HYDRAULICS

"Horsepower, Part 2"

Water and Wastewater

The following is from our text "MATH TEXT for WATER and
WASTEWATER TECHNOLOGY,
THIRD EDITION" by GROVER WRIGHT

(In the text we start with the "basic level" topic such as this, show an example, and then have you do several just like it. You are able to check your answers against ours for each problem. We then move to the intermediate level, and then to the advanced level. We show an example at each level, and have several additional problems for you to do on your own at each level... all with answers!)

This is meant to be printed out... that is the only way you will be able to properly view the math graphics.

GIVEN: We have a water flow rate of 50 gal/min that we wish to raise up a vertical distance of 150 feet.
PROBLEM: How much work will it take to raise the water?


PROBLEM: How much power will it take to pump the water?

PROBLEM: What is the water horsepower required to do this?

About now we are all wishing that there was a short-cut for this conversion of gallons into pounds of water. There is one! But as with all things, it is another number that we will have to memorize to use, and our memories do fail us, usually at "low-stress times" like certification exams! So it is very important to remember how we derived this math formula...now on to the short-cut...

Notice in the following formula the numerator always has the weight of water in it at 8.34 lbs. In the denominator there is always the definition of a horsepower of 33,000 ft-lb/min units. If we divide the 33,000 by 8.34 we get 3,956.8 as a constant now in the denominator, (that is rounded off to "3,960",) and nothing in the numerator as they will reduce each other......

to check this we invert the denominator and multiply:


this leaves us with the formula:

"Water Horsepower" is all theory. It does not account for the inefficiency of the pumps and the motors to pump the water. The equipment has flaws in the bearings, energy transfer, etc in design alone, much less considering that equipment does slowly wear out.
We need to correct for these inefficiencies......

We need to correct for the pumps inefficiency.....we do this by INCREASING the amount of horsepower APPLIED to the pump. By "over powering" the pump, we get it to do the amount of theoretical work we need it to do. This horsepower applied to the pump is called
"Brake Horsepower": denoted Bhp



In our example of 50 gal/min with a total head (lift) of 150 feet, and now adding a pump efficiency (Peff) of 62%:

Finally, we need to correct for the efficiency of the motor. The motor "loses" electrical power in the form of heat, bearings that have friction, magnetic flux that is lost in the transfer of power, etc. Lets say that the motor has an efficiency of 92%...and that we now label the "Motor Horsepower" as Mhp:

We have taken a theoretical water horsepower of 1.9. through a pump corrected Brake Horsepower of 3.05, to finally a Motor horsepower (also called "wire horsepower" in some circles) of 3.32 to do the job.


 

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