Sizing of relief valves for supercritical fluids

Sizing of relief valves for supercritical fluids

March 23rd, 2011

Alexis Torreele

Overview

� Jacobs – Introduction

�Relief Valve Study – An Engineering Approach

�Relief Calculation for Supercritical Fluids− Introduction− Theoretical Background− Example Case− Discussion & Evaluation

Jacobs

Introduction

Jacobs – Introduction: Who Are We

� Committed to BeyondZero® Safety as safety is our #1 priority

� Relationship based company� Global resource base – 57.500 employees in 25

countries on 4 continents� Fortune 500 #1 Engineering & Construction Company� Publicly 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

Jacobs – Introduction: Worldwide offices

Jacobs – Introduction: Europe

Jacobs – Introduction: BelgiumOil & Gas(Refining)

30

Others12

Chemicals & Polymers

45

Pharma& Bio

13

Process, 52Engineering &

Design, 316

Project Mgt., 48

G&A, 31

Constr. Mgt, 26

Project Serv. &Admin., 82

Procurement,14

Civil, 44Mechanical, 31Instrumentation, 88Piping, 127Electrical, 26

CAD/IT, 18

Jacobs – Introduction: Clients

Yara

Total

Solvay

Shell

SABIC

Client

30-60

40-80

20-80

15-60

15-60

Workload / People

2004

1985

2003

2004

2002

2003

Since

15-20GSK

200515-60ExxonMobil

200315-60Dow

200115-30BP Chembel

200725-50Borealis

200430-50BASF

SinceWorkload / People

Client

Relief Valve Study

An Engineering Approach

Relief Valve Study – An Engineering Approach

� Gather info:− P&ID’s− Equipment data − Etc.

� Define relief scenario’s:− E.g.: External fire, Blocked outlet, etc.− Use list API 521 as guidance− Use tools as HAZOP, PLANOP , client specific methods

to determine applicable scenarios


� Calculate relief scenario’s− Relief load− Relief valve orifice size

� Determine governing case− General approach:

Scenario requiring the largest orifice size =

Governing case


� Verify inlet and outlet conditions− Pressure drop over inlet (< 3% of set pressure)− Pressure at outlet (backpressure):

� Superimposed backpressure : static pressure (if variable: NO conventional type valve)

� Built-up backpressure : pressure increase as result of relief flow (< 10% for conventional, < ca. 50% for balanced & > 50% for pilot operated type valves)


� Determine safety valve type:− Conventional spring-loaded− Balanced bellows − Pilot operated

� Mechanical stress analysis

� Flare network study

Relief Calculation for Supercritical Fluids

Introduction

� Objective:Calculate mass relief flow , volume relief flow and required orifice size of heat-input driven relief cases on systems with supercritical relief 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 enthalpy:H = U + pV (1)

dH = dU + Vdp + pdV (2)

dU = δQ – pdV (3)

Combining (2) & (3)

dH = δQ + Vdp (4)

p is constant during relief; hence,

∆H = Q (5)

And,

∆∆∆∆H/∆∆∆∆t = Q (6)

Theoretical Background

� Heat input = Enthalpy change

Hi (∆H)p Hi+1

∆t * Q

Vi ∆t Vi+1

∆∆∆∆V////∆∆∆∆tH: Specific enthalpyV: Specific volumeQ: Heat inputt: Time

Example Case – Information

� Fire case for a Vessel� Process Data (normal operation):

− Content: Methane� Crit. Temp. -82,7 °C� Crit. Press. 45,96 bara

− Level: 60% Liquid− Pressure: 10 barg− Temperature: -122 °C− Volume: 10 m³− Area: 25 m²

Qfire

SP

50barg

Example Case – Relief Process Overview

� 1 → 2 Heating before Relief: ‘Isochoric’ processNo volume or mass change (no relief)

� 2 → 3 Relief: Isentropic flashAdiabatic & frictionless flow through relief valve

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

P-E Diagram of Methane

0.1

1

10

100

-100 100 300 500 700 900 1100 1300 1500

Enthalpy (kJ/kg)

Pressure (bar)

δ = 1kg/m3

δ = 0,1kg/m3

δ = 10kg/m3

δ = 100kg/m3

T = 10

0K

T = 20

0K

T = 15

0K

T = 30

0K

T = 40

0K

T = 50

0K

1

2 2'

3 3'

+ Qfire

+ Qfire

Density [kg/m³] - Temperature [K] - Entropy [kJ/(kgK)]

δ = 400kg/m3

Relief Press.

