How to Calculate Flow Resistance for Gases

GAS FLOW – LOHM RATE FORMULA

The Lohm Laws extend the definition of Lohms for gas flow at any pressure and temperature, and with any gas. The formulas work well for all gases because they are corrected for the specific gas and for the flow conditions caused by the compressibility of gases due to pressure.

A 100 Lohm restriction will permit a flow of 250 standard liters per minute of nitrogen at a temperature of 59°F, and an upstream pressure of 90 psia discharging to the atmosphere.

The steps for using the gas Lohm rate formula are below. For more information about sonic and subsonic flow conditions, refer to Restrictions in Series or in Parallel.

Sonic condition

i.e. P₁/P₂ ≥ 1.9

Subsonic condition

i.e. P₁/P₂ < 1.9

Where:

L = Lohm rate (Lohms)

K = Gas flow – Units Constant (see volumetric and gravimetric flow units in tables below)

f_T = Temperature correction factor (see “Temperature Correction Factor f_T” chart below)

P₁ = Upstream absolute pressure

P₂ = Downstream absolute pressure

Q = Gas flow rate

∆P = P₁ – P₂

Steps for using the gas Lohm Rate formula:

1. Compute the P₁ / P₂ pressure ratio.
2. Select the correct formula for the flow condition.
3. Look up the value of “K” for the gas.
4. Determine the temperature correction factor, “f_T“.
5. Use the formula to solve for the unknown.

TEMPERATURE CORRECTION FACTOR f_T

GAS FLOW – UNITS CONSTANT “K”

To eliminate the need to convert pressure and flow parameters into specific units such as psia and standard liters per minute, the tables below list the values of the units constant “K”, which is used in the gas flow Lohm formulas. Depending on your application, it may be more appropriate to use the volumetric or gravimetric flow units.

REFERENCE TABLE

REFERENCE TABLE

Gravimetric Example:

If a Lohm rate is being calculated for an Argon application using pounds per hour and bar as the flow and pressure units, the unit constant is 792.

GAS FLOW – EXAMPLE

EXAMPLE: What restriction will permit a flow of 1.00 SLPM of nitrogen at a temperature of 90°F with a supply pressure of 5 psig discharging to the atmosphere?

K = 276 (see table above)

T₁ = 90°F, f_T = 0.98 (see chart above)

P₁ = 5.0 + 14.7 = 19.7 psia, P₂ = 14.7 psia

P₁/P2 = 19.7/14.7 = 1.34 (subsonic)

∆P = 5.0 psid

Q = 1.00 SLPM