To evaluate the use of renewable resources, an experiment was carried out with r

Question

To evaluate the use of renewable resources, an experiment was carried out with rice hulls. After pyrolysis, the gas analyzed contained 6.4mol% CO2, 0.1% O2, 39% CO, 51.8% H2, 0.6% CH4, and 2.1% N2. It entered a combustion chamber at 90°F and a pressure of 35.0 inHg (abs), and was burned with 40% excess air at 70°F and an atmospheric pressure of 29.4 inHg; 10% of the CO remains; assume complete conversion of H2 and CH4. How many cubic feet of air were supplied per cubic foot of entering gas? How many cubic feet of product gas were produced per cubic foot of entering gas if the exit gas was at 29.4 inHg and 400°F? Assume the gas mixtures are ideal.

Explanation / Answer

Assume the gas mixtures are ideal

Basis: 100 moles of   gas entering combustion chamber

CO2 = 6.4mol%
O2 = 0.1%
CO = 39%
H2 = 51.8%
CH4 = 0.6%
N2 = 2.1%

Average molar mass of entering gas, MW = 44*0.064+32*0.001+28*0.39+2*0.518+16*0.006+28*0.021=15.488 g/mol
Mass of 100 moles of entering gas, W = 100*15.488 = 1548.8 g

Volume of gas at at 90°F and a pressure of 35.0 inHg (abs), V = n*R*T/P
n = 100 moles
R = 8.314 J/(mol*K)
P = 35 in Hg = 35*1.013E5/29.92 = 1.185E5 N/m2
T = 90 F = (90-32)*9/5+273 = 377.5 K
Vin = n*R*T/P = 100*8.314*377.5/1.185E5 = 2.65 m3 = 2.65/0.0283 = 93.64 ft3

Exit gas from combustion chamber analysis:

Assume complete conversion of H2 and CH4.
10% of the CO remains
90% conversion of CO
Moles of CO coming entered = 39 mol
Moles of CO unreacted = 3.9 mol
Moles of CO reacted = 39 - 3.9 = 35.1 mol

Reacting mixture:

O2 in mixture = 0.1 mol
(i)CO = 39 mol with 90% conversion
Moles of CO reacted = 39 - 3.9 = 35.1 mol
CO + 1/2O2 = CO2
Requirement of O2 = 35.1/2=17.55 mol

(ii)CH4 = 0.6 mol
CH4 + 2O2 = CO2 + 2H2O
Requirement of O2 = 0.6*2= 1.2 mol

(iii)H2 = 51.8 mol
H2 + 1/2O2 = H2O
Requirement of O2 = 51.8*1/2 = 25.9 mol

O2 available in feed = 0.1 mol
Total requirement of O2 = 17.55+1.2+25.9-0.1=44.55 mol
22 moles of O2 is supplied by 100 moles of air
Theoretical air requirement = 44.55*100/22= 202.50 moles of air
Actual air supplied is 40% excess air = 202.5*1.4=283.50 mol

Volume of air at 70°F and an atmospheric pressure of 29.4 inHg
n = 283.5 moles
R = 8.314 J/(mol*K)
P = 29.4 in Hg = 29.4*1.013E5/29.92 = 0.995E5 N/m2
T = 70 F = (70-32)*9/5+273 = 341.4 K
Vair = n*R*T/P = 283.5*8.314*341.4/0.995E5 = 8.09 m3 = 8.09/0.0283 = 285.77 ft3

cubic feet of air were supplied per cubic foot of entering gas = Vair/Vin

Vair/Vin = 285.77/93.64= 3.05

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