Superheated steam at 10 bar abs. and 300ºC admitted into the cylinder of a steam engine expands isentropically to a pressure of 0.7 bar. The pressure then falls at constant volume to a back pressure of 0.28 bar. Determine (a) modified Rankine cycle efficiency (b) steam consumption per kWhr (c) mean effective pressure (d) heat removed in the condenser per kg of steam.
Answers
Explanation:
Index Number: 150131A
2. Question 1 A steam power plant operates between a boiler pressure of 40 bar and a condenser pressure of 0.04 bar. Calculate the cycle efficiency, work ratio and specific stream consumption for I. Carnot cycle P1 = P2 = 40 bar = 4000 kPa P4 = P3 = 0.04 bar = 4 kPa T1 = T2 = 250.35℃ = 523.5 K T3 = T4 = 28.96℃ = 302.11 K h1 = hf @40 bar = 1087.4 kJ/kg h2 = hg @40 bar = 2800.8 kJ/kg hf @0.04 bar = 121.39 kJ/kg hfg @0.04 bar = 2432.3 kJ/kg s1 = s4 = sf @40 bar = 2.7966 kJ/kg.K s2 = s3 = sg @40 bar = 6.0696 kJ/kg.K sf @0.04 bar = 0.4224 kJ/kg.K sfg @0.04 bar = 8.0510 kJ/kg.K s4 = sf + xsfg x4 = s4−sf sfg x4 = 2.7966−0.4224 8.0510 x4 = 0.2949 s3 = sf + xsfg x3 = s3−sf sfg x3 = 6.0696−0.4224 8.0510 x3 = 0.7014
3. h3 = hf + xhfg h3 = 121.39 + 0.7014 × 2432.3 h3 = 1827.41 kJ/kg h4 = hf + xhfg h4 = 121.39 + 0.2949 × 2432.3 h4 = 838.68 kJ/kg Turbine work W23 = h3 − h2 W23 = 1827.41 − 2800.8 W23 = −973.39 kJ/kg Compressor work W41 = h1 − h4 W41 = 1087.4 − 838.68 W41 = 248.72 kJ/kg Heat transfer in the boiler Q12 = h2 − h1 Q12 = 2800.8 − 1087.4 Q12 = 1713.4 kJ/kg Heat transfer in the condenser Q34 = h4 − h3 Q34 = 838.8 − 1827.41 Q34 = −988.61 kJ/kg Net-work for the cycle W = W23 + W41 W = −973.39 + 248.72 W = −724.67 kJ/kg