QUESTION 7 A gas stream entering an absorber is 20 mol% CO2 and 80 mol% air. The

Question

QUESTION 7 A gas stream entering an absorber is 20 mol% CO2 and 80 mol% air. The flowrate is 1 m3/min at 1 bar and 360 K. When the gas stream exits the absorber, 98% of the incoming CO2 has been absorbed into a flowing liquid amine stream. a. What are the gas stream mass flow rates of the inlet and outlets in g/min? b. What is the volumetric flow rate of the gas outlet of the absorber if the stream is at 320 K and 1 bar? t ans "flow ins 1087 g/min, flow out-799.2 g/min, Volumetric flow rate=714|iters/min. o "flow ins 1087 g/min, flow out-1087 g/min, Volumetric flowrate=714 liters/min. o "flow ins 799.2 g/min, flow outs 1087 g/min, Volumetric flow rate:0.714 liters/min. o "flow ins 799.2 g/min, flow outs 1087 g/min, Volumetric flowrate=0.714 m3/min

Explanation / Answer

Pressure P = 1 bar = 10^5 Pa

Temperature T = 360K

Volumetric flow rate V = 1 m3/min

Moles of inlet gas stream n = PV/RT

= 10^5*1/8.314*360

= 33.41 mol/min

Part a

Average molecular weight of inlet gas stream

M = M1X1 + M2X2

= 0.2*44 + 0.8*29 = 32 g/mol

Mass flow rate of inlet = n x molecular weight

= 33.41*32

= 1069 mol/min

Flow rate of CO2 = 0.2*33.41 = 6.68 mol/min

Flow rate of inert gas = 0.8*33.41 = 26.73 mol/min

In the exit gas stream

Flow rate of CO2 = (1-0.98)*6.68 = 0.1336 mol/min

Flow rate of inert gas outlet = 26.73 mol/min

Total flow rate outlet n = 26.8636 mol/min

Part b

Mass flow rate outlet stream = 26.8636*29 = 780 g/min

Part c

Volumetric flow rate of outlet gas stream

V = nRT /P

= 26.8636 x 8.314 x 320/10^5

= 0.7147 m3/min x 1000L/m3

= 714.7 L/min

Option A is the correct answer

The exact values may vary depending on the exact molecular weight.

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