Page 136 - APPLIED PROCESS DESIGN FOR CHEMICAL AND PETROCHEMICAL PLANTS, Volume 1, 3rd Edition

122 Applied Process Design for Chemical and Petrochemical Plants

Complex Pipe Systems Handling Natural Example 2-14: Looped System
(or similar) Gas
Determine the equivalent length of 25 miles of 10-in.
The method suggested in the Bureau of Mines Mono- (10.136-h. I.D.) which has a parallel loop of 6 miles of 8-
graph No. 6 [43] has found wide usage, and is outlined in. (7.981-in. I.D.) pipe tied in near the midsection of the
here using the Weymouth Formula as a base. 10-in. line.
Figure the looped section as parallel lines with 6 miles
1. Equivalent lengths of pipe for different diameters
of 8-in. and 6 miles of 10-in. The equivalent diameter for
one line with the same carrying capacity is:
L1 = Lz (dl/dz)16/3 (2-105)

where L1 = the equivalent length of any pipe of length L, and do = [(7.981)8/3 + (10.136)8/3]3/8 = 11.9-in.
diameter, dz, in terms of diameter, dl.
This simplifies the system to one section 6 miles long of
dl = d2 (L1/Lz)3/16 (2-106) 11.9-in. I.D. (equivalent) pipe, plus one section of 25
minus 6, or 19 miles of 10-in. (10.136-in. I.D.) pipe.
where dl = the equivalent diameter of any pipe of a given
diameter, d2, and length, L2, in terms of any other Now convert the 11.9-in. pipe to a length equivalent to
length, L1. the 10-in. diameter.

2. Equivalent diameters of pipe for parallel lines L1 = 6(10.136/11.9)5.33 = 2.58 miles

do = (dlsI3 + dz8j3 . . . + d,8/3)3/s (2-107)
Total length of 10-in. pipe to use in calculating capaci-
ty is 19 + 2.58 = 21.58 miles.
where do is the diameter of a single line with the same
delivery capacity as that of the individual parallel lines of By the principles outlined in the examples, gas pipe
diameters dl, d2 . . . and d,. Lines of same length. line systems may be analyzed, paralleled, cross-tied, etc.
This value of do may be used directly in the Weymouth
formula. Example 2-15: Parallel System: Fraction Paralleled

Example 2-13: Series System
Determine the portion of a 30-mile, 18-in. (17.124in.
I.D.) line which must be paralleled with 20-in. (19.00-in.
Determine the equivalent length of a series of lines: 5
miles of 14in. (13.25-in. I.D.) connected to 3 miles of 10- I.D.) pipe to raise the total system capacity 1.5 times the
existing rate, keeping the system inlet and outlet condi-
in. (10.136-in. I.D.) connected to 12 miles of 8-in (7.981- tions the same.
in. I.D.).
Select 10-in. as the base reference size.
The five-mile section of 14in. pipe is equivalent to:
(2 - 108)
L1 = 5(10.136/13.25)5,33 1.195 miles of 10-in.
=

The 12 mile section of 8-in. is equivalent to:
For this example, qdb = 1.5 qda
L1 = 12(10.136/7.981)5.33 = 42.8 miles of 10-in.

Total equivalent length of line to use in calculations is: (1/1.5)n - 1 = 0.683
1
l2
1.195 + 3.0 + 42.8 = 46.995 miles of 10-in. (10.136in. I.D.). [l + (19.00/17.124)'~667 - I1
An alternate procedure is to calculate (1) the pressure
drop series-wise one section of the line at a time, or (2) This means 68.3 percent of the 30 miles must be paral-
capacity for a fixed inlet pressure, series-wise. lel with the new 19-in. I.D. pipe.