A system consisting of 2 Kg of water undergoes a cycle compo

A system consisting of 2 Kg of water undergoes a cycle composed of the following processes: 1-2 a constant volume process from P1=10 bar, x1=0.6533 to saturated vapor; 2-3 constant pressure process to P3=P1, Q23=328KJ 3-1 a constant pressure process Please tell me how to determine the net work for the cycle. I can tell that Wtotal=W12 +W23+W31 = 0 +(Q23-U23)+Pm(V1-V3); but what\'s next? I dont know how to find Q23 U23 V1 and V3 Thanks

Solution

Obtain the properties of water from the thermodynamic tables

State 1:

At 10 bar ,

u_f=761.39 kJ/kg

u_g=2582.8 kJ/kg

U1=761.39+0.6533(2582.8-761.39)=1951.317 kJ/kg

V1=mv_1=2x(0.001127+0.6533(0.19436-0.001127))=0.2547 m3

Process 2-3 must be constant temperature process

Check the question and correct it

State 2:

v2=v1=vg=0.12735 m3/kg

From staurated vapor steam tables

Find T2,u2

State 3:

T3=T2, and P3=P1

Find U3 and V3