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Microhidráulica Industrial

Introduction

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 region and incompressibility of low pressure gases.

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 atmosphere.

NOMENCLATURE
L = Lohms
K = Units Constant – Gas (click here)
fT = Temperature correction factor (click here)
P1 = Upstream absolute pressure
P2 = Downstream absolute pressure
Q = Gas flow rate
ΔP = P1 – P2
  1. Compute the P1/P2 pressure ratio.
  2. Select the correct formula for the flow region.
  3. Look up the value of “K” for the gas.
  4. Determine the temperature correction factor, “ fT”.
  5. Use the formula to solve for the unknown.

EXAMPLE: What restriction will permit a flow of 1.00 std L/min. of nitrogen at 90°F, with supply pressure at 5 psig, discharging to atmosphere?

K = 276 (click here)
T1 = 90°F, fT = 0.98 (see below)
P1 = 5.0 + 14.7 = 19.7 psia, P2 = 14.7 psia
P1/P2 = 19.7/14.7 = 1.34 (subsonic)
ΔP = 5.0 psid
Q = 1.00 std L/min.
L
= = 4640 Lohms

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 region and incompressibility of low pressure gases.

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 atmosphere.

NOMENCLATURE
L = Lohms
K = Units Constant – Gas (click here)
fT = Temperature correction factor (click here)
P1 = Upstream absolute pressure
P2 = Downstream absolute pressure
Q = Gas flow rate
ΔP = P1 – P2
  1. Compute the P1/P2 pressure ratio.
  2. Select the correct formula for the flow region.
  3. Look up the value of “K” for the gas.
  4. Determine the temperature correction factor, “ fT”.
  5. Use the formula to solve for the unknown.

EXAMPLE: What restriction will permit a flow of 1.00 std L/min. of nitrogen at 90°F, with supply pressure at 5 psig, discharging to atmosphere?

K = 276 (click here)
T1 = 90°F, fT = 0.98 (see below)
P1 = 5.0 + 14.7 = 19.7 psia, P2 = 14.7 psia
P1/P2 = 19.7/14.7 = 1.34 (subsonic)
ΔP = 5.0 psid
Q = 1.00 std L/min.
L
= = 4640 Lohms