Dry saturated steam at 30 bar expands to a pressure of 0.04 bar through a turbin

Question

Dry saturated steam at 30 bar expands to a pressure of 0.04 bar through a turbine whose [isentropic] efficiency is 80%. The steam is then condensed until its entropy is equal to that of saturated water at 30 bar. The wet steam is compressed to 30 bar in a compressor whose [isentropic] efficiency is 85%. Finally, the steam receives heat in a boiler at a constant pressure of 30 bar until it is dry saturated. Calculate the net work output, the heat input in the boiler and the thermal efficiency of the cycle. Sketch the cycle on the T-s and h-s diagrams for water.

Explanation / Answer

For steam properties see http://www.irc.wisc.edu/properties/

At 30 bar and quality = 1 (sat.vapor), we get, h1 = 2800 kJ/kg, s1 = 6.19 kJ-kg-K

At s2_is = s1 = 6.19 kJ/kg-K (isentropic) and P2 = 0.04 bar, we get, h2_is = 1860 kJ/kg

Entropy of sat.water(quality = 0) at 30 bar is = 2.65 kJ/kg-K

At s3 = 2.65 kJ/kg-K and P3 = 0.04 bar, we get h3 = 794 kJ/kg

At s4_is = s3 = 2.65 kJ/kg-K and P4 = 30 bar, we get h4_is = 1010 kJ/kg

Turb Eff = (h1 - h2)/(h1 - h2_is)

0.8 = (2800-h2)/(2800-1860)

h2 = 2048 kJ/kg

Comp Eff = (h4_is - h3)/(h4 - h3)

0.85 = (1010 - 794)/(h4 - 794)

h4 = 1048 kJ/kg

a) Net work output = Turb work - Compressor work

= (h1 - h2) - (h4 - h3)

= (2800-2048)-(1048-794)

= 498 kJ/kg

b) Heat supplied = h1 - h4 = 2800-1048 = 1752 kJ/kg

Efficiency = Work/heat supplied = .2842 = 28.4%

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