Example Case – Calculation Steps

� Step 1: Select Property Method� Step 2: Gather Relief Case Information� Step 3: Determine Heat Input� Step 4: Calculate Physical Properties� Step 5: Calculate Relief Flow Rate� Step 6: Determine Isentropic Choked Nozzle Flux� Step 7: Determine Required Orifice Size

Example Case – Step 1

Select Property Method

� Requirements:− Suitable for respective component(s)− Accurate for the relevant pressure and temperature range

(Pr > 1 // Tr > 1)− Accurate for both liquid and gas properties

� Important: Always verify property method with empirical property data!


� Selected Method: Lee Kesler− 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 ‘liquid’ to

supercritical ‘vapor’

− Integration of the thermal properties with the other physical properties→ Thermodynamic cohesiveness


Gather Relief Case Information

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

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

� Initial relief temperature:Considering an isochoric process:

(Tini(pini))ρini → (Trlf (prlf))ρini

(Tini(10barg))ρini → (Trlf(61,5barg))ρini

-122°C → -77°C


Determine Heat Input

� API 521 – external pool fire, heat absorption for liquids:

Qfire = 43.200 * f * αααα0,82

With f = 1 (no fireproof insulation / bare metal vessel)

α = 25 m²

Qfire = 605,05 kW= 2.178.196 kJ/h

αααα: Wetted surface [m²]f: Environment factor [-]Q: Heat input [W]


Calculate Physical 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.)

� Reiterate at increasing temperatures:− At relief pressure− Step size: ca. 3°C− # iterations: see later

P-E Diagram Methane

0.1

1

10

100

-100 100 300 500 700 900 1100 1300 1500

Etnhalpy (kJ/kg)

Pressure (bar)

δ = 1kg/m3

δ = 0,1kg/m3

δ = 10kg/m3

δ = 100kg/m3

T =

100K

T =

200K

T =

150K

T =

300K

T =

400K

T =

500K

1

2 2'

3 3'

+ Qfire

+ Qfire


δ = 400kg/m3


0,01459-8,710,079-38

0,01414-18,710,036-41

0,01303-43,79,927-47

0,01259-53,79,882-50

0,01193-68,79,814-53

0,01127-83,79,746-56

0,01062-98,79,676-59

0,00978-118,79,582-62

0,00896-138,79,487-65

0,00781-168,79,341-68

0,00662-203,79,169-71

0,00527-253,78,920-74

0,00455-288,78,742-77

V, m3/kgH, kJ/kgS, kJ/(kg.K)T, °C


Calculate Relief Flow Rate

� Volumetric flow rate:

� Mass flow rate:

HV

QV∆∆= &&

VV

m&

& =

H: Specific enthalpy [kJ/kg]V: Specific volume [m³/kg]V: Volume flow [m³/s]m: Mass [kg]m: Mass flow [kg/s]Q: Heat input [kW]


-

1,899

2,061

2,124

2,232

2,340

2,448

2,588

2,714

2,849

2,891

2,710

2,389

m, kg/s

0,01459

0,01414

0,01303

0,01259

0,01193

0,01127

0,01062

0,00978

0,00896

0,00781

0,00662

0,00527

0,00455

V, m3/kg

-

0,02686

0,02687

0,02674

0,02662

0,02638

0,02602

0,02532

0,02432

0,02227

0,01916

0,01427

0,01088

V, m3/s

-8,710,079-38

-18,710,036-41

Max. volume flow-43,79,927-47

-53,79,882-50

-68,79,814-53

-83,79,746-56

-98,79,676-59

-118,79,582-62

-138,79,487-65

-168,79,341-68

Max. mass flow-203,79,169-71

-253,78,920-74

-288,78,742-77

H, kJ/kgS, kJ/(kg.K)T, °C


Determine Isentropic Choked Nozzle Flux

� For ‘each’ relief temperature calculate the chokednozzle flux:− Iteratively, at decreasing

outlet pressure:

− And, along isentropic path:

