FORMULA'S AND MONOGRAPH FOR FLOW THROUGH VALVE FITTING AND PIPES

CACULATION VALVE FITTING AND PIPES

CHAPTER 3 CRANE FORMULAS AND NOMOGBAPHS FOR FLOW THROUGH VALV&S, FITTINGS AND PIPE 3-8

Reynolds Number for Liquid Flow Friction Factor for Clean Steel Pipe

887

.P])"bhooha24431143331093339933333/34338353

CHAPTER 3 FOR_FLOW THROUGH VALVES, FITTINGS-AND BIPE CRANE FORMULAS AND NOMOGRAPHS

CHAPTER 3 "

Pressure Drop in Liquid Linas for Laminar Flow

Pressure drop can be calculated from the formula below, or, from the nomograph on the opposite page, only when the flow

is laminar The nomograph is a graphical solution of the formula,

Flow is considered to be laminar at Reynolds number of 2000

or less; therefore, before using the formula or nomograph, determine the Reynolds number from the formula on page 3-2

or the nomograph on page 3-9

Given: A lubricating oil of density 897 ke/m* and Given: Oil having a density of 875 kg/m? and viscosity 95

viscosity 450 centipoise flows through a 6 inch Schedule centipoise flows through a steel pipe 79 mm inside

40 steel pipe at a rate of 3000 litres per minute diameter at a velocity of 2 metres per second

Find: The flow rate in litres per minute and the pressure Find: The pressure drop per 100 metres of pipe drop in 40 metres of pipe

4 Since Re< 2000, the flow is laminar and the 4 R= 1450 page 3-9

Romograph on the opposite page may be used 5 Since R,< 2000, the flow is laminar and the

