You have 5 kg of steam in a pistoncylinder system at 10 bar

You have 5 kg of steam in a piston/cylinder system at 10 bar, 280 degreeC. The water isentropic ally undergoes an adiabatic expansion to 1 bar. What will the final temperature of the water be? How much work will the water\'s expansion produce? If, Instead, the expansion is not isentropic, and has an isentropic efficiency of 65%, how much work will the system produce in expanding to 1 bar? What will the final temperature of the water in Part C be? How much entropy will be generated in Part C?

Solution

c)

From steam properties at T1 = 280 deg C and P1 = 10 bar, we get h1 = 3010 kJ/kg, s1 = 7.05 kJ/kg-K

For isentropic expansion, s2_s = s1 = 7.05 kJ/kg-K

From steam properties at P2 = 1 bar and s2_s = 7.05 kJ/kg-K, we get h2_s = 2560 kJ/kg

Ideal work done per kg = h1 - h2s

= 3010 - 2560

= 450 kJ/kg

Ideal work done = 5*450 = 2250 kJ

Actual work done = efficiency * Ideal work done

= 0.65*2250

= 1462.5 kJ

d) Isentropic efficiency = (h1 - h2) / (h1 - h2s)

0.65 = (3010 - h2) / (3010 - 2560)

h2 = 2717.5 kJ/kg

From steam properties at P2 = 1 bar and h2 = 2717.5 kJ/kg, we get T2 = 120 deg C