
For the ideal gas Û is a function of T only (not of p or V̂ ) so that
" ∂U # " ∂U #
$
% =$
% = Cv
& ∂T 'p & ∂T 'Vˆ
! ∂V̂ "
!R"
p$
% = p$ % = R
&p'
& ∂T ' p
ΔU = ∫ Cv̂dT = (3/2 R)(50) = 624 J
9–16
Solutions Chapter 9
9.2.33
Basis: 100 mol gas
Comp.
CO
H2
CO2
mol
68
30
2
100
ΔĤ (J/g mol) relative to 0°C and 1 atm
CO
H2
CO2
25°C =
1400°C =
298 K
1500 K
1673 K
1750 K
728
39,576
45,723
48,459
718
36,994
42,724
45,275
912
62,676
72,896
77,445
Basis: 1000 m3 at 101 kPa and 1673 K
1000 m3 273 K 1 kg mol
= 7.28 kg mol
1673 K 22.4 m3
Calculation of ΔH:
CO
H2
CO2
kg mol
ΔĤ (kJ/kg mol)
7.28 (.68)
7.28 (.30)
7.28 (.02)
-44,995
-42,010
-71,984
9–17
ΔH(kJ)
-222,743
-91,750
-10,481
Solutions Chapter 9
9.2.34
Basis: liquid water at 32°F and 1 atm
ΔH = 3 ΔH 300°F,1 atm – ΔH 32°F,1 atm = 3 (1192 – 0) = 3576 Btu
a.
b.
Basis: liquid water at 40°F and 60 psia
Δ H 32 °F,60 psia = 0
c.
Δ H = 3 (11192 – 0) =
3576 Btu
Basis: liquid water 40°F and 60 psia
ΔH = ΔH 300°F,60 psia – ΔH 40°F,60 psia = (1181.4 – 8.05)
= 1173.4 Btu
d.
Basis: water-steam mixture of 60% quality
ΔH, Btu/lb
State
1
60% steam - 40% water at 300°F; sat'd steam 1179.7
2
80% steam - 20% water at 300°F; water
ΔH = ΔH 2 – ΔH 1
ΔH 2 = (0.80) (1179.7) + (0.20) (269.6) = 998 Btu/lb
ΔH 1 = (0.60) (1179.7) + (0.40) (269.6) = 816 Btu/lb
Enthalpy change needed = ΔH
ΔH = ΔH 2 – ΔH 1 = 182 Btu / lb
e.
Basis: steam at 500°F and 120 psia
ΔH = ΔH 2 – ΔH 1 = 312.5 – 1276.7 = –964.2 Btu / lb
f.
ΔH = ΔH 2 – ΔH 1 = 487.8 – 1276.7 = 788.9 Btu / lb
9–18
269.6
Solutions Chapter 9
Data from text
ΔHA = 1205.0
ΔHB use tables in pocket
210o F
5 psia
200°F
1148.3
250°F
1171.1
7 psia
1146.7
1170.2
78.17
84.24
38.88
41.96
40 psia and 267.24°F
sat steam/sat liquid
70 psia/302°F
liquid
a. Enthalpy decrease
V̂A =1.0451 ft3 /lb
VB
(h)
b. volume goes up
70 psia/304°F
superheated steam
depending on heat supply
(i)
2.5 ft3 tank of water @ 160 psia and 363.5°F. @ 160 psia 6363.5 total volume is
occupied by sat steam
2.5
Vˆ steam = 2.834 ft 3 / lb ⇒ m steam =
= 0.88 lb
2.834
 water = 1-m steam
Vˆ Liq/ water = 0.0182 ft 3 / lb ⇒ m
(
V water =
)
2.5
3
× 0.0182 = 0.016 ft
2.834
Vsteam = 2.5 – 0.016 = 2.49 ft 3
9–19
Solutions Chapter 9
Of 5 lb H2 O assume x lb is sat steam (5 – x )lb is sat. water
x( 2.834) + ( 5 – x)(0.0182)
= 2.5
⇒ x(2.834 – 0.0182)
= 2.5 – 5(0.0182)
⇒
x
=
2.5 − .0910
= 0.86 lb
2.816
Yes, it is possible
(k)
Enthalpy of 10 lb steam @ 100 psia = 9000 Btu
Enthalpy/lb steam @ 100 psia = 900 Btu
ˆ
ΔH
steam @ 100 psia = 1187.3 Btu
ˆ
ΔH
water @ 100 psia = 298.43 Btu
Let x be fraction of steam
(1–x) is fraction of water
(1187.3)x + (1–x) 298.43 = 900
x(1187.3–298.43) = 900 – 298.43
or 68%
9–20
x=
601.57
= 0.68
888.87
Solutions Chapter 9
9.2.35
a-1:
V̂ ≅ 0.23 ft 3/lb
a-2:
gas
b-1:
42°F
b-2:
mixture of liquid and vapor
9.2.36
Basis: 1 g mol CO2
100o C
ΔH = m∫ o
50 C
(a + bT + cT + dT )dT
2
3
b
c
d
!
"
ΔH = m %a(100 − 50) + (1002 − 502 ) + (1003 − 503 ) + (100
− 4 50 4 ) &
2
3
4
'
(
"
#
4.233 ×10−2
2.887 ×10−5
2
2
(100 − 50 ) −
(1003 − 503 )+ &
%36.11(100 − 50) +
2
3
&
ΔH = 1 g mol %
−9
% 7.464 ×10
&
(1004 − 504 )
%(
&)
4
ΔH = 1956 J
From the CO2 chart ΔH ≅ 198-178 = 20 Btu/lb or 2040 J/g mol. From the enthalpy tables
(interpolating) ΔH = 1957 J/g mol. The CD gives 2038 J/g mol.
9–21
Solutions Chapter 9
9.2.37
ˆ − ΔH
ˆ ) = 10(86 − 288) = −
ΔH = 10(ΔH
2020 Btu . The CD gives –1962 J.
2
1
9.2.38
Assume the temperature is 80ºF, and the propane is saturated liquid and vapor. The
corresponding pressure is the saturation pressure, namely about 2.3 atm. Another
temperature would correspond to another pressure. After 80% of the propane is used, the
conditions are still saturated liquid-vapor mixture at 80ºF and 2.3 atm.
9.2.39
Basis: 10 lb mol ideal gas
(a)
Use pV = nRT
ˆ = Δ(RT)
Δ(pV)
3
359 (ft) 1 atm 500°R
3
(1) 10 lb mol
= 36.5( ft ) @ 100 atm, 40°F
lb mol 100 atm 492°R
(2) 100 atm
900°R
500°R
= 180 atm
(3)
ΔH = ( Hˆ 2 – Hˆ 1) 10 = 10 [300 + 8.0 (440) – (300 + 8.0 (40))] = 32, 000 Btu
(b)
The equation for H is based on the reference conditions where Ĥ = 0.
Ĥ = 0 when 300 = -8T or Tref = -37.5 ºF = -38.5ºC. U should have the same
reference temperature.
Δ Uˆ = Δ Hˆ – Δ(p Vˆ ) = Δ Hˆ – ΔRT for an ideal gas
Change units for Ĥ:
First term:
300 Btu 1 lb mol 1055 J
697 J
=
lb mol 454 g mol Btu
g mol
9–22
To F =1.8To C + 32
Solutions Chapter 9
o
8.0 Btu 1055 J lb mol (1.8To C + 32) F
J
Second term:
= (33.5To C +595)
o
(lb mol)( F) Btu 454 g mol
g mol
Ĥ ref T = 0
ΔH = H T − H ref T = H T
ΔRT = (RT)T − (RT)refT
ΔU = 697 + 595 + 33.5 To C − R(TABS − TABS
REF T
(T is absolute)
)
TABS =To C + 273
TASS1REF.T = − 38.5 + 273
ΔU = 1292 + 33.5 To C − 1.987 (To C + 38.5) = 1215 + 31.5 To C
9.3.1
All of them.
9.3.2
Closed unsteady state system
Q + W = ΔU
60 + W = 220
W = 160 Btu (work done on the system)
9–23
Solutions Chapter 9
9.3.3
(a)
pump
pump
system is the pump. Open, steady state.
Q
Very little
W
Yes
ΔU
No
ΔH
Yes(flow)
ΔPE
No
ΔKE
No
(b)
system is the pump and motor. Open, steady state
A little
Yes
No
Yes
No
No
(c)
System is the ice. Closed, unsteady state.
Yes
No
Yes
Yes
No
No
(d)
∞
Possibly Yes
system is the mixer and solution. The process is closed,
unsteady state.
Yes
Yes
9–24
Yes
No
No
Solutions Chapter 9
9.3.4
a.
boundary
Steam out
Energy out
Water in
Energy in
Q
b.
Q=0
Steam in
Energy in
Turbine
Steam out
Energy out
Work
c.
Work
Battery
Q=0
W ≠0
9.3.5
No. The energy balance is Q + W = ΔU, and for Q = 0 and since ΔU = 0 for an
isothermal process W = 0 .
9–25
Solutions Chapter 9
9.3.6
Closed unsteady state system
Q + W = ΔU
In kJ: (30-5) + 0.5 = Uf – 10
Uf = 35.5 kJ
9.3.7
Closed unsteady state system
Q + W = ΔU
Q=0
W = ΔU = CvΔT
W=
(100 W)(5 hr) 1 kW 3.6 × 103 kJ
= 1.8 ×103 kJ
1000 W 1 kW
mols of air = n =
pV
(100 kPa)(100 m3 ) (kg mol)(K)
=
= 3.97 kg mol
RT
303 K
(8.314)(kPa)(m3 )
1.8 ×103 = 30(3.97)(T − 30)
T = 45o C
9–26
Solutions Chapter 9
9.3.8
Closed, unsteady state process
Initial conditions
Final conditions
V = 1ft saturated dry satd. steam (the basis)
V = 1ft (know) + V2
Look up
At p = 1 psia T = 200ºF. The water
is vapor
V̂ = 33.610 ft3/lb
Ĥ = 1145.72 Btu/lb
Û = 1073.96 Btu/lb
From the steam tables:
Ĥ = 1150.15 Btu/lb
Û = 1077.48 Btu/lb
V̂ = 392.713 ft3/lb
p = 11.548 psia
Calculate the lb of steam
m=
1 lb
1 ft 3
= 0.02975 lb
33.610 ft 3
Q + W = ΔU
W = 0 (fixed boundary)
a.
Q = ΔU
b&c.
d.
ΔU = U2 – U1 = (1077.48) (0.02975) – (1073.96) (0.02975) = 0.105 Btu = Q
The volume of the second tank is Vfinal = (392.713) (0.02975) = 11.68
V2 = 11.68 − 1 10.68
= ft 3
9.3.9
No. For each path Q + W = ΔU, and for the cycle 1 to 2 and back ΔU = 0. Addition
gives
(QA + WA ) + (QB + WB ) = ΔU1−2 + ΔU 2−1 =0
(QA + QB ) = − (WA + WB )
Note: The equations in the problem used a different sign for W. A similar analysis
applies to the second equation in the problem.
9–27
Solutions Chapter 9
9.3.10
A closed steady-state system.
The cooling at the maximum efficiency is
0.695 kW 0.948 Btu 3600s
= 2372 Btu/hr
(1 kW)(s) 1 hr
not 5500 Btu/hr.
The main question with the advertisement is: where does the energy go that is removed
from the room if the unit does not require outside venting?
9.3.11
A closed steady-state system
 +W
 = ΔU
 =0
Q
 =−W

Q
 = 0.25 hp 0.7068 ⎛ Btu ⎞ 60 s = 10.6 Btu/min
W
1 hp ⎜⎝ s ⎟⎠ 1 min
 = − 10.6 Btu/min
Q
9.3.12
Open system, unsteady-state. The system is the volume containing 1 lb of H2O at 27ºC
(liquid). At the end of the process the system contains nothing. The water leaving is at
100ºC because the atmospheric pressure is 760 mm Hg.
Q + W = ΔH
W=0
ˆ (1) − H
ˆ (0)
Q = ΔH = H
out
in
The reference temperature for enthalpy is 0ºC but the water is at
ˆ =H
ˆ
ˆ
27ºC, H
−H
out
100 C
27 C
o
o
Q = (1 kg) (2675.6 – 111.7) kJ/kg = 2558 kJ
9–28
Solutions Chapter 9
9.3.13
Steps 1, 2, 3, and 4:
Figure P22.27 shows the known quantities. No reaction occurs. The process is clearly a
flow process (open system). Assume that the entering velocity of the air is zero.
Step 5:
Basis: 100 kg of air = 1 hr
Steps 6 and 7:
Simplify the energy balance (only one component exists):
ˆ + KE
ˆ + PE)m]
ˆ
ΔE = Q + W − Δ[(H
(1) The process is in the steady state, hence ΔE = 0.
(2) m1 = m2 = m.
ˆ
(3) Δ(PE)(m)
= 0.
(4) Q = 0 by assumption (Q would be small even if the system were not insulated).
(5) v1 = 0 (value is not known but would be small).
The result is
 )m] = ΔH + ΔKE
W = Δ[(Ĥ + KE
We have one equation and one unknown, W. ΔKE and ΔH can be calculated, hence the
problem has a unique solution.
Steps 7, 8, and 9
9–29
Solutions Chapter 9
ΔH =
(509 − 489)kJ 100 kg
= 2000 kJ
kg
ΔKE =
1
2
m(v22 − v12 )
3
1 kJ
! 1 " 100 kg (60 m )
=# $
= 180 kJ
2
1000(kg)(m 2 )
s
% 2&
(s) 2
W = (2000 + 180) = 2180 kJ
(Note: The positive sign indicates work is done on the air.)
To convert to power (work/time),
kW=
2180 kJ 1 kW 1 hr
= 0.61 kW
1 hr 1 kJ 3600 s
s
9.3.14
System: gas
 + PE
 m⎤+ Q + W
E t − E t = − Δ ⎡ Ĥ + KE
⎢⎣
⎥⎦
2
1
(
a.
)
no reaction
 + PE
 )m ⎤ = 0, no flow
-Δ ⎡(Ĥ + KE
⎣
⎦
ΔKE and ΔPE of gas = 0
Ut 2 − Ut1 = Q + W
b.
Here W = 0
Ut 2 − Ut1 = Q
c.
Here Q = 0,
9–30
Solutions Chapter 9
Ut 2 − Ut1 = W
d.
