F=ma make m the subject

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

Ilya [14]

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: $T_{2} =T_{sat} =151.83^{0}C$ .

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