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 '19 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 '19 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 '19 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 '19 at 4:09
  • $\begingroup$ The steam will cool as the pressure drops, but not enough to condense. $\endgroup$ – David White Apr 5 '19 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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.