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April 4, 2014, 05:43 |
Determine the torque of a turbine
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#1 |
New Member
Simon
Join Date: Apr 2014
Location: Stuttgart, Germany
Posts: 22
Rep Power: 12 |
Hello,
I have got some problems to determine the torque of an axial turbine. Because when I´m choosing the mesh goemetrie on the blade (pic 1), the power is about 310W. When I´m choosing the mesh geometrie for the blade (pic 2), hub and shroud, the power is about 90W. The Volume on the pictures is the fluid volume and between is the blade (invisible). Pic1.PNG Pic2.jpg Has somebody the solution? Thx error404 |
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April 4, 2014, 08:52 |
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#2 |
Senior Member
Bruno
Join Date: Mar 2009
Location: Brazil
Posts: 277
Rep Power: 21 |
If what you are modelling is an axial turbine and you pictures show your entire domain, then your simulation is incorrect, unless you have some really unconventional turbine. You should simulate the entire blade, including leading and trailing edges, plus some region up and downstream of the blade to allow the fluid to "settle" (I don't know what the correct word should be here).
Take a look at the tutorial 'Flow in an Axial Turbine Stage'. It covers everything related to your simulation. |
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April 4, 2014, 11:27 |
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#3 |
New Member
Simon
Join Date: Apr 2014
Location: Stuttgart, Germany
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Thank you for the information,
but the tutorial is done with a turbogrid mesh. I made the mesh with ICEM. I have got the "problem" that I haven´t got any gap between the stages. So I have to split it up, in a left and right fluid volumen?? |
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April 7, 2014, 04:24 |
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#4 |
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Simon
Join Date: Apr 2014
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April 7, 2014, 07:26 |
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#5 |
Senior Member
Bruno
Join Date: Mar 2009
Location: Brazil
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Then you might have to model the stator upstream and downstream to your rotor.
My guess is that the flow on your geometry will be very tightly coupled, so if you plan on using 'Frozen Rotor' for the interface, be sure to sweep across several different rotor positions. But a transient run with 'Transient Rotor Stator' would probably be a better idea. |
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April 7, 2014, 18:22 |
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#6 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
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This looks like a very unusual machine. The "blades" are big blocks which will produce torque when the passages align, but the flow will just about stop when they do not align as the flat front of the blade just about blocks the entire passage.
The result of this is I would expect this machine to have a very variable torque versus angle curve (high torque when the passages align, low torque when they do not) and a very pulsey flow rate (high flow when the passages align, low flow when the do not). Where did this design come from? For most applications this is a very poor machine design. But assuming you still want to simulate it, you are going to have to take account of this very pulsey flow. If you are using frozen rotor you are going to need to run lots of angles to get the full flow curve. |
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April 8, 2014, 06:53 |
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#7 |
New Member
Simon
Join Date: Apr 2014
Location: Stuttgart, Germany
Posts: 22
Rep Power: 12 |
Thank ghorrocks,
I´ll expect the same, regarding the torque plot. When I want to simulate transient and the frequenz is 100 000rpm, very fast. Which time step size should I choose? Is there any flashover value? I know some flashover value. It´s: 1/omega^2, it´s very small--> much calculationtime |
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April 8, 2014, 07:20 |
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#8 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
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Why is flashover relevant to this model? What gas is flowing through this thing?
CFD = long simulation time. Get used to it Yes, it is very small, but the time for each revolution is very small too. I should write an FAQ on this bit: To set the time step use adaptive time steps homing in on 3-5 coeff loops per iteration. Make sure the max and min time step size are wide enough that it never hits it. And that your initial time step is in the ball park (read the documentation for a guide to setting time step size in turbomachinery). Then the solver will find its own time step. |
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April 8, 2014, 07:28 |
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#9 |
New Member
Simon
Join Date: Apr 2014
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The fluid is ideal air, compressible.
600 000Pa (Inlet) 101 325Pa (Outlet) |
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April 8, 2014, 07:35 |
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#10 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
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So if it is air (ideal gas) then what does flashover got to do with it?
Your pressures mean you should be using a reference pressure of 101325Pa, an inlet pressure of 498675Pa and an outlet of 0Pa. |
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April 8, 2014, 07:38 |
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#11 |
New Member
Simon
Join Date: Apr 2014
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The flashover is from a student. He had seen this flashover in the www. for trubomachinery.
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