Question 14.6: Find out the heat generated by burning of n-Octane for a pro......
Find out the heat generated by burning of n-Octane for a product temperature of 257°C (530 K) without excess air supply.
| \overline{h}_{530K} kJ/ kmol |
\overline{h}_{298K} kJ/ kmol |
\overline{h^0_f} kJ/ kmol |
Substance |
| – – – | – – – | -208,450 | C_8 H_{18} (g) |
| 15,708 | 8682 | 0 | O_2 |
| 15,469 | 8669 | 0 | N_2 |
| 17,889 | 9904 | -241,820 | H_2O(g) |
| 19,029 | 9364 | -393,520 | CO_2 |
Given: Combustion reactants: Air and n-octane
Find out: Heat generated per kg of n-octane.
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.
Molar masses of propane and air are 44 kg/kmol and 29 kg/kmol, respectively.
Combustion reaction with no-excess air will be:
C_8H_{18} + (12.5 O_2 + 47.25 N_2) \longrightarrow 8CO_2 + 9H_2O + (47.25 N_2)
From the first law:
Q = W + \sum{N_P} (H^0_f + H – H_0)_P – \sum{N_R} (H^0_f + H – H_0)_R
With W = 0,
Q = \sum{N_P}(H^0_f + H – H_0)_P – \sum{N_R} (H^0_f + H – H_0)_R
From the given table data:
\sum{N_P} (H^0_f + H -H_0)_P = 8\times (-3,93,520 + 19,029 – 9,364) + 9 \times (-2,41,820 + 17,699 – 9,904) + 47.25 \times (0 + 15,469 – 8,669)
= –4,85,55,765 KJ/kmol
and
\sum{N_R} (H^0_f + H – H_0)_R = 1 \times (-2,08,450) + 12.5 \times 0 + 47.25 \times 0 = -2,08,450 kJ/kmol
So, Q = –48,555,765 – (–2,08,450) = –46,47,315 kJ/kmol.
Negative sign shows indicates heat outgoing.
Heat generated per kg:
= Heat generated / Mass of fuel = 46,47,315 (kJ/kmol)/114 (kg/kmol) = 40,765.92 kJ/kg