Ut 2 − Ut1 = Q + W
e.
Here, the system is the gas
Ut 2 − Ut1 = 0
Q = 0, W = 0
9.3.15
(a)
The system is the mixing point of the 1000°F “air” and water at 70°F
(b)
Open system
(c) and (d)
H2O at 70oF
H = 38.05 Btu/lb
Air at 1000oF
Products (100 lb)
System
H2O @ 400oF, H = 1201.2 Btu/lb
H = 6984 Btu/lb mol
air @ 400oF, H = 2576 Btu/lb mol
The enthalpies of water come from the steam tables. The properties of air come from the
tables in Appendix D7 (with interpolation). The reference temperature for the enthalpy is
32°F.
ΔE = [ΔU + ΔPE + ΔKE]inside = Q + W − ΔH − ΔPE − ΔKE



(e)
•
•
•
•
Thus
0 because steady state
Q is assumed to be 0 because time is short in transit
W = 0 (none specified)
ΔKE = ΔPE = 0 none with mass flow
ΔH = 0
or
ΔHout = ΔHin
(f) The units in the balance will be Btu and lb
( n air ) ΔHˆ air + ( n H O ) ΔHˆ H O (g) = ( n air ) ΔHˆ air + ( n H O ) ΔHˆ H2O
2
@ 400o F
2
@ 400o F
2
@ 1000o F
(a) 70o F
(100)(2576)
(6984)
+ (mH2O )(1238.9) = (100)
+ m H2O (38.05)
29
29
9–31
(e)
Solutions Chapter 9
9.3.16
General balance is
 + PE
 )m] + Q + W
ΔE = [ΔU + ΔKE + ΔPE]inside = − Δ[(Ĥ + KE
(1)
(2)
(3)
(4) (5)
(6)
(7) (8)
Assume all reactants are in bomb at start of reaction.
a.
Q
1
2
3
ΔU retain because changes
ΔKE delete - no change
ΔPE delete - no change
4
! ΔH
"
# ΔKE delete as no mass flow in or out
" ΔPE
$
Result: ΔU = Q
5
6
7
8
Q retain
W delete (fixed
boundary)
b.
Gases
Fuel
W
Q
ΔH = Q - W
1 ⎫⎪
5
delete as negligible
2 ⎬ΔE = 0 because steady state
3 ⎪⎭
6
delete not a factor
7
Q
keep
4
8
W
keep (electric work)
ΔH
keep
Result: ΔH = Q + W
9–32
Solutions Chapter 9
c.
1 ⎫⎪
2 ⎬ΔU = 0
3 ⎪⎭
4
steady state
ΔH is kept
5
ΔKE delete negligible
6
ΔPE delete not applicable
7
Q = 0 insulated applicable
8
W=0
no work done
Result: ΔH = 0
9.3.17
Unsteady state, closed, isothermal process at 25oC
Basis: 1 kg CO2
p1 = 550 kPa
T1 = 25°C
p2 = 3500 kPa
T2 = 25°C
V̂2 = 0.0125 m3 /kg
ˆ =170 kJ/kg
ΔH
From CO 2 !"V̂1 = 0.10 m 3 /kg
#
ˆ = 205 kJ/kg
chart "$ΔH
2
(converted to SI units)
Energy balance
ΔE = Δ[(U + PE + KE)]inside = Q + W − Δ(H + PE + KE)
ΔU = Q + W = ΔH − Δ(pV) so Q = ΔH − Δ(pV) − W
ΔH = (170 − 205)(1) = 35kJ
−
W = 4.106 kJ
! 3500 kPa 0.0125 m 3
Δ(pV) = %
−
kg
%'
Q = (−35) − (−11.25) − 4.106
=
! 3 N
"
550 kPa 0.10 m 3 " #10 m 2
1J $
&#
$ = −11.25 kJ
kg &( # 1 kPa 1(N)(m) $
&
'
−27.86 kJ / kg CO2
(removed)
9–33
Solutions Chapter 9
9.3.18
Pick the room as the system. The simplified form of the energy balance is (for no mass
flow):
ΔE = ΔU = Q + W = ΔU
where Q = 0 (room is insulated)
W = the electrical energy provided freezer through the system boundary
and is positive. (No volume change occurs). Hence ΔU = CvΔT is positive and ΔT is
positive; i.e. the temperature increases.
9.3.19
This is an unsteady state closed process
Basis: 1 lb water
Data from the CD
Initial Conditions
Final Conditions
T = 327.8ºF
T = 327.8ºF
p = 100 psia
V̂ = 0.018 ft 3 /lb for liquid and 4.435 for vapor
V̂ = 4.435
Ĥ = 298.475 Btu/lb
Û = 298.146 Btu/lb
V = 4.435 ft3
V = 4.435 ft3
At the final conditions the water is saturated vapor
9–34
Solutions Chapter 9
H = 1187.48 Btu/lb
U = 1105.46 Btu/lb
T = 327.75 ºF
p = 100 psia still
ΔH = 1187.48 – 298.475 = 889.0 Btu/lb
ΔU = 1105.46 – 298.146 = 807.3 Btu/lb
The energy balance is
ΔU = Q + W
W = 0 (fixed container)
Q = ΔU = 807.3 Btu/lb
9.3.20
A closed, unsteady state system comprised of 2 tanks. Get the steam properties from the
CD.
Basis: 1 lb steam.
State 1
State 2
m = 1 lb
p = 600 psia
V̂ = 0.795 ft3/lb
T = 500oF (960oR)
Û = 1128.19 Btu/lb
Ĥ = 1215.97 Btu/lb
V1 = 0.795 ft3
m = 1 lb*
p = 330.45 psia
V̂ = 1.590 ft3/lb
T = 500oF (960oR)*
Û = 1157.03 Btu/lb
Ĥ = 1253.98 Btu/lb
V2 = 2V1 = 1590 ft3*
* used to get state 2 properties
The closed system is Q + W = ΔU
ΔU = 1157.03 – 1128.19 = 28.84 Btu/lb
W = 0 (fixed boundary)
Q = ΔU = 28.84 Btu/lb
ΔH = 1253.98 – 1215.97 = 38.01 Btu/lb
9–35
Solutions Chapter 9
9.3.21
Basis: 1 lb H2O at 212ºF, 1 atm
Data from the CD
Û
Ĥ
V̂
Initial (liquid)
180.182 Btu/lb
180.226 Btu/lb
0.017
ft3/lb
Final (vapor)
1077.40 Btu/lb
1150.29 Btu/lb
26.773
ft3/lb
p = 1.00 atm
Non-flow process (unsteady-state)
Q + W = ΔU
ΔU = (1 lb)(1077.40 – 180.182) Btu/lb = 897.22 Btu
W = − ∫ pdV = (1 atm)(26.773
– 0.017) ft3/lb (2.72) Btu/(atm)(ft3)
−
= -72.78 Btu
Q = 897.22 – (-72.78) = 970.0 Btu
ΔH = 1150.29 − 180.226
=
970.0 Btu
Flow process (unsteady-state)
ΔU = Q + W –ΔH
Assume the volume of the system is fixed at V = 0.017 ft3 and that W = 0. Also assume
that nothing is left in the system after the water evaporates.
ˆ (0) − U
ˆ
ΔU = U
final
initial (1) = − 180.182 Btu
ˆ
ˆ (0) = 1150.29 Btu
ΔH = H
(1) − H
in
out 1212o F
Q = ΔU + ΔH = − 180.182 + 1150.29 = 970.1 Btu
9–36
Solutions Chapter 9
9.3.22
The tank is an open unsteady state system. f stands for final, i for initial, in for
in, m for mass in lb.
Data:
at 291°F and 50 psia
Ĥ Btu/lb
Û Btu/lb
V̂ ft3/lb
1179.39
1099.60
8.653
ΔU = Q + W – ΔH
at 14.7 psia saturated (212°F)
W=0
1150.26
1077.37
26.818
Q=0
ΔU = -ΔH
ˆ −m U
ˆ
ˆ
nf U
f
i i = (m f − m i )H in − 0
V = 50 ft 3 so mf =
50
= 5.78 lb
8.653
mi =
50
= 1.86 lb
26.818
ˆ ) − 1.86(1077.37) = (5.78 − 1.86) 1179.39
5.78(U
f
Û f = 1146.56 Btu/lb at 50 psia
Tf = 411o F
(superheated)
9–37
Hout = 0
Solutions Chapter 9
9.3.23
The general energy balance is
ˆ + KE
ˆ + PE)
ˆ m]
ΔE = Q + W − Δ[(H
(1)
(a)
(2) (3)
(4)
(5) (6)
(1)
ΔE = 0
steady state flow process, one overall system (You can also pick
two system, one the H2O and one the benzene).
(2)
Q is the net overall heat transfer to and from the heat exchanger with the
surroundings.
It is essentially 0 if exchanger is well insulated. If you pick 2 systems,
Q is the heat transfer between the benzene and the H2O, and is not 0.
(b)
(3)
W=0
no mechanical work in the process
(4)
ΔH ≠ 0
for all streams
(5)
ΔKE = 0 negligible KE change
(6)
(1)
ΔPE = 0
ΔE = 0
level exchanger assumed.
steady state process
(2)
Q
not zero as isothermal
9–38
Q = ΔH
Solutions Chapter 9
(3)
(4)
W
ΔH ≠ 0
retain
the pressure changes even if T is constant
(5)
ΔKE = 0 pipe diameter remains the same, and the flow rate is the same
(6)
ΔPE = 0
level pump lines
Q + W = ΔH
9.3.24
Unknowns
Relationships (R)
F
P
(1) F = P + V
(2) 0.05F = xPP
ˆ + SΔH = VH + PH
ˆ
(3) FH
F
S
V
P

(4) SΔH = Q

V
S

Q
S
 = UA(T − T )
(5) Q
S
V
TV
XP
Yes, two measurements must be made to balance the unknowns and the relationships. If
x P and TV are measured: solve (R5) for Q . (R4) for S substitute (R2) and (R1) in (R3)
 . Note that measuring
and solve for F. Use (R2) to calculate P and (R1) to calculate V
TV and S is not satisfactory because this combination eliminates either (R4) or (R5) as an
independent relationship. Also note that if you neglect the vapor pressure of the organic,
you can set Tv = 212ºF.
9–39
Solutions Chapter 9
9.3.25
Basis: 1 lb air
The energy balance is in the steady state. The system is the value.
 + PE
 )m ⎤ = Q + W
Δ ⎡(Ĥ + KE
⎣
⎦
if ΔKE ≅ 0
W=0
ΔPE = 0
Q ; 0
Hence ΔH = 0
ΔH = ∫ CpdT = 0
(a)
ΔT = 0 or
T2 = 250 K
Since ΔH = 0 (the kinetic energy term was negligible), the steady-state constraint requires
that the mass flow rate be constant:
 = ΔvS = Constant
m
where: ρ = density
v = velocity of the gas
S = cross-sectional area of pipe
Since the diameter of the pipe does not change and density is a function of temperature
and pressure, the gas velocity is also a function temperature and pressure.
vSTP =
100 lb Air 1 lb mol 359 ft 3 1 hr
1
= 7.0 ft/s
1 hr
29 lb Air 1 lb mol 3600 s π(1.5/12 ft)2
Therefore, the velocity downstream of the valve is:
v Outlet =
7.0 ft 250 K 1 atm
= 3.2 ft/s
s 273 K 2 atm
9–40
Solutions Chapter 9
9.3.26
Steady-state process. Basis: 1 second
 +W

 (Ĥ out − Ĥ in ) = Q
m
To get Ĥ in use the steam tables or a Cp equation. Using constant Cp is not the most
accurate method, but for liquids for small temperature changes, it is a good
approximation.
 +W


)(Tout − Tin ) = Q
(m)(C
p
Because the tank is well-mixed, the outlet water temperature is the same as the
temperature inside the tank T.
T = Tout
 (C p ) (T − Tin ) = 100 (20 − T) + 300
m
 =
m
1 kg 1 min
1 kg
=
min 60 s
60 s
Cp = 4180 J/(kg)(K)
4180
(T − 40) = 100 (20 − T) + 300
60
T = 30.0o C
9–41
Solutions Chapter 9
9.3.27
Assumptions
1.
This is a steady-flow process since there is no change with time at any point,
hence ΔE = 0.
2.
Air is an ideal gas since it is at a high temperature and low pressure relative to its
critical point.
3.
The kinetic and potential energy changes are negligible ΔKE ≅ ΔPE ≅ 0.
4.
Constant specific heats at room temperature can be used for air.
The simplified energy balance is
 +W
 = ΔH
 = m(Ĥ − Ĥ )
Q
out
in
 = − 200W
Q
 = 15 kW
W
Assume that with little error
ΔH = CpΔT
with Cp = 1.00 kJ/(kg)(ºC)
Otherwise you will need tables or a data base for the properties of air.
The ideal gas law gives the specific volume of air at the inlet of the duct
V̂1 =
RTm
[(0.287 kPa)(m 3 )/(kg)(K)](290 K)
=
= 0.832 m 3 /kg
pin
100 kPa
9–42
Solutions Chapter 9
The mass flow rate of the air through the duct is determined from
 =
m

V
150 m 3 /min ⎛ 1 min ⎞
1
=
= 3.0 kg/s
0.832 m 3 /kg ⎜⎝ 60 s ⎟⎠
V̂1
Substituting the known quantities, the exit temperature of the air is determined to be
(15 kJ/s) − (0.2 kJ/s) = (3 kg/s)[1.00 kJ/(kg)(oC)](Tout − 17)oC
Tout = 21.9o C
9.3.28
Open, unsteady state system – the adiabatic turbine
 =0
Q
 =0
ΔU
 =0
ΔPE
 = ΔH
 + ΔKE

W
Data from interpolating in the SI steam tables
At 600ºC and 100 kPa
3697.9
0.4011
At 400ºC and 100 kPa
3278.2
3.103
Ĥ(kJ/kg)
3
V̂(m /kg)
The velocities at the inlet and outlet to the turbine are calculated from the volumetric
flow rate = m V̂=
v1 =
d2
v where d is the pipe diameter. Therefore,
4
 1V̂1
4m
=
d 2in
⎛
kg ⎞ ⎛
m3 ⎞
(4) ⎜ 2.5 ⎟ ⎜ 0.4011
s ⎠⎝
kg ⎟⎠
⎝
(3.14159)(0.1 m)
2
= 127.7
9–43
m
,
s
Solutions Chapter 9
and
v2 =
 2 V̂2
4m
m
= 158.0
2
s
d out
⎡
⎛ 1⎞
⎛ 1 ⎞⎤
ΔKE = ⎜ ⎟ (2.5) ⎢158.02 − 127.7 2 ) ⎜
⎥ kJ/s = 10.82 kJ/s
⎝ 2⎠
⎝ 100 ⎟⎠ ⎦
⎣
 = (2.5)(3278.2 − 3697.9) kJ/s = − 1049.25 kJ/s
ΔH
 = − 1049.25 + 10.82 = 1038.4 kJ/s
W
(1038.4 kW)
9.3.29
This is an unsteady-state closed process. The energy balance reduces to
ΔE = ΔU + ΔPE + ΔKE = Q + W
Q = ΔU = U2 − U
=1
ΔPE = ΔKE = 0
ΔH − Δ(pV)
=
ΔH − (p2V2 − p1V2 )
where 2 (327.8ºF) is the final state and 1 (406ºF) is the initial state
Get the data from the CD in the back of the book. The final state is saturated conditions.
