School of Professional Eng Question of the Week June 26 1

[Blu1913]

Hello everyone. I am taking my PE in the Fall and found a great website that give you a new problem every week.

However, this weeks problem I believe may have an error, but I'm not Water/Distribution engineer so I thought I would ask you all....

Here is the website with the solution.

http://www.schoolofpe.com/answeroftheweek.asp

My question: in the answer arent they only solving for the motor cost of $44/week, not the pump and motor cost as the problem statement dictates? Why isnt the pump cost added to that? Because if the pump produces 2.5hp, then the cost per week is $25. GIving you a total of $69/week, not $44...right?? am i missing something?

Thanks for the help everyone!

 

[francesca]

It looks right to me. The pump gets its power from the motor. All you have to factor is the efficiencies of the motor and the pump and apply them to the 2.5 hp as the example shows.

 

[Maury]

The pump requires 2.5 hp, it does not produce any power.

 

[Blu1913]

So in the equation:

Total Efficieny = Hydraulic Horsepower
--------------------
Motor horsepower

Motor horsepower includes the hp of the pump? I thought the hydraulic hp was the horsepower of the pump, and the motor horsepower was separate.

In short, you're saying the Motor Horsepower includes the 2.5 pump hp? Is that correct?

 

[BRIS]

Hydraulic horesepower is what the system gives out, motor horsepower is what you have to put into the system to get out the hydraulic horsepower.

The overall system efficincy is what you put into the system divided by what you get out of the system.

The motor drives the pump so obviously the motor horsespower includes the pump horsepower as you put it. only the motor is burning electrical power and that is what you pay for.

 

[Blu1913]

Ah, I see it now. Geez, sometimes you have to beat it in, to make sense. Thank you everyone!!

 

[QQstuff]

the answer is very correct. note "hydraulic horsepower" of 2.5 hp. check PE reference manual to understand this question.

 

[25362]

Note that other costs such operators, controls, maintenance, capital costs, etc., are excluded.

 

[49merc]

A couple of references, besides the PE reference manual, go over this in a lot of detail. "The Pump Handbook", by Karassik, and "Groundwater and Wells", by Johnson Screen. (GPM X Head in Feet)/(3960 x pump eff. x motor eff.)= motor horsepower. .746 kW=1 HP. Leave out motor and pump efficiency, and you have water horsepower.