262 POWER PLANT ENGINEERINGTotal mass of air = 32 × 1.52 = 48.64 kg
Total mass of exhaust gases = mass of fuel + mass of air
=1.52 + 48.64 = 50.16 kg
Mass of steam formed = 1.4 × 1.52 = 2.13 kg
Mass of dry exhaust gases = 50.16 – 2.13 = 48.03 kg
(iii) Heat carried away by dry exhaust gases
= mg × Cpg × (tg – tr)
= 48.03 × 1.0 × (305 – 25) = 13448 kJ
(iv) Heat carried away by steam
= 2.13 [hf + hfg + cps (tsup – ts)]
At 1.013 bar pressure (atmospheric assumed):
hf = 417.5 kJ/kg
hfg = 2257.9 kJ/kg
= 2.13 [417.5 + 2257.9 + 2.09 (305 – 99.6)]= 6613 kJ/kg
[Neglecting sensible heat of water at room temperature]
Heat balance sheet (20 minute basis)Item kJ Percent
Heat supplied by fuel 66728 100(i) Heat equivalent of I.P. 19080 28.60
(ii) Heat carried away by cooling water 16929 25.40
(iii) Heat carried away by dry exhaust gases 13448 20.10
(iv) Heat carried away steam in exhaust gases 6613 9.90
(v) Heat unaccounted for (by difference) 10658 16.00Total 66728 100.00Example 8. The average indicated power developed in a C.I. engine is 13 kW/m^3 of free air
induced per minute. The engine is a three-liters four-stroke engine running at 3500 r.p.m., and has a
volumetric efficiency of 81%, referred to free air conditions of 1.013 bar and 15°C. It is proposed to fit
a blower, driven mechanically from the engine. The blower has an isentropic efficiency of 72% and
works through a pressure ratio of 1.72. Assume that at the end of induction the cylinders contain a
volume of charge equal to the swept volume, at the pressure and temperature of the delivery from the
blower. Calculate the increase in brake power to be expected from the engine.
Take all mechanical efficiencies as 78%.
Solution. Capacity of the engine = 3 liters = 0.003 m^3Swept volume =3500
2
× 0.003 = 5.25 m^3 /min.