CHEVRON - Flow Rates and Shock Pressures Through Tube Ruptures Heat Exchanger

Appendix F. Flow Rates and Shock Pressures Through Tube Ruptures

Methods to calculate tube rupture flow rates and shock pressures are given here for shell and tube exchangers containing gases, two-phase mixtures, and liquids.

Chevron Corporation F-1 April 1998

Appendix F Heat Exchanger and Cooling Tower Manual

F1.0 High-pressure Gases and Two-phase MixturesWhen the high-pressure fluid is part or all gas, the flow through the rupture may or may not be choked (limited by the gas reaching sonic velocity). If it is choked, flow through the rupture is independent of the low-pressure side. Otherwise, the response of the low-pressure side affects the rupture flow. The rupture flow may be choked on one side of the break and subsonic on the other side. Rupture flow from both sides of a break is maximum for a break at one end.

Rupture flow rates have been calculated for the usual range of conditions in heat exchangers and are given below in Equation F-1 and Figure F-1. For definitions, see the nomenclature section at the end of this appendix.

(Eq. F-1)

Equation F-1, with CF = 1, is the equation for isentropic (frictionless) choke flow from both sides of the break. CF is the ratio of actual flow with friction to the flow rate for choked isentropic flow with a break at one end and is given in Figure F-1.

Two-phase flow through a rupture is similar to all-gas flow. Initially, the two-phase rupture flow is mostly gas. The gas usually accelerates rapidly to near sonic speeds by virtue of its internal energy, whereas liquid is dragged along by viscous shear and accelerates much more slowly. The presence of liquid reduces the gas flow area and adds to frictional resistance of the gas flow. Treating liquid-gas mixtures on the high-pressure side as if they were all gas is conservative and is recommended.

The high-pressure fluid on the low-pressure side displaces and compresses the low-pressure fluid. The pressure rise on the low-pressure side depends on the volume flow rate into the low-pressure side, the compressibility of the low-pressure fluid and the flow area of the connected piping. The fluid compressibility and pipe sizes on the low-pressure side are usually the same for inlet and outlet lines for single phase fluids but may be significantly different for liquid-gas mixtures.

Equation F-2 below is a volume balance: volume flow rate of high-pressure fluid into the low-pressure side equals rate at which volume is made available by the compression waves in the inlet and outlet piping.

(Eq. F-2)

M·

R 2CF AT ρH g PH kH2

kH 1+----------------

kH 1+kH 1–---------------

⋅ ⋅ ⋅

0.5

⋅=

M·

R 2AL1 GL ρHFL PLR PL–( ) gρL1 KL1⋅------------------------

0.5⋅ ⋅ ⋅=

April 1998 F-2 Chevron Corporation

Heat Exchanger and Cooling Tower Manual Appendix F

where:

(Eq. F-3)

For single phase service on the low-pressure side, fluid properties and pipe sizes for inlet and outlet lines are usually equal. Then GL = 1 and Equation F-2 becomes:

(Eq. F-4)(From Equation F-3)

Solving Equations F-1 and F-2 simultaneously, with ρHFL = ρH (PLR/PH) for high-pressure gas, results in:

(Eq. F-5)

where:

(Eq. F-6)

Equation F-6 is plotted in Figure F-2.

Figure F-2 applies to any low-pressure fluid. Liquid cases are toward the right side of the figure, gas cases toward the left, and two-phase cases in the middle.

GL12--- 1

AL2

AL1----------

ρL1

ρL2---------

KL1

KL2----------

0.5

+=

M·

R 2AL ρHFL PLR PL–( ) gρL KL⋅------------------

0.5⋅=

PLR

PH----------

12---

PL

PH-------

PL

PH-------

2

4 X+ 0.5

+=

XCF

GL-------

AT

AL1----------

ρL1

ρH---------

KL1

PH----------

kH2

kH 1+----------------

kH 1+kH 1–---------------

0.5

=



Fig. F-1 Factor for Friction and Flow Regime



Fig. F-2 Rupture Pressure on Low-Pressure Side Due to High-Pressure Gas



F2.0 High-Pressure LiquidsTube rupture with high-pressure liquid involves decompression waves in the high-pressure system similar to the compression waves in the low-pressure system. Choked liquid flow does not occur at conditions encountered in heat exchangers.

The relationship between rupture flow and decompression is given by:

(Eq. F-7)

Equal pipe sizes and fluid properties are assumed for the high-pressure inlet and outlet lines.

