rupture disc sizing [PDF]

TECHNICAL BULLETIN RUPTURE DISC SIZING The objective of this bulletin is to provide detailed guidance for sizing rupture discs using standard methodologies found in ASME Section VIII Div. 1, API RP520, and Crane TP-410. To assist in the sizing process, Fike offers DisCalc™, a web based sizing program. See www.fike.com.

OVERPRESSURE ALLOWANCE When sizing pressure relief devices, the ASME Code defines the maximum pressure that may build up in the pressure vessel while the device is relieving. This pressure varies depending on the application of the device. The following table defines the various overpressure allowances. See technical bulletin TB8100 for ASME application requirements. Primary (Sole Relieving Device)

Secondary (Multiple Devices)

External Fire (Unexpected Source of External Heat)

External Fire (Storage Vessels Only)

Ref. UG-125(c)

Ref. UG-125(c)(1)

Ref. UG-125(c)(2)

Ref. UG-125(c)(3)

10% or 3 PSIG, whichever is greater, above the vessel MAWP

16% or 4 PSIG, whicever is greater, or above the vessel MAWP

21% above the vessel MAWP

20% above the vessel MAWP

RUPTURE DISC SIZING METHODOLOGIES Three basic methodologies for sizing rupture disc devices are described below. These methods assume single phase, non-reactive fluid flow. Resources such as API RP520 Part 1, the DIERS Project Manual, and CCPS Guidelines for Pressure Relief and Effluent Handling Systems provide other methods for two-phase, flashing, reactive, and otherwise non-steady state conditions. Coefficient of discharge method (KD) - The KD is the coefficient of discharge that is applied to the theoretical flow rate to arrive at a rated flow rate for simple systems. Resistant to flow method (KR) - The KR represents the velocity head loss due to the rupture disc device. This head loss is included in the overall system loss calculations to determine the size of the relief system. Combination capacity method - When a rupture disc device is installed in combination with a pressure relief valve (PRV), the valve capacity is derated by a default value of 0.9 or a tested value for the disc/valve combination. See technical bulletin TB8105 for specific application requirements when using rupture disc devices in combination with PRV’s. A listing of Fike certified combination factors can be found in technical bulletin TB8103.

COEFFICIENT OF DISCHARGE METHOD (KD)

Use this method for simple systems where the following conditions are true (8 & 5 Rule). This method takes into account the vessel entrance effects, 8 pipe diameters of inlet piping, 5 pipe diameters of discharge piping, and effects of discharging to atmosphere.

The inlet and outlet piping is at least the same nominal pipe sizes as the rupture disc device

The rupture disc device discharges directly to the atmosphere

The discharge piping does not exceed 5 pipe diameters

The rupture disc is installed within 8 pipe diameters of the vessel

Form No. TB8102-3 704 SW 10th Street  P.O. Box 610  Blue Springs, Missouri 64013-0610 U.S.A. Phone: 8162293405  www.fike.com

Pcf

Page 2:

If



P   k  1 

A C k KD

Pe  Pcf use critical flow equations

= = = =

minimum net flow area, (sq. in.) constant based on the ratio of specific heats k cp/cv coefficient of discharge 0.62 for rupture disc devices /k

 k 1   k  2 / k 1  r   r     GAS/VAPOR SIZING  k 1  1  r  Page 2: Page 2: k /( k  1 ) W = rated flow capacity, (lb/hr) Determination of Critical vs. Subcritical Flow per API RP520 M  2 W Pe capacity, (SCFM) V = rated flow r =flow Pcf  P  k /(k K1) D  C  A  P T  Z W = rated capacity, (lb/hr) A = minimum net flow area, (sq. in.) k /( k  1 ) Critical Pressure: W = rated flow capacity, (lb/hr)   k 1   2   k /( k 1) W = rated flow capacity, (lb/hr) V = rated flowPcapacity, P (SCFM)  2 P = set pressure plus allowance plus  2  P    V = rated flow capacity, (SCFM) onarea, the overpressure ratio of specific heats k V = rated flow capacity, (SCFM) C = constant based Pcfcf  P  k 1  Pcf  P  A = minimum net flow (sq. (psia) in.) atmospheric pressure A = minimum net flow area, (sq. in.)   A W= minimum (sq. in.)k /c = c  k  1  p v  k  1  T  Z net flow area, C = constant based on the ratio of specific heats k C = constant based on the ratio of specific heats k equations If Pe  Pcf use critical (psia) 0.62 for rupture disc devices Pek = exit pressure, C = constant based on the ratio of specific heats A  flow of discharge /cv = molecular kKv D == cpcoefficient k = cp/c K D  C=  Pcp/cv M M weight critical flow kequations IfIf PP PcfPuse e e  If P  P use critical flow equations K = coefficient of discharge 0.62 for rupture  kdevices for 1 / krupture disc devices cf use critical flow equations K = coefficient of discharge disc 0.62

