jacob fire evaluation

8/3/2019 Jacob Fire Evaluation

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Sizing of relief valves forsupercritical fluids

March 23rd, 2011

Overview

Relief Valve Study An Engineering Approach

Relief Calculation for Supercritical Fluids Introduction Theoretical Background Discussion & Evaluation


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Jacobs

Introduction

Jacobs Introduction: Who Are We

Committed to Be ondZero Safet as safet is our #1priority

Relationship based company

Global resource base 57.500 employees in 25countries on 4 continents

Fortune 500 #1 Engineering & Construction Company

Publicl traded on NYSE

Net income $65,8 Million 1Q FY11 ($246 Million FY10)

Revenues $2,4 Billion 1Q FY11 ($9,9 Billion FY10)

Backlog $13 Billion FY11

In business since 1947


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Jacobs Introduction: Worldwide offices

Jacobs Introduction: Europe


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Jacobs Introduction: BelgiumOil & Gas(Refining)

30

Chemicals &Polymers

45

Others12

Pharma& Bio

13

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Procurement,14

Process, 52

Engineering &Design, 316

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Jacobs Introduction: Clients

Client Workload Sinc Client Workload Since

/ People e / People

BASF 30-50 2003 SABIC 15-60 2004

Borealis 25-50 2002 Shell 15-60 2007

BPChembel

15-30 2004 Solvay 20-80 2001

ow - o a -

ExxonMobil

15-60 1985 Yara 30-60 2005

GSK 15-20 2004


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Relief Valve Study

An Engineering Approach


P&IDs

Equipment data

Etc.

Define relief scenarios: E.g.: External fire, Blocked outlet, etc.

Use list API 521 as guidance

Use tools as HAZOP, PLANOP, client specific methodsto determine applicable scenarios


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Relief load

Relief valve orifice size

Determine governing case General a roach:

Scenario requiring the largest orifice size

=

Governing case


Pressure drop over inlet (< 3% of set pressure)

Pressure at outlet (backpressure):

Superimposed backpressure: static pressure (if variable:NOconventional type valve)

Built-up backpressure: pressure increase as result of relief, .

100% for pilot operated type valves)


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Conventional spring-loaded

Balanced bellows

Pilot operated

Mechanical stress anal sis

Flare network study

Relief Calculation forSupercritical Fluids


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Introduction

Objective:Calculate mass relief flow, volume relief flow and required orificesize of heat-input driven relief cases on systems with supercriticalrelief temperature and/or pressure.

Examples:

Fire case for a Vessel

Blocked-in Heat Exchanger

References:

R. Ouderkirk, Rigorously Size Relief Valves for Supercritical Fluids,CEP magazine, pp. 34-43 (Aug. 2002).

L. L. Simpson, Estimate Two-Phase Flow in Safety Devices, Chem.Eng., pp. 98-102, (Aug. 1991).

Theoretical Background

Definition of enthal :H = U + pV (1)

dH = dU + Vdp + pdV (2)

dU = Q pdV (3)

Combining (2) & (3)

dH = Q + Vdp (4)

p is constant during relief; hence,

H = Q (5)

And,

H/t = Q (6)


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Theoretical Background

Heat input = Enthalpy change

Hi (H)p Hi+1

t * Q

i i+1

V/t H: Specific enthalpyV: Specific volumeQ: Heat inputt: Time

Example Case Information

Fire case for a Vessel SP

Process Data (normal operation):

Content: Methane

Crit. Temp. -82,7 C Crit. Press. 45,96 bara

Level: 60% Liquid

Pressure: 10 barg

arg

Temperature: -122 C

Volume: 10 m

Area: 25 mQ

fire


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Example Case Relief Process Overview

1 2 Heatin before Relief: Isochoric rocess

No volume or mass change (no relief)

2 3 Relief: Isentropic flash

Adiabatic & frictionless flow through relief valve

2 2 Relief Progression: Isobaric processSystem at constant pressure (i.e. relief pressure)

P-E Diagram of M ethane

100

= 10kg/m3

= 100kg/m3

2 2'

+ Qfire

+ Qfire

Density [kg/m] - Temperature [K] - Entropy [kJ/(kgK)]

= 400kg/m3

Relief

Press.

