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How to calculate the power generated by a turbine (steam)?

  1. Oct 28, 2014 #1
    Firstly, lets assume the expansion is isentropic.

    An example I found:
    Mass flow = 0.5kg/s
    Pi = 6000 kPa
    Ti = 500 Celcius

    Pf = 30kPa

    From this, how do we find the power generated and final steam quality?

    Do we look at steam tables, W = m(flow) (h_1 - h_2) --- is that helpful?
  2. jcsd
  3. Oct 28, 2014 #2


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    Unless you know the enthalpy of the inlet and exhaust steam, yes, you need to refer to steam tables.

    For your exhaust condition, knowing the pressure is often not enough information. However, since you have specified a turbine operating isentropically, then using a Mollier diagram (a special diagram of enthalpy versus entropy), the state line of this turbine will be a single vertical line running from the inlet point down to the exhaust pressure of 30 kPa.

    Since Mollier diagrams are hard to find on the web, you can also use steam tables.

    For P = 6000 kPa and T = 500 C,
    then h = 3423.1 kJ/kg and s = 6.8826 kJ/kg*K

    For P = 30 kPa, we need to find the properties for s = 6.8826 kJ/kg*K,

    sf = 0.9441 kJ/kg*K
    sfg = 6.8234 kJ/kg*K

    s = 6.8826 kJ/kg*K = sf + x * sfg, where x is the quality of the vapor

    doing the algebra and solving for x = 0.870, or a vapor quality of 87%

    The enthalpy of this mixture can be calculated using the saturated liquid properties:

    hf = 289.27 kJ/kg
    hfg = 2335.3 kJ/kg

    h = hf + x * hfg
    h = 2321.7 kJ/kg at the exit.

    The isentropic work (or more accurately, the power) is therefore W = m*(hi - ho) = 0.5 kg/s * (3423.1 - 2321.7) kJ/kg

    W = 550.7 kJ/s = 550.7 kW

    The property info comes from the attached tables:

    Table A-5, p.10 for the saturated water
    Table A-6, p. 14 for the superheated vapor

    Attached Files:

  4. Nov 4, 2014 #3
    SteamKing your calculation is correct, but the power you found is for a steam turbine 100% isoentropic efficiency.
    To calculate consumption, power and efficiency of steam turbines easily use the app from app store: https://itunes.apple.com/br/app/steam-turbine/id934240038?mt=8
  5. Nov 4, 2014 #4


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    Yes, I know the calculation is for an isentropic condition. That's what the OP specified.
  6. Nov 7, 2014 #5
    Thanks a lot!

    Do you recognise this formula...

    W = PnR(T2/P2-T1/P1)

    It's in the answers for a question, the question asked to calculate the work done by the air in a cylinder as it expands out. What's this equation, where is it derived from?
  7. Aug 20, 2015 #6
    Hi, everyone

    I have question about turbine and pressure.
    how to calculate Turbine pressure when
    N=20 MW
  8. Aug 20, 2015 #7


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    Hi, asdarw

    Unfortunately, you have not provided enough information to calculate pressure in the turbine. A turbine extracts work between an inlet pressure and an exhaust pressure, so you'll need to specify one or the other of these pressures.

    Also, if you want to ask a question, don't hijack another user's thread, especially one which is months, if not years, old. Make a thread for your own question. It's easy to do, and you can make as many different threads as you need.

    If your question is for home work or course work, be sure to fill out the HW Template and post in the appropriate HW forum.
  9. Aug 20, 2015 #8
    I know how to open new thread, but i didnt know where.

    Yes its about test in School. and that is all information.
    I know Q, N, and H, and the answer is 45.35
    and i dont get it, how they calculate it...

    The original text is not in english so i translated it. text is:

    whats pressure in turbin if: N=20 mgwt,
    Q=50 m^3/sec, h =0,9. and answer is 45.35
  10. Aug 20, 2015 #9


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    The answer is 45.35 what? Use units for all the quantities you write down. As I said before, you have not provided sufficient information to calculate "pressure", which I am assuming might be inlet pressure.

    The Rules for HW problems ask members to post the complete problem statement, not snippets, excerpts, abridged versions, etc. It gives others who read your post all the relevant facts which you possess, and it cuts down on this back-and-forth Q&A which just wastes time.

    At PF, the Mentors are perfectly happy to answer any questions you might have about posting threads, including in which forum they should be placed. It's their job to make sure that things run smoothly, so don't hesitate to contact them with any questions you might have about making a post.
  11. Aug 21, 2015 #10
    Good job on the first question, SteamKing. I would have done the problem 100% the same. Good old thermo...

    As for the second question, I'm thinking that perhaps the pressure is a change in pressure, or what I always call the delta P (of the flow work).

    That being said, there is enough to calculate the delta P (but not a static inlet or outlet pressure as you pointed out) without having all the other information.

    Let's make sure I understand the variables first:
    Q is volume flow rate
    h is isentropic efficiency
    N is power output?

