Question

Solution kg a. Wump(2088.9W kg kg h2 is from table C-4, and hi is from table C-1 at 20 °C. Alternatively, you can use the equation for pump work for an incompressible fluid. wpump = riv(P2-P) = (20-) ( 0.001 002 ) (5000 kPa-10 kPa) = 100 kW In the above equation, v is from table C-1 at 20 °C for liquid water b. On mChs- (343k865 k)66,93 Alternatively, if you didn't find h2 from table C-4, you can use a control volume that includes the pump and boiler as shown below. kg kJ kg kj wturbine-m(h-h-(2oY)(3433.8眚2464.96器)-19,377 kW c. (0.95) (2392.8訇-2464.96븝 kJ kg where h4 = 191.8 kg kg Qout=m(h4_h1)-(20坦)(2464.96 kg-83.9訇=47, 621 kW To check my work, check that the first law is obeyed: In Out-Wpump +Qin - Weurbine - out 9566903 19377- 47621 Energy is conserved! Now, to find the efficiency Method 1: -Wet 19377-95 d. e. 0 Qin = 66898 = 0.288 1--= 1-_ = 0.288 out Qin 47621 Method 2: η 66898

Answer 1

The solution shown above is absolutely correct. The enthalpy value at each point before entry and exit are taken from the saturated water enthalpy table from steam table.

The pump work is m.v( P2-P1). Here, v is specific volume at saturated water at 10 kPa.

The enthalpy at point 4 (exit of turbine) is calculated from formula h = hf + x( hg - hf) at pressure 10 kPa..

Since, from first law of thermodynamics, the net heat transfer is equal to net work transfer in the cycle. So the entropy change is zero.

And, thermal efficiency of the actual cycle is 28.8% and that of ideal cycle is 38.05%.

Thermal efficiency= 1 - T2/T1

= 1 - 293/473

= 38.05%.

So, the process is irreversible.