Question 3.1: Steam at 8.8 bar and 0.9 dry is supplied to an engine which ...
Steam at 8.8 bar and 0.9 dry is supplied to an engine which expands it adiabatically to the release pressure of 1.2 bar when the pressure falls at constant volume to the exhaust pressure of 0.2 bar. Calculate :
(i) Steam consumption in kg/kWh ;
(ii) Mean effective pressure ;
(iii) Heat to be removed by the condenser per kg of exhaust steam.
Steam pressure, p_{1} = 8.8 bar
Dryness fraction, x_{1} = 0.9
Pressure, p_{2} = 1.2 bar
Exhaust pressure, p_{3} = 0.2 bar
From steam tables :
At p_{1} = 8.8 bar. s_{f_1} = 2.0848 kJ/kg K, s_{g_1} = 6.6269 kJ/kg K
h_{f_1} = 738.5 kJ/kg, h_{fg_1} = 2032.8 kJ/kg
At p_{2} = 1.2 bar. s_{f_2} = 1.3609 kJ/kg K, s_{fg_2} = 7.2984 kJ/kg K
h_{f_2} = 439.4 kJ/kg, h_{fg_2} = 2244.1 kJ/kg,
v_{g_2} = 1.428 m³/kg
At p_{3} = 0.2 bar. h_{f_3} = 251.5 kJ/kg.
Equating entropy at points 1 and 2,
s_{1} = s_{2}
∴ s_{f_1} + x_{1} ( s_{g_1} – s_{f_1}) = s_{f_2} + x_{2} ( s_{g_2} – s_{f_2})2.0848 + 0.9 (6.6269 – 2.0848) = 1.3069 + x_{2} (7.2984 – 1.3609)
6.1727 = 1.3069 + 5.9375 x_{2}
∴ x_{2} = \frac { 6.1727 – 1.3069} {5.9375} = 0.819h_{1} = h_{f_1} + x_{1} h_{fg_1} = 738.5 + 0.9 × 2032.8 = 2568.02 kJ/kg
h_{2} = h_{f_2} + x_{2} h_{fg_2} = 439.4 + 0.819 × 2244.1 = 2277.3 kJ/kg
v_{2} = x_{2} v_{g_2} = 0.819 × 1.428 = 1.169 m³
Total work done = Area ‘451234’ = Area ‘512a5’ + area ‘a234a’
= (h_{1} – h_{2}) + (p_{2} – p_{3}) × v_{2}
= (2568.02 – 2277.3) + (1.2 – 0.2) × 10^{5} × 1.169 × 10^{–3}
= 290.72 + 116.9 = 407.62 kJ/kg.
(i) Steam consumption in kg/kWh :
= \frac {1 × 3600} {407.62} = 8.83 kg/kWh.
(ii) Mean effective pressure, p_{m} :
p_{m} × v_{2} = 407.62\frac{ p_{m} × 10^{5} × v_{2}}{10^{3}} = 407.62
p_{m} = \frac{407.62 × 10^{3}}{10^{5} × v_{2}} = \frac{407.62 × 10^{3}}{10^{5} × 1.169} (where p_{m} is in bar)= 3.48 bar.
(iii) Heat to be removed by the condenser :
Heat to be removed by the condenser
= Heat supplied – Work done = (h_{1} – h_{f_3} ) – 407.62
= (2568.02 – 251.5) – 407.62 = 1908.9 kJ/kg.