ˆ and H
ˆ are in Btu/lb and V
ˆ is ft3/lb.
U
Ĥ1,L = 381.65
Ĥ1,V = 1202.2
Ĥ 2,L = 298.5
Ĥ 2,V = 1187.5
Û1,L = 380.72
Û1,V = 1116.7
Û 2,L = 298.2
Û 2,V = 1105.5
V̂1,L = 0.019
V̂1,V = 1.743
V̂2,L = 0.0177
V̂2,V = 4.433
From the steam tables, get the initial amount of steam and use as the basis:
! 1 "
100 #
$ = 57.4 lb ignoring the liquid volume. (If you do not, you have to get the
% 1.743 &
average volume) and get the final state. Let y = the fraction of liquid at 100 psia
3
3
ˆ
ˆ
V
liquid = 0.0177 ft / lb, Vvapor = 4.433 ft / lb
Basis: 1 lb water
y(0.0177) + (1 − y)(4.433) = 1.74 y = 0.6095 lb (liquid) x = 0.3905 lb(vapor)
9–44
Solutions Chapter 9
Use values of Û from the CD to calculate Q. Otherwise you have to calculate Δ(pV).
Q = [(0.6095)(298.2) + (0.3905)(1187.5)]− [(1116.7)(1)]
Q = −471
=
Btu
or Q −2.71×104 Btu
lb
9.3.30
Steps 1, 2, 3 and 4
Get the needed properties from the CO2 chart or tables (Be sure to use the same reference
value). Let m CO2 be the mass of CO2 in the cylinder at the final state, V̂ is the specific
volume, Ûfinal is the specific internal energy of the CO2 at the end of the filling, and Ĥm
is the specific enthalpy of the CO2 in the pipeline.
At 200 psia, 40°F, Ĥ ≅ 153 Btu / lb and V ≅ 0.57 ft 3 / lb
Step 5
Basis: 3 ft3 of CO2 @ 200 psia (final conditions)
System: The cylinder (open system, unsteady state)
Steps 6 and 7:
Unknowns: V̂CO2 , mCO2 , TCO2
Equations: energy balance, enthalpy chart
ˆ chart to fix the conditions in the cylinder. The energy
We need two points on the p − H
balance reduces to (W = 0, Q assumed to be 0)
ΔU = −ΔH
which gives one point:
9–45
Solutions Chapter 9
ˆ ) − m (U
ˆ
ˆ
ˆ )
mCO2 (U
−(H
final
in itial
initial ) = min (Hin ) mout
out
minitial = 0
mout = 0
m CO2 = m in
ˆ
ˆ
U
final = Hin = 153 Btu / lb
ˆ and
The other point is the pressure p = 200 psia. Unfortunately the CO2 chart is p − H
ˆ , thus requiring a trial and error solution. Assume a T, lookup V̂ , and calculate
not p − U
ˆ =H
ˆ − pV
ˆ (ignore pV
ˆ at the reference sate of –40oF).
U
When Û = 153Btu / lb , then you have determined T. If you used tables of p vs Û in a
handbook, the calculations would be easier.
Ĥ
p
V̂
Û
Assume T = 160oF.
T = 140oF.
ˆ − pV
ˆ =
Then H
ˆ − pV
ˆ =
Then H
The calculates are quite approximate but
T = 140o F
mCO2 =
3 ft 3
= 4.3 lb
0.70 ft 3 / lb
9–46
182-(200)(0.68)(0.185) = 157
178-(200)(0.70)(0.185) = 153
Solutions Chapter 9
9.3.31
This is an unsteady state process in a closed system, or a closed system with air and
surroundings with water. From the latter viewpoint the system is the air, and ΔU = Q +
W where Q is the heat transfer to the water from the air.
Q
5 lb H2 O
air
W (positive)
For the system of the air
Q + W = ΔE = ΔU + ΔP + ΔK = ΔHΔ
− (pV)
For the water:
Q = ΔH – Δ( pV ) but Δ ( pV ) is negligible for liquid water so that
T +2.3
Q = ΔH = 5
∫ C pdT = 5 Cp (T + 2.3 – T ) for constant C p .
T
(or use the steam tables)
C p = 8.0
Btu
(lb mol )(°F)
Btu 2.3°F 1 lb mol H2 O 5 lb H2 O
= 5.11 Btu
(lb mol )
18 lb H2 O
Q is removed from the air, so for the system comprised of the air, Q = –5.11 Btu.
Q = 8.0
9–47
Solutions Chapter 9
For the system of the air
W=
12, 500(ft )(lb f ) 0.0012854 Btu
= 16.07 Btu
1(ft )(lb f )
W is positive when the surroundings do work on the system of the air.
3
Basis: 3 ft air (not essential)
∆U = –5.11 Btu + 16.07 Btu = 10.96 Btu
9.3.32
To make the process a steady state open process from (1) to (3) assume the system is the
coil plus the compressor. Then
0
0 0
ˆ
ˆ
ˆ )m + Q + W]
ΔE = –Δ[(H + P + K
or
W = ∆H – Q
Basis: 1 lb water evaporated
ˆ @ (1): (0.98) (1143.3) + (0.02) (161.17) = 1123.6 Btu/lb
Calculate Δ H
From (1) to (2): Balance on the compressor
ˆ – ΔH
ˆ = 1235.2 – 1126.6 = 111.6 Btu/lb
ΔH
2
1
9–48
Qˆ = –6 Btu / lb
Solutions Chapter 9
W = 111.6 –(–6) = 117.6 Btu/lb
From (2) to (3): Balance on the coil
ˆ – ΔH
ˆ = 168 – 1235.2 = – 1067.2 Btu/lb
ΔH
3
2
a.
Q
= 1067.2 = 9.1
W 117.6
ˆ = 0 so Q
ˆ = ΔH
ˆ = -1067.2 Btu/lb
W
(Note: the problem asks for the heat transfer
out as a + value)
! 38.46 ft 3 0.98 (0.016)ft 3 0.02 " lb steam
V1 = #
+
= 37.69 ft 3/lb
$
lb
lb
1067.2
Btu
%
&
Basis: 1,000,000 Btu/hr
b.
106 Btu 1hr 1 lb steam 37.69 ft 3
= 589 ft 3 / min
hr
60 min 1067.2 Btu lb steam
9–49
Solutions Chapter 9
9.3.33
Basis: 25 kg exit fluid
Q + W = ΔH
W=0
Q = ΔH = ΔH25 –ΔH15 –ΔH10 =ΔH25 – ( 2784.4 )(15) – (82.2 )(10 )
–50 J 25 × 103 g
= –1250 kJ
g exit
⎧ ΔHˆ L = 798.5 kJ / kg
For the exit fluid ⎨
ˆ
⎩ ΔH = 2784.4 kJ / kg
ˆ ~ 10 J )
( note ΔKE
–4
Q=
V
Let x = kg vapor, 25 – x = kg liquid
–1250 =
x kg 2784.4 kJ (25-x) kg 798.5 kJ 15 kg 2784.4 kJ 10 kg 82.2 kJ
–
–
+
kg
kg
kg
kg
– 1250 = 2784.4 x + 19,963 – 798.5 x – 41,766 – 822
x = 10.76
fraction vapor =
10.76
= 0.43
25.0
9–50
Solutions Chapter 9
9.3.34
Closed unsteady state process for each stage.
Q + W = ΔU overall
Calculate the moles of gas:
a.
n=
pV (1 atm)(0.387 ft 3 )(lb mol)(o R)
=
= 1×10-3 lb mol
RT
(530o R)(0.7302)(ft 3 )(atm)
Known values
p (atm)
T (ºF)
n (lb mol)
V
A
1
170 (630ºR)
1 × 10-3
?
B
C
1
70 (530ºR)
1 × 10-3
0.387
10
70 (530ºR)
1 × 10-3
?
(Basis)
D
10
823 (1283ºR)
1 × 10-3
?
For an ideal gas, C v = 5/2 R (has to be looked up or calculated) and U and H are functions
of T only
Also, Δ(pV) = Δ (nRT)
Calculate some of the missing values
b.
VA =
(1×10−3 )(0.7302)(630)
= 0.460 ft 3
1
VC =
(1×10−3 )(0.7302)(530)
= 0.0387 ft 3
10
VD =
(1×10−3 )(0.7302)(1283)
= 0.0937ft 3
10
The work for each stage will be assumed to be ∫ pdV (ideal process)
WAB (constant pressure process)
c.
W = − ∫ pdV = − p AB (VB − VA )
=−
1 atm (0.387 − 0.460)ft 3 2.72 Btu
= 0.199 Btu
(atm)(ft 3 )
9–51
Solutions Chapter 9
WBC (isothermal process)
d.
0.0387
W = − ∫=pdV − ∫=
0.387
0.0387 dV
nRT
dV −nRT
=− ∫
0.387
V
V
= −(1×10
= −3 )(1.987(530)(ln 0.1)
0.0387
nRT [ln V ]0.387
2.42 Btu
WCD (constant pressure process)
e.
W = − ∫ pdV = − pCD (VD − V
= c ) −(10)(0.0937 − 0.0387) = − 0.550 Btu
WDA (ideal adiabatic process)
W = − ∫ pdV = − ∫
nRT
dV However, both T and V vary.
V
Instead use
Q + W = ΔU
Q=O
R = 1.987 Btu/(lb mol) (ºR)
f.
" 5R #
W = ΔU = nC v ΔT = (10−3 ) %
& ( 630 − 1283) = − 3.243 Btu
' 2 (
g.
Woverall = 0.199 + 0.891 – 0.550 – 3.243 = – 1.74 Btu
h.
ΔH = 0 (a cycle)
i.
For the overall process Q + W = ΔU. Because of the cycle, ΔU = 0, and
thus Q = -W = 1.74 Btu
9–52
Solutions Chapter 9
9.3.35
Basis: 1 lb water/steam
Steady-state, open system for each unit
The data:
ˆ H
ˆ and quality or superheats at the numbered points:
Tabulation of p, T, V,
p (psia)
T (oF)
V̂(ft 3 / lb)
qual./sup.heat
1
14.7
65
0.026
0%
2
250
65
0.016
0%
3
250
401
1.844
100%
4
250
550
2.202
149º
5
40
268
10.14
96.5%
6
0.306
65
0.016
0%
Ĥ(Btu / lb)
33.09
33.79
1201.1
1291
1137
33.05
Calculate the specific volume at point 5:
V̂ = 0.965(10.506) + 0.035(0.01715) = 10.14
(a)
Heat to boiler:
Q = ΔH = H3 − H2 = 1201.1 − 33.8
=
1167.3 Btu / lb
(b)
Heat to superheater:
Q = ΔH = H4 − H3 = 1291 − 1201.1
=
90 Btu / lb
(c)
Heat removed in condenser = H6 − H5 = 33 −1137 = -1104 Btu/lb
(d)
Work delivered by turbine: Q + W = ΔH
Q=0
(e)
W = ΔH = H5 − H4 = 1137 − 1291
=−
154Btu / lb
Work required by the pump between 1 and 2:
Q + W = ΔH
Q=0
hence W = ΔH
Because we do not have values of ΔH for compressed water at 65ºF, we
will use
W = ΔU + Δ(pV)
65o F
ΔU = ∫ o Cv dT = 0
65 F
9–53
Solutions Chapter 9
Δ(pV) = p 2 V2 − p1V1 = (p 2 − p1 )V =
Efficiency =
(250 − 14.7) 0.016 144
= 0.70 Btu / lb
778
Net work delivered
154 − 0.70
=
Total heat Supplied 1167.3 + 90
= 0.121
For a water rate of 2000 lb/hr:
hp =
154 Btu 2000 lb steam 1 hr 1.415 hp
= 121
lb
hr
3600s 1 Btu/s
You can improve the efficiency by:
(a) Exhausting the turbine at a lower pressure.
(b) Use higher boiler pressure.
(c) Use higher superheat.
9.3.36
Basis: 100 lb/day liquid C12 feed at 8ºF
Assume
(1) Adiabatic operation of all units
(2) Flow process, steady-state
R
2.5 T/day, 0oF
R' )
P
D )
100 T/day, -30oF
F
100 T/day, 8oF
First find the enthalpy lost or gained by the C12 in the heat exchanger from which we can
get the work done by compressor A on the Freon.
