# What is the final pressure of steam in a tank, if we allow some steam to exit?

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?

• You should be able to find the final state in your steam tables. – Chet Miller Apr 4 at 22:46
• @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? – Drew Apr 4 at 23:12
• @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. – David White Apr 5 at 2:11
• @DavidWhite how does the JT effect influence the steam inside the tank? Only the steam downstream of the valve should be affected by that – Drew Apr 5 at 4:09
• The steam will cool as the pressure drops, but not enough to condense. – David White Apr 5 at 14:50

## 1 Answer

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.