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4.47 THE EXPANSION FACTOR, Y 219

4.47 THE EXP ANSION FACTOR, Y

The adiabatic flow Eq. (4-178) can be represented in a form:

The simultaneous solution of Eqs (4-173) and (4-174) eliminatesM a and yields a value for r , the critical pressure ratio. Table 4-25shows a wide variation in the critical values with respect to y (Le.,ratio of specific heats, Cp/C,) and the loss coefficient K .

4.46 ADIABATIC FLOW

If there is no heat transfer or energy dissipated in the gas when

traversing from state 1 to state 2, the process is adiabatic andreversible, that is isentropic. However, the actual flow conditionsare somewhere between isothermal and adiabatic and, as such,the flow behavior can be described by the isentropic equations,where the isentropic constant k replaced a polytropic constant y(e.g., 1 < y < k ) . For isothermal condition, y = 1, whereas trulyisentropic flow corresponds to y = k .

The density and temperature as a function of pressure are

(4-177)

The mass flow rate, G , by using Eq. (4-177) to eliminate p and Tand solving for G gives

L

where f =Fanning friction factor.

4 f L I Dof all loss coefficients in the system.

If the system contains fittings as well as straight pipe, the term(= Kf:pipe) can be replaced by CK, , that is the sum

(4- 179)

where

p1= P , M w / R T l , A P = P , - P2 and Y is the expansion factor.Note that Eq. (1.179) without the Y term is the Bernoulli equationfor an incompressible fluid of density p , . Therefore, the expansionfactor Y = Gadiabatic/Gincompressibles the ratio of the adiabatic massflux (Eq. (4-178)) to the corresponding incompressible mass flux,and is a unique function of P 2 / P , , k and K f. Figure 4-38a shows

values of Y fo r k = 1.3 and k = 1.4 as a function of A P / P , andKf (which is denoted by simply K on these plots). Figures 4-38b

and c show the expansion factor Y for compressible flow throughnozzles and orifices, and plots of the critical pressure ratio forcompressible flow through nozzles and venturi tubes respectively.The conditions corresponding to the lower ends of the lines onthe plots (e.g., the nought) represent the sonic (choked flow) statewhere P2 = P2* . These same conditions are shown in the tablesaccompanying the plots, thus allowing the relationships for chokedflow to be determined more accurately than is possible from readingthe plots.

Note: It is not possible to extrapolate beyond the nought

at the end of the lines in Figures 4-38a and b as this repre-sents the choked flow state, in which P2 = P2* (inside the pipe),and is independent of the external exit pressure. Figures 4-38a

EXAMPLE-11

From the table listed below determine the status of flow (e.g.,whether choking flow exists or not), (AP/P , ) , , , and the flow rate

WI.

Data Value

Upstream pressure, Po(kPaa)Downstream discharge pressure, PA (kPaa)Upstream specific volume, Vo(m3/kg)Isentropic coefficient, yInternal pipe diameter, D(mm)Length of pipe, rnNumber of elbowsLoss coefficient, K

6600200

0.017241.55

52.51004

45

SolutionFor known isentropic coefficient y and loss coefficient K, a guessedvalue of M a , is used in Eq. (4-17 3) until the left side of the equationapproximates to a value of zero. Otherwise, a new guess value ofM a , is used in Eq. (4-173). Once the required value is know n, Eqs.(4-174), (4-175), and (4-176) are subsequently used to determiner , P 2 ,A P, Y,,, and W respectively. This procedure involves the useof the Excel spreadsheet with the Goal Seek or Solver add-in fromthe Tools menu and is given in Example 4-1 1 . ~ 1 ~ .

Using the Ex cel sp readsheet Exam ple 4- 11 xls, the calculated

overall critical pressure ratio r is r = 0.04804The critical pressure P2 is

P2 = r x Po

= 0.04804 x 6600

= 317.06kPa

Test f o r choke f lowSince P2 > P A , the pipe will choke.The critical expansion factor Y,, from Eq. (4-175) is

45 (1 +0.04804)

'" = J 2 [ 4 5 + 2 ( 1 / 0 . 0 4 8 0 4 ) ]

= 0.6795

The critical mass flow rate from Eq. (4-176) is

6600-317.06

(4 5 x 0.01724)= 0.1265 (52.5)' (0.6795)

= 21,320.96kg/h

Figure 4-37 gives the Excel spreadsh eet snapshots of Example 4-1 1.

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220 FLUID FLOW

Figure 4-37 The Excel spreadsheet snapshot of Example 4-1 1

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4.47 T H E EXPANSION FACTOR. Y 221

Figure 4-37-(conrinued)

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222 FLUID FLOW

Figure 4-37-(conrinued)

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224 FLUID FLOW

3. Choked air flow at 50% pressure drop. An equation often used where p a is the critical static pressure at sonic velocity and p o isthe local stagnation pressure at the orifice/valve. Walters indicatedthat using Eq. (4-180) with y = 1.4 results in 47% pressure dropto obtain choking. Furthermore, he stated that Eq. (4-180) cannotbe used with the supply pressure if there is any significant pressuredrop from the supply to the orifice/valve.

to determine the likelihood of sonic choking is

yl (y-1)

(4-180)

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4.50 FRICTION DROP FOR FLOW OFVAPORS, GASES, AND STEAM 225

The entire pipe is solved in one lumped calculation instead ofcoupling the governing equations in marching order.

It is difficult to extend the equations to pipe networks.

Walters developed compressible flow equations for single pipe[35]:

Adiabatic flow equation and integrated solution are

k = cP/cv

Figure 4-38c Critical Pressure Ratio, rC, for compressible flow throughnozzles and venture tubes. (ReprintedA dapted with permission from “Flowof Fluids Through Valves, Fittings, and Pipe”, Technical Paper No. 410,1999, Crane Co. Al l rights reserved. Note: P‘ = psia, p = ratio of small-to-large diameter in orifices and nozzles, and contractions or enlargements inpipes).