10

Pressure

(ba

r)

= 1kg/m3

1

3 3'

0.1

-100 100 300 500 700 900 1100 1300 1500

Enthalpy (kJ/ kg)

= 0,1kg/m3

T=100K

T=200K

T=150K

T=300K

T=400K

T=500K


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Example Case Calculation Steps

Ste 1: Select Pro ert Method

Step 2: GatherRelief Case Information

Step 3: DetermineHeat Input

Step 4: CalculatePhysical Properties

Step 5: CalculateRelief Flow Rate

Step 6a: DetermineIsentropic Choked Nozzle Flux

Step 6b: DetermineRequired Orifice Size

Example Case Step 1

Requirements: Suitable for respective component(s)

Accurate for the relevant pressure and temperature rangeP > 1 // T > 1

Accurate for both liquid and gas properties

Important:Always verify property method with empirical property data!


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Example Case Step 1

Fit for light hydrocarbons

Application range

Pr: 0 to 10 (up to ca. 460 bara)

Tr: 0,3 to 4 (ca. -216 to 485 C)

One correlation for both liquid as well as vapor phase

No distinguishable transition from supercritical fluid tosupercritical vapor

Integration of the thermal properties with the otherphysical properties

Thermodynamic cohesiveness

Example Case Step 2

GatherRelief Case Information

Relief pressure:

PSV set press.: 50 bargFire case relief press.: 121 %of set pressure

Relief press.: 61,5 bara (Pr= 1,3)

n a re e empera ure:Considering an isochoric process:

(Tini(pini))ini (Trlf(prlf))ini(Tini(10barg))ini (Trlf(61,5barg))ini-122C -77C


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Example Case Step 3

DetermineHeat In ut

API 521 external pool fire, heat absorption forliquids:

Qfire = 43.200 * f * 0,82With f = 1 (no fireproof insulation / bare metal vessel)

= 25 m

Qfire = 605,05 kW

= 2.178.196 kJ/h: Wetted surface [m]f: Environment factor[-]Q: Heat input [W]

Example Case Step 4

CalculatePhysical Properties

Determine the specific volume (V), specific enthalpy (H) & entropy (S)at initial relief conditions:

Applying property method correlations in Excel spreadsheets

Using property models in Simulation Tools (Pro/II, Aspen Plus, etc.)

At relief pressure

Step size: ca. 3C

# iterations: see later


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P-E Diagram Methane

100

= 10kg/m3

= 100kg/m3

2 2'

+ Qfire

+ Qfire


= 400kg/m3

10

Pressure

(bar)

= 1kg/m3

1

33'

0.1

-100 100 300 500 700 900 1100 1300 1500

Etnhalpy (kJ/ kg)

= 0,1kg/m3

T=100K

T=200K

T=150K

T=300K

T=400K

T=500K

Example Case Step 4

T, C S, kJ/(kg.K) H, kJ/kg V, m3/kg

-77 8,742 -288,7 0,00455

-74 8,920 -253,7 0,00527

-71 9,169 -203,7 0,00662

-68 9,341 -168,7 0,00781-65 9,487 -138,7 0,00896

-62 9,582 -118,7 0,00978

-59 9,676 -98,7 0,01062

-56 9 746 -83 7 0 01127, , ,

-53 9,814 -68,7 0,01193

-50 9,882 -53,7 0,01259

-47 9,927 -43,7 0,01303

-41 10,036 -18,7 0,01414

-38 10,079 -8,7 0,01459


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Example Case Step 5

CalculateRelief Flow Rate

Volumetric flow rate:

Mass flow rate:

H

VQV

= &&

Vm

&

& =

H: Specific enthalpy [kJ/kg]V: Specific volume [m/kg]V: Volume flow [m/s]m: Mass [kg]m: Mass flow [kg/s]Q: Heat input [kW]

Example Case Step 5

T, C S, kJ/(kg.K) H, kJ/kg V, m3/kg V, m3/s m, kg/s

- , - , , , ,

-74 8,920 -253,7 0,00527 0,01427 2,710

-71 9,169 -203,7 0,00662 0,01916 2,891 Max. mass flow

-68 9,341 -168,7 0,00781 0,02227 2,849

-65 9,487 -138,7 0,00896 0,02432 2,714

-62 9,582 -118,7 0,00978 0,02532 2,588

-59 9,676 -98,7 0,01062 0,02602 2,448

- -- , - , , , ,

-53 9,814 -68,7 0,01193 0,02662 2,232

-50 9,882 -53,7 0,01259 0,02674 2,124

-47 9,927 -43,7 0,01303 0,02687 2,061 Max. volume flow

-41 10,036 -18,7 0,01414 0,02686 1,899

-38 10,079 -8,7 0,01459 - -


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Example Case Step 6

DetermineIsentro ic Choked Nozzle Flux

For each relief temperature calculate the chokednozzle flux:

