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

Fluid Flow 85

Table 2-4
Suggested Fluid Velocities in Pipe and Tubing: Liquids, Gases, and Vapors at low /moderate pressure to 50 psig and
50° to 100°F

The velocities are suggestive only and are to be used to approxi- The final line size should be such as to give an economical balance
mate line size as a starting point for pressure drop calculations. between pressure drop and reasonable velocity


Suggested Trial I Suggested Trial I
Fluid Velocity Pipe Material Fluid Velocity Pipe Material
Acetylene (Observe Sodium Hydroxide
pressure limitations) 4000 fpm Steel 0-30 Percent 6 fps Steel
Air, 0 to 30 psig 4000 fpm Steel 30-50 Percent 5 fps and
Ammonia 50-73 Percent 4 Nickel
Liquid 6 fps Steel Sodium Chloride Sol'n ,
Gas 6000 fpm Steel No Solids 5 fps Steel
Benzene 6 fps Steel With Solids (6Min.-·
Bromine I 15 Max.) Mone! or nickel
Liquid 4 fps Glass 7.5 fps
Gas 2000 fpm Glass Perchlorethylene 6 fps Steel
Calcium Chloride 4 fps Steel Steam
Carbon Tetrachloride 6 fps Steel 0-30 psi Saturated" 400Q-6000 fpm Steel
Chlorine (Dry) 30-150 psi Satu-
Liquid 5 fps Steel, Sch. 80 rated or super·
Gas 2000-5000 fpm Steel, Sch. 80 heated" 6000-10000 fpm
Chloroform 150 psi up
Liquid 6 fps Copper & Steel superheated 6500-15000 fpm
Gas I 2000 fpm Copper & Steel "Short lines 15,000 fpm
Ethylene Gas 6000 Ipm Steel (rnax.) I
Ethylene Dibromide 4 fps Glass Sulfuric Acid
Ethylene Dichloride 6 fps Steel 88-93 Percent 4 tps S. S.-316, Lead
Ethylene Glycol 6 fps Steel 93-100 Percent 4 fps Cast Iron & Steel,
Hydrogen 4000 fpm Steel Sch. 80
Hydrochloric Acid Sulfur Dioxide 4000 Iprn Steel
Liquid 5 fps Rubber Lined Styrene i 6 fps Steel
4000 fpm R. L., Saran, Trichlorethylene 6 fps Steel
Gas Ha veg Vinyl Chloride 6 fps Steel
Methyl Chloride Vinylidenc Chloride 6 fps Steel
Liquid 6 fp,s Steel \\later
Gas 4000 Iprn Steel Average service 3-8 (avg. 6) fps Steel
Natural Gas 6000 fpm Steel Boiler feed 4-12 fps Steel
Oils, lubricating 6 fps Steel Pump suction lines 1-5 fps Steel
Oxygen 1800 fpm Max. Steel (300 psig Max.) Maximum econorni-
(ambient temp.) 4000 fpm Type 304 SS cal (usual) 7-10 fps Steel
(Low temp.) Sea and brackish R. L., concrete,
Propylene Glycol .5 fps Steel water, lined pipe 5--8 fps 3 asphalt-line, saran-
Concrete 5-12 fps f (Min.) lined, transite

Note: R. L. = Rubber-lined steel.

From Figure 2-3, read, f = 0.0219 = fr 1 globe valve (open),�= 1: K = 340 fr= 340 (0.0219)
then, pipe only friction loss: = 7.446

2
hr = (f L/D; (v /2g) (2-2)
6 90° elbows, r/ d = 1.88/ 1.38 = l.36
D = pipe, l.D., in ft= 1.38/12 = 0.1150 fl

( 8 + 6 + 200) ( 4. 29) 2 K = 30 fT = 30 (0.0219) = 0.657
h =0.0219 ---------·
r 0.1150 (2)(32. ;2)
hr= 11.64 ft of kerosene flowing (pipe only)
For 6: 6 X 0.657 = 3.94
Loss through discharge fittings, valves, connections,
using K factors using Table 2-2:
2 check valves, swing, threaded, 100 fT = 100 (0.0219) 1 sharp edged entrance (sudden enlargement) = 1.0
= 2.19 = 4.38 1 gate valve, open,�= 1.0, K = 8 fT; K = 8 (0.0219)
(for 2) = 0.175

86 Applied Process Design for Chemical and Petrochemical Plants

Summation: Pressure drop = i:'.P = 0.00001799 Kp Q2/d 4 (2-52)
= 0.00001799 (52.48) ( 48.6)
K = [4.38 + 7.45* + 3.94 + 0.175* + l.O] = 16.941 (350)2/ ( 4.026) 4
i:'.P = 21.2 psi friction pressure loss only
*Threaded, from Table 2-2. (no elevation change)
For fittings:
Alternate Calculation Basis for Piping System Friction
Head Loss: Liquids
16.941 (4.292)
Lhen, h = Kv /2g = = 4.84 ft kerosene Pressure loss in a piping system (not including the
2
2(32. 2)
tanks, heat exchangers, distillation columns, etc.) is usu-
Total friction loss for discharge side pump due to fric- ally expressed in units of feet of flowing fluid, or the equiua-
tion: lent converted to pounds per square inch. Some published
pressure loss data is expressed as per 100 equioalent feet of
h = 11.64 + 4.84 = 16.48 ft fluid kerosene the size pipe being used or estimated.
hr= L'ip = 16.48/((2.31)/(0.81)) = 5.77 psi
Equivalent Feet Concept for Valves, Fittings, Etc.
Example 2-2: Laminar Flow Through Piping System
With pipe of any specified size as the basis, the total
A heavy weight oil, No. 5 fuel oil, is to be pumped footage of straight pipe in a system is just the measured
through 350 ft or existing 4-in. Schedule 40 pipe at 350 length (totaled).
gpm. Oil data: For fittings, valves, etc., in the same system, these can
be expressed as equivalent straight pipe, then added to the
Temperature = 100°F straight pipe described above, to arrive at a total equiva-
Viscosity = 150 cp lent straight length of pipe of the specific size in question.
Sp Gr= 0.78 = 48.6 lb/cu ft Figure 2-20 presents equivalent lengths of straight pipe
Pipe I.D. = 4.026 in. = 0.3355 ft (feet) for various pipe system components. For example,
a standard threaded 6-inch 90° elbow is equivalent to
Reynolds number= 50.6 Qp/(dµ) adding 17 feet of straight pipe to the system. This 17 feet
= 50.6 [ (350) ( 48.6) I ( 4.026) (150) l is additive to the lengths of nominal 6-inch straight pipe
= 1425 (2-50) in the system (dotted line). However, there is an impor-
tant consideration in the use of this chart, i.e., use only for
Flow is <2000, therefore, flow of viscous or laminar sys- threaded or screwed pipe/fittings, and only for sizes
tem consists or friction factor, fT, for 4-in. pipe = 0.017 under 2-inch nominal size. It is not practical in current
(Table 2-2). industry practice to thread a process or utility system
much greater in nominal diameter than 2 inches. For spe-
1 gate valve= K = 8 fr = 8 (0.017) = 0.136, (Table 2-2) cial situations, the larger sizes can be used, but from a
3 90° = K = 20 fr = 20 (0.017) = 0.345 handling standpoint, sizes greater than 3 inches or 4 inch-
1 90° = RID = 5; 5/D = 0.00045 (Figure 2-11); K = es are not practical.
0.1 (Figure 2-13A) For pipe sizes greater than 2 inches nominal, industry
1 pipe entrance to tank projecting inward, K = 0.78 practice is to weld the pipe and fittings into one continu-
(Table 2-2) ous system, and then use flanged or special bolted con-
For 350 ft pipe, K = f (L/D) = 0.0449 (350/0.3355) = nections for attaching the valves, orifices, and connec-
51.12 tions to vessels or other equipment: For special lethal,
[For f, see calculations below] high pressure, and steam power plant high tempera-
ture/high pressure utility systems, even the valves and
f = 64/Rc connections to vessels are welded into the system (See
f = 64/1425 = 0.0449 (2-17) ASME and Ai'\TSI Codes). For these situations of about 1 �-
inch to 2-inch nominal pipe size and larger, use Figure 2-
Total K values =51.12 + 0.78 + 0.1 + 0.136 + 0.345 = 21 to determine the equivalent pipe lengths for these fit-
52.48 tings, valves, etc. For example, a 45° welding elbow, or an
open 6-inch gate valve (see line on chart) have an equiva-
Velocity v, = 0.4.-08 Qld2 = 0.408 (350)/ ( 4.026) 2 lent length of 6-inch pipe of four feet (straight), which is
= 8.8 ft/sec (2-51) an addition to the actual straight pipe in the system. In

Fluid Flow 87













Gate Valve 3000
� Closed
,----'h Closed 2000
M Closed
Fully Open
1000
50
48
42
500
Angle Valve, Open 36
300 30---i-30
200
I Square Elbow 22 �
-: 20
100 18
Swing Check Valve, ... 6
�
fully Open � � 14--.
Borda Entrance 50 iz
@



Close Return Bend
tBJ



Standard Tee
Through Side Outlet





Standard Elbow or run of
Tee reduced Vz 2 2
I
l}h--j
I
I� i
0.5 1
Medium Sweep Elbow or
run of Tee reduced ;4
i 0.3 1-l-1
l-
¥J-E
�:,p@,J 45· Elbow ., Vi !-
0.2
r
'
run of Standard Tee to.s

Figure 2-20. Equivalent length resistance of valves and fittings to flow of fluids. Note: apply to 2 in. and smaller threaded pipe for process
applications (this author). By permission, Crane Co., Technical Paper #409, Engineering Div., 1942, also see [3).

88 Applied Process Design for Chemical and Petrochemical Plants

k e
0 20 40 60 80


20 Le
5,000-:
I 4POO "° d
;;L ,- 3POO�
,- 60-
:: t: ,-
� 2,000 i 50-:
...
=
�
Globe valve,open } 10 ..._' -: 40-::
9 c:,, l,OOOi �
.... '
8 ...... 800 � 30�
--
....
Cl
7 � 5 - -
8
680-
6 .,. 40 :;; 20 -,_ 24
c:,
�; 300{ _,_ 20
5 18
-
.
-
-
Angle valve, open }-- -"" �I 200 : _,_ 16
I
c:
"Ball check valve, open -� a 1 - -f- rn
Swing check valve, open ;;; = a,
c:
•45°-Y Globe valve, open <..> 100-= ·= to-- 10 c:
-
0
I 00°close screwed return 60� - 8 -
so'.§
.i:::.
� - .E 8-- -�
...
..
...
6 -
...
-
Screwed or fabricated tee 50- 0 'O 7�
),:
thru bronch ond }� 40i __. - 6 "O
90°single-miter elbow iii s�- 5 �
30-: .i:::. ..
..
...
a,
...
Welding tee lhru branch }---._ 20: c 'O 4-- 4 .�
-
.,
}
•sutterfly valve, open <, . - � 3 'tz
...
Submerged discharge \ I - c "' 3-- 3 -�
c
=
•3-Way plua cock, st. thru, open � : I 10..; ·- -=- 21/z a..
-
a
90 standard screwed or }--:-L � � 8� .� a -
D
60° single-miter elbow 0.:, - . 6-:: O' u 2-:- 2 ·e
::,
c:
- ::,
--
-
--
0
45"1olerol thru branch } 0.8 a,/ . 5- "' - l�z z
0.7-
-
90" long-sweep or }� 0 ' " A - 11/4
-
,•
90°double - miler elbow � \
=
-
•p1ug cock, full port, open o, �} \ \ / -;: '.;-
\
Screwed or fobricoled :..1?.-........ ..... \I\/,,- o.s-=- 314
tee or lateral thru run } o4 '� 0.1.::
submerged entrance 3- - 0.8-c: 0.6-= ,_. h
I
-
o.s-
90° triple- miler or 04 �] ; .. - "' 0.6-::
4 5°single miler elbow �� � Enlorging } Sudden·
x
180°welding return or � ;;. - Contracting chcnqe
rn section
4 5° screwed el bow 11 -o u I\ . rSA
-- -�
,....___
Welding tee thru run � \ ' Enlarg,n� reducer
Contract mg ( ta 12" size)
90" welding elbow } '
45°welding elbow or } •' Note: d = Small Pipe I dio. inc hes
\\
gate valve, open 0.1
0.09 d'= Lorge Pipe, dia. inc hes
0.08 - R = Bend Radius I inc he
0.07 "" For Eccentric Reducers I ncentric
Increase Values of Co
0.06 \, 'Enlarging } R�ducer O . Reducers by 33 1/3 o Yo.
20
\'
Contracting �,th
' ,.:.i, included ang re Values for Tees and Lal era ls
0.05 ,r \' Apply lo Full Size Fift
'\. l ings,
0.04 Losses Larger for Side Oullels.
" •Added by author 1-7 6
'
0.03 \ 3-Way plug cock op ening Is
80% of pipe size.
For partially closed globe-type valves,
For elbows 8 bends: R/d; 0 2 4 6 8
For confroctions 8r , Multiply open L 9 by:
enlargements: did: O 0.2 0.4 0.6 0.8 3/4 open 3.25
1/2open 12.0
1 /4 open 72.0
Figure 2-21. Equivalent length of fittings for pipe systems. Note: preferred use for 1 � in. and larger pipe butt-welded or socket-welded con-
nections (this author). By pennission, Tube Turns Div., Chemitron Corp. Bull. TI 725, 1952, reference now to Tube Turns Technologies, Inc.

Fluid Flow 89


summation, these equivalent lengths for all the compo- Table 2-5
nents determine the total pipe length to use in the pres- Typical Design Vapor Velocities* (ft/sec)
sure loss (pressure drop) equations to be described later.
Line Sizes
Fluid ::;6" 8"-12" � 14"
Friction Pressure Drop for Non-Viscous Liquids
Saturated Vapor
The only significance in differentiating between water Oto 50 psig 30-115 50-125 60-145
and liquids of different densities and viscosities is the con- Gas or Superheated Vapor
venience in having a separate simplified table for water. Oto 10 psig 50-140 90-190 110-250
11 to 100 psig 40-115 75-165 95-225
101 to 900 psig 30- 85 60-150 85-165
1. Using known flow rate in gallons per minute and a
suggested velocity from Tables 2-4 to 2-8 or Figure 2- * Values listed are guides, and final line sizes and flow velocities
must be determined by appropriate calculations to suit circum-
22, estimate first pipe size. Mean velocity of any liq- stances. Vacuum lines are not included in the table, but usually
uid flowing in a pipe [3] is given by Figure 2-22 and tolerate higher velocities. High vacuum conditions require careful
Equation 2-51. pressure drop evaluation.

2
v = 0.408 Qld = 0.0509 \V/(d 2) (p), ft/sec (2-51)
Table 2-6
1
d = (0.408 Q/v) 1 2 = (0.0509 W/vp) 1 2, in. (2-53) Usual Allowable Velocities for Duct and Piping Systems*
1
v = q/A = w,/Ap = 183.3 (q/d 2), ft/sec (2-54) Service/ Application Velocity, ft./min.
Forced draft ducts 2,500 - 3,500
2. Estimate or otherwise determine the linear feet of Induced-draft flues and breeching 2,000 - 3,000
straight pipe in the system, L. Chimneys and stacks 2,000
Water lines (max.) 600
3. Estimate (or use actual tabulation) number of fit- 10,000
tings, valves, etc. in system. Convert these to equiva- High pressure steam lines 12,000 - 15,000
Low pressure steam lines
lent straight pipe using Figures 2-20 or 2-21, Leg, or Vacuum steam lines 25,000
head by Figures 2-12 through 2-16 and Table 2-2. Compressed air lines 2,000
Refrigerant vapor lines
Note preferred pipe size type for charts. High pressure 1,000 - 3,000
Low pressure 2,000 - 5,000
4. Determine expansion or contraction losses, if any, Refrigerant liquid 200
including tank or vessel entrance or exit losses from Brine lines 400
Figures 2-l2A, 2-15, or 2-16. Convert units to psi, Ventilating duns 1,200 - 3,000
head loss in feet times 0.4331 = psi (for water), or Register grilles 500
adjust for Sp Gr of other liquids. �By permission, Chemical Engineer's Handbook, 3rd Ed., l\kGraw-Hill Book
Co., New York, N.Y., p. 1642.
5. Estimate pressure drop through orifices, control
valves, and other items in the system, but not equip-
ment. For control valves, estimate 11P from para-
graph to follow.
lish p1pmg system friction pressure drop (loss),
6. Determine pressure drop per unit of length. liquids (Figure 2-23):
a. Calculate Reynolds number [3]

R, = 50.6 Qp/(dµ) = 6.31 \V/(dµ) (2-16) For turbulent flow: �P/l 00 ft = 0.0216 f pQ / d 5 (2-55)
2
b. From Reynolds Number-Friction Factor Chart, = 0.000336 rw 2; ( d 5) (p) (2-55A)
Figure 2-3, read friction factor, f, at £/ d value
taken from Figure 2-11.
For laminar flow: �P/100 ft= 0.0668 (µ) v/d 2 (2-56)
c. Calculate presst,re drop per 100 feet of (straight
and/or equivalent) pipe [3] as psi/100 ft. Estab- = 0.0273 (µ) Q/d 4 (2-56A)

90 Applied Process Design for Chemical and Petrochemical Plants

rounded values to no more than one decimal place are
Table 2-7 limits for such head loss calculations.
Typical Design* Velocities for Process System The head losses calculated using K coefficients by these
Applications
figures can be added directly to the total friction head loss
for the straight pipe portions of a system. When equiva-
Service Velocity, ft./sec.
lent lengths are determined, they must be added to the
Average liquid process 4 - 6.5 straight pipe before determining the total head loss, as
Pump suction ( except boiling) 1 - 5 shown in the example calculations for a water system.
Pump suction, boiling 0.5 - 3
Boiler feed water ( disch., pressure) 4-8 Friction loss in rubber-lined pipe is usually considered
Drain lines 1.5 - 4 equivalent to that in new steel pipe of one-half to one
Liquid to re boiler ( no pump) 2-7 nominal size smaller, with little or no change due to
Vapor-liquid mixture out reboiler 15 - 30 aging, unless known conditions can be interpolated. For a
Vapor to condenser 15 - 80 given inside diameter, the friction loss is the same ( or
Gravity separator flows 0.5 - 1.5 slightly less) than clean steel pipe.
ln the turbulent flow range, friction loss in glass pipe is
*Tobe used as guide, pressure drop and system environment govern
final selection of pipe size. 70 to 85 percent of clean steel.
For heavy and viscous fluids, velocities should be reduced to about For 2-inch (nominal) and larger vinyl, saran, or hard
Y2 values shown.
Fluids not to contain suspended solid particles. rubber pipe, the friction loss does not exceed clean steel.
With saran and rubber-lined pipe the loss is about equal
to clean steel at the 2.5-inch size, increasing to 2 to 4 times
Table 2-8 the loss at the 1-inch size.
Suggested Steam Pipe Velocities in Pipe Connecting to
Steam Turbines Estimation of Pressure Loss across Control Valves:
Liquids, Vapors, and Gases
Service-Steam Typical range, h./sec.
Despite the need for good control in many process sys-
Inlet to turbine 100 - 150
Exhaust, non-condensing 175 - 200 tems, most engineers do not allow the proper pressure
Exhaust, condensing 500 - 400 drop for the control valves into their calculations. Many
literature sources ignore the problem, and many plant
operators and engineers wonder why the actual plant has
control problems.
7. Total pressure drop for system: Rather than assuming a pressure drop across the con-
trol as 25%, 33%, or 40% of the other friction losses in the
t.P, psi= (L + I:Leql (t.P/100 ft from 6 c above) system, a logical approach [9] is summarized here. The
+ 4 above + 5 above (2-57) control valve pressure drop has nothing to do with the

Note: L 0 q is from 3 above. valve size, but is determined by the pressure balance (See
Equation 2-59 [9]).
Control valve pressure drop:
If this pressure drop is too large or too small, recheck
the steps using larger or smaller pipes as may be indicat- (2-58)
ed. The tables in Cameron [57], Table 2-22, or Figure 2-
24 are very convenient to use, although they give much Available Af'c = (Ps - P cl - Fo, psi (2-59)
more conservative results (about twice unit head loss)
than the method outlined above. When using Figure 2-24, where Ps = total pressure at beginning (higher pressure) of
the results agree acceptably well with tests on 15 to 20 year system, psig, including any static heads to reach
old steel pipe. Also see Table 2-22. final pressure, P 0•
For brine, Table 2-9 gives multipliers to use with the Pe = pressure at lower end of system, psig
water unit losses of Figure 2-24. Figure 2-25 gives direct- F 0 = friction loss at design basis, total, for the system,
reading values with Dowtherm® liquid. psi, including equipment and piping, at Q 0 rate
Q� 1 = maximum anticipated flow rate for system, gpm,
It is important to note that comparison of results from orACFM
these charts does not yield exact checks on any particular F� 1 = friction pressure drop at maximum flow rate
fitting. Calculations should never be represented as being Q\1, psi
more accurate than the basic information. Therefore, Q 0 = design flow rate, gpm, or ACFM

