geoffman - You do not seem to be that confused.
Slightly over simplified explanation:
A) That referenced equation first takes the BHP (Total input power to the pump) and subtracts the WHP (The power used to lift the water to a certain head). The equation assumes that the lifted water then "falls off a cliff" and the WHP is lost and never recovered.
B) The result of this subtraction equals the *steady state* power dissipated due to all losses which include:
1) *"friction in the pump movement itself, from seals etc"*
2) Friction due to water flow restrictions/resistances
3) Losses due to pump electrical inefficiencies etc.
C) Last is the calculation of the temperature rise of the water due to this dissipated power.
To make a long story short - The water is heated by the total power into the pump minus the power used to lift the water. That calculation might be used for a large commercial non sealed system where the main purpose of the pump is to lift a lot of water to higher elevation.
But, here is the rub:
In the sealed closed loop PC cooling systems we are discussing, WHP IS ZERO
The power used to lift the water against gravity in the up path is fully recovered by gravity pulling the water down in the down path. (The "siphon effect")
ALL of the pump input power will heat the water!
In the case of an open reservoir PC system, the typical flow rates/head elevations will still only require a small amount of water "lifting power"
Summarized: Virtually all of the pump input power in PC cooling systems is required to overcome the friction of flow restrictions in the cooling system components and tubing and thus, heats the water.
And, this pump input power can easily measured (V X I)!
If you study the posts by the folks ThePimpulator recommended, you will not go wrong.
Slightly over simplified explanation:
A) That referenced equation first takes the BHP (Total input power to the pump) and subtracts the WHP (The power used to lift the water to a certain head). The equation assumes that the lifted water then "falls off a cliff" and the WHP is lost and never recovered.
B) The result of this subtraction equals the *steady state* power dissipated due to all losses which include:
1) *"friction in the pump movement itself, from seals etc"*
2) Friction due to water flow restrictions/resistances
3) Losses due to pump electrical inefficiencies etc.
C) Last is the calculation of the temperature rise of the water due to this dissipated power.
To make a long story short - The water is heated by the total power into the pump minus the power used to lift the water. That calculation might be used for a large commercial non sealed system where the main purpose of the pump is to lift a lot of water to higher elevation.
But, here is the rub:
In the sealed closed loop PC cooling systems we are discussing, WHP IS ZERO
The power used to lift the water against gravity in the up path is fully recovered by gravity pulling the water down in the down path. (The "siphon effect")
ALL of the pump input power will heat the water!
In the case of an open reservoir PC system, the typical flow rates/head elevations will still only require a small amount of water "lifting power"
Summarized: Virtually all of the pump input power in PC cooling systems is required to overcome the friction of flow restrictions in the cooling system components and tubing and thus, heats the water.
And, this pump input power can easily measured (V X I)!
If you study the posts by the folks ThePimpulator recommended, you will not go wrong.