WHITE PAPER
Equations and Example Benchmark Calculations within
the aspenONE® Engineering Safety Analysis Environment
V8.8: Two-phase Orifice Sizing, Vapor Noise and Reaction Forces
Craig Powers, Principal Software Developer, Aspen Technology, Inc.
Introduction
Introduced in Aspen HYSYS® V8.3, the Safety Analysis Environment provides a tool for adding pressure relief devices
and calculating relief loads inside Aspen HYSYS. Leveraging this tool within the rigorous Aspen HYSYS simulator, and in
combination with Aspen Flare System Analyzer, provides an integrated solution for pressure relief analysis (PRA) work.
This white paper contains hand calculations for two-phase orifice sizing and vapor noise and reaction forces inside the
Safety Analysis Environment, helping you to validate the calculations of this tool within Aspen HYSYS.
Leung Omega Method
There are two forms of the Leung Omega Method that are implemented in the Safety Analysis Environment. One form
is for cases that are saturated or two-phase at relief conditions, upstream of the relief valve. The other form is for cases
that are subcooled liquid upstream of the relief valve, with flashing occurring across the valve that produces a twophase stream at the valve outlet.
Non-Subcooled Omega Method
Equations
The governing equations for the application of the Omega Method are obtained from Appendix C of API Standard 520
Part I.1 It is a two-point method, meaning that physical properties are required at two thermodynamic state points. For
the non-subcooled method, the first state point is relief conditions. The second state point is obtained by an isentropic
flash to 90% of the relief pressure. Aspen HYSYS® uses an isenthalpic flash to 90% of the relief pressure in place of an
isentropic flash. With this information, the required orifice area may be obtained using the following equations below.
Eq. 1
𝜔𝜔 = 9×
𝑣𝑣!
𝜌𝜌!
− 1 = 9×
− 1 𝑣𝑣!
𝜌𝜌!
In the equation above, ω is the Omega parameter, v0 and v9 are specific volumes at upstream and 90% conditions,
respectively, and ρ0 and ρ9 are density at upstream and 90% conditions, respectively.
Eq. 2
Eq. 3
𝜂𝜂! =
𝑃𝑃!
≅ 1 + 1.0446 − 0.0093431 𝜔𝜔 ×𝜔𝜔 !!.!"#"$
𝑃𝑃!
!!.!"#$%!!.!"#$%& !" !
𝜂𝜂! = 𝑃𝑃! 𝑃𝑃! In the equation above, ηc is the critical pressure ratio, Pc is the pressure corresponding to critical flow
at the nozzle throat in the relief valve in absolute units, Pa is the total back pressure at the outlet of
the valve in absolute units, and P0 is the upstream relief pressure in absolute units.
If ηc exceeds ηa, then the flow through the relief valve is critical, and the following equation should be
used to calculate the maximum mass flux, where G is the maximum mass flux through the valve in
lb/s-ft² or kg/s-m², P0 is in psia or Pa, v0 is in ft³/lb or m³/kg, and ρ0 is in lb/ft³ or kg/m³. C1 = 68.09
in U.S. customary units, or 1.0 in SI units.
2
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Eq. 4
𝐺𝐺 = 𝐶𝐶! 𝜂𝜂!
𝑃𝑃!
𝑃𝑃! 𝜌𝜌!
= 𝜂𝜂!
𝑣𝑣! 𝜔𝜔
𝜔𝜔
If ηc does not exceed ηa, then the flow through the relief valve is subcritical, and the following
equation should be used to calculate the maximum mass flux, where units are as in (Eq. 4).
Eq. 5
𝐺𝐺 = 𝐶𝐶!
𝑃𝑃! 𝜌𝜌!
1
𝜔𝜔
−1 +1
𝜂𝜂!
−2 𝜔𝜔 ln 𝜂𝜂! + 𝜔𝜔 − 1
1 − 𝜂𝜂! Once the maximum mass flux has been obtained, the required effective discharge area may be
calculated by using Equation 6, where A is the required effective discharge area in mm², w is the
required relief load in kg/h, Kd is the discharge coefficient of the valve, Kb is the backpressure
correction factor (applicable to balanced bellows valves only), Kc is the combination capacity factor,
and Kv is the viscosity correction factor. C2 = 0.04 for U.S. customary units, 277.8 for SI units.