    [tex]N = h Q \Delta P[/tex]
    [tex]\Delta P = \frac{N}{h Q} [/tex]

    Could that be what he's asking? Granted, this only looks at the flow work, and I'm not getting the answer he gets, so this was only a guess. How did they arrive at 45.35? What units is that in? I get 444 kPa for the change in pressure as shown, but I'm somewhat perplexed what they're asking for. Given the variables you gave us, this is my only guess at how to even arrive at a pressure.
    Last edited: Aug 21, 2015
  12. Aug 21, 2015 #11


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    Turbines operate by extracting energy from the expansion of steam through the device. The change in energy is not determined solely by the difference in pressure; the temperature of the steam is also changing as it expands through the machine.

    You need to know the inlet conditions of the steam entering the turbine (pressure and temperature) and exit conditions (pressure and temperature) in order to calculate the amount of work extracted. Without this information, you're just guessing.
  13. Aug 21, 2015 #12
    For a compressible substance, yes. Did he (asdarw) specify that the substance was compressible? Otherwise, if you can assume an incompressible fluid, which based on his lack of supplying any other information on the problem, I was forced to make some kind of assumption, that's what I got.

    But then again, I guess if it were an incompressible substance, it wouldn't be using an isentropic efficiency, which begs the question, is h an enthalpy? Quite small for an enthalpy... Just not enough information to answer the question, so I made up the best solution I could that looked close to engineering equations I'm familiar with. But yeah, come to think of it, isentropic efficiency can't be used with incompressible substances, so, disregard my above explanation.

    Did they assume the outlet pressure at atmospheric or make some other kind of simplifying assumption? As SteamKing pointed out, without an inlet or exit pressure (and temperature or some other property to nail down the state), there really isn't enough information to solve that problem.
  14. Aug 21, 2015 #13
    Oh, is h the polytropic constant? In which case, you could use polytropic work to solve it.

    [tex]\dot{W}=\frac{Q\Delta P}{1-h}[/tex]
    [tex]\Delta P=\frac{\dot{W}(1-h)}{Q}[/tex]

    Naw, but that yields 360 kPa. Yeah, I'm at a loss on this one, and using pseudo-science trying to fill in the holes in the problem. Besides, if it's compressible, assuming the same volume flow rate at the entrance and exit is not realistic. Yeah, play with the equations as best I can, I can't figure out how they could make a homework problem out of the information he gave.
  15. Aug 21, 2015 #14


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    We don't really know what this guy wanted. :rolleyes:

    He hijacked another thread and posted a drive-by question. He hasn't responded with any additional information, so I assume that he no longer needs or wants any help with a solution. o_O
  16. Aug 22, 2015 #15
    Thank you guys to try solve this question.
    This "pressure" is not usual pressure i think, I didnt translated maybe properly. just knew that pressure means same.

    Pressure hydraulics - the linear size, which depicts a specific fluid flow (fluid weight unit correlated) energy given point. Design head will be selected according to the task. For example, in the upstream side of the dam will be the head of the water depth of the tailrace. Holes from the fluid or gas pressure at the time the resolution of problems relating to the matters gamodinebis Received hole cross-section area of the center of gravity sinking depth (fluid in the atmosphere of the leaves). HPP distinguish between gross head (upstream and downstream the difference in elevation) and net head. Net head equals the gross head hydraulic resistance caused by the losses. (google translate )

    the answer 45.35 is in metres i think, just 45.35m in answers and there is only 3 info for N, Q, H.. Becouse this misunderstanding, I posted here, maybe u know better why is 45.35m answer or enough or not the information. but the task stille exist :D

    Sorry for trouble, Im in trouble too
  17. Aug 22, 2015 #16


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    This is exactly why PF doesn't like thread hijacking.

    The thread started out originally talking about a steam turbine, and then someone comes along later and hijacks the thread with a post about a hydraulic turbine, but doesn't bother to tell any one about this crucial distinction. :mad:
  18. Aug 22, 2015 #17
    Aha! So, that will be pretty darn close to what I listed, only with the following variables:

    N = power output, units of W or MW in this case
    Q = volume flow rate, units of m^3/s
    h = turbine efficiency, unitless
    rho = fluid density (h20 at ~300 K), units of kg/m^3
    g = acceleration due to gravity (at sea level), units of m/s^2
    Delta h = head equivalent of power output of turbine, units of meters

    [tex] N = hQ\Delta P = hQ \rho g \Delta h[/tex]
    [tex] \Delta h = \frac{N}{hQ \rho g}[/tex]

    Assuming H = 20 MW, Q = 50 m^3/s, h = 0.9, rho = 999 kg/m^3, g = 9.81, then the head loss is 45.35 meters.

    Granted, those symbols are not what I'd use typically. Instead of h for efficiency, usually eta is used. I like to use an upside down A with a dot over it for volume flow, but Q is common. I prefer to use a W with a dot over it for power, I've never seen N used, but different textbooks use different symbols.
  19. Aug 22, 2015 #18
    Thank you for help, now I get it.

    Sorry for hijack, Just when I opened thread Staff deleted it, Ill try to get this forum rules :)
  20. Aug 22, 2015 #19


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    That's probably because you posted a HW problem without filling out the HW template or showing an attempt at a solution.
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