Finding power as a function of time in transferring water

[Kanda ryu]

Homework Statement

A completely filled cylindrical tank of height H contains water of mass M. At a height h above the top of the tank there is another wide container. The entire water from the tank is to be transferred into the container in time T such that level of water in tank decreases at a uniform rate. How will the power of the external agent vary with time.

Homework Equations

Change in kinetic energy = sum of work done by external forces

The Attempt at a Solution

As its given uniform rate, change in kinetic energy would be zero so I used work energy theorem.
W(ext force) = change in gravitational potential energy
= mgh'
I thought of taking water of mass dm at a height x below surface of water. So h'=(h+x).I am sort of confused how to proceed here to get power as a function of time. dW/dt would give power but how to proceed on RHS.

 

[TSny]

Welcome to PF!

You're thinking along the right lines.
Can you express dm in terms of M, T, and dt? (Hint: use the fact that water is leaving the cylindrical tank at a uniform rate.)

 

[Kanda ryu]

Density = M/A H (A is base area of cylinder
So M/AH = dm/A dx
So dm = Mdx/H
As water is decreasing at a uniform rate, Adx/dt = constant
dW/dt =(Mdx/Hdt)h'g
Even if I write Adx/dt = H/T, I don't get the answer as a function of time.

 

[TSny]

Density = M/A H (A is base area of cylinder
So M/AH = dm/A dx
So dm = Mdx/H
As water is decreasing at a uniform rate, Adx/dt = constant
dW/dt =(Mdx/Hdt)h'g

OK

Even if I write Adx/dt = H/T, I don't get the answer as a function of time.

This equation is not quite correct. Note the overall units on each side don't match. Once you fix that, you will have an expression for dx/dt in terms of H and T. Can you use that to find x as a function of time?

 

[Kanda ryu]

Yeah sorry I meant to type Adx/dt = AH/T
dx/dt = H/T
If I integrate dx from 0 to x and dt from 0 to t, I get
x= H t /T
dW/dT = M(dx/dt)g (h+x)
P= M H/T g (h+ Ht/T)
The answer is a function of a general time 't' so maybe this should be correct, thanks a lot for your support sir.
Am I allowed to confirm my final answer from physicsforum helpers as I currently do not have the means to access the correct answer.
Again thanks,
Cheers

 

[Kanda ryu]

Correction :
P(t)= MH/TH g(h+Ht/T)
= (M/T)g(h+Ht/T)

 

[TSny]

Correction :
P(t)= MH/TH g(h+Ht/T)
= (M/T)g(h+Ht/T)

That looks right to me. You can see if the answer makes sense by using P(t) to find the total work done during the time interval from t = 0 to t = T. Do you get what you expect?

 

[Kanda ryu]

Yes putting t=0,T/2 and T gives favourable results.

 

[TSny]

Did you use your result for the power P(t) to derive an expression for the total work W_tot done in emptying the cylindrical tank?

 

[Kanda ryu]

Yes, on integrating P(t)dt taking limits 0 to T, I got
Wtot= Mg(h+HT/2)

 

[Kanda ryu]

Mg(h+H/2)*
Which is like all the water is concentrated at the center of cylinder (by height) and finding work done to carry that to a height h. I hope this is correct

 

[TSny]

Mg(h+H/2)*
Which is like all the water is concentrated at the center of cylinder (by height) and finding work done to carry that to a height h. I hope this is correct

Yes, that looks very good.