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Say that we have an irreversible expansion process which extracts energy, like a turbine. Isentropic efficiency is commonly defined by the following relation, which applies in a similar fashion for pumps. Here, state 1 refers to the inlet conditions and state 2 refers to the outlet conditions, while the "s" indicates the isentropic pseudo-state. P, h, and s are pressure, enthalpy, and entropy.

$$ \eta = \frac{ h_1 - h_2 }{ h_1 - h_{2s} } $$

This has a monumentally huge, glaring problem, that has always bothered me - it can't be applied to sub-segments of a turbine. By that, I mean let's say we keep P1 and P2 the same, but instead of one turbine, we have two in series. Here is the basic situation on a hs diagram:

Isentropic Process

The general structure, isobaric lines, and 3 labeled states are cannon in the literature. The line should probably be curved slightly. But let's add another point somewhere in the middle of the 1->2 process. Let's label that midpoint "m". If you begin with the assumption that two irreversible expansion processes exist 1->m and m->2(by pressure) with the same isentropic efficiency as 1->2, then I'm saying that it follows that:

$$s_1 = s(P_1,h_1) \\ h_{2s} = h(P_2,s_1) \\ h_2 = h_1 - \eta ( h_1-h_{2s} ) \\ \text{Two-Step Process} \\ h_{ms} = h(P_m,s_1) \\ h_m = h_1 - \eta (h_1 - h_{ms} ) \\ h_{2s}' = h(P_2,s(P_m,h_m)) \\ h_2' = h_m - \eta (h_m - h_{2s}' ) \\ \text{The Problem:} \\ h_2' \ne h_2$$

In these equations, I intend for the 2-variable functions to be property lookups. I'm mostly using the steam tables.

A more correct way of saying this is that if you assigned values for $\eta_{1m}$ and $\eta_{m2}$ such that you wind up at the intended final point of 2, then both of these values will be different from the original $\eta$.

The topic has come up on the site before, considering what the derivative of some expansion line might mean. But that's already jumping the gun, because we don't even have any defined way to draw that line in the first place. I can write down $\frac{\Delta h}{\Delta s}$, but not $\frac{dh}{ds}$. Again, we could replace one turbine with two turbines, and assign them efficiencies such that point 2 has the same properties. But that leaves a free degree of freedom. You could make $\eta_{1m}$ and $\eta_{m2}$ the same, or you could have one a little higher and one a little lower, without point 2 being affected.

One simple way to phrase this question would be: how can I calculate P_m and h_m in a physically meaningful way?

My main interest is - in doing so, what metric defines the degree of irreversibility. Has anyone put forth a definition of a variable which will accomplish this in a truly differential sense along a non-ideal expansion line?

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I think the answer to your question is: Polytropic efficiency, it is differential isentropic efficiency and widely used in gas turbine literature. Let me know if you have trouble finding details.

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  • $\begingroup$ If I take some wet steam turbine parameters and force $P v^n$ to be constant, then I get some $n$ and I can draw an h-s line (mentioned in the question). Problem is, this line sees an initial decrease in entropy in the high pressure regions, which is clearly wrong. I think the correct interpretation is that $n$ must change over the different regions due to either non-ideal gas dynamics or phase change. As messy as it is, this looks like the right direction. Thanks. $\endgroup$ – Alan Rominger Jan 20 '15 at 21:41
  • $\begingroup$ Isentropic efficiency turbine = $\frac{\delta W_{actual}}{\delta W_{isentropic}}= \frac{h_2-h_1}{h_{2s}-h_1}$ and Polytropic efficiency is $\frac{dW_{actual}}{d W_{isentropic}}= \frac{dh}{dh_{s}}$. Since the isobars diverge from each other on a T-s plot along higher s, for each incremental pressure drop the denominator is changing as well. I found this online: take a look phad.cc.umanitoba.ca/~wang44/Teaching/MECH%204310/Ch_3/… $\endgroup$ – Sankaran Jan 21 '15 at 23:20
  • $\begingroup$ Right, but the point that your link gives a method to calculate intermediate states, it assumes a polytropic ideal gas. Do I have a polytropic ideal gas? Let's find out, calculate n=Cp/Cv. For my numbers this spans from 1.8 to 1.32 (inlet to outlet). So I probably shouldn't be surprised if application of the method gives unphysical results, no? That would be true even if I wasn't accounting for condensation, which also complicates the issue further. I think there's clearly no calculus-free approach. $\endgroup$ – Alan Rominger Jan 22 '15 at 12:39
  • $\begingroup$ Polytropic efficiency and the polytropic gas coefficient $\gamma$ are two different things all together. Polytropic efficiency is a machine (turbine) efficiency metric no what kind of gas/liquid it is. How you calculate different intermediate enthalpies from temperature pressure will depend on the equation of state of the gas, i.e., $h(T), s(T,P)$. And yes, you will have to integrate along the path of expansion in the steam turbine so no calculus free approach here. You can use computational tools that use lookup steam-table data (check out Chemkin, Cantera, also NASA gas properties) $\endgroup$ – Sankaran Jan 22 '15 at 23:16
  • $\begingroup$ That general method is what I'm asking about. Please correct me if I'm wrong, but the previous pdf you linked to does not employ such a general method, nor is it obviously generalizable to me. My best guess is to solve Pv^n=C for n and C, using the known inlet/outlet conditions, using the steam properties you reference. I did that and it gives an incorrect result. Do you want me to post the numbers from that? $\endgroup$ – Alan Rominger Jan 23 '15 at 12:48

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