Lohm Laws for Liquids – How to Calculate Flow Resistance for Liquids

A Simplified System of Defining Fluid Resistance

Over the years, The Lee Company has developed the Lohm system for defining and measuring resistance to fluid flow. Just as the "ohm" defines electrical resistance, the "Lohm" or "liquid ohm" can be used as a measure of fluid resistance.

The Lohm is defined such that 1 Lohm will flow 100 gallons per minute of water with a pressure drop of 25 psi at a temperature of 80°F. Since resistance is inversely proportional to flow, by definition:



By using Lohms, one can specify performance without concern for coefficients for discharge, passageway geometrics, physical dimensions or tolerances. The resistance of any flow can be expressed in Lohms and confirmed by actual flow tests.

Lohm Laws generalize the Lohm definition and allow the system designer to specify Lohm requirements for particular application based on the desired pressures and flow rates. Click here for a graph relating Lohms to hole diameter and flow coefficient, C_v.

Liquid Flow Lohm Laws (Gas Flow Lohm Laws can be found on another page.)

The Lohm Laws predict the actual performance of fluidic devices beyond the definition conditions of water at 25 psid and 80°F. In Liquid Flow several variables must be related, including:

I = Flow rate
H = Differential pressure
V = Viscosity correction factor. V factors compensate for the interaction of viscosity and device geometry and are unique to each class of device. See "V" factors graph for typical Lee orifices. Use 1.0 for water @ 80°F

S = Specific gravity. Use 1.0 for water @ 80°F
K = A constant to take care of units of measure. Use 20 for psi and gpm.

The Lohm Law for Liquid Flow is:



also:

d = orifice diameter (inches)
Cd = coefficient of discharge
Cv = flow coefficient

For special flow requirements, The Lee Company can determine the required Lohm rating.

Units Constant K

The Units Constant K eliminates the need to convert pressure and flow parameters to specific units such as PSI and GPM.

Example:
An orifice must flow 43in.3/min. of water at a head of 300 kPa. What Lohm rate is required?

Solution:
First, the appropriate K is selected from the table: K = 1760.
Second, the Lohm Formula is solved using the K value:


K for various flow and pressure units :

	psi	bar	kPa	N/m2	kg/cm2	ft.H2O	mm/Hg
gpm	20	76.2	7.62	.24	75.4	13.2	2.78
L/min.	75.7	288	28.8	.91	285	50	10.5
mL/min.	75 700	288 000	28 800	911	285 000	50 000	10 500
in3/min.	4620	17 600	1760	55.6	17 400	3040	642
ft3/min.	2.67	10.2	1.02	.032	10	1.76	.372