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PE Practice Exam Pump Problem
2

PE Practice Exam Pump Problem

PE Practice Exam Pump Problem

(OP)
Problem:
A centrifugal pump is used to transport 25 gpm of acetone at 78°F. The differential pressure across the pump is 50 psi. The vapor pressure of acetone at this temperature is 5.6 psia, viscosity is 0.3 cP, and density is 49 lbm/ft³ (sp gr = 0.79). The pump efficiency is 70%. The input shaft horsepower (hp) to drive the pump is most nearly:
(A) 0.12
(B) 0.73
(C) 1.04
(D) 1.35

Solution:
Shaft hp = W/eff
W=mass flow*weight
m=density * (volumetric flow)
v=1/density
bhp=density * Vol flow rate * specific volume * deltaP
W=0.73 hp
shaft hp = 0.73hp/0.7
=1.04 hp

SO MY QUESTION IS THIS:
Why can't I use HP=Q*h*SG/(3960*eff)?
It comes out to HP=0.73hp
But I shouldn't have to divide by efficiency again to get 1.04hp. Someone please explain. This has been bothering me.

Thanks in advance!

RE: PE Practice Exam Pump Problem

2
Check your math again.

Remember that the equivalent water head must be computed from pump discharge pressure and fluid weight of acetone.

When you determined head did you do
- 50psi/62.4lb^ft^3= 115 ft
or
-50psi/49 lb^ft3= 147 ft

If I work it out in mathcad using H=50psi/49lb/ft^3=147ft head

So
(25*147*0.79)/(3960*0.70)=1.0407hp





Jeff
Pipe Stress Analysis Engineer
www.xceed-eng.com

RE: PE Practice Exam Pump Problem

Totally agree, but all you need is head in the fluid you're using. The density term acts to give it the right units of work.

The head output of a centrifugal pump is the same at the same speed, but pressure and power vary by density for the same flow.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

RE: PE Practice Exam Pump Problem

Quote (ianertia)

W=mass flow*weight
m=density * (volumetric flow)
v=1/density
bhp=density * Vol flow rate * specific volume * deltaP

Something is incorrect about those equations. ponder

Yes, you can use
BHP(shaft) = (Q x H x SG) / (3960 x Eff)

where BHP(shaft) is brake horsepower being fed to the pump, Q is flow in GPM, H is (delta P * 144)/specific weight of acetone, SG is specific gravity and Eff is the hydraulic efficiency of the pump.

As stated above, you have made a math mistake somewhere.

"Where the spirit does not work with the hand, there is no art." - Leonardo da Vinci

RE: PE Practice Exam Pump Problem

Put more units into the calculations. Something seems off:

for example the equation W=mass flow*weight which, with units is

Work = (lbm/sec) * force ???


I would expect Power = volumetric rate * delta-P

= ((length^3)/time) * force/(length^2) -> force*length/time

(25 gallons/minute) * (231 inches^3/gallon)*(1 minute/60 seconds) = 96.25 inches^3/second

Power = 96.25 inches^3/second * 50 lbf/inches^2 = 4812 lbf-inches/second * 1 foot/12 inches = 401 lbf-foot/second; or .729 hp.

Since the hp output = 70% of the hp input, then divide by .7 to get 1.04

Density, specific gravity, and head don't matter for this calculation.

RE: PE Practice Exam Pump Problem

Quote:

SO MY QUESTION IS THIS:
Why can't I use HP=Q*h*SG/(3960*eff)?
It comes out to HP=0.73hp
But I shouldn't have to divide by efficiency again to get 1.04hp. Someone please explain. This has been bothering me.

You can absolutely use that formula, it's the easiest one to use.
Q=25 gpm
h = 50 psi x 2.31 / .79 = 146 ft
SG= .79
eff = .7

25 x 146 x .79 / (3960 x .7) = 1.04

RE: PE Practice Exam Pump Problem

1714 is a very useful HP conversion factor to memorize. Skip the head conversion and SG, it's already included in the pressure.

(gpm x psi)/ 1714 = HP

(25 gpm x 50 psi) / 1714 = 0.73 HP

Divide by efficiency

0.73/70% = 1.04 HP

I used to count sand. Now I don't count at all.

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