nomograph on the opposite page may be used

~~~

-GHAPTER 3 CRANE FORMULAS AND NOMOGRAFPHS FOR FLOW THROUGH VALVES, FITTINGS AND PIPE 3-13

Pressure Drop in Liquid Lines for Laminar Flow

b Ba Pressure, | bar = 10°Pa = 100kPa

B

a

CHAPTER 3

Flow of Liquids Through Noz zles and Orifices

The flow of liquids through nozzles and orifices can be

determined from the following formula, or, from the

nomograph on the opposite page The nomograph is a

graphical solution of the formula

9=348x10% df CVh, =3.51x10% a? C /Bp

a

Q= 0.209 a? CVh, = 21.07 d? € Ấp

0

d, = nozzle or orifice diameter

Head loss or pres-

sure drop is mea-

sured across taps lo-

cated 1 diameter up-

stream and 0.5 dia-

Given: A differential pressure of 0.2 bar is measured

across taps of a 50 mm inside diameter nozzle assembled

in 3-inch Schedule 80 steel pipe carrying water at 15°C

Find: The flow-rate in litres per minute

8 Calculate Re based on 1.D of pipe (73.66 mm)

di page A-3

10, R„= 220000 page 3-9

TỊ © = 1,12 correct for R¿ = 220 000; page A-20

12 When the C factor assumed in Step 3 is nat in

agreement with page A-20 for the Reynolds

number based on the calculated flow the factor

must be adjusted until reasonable agreement is

reached by repeating Steps 3 to }1 inclusive,

Example 2 Given: The flow of water, at 15°C through a 6-inch pipe, 150.7 mm LD., is to be restricted to 850 litres per minute

by means of a square edged orifice, across which there will be a differential head of 1.2 metres of water

Find: The size of the orifice opening

Solution:

3 Rạ= 1I0000 page 3-9

4, Assume a § ratio of say 0.50

2 d, (inlet diam) = 150.7

6 ad, = 0,50 d, = 0.50 x 150.7 = 75,35 Z5 C8 Ú62 page A-20

& | h,= 1.2 C = 0.62 | Index

9 | Index Q = 850 dc 17

10 Anorifice diameter of 77 mm will be satisfactory

since this is reasonably close to the assumed

value in Step 6

Hl H the value of a, determined from the nomo- graph is smaller than the assumed value used in Step 6 repeat Steps 6 to {0 inclusive using reduced assumed values for d, until it is in reasonable agreement with the value determined

in Step 9

Example 3

Given; A differential pressure of 3.5 kilopascals is measured across taps of a 25 mm inside diameter square edged orifice assembled in 1-1/2 inch Schedule 80 steel pipe carrying lubricating oil of 897 kg/m? density and

450 centipvise viscosity

Find: The flow rate in cubic metres per second

5 # = 450 suspect flow is laminar since viscosity

ishigh; page A-3

6 C= 085 assumed: page A-20

12 When the C factor assumed in Step 6 is not in agreement with page A-20 for the Reynolds number based on the calculated flow it must be adjusted until reasonable agreement is reached

by repeating Steps 6 to 12 inclusive Ñ\

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s P oO ete © NN Ñ se] HT N ‘gay rth ° iS, i HT aT i ‘i st be vi is LL > lóc + 8 LU a 5 °' # 3 pe AR 3, g Let TN = el Š - JJ„ aU, So = ® Lire TN N oil - 8 a 3 eee tA N ?, UR ho < š PT N > NANI he š es TaN la WEARS § s8 Lk aT A kì ` N AI VÀ BỊ" 3 Š NHẬNN GÌ > i ¢ 3 a i he 3 e ° | | -\ 2 io Clie - a N ih ở \ Š SG † | i | ~ | 8 shod Rị ET ved | + † 3 § § È |§|E |8|E 88 gg g9 || nh =2 |g|® Specific Volume, in cubic metres per kilogram MA aa na teip nN 8 fond go œ Qn 66 a 6 @ G&S np M 3 a On & tử từ 5” 56 ” ứ Ю b ® t bù ® Density, in kitograms per cubic metre s 2 * Velocity, in thousands of metres per minute Ð 8e -

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Rate of Flow, in thousands of kilograms per hour

internat Diameter of Pipe, in millimetres

20

40

BO

100

120

140

160

(penunuos)

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bere Bdid GNY SONILLIS “SBATWA HONOUHL MOTS HOS SHAVEDONON ONY SVININHOS SNVHO “cứ

£ H31dVH9

CHAPTER 3 3~22 FORMULAS AND NOMOGRAPHS FOR FLOW THROUGH VALVES, FITTINGS AND PIPE CRANE

Simplified Flow Formula for Compressible Fluids

Pressure Drop, Rate of Flow, and Pipe Size

The simplified flow formula for compressible fluids is accurate

for fully turbulent flow; in addition, its use provides a good

approximation in calculations involving compressible fluid Values of Ci (metric)

flow through commercial steel pipe for most normal flow

conditions

