Two tanks (Tank A and Tank B) are separated by a partition. Initially Tank A contains 2 kg of steam at 1 MPa and 300°C while Tank B contains 3 kg of saturated liquid–vapor mixture at 150°C with a vapor mass fraction of 50 percent. The partition is now removed and the two sides are allowed to mix until mechanical and thermal equilibrium are established. If the pressure at the final state is 300 kPa, determine (a) the temperature and quality of the steam (if mixture) at the final state and (b) the amount of heat lost from the tanks.

a) T2 = 133.5°C

x2 = 0.364

b) Qout = 3959.6 kJ

Explanation:

For this exercise we will make two assumptions: the tank is stationary, the kinetic and potential energy are equal to zero. The second assumption is that there are interactions between the two tanks. The contents of both tanks will be a closed system. The energy balance equals:

ΔEsystem = Ein - Eout

-Qout = ΔUA + ΔUB = (m*(u2-u1))A + (m*(u2-u1))B

The steam properties for the two tanks in the initial state can be found in Table A-4 to table A-6 of Cengel:

P1A = 1000 kPa

T1A = 300°C

v1A = 0.25799 m^3/kg

u1A = 2793.7 kJ/kg

T1B = 150°C

x1 = 0.5

vf = 0.001091 m^3/kg

uf = 631.66 kJ/kg

vg = 0.39248 m^3/kg

ufg = 1927.4 kJ/kg

v1B = vf + x1*vfg = 0.001091 + (0.5*(0.39248-0.001091)) = 0.19679 m^3/kg

u1B = uf + x1*ufg = 631.66 + (0.5*1927.4) = 1595.4 kJ/kg

V = VA + VB = mA*v1A + mB*v1B = (2*0.25799) + (3*0.19679) = 1.106 m^3

m = mA + mB = 3+2 = 5 kg

the specific volume will be equal to:

v2 = V/m = 1.106/5 = 0.2213 m^3/kg

With these calculations, we can looking the new properties in the same tables:

P2 = 300 kPa

v2 = 0.2213 m^3/kg

T2 = Tsat, 300 kPa = 133.5°C

x2 =(v2-vf)/(vg-vf) = (0.22127-0.001073)/(0.60582-0.001073) = 0.364

u2 =uf + x2*ufg = 561.11 + (0.364*1982.1) = 1282.8 kJ/kg

-Qout = (2*(1282.8-2793.7)) + (3*(1282.8-1595.4)) = -3959.6 kJ

Qout = 3959.6 kJ

a 2,000 calorie based diet.