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Homework Help: Actual Reheat Cycle

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Given:
    A reheat Rankine cycle

    - Steams enter the high-pressure turbine at 12.5 MPa and 550C
    - It leaves at 2 MPa
    - Steam is reheated at constant pressure to 450C before it expands in the low pressure turbine
    - Isentropic efficiency of turbine = 85%
    - Isentropic efficiency of pump = 90%
    - Steam leaves the condenser as saturated liquid
    - moisture content of the steam at the exit of the turbine is not to exceed 5%
    - mass flow rate is 7.7 kg/s

    Find the condenser pressure
    Net power output
    Thermal efficiency

    2. Relevant equations



    3. The attempt at a solution

    I already got the enthalpy of the inlet of high pressure turbine, at reheat, and at the inlet of the low pressure turbine. After that, I'm stock I do not know what should I do next. I do everything I know.

    The answers are 9.73 kPa, 10.2 MW, and 36.9%
     
  2. jcsd
  3. Jan 14, 2010 #2
    Ok lets take this problem in some steps.

    For ease of understanding the process lets use the diagram below.

    (The diagram in the attachment is taken from "Classical Thermodynamics" by Van Wylen and Sontag)

    From a First Law approach and considering just the high pressure turbine as the control volume and assuming the process has achieved steady state conditions

    The work of the high pressure turbine is

    Wt = h3 - h4

    where Wt = work of the turbine
    h3 = enthalpy of the turbine inlet
    h4 = enthalpy of the turbine outlet

    Now lets assume you have an isentropic turbine (constant entropy) you can say that (we will deal with the efficiencies later)

    s3 = s4

    where s3 is the turbine inlet entropy
    s4 is the turbine outlet entropy

    Now, the conditions of 12.5 MPa with 550 C specifies that the steam entering the high pressure turbine is in the superheat condition.

    Once the steam is expanded across the turbine the pressure is reduced to 2 MPa. when it enters the boiler to be reheated to 450 C.

    So what we need is the enthalpy at h4.

    Calculate that value and we can move onto the next step of considering the low pressure turbine.

    EDIT Sorry, I didn't see that you already have that value of h4. What did you calculate for h4?

    Thanks
    Matt
     

    Attached Files:

    Last edited: Jan 14, 2010
  4. Jan 14, 2010 #3
    Based on your diagram, I already have the h3, h4, and h5. After that, I do not know what to do. I got h4 by s3=s4(isentropic) then interpolate to get the temperature, then interpolate again to find h4(isentropic). After that I used the isentropic efficiency formula. Both the turbines have the isentropic efficiency of 85%. What should I do next after getting h3,h4, and h5?
     
  5. Jan 14, 2010 #4
    Why all of the interpolation? The steam is at 2 MPa when it leaves the high pressure turbine and enters the boiler. If you work out a quality for the steam you will find that it is at the saturation temperature, so h4 is simply the enthalpy at the saturation temperature at 2 MPa.

    Do you follow?

    Thanks
    Matt
     
  6. Jan 15, 2010 #5
    I have some questions, is the efficiency of both turbines are the same (85%)? Where is the moisture content pertaining to, the actual or the ideal? I cannot understand. How can you know that the point at the exit of the high pressure turbine is superheated or wet?
     
  7. Jan 15, 2010 #6
    The 85% efficiency is applied to both turbines. The moisture content is not dependent upon the actual or the ideal cycle.

    My reasoning for the steam to be at saturation temperature with a quality of 1 when it leaves the high pressure turbine is because most turbines are not condensing turbines (condensing turbines do exist but I am assuming that problem does not involve one.) and because you dont want moisture in the turbine.

    The enthalpy that I have for steam at 2 MPa at saturation temperature is 2799.5 kJ/kg

    So h4 = 2799.5

    From the equation I gave for the work of the high pressure turbine above is

    wt = h3 - h4

    h3 = 3475.2 kJ/kg (enthalpy at 12.5 MPa and 550 C)

    wt = (3475.2 - 2799.5) = 675.6 kJ/kg

    So the ideal work of the high pressure turbine is 675.7 kJ/kg

    Now the efficiency of 85% simply derates the ideal value by 15%

    So the actual work of the high pressure turbine is

    0.85*656.7 = 574.3 kJ/kg

    Now you need to calculate the work of the low pressure turbine. Now since it is specified that moisture content of the steam at the exit of the turbine is not to exceed 5% that means you have an exit quality of 95% so in this turbine there is moisture. You will need this value to calculate the exit enthalpy of the low pressure turbine. The sum of work of the two turbines is the total work of turbine system

    Do you follow?
     
  8. Jan 15, 2010 #7
    I know h5 (inlet of low pressure turbine), how can I get h6? to be able to determine work at the low pressure turbine?
     
  9. Jan 15, 2010 #8
    Do you understand what a quality value of the steam is?

    Please answer my questions instead of just asking for more values, I want you to gain an understanding of this process not just gain enthalpy values from me.

    Thanks
    Matt
     
  10. Jan 15, 2010 #9
    yes, from the given, the mixture of liquid-vapor contains a vapor of 95%. the enthalpy at the exit of the low pressure turbine is equal to hf + x(hf-hg). x is the quality (95%). At that point 6, what my problem is the condenser pressure where which will I get the hf and hg.
     
  11. Jan 15, 2010 #10
    how should i know the condenser pressure?
     
  12. Jan 16, 2010 #11
    That is how you calculate the value. Don't worry about the condensor pressure for now.

    Calculate the enthalpy of the exit of the low pressure turbine at 2 MPa and then calculate the total work of the turbine system with an efficiency of 85%

    Thanks
    Matt
     
  13. Jan 16, 2010 #12
    do you mean the enthalpy at low pressure turbine exit at 2Mpa? At 2 MPa, hf=908.79, hg=2799.5, so h at the LPT is 908.79+0.95(2799.5-908.79) = 2704.9645. Am I right thats how I understand it.
     
  14. Jan 16, 2010 #13
    Your are right.

    Now that you have the work of the turbine system, calculate the work of the pump.

    Thanks
    Matt
     
  15. Jan 16, 2010 #14
    work of the pump is v2(P2-P1) or h2-h1 but how? I dont have those values
     
  16. Jan 16, 2010 #15
    The pump, pumps water only. You NEVER want to pump a two-phase mixture. The pump supplies the pressure of 12.5 MPa. So h2 is the enthalpy of saturated liquid at 12.5 MPa.

    Apply some simple thinking on your part to the system and you will get the work of the pump.

    Thanks
    Matt
     
  17. Jan 16, 2010 #16
    why is the pressure at the low pressure turbine exit is 2Mpa? after getting the work of the pump whats next?
     
  18. Jan 17, 2010 #17
    Once you have the work of the pump, you can calculate the cycle efficiency. I used the pressure of 2 MPa at the exit of the low pressure turbine and calculated an efficiency of 36%

    Thanks
    Matt
     
  19. Jan 17, 2010 #18
    do mean you use 2Mpa as the condenser pressure? whats the reason?
     
  20. Jan 17, 2010 #19
    Yes, I mean to use 2 MPa because I am not sure how the condensor pressure is calculated yet.

    I left my thermodynamics textbook and notes in my office and I am at home until Monday. So for now think on your own and see if you can calculate the condenser pressure.

    Thanks
    Matt
     
  21. Jan 17, 2010 #20
    I have try solving this problem in several days, but I cannot get it, so I asked here. can you use pressure other than 2 MPa to solve the efficiency?
     
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