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Is this the right calculation?

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AHartman

Mechanical
Sep 17, 2010
32
Hi all,

I'm trying to find the power added to my working fluid (water @ 20 deg C) by a theoretical pump to see how much power I have to apply to the motor.

The pump draws from a bag, through flexible tubing, then out through more flexible tubing and a small plastic nozzle.

The conditions are as follows:

-P1 = 0 psi (ambient, the bag is not pressurized)
-P2 = 30 psi (my target output pressure)
-Q = 0.4 Gal/min (1.5 L/min)
-A1 = .051 sq. in. (tube area at inlet
-A2 = 0.0016 sq. in. (nozzle area at outlet)
-h1 = 0 ft
-h2 = 3 ft (total of +3 ft height change)

My equation for power is:

Power = rho * Q * wp (specific work)
wp = ((P2-P1)/rho)+(1/2)*(V2^2 - V1^2) + g*(h2-h1)
where V2 and V1 are the outlet and inlet velocities

Is this the right equation to use? And when I get 80W of power, does that seem right?
 
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Thanks for the input, bimr.

How do you get your flow resistance estimate?

I think the problem with my original calculation was that the output velocity V2 was 160 ft/s, but when I took my flow and area, that's what I got. That makes me think the application of my equation is incorrect.

Thanks,
Adam
 
Just made an estimate.

Size your tubing to obtain an 8 ft/sec velocity and it should be inconsequential in the calculations.
 
That makes sense. So maybe what I'm missing is that I took the area of the nozzle (very small, .0016 sq. in) and should instead look at the area of the outlet tube? That will certainly drop my output velocity, but it seems like I'll have a significant pressure drop across the nozzle which will jack up my power. I think I'm out of my depth here.
 
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