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Home Forums ANSYS Fluent Beginners to Advanced level Lesson 17 On Calculating efficiency of pump

  • Lesson 17 On Calculating efficiency of pump

    Posted by Raghu Karthik on December 19, 2023 at 8:48 am

    On Calculating efficiency.
    I read in book on turbomachinery by dixon and hall. That the efficiency of a pump is calculated by (isentropic specific work)/ (actual specific work). I can easily get actual work which is gH_actual. for isentropic specific work I need to completely define the state of fluid at both inlet and outlet, means i need to determine any 2 thermodynamic properties at inlet (point 1) and outlet(point 2). Then from second law of thermodynamics i can say isentropic specific work = h02s(s1,P02)-h01(T1,P01) since isentropic means s1=s02s. So i need temperature solution. in the case of temperature solution available. it should also be matching with the following expression Actual specific work (h02-h01) = gH = (P02-P01)/rho. only then i can proceed with isentropic specific work formula given above.

    For isentropic specific work without the thermal solution, I tried using T1 (say 20C) ,P01 =1e5 to get state at inlet which gives me s1/s01, h01 . Then after simulation i get P02. now i have s1 and P02 i can get h02s(s1,P02) but the thermodynamic property calculators online are very inaccurate with entropy inputs. because of that inaccuracy, for 80 kg/s outlet mass flow rate i am getting 98% pump efficiency and for 90 i am getting 102% . Although the exit pressures vary by 2148 the h02s values remain unchanged because of erroneous calculators online.

    I referred to mollier diagram of a compressor for reference.

    So for this i need an energy solution . Which is my earlier question (On why it is not getting solved and the solutions being uniform and constant.)

    MIDHUN replied 3 months ago 2 Members · 2 Replies
  • 2 Replies
  • Raghu Karthik

    Member
    January 11, 2024 at 3:21 am

    Hello, Can anyone go through this and tell me if this approach is correct or not.

  • MIDHUN

    Member
    January 12, 2024 at 7:43 am

    Hi Raghu,

    Sorry for not getting back to you sooner

    When modelling for incompressible flow we need not account for enthalpy. In devices such as liquid pumps the working fluid can be treated as incompressible. The efficiency of a pumping process with an incompressible working fluid is defined as the ratio of the head rise achieved by the fluid to the power supplied to the rotor/impeller called the “hydraulic efficiency”. In such situations we may need not account for temperature and enthalpies associated.

    You can refer to Ansys_Fluent_Users_Guide section (12.4.2.1.9. Efficiencies for Pumps and Compressors) for checking which comes under Modeling Turbomachinery Flows for details on the equations used under this topic. In case of compressors you can use these equations where temperature is also accounted.

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