If velocities are low, friction factors assumed in the simplified

formula may be too low; in such cases, the formula and

nomograph shown on pages 3-20 and 3-21 may be used to

provide greater accuracy

The Darcy formula can be written in the following form;

Ci = discharge factor, from chart at right

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

The limitations of the Darcy formula for compressible flow, as

outlined on page 3-3 apply also to the simplified flow formula Vatuas

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

8-inch Schedule 40 pipe at a rate of 100 000 kilograms per N

Given: Pressure drop is 1 bar with 7 bar gauge air at 30°C

flowing through 100 metres of 4 inch nominal size ISO steel

pipe, 6.3 mm wall thickness

Find: The flow rate in cubic metres per minute at metric

standard conditions (1.013 25 bar and 15°C)

Simplified Flow Formula for Comprassible Fluids (continued)

For steal pipes to ANSI 836.10: 1970 and BS1G00: Part 2: 1970

Values of C2 (metric)

80x 11.88 100 0.029 16 The letters, x, xx, in the columns

160 20.77 140 0.038 37 dard, Extra Strong, and Double Extra

Example 3

Given: A 6 bar gauge saturated Solution: Ap = 2.4 V= 0.273 page 3-17 or A-12

line with 9000 kil = 3

maximum pressure drop of 2.4

bar per 100 metres of pipe

Find: The smallest size of ISO

336 steel pipe suitable

Reference to the table of Cz values for ISO 336 pipes on page 3-24 shows that a

4 inch nominal size pipe with 7.1 mm wall thickness has the C2 value nearest to,

but less than, 10.85

The actual pressure drop per 100 metres of 4 inch, 7.1 mm wall thickness, pipe is;

Ap too = C1 C27 =0.81 x 10.22 x 0.273 = 2.26 bar

CHAPTER 3 3-24 FORMULAS AND NOMOGRAPHS FOR FLOW THROUGH VALVES, FITTINGS AND PIPE CRANE

Simplified Flow Formula for Compressible Fluids {continued}

Values of C, (metric) For steal pipes to ISO 336 — 1976

Pipe Size | Thickness of Cy Pipe Size | Thickness

of C2 Pipe Size | Thickness of Cz

CHAPTER 3

Simplified Flow Formula for Comprassible Fluids (continued)

Values of C, (matric) For steal pipes to ¡SO 336 — 1874

Pipe Size | Thickness of C, Pipe Size | Thickness of C; Pipe Size | Thickness of C,

$5.0 0.002 372 60.0 0,002 624

of the pipe sizes contained in BS 3600: 1973, within the same ranges of wall thicknesses

77ợnHHH37”000900909099/9123333333333390371333952

CHAPTER 3 3—26 FORMULAS AND NOMOGRAPHS FOR FLOW THROUGH VALVES, FITTINGS AND PIPE CRANE

Flow of Compressible Fluids Through Nozzles and Orifices

The flow of compressible fluids through nozzles and

orifices can be determined from the following formula,

or, by using the nomograph on the next page The

nomograph is a graphical solution of the formula

w= 3.512% 10% YduC/ App, =3.512x 104 Yd? C /#-

W= 1.265 YdyCJ Opp = 1.265 Yar [se

d, * nozzic or orifice diameter

(Pressure drop is measured across taps located |

diameter upstream and 0.5 diameter downstream

from the inlet face of the nozzle or orifice)

Example 1

Given: A differential pressure of 0.8 bar is measured

across taps located 1 diameter upstream and 0.5 diameter

downstream from the inlet face of a 25 mm inside

diameter nozzle assembled in a 2-inch Schedule 40 steel

pipe, in which, dry carbon dioxide (CO2) gas is flowing

Find: The flow rate in cubic metres per hour at metric

standard conditions (MSC = 1.013 25 bar and 15°C)

& Y2 093 page A-2]

9C = 1.02 turbulent flow assumed: page A-20

20 When the C factor assumed in Step 9 is not in

agreement with page A-20, for the Reynolds

number based on the calculated flow, it must be adjusted

until reasonable agreement is reached by repeating Steps

Find: The flow rate in kilograms per second and in cubic metres per minute at metric standard conditions (1.013 25 bar and 15°C)

FORMULAS AND NOMQGRAPHS FOR FLOW FRROUGH VALVES, FITTINGS AND PIPE — CRANE

mm

A few simple flow problems were presented in Chapter 3 to illus- trate the use of the nomographs

Other problems, both simple and

complex, are presented in this

chapter

Many of the exampies given in this chapter employ the basic formulas

of Chapters_1_and_2: these formulas were.rewritten.in.more commonly:

used terms for Chapter 3 Use of nomographs, when applicable, are indicated in the solution of these problems

