Answer:
a) 44.316 KJ
b) 110.3696 KJ
c) 151.83°C
Explanation:
Givens:
Substance: H20, m = 0.6 kg P1 = 1 MPa, T1 = 400°C v2 = 0.4 v1
First We Compute v1 as Following:
P1= 1 MPa, T1= 400°C
From TABLE we compare T(sat). corresponding to P1 with T1 determine the state phase
T(sat) = 179.88°C < T1
The state phase is Superheated
From TABLE we get v1 corresponding to P1 and T1
v1 = 0.30661 m^3/kg
(a) P2 = 1 MPa, T2 = 250°C
From TABLE we compare T(sat) corresponding to P2 with T2 determine the state phase
T(sat) = 179.88°C < T2
The state phase is Superheated
From we get v2 corresponding to P2 and T2
v1 = 0.23275 m^3/kg
P1 = P2 =1 MPa
The process is isobaric
W 1_2 = mP(v2 — v1) = 0.6 x 1000(0.23275 - 0.30661) = - 44.316 KJ
The compression work Wb = 44.316 KJ
(b) P2 = 500 kPa.
the piston reaches the stops so there are two processes
Process 1_2 is isobaric
W 1 _2 = MP(v2 —V1) = 0.6 x 1000(0.4 x 0.30661-0.30660) = -110.3796 KJ
Process 1_2 is isochoric
W2_3 = zero
The compression work Wb = W 1_2 + W 2_3 = 110.3796+ 0 = 110.3696 KJ
(c) v2 = 0.4v1 = 0.9 x 0.30661 = 0.122649 m^3/kg.
P2 = 500 kPa
From TABLE we compare vf and vg corresponding to P2 with v2 to determine the state phase
vf = 0.001093 m^3/kg < v2 < vg = 0.37483 m^3/kg
The state phase is Saturated Mixture
T2 = Tsat = 151.83°C