Fluid Flow 91


d
:!II .s
II" Q q
JOO
!0000 4IDlO .6
iDlO 80 V2
30000
!000 '° .7
200IIJ 40
«lOO l/4 .s
ll
mo .9
10000 lO
8000 1.0
1000
6000
p
1000 I', 37
800 LS
I'?
600 40
v
lOO
JOO
� llO 80 .... 0
60
0
0 .., M
� Cl v 45
"' 200 - Jl .., 2.5 ..c
�
0
u
.., :Ill c
u
c � � 15 O
0 im u v :... >
0.. .... u -------r=-10-'i---_:: ( 2
0 IOO � ---
.., 80 -=i2Do - s 6 - � ..
..
:
c
�
-
c
.,.·
"" ... Q) c - .3 .. - � �
0
u
c 40 �100 .2 ;;:: 0 55
� 0 80 0
..: JO u ..
.. co
u
.. 20 .I I
";; a: ""
u
"" .08 � 60
= 10 .06
.04
.I 65
.OJ
.02 1D 10
12
14
.006 IS IS
.004 IS
I
.003
.8 lO
2D
Figure 2-22. Velocity of liquid in pipe. By per- .6 .8 .002 24
mission, Crane Co., Technical Paper #410, .6
Engineering Div., 1957. Also see 1976 edition. ·.4 .5 .001 ·25
.3
L'i.P c = pressure drop across control valve Allowing 10% factor of safety, expected maximum
F�1 = friction pressure drop at maximum flow rate, psi increase in friction pressure drop allowance:
(2-61)
Friction Joss or drop at higher flow rates than design:
or r - l}F)
At maximum flow rate, Q:,.,i, the friction drop will
Increased pressu:;-e drop = [FD (QM /QD )2 - FD], become:
[
( :;:
(2-60)
(2-62)

92 Applied Process Design for Chemical and Petrochemical Plants

q Q
p
Index 2 65 l:iP, •
• 05
,06
.08
.I

Index l .2
!
.05

.04
.03 i
...
c ...
..
0 0
..
u ---.--� ·;;; 0
c
• 02 .._ c Q.
..
I = .. Q.
e
....... c
·;;
�
�
.015 "' ..
�
I
45 <I. ...
.. -
...
'· ;;
.OJ
<J
_., 40
O Q. .2
�ii:
Figure 2-23. Pressure drop in liquid ....
vi�
lines. By permission Crane Co., I -="'=i
Technical Paper #410, Engineering 0- E�
Div., 1957. Also see 1976 edition. I �� .I l 37
The friction loss or pressure drop, F0, is determined at where Pc = system end pressure = 22 + 15 = 37 psig (not
the design flow rate, Q 0, for the piping, valves, and fric- friction)
tion producing equipment (such as tubular heat exchang- Piping system pipe friction @ Q0 flow rate = 6 psi
ers, tubular furnaces/heaters), orifice or other meters, Heater, friction = 65 psi
and control valves. Because the system friction pressure Separator, friction = 1 psi
loss changes with flow rate through the system, recogni- Preheaters, 10 + 12 (friction) = 22 psi
tion must be given to the changes in flow rate (increase or Orifice, allow, friction = 2 psi
decrease) as it affects the pressure loss through the con- Total friction, excluding control valve, F0 = 96 psi
trol valves. For any design, the beginning and end points Assume pressure loss through control valve = 35 psi
of the system should be relatively constant for good
process operations. (2-59)
For good control by the valve, the pressure drop across
(or through) the valve must always be greater than the fric- 35 = (Ps - 37) - 96
tion losses of the system by perhaps 10% to 20% (see [9]).
Ps = 168 psi, at pump discharge, using assumed control
valve pressure drop of 35 psi
Example 2-3: Establishing Control Valve Estimated
Pressure Drop, using Connell's Method [9]. Note that P,. = 22 psig + 15 psi static Hd. = 37 psig
Assume that allowances must be made for a 10%
Refer Lo Figure 2-26 for an example to determine the increase in process flow rate, above design, Q 0. Pressure
pressure loss (drop) through the control valve. drop varies as the square of the flow rate.

Fluid Flow 93

Correction Factors Factors tO consider in evaluating the control valve pres-
Volur of C 60 70 80 90 100 110 120 130 140 sure drop are:
Multi lier to Correct Chart 2.57 1.93 1.50 1.22 1.0 .84 .71 .62 .54
A Allowance for increase in friction drop


Establish the ratio of the maximum antir:ipated flow rate
for system, QM, to the design basis rate, Q or QM/Q .
0
0
When Q:vr is not known, nor can it be anticipated, use:
Q.,1/Q0 of 1.1 for flow control and 1.25 for level pressure
and temperature control valves to anticipate the flow rate
transients as the control loop recovers from a distur-
bance [9J.
At the maximum flow rate QM, the friction drop will
become:

I (2-62)
IL
The increase in pressure drop will be:

(2-63)
1-- .... � <!l.,,. � '4-"°+-l-++l. � /:.-l c;,i.t.\ \ .� \
3 I/ [...,' \ �\� ' � . y (2-64)
2v � vv"" � K J,.t,. \�
1.q,L _ l___.::t: . 2.....L . 3.l:4..J . 5...:1.. . 7u.l.1.. . 0 � ..1.2£..L.3...14_J5...J....::7LLI- 1 0-2.i..o-30::':-'-'50 � 70'."" I OO F0 may not necessarily be very accurate at the design stage
where final drawing dimensions for the system are being
Pressure Loss in Feet of Water per 100 Feet
estimated. For this reason a 10% increase allowance is
Figure 2-24. Friction loss for flow of water in steel pipes. Note C = suggested to £\FM.
pipe roughness factor. See Tables 2-9 and 2-22. Courtesy of Carrier
Corp.
B. Allowance for possible falloff in: overall system pres-
sure drop, P5 - Pe
New flow rate = 110% (Q 0)
If there is an increase in system flow rate
F
Friction pressure drop will increase to 121 % of 0;
Overall system pressure drop = PF (all) = 0.05 PS (2-65)
1.21 (96) = 116 psi = FM C. Allowance for control valve (base pressure drop at
full-open position [9J
Friction increase =c 116 - 96 = 0 20 psi added for rela- This varies with the type and design of valve and can be
tively constant Ps and P.,
obtained from the manufacturer. It is identified here as
base pressure drop B for the valve itself. Using average
Available �pc= (168 - 37) - 116 line velocities and assuming that the control valve will be
�Pc = 15 psig through the control valve, which one pipe size smaller than the pipe line it is connected to,
means that the valve has to open more and using average B values over a range of sizes, the B values
reduce its sensitivity of response, from its for estimating purposes are [9]:
design �pc of 35 psig
Control Valve Type B,p_s_ i __
For design purposes, the assumed 35 psi for the control
valve could be used; however, decreasing the pipe friction Single Plug 11
Double Plug
7
of 6 psi to perhaps X or Y. by increasing the line size will Cage (unbalanced) 4
help the control of the valve. It would be better to have Cage (balanced) 4
the available valve pressure drop equal to or greater than Butterfly 0.2
the assumed. V-Ball l

94 Applied Process Design for Chemical and Petrochemical Plants

j-jllll I I I I I I I 11
20 0 I I
51.0 --..
15 - ... 4 ......_ J J I/ J I J J
=o.9 � I v I I v / I I I
01
10 ;:....uo.B -........... .............. � I I IJ I , I I/ , , I J
I
B � � -500 ,___ 600 � 700 I I I/ r I
0
TEMPERATURE- f J I I I I
I
I
Ii: 6 I I I I 'I I I I 'f I/ , I
J
8 J I I I I I I
<, 4 ,, I ,, , I .,., ,, , I '{ � ,
t ,, ... , ,, .J ,, J 'Of �I "-1-- ,, �� .... '\,, �
,
....
�J
d .,, 3 � J "' "'' I J I/ I
I
<, .:;;; I I I I I I I
.,; 2
"' .... � J J v IJ I , J I/ I
fy
.... I 1.5 v I v I I
0 J J � I/ i; J
� 1.0 , I , I I , I , ,
� 0.8 I , , ,
:::, I J I I , I I II
� 0.6 I I/ I/ I J , I J l
"' .... I j I I I
0.4 , I I r J , , I I '
v
0.3 I I , I I J
I I I I I
J I J J I I I I
0.2
I/ I J 'I J I I I I I
0.15 I I If I J IJ I I
10 20 30 40 60 BO 1 00 200 300 400 600 BOO 1 000 2000 4000 8000
OOWTHERM FLOW - GPM AT AVERAGE TEMPERATURE
Figure 2-25. Pressure drop for Dowtherm® liquid in schedule 40 pipe. By permission, Struthers Wells Corp. Bull. 4-45, 1956.

Then, incorporating the requirements of A, B, and C Assume, Q.1v1 = 120% ofQ
above, the estimated overall control valve drop is:
QM/Qo = 1.2
Required L'>Pc = 0.05 Ps + 1.1 [(QM/Qo)2 - I] Use cage type valve, B = 4
F 0 + B, psi (2-66)
From Equation 2-59,

B = base pressure drop for control valve with valve in wide- Available L'>Pc = (Ps - Pc) - Fo
open posit.ion, psi. (see list above). = (Ps - 37) - 96 = P5 - 133


From Equation 2-66,
Example 2-4: Using Figure 2-26, Determine Control
Valve Pressure Drop and System Start Pressure (See Required L'>P = 0.05Ps + 1.1 [QM/Qo)2 - l] F 0 + B
Example 2-3) = 0.05 (Pc) + 1.1 [(1.2) 2 - I] (96) + 4
= 0.05 Ps + 46.5 + 4 = 0.05Ps + 50
To determine Ps, the value of t!P through the control
valve must be known. Substituting:

Ps - 133 = 0.05Ps + 50
Pe= 37 psi 0.95Ps = 183
F 0 = 96 psi (all except control valve), psi Ps = 192 psi, start pressure at the pump

Fluid Flow 95

Table 2-9
Brine Pipe Friction Multiples
For Use With Water Friction Data, Figure 2-24
I BRINE TEMPERATURE, °F
Specific
BRINE Gravity 0 10 20 30 40 50 60 70
1.10 ... 1.23 1.20 1.18 1.16 I 1.14 I 1.13 1.12
Sodium Chloride .. ............ 1.15 1.43 1.33 1.29 1.26 1.24 1.22 1.21 1.20
1.20 1.53 1.44 1.38 1.3!5 1.32 1.30 1.28 1.27
- -----
1.05 . . . ... 1.15 1.12 1.10 1.08 1.07 1.06
1.10 ... 1.28 1.23 1.20 1.18 1.16 1.14 1.12
1.15 1.41 1.35 1.31 1.28 1.25 1.22 1.21 1.20
Calcium Chloride ........... .. ---- . ..
1.20 1.49 l.43 1.39 1.36 1.33 1.30 1.28 1.27
1.25 1.56 1.53 1.49 1.45 1.42 1.40 1.38 1.37
1.30 1.65 1.61 1.58 1.55 1.52 1.50 1.49 1.48
I i I
-
NOTE: To find brine friction loss, multiply loss from Fig. 2-10 by multiplier from above Table.
By permission, Crocker, S., Piping Handbook, McGraw-Hill Book Co.




Charge Nquid
Gate valve Flow orifice
2 psi 2 psi Preheater Preheat er
1777 ''' ,777 10 psi 12 psi
Charge pump
22 psi



To vent control system










Head = 15 psi (static)

Dlattllation column Fired healer 85 psi
Figure 2-26. Establishing control valve estimated pressure drop.


Control valve pressure drop: properly estimating the valve pressure drop. From Shin-
sky [10),
�Pc= 0.05 Ps + 50 = 0.05 (192) + 50 = 59.6 psi
GPM = a'(c,') � PJSpG--;- (2- 66)
�
Use this as estimated control valve pressure drop for
the system design.
where a' = fractional opening of control valve, generally
assume 60% == 0.60
The Direct Design of a Control Vafoe C's, == standard valve coefficient from manufacturer's
catalog
This does not require the system balance as outlined in 6Pc = pre�sure drop across valve, psi
A through C above; however, without first preparing a SpGr = specific gravity of fluid, relative to water at
pressure balance, the designer cannot be confident of same temperature

96 Applied Process Design for Chemical and Petrochemical Plants


or, from [11], for gases or vapors: b.P = (L + LLeq) (b.P/100' from Table 2-10)
+ Item (4) + Item (5) (2-57)
Flow, SCFH * t
�-----�
42.2 c: � (P 1 - P )(P + P 2) If this pressure drop is too large (or too small), recheck
1
2
(2-67) the steps using larger or smaller pipe as may be indicated.
� Table 2-22 [53] or Figure 2-24 are convenient to use,
although they give much more conservative results (about
Flow, SCFH ( temperature corrected) r twice unit head loss) than the method and figures just ref-
963C: � (P 1 -P 2 )(P 1 +P 2) erenced. When using Figure 2-24 the results agree accept-
(2-67A)
� SGT ably well with tests on 15-20-year-old steel pipe.

where Sg = specific gravity relative to air = 1.0 Example 2-5: Water Flow in Pipe System
P1 = inlet pressure ( 14. 7 + psig)
P2 = outlet pressure (14.7 + psig) The system of Figure 2-27 consists of 125 feet of
qh = flow rate, standard cu ft./hr (SCFI-1) unknown size schedule 40 steel pipe on the discharge side
T = flowing temperature, R abs, (°F + 460) of a centrifugal pump. The flow rate is 500 gallons per
0
c; = valve coefficient of flow, full open (from manufac- minute at 75°F. Although the tank is located above the
turer's tables) pump, note that this elevation difference does not enter
into the pipe size-friction drop calculations. However it
*The effect of flowing temperatures on gas flow can be will become a part of selection of the pump for the service
disregarded for temperatures between 30°F and 150°F. (see Chapter 3). For quick estimate follow these steps:
Corrections should apply to other temperatures above or
below [11]. 1. From Table 2-4, select 6 fps as a reasonable and usu-
tWhen outlet pressure P2 is less than J1 inlet pressure P 1 ally economical water rate.
the square root term becomes 0.87 P 1 [11].
From Table 2-10, a 6-inch pipe has a velocity of 5.55
Friction Loss For Water Flow
fps at 500 gpm and a head loss of 0. 720 psi/ I 00 ft.
The 5-inch pipe has a velocity of 8.02 fps and might
Table 2-10 is quite convenient for reading friction loss be considered; however 5-inch pipe is not common-
in standard schedule 40 pipe. It is based upon Darcy's ly stocked in many plants, and the velocity is above
rational analysis (equivalent to Fanning). usual economical pumping velocities. Use the 6--inch
Suggested procedure:
pipe (rough estimate).
1. Using known flow rate in gallons/minute, and a
suggested velocity from Tables 2-4, 2-5, 2-6, 2- 7 and 2- 2. Linear feet of straight pipe, L = 125 feet.
8 select an approximate line size. 3. From Figure 2-20, the equivalent length of fitting is:
2. Estimate (or use actual drawing or measured tabula- 6 inch-90° ell== 14 feet straight pipe (using medium
tion) total linear feet of pipe, L. sweep elbow to represent a welding ell). Note that
3. Estimate ( or use actual tabulation) number of this is given as 6.5 feet from Figure 2-21. This illus-
elbows, tees, crosses, globe valves, gate valves and trates the area of difference in attempting to obtain
other fittings in system. Convert these to equivalent close or exact values.
straight pipe using Figure 2-20 or 2-21, Leq, or to
head loss using Figures 2-12 through 2-16. Note pre- 3 90° ells= 3 (14) = Leq = 42 ft (conservative)
ferred pipe size/type for charts. 1 tee = 1 (12) = Leq = l2 (Run of std. tee)
4. Determine expansion and contraction losses (if any) 1 6" open Gate Valve = (1) (3.5) = Leq = 3.5
from Figures 2-12, 2-15, and 2-16. Convert units: 1 sudden enlargement in tank @ d/d' = O; = 10',
head loss in feet times 0.4331 = psi. (This term can Figure 2-21
usually be neglected for most liquids at reasonable
velocities < l O' I sec.) Total Leq = 67.5 feet
5. Estimate pressure drop through orifices, control 4. Neglect expansion loss at entrance to tank, since it
valves and other items that may be in system, per will be so small.
prior discussion. 5. No orifices or control valves in system.
6. Total pressure drop. 6. From Table 2-10, at 500 gpm, loss= 0.72 psi/100 eq ft.

