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!
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
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
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
Something is incorrect about those equations.
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
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
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
(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.