The rupture flow rate and pressure drop across the break for a tube break at one end (worst case) are related as follows:

(Eq. F-8)

where:

(Eq. F-9)

Solving Equations F-7, F-8, and F-3 simultaneously results in:

(Eq. F-10)

(Eq. F-11)

(Eq. F-12)

M·

R 2AH PH PHR–( )g ρH⋅KH

-------------- 0.5

=

M·

R 2CFL AT 4 3⁄( )g ρH PHR PLR–( )⋅( )0.5⋅=

CFL12--- 1

3 2⁄3 2⁄ F L D⁄⋅+------------------------------------

0.5+=

PLR PL–

PH PL–----------------------

Z

2---

1 SH+( )

SH2

--------------------- 14Z---

SH

1 SH+----------------

2

+0.5

1–

=

PH PHR–

PH PL–----------------------- SH

PLR PL–

PH PL–----------------------

=

M·

R

2CFL AT 4 3⁄( )g ρH PH PL–( )⋅[ ]0.5⋅-------------------------------------------------------------------------------------------- 1 1 SH+( )

PLR PL–

PH PL–----------------------

–0.5

=



where:

(Eq. F-13)

and

(Eq. F-14)

Equations F-10 and F-12 are plotted in Figures F-3 and F-4 respectively.

Fig. F-3 Rupture Pressure on Low-Pressure Side Due to High-Pressure Liquid

SH GL

AL1

AH----------

ρH

ρL1---------

KH

KL1----------

0.5

=

Z 4 3⁄( )CFL2

AT

AH--------

2 KH

PH PL–-------------------

=



F3.0 Effective Bulk ModulusEffective bulk modulus is the pressure change per unit of volume change.

(Eq. F-15)

Expansion or contraction of the pipe, as well as compression or decompression of fluid, contribute to the effective bulk modulus.

The effective bulk modulus, including pipe flexibility, is:

(Eq. F-16)

where C = 0.91 for restrained piping.

The effective bulk modulus is typically 10% less than the liquid bulk modulus. Equation F-16 is appropriate for liquid systems. Bulk modulus data for liquids are available in various reference books including the Fluid Flow Manual.

Fig. F-4 Liquid Rupture Flow Rate

K∆P

∆V V⁄----------------=

KKF

1KF D C⋅ ⋅

E tp⋅------------------------+

----------------------------------=



Bulk modulus for gases and two-phase flow may be calculated as follows:

(Eq. F-17)

where:

(Eq. F-18)

Only gas phase compressibility is included in Equation F-17. Liquid phase compressibility and pipe flexibility are insignificant in comparison to gas phase compressibility.

F4.0 Two-Phase DensityTwo-phase density of low-pressure fluid may be calculated as follows:

(Eq. F-19)

F5.0 ApplicationsThe use of the equations and graphs in this appendix is illustrated in the examples following the nomenclature section below.

The units indicated in the nomenclature section are consistent with the equations. The equations, however, are presented in terms of dimensionless parameters or ratios where practical. Any consistent units may be used in those cases. Bulk modulus, elastic modulus, and fluid pressure all have the same units. Where these variables appear as ratios in the examples, psi units are used; otherwise psf units are used.

Example 1 is for high-pressure gas and low-pressure liquid. Post-rupture pressure on the low-pressure side is 83% of the initial pressure of the high-pressure fluid and governs the design of the low-pressure side.

Example 2 is for high-pressure gas and low-pressure steam generation. Post-rupture pressure on the low-pressure side is insignificant and does not affect design pres-sure.

KL

PL-------

R 1+R

------------- PLR

PL---------- 1–

1PLR

PL----------

–1– kL⁄--------------------------------------------=

R ρ1 ρg⁄( ) Y1 Y–-------------

=

ρρg

Y ρg ρ1⁄( ) 1 Y–( )+--------------------------------------------------=



Example 3 is for high-pressure liquid and low-pressure liquid. Post-rupture pres-sure on the low-pressure side is about 40% of the initial pressure of the high-pres-sure liquid and governs the design of the low-pressure side.

F6.0 Nomenclature

A = Flow area of tube or pipe, ft2

C = Pipe restraint factor

CF = Actual tube rupture flow/isentropic choke flow

CFL = Actual tube rupture flow/frictionless liquid flow

D = Inside diameter of pipe, ft

E = Modulus of elasticity of pipe, psf

F = Darcy-Weisbach friction factor

g = 32.17 lb/slug

GL = Factor for dissimilar inlet and outlet piping

k = Ratio of specific heats (Cp/Cv)

K = Effective bulk modulus of fluid and pipe, psf

KF = Bulk modulus of fluid, psf

L = Flow length through broken tube, ft

= Rupture flow rate, lb/sec

P = Pressure, psfa

∆P = Increment of pressure, psf

R is defined by Equation F-18.

