Pump Heat Talk

[Johnseye]

Even if you had a pump that was 100% efficient and generated no heat while pumping water, thermodynamics dictates that you would still have a net increase in energy to the system; the energy extracted from the system can't be greater than that added.

Can you please explain pump efficiency and how that translates to less heat generation while pumping water?

I've seen numerous comments about not getting too technical, but since this discussion has been (correctly) pulled from the Vectra thread, and is to specifically discuss pump heat please get technical if necessary. No need to dumb it down.

 

[shred5]

Can you please explain pump efficiency and how that translates to less heat generation while pumping water?

I've seen numerous comments about not getting too technical, but since this discussion has been (correctly) pulled from the Vectra thread, and is to specifically discuss pump heat please get technical if necessary. No need to dumb it down.

That is a hard one. There are so many things that can cause something to be more efficient than another.

This is a made up scenario:
If you have two pumps that put out 3000 gph and pump one draws 30 watts and pump two draws 40 watts. Well pump one is more efficient and puts out less heat. 30 watts of heat is less than 40 watts.

There is allot that can make one pump more efficiency than another. Anything from pumps design, materials used, impeller design, controller design. It can go way deeper than that too.

I just did it simple.

 

[Johnseye]

That is a hard one. There are so many things that can cause something to be more efficient than another.

This is a made up scenario:
If you have two pumps that put out 3000 gph and pump one draws 30 watts and pump two draws 40 watts. Well pump one is more efficient and puts out less heat. 30 watts of heat is less than 40 watts.

There is allot that can make one pump more efficiency than another. Anything from pumps design, materials used, impeller design, controller design. It can go way deeper than that too.

I just did it simple.

Does it still all boil down to the wattage used by the pump? In your example if you have two 3000 gph pumps one at 30w and one at 40w the 30w pump is more efficient because for whatever factors it can pump 3000 gph using 30w? Or, if you have two pumps rated at 40w and can one of those pumps be more efficient than the other thus transferring less energy as heat?

Again, please don't keep it simple. Get technical and complicated if necessary.

 

[shred5]

If we are talking pump only minus the controller. If both pumps draw 40 watts and are completely surrounded by water it is 40 watts of heat. That has to go somewhere and that is the water, both would be transferring the exact amount of heat.

If you have two 40Watt pumps and one does 3000 gph and one does 4000 gph both are putting out the same heat but the 4000 gph is more efficient.

Efficiency for a water pump is based on GPH/watts consumed.

I have not had fluid dynamics in a long time but I think that is right.

Lighting would be the same except air. Watts is watts.

There are ways to make a unit disperse heat better but that heat has to go somewhere.

 

[africangrey]

tagging along

 

[sleepydoc]

Can you please explain pump efficiency and how that translates to less heat generation while pumping water?

I've seen numerous comments about not getting too technical, but since this discussion has been (correctly) pulled from the Vectra thread, and is to specifically discuss pump heat please get technical if necessary. No need to dumb it down.

When you pump water, you are adding energy to the water both in terms of velocity and height. Say, for example, that the actual movement of the water done by the pump takes 10 watts, but the pump itself actually draws 20 watts. The extra 19 watts is used by friction in the pump, heat generated by resistance in the wire windings, and other components in the pump that don't transfer energy perfectly as Klaus alludes to above. This pump would be 50% efficient (10W delivered power/20 watts consumed power.) i.e. 50% of the electrical power used by the pump goes towards the work you want to do and the other 50% is 'wasted.'

Now say that one of the bearings goes bad and the pump becomes harder to turn, so now you have to use 30 watts to keep the pump turning at the same speed. Now our pump would be 10/30 = 0.33, or 33% efficient, meaning that only ⅓ of the power used by the pump is the work you want while ⅔ is wasted.

To make an analogy, think of 2 bicycles - one is brand new with high pressure tires, perfect bearings and gears, oiled, lubricated and perfectly tuned up. The 2nd is 30 years old with rusty gears, flat tires and bad bearings. When you ride the good bike, you burn far fewer calories to ride a kilometer than you do with the old bike; the old bike is much less efficient because far more of the energy you use pedaling it is wasted on things other than forward motion.

Now remember that all energy gets converted to heat eventually. Going back to our example, your pump is using 20 watts do do the work you want it to do, so it's adding 20 watts to the tank. Now the inefficient pump with the bad bearings is adding 30 watts to the tank but still doing the same work. This means that an extra 10 watts of power is going into the tank as heat to move the same amount of water.

hope this makes more sense to you.

 

[Johnseye]

When you pump water, you are adding energy to the water both in terms of velocity and height. Say, for example, that the actual movement of the water done by the pump takes 10 watts, but the pump itself actually draws 20 watts. The extra 19 watts is used by friction in the pump, heat generated by resistance in the wire windings, and other components in the pump that don't transfer energy perfectly as Klaus alludes to above. This pump would be 50% efficient (10W delivered power/20 watts consumed power.) i.e. 50% of the electrical power used by the pump goes towards the work you want to do and the other 50% is 'wasted.'

Now say that one of the bearings goes bad and the pump becomes harder to turn, so now you have to use 30 watts to keep the pump turning at the same speed. Now our pump would be 10/30 = 0.33, or 33% efficient, meaning that only ⅓ of the power used by the pump is the work you want while ⅔ is wasted.

To make an analogy, think of 2 bicycles - one is brand new with high pressure tires, perfect bearings and gears, oiled, lubricated and perfectly tuned up. The 2nd is 30 years old with rusty gears, flat tires and bad bearings. When you ride the good bike, you burn far fewer calories to ride a kilometer than you do with the old bike; the old bike is much less efficient because far more of the energy you use pedaling it is wasted on things other than forward motion.

Now remember that all energy gets converted to heat eventually. Going back to our example, your pump is using 20 watts do do the work you want it to do, so it's adding 20 watts to the tank. Now the inefficient pump with the bad bearings is adding 30 watts to the tank but still doing the same work. This means that an extra 10 watts of power is going into the tank as heat to move the same amount of water.

hope this makes more sense to you.

Perfect sense. Thanks.

 

[Johnseye]

Can a pump overheat? I would assume it can. When it does can an 80w pump create more heat than the 80w it pulls from the wall or if it overheats would that be measured at the wall?

Is the measurement of a pump a the wall the end all be all of heat it can and will generate?

 

[sleepydoc]

Can a pump overheat? I would assume it can. When it does can an 80w pump create more heat than the 80w it pulls from the wall or if it overheats would that be measured at the wall?

Is the measurement of a pump a the wall the end all be all of heat it can and will generate?

Motors can and do overheat for a variety of reasons - overloading, worn bearings, snails jamming them, inadequate heat dissipation, etc.

Yes, the measurement at the wall (assuming it's accurate) is the end all be all of heat. Remember the 1st law of thermodynamics. Heat is energy. If you add more energy to the water than you take out of the wall, you are creating energy and the thermodynamics police will come and arrest you.

That said, it may be possible for a malfunctioning pump to draw more current than normal.