F2 =

Page 2:

D coefficient of discharge 0.62 for Drupture disc  of gas at standard  conditions, SGdevices = k specific  2gravity /k 1 r  k 1 / kr forair at  k 1 / k F2 k = 2 / k1  rSG=1.00  k 14.7 1 / k psia and 60°F    k  2 / k 1F2 r = 1  k r   absolute 1 r 2/ ktemperature 1  atr inlet  2k =1  r T =k    r    2 = F    (R=°F + 460°F) Calculations per ASME Section VIII (assumesFcritical flow)  1 r Z k factor for corresponding to P and T. = 1compressibility   k  1  T  Z  1  r   1 r W   M    use 1.0 if unknown. A M P P M     W K C A P e     W K C A P e D Critical D M F  KPDe M  PP  Pre  = 735 r = P  K D Flow: C  A P W T T Z r Z =2 eP Pr = T  Z W  KD  C  A P T  ZV P P = setP pressure plus overpressureplus allowance plus   T Z M P = set overpressure allowance plus allowance pressure plus pressure A  P = set pressure plus overpressureatmospheric (psia) pressure plus overpressure  K D PPpressure  Pe  (psia) P = set atmospheric pressure (psia) allowance plus W T  Z 4645  F2 atmospheric (psia) Pe = exit pressure,atmospheric W T Z A W T Z pressure (psia) (psia) M = molecular A exit pressure, (psia) Pe = weight A  K D  CW P M T  ZPe V= exit pressure, T  Z  SG KD  C  P M M = molecular weight SG = specific = exit pressure, (psia)conditions, P  A  A   K C P M gravity of gas at standard e M conditions, = molecular weight Calculation per API RP520 DK  C  P P  Peof gas at standard Pgravity SG F =2  K specific M864 D SG=1.00 air at 14.7 psia and 60°F M ==formolecular weight D SG specific gravity of+ gas at standard conditions, SG=1.00 for air at 14.7 and T psia = absolute atgravity inlet (R=°F 460°F) SG60°F =temperature specific of at gas14.7 at standard SG=1.00 for air andconditions, 60°F Subcritical Flow: Critical T Flow: = absolute temperatureZ at inlet (R=°F + 460°F) = compressibility factor for corresponding to Ppsia and SG=1.00 air at 14.7 psia andT.60°F W TZ  Z= compressibility factor for corresponding toabsolute P and T.for T = temperature at inlet (R=°F + 460°F) use 1.0 if unknown.  A W T Z W T  Z T = absolute temperature at inlet (R=°F + 460°F)  1.0 if unknown. 735  F2  K D AM  PP  Pe use A Z = compressibility factor for corresponding to P and T.

e

KD =

cf





735  F2  K D

A

V 4645  F2  K D

A

V 864  F2  K D

M  PP  Pe  WW K T C  PZ M A V T  ZD  MT  Z  A PZPMPPee  F 2F K T  Z AM735 V P D PM M 735  F2 2 K K PTP 4645 D  P DA e PP  Pe  6.32  K D  C  P  ZZ M T VVV T  ZT SG M A A864  F  K PP V P T  Z  SG T  Z ASG     F K P P P 4645 2 D e AD  PP  Pe  4645  2F2  K D 1.175  K D eC  P PP  Pe 

Z

=

compressibility factor for corresponding to P and T. use1.0 1.0if ifunknown. unknown. use

VV TT ZZ SG SG A A W   2F2 K  TKDDZ PPP 864 PPPee  F A 864 TABLE 1W TABLE 2 T Z KD  C  P M A  Constants Gas K C  P M D

V T Z M A  Gas or Vapor 6.32  K D  C  P

V Air T  Z  SG A  Acetic Acid 1.Acetylene 175  K D  C  P Ammonia Argon Benzene N-Butane ISO- Butane Butane Carbon Monoxide Carbon Disulfide Carbon Dioxide Chlorine Cyclohexane Ethane Ethyl Alcohol Ethyl Chloride Ethylene Helium Hydrochloric Acid Hydrogen Hydrogen Sulfide Methane Methyl Alcohol Methyl Chloride Natural Gas (Avg.) Nitric Acid Nitrogen Oxygen Pentane Propane Sulfur Dioxide Water Vapor

V T Z M A Molecular KD  C  PT  Z 6.32 W A  W k = cTp/c vZ  AWeight V T  ZP  P SG MM A K K DC C 28.971.D175  K  C 1.40 P D M 60 V T  Z1.15 M A  V T  Z 1.26 A26.04  6.32  K D  C  P  K D  1.33 CP 6.32 17.03 40 V T  Z1.67  SG A  V T  Z  SG A 78.1  1.175  K D1.12 C  P 58.12 1.175  K D1.094 C  P 58.12 1.094 56.1 1.10 28 1.40 76 1.21 44.01 1.30 70.9 1.36 84.16 1.09 30.07 1.22 46.07 1.13 64.5 1.19 28.05 1.26 4 1.66 36.5 1.41 2.016 1.41 34.07 1.32 16.04 1.31 32.04 1.20 50.48 1.20 19 1.27 30 1.40 28 1.404 32 1.40 72.15 1.07 44.09 1.13 64.06 1.29 18.02 1.324

