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
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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
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
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)
Relief Valve Study – An Engineering Approach 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 dH dU
= = =
U + pV dU + Vdp + pdV δQ – pdV
(1) (2) (3)
Combining (2) & (3)
dH
=
δQ + Vdp
(4)
p is constant during relief; hence,
∆H
=
Q
=
Q
(5)
And,
∆H/∆ ∆t
(6)
Theoretical Background Heat input Hi
=
Enthalpy change
(∆H)p
Hi+1
∆t * Q Vi
∆t
∆V//∆t
Vi+1
H: Specific enthalpy V: Specific volume Q: Heat input t: Time
Example Case – Information Fire case for a Vessel Process Data (normal operation): − Content: − − − − −
SP 50barg
Methane
Crit. Temp. Crit. Press.
-82,7 45,96
°C bara
Level: Pressure: Temperature: Volume: Area:
60% 10 -122 10 25
Liquid barg °C m³ m²
Qfire
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)
P-E Diagram of Methane Density [kg/m³] - Temperature [K] - Entropy [kJ/(kgK)]
100
δ = 400kg/m
3
δ = 100kg/m 2
Relief Press.
3
2'
+ Qfire
+ Qfire
δ = 10kg/m
3
δ = 1kg/m
3
δ = 0,1kg/m
3
1
Pressure (bar)
10
3
3'
1
Enthalpy (kJ/kg)
900
1100
T = 500K
700
T = 400K
500
T = 300K
300
T = 200K
100
T = 150K
T = 100K
0.1 -100
1300
1500
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!
Example Case – Step 1 Selected Method: Lee Kesler − Fit for light hydrocarbons − Application range Pr : 0 to 10 Tr : 0,3 to 4
(up to ca. 460 bara) (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
Example Case – Step 2 Gather Relief Case Information Relief pressure: PSV set press.: Fire case relief press.:
50 121
barg % of set pressure
Relief press.:
61,5
bara
Initial relief temperature: Considering an isochoric process: → (Trlf (prlf))ρini (Tini(pini))ρini (Tini(10barg))ρini → (Trlf(61,5barg))ρini -122°C → -77°C
(Pr = 1,3)
Example Case – Step 3 Determine Heat Input API 521 – external pool fire, heat absorption for liquids: = 43.200 * f * α0,82 Qfire With f = 1 (no fireproof insulation / bare metal vessel) α = 25 m² Qfire
= =
605,05 kW 2.178.196 kJ/h α: Wetted surface [m²] f: Environment factor [-] Q: Heat input [W]
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
P-E Diagram Methane Density [kg/m³] - Temperature [K] - Entropy [kJ/(kgK)]
100
δ = 400kg/m
3
δ = 100kg/m
3
2
2'
+ Qfire
δ = 10kg/m
+ Qfire
3
1
Pressure (bar)
10
δ = 1kg/m
3
δ = 0,1kg/m
3
3'
3
1
900
1100
T = 500K
700 Etnhalpy (kJ/kg)
T = 400K
500
T = 300K
300
T = 200K
100
T = 150K
T = 100K
0.1 -100
1300
1500
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
Example Case – Step 5 Calculate Relief Flow Rate Volumetric flow rate:
∆V & & V=Q ∆H
Mass flow rate:
& V & = m V 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]
Example Case – Step 5 T, °C
S, kJ/(kg.K) H, kJ/kg V, m3/kg V, m3/s m, kg/s
-77
8,742
-288,7
0,00455
0,01088
2,389
-74
8,920
-253,7
0,00527
0,01427
2,710
-71
9,169
-203,7
0,00662
0,01916
2,891
-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
-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
-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
-
-
Max. mass flow
Max. volume flow
Example Case – Step 6 Determine Isentropic Choked Nozzle Flux
For ‘each’ relief temperature calculate the choked nozzle flux: − Iteratively, at decreasing 2(H0 − Hb ) outlet pressure:
G=
Vb
− And, along isentropic path: S0 = Sb − Max. flux = Choked flux
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 Density [kg/m³] - Temperature [K] - Entropy [kJ/(kgK)]
100
δ = 400kg/m
3
δ = 100kg/m
3
2
2'
+ Qfire
δ = 10kg/m
+ Qfire
3
1
Pressure (bar)
10
δ = 1kg/m
3
δ = 0,1kg/m
3
3'
3
1
900
1100
T = 500K
700 Etnhalpy (kJ/kg)
T = 400K
500
T = 300K
300
T = 200K
100
T = 150K
T = 100K
0.1 -100
1300
1500
Example Case – Step 6 Relief temperature: -68 °C Tb, °C T0, p0:
pb, bara Vb, m³/kg Hb, kJ/kg G, kg/(m².s)
-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
-88
39,0
0,01134
-179,5
17931
-92
34,5
0,01309
-185,0
17479
: GChoked
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).
− Gchoked = η
p0 ω ⋅ V0
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: G = And
2(H0 − Hb ) Vb
H = xL.HL + (1-xL).HG V = xL.VL + (1-xL).VG
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
Example Case – Step 7 Determine Required Orifice Size •
API 521: & m A= GchokedK bK cK dK v
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 A: Effective orifice area [m²] m: Mass flow [kg/s] Gchoked: Choked mass flux [kg/(m².s)]
Example Case – Step 7 T, °C
S, kJ/(kg.K) H, kJ/kg V, m3/kg V, m3/s m, kg/s A, mm²
-77
8,742
-288,7
0,00455
0,01088
2,389
96
-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
-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
-
-
-
Req. Nozzle Size
Calculation Results
Volume Relief Rate 100%
90%
Orifice Area 80%
70%
M ass Relief Rate
60%
50%
40% 200
210
220
230 Temperature (K)
240
250
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
Pilot
Bellows
Conventional
Balanced Bellows
Pilot Operated
General flux equation
Pt ( ) ( ) − + − 2 xv 1 x ) v dp g f ∫P 2 r G = 2 xv g 2 + − ⋅ + − ( 1 x ) v xS 1 x f S t
(
)
