Question

A 10.0 m³ tank contains steam at 250°C and 10.0 bar. The tank and its contents are cooled until the pressure drops to 1.8 bar. Some of the steam condenses in the process.

(a) How much heat was transferred from the tank?

(b) What is the final temperature of the tank contents?

(c) How much steam condensed (kg)?

Answer 1

Initial state at 250°C and 10.0 bar (superheated steam)
Temperature T? = 250 °C
Pressure p? = 10 bar
From the steam table
specific volume v? = 0.23274 m³/kg
specific enthalpy h? = 2943.22 kJ/kg
specific volume = volume of the vessel / mass of water vapor

m = V/v? = 10m³ / 0.23274 m³/kg = 42.966 kg

At final state (saturated steam)
Pressure p? = 1.8 bar
Saturated temperature T? = 116.91 °C
specific volume of fluid vf = 0.001057 m³/kg
specific volume of gas vg = 0.977533 m³/kg
specific enthalpy of fluid hf = 490.67 kJ/kg
specific enthalpy of gas hg = 2701.42 kJ/kg
Specific volume balance

and enthalpy of the gas mixture are given by:
v? = vf + (vg - vf)?x

0.23274 = 0.001057 + (0.977533 - 0.001057) x

x = 0.2373
h? = hf + (hg - hf)?x
= 490.67 kJ/kg + (2701.42 kJ/kg - 490.67 kJ/kg) * 0.2373
= 1015.28 kJ/kg
Part a
heat was transferred from the tank
Q = m??h = m?(h? - h?)

= 42.966 kg * (2943.22 kJ/kg - 1015.28 kJ/kg)
= 82835.87 kJ

Part b
At final state (saturated steam)
Pressure p? = 1.8 bar
Saturated temperature T? = 116.91 °C


Part C
Let fraction of water vapor = x

fraction of liquid water = (1 - x)

mass of liquid water condensed
m = (1 - x) * m
= (1 - 0.2373) * 42.966 kg
= 32.770 kg