CHAPTER THREE 3.1
(a) m =
= (b) m
16 × 6 × 2 m3 1000 kg m 8 oz
1 0 3 i ≈ 2 × 10 5 kg ≈ b2 × 10gb5gb2 gd10
106 cm 3
1 qt
2 s
3
1g
32 oz 1056.68 qt
cm
3
4 × 106
≈
b3 × 10gd10 i 3
≈ 1 × 102 g / s
(c) Weight of a boxer ≈ 220 lb m
12 × 220 lb lb m
W max ≥
1 stone
≈ 220 stones
14 lb m
dictionary
(d) V =
≈
π D
2
L
4
=
2
3.14 4.5 ft 4
2
800 miles 5880 ft 1 mile
3 × 4 × 5 × d8 × 10 10 2 i × d5 × 10 3 i × 7 4 × 4 × 10
(e) (i) V ≈
6 ft × 1 ft × 0.5 ft
(ii) V ≈
28,317 cm 3 1 ft
1 ft 3
150 lb m
3
≈ 1 × 10 7 barrels
≈ 3 × 3 × 104 ≈ 1 × 105 cm3
28,317 cm3
62.4 lb m
1 ft
7.4805 gal 1 barrel 1 ft 3 42 gal
3
≈
150 × 3 × 10 4 60
(f) SG ≈ 1.05
3.2
995 kg
(a) (i)
m
(ii)
(b)
3.3
(a)
(b)
(c)
ρ
3
0.028317 m 3
1 lb m 0.45359 kg
995 kg / m3
1 ft
62.43 lb m / ft 3 1000 kg kg / m
3
3
. lb lb m / ft ft 3 = 6212
. lb lb m / ft ft 3 = 6212
= ρ H2O × SG = 62.43 lb m / ft 3 × 5.7 = 360 lb m / ft 3
50 L
0.70 × 10 3 kg
1 m3
m3
103 L m3
1150 kg min
0.7 × 1000 kg
= 35 kg
1000 L 1 min 1 m3
10 gal
1 ft 3
0.70 × 62 6 2.43 lb m
2 mi min
7.481 gal
1 ft 3
60 s
= 27 L s
lb m / mi m in ≅ 29 lb
≈ 1 × 105 cm3
3.3 (cont’d) 3 gasoline (d) Assuming that 1 cm kerosene was mixed with V g (cm3 ) gasoline
d
i d i 1dcm3 kerosenei ⇒ 0.82dg kerosene i d0.70V + 0.82idg blend i SG = = 0.78 ⇒ V + 1dcm3 blend i
g asoline ⇒ 0.70V g g gasoline Vg cm3ga
g
V g =
g
V gasoline
Volume Volumetr tric ic ratio ratio =
3.4
In France: In U.S.:
3.5
50.0 kg
V kerosene
0.82 − 0.78 0.78 − 0.70
3
= 0.5 0 cm
0.50 cm3 asoliine / cm3 ker kerosen osenee = = 0.50 cm3 gasol 3 1 cm
L
5 Fr $1
= $68.42 1L 5.22 Fr 1 gal $1.20 = $22.64 0.70 × 1.0 kg kg 3.7854 L 1 gal
0.7 × 1.0 kg kg 50.0 kg L
V B ( f t 3 / h ) , m B ( l b m / h ) 3 . V ( ft / h), SG = 0850
V H ( f t 3 / h ) , m H ( l b m / h )
= 700 lb m (a) V
h
0.850 × 62.43 lb m
. ft ft 3 / h = 1319
d i 0.879 × 62.43 lb m = 54.88V b kg / h g B ft 3 b hg bkg / h g = dV hb0.659 × 62.43g = 41.14V
B = m H m
ft 3
700 lb m / h
ft 3 V B
H
H
= 13.19 ft 3 / h V B + V H B + m H = 54.88VB + 41.14VH = 700 lb m m
⇒
B = 628 lb m / h benzene V B = 11.4 ft 3 / h ⇒ m H = 71.6 lb m / h hexane V H = 1.74 ft 3 / h ⇒ m
unit. (b) – No buildup of mass in unit. – ρ B and ρ H at inlet stream conditions are equal to their tabulated values (which are o
strictly valid at 20 C and 1 atm.) – Volumes of benzene and hexane hexane are additive. – Densitometer gives correct reading. reading.
