The engine uses n butane $(C_4H_{10})$ as liquid fuel is supplied with 40% excess air. Both fuel and air enter at 1 atmosphere pressure and 298 k. The products of combustion leave at 600K. Heat lost to the surroundings is 30% of power. The engine develops 60kW of power. - | Substance | $h^o$,(kJ/ kgmole) | $H_{298k}$ (kJ/kgmole) | $H_{600k}$(kJ/kgmole) | |-----------|----------------|-------------------|------------------| | $C_4H_{10}(I)$ | -126150 | 0 | - | | $O_2(g)$ | 0 | 8624 | 18260 | | $N_2(g)$ | 0 | 8660 | 17569 | | $CO_2(g)$ | -241830 | 8769 | 22285 | | $H_2O(g)$ | -393520 | 9856 | 20402 |

Mumbai university > MECH > SEM 3 > THERMO

Marks: 10M

Year: May 2015

Balanced reaction can be given as,

$C_4H_{10} + 7.5O_2 +7.5 × 3.76 N_2 → 4CO_2 + 5H_2O + 7.5 × 3.76 N_2$

When 140% $O_2$ is used

$C_4H_{10} + 1.4 × 7.5 O_2 + 1.4 × 7.5 × 3.76 N_2 → 4CO_2 + 5H_2O + 0.4 × 7.5 O_2 +1.4 × 7.5 × 3.76 N_2$

Qc = Hproducts – Hreactants

From table, we get the enthalpies of different chemicals at different temperatures

Hproducts= 5 (-393+ 20.4-9.86) + 4 (-241.8 + 22.3 – 8.77) + 3(18.26 -8.62) + 39.48 (17.57-8.66) = -2444.7 MJ/kmole

H reactants = 1 (-126.15) + 0 + 0 = -126.15 MJ/kmole

Qc = -2444.7 + 126.15 = -2318.55 MJ/kmole