Page 143 - APPLIED PROCESS DESIGN FOR CHEMICAL AND PETROCHEMICAL PLANTS, Volume 1, 3rd Edition

Fluid Flow 127

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

(qdM.65 )2 LS" TZ[ r, J . G/1c = 58,482/1.017 = 57,500
LiP= '_b_ (2-122)
TP 20 000 d 5 p E 2
� avg Reading Figure 2-40 type flow pattern is probably
annular, but could be wave or dispersed, depending
where E = l/i:)lcn on many undefined and unknown conditions.

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

(2-124)
[ � 11.07881 [ ; 0.5'.l94
q . = 0.43587 � - p 2 2
dJ<J.fo P ZTL
s m Determine R., for 3-in. pipe:
From Figure 2-11; £/d = 0.0006 for steel pipe
[ ;: ',:;� l (E) (2 123)
v = lOOO = 0.086 ft I sec
where E (Panhandle) = 0.9/<Jlcrr1.o 77 63 (3600) (0.0513)
u, = l cp/1488 = 0.000672 lbs/ft sec
Example 2-16: Two-phase Flow D = 3.068/12 = 0.2557 ft
p = 63.0
A liquid-vapor mixture is to flow in a line having 358 R 0 = D vp/µe = 0.2557 (0.086) (63.0)/0.000672
feet of level pipe and three vertical rises of 10 feet each R,, = 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
Density: lbs/cu ft; Vapor = 0.077 @L = 3.36 (10- 6) (0.0576) (1000)2 (1 foot)/(3.068) (63)
5
Liquid = 63.0 = l.l (10- 5) psi/foot
Viscosity, centipoise; Vapor = 0.00127
Liquid = 1.0 Gas pressure drop
Surface tension liquid = 15 dynes/ cm
Pipe to be schedule 40, steel v = 3 000 = 211 ft/ sec
Use maximum allowable vapor velocity= 15,000 ft/min. 0.077 (3600) (0.0513)

u, = 0.00127/1488 = 0.000000854 Ibs/ft sec
1. Determine probable types of flow:
R, = Dv p/µe = 0.2557 (211) (0.077)/0.000000854
r;·
77x63 = 4,900,000
'A= [(o /0.075)(0 /62.3)Jf15 = .o) � o
I
0.0
' g L [(o.o75 62.3 �
1c = l.017 Reading Figure 2-3, f = (l.0175
6
5
( � !11.0( � !: � r] 1/3 LiPc = 3.36 (10- )(0.0175)(1 foot)(3000)2/(3.068) (0.077)
= 0.0254 psi/foot
3
1j1 = (73/y)[µ L (62.3/pL )2 ]11 =
5
2
\JI= 4.86 3. X = (LiPdLiPg) 1 1 2 = (l.l (10- )/2.54 x 10- )11 2
= 2.10 (10- 2)
Try 3-in. pipe, 3.068-in. l.D., cross-section area =
0.0513 sq. ft. 4. For annular flow:
Wm= 1,000/0.0513 = 19,494 lbs/hr (sq ft) <[>GTI = ( 4.8 - 0.3 l 25d) X0.343 - 0.02ld
= [4.8 - 0.3125 (3.068) l (2.10 x 10··2)0.343 - 0021 (3.068)
G = 3,000/0.0513 = 58,482 lbs/hr (sq ft) = 1.31