Overall C12 balance through the heat exchanger
F+R=P
100 + 2.5 = 102.5
9–54
Solutions Chapter 9
Energy balance for C12 through the heat exchanger
For a flow process:
Q + W = ΔH
W=0
Q = ΔH
Choose as the reference state liquid C12 at -30ºF
ΔH = H Pout − H Fin − H R in
=
8.1 Btu lb mol 102.5 lb
!$ −30 − (−30)o F"%
o
(lb mol)( F) 71 lb 100 lb F
−
8.1 Btu lb mol 100 lb
!8 − (−30)o F"%
(lb mol)(o F) 71 lb 100 lb F $
−
8.1 Btu lb mol 2.5 lb
Btu
!$0 − (−30)o F"% = 0 − 433.5 − 8.55 = −442.1
o
(lb mol)( F) 71 lb 100 lb F
100 lb F
Consequently
Q = −442.1 Btu /100 lb F
This means heat is lost by the C12 and gained by the Freon
Energy balance for the Freon flowing through compressor A:
Q + W = ΔH
But since the flow of Freon is cyclical
ΔH = 0
Q Freon = −Q
=C12
−(−442.1)
=
442.1 Btu /100 lb F
WA = QFreon = 442.1
Btu
100 lb F
Energy balance for the C12 flowing through the compressor B:
2.5 lb/day @ -30ºF → 2.5 lb/day at 0ºF
Q + W = ΔH Q = 0
ΔH = Enthalpy change across compressor = H R − H R '
9–55
Solutions Chapter 9
Other units
W = 539 Btu/mole
= 1,520,000 Btu/day
ΔH =
+
H R = CpΔT + HΔvap
2.5 lb 8.1
[0 − (−30)]
100 lb F 71
4878 cal 1.8 Btu / lb 1 lb mol 2.5 lb
g mol cal / g mol 71 lb 100 lb F
= 8.55 + 309 = 317.5
Btu
100 lb F
Total work input needed to make the process operational:
hp =
317.5 + 442.1 Btu 2000 lb 100 ton F 778(ft)(lbf )
100 lb F
ton
day
Btu
1 day
(hp)(min)
(24) × (60) min 33,000(ft)(lb f ) 0.30
= 82.9 horsepower (actual power input)
9–56
Solutions Chapter 9
9.3.37
Basis: 1 day; Reference temp = 80ºF
a. Material and Energy Balances:
Overall Material:
In = 1,000,000 lb
Out : residue: 500,000 lb
naptha:
200,000 lb
gasoline:
300,000 lb
Total out
1,000,000 lb
Tower 1:
Material
feed
reflux
total
In
1,000,000
1,500,000
2,500,000
Out
2,000,000
500,000
2,500,000
vapor
residue
total
Energy:
In: feed: ( 1×106 )(0.53)(480 − 90) 106 (100) +
+(106 )(0.45)(500 − 480)
=
6
reflux: (1.5 ×10 )(0.59)(180 − 90)
=
steam:
(by difference)
total:
315,000,000 Btu
79,700,000 Btu
114,500,000 Btu
509,700,000 Btu
Out: residue: (5 ×105 )(0.51)(480 − 90)
vapor:
(2 ×106 )(111)
liquid:
(2 ×106 )(0.59)(250 − 90)
total (Btu):
99,200,000
222,000,000
188,500,000
509,700,000
=
=
=
=
Furnace:
Material
In
1,000,000 lb
Energy
In
6
feed: (10 )(0.53)(200 − 90) = 58,300,000
furnace ht:
257,200,000
total: (Btu)
315,500,000
Condenser I:
Material
In: vapor
H2O: 304,500,000/(1)(120-70)
9–57
Out
1,000,000 lb
Out
315,500,000
=
=
2.0 ×106 lb
6.1×106 lb
Solutions Chapter 9
2.0 ×106 lb
Out: liquid:
Energy
H2 O
In: vapor: 222 ×106 + 188.5
× 106
Out: liquid: (2 ×106 )(0.59)(180 − 90)
Removed by water
Total out:
Preheater I:
Material
In:
liquid = 500 ×106 lb ; Out: vapor
Energy
In: liquid: (5 ×105 )(0.59)(180 − 90)
steam:
Total
Out: vapor:
(5 ×105 )(0.59)(250 − 90)
+(5 ×105 )(111)
+ (5 ×105 )(0.51)(200 − 250)
Tower II
Material
In:
feed = 500,000 lb Out:
reflux = 600,000 lb
Total = 1,100,000 lb
=
=
=
=
6.1×106 lb
410.5 ×106 Btu
106 ×106 Btu
304.5 ×106 Btu
=
410.5 ×106 Btu
=
500 ×106 lb
=
=
26.5 ×106 Btu
78.65 ×106 Btu
=
105.15 ×106 Btu
=
105.15 ×106 Btu
vapor =
residue =
total =
Energy
In:
feed:
reflux: (6 ×105 )(0.63)(120 − 90)
steam: (by difference)
Total: (Btu):
=
=
=
=
900,000
200,000
1,100,000
105,150,000
11,300,000
42,700,000
159,150,000
Out: vapor: (9 ×105 )(0.63)(150 − 90) + 9 105 (118) =×140.05 ×106
residue: (2 ×105 )(0.58)(255 − 90)
= 19.1× 106
= 159.15 ×106
Total (Btu):
Condenser II
Material
In:
vapor:
H2O:
Out:
liquid
= 0.9 ×106 lb
123.0.5 ×106
(1)(110 − 70)
= 3.08 ×106 lb
= 0.9 ×106 lb
9–58
Solutions Chapter 9
H2 O
= 3.08 ×106 lb
Energy
In:
(Btu)
Out: liquid: (9 ×105 )(0.63)(120 − 90)
removed by H2O:
Total (Btu):
140.05 ×106
17 ×106
123.05 ×106
140.05 ×106
=
=
=
=
Heat Exchanger II (Assume Tout = 90ºF)
Material
In
liquid:
300,000 lb
H2O: 5,670,000/(1)(80-70) = 567,000 lb
Energy
In: (300,000)(0.63)(120-90)
Out: liquid:
removed by H2O
liquid:
H2O:
= 5.67 ×106 Btu
= 0.00
= 5.67 ×106 Btu
Heat Exchanger III
Material
In = Out:
Bottoms
200,000 lb
Charge
1,000,000 lb
Energy
In:
(200,000)(0.58)(255-90)
(1 × 106)(0.53)(90-90)
=
=
19.15 ×106
0.0
Total (Btu):
Out: (2 × 105)(0.58)(26.3-90)
(1 × 106)(0.53)(140-90)
Total (Btu):
=
=
=
=
19.15 ×106
−7.35 ×106
26.5 ×106
19.15 × 106
Tout of 26.3ºF does not seem reasonable
Heat Exchanger IV
Material
In = Out
Bottoms
0.5 ×106 lb
Charge
1.0 ×106 lb
9–59
Out
300,000
567,000
Solutions Chapter 9
Energy
In:
(5 × 105)(0.51)(480-90)
(1 × 106)(0.53)(140-90)
=
=
99.3 ×106
26.5 ×106
Total (Btu):
=
125.8 ×106
Out: (5 × 105)(0.51)(257-90)
(1 × 106)(0.53)(200-90)
Total (Btu):
=
=
=
67.5 ×106
58.3 ×106
125.8 × 106
Heat load of furnace = 315.5 × 106 - 58.3 × 106 = 257.2 ×106 Btu
b.
Additional heat if charge to furnace is at 90ºF
Total heat load = 315.5 × 106 Btu
Additional heat = 315.5 × 106 – 257.2 × 106 = 58.3 ×106 Btu
9.3.38
Basis: one hour operation open, steady-state process.
A material balance initially is necessary to determine amounts and compositions of
streams.
A.
Feed
Component
C2H6
C2H4
Wt.%
65
35
lb
650
350
9–60
Solutions Chapter 9
B.
Product
Have 97% recovery of ethylene
lb C2H4 = (0.97) (350) = 340
Now, since the product is 98% C2H4
340 lb C2 H 4 2 lb C2 H6
= 6.95 lb C2 H6
98 lb C2 H 4
C.
Component
C2H6
Wt.%
2
lb
6.95
C2H4
98
340
Bottoms
lb
lb
C2H4 = 350 – 340 = 10
C2H6 = 650 – 6.95 = 643.05
Component
Wt.%
lb
C2H6
C2H4
98.47
1.53
643.05
10
We are now ready to determine, by a series of energy balances, the quantities
desired.
Energy Balance on System “A” (see Fig.)
!Energy out with Product +
"
Energy in with Feed + energy in with steam = #Energy out with bottoms +
"Energy out with refrigerant
$
We will assign an arbitrary enthalpy of zero to the liquid feed as it enters the Pre-heater at
–100ºF. Actually, any reference condition could be chosen since we are only concerned
with enthalpy changes in the balances.
------------------------------------------------------------The feed undergoes at 20ºF rise before it enters the still.
T = -80ºF and
Cp feed = (0.65) (0.65) + (0.35) (0.55) = 0.62 Btu/(lb) (ºF)
9–61
Solutions Chapter 9
Enthalpy in with feed = (lb feed) (CpΔT) = (1000) (0.62) (-80-(-100))
= 12,400 Btu/hr
------------------------------------------------------------Enthalpy in with steam = (lb steam) (ΔHv) at 30 psig,
ΔHv = 945 Btu/lb
Enthalpy in with steam = (S) (945) btu/hr
------------------------------------------------------------From the data given, the temperature of the saturated liquid product is =30ºF since it is
essentially pure ethylene.
Cp = 0.55 Btu/(lb) (ºF)
Enthalpy out with product = (346.95) (0.55) [-30-(-100)] = 13,360 Btu/hr
The bottoms are practically pure ethane so, T = 10ºF,
Cp = 0.65 Btu/(lb) (ºF)
Enthalpy out with bottoms = (653.05) (0.65) [10-(-100)] = 46,700 Btu/hr
------------------------------------------------------------The refrigerant experiences a temperature rise of 25°F as it passes through the condenser.
The heat capacity may be assumed to be 1.0 Btu/(lb) (°F)
Enthalpy out with refrigerant = (lb refrig.) (1.0) (25) m Btu/hr
Now, rewriting the energy balance “A” with a substitution of known terms:
12,400 + 945S = 13,360 + 46,700 + 25m
(1)
It is evident from this first balance that a second balance will be necessary to determine
the unknown quantities.
Energy Balance “B”
Energy in with Overhead =
Energy out with refrigerant +
Energy out with Product +
Energy out with Reflux
Since we are using a reflux ration of 6.1
lb reflux = (6.1) (346.95) = 2,120 lb/hr.
Now, the specific enthalpy of the reflux = the specific enthalpy of the product, since they
9–62
Solutions Chapter 9
are from the same stream, and the differences in enthalpy between the overhead and the
product, or overhead and reflux, is merely the heat of vaporization of ethylene
(neglecting the small amount of C2H6 present), therefore, we can rewrite Energy Balance
“B” as:
(lb Overhead) (ΔHv) = 25 m
(2,120 + 346.95) (135) = 25 m
or m = 13,320 lb/hr.
Now substituting this value into equation (1) we have
12,400 + 945 S = 13,360 + 46,700 + (25) (13,320)
so that S = 403 lb / hr
a.
403 lb steam
hr
= 0.403 lb steam / lb feed
hr
1000 lb feed
b.
13,320 lb refrigerant 1 ft 3 refrig. 7.48 gallons
= 1,993 gallons of refrigerant/hr
hr
50 lb refrig.
ft 3
The bottoms leave the still and enter the Pre-heater at approximately 10ºF, the saturation
temperature of pure C2H6. The final temperature of the bottoms as it leaves the Preheater can be calculated from a simple energy balance around the Pre-heater.
c.
Enthalpy Balance “C”
(lb bottoms) [0.65 Btu/(lb) (°F)] (10 – Tf) = (lb feed)
×[0.62 Btu /(lb)(o F)][−80 − (−100)]
(653.05) (0.54) (10 – Tf) = (1000) (0.62) (20)
Tf = 19.2o F
9–63
Solutions Chapter 10
10.1.1
(a)
10.1.2
(b), (d), (f)
10.1.3
(c)
10.1.4
ˆ ° of Fe O is calculated as follows.
ΔH
f
2 3
(1)
ˆ ° – Σn ΔH
ˆ°
ΔHRxn = Σni ΔH
fi
i
fi
products
reactants
°
ˆ ° (2 ) – ΔHˆ ° ( 3 )
–822.200 kJ = ΔHˆ fFe2 O 3 – Δ H
fFe
fO 2 2
ˆ° –0–0
= ΔH
fFe O
2 3
(2)
ΔHRxn = –284.100 kJ
ˆ ° – ΔH
ˆ ° (2) – ΔH
ˆ ° (1 )
= ΔH
fFe O
fFeO
fO 2
2 3
2
ˆ ° (2) – 0( 1 )
–284.100 = –822.200 – Δ H
fFeO
2
ˆ
ΔH
fFeO =
°
–538.10
kJ = –269.05 kJ
2
vs – 267 in Appendix F
10–1
Solutions Chapter 10
10.1.5
ΔH°(kJ)
+123.8
+2220.0
C3H8 (g) → C3H6 (g) + H 2 (g)
3CO2 (g) + 4H2O(1) → C3H8 (g) + 5O2 (g)
4 × (−285.8)
4[H 2 (g) + 1/ 2O2 (g) → H 2O(1)]
3 × (−393.4)
3[C(s) + O2 (g) → CO2 (g)]
________________________________________________
+20.1 kJ/g mol
3C(s) + 3H2 (g) → C3H6 (g)
ΔHof of C3H6 = 20.1 kJ/g mol
10.1.6
Basis: 1 g mol C5H2
ΔH (kJ/g mol)
5 CO2 (g) + H 2O(l )
→
add : 5[C(s) + O 2 (g)] →
1
1
C5 H 2 (s)+ 5 O 2 (g)
2110.5
5[CO 2 (g)]
5[-394.1]
2
add: H 2 (g)+ O 2 (g)
→
H 2O(g)
-241.826
add: H 2O(g)
→
H 2 O(l )
-43.911
net: 5C(s)+H 2 (g)
→
C5 H 2 (s)
ΔH o f = -143.737 kJ/g mol
2
10–2
Solutions Chapter 10
10.2.1
a.
Basis: 1 g mol NH3 ( g)
NH3 ( g)
ˆ °(kJ / g mol ) :
ΔH
f
°
ΔHrxn =
b.
HCl( g)
+
–46.191
ˆ –
∑ ni ΔH
i
products
→
–92.311
–315.4
ˆ
∑ ni ΔH
i
reac tants
=
(1)( –315.4) – [(1)( –46.191) + (1)( –92.311)]
=
–176.9 kJ / g mol NH3 ( g)
Basis: 1 g mol CH4 (g )
CH4 (g ) + 2 O2 (g ) → CO2 (g)
ˆ (kJ / g mol ) :
ΔH
f
°
ΔHrxn =
c.
