Moran and Shapiro Fundamentals of Engineering Thermodynamics (p147) has a worked example:

Steam at a pressure of $15$ bars and a temperature of $320^{\circ}C$ is contained in a large tank. Connected to the tank through a valve is a turbine followed by a small initially evacuated vessel with a volume of $0.6m^3$. The valve is opened and the vessel; fills with steam until the pressure is $15$ bars and the temperature $400^{\circ}C$. The valve is then closed. The filling process takes place adiabatically and kinetc and potential energy effects are negligible. Determine the amount of work developed by the turbine in kJ ...

The solution is then given but in a comment at the end they add:

If the turbine were removed and steam allowed to flow adiabatically into the smaller vessel without doing work, the final steam temperature in the vessel would be $477^{\circ}C$, as may be verified

How do you find the temperature without the turbine?

They start by observing that with their assumptions the energy rate balance reduces to:

$$\frac{dU_{cv}}{dt} = -\dot{W} + \dot{m_i}h_i$$

and integrating this and re-arranging they have:

$$ W_{cv} = h_i\Delta m_{cv} - \Delta U_{cv}$$

Because the smaller vessel is initially evacuated the changes are just the final values which they indicate with the subscript 2:

\begin{eqnarray*} W_{cv}&=& h_i\Delta m_{cv} - \Delta U_{cv}\\ &=& h_i m_2 - m_2 u_2\\ &=& m_2(h_i - u_2)\\ &=& 386.6 kJ \end{eqnarray*}

They have:

steamtables(400C, 15bar, v) = $0.203\,m^3/kg$ (use for mass)
steamtables(400C, 15bar, $u_2$) = $2951.3\,kJ/kg$
steamtables(320C, 15bar, $h_i$) = $3081.9\,kJ/kg$

What is the temperature of the steam in the second question

  • $\begingroup$ I have edited your question to help bring it in line with the site's expectations regarding quality. Homework questions are allowed, so your earlier statement regarding "this is not homework" is not necessary. That said, you do need to demonstrate some level of effort at solving the problem yourself. Please edit your question and show what you've attempted and why you're not able to verify their statement. $\endgroup$
    – user16
    Nov 1, 2015 at 14:02
  • $\begingroup$ Hints: the pressure in both chambers will be the same after the exchange, and you can write the first law of thermodynamics for an open system on the smaller chamber. The solution to the first law will give you another state variable which you can use (in conjunction with pressure) to look up the temperature of the steam in chamber two. $\endgroup$
    – Carlton
    Nov 1, 2015 at 15:04
  • $\begingroup$ The state of the steam is determined by any two properties and we have the pressure and the internal energy. Bingo. Here's a link to some free steam tables that have internal energy, you have to do some interpolation for your problem though. $\endgroup$
    – Carlton
    Nov 1, 2015 at 20:19
  • $\begingroup$ The steam heats up as it expands? That doesn't seem right - what am I missing? $\endgroup$
    – user253751
    Nov 12, 2015 at 8:20
  • $\begingroup$ @immibis, the steam does in fact heat up, because it's being compressed by the steam flowing in after it. Very counter-intuitive, but true. $\endgroup$
    – Carlton
    Nov 12, 2015 at 18:32

1 Answer 1


For the second question we have only the third value

But Carlton gives hints. The work is set to zero so the equations become:

\begin{eqnarray*} 0 &=& h_i\Delta m_{cv} - \Delta U_{cv}\\ &=& h_i m_2 - m_2 u_2\\ h_i &=& u_2\\ \end{eqnarray*}

The state of the steam is determined by any two properties and we have the pressure and the internal energy. So we just need go to the steam tables to get the answer ... unfortunatlely my free steam tables at http://www.spiraxsarco.com/ do not provide this combination and the other place does but wants $30.00 to tell me the answer. But my printed tables tell me that the answer is about 480C.


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