3.6
(a) V =
1955 . kg H 2SO 4
1 kg solution
L 12563 . . kg × 1000
0.35kg H 2SO 4
= 445 L
(b) L 1.8255 × 1. 00 kg 195.5 kg H 2 SO 4 0.65 kg H 2 O + 0.35 kg H 2 SO 4 470 − 445 × 100% = 5.6% % error = 44 5
V ideal =
3.7
1955 . kg H 2 SO 4
L = 470 L 1.000 kg
Buoyant force b upg = Weight of block bdown g
E
Mass of oil displaced + Mass of water displaced = Mass of block ρ oil
b0.542gV + ρ H O b1 − 0.542gV = ρ c V 2
From Table B.1:
= 2.26 g / cm3 ,
ρ c
ρw
= 1.00 g / cm3 ⇒
ρ oil
= 3.325 g / cm3
moil = ρ oil × V = 3.325 g / cm3 × 35.3 cm3 = 117.4 g moil + flask = 117.4 g + 124.8 g = 242 g
3.8
Buoyant force b upg = Weight of block bdown g
⇒ Wdisplaced liquid = Wblock ⇒ ( ρVg)disp. Liq = ( ρ Vg) block Expt. 1:
ρ w
b15. Ag g = ρ b2 Ag g ⇒ ρ B
ρ w =1.00 g/cm
Expt. 2:
ρ soln
B
= ρ w ×
15 . 2
3
ρ B
= 0.75 g / c m3 ⇒ b SGg B = 0.75
b Ag g = ρ B b2 Ag g ⇒ ρ soln = 2 ρ B = 15. g / c m3 ⇒ b SGgsoln = 15.
3.9
Let W A + W B
= density of water. Note:
ρ A
> ρ w (object sinks)
Volume displaced: Vd 1 = Ab hsi = Ab d hp1 − hb1 i (1)
hs 1
Archimedes ⇒
h ρ 1
hb1
ρ w
= WA + WB
ρ wVd 1 g
weight of displaced water Subst. (1) for V d 1 , solve for dh p1 − hb1 i
Before object is jettisoned
h p1 − hb1 =
W A + W B pw gAb
(2)
bi g Volume of pond water: Vw = Ap h p1 − Vd 1 ⇒ Vw = Ap hp1 − Ab d hp1 − hb1 i subst. b 2 g for b p 1 − hb1
Vw = Ap hp1 −
subst. b 3g for h p 1 in
b 2 g , solve for hb1
hb1 =
W A + W B
V w A p
pw g
+
bW
A
⇒ h p1 =
V w A p
+ W B g L 1
pw g
MA MN
p
−
+
W A + W B pw gAp
1 Ab
O P PQ
(3)
(4)
3.9 (cont’d) s2
W B
W A
Let V A = volume of jettisoned object = h ρ 2
h b2
(5)
Volume displaced by boat: Vd 2 = Ab dhp 2 − hb 2 i
(6)
E Subst. for Vd 2 , solve for dh 2 − h 2 i p
W B
(7)
Volume of pond water: Vw = A ph p2 − Vd 2 − VA solve for
V w
h p 2
A p
⇒ h p 2 =
subst. b 8g
⇒
for h p 2 in b 7 g, solve for hb 2
(a)
b
pw gAb
ρ A g
Archimedes ⇒ ρ WVd 2 g = W B
After object is jettisoned
h p 2 − hb 2 =
W A
+
hb 2 =
WB pw gAp V w A p
+
+
W A
pw gAp
+
Vw = A ph p2 −
p A gAp
W B
b 5g , b 6 g & b 7 g
W B pw g
−
W A p A g
(8)
W A p A gAp
−
W B pw gAb
(9)
Change in pond level
⎡ 1 1 ⎤ W A ( pW − p A ) ρW < ρ A − ⎯⎯⎯⎯ →<0 ⎢ ⎥= A p g ⎣ p A pW ⎦ pA pW gAp
( 8)−( 3) W A
h p 2 − h p1 =
⇒ the pond level falls (b)
