Low torque values on Screw Turbine

ghorrocks · July 19, 2015, 19:53

It looks like it has converged a bit but not very far. This is an FAQ: http://www.cfd-online.com/Wiki/Ansys...gence_criteria

When you assess the inlet and outlet pressures, you should use areaAve, not massflowAve. It does not make much sense to massflow average pressure.

Can you post an image of the flow you are getting? Have you checked there is not soemthing funny going on and the flow is not as you expect?

Shaun Waters · July 21, 2015, 18:35

I've attached a picture of the velocity flow. To me it looks as it should. Again, it's just that torque comes out in the region of 1e+7.

I have also changed to areaAve as suggested.

Thanks

ghorrocks · July 21, 2015, 21:35

Can you:
* Show some images of the detailed flow in the passage? Is there separations? What does the pressure distribution look like?
* What does areaInt(p)@rotor give? Does it match the force()@rotor? Also check areaInt(p*x)@rotor and compare to torque.

Shaun Waters · July 22, 2015, 07:15

When using the setup as suggested, with inlet: mass flow rate =297,000kg/s and outlet pressure and direction = 0[pa] with a reference pressure set as 39KPa, the velocity streamline is not as it should be. The convergence is also not as smooth.

For this setup, the values for the expressions are as follows:
areaInt(p)@Turbine = 4.26533e+010N
force_z()@Turbine = -3.05309e+008N (the force expressions required a coordinate direction, so I chose the axis of rotation)
areaInt(p*x)@Turbine = 1.14151e+008 [J]
torque_z()@Turbine = -1.382e+008 [N m]

For this setup, the pressure contour over the turbine is of a high value.

For the setup where opening pressure and direction outlet has the 39KPa value and there is no reference pressure the values are as follows:
areaINt(p)@Turbine = 6.12118e+007 [N]
force_z()@Turbine = -2.5187e+006 [N]
areaInt(p*x)@Turbine = -1.36444e+006 [J]
Torque_z()@Turbine = 1.79003e+006[N m]

The first three images are from the 1st setup, the following images are from the second setup.
For both:
1st image = convergence
2nd image= velocity streamline
3rd image = pressure contour
4th image = wall shear contour

Shaun Waters · July 22, 2015, 07:17

Here are the images for the second setup as mentioned in the previous post.

Thanks.

ghorrocks · July 22, 2015, 07:27

They are seriously big pressures. Are you sure they are in the right range? Are you sure the flow rate you are pushing through this thing is correct?

Have a look at the negative pressures. You are getting pressures below absolute zero which means you are cavitating. Would you expect this thing to cavitate? If cavitation is significant then that would completely change things.

Can you post an image of your mesh.

Shaun Waters · July 22, 2015, 07:49

The flow rate is based on the values of a bulb turbine with the same diameter. I was asked to simulate the same conditions as this turbine. Is it a possibility that I am trying to force a mass flow rate that is impossible for this turbine?

The pressures that I have set are based on the amount of water pressure at the inlet and outlet. At the inlet side there is 6.5m of water above the inlet, and the water is flowing through the turbine into a basin already with 4m of water above the outlet. Outlet pressure based on density x gravity x 4m = 39kpa

I've attached images of my mesh.

I forgot to mention, I would not expect any cavitation with this turbine. Also, I agree the pressures encountered here, are very high also, I do not know why.

Thanks

ghorrocks · July 22, 2015, 08:18

If your expected inlet pressure is 6.5m = 65kPa and outlet pressure = 39kPa then I would be seriously worried about the pressure distribution you have modelled - it ranges from 1.6MPa to -1.1MPa. Clearly you have not modelled what you have in mind.

Also the fact that changing the reference pressure significantly changes the result is a worry. This means something is quite wrong with your simulation.

Are you saying that you are modelling a variation on a bulb turbine of the same diameter? Why do you expect it to perform as you say? See my post #1

- this device will not perform any where near a bulb turbine.

So yes, I am suspicious that the flow rate you are specifying through this device will require enormous pressure head to drive it through (like what you are getting) and will result is crazy torque values on the rotor (like what you are getting).

Shaun Waters · July 22, 2015, 08:37

My supervisor wants me to understand how an archimedes screw would perform in a tidal range situation, instead of a bulb turbine. It is supposed to be for a theoretical location where the environmental effects and cost of the bulb turbine, mean it cannot be used. I know that the screw turbine will produce only a fraction of what a bulb turbine would produce, which is why I am expecting a far, far lower value for torque and power.

If, as you say it would not be possible to force the amount of mass flow rate through the turbine as I have specified, is there a way that I could just use the pressure boundary conditions at the inlet and outlet instead?

