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Temperature drop over hydraulic turbine

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fuentes303

Chemical
Mar 2, 2010
3
Assuming an hydraulic turbine of an hydro electric plant. Is it correct to assume that the water temperature at the outlet of the turbine can be calculated from the balance:

enthalpy of water at outlet = (enthalpy of water at inlet) minus (shaft power hydraulic turbine)?

Thanks

 
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Are we really speaking about hydraulic turbines? I mean those devices which convert the energy (potential or kinetic) of a fluid into mechanical energy to generate electric power. If so I think your equation is not valid
 
Ione

I really mean a Hydraulic (water) turbine (Francis / Kaplan / Pelton turbine). If above mentioned heat balance is not applicable, what would be applicable?
 
The answer would be yes, but you should also subtract any losses (heat exchange) to the external environment.
 
Oh, but re-reading your post title: the temperature will rise across the turbine, not drop, assuming the turbine is actually removing energy from the fluid.
 
You need the energy equation:

V1^2/2g+h1=V2^2/2g+h2+ Work done on turbine.
So, if you can get V1,V2 and the work done, you can get delta h, the change of enthalpy. In an incompressible fluid, h is not a function of temperature alone, so you will have to get delta pv of the water before you get delta u, the internal energy and then delta T.
 
Hummmm... Now that That looks suspiciously like a heavily modified Bernoulli equation.

Is there any heat gain by the water from the turbines inefficiency, assuming most of the heat lost to friction is carried off by the water?

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
 
There shouldn’t be a change in temperature for incompressible fluid.
 
Ok, if from an engineering point of view we have to consider water as a compressible fluid, I think I have nothing to add to this thread.
 
"There shouldn't be a change in temperature for incompressible fluid. "

There will be a temperature change due to losses (friction>> turbulence >> heating, even for a theoretically incompressible fluid.

And, yeah, velocities matter. So does vertical (head) change.
 
No argument from me. The friction loss just from pipe flow can add a couple of degrees to a fluid in a long pipeline, if friction energy is high in relation to the heat that can be conducted off laterally. Most all of heat generated in a pump is picked up and carried off by the the fluid being pumped. I assumed power turbines might be the same, but I don't have any experience with those.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
 
Hydraulic turbines are not thermal devices. They just convert potential energy into kinetic energy and then into mechanical energy to the shaft. The equation proposed by the OP is meaningless for this stuff.
 
If the turbine does not have a 100% efficiency then you will experience a temperature increase (although it may be well below detection threshold).

I know this is a little hard to accept - but its the same with liquid through valves - the temperature increases!

This about it: Where will the potential energy go? It can only go into the fluid - thus increasing temperature.

Best regards

Morten
 
Right. A pump isn't a thermal device either, but they get pretty hot when you try to run them on min flow where there isn't enough flow leaving the pump to take the heat away. During normal flows a small amount of heat reaches the outer surface of the pump and is convected off, however the bulk of the heat goes to heating the fluid and is pumped into the discharge pipeline. A pipeline simulator that does heat transfer, such as Stoner PL Simulator, which I use, or Pipeline Studio (I think MortonA uses that one) will clearly show a temperature increase even on normal flows. If I run it in full thermal mode, it will calculate the heat transfer along the pipeline too. Typically at less than 10% BEP flows, the pump efficiency is so poor and outflow so low that no heat is carried off and they clearly show pump overheating. Stoner will automatically shut down a pump that reaches 350F.

While the SPL doesn't have ready-made hydro-power turbines, I suppose I could make a turbine curve, call it a pump and spin it backwards to see what happens. Not today though. I'm not sure I could handle all the "parameter out of bounds" warnings on a Friday.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
 
Hydraulic turbines are devices which extract energy from a flow converting its kinetic energy into mechanic energy. Let’s consider the Pelton wheel (impulse hydraulic turbine). The whole energy available at the nozzle outlet (I mean already diminished of the head loss in the conduit to reach the nozzle) could not be entirely converted to mechanical shaft power since:

1. Water still possesses kinetic energy when it leaves the bucket, related to its leaving velocity
2. Part of the energy is dissipated by friction in shaft bearing system.

How could anyway be the temperature of the water at the exit of nozzle, different from that which leaves the bucket after impact?
 
Since water can't grab and spin the wheel, the only mechanism of transfer of energy from the water to the wheel face is via impact and subsequent diversion of the flow around the curvature of the cup before splattering away. There's a lot of friction there as it flows on the cup surface, so there is the inefficiency and consequent heat source. I would suspect some of the bearing friction heat is conducted along the shaft and is exposed to conduction by the water too.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
 
Maybe this is the culprit of the water overheating: we can get fishes well done downstream the turbine. [wink]

Sincerely I think inefficiency is due to the fact the wheel rotates in the air which offers resitance to the movement and it is responsible of the energy loss.
 
Fish love the warm water coming off the outlets of fossile fuel and nuclear plants. That's where the biggest bass are found in Texas; at the cooling water spillways going into the lakes. They will grow around twice as fast as those in normal lakes. Just ignore the ghostly green glow from those bass caught at Commanche Point, N32°17?54? W97°47?6?
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Nope, because in a vacuum that would mean turbines could be 100% efficient. Friction is from atoms rubbing against atoms, whether they form a solid, a gas or a liquid. Impact and friction both make heat. Ever touch a penny just after it got hit by a bullet? That wasn't the heat from the gunpowder. And the bullet was hot too. If it was a water bullet, do you think the water and the penny would both get hot? I do.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
 
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