For gases with different specific heat ratios, the pressure dropratio will differ somew hat, in accordance with E q. (4-180). In addi-tion, Eq. 4-180) breaks down for pipe-system analysis when pipefriction becomes a factor. This is because the stagnation pressure inthe equation is the pressure at the upstream side of the shock wave.However, if there is any pressure drop in the pipe from the supplypressure to the shock wave, then the supply pressure cannot be usedin Eq. (4-180). Instead, the local stagnation pressure at the shockwave must be used, which is unknown unless the pressure drop isdetermined from alternative means. Therefore, Eq. (4-180) cannotbe used to predict the supply and discharge pressures necessary forsonic choking unless the piping has negligible friction loss.

4.49 OTHER SIMPLIFIED COMPRESSIBLE FLOW

As shown earlier, most gases are not isothermal and, therefore,it is impossible to know how much error is introduced by theassumption of constant temperature.

Simplified equations typically do not address sonic-choking condi-

These equations break down at high M ach numbers.

METHODS

tions, and cannot address the delivery temperature.

Ma2 1- M a2

soL k d x = / M a I y Ma4 [1+ Ma2] dMa2

Isothermal flow equation and integrated solution are

Ma2 (1 - Ma2)lLTd x = lMa dMa2I YMa4

(4-181)

(4- 182)

(4- 83 )

(4- 184)

Computer software has been developed that models pipe systemsof compressible fluids and this can be obtained from the websitewww.aft.com.

4.50 FRICTION DROP FOR FLOW OF VAPORS, GASES,AND STEAM

See Figures 4-39a and 4-3913

A. D ARCY RATIONA L RELATION FOR COMPRESSIBLEFLOW FOR ISOTHERMAL PROCESS [41

(4-65)P 0 . 0 0 0 3 3 6 f w 2 v - .000336 f W2

-- -lOOft d5 d5P

or

A P 0.000001959f (qL)2Sg2-- -10 0 t d5P

In S I units,

62,530f W 2 v 62,530f W2-

d5 d5A P/ 1 0 0 m =

or

(4-185)

(4-66)

(4- 186)

The general procedures outlined previously for handling fluidsinvolving the friction factor, f , and the Re chart are used with theabove relations. This is applicable to compressible flow systemsunder the following conditions [4].

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226 FLUID FLOW

AP I 100 ft = 0.000336 WPld6p p V

40

30

.04

Ind

2

APnm

.4

.5

.S

‘.l

.8

.9

1.0d

307

Figure 4-39a Pressure drop in compressible flow lines. (Reprinted/Adaptedwith permission from “Flow of Fluids Through Valves, Fittings, and Pipe”,

Technical Paper No . 410, 1999,Crane Co. All rights reserved.)

where

W = rate of flow, Ib/h (kgh)

V = specific volume of fluid, Ib/ft3(m3/kg)f = friction factord = internal pipe diameter, in. (mm)

p = fluid density, Ib/ft3(kg/m3)

-

K = loss coefficient for all valves, fittings, and pipe (resistancecoefficient)

r=fittings+valves

-S, = specific gravity of gas relative to air= he ratio of

molecular weight of the gas to that of air

q; = rate of flow, ft3 /h( m3/ h) at standard conditions (14.7psiaand 60” F), SCFH (m3/s at metric standard conditions(MSC) -1.01325bara and 15°C).

q i = flow rate, ft3/h at 14.7 psia and 60” FS, = specific gravity of a gas relative to air = the ratio of the

P’ = pressure, lb,/in.2 abs

T , = inlet temperature, abs (” R = F+460)

p1 = upstream density of steam, lb/ft3.

molecular weight of the gas to that of air

A P = pressure drop, psi

Babcock formula for steam flow at isothermal condition is

q; = 24,700 [Yd’/S,] ( A P p , / K ) ’ ’ ’ , CFH at 14.7psia and 60”F In SI units,

(4- 187) qh = 1.0312 [Yd’lS,] ( A p p I / K ) ’ I 2 (4-1 89)

or or

qh = 40,700Yd’ [ ( A P ) P i ) / ( K T I S g ) ]” * (4-188) qi = 19.31 Yd2[ ( A p ) ( p ; ) / ( K T 1 S , ) ] ’ / 2 (4- 190)

where where

d = internal pipe diameter, in.Y = net expansion factor for compressible flow through

orifices, nozzles, or pipe (see Figures 4-38a-c)

d = internal pipe diameter, mmY = net expansion factor for compressible flow through

orifices, nozzles, or pipe

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4.50 FRICTION DROP FOR FLOW OFVAPORS, GASES, AND STEAM 227

-- 0

r 3 0 2

100.0150 .01

Pressure Equivalent: Ibar =-

Index 2

1 0 5 p a

IO0 Pa

d

f

W

T o o

[ w

Figure 4-39b Pressure drop in compressible flow lines (Metric units). (ReprintedAdapted with permission from “Flow of Fluids Through Valves,Fittings, and Pipe”, Technical Paper No . 410, 1999,Crane Co. Al l rights reserved.)

K = loss coefficient (resistance coefficient)p’ = pressure, baraqi, = flow rate, m3/h at MSC (metric standard conditions -

S, = specific gravity of a gas relative to air = he ratio of the1.01325bar at 1 5 ° C )

molecular weight of the gas to that of airAp = pressure drop, barT , = inlet temperature, abs ( K = “C +273)

p 1 = upstream density of steam, kg/m3.

Figures 4-40a and b (SI) are useful in solving the usual steamor any vapor flow problem for turbulent flow based on the modifiedDarcy friction factors. At low vapor velocities the results may below; then use Figure 4-39a or b (SI). For steel pipe the limitations

listed in (A) above apply.

