Question

Benzene gas (C6H6) at 25° C and 1 atm, enters a combustion chamber operating at steady state and burns with 95% theoretical air entering at 25° C and 1 atm. The combustion products exit at 1000 K and include only CO2 , CO , H2O , and N2.
Determine:
The mass flow rate of the fuel, in kg/s, to provide heat transfer at a rate of 1000 kW

Answer 1

Answer:

The mass flow rate of the fuel, in kg/s = $0.037 kgs^{-1}$

Explanation:

The actual balanced equation for the combustion reaction can be written as :

$C_6H_6 +(7.5)(O_2+3.76N_2) ------> 6CO_2 +3H_2O+28.2N_2$

At 95% theoretical air entering at 25° C and 1 atm, the equation of the reaction can be represented as :

$C_6H_6 + 7.125(O_2+3.76N_2) ------> 5.25 CO_2+ 0.75CO+3H_2O+26.79N_2$

The Energy rate formula can be use to determine the mass flow rate of the fuel and which is given as:

$\frac{Q_{Cv}}{n_{fuel}}} = 5.25( \bar h_f^0 + \delta \bar h)_{co_2} + 0.75 (\bar h^0_f +\delta \bar h )_{co} +3( \bar h_f^0 + \delta \bar h)_{H_2O} + 26.79( \bar h_f^0 + \delta \bar h)_{N_2}- (\bar h_f^0)_{fuel}$

from ideal gas table at respective amount for each compound; we have:

$\frac{Q_{Cv}}{n_{fuel}}} = 5.25 (-393520 + 42769-9364) + 0.75 (-110530+30355-8664 ) +3(-241820+35882-9904) + 26.79( 30129-8669)-82930$

$\frac{Q_{Cv}}{n_{fuel}}} = -2112780 kJ/mol.k$

$n_{fuel} = \frac{Q_{cv}}{-2112780kJ/kmol}$

Given that;

The combustion products exit at  1000 K = $Q_{cv}$

$n_{fuel}= \frac{-1000}{-2112780}$

$n_{fuel}= 4.73*10^{-3} kmol/s$

From the ideal gas tables M = 78.11 kg/kmol

∴

$n_{fuel}= 4.73*10^{-3} kmol/s * 78.11 kg/kmol$

$n_{fuel} = 0.037 kgs^{-1}$

∴ The mass flow rate of the fuel, in kg/s = $0.037 kgs^{-1}$

Answer 2

Answer:

Explanation:

Benzene gas is burned with 95 percentage theoretical air during a steady â flow combustion process. The mole fraction of the CO in the products and the heat transfer from the combustion chamber are to be determined

Assumption

steady operating conditions exit .

Air and combustion gases are gases

Kinetic and potential energies are negligible

The fuel is burned with insufficient amount of air and thus the products will contain some CO as well as CO2, H2O and H2

Combustion equation is

C6H6 + ath (O2 + 3.76N2) â 6CO2 +3H2O +3.76ath N2

Where ath is the stoichiometric coefficient and is determined from the O2 balance

ath = 6+1.5 = 7.5

then the actual combustion equation can be written as

C6H6 + 0.95 * 7.5(O2 + 3.76N2) â xCO2 + (6-x) CO + 3H2O + 26.79N2

O2 balance: 0.95 * 7.5 = x +(6-x)/2 +1.5 â x=5.25

Thus C6H6 + 7.125(O2 +3.76 N2) â 5.25 CO2 + 0.75CO + 3H2O + 26.79N2

The mole fraction of CO in the products is

yCO = N CO/ N total = 0.75/5.25 + 0.75+3+26.79

=0.021 or 2.1% mole fraction of the CO in the products