Eq. 6
𝐴𝐴 =
𝐶𝐶! 𝑤𝑤
𝐾𝐾! 𝐾𝐾! 𝐾𝐾! 𝐾𝐾! 𝐺𝐺
When an orifice area has been selected, Equation 6 may be rearranged to find the rated capacity of
the valve, shown in Equation 7:
Eq. 7
𝑤𝑤!"# = 𝐴𝐴 𝐾𝐾! 𝐾𝐾! 𝐾𝐾! 𝐾𝐾! 𝐺𝐺 277.8 Example with Critical Flow
The example is based on the following conditions:
Composition
15% propane, 25% n-butane, 30% n-pentane,
30% n-heptane using the Aspen HYSYS PR
package for physical properties
Relief conditions
8.03 barg / 101.2 C
Required relief load 63,000 kg/h
63,000 kg/h
Based on the relief pressure of 8.03 barg, the 90% intermediate flash pressure is at 7.227 barg.
Setting up a stream in Aspen HYSYS at the relief conditions and running it through an isenthalpic
flash will yield the following properties:
ρ0 = 112.0 kg/m³
ρ9 = 90.73 kg/m³
3
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The Omega parameter may be calculated using (Eq. 1):
112.0
− 1 = 2.110 90.73
𝜔𝜔 = 9×
The critical pressure ratio and backpressure ratio may be calculated using (Eq. 2) and (Eq. 3):
𝜂𝜂! =
𝜂𝜂! = 0.6989 1.01325
= 0.1120 8.03 + 1.01325
Figure 1: Two-phase critical flow case calculated in Aspen HYSYS
Because ηc is larger than ηa, flow across the relief valve is choked; therefore, the maximum mass flux
is calculated using (Eq. 4).
𝐺𝐺 = 0.6989
112 9.043 ×100,000 𝑃𝑃𝑃𝑃/𝑏𝑏𝑏𝑏𝑏𝑏
= 4,844 kg/s-­‐m² 2.110
Assuming that all capacity coefficients are 1.0, aside from the discharge coefficient which is 0.85 in
accordance with guidance from API, the required area is calculated using (Eq. 6):
𝐴𝐴 =
4
277.8 63,000
= 4,250 mm² 0.85 4,844 Equations and Example Benchmark Calculations within the
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For a selected Q orifice with an effective area of 7129 mm², the rated capacity is calculated using
(Eq. 7):
𝑤𝑤!"# =
0.85 4,844 7,129
= 105,700 kg/h 277.8
The results calculated above are compared to results obtained in Aspen HYSYS in Table 1.
Variable
Units
Example Calculation
Aspen HYSYS
kg/h
63,000
63,000
Relief Pressure (P0)
barg
8.03
8.03
Mass Density at 90% (ρ9)
Required Relief Load (w)
90% Pressure (P9)
Mass Density at Relief (ρ0)
7.227
barg
kg/m
3
112.0
112.0
kg/m
3
90.73
88.34
Omega Parameter (ω)
Critical Flow Ratio (ηc)
2.110
0.6989
Flow Type
Maximum Mass Flux (G)
Required Area (A)
Rated Capacity (wmax)
Critical
kg/s-m
mm
2
2
kg/h
Blue = Calculation input
Critical
4,844
4,250
4,448
105,700
101,000
Gray = Calculated value
Table 1: Comparison of example calculation and Aspen HYSYS calculation for critical two-phase flow
Example with Subcritical Flow
The example is based on the following conditions:
Composition
15% propane, 25% n-butane, 30% n-pentane,
30% n-heptane using the Aspen HYSYS PR
package for physical properties
Relief conditions
154 kPag / 85 C
Required relief load
20,000 kg/h
Back pressure
60 kPa superimposed
Based on the relief pressure of 1.54 barg, the 90% intermediate flash pressure is at 1.28 barg. Setting
up a stream in Aspen HYSYS at the relief conditions and running it through an isenthalpic flash will
yield the following properties:
ρ0 = 8.264 kg/m³
ρ9 = 7.335 kg/m³
5
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The Omega parameter may be calculated using (Eq. 1):
8.264
− 1 = 1.140 7.335
𝜔𝜔 = 9×
The critical pressure ratio and backpressure ratio may be calculated using (Eq. 2) and (Eq. 3):
𝜂𝜂! =
𝜂𝜂! = 0.6234 1.01325 + 0.6
= 0.6318 1.54 + 1.01325
Figure 2: Two-phase subcritical flow case calculated in Aspen HYSYS
Because ηc is smaller than ηa, flow across the relief valve is unchoked, and the maximum mass flux is
calculated using (Eq. 5), yielding a maximum mass flux G = 830 kg/m²-s.
