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.
|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|
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.|
To eliminate the need to convert pressure and flow parameters into specific units such as “psia" and “std L/min.", the table below lists values of the Units Constant “K", which is used in the Gas Flow Lohm Formulas:
|VOLUMETRIC FLOW UNITS|
|Abs. Pres||psia||bar||kPa||mm. Hg|
|H2||1030||36.3||62 700||14 900||526||149||19 900|
|He||771||27.2||47 100||11 200||395||112||14 900|
|Neon||343||12.1||20 900||4 980||176||49.8||6 640|
|Nat. Gas||319||11.3||19 400||4 620||163||46.2||6 160|
|N2||276||9.73||16 800||4 000||141||40.0||5 330|
|CO||274||9.69||16 700||3 980||141||39.8||5 300|
|Air||271||9.56||16 500||3 930||139||39.3||5 230|
|Ethane||251||8.86||15 300||3 640||129||364||4 850|
|O2||257||9.08||15 700||3 730||132||37.3||4 970|
|Argon||245||8.65||14 900||3 550||125||35.5||4 730|
|CO2||213||7.52||13 000||3 090||109||30.9||4 110|
|N2O||214||7.56||13 100||3 100||110||31.0||4 140|
|SO2||176||6.21||10 700||2 550||90.1||25.5||3 400|
|Freon-12||123||4.34||7 510||1 780||63.0||17.8||2 380|
|GRAVIMETRIC FLOW UNITS|
|Abs. Pres||psia||bar||kPa||mm. Hg|
When selecting components for use in a gas system, certain factors must be considered which arise only because of the compressibility of the gaseous medium. The nature of gas compressibility is defined by the following two rules.
|Boyle's Law –||The pressure and specific volume of a gas are inversely proportional to each other under conditions of constant temperature.|
|Charles' Law –||The pressure and temperature of a gas are directly proportional to each other when the volume is held constant, and the volume and temperature are directly proportional when the pressure is held constant.|
Thus, a gas will expand to fill any container, and pressure and temperature will adjust to values consistent with the above rules. Gas flowing through valves and restrictors will be subject to an increasing specific volume as pressure drops take place, and temperatures will change as determined by the Joule-Thompson effect.
The combination of the above rules forms the basis for the “Equation of State" for perfect gases. This allows either pressure, temperature, or volume to be calculated for a known quantity of gas when the other two variables are known.
|i.e.||p V = m R T||(click here for values of the Gas Constant, R)|
In general, the following comments apply to gas flow.
The Rule of Forbidden Signals:*
“The effect of pressure changes produced by a body moving at a speed faster than the speed of sound cannot reach points ahead of the body."
This rule can be applied to pneumatic flow restrictors where the body is not moving, but the flow velocity relative to the body can reach, or exceed, the speed of sound. Whenever the downstream pressure is low enough to produce Mach 1 at the restrictor throat, any effect of changes in the downstream pressure cannot reach points upstream of the throat. Thus, flow rate will be independent of downstream pressure. This situation applies to a single orifice restrictor flowing air when the overall pressure ratio exceeds 1.89/1.
*von Kármán, Jour. Aero. Sci., Vol. 14, No. 7 (1947)
Gas flow is a function of upstream absolute pressure, and of the ratio of upstream to downstream pressures. Lohm testing done at The Lee Company is performed at an upstream pressure which is high enough so that downstream pressure does not affect the flow rate. To accurately determine the upstream absolute pressure, it is necessary to measure atmospheric pressure with a suitable barometer. This measurement will normally be in units of in. Hg, while the gauge pressure reading is in units of psig. Thus, the barometer reading must be converted to psia, and added to the gauge reading to get the value of pressure in psia.
Pres. (psia) = Pres. (psig) + 0.4912 x Pres. (in. Hg)
EXAMPLE: What single-orifice restriction will permit a flow of 2.00 std L/min. of nitrogen at 70°F, with supply pressure at 10 psig, discharging to an atmospheric pressure of 29.5 in. Hg.