NH4 Cl (s)
–74.84
ˆ
0
+ 2 H 2 O ( )
–393.51
–285.840
ˆ
∑ ni ΔHi – ∑ ni ΔHi
products
reac tants
=
!#( 2)( –285.840) + (1)( –393.51)"$ – !#(1)( –74.84) + 0"$
=
–890.4 kJ / g mol CH4 ( g )
Basis: 1 g mol C6 H 2 (g )
C6 H1 2(g )
ˆ °(kJ / g mol ) :
ΔH
f
–123.1
→
C6 H 6 ( )
+48.66
+
3H2 (g)
0
ΔH°rxn = [ 0 + (1)(+48.66)] – [(1)(–123.1)] = 171.76 kJ / g mol C 6 H12 (g)
10–3
Solutions Chapter 10
10.2.2
a. Basis: 1 g mol CO2 (g)
CO2 (g) + H2 (g) → CO (g) + H2O (1)
ΔH°rxn = (– 110.52 – 285.84) – (– 393.51) = –2.849 kJ
b.
Basis: 1 g mol CaO (s)
ΔH°rxn = (–986.59 – 924.66) = [(–635.55 – 601.83) - 2 (285.84)] =
−102.19 kJ per mol CaO(s)
Multiply by 2 for the reaction listed −204.38 kJ
c.
Basis: 1 g mol Na2SO4 (s)
ΔH°rxn = (–1090.35 – 110.54) – (–1384.49) = 183.59 kJ
d.
Basis: 1 g mol NaCl (s)
ΔH°rxn = (–92.31 – 1126.33) – (–811.32 – 411.01) = 3.67 kJ
e.
Basis: 1 g mol O2
ΔH°rxn = [2 (–1384.5) + 4 (–92.31)] – [(–411.00) + 2 (–296.90)
+ 2 (–285.84)] = –1561.76
f.
Basis: 1 g mol SO2 (g)
ΔH°rxn = (–811.32) – (–285.84 – 296.90) = –228.58 kJ
g.
Basis: 1 g mol N2
ΔHºrxn = 2 (90.374) = 180.75 kJ
h.
Basis: 1 g mol Na2CO3 (s)
ΔH°rxn = [3 (–1117.13) + (–393.51)] – [–1130.94 + 2 (–373.21)
+ 4 (–296.90)] = –1867.54 kJ
10–4
Solutions Chapter 10
i.
Basis: 1 g mol CS2 (1)
ΔH°rxn = (–1394.69 – 60.25) – (+87.86) = –287.61 kJ
j.
Basis: 1 g mol C2H2 (g)
ΔH°rxn = (+105.02) – (52.28 – 92.31) = –145.05 kJ
k.
Basis: 1 g mol CH3OH (g)
ΔH°rxn = (–115.90 – 241.82) – (–201.25) = –156.48 kJ
l.
Basis: 1 g mol C2H2 (g)
ΔH°rxn = (–140.7) – (226.75 – 285.84) = −81.61 kJ
m.
Basis: 1 g mol C4H10 (g)
ΔH°rxn = (52.28 – 84.67) – (–124.73) = 92.34 kJ
10.2.3
Basis: 1 g mol Fe S2
ΔH o rxn = 567.4
− kJ / g mol FeS2
( − 567.4) kJ 1 g mol FeS2 1000 g
kJ
= −4728.3
g mol FeS2 120 g FeS2 1 kg
kg FeS2
Note: The information about conversion does not affect the value of the standard heat of
reaction. It will affect the values in the energy balance.
10–5
Solutions Chapter 10
10.2.4
G+E →
← E+F
ˆ
ΔH
Rxn is at 25°C, 1 atm
°
ˆ ° = ΣΔ H
ˆ ° – ΣΔH
ˆ ° = 1.040 × 10 9 – 0.990 × 109 J / g mol G
ΔH
Rxn
f
f
prod.
react.
(
)
6
= 50 × 10 J / g mol G converted
Since 0.48 reacts, ΔH°Rxn = 0.48( 50) × 10 6 J / g mol G
6
= 24 × 10 J / g mol G fed (not asked for)
10.2.5
No. The value reported is the heat transfer for constant volume, Qv. The heating value at
constant pressure should be reported for the standard heat of reaction (an enthalpy
change). Assume the reaction is at 25ºC.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l )
Basis: 1 g mol CH4
Q = ΔU = ΔH – Δ(pV)
If the gases can be treated as ideal Δ(pV) = Δ(nRT) = Δn (RT).
Δn = 1 – 2 – 1 = -2
The correction is
ΔnRΤ = -2 (8.314) (298) = -4958 J/g mol
In SI units of 273.15 K and 101.3 kPa, the volume is 22.415 m3 g mol.
The correction is (ignoring the pressure change from 1000 kPa to p2)
−4958
= 221.2J / m3 or 0.22 kJ / m3
22.415
Otherwise you have to calculate p2 V − p1V where p2 = 1000 kPa and p1 = p*H2O + pCO2 .
The value to be reported should be 39.97 – 0.22 = 39.75 kJ / m3
10–6
Solutions Chapter 10
10.2.6
Yes. Calculate the sensible heats using the tables and using a common reference
temperature.
10.2.7
a.
If the exit temperature is > 25oC, and the dilutent is at the entering temperature of
the reactants, then less heat has to be removed from the process.
b.
Remains the same as the presumption is in calculating ΔHºrxn that stoichiometric
quantities react to completion.
c.
ˆ o = ΔH
ˆ o with H O(g) = 241.826
At 25ºC, ΔH
−
kJ / g mol H2O.
rxn
f
2
At 500 K, the sensible heats have to be considered, and ½ O2 and H2 together have
a larger ΔH than does H2O so that ΔHrxn at 500 K is less than ΔHrxn at 25ºC.
10.2.8
S (s) + O2 (g)
SO2 (g)
ΔHRxn
600K (327oC)
ΔHReact.
ΔHProd.
298K (25oC)
ΔHoRxn
Basis: 1 g mol S(s) =1 g mol SO2 (g)
ΔHrxn, 600 K = ΔHorxn + ∑ ΔHprod, 600 K − ∑ ΔHreact, 600 K
ΔHo rxn, 298 = ∑ ΔHo f − ∑ ΔHo f
prod
react
= [−296.90) − [0 + 0] = −296.90 kJ / g mol S
= − 296,900 J/g mol S
∑ ΔH
prod,600 K
10–7
Solutions Chapter 10
ΔHSO2 ,600 K = ∫
327
25
(38.91 + 3.904 ×10−2 T − 3.104 ×10−5 T 2 + 8.606 10×−9 T3 )dT
= 13,490 J/g mol
∑ ΔH
(13,530 from Enthalpy Tables)
react,600 K
This assumes s is really a liquid at 600ºC, not a solid as in the equation
ΔHS,600K = ΔH 298−386 + ΔH fusion + H 386−600
Δ
solid
=∫
386
298
liquid
(15.2 + 2.68 ×10−2 T)dT + 10, 000 + 1/ 2∫
600
386
(35.90 + 1.26 10−3×T)dT
½ (S from Perry)
2
= 2144 + 10,000 + 3975
= 16,119 J/g mol
ΔHO2 ,600K = ∫
327
25
(29.10 + 1.158 ×10−2 T − 0.6076 ×10−5 T 2 + 1.311 10
× −9 T3 )dT
= 9,340 J/g mol (8899 from the enthalpy tables)
ΔH rxn,600K = −296,900 + (13, 490 − 25, 429)
= 308,840 J / g mol S
10–8
Solutions Chapter 10
10.2.9
C3H8 (g) → C2 H2 (g) + CH4 (g) + H 2 (g)
Basis: 1 g mol C3H8
ΔH500 = ΔH25 + ∑ ΔHprod − ∑ ΔHreact
ΔH o 25 = 226.75 + (−74.84) + 0 − (−103.85)
= 255.76 kJ / g mol
ΔH = ∫ CpdT = aT + b / 2 T2 + c / 3 T3 + d / 4 T4
ΔHC3H8 = 68.032T + 1.130 ×10−1 T2 − 4.37 ×10−5 T3 + 7.928× 10−9 T4 500
25
= 55.53 kJ / g mol
ΔHC2H2 = 42.43 T + 3.027 ×10−2 T2 − 1.678 ×10−5 T3 + 4.55 ×10−9 T4 500
25
= 25.89 kJ / g mol
ΔHCH4 = 34.31 T + 2.735 ×10−2 T2 + 1.220 ×10−6 T3 + 2.75 ×10−9 T4 500
25
= 23.10 kJ / g mol
ΔHH2 = 28.84 T + 3.825 ×10−5 T2 + 1.096 ×10−6 T3 + 2.175
× 10−10 T4 500
25
= 13.83 kJ / g mol
ΔH500 = (255.76 + 25.89 + 23.10 + 13.83) − 55.53=
10–9
263.05 kJ / g mol
Solutions Chapter 10
10.2.10
All data is from Tables E.1 and F.1 of the Appendix.
a.
o
CH 3OH(g) + 1/ 2O 2 (g) 200
C H 2CO(g) + H 2O(g)
Basis: 1 g mol CH3OH
Comp.
Cp (J /(g mol)(K or oC))
T
ΔĤof (kJ / g mol)
CH3OH(g)
42.39 + 8.301× 10−2 T − 1.87 × 10−5 T 2 − 8.03 × 10−9 T 3
ºC
-201.25
O2(g)
29.10 + 1.158 × 10−2 T − 0.6076 × 10−5 T 2 + 1.311 ×10−9 T 3 ºC
H2CO(g)
34.28 + 4.268 × 10−2 T − 8.694 ×10−9 T 3
ºC
-115.89
H2O(g)
33.46 + 0.6880 × 10−2 T +0.7604 × 10−5 T 2 − 3.593 × 10−9 T 3 ºC
-241.826
0
25ºC = 298 K
200ºC = 473 K
ΔHo rxn ,25o C = [−115.89 + (−241.826)] − (−201.25) = 156.47 −kJ / g mol
= −156, 470 J / g mol
Products
Add the Cp equations for the two products and integrate (each involves 1 mole).
∑ ΔH prod,200o C = ∫
200
25
(67.74 + 4.95 ×10−2 T + 0.7604 ×10−5 T 2 − 1.228 ×10−8 T 3 )dT
= 12,850 J/g mol
Reactants
ΔHCH3OH = ∫
200
25
(42.39 + 8.301×10−2 T − 1.87 ×10−5 T 2 − 8.03 ×10−9 T3 )dT
= 9000 J/g mol
10–10
Solutions Chapter 10
ΔHO2 =
1 200
(29.10 + 1.158 ×10−2 T − 0.6076 ×10 −5 T 2 + 1.311× 10 −9 T 3 )dT
∫
25
2
= 2,650
ΔHrxn,200o C = −156, 470 + 12,850 − 2,650
146,
= 270 J / g mol
o
SO 2 (g) + 1/ 2 O 2 (g)300
C SO3 (g)
b.
Basis: 1 g mol SO2
DATA:
Comp.
Cp (J /(g mol)(K or oC))
O2(g)
29.10 + 1.158 × 10−2 T − 0.6076 × 10−5 T 2 + 1.311 ×10−9 T 3 ºC
0
SO2(g)
38.91 + 3.904 × 10−2 T − 3.104 × 10−5 T 2 + 8.606 × 10−9 T 3 ºC
-296.90
SO3
48.50 + 9.188 × 10−2 T − 8.540 ×10 −5 T 2 + 32.40 ×10 −9 T 3 ºC
-395.18
T
ΔĤof (kJ / g mol)
ΔHo rxn ,25o C = −395.18 + 296.90 = −98.28 kJ / g mol = 98, 280 −J / g mol
Products
∑ ΔH prod,300o C = ∫
300
25
(48.50 + 9.188 ×10−2 T − 8.540 ×10−5 T 2 + ×
32.40 10−9 T3 )dT
= 16, 740 J/g mol
Reactants
ΔHSO2 = ∫
300
25
(38.91 + 3.904 ×10−2 T − 3.104 ×10−5 T 2 + 8.606× 10−9 T3 )dT
= 12, 180 J/g mol
ΔH O2 = 8, 470 J / g mol
1
ΔH rxn,300o C = −98, 280 + 16, 740 − 12,180 − (8, 470) = 97,960J
− / g mol SO2
2
10–11
Solutions Chapter 10
10.2.11
Basis: 1 g mol yeast
The heat of formation is
3.92 C + 6.5 H + 1.94 O2 → C3.92 H6.5 O1.94
The heat of combustion is for the reaction
C3.92 H6.5 O1.94 + 5.60 O2 → 3.92 CO2 (g) + 3.25 H2 O(l )
ˆ 0 − ∑ n ΔH
ˆ0 "
−1,518 kJ = − !& ∑ n i ΔH
f ,i
i
f ,i '
product
reactants
ˆ0
"
−1,518 = − !%(3.92)(−393.51) + (3.25)(−285.840) − 5.60(0) − (1)ΔH
f ,yeast &
ΔĤ 0f,yeast = 963.54
− kJ/g mol
The molecular weight of C3.92 H 6.5O1.94 is 84.63. Per 100 g yeast
ΔĤ0f,yeast = 1139−kJ/100 g yeast
10.2.12
Basis: 100 g mol CO
Assume pressure is 1 atm, and an open, steady state process.
CO +
1
2
O 2 → CO 2
The energy balance reduces to Q = ΔH. The reference temperature is 25°C.
100 g mol
CO
CO2
400oC
300oC
100 g mol O2
100oC
10–12
Solutions Chapter 10
Data:
Sensible heats
ˆ
from CD : ΔH(kJ
/ g mol)* ΔH(kJ)
mol
o
T( C)
ΔĤ (kJ / g mol)
O2
CO
100
100
100
300
-110.541
2.235
8.294
223.5
-10,225
Out
O2
CO2
50
100
400
400
-393.505
11.715
16.245
585.8
-37,726
Comp.
o
f
In
*You can calculate ΔĤsensible using the tables in the appendix, but the process requires
interpolation.
Q = ΔH = (−37,726 + 585.8) − (−10, 225 + 223.5) = − 27,139 kJ
10.2.13
The problem requires more information to solve. To make the solution easier but
approximate, assume the reaction takes place at 25°C and 1 atm. Assume that the energy
balance reduces to Q = ∆H, and that Q represents the desired kJ. Data:
°
ˆ°
ΔH
f , H 2 O = –285.840 ΔĤ f, CO2 = –393.51 , both in kJ/g mol. Ref. temp. = 25°C. Basis: 1
g mol glucose. MW of glucose = 180.
(C6 H12O6 )glucose + 6O2 (g) → 6H 2O(l ) + 6CO2 (g)
ΔH° =ΔH°products –ΔH°reactants
= [6( –285.840) + 6(–393.51)] – [1( –1260) + 6( 0)] = –2816 kJ
0.90 ( –2816 )
= 14.0 kJ/g glucose
180
At 37°C (body temperature), assume only the O2 is used in a reaction
6 g mol O2 22.4L at SC 310K 1 g mol suc.