It would be much easier to achieve higher flow rates with a bulb turbine than a screw turbine as the water is able to pass through a bulb turbine much easier, than being trapped inside the screw for a number of turns.

The pressures are based on the tide coming in, creating a head difference between the sea and a closed off basin. The water then passes through the turbine and filling the separate basin. Pressure on the inlet is the head of the higher sea water, and the pressure on the outlet is the lower level of water inside the basin that is filled by the water passing through the turbine.

ghorrocks · July 22, 2015, 09:03

Now things are clear, and the reason for your crazy result is apparent. You have not chosen appropriate boundary conditions. You have a mass flow at the inlet and a pressure at the outlet. This is wrong as the mass flow is a function of the system and the mass flow you chose is what you get with a bulb turbine, not the screw you have modelled. You should have pressure inlet and outlet - because that is what is externally driving the flow, the pressure head difference from inlet to outlet.

If you repeat this analysis using a total pressure inlet condition and the pressure outlet you currently are using then you should get something close to the result you expect.

singer1812 · July 22, 2015, 10:04

Glenn is correct and it harks back to my posting.

If you fix the flowrate and fix the screw turning, then the only way you will get a correct answer is if you guess the exact rpm associtated with the flow and geometery. I still think that if you do this, your monitored torque (assuming you guess the right rpm for the flow) will be near zero because you are putting in work to turn the screw, outside of the flow.

In reality, you should either fix the RPM and let the solver determine the mass flow, as Glenn suggests doing, or fix the flow rate and put in CEL to turn the screw according to the torque applied and the interia of the screw.

Shaun Waters · July 22, 2015, 18:06

I have now run the setup, as suggested with total pressure inlet of 64kpa and a pressure outlet of 39kpa.

I have attached some images of the flow from this run, including both the total pressure and pressure contours.

areaInt(p)@Turbine = 3.92589e+007 [N]
force_z()@Turbine = -414675 [N]
areaInt(p*x)@Turbine = -750455 [J]
torque_z()@Turbine = 1.26804e+006

The values for torque are less, which is good. However, when combined with a 100RPM, the power comes out at approximately 13MW, which is very high compared to what I expected, as the bulb turbine of the same diameter and exposed to the same pressure conditions, would produce approximately 16MW.

Could this power/torque values for my screw turbine be high because losses such as friction are not included, or is there still an issue with my setup?

Thanks for all of your help with this.

ghorrocks · July 22, 2015, 20:21

It looks like you have the boundary conditions sorted - now you have to address the rotation speed as suggested by Edmund.

The screw device will not rotate at the same speed as the bulb device. The flow rates are different for a start. So what you need to do is to establish what is the speed this device will turn at. Assuming you are only interested in the steady state speed, the speed will be constant when there is zero net torque acting on the system. This occurs when the fluid torque on the rotor equals the torque on the load (with allowance for losses if they are important). My preferred way of working this out is manual (there are other ways but they tend to be more difficult for simple applications like this).

So you need to know your load torque versus speed curve. You then do a series of simulations on your rotor at a range of speeds, which gives you a rotor torque versus speed curve. Where those curves intersect is where the rotor torque equals the load torque, and that is the speed this device will operate at.

And before you ask - yes, you need to know the load torque versus speed curve. You can assume something if you like, but be aware that you are generating a load torque versus speed curve.

Shaun Waters · July 23, 2015, 08:19

Ok, thank you.
So to begin with, I need to run the setup at a range of RPM values to get a rotor torque vs RPM curve using the torque monitor point for the turbine (torque_z()@Turbine).

Then I need to compare this to a load torque vs RPM curve. Where these two curves intersect, is the RPM this device will operate at for the pressures given?

How do I be able to work out the load torque at the range of RPM's to create this second graph?

ghorrocks · July 23, 2015, 09:16

The rotor will obviously turn at a different speed if there is no load on it (so it freewheels) or has a heavy load on it (so it turns slowly). This shows that the rotation speed depends on the load you apply to it.

So your bulb turbine which operates at 100RPM (I think that is what you said) does so because these torques balance for this system at that speed for that pressure head. You say the bulb turbine generates 16MW, so you know the torque at that speed. That is one point on the curve. For speeds higher and lower than this you will have to define a load curve somehow.

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July 19, 2015, 19:53		#21
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,870 Rep Power: 144	It looks like it has converged a bit but not very far. This is an FAQ: http://www.cfd-online.com/Wiki/Ansys...gence_criteria When you assess the inlet and outlet pressures, you should use areaAve, not massflowAve. It does not make much sense to massflow average pressure. Can you post an image of the flow you are getting? Have you checked there is not soemthing funny going on and the flow is not as you expect?