Assuming that all capacity coefficients are 1.0, aside from the discharge coefficient which is 0.85 in
accordance with guidance from API, the required area is calculated using (Eq. 6):
𝐴𝐴 =
277.8 20,000
= 7,870 mm² 0.85 830
For a selected R orifice with an effective area of 10322.5 mm², the rated capacity is calculated using
(Eq. 7):
𝑤𝑤!"# =
0.85 830 10322.5
= 26,230 kg/h 277.8
The results calculated above are compared to results obtained in Aspen HYSYS in Table 2.
6
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Variable
Units
Example Calculation
Aspen HYSYS
Required Relief Load (w)
kg/h
20,000
20,000
Relief Pressure (P0)
barg
1.54
1.54
Mass Density at 90% (ρ9)
kg/m3
90% Pressure (P9)
Mass Density at Relief (ρ0)
1.28
barg
kg/m
3
Omega Parameter (ω)
Critical Flow Ratio (ηc)
8.264
8.264
7.335
7.337
1.140
0.6234
Flow Type
Subcritical
Maximum Mass Flux (G)
Required Area (A)
Rated Capacity (wmax)
kg/s-m
mm
2
2
kg/h
Blue = Calculation input
Subcritical
830
7,870
7,710
26,230
26,780
Gray = Calculated value
Table 2: Comparison of example calculation and Aspen HYSYS calculation for subcritical two-phase flow
Subcooled Omega Method
Equations
As with the non-subcooled Omega Method, the subcooled Omega Method is a two-point method.
The first point remains the same with the conditions at the PRV inlet. For the second point, the relief
stream is flashed isentropically to 90% of its saturation pressure corresponding to the PRV inlet
temperature T0. As with the non-subcooled method, flow may be critical or subcritical; however,
the subcooled Omega Method also requires a determination as to whether the relief conditions
correspond to “high” or “low” subcooling; that is, whether the fluid will begin to flash in the inlet line,
or whether it will flash at the valve throat.
The omega parameter is calculated using a similar equation to (Eq. 1), where ρℓ0 is the liquid density
at the PRV inlet in kg/m³.
Eq. 8
𝜔𝜔! = 9×
𝜌𝜌ℓ𝓁𝓁"
− 1 𝜌𝜌!
“High” or “low” subcooling is determined via the transition saturation pressure ratio, which is
calculated below in Equation 9 and 10, where ηst is the transition saturation pressure ratio, ηs is
the saturation pressure ratio, Ps is the saturation pressure in absolute units, and P0 is the upstream
pressure in absolute units.
Eq. 9
Eq. 10
7
𝜂𝜂!" =
𝜂𝜂! =
2𝜔𝜔!
1 + 2𝜔𝜔!
𝑃𝑃!
𝑃𝑃!
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If ηst exceeds ηs, then relief is considered to lie in the “high” subcooling region, and flow will
necessarily be critical unless the total back pressure exceeds the saturation pressure (but in this
case, the fluid is liquid across the valve, and the API liquid sizing will be sufficient to calculate the
orifice size). Otherwise, relief is considered to lie in the “low” subcooling region, and a check must be
made to determine if the flow across the PRV is critical or subcritical. The critical pressure ratio may
be estimated as shown below in Equation 11.
Eq. 11
𝜂𝜂! ≅ 𝜂𝜂!
2𝜔𝜔!
2𝜔𝜔! − 1
1−
1−
1 2𝜔𝜔! − 1
𝜂𝜂!
2𝜔𝜔!
The back pressure ratio ηa is calculated using (Eq. 3) as with the non-subcooled method.
For “low” subcooling, the maximum mass flux is calculated using Equation 12 below.
