# power of a water pump

• Sep 5th 2011, 01:38 AM
furor celtica
power of a water pump
A pump, taking water from a large reservoir, is used to spray a jet of water with speed 20 ms^1 and radius 0.05 metres, from a nozzle level with the surface of the reservoir. Calculate the power of the pump.

Alright I'll admit I'm pretty lost here and haven't been able to mount a very convincing attempt at a solution. I'm used to force-work-power problems based on the motion of an object so I'm not sure how to model this exercise. Even the wording is unclear; should I take initial speed 20 ms^-1, ending speed 0, distance 0.05? Where is the jet directed, horizontally or vertically?
Any suggestions would be appreciated.
• Sep 5th 2011, 05:18 AM
skeeter
Re: power of a water pump
Quote:

Originally Posted by furor celtica
A pump, taking water from a large reservoir, is used to spray a jet of water with speed 20 ms^1 and radius 0.05 metres, from a nozzle level with the surface of the reservoir. Calculate the power of the pump.

water density is 1 kg/L

1 L = 0.001 m^3

In a pipe of radius 0.05 m , 1 kg of water occupies a length approximately 0.127 m.

Traveling at 20 m/s , 1 kg of water takes about 0.006 sec to travel that distance.

$\displaystyle P = \frac{W}{t} = \frac{\Delta E_K}{t} = \frac{\frac{1}{2}mv^2}{t} = \frac{\frac{1}{2}(1)(20^2)}{0.006} \approx 31.4 kW$
• Sep 5th 2011, 06:22 AM
furor celtica
Re: power of a water pump
thanks bro
• Sep 6th 2011, 07:42 PM
bjhopper
Re: power of a water pump
Hello Skeeter,
Excuse me for using US Eng units

Velocity of jet = 20mps*3.281=65.6fps
Jet dia =.05m * 39.37 = 1.97 in (.164 ft)
Cross-section of jet (,164)^2*pi/4=.0211 ft^2
Q=V*A =65.6*.0211=1.38cfs
Q gpm = 1.38*7.5*60 =623 gpm
Velocity head of jet =(65.6)^2/64.4 =66.8 ft
Hydraulic Hp = gpm *8.33* 66.8/33000 =10.5Hp 7.84 KW
No pipe friction loss and no potential head loss and no pump eff loss
• Sep 7th 2011, 03:11 AM
skeeter
Re: power of a water pump
Quote:

Originally Posted by bjhopper
Hello Skeeter,
Excuse me for using US Eng units

Velocity of jet = 20mps*3.281=65.6fps
Jet dia =.05m * 39.37 = 1.97 in (.164 ft) ... radius?
Cross-section of jet (,164)^2*pi/4=.0211 ft^2
Q=V*A =65.6*.0211=1.38cfs
Q gpm = 1.38*7.5*60 =623 gpm
Velocity head of jet =(65.6)^2/64.4 =66.8 ft
Hydraulic Hp = gpm *8.33* 66.8/33000 =10.5Hp 7.84 KW
No pipe friction loss and no potential head loss and no pump eff loss

...
• Sep 7th 2011, 03:37 AM
bjhopper
Re: power of a water pump
Sorry Skeeter I used .05 m as a dia not a radius.Flow increases to 2494 gpm and Hp to 42 same as your answer in KW
• Sep 7th 2011, 04:37 AM
skeeter
Re: power of a water pump
Thanks for the confirmation from an engineer's perspective. ;)