Gas Flow Constant C for Sonic Flow STEAM SIZING k

C

k

C

1.00

315

1.40

356

1.02

318

1.42

358

1.04

320

1.44

360

1.06

322

1.46

361

1.08

325

1.48

363

1.10

327

1.50

365

1.12

329

1.52

366

1.14

331

1.54

368

1.16

333

1.56

369

1.18

335

1.58

371

1.20

337

1.60

373

1.22

339

1.62

374

1.24

341

1.64

376

1.26

343

1.66

377

1.28

345

1.68

379

1.30

347

1.70

380

1.32

349

2.00

400

1.34

351

2.10

406

1.36

352

2.20

412

1.38

354

2 of 9

STEAM SIZING Calculation per ASME Page 3: Section VIII Steam:

Page 3: .5 ⋅ A ⋅ P ⋅ K D ⋅ K N W = 51

A=

W 51.5 ⋅ P ⋅ K D ⋅ K N

A=

W 51.5 ⋅ P ⋅ K D ⋅ K N ⋅ K SH

KN =

Correction factor for steam

W K N = when P ≤ 1500 psia W A == 51.5 ⋅ A ⋅ P ⋅ K D ⋅ K N 51.5 ⋅ P ⋅ K D ⋅ K N ⎛ 0.1906 P − 1000 ⎞ W KN = ⎜ ⎟ when P > 1500 psia and P ≤ 3200 psia A= ⎝ 0.2292 P − 1061 ⎠ Calculation per API 51.RP520 5 ⋅ P ⋅ KWD ⋅ K N A= K SH = See Table 3 for superheat steam correction factors. For saturated steam use 1.0. Steam: 51.5 ⋅ P ⋅ K D ⋅ K N ⋅ K SH W A= 51.5 ⋅ P ⋅ K D ⋅ K N ⋅ K SH K N = Correction factor for steam K = when P ≤ 1500 psia K NN = Correction factor for steam TABLE 3 P − 1000 ⎞ ⎛ 0.1906 Superheat Correction (API RP520 Part P 1 Table > 15009)psia and P ≤ 3200 psia ⎜ K NN = Factors, when P K≤SH1500 psia⎟ when ⎝ 0.2292 P − 1061 ⎠ 19063Pfor − 1000 ⎛ 0.Table ⎞ steam correction factors. Temperature °F Burst K = See superheat saturated ≤ 3200 psia steam use 1.0. SH = ⎜ ⎟ when P > 1500 psia and PFor Pressure N ⎝ 0.2292 P − 1061 ⎠ 300 400 500 600 700 800 900 1000 1100 1200 (psig) K SH = See Table 3 for superheat steam correction factors. For saturated steam use 1.0. 15

1.00

.98

.93

.88

.84

.80

.77

.74

.72

.70

20

1.00

.98

.93

.88

.84

.80

.77

.74

.72

.70

40

1.00

.99

.93

.88

.84

.81

.77

.74

.72

.70

60

1.00

.99

.93

.88

.84

.81

.77

.75

.72

.70

80

1.00

.99

.93

.88

.84

.81

.77

.75

.72

.70

100

1.00

.99

.94

.89

.84

.81

.77

.75

.72

.70

120

1.00

.99

.94

.89

.84

.81

.78

.75

.72

.70

140

1.00

.99

.94

.89

.85

.81

.78

.75

.72

.70

160

1.00

.99

.94

.89

.85

.81

.78

.75

.72

.70

180

1.00

.99

.94

.89

.85

.81

.78

.75

.72

.70

200

1.00

.99

.95

.89

.85

.81

.78

.75

.72

.70

220

1.00

.99

.95

.89

.85

.81

.78

.75

.72

.70

240

-

1.00

.95

.90

.85

.81

.78

.75

.72

.70

260

-

1.00

.95

.90

.85

.81

.78

.75

.72

.70

280

-

1.00

.96

.90

.85

.81

.78

.75

.72

.70

300

-

1.00

.96

.90

.85

.81

.78

.75

.72

.70

350

-

1.00

.96

.90

.86

.82

.78

.75

.72

.70

400

-

1.00

.96

.91

.86

.82

.78

.75

.72

.70

500

-

1.00

.96

.92

.86

.82

.78

.75

.73

.70

600

-

1.00

.97

.92

.87

.82

.79

.75

.73

.70

800

-

-

1.00

.95

.88

.83

.79

.76

.73

.70

1000

-

-

1.00

.96

.89

.84

.78

.76

.73

.71

1250

-

-

1.00

.97

.91

.85

.80

.77

.74

.71

1500

-

-

-

1.00

.93

.86

.81

.77

.74

.71

1750

-

-

-

1.00

.94

.86

.81

.77

.73

.70

2000

-

-

-

1.00

.95

.86

.80

.76

.72

.69

2500

-

-

-

1.00

.95

.85

.78

.73

.69

.66

3000

-

-

-

-

1.00

.82

.74

.69

.65

.62

3 of 9

AV =

Page Page 4: 4:

