Page 141 - APPLIED PROCESS DESIGN FOR CHEMICAL AND PETROCHEMICAL PLANTS, Volume 1, 3rd Edition

Fluid Flow 127

or W,hyr/G = 19,494 (1.017) (4.86)/58,482 = 1.641

G/h = 58,482/1.017 = 57,500
(2 - 122)
Reading Figure 2-40 type flow pattern is probably
annular, but could be wave or dispersed, depending
on many undefined and unknown conditions.

(b) For the Panha-Pde equation, Baker [33] summarizes: 2. Liquid Pressure drop

APL = 3.36 fLW (1W)/d5p (2-1 24)

Determine & for %in. pipe:
From Figure 2-11; E/d = 0.0006 for steel pipe
(2-123)
1000
v= = 0.Q86 ft / sec
where E (Panhanldle) = 0.9/@GTT1.077 63 (3600) (0.0513)
ke = 1 cp/1488 = 0.000672 lbs/ft sec
D = 3.068/12 = 0.2557 ft
p = 63.0
A liquid-vapor mixture is to Row in a line having 358 = D vp/y, = 0.2557 (0.086) (63.0)/0.000672
feet of level pipe and three vertical rises of 10 feet each & = 2060 (this is borderline, and in critical region)
plus one vertical rise of 50 feet. Evaluate the type of flow
and expected pressure drop. Reading Figure 2-3, approximate f = 0.0576
Vapor = 3,000 lbs/hr Substituting:
Liquid = 1,000 lbs/hr
ensity: Ibs/cu ft; Vapor = 0.077 APL = 3.36 (lo-‘) (0.0576) (1000)* (1 f00t)/(3.068)~ (63)
Liquid = 63.0 = 1.1 (1O-j) psi/foot
Viscosity, centipoise; Vapor = 0.00127
Liquid = 1.0 Gas pressure drop
Surface tension liquid = I3 dynes/cm
Pipe to be schedule 40, steel v= 3000 = 211 ft/sec
Use maximum allowable vapor velocity = 15,000 ft/min. 0.077 (3600) (0.0513)

pe = 0.00127/1488 = 0.000000854 ?.bs/ft sec
etermine probable types of flow:
I& = Dv p/ke = 0.2537 (211) (0.077)/0.000000854
= 4,900,000

h= 1.017 Reading Figure 2-3, f = 0.0175

APG = 3.36 (lo-‘) (0.0175) (1 foot) (3000)n/(3.068)5(0.077)
= 0.0254 psi/foot

xy = 4.86

Try 3-in. pipe, 3.068411. I.D., cross-section area =
0.0513 sqq. ft. 4. For annular flow:

Wm = 1,000/0.0513 = 19,494 Ibs/hr (sq ft) @Gm = (4.8 - 0.3125d) X0.343 - 0.021d
= [4.8 - 0.3125 (3.068)] (2.10 X 10-‘)0.343-0.021 (3.068)
G = 3,000/0.0513 = 58,482 Ibs/hr (sq ft) = 1.31