= 0.85 L/g at 37o C and 1 atm
1 g mol suc. g mol O2 273K 180g suc.
10–13
Solutions Chapter 10
10.2.14
a.
Presumably H2, O2, C1, HC1, and H2 are gases.
(1)
6 g mol FeC13(-403.34 kJ/g mol FeC13) + 9 g mol H2O (-241.826 kJ/g mol H2O)
-3 g mol Fe2O3 (-822.156 kJ/g mol Fe2O3)-18 g mol HC1(-92.312 kJ/g mol HC1)
= −468.39 kJ
(2)
6 g mol FeC12(-342.67 kJ/g mol FeC12) + 3 g mol C12 (0.0 kJ/g mol C12)
-6 g mol FeC13 (-403.34 kJ/g mol FeC13) = 364.02 kJ
(3)
2 g mol Fe3O4 (-1116.7 kJ/g mol Fe3O4) + 12 g mol HC1 (-92.312 kJ/g mol HC1)
+ 2 g mol H2 (0.0 kJ/g mol H2) – 6 g mol FeC12 (-342.67 kJ/g mol FeC12)
-8 g mol H2O (-241.826 kJ/g mol H2O) = 649.48 kJ
(4)
3 g mol Fe2O3 (-822.156 kJ/g mol Fe2O3) – 2 g mol Fe3O4 (-1116.7 kJ/g mol
Fe3O4) – ½ g mol O2 (0.0 kJ/g mol O2) = −233.07 kJ
(5)
6 g mol HC1 (-92.312 kJ/g mol HC1) + 3/2 g mol O2 (0.0 kJ/g mol O2)
-3 g mol H2O (-241.826 kJ/g mol H2O) – 3 C12 (0.0 kJ/g mol C12) = 171.61 kJ
b.
2H2O
2H2 + O2
∑ ΔH
o
n,i
= 483.65 kJ
10–14
Solutions Chapter 10
10.2.15
Basis: 1 g gasoline
ΔH =
40 kJ 0.84
= 33.6 kJ / cm3
3
g cm
Methanol:
Assume the reaction is:
ΔĤof (kJ / g mol)
1
CH3OH(l ) + 1
2
O2 (g) → CO2 (g) 2H
+ 2O(g)
0
− 238.64
-393.51
-241.826
ΔHRxn = 638.52
− kJ / g mol
638.52 kJ 1 g mol 0.792 g
= 15.79 kJ / cm3
3
g mol 32.04 g cm
33.6 kJ cm3 methanol
cm3 methanol
=
2.13
cm3 gas 15.79 kJ
cm3 gas
a. (2.13-1) 100 = 113% larg er tan k
Ethanol:
C2 H5OH(l ) + O2 (g) → 2CO2 (g)
+ 3H2O(g)
ΔĤof (kJ / g mol)
− 277.63
0
-393.51 -241.828
ΔHRxn = 1235
− kJ / g mol
1235
1 g mol 0.789 g
= 21.15 kJ / cm3
g mol EtOH 46.07 g cm3
33.6 kJ cm3 EtOH
= 1.59
cm3 gas 21.15 kJ
b. (1.59) 100 -100 = 59% larg er
10–15
Solutions Chapter 10
10.2.16
Basis: 1 g mol Cu and 1 g mol H2SO4
Data from Perry
Cu(s) + H2SO4 (l ) → H2 (g) CuSO
+ 4 (s)
ΔH o rxn = (0 − 772.78) − (0 − 810.40)
= 37.61 kJ
For a constant volume process:
Q = ΔU = ΔH – Δ(pV) = ΔH –Δn(RT)
Δn = 1
Q = 37.61 – (1) (8.314 × 10-3) (298) = 35.13 kJ/g mol
Basis: 1 lb mol of each reactant
(37.13)(1000)J 454 g 1 Btu
= 15,978 Btu / lb mol
g mol
1 lb 1055 J
10.2.17
Assume steady state flow process with reaction.
N2(g)
H2(g)
C6H12(g)
C6H6(g)
Q
Ref. temp. = 25°C
C6 H12 (g) → C6 H6 (g) + 3H2 (g)
The standard heat of reactions is
ΔĤoRxn = [1(82,927) + 3(0)] − [1(−123,100)] = 2.06 ×105 J / g mol C6 H12
Material balance:
Step 5:
Basis: 1 g mol C6H12 (g) in
Steps 6 to 9:
10–16
Solutions Chapter 10
Make balances for each species so that the unknowns H2 and C6H6 can be calculated.
Results:
n H2
out
=
0.70(1)(3)
=
n C6 H 6
out
=
0.70(1)(1)
= 0.70
n C6H12
out
=
0.30(1)
= 0.30
out
=
0.50
n N2
Energy balance
Out:
In:
Component
ˆ o f (J/gmol)
ΔH
H2(g)
C6H6(g)
C6H12(g)
N2(g)
0
+ 82.927 × 103
- 123.1 × 103
0
C6H12(g)
N2 (g)
-123.1 × 103
0
(a)
2.10
n C6H12 in = 1.0
n N2 in
T
ni
ΔH 300
(kJ)
25
0
0
0
0
2.10
0.70
0.30
0.50
0
0
1.00
0.50
16.865
71.226
-47.552
4.016
44.556
-123.1
0
-123.1
∫ C dT(or use tables)*; T = 300°C
25
Q = 1.44 × 105 J / gmol
p
8031
18,824
35,408
8,031
or
0
0
or
(b)
= 0.50
Q = + 1.68 × 105 J / gmol
*at 25°C exit temp. and entrance temp. these terms are 0.
10–17
Solutions Chapter 10
10.2.18
At 25°C and 1 atm H2O is a liquid.
1
H 2 (g) + O2 (g) → H 2 O(l )
2
o
ΔH c (kJ/g mol): − 285.84
0
0
ΔHRxn25oC
ΔHReact
ΔHProd
ΔHRxn0oC
Basis: 1 g mol H2 (g)
ΔHo rxn = −[∑ ΔHo c,Prod − ∑ ΔHo c,React ]
1
= −[(1)(0) − (1)(−285.84) − (= −
)(0)]
285.84 kJ / g mol H 2
2
H2O at 25ºC is a liquid
ΔH Pr od = ∫
0o C
Cp dT =
o
25 C
ΔH React = (1) ∫
0o C
4.184 J 18 g (−25o C)
= 1.883
− kJ
(g)(o C) 1 g mol
1
0o C
2
25
CpH2 (g) dT + ( ) ∫
o
25 C
1
CpO2 (g) dT = −(718)(1) − (727)( ) = 1.081
− kJ
2
ΔHrxn (0o C) = (−1.883) − (−1.081) + (−285.84) = 286.64 kJ
− / g mol H2
10–18
Solutions Chapter 10
10.2.19
Basis: 1 g mol dry cells
The heat of reaction is
ˆ 0 − ∑ n ΔH
ˆ0 "
ΔH orxn = − !& ∑ n i ΔH
i,c
i
i,c '
products
reactants
= −[(1)(−1,517) + 2.75(−393.51) + 3.42(
−
285.840)
−6.67(−2,817) − 2.10(0)] = 15, −
213 kJ/g mol cells
For 100 g dry cells
-15,213 kJ
1g mol
100 g dry cells
= -1.800×104 kJ
g mol dry cells 84.58 g dry cells
10.3.1
Steady State, open process. Ref T = 25ºC
Basis: 100 g mol feed
The moles out come from the material balance, and the other data from the CD.
Q = ΔH
ΔH data from the CD.
Comp.
% = g mol
T(ºC)
ΔĤof (kJ / g mol)
IN with gas
10–19
ΔĤsensible (kJ)
ΔH(kJ)
Solutions Chapter 10
CO2
6.4
500
-393.505
20.996
-2384.06
O2
0.2
500
-
15.034
3.04
CO
40.0
500
-110.541
14.690
-3834.04
H2
50.8
500
-
13.834
702.77
N2
2.6
500
-
14.242
37.03
100.0
-5,475.29
In with air
O2
63.28
25
-
0
0
N2
240.65
25
-
0
0
303.93
Out with fg
CO2
46.4
720
-393.505
32.006
-16,773.55
H2O(g)
50.8
720
-241.835
26.133
-10,957.66
N2
240.65
720
-
21.242
5,111.89
O2
18.1
720
-
22.555
408.25
356.0
-22,211.08
Q = ΔH = (−22, 211.08) − (−5, 475.29) = − 16,736 kJ
(Heat leaving)
10–20
Solutions Chapter 10
10.3.2
Steady state, open process. Ref T = 25°C
Basis: 100 g mol P
W
H2O
coke F 40oC
Furnace
P 1100oC
CO2
CO
O2
N2
mol %
16.1
0.8
4.3
78.8
1000
R Refuse 200oC
mass fr.
C
0.90
H
0.06
inert 0.04
1.00
A
Air 40oC
mol fr.
O2 0.21
N2 0.79
1.00
C
inert
mass fr.
0.10
0.90
1.00
Data come from the material balances and the CD.
C p (J / g mol)( o C)
H 20.8
C 14.3
Comp. Percent (g)
g mol
T(ºC) ΔĤof (kJ / g mol)
ΔĤsensible (kJ / gmol) ΔH(kJ)
In F (214.8 g) solid
C
90.0
16.11
40
-
0.214
3.456
H
6.0
12.79
40
-
0.312
3.991
Inert
4.0
-
40
-
ignore
-
100.0
28.90
7.447
In air (99.7 g mol)
O2
20.94
40
-
0.442
9.255
N2
78.76
40
-
0.436
34.339
99.70
43.595
10–21
Solutions Chapter 10
Out in P (g)
CO2
16.1
1100
-393.505
24.744
-5937.05
CO
0.8
1100
-110.541
25.032
-68.41
O2
4.3
1100
-
26.209
112.70
N2
78.8
1100
-
38.891
3064.59
Water out (g)
100.0
6.80 1100
30.174
–2828.17
205.18
Refuse
10 g
200
-241.826
-
1.488 per g
14.88
Q = ΔH = (−2828.17 + 205.18 + 14.88) − (7.447 + 43.595) = − 2659.2 kJ
(heat exiting)
10.4.1
If the fuel cannot produce H 2O as a product, HHV = LHV, but the concept really refers
to cases in which water is produced so that HHV ≠ LHV.
10.4.2
(a)
0.005⎞
⎛
+ 4050(0.006)
HHV = 14,544 (.80) + 62,028 ⎝ 0.0003 –
8 ⎠
= 11,640 Btu/lb
LHV = 11,800 – 91.23(0.3) = 11,770 Btu/lb
(b)
HHV = 17,887 + 57.5(30) – 102.2 (0.5) = 19,560 Btu/lb
LHV = 19,560 – 91.23 (12.05) = 18,460 Btu/lb
10–22
Solutions Chapter 10
10.4.3
Basis: 100 lb mol gas
The reaction for each compound is the stoichiometric amount of O2 going to CO2 gas and
H2 O ( l )
Comp.
lb mol
CO2
C2 H4
CO
H2
CH4
N2
9.2
0.4
20.9
15.6
1.9
52.0
100.0
ΔĤ(Btu/lb mol)
ΔH (Btu)
0
598,000
122,000
123,000
383,000
0
0
239,000
2,544,000
1,918,000
728,000
0
5,429,000
Note: NGI standard state is 60°F
Btu 5,429,000 Btu
1 lb mol 520°R
3 =
100 lb mol 359 ft 3 492°R
ft
3
= 143 Btu / ft at std. conditions of NGI
10–23
Solutions Chapter 10
10.4.4
Basis: 100 mol of gas
(a1)
Comp.
mol.wt.
CH4
C2 H6
C3H8
C4H10
Total
16
30
44
58
Mol%=
Vol.%
lb
88
6
4
2
100
1408
180
176
116
1880
–ΔH°c1
kJ/g mol
–ΔH°c2
kJ/g mol
890.35
1559.88
2220.05
2878.52
802.32
1427.84
2044.00
2658.45
mol. wt. mixture = 18.80 lb/lb mol
−ΔHo c1 =H2O(1) to CO2 (g) products;
− ΔHo c2 = H 2O (g) + CO2 (g) products
ΔHº1 rxn = – [ΔHc prod – ΔHºc react] = 1022.90 kJ/g mol
(a1)
ΔH o rxn =
−1022.90 454 g mol Btu
lb mol
Btu
= − 23,600
g mol
lb mol 1.055 kJ 18.80 lb
lb
HHV = 23,600 Btu/lb
(a2)
ΔHºrxn = (–23,600) (18.80) = -439,000 Btu / lb mol
HHV = 439,000 Btu/lb mol
(a3)
ΔHo rxn =
−439,000 Btu
1 lb mol
= − 1160
3
lb mol
379.2 ft at 60o F, 760 mm Hg
HHV
3
= 1160 Btu / ft at 60°F, 760 mm Hg
(b1) ΔH°2 rxn = –925.71 kJ/g mol
ΔH°2 rxn =
–925.71 454
= –21,200 Btu/lb
1.055 18.80
LHV = 21,200 Btu/lb
(b2) ΔH°2 rxn = (–21,200) (18.80) = -398,000 Btu/lb mol
10–24
Solutions Chapter 10
LHV = 398, 000 Btu / lb mol
(b3) ΔH°2 rxn = (–398,000)/(379.2) = -1051
LHV = 1051 Btu/ft 3 at 60°F and 760 mm Hg
10.4.5
Write the equation for the combustion for 1 gram as:
1 g of Fuel or food + Oxygen → CO2 (g) + H2O(1) + N2 (g)
Both the reactants and the products are at 25ºC and 1 atm. The chemical energy released
by the complete combustion of 1 g of a food or fuel according to the above equation at
standard conditions is the heat of combustion. Assume the ΔHc are additive. Then
ΔHC = (17.1)(28) + (39.5)(10) + (14)(4) = 930 kJ
The High Energy bar has (capital C for calorie)
220 Calories 1000 calories 4.184 J 1 kJ
= 921 kJ
1 Calorie 1 calorie 1000 J
The values are close enough.
10.4.6
When the combustion results in a single component.
10–25
Solutions Chapter 10
10.4.7
Basis: 1 lb mixture
Get the data for the ΔHc in Btu/lb, not Btu/mol.
