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).

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

answer:

true

explanation: