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Steam Engine Question - Help Understanding Needed

  1. Jan 5, 2013 #1
    Ok, so I have complete faith in the tried and proven laws of thermodynamics, including the part of the second law which states that it is impossible to create a heat engine working in cycle that takes in a certain amount of heat energy, and converts all the heat energy it takes in, into Mechanical work.

    But I am counfused, with regards to how a steam engine would not violate that law.

    In particular, this is what I am confused about: Couldn't you build a steam engine, where the steam is ran through just enough rotors that the temperature gained from heating the steam is in effect all lost to turning the rotors in the turbine, and thereby is returend to the boiler in the state that it left it in, thereby making no need for a cold sink for the heat to be dumped into?

    I know that my reasoning is flawed somewhere, but I'm not sure where . . . . .
  2. jcsd
  3. Jan 5, 2013 #2
    There are two problems one is mechanical losses, but even assuming a perfect world with frictionless bearings and perfect seals, for a steam engine to work there has to be delta T somewhere, (between the boiler flue gas and the boiler water, or between the condenser cooling water and the LP steam being condensed) those delta Ts are where you loose efficiency.
  4. Jan 5, 2013 #3


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    Welcome to PF.

    The problem is that "just enough rotors" (turbine stages) is an infinite number of stages. Each stage only extracts some fraction of the remaining energy, so it might look something like this, if each stage extracts half the remaining energy:

    1 stage = 1/2 the energy
    2 stages= 1/2 + 1/4 = 3/4
    3 stages= 1/2 + 1/4 + 1/8 = 7/8
    4 stages= 1/2 + 1/4 + 1/8 + 1/16 = 15/16

    Etc. And each stage is dealing with lower pressure steam, which means each successive stage is larger than the last. So you see, there is a quickly diminishing return.
  5. Jan 5, 2013 #4

    jim hardy

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    I hope i'm not off topic here -

    Note your steam exits the turbine at quite low pressure.
    To get it back into boiler its pressure must be raised back up to that of the boiler.
    To that end a phase change is employed - because to reach any given pressure it takes far less work to pump an incompressible fluid than a compressible one,,,

    look at your vdp + pdv terms


    The condenser in a steam power plant ideally just discards the heat of vaporization.
    They intentionally mix steam with the condensate to minimize subcooling.


    old jim
  6. Jan 5, 2013 #5


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    Staff: Mentor

    Lol, good point -- you need both the condenser and the pump!
  7. Jan 5, 2013 #6
    What you are asking for is to adiabatically expand steam until it turns back into liquid - it is impossible.
    When you expand steam it cools, yes, but it will not condense because you have increased volume. That's why you need the cold sink. Sorry.
  8. Jan 7, 2013 #7
    Ok, I think I understand it now. I didn't realize that in order to get the steam back into the boiler, you would have to raise it to the same pressure as the boiler. So I assume, if you didn't use the condenser, you probably would have to use the same amount of work to adiabaticaly compress the steam, as the steam exerted on the rotors while adiabaticaly expanding, thereby resulting in zero net work. Haha, and yes, I guess I was expecting the steam to adiabatically expand into a liquid. Thanks!
  9. Jan 8, 2013 #8
    This makes sense to me. All you want to do is turn the fluid back into a liquid to decrease the volume to make VdP smaller. Why cool the liquid further if you're just going to heat it up again?

    My question is given two incompressible fluids with comparable boiling points, will the fluid with the smaller heat of vaporization produce a more efficient cycle? If this is the case, could we possibly use a liquid other than water to avoid it's relatively large heat of vaporization?

  10. Jan 8, 2013 #9

    jim hardy

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    I would think so, but that's just my intuitive answer. I do better at math in the mornings
    and i'm an electrical type who learned enough thermo to not appear totally ignorant in presence of mechanical types.

    Probably there's somebody here who can give you a good answer with numbers quicker than i.
    But i'll take a look.
    Here's wher i'll start: http://en.wikipedia.org/wiki/Rankine_cycle

    Here is an interesting anecdote - well to me any way -

    in the 1930's when they were still trying to figure out what water chemistry would allow high temperature and pressure in steel boilers,
    somebody built a two stage plant in downtown Boston.
    The high temperature turbine used mercury vapor
    and its exhaust heated water for a steam turbine.


    9800 was still a respectable heat rate thirty years later after they'd solved the chemisty problems and had steam pressures 10X higher...

    Can you imagine permitting that thing today ?
  11. Jan 9, 2013 #10

    jim hardy

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    Well i did take a further look

    and as so often happens, my "off the cuff" answer was dead wrong.

    Here's a scholarly article i found with Google:

    Around page 3063 is the answer to your question, see section 3.1.2

    And it's obvious in hindsight:
    Latent heat defines the length of the horizontal lines on T-S diagram.
    That length determines the width of the polygon representing the cycle.
    Width X height is area, and area enclosed by the polygon is the work output.
    Obviously the wider polygon delivers more work.

    So i apologize for the mis-information. I shoulda had more faith in my limited appreciation of basic thermo.

    It's a pretty good looking article - check it out!

    And thanks for the question - obviously i needed a refresher.

    old jim
  12. Jan 9, 2013 #11
    Interesting. I haven't read it that closely, but I'm guessing that the "length of the horizontal lines on T-S diagram" is directly related to the expansion ratio of the working fluid from it's liquid state volume to it's largest gas volume.

    There is another thing I'm finding confusing when trying to figure this out. Most of the graphs are temp vs. entropy. Why don't they look at the good old Pressure vs. Volume graph or "specific volume" and "specific pressure". Pressure and volume are much more intuitive that entropy.
    Last edited: Jan 9, 2013
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