1. Determine C1 and C from Figure 4-40a or b (SI units) andTable 4-26 for the steam flow rate and assumed pipe size respec-tively. Use Table 4-6 or 4-19 to select steam velocity for line

size estimate.2. Read the specific volume of steam at known temperature and

3. Calculate pressure drop (Figure 4-40a or b) per lOOft of pipe

B. ALTERNATE VAPOR/GAS FLOW METHODS

Note that all specialized or alternate methods for solving arebased on simplified assumptions or empirical procedures presented

pressure from steam tables.

from

earlier. They-are not presented as better approaches to solving thespecific problem. AP/lOOft = CIC,v (4- 191)

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228 FLUID FLOW

7

64

2 5

G

B

0

'X

v)

-VI

f 3.5

= 3

2-- 25

a

+.-

0

01

mCT

-' 2

-

L5

la

.9

.8

Th e Darcy formula can be written in the following form :

-- .07

.06

.35

.04

.03

,025

.02

.015

.o 1.no9

.008

.@I6

.005

.004

.a3

.0025

.no2

.0015

.m7

.001

.In09

.owe

.COO7

.OW6

c l = - = -Pioo APioop c * = - = -Pwo A P m P

C1 = discharge factor from chart at right.C, = size factor, from table on next page.

Czv CZ ClV C1

The limitations of the Darcy formLila for compressible

flow, as outlined on page 3 - 3 , apply also to the simplifiedflow formula.

Example 1

Given: Steam a t 34 5 psig and 500 F flows through 8-inchSchedule 40 pipe at a rate of 240,000 pounds per hour.

Find: The pressure drop per io0 feet of pipe.

Soluiion: CI = 57

Ce = 0.146

V = 1 .4 j . . . . _ . . . . pgc 3-17cr A-16-

APioo = 57 x c.146 x 1.4j = 12

Example 2

Given: Pressure drop is 5 psi with 100 psig air at go Fflowing through 1 0 0 feet of 4-inch Schedule 40 pipe.

Find: The flow rate in standard cubic feet per minute.

Solu t ion : APloo = 5.0

Cz = j . 1 7

p = 0 . j64

W = 2 3 OCO

. . . . . . . . . . . . .page .<%lo

( 5 . 0 x 0.564) + j . 1 7 = 0 . j 4 ji

q f m = w f (4.58 s,) . . . . . . . . . . . . . .page B-2

q f m = 23 ooo + (4.58 x 1 . 0 ) = 5000 scfm

1i - ii'

.I---u)d

---

- .7

>

II ' CI

For C? alues and an example on "determining p ipe dxe",

see th e opposite page.

Figure 4-40aPipe". Technical Paper N o . 410, 1999, Crane Co. All rights reserved.)

Simplified flow formula fo r compressible fluids. (Reprinted/adapted with permission from "Flow of Fluids Through Valves, Fittings, and

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4.50 FRICTION DROP FOR FLOW OFVAPORS. GASES, AN D STEAM 229

The Darcy formula can be written in the following form;

ktcl = E and c2= 62 530 x 10’ L10 d5

The simplified flow formula can then be written:

Ca = disduuge factor , from chart at right

Cz =size factor from tables on pages 3-23 to 3-25

The limitations of the &cy formula for compressible flaw, asoutlined on page 3-3 apply also to the simplified flow formula.

Example 1

Given: Steam at 24 bar absolute and 250’C flows through an

&inch Schedule 40 pipe at a rate of 100 000 kilograms per

hour.Fmd: The pressure drop per 1Do met= of pipe.

Solufion: CI = 100

C2

vAp ,oo = 100x 0.257 x 0.091

Ap = 2.34bar

= 0.257 . . . . . . . . . . . . . . . . acing page

= 0.091 m3 kg. . . . . . . page 3-17 or A 4 5

Exampte 2

Given: Pressure drop is 1 bar with 7 bar gauge air at 3OoCflowing through 100 metres of 4 inch nominal size IS 0 steelpipe, 6.3 mm wall thickness.

Fd: he flow rate in cubic metres per minute at metricstandard con ditions(1.01’3 5 bar and 15°C).

SblUtfoR: &,a, = 1Cz = 9 . 4 2 . . . . . . . . . . . . . . . . age 3-24

P = 9.21 . . . . . . . . . . . . . . . p g e A - 1 0

w = 9 9 0 0

q‘,,, = W + 73.5 Sg) . . . . . . . . . . . age B-2

q’,,, = 9 9 0 0 + ( 7 3 . 5 ~ 1 ) = 134.7m3/min

Fo r C valuesse opposite pagean d pager 3-24, 3-23.

Fo r example on determiningpipe

me see oppomtc page.

Figure 4-40bFittings,an d Pipe”, Technical Paper No. 410, 1999, Crane Co. Al l rights reserved.)

Simplified flow formula for compressible fluids (Metricunits). (Reprintedadapted with permission from “Flow of Fluids Through Valves,

4.

5 .

6.

7.

8.

Determine the loss coefficient K of all fittings, valves, and so on,

and hence the equivalent length ( K = f L,,/D) or, alternatively,from Figure 4-24 or 4-25.Determine expansion and contraction losses, fittings and atvessel connections.Determine pressure drops through orifices and control valves.Total system pressure drop

APTOTAL= (L+L,,) (AP/lOO)+Item5+Item6 (4-192)

If pressure drop is too large (or greater than a percentage of theinlet system pressure), re-estimate line size and repeat calcula-tions (see paragraph (A ) above) and also examine pressure dropassumption for orifices and control valves.

C. AIR

For quick estimates for air line pressure drop and through an orifice,see Tables 4-21a and b.

D. BABCOCK EMPIRICAL FORM ULA FOR STEAM

Comparison of results between the various empirical steam flowformulas suggests the Babcock equation is a good average for mostdesign purposes at pressure 500 psia and below. For pipe linessmaller than 4in., this relation may be 0 4 0 % high [38].

