Question

A well insulated rigid tank with a volume of 0.25 m3 contains saturated water vapor at 100˚C. Water is rapidly mixed until the pressure reaches 150 kPa. Determine the final temperature (˚C) and the work (kJ) during this process. (Draw the P-v and T-v graphics.) Note: Use the provided feature table.

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