# Pump and nozzle

1. Apr 9, 2006

### JSBeckton

Consider a water pump that recieves liquid water at 15C and 100kPa, and delevers it to a same diameter short pipe having a nozzle with exit diameter of .01m to the atmosphere at 100 kPa. Neglect KE in pipes and assume constant internal energy u for water. Find exit velocity and mass flow rate if the pump draws 1kW of power.

I started with the energy equation but since the pressure and internal energy is the same, wont the he and hi also be the same? If I reduce I get

$$\begin{array}{l} \omega = - \frac{1}{2}V^2 \\ V = \sqrt { - 2\omega } \\ V = \sqrt {( - 2)( - 1kW)} \\ \end{array}$$

Which isn't giving me the right velocity of 29.43 m/s. I know it has to do with the exit area but its not comming to me, can anyone point me in the right direction? Thanks.

2. Apr 9, 2006

### lightgrav

are you trying to use "w" as (KE density)/(mass density) ?
you've ended up substituting Power for it.

While the intake pressure is 1 atm and outside the nozzle it is 1 atm,
the pressure in the pipe, just INside the nozzle, is MORE than 1 atm.

Use Power = F . v ... and F = P . A

3. Apr 9, 2006

### JSBeckton

I'm using "w" as specific work

F=PA
=(100kPa)(.00007853)
.007853N

W=FV (V=velocity?)
V=W/F
=(1kW)/(.007853N)
=127.324

The units aren't comming out right and the anwser is much too high, can you see what I'm doing wrong?

4. Apr 9, 2006

### lightgrav

You're just "plugging and chugging"!

Why are you using 1 atm for the Pressure inside the pipe ?
If the pressure inside the nozzle opening = P outside the opening,
there's no Force to accelerate the water!
But the water has to speed up, so that dm/dt = rho.A.v
is the same outside the nozzle as inside the pipe.

Units?
Power[Watts] = E/t [Joule/sec = N.m/s] = F.v [N.m/s]

[J/kg] is not the same as [J/s], if you want to check your first post.

Last edited: Apr 9, 2006
5. Apr 9, 2006

### Cyrus

That is wrong, dm\dt = 0 for the control volume.

6. Apr 9, 2006

### JSBeckton

I do not know how to find the pressure inside the pipe before it exits w/o knowing the diameter of the short pipe. I makes sense that the mass flow rate must be the same at the pump and at the nozzle, but I cannot find the mass flow rate w/o the dia of the pipe, is there another way?

7. Apr 9, 2006

### lightgrav

yes.
the power from the pump provides KE to the water as it exits the hole.
... what's Pressure times velocity ?

Cyrus,
dm/dt is the mass flow rate past any point (say, the nozzle)
- which was asked for -
= the mass current = mass current density integrated thru(dot) the Area
= the momentum density integrated thru(dot) the Area ...
.
no one implied (and no one inferred) a change in mass density.

Last edited: Apr 9, 2006
8. Apr 9, 2006

### JSBeckton

Power.

I guess i'm stuck because I don't know the velocity or the pressure at the exit point and I can't seem to figure out what you are hinting at for whatever reason. Can you be more specific or try to explain in another way? Thank you for your help by the way.

9. Apr 9, 2006

### Cyrus

Man, I have no clue now. He says ignore KE in the pipes and then asks right after wards to find the velocity of the flow.

To Lightgrav,

usually, dm/dt is used for the control volume, and m-dot for the flow. That's why I said dm/dt is zero. But we are in agreement.

10. Apr 9, 2006

### lightgrav

Your first post is in the right direction ... Pump Power becomes KE ...
but the mass is important, which you got rid of (intentionally).

in 1 sec, 1kJ of Work is done => KE = 1/2 m v^2 .
but dm/dt = rho.A.v , so in 1 sec , m = rho.A.v.1 exits the nozzle.
Replace the "m" in KE with this formula, and solve for v at the nozzle.

(KE in pipe is small, becomes large as iot accelerates thru nozzle)

11. Apr 9, 2006

### Cyrus

What a poorly, poorly written problem.

12. Apr 9, 2006

### JSBeckton

KE=1/2 mv^2

but m=rho.A.v.1s

KE=1/2 rho.A.v^3

v=cube rt [(2KE)/(rho.A)]

V= 29.43 m/s

m dot=rho v A
=(997)(29.43)(.00007853)
= 2.13 kg/s

I guess it was a big mistake to cancel out the mass. Thanks both of you for all of your help, it is greatly appreciated!

13. Apr 9, 2006

### Cyrus

I was of no help to you.