SH is defined by Equation F-13.

tp = Pipe wall thickness, ft

V = Volume, ft3

∆V = Increment of volume, ft3

X is defined by Equation F-6.

Y = Weight fraction vapor

Z is defined by Equation F-14.

ρ = density, lb/ft3

M·

R



Subscripts:

Example 1

High-Pressure Hydrogen Gas and Low-pressure Cooling Water

Gas

Cooling Water

Tube

Low-Pressure Piping

1 - Inlet piping, low pressure side (LPS)

2 - Outlet piping, LPS

g - Gas, LPS

H - High-pressure side, normal conditions

HR - High-pressure side, rupture conditions

HFL - High-pressure fluid on low pressure side

l - Liquid

L - Low-pressure side, normal conditions

LR - Low-pressure side, rupture conditions

R - Rupture conditions

T - Tube

PH = Pressure = 2000 psia = 288,000 psfa

ρH = Density = 0.459 lb/ft3

kH = Ratio of specific heats = 1.4

PL = Pressure = 65 psia = 9360 psfa

ρL = Density = 62 lb/ft3

KF = Bulk modulus = 315,000 psi = 45,360,000 psf

Inside diameter = 0.482 in. = 0.0402 ft

Inside Diameter = 4.026 in.

tp = Wall thickness = 0.237 in.

E = Elastic modulus = 29 × 106 psi = 4.2 × 109 psf

GL = 1 (similar inlet and outlet)



(Eq. F-20)

Assume CF = 0.5

(Eq. F-21)

=

(Eq. F-22)

(Eq. F-23)

PLR = 0.83 (2000) = 1660 psia

CF from Figure F-1 is 0.5; therefore assumed value above is okay.

(Eq. F-24)

=

= 1.79lb/sec Total Flow from Rupture

KL

KF

1KF D C⋅ ⋅

E tp⋅------------------------+

----------------------------------315,000

1315,000

29,000,000---------------------------

4.0260.237-------------

0.91+

------------------------------------------------------------------------ 269,700psi= = =

XCF

GL-------

AT

AL1----------

ρL1

ρH---------

KL1

PH----------

kH2

kH 1+----------------

kH 1+kH 1–---------------

0.5

=

0.51

------- .4824.026-------------

2 620.459-------------

269,7002000

------------------- 1.4

22.4-------

2.40.4-------

0.5

0.662=

PLR

PH----------

12---

PL

PH-------

PL

PH-------

2

4X+ 0.5

+12---

652000------------

652000------------

24 0.662( )+

0.5+ 0.83= = =

M·

R 2CF AT ρH g PH kH2

kH 1+----------------

kH 1+kH 1–---------------

⋅ ⋅

0.5

⋅=

2 0.5( ) 3.14( ) 0.0402( )2[ ]4

-------------------------------------------- 0.459 32.17( ) 288,000( ) 1.4( ) 22.4-------

2.40.4-------

0.5



Example 2

High-Pressure Hydrogen Gas and Low-Pressure Steam Generator

Gas

BFW/Steam

Tube

BFW Piping

(Eq. F-25)

Steam PipingInside diameter = 7.981 in.

Y2 = Weight fraction vapor = 1 (1/R2 = 0)

Assume PLR/PL = 1.13


ρH = Density = 0.459 lb/ft3

kH = Ratio of specific heats = 1.4

PL = Pressure = 165 psia = 23,760 psfa

ρl = Liquid Density = 55 lb/ft3

ρg = Vapor density = 0.36 lb/ft3

kL = Ratio of vapor specific heats = 1.28

KF = Liquid bulk modulus = 183,000 psi = 26.4 × 107 psf

Inside diameter = 0.482 in. = 0.0402 ft

Inside diameter = 2.067 in.



Y1 = Weight fraction vapor = 0 (R1 = 0)

KL1

KF

1KFDC

Etp----------------+

-------------------------183,000

1183,000

29,000,000---------------------------

2.0670.154-------------

0.91+

------------------------------------------------------------------------ 169,000 psi= = =



(Eq. F-26)

KL2 = 1.43 (165) = 236 psi

(Eq. F-27)

Assume CF = 0.675

(Eq. F-28)

=

(Eq. F-29)

(Eq. F-30)

PLR = 0.0935 (2000) = 187 psia

PLR/PL = 187/165 = 1.13; therefore assumed value above is okay.

