Question

Exhaust gas from a combustion process is 70% Nitrogen, 15% Carbon Dioxide, 10% Carbon monoxide, and 5% Oxygen. The absolute pressure is 0.8 atmospheres. The temperature of the exhaust gas is T=240^oF. Find:

The specific gas constant R of the exhaust gas in units of ft*lbf/slug*R

Answer 1

Basis -

1 mol/m3 of gas

Temperature T(K) = (240°F - 32) *(5/9) + 273.15 = 388.7K

Pressure P = 0.8 atm x 101325 Pa/atm = 81060 Pa

From the ideal gas equation

PV = nRT

R = PV/nT

= 81060 Pa x 1m3/1 mol x 388.7K

R = 208.5412 Pa-m3/mol·K

R = 208.5412 N-m/mol·K

Average molecular weight

M avg = M1 * X1 + M2 * X2 + M3 * X3 + M4 * X4

= 0.7 x 28 + 0.15 x 44 + 0.10 x 28 + 0.05 x 32

= 30.6 g/mol

R = (208.5412 N-m/mol·K) x (1 mol/30.6g)

= (6.815 N-m/g-K) x (14593.9 g/slug)

= 99458.477 N-m/slug-K x (1K/1.8R)

= (55254.7098 N-m/slug-R) x (3.28ft/m)

= 181235.44 N-ft/slug-R x (0.2248lbf/N)

= 40741.728 ft-lbf/slug-R