Power to raise water from point A to B

[zeralda21]

Homework Statement

Water is pumped from a well into a home. The height of the water to be lifted is 20 m, and they want to reach a water flow of up to 3.6 m3/h. The effective efficiency, when one takes into account the energy losses in both the pump in the water flowing in the pipes, is estimated at 40%. How much power should the pump be? (It is assumed that the kinetic energy of the water can be completely neglected in this context.)

Homework Equations

W=mgh
Efficiency=W_{applied}/(W_{quarried})
And I am sure that I am missing an equation(don't know which).

The Attempt at a Solution

It is given that the desired flow is up to 3.6 m^3/h = 3600 dm^3/h = 3600 kg/h. And the energy required to transport the water from the well to the home per hour is

W=mgH/h=3600*9.81*20/h≈ 700 000 W/h. Now the efficiency is 40% which means that the minimum power P form the pump must be:

0.4P = 700 000 W/h. Hence, P = 1.8*10^6 W/h. Very far from the correct answer(0.5kW). What am I missing?

 

[Sunil Simha]

Check your units. That is the only mistake you have committed.

 

[zeralda21]

Check your units. That is the only mistake you have committed.

Oh. (1.8*10^6)/(3600s)=500 W/second. But in that case I transform from W/hour to W/second. Confusing since it should only be W. Also the answer is in only W

 

[zeralda21]

I also have an other concern that I don't understand from the problem statement:

It is assumed that the kinetic energy of the water can be completely neglected in this context.

I don't understand the relevancy of this at all. How is this relevant? Or in what context could it be relevant?

 

[Sunil Simha]

Oh. (1.8*10^6)/(3600s)=500 W/second. But in that case I transform from W/hour to W/second. Confusing since it should only be W. Also the answer is in only W

Are you sue it is W/h. Or is it J/h?

You can neglect the kinetic energy of the water pumped on the assumption that the water was pumped quite slowly ((i.e. reversibly;like in a reversible thermodynamic process). This assumption is just an ideal scenario.

 

[zeralda21]

Are you sue it is W/h. Or is it J/h?

You can neglect the kinetic energy of the water pumped on the assumption that the water was pumped quite slowly ((i.e. reversibly;like in a reversible thermodynamic process). This assumption is just an ideal scenario.

I am not sure but I believe so since the answer is 0.5 kW. Does anyone know?

 

[sankalpmittal]

I am not sure but I believe so since the answer is 0.5 kW. Does anyone know?

Yes, 0.5 kW is the correct answer. Remember , Efficiency= Power output/Power input...
You had to calculate power input in this case which comes out to be 0.5 kW.

 

[Sunil Simha]

I am not sure but I believe so since the answer is 0.5 kW. Does anyone know?

Well, check the dimensions of power. Which one of J/h and W/h have the same dimensions?

 

[zeralda21]

Yes, 0.5 kW is the correct answer. Remember , Efficiency= Power output/Power input...
You had to calculate power input in this case which comes out to be 0.5 kW.

Forgive me for not being smooth here. But I still don't understand. It is something with the units I think.

The amount of work required from A to B is 700 000 W/h and η=40%=0.4 Therefore

0.4*Power input = 700 000. Hence power input = 1.8*10^6 W/h. The answer is 500 W = 0.5kW which is given if 1.8*10^6 W/h is divided by 3600 seconds which would yield W/second. So what am I doing wrong?

 

[Sunil Simha]

Forgive me for not being smooth here. But I still don't understand. It is something with the units I think.

The amount of work required from A to B is 700 000 W/h and η=40%=0.4 Therefore

0.4*Power input = 700 000. Hence power input = 1.8*10^6 W/h. The answer is 500 W = 0.5kW which is given if 1.8*10^6 W/h is divided by 3600 seconds which would yield W/second. So what am I doing wrong?

Okay, back to the basics. What is the unit of power? (consult your textbook,if you are confused)
What are its dimensions?
What are those of W/h? Does it have the same dimensions as those of power?

 

[sankalpmittal]

Forgive me for not being smooth here. But I still don't understand. It is something with the units I think.

The amount of work required from A to B is 700 000 W/h and η=40%=0.4 Therefore

0.4*Power input = 700 000. Hence power input = 1.8*10^6 W/h. The answer is 500 W = 0.5kW which is given if 1.8*10^6 W/h is divided by 3600 seconds which would yield W/second. So what am I doing wrong?

Again, kW is the dimension of power. Watt=Joule per second. KiloWatt= Joule *1000 per second.

Watt per hour as you state is not the dimension of power, watt is.

The question specifically stated that water "flow up" rate is 3.6 m³ per hour. What will be its rate in m³ per second ? How will you find mass flow up per second then ? How will you get gain in potential energy per second ?

CHECK YOUR UNITS !