(4-193)2 L

p l - p z =AP=0.000131(1+3.6/d)-Pd5

A P / 1 0 0 ft = w2F/p (4-194)

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230 FLUID FLOW

TABLE 4-26 Simp lified Flow Form ula For Compressible Fluids Pressure Drop, Rate of flow, and pipe sizes(use wit h figure 4-40a)

Valuesof C,

Nominal

Pipe Size

Inches

18

14

38

12

-

-

-

-

34-

1

I4

I2

2

2_!2

3

3_12

4

Schedule

Number

40 s80x

40 s80x

40 s

80 x

40 s80 x

160

.. .xx

40 s80 x

160

40 s80 x

...xx

160

.. . xx

40 s

80 x160

...xx

40 s80 x

160

...xx

40s80x

160...xx

40 s

80x160

.. . xx

40 s

80x160

...xx

40s

80 x

40 s

80x120

160

...xx

Value

1f C, Pipe Size

26,200,000.00

1,590,000.004,290,000.00

319,000.00

718,000.00

93,500.00

186,100.00

4,300,000.00

11,180,000.00

21,200.00

36,900.00

100,100.00

627,000.00

5,950.009,640.00

22,500.00

114,100.00

1,408.00

2,110.00

3,490.00

13,640.00

627.00

904.00

1,656.00

4,630.00

169.00

236.00

488.00899.00

66.70

91.80

146.30

380.00

21.40

28.70

48.30

96.60

10.00

37.70

5.17

6.75

8.94

11.80

18.59I1

Schedule

Number

40 s80 x

120

160...xx

40 s

80x120

160

.. . xx

2030

40 s

60

80x

100

120140

160

2030

40 s60x

80

100120

140

160

2030

40

60

80

100

120

140160

10

20

30 s40

60

80

100

120140

160

...xx

...s

...x

.. .x

Inches

2.042.69

3.594.93

0.610

0.798

1.015

1.376

1.861

0.1330.135

0.146

0.163

0.185

0.21 1

0.2520.289

0.317

0.333

0.03970.0421

0.0447

0.0514

0.0569

0.06610.0753

0.0905

0.1052

0.01570.0168

0.0175

0.01800.0195

0.0206

0.0231

0.0267

0.0310

0.03500.0423

0.00949

0.00996

0.01046

0.010990.01155

0.01244

16

18

20

24

0.01416

of Schedule Numbers indicate

jchedulc

Number

102030s40x60

80

100

120

140

160

10

20

...s30

...x40

6080

100

120140160

10

20 s30 x40

60

80100120

140

160

10

20s

30

4060

80

100

120

140160

...x

Note:

Value

of c,

0.004630.00421

0.00504

0.005490.00612

0.00700

0.00804

0.00926

0.01099

0.01244

0.00247

0.00256

0.0026 6

0.00276

0.00287

0.00298

0.00335

0.00376

0.00435

0.005040.005730.00669

0.00141

0.00150

0.00161

0.00169

0.00191

0.002170.002510.00287

0.00335

0.00385

0.000534

0.000565

3.000597

3.000614

3.0006513.000741

1.000835

1.000972

1.001119

3.0012741.001 478

Source: By permission from Crane Co., Technical Paper No. 410, Engineering Div., 1957. See author’s note at Figure 2-31.

4.51 DARCY RATIONAL RELATION FORCOMPRESSIBLE VAPORS AN D GASES

Figure 4-41 is a convenient chart for handling most in-plantsteam line problems. For long transmission lines over 200 ft. the

line should- be calculated in sections in order to re-establish thesteam-specific density. Normally an estimated average p should beselected for each line increment to obtain good results.

Table 4-28 is used to obtain the value for “F’ n Eq. (4-194).

1. Determine first estimate of line size by using suggested velocity

2. Calculate Reynolds number R e and determine the friction factor,from Table 4-6.

f, sing Figure 4-5, 4-42a, or b (SI) (for steel pipe).

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4.51 DARCY RATIONAL RELATION FOR COMPRESSIBLE VAPORS AND GASES 231

TABLE 4-27a Flow of Air Through Sch. 40 Pipe' (use for estimating; for detailed calculations use friction factors f)

For lengths of pipe other than100 feet, the pressure drop is

hUSfor 50 feet Of pipe, theP m rop isa F C " ~ ~ l Yone-half thevaluegiveninthetable . . for 30 0 feet, threetimes the given value, etc.

Airroportional to the length. Q'n

QbiC FeetPer Minuteat 60F nd

14,7psia

The ressuredropisalsoin-verseyy proportional to theabsolute pressure and directly.proportional to the ahsolutetemperature.

Therefore, to determine thepressure drop for inletoraver-age pressures other than 100psi and at temperatures otherthanMF, multiply thevaluesgivenin the table by the ratio:

100+14.7 460+1( P+14.7 )( 520 )

where:

"P" is the inlet or averagegauge pressure in pounds per

square inch, and,"t " is the temperature indegrees Fahrenheit underconsideration.

The cubic feet per minute ofcompressed air a t any pres-sure is inversely pmportiohalto the absolute pressure anddirectly proportional to theabsolute temhrature.

To determine the cubic feetper minute of compressed airat any temperature and pres-sure other than standard con-ditions, multiply the value ofcubic feet per minute of freeair by the ratio:

Calculations for PlprOlh w than Schodulo 40

To determine the velocit Of

water, or the pressure &pof water or air, through p~peother than Schedule 40,usethe following formulas:

Subscript "a" refers to theSchedule of pipe throughwhich velocity or pressuredrop is desired.

Subscript "40" refers to t h evelocity or ressure dropth ro ue Schelule 40 pipe, a3given in the tables on thesefacing pages.

