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Start by
Malcolm Heys
01-19-2014 03:04 AM

How can I calculate power to the fluid?

I have been calculating the power that a pump (or fan) imparts on the fluid by:
Power[W] = flow [m3/s]x Static pressure [Pa] I call it hydraulic power.

I then divide this by the pump (or fan) efficiency and the motor efficiency to get the power required to run the pump (or fan).

My problem is when I search on the internet for hydraulic or fluid power I can't find this formula. Does anyone know what it's called and a website that confirms this formula?
01-19-2014 05:18 AM
Top #2
José Miguel Victoria Martínez
01-19-2014 05:18 AM
Power (W) = [ Flow (m3/h) * Static Pressure (liquid column meters) * Spc.gravity ] / [ Pump Efficiency * 270 ] * 735,48

Static Pressure (liquid column meters <> mcl <> m) = [Static Pressure (kg/cm2) / Density (kg/cm3)] *10

* 270 and 735,48 are conversion factors to correct units.

* This Power is absorved power for the pump, it's not considered driver efficiency.

3) If you want to calculate Hydraulic power you have not to consider the pump efficiency

I can send you a scaned sheet from Hydraulic Institue aobut the formula you are looking for.
01-19-2014 07:31 AM
Top #3
Brian Parker
01-19-2014 07:31 AM
You can also find it at http://www.mcnallyinstitute.com/16-html/16-12.htm and http://www.engineeringtoolbox.com/pumps-power-d_505.html
01-19-2014 10:03 AM
Top #4
Malcolm Heys
01-19-2014 10:03 AM
Thank you very much for your feedback. I compared the different methods:
1. W=Pa x m3/s
3. McNally Institute example
4. Engineering Toolbox

They all gave the same results. Only slightly different due to the conversion factors, so I'm happy with my original equation. I can use it for fans and pumps. I'm surprised it's not given a name like "Hydraulic Power" or "Fluid Power" which is universally recognised.

I've recently had a fan supplier give data of a fan that only converted 10% of electrical power to "hydraulic power". He didn't understand what I meant by "hydraulic power" so I searched for a universal expression, but haven't yet found one.

I have also been talking to a heat exchanger supplier about optimising size to use the least amount of "hydraulic Power". ie balance the pressure drop and flow to reject kW with the minimum fan/ pump "hydraulic Power". But, Hydraulic power is not recognised so again I think I must have the name wrong.
01-19-2014 12:50 PM
Top #5
Merritt Elmore
01-19-2014 12:50 PM
I am a current Mechanical Engineering student, we call it "Hydraulic Power" even if that isn't the name of the equation. Hydraulic Power=Shaft Power * Pump Efficiency; Shaft Power= Electrical Power * Motor Efficiency.
01-19-2014 03:17 PM
Top #6
Ned Hoag
01-19-2014 03:17 PM
Jose is correct, the term hydraulic power or also referred to as hydraulic horsepower refers only to the work performed. Brake horsepower takes into consideration the losses incurred within the pump or fan through friction, leakage, etc. which represents its efficiency. So brake horsepower is the input power required at the shaft.
01-19-2014 05:59 PM
Top #7
Malcolm Heys
01-19-2014 05:59 PM
Thank you. Seems like Hydraulic Power is the recognised term for pumps. Has anyone heard if it refers to air too, or do fan engineers use another term?
01-19-2014 08:43 PM
Top #8
01-19-2014 08:43 PM
Dear Mr. Malcolm Heys,
In Hydraulics, the term Fluid covers all matters that flow. So, Fluid Power could be a better term. Hydraulic Power is a term more popular in Power Hydraulics. Power Hydraulics is more common in heavy earth moving machineries. So, Hydraulic Power may be misunderstood. Fluid Power also becomes more general, since it would take cognisance of all characteristics of the fluid. General Formula for Fluid Power is -
Power (kW) = (Rho) * (Q in l/s) * (H in m) / 102 / (Pump Efficiency) / (Motor Efficiency)
kW i.e. 1000 W is the commonly used unit of power. 736 W make one metric HP. 746 W make one British HP
Rho = density of the fluid in kg per liter
Q in l/s is rate of flow in liters per second. Change of units will need corresponding change in the value of constant 102
H in m is the head of pumping (or of fans). This is total of Static head and frictional head and also head-equivalent of pressure, if any in the vessel where the pump has to deliver. (This is commonly the case in boiler feed pumping)
102 is the constant as mentioned above. The value will change if units for flow-rate and head are not l/s and m.
When pumping viscous fluids, one needs to take into account viscosity correction factors. To make the formula more general I would put it as -
Power (kW) = (Rho) * (Q in l/s) / (Visc. Corr. for Q) * (H in m) / (Vis. corr. for H) / 102 / (Pump Efficiency) / (Visc. Corr. for Pump Effy.) / (Motor Efficiency)
As can be seen Viscosity correction factors are to be taken into account for Q, H and also for Pump Efficiency.
Hope, this helps.
S. L. Abhyankar <engg.sla@gmail.com>, <pumpquest@gmail.com>
01-19-2014 11:11 PM
Top #9
Malcolm Heys
01-19-2014 11:11 PM
I've tried "fluid power" into google and it seems to be a term for the power derived from fluids (including turbine efficiencies) rather than the power to the fluid. Hydraulic power, I agree, is not universal and could be misunderstood as it may only apply to liquids. I've tried both terms with suppliers and I would like to say "fluid power is a known term, you can google it" which doesn't work.

