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) 
P_{1}  = Upstream absolute pressure 
P_{2}  = Downstream absolute pressure 
Q  = Gas flow rate 
ΔP  = P_{1} – P_{2} 
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)  

T_{1}  = 90°F, fT = 0.98 (see below)  
P_{1}  = 5.0 + 14.7 = 19.7 psia, P2 = 14.7 psia  
P_{1}/P_{2}  = 19.7/14.7 = 1.34 (subsonic)  
ΔP  = 5.0 psid  
Q  = 1.00 std L/min.  
L 

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  
Flow  SLPM  SCFM  in^{3}/min  SLPM  SCFM  SLPM  mL/min 
H_{2}  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 
N_{2}  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 
O_{2}  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 
CO_{2}  213  7.52  13 000  3 090  109  30.9  4 110 
N_{2}O  214  7.56  13 100  3 100  110  31.0  4 140 
SO_{2}  176  6.21  10 700  2 550  90.1  25.5  3 400 
Freon12  123  4.34  7 510  1 780  63.0  17.8  2 380 
GRAVIMETRIC FLOW UNITS  

Abs. Pres  psia  bar  kPa  mm. Hg  
Flow  PPH  lbm/s  kg/min  PPH  kg/min  kg/min  gm/min 
H_{2}  11.6  0.00322  0.0876  168  1.27  0.0127  1.69 
He  17.3  0.00479  0.131  250  1.89  0.0189  2.25 
Neon  38.7  0.0108  0.293  561  4.25  0.0425  5.66 
Nat. Gas  34.8  0.00966  0.263  505  3.82  0.0382  5.09 
N_{2}  43.2  0.0120  0.326  626  4.73  0.0473  6.31 
CO  43.0  0.0119  0.325  623  4.71  0.0471  6.28 
Air  43.8  0.0122  0.331  636  4.81  0.0481  6.41 
Ethane  42.2  0.0117  0.319  611  4.62  0.0462  6.16 
O_{2}  46.0  0.0128  0.348  667  5.04  0.0504  6.72 
Argon  54.6  0.0152  0.413  792  5.99  0.0599  7.99 
CO_{2}  46.0  0.0145  0.396  759  5.74  0.0574  7.65 
N_{2}O  52.7  0.0146  0.398  764  5.77  0.0577  7.70 
SO_{2}  63.0  0.0175  0.476  914  6.91  0.0691  9.21 
Freon12  83.2  0.0231  0.629  1210  9.12  0.0912  12.2 
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 JouleThompson 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 singleorifice 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.
K  = 276 (click here)  

T_{1}  = 70°F, fT = 1.00 (click here)  
P_{2}  = 0.4912 x 29.5 = 14.5 psia  
P_{1}  = 10.0 + 14.5 = 24.5 psia  
P_{1}/P_{2}  = 24.5/14.5 = 1.69 (subsonic)  
ΔP  = 24.5  14.5 = 10.0 psid  
Q  = 2.00 std L/min  
L 

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.
UNITS:
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?
SOLUTION:
When multiple orifices appear in series or when a restrictor has several stages, there is a nonuniform 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 multiorifice device is higher than would be expected from a singleorifice 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 multiorifice restrictors.
EXAMPLE: What multiorifice 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.
K  = 1030 (click here) 

T_{1}  = 70°F, fT = 1.0 (click here) 
P_{1}  = 40.0 + 14.7 = 54.7 psia 
P_{2}  = 14.7 psia 
P_{1}/P_{2}  = 54.7 / 14.7 = 3.72 
f_{M}  = 1.30 (click here) 
Q  = 0.50 std L/min 
Sonic flow:
for P_{1} / P_{2} > 3, use f_{M} = 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. P_{1} / P_{2} ≥ 1.9) to be in the sonic region.
1.) L_{1} and L_{2} are both sonic (L_{1} > L_{2} ):
2.) L_{1} is subsonic, and L_{2} is sonic (L_{1} ≠ L_{2} ):
3.) L_{1} is subsonic, and L_{2} is sonic (L_{1} = L_{2} ):
P_{2} = 0.8 x P_{1}
4.) L_{1} is sonic, and L_{2} is subsonic (L_{1} > L_{2} ):
5.) L_{1} is subsonic, and L_{2} is subsonic (L_{1} ≠ L_{2} ):
6.) L_{1} is subsonic, and L_{2} is subsonic (L_{1} = L_{2} ):
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.
L_{1} = 1500 Lo.  L_{2} = 1500 Lo. 
Use solution procedure from here to determine approximate value of intermediate pressure, P_{2} :
L_{1} / L_{2} = 1500 / 1500 = 1.0 , P_{1} / P3 = 30.0 / 14.7 = 2.04
P_{2} = 0.81 x 30.0 = 24 psia. (approx.)
P_{1} / P_{2} = 30.0 / 24.0 = 1.25 , P_{2} / P3 = 24.0 / 14.7 = 1.63
(L_{1} and L_{2} 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.
EXAMPLE:
Therefore, L_{T} = 970 Lohms