12346

68

101620

1303s404

8708090

10012S160176lW

12Slso2763003%

3503764w4%480

476600660600650

700

SKI900

9110100011001 l W1300

11001600160018001000

16003000360040004600

60006000

700080009000

1000011000120001300014000

1600016000180002000022000

u000160002800030ooo

a

corn-arcdAil

ubic Feel:r Minutc60Fanc

100 psig

0.12s0.m0.3840.6130.641

0.7691.0251.2821.9222.663

3.2043.8454.-5.1266.767

6.4087.6908.971

1o.Zs11.63

12.8216.0219.2222.4325.63

28.8432.0436.2438.4541.65

44.8748.0651.2654.4757.67

60.8864.0870.4976.9083.30

89.71%.I2102.5108.9115.3

121.8128.2141.0163.8166.6

179.4192.2205. 1230.7256.3

320.4384.5448.6512.6676.7

W 8769:O

897.1IO2511 8

12821410163816661794

19222051230725632820

307633323688384s

-0.3611.313.064.837.46

.0.6

.8.6s.7........ .*..*....... .

234'

0.0190.023

0.0290.0440,0620.0630.107

0.1340.1640.1910.2320.270

0.3130.3560.4020.4620.507

0.5620.6230.7490.8871.04

1.191.361.551.741.96

2.182.402.893.444.01

4.655.316.047.659.44

14.721.128.837.647.6

. ..

.. .

.. ... ....

.. ...... .

. . .

.. .

.. ... .

.. ... .

.. .

...

.. .

. . ....-

-1/11

0.0830.2850.6051.041.68

2.233.896.963.02.8

5.6. . ...... ... .

3'

0.0110.0280.036

0.0450.0550.0640.0780.090

0.1040.1190.1340.1510.168

0.1870.2060 .240.2930.341

0.3950.4510.5130.5760.641

0.7180.7880.9M1.131.32

1.521.741.972.503.06

4.766.829.23

12.115.3

18.827.1

36.9.. ... ... .. ... . ... .. . .

.. .

.. .

. . .

.. .

. . .

.. .

. ..

. . .

.. .-

Pre-ure Drop of AirIn Pounds per Square Inch

Per 100Feet of Schedule40P i p

For Air at 100 Pounds pe rSquare Inch Gaul %ss.me

and 60 F Teml ature-/8'

0.0180.0640.1330.2260.343

0.4080.8481.262.734.76

7.34L0.514.218.413.1

L8.5w.7. . .. . .. . .

3lh"

0.0220.0270.0320.0370.043

0.0500.0570.0640.0720.081

0.0890.0990.1180.1390.163

0.1880.2140.2440.2740.305

0.3400.3750.4510.5330.626

0.7180.8240.9311.181.45

2.253.204.335.667.16

8.8512.7

17.222.628.5

36.2. . .. . .. . .. . .. . .. . .. . .. . .. . .

. . .

. . .

.. .

. . .-

-M

0.0200.0420.0710.106

0.1480.2550.3560.8341.43

2.213.154.245.496.90

8.492.26.5.1.4,7.0

13.2. . ... .t..

.. .

4'

0.0300.0340.0380.042

0.0470.0520.0620.0730.086

0.0990.1130.1270.1440.160

0.1780.1970.9360.2790.327

0.3770.4310.4900.6160.757

1.171.672.262.943.69

4.566.57

8.9411.714.9

18.422.226.431O36.0

. . .

...

. . .

. . .

. . .

. ..

. . .

. . .

.. .

-3A'

0.027

0.0370.0620.0940.2010.346

0.5260.7481.001.301.62

1.992.853.834.966.25

7.691.97.03. I0.0

7.9.. .. . .. . .. . .. . .. ... . .. . .. . .

5"

0.0320.0360.0410.0460.051

0.0570.0630.0750.0890.103

0.1190.1%0.1540.1980.239

0.m0.5240.7s0.91!1.16

1.422.03

2.763.594.54

5.606-788.079.47

11.0

12.614.318.222.427.1

32.337.9

.. ....

-1"

3.0191.0291.0623.102

0.156D.219D ,293D.379D.474

0.578Q.8191.101.431.80

2.213.394.876.608.54

0.83.36.09.02.3

5.89.63.67.9. . .

. . .

. . .

. . .

. .

. . .

6"

0.02:0.0250 03(0: 03!0.041

0.04:0 . 0 5 ~0 . 06 '0.07!0.09,

0.14.0.20.0.2710.350.45

0.550.79

1.071.391.76

2.162 . a3.093.634.21

4.845.506.968.66

10.4

12.414.516.919.3-

-%*

), 026

).039). 55).0731.0951.116

1.1491.2001.2701.3501.437

1.5341.8251.171.581.05

1.59?.83.831.565.39

5.171.055.029.01D.2

1.31.55.18 .01.1

4.37.91.85 . 90 . 2

. . .

. . .

. . .

. . .

. . .

0"

0.02;

0.03I0.0510.0610.W0.11:

0.13(0.191

0.26'0.33'0.42'

0.52'0.63.0.75,0.881.02

1.171.331.682.012.50

2.973.494.044.64-

-1H'

0.0190.0260.0360.0440.056

0.0670.0940.1260.1620.203

0.2470.3800.5370.7270.937

1.191.451.752.072-42

2.803.203.644.094.59

5.095.616.798.049.43

0.92.64.26.08. 0

!O.O12.116.711.817.3

10"

0.0160.0210.0240.W

0.04:

0.08,0. lb;0.13

0.160.19;0.230.27;0.31(

0.410.5210.640.77

0.911.121,51.42

0.06!

0 . F

-

-2.

0.0190.0270.0360.0460.068

0.0700.1070.1610.2080.264

0.3310.4040.4840.6730.673

0.7760.8871.001.131.26

1.401.681.872.212.60

3.003.443.904.404.91

6.476.067.298.630.1

1.83.65.3.9.3u. 9

17.3

12"

0.0180.-0.0340.0440.W

0.0670.w10.M0.1120.129

0.1480.1670.2130.2600.314

0 371

0.6050.620

0:u5-*By permission TechnicalPaper No. 410, Crane Co., EngineeringDiv., Chicago,1957.