The controversial subject regarding the selection of a formula most applicable to the flow of gas through long pipe lines is analyzed in Chapter 1 It is shown that the three commonly used formulas are basically identical, the only difference being in the selection of friction factors A comparison of results obtained, using the three formulas, is presented in this chapter

An original method has been developed for the solution of problems involving the discharge of compressible fluids from pipe systems

Mlustrative examples applying this method demonstrate the simplicity

of handling these, heretofore complex, problems

Reynolds Number and Friction Factor For Pipe Other Than Steet The example below shows the procedure in obtaining the Reynolds number and friction factor for smooth pipe (plastic) The same pro- cedure applies for any pipe other than steel or wrought iron, such as

concrete, wood stave, riveted steel, etc For relative roughness of these

and other piping materials, see page A-23

Example 4-1 Smooth Pipe (Ptastic}

Determination of Valve Resistance

In L, L/D, K, and Flow Coefficient C, Example 4-2 4, 24/B, and K from C, for

Conventional Type Valves Given:

A 150 mm (6-inch) Class 125 iron Y-pattern globe vaive

has a flow coefficient, C,, of 600 (US gal/min)

Find: Resistance coefficient K and equivalent lengths

L/D and L for flow in zone of complete turbulence

Solution;

t K, L/D, and L should be given in terms of 6-inch

Schedule 40 pipe; see page 2-10,

2 C= Se or VR K eee (Cu? me age 3-4

In this equation đ is in inches (1 inch = 25.4 mm)

- for 154 mm LD pipe in fully

6 f= 0.015 turbulent flow range; page A-25

LK 335

% DF O03 773

D= 154.1 +1000 = 0.1541 metres

& oL -(5)o = 223 x 0.1541 = 34.4 metres

Example 4-3 £4, L/D and K for

Conventional Type Valves

Given:

A 100 mm (4-inch) Class 600 steel conventional angle

valve with full area seat

Find: Resistance coefficient K and equivalent lengths

L/D and L for flow in zone of complete turbulence

Solution:

1 K, L/D, and L should be given in terms of 4-inch

Schedule 80 pipe; see page 2-10

for graphical solutions

=180 of steps 5 and 6, use

of body rings to vaive ends Face-to-face dimension is

560 mm and back of seat ring to back of seat ting is about 150 mm

Find: Kz for any flow condition, and L/D and L for flow in zone of complete turbulence

i Kz, L/D, and L should be given in terms of 6-inch

Schedule 80 pipe; see page 2-10

beeen ates page A-26

ath

d,= 146.4 6” Sched 80 pipe; page B-16

#r=0.015 for 6” size; page A-26

8° = 0.48 B° = 0.23 tan Ê „85464 -101.6@

2 0.5560 - 150) tang =0.i11= sing approx,

7yH0AE005nPPDPPPSTPoballtslddllli434339330A4333053

Example 4-5 Lift Check Valves Example 4-6 Reduced Port Ball Valve

Given: A slobe type lift check valve with a wing-guided

disc is required in a 3-inch Schedule 40 horizontal

pipe carrying 20 C water at the rate of 300 litres per

minute,

Find: The proper size check valve and the pressure

drop The valve should be sized so that the disc is fully

lifted at the specified flow; see page 2-7 for discussion

2 dị =62.7

d; =719

V = 0.001 002 a=998.2

fr = 0.018

3 Unpin = 50S 0.001 = 1.585

= 21-22 x 300 77x92

In as much as v js less than vmin, 4 3-inch valve will be too large Try a 2'4-inch size

21.22 x 300 62.77 Based on above, a 2%-inch valve installed in 3-inch Schedule 40 pipe with reducers is advisable

- for 2%” Sched 40 pipe; page B-16 for 3” Sched 40 pipe; page B-16 bec eeeeee 20 C water: page A-6

""——— 20 C water; page A-G for 21⁄2” or 3” size; page A-26

in litres per minute

2 K=05 entrance; page A-29

3 For K (ball valve), page A-28 indicates use of Formula 5 However, when inlet and outlet angles (8) differ, Formule 5 must be expanded to:

10 Calculate Reynolds number to verify that friction factor of 0.018 (zone of complete turbulence) is correct for flow condition or, use “vd” scale

at top of Friction Factor chart on page A-25

ud = 2.675 x 77.9 = 208

11 Enter chart on page A-25 at ud = 208 Note ƒ for 3-inch pipe is less than 0.02 Therefore, flow is in the transition zone (slightly less than fully tur- bulent) but the difference is small enough to forego any correction of K for the pipe

