Question

A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the steam is cooled. Determine the compression work if the final state is (a) 1.0 MPa and 250°C and (b) 500 kPa. (c) Also determine the temperature at the final state in part (b).

Answer 1

Answer:

(a) Compression work at the final state with a pressure of 1(MPa) is: 44.32(KJ), (b) Compression work at the final state with a pressure of 500(KPa): 110.37(KJ) and (c) temperaure of the final state in part b: T=151.83(°C).  

Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are $P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg$ and$v_{2} =0.4v_{1}$ from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the $T_{sat}=179.88^{0} C \leq T_{1}$ from the table 1 we get:$v_{1} =0.30661(m^{3}/Kg)$. Now we have second conditions as: $P_{2}=1(MPa), T_{2}=250^{0}C$ so from the same table we can see the state still superheated and we get$v_{2}=0.23275(m^{3}/Kg)$, knowing that it's a isobaric process we can find the compression's work as:$W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ)$ so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and:$W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ)$ and the second process is isochoric:$W_{2}=zero$,now$W_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ)$. Finally to get the temperarure at the final state in part (b) we get:$v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa)$ from table 2 (see attached) we compare$v_{f}$ and$v_{g}$ at the saturated water table and find the following:$v_{f}=0.001093(m^{3}/Kg)$, so we know that the final state phase is a satured mixture and we get the temperature at the final state as:$T_{2} =T_{sat} =151.83^{0}C$.