Question

Steam at 275ºC and 15.0 bar is contained in a rigid tank with the volume of 10.0 m3 . In a constantvolume cooling process, the pressure inside the tank drops to 1.8 bar and some of the steam condenses. Calculate 1) the temperature of the tank contents after the process, 2) the amount of steam condensed during the process (in kg), and 3) the amount of heat removed from the tank during the process.

Answer 1

At initial conditions (superheated steam)

Temperature T1 = 275 °C

Pressure P1 = 25 bar

specific volume v1 = 0.16099 m3/kg

specific enthalpy h1 = 2982 kJ/kg

specific volume
v1 = mass/Volume
Mass of water vapor

m = V/v? = 10m3/ 0.16099 m3/kg

= 62.115 kg

at final conditions (saturated steam)
Pressure P2 = 1.8 bar
T2 = 116.91 °C
specific volume of fluid vf = 0.0010575 m3/kg
specific volume of gas vg = 0.97753 m3/kg

specific enthalpy of fluid hf = 490.668 kJ/kg

specific enthalpy of gas hg = 2701.42 kJ/kg

Quality of steam
v1 = vf + (vg - vf)*x
x = (v1 - vf) / (vg - vf)
= (0.16099 - 0.0010575) / (0.97753 - 0.0010575)
= 0.16378
specific enthalpy of at state 2
h2 = hf + (hg - hf)*x
= 490.668 + (2701.42 - 490.668) * 0.16378
= 852.744 kJ/kg
Part 3
The heat removed
Q = m??h

= m?(h? - h?)
= 62.115 kg * (2982 kJ/kg - 852.744 kJ/kg)
= 132258.73 kJ

Part 1
T2 = 116.91 °C

Part 2
Mass of steam condensed
m = (1 - x) * m
= (1 - 0.16378) * 62.115 kg
= 51.94 kg