8/6/2019 7766X_04e[1]


232 FLUID FLOW

TABLE 4-27b Discharge of Air Th roug h a n Or ifice" in Cubic Feet of Free Air per m inute at Standard A tmosphericPressure of 14. 7 Ib in.2 ab s and 700F

GaugePressureBeforeOrifice

in PoundsPer square inch

Diameter of Orifice (in.)

7

32 16 8 4 8 2 8 4 81-------

6 4

-1 0.028

2 0.040

3 0.048

4 0.056

5 0.062

6 0.068

7 0.073

9 0.083

12 0.095

15 0.105

20 0.123

25 0.140

30 0.158

35 0.176

40 0.194

45 0.211

50 0.229

60

70

80

90

100

110

120

125

0.264

0.300

0.335

0.370

0.406

0.441

0.476

0.494

0,112 0.450

0.158 0.633

0.194 0.775

0.223 0.892

0.248 0.993

0.272 1.09

0.293 1.17

0.331 1.32

0.379 1.52

0.420 1.68

0.491 1.96

0.562 2.25

0.633 2.53

0.703 2.81

0.774 3.10

0.845 3.38

0.916 3.66

1.06 4.23

1.20 4.79

1.34 5.36

1.48 5.92

1.62 6.49

1.76 7.05

1.91 7.62

1.98 7.90

1.80 7.18

2.53 10.1

3.10 12.4

3.56 14.3

3.97 15.9

4.34 1 17.4

4.68 18.7

5.30 21'2

6.07 24.3

6.72 26.9

7.86 31.4

8.98 35.9

10.1 40.5

11.3 45.0

12.4 49.6

16.2 1 28.7

22.8 1 40.5

27.8 1 49.5

39.1 69.5

42.2 75.0

47.7 84.7

54.6 97.0

60.5 108

70.7 126

80.9 144

91.1 162

101 180

112 198

122 216

132 235

152 271

173 307

193 343

213 379

234 415

254 452

274 488

284 506

45.0

63.3

64.7

91.2

77.5 I 111

89.2 128

99.3 143

109 156

117 168

132 191

152 218

168 242

196 283

225 323

366 528

423 609

649 934

1016

762 1097

1138

175 228

213 278

230 300

329 430

50 3

440 575

49 6

66 2

718 938

1082

939 1227

1050 1371

1161 1516

1272 1661

1383 1806

1494 1951

1549 2023

Notes: Table is based on 100% coef f ic ient of f low. For wel l -roun ded entrance mul t ip ly values by 0.97. For sharp-edged or i fices a

Values for pressures f rom 1 to 15 Ibs gauge calculated by s tandard adiabat ic formula.Values for pressures above 15 Ib. gauge calculated by approx imate formu la p roposed b y S.A. Moss.

mult ipl ier of 0.65 ma y be used for approx ima te resul ts .

aCPlW, = 0.5303-

l

where

W,=discharge in Ibis

a =area o f o r i f ice in.2

C =Coef f i c ien t o f f l ow

P =Up stream total pressure in Ib/in.2 abs

Tl =Ups t ream temperature in O F , ab s

Values used in calculating ab ove table were: C = 1.0, P, =gauge pressure + 14.7 Ib/in2.

Weights ( W )were converted to volu me us ing dens i ty of fac tor of 0.07494 Ib/ ft . This is correct for dry air at 14.7 Ib/ in2.

Formula cannot be used where P is less than tw o t imes the baro metr ic pressure.

(Source: By permiss ion f rom F.W. O'Neil, ed., "Comp ressed Ai r Data", Compre ssed Air Magazine, 5th Ed. N ew York, 1939 [371.)

abso lute pressure an d 70°F.

8/6/2019 7766X_04e[1]


:LIJlDS IN PIPE 233

PRESSURE LOSS IN LBPER SQ NCH PE R I00FEET

Based on Babcock Formula: .P~O.000131 I t x )dLd P&

Figure 4-41 Steam flow chart. (B y permission from Walworth Co . Note: Use for estimating only (Ludwig [19].)

3.4.5.

6 .

7.

8.

Determine total straight pipe length, L .

Determine equivalent pipe length for fittings and valves, Leq.

Determine or assume losses through orifice plates, controlvalves, equipment, contraction and expansion, and so on .Calculate pressure drop, AP/lO Oft (or use Figure 4-39a or b

(SI)).

0.000336 f W2AP/100ft =

P d5

0.000000726f T S, (q;)’

P’d5

--

(4-65)

(4-65a)

Total pressure drop, A P total

= ( L + L e g ) APllOO) + tem5 (4-195)

If total line or system pressure drop is excessive (or greater thana percentage of the inlet system pressure), examine the portionof pressure drop due to pipe friction and that due to other factorsin the system. If the line pressure drop is a small portion of

the total, little will be gained by increasing pipe size. Considerreducing losses through items in step 5 above. Recheck otherpipes sizes as may be indicated.

4.52 VELOCITY OF COMPRESSIBLE FLUIDS IN PIPE

See Figures 4-43a and 4-43b

2.40Wv 3.06Wv 3.06Wurn=-.---- - -

a d 2 d 2P

In SI units,

16,670Wv 21,220Wv 21,220W--

a d 2 d 2=

(4-196)

(4- 196a)

where

v, = mean velocity in pipe, at conditions stated for v, tlmin

( d s )a = cross-sectional area of pipe, in.2 (m2)

W = f low rate, Ib h (k gh)v = fluid specific volume, ft3/lb (m3/kg)d = inside pipe diameter, in. (mm)p = fluid density, lb/ft3 (kg /m3 ) at T and P.

