Q2 A Ans
Q2 A Ans
Q2 A Ans
E L
Find: E/L
L = 0.1111 + 1 = 1.1111
E = 0.0101 + 1 = 1.0101
Pp = 0.3(760) = 228 mm Hg
At 30 oC
Pv = 281.8 mm Hg
c. %S
= 72.73 %
d. Sm
= 0.429
e. S abs
= 0.429(58)/29)
= 0.858
3. A waste gas containing SO2 (a precursor of acid rain) and several other species
(collectively designated as A) is fed to a scrubbing tower where it contacts a
solvent (B) that absorbs SO2.The solvent feed rate to the tower is 1000L/min.
The specific gravity of the solvent is 1.30. Absorption of A and evaporation of B
in the scrubber may be neglected. Determine the (a) mass fraction of SO2 in the
liquid effluent stream and (b) the rate at which SO2 is absorbed in the liquid
effluent stream. Mole fraction of SO2 in the exit gas stream is 0.001 and 0.07 in
the feed gas. The feed gas enters 400 oC and 10 atm with molar density
(mol/L) of 0.185. Volumetric flowrate of feed gas is 207 m3/min.
SO2 (nf = 0.001) L Solvent, S
A 1000L/min
Sp. Gravity = 1.3
Scrubber/absorber
207 m3/min
SO2 (nf = 0.07) S
A E SO2
o
400 C
10 atm
0.185 mol/L
Basis: 1 mol of A
SmE = 0.07/(1-0.07) = 0.075
SmL = 0.001/(1-0.001) = 0.001
Rate SO2 absorbed = 0.0736 mol / 2.79 x 10-5 min = 2616 mol / min
= 167 kg/min
4. a) Calculate the average molal heat capacity (J/mol-oC) of methane from 0 to
1200 oC. Using Table E.1
34.31
1200
Cp (J/mol) = T + 5.469 x 10-2T + 0.3661 x 10-5 T2 - 11.00 x 10 -9
T3) dT
0
= 76955 J/mol
b) Using the average molal heat capacity calculated, calculate the heat needed
to cool 1 kmol methane from 500 oC to 40 oC .
Q = nCpT = (1kmol)(1000 mol/kmol)(64.13)(40-500)
= - 29499800 J
= - 29,500 kJ
5. Moist air at 38 oC and 1 atm with volumetric flowrate of 50 L/min has an
absolute humidity of 0.003. Part of the water is condensed to have a final
absolute humidity of 0.002 What will be the temperature of temperature of
the gases leaving the unit. Calculate the rate of cooling (kW).
38 oC, 1 atm
50 L/min
E Condenser/dehumidifier L
Habs = 0.003 Habs = 0.001
nE = PV/RT = (1 atm)(50L)/(0.0821)(38+273)
= 1.96 mol
HmE = 0.003(29/18) = 0.0048
HmL = 0.002(29/18) = 0.0032
Pv = 2.43 mm Hg
Reference: air at 38 oC
Water vapor at 38 oC
Inlet enthalpy = 0
Outlet
n H
=1.95 H
= 0.0063 H
H2O (s) = 3
[1.96(0.0048/1.00 H
48] 0.0063
= 0.003
28.94
= Cp (J/mol) = + 0.4147 x 10-2T + 0.3191 x 10-5 T2 - 1.965 x 10 -9
8
1
H
38
T3) dT
= -1334
33.46
= Cp (J/mol) = + 0.668 x 10-2T + 0.76044 x 10-5 T2 - 3.593 x 10 -9
8
2
H
38
T3) dT
= -1544
3 = SH H2O (g) ( 38 to 100 oC) + LHc H2O (at 100 oC)
H
+ SH H2O(l) [100 oC(373K) to 0 oC(273K)
+ LH freezing H2O (at 0oC) + SH H2O (s) 0oC to -8 oC
33.46
100
SH H2O (g) = + 0.668 x 10-2T + 0.76044 x 10-5 T2 - 3.593 x 10 -9
T3)
38
dT
=2105 J/mol
= - 8105 J/mol
LH freezing = - 6 kJ/mol
SH H2O (s) = neglect ( no available Cp in Table E.1)
3 = 2105 - 40.6(1000) -6(1000) - 8105 = -52600 J/mol
H
Q = -2769 0 = -2769 J
= -0.05