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January 25, 2017, 09:38 |
Input power calculation for pump in CFD post
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#1 |
New Member
Dhairyasheel deshmukh
Join Date: Jan 2017
Posts: 4
Rep Power: 9 |
hello,
I am simulating fluid flow through centrifugal pump using cfx. after simulation, I am generating pump report using report template. the head value matches exactly with the experimental data but the efficiency is not matching. I think I am not using correct input power value from the expression. can someone help me how to find the exact input power value for a centrifugal pump in CFD-post? thanks, regards, |
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January 29, 2017, 09:10 |
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#2 |
Senior Member
Join Date: Jun 2009
Posts: 174
Rep Power: 17 |
See below.
See below my old post related with your question : find how many are missing in your CFD to evaluate torque comparable to test values. Let me clarify "Pump Efficiency in CFD" for you (and also for many people who have misunderstood it). Below are what you defined, "head (massFlowAve(Total Pressure in Stn Frame )@R1 Outlet-massFlowAve(Total Pressure in Stn Frame )@R1 Inlet)/(ave(Density )@R1 Outlet*g) shaft power abs(sum(Velocity in Stn Frame u * Force X + Velocity in Stn Frame v * Force Y + Velocity in Stn Frame w * Force Z)@Calc Wall Region) * numComponents" The bottom part, what you named "shaft power", is however not the real shaft power because your CFD domain excludes the followings : - Disk friction on the front and back cavities of the impeller - Leakage flow power loss through the front and rear seals - Any power loss due to recirculation flows returning back to the impeller - Bearing windage power losses Therefore, any well-designed pump impeller will show over 90% efficiency from CFD if all of them are excluded in the analysis, and if the bottom is a pure calculation of (Torque x omega) from CFX-Post. No surprises ! The CFD pump efficiency cannot be compared to test efficiency unless all of external parasitic losses are included in CFD (but so much rare cases). In other words, what CFX-Post displays "Pump hydraulic efficiency = (rho * g * Q * dH) / (Torque x omega)" is not a true one, only confusing users. Be careful. |
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February 2, 2017, 11:05 |
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#3 |
New Member
Dhairyasheel deshmukh
Join Date: Jan 2017
Posts: 4
Rep Power: 9 |
Thanks for the valuable information. The head value is close to the experimental for a wide range but, efficiency is deviating except one point at all other points.
Yes, CFD don't consider the losses you mentioned. but it is giving perfect results for head calculation. if I want to find the input power value close enough to match with experimental can you tell me the procedure to do it in CFD post. I have followed one method of calculating (torque on blade*no. obf blades*omega). but that value also don't matches. thank you, Regards, |
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April 27, 2017, 09:06 |
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#4 | |
New Member
Hazem
Join Date: Mar 2017
Posts: 3
Rep Power: 9 |
Quote:
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