Problem 1: Let K denote K2CrO4 and W denote water. The process flow chart is shown
below:
M1
100% W (vapor)
Evaporator
4500 kg/h
33.3% K
66.7% W
M2
49.4% K
50.6% W
Crystallizer
& Filter
M5
36.4% K
63.6% W
Two Streams Combined
M3 (crystals)
100% K
M4 (solution M5)
36.4% K
63.6% W
The filter cake is a mixture of solid K crystals and a liquid solution with known K and W
mass fractions. The crystals constitute 95% by mass of the filter cake stream (M3+M4).
Please determine
a) Flow of M2.
b) Flow of M3.
c) Flow of M5.
Solution
Evaporator Balance In-Out=Acc
K balance
4500[kg/h]*0.333-M2*0.494=0
M2=4500*0.333/0.494= 3033.4 kg/h
Crystallizer Balance In-Out=Acc
Overall Balance
M2-M3-M4-M5=0
3033.4 kg/h-M3-M4-M5=0
K Balance
3033.4 kg/h *0.494-M3*1.0-M4*0.364-M5*0.364=0
W Balance
3033.4 kg/h *0.506-M3*0 -M4*0.636-M5*0.636=0
Define a quantity M6=M4+M5 since it has the same composition as both stream 3
and 4. Use this substitution on Evaporator W Balance.
3033.4 kg/h *0.506-M3*0 -M6*0.636 =0
M6=3033.4 kg/h *0.506/0.636=2413.4 kg/h
Use overall balance to determine M3
3033.4 kg/h-M3-M6=0
M3=3033.4 kg/h-M6=3033.4 kg/h-2413.4 kg/h = 620 kg/h crystals
M3/(M3+M4)=0.95
gives 0.05*M3=0.95*M4
0.05*620 kg/h/0.95=M4=32.6 kg/h
Evaporator Overall Balance M2-M3-M4-M5=0
3033.4 kg/h-620 kg/h-32.6 kg/h-M5=0
M5= 2380.8 kg/h
Problem 2: Perform a degrees of freedom analysis for the splitter in the following process
flow sheet.
N5
0.983 DA
0.017 W(v)
N1 (mole)
0.96 DA
0.04 W(v)
Mix
N2
0.977 DA
0.023 W(v)
DA = dry air
W(v) = water vapor
W(l) = liquid water
Air
Conditioner
N4
0.983 DA
0.017 W(v)
Split
N6=100 mole
0.983 DA
0.017 W(v)
N3
1.0 W(l)
Solution: DoF = 1
Problem 3: Calculate the theoretical air for the combustion of 100 kg of dried coal with
ultimate analysis.
Element
C
H
S
O
Ash
Ultimate Analysis
71.2%
4.8%
4.3%
9.5%
10.2%
Mole Weight (gm/mole)
12
1
32
16
-
Solution:
Basis: 100 kg coal
Element
C
H
S
O
Ash
Ultimate Analysis
71.2%
4.8%
4.3%
9.5%
10.2%
Mole Weight (kg/kmol)
12
1
32
16
-
kmol
5.933
4.8
0.134
Reactions
C + 02  CO2
H + ¼ O2  ½ H2O
S + O2  SO2
Total O2 required for complete combustion= 5.933+4.8/4+0.134=7.267 kmol
Minus the amount of O in the coal = ½ * (9.5/16)=0.297*kmol
Equals 7.267-0.297=6.970 kmol O2/100 kg coal.
Theoretical Air = Theoretical O2/0.21= 6.970 kmol O2/0.21=33.19 kmol Air