Table 2-10
Flow of Water Through Schedule 40 Steel Pipe*
Note: This Table closely matches results of Hydraulic Institute
f W t er a
P ressure rop per 100 f t an d V l it · S h d l 40 p· 1pe or a t 60F
e OCI y m c e ue
ee
D
Discharge
Veloc- Press. Veloc- Press. Veloc- Press. Veloc- Press. Veloc- Press. Veloc- Press. Veloc- Press. Veloc- Press.
lty Drop ity Drop ity Drop ity Drop ity Drop ity Drop ity Drop ity Drop
Gallons Cubic Ft. Feet Lbs. Feet Lbs. Feet Lbs. Feet Lbs. Feet Lbs. Feet Lbs. Feet Lbs. Feet Lbs.
per per per per per per per per per per per per per per per per per per
Minute I Second Second Sq. In. Second Sq. In. Second Sq. In. Second Sq. In. Second Sq. In. Second Sq. In. Second Sq. In. Second Sq. In.
1/s" %" %" l/2"
.2 o.000446 1.13 l.86 0.616 0.359
.3 0.000668 l.b9 4.22 0.924 0.903 0.504 0.159 0.317 0.061 3A"
.4 0.000891 2.26 6.98 1.23 l.61 0.672 0.345 0.422 0.086
.5 0.00111 2.82 10.S I. 54 2,39 0.840 0.539 0.528 0.167 O.JOI 0.033
.6 0,00134 J .39 14.7 1.85 3.29 1.01 0,751 0.6JJ 0.240 0.361 0.041
.8 0.00178 4.52 25.0 2.46 S.44 1.34 l.25 0.844 0.408 0.481 0.102 1" 11,4"
I 0.00223 5 .65 37.2 J .08 8.28 1.68 l.85 1.06 0.600 0.602 0.155 0.371 0.048 1%"
2 0.00446 11.29 134.4 6.16 30.1 J .36 6.58 2.1 i 2.10 1.20 0.526 0 .743 0.164 0.429 0.044
3 0.00668 9.25 64.1 5 .04 13.9 J .17 4.33 1.81 1.09 1.114 0.336 0.644 0.090 0 .473 0.043
4 0,00891 12.JJ 111.2 6.72 23.9 4.22 7.4212.41 1.83 1.49 0. 565 0.858 0.150 0.630 0.071
5 0.01114 2" 8.40 36.7 5 .28 11.2 J .01 2.75 1.86 0.835 i ,073 0.223 0.788 0.104
6 0.01337 0.574 0.044 21/z" 10.08 51.9 6.JJ 15.8 J .61 3.84 2.23 1.17 1.29 0.309 0.946 0.145
8 0.01782 0.765 0.073 IJ .44 91.1 8.45 27.7 4.81 6.60 2.97 1.99 1.72 0.518 1.26 0.241
10 0.02228 0.956 0.108 0.670 0.046 10.56 42.4 6.02 9.99 3 .71 2.99 2.15 0.774 I. 58 0.361
15 0.03342 1.43 0.224 I.Of 0.094 3" 9.03 21.6 5. 57 6.36 3 .22 1.63 2.37 0.755
20 0.04456 1.91 0.375 l .34 0.158 0.868 0.056 31/z'' 12.0J 37.8 7 .43 10.9 4.29 l.78 3 .16 1.28
25 0.05570 2.39 0.561 l.68 0.234 1.09 0.083 0.812 0.041 4" 9.28 16.7 5 .37 4.22 J .94 1.93
30 0.06684 2.87 0.786 2.01 0.327 I. JO 0.114 0.974 0.056 11.14 23.8 6.44 5.92 4.73 2.72
35 0.07798 3 .35 I.OS 2.35 0.436 I. 52 0.151 1.14 0.704 0.882 0.041 12.99 32.2 7. 51 7.90 5. 52 3.64
40 0.08912 J .83 1.35 2.68 0.556 1.74 0.192 l. JO 0.095 1.01 0.052 14.85 41.5 8.59 10.24 6.JO 4.65
45 0.1003 4.JO 1.67 J .02 0,668 1.95 0.239 1.46 0.117 I. I J 0,064 9.67 12.80 7 .09 5.85
50 0, 1114 4.78 2.03 J .35 0.839 2.17 0.288 1.62 0.142 1.26 0.076 5" 10.74 15.66 7 .88 7,15
60 0.1337 5 .74 2.87 4.02 l.18 2.60 0.406 1.95 0.204 l. 51 0.107 12 .89 22.2 9.47 10.21
70 0,1560 6.70 3.84 4 .69 1.59 J.04 0.540 2.27 0.261 1.76 0.143 1.12 0.047 i l .05 13.71
80 O,li82 7 .65 4.97 5 .36 2.03 3 .47 0.687 2.60 0.334 2.02 0.180 1.28 0.060 12.62 17.59
90 0.2005 8.60 6.20 6.0J 2.53 3 .91 0.861 2.92 0,416 2.27 0.224 l.44 0,074 6" 14.20 22.0
100 0.2228 9.56 7.59 6.70 3.09 4.34 I.OS J .25 0.509 2. 52 0.272 l.60 0.090 1.11 0.036 15 .78 26.9
125 0.2785 11.97 11.76 8 .38 4.71 5 .43 l.61 4.06 0.769 J .15 0.415 2.01 0.135 1.39 0.055 19.72 41.4
150 0.3342 14.36 16.70 10 .05 6.69 6. 51 2.24 4.87 1.08 3.78 0.580 2.41 0.190 1.67 0.077
175 0.3899 116.75 22.3 11 .73 8.97 7 .eo 3.00 5 .68 1.44 4.41 0.774 2.81 0.253 1.94 0.102
200 0,4456 19.14 28.8 13 .42 11.68 8.68 3.87 6.49 1.85 5 .04 0.985 J .21 0,323 2.22 0.130 8"
225 0.5013 ... . .. 15 .09 14.63 9.77 4.83 7.JO 2.32 5 .67 1.23 3 .61 0.401 2.50 0.162 l.44 0.043
250 0.557 ... ... ... . .. 10.85 5.93 8.12 2.84 6.JO 1.46 4.01 0.495 2.78 0.195 1.60 0.051
7,75 0.6127 I ... ... . .. . .. 11.94 7 .14 8.93 3.40 6.93 1.79 4.41 0.583 3 .05 0.234 I.76 0.061
300 0.6684 ... . .. ... . .. IJ .00 8.36 9.74 4.02 7. 56 2.11 4.81 0.683 J.33 0.275 I.92 0.072
325 0.7241 ... ... ... . .. 14.12 9.89 10. 53 4.09 8.19 2.47 5 .21 0.797 J .61 0.320 2.08 0,083
350 0.7798 ... ... ... . .. I I.Jo 5.41 8.82 2.84 5 .62 0.919 J .89 0.367 2 .24 0.095
375 0.8355 ... ... ... . .. 12.17 6, 18 9.45 3.25 6.02 I.OS 4 .16 0.416 2 .40 0.108
400 0.8912 ... ... . .. . .. 12.98 7,03 10.08 3.68 6.42 1.19 4.44 0.471 2.56 0.121
HS 0.9469 ... ... ... . .. 13 .80 7,89 10.71 4.12 6.82 1.33 4.72 0.529 2.73 0.136
450 1.003 10" ... . .. ... . .. 14.61 8.80 l I. 34 4.60 7 .22 1.48 5 .00 0.590 2.89 0.151
475 1.059 1.93 0.054 ... ... ... . .. 11 .97 5.12 7 .62 1.64 5 .27 0.653 J .04 0.166
soc 1,114 2.0J 0.059 ... ... ... . .. 12.60 5.65 8.02 1.81 5. 55 0.720 3 .21 0.182
550 1.225 2.24 0.071 ... ... ... . .. n .85 6.79 8.82 2.17 6.11 0.861 3. 53 0.219
600 1.337 2.44 0.083 ... ... ... . .. 15 .12 8.04 9.63 2.55 6.66 1.02 3.85 0.258
650 1.448 2.M 0.097 12" ... ... . .. . .. ... ... 10.43 2.98 7 .22 1.18 4.17 0.301
700 1.560 2.85 0.112 2.01 0.047 ... . .. ... . .. l l.23 3.43 7.78 1.35 4.49 0.343
750 1.671 J.05 0.127 2.15 0.054 14" ... ... ... ... 12.0J 3.92 8.33 1.55 4.81 0.392
800 1.782 J.25 0.143 2.29 0.061 ... ... ... ... 12.83 4.43 8.88 1.75 5 .13 0.443
850 1.894 3 ,4<, 0.160 2.44 0.068 2.02 0.042 ... ... ... ... IJ .64 s.oo 9,44 1.96 5 .45 0.497
900 2.005 J .66 0.179 2.58 0.075 2.13 0.047 ... ... ... . .. 14 .44 5.58 9.99 2.18 5 .77 0.554
950 2.117 J .86 0.198 2.72 0.083 2.25 0.052 ... ... 15 .24 6.21 IO. 55 2.4216.09 0.613
1 000 2.228 4.07 0.218 2.87 0.091 2 .37 0.057 16" ... ... 16.04 6.84 11.10 2.68 6.41 0.675
1100 2.451 4 .48 0.260 3 .15 0.110 2.61 0.068 ... ... 17 .65 8.23 12.22 3.22 7.05 0.807
1 200 2.674 4.88 0.306 J.44 0.128 2.85 0.080 2.18 0.042 ... ... ... ... 13 .JJ 3.81 7.70 0.948
1300 2.896 ;_29 0.355 J .73 0.150 3 .08 0.093 2.36 0.048 ... ... ... ... 14 .43 4.45 . 8.33 1.11
1 400 3.119 5 .70 0.409 4.01 0.171 J .32 0.10712. 54 0.055 I 15. 55 5.13 8.98 1.28
1 500 3.342 6.10 0.466 4 .JO 0.195 J. 56 0.122 2.72 0.063 18'' 16.66 5.85 9.62 1,46
1600 3.565 6.51 0.527 4.59 0.119 J .79 0.138 2.90 0.071 17.77 6.61 10.26 1.65
1800 4.010 7 .32 0.663 5. Io 0.276 4.27 0.172 J .27 0.088 2.58 0.050 19.99 8.37 I I.54 2,08
2000 4.456 8.14 0.808 5 .73 0.339 4 .74 0.209 3 .63 0.107 2.87 0.060 22.21 10.3 12.82 l.55
20"
2500 5.570 10.17 1.24 7 .17 0.515 5 .93 0.321 4.54 0.163 3. 59 0.091 16.03 3.94
3 000 6.684 12.20 1.76 8.60 0.731 7.11 0.451 5 .45 0,232 4 .30 0.129 3 .46 0.075 24" 19.24 5.59
3 500 7.798 14.24 2.38 10.0J 0.982 8.30 0.607 6. 35 0.312 5 .02 0.173 4.04 0.101 22.44 7,56
4 000 8.912 16.27 3.08 11.47 1.27 9.48 0.787 7.26 0.401 5 .74 0.222 4.62 0.129 J .19 0.052 25 .65 9,80
4 500 !0.03 18.Jl 3.87 12.90 1.60 10.67 0.990 8.17 0.503 6.46 0.280 5 .20 0.162 J. 59 0.065 28.87 12.2
5000 n.rs 20.35 4.71 14.33 1.95 i I .85 J.21 9.08 0.617 7 .17 0.340 5 .77 0.199 3 .99 0.079 ... ...
".
6000 13.37 24.41 6.74 17 .20 2.77 14.23 J.71 10.8g 0.877 8.61 0.483 6.93 0.280 4.79 C.111 ... ...
0.192 ...
7000 15.60 28.49 9.11 20.07 3.74 16.60 2.31 l2.71 1.18 10.04 0.652 8.08 0.376 5. 59 0.150 ...
8 000 !7.82 ... ... 22.93 4.84 18.% 2.99 14. 52 I.SI 11.47 0.839 9.23 0.488 6.38 ...
. "
9000 20.05 ... . .. 25 .79 6.09 21. 34 3.76 16.34 1.90 12.91 I.OS 10.39 0.608 7.18 0.242 ...
I
10 000 22.28 ... ... 28.66 7.46 23 .71 4,61 18 .15 2,34 14.34 1.28 11.54 0.739 7.98 0.294 ... ...
. "
12000 26.74 I ... . .. 34.40 10.7 28.45 6.59 2} .zo 3.33 17.21 1.83 IJ.85 1.0619.58 0.416 ... ...
0.562 ...
14 000 31.19 ... ... . .. . .. 33 .19 8.89 25 .42 4.49 20.08 2.45 16. i6 1.43 11.17 ...
16 000 35.65 ... ... . .. ... .. . . 29.05 5.83 22 .95 3.18 18.47 1.85 12.77 0.723 ...
18 000 40.10 ... ... ... ... .. . .. 32.68 7.31 25 .82 4.03 20.77 2.32 i4.36 0.907 ... . ..
20000 44.56 I ... ... ... .. . .. . . ,36.31 9.03 28.69 4.93 23 .08 2.86 15.96 J.12 ... ...
For pipe lengths other than 100 feet, the pressure drop is proportional to the length. Velocity is a function of the cross sectional flow
Thus, for 50 feet of pipe, the pressure drop is approximately one-half the value given area; thus, it is constant for a given fl.ow r ate
in the table , .. for 300 feet, three times the given value, etc. and is independent o[ pipe lenath.
For pipe other than Schedule 40: v = v1-0 (d10/d)' and t.P = ,lP10 (d10/d)' where subscript
40 refers to the conditions for schedule 40 pipe.
*By permission, "Technical Paper No. 410," Crane Co., Engineering Div., Chicago (1957)

98 Applied Process Design for Chemical and Petrochemical Plants

The pressure that can develop from the shock wave can
be destructive to the containing system hardware, partic-
ularly in long pipe. Examples of conditions that can devel-
Tonk op water hammer are:


125 feet 1. start, stop, or an abrupt change in a pump's speed
Straight pipe 90° Elbow 2. power failure
(Welding Type) 3. rapid closing of a valve (usually a control valve,
which can slam shut in one or two seconds)

__ e" Gate Valve The magnitude of this shock wave can be expressed
Full Open [19, 20]:
Suction
Line
Centrifugal (2-69)
Pump
Figure 2-27. Example 2-5, pipe system for pipe sizing calculations.
For water:

Total pressure drop from face of discharge flange on a,.= 4660/(l + Ki,, Br)ll ft/sec (2-70)
2,
pump to nozzle connection on tank:
where h,..h = maximum pressure developed by hydraulic
shock, ft of water
�p = (125 + 67.5) [(0.720)/100] + 0 v,.. = reduction in velocity, ft/sec (actual flowing veloc-
�p = 1.386 psi ity, ft/ sec)
�p = 1.386 psi (2.31 feet/ psi) = 3.20 feet water
g = gravitational constant, 32.2 ft/sec
Ki,, = ratio of elastic modulus of water to that of the
Note that a somewhat more accurate result may be pipe material (See list below)
obtained by following the detailed loss coefficients given Br = ratio of pipe diameter (I.D.) to wall thickness
in Figures 2-12 through 2-16. However, most preliminary a; = velocity of propagation of elastic vibration in the
engineering design calculations for this type of water sys- discharge pipe, fl/sec
tem do not warrant the extra detail.
Some typical Ki,s values for water/metal are [19]:
Flow of Water from Open-End Horizontal Pipe

The equation of Brooke [36] is useful in estimating Metal Ki...
water or similar fluids flow from the end of open pipes: Copper 0.017
Steel 0.010
GPM = 1.04 a (1) (2-68) Brass O.Ql7
Wrought iron 0.012
Malleable cast iron 0.012
where GPM = flow rate, gallons per minute Aluminum 0.030
a = internal cross-sectional area for flow in pipe, sq in.
I = horizontal distance from pipe opening to point
where flow stream has fallen one ft, in. The time interval t 5, required for the pressure wave to
travel back and forth in the pipe is:
Water Hammer [19]
t,. = 2 L/a,., sec (2-71)
Water hammer is an important problem that occurs in
some liquid control systems. It is defined as hydraulic L = length of pipe, ft (not equivalent ft)
shock that occurs when a non-viscous liquid flowing in a
pipe experiences a sudden change in velocity, such as the
fast closing of a valve. The kinetic energy of the moving When the actual abrupt closing of a device to stop the
mass of liquid upon sudden stoppage or abrupt change of flow has a time shorter than ts, then the maximum pres-
direction is transformed into pressure energy, thereby sure, hwh, will be exerted on the closed device and line.
causing an abrupt pressure rise in the system, often result- Note that the value of, hwh• is added to the existing static
ing in severe mechanical damage [53]. pressure in the system.

Fluid Flow 99

Example 2-6: Water Hammer Pressure Development 1. From Table 2-4, selected velocity = 6 fps.

An 8-inch process pipe for transferring 2000 GPM of Estimated pipe diameter, d = (0.408 Q!v) 112
2
methanol of Sp Gr = 0.75 from the manufacturing plant = [ (0.4.-08) 25/6] 11 = 1.3 inch
site to a user plant location is 2,000 feet long, and the liq-
uid is flowing at 10.8 ft/ sec. Try l�inch (i.d. = 1.61), since lX-inch (i.d. =
Maximum pressure developed (preliminary solution) 1.38) is not stocked in every plant. If it is an accept-
when an emergency control valve suddenly closes: able plant pipe size, then it should be considered
and checked, as it would probably be as good pres-
sure drop-wise as the lY,-inch. The support of lX-
(2-69)
inch pipe may require shorter support spans than
the lY,-inch. Most plants prefer a minimum of l�
Since methanol has many properties similar to water: inch valves on pressure vessels, tanks, etc. The valves
at the vessels should be 11.! inch even though the
1 2
aw = 4660/ (I + Ki,, Br) 1 pipe might be IX inch The control valve system of
= 4660/(1-:-- 0.01 (24.7"')] 112 = 4175 ft/sec gate and globe valves could very well be iX inch. For
this example, use lY,-inch pipe, Schedule 40:
"For 8-inch std pipe, Br= 7.981/0.322 = 24.78
2. Linear length of straight pipe, L = 254 ft.
Time interval for pressure wave travel: 3. Equivalent lengths of fittings, valves, etc.

ts= 2L/aw = 2 (2000)/4175 = 0.95 sec (2-71)
Estimated Eq. Feet
Fittings Type (from Figure 2-20)
If the shutoff time for the valve (or a pump) is less than
0.95 seconds, the water hammer pressure will be: 10 lW'-90° Elbows 4' (10) = 40
8 L;,5"-Tees 3' (8) = 24
4 l W'-Gale Valves l ' (4) = 4
h" 11 = 4175 ( 10.8) /32.2 = 1400 ft of methanol 68 ft. Use 75 ft.
�============��==�==��=====�
= (1400)/((2.�\l)/0.75)] = 454 psi hydraulic shock

Then total pressure on the pipe system 4. No expansion or contraction losses ( except control
valve).
5. Pressure drop allowance assumed for orifice plate =
= 454 + (existing pressure from process/or pump)
5 psig.
This pressure level would most likely rupture an 8-inch Control valve loss will be by difference, trying to
Sch. 40 pipe. For a more exact solution, refer to specialty maintain minimum 60% of pipe friction loss as min-
articles on the subject. imum drop through valve, but usually not less than
10 psi.
Example 2-7: Pipe flow System With Liquid of Specific
Gravity Other Than Water 6. Reynolds number, R., = 50.6 Qp/dµ (2-49)
== so.e (25) [0.93 (62.3)]/
This is illustrated by line size sheet, Figure 2-28. (1.61) (0.91)
= 50,025 (turbulent)
Figure 2-29 represents a liquid reactor system discharg-
ing crude product similar to glycol through a flow control 7. From Figure 2-11, £/d = 0.0012 for lY,-inch steel pipe.
valve and orifice into a storage tank. The reactor is at 350
psig and 280°F with the liquid of 0.93 specific gravity and From Figure 2-3, at R,, = 50,025, read f = 0.021
0.91 centipoise viscosity. There is essentially no flashing of
liquid across the control valve.
8. Pressure drop per 100 feet of pipe:
Flow rate: 11,000 lbs/hr �P/100' = 0.0216 fp Q /d 5 (2-72)
2
2
GPM actual= 11,000/(60) (8.33) (0.93) = 23.7 = 0.0216 (0.021) (62.3) (0.93) (25) /(1.61) 5
= 1.52 psi/ 100 ft equivalent
Design rate = 23.7 (1.05) = 25 gpm

100 Applied Process Design for Chemical and Petrochemical Plants


I SHEET NO.---------
LUDWIG CONSULTING ENGINEERS


LINE SIZE SHEET
Joi, Ho. ----------
Date _ Charge Ha. _

Line Ha. LP - 51 _ Flow ShHt Drowi.ng Ha. _
Line Description Reactor Discharge


280
Fluid In line Crude Product Temperature -------'--'---- F
0
GPM (Cale.) 23,7 GPM (des.) 25 Pressur"'t ------�3=-5_0_psig
CFM (Cale.) CFM (des.) Sp. Gr.0. 93

L bs./hr. (Cale.) Lbs./hr.(des.) Sp. Vol. -------- cu.ft./11,.
0.91
Recommended V eloclty 6,o fp_s __ Viscosity __________ cp
Straight pipe, fittings, valves Pr� 1 sure. Drop
expon1(on, contraction, etc. Item 1n, ps I
Item Mo. Unit E q. Ft. Tata I E q. Ft. Pipe & Equivole"t c: )
Pipe 254 I 254 Orifice c ) - In
90°Elbow 10 4 40 ,': Motor Valve (��"�rr.1) �l,n
Tee 8 3 24 Mi scel raneous
Gate Va. 4 I 4

Total ':!c;n
Total 32g 1,
Estimated line size H:" (ye r j f j ed)
* Rounded total to 75 feet
** By difference Actual Velocity fp s
Inlet= 350 psig; Outlet= 0 psig Unit Loss per 100 ft. 1. si ss i (see belQ\:l)
Friction Loss= 10 (includes orifice loss)
Balance for Valve= 340 psi Total head loss 869
in feet of liquid
Total pressure
drop in psi 350


Selected pipe ,I�• _ _._ ) 4 · '- ' -------- Material & Weight Schedule 40 steel

Calculatlons: Re = so.6 QPl<y«= 50,6(25) (0.93 x 62.3)/(1.61) (0.91) 50,025
t
�Id= 0.0012; f = 0.021 (Figures 2-3 and 2-11)
l:a.P/100' 0.0216 f/J Q2 / ds = 0.0216(0.021) (62.3) (0.93) (25)2/(1.65)5
J
1.52 psi/100 ft.
Total Pipe System Friction AP =((329) (l.52/100)) + 5,'< = 10 psi for friction; >'<Orifice

Total Loss, Feet Liquid= 350(2.31ft./psi)(l/0.93) = 869 feet of Liquid

Checked by: _ Date: _
Figure 2-28. Line sizing sheet for example problem, Example 2-7.

Fluid Flow 101
,-u--- Vo Ive Valve Valve a piping or process system, there may be ( 1) adiabatic
Flaw
Orifice
Gate Control Gale
Dow where for practical purposes there is no exchange of
Plate
heal into or from the pipe. This is expressed by:
*Gale
------ Valve Globe P' V/ = constant (adiabatic) (2-73)
Valve
Dip,
Pipe I
Reactor Ill 350 psig Crude Product Storage or, (2) isothermal flow, which is flow at constant tempera-
Tank at Atmospheric ture ( often close to practical experience) and:
Pressure

Figure 2-29. Liquid flow system, Example 2-7.
P' Va= constant (isothermal) (2-74)

9. Total Pressure Drop Often for a large variety of process gases, some relation-
ship in between expresses the pressure-volume relation-
The control valve must be sized to take the residual ship by:
pressure drop, as long as it is an acceptable minimum.
Pressure drop accounted for:
P' Va" = constant (polytropic) (2-75)

Total psi drop= (245 + 75) (1.52/100) + 5 = 10 psi
For gases/vapors flowing in a pipe system from point 1
Drop required across control valve: with pressure P 1 and point 2 with pressure P 2, the P 1 - P 2
is the pressure drop, �P, between the points [3].
Reactor = 350 psig
Storage O psig Velocity of Compressible Fluids in Pipe
Differential = 350 psi
6.P = 10 psi (sys. friction)
Control Valve L'.P = 340 psi
3.06 WV 3.06\,\1 (2- 76)
d ? d2p
Note that this control valve loss exceeds 60 per-
cent of this system loss, since the valve must take the
difference. For- other systems where this is not the sit- where vm = mean velocity in pipe, at conditions stated for V,
uation, the system loss must be so adjusted as to ft/min.
assign a value (see earlier section on control valves)
of approximately 10 to 20 psi or 25 to 60 percent of \V = flow rate, lbs/hr
the system other than friction losses through the V = fluid specific volume, cu ft/lb, at T and P
valve. For very low pressure systems, this minimum
value of control valve drop may be lowered at the sac- d = inside pipe diameter, in.
rifice of sensitive control. p = fluid density, lbs/ cu ft, at T and P

P' = pressure, pounds per sq foot absolute
Friction Pressure Drop For Compressible Fluid Flow
k = ratio of specific heats, cp/c"
Vapors and Gases
Note that determining the velocity at the inlet condi-
The flow of compressible fluids such as gas, vapor, tions to a pipe may create significant error when results
steam, etc., is considered in general the same as for liq- are concerned with the outlet conditions, particularly if
uids or non-compressible fluids. Specific semi-empirical the pressure drop is high. Even the average of inlet and
formulas have been developed which fit particular sys- outlet conditions is not sufficiently accurate for some sys-
tems and have been shown to be acceptable within engi- tems; therefore conditions influenced by pressure drop
neering accuracy. can produce more accurate results when calculations are
Because of the importance of the relationship between prepared for successive sections of the pipe system (long
pressure and volume for gases and vapors as they flow in or high pressure).

102 Applied Process Design for Chemical and Petrochemical Plants

Friction Drop for Flow of Vapors, Gases, and Steam q\ = rate of How, cu fl/hr at standard conditions (14. 7
Figure 2-30 psia and 60°F), SCFH.