Eq. 12
𝐺𝐺 = 𝐶𝐶!
𝜔𝜔!
𝜂𝜂!
𝑃𝑃! 𝜌𝜌ℓ𝓁𝓁"
2 1 − 𝜂𝜂! + 2 𝜔𝜔! 𝜂𝜂! ln − 𝜔𝜔! − 1 𝜂𝜂! − 𝜂𝜂 𝜂𝜂!
𝜂𝜂
−1 +1
𝜂𝜂
For critical flow, η=ηc; for subcritical flow, η=ηa. Pressure is in psia or Paa. C1 = 68.09 for U.S.
customary units, 1.0 for SI units.
For “high” subcooling, the maximum mass flux is calculated using Equation 13 below.
Eq. 13
𝐺𝐺 = 𝐶𝐶! 2 𝜌𝜌ℓ𝓁𝓁" 𝑃𝑃! − 𝑃𝑃! The pressure is measured in psia or Paa.
Given a required relief load in lb/h or kg/h, the required area and rated capacity may be calculated
using (Eq. 6) and (Eq. 7).
Example with Low Subcooling, Critical Flow
The example is based on the following conditions:
Composition
15% propane, 30% isobutane, 30% n-butane,
25% isopentane using the Aspen HYSYS PRSV
package for physical properties
Relief conditions
242 psig, 205 F
Required relief load
92,590 lb/h
Setting up a stream in Aspen HYSYS at the relief conditions and flashing appropriately, yields the
following properties:
ρℓ0 = 29.23 lb/ft³
Ps = 241.4 psia
P9 = 217.3 psia
ρ9 = 17.53 lb/ft³
8
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The omega parameter is calculated using (Eq. 8):
𝜔𝜔! = 9×
29.23
− 1 = 6.011 17.53
Figure 3: Low subcooled critical flow case calculated in Aspen HYSYS
The transition saturation pressure ratio is calculated using (Eq. 9):
𝜂𝜂!" =
2 6.011
= 0.923 1 + 2 6.011
The saturation pressure ratio is 0.956. Because this is greater than the transition ratio, this case
involves “low” subcooling, and the critical pressure ratio is calculated using (Eq. 11):
𝜂𝜂! ≅ 0.923
2 6.011
2 6.011 − 1
1−
1−
1
2 6.011 − 1
0.923
2 6.011
= 0.8633 The back pressure ratio of 0.0573 is much lower than the critical pressure ratio, so the flow is critical.
The maximum mass flux is calculated using (Eq. 12) with η=ηc. The result shows that G = 2,150
lb/s-ft².
Assuming that all capacity coefficients are 1.0, aside from the discharge coefficient which is 0.65 in
accordance with guidance from API, the required area is calculated using (Eq. 6):
𝐴𝐴 =
9
0.04 92,590
= 2.651 in² 0.65 2,150
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For a selected L orifice with an effective area of 2.853 in², the rated capacity is calculated using
(Eq. 7):
𝑤𝑤!"# =
0.65 2,150 2.853
= 99,640 lb/h 0.04
The results calculated above are compared to results obtained in Aspen HYSYS in Table 3.
