
Lesson 17 On Calculating efficiency of pump
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 (h02h01) = gH = (P02P01)/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.)
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