Combustion products with molar analysis of 10 CO2 20 H2 O 70

Combustion products with molar analysis of 10% CO_2, 20% H_2 O, 70% N_2 enters an engine exhaust pipe at 827 degree C and are cooled as they pass through the pipe exiting at 43 degree C and 1 atm. Determine if there will be any condensate; if there is then determine the rate of condensation, in kg, per kg of the mixture. Also, determine the heat transfer at steady state, in kJ per kg of the mixture.

Solution

solution:

1) here molar mass of mixture of CO2,H2O and N2 is m=44+18+28=90 kg/kmol

2) here specific heat of exhaust assume Cp=1.005 kj/kg k

3) here total energy lost by gas is

q=m*cp*(827-43)=90*1.005*(827-43)=70912.8 kj/kmol

4) here total energy lost by vapor is 20%

Qwater=.2*Q=.2*70912.8=14182.56 kj

5) this energy is lost by vapor and water

Qwater=Qgas+Qliquid

14182.56=mwg*1.005*(827-100)+Mw*4.187*(100-43)

mwg=18 kg/kmol -it is mass of water vapour

14182.26=18*1.005*(827-100)+mw*4.187*(100-43)

mw=mass of condensate per kmol=4.3205 kg/kmol

5) mass of condensate produce per kg of mixture is

for 90 kg of mixture-4.3205 kg of condensate

for 1 kg mixture- x kg of condensate

x=4.3205*1/90=.0480057 kg/kg of mixture

6) heat transfer per kg of mixture is given by

heat transfer 70912.8 kj-90 kg of mixture

heat transfer is x-1 kg of mixture

x=70912.8*1/90=787.92 kj/kg of mixture