A. The Darcy rational relation for compressible flow [3 J is: 1. When calculated Af> total < 10 percent inlet pres-
sure, use p or V based on inlet or outlet conditions.
o. 000335 nv v 2. When calculated LiP total > 10 percent inlet pres-
2
�PI 100 rt = ------ (2-77)
d 5 sure, but< 40 percent, use average p or V based on
inlet and outlet conditions.
2
0.000001959f(q� ) Sg 3. When calculated iiP total, P1 to P2 is> 40% of inlet
2
or, �p I 100 ft= ------------'-- (2-78)
d"p pressure, primarily for long lines, use the following
choices, or break the line into segments and calcu-
The general procedures outlined previously for han-
dling Iluids involving the friction factor, f, and the Re late LiP for each as above.
chart are used with the above relations. This is applicable
to compressible flow systems under the following condi- Also use Babcock formula given in another paragraph
tions [3]. for steam flow.

where Sg = specific gravity of gas relative to air = the ratio of q' h = 24,700 [Yd /Sg] (�P p 1/K) 12, CFH @ 14.7 psia
2
1
molecular weight of the gas to that of air. and 60°F (2-79)
w
1600
p v
AP/100 ft. = 0.000 336 f W2/d5p
1000
50 800
40 600
500
30 400
300
d
20 200
15 2!
s: 100
:::,
0 80
..... 10 1g_ 60 g
0
'-'
:.0 9 - 50 ::: �<>..
8 gi
:::,
c..., f 40 � �
:;; 7�
c,. 6 :.0 .05 30 � '\;
:::,
"' c..., :::,
"C 5 ·= .04 20 &. �
c:
s: 4 ·:; c::, �
:::,
-c"
c::,
...
;::;::: .03 � c::, � -
,,
c:
'ii: � 8 10 ·= es
':I:,"
,S! .....
c
� :c 6 ..... ...
0
0 .OH� 5 0 �
.... ..... 4 .!! �
I
1.s1 .... 3 &l �
1i, .015 I �
u :t: <I
""
1.0·2
.9.,.�
.8 I .01 1
.7 c::,.. .8
.6
.5 .6
.5
.4 .4
.3
.3
.2
.2
.I
Figure 2-30. Pressure drop in compressible flow lines. By permission, Crane Co., Technical Paper #410, Engineering Div. 1957. Also see 1976
edition.

Fluid Flow 103

2
q\ = 40,700Yd [(L\P) (P' 1 )/(KT 1 Sg)J 112 (2-80) 5. Determine expansion and contraction losses, fittings
same units as Equation 2-79 above and at vessel connections.
6. Determine pressure drops through orifices and con-
where Y = net expansion factor for compressible flow through trol valves.
orifices, nozzles, or pipe 7. Total system pressure drop
K = resistence coefficient, ft
P' = pressure, lbs/sq in. absolute L'i.P Total= (L + Leq) (L'i.P/100) + Item 5 + Item 6 (2-57)
w, = flow rate, lbs/sec.
8. If pressure drop is too large, re-estimate line size and
Isothermal conditions, usually Jong pipe lines [3]: repeat calculations (see paragraph (A) above) and
also examine pressure drop assumptions for orifices
I and control valves.
I 2
" � 144 gA C. Air
v,
v ( fL For quick estimates for air line pressure drop, see
i D
Tables 2-12A and 2-l 2B.
lbs/sec (2- 3)
D. Babcock Empirical Formula for Steam
plus the conditions listed. The equation is based on steady
flow, perfect gas laws, average velocity at a cross section, Comparison of results between the various empirical
constant friction factor, and the pipe is straight and hori- steam flow formulas suggests the Babcock equation as a
zontal between end points. good average for most design purposes at pressure 500
psia and below. For lines smaller than 4 inches, this rela-
D = pipe ID, ft tion may be 0-40 percent high [56].
L = pipe length, ft
A = cross-sectional area for flow for pipe, sq ft
w 2 �
p 1 - Ps = L'i.P = 0.000131 (1 + 3.6/d) (2-82)
B. Alternate Vapor/Gas Flow Methods pd"
L'i.P/100 feet= w2 F/p (2-83)
Note that all specialized or alternate methods for solv-
ing are convenient simplifications or empirical proce- Figure 2-32 is a convenient chart for handling most in-
dures of the fundamental techniques presented earlier. plant steam line problems. For long transmission Jines
They are not presented as better approaches to solving over 200 feet, the line should be calculated in sections in
the specific problem. order to re-establish the steam specific density. Normally
Figure 2-31 is useful in solving the usual steam or any an estimated average p should be selected for each line
vapor flow problem for turbulent flow based on the mod- increment to obtain good results.
ified Darcy relation with fixed friction factors. At low Table 2-13 for "F" is convenient to use in conjunction
vapor velocities the results may be low; then use Figure 2- with the equations.
30. For steel pipe the limitations listed in (A) above apply.
Darcy Rational Relation for Compressible Vapors and
1. Determine C 1 and C 2 from Figure 2-31 and Table 2-11 Gases
for the steam flow rate and assumed pipe size respec-
tively. Use Table 2-4 or Table 2-8 to select steam veloc- 1. Determine first estimate of line size by using sug-
ity for line size estimate. gesLed velocity from Table 2-4.
2. Read the specific volume of steam at conditions, 2. Calculate Reynolds number Re and determine fric-
from steam tables. tion factor, f, using Figure 2-3 or Figure 2-33 (for
3. Calculate pressure drop (Figure 2-31) per 100 feet of steel pipe).
pipe from 3. Determine total straight pipe length, L.
4. Determine equivalent pipe length for fittings, valves,
Lcq·
LiP/100feet = C C V (2- 81)
2
1
5. Determine or assume losses through orifice plates,
4. From Figure 2-20 or 2-21 determine the equivalent control valves, equipment, contraction and expan-
lengths of all fittings, valves, etc. sion, etc.

104 Applied Process Design for Chemical and Petrochemical Plants

YaluesefCa
-
w c,
2500
191

19111
w w
II ,1 ,1 II 1IIOO m
.OS ,m
' ,DI !IOI IOO
• .117 ,15 IOO a
7111
JS
1 � ,2 111 !!ell
15
.14 ,25 a
Pressure Drop per 100 feet Pipe: i ' .03 .3 I IIO DI
...
ilP100 = C1C2V = Ci Cz • ... .D2S .4 a l 5111 250
p I .112 .s I 2QJ
.,
a.
C1 = � DO = Jl.P100P C2= � 0 =ilP100P - • .1115 a .7 25 0 400 1!11 -s
...
..
..
0
..
C2V C2 C1V C1 0 .I 'O 0
J 3,5 ., JI Ii
:::,
!I
C1 = Discharge Factor from Chart .01 ii LO � 1111 '11
:::,
.,. .
C 2 = Size Factor, from Table 2-11 I- 3 .00, > I- D) 90 >
.!:
.,..
.!:
IQ
,OOI
.a11
�
"'"
For llP>40% P1, do not use this method. - 2.5 .ms LS 40 .9 250 10
ii
For Jl.P between 10% and 40% of P1, use average for V. j .Dll9 0 50
0
�
For Jl.P<10% P1, use Vat P1 or Pz. c,:
ilP100 = Psi, pressure drop per 100' pipe. .OOt 2.5 50 = ZIO 40
v = Specific Volume, cu ft/lb. = ,1I03 3 l)
IO
.1»25 25
Note: For quick estimates; not as accurate as friction LS ISO
loss calculations JI02 70 10
.ocns II 15
a 90
9
LI .001 10 m 100 10
.0009
J .Gall
.GJ07
,I
JI006
Figure 2-31. Simplified flow formula for compressible fluids. By permission, Crane Co., Technical Paper #410, 1957. Also see 1976 edition.
6. Calculate pressure drop, LiP/100 ft ( or use Figure 2-34). ing pipe size. Consider reducing losses through
items in step 5 above. Recheck other pipe sizes as
may be indicated.
0.000336 fW 2
�P/100 feet = (2-77)
pd5
0.000000726 fTSs ( (b ) 2 Example 2-8 Pressure Drop for Vapor System
1
P d 5 (2-77A) The calculations are presented in Figure 2-35, Line
7. Total pressure drop, LiP total Size Specification Sheet.
Figure 2-36 is convenient when using Dowtherm vapor.
= (L + Leq) (�PI 100) + Item 5 (2-57)
8. If total line or system drop is excessive, examine the Alternate Solution to Compressible Flow Problems
portion of drop due to pipe friction and that due to
other factors in the system. If the line drop is a small There are several good approaches to recognizing the
portion of the total, little will be gained by increas- effects of changing conditions on compressible flow [ 44, 47].

Fluid F!ow 105

Table 2-11
Simplified Flow Formula For Compressible Fluids Pressure Drop, Rate of Flow and Pipe Sizes*
(Use With Figure 2-31)
Values of C2
I
Nominal Schedule Value Nominal I Schedule I Value f;!i�:k�:! tt�bu1: Value
Pipe Size I Number of C, Pipe Size Number of C, of C2
Inches II Inches 11 Inches -
Ya 40 s 7920 000. I 5 40 s 1.59 16 10 0.004 63
80x 26 200 000. 80 x 2.04 20 0.004 21
120 2.69 30s 0.005 04
tA 40 s 1590000. 160 3.59 40x 0.005 49
80x 4 290 000. ... xx 4.93 60 0.006 12
% 40 s 319 000. 6 ! 40s 0.610 80 0.00700
80 x 718 000. 80x 0.798 100 0.008 04
120 1.015 120 0.00926
1h 40s 93 500. 160 1.376 140 0.01099
80 x 186 100. ... xx 1.861 160 0.01244
160 4 300 000.
... xx 11180 000. 8 20 0.133
30 0.135 18 10 0.002 47
40 s 0.146 20 0.002 56
3,4 40 s 21 200. 60 0.163 .. s 0.00266
80x 36 900. 80 x 0.185 30 0.002 76
160 100 100. .. x 0.002 87
... xx 627000. 100 0.211 40 0.002 98
120 0.252
1 40 s 5950. 140 0.289 I 60 0.003 35
80x 9'640. ... xx 0.317 80 0.003 76
160 22 500. 160 0.333 100 0.004 35
... xx 114100. 120 0.005 04
10 20 0.039 7 140 0.00573
11/.a 40s 1408. 30 0.042 1 160 0.00669
80 x 2110. 40 s 0.0447
160 3 490. 60x 0.0514
·.·. xx 13 640. 80 0.056 9 20 10 0.001 41
20 s 0.001 50
100 0.066 1 30x 0.00161
11h 40 s 627. 120 0.075 3 40 0.00169
80x 904. 140 0.090 5 60 0.001 91
160 1656. 160 0.105 2
... xx 4630. 80 0.00217
12 20 0.015 7 100 0.002 51
2 40 s 169. 30 0.016 8 120 0.002 87
80 x 236. ... s 0.017 5 140 0.003 35
160 488. 40 0.018 0 160 0.003 85
.. . xx 899 . ... x 0.019 5
60 0.020 6
21h 40s 66.7 24 IO 0.000 534
80x 91.8 80 0.023 1 20s 0.000 565
160 146.3 100 0.026 7 .. x 0.000 597
... xx 380.0 120 0.031 0 30 0.000614
140 0.035 0 40 0.000651
160 0.042 3 60 0.000741
J 40 s 21.4
80x .28.7 14 10 0.009 49 80 0.000 835
160 48.3 20 0.009 96 100 0.000 972
... xx 96.6 30 s 0.01046 120 0.001119
40 0.010 99 140 0.001274
31/z 40 s 10.0 ... x o.ou 55 160 0.001478
80x 37.7 60 0.012 44
4 40 s 5.17 80 0.014 16 Note
80 x 6.75 100 0.016 57 The letters s, x, and xx in the col-
120 8.94 120 0.018 98 umns of Schedule Numbers indicate
Standard. Extra Strong. and Double
11.80
0.021 8
140
I 160 18.59 160 0.025 2 Extra Strong pipe respectively.
... xx
-
By permission, Crane Co., Technical Paper#4IO, Engineering Div., 1957. See author's note al Figure 2-31.

Table 2-12A
Flow of Air Through Schedule 40 Pipe*
(Use for estimating; for detailed calculations use friction factors, f)
For lengths of pipe oth er than Free Air ·- ·- ·-·--- Pressure Drop of Air
Com-
100 feet, the pressure drop is q'm
proportional to the length. pressed Air In Pounds per Square Inch
Thus, for 50 feet of p ipe, the Cubic Feet Cubic Feet Per 100 Feet of Schedule 40 Pipe
pressure drop is approx imately Per Minute Per Minute For Air at I 00 Pounds per
one-half the value give nin the at 60 F and at 60 Fand Square Inch Gauge Pressure
table ... for 300 fee t, three 14.7 psia 100 psig and 60 F Temperature
times the given value , etc.
The pressure drop is also in- Ys' 1.1c• :Ya" 112' I
versely proportional to the 1 0,118 0,361 0.083 0.018
absolute pressure and directly l 0.156 1.31 0.185 0.064 0.020
.proportional to the absolute 3 0.384 3,06 0.605 0.133 0.042 3,4"
temperature. 4 0.513 4.83 1.04 O.ll6 0.071 0.017
0.641
0.343
0.106
1.58
7.45
5
Therefore, to determ ine the 6 0.769 10.6 2.13 0.408 0.148 0.037 1 ...
pressure drop for inlet or aver- 8 l.OlS 18,6 3.89 0.848 0.155 0.061 0.019
age pressures other t han 100 10 l.l8l 28.7 5.96 1.16 0.356 0.094 0.029 11/c" 1112·
...
l.9ll
15
13.0
psi and at temperatur es other 20 l.563 ... za.s l.73 0.834 O,lOI 0.062 0.026
1.43
4.76
0.345
0.101
than 60 F, multiply th evalues
given in the table by t he ratio: 25 3,204 ... 35.6 10.5 l.ll o.748 0.156 0.039 0.019
7.34
0.516
...
...
30
0.119
3.15
3.845
0.055
0.026
+t)
4.486
(JOO+ 14.7)( � 35 5,116 ... ... 14.l 4.14 1.00 0.293 0.073 0.035
...
...
18.4
0.379
40
5.49
P+ 14.7 5 20 45 5.767 ... ... 23.1 6.90 1.30 0.474 0.095 0.044 2'
1.61
0.055
0.116
where: 50 6.408 28.5 8.49 1.99 0.578 0.149 0.067 0.019
60
0.094
"P" is the inlet or average 70 7.690 2W 40.7 12.2 l.85 0.819 0.200 0.116 0.027
...
8.971
1.10
3.&3
16.5
0.036
0.270
gauge pressure in pound s per 80 10,25 0.019 ... ll.4 4.96 1.43 0.350 0.162 0.046
square inch, and, 90 11,53 0.013 ... 17.0 6.15 1.80 0.437 0.103 0.058
100
"r" is the tempera ture in 115 12.81 0.029 3" 33.2 7.69 l.21 0.534 0.247 0.070
...
16.02
degrees Fahrenhei t under 150 19.21 0.044 0.021 ... 11.9 3.39 0.815 0.380 0.107
17.0
0.151
1.17
o.537
0.062
4.87
consideration. 175 22.43 0.083 0.028 ... 23,1 6.60 1.58 0.727 0.205
200 lS.63 0.107 0,036 3lf2" ... 30,0 8.64 l.05 0.937 0.164
The cubic feet perm lnute of llS 28.84 0.134 0.045 0.022 37.9 10.8 2.59 1.19 0.331
compressed air at an y pres- :2SO 31.04 0.164 0.055 0.017 ... 13.3 3.18 1.45 0.404
...
1.75
sure is inversely propo rtional :275 35.14 0.191 0.066 O,OJl ... 16.0 3.83 l.07 0.484
38,45
0.078
O,l3l
300
0.037
19.0
4.56
0.573
to the absolute press ure and 315 41.65 0.170 0.090 0.043 ... 12.3 5.32 l.4l 0.673
directly proportional to the 4"
absolute temperature 350 44.87 0.313 0.104 0.050 ... 15.8 6.17 l.80 0.776
...
375 48.06 0.356 0.119 0.057 0.030 29.6 7.05 3.lO 0,887
To determine the cu bic feet 400 51.16 0.401 0.134 0.064 0.034 ... 33.6 8.02 3.64 1.00
...
425
54.47
0.451
37.9
0.038
9.01
1.13
4.09
0.151
0.072
per minute of compressed air 450 57.67 0.507 0.168 0.081 0.042 ... . .. 10.2 4.59 1.26
at any temperature an d pres-
sure other than stand ard con- 415 60.88 0.561 0.187 0.089 0.047 ... 11.3 5.09 1.40
...
O.l06
0.051
1.55
12.5
0.099
5.61
0.613
64.08
500
ditions, multiply the value of 550 70.49 0.749 0.148 0.118 0.061 ... 6.79 1.87
15.1
cubic feet per minut e of free 600 76.90 0.887 0.193 0.139 0.073 ... I 18.0 8.04 l.ll
air by the ratio: 650 83.30 1.04 0.341 0.163 0.086 5" ... 21.1 9.43 l.60
( 14.7 )(460 t) 700 89.71 1.19 0.395 0.188 0.099 0,032 24.3 10.9 3.00
ss.n
14.7+P -n- 750 101.5 1.36 0.451 0.114 0.113 0.036 27.9 ll.6 3.44
0.513
0.144
0.127
800
14.l
31.8
3.90
I.55
0.041
850 108.9 1.74 0.576 0.274 0.144 0.046 35.9 16.0 4.40
900 115,3 1.95 0.641 0.305 0.160 0.051 6" 40.2 18.0 4.91
950 1:21.8 l.18 0.715 0.340 0.178 0.057 0.023 ... 20.0 5.47
l 000 118.l l.40 0.788 0.375 0.197 0.063 0.025 ... 22.1 6.06
1100 141.0 l.89 0.948 0.451 0.136 0.075 0,030 ... 16.7 7.19
1 lOO 153.8 3.44 1.13 0.533 0.179 0.089 0.0351 ... 31.8 8.63
1300 166.6 4.01 1.31 0.616 0.317 0.103 0.041 ... 37.3 10,1
1400 179.4 4.65 1.51 0.718 0.377 0.119 0.047 11.8
,,. 1500 105.1 5.31 1.74 0.8l4 0.431 0.136 0.054 8" 13.5
192.2
6,04
1.97
1600
0.490
0.931
0.061
0.154
15.3
1800
Calculatlon1 for P :2000 130.7 7.65 l.50 1.18 0.616 0.193 0.075 19.3
256.3
9.44
Other than Schedu ,. 40 3.06 1.45 0.757 0.237 0.094 0.023 10" 13.9
:2500 310.4 14.7 4,76 l.25 1.17 0,366 0.143 0.035 37.3
384.5
To determine the ve loci% of 3000 448,6 21. l 6.81 3.lO 1.67 0.5l4 0.104 0.051 0.016
9,23
3 500
l.26
0.709
water, or the pressu re rop 4000 511.6 28.8 ll.l 4.33 l.94 0.919 0.176 0.068 0.021
0.358
0.028
5.66
0.088
37.6
of water or air, throu gh pipe 4500 576.7 47.6 15.3 7.16 3.69 1.16 0.450 0.111 0.035 12#
other than Schedule 40, use ...
8.85
the following formula s: 5000 640,8 ... 18.8 12,7 4.56 1.41 0.552 0.136 0.043 0.018
27.1
0.061
769,0
2.03
6.57
0.794
0.195
6000
0.025
...
8.94
v4 = v40 (_�o ) 2 7000 1025 ... 36.9 17.2 14.9 2.76 1.07 0.262 0.082 0.034
897.1
. ..
8000
0.339
0.107
1.39
0.044
11.7
3.59
22.5
...
. ..
1153
9000
0.055
0.427
1.76
0.134
4.54
28.5
or
2.16
0.526
...
AP,.= AP,o( i 10000 1282 ... ... 35.2 18.4 5.60 2.62 0.633 0.164 0.067
11000
. ..
0.197
0.081
1410
22.2
6.78
12000 1538 ... ... . .. 26.4 8.07 3.09 0.753 0.234 0.096
0.273 e.m
...
...
...
9.47
Subscript "a" refers to the 13 000 1666 ... . .. ... 31,0 11.0 3.63 0.884 0.316 0.129
HODO
4.21
36.0
1.02
1794
Schedule of pipe through 15000 19l2 ... ... ... .. .
which velocity or p ressure 16 000 2051 ... . .. ... . .. ll.6 4.84 1.17 0.364 0.148
0.167
5.50
14.3
1.33
0.411
drop is desired. 18 000 2307 ... ... ... . .. 18.2 6.96 1.68 0.520 0.213
:20000 2563 ... ... ... ... 22.4 8.60 2.01 0.642 0.160
Subscript "40" refer s to the l:2000 2820 ... ... ... . .. 27.1 10.4 2.50 0.771 0.314
velocity or pressu re drop ... ... ... ...
32.3
through Schedule 40 pipe, as :24000 3076 ... ... ... ... 37.9 12.4 2.97, 0.918 0.371
:?6000
3332
1.12
14.5
0,435
3.49
given in the tables on these 28 000 3588 ... ... ... . .. ... 16,9 4.04 1.25 o.505
facing pages. 30000 3845 ... ... I ... . .. ... 19.3 4.64 1.42 0.520
*By permission Technical Paper No. 410, Crane Co., Engineering Div., Chicago, 1957.