4-4 Na CHAPTER 4 ~ EXAMPLES OF FLOW PROBLEMS CRANE

Laminar Flow in Valves, Fittings, and Pipe

in flow problems where viscosity is high, calculate the Reynolds Number to determine whether the flow is laminar or turbulent

Example 4-7

Given: S.A.E, 10 Lube Oil at 15 C flows through the

system described in Example 4-6 at the same differential

f= 64 Re ky ky pipe, laminar flow; page 3-2

“Note: This problem has two unknowns and, therefore, requires

a triakund-error solution Two or three trial assumptions will

usually bring the solution and final assumption into agreement

within desired limits

Exampie 4-8

Given: S.A.E 70 Lube Oil at 40 C is flowing at the rate of 600 barrels per hour through 60 metres of 8- inch Schedule 40 pipe, in which an 8-inch conventional globe valve with full area seat is installed,

Find: The pressure drop due to flow through the pipe and valve,

_ 64

2 $=0916a60F (S60) page A-7 Š=0.90at40C page A-7 đ=2027 8” Sched 40 pipe; page B-16

———

CRANE

Laminar Flow in Valves, Fittings, and Pipe — continued

In flow problems where viscosity is high, calculate the Reynolds Number to determine whether the flow is laminar or turbulent

Exaraple 4-8

Given: S.A.E 70 Lube Oil at 40 C is flowing through S-inch Schedule 40 pipe at a rate of 2300 litres per minute, as shown in the following sketch

5°’ Class 156 Steet Angie Valve with full-area Py 8° Class 180 Stee} Gate

Valve with fult area

1 seat ~ wide open

seat ~ wide open Oo

—T Elevatian 18m 5" Weld 185m

Elbow

| Elevation O

d=1282 5” Sched 40 pipe: page B-16 S*0.916 at60F(IS6C) page A-7 S=0.90at40C page A-7 u=450 page A-3 p=999x0.9=899 , page A-6, A-7

21.22 x 2300 x 899 128.2 x 450

Re < 2000; therefore flow is laminar

= 6t «- f“ ao = 0.084

Re= = 760

Summarizing X for the entire system (gate valve, angle valve, elbow, and pipe),

K= (8 x 0.016) + (150 x 0.016) + (20 x 0.016) + (0.084 x 85 x 1000) _ 58,54

:¬m „ CHAETER 4 nm EXAMPLES OF FLOW PROBLEMS: “CRANE

Pressure Drop and Velocity in Piping Systems Example 4-10 Piping Systems — Steam

Given: 40 bar abs steam at 460 C flows through 120

metres of horizontal 6-inch Schedule 80 pipe at a rate

of 40 000 kilograms per hour

The system contains three 90 degree weld elbows having

a relative radius of 1.5, one fully-open 6 x 4-inch

Class 600 venturi gate valve as described in Example 4-4,

and one 6-inch Class 600 y-pattern globe valve Latter

has a seat diameter equal to 0.9 of the inside diameter

of Schedule 80 pipe, disc fully lifted

Find: The pressure drop through the system

4 d=1464 6” Sched 80 pipe; page B-16

P=0.081 _ #0 bar steam 460 C; page A-16

K=3x14x0015 = 0.63 3 elbows; page A-29

K£LA44 6 x 4” gate valve; Example 4-4

7, Summarizing K for the entire system (globe valve,

pipe, venturi gate valve,-and elbows),

K* 144+ 12.3 +0.63+ 1.44 = 15.8

Ap = 9:6253 x 15.8x 40 000? x 0.081 Pp =

a 146.4

Ap = 2.8 bar

Example 4-17 Flat Heating Coils — Water

Given: Water at 80 C is flowing through a flat heating coil, shown in the sketch below, at a rate of 60 litres per minute

— eee water 40 C; page A-6

¬ water 40 C; page A-3 3=266 1” Sched 40 pipe; page B-16 /r0023 1” Sched 40 pipe; page A-26 21.22 x 60 x 971.8

f= 0.024 0.024 x 5.4 x 1000 26.6

yy

ri

CC

P000900909020)02920232939203363332143839535

Pressure Drop and Velocity in Piping Systems — continued Example 4-12 Orifice Size for Given

Pressure Drop and Velocity

Given: A 12 inch nominal size, ISO 336 steel pipe,

1] mm wall thickness, 18 metres long containing a

standard gate valve discharges 15 C water to atmosphere

from a reservoir The entrance projects inward into the

reservoir and its centre line is 3.5 metres below the water

level in the reservoir

Find: The diameter of thin-plate orifice that must be

installed in the pipe to restrict the velocity of flow to

3 thetres per second when the gate valve is wide open

2 K=078 entrance: page A-29

Ke 8fr wee ee eee gate valve; page A-27

8 Assume § = 0.7 C=0,7 puge A-20

then K x 4.3 B is too large

9 Assume B = 0.65 C=0.67 page A-20

then K x 7.1 “ Bis tov small

Example 4-13 Flow Given in Traditional Units

