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Hydraulic Power
Fluid Mechanics for Liquids

Hydraulic Power

Whenever there is flow through an orifice, there is power consumption (or loss) which is a function of the pressure drop and the flow rate. The following data is useful in calculating the hydraulic power requirements of a system.

\[\ \text{HP}=\frac{\text{H}\times\text{I}\ }{1714}\]

Where:

H = Pressure drop (psi)

I = Flow rate (GPM)

HP = Hydraulic power (horsepower)

The hydraulic power can also be expressed in another convenient form.

For Water Applications:

\[\text{HP}=\frac{0.0117\text{H}^{\frac{3}{2}}\ }{L}\quad\ \text{or}\quad\ \frac{0.0117\text{H}\ \sqrt{\text{H}}\ }{L}\]

Since 1 HP = 746 watts, the above formula can be:

\[\text{Watts}=\frac{8.70\text{H}^{\frac{3}{2}}\ }{L}\quad\ \text{or}\quad\ \frac{8.70\text{H}\ \sqrt{\text{H}}\ }{L}\]

The nomogram below shows this relationship.

For Liquid Applications:

\[\text{HP}=\frac{0.0117\text{HV}\ \sqrt{\frac{\text{H}\ }{\text{SG}\ }}\ }{L}\]

Where:

V = Viscosity compensation factor

SG = Specific gravity

L = Lohm rate (lohms)

Example

A Lee Chek® Valve such as CKRA1876005 with 400 Lohms will flow 0.35 GPM of water at 50 psid. At those conditions, what horsepower is lost?

\[\text{HP}=\frac{\text{H}\times\ \text{I}\ }{1714}=\frac{50\times\ 0.35\ }{1714}\]
\[\text{HP}=0.010\;\text{Horsepower}\]

 

NOMOGRAM FOR HYDRAULIC POWER – WATER APPLICATIONS