Doug's farm has a tank to store water for the livestock. The top of the tank is 19 ft tall and it sits on a hill that is 34 ft above the pump that fills it from a deep well. If the water from the pump pours into the tank from the top, How much pressure must the pump overcome to fill the tank?

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Shouldn't the answer be equal to the difference in atmospheric pressure that rises in a height difference of 34 feet?
Any expert , please tell me if I am right or not.

Shouldn't the answer be equal to the difference in atmospheric pressure that rises in a height difference of 34 feet?
Any expert , please tell me if I am right or not.

You need the water volume. 1/8" pipe vs 4 ft. diameter. Density, volume and height all factor in.

Shouldn't the answer be equal to the difference in atmospheric pressure that rises in a height difference of 34 feet?
Any expert , please tell me if I am right or not.

No, not right.

You need the water volume. 1/8" pipe vs 4 ft. diameter. Density, volume and height all factor in.

You speak from practical experience, where issues such as the pipe cross-sectional area and the material it's made from can have an impact on the resistance the water experiences as it flows through the pipe. But for high school physics problems it's typical to ignore these complicating factors. Hence the pipe diameter, and water volume don't enter into the calculation. All that's needed is the height the water is being transported, the density of the water, and the gravitational constant g.

Yes, NFPA 13. Used to do it manually but now the computer does it for you. Although retired 11 years I still have some tables around.

No, not right.

Okay, assume I correct the height to 34+19=53.
So the maximum pressure P the pump has to take is (53*30/100 meters) *density of water*g. But again, atmospheric pressure is pushing on the water column. So shouldn't the total maximum pressure be pressure caused by the water column + atmospheric pressure?

Okay, assume I correct the height to 34+19=53.
So the maximum pressure P the pump has to take is (53*30/100 meters) *density of water*g. .

Where did the 30/100 come from?


But again, atmospheric pressure is pushing on the water column. So shouldn't the total maximum pressure be pressure caused by the water column + atmospheric pressure?

Sure, but remenber that the difference in air pressure assists the pump. So you have:

Pump pressure = (\rho_{water} - \rho_{air}) \ \times \ 53m \ \times\ g

But if you consider that the data provided is good to only 2 decimal places, and the density of air is so much less than the density of water, I think you'll find the impact of air pressure is negligible.

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Where did the 30/100 come from?



Sure, but remenber that the difference in air pressure assists the pump. So you have:

Pump pressure = (\rho_{water} - \rho_{air}) \ \times \ 53m \ \times\ g

But if you consider that the data provided is good to only 2 decimal places, and the density of air is so much less than the density of water, I think you'll find the impact of air pressure is negligible.

Well, that 30/100 is converting feet to meters...

Got it. Using 3 ft/meter is a pretty rough approximation, but OK. My formula mistakenly said 53 m, where it should have been 53 ft:

Pump pressure = ( \rho _{water} - \rho_{air}) \ \times \ 53 \mathbf{ft} \ \times \ g

where densities and the factor g are in Imperial units.