LIQUID SIZING

W )w 2407⋅⋅AA⋅⋅KKDD ((PP−−PP W == 2407 e e)w Page 4:

== Calculation per ASME SectionA AVIII Water:

DD W = 2407 ⋅ A ⋅ K D

A= Calculation per API RP520 Non-viscous liquid:

Viscous liquid:

2407 2407⋅⋅KK

W W

W

2407 ⋅ K D

((PP−−PPe e)w)w (P − P )w e

KKV

V

rated capacity, (gal.min) required Area without viscosity corrections (in2)

Av =

required Area with viscosity corrections (in2) Specific weight of water, (lb/ft3)

2.878 342.75 ⎞ ⎛ Kv = ⎜ 0.9935 + + ⎟ R 0.5 R1.5 ⎠ ⎝

QQ SG SG AAR == R 38 ⋅ K ⋅ K P − D V e 38 ⋅ K D ⋅ KV P −PP e

Q AR = AAR AAV ==38 ⋅RK D ⋅ KV

Q = AR = W =

(P − Pe )w

AR KV

Re =

SG P − Pe

−1.0

, viscosity correction factor

Q(2800 ⋅ SG ) Reynolds Number (u is in centipoises) u A

or

A V 12700 ⋅ Q AV = R (U is in Saybolt Universal Seconds, SSU) Re = KV A using K of 1.0. U A For viscous liquid sizing, first calculate R V Apply the area A of the next larger disccapacity, to the Reynolds Q = size rated (gal.min) rated capacity, (gal.min) number calculations to arriveQat K= . Then re-calculate V = required Area without viscosity corrections (in2) A R required area AV using the derived . required Area without viscosity corrections (in2) AR ==KVrated Q capacity, (gal.min) A

=

2 2 required Area with viscosity corrections (in )

Area without corrections (in )(in 2) AARv == required Area withviscosity viscosity corrections v RESISTANCE TO FLOW METHOD (Krequired ) 3 R

W == required SpecificArea weight water,corrections (lb/ft )3 (in2) with of viscosity Wv 5 Rule = Specific of water, (lb/ft )disc is not installed in combination with a pressure relief valve. Use this method when the 8 A& does notweight apply and the rupture −1.0 TM ) ⎞DisCalc = Specific weight water,designer. (lb/ft.375 This type of calculation is theWresponsibility of the2system does not perform this type of calculation. .of878 342 ⎛ −1.0 Kv = ⎜⎛0.9935 + 2.0878 + , viscosity correction factor 342 . 75 ⎟ ⎞ − 1 . 0 .5 1.5 2+.878 Kv to + .75 = ⎛Flow R 0.5 342 R ⎞1.5 ⎠ ⎟ , viscosity correction factor Characteristics of the Resistance Method ⎝⎜ 0.9935 Kv =

⎜ 0⎝.9935 +

R+

R⎟

,⎠ viscosity correction factor

R individual ⎝ ⎠ • Sizing is done on a relief system basis not by R capacity of components • Rupture disc is treated as another component in the relief system ) Reynoldsvalue Q(a2800 ⋅ SGresistance • Each device or family of devices has unit-less (KR) that represents the expected resistance to flow that is independent Number (u is in centipoises) Re = QQ 2800 ⋅ SG ( ) 2800 ⋅ SG of the fluid flowing Reynolds Number (u is in(u centipoises) RRee == u A Reynolds Number is in centipoises) • System relief capacity must be multiplied u uAby Aa factor of 0.90 or or Types of KR or 12700Q⋅ Q 12700 ⋅opening Because many rupture discs have characteristics depending on whether (U in is Saybolt in Saybolt Universal Seconds, SSU) they are opened with a compressed vapor or R e ==different 12700values ⋅Q (U isthat Universal SSU) service media. The K values for different media R e incompressible liquid, there are areSaybolt denoted bySeconds, the applicable = UUKA (U is in Universal Seconds, SSU) Re certified RA R are a result of differences in how the rupture U Adisc opens with different media and test methods that have been standardized in ASME PTC25. A list of Fike certified KR factors can be found in technical bulletin TB8104.

