Power of Water Pump (from continuity?)

[mbrmbrg]

Homework Statement

Water is pumped steadily out of a flooded basement at a speed of 5.5 m/s through a uniform hose of radius 1.2 cm. The hose passes out through a window 2.9 m above the waterline. What is the power of the pump?

Homework Equations

R_v=Av
R_m=\rho Av

where P is power, W is work, and K is kinetic engery:
P=\frac{W}{\Delta t}
W=\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mv_0^2

The Attempt at a Solution

R_v=Av=\pi r^2v=(\pi)(.012 m)^2(5.5 m/s)=0.002488 m^3/s

R_m=\rho Av=(1000 kg/m^3)(0.002488m^3/s)=2.488kg/s

Power=\frac{W}{\Delta t}=\frac{\Delta K}{\Delta t}=(\frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2)/(\Delta t)

Take delta time to be 1s, initial velocity (water sitting in basement) to be 0m/s, final velocity (water in pipe) to be 5.5m/s, mass dealt with in one second is 2.488kg.
NOTE: I think my error lies here; something was assigned a totally evil value. Probably my choices of velocity.

But going with the values I picked, I got P=(\frac{1}{2})(2.488 kg/s)(5.5 m/s)^2=37.63W

This be wrong. I be sad.

 

[FredGarvin]

As a first thought, use Bernoulli to calculate the pressure at the pump knowing the velocities and elevation changes. Use the pressure to calculate power via
P = pQ where

P= Power
p= pressure
Q = volumetric flow rate

 

[Andrew Mason]

You seem to be overlooking the \rho gh term.

AM

 

[FredGarvin]

The \rho g h is taken care of in Bernoulli in the \gamma \Delta z terms.

 

[mbrmbrg]

I don't think I know that much Bernoulli... the only Bernoulli equation that we've covered in class so far isp+\frac{1}{2}\rho v^2+\rho gy=constant. And the problem claims that it can be solved without even that. Sigh.

 

[FredGarvin]

That's just a different form of it. That is correct. Like I mentioned, you can calculate the pressure at the pump, i.e. P1 from Bernoulli. You know the constant velocity and density, so the two terms with velocity will drop out.

Once you get the pressure at the pump from the Bernoulli equation, use the second equation I mentioned to calculate the power.

A quck run through gets about 71 W. Is that close to your answer?

 

[Andrew Mason]

I don't think I know that much Bernoulli... the only Bernoulli equation that we've covered in class so far isp+\frac{1}{2}\rho v^2+\rho gy=constant. And the problem claims that it can be solved without even that. Sigh.

The power is the kinetic + potential energy added to the water per unit time. You have figured out the kinetic energy/time at 37.6 W. What is the potential energy added per unit time?

AM

 

[civil_dude]

A simple equation for power, P, is:

P = Q * ((gamma)force density) * h

I get 70.7 W

Of course this doesn't account for any parasitic losses.

 

[mbrmbrg]

No matter what I do, I keep getting Power in the hundreds of Watts.

Here's the latest work (and thank you for putting up with me!)

Where the 1's are the pipe at the waterline, and the 2's are the top of the pipe:
p_1+\frac{1}{2}\rho {v_1}^2+\rho gy_i=p_2+\frac{1}{2}\rho {v_2}^2+\rho gy_2

Since v is constant along the pipe, the 0.5*rho*v^2 terms cancel.
Let the waterline be 0m.

p_1=p_2+\rho gy_2=10^5 Pa + (1000kg/m^3)(9.81m/s^2)(2.9m) = 128449 Pa

OK. Now to find the force that the pipe exerts.

p=\frac{F}{A} so F=pA=(p)(\pi r^2)=(128449 Pa)(\pi)(.012 m)^2=58.11N

Lastly (*gasp*) Power=Fv=(58.11 N)(5.5 m/s)=319.6 W

And I lose yet again...

 

[Andrew Mason]

No matter what I do, I keep getting Power in the hundreds of Watts.

Here's the latest work (and thank you for putting up with me!)

Where the 1's are the pipe at the waterline, and the 2's are the top of the pipe:
p_1+\frac{1}{2}\rho {v_1}^2+\rho gy_i=p_2+\frac{1}{2}\rho {v_2}^2+\rho gy_2

Since v is constant along the pipe, the 0.5*rho*v^2 terms cancel.
Let the waterline be 0m.

p_1=p_2+\rho gy_2=10^5 Pa + (1000kg/m^3)(9.81m/s^2)(2.9m) = 128449 Pa

OK. Now to find the force that the pipe exerts.

p=\frac{F}{A} so F=pA=(p)(\pi r^2)=(128449 Pa)(\pi)(.012 m)^2=58.11N

Lastly (*gasp*) Power=Fv=(58.11 N)(5.5 m/s)=319.6 W

And I lose yet again...

One kg of stationary water in the basement is lifted 2.9 m. and moved to a speed of 5.5 m/s. This is an increase of mgh + .5mv^2 = 1*9.8*2.9 + .5*1*5.5^2 = 28 + 15 = 43 Joules. Since the flow rate is 2.5 kg/sec., the power is 43*2.5 = 107.5 Joules/second.

AM

 

[FredGarvin]

No matter what I do, I keep getting Power in the hundreds of Watts.

Here's the latest work (and thank you for putting up with me!)

Where the 1's are the pipe at the waterline, and the 2's are the top of the pipe:
p_1+\frac{1}{2}\rho {v_1}^2+\rho gy_i=p_2+\frac{1}{2}\rho {v_2}^2+\rho gy_2

Since v is constant along the pipe, the 0.5*rho*v^2 terms cancel.
Let the waterline be 0m.

p_1=p_2+\rho gy_2=10^5 Pa + (1000kg/m^3)(9.81m/s^2)(2.9m) = 128449 Pa

OK. Now to find the force that the pipe exerts.

p=\frac{F}{A} so F=pA=(p)(\pi r^2)=(128449 Pa)(\pi)(.012 m)^2=58.11N

Lastly (*gasp*) Power=Fv=(58.11 N)(5.5 m/s)=319.6 W

And I lose yet again...

You're close. Bernoulli will reduce to p_1-p_2 = \rho g(y_2-y_1)

The p_1-p_2 term is equal to the pressure that the pump is putting out since the pipe is exiting to atmospheric pressure.

Your next step is where you went wrong. That is not the way to calculate pump power. The power from the pump is a function of pressure and flow rate. Look at the equation that has been given to you twice:

P = p Q

 

[mbrmbrg]

Your next step is where you went wrong. That is not the way to calculate pump power. The power from the pump is a function of pressure and flow rate. Look at the equation that has been given to you twice:

P = p Q

Yessir. Thank you, sir. Sorry for disregarding you, sir.
Now that the final's over, I no longer have any compunctions about using equations not covered in class or the textbook

 

[Hurkyl]

I agree with Andrew's calculation:

power = (GPE created per second) + (KE created per second)
= 108.39... Joules per second

(+ losses due to friction, etc)