Iteratively, at decreasing

outlet pressure:( )

b

b0

V

HH2G

=

And, along isentropic path:

Max. flux = Choked flux

b0 SS =

H: Specific enthalpy [J/kg]V: Specific volume [m/kg]G: Mass flux [kg/(m.s)]S: Entropy [kJ/(kg.K)]

0: Inlet condition

b: Outlet condition

P-E Diagram Methane

100

= 10kg/m3

= 100kg/m3

2 2'

+ Qfire

+ Qfire


= 400kg/m3

10

Pressure

(ba

r)

= 1kg/m3

1

33'

0.1

-100 100 300 500 700 900 1100 1300 1500

Etnhalpy (kJ/ kg)

= 0,1kg/m3

T=100K

T=200K

T=150K

T=300K

T=400K

T=500K


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Example Case Step 6

Relief tem erature: -68 C

Tb, C pb, bara Vb, m/kg Hb, kJ/kg G, kg/(m.s)

T0, p0: -68 61,5 0,00808 -158,8 -

-72 57,0 0,00840 -162,5 10248

-76 52,5 0,00878 -166,4 14009

-80 48,0 0,00924 -170,4 16496

:

-85 43,5 0,00988 -174,7 18058 GChoked-88 39,0 0,01134 -179,5 17931

-92 34,5 0,01309 -185,0 17479

Example Case Step 6

Iteration = time consumin rocess!!

Alternative method: use simplified correlations to

determine isentropic choked flux J.C. Leung, A Generalized Correlation for One-component

Homogeneous Equilibrium Flashing Choked Flow, AIChE Journal,pp. 1743-1746 (Oct. 1986).

0

0choked

V

pG

=


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ATTENTION: 2-phase flow

Relief of su ercritical fluids can lead to 2- hase flow!

Homogenous Equilibrium Model (HEM)

Assumptions

1. Velocities of phases are equal

2. Phases are at thermodynamic equilibrium

Formula applies:

And H = xL.HL + (1-xL).HGV = xL.VL + (1-xL).VG

( )b

b0

VHH2G =

H: Specific enthalpy [J/kg]V: Specific volume [m/kg]G: Mass flux [kg/(m.s)]

0: Inlet condition

b: Outlet condition

L: Liquid phase

G: Gas phase

Example Case Step 7

DetermineRe uired Orifice Size

API 521:

With backpressure correction, Kb = 1 (backpressure


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Example Case Step 7

T, C S, kJ/(kg.K) H, kJ/kg V, m3/kg V, m3/s m, kg/s A, mm

- -- , - , , , ,

-74 8,920 -253,7 0,00527 0,01427 2,710 -

-71 9,169 -203,7 0,00662 0,01916 2,891 153

-68 9,341 -168,7 0,00781 0,02227 2,849 155 Req. Nozzle Size

-65 9,487 -138,7 0,00896 0,02432 2,714 152

-62 9,582 -118,7 0,00978 0,02532 2,588 -

-59 9,676 -98,7 0,01062 0,02602 2,448 -

-56 9 746 -83 7 0 01127 0 02638 2 340 -, , , , ,

-53 9,814 -68,7 0,01193 0,02662 2,232 -

-50 9,882 -53,7 0,01259 0,02674 2,124 141

-47 9,927 -43,7 0,01303 0,02687 2,061 -

-41 10,036 -18,7 0,01414 0,02686 1,899 -

-38 10,079 -8,7 0,01459 - - -

Example Case Results

When all values (relief volume flow, mass flow and nozzle size)decrease with increasing relief temperature: stop iterations.

Determine selected effective orifice (API 526) based on maximum

calculated nozzle size value: Max. nozzle size value: 155 mm

Selected standard orifice: 198 mm (F)

a cu ate pressure rop over n et an sc arge

Determine safety valve type (conventional, balanced bellows, pilotoperated)


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Calculation Results

100%Volume Relief Rate

70%

80%

90%

Orifice Area

40%

50%

60%

200 210 220 230 240 250

Temperature (K)

Mass Relief Rate

Example Case Conclusions

S ecific calculation method is re uired:

Fluids that are below critical conditions in normal operation

can have super critical relief

Max. mass flow Max. volume flow Min. required nozzlesize

Required nozzle size determined using a simplified method(API 521 5.15.2.2.2): 58 mm vs. 155 mm

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