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myv65 · 07-14-2002, 09:40 PM

"Useful work" is defined by the task at hand. In the case of a pump, I define it as flow rate multiplied by pressure rise. Another way to look at it is it is the energy input devoted to the desired task. If the motor and pump each worked at 100% efficiency, the motor input would equal the useful work.

I'm not so sure that skulemate's question is easily answered. Total energy of a fluid (ignoring thermal considerations for a moment) is defined by Bernoulli's equation. It basically says that energy = p + v^2 + z. I've left out a few pertinent terms like density and gravity to simplify. "p" is pressure, "v" is velocity, and "z" is elevation. "p" and "z" are what I would consider as potential energy while "v" is what I would consider kinetic energy.

Does a pump add kinetic or potential? Well, the fluid moves the same speed on each side of the pump, so one could argue that the kinetic energy is a constant. In this case, the pump couldn't add kinetic energy, right? However, without the pump there would be no flow, so the pump must add kinetic energy, right?

Like Sirpent said, it really doesn't matter. One energy form or another, it's still energy and must balance out at the end of the day.

07-14-2002, 09:40 PM	#40
myv65 Cooling Savant Join Date: May 2002 Location: home Posts: 365	"Useful work" is defined by the task at hand. In the case of a pump, I define it as flow rate multiplied by pressure rise. Another way to look at it is it is the energy input devoted to the desired task. If the motor and pump each worked at 100% efficiency, the motor input would equal the useful work. I'm not so sure that skulemate's question is easily answered. Total energy of a fluid (ignoring thermal considerations for a moment) is defined by Bernoulli's equation. It basically says that energy = p + v^2 + z. I've left out a few pertinent terms like density and gravity to simplify. "p" is pressure, "v" is velocity, and "z" is elevation. "p" and "z" are what I would consider as potential energy while "v" is what I would consider kinetic energy. Does a pump add kinetic or potential? Well, the fluid moves the same speed on each side of the pump, so one could argue that the kinetic energy is a constant. In this case, the pump couldn't add kinetic energy, right? However, without the pump there would be no flow, so the pump must add kinetic energy, right? Like Sirpent said, it really doesn't matter. One energy form or another, it's still energy and must balance out at the end of the day. Last edited by myv65; 07-14-2002 at 09:46 PM.