Question

The air quantity at 25 ^oC and 80% RH required for complete combustion for refuse (dry basis) is 7.0 lb/10000 Btu. Air supplied for complete combustion is 30% in excess.

please find the volume of air at 25 ^oC and 80% RH needed for complete combustion (30% excess)

Answer 1

Air needed for complete combustion of refuse = 7 lb/10000 BTU

R. H = 80%

P = 1 atm

From chemical engineer handbook

Saturation pressure of water at 25°C

= 0.0419 atm

R. H = PA/Pas

PA = 0.80(0.0419) = 0.03352 atm

Mole fraction of water vapor in air = 0.03352/1 = 0.03352

Mole fraction of oxygen in air = 0.21(1-0.03352) = 0.2029

Mole fraction of nitrogen in air = 0.79(1-0.03352) = 0.7635

M. W of water vapor = 18 lb/lbmole

M. W of oxygen gas = 32 lb/lbmole

M. W of nitrogen gas = 28 lb/lbmole.

Substuting for water vapor we get

x1= 0.0211

Similarly for oxygen

x2 = 0.2280

x3 = 1-(0.2280-0.0211) = 0.7509

Inlet air analysis

Component	mass%	mass(lb)	M.W	lbmole	mole%
Water vapor	2.11	0.1477	18	0.008205	3.35
Nitrogen gas	75.09	5.2563	28	0.1877	20.29
Oxygen gas	22.80	1.596	32	0.04987	76.36
Total	100	7		0.24577	100

Total moles = 0.24577 lbmoles

Air is 30% excess

Air supplied =0.24577(1.3) = 0.319501 lbmole

1 kg = 2.205 lb

Air supplied = (0.319501/2.205) = 0.144898 kmol

= 144.898 moles

According to ideal gas law

n = PV/RT

V = nRT/P

T = 25°C = 298 K

P = 1 atm = 1.013×105 Pa

V = (144.898×8.314×298) /(1.013×105)

V = 3.5438 m3

Volume of air needed = 3.543 m3

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