(

0.5

1.5

)

• Air or gas service – KRG Use KRG when the media is a gas or vapor, or when the media is liquid but there is a significant vapor volume directly in contact with the disc at the time of rupture • Liquid service – KRL Use KRL when the media is liquid and the liquid is against the disc at the time of rupture • Air or gas and liquid service – KRGL KRGL can be used for any service conditions The following examples will illustrate how KR values are used to establish the flow capacity of a pressure relief piping system. Vapor Sizing The following example, see Figure 1, assumes that k = cp/cv = 1.4 which results in a conservative calculation. The example shown is based on Crane TP-410 methods. It also assumes a steady state relieving condition where the vessel volume is large relative to the relieving capacity. Given information: • Pressure vessel MAWP = 1000 psig • Relieving pressure as allowed by ASME Section VIII Div. 1 = 110% x MAWP = 1114.7 psia = P’1 • Back pressure (outlet pressure) = 14.7 psia • Working fluid - air (k = cp/cv = 1.4) • Air temperature at disc rupture = 500°F = 960R = T1 • Maximum flow rate into the vessel = 20,000 SCFM • Rupture Disc - Fike 3” SRX-GI g KRG = 0.99

  



   

 



 

Figure 1

4 of 9

DETERMINE THE TOTAL PIPING SYSTEM RESISTANCE FACTOR: Piping Component or Feature

Flow Resistance Value (K)

Reference

Entrance - Sharp Edged

K1 = .50

Crane 410 pg A-29

1 ft of 3” Sch. 40 Pipe

K2 = .07

Fike 3” SRX-GI Rupture Disc

KRG = 0.99

20 ft or 3” Sch. 40 Pipe

K3 = 1.41

3” Sch. 40 Standard 90° Elbow

K4 = 0.54

Crane 410 Pg A-29

40 ft of 3” Sch. 40 Pipe

K5 = 2.82

K=fL/D: f = .018 (Crane 410 Pg A-26 L= 1 ft. ID = 3.068/12 ft

Pipe exit - Sharp Edged

K6 = 1.00



National Board Cert. No. FIK-M80277

Page 5:

Total System Flow Resistance

KT

K=fL/D: f = .018 (Crane 410 Pg A-26 L= 1 ft. ID = 3.068/12 ft

q' = 678 ⋅ Y ⋅ d 2 = 7.33 m

K=fL/D: f = .018 (Crane 410 Pg A-26 L= 1 ft. ID = 3.068/12 ft

'1 410 Pg A-29 ΔP ⋅ PCrane K2 + KRG+ K3 + K4 + K5 + K6 KK⋅T T=1K⋅1 +SG

The Darcy Equation defines the discharge of compressible fluids through valves, fittings and pipes. Since the flow rate into the example vessel is defined in SCFM, the following form of the Darcy equation is used: Page 5:

Crane Equation 3-20

q’m = Y

=

ΔP ⋅ P'1 K ⋅ T1 ⋅ SG

q’m =

rate of flow in cubic feet per minute at standard conditions, (SCFM) (14.7 psia and 60°F) Y = net expansion factor for compressible flow through orifices, nozzles and pipes (Crane 410 Pg A-22) d = internal diameter of pipe, (in) ΔP = change in pressure entrance to exit, (psia) rate of flow in cubic feet per minute at standard P’1 = pressure at entrance, (psia) conditions, (SCFM) (14.7 psia and 60°F)K = loss coefficient net expansion factor for compressible flow = absolute temperature at entrance, (R) T1through

q'm = 678 ⋅ Y ⋅ d 2

orifices, nozzles and pipes (Crane 410 Pg A-22) = internal diameter of pipe, (in) ΔP = change in pressure entrance to exit, (psia) P’1 = pressure at entrance, (psia) K = loss coefficient T1 = absolute temperature at entrance, (R)

d

5 of 9

To determine Y, first it must be determined if the flow will be sonic or subsonic. This is determined by comparing the actual DP/P’1 to the limiting DP/P’1 for sonic flow. Crane Table A-22 shows limiting factors for k=1.4 for sonic flow at the known value of KT. If (DP/P’1)sonic < (DP/P’1)actual, then the flow will be sonic. K

DP/P’1

Y

1.2

.552

.588

1.5

.576

.606

2.0

.612

.622

3

.662

.639

4

.697

.649

6

.737

.671

8

.762

.685

10

.784

15

.818

20

.839

.695 Page 6 .702 Page 6 .710

40

.883

100

.926

Limiting Factors for Sonic Velocity (k=1.4) Excerpt from Crane 410, Pg A-22

1114.7 − 14.7 ⎛ ΔP ⎞ .710Page = = 0.9868 ⎜ Page 66⎞' ⎟ 1114 .7 − 14 ⎛⎝ΔP P 1114 .7.7 = 0.9868 1 ⎠ actual= .710⎜ '⎟ P1 ⎠ actual

