Chemical Engineering

Multiple Component Systems

Question 1
Marks : +2 | -2
Pass Ratio : 100%
Waste water flowing with TSS concentration 40 mg/L at the rate of 10 Liter/min is discharged into the river if the river is flowing at the rate of 15 Liter/min, what is the concentration of TSS in the river?
10 mg/L
30 mg/L
45 mg/L
60 mg/L
Explanation:
Concentration of TSS in river = 40*15/10 = 60 mg/L.
Question 2
Marks : +2 | -2
Pass Ratio : 100%
Feed of a reactor has 0.25 mass fraction of H2O and 0.75 mass fraction of CO2 with rate 4 g/hr, if the product rate is 20 g/hr what is the mass of CO2 in the product side?
3 grams
9 grams
15 grams
21 grams
Explanation:
Mass fraction of CO2 will be same on both sides, => mass of CO2 in the product side = 20*0.75 = 15 grams.
Question 3
Marks : +2 | -2
Pass Ratio : 100%
A reactor with efficiency 50%, has the feed CO2 and SO2 with mass fraction 0.4 and 0.6 with the rate 10 Kg/hr, what is the rate of CO2 in product side?
0.5 Kg/hr
1 Kg/hr
1.5 Kg/hr
2 Kg/hr
Explanation:
Rate at product side = 10*50/100 = 5 Kg/hr, => rate of CO2 = 5*0.4 = 2 Kg/hr.
Question 4
Marks : +2 | -2
Pass Ratio : 100%
A reactor has feed of two streams, stream 1 has CO2 and H2O with mass fraction of 0.4 and 0.6 and rate 10 kg/hr, and stream 2 has CO2 and H2O with mass fraction 0.1 and 0.9 and rate 40 Kg/hr, what the mass fraction of CO2 in the product stream?
0.08
0.16
0.24
0.32
Explanation:
Mass fraction of CO2 in product stream = (0.4*10 + 0.1*40)/(10 + 40) = 0.16.
Question 5
Marks : +2 | -2
Pass Ratio : 100%
Water flowing with chlorine concentration 10 mg/L at the rate of 20 Liter/min, is discharged into a river if the concentration of chlorine in a river is 5 mg/L what is the flow rate of river?
10 Liter/min
20 Liter/min
30 Liter/min
40 Liter/min
Explanation:
Rate of flow of river = 20*5/10 = 10 Liter/min.
Question 6
Marks : +2 | -2
Pass Ratio : 100%
Feed of a reactor has 0.75 mass fraction of O2 and some SO2, if the product rate is 1000 Kg/hr then what is the rate of SO2 in the feed?
100 Kg/hr
250 Kg/hr
750 Kg/hr
1000 Kg/hr
Explanation:
Feed rate = Product rate = 1000 Kg/hr, => Rate of SO2 = (1 – 0.75)*1000 = 250 Kg/hr.
Question 7
Marks : +2 | -2
Pass Ratio : 100%
Water flowing with TSS at the rate of 40 Liter/min, is discharged into a river flowing at the rate of 20 Liter/min, if the concentration of TSS in the river is 5 mg/L, what is the concentration of TSS in water?
5 mg/L
10 mg/L
15 mg/L
20 mg/L
Explanation:
Concentration of TSS in water = 5*40/20 = 10 mg/L.
Question 8
Marks : +2 | -2
Pass Ratio : 100%
A reactor has two feed streams with the ratio of their rate as 1:9 and has O2 and O3 with the mass fraction of 0.6 and 0.4 in stream 1 and 0.2 and 0.8 in stream 2, what is the mass fraction of O2 in the product side?
0.24
0.48
0.72
0.96
Explanation:
Let the stream rates be 1 Kg/hr and 9 Kg/hr, => mass fraction of O¬2 in product side = (1*0.6 + 9*0.2)/(1 + 9) = 0.24.
Question 9
Marks : +2 | -2
Pass Ratio : 100%
A reactor has feed from 2 streams, one with rate 10 Kg/hr and other with 30 Kg/hr, and product side also has 2 streams one of which is at the rate of 15 Kg/hr, what is the rate of other product stream?
5 Kg/hr
15 Kg/hr
25 Kg/hr
35 Kg/hr
Explanation:
Total feed rate = Total product rate, => 10 + 30 = 15 + x, => x = 25 Kg/hr.
Question 10
Marks : +2 | -2
Pass Ratio : 100%
A reactor has feed rate of 10 Kg/ hr and 5 Kg/hr and product rate of 9 Kg/hr, what is the efficiency of the reactor?
10%
30%
60%
100%
Explanation:
Efficiency = 9/(10 + 5)*100 = 60%.