Component
Mass fr., ωi
ΔHc (Btu / lb)
ΔH(Btu)
Basis
1200 lb
Hexachloroethane, C2C16
Tetrachloroethene, C2C14
Chlorobenzene, C6H5C1
Toluene, C7H8
0.0487
0.0503
0.2952
0.6058
828
2141
11,876
18,246
40
108
3,506
11,053
58.44
60.36
354,24
726.96
Total
1.000
14,707
1200.00
_______________________________________________________________________
The Btu/lb are 14, 707 from the heat of combustion data, a deviation of 3.2% from
15,200. For DuLong’s equation, the component elements are needed.
Basis: 1200 lb
_______________________________________________________________________
C
H
C1
O
MW Total (lb)
_______________________________________________________________________
C2C16
24
0
213
0
237.00
ωi
0.10
0.00
0.90
0.00
lb
5.84
0.00
52.60
0.00
C2C14
24
0
142
0
ωi
0.14
0.00
0.86
0.00
lb
8.45
0.00
51.91
0.00
C6H5C1
72
5
35.5
0
10–26
58.44
166.00
60.36
112.50
Solutions Chapter 10
ωi
0.64
0.04
0.32
0.00
lb
226.71
14.17
113.36
0.00
C7H8
84
8
0
0
ωi
0.91
0.09
0.00
0.00
lb
661.53
65.43
0.00
0.00
726.91
Total
902.53
79.60
217.87
0
1200
ωi
0.75
0.07
0.18
0
1.00
354.24
9200
HHV = 14,455 C + 62,028 H – 7753.5O2 + 4050 S
= (14,455)(0.75) + (62,028)(0.07)
= 15,183 Btu/lb (the higher heating value) a deviation of 0.11%
10.4.8
The reactions are:
C2 H5OH(1) + 3 O2 (g) → 2 CO2 (g)
+ 3 H 2O(1)
(1)
C8H18 (1) + 12.5 O2 (g) → 8 CO2 (g) +9 H2O(1)
(2)
!
ˆ o − ∑ n ΔH
ˆ o "&
ΔH rxn = − % ∑ n i ΔH
c,i
i
c,i
reac tan ts
' products
(
For (1): −[(2)(0) + (3)(0) − (3)(0) − (1)(−1366.91) = 1366.91
− kJ / g mol C2 H5OH
For (2): −[(8)(0) + (9)(0) − (12.5)(0) − (−1307.53)(4.184)] = 5470.71
− kJ / g mol C8H18
Note: the ΔĤc = 1307.53 k cal / g mol for n-octane liquid is from Perry.
10–27
Solutions Chapter 10
Basis: 1 kg gasahol
Component
mass.fr. = kg
MW
g mol
ΔĤ(kJ)
C2H5OH
0.10
46.05
2.17
-2,966
C8H18
0.90
114.14
7.89
-43,137
Total
1.00
−46,103
If the fuel were all octane
1 kg oc tan e 1 kg mol octane 1000 g −5470.71 kJ
= −47,930 kJ
114.14 kg octane 1 kg g mol octane
−47,930 − (−46,103)
(100) = 3.8%
−47,930
10–28
Solutions Chapter 11
11.1.1
Basis: gas as given in problem statement
(a) 0.030 =
p H2O
pTot − pH2O
=
p H 2O
101.6 − p H2O
p* H2 O at 60 °C
% humidity
so, pH 2 O = 2.96 kPa
= 19.9 kPa
⎛ 2.96 ⎞ ⎡ 101.6 −19.9 ⎤
= 100 ⎝
= 12.2%
19.9 ⎠ ⎢⎣ 101.6 − 2.96 ⎥⎦
(b) Relative humidity
(100)
2.96
19.9
=
14.9%
*
(c) Dewpoint occurs where pH 2 O = 2.96 kPa, or T = 24oC from the steam tables.
11.1.2
*
At 27 °C, pH 2 O = 3.52 kPa. The actual pressure of the water vapor is obtained from
p=
nRT ⎛ 0.636 ⎞ ⎛⎜ 8.314 ⎞⎟ ⎛⎜ 300 K ⎞⎟ ⎛
⎞ = 3.15 kPa
=⎝
⎠
⎝
⎠⎝
⎠ 28.0 ⎠
V
18.01 ⎝
⎛ 3.15 ⎞
100 ⎝
3.52 ⎠
= 89%RH
11–1
Solutions Chapter 11
11.1.3
.5067
---------p
.1217
-----
Initial state
40 80 T °F
*
H 2O
At 80 °F, p
= 0.5067 psia from the steam tables
*
At 40 °F, pH 2 O = 0.1217 psia
*
At 58 °F, pH 2 O = 0.2384 psia
(a) 100
0.1217 psia
=
0.5067psia
24%
(b) The mol fraction of H2O vapor is constant on compression unless saturation is
reached. At the
initial conditions
y H 2O =
0.1217
−3
= 8.28 × 10
14.696
⎛ 0.1217⎞
= 0.2437 psia
At 2 atm, pH 2 O = p Tot y H2 O = (14.692)(2) ⎝
14.696 ⎠
0.2437
> 1 hence water condenses and the relative humidity is 100%
0.2384
11–2
Solutions Chapter 11
11.1.4
P* H 2 O = 5.878 in Hg
T = 140°F
p = 30 in Hg
H = 0.03 mol H2O/mol BDA ⇒ Basis: 1 mol BDA
mol H 2O
0.03
=
= 0.0291 = y H 2O
mol H 2O + BDA 1 + .03
p H 2O = 30 (.0291) = 0.874 in Hg
pair = 30 (1 – 0.0291) = 29.13 in Hg
(a)
% rel. humidity = 100
p H2O
! 0.874 "
= 100 #
$ = 14.9%
p*H2O
% 5.878 &
(b)
Dew point is the temperature at which vapor first condenses on cooling at
constant humidity, hence p* = 0.874 in Hg ~ 75°F.
11–3
Solutions Chapter 11
11.1.5
(a)
Entering
Leaving
80°F = 26.8°C
70°F = 21.0°C
p* at 26.8°C = 26 mm Hg
p* at 21.0°C = 19 mm Hg
Rel. Humidity =
(b)
p
5 mm Hg
100
=
p* 26 mm Hg
(c)
= 94.8%
Entering
Leaving
V
p
=
Vtot p tot
Vol fr. =
5 mm Hg
100 = 0.676%
740 mm Hg
H 2O:
p
p tot
18 mm Hg
100 = 2.43%
740 mm Hg
Air:100 - 0.0676 = 99.324%
Air: 100 - 2.43 = 97.57%
Entering
Leaving
Basis: 1 lb mol at 80°F, 740 mm Hg
Vol % = mol %
lb
wt %
H 2O: (0.00676)(18)
0.12
0.4
Air: (99.324)(29)
28.80
28.92
99.6
100.0
(d)
p 18 mm Hg
100
=
p* 19 mm Hg
= 19.2%
Vol fr. =
H 2O:
Rel. Humidity =
Basis: 1 lb mol at 70°F, 740 mm Hg
lb
wt %
H 2O: (0.0243)(18)
0.44
1.53
Air: (0.9757)(29)
28.30
28.74
98.47
100.00
Entering
Leaving
Basis: 1 lb mol at 80°F, 740 mm Hg
Basis: 1 lb mol at 70°F, 740 mm Hg
Actual: pvap = 5 mm Hg
Actual: pvap = 18 mm Hg
pvapor-free gas = 740 – 5 = 735 mm Hg
pvapor-free air = 740 – 18 = 722 mm Hg
11–4
Solutions Chapter 11
Humidity =
n H 2O
n dry air
=
g mol H 2O
18 g
1 g mol dry air
g mol dry air 1 g mol H 2O
29 g
p H 2O
p dry air
Entering
Humidity =
(e)
Leaving
5 18
g H 2O
= 4.22×10-3
735 29
g dry air
18 18
g H 2O
= 0.0155
722 29
g dry air
Basis: 1000 ft3 of mixture at 740 mm Hg and T
Entering
Leaving
Vol % H2O = 0.676 %
Vol % H2O = 2.43 %
Vol H2O = (0.00676)(1000 ft3) = 6.76 ft3
Vol H2O = (0.0243)(1000 ft3) = 24.3 ft3
@ 26.8 °C and 740 mm Hg
@ 21.0°C and 740 mm HG
3
3
6.76 ft 1 lb mol 492°R 740 mm Hg 18 lb
3 540°R 760 mm Hg lb mol
359 ft
24.3 ft 1 lb mol 740 mm Hg 492°R 18 lb
3 760 mm Hg 530°R lb mol
359 ft
= 0.30 lbH 2 O / 1000 ft 3
(f)
at 80 F and 740 mm Hg
= 1.10 lb H2 O / 1000 ft 3
at 70°F and 740 mm Hg
Entering
Leaving
0.30
0.99324
1.10
0.9757
3
3
= 0.302 lbH 2 O / 1000 ft vapor free air
(g)
= 1.127lb H2 O / 1000 ft vapor freeair
Basis: 800,000 ft3/day at 80°F, 740 mm Hg
Entering
Leaving
Vol dry air = 794, 600 (
Vol % air = 99.324 %
530
)
540
= 779,900 ft3 at 70°F
Vol dry air = (800,000)(0.99324)
11–5
Solutions Chapter 11
= 794,600 ft3
Vol % = 97.57 %
Total vol =
Vol H2O vapor = (800,000)(0.00676)
= 5400 ft3 at 80°F, 740 mm Hg
5400 ft
779, 900
3
= 799, 300 ft
0.9757
Vol H2O vap = 799,300 – 779,900
3 740
492
760 540
= 19,400 ft3 at 70°F, 740 mm Hg
= 4790 ft3 at SC
19, 400
492 740
= 17, 535 at SC
530 760
Water evaporated = 17,535 – 4,790 = 12,745 ft3 at SC
3
12, 745 ft 1 lb mol 18 lb
= 639 lb H 2O/day
3 lb mol
359 ft
Part g could also be worked using part f and the volume of dry air:
Entering
Leaving
794,600 ft3 dry air
779,900 ft3 dry air
0.302 lb H2O/1000 ft3 dry air (see f)
1.127 lb H2O/1000 ft3 dry air (see f)
794, 600 0.302
= 240 lb H 2O
1000
779, 900 1.127
= 879 lb H 2O
1000
Water evaporated = 879 – 240 = 639lb H2 O / day
11–6
Solutions Chapter 11
11.1.6
Basis: Air at T = 66 °F and 21.2 psia; assume V is constant.
(a) Initial
pair ,i V = n air RT 6 6 F
Final
Assume all of the water evaporates, and check later on to see if the assumption is true.
pair , f + pH 2 O = p t so pair , f = pt − p H 2 O
pair , f V = n air RT 1 8 0  F
The material balance is simple: all the initial air = final air.
pair , i
pt − p H 2 O
=
66 + 460 526
21.2
=
=
180+ 460 640 33.0− pH 2 O
pH 2 O = 7.20 psia
Since p* = 7.51 psia, all of the water can evaporate and the air will not be saturated.
(b) Basis: 1 lb H 2O (0.0555 lb mol H 2O )
At the final condition
pH 2 O
pt
=
7.20 n H 2 O 0.0555
=
=
33.0
nt
nt
so n = 0.254 lb mol
nRT ⎛⎜ 0.254 ⎞⎟ ⎛ 10.73 ⎞ ⎛⎜ 640 ⎞⎟
53.0 ft 3
=
=
⎝
⎠ ⎝ 33.0 ⎠ ⎝
⎠
p
0.17 lb H2 O
1lbH 2O
(c) n air = 0.254 - 0.0555 = 0.1985 lb mol
=
lb air
0.1985(29)
V=
11–7
Solutions Chapter 11
11.2.1
a.
Dew point = 10ºC
b.
%RH =
c.
H =
p H2O
p H2O
pair
psat
=
100 =
p10o C
p27o C
100 =
1.27 kPa
100 = 38%
3.356 kPa
1.27
18
= 0.79 kg H 2O / kg air
101.3 − 1.27 29
11.2.2
When the air is saturated.
11.2.3
When the air is saturated (100% relative humidity).
11.2.4
The two temperatures are approximately equal at atmospheric temperatures and
pressure.
11.2.5
If by “air” is meant wet air, H =
0.02
lb H 2O
= 0.204
0.98
lb dry air
From the SI humidity chart
a.
Dew point = 25.2ºC
b.
%RH ≅ 59%
11–8
Solutions Chapter 11
11.2.6
Let A = alcohol and C = CO2. MW of A = 46; MW of C = 44.
At 40ºC, pA* = 134.26 mm Hg or 17.90 kPa
a.
pA = 0.10 (100) = 10 kPa
H =
(46)(10)
= 0.116 g A/g C
(44)(100 − 10)
p
10
(100) =
(100) = 55.9%
p*
17.90
b.
%RS =
c.
Cs = 1.00 + 1.88(H) = 1.00 + 1.88 (0.116) = 1.22 J/(K) (g C)
d.
V̂' = 2.83 ×10−3 TK + 4.56 ×10−3 (H )
= (2.80)(10-3)(313.15) + (4.56)(10-3)(0.116) = 0.877 m3 / kg C
c.
At saturation at 40ºC
H =
(46)(17.90)
= 0.228 g A / g C
(44)(100 − 17.90)
V̂' = (2.80)(10-3) (313.15) + 4.56 × 10-3 (0.228) = 0.878 m 3 / kg C
11.2.7
From the SI chart at the intersection of TDB = 30o C and RH = 65%
H = 0.0174 kg H2O / kg dry air
11–9
Solutions Chapter 11
11.2.8
They are almost parallel to each other.
11.2.9
a.
From Humidity Chart, where tDB = 90ºC (194ºF) and tWB = 46ºC (115ºF) H =
0.049 kg H2O/kg air. Upon cooling to 43ºC (109ºF), no condensation occurs,
therefore H is constant.
0.049 kg H 2O 29 kg air 1 kg mol H 2O
kg mol H 2O
= 0.079
1 kg air
1 kg mol air 18 kg H 2O
kg mol air
b.
c.
⎛ 273 + 43 ⎞
Final pressure = 100 ⎝
= 87.1 kPa
273 + 90 ⎠
At saturation: 0.079 =
p*H2O
*
87.1 − p H2O
; solving p*H2O = 6.38 kPa
At the dew point, the vapor pressure of pure water is equal to 6.38 kPa. Dew
point = 37°C (99°F)
The same answer can be obtained by proceeding to the dew point at constant H on
a humidity chart for the correct pressure.
