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l.govUsing RELAP5 to Analyze Pressure Relief Systems for Noncondensable Gases

2011 IRUG Meeting

Joe Palmer

Presentation Outline

• Test Cases

• Typical Problems

• Modeling Details

• Energy Conservation and Heat Transfer Details

Test Case 1 - Measured Flow Through an Orifice

0 10 20 30 40 50 60 70 80 90

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

RELAP Vs. Measured Orifice FlowDischarge Coefficient = 0.843

RELAP Prediction

Measured Data

Pressure (psig)

Flo

w R

ate

(lb

m/s

ec

)

Measured data is from O’Keefe Orifice Size 63 (.063 inch dia)

Fluid is air

Test Case 2 – Tank BlowdownComparison of RELAP5 to Two Other Compressible Flow Codes

0 1 2 3 4 5 60

200

400

600

800

1000

1200

Tank Blowdown

RELAP5COMPFLOCOMPG5

Time (sec)

Tank

Pre

ssur

e (p

sia)

Fluid is air

Case 3 – Sonic Flow in a PipeComparison of RELAP5 to Crane Tech Paper 410 Example Problem 4-21

Test Case 3 (cont)

Case 4 – Subsonic Flow in a PipeComparison of RELAP5 to Crane Example Problem 4-22

Would like more test cases with measured data

• I would be interested in doing more comparisons to measured data – especially flow down pipes – sonic and subsonic. This would be for noncondensable gases, not steam.

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ATR Temperature Controlled Fuel Experiment

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Typical Relief Valve Problem

Basic Problem - Modeling Approach

The components being protected by the relief valve are downstream

Relief Valve Modeling• We can not simply treat a RV as an orifice unless the

discharge is directly to atmosphere. • This is because the RV is a differential pressure

device. Most RVs open and stay open based on the differential between upstream and downstream pressure rather than the difference between upstream and atmospheric pressure.

• Therefore the back pressure from the relief exhaust line must be taken into account. This is especially important for the Advanced Test Reactor (ATR) installations where the exhausts are routed long distances to HVAC ducts. Backpressure is roughly translated into an increase in pressure at the RV inlet.

• “Roughly” because the relief valves do have a blowdown pressure, which keeps the valve open at a certain percentage below the set pressure – typically 10%.

Relief Valve Modeling (cont)

• So, once open, a 100 psi relief valve would require at least 90 psi differential to stay open. This is important because a RV can be massively oversized but if the exhaust line is too small or too long, it will not properly protect the system.

Relief Valve Modeling (cont)

Valve is modeled in RELAP5 using vendor supplied orifice area and then discharge coefficient is adjusted to achieve vendor’s published flow capacity

Relief valve is modeled as a Motor Operated Valve (MOV) with trip open at 5% over set pressure and trip closed at 5% under set pressure (i.e., 5% blowdown which is conservative)

ATR Experiments – Relief Protection of Downstream Low-Pressure Components

Effect of heat transfer assumption

Case Relief valve

Inlet temp.

(°F)

Heat transfer?(isothermal)

Flow rate

(SCFM)

Pressure

(psig)

3 118CSS 30 Yes 58.0 38.3

4 118CSS 100 Yes 57.7 38.1

5 118CSS 30 No 60.4 38.2

Table 3. Calculated results for the supply line to the actuator

Taken from C. Davis calculations supporting INL internal doc, ECAR-1464

Relief valvePressure regulator

2 inch of1/2-inch pipe

AtmosphereP = 0.0 psig

125 175115105

Air supplyP = 150 psigT = 30 F

100 110 120 170 190

16 ft of 3/8-inch tube 4 inch of 1/4-inch tube100F

Almost no effect

Effect of RELAP junction e flag

• At large discontinuities the user manual recommends setting the e flag equal to “1”.

• * from to area f loss r loss• 1050101 100010000 110000000 1.576e-6 20800. 20800. 1000000

• This turns out to be very important for the restriction orifice in a typical ATR experiment supply system

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Effect of RELAP junction e flag

• With e flag set to 0 temperature downstream of orifice is much too low.

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This does not seem to be a result of Joule-Thompson cooling since RELAP calculates a deltaT about the same for both helium and argon which have dramatically different J-T coefficients

Helium is the fluid

Effect of e flag on flow through straight pipe

Another look at test Case 4

.0858 lb/sec

.0793 lb/secMatches Crane almost exactly

.0851 lb/sec

Air, 10 ft of ½” Pipe

A look at temperature change across individual nodes – Case 4

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First Lawe = 0e = 1

-.464oF-.387oF

-6.97oF

-32.9oF

First Lawe = 0e = 1 +.37oF

+2.54oF

First Law:

Air, 10 ft of ½” Pipe

Isothermal Modeling

• RELAP5-3D does not have an isothermal flag• However, noncondensable gas models can be made to run

isothermally by incorporating heat structures, i.e., by setting the pipe wall to a fixed temperature and using it to keep the fluid at approximately the set temperature

• Another (less elegant) way is to switch the e flag to 1 judiciously, i.e., turn the e flag on and off at various junctions down the length of a pipe. This is done by trial and error, but for small problems it may be easier than the heat structure approach

• Suggest incorporating an isothermal option in RELAP5. COMPFLO has this and it is handy for noncondensable gas problems such as those described in this presentation

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Conclusions

• RELAP5-3D is an effective tool for modeling pressure relief systems for noncondensable gases

• Care must be taken in modeling relief valves to ensure sufficient pressure drop is taken across the valve

• If choking is possible at the end of any pipe the node density should be increased at that end

• The thermodynamics behind the junction e flag are not clear to this user

• Suggest incorporating an isothermal option in RELAP5.

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