Page 138 - 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-in. 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 '\Veymouth 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:
(2-105)

0
where L 1 = the equivalent length of any pipe of length L 2 and d = [(7.981) 813 + (10.136) 813]318 = 11.9-in.
diameter, d2, in terms of diameter, d 1.
This simplifies the system to one section 6 miles long of
(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 d 1 = the equivalent diameter of any pipe of a given
diameter, d2, and length, L 2, 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) 533 = 2.58 miles

(2-107)
Total length of 10-in. pipe to use in calculating capaci-
ty is 19 + 2.58 = 21.58 miles.
where d., 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 dj, d 2 ... and dn- Lines of same length. line systems may be analyzed, paralleled, cross-tied, etc.
This value of d., may be used directly in the Weymouth
formula.
Example 2-15: Parallel System: Fraction Paralleled
Example 2-13: Series System
Determine the portion ofa 30-mile, 18-in. (17.124-in.
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 14-in. (13.25-in. l.D.) connected to 3 miles of 10- I.D.) pipe to raise the total system capacity 1.5 times the
in. (10.136-in. I.D.) connected to 12 miles of 8-in (7.981- existing rate, keeping the system inlet and outlet condi-
in. I.D.). tions the same.
Select 10-in. as the base reference size.
The five-mile section of 14-in. pipe is equivalent to: ( q da I q db ) 2 - 1
x= --------- 1 (2- 108)
1
L 1 = 5(10.136/13.25) 533 == 1.195 milesoflO-in.

The 12 mile section of 8-in. is equivalent to:
For this example, qdb = 1.5 qda
L1 == 12(10.136/7.981) � 3 = 42.8 miles of 10-in.
5·
Total equivalent length of line to use in calculations is: (1/1.5) 2 - 1
x= = 0.683
[ (19.0(VJ\124)2 667 ]
1.195 + 3.0 + 42.8 = 46.995 miles of 10-in. (10.136-in. I.D.). [1 + ]2 - 1

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.