Given: Fuel oil, with a specific gravity of 0.815 anda kinematic viscosity of 2.7 centistokes flows through a 2-inch Schedule 40 steel pipe, 100 feet long, at a rate of

2 US gallons per second

Find: The pressure drop in bars and in pounds force per square inch

(

Pressure Drop and Velocity in Piping Systems — continued

Example 4-14 ,,, Bernoulli's Theorem — Water

Given: Water at 15 C is flowing through the piping dz =128.2 5” Sched 40 pipe; page B-16

system, shown in the sketch below, at a rate of 1500 FpEOOG cece, 5” sizes page A-26

litres per minute

102.3

4 = =e = 0.80 5” Waiding Eihow 5" Schaduta 40 Pipe Py 8 128.2 0

34m——Ê 5" x 4” Reducing Welding Elbow Ep “v88 TỔ 0.28 metres

_ Find: The velocity in both the 4 and 5-inch pipe sizes ý For Schedule 40 pipe,

and the pressure differential between gauges p, and p; R= 21.22 x 1500x999 _ 2/83 x 105

1 — Use Bernoull?s theorem Gee page32) z,+ Wm, ilz, 4 — l0 ở C77777 R.- 2L22x150014999 _ 2.2 495

ƒ#0018 4or 5” pipe

0# 2-24 6K 128.2

Pro Py * 4 [as ~2,)+ 2a=H +4

= page 3-4 k= 2018 x 34 x 1000

102.3

K =f 5 ăn page 3⁄4 K¿ =60+0/82 #146 page 2-11

a- xy reducing 90° ? Then, in terms of S-inch pipe,

* Page Ae KTOTAL = 9.4 + 14.6 + 0.22 + 0.58 = 24.8

Note: In the absence of test data for increasing elbows, th

resistance is conservatively estimated 10 be const to the wir 8 hạ = 2226 x 24.8 x 1500" = 4.74

mation of the resistance due to a straight size §" elbow and a 128.2

BFE ieee ee cee eee eee page A-3

d, = 102.3 4” Sched 40 pipe; page B-16

Hư

cgg8099090992393953201923335331333333371323333

Pressure Drop and Velocity in Piping Systems ~ continued Example 4-16 Power Required for Pumping

Given: Water at 20 C is pumped through the piping

system below ut a rate of 400 litres per minute

4 2%" Globe Lift Check Vaive with wing guided

disc instalted with reducers in 3” Pree

Elevation Z, = 0

Find: The total discharge head (H) at flowing conditions

and the power demand (brake power) required for a

pump having an efficiency (@p) of 70 per cent

Solution: 1, Use Bernoulli's theorem (see page 3-2):

Điển — By

2 Since p, = pz and x, = 0¿, the equation can be

rewritten to establish the pump head, H:

4 K=30fp 90° elbow; page A-29

Ke 8fp gate valve: page A-27 K=f 5 "———- Straight pipe; page 3-4

3 d= 779 3” Sched 40 pipe: page B-16

779

9 4j= 120 + 7 = 127 metres

400 x 127 x 998.2

6116 x 10” x0.7 Example 4-16 Air Lines

Given: Air at 5 bar gauge and 40 C is flowing through

25 metres of f-inch Schedule 40 pipe at a rate of 3 standard (MSC) cubic metres per minute (see page B-12)

3 Correction for length pressure and temperature (page B-15):

3 B= O66 cu page B-16

6 A= 0.7854 (48 * 0,000 556

3 Ve ng = 987 m/min (upsteam) V=z ñngg siÿ = 1023 m/min (downstream)

Note: Example 4-16 may also be solved by use of the pressure

drop formula and nomograph shown on pages 3-2 and 3-21 respectively or the velocity formula and nomograph shown on pages 3-2 and 3-17 respectively.

Pipe Line Flow Problems

Example 4-17 Sizing of Pump for Oil Pipe Lines Given: Crude oil 30 degree API at 15.6 C with a viscosity of

75 Universal Saybolt seconds is flowing through a BS 1600, 12 inch, Schedule 30 steel pipe at a rate of 1900 barrels per hour

The pipe line is 80 kilometres long with discharge at an elevation

of 600 metres above the pump inlet Assume the pump has an efficiency of 67 per cent

Find: The power demand of the pump

Ap = 15.81 fag or, after converting B to Q,

use nomograph on page 3-11

Re = 56.23 Be Gece cece eaee page 3-2 or 3-8

Then the power demand is:

3035 x 1026.3 x 875.3

6116x 10° x 0.67 1104, say 1110 kW

POCROS RRR

Pipe Line Flow Problems — continuad

Given: A natural gas pipe line made of BS 3600 14inh 9-9 y= (Pn (2) = 2.938

The gas consists of 75% methane (CH), 21% ethane Ih H=0011 estimated; page A-5

Find: The flow rate in millions of cubic metres per day 12, Re= 333.6 x 0.011

Solutions: Three solutions to this example are presented Re = 9 986 000 or 9.986 x 10

for the purpose of illustrating the variations in results 1 f=0.0128 oo ey page A-25

formula, the Weymouth formula and the Panhandle, correct, the flow rate is 2.938 million m?/d at

it would have to be adjusted and Steps 8, 9, 12, and 13 repeated until the assumed friction factor

