Question 14.7: A mixture of benzene gas with 60 % excess air contained in a......
A mixture of benzene gas with 60 % excess air contained in a constant-volume tank undergoes the combustion. Determine the net heat transfer. Product-temperature is 1,000 K. Use the data given below:
| \overline{h}_{1000K} kJ/ kmol |
\overline{h}_{298K} kJ/ kmol |
\overline{h^0_f} kJ/ kmol |
Substance |
| – – – | – – – | 82,930 | C_6 H_{6} (g) |
| 31,389 | 8682 | 0 | O_2 |
| 30,129 | 8669 | 0 | N_2 |
| 35,882 | 9904 | -241,820 | H_2O(g) |
| 30,355 | 8669 | -110,530 | CO |
| 42,769 | 9364 | -393,520 | CO_2 |
Given: (1) Combustion reactants: Air and benzene, (2) Excess air: 60%
Find out: heat transferred per kmol of benzene.
Properties: Molar mass of C, H_2 and O_2 and air are: 12 kg/kmol, 2 kg/kmol, 32 kg/kmol and 29 kg/kmol
respectively.
Assumption: (1) Combustion is complete, (2) Air and combustion gases are ideal (3) Kinetic and potential energies are negligible (4) Combustion products are H_2O, CO_2 and N_2 only.
Combustion reaction with no-excess air will be:
C_6H_6 + (7.5 O_2 + 28.2 N_2) \longrightarrow 6 CO_2 + 3 H_2O (28.2 N_2)
From the first law:
For constant volume,
Q – W = \sum{N_R} (H^0_f + H – H_0 – pV)_R – \sum{N_P (H^0_f + H – H_0 – pV)_p}
With W = 0,
Q = \sum{N_R} (H^0_f + H – H_0 – pV)_R – \sum{N_P (H^0_f + H – H_0 – pV)_p}
Assuming ideal gas condition:
Q = \sum{N_R} (H^0_f + H – H_0 – RT)_R – \sum{N_P (H^0_f + H – H_0 – RT)_p}
From the given table data:
\sum{N_P} (H^0_f + H – H_0 – RT)_p = 6 \times (- 3,93,520 + 42,769 – 9,364 – 8.314 \times 1,000) + 3\times (-2,41,820 + 35,882 – 9,904 – 8.314 \times 1,000) + 28.2 \times (0 + 30,159 – 8,669 – 8.314 \times 1,000)
= –24,81,766 kJ/kmol
\sum{N_R} (H^0_f + H – H_0 – RT)_R = 1 \times (82,930 -8.314 \times 1,000) + 7.5 \times (- 8.314 \times 1,000) + 28.2 \times (- 8.314 \times 1,000)
= –7,996.89 kJ/kmol.
So, Q = –24,81,766 – (–7,996.89) = –24,73,769.11 kJ/kmol.
Negative sign indicates heat outgoing.