Change in boat level h p 2 − hp1
L 1 M g MN p A
b 9 g− b 4 g W A
=
A p
A
p
−
1 pW Ap
+
⇒ the boat rises
3.10 (a)
ρ bulk =
2.93 kg CaCO 3 L CaCO 3
(b) Wbag = ρ bulk Vg =
> 0 O L 5gF M b O I F p F A I I P 1 V M 1 = + P G J G − 1J K J K PP > 0 p A PQ H A K M GH p H A MN PQ > 0
W
0.70 L CaCO 3 L total
b
A
A
p
p
W
b
= 2.05 kg / L
2.05 kg 50 L 9.807 m / s 2
1N
. × 10 3 N = 100
L 1 kg ⋅ m / s Neglected the weight of the bag itself and of the air in the filled bag. 2
(c) The limestone would fall short of filling three bags, because – the powder would pack tighter than the original particles. – you could never recover 100% of what you fed to the mill.
V b = ρ b
=
ρ w g
mb V b
=
1N 1 kg ⋅ m / s 2
Wb − W I
=
9.807 m / s2
1225 . kg
3.11 (a) Wb = mb g =
= 1202 N
(1202 N - 44.0 N)
1 kg ⋅ m / s2
0.996 kg / L × 9.807 m / s 2
1N
1225 . kg 119 L
. kg / L = 103
m f + mnf = mb
(b)
x f =
m f mb
(1)
⇒ m f = mb x f
(2)
(1),(2) ⇒ mnf = mb d1 − x f i m f
V f + Vnf = V b ⇒
F x H ρ
b 2 g,b 3g
f
⇒ mb G
(c) x f =
+
+
ρ f
J = K
ρ nf
1 / ρ b − 1 / ρ nf 1 / ρ f − 1 / ρ nf
mnf
(3)
=
ρ nf
1 − x f I mb
f
= 119 L
mb ρ b
F 1 H ρ
⇒ x f G
ρb
1 / 0.9 − 1 /1.1
1 ρ nf
f
1 / 103 . − 1/1.1
=
−
I 1 J K = ρ
b
−
1
⇒ x f =
ρ nf
1 / ρ b − 1 / ρ nf 1 / ρ f − 1 / ρ nf
= 0.31
(d) V f + Vnf + Vlungs + Vother = Vb m f
+
ρ f
mnf ρ nf
+ Vlungs + V other =
m = m x f b f m = m ( 1− x ) nf b f
mb
F x GH ρ
f
−
mb ρ b
1 − x f I ρ nf
f
K
b
−
1 ρ nf
I J K
F 1 1 I 1 1 V + V − = − − m H ρ ρ J K ρ ρ F 1 1 I F V + V I F 1 1 I F 1.2 + 01. I J K GH − J K − GH GH ρ − ρ J K − GH m J 103 . 11 . 1225 . K = = = 0.25 F 1 1 I F G 1 − 1 J I GH ρ − ρ J K H 0.9 11. K
⇒ x f G
lungs
f
nf
b
nf
b
nf
other
b
f
nf
other
b
lungs
⇒ x f
F 1 H ρ
J + (Vlungs + Vother ) = mb G
3.12 (a) 4.5
) O 4 2 H 3.5 g 0 3 0 2.5 1 / e 2 l I g 1.5 ( . c 1 n o 0.5 C
y = 545.5x - 539.03 R2 = 0.9992
0
0.987
0.989
0.991
0.993
0.995
0.997
Dens ity (g/cm 3)
From the plot above, r = 5455 . ρ − 539.03 r = 3.197 g Ile / 100g H 2 O
(b) For ρ = 0.9940 g / cm3 , Ile = m
150 L
0.994 g
h
cm3
1000 cm 3
3.197 g Ile
L
1 kg
103.197 g sol 1000 g
= 4.6 kg Ile / h
