## Restrictors in Series or in Parallel

### IN SERIES

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 below may be used when the Lohm rates of the applicable restrictors are known.

### Two Restrictors

The chart below 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 1: 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.

Swipe to the right for more table information

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

### Two Restrictors – General

The following will allow solutions to be obtained for 2 restrictor problems even when Lohm or pressure ratios are off – scale:

1. When Lohm ratio is less than 0.1, then P2 = P1.
2. When Lohm ratio is less than 8.0, then solution for pressure ratio greater than 10, is the same as at 10.
3. When Lohm ratio is greater than 1.5, then solution at high values of pressure ratio is such that ratio P2 / P1 is equal to the reciprocal of the Lohm ratio.

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 above. 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 ):

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 2: Find the intermediate pressure in the example problem above with more precision.

Swipe to the right for more table information

 L1 = 2000 Lo. L2 = 500 Lo.

P1 = 72 psia, P2 = 20 psia, P3 = 14.7 psia
P1 / P2 = 72 / 20 = 3.60 (Sonic)
P2 / P3 = 20 / 14.7 = 1.36 (Subsonic)

EXAMPLE 3: Find the intermediate pressure between two restrictors with an upstream pressure of 30 psia, exhausting to atmosphere at 14.7 psia.

Swipe to the right for more table information

 L1 = 1500 Lo. L2 = 1500 Lo.

Use solution procedure from example 1, above, 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)

### IN PARALLEL

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: