Question

A mixture of hydrocarbon gaseous, containing on a volume basis: 80% methane, 10% ethane and 10% n-propane are burned in a boiler where saturated steam are produced at 1 atm. The gas mixtures enters the boiler at a rate of 100 m 3 /hr (STP) and 25oC and the water enters the boiler at 25oC. The percentage conversion of methane is 90%; of the methane burned, 75% reacts to form carbon dioxide and the balance reacts to form carbon monoxide; while ethane and n-propane burned completely to form carbon dioxide and water. The exit flue gas temperature from the boiler is 400oC and the thermal efficiency of the boiler is approximately 83%. The net heating values of methane, ethane and n-propane are 802.34 kJ/mol, 1437.2 kJ/mol and 2044.2 kJ/mol, respectively. Based on the given information,

(a) estimate the percentage of excess air required.

(b) estimate the flow rate (kg/hr) of steam produced.

Answer 1

Vol% = mol%

V = 100 m³/h

STP conditions

T = 273 K

P = 1 atm = 1.013×10⁵ Pa

n = PV/(RT)

n = (1.013×10⁵×100) /(8.314×273)

n = 4463.101 mol/h

Methane in feed = 4463.101(0.80)

= 3570.48 mol/h

Ethane in feed = 4463.101(0.10) =446.310 mol/h

Propane in feed = 4463.101(0.10/= 446.310 mol/h

All of ethane and propane reacts

90% of methane reacts out of which 75% form CO2 and 25% CO

The reactions occuring are

Methane reacted = 3570.48(0.90) =

3213.432 mol/h

Methane undergoing first reaction = 3213.432(0.75) = 2410.074 mol/h

Methane undergoing second reaction = 803.358 mol/h

All of ethane and propane completely burns

According to stiochiometry of four reactions O₂consumed = (2410.074)(2) +(803.358) (1.5) +(446.310) (3.5) +(446.310) (5) = 9818.82 mol/h

Heating value of methane for first reaction = 802.3 KJ/mol

Heating value of ethane = 1437.2 KJ/mol

Heating value of propane = 2044.2 KJ/mol

For second reaction

At T = 25°C

∆H(CO) = -110. 3 KJ/mol

∆H(CH4) = -74.86 KJ/mol

∆H(H2O) = -241. 83 KJ/mo

∆Hr (2) =∆H_products - ∆H_reactants

∆Hr(2) = -110. 3-(2×241.83) -(-74.86) =

-519. 1 KJ/mol

Heating values are heat liberated when fuel burns in complete combustion

Heat released due to methane = (2410.074) (-802.3) + (803.358(-519.1)) =

-2350625. 508 KJ/h

Heat released due to ethane = 446.310(-1437.2) = -641436. 732 KJ/h

Heat released due to propane = 446.310(-2044.2) = -912346. 902 KJ/h

Total heat released = ∆H_r =

-2350625. 508-641436.732-912346.902 = -3904409. 142 KJ/h

The process is essentially adiabatic

Hence

∆H_r = H_products- H_reactants

Product analysis

Reference enthaply temperature = 25°C

Product temperature = 400°C

Feed temperature = 25°C

The Cp is taken from handbook at average temperature of (25+400) /2 = 212.5°C

Let x be supplied air

Component	inlet moles	Cp(KJ/mol°C)	Hi(KJ)	outlet moles	Cp(KJ/mol°C)	Ho(KJ)
CH4	3570.48	0.04576	61269.375	357.048	0.04576	6126.75
CO2	-	-	-	4641.624	0.04413	76813.075
CO	-	-	-	803.358	0.029744	8960.655
H2O	-	-	-	9551.034	0.03506	125572.21
N2	x(0.79)	0.029532	8.7488(x)	x(0.79)	0.029532	8.7488(x)
O2	x(0.21)	0.03093	78.75(x)	x(0.21) -9818. 82	0.03093	(x(0.21)-303.696) (400-25)
Total

∆Hr = ∆H(products) - ∆H(reactants)

∆Hr = -3904409. 142

-3904409. 142 = 217472.690+(0.21x -303. 696) (400-25) -(61269.375) -(78.75x)

-4060612. 132=(0.21x-303.696) (400-25) -78. 75x

x = 50251.160 mol/h

O2 in feed = 50251.160(0.21) = 10552.743 mol/h

Theoretical O2 required = 9818.82 mol/h

% excess O2 =

(10552.743-9818.82) /(9818.82)) (100) = 7.474%

% excess air = 7.474%

B)

Heat released from reactor = -3904409. 142 KJ/h

Boiler efficiency =83%

Useful heat = -3904409.142(0.83) =

-3240659. 588 KJ/h

From steam tables

Enthaply of saturated steam at 1atm = 2674.9 KJ/Kg

Mass rate of steam produced

= 3240659.588/(2674.9) = 1211.506 Kg/h

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