PE Practice Exam Pump Problem 2

[ianertia]

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!

 

[JGard1985]

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

http://www.xceed-eng.com

 

[LittleInch]

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.

 

[Lnewqban]

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.

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

 

[3DDave]

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.

 

[TenPenny]

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

 

[SandCounter]

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.