I still like the formula below because there are no correction factors required.
Interestingly, there is also no density [kg/m3].

So, I'm surpised the formula from S. L. Abhyankar includes density. I may be wrong.
01-20-2014 01:32 AM
Top #10
Roumen Naidenov MBA
01-20-2014 01:32 AM

Hydraulic power is a recognized term for pumps. I would like to add some precisions:
* if you talk about pumps, you need to take into account the total head (static head + friction head), not only the static head. The pump needs to give enough energy to the fluid to compensate not only the difference in height but the friction losses in the pipes.

* the density shall be taken into account in the hydraulic power calculation. I guess in the last formula you cite (Power[W] = flow [m3/s]x Static pressure [Pa] ) it is understood the liquid is water. Most of the technical data sheets of pumps (curves, etc.) are done for water.

* S. L. Abhyankar is correct, viscosity shall also be taken into account if you pump liquids with viscosity different than the one of water.

* there are also non-Newtonian liquids (paper pulp, ketchup, mayonnaise...) whose properties modify the pumps characteristics/curves. To simplify, non-Newtonian liquids are the one whose viscosity changes not only with temperature, but with shear stress as well.

At the end, the formula of the Hydraulic power for water will be:

Pi = rho (density) x g (acceleration rate of Earth) x Hst (static head) + Hi (friction losses) x Q (flow rate)

(for viscous fluids cf. the corrections done by S. L. Abhyankar)

This is the pure instantaneous hydraulic power needed, in order to define the real energy needed, it shall be divided by the pump and the motor efficiency (and additionally by the efficiency of the transmission or the variable speed drive if there are such).

For viscous liquids the corrections are somehow more complicated, you can have a look on the KSB method or the Hydraulic Institute method. Very little technical info is available to adapt the pumps for Non-Newtonian liquids, I know only the Technical Data Sheets of TAPI as a source for the paper pulp.
01-20-2014 04:09 AM
Top #11
Johnny Qing Sheng Ke
01-20-2014 04:09 AM
Power is the energy used in a time unit. To lift a object (e.g a water block )to a certain height (TDH= static head+ friction losses+minor losses), the energy consumed is : Mass (in kg) X g ( gravity constant) X Height ( in meters) = Joule. Divide the energy with time, you then have Power (Watt=Joule/second). So for a liquid pumping problem, you hydraulic power : Q ( Mass/ unit time) X g X TDH. If the liquid is water : 1 liter water = 1 kg Mass. So you can use : Q (liter/second)X g X TDH to calculate hydraulic power
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