Hydraulic Power from Alfa Laval Book

[MIANCH]

Hi there,
Calculating the hydraulic power of a pump from Alfa Laval book is very simple and it dose not take in account any critical parameters like, piping size, elevation, viscosity.
How you can rate the authenticity of this equation. the equation is Kw=QxH/K where is constant 1715.
Can any one explain about this equation and how far I can use it for pump specification.
Thanks
MIANCH

 

[ione]

How can you state this?

Flow rate (Q) and head (H) for each working point are strictly related to the system curve, and so to the pipe size, roughness, elevation, fittings. Performance curves are also affected by viscosity of the fluid and so the working points vary accordingly.

 

[BigInch]

Hydraulic power does not even include the efficiency of the pump, or the motor, not to mention anything else such as pipe, elevation or viscosity. You need NO more than the physical definition of Power = Work per unit Time to prove the equation, Work = Force x Distance, time is time.

Hydraulic power is the power required to do the job. The power required to do whatever it is that you want to do, and it considers how you want to do it. If you need pipe to do it, the hydraulic power required must include the loss of that pipe. If you have a waterfall available, the power required from a pump can be that much less. If you have a viscous fluid, the hydraulic power required to use that fluid may be more, but only if you are trying to use its available power by flowing it. If your viscous fluid will deliver power simply by its weight moving through a distance not enclosed by a pipe, such as down through a water wheel, viscosity will have no or little effect.

Now, what do you suppose Brake Power is.

 

[micalbrch]

The confusion comes from the bad explanation in the Alfa Laval handbook. Is says:

Hydraulic Power = Q x H x ? x g

and

Hydraulic Power = Q x H / k (k = unit constant)

They use "Head" and "Pressure" as identical terms and units.

 

[BigInch]

The problem is that those are equations, not definitions and, as long as k = ? x g, that's going to be correct because that is what hydraulic power equals in mathematical terms. It would seem the real problem is that someone doesn't know what hydraulic power equals "in words".

 

[micalbrch]

I agree with you, BigInch. But have a look at the corresponding page of the handbook. Their "k" explanation is bad (wrong).

 

[MIANCH]

Biginch,
Ok, I agree with you Hydraulic power is to move the fluid by any mean, either used water fall or pump the viscous fluid with force.
Now if you go through engineering books, there is including every parameter and my question was about Alfa Laval book.
In this book they used equations (short cut method).
Thanks
MIANCH

 

[hydtools]

Power is force x velocity. units lb-ft/sec
Force is pressure x area. units lb.
Velocity is volume rate / area. units ft/sec
power = p x A x Q/A = p x Q(231/720). Q in gpm
horsepower = power/550 = p x Q(231/720)/550 = p x Q/1714

The hydraulic horsepower is just the power state of the fluid. It does not indicate how much input created it or how much will remain after flow through the system.

Ted

 

[BigInch]

"Power state of the fluid." OK, I buy that one.

"It does not indicate how much input created it"
It would seem that work (ft-lbf) must have been the input, and you would know how much, but only over an uncertain amount of time.

"How much will remain after flow through the system." I don't think work can remain. Wouldn't what "remained" have to be the consequences of a change in work done, plus or minus, on the actual system itself, and again, plus or minus the change in potential energy stored within the system.

In other words, don't mind me. I just quit smoking a couple of days ago and this is probably the least of what's bothering me.

 

[hydtools]

I never smoked so I really can't relate to quitting. Caffine, now that's a different matter. Good luck with quitting.

Let's say you have analyzed the system and you have determined that you need pressure p to drive flow rate gpm through that system. Now you can calculate the hydraulic power hp to drive the system. If you know the efficiency of the pump at that flow and pressure, you can then determine the shaft horsepower required to drive the pump. If you know the efficiency of the prime mover, then you can calculate the input power required to drive the prime mover to drive the pump to drive gpm through the system.

Ted

 

[btrueblood]

"I just quit smoking a couple of days ago"

I've never been on fire either, hope the burns are all less than 1st degree....how many gpm to put the fire out (just to keep the topic remotely related to pumps...

Seriously, you are all saying the same thing, and yes, the handbook of the OP is technically correct without any of the caveats and exceptions that would allow anybody to utilize that equation to size a pump and motor combination. But, somewhere, a marketing type wanted the handbook to "look" simpler than the competition's handbook.

 

[MIANCH]

hydtools,
by solving eqaution answer is not coming 1714, how do you have bring it.
Thanks
MIANCH

 

[quark]

The book says k = 600 and you would get it if you use SI uiints.

Hydtools already gave the equation.

Power = Pressure (total head here) in N/sq.mtr * Area in sq.mtr * (Flow in cu.mtr/sec/Area in sq.mtr)

This gives you power unit in N-m/sec which is J/s or W.

If you use pressure (or total head) in bar and flowrate in lit/min then,

Q l/min = Q/(60*1000)cu.mtr/sec
H Bar = 100000 N/sq.mtr (or 100000 Pa)
So, P (W) = Q*H*100000/(60*1000) = Q*H*10/6
If you want P to be in kW then
P(kW) = Q*H*10/(6*1000) = Q*H/600

 

[hydtools]

231/(12*60*550) = 1/1714 It is a constant to make equation units consistent. And it is rounded down from 1714.2857

231 cu.in./gal
12 in/ft
60 sec/min
550 ft-lb/sec/hp
p, lb/in^2
Q, gal/min

Ted