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?

  • $\begingroup$ You should be able to find the final state in your steam tables. $\endgroup$ – Chet Miller Apr 4 at 22:46
  • $\begingroup$ @ChetMiller I found values in my steam tables that matched my calculated $(v_2, s_2)$, but I think I got lucky because this means that it's still steam. What if we kept rejecting steam until we got a liquid vapor mixture? Then it seems like $(v_2, s_2)$ wouldn't be enough info to define the state, since we don't have the quality. Is this correct? $\endgroup$ – Drew Apr 4 at 23:12
  • $\begingroup$ @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. $\endgroup$ – David White Apr 5 at 2:11
  • $\begingroup$ @DavidWhite how does the JT effect influence the steam inside the tank? Only the steam downstream of the valve should be affected by that $\endgroup$ – Drew Apr 5 at 4:09
  • $\begingroup$ The steam will cool as the pressure drops, but not enough to condense. $\endgroup$ – David White Apr 5 at 14:50

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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