(c) The measured solution density is 0.9940 g ILE/cm3 solution at 50oC . For the calculation of Part (b) to be correct, the density would have to be changed to its equivalent at 47 oC. Presuming that the dependence of solution density on T is the same as that of pure water, the solution density at 47 oC would be higher than 0.9940 g ILE/cm 3. The ILE mass flow rate calculated in Part (b) is therefore too low. 3.13 (a)
1.20 ) n i m / g k ( e t a R w o l F s s a M
1.00
y = 0.0743x + 0.1523
0.80
R = 0.9989
2
0.60 0.40 0.20 0.00 0.0
2.0
4.0
6.0
8.0
Rotameter Reading
10.0
12.0
3.13 (cont’d) = 0.0743 b5.3g + 01523 From the plot, R = 5.3 ⇒ m . = 0.55 kg / min
(b)
Rotameter Collection Collected Volume Reading Time (min) (cm3) 2 1 297 2 1 301 4 1 454 4 1 448 6 0.5 300 6 0.5 298 8 0.5 371 8 0.5 377 10 0.5 440 10 0.5 453
Mass Flow Rate (kg/min) 0.297 0.301 0.454 0.448 0.600 0.596 0.742 0.754 0.880 0.906
Difference Duplicate (Di)
Mean Di
0.004 0.006 0.004
0.0104
0.012 0.026
1
b0.004 + 0.006 + 0.004 + 0 .012 + 0.026 g = 0 .0104 kg / min 5 95% confidence limits: ( 0.610 ± 174 . Di ) kg / min = 0.610 ± 0.018 kg / min
Di =
There is roughly a 95% probability that the true flow rate is between 0.592 kg / min and 0.628 kg / min .
3.14 (a) (b)
(c)
(d)
(e)
(f)
(g)
(h)
15.0 kmol C 6 H 6
78.114 kg C 6 H 6 kmol C6 H 6
15.0 kmol C 6 H 6
1000 mol kmol
15,000 mol C 6 H 6
. × 10 4 mol C 6 H 6 = 15
lb - mole 453.6 mol
15,000 mol C 6H 6
. lb - mole C 6 H 6 = 3307
6 mol C 1 mol C 6 H 6
15,000 mol C 6H 6
6 mol H 1 mol C 6 H 6
90,000 mol C 12.011 g C mol C 90,000 mol H 1.008 g H mol H 15,000 mol C 6 H 6
. × 10 3 kg C 6 H 6 = 117
= 90,000 mol C = 90,000 mol H
= 1.08 × 10 6 g C
= 9.07 × 10 4 g H
6.022 × 10 23 mol
= 9.03 × 10 27 molecules of C 6 H 6
= 3.15 (a) m
(b) n =
175 m3
1000 L
0.866 kg
m3
h 2526 kg min
L
1000 mol 1 min 92.13 kg
60 s
1h 60 min
= 2526 kg / min
= 457 mol / s
o (c) Assumed density (SG) at T, P of stream is the same as the density at 20 C and 1 atm
3.16 (a)
200.0 kg mix 0150 . kg CH 3OH kg mix
mix = (b) m
M =
3.17
kmol CH 3OH 32.04 kg CH 3OH
1 kmol
100.0 lb - mole MA
74.08 lb m MA
1 lb m mix
h
1 lb - mole MA
0.850 lb m MA
0.25 mol N 2
N = m 2
1000 mol
3000 kg h
28.02 g N 2
+
0.75 mol H 2
mol N 2 kmol 0.25 kmol N 2 8.52 kg
kmol feed
= 936 mol CH3OH
= 8715 lb m / h
2 .02 g H 2