M = 20.06, or say 20.1 20 Assume average operation conditions; then ef-

ficiency is 92 per cent:

4 —~ 12

Discharge of Fluids from Piping Systems

Example 4-19 Water

Given Water at 20 C is flowing from a reservoir through

the piping system below The reservoir has a constant

tances page A-26

K=60f mitre bend: page A-29

KEBfr gate valve; page A-27

ø +++ sudden contraction: page A-26

3 d#525 2” Sched 40 pipe: page B-16

d2779 1.0 3” Sched 40 pipe: page B-16

#r=0.019 2” pipe: page A-26

/#r=z0.018 3” pipe: page A-26

and KTOTAL = 0.5 + 1.08 + 0.14 +0.69 4

10.8 + 5.0 + 1.37 = 19.58

= 0.2087 x 77 STS FT TTAE = $35 (this solution assumes How in fully turbulent zone) Calculate Reynolds numbers and check friction factors for flow in straight pipe of the 2-inch size:

21.22 x 535_x 998.2

and for flow in steaight pipe of the 3-inch size:

= 21.22 x $35 x 998.2

=1,32 $

Since assumed friction factors used for straight pipe in Step 4 are not in agreement with those based on the approximate flow rate, the K factors for these items and the total system should be corrected accordingly,

-peer

ỷ 7”-ooo»200909933830/329332339922932303?7%3323539

Discharge of Fluids from Piping Systems — continued

Example 4-20 Steam at Sonic Velocity

Given: A header with 12 bar absolute saturated steam is feeding

a pulp stock digester through 10 metres of 2-inch, ISO 336 steel pipe, 4 mm wall thickness, which includes one standard 90 degree elbow and a fully-open conventional plug type disc globe valve

‘The initial pressure in the digester is atmospheric

Find: The initial flow rate in kilograms per hour, using both the modified Darcy formula and the sonic velocity and continuity

equations

Solutions — for theory, see page 1-9:

L w= 2e Equation 3-2; page 3-2

+ Kr3M0fr globe valve; page A-27 10 p'= pi Ap

K=05 — entrance from header; page A-29 Ap determined in Step 6

XỔ eo exit to digester: iu dl Ag = 2782.7 12 bar abs sat steam; page A-13

6 Using the chart on page A-22 for y = 1.3 it is

found that for K = 12.16 the maximum Ap/p' is 0.786 (interpolated from table on page A-22)

Since Ap/p; is less than indicated in Step 5, sonic

velocity occurs at the end of the pipe, und Ap in

the equation of Step / is:

NOTE

ap = 0.786 x 12 = 9.432 say 9.43

7 Y=0210 interpolated from But the increase in specific volume from inlet to outlet

92 Kinetic energy increase is the internal heat energy of the

8 W=4,265 x 0.71 x $2.3? STET OTR fluid Consequently, the heat energy actually decreases

ve xÙ - toward the outlet Calculation of the correct hạ at the

outlet wil yield a flow rate commensurate to the answer

4 =Mâ$ ne GHAPTER 4 ~ EXAMPLES OF FLOW PROBLEMS: “CRANE

Discharge of Fluids from Piping Systems ~ continued Example 4-27 Gases at

Sonie Velocity Given: Coke oven gas having a specific gravity of 0.42,

a header pressure of 8.0 bar gauge and a temperature of

60 C is flowing through 6 metres of 3-inch Schedule 40

pipe before discharging to atmosphere Assume ratio of

Note: The Reynolds number need not be calculated since

gas discharged to atmosphere through a short pipe will

have a high Re and flow wilt always be in a fully

turbulent range, in which the friction factor is constant

4 4=719, D=00719 page B-16

K=0§ for entrance; page A-29

KF10 0 oo cece for exit; page A-29

Ap

oo 9.013

7 Using the chart on page A-22 for y = 1.4, itis

found that for K = 2.85, the maximum Apip, is

0.655 (interpolated from table on page A-22)

Since App; is less than indicated in Step 6, sonic

velocity occurs at the end of the pipe and Ap in

Step / is:

Ap =0.655p), = 0.655 x 9.013 = 5.9

& 1, =60+273 = 333

interpolated from

10 dạy is equal to:

3.9.x 9013 19.31 331% x0.636x77.9? (= 27S * V2-8Sx333 x0.42

Find: The flow rate in cubic metres per minute at Metric Standard Conditions

K=#i0 , for exit: page A-29

Flaw Through Orifice Meters Example 4-23 Liquid Service

Given: A square edged orifice of 50 mm diameter is

installed in a 102.5 mm inside diameter pipe having a

mercury manometer connected between the pipe taps |

diameter upstream and 0.5 diameter downstream

Find: (a) The theoretical calibration constant for the

meter when used on 15 C water and for the flow range

