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# Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

## Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

(OP)
Hi All,

I calculated the flow rate of steam due to tube rupture from tube side (high pressure side) to shell side( low pressure side using the below equation taken from Wing Wong's paper "PRV Sizing for exchanger Tube Rupture" attached with this thread.

Mass flow rate considering one orifice ( one tube diameter ) in lb/hr is given by Wv as below.

Wv = 1891 Y d2 C (DP. Rho)^0.5

where Y = Net Expansion Factor = 1-0.317 (DP/P1)

d is the tube diameter , inches

C = orifice coefficient = 0.6

DP = P1 - P2

where P1 = High pressure side operating pressure , psia
P2 = greater of critical pressure PCF and relief pressure ( Prelief), psia

I know that this equation is a modification on Page 3-26 in Crane TP 410 M. My question are the units that I mentioned correct. If anyone has used this equation kindly confirm.

Thanks and Regards,
Pavan Kumar

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

there are corrections (orifice area equations), see Hydrocarbon Processing May 1992,
for this topic see also the papers / methods proposed by Leung and others...

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

Pierre,
in case of vapor+liquid I would suggest to follow Leung "Two-Phase Flow Discharge in Nozzle and Pipes-A Unified Approach" and similar methods,
see for example "Using DIERS Two-phase Equations to Estimate Tube Rupture Flowrates" Hydrocarbon Processing August 2001 (but several more recent available)
a simple but rigorous alternative could be to calculate speed of sound / critical flow with one validated (HEM, HNE etc.) model, see also DIERS papers

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

(OP)
Hi PaoloPemi,

Request you to kindly share the paper by Leung if you have access and if it is ok to post here.

Thanks and Regards,
Pavan Kumar

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

Pavan,
in the past many works of Leung were available for download from Fauske but I do not know if it is still true...
anyway you should be able to find a copy of "Using DIERS Two-phase Equations to Estimate Tube Rupture Flowrates" Hydrocarbon Processing August 2001

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

(OP)
Hi PaoloPemi,

I donot have subscription to Hydrocarbon prcoessing. Since my case involves just vapor flow I good with the equation given by Wing Wong's paper. Would you be able to share his May 1992 paper which has the corrections. I know pierreick shared the correction document but I want to verify in the paper itself.

Thanks and Regards,
Pavan Kumar

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

for vapor the paper defines relieving flowrates for tube rupture cases according Crane 3.22 (I presume you have a copy),
the corrections are those in Pierre post, see Letters to the editor, Corrections needed, original copy in
https://www.cheresources.com/invision/topic/29325-...

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

(OP)
Hi PaoloPemi,

I checked the Aspen document sent by pierreick has the corrected equations of Wing Wong paper. These equations match with the corrections mentioned in the Letters to Editor provided by pierreick. I got all the information I need to complete my calculation.

Thanks and Regards,
Pavan Kumar

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

Hi Paolo,
Noted with thanks
Pierre

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

(OP)
Hi pierreick and PaoloPemi,

Thanks for the documents provided. Your help is invaluable to me.

Thanks and Regards,
Pavan Kumar

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

Pavan, please note that the Cd=0.6 you are stating is only for non choked vapor or liquid relief flow across the orifice and this would give you less relieving rate across the tube rupture guillitone leak. please consider using cummingham paper which considers more conservative Cd in case you have choked conditions for square edge orifice as well as Ward Smith for Cylindrical nozzles critical flow demonstrating maximum Cd=0.84 depending on the ratio of the thickness to diameter.
Considering that phase equilibrium requires minimum 100 mm to be attained, flashing is usually expected past the tube leak point and not within the orifice and hence if you have liquid at the inlet (and not already two phase), you can consider non flashing liquid rather than two phase.

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

(OP)
Hi Sawsan311,

I have used a Cd of 0.74 for tube to shell flow and used the correct equation for Expansion factor Y = 1-0.48 DP/P1. Do you think this is less conservative for choked flow conditions?. Also in my case the tube to shell flow is saturated steam and I think will condense for flashing flow to be possible. Would it be possible for you to point me where I can obtain Cunningham's paper?.

Thanks and Regards,
Pavan Kumar

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

(OP)
Thank so much Pierre. I have AFT Arrow software with me and this document would be very valuable to me. Besides there are so many good references mentioned in the document which are very valuable.

Do you have a good reference for calculating Two phase pressure drop for flashing flow?

Thanks and Regards,
Pavan Kumar

### RE: Tube Rupture Vapor Flow Rate Calculation - Wing Wong's Paper

(OP)
Hi All,

I am writing this message to determine the correct basis for Tube rupture relief rate calculation. The heat exchanger is steam reboiler ( sketch attached with this message). The tube side is LP steam with normal operating pressure of 70 psig while the shell side is condensate which is being boiled to generate steam. The MAWPs and operating condition of shell and tube sides are shown as below:

A) Shell Side

Fluid = Steam Condensate
Normal Operating Pressure = 10 psig
Normal Operating Temperature = 248 Deg F
MAWP = 210 psig at 329 Deg F

B) Tube Side

Fluid = LP Steam
Normal Operating Pressure = 70 psig
Normal Operating Temperature = 315.98 DEg F ( saturation temperature of 70 psig steam)
MAWP = 247 psig at 554 Deg F

The shell side is the low pressure side. Applying the 2/3 rd rule

MAWP of Shell Side / MAWP of tube Side = 210 / 247 = 0.85 > 2/3

Hence tube rupture will not result in loss of containment on shell side because. This however does not mean tube rupture is not feasible. There is a PSV on Shell side which is set at 72.5 psig lower than the MAWP if shell side of 210 psig.

Assuming that complete tube failure occurs due to pressure difference between tube and shell sides and due to corrosive nature of water, I now try to calculate the tube rupture flow rate.

Per Wing Wong's paper the tube rupture relief rate is calculated a below

Wv =2407.7 C A Y SQRT( DP. RHO)

where C= 0.74 for tube to shell flow
A = T/4(d^2), where d is the inside diameter of the tube in inches,

Y = 1 - 0.4 (DP/ P1)

where P1 is the pressure upstream of tube break, psia
P2 = pressure downstream of tube break, psia

P2 = Greater of PCF and PRel

where PCF is critical flow pressure calculated as PCF = P1 . (2/(k+1))^(k/k-1)

where k is the ratio of specific heats of steam

PRel is the relieving pressure = 1.1* PSV set pressure + 14.7 , psia

DP = P1-P2, psi

With P1 = 70 psig = 84.7 psia
k = 1.378 (for 70 psig steam)

PCF/P1 = 0.53

PCF = 0.53 *84.7 = 45 psia

PRel = 1.1*72.5+14.7 = 94.45 psia

Since PRel > PCF , P2 = PRel = 94.45 psia

DP = 84.7 - 94.45 = -9.75 psi

RHO = density at P1.

Notice that the DP is negative.

Now my question is I considered the normal operating pressure as P1. Should P1 be taken as the MAWP of tube side which is equal to 247 psig as this is the maximum possible pressure that tube side can ever see for conservative purposes?. Can I consider this?. In the P&IDs I have I did not see a PSV on the LP Steam line. But there will be surely a PSV downstream of the pressure reducing station that reduced the pressure from HP steam to LP steam. Is it acceptable to consider MAWP as the upstream pressure for the lack of maximum possible pressure information?.

If this is so DP = 247+14.7 - 94.45 = 167.25 psi

With this the calculated relief rate = 6465.44 lb/hr

Thanks and Regards,
Pavan Kumar

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