Please see the attached file.

Steam at 800 kPa, 600°C enters a steady flow turbine and leaves at 100 kPa. The steam undergoes an expansion process through the turbine, does work in the amount of 600 kJ/kg, and loses 50 kJ/kg of energy by heat transfer to the surroundings.

(a) Determine the exit temperature of the steam, in °C.
(b) If the surroundings have a temperature of 25°, what is the total entropy change for this process?
(c) Comment on your answer to part (b). Is this process possible?
(d) Determine the maximum work that the steam could have done and calculate the isentropic efficiency of the process.

SOLUTION:

a) Exit tmperature:

From first law of thermodynamics, for flow process

H2-H1 = Q-WS

In the given problem Q = 50 KJ / Kg of stem (Note the negative sign. As heat is transferred from system to surroundings Q is negative). WS = 60 KJ / Kg of steam.

Fro steam tables enthalpy H1 of steam at inlet conditions (P=800 K Pa and T= 600 C) is 3700 KJ / Kg and Entropy S1 = 8.17 KJ /Kg-K.

H2 = H1 +Q-WS = 3700-50-600 = 3050 KJ / Kg of steam.

Thus the exit conditions of steam (State 2) are P=100 K Pa (given) and H = 3050 KJ / Kg. At these conditions from steam tables the temperature is 288 C and entropy = 8.17 KJ / Kg.

b) Total Entropy change:

Entropy change for system ( steam ) = S2-S1= 8.17 - 8.14 = 0.03 KJ /Kg.
Entropy change for surroundings = Qsur/Tsur = 50/298.15 = 0.1677 KJ /Kg
( Note that here Qsur the amount of heat transferd to surroundings is positive)
Total entropy change = 0.03 + 0.1667 = 0.1997 KJ / Kg

c) The process is feasible because the total entropy change is > 0 (Zero)
d) Maximum work:

Lost work = ?Stotal Tsur = 0.1977 x298.15 = 58.94 KJ / Kg

Wmax = WS + Wlost = 600+58.94 =658.94 KJ Kg

Isentropic Efficiency = WS / Wlost = 600/658.94 = 0.9105

Note: There is a good excel software freely available for steam properties. This can be used for getting the steam properties. This can be downloaded from the site http://www.x-eng.com