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
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
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)
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
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]
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
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)]
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