11–10
Solutions Chapter 11
11.2.10
Basis: 1 lb dry air
Data from the humidity chart.
Initial state:
TDB = 180o F and TWB = 120o F
H = 0.0637 lb H2O/lb dry air
Hsaturated ≅ 120 Btu / lb dry air
δHdeviation ≅ −1.5 Btu / lb dry air
H = 118.5 Btu/lb dry air
Final state
TDB = 115o F,
TWB = ?, H = 0.0657 lb H 2O / lb dry air
Hsaturated ≅ 101 Btu / lb dry air
ΔH = 101 – 118.5 = −17.5 Btu / lb dry air
11–11
Solutions Chapter 11
11.2.11
Basis: 1 lb bone dry air (BDA)
Basis: 100 m3
Data from psychometric chart (BDA = bone dry air).
a.
@ 1, Dew point = 23°C
b.
@ 1, humidity = 0.018 kg H 2O / kg dry air
c.
d.
@1, Relative humidity =
ΔH = Q + W
H 0.018
=
(100 = ) 54.5%
H 2 0.033
W=0
Q = ΔH 2 – ΔH 1 = (141.5 – 3.6)
kJ
kJ
kJ
– (79.0 – 0.3)
= 59.2
kg BDA
kg BDA
kg BDA
V 100 m 3 entering air kg BDA
m= ˆ =
= 112.36 kg BDA
V
0.89 m3
Q=
e.
59.2 kJ 112.36 kg BDA
= 6, 651.7 kJ
kg BDA
Adiabatic cooling by evaporation yields a saturated humidity of 0.041 kg
H2O/kg BDA:
(0.041 − 0.018) kg H 2O 112.36 kg BDA 2.58 kg H2 O
=
3
100 m
kg BDA
110 m3 air
f.
Texit = 37°C
11–12
Solutions Chapter 11
11.3.1
a.
The humidity of the entering air at 225ºF DB and 110ºF WB is obtained from the
humidity chart.
Humidity = 0.031
lb H 2O
lb dry air
Assume the exit air to be saturated at 125ºF.
Humidity = 0.0955
b.
lb H 2 O
lb dry air
Basis: 1 hr
10 tons 1 day 2000 lb
= 835 lb/hr
day 24 hr ton
Water in = (0.1) (835) = 83.5 lb/hr
Water out =
(0.9) (835) lb dry grain
1 lb H 2O
= 7.59 lb/hr
99 lb dry grain
lb H2O removed/hr = water in – water out = 83.5 – 7.59 = 75.9 lb H 2 O / hr
c. Product output = (0.9) (835)
d.
lb H 2 O
lb dry grain
+ 7.59
hr
hr
24 hr = 18, 200 lb / day
1 day
7.59 lb H 2O/hr
lb BDA
= 1175 lb BDA/hr
=
(0.0955 – 0.0310) lb H 2O/lb BDA
hr
Q = ΔH = Enthalpy out – Enthalpy in
= ΔHair + ΔHdry grain + ΔHwater
= 1175
+
lb BDA
Btu
Btu
136.5
– (92.25 – 2.25)
hr
lb BDA
lb BDA
(0.9) (835) lb dry grain 0.18 Btu (110 – 70) °F
hr
(lb) (°F)
11–13
Solutions Chapter 11
+
–
7.59 lb H 2O 1.0 Btu (110 – 32) °F
hr
(lb) (°F)
83.5 lb H 2O 1.0 Btu 2 (70 – 32) °F
hr
(lb) (°F)
4
= 5.46 × 104 + 0.54 × 104 + 0.059 × 104 – 0.32 × 104 = 5.74 × 10 Btu / hr
11.3.2
Assume in this problem
1.
2.
3.
4.
Steady operating conditions
Dry air and water vapor are ideal gases
ΔKE = ΔPE = W = 0
The mixing is adiabatic (Q = 0)
Data from the humidity chart:
Stream 1:
H1 = 110.3 kJ/kg dry air
H = 0.0272 kg H2O/kg dry air
Stream 2:
H2 = 50.9 kJ/kg dry air
H 2 = 0.0130 kg H2O/kg dry air
The specific humidity and the enthalpy of the mixture can be determined from mass and
energy balances for the adiabatic mixing of the two streams:
Basis: 1 kg dry air
Total Mass balance: 8 + 6 = 14 kg total
Water mass balance:
8(0.0272) + 6(0.0130) = 14 (Hmixture)
b.
H = 0.0211 kg H2O/kg dry air
11–14
Solutions Chapter 11
Energy balance (ΔH = 0)
8(110.3) + 6(50.9) = 14 (Hmixture)
H = 84.8 kJ/kg dry air
These two properties fix the state of the mixture. Other properties of the mixture are
determined from the psychometric chart.
a.
T = 30.7o C
c.
RH = 75.1%
11.3.3
Steps 1, 2, 3, and 4
The process will be assumed to be a steady state, open, continuous one with 5L
flowing in (in the material balance the air is invariant). All of the data for the stream
flows and compositions have been placed on the figure. The lungs will be the system.
P L air, 37oC
5 L air, 25oC
p
H2O
pair
ttot
Lungs
0.95 kPa
96.05 kPa
97.0 kPa
p
H2O
pair
ttot
= p*
H2O
6.27 kPa
90.73 kPa
97.0 kPa
p*25o C = 3.17 kPa
p H2O = 3.17(0.3) = 0.95 kPa
Step 5:
Basis: 5 L air → 1 min
Steps 6, 7, 8, and 9:
Material balances
In: n air =
96.05 kPa
1 atm
5L
1(g mol)(K)
= 0.194 g mol
101.3 kPa 298 K 0.08206(L)(atm)
11–15
Solutions Chapter 11
" 0.95 #
−3
n H2O %
& (0.194) = 1.92 ×10 g mol
' 95.05 (
Out:
n air = 0.194 g mol
! 6.27 "
n H2O = 0.194 #
$ = 0.0134 g mol
% 90.73 &
Energy balance
The energy balance reduces to Q = ΔH. The reference temperature will be 25ºC.
Assume the increase in water vapor comes from water vaporized at 37ºC
ˆ = 2414.3 kJ / kg).
(ΔH
vap
ΔHin: Both the air and water enter at 25ºC so ΔH = 0 for both the input streams.
ΔHout:
ΔHair = 0.194∫
310
298
(27.2 + 0.0041 TK )dT = 66.2 J
310
ΔH H2O = 1.92 ×10−3 ∫+ (34.4 0.0063 TK )dT
298
+ (0.0134 – 1.92 × 10-3) (2414.3)(18) = 499 J
Q = ΔH = 499 + 66.2 = 565 J/min or 33.9 kJ / hr
11–16
Solutions Chapter 11
11.3.4
Initial air: TDB = 38ºC, TWB = 27ºC,
H1 = 0.0175 kg/kg dry air
Air from scrubber: T = 24ºC, RH = 100%, RH2 = 0.0188 kg/kg dry air
Heated to 93ºC:
H3 = 0.0188
From drier: TDB = 49ºC,
H4 = 0.0377
(1000 kg/hr) (0.05) = 50.0 kg H2O to be evaporated
50.0/(0.0377 – 0.0188) = 2650 kg dry air/hr
2650 (0.0377) = 100 kg H2O/hr
! 2650 100 " 22.4 273 + 49
V4 #
+
= 2560 m3 at 49ºC and 1 atm
$
29
18
273
%
&
Heat supplied:
Q = [(2650) (1.00) + 100 (0.200] (93 – 24) = 1.84 × 105 kJ/hr
Water at 93ºC has p* = 79.4 kPa
H3 =
(18) ( 79.4) = 2.25 kg H O/kg dry air
2
( 29) (101.3–79.4 )
[(0.0188/2.25)] (100) = 0.83%
RH air from heater
Answers:
a.
1. H = 0.0175
2. H = 0.0188
3. H = 0.0188
4. H = 0.0377
b. 1. 42%
2. 100%
3. 0.83%
4. 47%
c. 2650 kg dry air/hr
d. 2560 m3/hr
e. 1.84 × 105 kJ/hr
11–17
Solutions Chapter 11
11.3.5
Basis: 180 kg/hr of product
Material Balance
Dry product =
a.
92 kg dry P 180 kg P
= 166 kg dry P/hr
100 kg P
1 hr
Water removed from solid:
In =
1.25 kg H 2O 166 kg/hr dry P
– (0.08) (180) kg/hr
1.00 kg dry P
= 207.5 – 14.4 = 193 kg H 2 O / hr
Basis: 1 BDA
Water picked up in dryer
out
in
0.0571 kg H 2O
0.0083 kg H 2O
kg H 2O
–
= 0.0488
kg BDA
kg BDA
kg BDA
out, 53°
in, 21°
193 kg H 2 O
kg BDA
=
= 3955 kg BDA / hr
hr
0.0488 kg H 2O / kg BDA
b.
Energy balance
Air:
ΔH (kJ/kg BDA)
air in (21ºC, 52% RH ):
58.2
air out (53ºC, 60% RH):
219.7
ΔH =
161.5 kJ/kg BDA
Basis: 1 hr
ΔH = (161.6) (3955) = 6.387 × 105 kJ
Solid (ref. 21ºC)
11–18
Solutions Chapter 11
3
0.18 cal 4.184 J 10 (43 – 21) °C
solid out: ΔH =
C pdT = =
= 16.57 kJ/kg P
21
(g) (°C) cal 10 3
43
solid in: ΔH = 0 (because of reference temperature)
Basis: 1 hr
ΔH = (16.57) (180) = 2983 kJ
If the dryer and reheater are insulated, then for the system
Q – W = ΔH and
W=0
5
Qreheater = 2983 + 6.387 × 105 = 6.42 × 10 kJ / hr
11.3.6
Step 5:
Basis: 1 hr
Assume:
1.
2.
3.
4.
ΔPE = ΔKE = W = 0,
No reaction occurs,
open, steady state process,
ideal gas behavior
Data:
Entrance air (in)
Ĥ (kJ/kg dry air)
H (kg H2O/kg dry air)
V̂(m3 / kg dry air)
50.5
0.00587
0.88
Assume the properties of the wet penicillin are the same as those of water. ΔHvap at 34ºC
= 2420.25 kJ/kg water. Let P = kg dry penicillin per hour, and F = kg dry air/hr.
Steps 3 and 4
11–19
Solutions Chapter 11
Moist air
TDB= 34oC
TWB = ?
Penicillin
Dryer
34oC
Penicillin
P
in
ω = 0.80
34oC
ωPout = 0.50
Moist air
3300 m3/hr
TDB=34oC
TWB = 17oC
air in =
4500 m3 1 kg dry air
= 5114 kg dry air
0.88 m3
Steps 6 and 7:
The exit conditions for the air are not known, but the H and H are related on the
humidity chart hence only one is unknown. P (dry penicillin is unknown).
The balances are water, dry air, and dry penicillin.
Steps 8 and 9
TWB ( o C)
Assume:
20
21
22
! kg H 2O "
H#
$
% kg dry air &
0.009
0.010
0.0115
Ĥ(kJ / kg dry air)
57.0
60.6
64.0
Water balance:
5114 (0.009 – 0.00587) = water evaporated = 16.0 kg
5114 (0.010 – 0.00587) = 21.1 kg
5114 (0.0115 – 0.00587) = 28.8 kg
Energy balance:
5114 (57.0 – 50.5) / (2420.25) = 13.73 kg
5114 (60.5 – 50.5) / (2420.25) = 21.1 kg
5114 (64.0 – 50.5) / (2420.25) = 28.5 kg
The solution is very sensitive to the values read from the psychometric chart. Assume the
final TWB ; 21 or 22o C .
a.
Water evaporated = 28.5 kg
11–20
Solutions Chapter 11
ΔH (21ºC) = 51,140 kJ/kg dry air ⇒ 51,140 kJ / hr
b.
c.
Steps 6 and 7
Unknowns
P, exit TWB
Equations
air, water
Steps 8 and 9
Water balance on air gives water evaporated (5114) (0.010-0.00587) = 21.1 kg H2O
Energy balance: (60.5 – 50.5) (5114) + 21.1 (2420.25) = 0
0.80 → 0.60
Use 21ºC
! 0.80 0.50 "
21.1 kg H2O evap = P $
−
3P
% =
& 0.20 0.50 '
P=
21.1
= 7 kg / hr
3
11.3.7
R
80oF
120oF
Hin=0.0031
lb H2O/lb DA
H out=0.0292
lb H2O/lb DA
H = 0.0075
lb H2O/lb DA
Basis = 2000 lb dry waste (DW)
Comp.
in %
out
waste in =
H2 O
63.4
22.7
2000 0.773 100
= 4, 220 lb
36.6
DW
36.6
77.3
DW out = 2,000 lb
→
11–21
Solutions Chapter 11
100.0
100.0
H2O evap.
= 2,220 lb
Basis: 1 lb dry air
Air in, 80ºF, 29.92 in. Hg, and a wet bulb temperature = 54ºF:
Hin = 0.0031 lb H2O/lb dry air Ĥ'in = 22.62 − 0.18 =22.54 Btu / lb dry air
Air out, 120ºF, 29.92 Hg, and a wet bulb temperature = 94ºF:
Ĥ ' = 61.77-0.4 = 61.37 Btu/lb dry air
Hout = 0.0292 lb H2O/lb dry air
Overall
b.
Hout
=
0.0292
Hin
=
0.0031
H2O evaporated =
0.0261
air used per ton dust =
lb H2O/lb dry air
lb H2O/lb dry air
(1 + 0.0031)
2220
0.0261
= 85, 400 lb moist inlet air
per ton DW
Air out =
1.00
2220
0.0261
0.0292
(29)
= 2,930 lb mol dry air out
2220
0.0261
(18)
= 138 mol H2O out
3068 total mol out
c.
3068 359 580
= 1,300, 000 ft3 wet air exiting/ton DW
492
Air Recirculated
R = lb air added at mixing point before kiln
Water balance:
(0.0031) (1) + (0.0292) (W) = (0.0075) (1+W)
0.0292 W
R=
a.
=
0.0075
-0.0075 W
-0.0031
0.0217 W
0.0044
0.0044
= 0.203 lb
0.0217
Recirculation =
0.203 100
= 16.8% of gas out is recirculated
1.203
11–22