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Efficiency calculation in a francis turbine (CFX) |
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April 3, 2014, 09:32 |
Efficiency calculation in a francis turbine (CFX)
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#1 |
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Hi,
I've been experiencing some problems when calculating the efficiency in a Francis turbine runner. The simulation is runned in steady state. The runner is created from a single blade with a fluid passage and then copied 10 times into a full circle. I use the following equation in CFX-Pre: torque_z()@Blade*omega/(massFlow()@Inlet*(dP/roh+dV/2)) dV=(massFlow()@Inlet/(area()@Inlet*roh))^2-(massFlow()@Outlet/(area()@Outlet*roh))^2 dP=(massFlowAve(Pressure)@Inlet-massFlowAve(Pressure)@Outlet) roh=massFlowAve(Density)@Inlet omega=2300*pi/30 The efficiency is calculated to 1.10 which is obviously wrong. The boundary conditions are set to: Inlet: cyl.vel.components. (0 -2.02 -7.2682) Outlet:static pressure (1 atm) The Domain is set to "Rotating" with 2300revmin^-1. The interface between the passage is set to Rotational Periodicity. Does anyone have any idea of what could be wrong? All answers are appreciated. |
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April 3, 2014, 17:46 |
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#2 |
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Bruno
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Try this:
dPtot = massFlowAve(Total Pressure in Stn Frame)@Inlet - massFlowAve(Total Pressure in Stn Frame)@Outlet volFlow = massFlow()@Inlet / areaAve(Density)@Inlet omega = 2300 [rev/min] eff = (torque_z()@Blade * omega) / (volFlow * dPtot) This is pretty much what you have, only rewritten to use the total pressure directly. If you still have an efficiency > 1, then something is wrong with your setup. Check the mass flow, the velocity vector at the inlet, the rotation direction (from your velocity vector, it should be negative), if the interfaces between the copies are ok (that is, if CFX identified them correctly and didn't create walls) and if your simulation has fully converged (monitor the torque during the simulation and ensure that it too achieved steady state). If you can't find anything, post the CCL from your model here. |
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April 3, 2014, 18:11 |
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#3 |
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Glenn Horrocks
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Have you done the basic accuracy checks? http://www.cfd-online.com/Wiki/Ansys..._inaccurate.3F
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April 4, 2014, 08:53 |
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#4 |
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Thanks for the help, it is highly appreciated. I tried your equation brunoc. The velocity wasnt negativ so I rotated the runner in opposity direction (-2300). Theh the result was eff= -76.8503 [degree]. The negative value comes from the torque around the z-axis. And when i multiplied it with pi/180 to transform it into radians the eff= -0.971969 [degree] which seems reasonable. Do you know how to change CFX-pre to calculate in radians? And again thanks for the help!
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April 4, 2014, 09:39 |
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#5 |
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Bruno
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You can change the units used in the options. Or just divide your expression by '1 [rad]':
eff = (torque_z()@Blade * omega / 1 [rad]) / (volFlow * dPtot) Also, if you set omega to -2300 [rev/min] and use this expression to set the domain rotation speed, you won't have to play around with signals. BTW, did your previous results converge well? Usually when you have the rotating signal wrong the solver complains quite a lot. Take a look at the streamlines on your old and new results. If your new efficiency of ~0.97 is right (it's almost too good to be right, if I may add) then you'll have quite smooth streamlines. If not, then something is still wrong. |
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April 4, 2014, 13:00 |
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#6 |
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I compared the streamlines of the old and the new calculations. The streamlines from the old calculation were quite chaotic, with a swirl at the inlet by each passage. The new calculation has very smooth streamlines all the way through. You have been very helpful, thank you
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April 5, 2014, 12:39 |
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#7 |
New Member
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I presume, your simulation is perfectly converged with acceptable accuracy as well as you have performed mesh scaling test before the final results.
----- I would suggest, Start with simple math.. Perform manual calculations by extracting the variable values using function calculator in CFD Post. Compute hydraulic efficiency. Please remember the torque may be negative in z direction if the runner rotation is counter clockwise. One you are sure about the value obtained from function calculator then go for long expressions. If you find will find wrong value then something goes adverse in the simulation. |
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April 10, 2014, 04:26 |
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#8 |
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Thanks for the input. I calculated the efficiency by hand based on eff=M*omega/(roh*g*H*Q). Where M is derived from the vorticity gamma=pi*diameter*Cu(tangential absolute velocity), by M=roh*gamma*Q/(2*pi).
I also used an eff. calc. based on the inlet and outlet velocity triangles. eff=((C1^2-C2^2)+(U1^2-U2^2)+(W2^2-W1^2))/(2*g*H). Where C=absolute velocity, U=tangential velocity and W=relative velocity. They both end up around 0.50. The problem, as I see it, is that calculations by hand are defined by the estimated omega, and can then only be indicative of the true efficiency. I'm a beginner in runner design so correct me if I'm wrong. Probably there is a more accurate way to calculate it. Where the force from the pressure and suction side of the blades come into count. The runner will soon be 3D printed in plastic and tested in a hydro-power prototype lab. Any advice or suggestion is appreciated. |
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April 10, 2014, 14:45 |
Bad interface model
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#9 | |
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Quote:
If you use the full turbine geometry, you must use the General Connection interface model between passages not a rotational periodicity. |
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April 11, 2014, 05:24 |
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#10 |
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Hi,
the rotational periodicity is set between the periodic sides of the fluid passage around one blade. I think my former explanation was a bit sloppy. So the simulation is based on one blade. |
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April 11, 2014, 10:39 |
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#11 |
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The Francis turbine is a radial turbine, the axial velocity component is zero. At the rotor inlet, the velocity vector has two components tangential and radial.
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February 6, 2017, 18:17 |
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#12 |
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Have a nice time!
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Hello, how can I calculate the torque? I cannot write its expression..
Thanks |
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February 6, 2017, 18:49 |
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#13 |
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Glenn Horrocks
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torque_x()@Locator
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February 6, 2017, 18:51 |
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#14 |
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Have a nice time!
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February 6, 2017, 19:01 |
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#15 |
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Have a nice time!
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What should I do to complete defining the torque?
Thanks |
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February 6, 2017, 20:57 |
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#16 |
Super Moderator
Glenn Horrocks
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"Locator" should be replaced with the name of the wall object you want the torque from. Also make sure you are getting torque about the right axis.
You need to read the CFX reference manual on CEL expressions and the functions in particular. You need to understand what these functions do before you can use them correctly. |
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