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myv65 · 07-18-2002, 11:42 AM

BigBen2k,

I think you're confusing my intent a little. It isn't so much that a given portion of the energy input to the pump immediately shows itself as heat. The point I am making is that all the energy consumed by the pump must at some point escape from the system. It does this as thermal energy.

As to how much does immediately show itself as thermal energy, one must either measure this directly or believe the data provided by the pump manufacturer (assuming they provide it). The best industrial centrifugal pumps peak around 80% efficient. These pumps have very closely held tolerances while dealing with flows in excess of 1000 gpm. These comparatively dinky pumps we use have tolerances on the same order of magnitude, but pump less than 1000 gph. This means even the best of our pumps probably does no better than 70% efficiency.

The efficiency peaks at only one specific flow/discharge head. As flow nears zero, efficiency nears zero. As discharge head nears zero, efficiency also nears zero. Ya gotta remember that pump work is delta-P * flow rate. If either of these terms approaches zero, so does work hence efficiency.

So that I am clear, I'll rehash the various energies and efficiencies involved here. A pump's work output is delta-P * flow rate. Energy required to generate this flow is work divided by pump efficiency. Assuming a "best case" scenario, the shaft power must be about work divided by 0.7 (more realistically, probably 0.5 to 0.6). Shaft power is the electric motor's output power. The motor's input power is work divided by motor efficiency. Motor efficiency is probably between 80 and 85%.

So, line power to the motor should roughly equal delta-P * flow rate / ~0.7 (pump efficiency) / ~ 0.8 (motor efficiency). I'll leave it to anyone interested to apply the proper unit conversions.

The motor's inefficiency mainly goes off as convection for an inline pump. It obviously goes into the water for a submerged pump. The pump's inefficiency goes into thermal energy in the water. The pump's useful energy goes into circulating flow. All of the energy put into the pump's motor must eventually leave the box. Under steady-state conditions, energy leaving equals energy entering. For a while after starting up, energy leaving is less than entering with the excess showing up as thermal energy in the fluid (and kinetic energy of the moving fluid). After powering off, flow will slow to a halt and the fluid will cool to ambient. This is the release of stored energy and results in an energy output with no energy input.

07-18-2002, 11:42 AM	#49
myv65 Cooling Savant Join Date: May 2002 Location: home Posts: 365	BigBen2k, I think you're confusing my intent a little. It isn't so much that a given portion of the energy input to the pump immediately shows itself as heat. The point I am making is that all the energy consumed by the pump must at some point escape from the system. It does this as thermal energy. As to how much does immediately show itself as thermal energy, one must either measure this directly or believe the data provided by the pump manufacturer (assuming they provide it). The best industrial centrifugal pumps peak around 80% efficient. These pumps have very closely held tolerances while dealing with flows in excess of 1000 gpm. These comparatively dinky pumps we use have tolerances on the same order of magnitude, but pump less than 1000 gph. This means even the best of our pumps probably does no better than 70% efficiency. The efficiency peaks at only one specific flow/discharge head. As flow nears zero, efficiency nears zero. As discharge head nears zero, efficiency also nears zero. Ya gotta remember that pump work is delta-P * flow rate. If either of these terms approaches zero, so does work hence efficiency. So that I am clear, I'll rehash the various energies and efficiencies involved here. A pump's work output is delta-P * flow rate. Energy required to generate this flow is work divided by pump efficiency. Assuming a "best case" scenario, the shaft power must be about work divided by 0.7 (more realistically, probably 0.5 to 0.6). Shaft power is the electric motor's output power. The motor's input power is work divided by motor efficiency. Motor efficiency is probably between 80 and 85%. So, line power to the motor should roughly equal delta-P * flow rate / ~0.7 (pump efficiency) / ~ 0.8 (motor efficiency). I'll leave it to anyone interested to apply the proper unit conversions. The motor's inefficiency mainly goes off as convection for an inline pump. It obviously goes into the water for a submerged pump. The pump's inefficiency goes into thermal energy in the water. The pump's useful energy goes into circulating flow. All of the energy put into the pump's motor must eventually leave the box. Under steady-state conditions, energy leaving equals energy entering. For a while after starting up, energy leaving is less than entering with the excess showing up as thermal energy in the fluid (and kinetic energy of the moving fluid). After powering off, flow will slow to a halt and the fluid will cool to ambient. This is the release of stored energy and results in an energy output with no energy input.