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February 9, 2016, 09:25 |
Turbine Efficiency
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#1 |
Member
Alex
Join Date: Feb 2016
Posts: 81
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Cfx provides polytropic and Isoentropic efficiencies.
As well turbine efficiency can be presented as eff = torque * / massflow*Specific Heat Capacity*Temperature*(1- ^((1-gamma)/gamma))) Torque, massflow and omega are simple to determine. What about other variables? 1. Specific Heat Capacity (Cp) depends on static temperature. Is temperature should be calculated at inlet face? 2. Temperature - total one in stationary frame at inlet face? 3. - total pressure in stationary frame at inlet face divided by total pressure in stationary frame at outlet face? 4. gamma = Cp/(Cp - R) - well, what surface static temperature should be averaged on? Or may be volume average function should be used? I'd like to avoid constant values like 1.4 for air lest simplify the task. |
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February 9, 2016, 09:58 |
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#2 |
Senior Member
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Such formula is ONLY valid for constant properties and ideal gas equation of state. Their use for variables properties is questionable at best because of many factors.
For example, how strong is the dependence within the range of operation of the machine? What is a function of temperature and pressure ? Is the equation of state non-ideal ? If you want to account for variable properties and everything else, use the efficiency definition, i.e ratio of "real work|power produced" / "ideal work|power possible". Efficiency = ( omega * torque ) / ( massFlow * [Inlet Total Enthalpy - Outlet Total Enthalpy] ) Hope the above helps, |
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February 9, 2016, 17:15 |
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#3 |
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I assume your CFX setting is based on an ideal gas (air) for the turbine. Because you had no real gas model, you need to stick to the isentropic efficiency equation for a perfect gas. Take an average gamma and Cp between inlet and outlet (based on Stn total temperature), and the inlet Stn total (mass-averaged) temperature.
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February 10, 2016, 01:38 |
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#4 |
Member
Alex
Join Date: Feb 2016
Posts: 81
Rep Power: 10 |
turbo, you are right, I just took default Air Ideal Gas (constant Cp) that's represented in CFX and just changed SHC to Zero Pressure Polynomial Option, default Dynamic Viscosity to Sutherlands formula option, Thermal Conductivity to polynomial founded somewhere in the www. But! These parameters are based on static temperature, ain't it? So Cp (and gamma as a consequence) should be based on static temperature, not Stn total temperature?
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February 10, 2016, 02:01 |
P.S.
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#5 |
Member
Alex
Join Date: Feb 2016
Posts: 81
Rep Power: 10 |
Actually I have to estimate geometry change effect on turbine blade wheel somehow. I have both blade wheel geometry editions (3d models) and all necessary conditions. I have no stator models. So I simulate just one rotating domain in CFX (I can afford steady state mode only). I suppose efficiency is main criterion in my case?
Taking the opportunity, I'd like to ask: what boundary conditions are correct for the second simulation simulation? For the original edition I used profile (Total Pressure and Temperature in Stn Frame and flow direction angles) at inlet and averaged static pressure at outlet. I suppose that I should take the same profile data at inlet and massflow at outlet? But one point makes me hesitate. If turbine power gets lower than its rotating becomes slow, moreover, compressor yields less massflow? How should I treat this case? |
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February 10, 2016, 09:19 |
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#6 |
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Specific heat does not care about total or static temperature because dh = Cp x dT or dho = Cp x dTo, all the same Cp. Your model is with a dynamic flow, therefore total temperature would be right to say.
If an ideal gas is concerned, the isentropic efficiency definition reduces to the ratio of temperature only, leading to requiring the pressure ratio, temperature ratio and gamma only. You do not have to worry about Cp or T1. Original BC combinations are correct for turbomachinery CFD. Do not use the mass flow exit BC unless you are in troubles. If compressible viscous flow (of hyperbolic equation types) is simulated, specifying static pressure at exit is making sense in real physics. |
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February 11, 2016, 02:24 |
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#7 |
Member
Alex
Join Date: Feb 2016
Posts: 81
Rep Power: 10 |
turbo,
I prefer to use total pressure at in and static one at out. Tried to use massflow at out, but convergence (I mean monitor points) proved about dozen! times slower. So, you suggest me to use the same pressure and temperature values at inlet and outlet for both cases, original geometry and modified one? And pay attention to isoentropic and polytropic efficiency differences as well as ratio of expansion? |
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February 11, 2016, 15:40 |
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#8 |
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Always apply (Po, To, angle) at inlet and p at exit.
The efficiency definition is called "isentropic efficiency", not polytropic. |
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February 12, 2016, 12:07 |
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#9 |
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There are multiple types of efficiencies in turbomachinery, among them are the overall efficiency, isentropic efficiency and not least the polytropic efficiency.
Each one has its usage. A good reference for definitions of the 3 above is the classic by Dixon, http://www.ewp.rpi.edu/hartford/~ern.../Dixon-Ch2.pdf |
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October 5, 2016, 07:07 |
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#10 | |
Member
AHMAD
Join Date: Oct 2013
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Quote:
If let say I have two cases(Case 1 and Case 2) to be simulated. Case 1 and Case 2 do not share the same blade profile but both Cases 1 and 2 have the same boundary conditions. Slight modifications to be performed on the blades of Case 2 but the operating condition of the turbine is the same as Case 1. Then we have the equation of isentropic efficiency where the values at the inlet and outlet can be used to calcuate the isentropic efficiency. Since, the values at the inlet and the outlet are prescribed as the boundary condition, does it mean that we can calculate the isentropic efficiency before we begin the simulation? Or shall I take the values at a plane somewhere close to the boundary condition but not on the boundary location itself? P/S: My SV is an experimentalist and gave this question to me. |
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October 10, 2016, 12:10 |
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#11 | |
Senior Member
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Quote:
Different turbine geometry will provide different outlet performance. Though you fix the static p at the exit boundary, you will get a different mass flow from the other case. That means you will have two different performance of (mflow1, PR, eta1) and (mflow2, PR, eta2) for the same total-to-static pressure ratio (PR). You need to find the outlet performance from CFX-post. |
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