Water vapor is cooled in a closed rigid tank from 793 K and

Water vapor is cooled in a closed, rigid tank from 793 K and 10^7 Pa to a final temperature of 543 K Determine the final pressure 1 bar = 10^5 Pa K = degree C + 273

Solution

Now that you\'ve given the initial pressure, I\'m using steam tables from a book.

At 793K and 100 bars, the specific volume is 0.0424 m^3/kg. The specific volume must remain the same, since the tank is closed and rigid. Now I seek some point where T = 793K and the specific volume is still 0.0424 m^3/kg. This is tricky, because if the pressure were now 55 bars (the saturation pressure), the specific volume would be just a bit too small (0.036 m^3/kg), but in a superheated condition at 40 bars, the specific volume at 553K is a bit too large; I have to extrapolate to find that v is about 0.054 m^3/kg at 40 bars. The best I can after that is to interpolate pressure by observing that 42 is 1/3 of the way from 36 to 54...which leads to the conclusion that the pressure where v = 0.0424 m^3/kg and T = 543K will be around 50 bars.

I hope this resolve your query