Fluid Flow 107

Table 2-12B
Discharge of Air Through an Orifice*
In cubic feet of free air per minute at standard atmospheric pressure of 14.7 lb. per sq. in. ahsolute and 70° F.
Gauge DIAMETER OF ORIFICE
Pressure
.i:«
before Orifice .v.>. n" %" %:" %" Y2" %" %'" Ys" 1.
1,,
64
32
in Pounds ---- I I I I I I I I
per sq. in. Discharge in Cubic feet of free air per minute
-,
1. ........... .028 .112 .450 1.80 7.18 16.2 28.7 45.0 64.7 88.1 115
2 ............ .040 .158 .633 2.53 10.1 22.8 40.5 63.3 91.2 124 162
3 ............ .048 .194 .775 3.10 12.4 27.8 49.5 77.5 111 152 198
4 .......•.... .056 .223 .892 3.56 14.3 32.1 57.0 89.2 128 175 228
5 ............ .062 .248 .993 3.97 15.9 35.7 63.5 99.3 143 195 254
- -··
6 .068 .272 1.09 4.34 17.4 39.1 69.5 109 156 213 278
7 .073 .293 1.17 4.68 18.7 42.2 75.0 117 168 230 300
g .083 .331 1.32 5.30 21.2 47.7 84.7 132 191 260 339
12 .095 .379 1.52 6.07 24.3 54.6 97.0 152 218 297 388
15 .105 .420 1.68 6.72 26.9 60.5 108 168 242 329 430
- - ----
20 .123 .491 1.96 7.86 31.4 70.7 126 196 283 385 503
25 .140 .562 2.25 8.98 35.9 80.9 144 225 323 440 575
30 .158 .633 2.53 10.1 40.5 91.1 162 253 365 496 648
35 .176 .703 2.81 11.3 45.0 101 180 281 405 551 720
40 .194 .774 3.10 12.4 49.6 112 198 310 446 607 793
-
45 .211 .845 3.38 13.5 54.1 122 216 338 487 662 865
50 .229 .916 3.66 14.7 58.6 132 235 366 528 718 938
60 .264 1.06 4.23 16.9 67.6 152 271 423 609 828 1082
70 .300 1.20 4.79 19.2 76.7 173 307 479 690 939 1227
80 .335 1.34 5.36 21.4 85.7 193 343 536 771 1050 1371
- -
90 .370 1.48 5.92 23.7 94.8 213 379 592 853 1161 1516
100 .406 1.62 6.49 26.0 104 234 415 649 934 1272 1661
110 .441 1.76 7.05 28.2 113 254 452 705 1016 1383 1806
120 .476 1.91 7.62 30.5 122 274 488 762 1097 1494 1951
125 .494 1.98 7.90 31.6 126 284 506 790 1138 1549 2023
Table is based on 100% coefficient of flow. For well rounded entrance multiply values by 0.97. For sharp edged orifices a multiplier of
0.65 may be used for approximate results.
Values for pressures from 1 to 15 lbs. gauge calculated by standard adiabatic formula.
Values for pressures above 15 lb. gauge calculated by approximate formula proposed by S. A. Moss.
aCP1 Where:
W. = .5303 ,,. -- \V 8 = discharge in lbs. per sec.
1
V T 1 a = area of orifice in sq. in.
C = Coefficient of flow
P1 = Upstream total pressure in lbs. per sq. in. absolute
T 1 = Upstream temperature in °F. abs.
Values used in calculating above table were; C = 1.0, P 1 = gauge pressure + 14.7 lbs./sq. in. T1 = 530° F. abs.
Weights (W) were converted to volumes using density factor of 0.07494 lbs./cu. ft. This is correct for dry air at 14.7 lbs. per sq. in
absolute pressure and 70° F.
Formula cannot be used where P1 is less than two times the barometric pressure.
*By permission "Compressed Air Data," F. W. O'Neil, Editor, Compressed Air ,Hagazine, 5th Edition, New York, 1939 [49].




Friction Drop for Air Example 2-9: Stearn Flow Using Babcock Formula

Table 2-12A is convenient for most air problems, not- Determine the pressure loss in 138 feet of 8-inch
ing that both free air (60°F and 14.7 psia) and com- Schedule 40 steel pipe, flowing 86,000 pounds per hour
pressed air at 100 psig and 60°F are indicated. The cor- of 150 psig steam (saturated).
rections for other temperatui·es and pressures are also
indicated. Figure 2-37 is useful for quick checking. How- Use Figure 2-32, w = 86,000/60 = 1432 lbs/min
ever, its values are slightly higher (about 10 percent) than Reading from top at 150 psig, no superheat, down ver-
the rational values of Table 2-11, above about 1000 cfm of tically to intersect the horizontal steam flow of 1432
free air. Use for estimating only. lbs/min, follow diagonal line to the horizontal pipe size

108 Applied Process Design for Chemical and Petrochemical Plants
Table 2-13 Sonic velocity will be established at a restricted point in
Factor "F'' For Babcock Steam Formula* the pipe, or at the outlet, if the pressure drop is great
Nominal Pipe Size *Standard Weight #Extra Strong enough to establish the required velocity. Once the sonic
Inches Pipe Pipe velocity has been reached, the pressure drop in the system
V2 955.1 x 10- 3 2.051 x
% 184.7 x J0-3 340.8 x I 0-3 will not increase, as the velocity will remain at this value
l 45.7 x 10-3 77.71 x 10- 3 even though the fluid may be discharging into a vessel at
l v.. 9.432 x 10- 3 14.67 x 10- 3 a lower pressure than that existing at the point where
IY2 3.914 x 10- 3 5.865 x 10-s sonic velocity is established.
2 951.9 x J0-6 1.365 x 10- 3
2V2 351.0 x 10- 6 493.8 x 10-6 In general, the sonic or critical velocity is attained for an
3 104.7 x J0-6 143.2 x 10-6 outlet or downstream pressure equal to or less than one
3V2 46.94 x J0-6 62.95 x 10- 6 half the upstream or inlet absolute pressure condition of a
4 23.46 x J0-6 31.01 x 10- 6 system. The discharge through an orifice or nozzle is usu-
5 6.854 x 10- 6 8.866 x 10-6
6 2.544 x 10- 6 3.354 x 10- 6 ally a limiting condition for the flow through the end of a
8 587.1 x 10-9 748.2 x 10·-9 pipe. The usual pressure drop equations do not hold at
10 176.3 x 10- 9 225.3 x J0-9 the sonic velocity, as in an orifice. Conditions or systems
12 70.32 x (0-9 90.52 x 10- 0 exhausting to atmosphere (or vacuum) from medium to
140.D. 42.84 x 10- 9 55.29 x J0-9 high pressures should be examined for critical flow, other-
160.D. 21.39 x 10-9 27.28 x J0-9
180.D. 11.61 x 10- 9 14.69 x J0-9 wise the calculated pressure drop may be in error.
200.D. 6.621 x 10-u 8.469 x 10- 9 All flowing gases and vapors (compressible fluids)
240.D. 2.561 x 10- 9 3.278 x 10- 9
including steam (which is a vapor) are limited or
*Factors are based upon 1.0. listed as Schedule 40. approach a maximum in mass flow velocity or rate, i.e.,
#Factors are based upon I.D. listed as Schedule 80. lbs/sec or lbs/hr through a pipe depending upon the
tBy permission The Walworth Co.
specific upstream or starting pressure. This maximum
rate of flow cannot be exceeded regardless of how much
the downstream pressure is further reduced [3]. To
of 8 inches, and then vertically down to the pressure drop determine the actual velocity in a pipe, calculate by
loss of 3.5 psi/100 feet.
For 138 feet (no fittings or valves), total .1P is 138 (3.5/ 3.06W V
100) = 4.82 psi. v = d 2 or use Figure 2-34.
For comparison, solve by equation, using value of F =
587.l X 10- 9 from Table 2-13. This maximum velocity of a compressible fluid in a
pipe is limited by the velocity of propagation of a pressure
LlP/100 ft= (1432) (587.1 x 10- )/0.364 wave that travels at the speed of sound in the fluid [3].
9
2
= 3.32 psi/ I 00 ft This speed of sound is specific for each individual gas or
LlP total= (3.32/100) (138) = 4.75 psi
vapor or liquid and is a function of the ratio of specific
heats of the fluid. The pressure reduces and the velocity
These values are within graphical accuracy. increases as the fluid flows downstream through the pipe,
with the maximum velocity occurring al the downstream
Sonic Conditions limiting Flow of Gases and Vapors end of the pipe. When, or if, the pressure drop is great
enough, the discharge or exit or outlet velocity will reach
The sonic or critical velocity (speed of sound in the
fluid) is the maximum velocity which a compressible fluid the velocity of sound for that fluid.
can attain in a pipe [3]. If the outlet or discharge pressure is lowered further,
the pressure upstream at the origin will not detect it
v, = [ ( cp/ c.) (32.2) (1544/MW) ( 460 + t) ]112 (2-84) because the pressure wave can only travel al sonic veloci-
= 68.1 [(cp/cv) P'/p] 112, ft/sec ty. Therefore, the change in pressure downstream will not
be detected upstream. The excess pressure drop obtained
where the properties are evaluated at the condition of by lowering the outlet pressure after the maximum dis-
sonic flow. charge has been reached takes place beyond the end of
This applies regardless of the downstream pressure for the pipe [3]. This pressure is lost in shock waves and tur-
a fixed upstream pressure. This limitation must be evalu- bulence of the jetting fluid. See References 12, 13, 24, and
ated separately from pressure drop relations, as it is not 15 for further expansion of shock waves and detonation
included as a built in limitation. waves through compressible fluids.

Fluid Flow 109

AVERAGE PRESSURE -LB. PER SQ. IN. ABSOLUTE
·'-" '�� �" \"\ ""� ,,, -& \\ t,\' Cl\�,\'\'\\ .. ,
' .x
� -� " � ....:i� � �I\ \ " ....... .\ '- '\ 0
,._� � r-� � L \ ' .... - \ ' - \ '·" ., \ IOO �
l
I
-14i
-f.,\ 0
- " 1-,.l \ ,-.... ,� - \ u I\ I\ so� ..L ., . ' � �- \ \ 1 !00!
�
l..l
·oo a::
..... \...
_,
l "to u, ..... 11 o-•., ,...
-"\, "l"""' \ 1 � .... L_1 � t-- u \ \ { \ = ·- �
\ 1 \ II � I = --,, 1 I \ \ = ICO II>
700
I \ \ \ ,-.......1.. ,\ ;, M.� --i 1 c
-r- t-- t-- t-- r- r- i---.. t-- ' r- t--
r--, i---.. -i-- - ...... -i--. ........... - ....... ..... i--. .... t-- .... t--
1()- ...... - ........... ,___ 40000
l!O- r- r--. t--.. r--. -i---.. -t-- r- t--
- -i--. -r-.. t--.. 1--... ,-..... ...... ....... -i- 't..... -t..... - _oooo
-
40 -,-..... -i- .._
- -i--.- t--.. !-.... t-... r--. t--.. t..... �
r--, ,-..... t--.. :-.... t-- 1--...� t..... - ...... 't..... -t-- iliilEIOOOO
t..... .=-8000
:,0 - -r- -t..... - :-.... t-- t-- - -i- -
t--
i-�
,-.....
t--
t--
20 -- -r-� ,-..... r- - t-- t-- - ...... .......... :-.... c::-4000
,-.....
-
,-.....
.......
t--
t--
�
t--
16- -t-- t-- -- ....... ..... t-- t-- ...... t-- t-- - 2000
14-.: r- r- ,-..... t--.. t-- <, t---.. t-- I--... =--1000
1..-_ r--... r-_ - i,... I ,-..... t--.. - ....... -.._ --i-- -t-- §= ___
.......
:-.... .... ---r-. ..
-
10- r--... - ,-..... - ....... �- - ,-..... � ....... -t---.. .... i--. t..... �-�o
t--
.. _
t-- t-- ,-..... .... �- -- .....
I
- -i---.. ..... t-- I-..._ :--..._ --·� :-,... t-- 1--. -IQO
..._
·- t--
-
6 - - ..... � - '""· !-.... t--
....... �
,- r- t-- -- -- t--� t-- -...__ -i- t--
:-...._
-
- r--... t-- t-- r- -...__ -""- i- ,___ -
�
4 r- -r- -i---.. ""- ,-...._ I-.... 1--... t.....
.......
- - - .......
.....
.....
.....
:, :--- -- -.... :--... - i--... .....""- i- -=-r.o
.......
- ...... -- -- t-- t-- - -i---.. �.
�
-1--...
.......
- -:-.... - ""- t-- ....... ""- 1--... ==:10
.......
z- ...... ....... ....... ...... ...... ....... ....... t--
- ....... :--... -- ,..._ ...... -i,...J I --- ....._ i--. -- E-4
-....
.......
1--...
.......
- - - -.... -� i-,......_ !-.... ....... --- -i- i- - r.
.......
.......
.......
-
.......
t--
1--.
i--...
-i--.. ....... _ -- i-,... :--... ...... ....... i- :-.... �=
I t-- i-,... ...... ...... r---.. ...... -.... ...... i--. i--.
-....
......
o.a- !'-.. - -- -- r- t-- - ..... _ t---.. -i- -i---.. - ..
......
.75 -- -- ....... _ t-- i--... ....... ..... - ..... i- -
r--...
0.6 - -.... r- :-.... ..... 1--... -1--... ""- r-:t
.5 r---.. :--... ...... ..... t-- 1--... COPYRIGHT-WALWORTH COIINH'l'-1157 !§ -
11111p1111 I 1111 Iii' 111111 '11 11� � 13111111111111111r1Tl1""1'"l 111
JI
..
0. 2! s • q'° � ci Of i
• q q
PRESSURE LOSS IN Le. PER SQ. INCH PER 100 FEET
Based on Babcock Formula: P = 0.000131 (I+ d ;!}
3 6)
Figure 2-32. Steam flow chart. (By permission, Walworth Co. Note: use for estimating only (this author).)
The maximum possible velocity of a compressible fluid P' = pressure, Psi abs (Psia)
in a pipe is sonic (speed of sound) velocity, as: �· = sonic or critical velocity of flow of a gas, ft/sec
V 1 = specific volume of fluid, cu ft/lb at T and P'
0.000001959f (q'h) 2Sg 2 g = acceleration of gravity= 32.2 ft/per/sec
or, �p I l 00 ft = (2- 78)
d''p Thus the maximum flow in a pipe occurs when the veloc-
ity at the exit becomes sonic. The sonic location may be
where k = ratio of specific heat for gas or vapor al constant
pressure other than the exit, can be at restrictive points in the sys-
R = individual gas constant = MR/M = 1544/M tem, or at control/ safety relief valves.
M = molecular weight
MR = universal gas constant = 1544 Shock waves travel at supersonic velocities and exhibit
T = temperature of gas, R, = ( 460 + °F) a near discontinuity in pressure, density, and tempera-

110 Applied Process Design for Chemical and Petrochemical Plants

w d
�000 Index Re -= 6.31 W/d/1, JL 0.3 ·-
h
800 = 0.-482q' Sg Id/A
0.008- -
-
600 - 0.4-
500 : -
400 0.009- 0.5-
Internal pipe diameter, in. - -
300 0.6-
-
-
�� 6 � 0.7-:_
200 �:: O.ol I - 0.8--: '--
-
0.9-:
I -----12 O.o12- 1.0- '--
-
-
100 '\) � 10 0.013- -
-
80 0.014- - 1�4
I
- '-
-
11 :=------: .;
I
60 NR, I :::=-----::: 8 0015- 1.5-: '-- 1�2 .5
-
-
-
50 IQ,000 ... 0.016- - .;
a.
2-
c:
40 ..: <) 8,000 -. .'!! - ·- - � 2 ·a.
.....
.c 6,000 4f:: '- .� 0.017- ... - 0
'-
- ,_
a.
30 ..,; 4,000 -i 3- '- - - 'a. - 2h ""' ...
'- a 0.(Jl8-
.,; 3,000 '.:, = '- u - - 3 ..
:::,
:::, 1�2 '- :E- 0.019- - 3- "D
0
L---
...
20 0 2,0� I • >- ·;;; - ... - - .c
0 0.020-
u
= -; (3) 1\ 1· 174- - u u -- 312 -
"'
"'
-
·;;
- I, � 3/4= = � - .� 4- - 4 ..
E
I
0
-
;:
0
...
-
""
-
..
-
0
- :::,
"'
10 ... "D 600 ' h= - - ;; s-= - 5 ·;;;
a c: 400 ' 0 - c
"'
a
a: "' .D c - ·e
0
c
ct
::,
... -
-
0
-
.c 300 ' .. 0.025- - 6- - 6
6 - 200 I\\ \ 7- z
0
x
5 .D '� 8- 8
- ,_
"'
-
4 E 100 � I'� 9-
:::,
c
80
....
10
"" 60 N
0
c: 40 - '- 12
"" ...
a: 30 - -
14
-
.......
20 " 0.035- 15- 16
-
'I\ - -
1\1' - -'- 18
-
10 0.040- - '-- 20
8 - 20-
6 - -� 24
(4) - -
4 ' -
-
-
3 0.045-: 30-
-
2 -
0.01 . 0.02 0.03 0.04 005 0.050....: -
f -frrct,on factor for clean-- 40 ,_
steel and wrought iron pipe
Figure 2-33. Reynolds number for compressible flow, steel pipe. By permission, Crane Co., Technical Paper #410, Engineering Div., 1957. Also
see 1976 edition.
ture, and a great potential exists for damage from such the speed of sound. The ratio of the actual fluid velocity
waves [15]. A discussion of shock waves is beyond the to its speed of sound is called the Mach number [38].
scope of this chapter.
The velocity of sound at 68°F in air is 1126 ft/sec.
Velocity considerations are important in rotating or For any gas, the speed of sound is:
reciprocating machinery systems, because, if the com-
pressible fluid velocity exceeds the speed of sound in the
fluid, shock waves can be set up and the results of such », = � kgp "/p, ft/ sec (2- 86)
conditions are much different than the velocities below (equation continued on page 113)

Fluid Flow 111



3.06WV 3.06W
v= ----
d 2 d2p

OTT'T...,........,....,..........,......,...,....,.� .......................... ���,........,..,.,..�.....- ......... �������.,...,.���,�
H-H+H-1-1.\+--+-l-\-l-+i---f\r+-+-lf-\+-+++-!-+l++-++++t ,� ;;;
111-U-U+l��I-Al,++<���l+l+,��,l---l���-l-4,...j....l-4���++1,�j
�H-H-�++-1,\Si z
...++sl..+-1--+-l4-i-��++++++l-lll-l--+-l'-M...j...j.-!-+li-1+1+,1.l+M������+l-l-l-l,1.1�1+4,,+--...j.+��:\.+4-+--'--+44\!-1-+-I-I-�� �
H+l+++-+\+---+--\+ ....... -t-\+-+-\-+-*+-++-fr+++-+++-H� �
1-+-+-1�>-i,..;�...j...j.��+1-'+l+µj....�lll--l-++141!--l-i+.f,l.j.j�����l++,l�-l+IJI-U+U-l-+--4--4-....\-1---1-,...j....�..ij.....j��-+-+�� �
l--'-+-'4-J4-4-l+.L+4-4J�J.µ,J-J,-u...J4-I..U..4..L.L.1.L�u.µ.LJ..4.J..LL.4u....L.1J.I.J.l.Jl.l.j.lJ.LLjlLLL4,-l-4.l.,....L....ll....a..i.......+,1o..1-..ll..L�.....L-J�+,l,.i.J..l.jt� �
� �

sa4�u1 U! 'ad!d 10 1 1aweio 1eu1a1u1 - p


JROH 1ad spunod io spu sno41 U! 'MOl.::I IO a1e� - M
.,,
N ...I
I I I I I I I I I I I , I I I I I , I I I I 1111 I , I I I I I I I I I I l I I I I I I I I Ill I I I I I I I I I I l1i1 I, 1 •••• l •• 11111111

I
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0
a1nuii, 1ad 1aa.::1 7. spuesno41 ui '.<;1 � 0:a11 ·-'1
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a,
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-=
N
100.::1 �,qn:> 1a spunod u, '.<1tsuao 1481.1M - d
.,,
.... ..,.,
s
I
N ('I') .... ,.,., "'
q
q <=! q
q
U"? �
""!
Q.. "" I I 11 <D I "" I I � I I "": I I ii I I I I c, I I I 11 I I a:, s
�
N
...J
I
I
,� I� I � !:!:! s I QC) I I C.O�V(¥') I I. l"'I . .., "'"' .... "'! I I ""! !:!? I I
I
I
I
I
I �
punod rad a.::1 �,qn:, ui •awn1011 �!J!�ads - ,:!
I
-·
•••
N....f
\ �
I §
\ ' ' 1 \ ' � I\ \ \ ' I � ·-
I I \
...
.s;::
c:
' \ ' \ I \ \ ,v .� .s;::
e
...
"' ...
"' � \ � 1, I\ 11 � u,
aO
� 11 'i I� \�� � � 8 ...
�
I ' \ ' a:, ...
11� I\ �� �,.
.':: l � � ' \ ' \ 11 \ \ � � � � "'i-.. c
c:
"' -- II II -� (� \ :\ -- � .,;
-i:
I�
\\
oft � I � \\ \ ' I� '' ' � � :::,
.,\ \ \ � .;;
p. � ) § ...
:,, \ \I\ \ I\ Q.
.........
\
�
e
!\
�\ ' \ \\\ ' ' \ �-- ' \. 8 "'
.,,
\
I-
� <g,,
�
� ,, \. \ ' ':,\ea\\\ � 1; -
I
\
l( � �
\ \ \ \�)... ':,a\111a\e � f:i
,iii
.... �
I 111111,
Figure 2-34. Pressure drop in compressible flow lines. By permission, Crane Co., Technical Paper #410, Engineering Div., 1957. Also see 1976
edition.

