# water flow rate

Here’s a problem putting together some ideas about the first law of thermodynamics, work, and efficiency. We have a reservoir at the top of a hill. At the bottom of the hill , 20 m below, is a 275 kW pump with an efficiency of 77 percent moving water from the reservoir to the bottom of the hill at a rate of 1000 kg/s. If the pump is working at full load, how much of the work it’s doing is going towards overcoming frictional losses?

If we know the pump is operating at full load, we can figure out how much power it’s actually putting out by looking at the power put into it and its efficiency. In this case, it puts out about 211,750 W.

Of that 211,750 W it’s putting out, only some of it’s actually doing the work of moving the water. The rest is going towards overcoming frictional losses in the piping. To figure out how much work is required for the latter, we have to figure out how much energy you would need to move this much water in a perfect world without friction at play.

Our life is also complicated by the fact that we’re looking at a constant flow of water, not a single discreet unit moved a definite difference. But we can work with that.

Work is just energy spent. If you remember your physics, you know that there’s a change in gravitational potential energy when you move an object through a gravitational field. That change in potential energy is the mass of the object multiplied by the gravitational acceleration and the change in its distance from the source of the gravitational field - its height. That’s exactly the kind of change in energy that’s going on here. Let’s look at how much of a change of potential energy there is on a per kilogram basis for the water in this scenario.

In this case, it’s about 196 J/kg of water. What we’re interested in, though, is a rate. We need to know the power this takes up - how many Joules per second. Fortunately, we know the mass flow rate of the water. Multiplying that mass flow rate by the work per unit mass, we get the actual power required to move this much water. Here, it’s about 196,000 W.

To find how much power actually goes to frictional losses, we just subtract the power required to move the water from the power the pump is actually putting out.

About 15,750 W of losses in this case.