Power Requirements of a Pump - Solved Example

A solved example of how to calculate power requirements of a pump.
Please provide feedback on this tutorial by selecting "Like" or "Dislike". Your feedback including any comments ([e-mail]) will be used in developing new tutorials.
For additional contents related to the topic of this tutorial, you may refer to Singh, R.P. and D.R. Heldman. 2014. Introduction to Food Engineering. 5th edition. Elsevier Publ. London.

[@leondelgadop]

Excellent explanation. Very concise and useful. Thanks.

[@jarleerlandherstad5292]

Hello, Thanks for the video. I have 2 questions and 1 comment :)
1. In line 7) of your calculation (approx. 09:35 - :09:50): Why do you use a factor of 2 i.e. 2 x 0.0046 x ... x 35 /diameter. You are calculating the loss in the pipe which there is 1 off. (There are 2 bends, hence a factor of 2, OK)
2. In line 7) of your calculation (approx. 10:06): Why do you use 0.00291 m for the pipe diameter when you at the start says it is: 0.0291 m (a decade in difference).
3. Also, the friction factor of 0.0046 ...? Does such a smooth pipe exist? (The value is outside my Moody diagram).

[@RPaulSingh]

In line 7, the factor "2" is part of the equation, please see the derivation. Ef =2 x f x u^2 x L/D
The denominator term for Diameter should be 0.02291. In fact, in the calculation, the correct diameter is used. The resulting value of 187.2 is correct. Thank you for noting this, I will make the correction in the video.
The friction factor of 0.0046 is correct. Please recheck the Moody diagram.

[@araieskey2183]

thank you so much for the explanations, but why in Ef pipe length of 35 m, the D become 0,00291 m? is it the D or not? in the problem the D is 0,02291 m

[@RPaulSingh]

Thank you for identifying the correction. In step 7, the value of D is incorrectly written as 0.00291, it should be 0.02291. The calculated value of 187.2 is correct.

[@alessandrabyrrorodrigues9311]

Hi. In line 6) two elbows of 90 degrees are 2*0.75 instead of 2*1.5. The result in line 6) should be 3.5. So, in line 8) your total friction loss is 213.9 J/kg. Am I correct?

[@RPaulSingh]

The friction loss for a regular 90 degrees threaded elbow is 1.5, therefore, for two elbows it is 2*1.5. The calculation shown in the video is correct!