112 Applied Process Design for Chemical and Petrochemical Plants

I SHEET NO.--------
LUDWIG CONSULTING ENGINEERS


By __ LINE SIZE SHEET

Date----------- Char9e Ha, _
Line Ha, LP - 61 Flciw Shut Drawl,119 Na. _
L In• Dea cri ptl on __ V_e:...n_t.:._t:...h_r_o_u...::g:...h_E_x_c_ha---' n g � e-r_f_o_r_T_o_w_e_r_T_-_...:3:.__ _


N2 + Hydrocarbon
140 __ F
Fluid in lin•--------------------- Temperature _;_.:....:.._ 0
GPM (Cale,) GPM (du,)------- Pressur• 5_ . :... 3 __ p1i9
CFM (Calc,) __ 2_0_.6-=0 CFM (du,) _2_2_7_0 _ Sp. Gr. 0.975
11.3
Lbs,/hr.(Calc.) __ 1..:.0..i...;:...84...c...;.. 1 Lb1,/hr.(des,) 12, 000 Sp, Val, cu.lt,/11,,
Recammended Velocity------------- fp --- Vi 1ca1 itY._o.,:. . _0_ 1 9:::__ cp

Straight pipe, fitting•, valve• P re1aure Drop
exponslon, contraction, etc. Item In•
Item Mo, Unit E q, Ft. Tata I E q, Ft. Pipe & Equivalent n n&.,17
C,t- I in<> i; Orifice
,-�,-., \I 1 11 11 Matar Valve
1 6 6 Ml 1cellaneou1
T<><> - 4-n 1 en so l'v.-h�"�"'r rlrnn 1 i:;n

Total 1- t;F,
Tatol 72
E atlmated lln• ai:ae 10 11 (existing) _

Cross-sect, area, 10 11 pipe=0,547 sq.ft. Actuol V eloclty _ _,41.JJ�SwDc__ fp

__
Velocity= 2270/0.547 = 4150 Feet/Min, Unit Lass per 100 ft._o_._0_8_57 p_s_i _
Tatal head 1011
In fHt af liquid -----------
Total pre, sure
drap in psi __ -..:cl...:c•..::.5-=.6 _


Selected pipe alu 1 O" Moterlal & Weight Schedu 1 e 40, S tee 1
6.31 W (6.31) (12,000) 3,98 x 10
,,
Calculatlana:Re = d � (I0,02)l0.019)
f = o. o 1 58 I'? = 1 ;V
(o.000336)(f)(w)2 - (o.000336)(0.0158)(12,000)2

l
= 0.0857 Psi/100 equivalent feet of plpe(as pipe,fittings,valves, etc)

AP Total (friction)= (0.0857/100)(72) 0.0617 Psi



Checked by: _ Cote:-----------------

Figure 2-35. Example of pressure drop for a vapor system, Example 2-8.

Fluid Flow 113



80 , ,
J IJ J J "
50 I/ [/ I II IJ I f
40 1, I, J , I, C7 7 , I/ I/ , I '
'
1, f
30 ' I I, IJ I 11 11 I
I/ I I/ II ' I II J I J
20 � ' ,
) , J ) I/ J I I II
ti: 1 S v � v v v 11
0 I ll ll � I/
0 10 1, LJ ' J ,
\\ - Ii ,, ,
I I ,,--1 --,, , I I I I
� '
i 8.0 :.J - - t-t- � L- ;.. '/- ;;."l- -:,r ._,.i l7 �.t'.J � �J -� 7
'J
�I
g 5.0 .::.1 - .... v I 17 I I
I
�
.,, � . / r J " , , J , , J -. , ' I,
': 4.0
--..
s 3.0 � J v J I I/ ,, I I 17 IJ
I � I/ I I II' I I I
'
e, 2.0 II' II ' J
0 ) II I I J J l
� i.s II' � v 'I'
w I I j I/ II I I J II I
� 1.? ' I / TEMP. •F CORRECTION
� 0.8 I , I I , I I I
w J IJ J 500 8.95:
"' 0.5 J I I J 550 5.77-
.....
3.72 -
600
0.4 J J I I/ I r , J J I I/ 650 2.40 �
I r J J J ,
0.3 I I r I II' I IJ I I , I I 700 1.55
750
1.00
I
J
0.2 v J I I r I/ � I/ I) 1, , I I/ I 11 I I ....
All sizes are schedule
J
J
40 except 1" & 1 'A" ....
.
which are schedule 80
2 3 8 10 20 30 50 80 100 200 300 500 800 1000 2000 5000 10,000
DOWTHERM VAPOR FLOW - LBS./HR. x 10 2
Figure 2-36. Pressure drop, Dowtherm "A"® vapor in steel pipe. By permission, Struthers Wells Corp., Bull. D-45.
(equation continuedfrom pag,: 110) For nozzles and orifices (vapors/gases):
k = ratio of specific heat of gas, at constant pressure to that at
constant volume, = cp/c, .. See Table 2-14
g = 32.2 ft/sec squared w, = 0.525Yd c' � (2-88)
'
2
' {V:-
p" = pressure, pounds per sq ft, abs (Psf abs) (note units)
p = the specific weight, lb/cu ft (see Appendix) at T and p"
For valves, fittings, and pipe (liquids):
This sonic velocity occurs in a pipe system in a restrict-
ed area (for example, valve, orifice, venturi) or at the out-
let end of pipe (open-ended), as long as the upstream . = 0 52- d2 /!lP (P1)
pressure is high enough. The physical properties in the '\ s . !J i I K (2-89)
above equations are at the point of maximum velocity.
For the discharge of compressible fluids from the end For nozzles and orifices (liquids):
ofa short piping length into a larger cross section, such as
a larger pipe, vessel, or atmosphere, the flow is considered
adiabatic. Corrections are applied to the Darcy equation w, = o.s2s d� c' � lip (p 1) (2-90)
to compensate for fluid property changes due lo the
expansion of the fluid, and these are known as Y net where V 1 = upstream specific volume of fluid, cu ft/lbs
expansion factors [3]. The corrected Darcy equation is: w, = rate of flow, lbs/sec
For valves, fittings, and pipe (vapors/gases): Af> = pressure drop across the system, psi (inlet-dis-
charge)
K = total resistance coefficient of pipe, valves, fittings,
w, = 0.525Ydf � P/(KYi), lbs/sec (2- 87) and entrance and exit losses in the line

114 Applied Process Design for Chemical and Petrochemical Plants

Fluld Flow
- 8AIE l'IIUUt • L&. PEii iQ. It. ,,,
......... _,. .. ,,
-
/41 D ! llD " llD - U) - e.o 1!0 ,.... - - l>I T 1r
'
IP
limv I
l90,
.....
--
·-
- ......
- .....
- ......
- ........ _ --- -- r-c, ..... - -
..... ..... _ "' -- , ...... - ....._ ......... ..... -
--
-
ID- ....... ........ - --- -- ..... - .......... ....... :;;
-
........ _
='°°°
- r-. .... - ..... i - - r-..._ r--,._ ..... ....... _ I"'-, -
-
-
....... -- .....
-
........ ·---
--- .....
-
--
I"'-,..._
- ....
......
.......
....._ 3000
I"'-,
-- - .....
--
.............
&- -- - -- - - ..... - ..... - -- ....... �
- ....
----
- .....
....... _ e,
r- ......
.......
-
&- - ...... ...... - ....... -- - r-- ..... ...... .......
__
. --- -.. - ,..., ....... r-..._ ....... .... � IOOO
- ......
-· .......
-- .......
r-_
-
<,
-· ..... r- -,.... ....... - - ....... ..... - -- ;a
.....
-
- ..... =-
- --
- ...........
--
--- ......
• - -- -- - ...... - - ..... __ ....... ....... ....... ,-200
,...... ,...... - .......
.......
3'-
-
.......
- ......
......
......
.....
-- .......
r- .....
,...... ,....
.......
....... ....
I"'-, ....
z- - - ---- ........ _ - ......... -r- - - ....... c-, r--.. I
.......
- ....... - -- ........... ....... ..... ..... -- - ,...., .... ;:: IDO
........
Ir 1-. -- r- -.. -- - ....... ........... ......... - ...... =-IO
-
............
......
....... _
r- ..... - ......
r-
1t - -.. ..... _ ..... - ....... r-.. -- r--. ,__ -"""' - ..... =- 40
-
..... _
..... ........
--,�
-
............
IO
.......
r-..
I'-
--
.....
.......
....... _
......
I - - ,...... r- -- ....... .... ..... ....... ........... ......_ ..... -�
-
-- -- - --
......
.......
....... _
.. ..... ..... - - -- -- - -- - ..... -
-
-.. ...... - .......
,......
. r- ..... - .... -- - -
-
--
- .......
-'- - ....... -- -...._ --- ..... ....... ..... ,...... -- - ......
- .......
-- -- -- r-.. ...... - ....... - ...... ..... ..............
..... _
.......
1--.
-- ...........
........
--
...........
......
r-......_
r-,,.
..... ,....
......
I ,,
r-..
.J. I '.l.'I.J'J'I' . J ]""I ! 1'/j i1'io ..... 'l
I .04 Ml. IF.al
..._. DIQI Lii PIR IQ. INCH Piil lOO RET LI
.
• 0.1025 LY
•
A
I
d
B ase on H ams F ormu a t Qp - da.:s, Re , psi
L : Pipe Length , feet
Re= Ratio of Compression !from free air) at Entrance of Pipe
d = I. D. , Inches
V : Air Flow, cu. ft./sec. ( free air)
Figure 2-37. Compressed air flow chart. By permission, Walworth Co. Note: use for estimating only (this author).
Y = net expansion factor for compressible flow For example, for a line discharging a compressible fluid
through orifices, nozzles, and pipe [3] (see Fig- to atmosphere, the b.P is the inlet gauge pressure or the
ures 2-38A and 2·38B) difference between the absolute inlet pressure and atmos-
�p = pressure drop ratJo in M /P', used to determine pheric pressure absolute. '\,\,'hen �p /Pi' falls outside the
Y from Figures 2-38A and 2-38B. The �p is the limits of the K curves on the charts, sonic velocity occurs
difference between the inlet pressure and the
pressure in the area of larger cross section. at the point of discharge or at some restriction within the
di = pipe inside diameter, in. pipe, and the limiting value for Y and �p must be deter-
C' = flow coefficient for orifices and nozzles (Figures mined from the tables on Figure 2-38A, and used in the
2-17 and 2-18) velocity equation, v 5, above [3].

Fluid Flow 115

Table 2-14 For flow of gases and vapors through nozzles and orifices:
Typical Ratios of Specific Heats, k
Compound k = <;,/ Cv __ � (2-48)
Air 1.40
Ammonia 1.29
Argon 1.67 where � = ratio of orifice throat diameter to inlet diameter
Carbon Dioxide 1.28 C' = flow coefficient for nozzles and orifices (see Fig-
Carbon Monoxide 1.41 ures 2-17 and 2-18), when used as per ASME speci-
Ethylene 1.22 fication for differential pressure
Hydrochloric acid 1.40 p = fluid density, lb/ cu ft
Hydrogen 1.40 A = cross-sectional flow area, sq ft
Methane 1.26
Methyl Chloride 1.20 Note: .jhe use ofC' eliminates the calculation of velocity of
Nitrogen 1.40
4
Oxygen 1.40 approach. The flow coefficient C' is C' = Cd/ � 1 - � > 'f/
Sulfur dioxide 1.25 Cd = discharge coefficient for orifices or nozzles [3].
For compressible fluids flowing through nozzles and ori-
fices use Figures 2-17 and 2-18, using hL or f1P as differen-
tial static head or pressure differential across taps located
one diameter upstream at 0.5 diameters downstream from
Figures 2-38A and 2-388 are based on the perfect gas the inlet face of orifice plate or nozzle, when values of Care
laws and for sonic conditions at the outlet end of a pipe. taken from Figures 2-17 and 2-18 [3]. For any fluid:
For gases/vapors that deviate from these laws, such as
steam, the same application will yield about 5% greater
1
flow rate. For improved accuracy, use the charts in Figures q = C'A ([2g (144) l1P]/p) 1 2, cu ft/sec flow (2-48)
2-38A and 2-38B to determine the downstream pressure
when sonic velocity occurs. Then use the fluid properties Note for liquids f1P is upstream gauge pressure.
at this condition of pressure and temperature in: For estimating purposes for liquid flow with viscosity
similar to water through orifices and nozzles, the follow-
ing can be used [53]:
v, = � kgRT, ft/sec= (kg (144)P'V)l/ 2 (2-85)

Q = 19.636 c' d / .i; /----
to determine the flow rate at this condition from: ( : �
� I
1
v = q/A = 183.3q/d 2 = 0.0509W/(d 2 )(p) (2-91)
do
where - is greater than 0.3 (2- 92)
d = internal diameter of pipe, in. d
A= cross section of pipe, sq ft
q = cu ft/sec at flowing- conditions d
02
T = temperature, R Q = 19.636 c' d {h where - is less than 0.3 (2 - 93)
0
k = ratio of specific heats di
P' = pressure, psi abs
02
W = flow, lbs/hr or [3], '"' = 157.6d C' � hLp 2
v = velocity, mean or average, ft./sec = 1891 d 0 C -y l1Pp (2-94)
2
,�
These conditions are similar to flow through orifices, where Q = liquid flow, gpm
nozzles, and venturi tubes. Flow through nozzles and ven- d, = diameter of orifice or nozzle opening, in.
turi devices is limited by the critical pressure ratio, re = di = pipe inside diameter in which orifice or nozzle is
downstream pressure/upstream pressure at sonic condi- installed, in.
tions (see Figure 2-38C). h' L = differential head at orifice, ft liquid
C' = flow coefficient (see Figure 2-39 for water and
For nozzles and venturi meters, the flow is limited by crit- Figure 2-18 and 2-19 for vapors or liquids)
ical pressure ratio and the minimum value of Y to be used. (text continued on page 118)

116 Applied Process Design for Chemical and Petrochemical Plants

le = 1.3
1k =arrn•ximati:ly 1.3 for CO,. 502, H20. 11:z..'i. NI 13, Np, Cl2, CH�. C2H2. and C2H�)
1.0 �-,---,-.---,--�-,---,r--..,....�-.---,--�-,---,r--...,..---y--r-.....---r--. limiting Factors
For Sonic Velocity
le = 1.3


K I��, y


1.2 .525 .612
o.ao 1--+--+--+---+--1��..P.c:-f"',,:����,e,,..i�--+--+-+-+--+--+-� 1.5 .550 .631
r 2.0 .593 .635
3 .642 .658
4 .678 .670
6 .722 .685
8 .750 .698
10 .773 .705
15 .807 .718

20 .831 .718
40 .877 .718
100 .920 .718
o.ss':"-- ........ __.-��-�""="'""-....__...,.__._,,...__....__,,..__.__..__ .......... _.,.....,._.....___.�...._....J
0 0.1 0.2 0.3 0.4 0.5 0,6 0.7 0,8 0.9 1.0
f}.P
p�



le = 1.4
(k =approximately 1.4 for Air, H2. 02, N2, CO, NO, and HCI)
l,Qs:c--,--.---.--,--�---.-.,--.---,.-,---.----"T-,---,--.--...---,---r--,r---, limiting Fadors
For Sonic Velocity
le = 1.4


K I �(I y


1.2 .552 .588
1.5 .576 .606
0.80 r--+--+--+--+--�:>o.l--'"c-+�'l<-"'lc-'""d'-�..::,,....�....ic--+--1--+---+--4-�---I 2.0 .612 .622
1·
3 .662 .639
4 .697 .649
6 .737 .671
8 .762 .685
10 .784 .695
15 .818 .702

20 .839 .710
40 .883 .710
100 .926 .710
0.9 1.0




Figure 2-38A. Net expansion factor, Y, for compressible flow through pipe to a larger flow area. By permission, Crane Co., Technical Paper
#410, Engineering Div., 1957. Also see 1976 edition.

Fluid Flow 117

Expansion
Factor
y
1.00�--�---,.----,----r----,----r----,,----..,.....----------...P-'".-.--.--..



Square edge orifice
!!R. -
d, -












o. 75

































0.35 _. __ , __ .._ _,_ __ �.._ __ ..._ __ ........ '-----'--'--L-L...L�

I= I .45 � · • e I I I e I el e e e I I e I e I,, I I I,, e I I l' I I I,, I I I,,• I e,, I I I I
. .4 . . 1.0
I = ! . 40 � I e I I I • I e • l ' f e e I • 1 e e I e I e e , , e I e l ' e I e I • I • • l ' e • • I e • • • I
.4
•
•
1.0
,.
.
L
I
II= 1.35 0' ' ', ' , , '! I' I I I ', ' 'l' , I I I I I ' I I I I I 'I I I ' 'l' ' , 'I I I ' 1 I
1.0
.
•
.6
.
!
�
I= I . 30 • • • I C • e 1 1 • • • e I 1 1 • • I • • I t I • • • t I • , • • I • • , , I , , , e I , , · , , i
to
.4
.
.8
.6
I= 1.25 � • I I I t t ! • t • • I • • • t I e • • • I t t t I I• • 1 • I • • I • I •II • I • I • • I
.4
.8
1.0
.6
.
Pressure ratio, A P/P,'
Figure 2-386. Net expansion factor, Y, for compressible flow through nozzles and orifices. By permission, Crane Co., Technical Paper #410,
Engineering Div., 1957. Also see 1976 edition and Fluid Meters, Their Theory and Application, Part 1, 5th Ed., 1959 and R. G. Cunningham,
Paper #50-A-45, American Society of Mechanical Engineers.

118 Applied Process Design for Chemical and Petrochemical Plants

(text continued from page 115) Flow of gases and vapors ( compressible fluids) through
q = cu ft/sec at flowing conditions (Figure 2-37) Coef- nozzles and orifices. (For flow field importance see Refer-
ficient from Reference [22] for liquids discharge ences [31]). From [3]:

r0 = critical pressure ratio for compressible flow, =
P'2/P'1 , 2g ( l 44)6P
q = YC A ' , cu ft I sec
p
(at flowing conditions) (2- 48)

Y = net expansion factor from Figures 2-38A or 2-38B
llP = differential pressure (equal Lo inlet gauge pressure when
.14 discharging to atmosphere)
................ ..... p = weight density of fluid, lbs/cu ft at flowing conditions
-.............. r--..... A = cross section of orifice or nozzle, sq ft
.II C' = flow coefficient from Figures 2-38A or 2-38B
<, ,....__ ............. ... r----.J. ...
........._
r-,...... ! i
---·'° ........ ............. +-, or, W = 1891 Yd 1 C , I =-, lbs/hr (2- 95)
!lP
2
Q.
..... !'-.... ........._ t-r " I vi
..........
Q. ............ � -.. ............. ....... """<
II .!ol ............ r-; -.............. /� " where d, = internal diameter of orifice, in
...... .... .... ........._ ........... i""---.... ....... V1 = specific volume of fluid, cu ft/lb
...........
� ............. � r-; ........... <, 1 I I ..
. 16 r-, -...... ............ - i""'-<.,,
� � <; r-; .. , /ilP P/
•
O f- : -,
.......... .......... , 2
:::::::: � r-....... <, � or, q = 11. 30 Yd C cu ft/sec
___.....
.!>< .... I T1 Sg
""""R � � v' II at 14. 7 psia and 60° F (2- 96)
--- L----"_:
� �
.....,.., _,
.ail where Sg = Sp Gr gas relative Lo air, = mo! wt, gas/29
0
us I.I l.lS l4 I'S Ti= absolute temperature, R
k=Cp/C� P' 1 = pressure, psi abs
P' = psia Procedure
B = ratio small-to-large diameter in orifices and nozzles, A. How to determine fripe size for giuen capacity and pressure
and contractions or enlargements in pipes
drop.
Figure 2-38C. Critical Pressure Ratio, re, for compressible flow
through nozzles and venturi tubes. By permission, Crane Co., Tech- 1. Assume a pipe diameter, and calculate velocity in
nical Paper #410, 1957. Also see 1976 edition. See note at Figure 2- feet/ second using the given flow.
18 explaining details of data source for chart. Note: P' = psia �= ratio
of small-to-large diameter in orifices and nozzles, and contractions or 2. Calculate sonic velocity for fluid using Equations 2-
enlargements in pipes. 84 or 2-85.
,
lt6·INUANT SHARP- $QVAltE RE·ENTRANT 5QtlAllE WELL
TV61 IDtilD IDG£D _J __ w __ ROVNDED
TV/JI
EDGED
! __
J
-,= 1-- -,,- -,-- ....... -+ ___
�-
n- i---
�
1DCrJ1 1/•t.tDIA. ITIIIAM Cl.Ult lll!U UIICT!I • 'L-tl'a DIA. Tlllt IUlnl Fllt
1
C= .52 C= .61 C= .61 C= .73 C= .82 C= .98

Figure 2-39. Discharge coefficients for liquid flow. By permission, Cameron Hydraulic Data, Ingersoll-Rand Co., Washington, N.J., 1979.