⎝

1114.7

1114..77−−14 14..77 1114 ⎛⎛⎜ΔΔPP '⎞⎞⎟ 9868 ⎜K =P7' .⎟33 == 1114.7 ==00..9868 actual 1114.7 ⎝⎝ T P11 ⎠⎠actual KT = 7.33 ⎛ ΔP ⎞ = 0.754 From table A-22 at KT=7.33 ⎛⎜ΔP P⎞' ⎟ ⎜⎝ ' 1⎟ ⎠ sonic= 0.754 33 P1 ⎠ sonic KKT == 77.⎝.33 For this example:

T

((

⎛⎛ΔΔPP' ⎞⎞ )

= 0.754

⎜ P ' '⎟⎟ = 7=.62 0.754 Since Δ⎜P , then ΔP = 0.754 ⋅ P = 0.754 ⋅ 1114.7 = 840.5 psi Pactual '1 P sonic 11 ⎠⎠sonic Since ΔP⎝⎝ P1 actual = 7.62, then ΔP = 0.754 ⋅ P1' =1 0.754 ⋅ 1114.7 = 840.5 psi

)

'

Since (DP/P’1)sonic = 0.754, then DP = 0.754 * P’1 = 0.754 * 1114.7 = 840.5 psig Calculating the system capacity is completed by substituting the known values into Crane 410 Equation 3-20.

((

))

Since P' actual =Δ7P.62 ⋅ P, 'then ΔP = 0.754 ⋅ P' = 0.754 ⋅ 1114.7 = 840.5 psi Since ΔΔPP⋅P P,' 1then ΔP = 0.754 ⋅ P11 = 0.754 ⋅ 1114.7 = 840.5 psi q ' = 678 Y11⋅actual d 2 =Δ7P.⋅62 '

'

q = 678 ⋅ Y ⋅ d 'm m

2

⋅ SG KK⋅ T⋅1T⋅1 SG 2 840.5 ⋅ 1114.7 ' 840.5 ⋅ 1114.7 678⋅ ⋅00.680 .680⋅ (⋅3(.3068 .068 ΔPP)⋅2⋅)PP11' ' 7.33 ⋅ 960 ⋅ 1 qqm''m' ==678 2 Δ 2 678⋅⋅YY ⋅⋅dd 7.33 ⋅ 960 ⋅ 1 qqmm' ==678 SG KK ⋅⋅TT11⋅⋅SG SCFM 50,074 ,074SCFM qq' m ==50 1

m

840..55⋅⋅1114 1114..77 840

678 680 068))2 1, paragraph UG-127(a)(2), also requires that the calculated system capacity using qm' m ==678 The ASME Pressure Vessel qCode, Section VIII,⋅⋅((Division ⋅⋅00..680 33..068 33 960 1 or less to account for uncertainties inherent with this method. ⋅⋅960 the resistance to flow method must also be multiplied by77a..33 factor of⋅⋅10.90 '

2

' SCFM 50,,074 074SCFM qq' m ==50

' 50,074 ,074⋅ 0⋅ .090 .90= =4545 ,066 SCFM qqm'mm−−ASME ,066 SCFM ASME ==50

Thus, the system capacity is greater than the required process capacity (20,000 SCFM)

=50 50,,074 074⋅⋅00..90 90==45 45,,066 066SCFM SCFM qqmm−−ASME Subsonic Flow Case ASME = In the case where the flow is subsonic, or (DP/P’1)sonic > (DP/P’1)actual , simply read the value of Yactual from Crane 410 chart A-22, Substitute (DP/P’1)actual and Yactual into the calculations ''

6 of 9

LIQUID SIZING For this example Figure 2 is assumed, water will be considered the flow media. The example shown is based on Crane TP-410 methods. It also assumes a steady state relieving condition where the vessel volume is large relative to the relieving capacity. Given information: • Pressure vessel MAWP = 500 psig • Relieving pressure as allowed by ASME Section VIII Div. 1 = 110% x MAWP = 550 psig = P1 • Back pressure (outlet pressure) = 1 psig = P2 • Working fluid - water • Temperature = 70°F • Maximum flow rate into the vessel = 50 ft3/min • Rupture disc - Fike 2” SRL-GI → KRGL = 0.59 From Crane 410: “Bernoulli’s Theorem is a means of expressing the application of the law of conservation of energy to the flow of fluids in a conduit (piping). The total energy at any particular point, above some arbitrary horizontal datum plane, is equal to the sum of the elevation head (Z), the pressure head (P), the velocity head (V). In real applications, there are energy losses in piping systems between states (or location) 1 and 2. Those losses are accounted for in the term Page hL, which 77 are predominately frictional head losses. Page The energy balance is then expressed:

   



   

 



 

Page 7

Z1 + 144 ⋅⋅PP1 VV122 144 ⋅ ⋅PP VV22 2 2 144 144 1+ 1 = Z + 2+ ZZ1 ++ +2 ⋅ g2 ++hhL L 1 ρρ1 +22⋅ ⋅gg = Z2 2 + ρρ2 2⋅ g 1 2 Crane Equation Page 71-3

