1 m3 of saturated liquid water at 200°C is expanded isothermally in a closed system until its quality is 80 percent. Determine the total work produced by this expansion, in kJ

Total work produced = 1.355 × 10^5 KJ

Explanation:

Substance :H2O, V1=1m3

State1= sat. liquid, T1=200°C, x1=0

State2= sat.mixture, T1=200°C, x2= 0.8

v1=vf=0.001157m3/kg, vg=0.12721m3/kg

P1=P2=Psat. =1554.9 KPa

v2= vf+x2 (vg-vf) =0.001157 +0.8(0.12721-0.001157)=0.102m3/kg

V2= V1×v2/v1 = 1× 0.102/0.001157 =88.159m3

W= P(V1-V2) =1554.9(88.152- 1) = 1.355x10^5KJ

its lower due to the doppler effect.: apex