Fluid Flow 119

3. If sonic velocity of step 2 is greater than calculated From Figure 2-38, Y = 0.97; from Figure 2-18.
velocity of step 1, calculate line pressure drop using
usual flow equations. If these velocities are equal,
then the pressure drop calculated will be the maxi- c' (assumed turbulent) (2-47)
mum for the line, using usual flow equations. If [l - (dof a, )-1 ]1;2
sonic velocity is less than the velocity of step 1, reas-
sume line size and repeat calculations. where Cd = orifice discharge coefficient, uncorrected for
velocity of approach
B. How to determine flow rate ( capacity) for a given line size
and fixed pressure drop. C' = 0.74 at est. Re� 2000
Temperature = 460 + 50 = 510°F
This is also a trial and error solution following the pat-
tern of (A), except capacities are assumed and the pres-
sure drops are calculated to find a match for the given D . 144P 144 (54.7)
conditions of inlet pressure, calculating back from the ensity = P = RT = (96.4) (510)
outlet pressure. = 0.1602 lb/cu ft

C. How to determine pressure at inlet of pipe system for fixed W = 1891 Y d/C (Af>p)1;2 (2-95)
pipe size and flow rate. W = 1891 (0.97) (0.750) 2 0.74 [(3) (0.1602)]112
W = 529.2 lbs/hr methane
1. Determine sonic velocity at outlet conditions and
check against a calculated velocity using flow rate. If Check assumed R., to verify turbulence; if not in rea-
sonic is the lower, it must be used as limiting, and capac- sonable agreement, recalculate C' and balance of solu-
ity is limited to that corresponding to this velocity.
2. Using the lower velocity, and corresponding capaci- tion, checking:
ty, calculate pressure drop by the usual equations.
For greater accuracy start at the outlet end of the Viscosity of methane = 0.0123 centipoise
line, divide it in sections using the physical proper- = 6.31 W/dµ
ties of the system at these points, backing up to the = 6.31 (502)/ (0.750) (0.0123)
inlet end of the line for the friction loss calculations. Re= 343,373
This procedure is recommended particularly for
steam turbine and similar equipment exhausting to This is turbulent and satisfactory for the assumption.
atmosphere or vacuum. The pressure at the inlet of For helpful quick reference for discharge of air through
the line is then the sum of the discharge or outlet an orifice, see Table 2-12B.
line pressure and all the incremental section pres-
sure losses. In the case of a turbine, this would set its Example 2-11: Sonic Velocity
outlet pressure, which would be higher than the
pressure in the condenser or exhaust system. Water vapor ( 4930 lbs/hr) is flowing in a 3-inch line at
730°F. The outlet pressure is less than one half the inlet
Example 2-10: Gas Flow Through Sharp-edged Orifice absolute pressure. What is maximum flow that can be
expected?
A l"-Schedule 40 pipe is flowing methane at 40 psig
and 50°F. The flange taps across the orifice (0.750 inch s,I c, = 1.30
diameter) show a 3 psi pressure differential. Determine
the flow rate through the orifice. MW vapor = 18.02
Solution:
CH 4; Sp Gr = Sg = 0.553 v, = [ ( 1.30) (32.2) (1544/18.02) (730 + 460)] 1
1 2
= 2,065 ft/sec
Gas constant = R = 96.4
Ratio Sp. ht. = k = 1.26 Cross section of 3-inch pipe = 0.0513 sq ft
Absolute system pressure = P = 40 + 14.7 = 54.7 psia
b.P/P 1 = 3.0/54.7 = 0.0549 Volume flow = (2,065) (0.0513) = 105.7 cu ft/sec
Pipe ID = 1.049 in.
d /d1 = 0.750/1.049 = 0.7149 Vapor density= 4930/(3600) (105.7) = 0.01295 lb/cu ft
0

120 Applied Process Design for Chemical and Petrochemical Plants

Pressure at end of line Table 2-15
= 0.01295 (379/ 18.02) (14.7) ( 1190/520) Dry-Gas Flow Transmission Factors
= 9.16 psia (below atmos.)
Title Transmission Factor (Yl/f) Ref.*
Weymouth l l.2Do.167
0 125
Blasius 3.56Re ·
Friction Drop for Compressible Natural Gas Panhandle A 6.87Re 0 · 073
in Long Pipe Lines Modified Panhandle 16.5Reo.01g5
Smooth pipe law 4 log (Re ..ff) - 0.4
Tests of the U.S. Department of the Interior, Bureau (Nikuradse)
of Mines, reported in Monograph 6 Flow of Natural Gas Rough pipe law
(D)
Through High-Pressure Transmission Lines [ 43] indicate (Nikuradse) 4 log-- + 3.48
that the Weymouth formula gives good results on flow (2E)
measurements on lines 6 inches in diameter and larger [ ]
when operating under steady flow conditions of 30 lo Colebrook 1 log � + 3.48 - 4 log 1 + 9.35 2R �
600 psig. .[j
Note: D = inches
Long gas transmission lines of several miles length *See listing of source references in Reference (15]. By permission,
are not considered the same as process lines inside Hope, P. M. and Nelson, R. G., "Fluid Flow, Natural Gas," McKetta,J.J.
plant connecting process equipment where the lengths Ed., Encyclopedia of Chemical Processing and Design, vol. 22, 1985, M.
Dekker, p. 304 [15].
usually are measured in feet or hundreds of feet. Some
plants will transfer a manufactured gas, such as oxygen,
carbon dioxide, or hydrogen, from one plant to an adja- sure base of 14.4 psia is to be used with the Bureau of
cent plant. Here the distance can be from one to fifteen Mines multipliers [ 43].
miles. In such cases, the previously discussed flow rela-
tions for compressible gases can be applied in incre-
2
mental segments, recalculating each segment, and then [ P1 - P9 2 ]1/2
the results can be checked using one of the formulas % (aL14.4psia & 60°F) = 36.926d 2·667 Lm -
that follow. However, there are many variables to evalu-
ate and understand in the Weymouth, Panhandle, Pan- scfh (2- 97)
handle-A and modifications as well as other flow rela- q\ (al 14.4 psia and 60°F)
tionships. Therefore, they will be presented for
reference. However, the engineer should seek out the
specialized flow discussions on this type of flow condi- [ 520) � 1/2,
tion. The above mentioned equations are derived some- = 28.0 d 2 ,667 P/ - P/ ( scfh (Ref. 8) (2-98)
what empirically for the flow of a natural gas containing Sg i., T J
some entrained liquid (perhaps 5% to 12%), and the
results vary accordingly, even though they are not two-
phase flow equations. Weymouth's formula [57] has friction established as a
function of diameter and may be solved by using align-
ment charts.
Table 2-15 [15] tabulates the transmission factors of
the various equations. Most of these are established as The Weymouth formula is also expressed (at standard
correction factors to the correlation of various test data. conditions) as:


Dunning [ 40] recommends this formula (from Ref-
erence [ 43]) for 4 to 24-inch diameter lines with specif- E = transmission factor, usually taken as: 1.10 X 11.2 d 0·167
ic gravity of gas near 0.60, and actual mean velocities (omit for pipe sizes smaller than 24 in.)
from 15 to 30 feet per second at temperature near 60°F. d = pipe I.D., in.
Ts= 520°R
The Bureau of Mines report states that minor cor- P, = I 4.7 psia
rections for bends, tees, and even compressibility are T 1 = flowing temperature of gas, R
0
unnecessary due to the greater uncertainties in actual qd = cu ft/day gas at std conditions of P, and T,
line conditions. Their checks with the Weymouth rela- P'i = inlet pressure, psia
tion omitted these corrections. The relation with pres- P' 2 = outlet pressure, psia

Fluid Flow 121


Z = compressibility factor For bends in pipe add to length [38]:
Lm = pipe length, miles

or frcm Reference [3]: Bend Radius Add", as pipe diameters, de
1 Pipe dia. 17.5
1.5 Pipe dia, 10.4
2 Pipe dia. 9.0
Example 2-12: Use of Base Correction Multipliers 3 Pipe dia. 8.2
*These must be converted to the unit of length used in the formula.
Tables 2-16, 2-17, 2-18, and 2-19 are set up with base ref-
erence conditions. In order to correct or change any base
condition, the appropriate multiplier(s) must be used. If a line is made up of several different sizes, these may
A flow of 5.6 million cu ft/day has been calculated be resolved to one, and then the equation solved once for
using ·weymouth 's formula [57], with these conditions: this total equivalent length. If these are handled on a per
measuring base of 60°F and 14.4 psia; flowing tempera- size basis, and totaled on the basis of the iongest length of
ture of 60°F, and specific gravity of 0.60. Suppose for com- one size of line, then the equivalent length, Le, for any
parison purposes the base conditions must be changed to: size d, referenced to a basic diameter, de.
measuring base of 70°F and 14. 7 psia; flowing tempera-
ture of80°F, and specific gravity of0.74. (2-103)

Multipliers from the tables are:
Pressure base: 0.9796 where Lm is the length of pipe of size d to be used.
Le is the equivalent length of pipe size cl, length Lm
Temperature base: 1.0192 after conversion to basis of reference diameter, d.,
Specific gravity base: 0.9005
Flowing temperature base: 0.9813
New base flow The calculations can be based on diameter de and a
= (5,600,000) (0.9796) ( 1.0192) (0.9005) (0.9813) length of all the various Le values in the line plus the
length of line of size de, giving a total equivalent length
= 4,940,000 cu ft/day for the line system.
Panhandle-A Gas Flow Formula [3]
Modified Panhandle Flow Formula [15]
This formula is considered to be slightly better than
0
the ±10 percent accuracy of the Weymouth formula. %s = 737.2 E (T /P 0) 102 [[P/ (1 + 0.67 ZP1)
- P22 (1 + 0.67 ZP2)J/T Lm G0.961]051 (d)253 (2-104)

)1.01ss1[ lo.;394 where Lm = miles length
= P/- P/
q 435.87E(T ip d = inside diameter, in.
d, S I S s 0.8539TI z T = flowing temperature, R
g "rn
d 2.6182 (2- 101) Z = gas deviation, compressibility factor
T = base temperature, (520 R)
O
or E = 0.92, usually 0.8539 G = gas specific gravity
Z = compressibility correction term
13539 P = pressure, psi, absolute
[
S 0.4606
2 435.8_7_(T_,- . 1""" g r_ , _ ) P 0 = base pressure, (14.73 psi, absolute)
P/ - p 2 = l- . O- i 8_8_-<l-2- . 6- 1 8_ 2 _ l TLm q l.B 539 E = "efficiency factor," which is really an adjustment
l
Lo fit the data
f = fanning friction factor
(2 - 102)
qos = flow rate, SCF I day
0
where T = gas flowing temperature, R = 460°F + t
E = efficiency factor for flow, use 1.00 for new pipe
without bends, elbows, valves and change of pipe American Gas Association (AGA) Dry Gas Method
diameter or elevation
0.95 for very good operating conditions See Reference [16] A.GA, Dry Gas Manual. Some tests
0.92 for average operating conditions indicate that this method is one of the most reliable above
0.85 for poor operating conditions a fixed Reynolds number.

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.

Fluid Flow 123

Table 2-16 Table 2-17
Pressure-Base Multipliers For Quantity* Temperature-Base Multipliers For Quantity*

14.4 . 460 + new temperature base, ° F.
.
Multiplier= �-----------cc---�----,-- Mu I tip ier = ------�------
1
New pressure base, Lbs./sq. in. abs. 460 + 60
New Pressure Base, Lbs./sq.in. abs. Multiplier New Temperature Base, ° F. Multiplier
12.00 1.2000 45 0.9712
13.00 1.1077 50 0.9808
14.00 1.0286 55 0.9904
14.40 1.0000 60 1.0000
14.65 0.9829 65 1.0096
14.7 0.9796 70 1.0192
14.9 0.9664 75 1.0288
15.4 0.9351 80 1.0385
16.4 0.8780 85 1.0481
90 1.0577
*By permission, Johnson, T. W. and Berwald, W. B., Flow of Natural Gas 95 1.0673
Through High Pressure Transmission Lines, Monograph No. 6, U.S. Dept. of 100 1.8769
Interior, Bureau of Mines, Washington, DC.
*By permission, Johnson, T. W. and Berwald, W. B., How of Natural Gas
Through High Pressure Transmission Lines, Monoprapl: No. 6, U.S. Dept. of
Interior, Bureau of Mines, 'Washington, D.C.
Table 2-18
Specific Gravity Multipliers For Quantity*

[ 0.600 J Y2
Multiplier =
actual Specific Gravity


Specific Gravity 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
----- -----
0.5 1.0954 1.0847 1.0742 1.0640 1.0541 1.0445 1.0351 1.0260 1.0171 1.0084
0.6 1.0000 0.9918 0.9837 09759 0.9682 0.9608 0.9535 0.9463 0.9393 0.9325
0.7 0.9258 0.9193 0.9129 0.9066 0.9005 0.8944 0.8885 0.8827 0.8771 0.8715
0.8 0.8660 0.8607 0.8554 0.8502 0.8452 0.8402 0.8353 0.8305 0.8257 0.8211
0.9 0.8165 0.8120 0.8076 0.8032 0.7989 0.7947 0.7906 0.7865 0.7825 0.7785
1.0 0.7746 0.7708 0.7670 0.7632 0.7596 0.7559 0.7524 0.7488 0.7454 0.7419
1.1 0.7385 0.7352 0.7319 0.7287 0.7255 0.7223 0.7192 0.7161 0.7131 0.7101
-- I
,;,By permission.Tohnson, T. W. and Berwald, W. B., Flow of Natural Gas Through High Pressure Transmission Lines, Monograph No. 6, U.S. Dept. of Inte-
rior, Bureau of Mines, Washington, DC.
Table 2-19
F1owing-Temperature Multipliers For Quantity*


[ 460 + 60 J Y2
Multiplier =
460 + actual flowing temperature

Temp. °F. 0 1 2 3 4 5 6 7 I ---- -----
8
9
.. 1.0632 1.0621 1.0609 1.0598 1.0586 1.0575 1.0564 1.0552 1.0541 1.0530
10 1.0518 1.0507 1.0496 1.0485 1.0474 1.0463 1.0452 1.0441 1.0430 1.0419
20 i.0408 1.0398 1.0387 1.0376 1.0365 1.0355 1.0344 1.0333 1.0323 1.0312
30 1.0302 1.0291 1.0281 1.0270 1.0260 1.0249 1.0239 1.0229 1.0219 1.0208
40 1.0198 1.0188 1.0178 1.0167 1.0157 1.0147 1.0137 l.0127 1.0117 1 0107
50 1.0098 1.0088 1.0078 1.0068 1.0058 1.0048 1.0039 1.0029 1.0019 1.0010
60 1.0000 0.9990 0.9981 0.9971 0.9962 0.9952 0.9943 0.9933 0.9924 0.9915
70 0.9905 0.9896 0.9887 0.9877 0.9868 0.9859 0.9850 0.9841 0.9831 0.9822
80 0.9813 0.9804 0.9795 0.9786 0.9777 0.9768 0.9759 0.9750 0.9741 0.9732
90 0.9723 0.9715 0.9706 0.9697 0.9688 0.9680 0.9671 0.9662 0.9653 0.9645
*By permission,Johnson, T. W. and Berwald. W. B., Flow of Natural Gas Through High Pressure Transmission Lines, Monograph No. 6, C.S. Dept. oflnteri-
or, Bureau of Mines, Washington, DC.

124 Applied Process Design for Chemical and Petrochemical Plants

Parallel System: New Capa<:ity after Paralleling If two-phase flow situations are not recognized, pres-
sure drop problems may develop which can prevent sys-
Solve this relation, rearranged conveniently to [ 43): tems from operating. It requires very little percentage of
vapor, generally above 7% to 8%, to establish volumes and
flow velocities that must be solved by two-phase flow analy-
"'" � H-[1-+-( d- b ....;):....d-.--)-2- . 66_7_]_ 2 sis. The discharge flow through a pressure relief valve on
a process reactor is often an important example where
two-phase flow exists, and must be recognized for its back
pressure impact.

Two-phase Liquid and Gas Flow
Flow Patterns
The concurrent flow of liquid and gas in pipe lines has
received considerable study [33), [35], [37], [ 41]. How- Six or seven types of flow patterns (Figure 2-40) are
ever, pressure drop prediction is not extremely reliable usually considered in evaluating two-phase flow. Only one
except for several gas pipe line conditions. The general type can exist in a line at a time, but as conditions change
determinations of pressure drop for plant process lines (velocity, roughness, elevation, etc.) the type may also
can only be approximated. change. The unit pressure drop varies significantly
The latest two-phase flow research and design studies between the types. Figure 2-40 illustrates the typical flow
have broadened the interpretation of some of the earlier regimes recognized in two-phase flow.
flow patterns and refined some design accuracy for select-
ed situations. The method presented here serves as a fun- Figure 2-41 [17] typically represents a graphical illustra-
damental reference source for further studies. It is sug- tion of the various flow patterns of Figure 2-40 as the two-
gested that the designer compare several design concept phase mixture flows through the piping. Long gas trans-
results and interpret which best encompasses the design port lines may have hydrocarbon or other liquids form
problem under consideration. Some of the latest refer- (condense) as the fluid flows, and this becomes a real prob-
ences are included in the Reference Section. No one ref- lem for offshore or long buried onshore raw gas transmis-
erence has a solution to all two-phase flow problems. sion (see section dealing with calculation methods).


y=G/X
100,000 I I I
......... " ........... , I
....
-, ............. Dispersed Flow
.......... ......_
1, . -- - ..... I
:-...
Wave Flow Annular Flow A Bubble or Froth Flow
10,000 '· ........... 1 ...
.... � � -- I,.,; "" '\..
---..... .... .... L_.
""'""' � .. " � �

<, r-.
Stratified Flow i"-.. Slug Flow
�-
1,000 ....
----
'
............ ""'i--- � '
<, Plug Flow '�
' n-. ...
I
1000.1 0.2 0.4 0.60.81.0 2 4 6 8 10 2 4 6 8 100 2 4 6 B 1,000 2 4 6 810,000
x = WmXy,/G

Figure 2-40. Flow patterns for horizontal two-phase flow. (Based on data from 1-in., 2-in., and 4-in., pipe). By permission, 0. Baker, Oil and
Gas Journal, Nov. 10, 1958, p. 156.

Fluid Flow 125


Type of Flow For Horizontal Pipes Total System Pressure Drop

Bubble or Froth: Bubbles dispersed in liquid The pressure drop for a system of horizontal and verti-
Stratified: Liquid and gas flow in stratified layers cal (or inclined) pipe is the sum of the horizontal pres-
Wave: Gas flows in top of pipe section, liquid sure drop plus the additional drop attributed to each ver-
in waves in lower section tical rise, regardless of initial and final elevations of the
Slug: Slugs of gas bubbles flowing through line [33].
the liquid
Annular: Liquid flows in continuous annular L'i.PTPh = L'i.Ppy (horizontal pipe) + nhf', PL/144 (2-110)
ring on pipe wall, gas flows through
center of pipe A. To determine most probable type of two-phase flow
Plug: Plugs of liquid flow followed by plugs using Figure 2-40.
of gas
Dispersed: Gas and liquid dispersed
1. Calculate Wm A.\jl/G
2. Calculate G/t..
3. Read intersection of ordinate and abscissa to identi-
Segregated
fy probable type of flow. Since this is not an exact,
clear-cut position, it is recommended that the adja-
cent flow types be recorded also. Note: See Example
2-16 for definitions of A. and 'I'·
Stlatllled
B. Calculate the separate liquid and gas flow pressure
drops.