144 ⋅ P1

ρ1

Z1 and Z2 144 ⋅ P V12 144 ⋅ P2 V22 and P2 Z1 + Z and1 Z + ==Z 2 + elevation + head at+ states hL 1 and P21(ft) ρ2 2head ⋅ g at states 1 andV12and Z1ρ1 1and Z222 ⋅ g = elevation (ft) V2

+

V12 144 ⋅ P2 V22 = Z2 + + + hL 2⋅ g 2⋅ g Figureρ22

= = = = = =

elevation head at states 1 and 2 (ft) pressure at states 1 and 2 (psig) velocity at states 1 and 2 (ft/sec) fluid density at states 1 and 2 (lb/ft3) acceleration due to gravity (32.2 ft/sec2) frictional head loss (ft)

P2 == pressure PP1 and pressureatatstates states11and and22(psig) (psig) 1 and P2 ρ1 and ρ2 V = velocity at states 1 and VV1 and 2 = velocity at states 1 and22(ft/sec) (ft/sec) 1 and V2 g 3 ρ2 == fluid )3 ρρ1 and fluiddensity densityatatstates states11and and22h(lb/ft 1 and ρ2 L(lb/ft ) 2 g = acceleration due to gravity (32.2 ft/sec )2) g = acceleration due to gravity (32.2 ft/sec Z1 and h ZL2 = elevation head at states 1 and 2 (ft) = frictional head loss (ft) = at frictional loss (ft) = pressure states 1 andhead 2 (psig) P1 and Ph2L = velocity at states 1 and 2 (ft/sec) As in theVprevious head losses due to friction in the piping and the head losses due to fittings are proportional to the sum of 1 and V2 example, hL = ∑ K the flow ρresistances: = fluid density at states 1 and 2 (lb/ft3) 1 and ρ2 g hL

hhL ==∑∑=KK = L

acceleration due to gravity (32.2 ft/sec2) frictional head loss (ft)

Since the acutal head loss is velocity dependent,

⎛ V 22 ⎞ hL = ∑ K ⎜⎜⎛ V ⎟⎟⎞ hL = ∑ K⎝⎜⎜2 ⋅ g ⎠⎟⎟ ⎝ 2⋅ g ⎠

hL = ∑ K

⎛ V2 ⎞ ⎟⎟ hL = ∑ K ⎜⎜ ⎝ 2⋅ g ⎠

⎛ V2 ⎞ ⎟⎟ hL = ∑ K ⎜⎜ ⎝ 2⋅ g ⎠

7 of 9

Frictional loss coefficients and fitting loss coefficients for the example are as follows: Piping Component or Feature

Flow Resistance Value (K)

Reference

Piping Frictional Losses 1 ft of 2” Sch. 40 Pipe

K1’ pipe = 0.11

K=fL/D; f = .019 (Crane 410 Pg A-26) L = 1 ft, ID = 2.067/12 ft

20 ft of 2: Sch. 40 Pipe

K20’ pipe = 2.21

K=fL/D; f = .019 (Crane 410 Pg A-26) L = 20 ft, ID = 2.067/12 ft

40 ft of 2” Sch. 40 Pipe

K40’ pipe = 4.41

K=fL/D; f = .019 (Crane 410 Pg A-26) L = 40 ft, ID = 2.067/12 ft

Entrance - r/d = 0.10

Kent - 0.09

Crane 410 Pg A-29

Fitting Losses Fike 2” SRL - GI Rupture Disc

KRGL = 0.59

National Board Cert. No. FIK-M80031

2” Sch. 40 Standard 90° Elbow

Kel = 0.57

Crane 410 Pg A-29

Pipe Page exit - Sharp 8 Edged

Kexit = 1.00

Crane 410 Pg A-29

Total Losses

KT = 8.98

Page 8

Page 8 Thus,

Page 8

⎛ V2 ⎞ hL = 8.98⎜⎜ ⎛ V 2⎟⎟ ⎞ hL = 8.98 ⎛⎝ ⎜2V⋅ 2g ⎞⎠ ⎟⎟ hL = 8.98⎜⎜⎛⎜⎝ V 2 ⋅2 g⎟⎟ ⎞⎠ ⋅ g ⎠⎟ = =8.98⎝0⎜2ft/sec Vh conditions: Other known vessel L

Zvessel = Vvessel = Vvessel = Z Z vessel == ZVvessel P exit vessel = vessel == ρZZ Z1vessel vessel = exit == PP exit = Zvessel == ρρ11

