Question

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.

Answer 1

a) the final mass is mf= 0.216 Kg

b) the final pressure is P= 2.06 bar

Step-by-step 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