1. For general process application both L'i.PL and L'i.P g
may be calculated by the general flow equation:
L'i.PL or L'i.P ( using proper values respectively)
g

�m� where f is obtained from Reynolds-Friction Factor
2
3.36fLW (l O -6 )
Intermittent
(2-111)
d5p
chart (Figure 2-3) for an assumed line size, d.
Plug
--2!:3 2. For gas transmission, in general form [33]



Slug (q d 14.65 ) LS g TZf
L'i.P = (2-112)
Distributed g 20 000 d 5 P avg
'
is the thousands cf standard cubic feet
where qd 14 _ 65
of gas per day, measured at 60°F and 14.65 psia, and
Pavg is the average absolute pressure in the pipe sys-
tem between inlet and outlet. This is an estimated
value and may require correction and recalculation
of the final pressure drop if it is very far off.
For oil flow .in natural gas transmission lines [33]


Figure 2·41. Representative forms of horizontal two-phase flow pat- fLQ21,p
terns; same as indicated !n Figure 2-40. By permission, Heim, H., Oil (2-113)
and Gas Journal, Aug. 2, 1982, p.132. 181,916d 5

126 Applied Process Design for Chemical and Petrochemical Plants

3. Calculate value times 1.1 to 2.0, depending upon critical
nature of application.
(2-114)
(2-118)
4. Calculate e for types of flow selected from Figure
2-40 [33].
where PL is the density, lb/ cu ft, of the liquid flowing
in the line, and Fe, elevation factor using gas veloci-
Type Flow Equation for 4>GTT ty, v.
Froth or Bubble <I>= 14.2 xo.75/WmO.l
x
Plug cf>= 27.315 0·855/Wmo.17 (2-ll9)
Stratified cf> = 15,400 X/V. 1m0·8
Slug ¢ = 1,190 xo.1s,,;wmo.5 or as an alternate: F = 1. 7156 -o.102 (2-120)
0
Annular+ <I> = (4.8 - 0.3125d) X0.348 - 0.02Id Vg
*Set d = 10 for any pipe larger than 10-in.
X = [LlPL;q/ LlP ga,] V 2 Use Figure 2-42 for v less than 10. Most gas trans-
mission lines flow at from 1-15 ft/sec.
5. Calculate two-phase pressure drop, horizontal por- For fog or spray type flow, Baker [33] suggests using
tions of lines. For all types of flow, except wave and Martinell i's correlation and multiplying results by two [ 46].
fog or spray:
(a) For gas pipe line flow, the values of <l>crr may be
�PTP = �pc$2 err, psi per foot (2-115) converted to "efficiency E" values and used to cal-
culate the flow for the horizontal portion using a
For wave [52]. fixed allowable pressure drop in the general flow
equation [33]. The effect of the vertical compo-
�P'lr = fTP (G'g)2 /193.2 dPg, psi/foot (2-116) nent must be added to establish the total pressure
drop for the pumping system.
where

(2-117) = [38.7744T, (P/ - P/ )d 5 ]0'5
( -
14 65
f
qd ' lOOOP L S TZ (..£)o.s 2 121 )
s m g S
6. Total two-phase pressure drop, including horizon-
tal and vertical sections of line. Use calculated where 14.65 refers to reference pressure P 5•



Liquid Head
Factor, Fe
1.0 T l T, I I I I II II I I I I I !
0.9 I 'r T I Tl, I I I I I I I I I I I I ) I
I I I I
I 11 11 I
I
I
I I
I I
0.8 \ • Natural Gos Condensate in 16" f!i11eline
0.7 , o Natural Gos,Oil and Water in 2 011 Well Tubing
e Air and Water in I" Ver;.tical Tubing
o Air and Lube Oil in 2 Inclined Tubing
0.6
'
..
0.5 ..... r •
0.4
0.3
·� In �
0.2 ..... _ c
0.1 • ' b
.
I"'
0·0 o 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50
Superficial Gos Velocity, fl./sec.
Figure 2-42. Estimating pressure drop in uphill sections of pipeline for two-phase flow. By permission, 0. Flanigan, Oil and Gas Journal, Mar.
10, 1958, p. 132.

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

128 Applied Process Design for Chemical and Petrochemical Plants

5. Two-phase flow for horizontal flows: lent flow, (b) sub-atmosphere pressure, (c) pressure drop
is limited to 10% of the final pressure (see comment to
2
LlPTr = LlPc<I> crr = (0.0254) (1.31) 2 = 0.0438 psi/ft follow), and ( d) the lower limit for application of the
method is
6. Fe= 0.00967 (Wm) 05 /v1'· 7
= 0.00967 (19,494)0.5/(2ll)0.7 W/d 7 20 (2-125)
= 0.032

Vertical elevation pressure drop component: where Wis the flow rate in lbs/hr and d is the inside pipe
diameter in inches. If the above ratio is less than 20, the
= n h fcpdl44 = [(3) (10) + (]) (50)] (0.032) (63)/144 flow is "streamlined" and the data does not apply.
= 1.125 psi total If the pressure drop is greater than 10% of the final
pressure, the pipe length can be divided into sections and
Total: the calculations made for each section, maintaining the
same criteria of (c) and (d) above.
LlPTPh = (0.0438) (358) + 1.125
= 16.7 psi, total for pipe line
Method [54]
Because these calculations are somewhat uncertain
due to lack of exact correlations, it is best to calculate The method solves the equation (see Figure 2-43)
pressure drop for other flow patterns, and apply a gener-
ous safety factor to the results.
Table 2-20 gives calculated results for other flow pat- LlP vac (FICDICTI) + (F2CD2CT2) (2- 126)
terns in several different sizes of lines. P1

where LlP,,.c = pressure drop, in. water/100 ft of pipe
Table 2-20
Two-Phase Flow Example Pt = initial pressure, inches mercury absolute
F 1 = base friction factor, Figure 2-43
Horizontal Flow Pattern F 2 = base friction factor, Figure 2-43
Eleva-
Strati- tion Gy, 1 = temperature correction factor, Figure 2-43
Pipe I.D. Annular tied Wave Factor, Ft./sec ..
Inches Psi/Ft. Psi/Ft. Psi/Ft. F. Gas Ve!. � = temperature correction factor, Figure 2-43
3.068 0.0438 0.000367 0.131 0.032 210.9 CD! = diameter correction factor, Figure 2-43
4.026 0.0110 0.000243 0.0336 0.0465 122.5 Cn 2 = diameter correction factor, Figure 2-43
6.065 0.00128 0.000131 0.00434 0.0826 53.9
7.981 0.00027 0.000087 0.00110 0.121 31.1
10.020 0.000062 0.000062 0.00035 0.166 19.7
Example 2-17: Line Sizing for Vacuum Conditions

Determine the proper line size for a 350 equivalent
feet vacuum jet suction line drawing air at 350°F, at a rate
Pressure Drop in Vacuum Systems of 255 lbs/hr with an initial pressure at the source of 0.6
in. Hg. Abs. Assume 10-in. pipe reading Figure 2-43. Note:
Vacuum in process systems refers to an absolute pressure watch scales carefully.
that is less than or below the local barometric pressure at
the location, It is a measure of the degree of removal of
atmospheric pressure to some level between atmospheric F1 = 0.0155
barometer and absolute vacuum (which cannot be F2 = 0.07l
attained in an absolute value in the real world), but is CDI = 0.96
used for a reference of measurement. In most situations, C02 = 0.96
a vacuum is created by pumping air out of the container Cn = 1.5
(pipe, vessels) and thereby lowering the pressure. See Fig-
ure 2-1 to distinguish between vacuum gauge and vacuum � = 1.67
absolute. LlP,.,.c = [(0.0155) (0.96) (1.5) + (0.071) (0.96) (l.67)]/0.6
This method [54] is for applications involving air or = (0.02232 + 0.1138)/0.6
steam in cylindrical piping under conditions of (a) turbu- = 0.2269 in. water/100 ft.

Fluid Flow 129

Total line pressure drop: The majority of industrial chemical and petrochemical
plants' vacuum operations are in the range of 100
( o.2259) microns to 760 torr. This is practically speaking the rough
/'I.P vac (350) = 0.794 in. water (for 350') vacuum range noted above. For reference:
100
(0.794/13.6) = 0.0584 in. Hg l torr = l mm mercury (mmHg)
1 in. mercury (in. Hg) = 25.4 torr
l micron (urn Hg) = 0.0010 torr
Final calculated pressure = 0.6 + 0.0584 = 0.6584 in. Hg For other conversions, see Appendix.
10% of 0.658 = 0.0658 in. Hg In general, partially due to the size and cost of maintain-
Therefore the system is applicable to the basis of the ing vacuum in a piping system, the lines are not long (cer-
method, since the calculated pressure drop is less than tainly not transmissions lines), and there is a minimum of
10% of the final pressure, and w/d = 25.5, which >20. valves, fittings, and bends to keep the resistance to flow low.
The procedure recommended by Reference [18] is
Low Absolute Pressure Systems for Air [54] based on the conventional gas flow equations, with some
slight modifications. The importance in final line size
Fer piping with air in streamline flow at absolute pres- determination is Lo determine what is a reasonable pres-
sures in the range between 50 microns and l millimeter sure loss at the absolute pressure required and the corre-
of mercury, the following is a recommended method. Cal- sponding pipe size to balance these. In some cases a
culation procedures in pressure regions below atmos- trial/ error approach is necessary.
pheric are very limited and often not generally applicable Method [18], by permission:
to broad interpretations.
For this method to be applicable, the pressure drop is
limited to 10% of the final pressure. l. Convert mass flow rate to volumetric flow rate, gm.
qm = W (359/M) (760/P,) (TI (32 + 460) (1/60),
ivlethod [54] cu ft/ min (2-128)

Refer to Figure 2-44 for iow pressure friction factor and where P, = pressure, torr
air viscosity of Figure 2-45 to correspond to Figure 2-44.
T = temperature, R
0
W = mass flow, lbs/hr
4fLpv 2 . (2-127) M = molecular weight
2gD(144)' P SI
2. Calculate section by section from the process vessel to
the vacuum pump (point of lowest absolute pressure).
where P' 1 = upstream static pressure, psi abs. 3. Assume a velocity, v, ft/sec consistent with Figure 2-
P' 2 = downstream static pressure, psi abs. 46. Use Table 2-21 for short, direct connected con-
f = friction factor, from Figure 2-44.
L = length of pipe (total equivalent), ft, incl. valves nections to the vacuum pump. Base the final specifi-
and fittings cations for the line on pump specifications. Also the
p = average density, lbs/ cu ft diameter of the line should match the inlet connec-
v = average velocity, ft/ sec tion for the pump. General good practice indicates
g = acceleration due to gravity, 32.17 ft/sec-sec that velocities of100 to 200 ft/sec are used, with 300
D = inside diameter of pipe, ft to 400 ft/ sec being the upper limit for the rough vac-
µ = abs. viscosity of air, lbs/ft-sec uum classification.

Vacuum for other Gases and Vapors Sonic velocity, v5 = (kg [ 1544/ M] T) 112, ft/ sec.
Use v from Figure 2-46, and qm from Equation 2-128.
Ryans and Roper categorize [18] vacuum in process
systems as: 4. Determine pipe diameter, D, ft,


Category Absolute Vacuum (Absolute Pressure) (2-129)
Rough vacuum 760 torr to 1 torr
Medium vacuum l to 10- torr Round this to the nearest standard pipe size. Recal-
3
High vacuum 10- to l o-i torr culate v based on actual internal diameter of the line.
3
7
Ultra high vacuum 10-- torr and below
,======� (text continued on page 132)

130 Applied Process Design for Chemical and Petrochemical Plants














STANDARDS OF THE HEAT EXCHANGE INSTITUTE, INC.
1000
800
600

400


200
""'
<,J
a:
;=,
E- 100
<
a:
<,J 80
Cl.
::; 60
<,J
E-
40
1.0 1.4 1.8 2.2 2.6 3.0 3.4 3.8 4.2 4.6 5.0
CrC'.,. TEMPERATURE CORRECTION FACTORS




60
40

u:
<,J
:r:
u 20
�
0::
<,J
E-
<,J 10
::;;
<
s 8
<,J 6
Cl.
5::
..J
-e 4
;=,
E-
u
-e
2
10-• IO__. 10-• 10- 1 10--'
Note: friction Factors FI and F 2 are based on rate of flow, while Factors Co I and Co2 are based on actual pipe diameter.


Figure 2-43. Evaluation curves for friction losses of air and steam flowing turbulently in commercial pipe at low pressures. By permission,
Standards for Steam Jet Ejectors, 4th Ed., Heat Exchange Institute, 1988.

Fluid Flow 131














STEAM JET VACUUM SYSTEMS
10 5











104













For turbulent flow "!... must be greater than 20
d
F = (F, X Cr" X C-ri) + (F2 X C.,2 X CTZ)
P,
F = Pressure Drop, inches of water in 100 feet of pipe
P 1 == Initial Pressure, inches of mercury absolute




IO'











IO
10 10 2 10' I ()4

132 Applied Process Design for Chemical and Petrochemical Plants

ABSOLtrrE VISCOSITI' X ICY· POUNDS PER l'OOT-SECOND

5.00
4.00 I\
3.00
' \
2.00
'
� I 12 AND 18 INCH PfPE I I\
1.00
8:'8
0.60 \
0.50 '
0.40 r-..
' ... \
'- '\. 6 AND I INCH PIPE � - I\ \
" ' \
1,
0.10 "� r\
0.01 -
0.07 '
0.06
0.05 \
0.04 \.
0.03
:\.
" '
0.02
'\
o.o I '\
.
40 5060 80100 200 300 500 1000 2000
D,,p I'\.
REYNOLDS NUMBER Ro = - 0
Figure 2-44. Friction factor for streamlined flow of air at absolute
pressures from 50 microns Hg. to 1 mm Hg. By permission, Stan- ABSOurrE VISCOSITY OF AIR
dards for Steam Jet Ejectors, 3rd. Ed., Heat Exchange Institute, 1956
[54] and Standards for Steam Jet Vacuum Systems, 4th Ed., 1988. Figure 2-45. Absolute viscosity of air. By permission, Standards for
Note: f on same basis as Figure 2-3 [58]. Steam Jet Ejectors, 3rd Ed., Heat Exchange Institute, 1956 (54]; also,
Standards for Steam Jet Vacuum Systems, 4th Ed., 1988 [58].
(text continued from page 129)
5. Determine Reynolds Number, R,,. 8. Calculate the pressure drop for the specific line sec-
tion ( or total line) from:
(2-15)
(2-130)

p = density, lb/ cu ft at flowing conditions
2
D = pipe inside diameter, ft or, = 4.31 p.f Lv /2gd, Lorr (2-l 30A)
v = vapor velocity (actual), ft/sec
µ 0 = viscosity of vapor, lb/ft-sec where p = density, lb/cu ft
d = pipe inside diameter, in.
qm = volumetric flowrate, cu ft/min
6. Determine friction factor, f, from Moody Friction f = friction factor, (Moody) Figure 2-3
Factor Charts, Figure 2-3. .1.PT = pressure drop, torr


or, calculate for turbulent flow using Blausius' equa- Calculate: P; = P;M/555Ti, lb/cu ft (2-131)
tion [18]:
P; = pressure, torr
M = average molecular weight of mixture flowing
1
f = 0.316/ (R,,) 11, for R,, < 2.0 X 10 5
T; = temperature, R
0
7. Tabulate the summation of equivalent lengths of 9. If the calculated pressure drop does not exceed the
straight pipe, valves, fittings, entrance/exit losses as maximum given in Figure 2-47, use this calculated
presented in earlier sections of this chapter. value to specify the line. If the �p exceeds the limit

Fluid Flow 133


Table 2-21 The suction pressure required at the vacuum pump (in
Criteria for Sizing Connecting Lines in Vacuum Service absolute pressure) is the actual process equipment operat-
ing pressure minus the pressure loss between the process
Vacuum pump Assumed flow velocity, ft/ s
equipment and the source of the vacuum. Note that absolute
Steam.jet: pressures must be used for these determinations and not
System pressure, torr gauge pressures. Also keep in mind that the absolute pres-
0.5-5 300 sure at the vacuum pump must always be a lower absolute
5-25 250
25-150 200 pressure than the absolute pressure at the process.
150--760 150
Liquid ring pump: Pipe Sizing for Non-Newtonian Flow
Single-stage" 100
Two-stage 150 Non-Newtonian fluids vary significantly in their prop-
Rotary piston: erties that control flow and pressure loss during flow from
Single-stage 50 the properties of Newtonian fluids. The key factors influ-
Two-stage 25 encing non-Newtonian fluids are their shear thinning or
Rotary vane.j thickening characteristics and time dependency of viscos-
Single-stage 200 ity on the stress in the fluid.
Two-stage 400 Most conventional chemical and petrochemical plants
Rotary blowers: do not process many, if any, non-Newtonian fluids. How-
Atmospheric discharge 50 ever, polymers, grease, heavy oils, cellulose compounds,
Discharging to backing pump 100
paints, fine chalk suspensions in water, some asphalts, and
�Assumes the pump features dual inlet connections and that an inlet other materials do exhibit one type or another of the
manifold will be used. characteristics of non-Newtonians, classified as:
+Based on rough vacuum process pumps. Use 25 ft/s for high vacuum
pumps.
By permission, Ryans.]. L. and Roper, D. L., Process Vacuum System Design • Bingham plastics
and Operation, McGraw-Hill Book Co. Inc., 1986 [18]. • Dilatant
• Pseudoplastic
• Yield pseudoplastics
of Figure 2-47, increase the pipe size and repeat the cal-
culations until an acceptable balance is obtained. For ini- Solving these classes of flow problems requires specific
tial estimates, the authors [ 18] recommend using 0.6 data on the fluid, which is often not in the public litera-
times the value obtained from Figure 2-4 7 for an accept- ture, or requires laboratory determinations using a rota-
able pressure loss between vessel and the pump. tional viscometer. The results do not allow use of the usual




1000
8
6

4
3 - ... -�
.. 2 - -Wui

� -�
ci J I � .
i �i---... es1gn as,s ""'"'�
·g 100
"ii 8 .
Miniin�rri
> 6 � � .......
..---
4 -�
3
Figure 2-46. Typical flow velocities for vacuum lines. 2 .
Note: 1 torr = 1.33 mb = 133.3 Pa. 1.0 ft/sec = 0.3048
rn/sec, By permission, Ryans, J. L. and Roper, D. L., 10
Process Vacuum System Design & Operation, McGraw- 1 2 3 466789 2 3 456789100 2 3 4 567891000
Hill Book Co., Inc., 1986 [18]. Pressure, torr

134 Applied Process Design for Chemical and Petrochemical Plants

1. 0
8
6
-
4
.,'."'rimary condenser·'-
3
Vent condenser
2- ._ t,.,... -...... K. ...._
-Section
,Seftij>']. -- '"' �p,, r-p,, �k ... ��""�
Figure 2-47. Acceptable pressure losses """" """ ""- """' ...._
between the vacuum vessel and the vacuum .1 Sections enc 4
pump. Note: reference sections on figure to 8 � ..
system diagram to illustrate the sectional type 6 ,-p,, ....
5
hook-ups for connecting lines. Use 60% of the 4
pressure loss read as acceptable loss for the
system from process to vacuum pump, for ini- 3
tial estimate. P = pressure drop (torr) of line in 2
question; P 0 = operating pressure of vacuum
process equipment, absolute, torr. By permis-
sion, Ryans, J. L. and Roper, D. L., Process 0.0 1
Vacuum System Design & Operation, McGraw- 1 2 3 4 6 6 7 8 10 2 3 4 6 6 78 100 2 3 4 6 6 7 8 1000
Hill Book Co., Inc., 1986 [18]. Pressure in vacuum ve.l P0


Fanning or Moody friction charts and are beyond the used as the viscosity factor in the pressure drop calcula-
scope of this chapter. Design literature is very limited, tions. The two principal classifications are [25]:
with some of the current available references being Sul-
tan [21], Bird et al. [22], Cheremisinoff, N. P. and Gupta 1. Newtonian slurries are simple rheological property
[14], Perry et al. [5], and Brodkey and Hershey [23]. viscosities, and can be treated as true fluids as long
as the flowing velocity is sufficient to prevent the
Slurry Flow in Process Plant Piping dropout of solids. For this type of slurry, the viscosi-
ty= µ.
Most industrial process plants have from none to a few 2. Bingham-plastic slurries require a shear stress dia-
slurry flow lines to transport process fluids. The more gram showing shear rate vs. shear stress for the slurry
common slurry lines discussed in the literature deal with in order to determine the coefficient of rigidity, T],
long transmission lines for coal/water, mine tailings/ which is the slope of the plot at a particular concen-
water, limestone/water, wood pulp-fibers/water, gravel/ tration. This is laboratory data requiring a rheometer,
water, and others. These lines usually can be expected Lo These are usually fine solids at high concentrations.
have flow characteristics somewhat different than in-plant
process slurries. Considerable study has been made of the Reference [25] has two practical in-plant design exam-
subject, with the result that the complexity of the variables ples worked out.
make correlation of all data difficult, especially when The pressure drop design method of Turian and Yuan
dealing with short transfer lines. For this reason, no single [24] is the development of the analysis of a major litera-
design method is summarized here, but rather reference ture data review. The method categorizes slurry flow
is given to the methods that appear most promising (also regimes similar in concept to the conventional multi-
see Reference [30]). regime diagram for two-phase flow, Figure 2-50. Their fric-
Derammelaere and Wasp [25] present a design tech- tion factor correlations are specific to the calculated flow
nique that ties into their classification of slurries as het- regime. See Figure 2-51 for one of four typical plots in the
erogeneous and homogeneous (Figures 2-48 and 2-49). original reference.
This method uses the Fanning friction factor and conven- Example calculations are included, and Figure 2-52
tional equations for pressure drop. The recommended illustrates the effect of pipe size on the placement of the
design slurry velocities range from 4 to 7 fl/sec. Pipe abra- flow regime.
sion can be a problem for some types of solids when the
velocity approaches 10 ft/sec. For velocities below 4 ft/sec Pressure Drop for Flashing Llquicls
there can be a tendency for solids to settle and create
blockage and plugging of the line. Steam is the most common liquid that is flashed in
The concentration of the solids in the slurry deter- process plants, but of course, it is not the only one as
mines the slurry rheology or viscosity. This property is many processes utilize flash operations of pure com-