3.
Feed: 100 kg
18.6 kg oil
69 kg solid
12.4 kg is moisture
69 kg= (87.7/100)*S
S=78.67 kg
Amount of oil in the exit stream: 0.008*78.67 kg = 0.629 kg
Amount of oil transferred is: 18.6-0.629 = 17.971 kg
% recovery= (17.971/18.6)*100= 96.6%
4.
Inlet (F) Outlet (O)
66% H2 98% H2
33% CH4 2% CH4
1% other Vent=? (V)
Basis: 100 kmol flue gas
Inlet:
66 kmol H2
33 kmol CH4
1 kmol other component
Outlet:
H2: 0.98*O=0.85*66=56.1 kmol
O= 57.24 kmol
CH4 in outlet= 0.02*57.24 = 1.145 kmol
V= 100-57.24 kmol = 42.76 kmol
Composition of V:
H2: 66-56.1 = 9.9 kmol
CH4= 33-1.145 kmol = 31.855
Other= 1 kmol
% H2: (9.9/42.76)*100=23.15%
%CH4: (31.855/42.76)*100= 74.49%
Other: 2.36%
5. x+y = 100
(MW CuSO4 / MW CuSO4.5H2O )x+(MW FeSO4 / MW FeSO4.7H2O )y=59.78
MW CuSO4 : 159.5
MW CuSO4.5H2O : 249.5
MW FeSO4 : 152
MW FeSO4.7H2O : 278
0.639x+0.547y=59.78
0.639x+0.547(100-x)=59.78
0.092x=5.08
X=55.21 kg
X=55.21 kg
Y=44.79 kg
6. Let the mass of crystal be x
That of ethanol be y
X+y = 1000
Mass of crystal/Mass of ethanol= 190/100
x/y = 1.9
x=655.17 kg
y= 344.83 kg
1 mole of Crystal =MW of MgCl2. 6H2O
MW of MgCl2 in one mole of crystal
Amount of MgCl2 = (MW of MgCl2/ MW of crystal)*x
Rest will be water of hydration
7.
Top (T)
Inlet (F) = 100 kg
81% C6H6
6.5% H2O = 6.5 kg
3.4% H2O
74% C6H6 = 74 kg Bottom (B)
15.6% C2H5OH
19.5% C2H5OH = 10% C6H6
19.5 kg
34.7% H2O
55.3% C2H5OH
100 = T+B
74=0.81T+0.1B
T = 90.17 kg
B = 9.83 kg
8.
Inlet (F) Outlet (S)
9.6% glycerol 80% glycerol
10.3% salt Vent (V) 6% salt
Rest water 4.5% of F glycerol Rest water
X% salt
Rest water
Percentage evaporation= ? 78.93%
Percentage Crystallization= ? 9.61%
Basis: 100 kg spent lye
Inlet: 9.6 kg glycerol; 10.3 kg salt
In the vent: glycerol: 0.045*9.6 kg = 0.432 kg
In the outlet: (9.6-0.432)= 9.168=0.8*S
S= 11.46 kg
Salt in outlet: 0.06*11.46 kg = 0.69 kg
Salt in vent: 10.3-0.69 kg = 9.61 kg
V= 88.54 kg
Inlet water: 80.1 kg
Outlet water: 0.14*S = 1.6 kg
Water evaporated = 80.1-1.6 kg = 78.5
% evaporation: 78.5%
% crystallization of salt: 9.61%
9.
Fresh air
0.02 kg water vapour
per kg dry air
Outlet (S) Dry board
Inlet (F) (1000 kg)
16% H2O 0.5% H2O
Wet air
0.09 kg water vapour
per kg dry air
Amount of water taken up by 1 kg of dry air= 0.07 kg water vapour per kg dry air
Amount of DRY board entering the system= Amount of DRY board leaving the system= 1000 kg
INLET:
16% of the total mass is accounted by the amount of moisture that is present
84% of the total mass will be attributed by the mass of the dry board
0.84*F=1000 kg
F= 1190.47 kg
OUTLET:
0.5% of the total mass of the outlet is attributed by the water present
99.5% of the mass will be because of the dry board= 1000 kg
0.995*S = 1000
S= 1005 kg
Water uptake by the air is the weight difference between F and S= 1190.47 – 1005 kg
=185.47 kg
Therefore, air required for drying: 185.47/0.07 = 2649.57 kg/h
10. A spent solution of chloroacetic acid in ether contains 20 mole% chloroacetic acid. It is desired to
make 500 kg of a saturated solution at 25°C. Find the quantities of spent solution and chloroacetic acid
required to make the above solution.
(Solubility of chloroacetic acid in ether is 190 g/100 g ether at 25°C).
Molecular weight of chloroacetic acid (CAA) = 94.5 g/mol
Molecular weight of ether = 74 g/mol
Mass % CAA= 24.2%
Mass % Ether= 75.8%
At saturation,
Mass % CAA = 65.5%
Mass % Ether = 34.5%
Let us assume that in the saturated solution, x is the amount of spent solution and y is the amount of
pure CAA.
x+y = 500
CAA Balance:
0.242*x+y=0.655*500
x= 227.43 kg
y = 272.57 kg
Ether Balance:
0.758*x=0.345*500
x = 227.57 kg
y = 500 – 227.57 kg
11. A mixed fertilizer, having the NPK composition 10:26:26 as %N, %P 2 O 5 and % K 2 O by mass,
respectively, is to be formulated by mixing ammonia, phosphoric acid, potassium chloride and/or urea.
a) If anhydrous ammonia, anhydrous phosphoric acid and pure potassium chloride are used
for mixing, calculate amount of each of them for formulating 100 kg fertilizer.
b) If pure urea is used instead of anhydrous ammonia, calculate the amount of urea required
for formulating 100 kg fertilizer.
2H 3 PO 4 P2O5 + 3 H2O
2KCl + H 2 O K 2 O + 2HCl
Amount of NH 3 required: 12.14 kg
Amount of H 3 PO 4 required: 35.89 kg
Amount of KCl required: 41.2 kg
CH 4 N 2 O = 60.06 g/mol
Amount of Urea required 21.43 kg
12. You have a tank that initially holds solution containing 40% of salt in water.
i. You add 20 kg of salt to the tank and allow it to dissolve. How much salt and water
do you now have in the tank?
ii. Further, you withdraw 40 kg solution from the tank. How much salt and water do
you now have in the tank?
(ii) Salt in the withdrawn 40 kg solution = 15 kg
Water in the withdrawn solution = 25 kg
If water is 100 kg then salt would be 40 kg
Total mass of the solution will be = 100 + 40 = 140 kg
Salt in the tank after adding 20 kg = 40 +20 = 60 kg
Water in the tank = 100 kg
Mass fraction of salt in the modified solution = 60/160 = 0.375
Mass fraction of water = 0.635
Amount of salt that is withdrawn = 0.375 * 40 = 15 kg
Water = 25 kg
Salt left in the tank = 45 kg
Water left in the tank = 75 kg