Note that determining the velocity at the inlet conditions to a pipemay create significant error when results are concerned with theoutlet conditions, particularly if the pressure drop is high. Even theaverage of inlet and outlet conditions is not sufficiently accuratefor some systems; therefore conditions influenced by pressure drop

8/6/2019 7766X_04e[1]


234 FLUID FLOW

TABLE 4-28 Factor”F” For Babcock Stea m Formula

Nominal Pipe Standard Extra StrongSize (in.) Weight Pipe’ Pipe+

955.1x 2.051x

184.7 x I 0-3

1 114 9.432 x 10-3 14.67x I 0-3

2 951.9 x 1.365x

2 ’I 2 351.0 x 493.8 x3 104.7x 143.2x

46.94 x 62.95x3 ’ I 2

4 23.46x 31.01 x

5 6.854 x 10-6 8.866x6 2.544 x 3.354 x 10-6

8 587.1 x IO-’ 748.2 x 10-910 176.3x 225.3 10-912 70.32 10-9 90.52 10-9

14 OD 42.84 x 10-9 55.29 x 10-916 O D 21.39 x IO-’ 27.28 10-9

20 O D 6.621 x IO-’ 8.469 10-924 OD 2.561 x IO-’ 3.278 10-9

340.8 10-3I23i 4

1 45.7 x 10-3

1 12 3.914 x 5.865 10-3

77.71 x

18 OD 11.61x IO-’ 14.69x I 0- 9

(Source:By permiss ion f rom The Wa lwor th C o . )* Factors are based up on ID l isted as Schedule 40

Factors are based upo n ID l isted as Schedule 80

can produce more accurate results when calculations are prepared

for successive sections of the pipe system (long or high pressure).

~~

EXAMPLE-12Pressure Drop for Vapor SystemThe calculations are presented in Figure 4-44, Line Size Spec-

Figure 4-45 is convenient when using Dowtherm vapor.

ification Sheet.

4.53 ALTERNATE SOLUTION TO COMPRESSIBLEFLOW PROBLEMS

There are several good approaches to recognizing the effects ofchanging conditions on compressible flow [39, 401.

FRICTION DROP FOR AIR

Table 4-27a is convenient for most air problems, noting that bothfree air (60 °F and 14.7psia) and compressed air at 1OOpsig and

60”F are indicated. The corrections for other temperatures and pres-

sures are also indicated. Figure 4-46 is useful for quick checking.

However, its values are slightly higher (about 10%) than the rational

values of Table 4-26, above about 1OOOcfm of free air. Use for

estimating only.

EXAMPLE-1 3Steam Flow Usine Babcock Formula

w = 1891Yd i2JAPpl/K, Ib/hY

Determine the pressure loss in 138ft of 8-in. Sch. 40 steel pipe,

flowing 86,000 lblh of 150 psig steam (saturated).In units,

Use Figure 4-41, w = 86,000/60 = 14321b/minReading from top at 150 psig, no superheat, down vertically to

intersect the horizontal steam flow of 1432 Ib/min, follow diagonal

line to the horizontal pipe size of 8 in., and then vertically down to

the pressure drop loss of 3.5 psi/lOOft.

For 138ft (no fittings or valves), total A P is 138(3.5/100) =

4.82psi.

w = 1.111 x 10-6Yd1?/g>g/s

w = 1.265 Y d 1 2 / F , g/h

For comparison, solve by equation, using value of F = 587.1 x

from Table 4-28.For nozzles and orifices (vapors/gases):

AP/lOOft = (1432)* (587.1 x 10-9)/0.364

= 3.31 psi/lOOft

AP,,,,, = (3.31/100) (138) =4.57psi

u3 = 0.525 Y dI2 ’,/?, Ib/s

In SI units,

These values are within graphical accuracy.For the discharge of compressible fluids from the end of a

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 to compensate for fluid property

changes due to the expansion of the fluid, and these are known as

Y net expansion factors [4]. The corrected Darcy equation is:

w=1.111x10-6Yd,2C’{g>g /s

For

English Engineering units

fittings$and Pipe (liquids):

For valves, fittings, and pipe (vapordgases):

w = 0.525 Y d i 2 d m ,b/s (4-197)

(4-198)

(4-199)

(4-200)

(4-201)

(4-202)

(4-203)

(continued)

8/6/2019 7766X_04e[1]


4.53 ALTERNATE SOLUTIONTO COMPRESSIBLE FLOW PROBLEMS 235

EXAMPLE-13-(continued)In SI units,

W = 1.111 x 10-‘dl2/;,kg/s AP(P1)

English Engineering unitsFor nozzles and orifices (liquids):

w = 0.525 d I 2C ’ dm ,b/s

In SI units,

(4-204)

(4-205)

w = 1.111 x lo-‘ d i2C ’ J m , g/s (4-206)

where

d , = pipe inside diameter, in. (mm)

p 1 = upstream fluid density, lb/ft3(kg /m3 )w = rate of flow, lb/s (kg/s)

AP = pressure drop across the system, psi (bar) (inlet-discharge)K = total resistance (loss) coefficient of pipe, valves, fittings,

Y = net expansion factor for compressible flow throughand entrance and exit losses in the line

orifices, nozzles, and pipes [4] (see Figure 4-38a, b, or c)A P = pressure drop ratio in AP/P’ , used to determine Y from

Figure 4-38a and b. The AP is the difference between theinlet pressure and the pressure in th e area of larger crosssection

C’ = flow coefficient for orifices and nozzles (Figures 4-19 and

For example, for a line discharging a compressible fluid to atmo-sphere, the A P is the inlet gauge pressure or the difference betweenthe absolute inlet pressure and the atmospheric pressure absolute.When AP/P‘ falls outside the limits of the K curves on the charts,sonic velocity occurs at the point of discharge or at some restric-

tion within the pipe, and the limiting value for Y and A P must bedetermined from the tables in Figures 4-38a and b and used in thevelocity equation, us,above [4].

