Question

9.77

Methane is burned with 25% excess air in a continuous adiabatic reactor. The methane enters the reactor at 25 C and 1.10 atm at a rate of 550 L/s, and the entering air is at 150 C and 1.1 atm. Combustion in the reactor is complete, and the reactor effluent gas emerges at 1.05 atm.

a) Calculate the temperature and the degrees of superheat of the reactor effluent. (Consider water to be the only condensable species in the effluent).

b) Suppose only 15% excess air is supplied. Without doing any additional calculations, state how the temperature and degrees of superheat of the reactor effluent would be affected [increase, decrease, remain the same, cannot tell without more information] and explain your reasoning. What risk is involved in lowering the percent excess air?

Answer 1

Methane is burned with 25% excess air

P = 1.10 atm = 1.10×1.013×105 = 111430 Pa

V = 550 L/s = 0.550 m3/s

T = 25°C = 298 K

n = PV/RT

n = (111430) (0.550) /(8.314×298)

n = 24.736 moles

The reaction is complete so no CO is formed and mthane is completely consumed

According to stiochiometry

For 24.736 moles methame O2 required is 24.736(2) = 49.472 moles

Oxygen is 25% in excess

Oxygen supplied = 49.472(1.25) = 61.84 moles

Air supplied = 61.84/0.21 = 294.476 moles

Air is supplied at 150°C and 1.1 atm

At 25°C from Perry handbook

∆Hf(CO2) = -393. 15 KJ/mol

∆Hf(H2O) (g) = -241. 82 KJ/mol

∆Hf(CH4) (g) = -74. 87 KJ/mol

∆Hrnx = ∆Hf products - ∆Hf reactants

∆Hrnx = (-393.15) +(2) (-241.82) -(-74.87)

∆Hrnx = -801. 92 KJ/mol at 25°C

The Cp for exit gases is calculated at average temperature of 1000 K

Component	moles	mol%	Cp(J/mol K) (1000 K)
O2	61.84-49.472 = 12.368	3.87	34.88
N2	294.47(0.79) = 232.636	72.87	32.676
CO2	24.736(1) = 24.736	7.749	54.296
H2O	24.736(2) = 49.472	15.49	41.184
Total	319.212	100

Cp of O2 at 150°C = 30.3328 J/mol K

Cp of N2 at 150°C = 29.2964 J/mol K

∆H =∆Hproducts - ∆Hreactants

∆Hproducts = 24.736(54.296) (T- 25) +(49.472)(41.184) (T-25) +(12.368) (34.88) (T-150) + 232.636(32.676) (T-150)

∆Hproducts = 3380.52(T-25) +(8033) (T-150)

∆Hreactants  = 12.368(30.3328) (150-0) +(232.636) (29.2965) (150-0)

∆Hreacatants = 1078586.497

∆Hrnx = -801. 92(24.736)

Substituting we get

-801.92(24.736)(1000) = 3380.52(T-25) +(8033) (T-150) -1078586. 497

Solving we get

T = 1530.48 °C

The temperature of outlet gas is 1530.48°C

The pressure of outlet gas = 1.05 atm

Partial pressure of water vapor in air = 1.05(0.41184) = 0.4324 atm

The water condenses once partial pressure equals saturation pressure

From perry handbook

At PA = 0.4324 atm , the saturation temperature = 77.77°C

The degrees of superheat = 1530.48-77.77 = 1452.71 °C

B) If the percent excess air is decreased to 15%

The temperature of outlet gas increases and also superheat increases

The more the amount of excess air the more heat the components hold and adiabatic temperature decreases

The risk of lower excess air is incomplete combustion , overshoot of temperature and also high cooling costs

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