An electric pump raise 9.1m^3 of water from a reservoir

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  • #1
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1. An electric pump raise 9.1m^3 of water from a reservoir whose water-level is 4m
below ground level to a storage tank above ground. If the discharge pipe outlet is
32m above ground and the operation takes 1hour,find the minimum power rating of
the pump if it's efficiency is 70%.(1m^3 of water has a mass of 1000kg)

Homework Equations





The Attempt at a Solution



An Idea suddenly came to me I need to find the energy but using Potential Energy formula? mgh?

I'm sorry I have no Idea how to attempt it, please help me, give me a hint on what to do.
 

Answers and Replies

  • #2
688
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Yes, potential energy = mgh.

And power = Energy / Time = mgh / time

Or the ideal power is expressed as:

Power (ideal) = (Δm / Δt) gh

And we divide this by the pump efficiency [itex]\eta[/itex] to get the real power:

Power (ideal) = (Δm / Δt) X gh / [itex]\eta[/itex]


Notice that Δm can be figured our from your volume and density and Δt is given.
 
  • #3
576
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Sorry I'm not a pro physics student that triangle means change right? so it's change in mass/change in time x gravity x height/ 70%?
 
  • #4
688
1


Sorry I'm not a pro physics student that triangle means change right? so it's change in mass/change in time x gravity x height/ 70%?
Yes ! And you are getting the idea, so maybe one day soon you will be a pro physics student.
 
  • #5
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But the thing is I don't see a change in mass or density. :S
 
  • #6
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Ohh wait I should find the density by doing 9100KG x 9.1? because Density = Mass x volume?
 
  • #7
688
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But the thing is I don't see a change in mass or density. :S
Think of the change is mass as follows:

First note: mass = density X volume = 1000 kg / m^3 X 9.1 m^3

After 1 hour, all this mass will be pumped out, so this is your "change" in mass.
 
  • #8
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Ohh wait I should find the density by doing 9100KG x 9.1? because Density = Mass x volume?
Close. But, Density = Mass / Volume
 
  • #9
576
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Alright should I do this 9100 x 10 x 32( I believe this is the height) to get the energy then divide it by the efficiency?
 
  • #10
688
1


Alright should I do this 9100 x 10 x 32( I believe this is the height) to get the energy then divide it by the efficiency?
Close. Notice that the water is 4 m underground, so work is done over a total height change of: 32m + 4m. And you have to introduce your time of 1 hour in the equation to get power.
 
  • #11
576
2


Alright so it's Power= Energy/Time = mgh/time = 9100kg x 10 x 36/ 60 or convert the 36m to km?
 
  • #12
688
1


Alright so it's Power= Energy/Time = mgh/time = 9100kg x 10 x 36/ 60 or convert the 36m to km?
With some practice, you can learn to manipulate units: (kg/sec) X (m/s^2) X m = watt. For now, if you stay in base SI units of kg, m, s, then energy will be in joules, and power will be in watts. So your time should be noted as: 1 hr = 3600 sec.
 
  • #13
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I can't seem to get the answer in the back of the book it says 1.3kW
 
  • #14
688
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I can't seem to get the answer in the back of the book it says 1.3kW
Show me the numbers you used.
 
  • #15
576
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9100kg x 10m/s2 x 36m / 60 seconds = 54,600J/70%
 
  • #16
688
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9100kg x 10m/s2 x 36m / 60 seconds = 54,600J/70%
The elapsed time is 1 hr:

1 hr = 60 minutes = 3600 sec.
 
  • #17
576
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LOL I didn't even realize LOL I'm STUPID
 
  • #18
688
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LOL I didn't even realize LOL I'm STUPID
Honest mistake.

This is a good "real world" problem. Try to understand all we did.
 
  • #19
576
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I will this text book I'm using is full of challenging questions, Physics by Abbott you know of it?
 
  • #20
688
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I will this text book I'm using is full of challenging questions, Physics by Abbott you know of it?
I used Modern Physics by Tipler back in my college days.
 
  • #21
576
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Oh, I'm in grade 11.
 

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