Graystar said:
The purpose of the pump is to move the water. The pump doesn’t care if you’re watercooling, watering the lawn, or pumping water out of your basement...its job is to load up water with potiential energy, bringing it to a higher energy state. How that water gets back down to a lower energy state is your business.
I don't beleive you addressed the question.
Actually, it’s 100% heat. Sound heats the air, and deforming a solid heats that solid.
The object applies a force to the table, it is deformed; something moves - this requires a force, kinetic energy, to do so. It is not 100% heat.
Take a piece of metal and bend it back and forth for a bit then touch the bend. It will be hot. It all turns into heat.
The vast majority of the force is used to break the metallic bonds.
Lets say the pump was filling a tub 20 feet in the air. The action of the pump is kinetic. More than simply being in motion, it is transferring energy. So, the Iwaki MD-20RZ is transferring its 20 watts into the water as it pushes it up to the tub. But remember, this pump is rated at 50 watts. So the pump consumed 30 watts of power, turning it into heat, in order to impart 20 watts worth of energy into the water.
Yep. I am in agreeance with you here. Pumps are not 100% efficient. My MD-30rz get exceptionally hot.
At this point the pump is done and is out of the picture. If you open a drain in the tub then the water will flow out, returning to its previous state, and being heated on impact. Or, the water can turn a waterwheel, have a nice cool trip down to its previous state, and perform the same type of transfer the pump just made, but to some other device...such as a grindstone. As with the pump, there will be a cost associated with the transfer in terms of friction of the moving parts. So maybe you end up with 10 watts of power at the grindstone’s business end.
The water applies a force to the water wheel to make it turn, friction saps some more energy too. Yes, when it impacts there will be heat, but again, not 100%.
Ein = Eout.
Ein = EKpumpoutput
Eout = (mass_of_waterwheel * acceleration_of_waterwheel) + sound (which is really EK) + heat (and so is this if you want to be pedantic).
As you said, you have a restrictive loop. There is an energy cost associated with pushing water through small holes at a certain speed. That energy is turned into heat through friction.
Some of it is, sure. But the system is pressurised (by the pump), and thus exerts a force on the walls of the loop, which because of newtons laws, is applied back to it (by the atmosphere).
The faster the water, the more heat is generated.
Friction sucks hey.
The speed of water through your system is determined by the number of watts available to push it, and turn into heat. More watts equals faster flow.
Phextwin said:
So if all the output power is used to heat up the water, what is the point of a pump? What moves the water? Where does the kinetic energy come from?
Some time ago where was a discussion as to how much more water a Mag 5 pump will push than a Mag 3. People thought the Mag 5 will push more because it had a zero-head rate of 500 gph, as opposed to the Mag 3 which had a zero-head rate of 350 gph. I, however, said that the two pumps would have the same rate of flow...and when measurements were finally made and charted, the two pumps did indeed have the same flow rate in a watercooling system. Why?
Because the restriction curve of the test system was to steep to take advantage of the MAG5's extra flow capacity and/or the accuracy of the flowmeters was not good enough.
Because both pumps had the same head...and the head rating (a measure of fluidic energy) is directly related to the BHP of the pumps.
Explain. I would have thought it would be more due to the impellor design. All the high head pumps i have seen use a closed impellor design, ie Iwaki MD-XXz & RD-XX, panworld PI-XX, laing DDC, D4 & D5 all are considered high head pumps, all have closed impellors.
The Iwaki MD-30 R, RX & RZ all have the same output power, yet their head ratings are vastly different.