Figures 4-38a and b are based on the perfect gas laws and forsonic conditions at the outlet end of a pipe. For gaseshapors thatdeviate from these laws, such as steam, the same application willyield about 5% greater flow rate. For improved accuracy, use thecharts in Figures 4-38a and b (SI) to determine the downstreampressure when sonic velocity occurs. Then use the fluid propertiesat this condition of pressure and temp erature in:

4-20).

us = JksRr= /kg144Ptv, f t /s (4-207)

or v , = = / ~ P I v , /s (4-208)

to determine the flow rate at this condition from

u = q / A = 183.3 q/d 2 = 0.0509 W / ( p ) d’) , f t /s (4-30)

In SI units,

Q Wu = q / A = 1.273 x 10‘ q / d 2= 21.22- = 354 -, m /s (4-31)

d 2 p d 2

where

d = nternal diameter of pipe, in. (mm)

A = cross section of pipe, ft2 (m2 )

q = ft3/sc (m 3/s ) at flowing conditions

T = temperature, ’ (K = 273 + )t = fluid temperature, Ck = y = ratio of specific heats (C,/C,)

P’ = pressure, psia (N/ m2 abs)W = flow, lb/h (kg/h)

v = velocity, mean or average, ft/s ( d s ) .

These conditions are similar to flow through orifices, nozzles, andventuri tubes. Flow through nozzles and venturi devices is limitedby the critical pressure ratio, r, = downstream pressurehpstreampressure, at sonic conditions (see Figure 4-38c).

For nozzles and venturi meters, the flow is limited by criticalpressure ratio and the minimum value of Y is to be used. For flowof gases and vapors through nozzles and orifices:

o r q = Y C’ A - - - , m 3 / sJ2,,

(4-209)

(4-210)

where

p = ratio of orifice throat diameter to inlet diameterC’ = flow coefficient for nozzles and orifices

(see Figures 4-19 and 4-20), when used as perASME specification for differential pressure

p = fluid density, lb/ft3 (kg/m3)A = cross-sectional flow area, ft2 (m 2).

Note: The use of C’ eliminates the calculation of velocity of

approach. The flow coefficient C‘ is C‘ = C , / m , C =

discharge coefficient for orifices and nozzles [4].For compressible fluids flowing through nozzles and orifices

use Figures 4-19 and 4-20, using h , or A P as differential static heador pressure differential across taps located one diameter upstream,and 0.5 diameters downstream from the inlet face of orifice plate ornozzles, when values of C are taken from Figures 4-19 and 4-20 [4 ].For any fluid:

In SI units,

(4-21 1)

q = CIA/= , m3/sf lowP

(4-2 12)

(continued)

8/6/2019 7766X_04e[1]


236 FLUID FLOW

In SI units,EXAMPLE-13-(continued)For estimating purposes in liquid flow with viscosity similar towater through orifices and nozzles, the following can be used [7]:

q = Y m'/s (4-222)

(4-213)where

Y = net expansion factor from F igure 4-38a or b

A P = differential pressure (equal to inlet gauge pressure whendischarging to atmosphere)

conditions

(4-214) p = weight density of fluid, lb/ft3 (kg/m3) at flowing

A = cross-sectional area of orifice or nozzle, ft3 (m2)

In S I units,

1Q = 0.2087C'd12& 4 , V m i n1 - ( $ )

do .

diwhere - s greater than 0.3

Q = 19.636C'dO2& where -o is less than 0.3

di

In SI units,

Q = 0 . 2 0 8 7 C ' d O 2 h , l m i n

or [4], W = 1 5 7 . 6 d O 2 C ' m

= 1891 do2C ' m

In SI units,

W = 0 . 0 1 2 5 2 d O 2 C m 1.265dO2C&

where

Q = liquid flow, gpm (Ymin)C =

flow coefficient for orifices and nozzles=

dischargecoefficient corrected for velocity of

C' = flow coefficient from Figure 4-38a or 4-38b.(4-215)

(4-216) W = 1 8 9 1 Y d O 2 C ' / ~ , l b / h (4-223)

or W = 1.265 do2C'&, kg/h (4-224)

(4-217)

(4-21 8) where

(4-219)do= internal diameter of orifice, in. (mm)

VI = specific volume of fluid, ft3/lb (m3/kg)

p1 = density of fluid, Ib/ft3 (kg/m3)A p = pressure drop, psi (bar).

(4-220)

q ' = l l . 3 0 Y d ~ C ' & , f t 3 / s , a tP P ; 14.7 psia and 60 °F

(4-225)

approach = C, /-do = diameter of orifice o r nozzle opening, in. (mm)di = pipe inside diam eter in which orifice or nozzle is installed,

h = static pressure head existing at a point, in. (m) of fluid.h, = loss of static pressure head due to fluid flow, m of fluid.C' = flow coefficient (see Figure 4-47 for water and

in . (mm)

Figures 4-20 and 4-21 fo r vapors or liquids)

q = f t3/s (m3/s) at flowing conditionsrc = critical pressure ratio for compressible flow, = P ; / P i

A P = pressure drop, psiA p = pressure drop, bar (hLan d Ap measured across taps at 1

W = flow rate, lb/h (kg/h).diameter and 0.5 diameter)

Flow of gases and vapors (compressible fluids) through nozzlesand orifices (for flow field importance, see [41]). From [4]:

where

S, = sp gr of gas relative to air = mol wt of gad29T I = absolute temperature, ' = (460 +' F)Pi = pressure, psia.

In SI units,

q = 0.005363YdO2I/""";. m 3 / sTI s,

(4-226)

where

T , = 2 7 3 . 1 5 + t

p i = pressure, baraA p = pressure drop, barS, = sp gr of gas relative to air = mol wt of gad29Y = net expansion factor compressibility flow through orifices,

t = fluid temperature, a C

nozzles, or pipe.

(4-221)

(at flowing conditions)

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