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A fuel gas containing 45.00 mole% methane and the balance ethane is burned completely with pure oxygen at 25.00 degree C, and the products are cooled to 25.00 degree C. Suppose the reactor is continuous. Take a basis of calculation of 1.000 mol/s of the fuel gas, assume some value for the percent excess oxygen fed to the reactor (the value you choose will not affect the results), and calculate - Q(kW), the rate at which heat must be transferred from the reactor if the water vapor condenses before leaving the reactor and if the water remains as a vapor. Now suppose the combustion takes place in a constant-volume batch reactor. Take a basis of calculation 1.000 mol of the fuel gas charged into the reactor, assume any percent excess oxygen, and calculate -Q(kJ) for the cases of liquid water and water vapor as products.

User Vectran
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1 Answer

4 votes

Answer:

A)

- Q ( kw ) for vapor = -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B )

- Q ( kj ) for vapor = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

Step-by-step explanation:

Given data :

45.00 % mole of methane

55.00 % of ethane

attached below is a detailed solution

A) calculate - Q(kw)

- Q ( kw ) for vapor = -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B ) calculate - Q ( KJ )

- Q ( kj ) for vapor = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

since combustion takes place in a constant-volume batch reactor

A fuel gas containing 45.00 mole% methane and the balance ethane is burned completely-example-1
User Lukegf
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