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Low temperature production with compressed air

  1. Aug 7, 2016 #1
    Suppose, there are two cylinders of compressed gas, say air. Both are at same temperature and pressure and the amount too is same.
    Now, contents of both are released but in a different way. One has been released directly and the other has been released through compressed air turbine having a generator attached to it. Therefore, some amount of power/electricity has been generated by the second. I want to know which one will be able to produce more cold.
    Common sense tells that it's the second because by this process that dynamic energy of the compressed air has been converted into power/electricity and it should has lesser energy than the first where the velocity of release has been converted into random motion of molecules i.e. heat.
    Am I right or wrong?
     
  2. jcsd
  3. Aug 7, 2016 #2
    Good question ! ...I have some thoughts ...

    If a cylinder is inside a room and contents are emptied into this closed room , the cylinder will be cooler but the room would have heated up , no net temperature change.

    If a turbine is connected and electricity (heat) is taken out of the room by wires then surely this room will be cooler than the first .
     
  4. Aug 7, 2016 #3
    Well, that's the same as mine. Glad to know that I have been supported by at least one. But, I want more to be sure.
    Actually, I have come up with an idea of increasing the efficiency of thermal power plants and it suddenly struck my mind if the temperature at the Condenser section is lowered, the efficiency and output can be increased. Now, if we use compressed air to rotate a turbine and then use that colder air to create lower temperature at the Condenser; I am sure that the increase in output will outrun the power needed to compress air or other gaseous fluid.
     
    Last edited: Aug 7, 2016
  5. Aug 7, 2016 #4
    I'm not sure how lowering the temperature of the condenser would increase efficiency , unless it allowed use of a different working fluid with a lower boiling point ( practically impossible)....after all , we only need to condense steam to water at 100C and pump it back to the boiler , very easy.

    Raising the the max temp.would increase efficiency.
     
  6. Aug 8, 2016 #5
    That's very basic physics. You can increase the efficiency of thermal based system by either increasing temperature at the Boiler or by decreasing temperature at the Condenser. Pressure at the Condenser can be varied and therefore we can control the temperature at the Condenser and engineers often try to reduce the pressure as much as possible so that efficiency can be increased.
     
  7. Aug 8, 2016 #6

    Mech_Engineer

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    Turbines don't convert heat to work, they convert pressure to work. It seems to me the best an expansion cycle can do (regardless of whether it's a turbine or expansion valve) is follow an isentropic expansion curve. In that case from the same starting point they will both result in the same end state. In practice the turbine will have an efficiency associated with it that will result in a state slightly off of isentropic, but there isn't really a way to calculate the efficiency of a valve since it doesn't produce work. In any case, they will best-case result in the same end-state.
     
  8. Aug 13, 2016 #7
    At this wiki page one steam turbine, it has been clearly said that the steam gains velocity by passing through a nozzle shaped path inside the turbine that is called the stator. That means not only the pressure, but also the heat of the steam itself has also been converted into work inside the turbine, to precise at the stator section.
    If just pressure will be enough, then why steam is superheated in thermal power plants to get higher efficiency?
     
  9. Aug 14, 2016 #8
    You're right ....I had assumed the final exhaust steam pressure was atmospheric ,... but from wikepedia .."The exhausted steam is at a pressure well below atmospheric,"....
     
  10. Aug 14, 2016 #9

    Mech_Engineer

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    This answer is easily answered with a Google search:

    http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node66.html
     
  11. Aug 15, 2016 #10
    Thank you Mech_Engineer, the part that you have posted clearly supports my claim. Increasing temperature i.e. superheating steam means increasing efficiency and that will lead to higher output. In short, by increasing temperature we can increase output and that isn't possible until and unless a part of the extra heat will be converted into power.
     
  12. Aug 15, 2016 #11

    Mech_Engineer

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    It is well-know that superheating in a Rankine Cycle increases efficiency. There are also more complex "superheat with regeneration" Rankine cycles which re-heat the fluid a little in between stages of a multi-stage turbine. The goal is to follow the saturation curve and extract as much energy as possible out of the fluid.

    http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/fig6RankineReheat_web.jpg

    Keep in mind (as I said before) turbines convert pressure into mechanical work. A turbine does not convert heat directly to work, the advantage of superheating the steam is to increase its pressure (and to increase steam quality after expansion through the turbine). When you think practically about a turbine, it is a set of blades which must be pushed by the working fluid; this net force doesn't come from the fluid's temperature, it comes from pressure.
     
  13. Aug 15, 2016 #12
    In Boilers of thermal power plants, the mode of heating is isobaric i.e. pressure remains constant during compression. Inside the turbine at the stator section, this additional heat will be converted into dynamic pressure and that will drive the blades of the rotors of the next stage. Overall, heat will certainly be converted into energy by a turbine but in the stator section, it has been converted into velocity i.e. dynamic pressure and then into energy by rotating the rotor.
     
  14. Aug 15, 2016 #13

    Mech_Engineer

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    I'm not aware of any process step which directly converts "additional heat into dynamic pressure." You might read through this section of my previous link to gain some more insight though:

    http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node66.html
     
  15. Aug 15, 2016 #14
    Some excerpts from wikipedia:
    Source: https://en.wikipedia.org/wiki/Steam_turbine
     
  16. Aug 15, 2016 #15

    Mech_Engineer

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    By definition a nozzle uses a pressure differential to create velocity. In your provided reference:

    The temperature of the fluid drops as a result of the pressure drop within the turbine as a system as work is generated.
     
  17. Aug 15, 2016 #16
  18. Aug 15, 2016 #17

    Mech_Engineer

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    I'm aware of that thread, I see no useful information in that thread w.r.t. the posted question in this thread.

    A nozzle by itself will not convert temperature to velocity (a.k.a. kinetic energy). The purpose of a thermodynamic process like the Rankine Cycle as a whole is the conversion of heat (combustion source) to mechanical work.
     
    Last edited: Aug 15, 2016
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