July 21, 2015, 21:35		#23
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,870 Rep Power: 144	Can you: * Show some images of the detailed flow in the passage? Is there separations? What does the pressure distribution look like? * What does areaInt(p)@rotor give? Does it match the force()@rotor? Also check areaInt(p*x)@rotor and compare to torque.

July 22, 2015, 07:27		#26
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,870 Rep Power: 144	They are seriously big pressures. Are you sure they are in the right range? Are you sure the flow rate you are pushing through this thing is correct? Have a look at the negative pressures. You are getting pressures below absolute zero which means you are cavitating. Would you expect this thing to cavitate? If cavitation is significant then that would completely change things. Can you post an image of your mesh.

July 22, 2015, 08:18		#28
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,870 Rep Power: 144	If your expected inlet pressure is 6.5m = 65kPa and outlet pressure = 39kPa then I would be seriously worried about the pressure distribution you have modelled - it ranges from 1.6MPa to -1.1MPa. Clearly you have not modelled what you have in mind. Also the fact that changing the reference pressure significantly changes the result is a worry. This means something is quite wrong with your simulation. Are you saying that you are modelling a variation on a bulb turbine of the same diameter? Why do you expect it to perform as you say? See my post #1 - this device will not perform any where near a bulb turbine. So yes, I am suspicious that the flow rate you are specifying through this device will require enormous pressure head to drive it through (like what you are getting) and will result is crazy torque values on the rotor (like what you are getting).

July 22, 2015, 09:03		#30
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,870 Rep Power: 144	Now things are clear, and the reason for your crazy result is apparent. You have not chosen appropriate boundary conditions. You have a mass flow at the inlet and a pressure at the outlet. This is wrong as the mass flow is a function of the system and the mass flow you chose is what you get with a bulb turbine, not the screw you have modelled. You should have pressure inlet and outlet - because that is what is externally driving the flow, the pressure head difference from inlet to outlet. If you repeat this analysis using a total pressure inlet condition and the pressure outlet you currently are using then you should get something close to the result you expect.

July 22, 2015, 10:04		#31
singer1812 Senior Member Edmund Singer P.E. Join Date: Aug 2010 Location: Minneapolis, MN Posts: 511 Rep Power: 21	Glenn is correct and it harks back to my posting. If you fix the flowrate and fix the screw turning, then the only way you will get a correct answer is if you guess the exact rpm associtated with the flow and geometery. I still think that if you do this, your monitored torque (assuming you guess the right rpm for the flow) will be near zero because you are putting in work to turn the screw, outside of the flow. In reality, you should either fix the RPM and let the solver determine the mass flow, as Glenn suggests doing, or fix the flow rate and put in CEL to turn the screw according to the torque applied and the interia of the screw.

July 22, 2015, 20:21		#33
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,870 Rep Power: 144	It looks like you have the boundary conditions sorted - now you have to address the rotation speed as suggested by Edmund. The screw device will not rotate at the same speed as the bulb device. The flow rates are different for a start. So what you need to do is to establish what is the speed this device will turn at. Assuming you are only interested in the steady state speed, the speed will be constant when there is zero net torque acting on the system. This occurs when the fluid torque on the rotor equals the torque on the load (with allowance for losses if they are important). My preferred way of working this out is manual (there are other ways but they tend to be more difficult for simple applications like this). So you need to know your load torque versus speed curve. You then do a series of simulations on your rotor at a range of speeds, which gives you a rotor torque versus speed curve. Where those curves intersect is where the rotor torque equals the load torque, and that is the speed this device will operate at. And before you ask - yes, you need to know the load torque versus speed curve. You can assume something if you like, but be aware that you are generating a load torque versus speed curve.

July 23, 2015, 08:19		#34
Shaun Waters New Member Shaun Waters Join Date: Mar 2015 Posts: 19 Rep Power: 11	Ok, thank you. So to begin with, I need to run the setup at a range of RPM values to get a rotor torque vs RPM curve using the torque monitor point for the turbine (torque_z()@Turbine). Then I need to compare this to a load torque vs RPM curve. Where these two curves intersect, is the RPM this device will operate at for the pressures given? How do I be able to work out the load torque at the range of RPM's to create this second graph?

July 23, 2015, 09:16		#35
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,870 Rep Power: 144	The rotor will obviously turn at a different speed if there is no load on it (so it freewheels) or has a heavy load on it (so it turns slowly). This shows that the rotation speed depends on the load you apply to it. So your bulb turbine which operates at 100RPM (I think that is what you said) does so because these torques balance for this system at that speed for that pressure head. You say the bulb turbine generates 16MW, so you know the torque at that speed. That is one point on the curve. For speeds higher and lower than this you will have to define a load curve somehow.