Question

A well-insulated rigid tank having a volume of 0.25 m3 contains saturated water vapor at 100ᴼc. the water is rapidly stirred until the pressure is 1.5 bars. determine the temperature at the final state, in ᴼc, and the work during the process, in kj.

Answer 1

State 1:
From steam tables;
T1 = 100°C
P1 = 1.01325 Bar
S1=Sg = 7.355 kJ/kg.K

State 2:
P= 1.5 Bar
T2 = ?

This point is super-heated but entropies should be equal.

S1=Sg=S2 = 7.355 kJ/kg.K

This point lies between saturation point at 1.5 Bar and 150 °C super-heat state. By interpolation,
T2 = 111.4+ (7.355-7.223)/(7.420-7.223)*(150-111.2) = 137.26 °C

Work done:

W=ΔU = m(u2-u1)
u1= 2506 kJ/kg
u2 = 2519 + (7.355-7.223)/(7.420-7.223)(2580-2519) = 2559.87 kJ/kg
m= volume/specific volume, where specific volume = 1.159+(7.355-7.223)/(7.420-7.223)(1.286-1.159) = 1.244 m^3/kg

Then, m=0.25/1.244 = 0.2 kg

Therefore,

W= 0.2*(2559.87-2506) = 10.77 kJ