− Max. flux = Choked flux

( )b

b0

V

HH2G

−=

b0 SS =H: Specific enthalpy [J/kg]V: Specific volume [m³/kg]G: Mass flux [kg/(m².s)]S: Entropy [kJ/(kg.K)]

0: Inlet condition

b: Outlet condition

P-E Diagram Methane

0.1

1

10

100

-100 100 300 500 700 900 1100 1300 1500

Etnhalpy (kJ/kg)

Pressure (bar)

δ = 1kg/m3

δ = 0,1kg/m3

δ = 10kg/m3

δ = 100kg/m3

T = 1

00K

T = 2

00K

T = 1

50K

T = 3

00K

T = 4

00K

T = 5

00K

1

2 2'

3 3'

+ Qfire

+ Qfire


δ = 400kg/m3


� Relief temperature: -68 °C

17479

17931

18058

16496

14009

10248

-

G, kg/(m².s)T0, p0:

-185,00,0130934,5-92

-179,50,0113439,0-88: GChoked

-174,70,0098843,5-85

-170,40,0092448,0-80

-166,40,0087852,5-76

-162,50,0084057,0-72

-158,80,0080861,5-68

Hb, kJ/kgVb, m³/kgpb, baraTb, °C


� Iteration = time consuming process!!

� 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

⋅=

ωη

ATTENTION: 2-phase flow

� Relief of supercritical fluids can lead to 2-phase flow!

� Homogenous Equilibrium Model (HEM)Assumptions1. Velocities of phases are equal2. Phases are at thermodynamic equilibrium

� Formula applies:

And H = xL.HL + (1-xL).HG

V = xL.VL + (1-xL).VG

( )b

b0

V

HH2G

−=

H: Specific enthalpy [J/kg]V: Specific volume [m³/kg]G: Mass flux [kg/(m².s)]

0: Inlet condition

b: Outlet condition

L: Liquid phase

G: Gas phase


Determine Required Orifice Size

• API 521:

With backpressure correction, Kb = 1 (backpressure << 10%)

combination correction, Kc = 1 (no rupture disk)

discharge coefficient, Kd = 0,975 (assuming vapor)

viscosity correction, Kv = 1

vdcbchoked KKKKGm

A&

=

A: Effective orifice area [m²]m: Mass flow [kg/s]Gchoked : Choked mass flux [kg/(m².s)]


-

1,899

2,061

2,124

2,232

2,340

2,448

2,588

2,714

2,849

2,891

2,710

2,389

m, kg/s

-

-

-

141

-

-

-

-

152

155

153

-

96

A, mm²

0,01459

0,01414

0,01303

0,01259

0,01193

0,01127

0,01062

0,00978

0,00896

0,00781

0,00662

0,00527

0,00455

V, m3/kg

-

0,02686

0,02687

0,02674

0,02662

0,02638

0,02602

0,02532

0,02432

0,02227

0,01916

0,01427

0,01088

V, m3/s

-8,710,079-38

-18,710,036-41

-43,79,927-47

-53,79,882-50

-68,79,814-53

-83,79,746-56

-98,79,676-59

-118,79,582-62

-138,79,487-65

Req. Nozzle Size-168,79,341-68

-203,79,169-71

-253,78,920-74

-288,78,742-77

H, kJ/kgS, kJ/(kg.K)T, °C

Calculation Results

40%

50%

60%

70%

80%

90%

100%

200 210 220 230 240 250

Temperature (K)

Orifice Area

Volume Relief Rate

Mass Relief Rate

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 mm²− Selected standard orifice: 198 mm² (‘F’ - orifice)

� Calculate pressure drop over inlet and discharge

� Determine safety valve type (conventional, balanced bellows, pilot operated…)

� …

Example Case – Conclusions

� Specific calculation method is required:

− Fluids that are below critical conditions in normal operation can have super critical relief

− Max. mass flow ≠ Max. volume flow ≠ Min. required nozzle size

− Required nozzle size determined using a simplified method (API 521 §5.15.2.2.2): 254 mm² vs. 155 mm²

Extra Slides

Safety Valve Types

Bellows

Pilot

Conventional BalancedBellows

Pilot Operated

General flux equation

( )( )

( )

−+⋅

−+

−+−=

∫

t

2

2

fg

P

Pfg

2

x1xSv)x1(S

xv

dpv)x1xv2

G

t

r