Energy equation modeling in turbine rotor?

mosi · May 31, 2014, 20:48

Hi everyone,

I have faced a very fundamental question in modeling energy equation in turbine rotor which I can not find the answer to.

In a real turbine, the power used to rotate the rotor blades, is extracted from the hot fluid. So the fluid loses temperature and enthalpy in the rotor. But in CFX when we are modeling the rotating stage, we just enter the rotation speed and it seems the solver does not ask you where the power to rotate the blades is coming from. so the results show that the temperature and enthalpy is not decreased in the rotor stage. so the power to rotate the rotor is not extracted from the fluid which should be wrong. Anybody has any explanation for this problem?

Thanks,
Mosi

mosi · June 1, 2014, 00:29

I think I might have figured out the problem. It is likely that it is caused by my boundary conditions. Because when I checked the results of CFX Tutorial #14 it showed a reasonable decrease in temperature and enthalpy in rotor. I will change my boundary conditions and update the results here.

Thanks

ghorrocks · June 1, 2014, 02:53

If your simulation is accurate and there is no enthalpy drop then I bet there is no torque on the rotor either. Then you have no energy drop in the fluid and no energy extracted. Hey presto energy is conserved.

mosi · June 1, 2014, 11:55

OK, as I guessed the problem was fixed by changing the boundary conditions. Before, I was using mass flow inlet=0.11 kg/s and static pressure outlet=0 atm, and I saw no enthalpy drop in rotor. Now I've changed it to P_tot inlet=0 atm and mass flow outlet=0.11 kg/s and I see a 22000 J/kg total enthalpy drop in rotor which is reasonable.

@Glenn, thank you for your reply. You are right, the torque was zero and it's non-zero now. But I still don't know why the former boundary conditions created no power. Do you have any idea?

ghorrocks · June 1, 2014, 20:50

At the operating point the rotor will generate torque. If the rotor rotational speed is too fast it will generate negative torque (ie the rotor will want to slow down). At some point between it will generate zero torque.

mosi · June 1, 2014, 22:37

That is absolutely true. Thank you.

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May 31, 2014, 20:48	Energy equation modeling in turbine rotor?	#1
mosi New Member mosi Join Date: May 2014 Location: College Station, Texas Posts: 10 Rep Power: 11	Hi everyone, I have faced a very fundamental question in modeling energy equation in turbine rotor which I can not find the answer to. In a real turbine, the power used to rotate the rotor blades, is extracted from the hot fluid. So the fluid loses temperature and enthalpy in the rotor. But in CFX when we are modeling the rotating stage, we just enter the rotation speed and it seems the solver does not ask you where the power to rotate the blades is coming from. so the results show that the temperature and enthalpy is not decreased in the rotor stage. so the power to rotate the rotor is not extracted from the fluid which should be wrong. Anybody has any explanation for this problem? Thanks, Mosi

June 1, 2014, 00:29		#2
mosi New Member mosi Join Date: May 2014 Location: College Station, Texas Posts: 10 Rep Power: 11	I think I might have figured out the problem. It is likely that it is caused by my boundary conditions. Because when I checked the results of CFX Tutorial #14 it showed a reasonable decrease in temperature and enthalpy in rotor. I will change my boundary conditions and update the results here. Thanks

June 1, 2014, 02:53		#3
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,701 Rep Power: 143	If your simulation is accurate and there is no enthalpy drop then I bet there is no torque on the rotor either. Then you have no energy drop in the fluid and no energy extracted. Hey presto energy is conserved.

June 1, 2014, 11:55		#4
mosi New Member mosi Join Date: May 2014 Location: College Station, Texas Posts: 10 Rep Power: 11	OK, as I guessed the problem was fixed by changing the boundary conditions. Before, I was using mass flow inlet=0.11 kg/s and static pressure outlet=0 atm, and I saw no enthalpy drop in rotor. Now I've changed it to P_tot inlet=0 atm and mass flow outlet=0.11 kg/s and I see a 22000 J/kg total enthalpy drop in rotor which is reasonable. @Glenn, thank you for your reply. You are right, the torque was zero and it's non-zero now. But I still don't know why the former boundary conditions created no power. Do you have any idea?

June 1, 2014, 20:50		#5
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,701 Rep Power: 143	At the operating point the rotor will generate torque. If the rotor rotational speed is too fast it will generate negative torque (ie the rotor will want to slow down). At some point between it will generate zero torque.

June 1, 2014, 22:37		#6
mosi New Member mosi Join Date: May 2014 Location: College Station, Texas Posts: 10 Rep Power: 11	That is absolutely true. Thank you.