CF from Figure F-1 is 0.675; therefore assumed value above is okay.

(Eq. F-31)

KL2

PL----------

11

R2------+

PLR

PL---------- 1–

1PLR

PL----------

–1– kL⁄----------------------------------------------

1( ) 1.13 1–( )1 1.13( )– 1– 1.28⁄----------------------------------------- 1.43= = =

GL12--- 1

AL2

AL1----------

ρL1

ρL2---------

KL1

KL2----------

0.5

+12--- 1

7.9812.067-------------

2 550.36----------

169,900236

------------------- 0.5

+ 2472= = =

XCF

GL-------

AT

AL1----------

ρL1

ρH---------

KL1

PH----------

kH2

kH 1+----------------

kH 1+kH 1–---------------

0.5

=

0.6752472-------------

0.4822.067-------------

2 550.459-------------

169,9002000

------------------- 1.4

22.4-------

2.40.4-------

0.5

0.001025=

PLR

PH---------

12---

PL

PH------

PL

PH------

2

4X+0.5

+12--- 165

2000------------

1652000------------

24 0.001025( )+

0.5+ 0.0935= = =

M·

R 2CF AT ρH g PHkH2

kH 1+----------------

kH 1+kH 1–---------------

⋅ ⋅

0.5

⋅=



=

(Eq. F-32)

= 2.42 lb/sec Total Flow from Rupture

Example 3

High-Pressure Liquid and Low-Pressure Cooling Water

High-Pressure Liquid

Cooling Water

Tube

High-Pressure Piping

(Eq. F-33)


ρH = Density = 50 lb/ft3

KFH = Bulk modulus = 200,000 psi = 28,800,000 psf

Viscosity = 0.000672 lb/ft⋅sec = 1 CP

PL = Pressure = 65 psia = 9,360 psfa

ρL = Density = 62 lb/ft3

KFL = Bulk modulus = 315,000 psi = 45,360,000 psf

D = Inside Diameter = 0.56 in.

L = Length = 480 in. = 40 ft

Inside diameter = 4.026 in.



2 0.675( ) 3.14( ) 0.0402( )2[ ]4

-------------------------------------------- 0.459 32.17( ) 288,000( ) 1.4( ) 22.4-------

2.40.4-------

0.5

KH

KFH

1KFHDC

Etp--------------------+

-----------------------------200,000

1200,000

29,000,000---------------------------

4.0260.237-------------

0.91+

------------------------------------------------------------------------ 180,700 psi= = =



Low-Pressure Piping

(Eq. F-34)

(Eq. F-35)

Assume Friction Factor F = 0.0126

(Eq. F-36)

(Eq. F-37)

from Figure F-3 (0.318 from Equation F-10)

from Figure F-4 (0.669 from Equation F-12)(Eq. F-38)

Inside Diameter = 4.026 in.



GL = 1 (Similar Inlet and Outlet)

PLR = 65 + 0.32 (400 - 65) = 172 psia

KH

KFL

1KFLDC

Etp-------------------+

-----------------------------315,000

1315,000

29,000,000---------------------------

4.0260.237-------------

0.91+

------------------------------------------------------------------------ 269,700 psi= = =

SH GL

AL1

AH----------

ρH

ρL1---------

KH

KL1----------

0.5

14.0264.026-------------

2 5062------

180,700269,700-------------------

0.50.735= = =

CFL12--- 1

3 2⁄3 2⁄ FL D⁄+--------------------------------

0.5+

12--- 1

3 2⁄3 2⁄ 0.0126 480( ) 0.56⁄+-------------------------------------------------------------

0.5+ 0.675= = =

Z 4 3⁄( )CFL2

AT

AH--------

2 KH

PH PL–-------------------

4 3⁄( ) 0.675( )2 0.564.026-------------

4 180,700400 65–---------------------

0.123= = =

PLR PL–

PH PL–---------------------- 0.32=

M·

R

2CFLAT 4 3⁄( )g ρH PH PL–( )⋅[ ]0.5--------------------------------------------------------------------------------------- 0.67=



(Eq. F-39)

= 15.7 lb/sec Total Flow from Rupture

Check assumed friction factor (above value is okay).

M·

R 0.67 2( ) 0.675( )3.14

0.5612

---------- 2

4----------------------------- 4 3⁄( ) 32.17( ) 50( ) 400 65–( )144[ ]0.5=


CHEVRON - Flow Rates and Shock Pressures Through Tube Ruptures Heat Exchanger

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