Consider steady flow combustion of methane gas. Fuel and 30 % excess air enter e

Question

Consider steady flow combustion of methane gas. Fuel and 30 % excess air enter enter the combustor at 101 kPa and 25 oC. The combustion products leave the chamber at 1600 K and the same pressure, and they consist of CO2, H2O, NO, O2, and N2. If the mole fraction of NO in the combustion products is written as y * 10-4, what is y?

(Hint: The stoichiometric reaction you want to consider from the table is the forming of NO:   N2 + O2 <---> 2 NO, since the combustion is considered complete there is no methane left in the products. The actual reaction is the full reaction including combustion.)

Explanation / Answer

1 moL CH4(g) + 2 moL O2(g) ----> 1 moL CO2(g) + 2 moL H2O(g)

1 moL N2(g) + 1 moL O2(g) ----> 2 moL NO(g)

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1 moL N2(g) +1 moL CH4(g) + 3 moL O2(g) ----> 1 moL CO2(g) + 2 moL H2O(g) +2 moL NO(g)

If:

Molar fraction of NO is XNO= Yx10-4, what is Y?

And if:

XNO= (2 moL / 10 moL) = 0.2

Then:

XNO/10-4 = Y

Y = 0.2/10-4

Y = 2000 Pa = 2 Kpa

"Y" Is the partial pressure corresponding to hiponitroso anhydride

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