Pumping Head

[cougarfan]

I need some general help checking what head the pump will be up against for a fluid with a sg of 1.55 through a 3/4" line - 325 ft long, 10 ft elevation. Can someone give me a formula for doing this?

I have done this a long time ago but forgot how.....

 

[Scipio]

You'll also need to determine the static pressure at the downstream end of the piping, and in order to calculate the line losses, the pipe schedule/wall thickness along with the expected flowrate to calculate the velocity of the liquid. That will give you a ballpark - the presence of elbows, tees, valves, etc. in the piping will also increase pressure drop. There are other considerations, like the surface roughness of the interior wall of the pipe, but depending on the installation the impact is minor and not always considered.

 

[25362]

The frictional losses for the flow of any incompressible fluid is given by:

h_f = f (L/D) V²/2g

h_f = frictional resistance in ft of fluid
L = length of pipe in ft
D = average internal diameter of pipe in ft
V = average fluid velocity in f/s
g = acceleration of gravity, 32.17 ft/s²
f = dimensionless friction factor, there are graphs for its estimation based on the dimensionless Reynolds' number. Re = (density)(velocity)(diameter)/(absolute aka dynamic viscosity) in consistent units.

Losses on valves and fittings are expressed either as equivalent pipe lenghts, or by h = k.V²/2g

k values are tabulated.

To these friction head losses, you must add any static pressure differences as well as any difference in heights.

 

[cougarfan]

25362 - QUESTION - DOESN'T THE DENSITY OF THE PRODUCT IN QUESTION PLAY INTO THIS EQUATION AT ALL - IN OTHER WORDS, IF i AM PUMPING WATER VS PUMPING MOLOSSES, ISN'T THE HEAD GOING TO BE DIFFERENT?

 

[briand2]

Dideyjohn,

If you look at the last line of 25362's response, you'll see that density is catered for thus:

"To these friction head losses, you must add any static pressure differences as well as any difference in heights."

Regards,

Brian

 

[cougarfan]

25362 & Briand2 - let me explain my question - isn't the pressure drop of a 100 ft of pipe different for a water solution compared to a molasses solution? If I had a 10 ft of head drop for a water line for 100 ft, my pressure drop for molosses would be much greater wouldn't it? How does one factor that in?

 

[CanuckMiner]

Dideyjohn,

The friction head will be more for pumping molasses than it would be for water due to the fact that molasses has a higher viscosity than water (at ambient temperatures). This would be accounted for in the determination of the Reynold's number, which in turn is used to determine the friction factor f as outlined in 25362's post.

However, an increased density of the fluid does not necessarily mean an increase in head loss, provided the "kinematic" viscosity (equal to the absolute viscosity divided by the density) of the two fluids is equal.

In short, you have to determine the friction factor for your fluid. To do this, you'll need to find/determine the viscosity.

Cheers,
CanuckMiner

 

[Scipio]

The Darcy-Weisbach equation 25362 posted actually does handle that, different liquids are addressed with the kinematic viscosity, which is used in calculating the dimensionless friction factor f in that equation.

There's another equation, simpler, but valid only for laminar flow (Reynolds number < 2000) under pressure in circular pipes, called the Hagen-Poiseuille law.

Head loss h_L=32v(L/gD²)V

Where
v = kinematic viscosity
L = length of pipe
g = gravitational acceleration
D = inner diameter of pipe
V = average velocity of fluid

Everything else, valves, fittings, static head, etc. still has to be considered the same as if you were using Darcy Weisbach. That's digging back to my university fluid dynamics though, these days I usually just calculate friction losses with software.

 

[DLANDISSR]

dideyjohn:
Get a copy of &quot;Crane's Flow of Fluids&quot; and use the formulas and graphs therein.