# Thread: power of a water pump

1. ## 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.

2. ## Re: power of a water pump

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.

$\displaystyle P = \frac{W}{t} = \frac{\Delta E_K}{t} = \frac{\frac{1}{2}mv^2}{t} = \frac{\frac{1}{2}(1)(20^2)}{0.006} \approx 31.4 kW$

thanks bro

4. ## 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

5. ## Re: power of a water pump

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
...

6. ## 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

7. ## Re: power of a water pump

Thanks for the confirmation from an engineer's perspective.

### a pump, taking water 0.05 metres

Click on a term to search for related topics.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/math-topics/187305-power-water-pump.html", "fetch_time": 1539807592000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-43/segments/1539583511216.45/warc/CC-MAIN-20181017195553-20181017221053-00425.warc.gz", "warc_record_offset": 230983943, "warc_record_length": 11476, "token_count": 838, "char_count": 2562, "score": 4.125, "int_score": 4, "crawl": "CC-MAIN-2018-43", "snapshot_type": "latest", "language": "en", "language_score": 0.874426007270813, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00024-of-00064.parquet"}