Question.NoA 19 A stream of air (21 mole%o, the rest N2) flowing at a rate of 10

Question

Question.NoA 19 A stream of air (21 mole%o, the rest N2) flowing at a rate of 10.0 kg/h is mixed with a stream of CO.. The CO, enters the mixer at a rate of 20.0 m'/h at 150 °C and a pressure 1.5 bar. Assume ideal gas behavior. (a) What is the mole percent of CO; in the product stream? (b) Express the flowrate of the exit stream in SCMH. (c) Determine the molar flowrate of CO2 that enters the mixer using the /9 I 3 13 compressibility factor equation of state. Is /5 (d) Assuming the answer in part (c) to be correct, what is the percentage error in the calculation if ideal gas behavior is assumed? I A /21

Explanation / Answer

Ideal gas equation => PV = nRT

or, n = PV/RT where n = no. of moles ;

150 degrees = 423.15 K ; mol. wt of O2 = 32 gm/mol ; mol. wt of N2 = 28gm/mol ; mol. wt of CO2 = 44 gm/mol

In 1 hour,

Moles of CO2 that got into the mixer = [(1.48 atm)*(20*103 L)]/[(0.082 Latmmol-1K-1)*(423.15K)] = 853.07 mol

Let n moles of O2 + N2 was given.

Therefore, for 1 hour,

(21n/100 mol)*(32 gm/mol) + (79n/100 mol)*(28 gm/mol) = 10000 gm

=>n = 346.74

Therefore, mole fraction of CO2 = (853.07 mol)/[(853.07 mol + 346.74 mol) = 0.711

therefore, mole% of CO2 = 71.1% (Ans to a)

SCMH = standard cubic meters per hour

In 1 hour,

from ideal gas equation, we have V = nRT/P = ({346.74 + 853.07} mol)*(0.082 Latmmol-1K-1)*(423.15 K)/(1.48 atm) = 28129 L = 28.129 m3

Therefore, flowrate of exit stream = 28.129 SCMH (Ans to b)

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