where the orifice flow coefficient C is constant and

(b), the flow rate of 15 C water when the mercury

3 The weight density of mercury under water equals

Pw (Sug ~ Sy), where (at 15 C):

Pw = density of water=999.0 page A-6

Sag = specific gravity of mercury = 13.57

ves page A-7

Sy — specific gravity of water = 1.00

Re = 72 000 or 7.2 x 108

12, Rẹ=

13 C= 0,625 is correct for Re = 7.2 x 104, per page 4-20; therefore, the flow rate through the pipe is

383 litres per minute

i4 When the C factor on page A-20 is incorrect, for the Reynolds number based on calculated flow,

it must be adjusted until reasonable agreement is reached by repeating Steps 9, 10, and 12

Example 4-24 Laminar Flow

in flow problems where the viscosity is high, calculate the Reynolds number to determine the

type of flow, Given: SAE 10 Lube Oil at 32 C is flowing through a 3- inch Schedule 40 pipe and produces 2.8 kPa pressure differential between the pipe taps of a 55 mm LD square edged orifice

Find.The flow.rate.in litres-per minute:

Solution:

L Q=2107d? C / oe ch page 3-5 or 3-15

„ 21.220p

3 H=38 Suspect laminar flow; page A-3

4 d, (larger diam.) #77.9 2.0.0.2 page B-16

a, 32v

S de F795 0.706

based on laminar flow

% 6§=0.876atlSC page A-7 SFFO0.87at32C0 page A-7

& p=999x0.87=869 page A-7

9 g=2.07x557x08 (9228 = 289.5

869

„ 21.22 x 289.5 x 869 _

1l C=0.9forReg=1803 page A-20

Since the assumed C value of 0.8 is not correct, it

must be adjusted by repeating Steps 6, 7, 8 and 9

0.028 l2 @#21.07x 557x087 / 2 sàng

9 869

21.22 x 315 x 869

779 x 38

Since C = 0.87 is correct for the flow, the flow through the meter is 315 litres per minute

Application of Hydraulic Radius to Flow Problems

Example 4-25 Rectangular Duct Given: A rectangular concrete overflow aqueduct 7.6 metres high and 5 metres wide, has an absolute roughness (e) of 3 millimetres

assumed; page A-23 4*4.428A f— Lg

hydraulic radius is used; refer to page 3-5 Me E999 eee eeee eee eee page A6

Re = 318.3 Ru bbe nena eaees page 3-2 Re = 162 000 000 or 1.62 x 108

4 Assuming a sharp edged entrance, 45 Since the friction factor assumed in Step 8 and that

Then, resistance of entrance and exit, Kg=0.5 + 10= 1.5

factor based on the calculated Reynolds number were not in reasonable agreement, the former should be adjusted and calculations repeated until reasonable agreement is reached

Application of Hydraulic Radius to Flow Problems — continued

Example 4-26 Pipe Partially Filled

With Fiowing Water

Given: A cast iron pipe is two-thirds full of steady 10 The cross sectional flow area equals:

uniform flowing water (15 €) The pipe has an inside A+B+C = (2 x 14 150) + 172 000

' 12, L* Ah = 16" 0.0625 metre per metre

Since pipe is flowing partially full an equivalent 16, Relative roughness 7 0.00036 +++ page A-23

friction factor assumed in Step 17

5 Depth of flowing water equais:

x = 90° ~ 10°32’ = 19°28" = 19.47° 24, Since the friction factor assumed in Step 17 and

Area C= nd 180 + (2.x 19.47) flow rate will be 89 000 litres/min

should be adjusted and the calculations repeated

8 b=JS 3007 7100? = 283mm until reasonable agreement is reached

9 Area A = Arca B= % (100 x 283)

Area Aor B= 14 150 mm?

GRANGE CHAPTER 4 ~ EXAMPLES OF FLOW PROBLEMS

Application of Hydraulic Radius to Flow Problems — continued

Example 4-26 Pipe Partially Filled

With Flowing Water Given: A cast iron pipe is two-thirds full of steady 10 The cross sectional flow area equals:

uniform flowing water (15 €) The pipe has an inside AtB4C = (2.x 14.150) + 172.000

wow Hg? = 48 = 44200300 2 55 099

12 te Ah = i = 0.0625 metre per metre

18: m wso0( 2182 ty ) = 1146 mm

diameter based upon hydraulic radius is substituted In, f= 0.0156 assuming fully turbulent

2 Q@=0.2087d* T7 0.4174d2 — " 0.0156

‘ @= 89000 litres/min

friction factor assumed in Step 17

« = 90° ~ 70932! = 19°38" = 19.47° 24 Since the friction factor assumed in Step 17 and

ce ee 180 + (2 x 19.47) flow rate will be 89 000 litres/min

Area C = T0 172 000 mm? were not in reasonable agreement, the former

should be adjusted and the calculations repeated

9 Area A= Arca B = 4 (100 x 283)

Area Aor B= 14 150 mm?

CHAPTEB.4~.EXAMPL68 GE FLOW PROGLEMS CRANE