Microhidráulica Industrial

# 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 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) 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 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) 