Answer:
The mass flow rate of the fuel, in kg/s = [tex]0.037 kgs^{-1}[/tex]
Explanation:
The actual balanced equation for the combustion reaction can be written as :
[tex]C_6H_6 +(7.5)(O_2+3.76N_2) ------> 6CO_2 +3H_2O+28.2N_2[/tex]
At 95% theoretical air entering at 25° C and 1 atm, the equation of the reaction can be represented as :
[tex]C_6H_6 + 7.125(O_2+3.76N_2) ------> 5.25 CO_2+ 0.75CO+3H_2O+26.79N_2[/tex]
The Energy rate formula can be use to determine the mass flow rate of the fuel and which is given as:
[tex]\frac{Q_{Cv}}{n_{fuel}}} = 5.25( \bar h_f^0 + \delta \bar h)_{co_2} + 0.75 (\bar h^0_f +\delta \bar h )_{co} +3( \bar h_f^0 + \delta \bar h)_{H_2O} + 26.79( \bar h_f^0 + \delta \bar h)_{N_2}- (\bar h_f^0)_{fuel}[/tex]
from ideal gas table at respective amount for each compound; we have:
[tex]\frac{Q_{Cv}}{n_{fuel}}} = 5.25 (-393520 + 42769-9364) + 0.75 (-110530+30355-8664 ) +3(-241820+35882-9904) + 26.79( 30129-8669)-82930[/tex]
[tex]\frac{Q_{Cv}}{n_{fuel}}} = -2112780 kJ/mol.k[/tex]
[tex]n_{fuel} = \frac{Q_{cv}}{-2112780kJ/kmol}[/tex]
Given that;
The combustion products exit at 1000 K = [tex]Q_{cv}[/tex]
[tex]n_{fuel}= \frac{-1000}{-2112780}[/tex]
[tex]n_{fuel}= 4.73*10^{-3} kmol/s[/tex]
From the ideal gas tables M = 78.11 kg/kmol
∴
[tex]n_{fuel}= 4.73*10^{-3} kmol/s * 78.11 kg/kmol[/tex]
[tex]n_{fuel} = 0.037 kgs^{-1}[/tex]
∴ The mass flow rate of the fuel, in kg/s = [tex]0.037 kgs^{-1}[/tex]
