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

Answer: 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

work done = 1.36 x 10^5 KJ

Explanation:

liquid water, volume (v) = 1m^3

from the table; at Temperature of 200oC

v1 = vf = 0.001157m^3/Kg and vg = 0.12721 m^3/Kg

for saturated liq water T1 = 200oC, percentage (x1) = O

for saturated mixture T2 = 200oC, percentage (x2) = 0.8

p1 = p2 = p (saturated solution) = 1554.9 KPa (this shows that the process is isobaric)

how ever,

v2 = vf + x2 (vg - vf)

v2 = 0.001157m^3/Kg + 0.8 (0.12721 m^3/Kg - 0.001157m^3/Kg)

v2 = 0.102m^3

v2 = v1 x v2/vf

v2 = 1 x 0.102/0.001157

v2 = 88.159m^3

work done = p (v2 - v1)

work done = 1554.9 (88.159 - 1)

work done = 1.36 x 10^5 KJ