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How much work can you extract from a turbine connected to a pressure cooker?

The situation is modeled as a rigid adiabatic tank of volume $V_{tank}$, which initially contains water at a high pressure $P_1$ and quality $x_1$.

A tube connects the tank to a valve that only allows the flow of vapor to an isentropic turbine, which discharges to the atmosphere and produces work $W$. We know that the discharged vapor is a saturated vapor at $P_{atm}$.

The schematic is shown below:

enter image description here

Now, the valve is opened and vapor flows out until all the liquid phase in the tank is evaporated. Then the valve is closed.

Known parameters: $V_{tank}$, $P_1$, $x_1$, $x_2 = 0$, $P_{outlet}$

Unknown parameters: $T_2$, $P_2$, $W$


I begin by performing an energy balance on the whole system:

$$ \Delta U = m(u_{f,1} + x_1 u_{fg,1}) - (m-\Delta m)(u_{f,2})= -W -\Delta m h_{out}$$

where $h_{out}=h_g(P_{atm})$, the mass of water (vapor) loss is given by $\Delta m$, and $u_{f,1}$, $u_{fg,1}$ are tabulated values for water.

Initially, we have mass $m = \frac{V_{tank}}{v_1}$ and $v_1 = v_f + x_1v_{fg}$ is defined from steam tables since state 1 is fully defined.

I think I can assume that the final pressure is $P_{atm}$, once all the steam runs through the turbine, therefore state 2 is fully defined and $u_{f,2} = u_f(P_{atm})$.

Now we have 2 unknowns: $\Delta m$, and $W$.

The amount of mass that we lose, $\Delta m$, is the only other parameter needed to calculate work.

Entropy balance on the entire system adds another equation:

$$ \Delta S = m(s_{f,1} + x_1s_{fg,1}) - (m-\Delta m)s_{f,2} = -\Delta m s_{out} + S_{gen}$$

where $s_{out} = s_g(P_{atm})$, and $s_{f,2} = s_f(P_{atm})$. This adds entropy generation $S_{gen}$ as another unknown, however, and therefore I have 3 remaining unknowns ($\Delta m$, $W$, $S_{gen}$). If I could assume zero entropy generation, I could solve this problem, but I think the steam extraction is irreversible.

How else can I go about this?

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The trick to this problem is finding the final state in the tank, which is not when the pressure in the tank gets to 1 atm. It is when no liquid remains in the tank. To do this, you need to recognize that, at any time during the expulsion process, the liquid and vapor remaining inside the tank has experienced an adiabatic reversible expansion (in pushing the vapor ahead of it out of the tank through the valve). This means that the entropy per unit mass of the vapor remaining in the tank at the final state is the same as the entropy per unit mass in the tank initially: $$s_{f,g}(P_f)=s_{1,l}(1-x_1)+s_{1,g}x_1$$ This tells you the final saturation vapor pressure $P_f$ in the tank. In addition to this, you have the following equation to get the final mass of vapor in the tank: $$m_fv_{f,g}(P_f)=V_{tank}$$where $v_{v,g}$ is the specific volume of the vapor at saturation vapor pressure $P_f$. The final internal energy per unit mass in the tank is just $u_v(P_f)$. This is all you need to provide closure on the problem.

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  • $\begingroup$ I considered the isentropic expansion, but it doesn't make sense when looking at a $T-s$ diagram. The tank depressurizes, so the temperature of the vapor isentropically decreases. This means that the final state must have a quality $x_f < 1.0$ (since the width of the vapor dome increases as $T$ decreases). The problem statement, however, says that $x_f = 1.0$. $\endgroup$ – Drew Jan 15 at 14:08
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    $\begingroup$ I agree. So there must be something wrong with the problem statement. $\endgroup$ – Chet Miller Jan 15 at 14:22
  • $\begingroup$ Is it possible that the initial tank contents are meant to be a compressed liquid? That would allow isentropic expansion until the tank is filled with saturated liquid... although the mass that would leave would be minuscule. $\endgroup$ – user1476176 Jan 15 at 15:35
  • $\begingroup$ @user1476176 I was just stating what the problem statement said. Perhaps the problem is meant to be confusing or not make sense, or maybe we aren't supposed to wonder about "how" the remaining water evaporated in an externally adiabatic tank. $\endgroup$ – Drew Jan 16 at 16:45
  • $\begingroup$ Frustrating either way. I guess that, if we want to get technical, the only process compatible with the given constraints is the null process (nothing happens), so the "correct" answer is that the maximum available work is 0. If it's meant as a trick question, I'd say it's a bit mean (of the question-writer, not of you). $\endgroup$ – user1476176 Jan 16 at 21:12
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The extra constraint that you've forgotten is the Mass Balance. You should be able to figure out the change in system mass by considering the initial and final states of the tank, which are both well-defined. Note that the question states that the process ends as soon as all of the liquid in the tank evaporates, so the final tank pressure is actually $P_1$ rather than $P_\text{atm}$ (the tank can continue to vent, but that is not considered part of the process under consideration). This makes the math significantly simpler.

It's true that $S_\text{gen}$ can't be zero for the entire system (the throttling process inside the valve is irreversible), but you can deal with this by dividing the entropy generation into a component associated with the valve (which you can calculate) and a component associated with the turbine (which must be zero, if work output is to be a maximum).

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  • $\begingroup$ I don't see how pressure remains at $P_1$, especially after losing the steam. $\endgroup$ – Drew Jan 15 at 11:16
  • $\begingroup$ Yes, I agree with that comment Drew. You corrected me on this same point yesterday evening. The problem statement clearly says that the tank is adiabatic, so the final pressure can't be $P_1$. $\endgroup$ – Chet Miller Jan 15 at 11:45
  • $\begingroup$ Ah you're right - I was preoccupied with the idea of a real pressure cooker, in which heat is continuously being added and a valve lets steam escape to maintain a specified pressure. I have removed the offending sentence. $\endgroup$ – user1476176 Jan 15 at 15:31
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    $\begingroup$ It would be nice if the problem statement was more practical. Every real world application of this system will involve a boiler, where heat is added to the "pressure cooker" in order to generate a constant pressure steam to the turbine. $\endgroup$ – David White Jan 15 at 15:53

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