The document provides an overview of Jacobs, an engineering company, and discusses their approach to sizing relief valves for supercritical fluids. It then presents a case study example of calculating the relief requirements for a vessel containing methane undergoing an external fire. The key steps involve: (1) gathering process data; (2) determining heat input from the fire; (3) calculating fluid properties as temperature increases; (4) determining mass and volume relief rates; (5) calculating choked flow rates; and (6) sizing the required relief valve orifice. The example demonstrates that relief of supercritical fluids can involve complex two-phase flow that requires specialized modeling approaches.

1. Sizing of relief valves for
supercritical fluids
March 23rd, 2011
Alexis Torreele

2. Overview
Jacobs – Introduction
Relief Valve Study – An Engineering Approach
Relief Calculation for Supercritical Fluids
− Introduction
− Theoretical Background
− Example Case
− Discussion & Evaluation

3. Jacobs
Introduction

4. 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

5. Jacobs – Introduction: Worldwide offices

6. Jacobs – Introduction: Europe

7. Jacobs – Introduction: Belgium
Oil & Gas
(Refining)
30
Others
12
Chemicals &
Polymers
45
Pharma
& Bio
13
Process, 52
Engineering &
Design, 316
Project Mgt., 48
G&A, 31
Constr. Mgt, 26
Project Serv. &
Admin., 82
Procurement,
14
Civil, 44
Mechanical, 31
Instrumentation,
88
Piping, 127
Electrical, 26
CAD/IT, 18

8. 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
200515-60ExxonMob
il
200315-60Dow
200115-30BP
Chembel
200725-50Borealis
200430-50BASF
SinceWorkload
/ People
Client

9. Relief Valve Study
An Engineering Approach

10. 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

11. Relief Valve Study – An Engineering Approach
Calculate relief scenario’s
− Relief load
− Relief valve orifice size
Determine governing case
− General approach:
Scenario requiring the largest orifice size
=
Governing case

12. Relief Valve Study – An Engineering Approach
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)

13. Relief Valve Study – An Engineering Approach
Determine safety valve type:
− Conventional spring-loaded
− Balanced bellows
− Pilot operated
Mechanical stress analysis
Flare network study

14. Relief Calculation for
Supercritical Fluids

15. 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).

16. 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)

17. Theoretical Background
Heat input = Enthalpy change
Hi (∆H)p Hi+1
∆t * Q
Vi ∆t Vi+1
∆∆∆∆V////∆∆∆∆t
H: Specific enthalpy
V: Specific volume
Q: Heat input
t: Time

18. 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

19. Example Case – Relief Process Overview
1 → 2 Heating before Relief: ‘Isochoric’ process
No volume or mass change (no relief)
2 → 3 Relief: Isentropic flash
Adiabatic & frictionless flow through relief valve
2 → 2’ Relief Progression: Isobaric process
System at constant pressure (i.e. relief pressure)

20. 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=100K
T=200K
T=150K
T=300K
T=400K
T=500K
1
2 2'
3 3'
+ Qfire
+ Qfire
Density [kg/m³] - Temperature [K] - Entropy [kJ/(kgK)]
δ = 400kg/m3
Relief
Press.

21. 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

22. 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!

23. Example Case – Step 1
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

24. Example Case – Step 2
Gather Relief Case Information
Relief pressure:
PSV set press.: 50 barg
Fire 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

25. Example Case – Step 3
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]

26. Example Case – Step 4
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

27. 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
Density [kg/m³] - Temperature [K] - Entropy [kJ/(kgK)]
δ = 400kg/m3

28. Example Case – Step 4
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

29. Example Case – Step 5
Calculate Relief Flow Rate
Volumetric flow rate:
Mass flow rate:
H
V
QV
∆
∆
= &&
V
V
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]

30. Example Case – Step 5
-
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

31. Example Case – Step 6
Determine Isentropic Choked Nozzle Flux
For ‘each’ relief temperature calculate the choked
nozzle flux:
− Iteratively, at decreasing
outlet pressure:
− And, along isentropic path:
− Max. flux = Choked flux
( )
b
b0
V
HH2
G
−
=
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

32. 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
Density [kg/m³] - Temperature [K] - Entropy [kJ/(kgK)]
δ = 400kg/m3

33. Example Case – Step 6
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

34. Example Case – Step 6
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
0
choked
V
p
G
⋅
=
ω
η

35. ATTENTION: 2-phase flow
Relief of supercritical fluids can lead to 2-phase 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).HG
V = xL.VL + (1-xL).VG
( )
b
b0
V
HH2
G
−
= 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

36. Example Case – Step 7
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 KKKKG
m
A
&
=
A: Effective orifice area [m²]
m: Mass flow [kg/s]
Gchoked: Choked mass flux [kg/(m².s)]

37. Example Case – Step 7
-
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

38. 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

39. 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…)
…

40. 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

41. Extra Slides

42. Safety Valve Types
Bellows
Pilot
Conventional Balanced
Bellows
Pilot Operated

43. General flux equation
( )( )
( ) 























−+⋅





−+
−+−
=
∫
t
2
2
f
g
P
P
fg
2
x1xSv)x1(
S
xv
dpv)x1xv2
G
t
r