Fluid_Flow_Rules_of_Thumb_for_Chemical

20

Rules of Thumb for Chemical Engineers

Equivalent Lengths for Multiple Lines Based on Panhandle A

dl, d2, d3 & dn - internal diameter of individual line corresponding to lengths L1, L2, L3 & Ln

Condition I. A single pipe line which consists of two or more dif- ferent diameter lines. LE -- equivalent length L1, L2 , . . . Ln - length of each diameter D1, D2 , . . . Dn = internal diameter of each separate line corresponding to L1, L2 , . . . Ln DE = equivalent internal diameter IDE14"8539 "DE -'4"8539 ,-DE --4.8539 Let

1.8539

2.6182

dE 2.6182+ d22.6182+d326182+. . . dn2"6182

LE - L l [ d 1

i]18539

90 9

2.6182

dE

2.6182

2.6182

2.6182

E Ln d12.6182-k-d2

+d3

+. . .dn

+. . . gnl-~nnI

t e - gl[_-~-ij

+ L2 [-~2 ]

when L1 - length of unlooped section

L2 - length of single looped section L3 - length of double looped section dE - dl - d2

Example. A single pipe line, 100 miles in length con- sists of 10 miles 10~4-in. OD; 40 miles 123/4-in. OD and 50 miles of 22-in. OD lines. Find equivalent length (LE) in terms of 22-in. OD pipe.

then"

.8539

i ]1 LE - L1 + 0.27664 L2 + L3 2dl 2"6182-k-d326182 d12.6182

L- - 50+E 40[ 215 -'4"8 3912.;51 +1 01215 -'4"853910.;5J

= 50 + 614 + 364 = 1,028 miles equivalent length of 22-in. OD

when dE- d l - d2 - d3 then LE- L1 + 0.27664 L2 + 0.1305 L3

Example. A multiple system consisting of a 15 mile section of 3-85/8-in. OD lines and 1-103/4-in. OD line, and a 30 mile section of 2-85/8-in. lines and 1-103/4-in. OD line. Find the equivalent length in terms of single 12-in. ID line.

Condition II. A multiple pipe line system consisting of two or more parallel lines of different diameters and different lengths. LE= equivalent length L1, L2, L3, . . . Ln- length of various looped sections dl, d2, d3, 9 9 9 = internal diameter of the individ- ual line corresponding to length L1, L2, L3 82 Ln i dE2.6182 ]1.8539 LE -- L1 dl2"6182+ d226182+d326182-t-... dn2"6182 Let

122.6182

11"8539

LE -- 15 3(7.981)2.6182+ 10.022"6182 I 22.6182

]1.8539

+ 30 2(7.981)2.16182 + 10.022"6182 = 5 . 9 + 18.1 = 24.0 miles equivalent of 12-in. ID pipe

0 9 9

dE2.6182

11"8539

Example. A multiple system consisting of a single 12-in. ID line 5 miles in length and a 30 mile section of 3-12-in. ID lines. Find equivalent length in terms of a single 12-in. ID line.

Ln dl2"6182-k-8226182+d326182+. . . dn2"6182

LE - - equivalent length L1, L2, L3 82 Ln- length of various looped sections

Let