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 [tex]P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg[/tex] and[tex]v_{2} =0.4v_{1}[/tex] from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the [tex]T_{sat}=179.88^{0} C \leq T_{1}[/tex] from the table 1 we get:[tex]v_{1} =0.30661(m^{3}/Kg)[/tex]. Now we have second conditions as: [tex]P_{2}=1(MPa), T_{2}=250^{0}C[/tex] so from the same table we can see the state still superheated and we get[tex]v_{2}=0.23275(m^{3}/Kg)[/tex], knowing that it's a isobaric process we can find the compression's work as:[tex]W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ)[/tex] 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:[tex]W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ)[/tex] and the second process is isochoric:[tex]W_{2}=zero[/tex],now[tex]W_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ)[/tex]. Finally to get the temperarure at the final state in part (b) we get:[tex]v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa)[/tex] from table 2 (see attached) we compare[tex]v_{f}[/tex] and[tex]v_{g}[/tex] at the saturated water table and find the following:[tex]v_{f}=0.001093(m^{3}/Kg)<v_{2}<v_{g} =0.37483(m^{3}/Kg)[/tex], so we know that the final state phase is a satured mixture and we get the temperature at the final state as:[tex]T_{2} =T_{sat} =151.83^{0}C[/tex].
a) Compression work at the final state is 44.32 KJ,
(b) Compression work at the final state with a pressure of 500 KPa is 110.37KJ
The piston work is defined as the amount of work done on the piston in order to do its reciprocating motion. It is found by multiplying the net force with the displacement of the piston.
(a) Compression work at the final state with pressure is 44.32(KJ),
The given condition is;
P₁= 1.0 MPa
T₁ = 250°C
M=0.6 Kg
v₂=0.4v₁
v₁=0.30661 m/sec
P₂=1 Mpa
T₂=250°C
Work done on the piston;
[tex]\rm W= mP(V_2-V_1)\\\\ \rm W= 0.6 \times 1000(0.232-0.30661) \\\\ \rm W=-44.32 \J[/tex]
(- ve )shows the compression work
Hence compression work at the final state with pressure is 44.32(KJ),
(b) Compression work at the final state with a pressure of 500 KPa is 110.37 KJ.
The piston reaches the stop and there are two processes in this stage,
Process 1 is isobaric
The work done during the isobaric process;
[tex]\rm W= mP(V_2-V_1)\\\\ \rm W= 0.6 \times 1000(0.40 \times 0.3066-0.30661) \\\\ \rm W=-110.38 \ KJ[/tex]
The isochoric work done is zero.
[tex]\rm W_{net} = W_1+W_2 \\\\ \rm W_{net} =-110.38+0 \\\\ \rm W_{net} =-110.38[/tex]
Hence compression work at the final state with a pressure of 500 KPa is 110.37(KJ)
(c)Temperature of the final state in part b will be T=151.83(°C)
Volume at stage 2
[tex]\rm V_2= 0.4 \TIMES 0.30661 \\\\ \rm V_2= 0.1022644 \ m^3[/tex]
P₂ = 500 Kpa
The final state temperature is the temperature of the saturation stage.
[tex]T_{2}=T_{sat}=151.83[/tex]
Hence temperature of the final state in part b will be 151.83(°C).
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https://brainly.com/question/3406116
