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Consider $m_1 = $ 100 kg of steam at pressure $P_1$ and temperature $T_1$ in a 3 $m^3$ insulated tank, closed off by a valve. We open the valve and allow 50 kg of steam to exit. What is the final pressure?

Here are known values:

  • $m_1 = $ 100 kg
  • $P_1 = 10^7$ Pa
  • $T_1 = 500$ ° C
  • $m_2 = $ 50 kg

My solution starts with assuming the remaining steam undergoes an isentropic process (adiabatic + reversible expansion), so that $s_1=s_2=6.58$ kJ/kgK from steam tables. Now, from the final mass we can get the specific volume $v_2=\frac{V}{m_2} = 3/50 = 0.06 m^3/kg$.

Since we know $(s_2,v_2)$, doesn't this mean that state 2 is fully defined? How can we get the pressure with this knowledge?

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  • $\begingroup$ You should be able to find the final state in your steam tables. $\endgroup$ Commented Apr 4, 2019 at 22:46
  • $\begingroup$ @Drew, I seriously doubt that steam would condense as you went to lower and lower pressure, unless you used a condenser. A Joule-Thompson effect would occur as the pressure fell, causing the temperature to also fall, but not enough to condense any of the low pressure steam. $\endgroup$ Commented Apr 5, 2019 at 2:11
  • $\begingroup$ The steam will cool as the pressure drops, but not enough to condense. $\endgroup$ Commented Apr 5, 2019 at 14:50
  • $\begingroup$ @Drew, over short time periods, if you have boiling water in the vessel with steam and you relieve pressure, some of the water will boil, but the steam will not condense. Over longer time periods, if you allow time for heat transfer to the environment (e.g., several minutes or longer), then steam will begin to condense. $\endgroup$ Commented Apr 5, 2019 at 18:18
  • $\begingroup$ @Drew, I'll add to my last comment. For your high pressure steam, look up the enthalpy per pound (or kg) in a steam table. Based on conservation of energy, as you reduce the pressure on that steam, it will retain the same enthalpy per pound even though it is brought to a lower pressure. Look up the same enthalpy at that lower pressure, and note that the lower pressure steam is more superheated (farther away from its condensation temperature), not closer to condensing. $\endgroup$ Commented Apr 5, 2019 at 19:17

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If the final state in the tank is a mixture of liquid and vapor, then you have $$v_{2L}(1-x)+v_{2V}x=v_2$$and $$s_{2L}(1-x)+s_{2V}x=s_2$$where x is the steam quality. Eliminating the quality x, we have $$\frac{(v_2-v_{2L})}{(v_{2V}-v_{2L})}=\frac{(s_2-s_{2L})}{(s_{2V}-s_{2L})}$$ All you need to do is find the temperature at which this relationship is satisfied.

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