= 8.52 g mol mol H 2 28.02 kg N 2 = 2470 kg N 2 h kmol N 2
3.18 M suspension = 565 g − 65 g = 500 g , M CaCO 3 = 215 g − 65 g = 150 g = 455 mL min , m = 500 g min (a) V
(b)
ρ =
= 500 g / 455 mL = 110 / V . g mL m
(c) 150 g CaCO 3 / 500 g suspension = 0.300 g CaCO 3 g suspension 3.19
Assume 100 mol mix. mC2 H5OH =
mC4 H8O2 =
10.0 mol C 2 H 5OH
mol C 2 H 5OH 75.0 mol C 4 H 8O 2
xC4 H8O2 =
88.1 g C 4 H 8O 2 mol C 4 H 8 O 2
mCH3COOH =
xC2 H5OH =
46.07 g C 2 H 5OH
15.0 mol CH 3COOH
= 6608 g C 4 H 8 O 2
60.05 g CH 3COOH mol CH 3COOH
461 g 461 g + 6608 g + 901 g 6608 g 461 g + 6608 g + 901 g 901 g
xCH 3COOH =
= 461 g C 2 H 5OH
= 901 g CH3COOH
= 0.0578 g C2 H 5OH / g mix . g C 4 H 8 O 2 / g mix = 08291
. g CH 3COOH / g mix = 0113 461 g + 6608 g + 901 g 461 g + 6608 g + 901 g = 79.7 g / mol MW = 100 mol 25 kmol EA 100 kmol mix 79.7 kg mix m= = 2660 kg mix 75 kmol EA 1 kmol mix
3.20 (a) Unit Crystallizer Filter Dryer
(b) m gypsum = V gypsum =
Function Form solid gypsum particles from a solution Separate particles from solution Remove water from filter cake 0.35 kg C aSO 4 ⋅ 2 H 2 O
1 L slurry
L slurry
0.35 kg CaSO4 ⋅ 2H2O
= 0.35 kg CaS O 4 ⋅ 2 H 2 O
L CaSO4 ⋅ 2H2O
= 0151 . L CaSO4 ⋅ 2H2O 2.32 kg CaSO4 ⋅ 2H2O 0.35 kg g ypsum 136.15 kg CaSO 4 = 0277 CaSO 4 in gypsum: m = . kg CaSO 4 172.18 kg g ypsum CaSO4 in soln.: m =
. g L sol b1 − 0151
L
0.35 kg gypsum
(c) m =
% recovery =
1.05 kg 0.209 kg CaSO 4 100.209 kg sol
0.05 kg sol
0.209 g CaSO 4
0.95 kg gypsum
100.209 g sol
0.277 g + 3.84 × 10 -5 g
. kg CaSO 4 = 000186
= 3.84 × 10-5 kg CaSO 4
× 100% = 99.3%
0.277 g + 0.00186 g
3.21 CSA: FB:
45.8 L
0.90 kg
kmol
min L 55.2 L 0.75 kg
75 kg kmol
min
L
90 kg
kmol U | 05496 . mol CSA min | ⇒ = 1.2 V kmol | 0.4600 mol FB 0.4600 min | W
= 05496 . =
She was wrong. The mixer would come to a grinding halt and the motor would overheat. 3.22 (a)
150 m ol E tO H
4 6.0 7 g E tO H mol EtOH
6910 g EtO H
0.600 g H 2 O
= 10365 g H 2 O
0.400 g EtOH
V =
6910 g EtOH
789 g EtOH
SG =
(b) V ′ =
L
(6910 +10365) g 19.1 L
% error =
+
L 1000 g
( 6910 + 10365) g mix
= 6910 g EtOH
10365 g H 2 O
1000 g H 2 O
. = 0903
L 935.18 g
(19.123 − 18.4 72 ) L 18.472 L
L
= 18472 . L ⇒ 18.5 L
× 100% = 3.5%
= 19.123 L ⇒ 19.1 L
