Let's say I have a steam turbine that I have modified to increase its isentropic efficiency. As a specific case, consider the modification outlined in the Mollier diagram below. The arrows represent the before and after modification paths of the turbine on the Mollier diagram.

The slope of the arrows represent how efficient the turbine is. A horizontal slope is a throttling process whereas a vertical slope is 100% isentropic efficiency. Slopes in between obviously are something less than 100%.

The slope is $\Delta{H} \over \Delta{S}$ which has units of temperature (degR).

Does the fact that the units of the slope are temperature hold any special meaning? Why would the value of a temperature inherently represent a performance level of isentropic efficiency? Does the fact that the values are always negative mean anything? Is the reason 100% efficiency is impossible related to the fact that the value of the temperature is $-\infty$?

turbine mod


Yes, the fact that the slope is infinite means 100% efficiency is not possible. This is because $\Delta S$ is zero while $\Delta H$ is non-zero, which is not possible. Entropy will be created when heat is released.

In other situations, for example when considering Mollier diagrams of pure substances, the slope of the isobaric line is the saturation temperature. In this case, it will likely be the absolute temperature of the system because:

$Tds = dh$

at constant pressure from the second law of thermodynamics. So, the slope on the diagram is the temperature of the system.

In your figure, I don't understand why the slope is negative because the curves clearly have positive slopes.

  • $\begingroup$ I don't fully follow your comments. Ok on infinite slope. My lines aren't isobaric since I'm reducing the pressure to internally generate the kinetic energy to turn the turbine blades. As you say this will always be done w/ some irreversibility and so net P will have to decrease. Just from the math the slope has to be negative since $\Delta H$ is negative whereas &\Delta S$ is positive. $\endgroup$ – Jason Waldrop Mar 15 '12 at 16:26
  • $\begingroup$ I'm just looking at your plot you put in the post. As the lines move right, they also move up which is a positive slope. Also, isn't it isobaric -- the lines in the figure are labeled with a single pressure? Typically the diagrams consist of constant pressure, constant temperature, and constant volume lines. What is your plot showing? $\endgroup$ – tpg2114 Mar 15 '12 at 19:46
  • $\begingroup$ Sorry I was talking about the straight lines which represent the paths taken by the two turbines. I understand your original comments on the isobaric and isothermal lines now. I was talking about the path lines slope. $\endgroup$ – Jason Waldrop Mar 15 '12 at 20:53

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