A 380-L tank contains steam, initially at 500oC, 3 bar. A valve is opened for 10 seconds and steam flows out of the tank at a constant mass flow rate of 0.005 kg/s. During steam removal, a heater maintains the temperature within the tank constant. Determine final mass remaining in the tank, in kg, and the final pressure in the tank, in bar.

a) the final mass is mf = 0.216 Kg

b) the final pressure is P = 2.06 bar

Explanation:

Since the steam does not behave as ideal gas we should use the thermodynamic property tables for steam:

- at 500°C and 2 bar, the specific volume is v = 1781,4*10^-3 m3/Kg

- at 500°C and 5 bar, the specific volume is v = 710.9*^-3 m3/Kg

therefore, using interpolation (v=a+b*P)

- at 500 °C and 3 bar, the specific volume is vi = 1424.5*10^-3 m3/Kg = 1424.5 L/Kg

therefore the initial mass is

mi = V / vi = 380 L / 1424.3 L/Kg = 0.266 Kg

the final mass is

mf = 0.266 Kg - 0.005 Kg/s * 10 seg = 0.216 Kg

since the tank volume is constant the final specific volume is

vf = V / mf = 380 L / 0.216 Kg = 1759.25 L/Kg = 1759.25*10^-3 m3/Kg

and using previous v values along with v=a+b*P

- at 500°C and vf = 1759.25*10^-3, the pressure should be P = 2.06 bar