Variable
Units
Example Calculation
Aspen HYSYS
Required Relief Load (w)
lb/h
92,590
92,590
Relief Pressure (P0)
psig
242
242.0
226.7 psig
241.4 psia
psig
202.6
Mass Density at Relief (ρℓ0)
lb/ft3
29.23
lb/ft
17.53
Saturation Pressure (Ps)
90% Pressure (P9)
Mass Density at 90% (ρ9)
3
Omega Parameter (ω)
6.011
Transition Subcool Ratio (ηst)
Critical Flow Ratio (ηc)
0.923
0.8633
Flow Type
Maximum Mass Flux (G)
Required Area (A)
Rated Capacity (wmax)
17.51
Low Subcool, Critical
lb/s-ft
in
2
2
lb/h
Blue = Calculation input
Critical
2,150
2.651
2.663
99,640
99,180
Gray = Calculated value
Table 3: Comparison of example calculation and Aspen HYSYS calculation for low subcooled critical flow
Example with Low Subcooling, Subcritical Flow
The example is based on the following conditions:
Composition
15% propane, 30% isobutane, 30% n-butane,
25% isopentane using the Aspen HYSYS PRSV
package for physical properties
Relief conditions
44 psig, 87.5 F
Required relief load
92,590 lb/h
Back pressure
40 psi superimposed
Setting up a stream in Aspen HYSYS at the relief conditions and flashing appropriately yields the
following properties:
ρℓ0 = 35.17 lb/ft³
Ps = 57.13 psia
P9 = 51.4 psia
ρ9 = 13.72 lb/ft³
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Figure 4: Low subcooled subcritical flow case calculated in Aspen HYSYS
The omega parameter is calculated
using (Eq. 8):
𝜔𝜔! = 9×
35.17
− 1 = 14.07 13.72
The transition saturation pressure ratio is calculated using (Eq. 9):
𝜂𝜂!! =
2 14.07
= 0.9657 1 + 2 14.07
The saturation pressure ratio is 0.9733. Because this is greater than the transition ratio, this case
involves “low” subcooling. The critical pressure ratio is calculated using (Eq. 11):
𝜂𝜂! ≅ 0.9733
2 14.07
2 14.07 − 1
1−
1−
1
2 14.07 − 1
0.9733
2 14.07
= 0.9130 The back pressure ratio of 0.9319 is higher than the critical pressure ratio, so the flow is subcritical.
The maximum mass flux is calculated using (Eq. 12) with η=ηa=0.9319. The result is that G = 765.4
lb/s-ft².
Assuming that all capacity coefficients are 1.0, aside from the discharge coefficient which is 0.65 in
accordance with guidance from API, the required area is calculated using (Eq. 6):
𝐴𝐴 =
11
0.04 92,590
= 7.444 in² 0.65 765.4
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For a selected Q orifice with an effective area of 11.05 in², the rated capacity is calculated using
(Eq. 7):
𝑤𝑤!"# =
0.65 765.4 11.05
= 137,400 lb/h 0.04
The results calculated above are compared to results obtained in Aspen HYSYS in Table 4.
Variable
Units
Example Calculation
Aspen HYSYS
Required Relief Load (w)
lb/h
92,590
92,590
Relief Pressure (P0)
psig
44
44.00
42.43 psig
57.13 psia
psig
36.7
Mass Density at Relief (ρℓ0)
lb/ft3
35.17
lb/ft3
13.72
Saturation Pressure (Ps)
90% Pressure (P9)
Mass Density at 90% (ρ9)
Omega Parameter (ω)
14.07
Transition Subcool Ratio (ηst)
Critical Flow Ratio (ηc)
0.9657
0.9130
Flow Type
Maximum Mass Flux (G)
Required Area (A)
Rated Capacity (wmax)
13.75
Low Subcool, Critical
lb/s-ft
in
2
2
lb/h
Blue = Calculation input
Critical
765.4
7.444
7.450
137,400
137,300
Gray = Calculated value
Table 4: Comparison of example calculation and Aspen HYSYS calculation for low subcooled, subcritical
flow
Example with High Subcooling
The example is based on the following conditions:
Composition
15% propane, 30% isobutane, 30% n-butane,
25% isopentane using the Aspen HYSYS PRSV
package for physical properties
Relief conditions
242 psig, 188.3 F
Required relief load
92,590 lb/h
Setting up a stream in Aspen HYSYS at the relief conditions and flashing appropriately, yields the
following properties:
ρℓ0 = 30.34 lb/ft³
Ps = 203.1 psia
P9 = 182.8 psia
ρ9 = 17.40 lb/ft³
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Figure 5: High subcooled critical flow case calculated in Aspen HYSYS
The omega parameter is calculated using (Eq. 8):
𝜔𝜔! = 9×
30.34
− 1 = 6.693 17.4
The transition saturation pressure ratio is calculated using (Eq. 9):
𝜂𝜂!" =
2 6.693
= 0.9305 1 + 2 6.693
The saturation pressure ratio is 0.7912. Because this is less than the transition ratio, this case
involves “high” subcooling. The critical pressure ratio is equal to the saturation pressure ratio. The
maximum mass flux is calculated using (Eq. 13):
𝐺𝐺 = 68.09 2 30.34 256.7 − 203.1 = 3,883 lb/s-­‐ft² Assuming that all capacity coefficients are 1.0, aside from the discharge coefficient which is 0.65 in
accordance with guidance from API, the required area is calculated using (Eq. 6):
𝐴𝐴 =
0.04 92,590
= 1.467 in² 0.65 3,883
For a selected K orifice with an effective area of 1.838 in², the rated capacity is calculated using
(Eq. 7):
𝑤𝑤!"# =
0.65 3,883 1.838
= 116,000 lb/h 0.04
The results calculated above are compared to results obtained in Aspen HYSYS in Table 5.