⎟

⎜

0⎝ft2 ⋅ g ⎠ 0 ft/sec 01 ft/sec ft + 20 ft = 21 ft = elevation change of piping ft 000ft ft/sec ft/sec 3= 21 ft = elevation change of piping + 20 1ρ021ft=ftft +62.3 20 ftlb/ft =ft21 ft =water elevation change of piping for at room temperature ft/sec 010ft/sec ft + 20 ft =3 213 ft = elevation change of piping 62.3 lb/ft water at room temperature ρ2ρ=2 = 62.3 lb/ft for for water at room temperature

Pexit = 0 ft/sec = ρ2 = 62.3 lb/ft3 for water2 at room ρ1 ⎡ temperature ⎛ V22 V2 ⋅into 144 550Equation 1-3, Substituting0 values + + 0 = 21 + 0 + + ⎢8.98 ⋅ ⎜

⎞⎤ ⎟⎟⎥ ⎜ 62⋅ .550 3 2 ⋅V32 2 ⋅V32 2 .2 2 .2 ⎠ ⎝ ⎤⎦2 2 ⎡⎣ ⎞ ⎛ 144 ⎡ 0 =0 21 + 0++0 + 2 V2+ ⎢8.+98 8 ⋅ ⎜⎜.98 ⋅2⎛⎜ ⎟⎟V⎥2 ⎞⎟⎤ 00++ 144 ⋅ 550+ + = 21 2 2 ⎢ 62 . 3 2 ⋅ 32 . 2 2 ⋅ 32 . 2 ⋅ 550 ⎦ ⎞.2⎤ ⎟⎠⎥ .3ft/sec 2V⋅232.⎣2 ⎡ ⎝ ⎛ ⎜⎝ V 22⋅⎠32 V =144 8962 .82

02 +

+ 0 = 21 + 0 +

62.3

Solving for VV22 (exit = 89.velocity), 82 ft/sec V2 = 89.82 ft/sec V2 = 89.82 ft/sec

2 ⋅ 32.2

⎟⎟⎥ ⎦ + ⎢8⎣.98 ⋅ ⎜⎜ ⎝ 2 ⋅ 32.2 ⎠⎦ ⎣

9 losses assumed that the flow in the pipes was fully turbulent flow. The friction factor used earlier in the calculations for pipingPage frictional The value of the friction factor is related to the Reynolds Number (Re) of the resulting flow (Ref: Crane 410 pg 1-1). For Re< 2000, the flow is laminar and the friction factor is a function of Reynolds Number, only. For Re >4000, the flow is fully turbulent and the friction factor is also a function of the character of the piping wall (relative roughness). 89.82 ⋅ 2.067 1

V ⋅d v

( 12)

= Reynolds = Number: Re the Page 9 used earlier must be verified. First calculate The friction factor Re = V d v

( )

V ⋅ d 89.82 ⋅ 2.067 112 = v .000011

= = =

fluid velocity = 89.82 ft/sec pipe diameter = 2.067 in/12 in/ft kinematic viscosity = 0.000011 ft2/sec

Q = A ⋅V Q A V Q

volumetric flow rate (ft3/sec) area of pipe (ft2) - πd2/4 fluid velocity (ft/sec)

= = = =

Qcalc =

(

π 2.067 ⋅

) ⋅ 89.82 12 2

4 2.09 ft3/sec = 125.6 ft3/min

V d v

= = =

.000011

fluid velocity = 89.82 ft/sec pipe diameter = 2.067 in/12 in/ft kinematic viscosity = 0.000011 ft2/sec

Q = A ⋅V Q A V Q

volumetric flow rate (ft3/sec) area of pipe (ft2) - πd2/4 fluid velocity (ft/sec)

= = = =

Qcalc =

(

π 2.067 ⋅

) ⋅ 89.82 12 2

4 2.09 ft3/sec = 125.6 ft3/min 8 of 9

If the flow had been laminar, Re < 2000 the friction factor is calculated as:

If the flow had been laminar, Re < 2000 the friction factor is calculated as:

A = area of pipe (ft2) - πd2/4 VPage = 9fluid velocity (ft/sec) Q

=V

(

( )

)

π .067 1 .82 ⋅ 22.⋅067 ⋅ d ⋅ 289 89.8212

12 the flow is turbulent, and the friction factor is now a function of the relative roughness of the pipe. Page 9 4 = is >4000, Re = Number Since the Reynolds v2.09 000011 3the .friction From Crane Q 410 Figure A-23, factor, f, for 2” commercial steel pipe in fully turbulent flow is 0.019. This verifies the original = ft /sec = 125.6 ft3/min calc assumption for friction factor. V = Vfluid = 89.82 ft/sec 1 Page 9 89.82 ⋅ 2.067 ⋅ d velocity 12 d = pipe diameter = 2.067 in/12 in/ft = = R Page e 9 LaminarIfvFlow Considerations /sec factor is calculated as:1 viscosity 0.000011 ft2friction v been laminar, .000011 the = flow kinematic had R=e < 2000 the V ⋅ d 89.82 ⋅ 2.067 If the flow had been laminar, Re