It is frequently convenient to express gas flow in terms of flow at standard conditions. This is useful for calculation purposes, or for application to flow measuring instruments.
|T||= Gas temperature, °R = 460 + °F|
|P||= Gas pressure, psia|
|ACFM||= Gas flow, actual cubic feet/minute|
|SCFM||= Gas flow, standard cubic feet/minute|
EXAMPLE: What is SCFM corresponding to 0.032 ACFM at 300 psia and at 240°F?
When multiple orifices appear in series or when a restrictor has several stages, there is a non-uniform distribution of the overall pressure drop through the restrictor. Click here for additional discussion of series gas flow.
The effect of the above flow behavior is that the gas flow rate of a multi-orifice device is higher than would be expected from a single-orifice device of the same lohm rate, and at the same pressure conditions. This characteristic is reflected in the flow factor, “ fM", which reaches a maximum value of 1.3 at a pressure ratio of 3/1. See the graph on page 171 for values of “ fM", at any pressure ratio for multi-orifice restrictors.
EXAMPLE: What multi-orifice restriction will permit a flow of 0.5 std L/min. of hydrogen at 70°F, with supply pressure at 40 psig, discharging to atmosphere.
for P1 / P2 > 3, use fM = 1.3
When gas flow passes through orifices in series, the pressure drops are not evenly distributed. This is caused by the compressibility of the gas, and generally results in higher pressure drops at the downstream orifices. Thus, it becomes difficult to calculate the intermediate pressure between series restrictors flowing gas without using a trial and error process. To simplify this calculation, the chart on the following page may be used when the Lohm rates of the applicable restrictors are known.
The chart on the adjacent page solves for the absolute pressure between two orifices as a percentage of the supply pressure. To solve a problem, simply follow the graph line corresponding to the Lohm ratio, L1/L2 , until it crosses the overall pressure ratio, P1/P3. Then read horizontally across to the left hand scale to obtain the value of P2 as a percentage of the upstream absolute pressure, P1.
EXAMPLE: Find the intermediate pressure between two restrictors with an upstream pressure 72 psia, exhausting to atmosphere at 14.7 psia.
|L1 = 2000 Lo.||L2 = 500 Lo.|
Calculate the Lohm ratio: L1/L2 = 2000/500 = 4.0
Calculate the overall pressure ratio: P1/P3 = 72.0/14.7 = 4.9
Read 28% from left hand scale of graph.
The upstream pressure is known, thus:
|P2 = 0.28 x 72.0 = 20 psia|
The following will allow solutions to be obtained for 2 restrictor problems even when Lohm or pressure ratios are off – scale:
The following formulas provide solutions to series gas flow problems which must be solved with more precision than can be obtained by use of the graph here. In each case, the graph may be used to determine whether or not each restrictor has a high enough pressure ratio (i.e. P1 / P2 ≥ 1.9) to be in the sonic region.
1.) L1 and L2 are both sonic (L1 > L2 ):
2.) L1 is subsonic, and L2 is sonic (L1 ≠ L2 ):
3.) L1 is subsonic, and L2 is sonic (L1 = L2 ):
P2 = 0.8 x P1
4.) L1 is sonic, and L2 is subsonic (L1 > L2 ):
5.) L1 is subsonic, and L2 is subsonic (L1 ≠ L2 ):
6.) L1 is subsonic, and L2 is subsonic (L1 = L2 ):
EXAMPLE: Find the intermediate pressure in the example problem here with more precision.
EXAMPLE: Find the intermediate pressure between two restrictors with an upstream pressure of 30 psia, exhausting to atmosphere at 14.7 psia.
|L1 = 1500 Lo.||L2 = 1500 Lo.|
Use solution procedure from here to determine approximate value of intermediate pressure, P2 :
L1 / L2 = 1500 / 1500 = 1.0 , P1 / P3 = 30.0 / 14.7 = 2.04
P2 = 0.81 x 30.0 = 24 psia. (approx.)
P1 / P2 = 30.0 / 24.0 = 1.25 , P2 / P3 = 24.0 / 14.7 = 1.63
(L1 and L2 are both subsonic)
For parallel flow, the total Lohm rating is:
Note that this relationship is identical to that for hydraulic flow, and to the electrical equation.
Therefore, LT = 970 Lohms