Physics Question Re: Power

[jumpulas]

Hey all, Can you please help me out with this question...

Water is pumped up to a water tower which is 92.0 m high. The flow rate up to the top of the tower is 75 L/s, and each litre of water has a mass of 1.0 kg. What power is required to keep up this flow rate to the tower?

Thanks.

[UncleAl]

Power is energy/time. What is the difference in energy between water at baseline and 92 meters higher, potential and kinetic energies? Are we talking in-process or a cold start? ** *** ***** ***'* *** **, ****** * ********* *****.

 

http://www.mazepath.com/uncleal/horse.htm

[infamous]

Hey all, Can you please help me out with this question...

Water is pumped up to a water tower which is 92.0 m high. The flow rate up to the top of the tower is 75 L/s, and each litre of water has a mass of 1.0 kg. What power is required to keep up this flow rate to the tower?

Thanks.

Ok let me see if I can help with this?

 

1 HP = 550 ft. lb. sec.

92m = 302 ft.

75kg/sec. = 34.02 lb./sec.

302 times 34.02 = 10274.04

10274.04/550 = 18.68 HP

 

One must figure in the weight of the water in the 92 meter tall plumbing. And that woud depend upon the dia. of the pipe.

 

So adding a couple HP to the figure 18.68 should give us a figure of around 20 HP. Not knowing the total weight of the water in the pipe makes this calculation only an approximation.

[Erasmus00]

Hey all, Can you please help me out with this question...

Water is pumped up to a water tower which is 92.0 m high. The flow rate up to the top of the tower is 75 L/s, and each litre of water has a mass of 1.0 kg. What power is required to keep up this flow rate to the tower?

Thanks.

 

How much energy is required to lift 75 liters of water 92 meters? This is the energy input required every second, which is, in fact, the power.

-Will

[C1ay]

Wow, that's about 1200 gallons/minute. That'll take about a 125mm pipe and the pressure will be just short of 900 kpascals. Hopefully that'll help.:shrug:

[CraigD]

… Water is pumped up to a water tower which is 92.0 m high. The flow rate up to the top of the tower is 75 L/s, and each litre of water has a mass of 1.0 kg. What power is required to keep up this flow rate to the tower? …

You have a distance, a volume, a time, and a density. You want a power. Since there’s no mention of that complicating factor, friction, we can work the problem assuming there is none (a good thing, because water friction is complicated). Our answer, therefore, is the minimum power that the water tower pump would require.

 

Start with the appropriate basic definitions:

Power = Work / time

Work = Force * Distance

Force = Mass * Acceleration

Mass = Volume * Density

 

Replace the variable with your given quantities, and calculate power. Don’t assume infamous’s answer is right, or wrong – work out your own.

 

It will be easiest to use metric units – meters, litres, seconds, kilograms, newtons, jouls, and watts. If you like your power in English units, note that 1 horsepower = 750 watts.

 

Don’t see a time anywhere? Look at the “flow rate” of “75 L/s”. “75 L” is a volume. “s” is a time. When expressed as a ratio, like this it means you can use any volume and time you want, so long as volume / time = 75. It’s easiest to just use a time of “1 s”, and a volume of “75 L”.

 

What acceleration to use? Assuming this water tower is on the surface of the earth, that would be gravity, about 9.8 (m/s)/s.

 

Notice how little information this problem actually needs. We don’t care if the water is pumped slowly through a large pipe, quickly though a small one, shot from a cannon, or carried in buckets attached to a wheel – ignoring friction, they all require the same power! Physicists love ignoring friction.

 

Post your results back here, and we’ll see if they agree with mine.

[C1ay]

One must figure in the weight of the water in the 92 meter tall plumbing. And that woud depend upon the dia. of the pipe.

Not really. The weight translates to a pressure per square unit regardless of the square units in the cross section of the pipe. The pipe does need to be sized for the flow to minimize friction. At high flow velocity the flow would be non-laminar and reynolds number would climb exponentially and friction would become a significant part of the equation. I would keep the flow for this circuit at 4 to 5 m/s.

[Little Bang]

Ok Infamous, what did I do wrong.

 

After the first 92 seconds the pipe would be full of water.

 

V = 75 * 92

V = 6900 L

Weight of water = 6900kg * 2.2lbs = 15180 lbs/550 =27.7 horse power

[CraigD]

... 75kg/sec. = 34.02 lb./sec. ...

Your conversion is in error. 75 kg = 75 kg * ~2.2 lb/kg =~ 165 lb, not 75 / 2.2 = 34. Your results are about 5 times understated.

 

You don’t need any other data, such as pipe size.

 

English units are anoying.

[Qfwfq]

75 kg/s times 92 m times acceleration of gravity 9.8 m/s^2 = 67620 W, equivalent to 92 HP.

[C1ay]

75 kg/s times 92 m times acceleration of gravity 9.8 m/s^2 = 67620 W, equivalent to 92 HP.

:shrug:

[infamous]

Your conversion is in error. 75 kg = 75 kg * ~2.2 lb/kg =~ 165 lb, not 75 / 2.2 = 34. Your results are about 5 times understated.

 

You don’t need any other data, such as pipe size.

 

English units are anoying.

I stand corrected CraigD.

[Qfwfq]

:wave:

:shrug:

[C1ay]

:shrug:

IMO, we shouldn't give people the answer to their homework, only help in finding the answer themselves.

[infamous]

IMO, we shouldn't give people the answer to their homework, only help in finding the answer themselves.

You make a very good point there C1ay, you won't catch me doing it again, Ha,ha,ha.

[Qfwfq]

IMO, we shouldn't give people the answer to their homework, only help in finding the answer themselves.

That depends on what his teacher calls "the answer". :shrug:

[infamous]

Your conversion is in error. 75 kg = 75 kg * ~2.2 lb/kg =~ 165 lb, not 75 / 2.2 = 34. Your results are about 5 times understated.

 

You don’t need any other data, such as pipe size.

 

English units are anoying.

Correct me if I'm wrong here CraigD, but my physics reference book gives the conversion for kg to lbs as .45359kg = 1 lb. Therefore, 75 times .45359 would equal: 34.01925 lbs. Where am I going wrong here?