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?
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?