Equations and Example Benchmark Calculations within the
13 aspenONE® Engineering Safety Analysis Environment
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Variable
Units
Example Calculation
Aspen HYSYS
Required Relief Load (w)
lb/h
92,590
92,590
Relief Pressure (P0)
psig
242
242.0
188.4 psig
203.1 psia
psig
168.1
Mass Density at Relief (ρℓ0)
lb/ft3
30.34
lb/ft3
17.40
Saturation Pressure (Ps)
90% Pressure (P9)
Mass Density at 90% (ρ9)
Omega Parameter (ω)
6.693
Transition Subcool Ratio (ηst)
0.9305
Flow Type
Maximum Mass Flux (G)
Required Area (A)
Rated Capacity (wmax)
17.38
High Subcool, Critical
lb/s-ft
in
2
2
lb/h
Blue = Calculation input
Critical
3,883
1.467
1.473
116,000
115,600
Gray = Calculated value
Table 5: Comparison of example calculation and Aspen HYSYS calculation for high subcooled, critical flow
Direct Integration Method
The direct integration method involves evaluating thermodynamic properties of the relieving stream
at a number of state points and then numerically integrating the nozzle equation to determine the
maximum mass flux. This equation is the theoretical basis for every orifice method, so the direct
integration method is applicable to any type of flow, single or multi-phase.
Equations
The nozzle equation is shown below, where G is the mass flux in lb/ft²-s or kg/m²-s, v is the specific
volume in ft³/lb or m³/kg, and dp is the differential pressure increment in lbf/ft² or Pa.
Eq. 14
𝐺𝐺 ! =
−2𝑔𝑔!
𝑣𝑣 !
𝑣𝑣 𝑑𝑑𝑑𝑑 This is approximated as shown below, where v̅ is the average specific volume over the pressure
increment Δp.
Eq. 15
𝐺𝐺 ! ≅
−2𝑔𝑔!
𝑣𝑣 !
𝑣𝑣 ∆𝑝𝑝 The sum is performed from the relief pressure until either a maximum value of G is reached or the
back pressure Pa is reached. If the maximum in G occurs above Pa, then the flow is choked.
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Example
The example is based on the following conditions:
Composition
25% n-octane, 35% n-decane, 30% n-C11, 10%
nitrogen using the Aspen HYSYS SRK package
for physical properties
Relief conditions
176 psig, 122 F
Required relief load
138,900 lb/h
A property table may be used to obtain the mass density along an isentropic path for numerical
integration, as shown below.
Pressure
psia
Mass Density
lb/ft3
Specific Volume
ft3/lb
gcv̅ ∆P
ft²/s²
Mass Flux
lb/ft2-s
190.7
23.21
0.0431
0.0
0.0
180.7
22.48
0.0445
2028
1432
170.7
21.71
0.0461
2098
1972
160.7
20.90
0.0478
2175
2347
150.7
20.06
0.0499
2263
2625
140.7
19.18
0.0521
2363
2835
130.7
18.25
0.0548
2477
2988
120.7
17.28
0.0579
2610
3092
110.7
16.25
0.0615
2766
3150
100.7
15.17
0.0659
2952
3164
90.7
14.04
0.0712
3177
3133
Note that the mass flux reaches a maximum at a pressure of 100.7 psia, indicating that flow is
choked and the relief mass flux is 3164 lb/ft²-s.
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Figure 6: Direct integration case calculated in Aspen HYSYS
Assuming that all capacity coefficients are 1.0, aside from the discharge coefficient which is 0.85
in accordance with guidance from API for two-phase flow at relief conditions, the required area is
calculated using (Eq. 6):
𝐴𝐴 =
0.04 138,900
= 2.066 in² 0.85 3,164
For a selected L orifice with an effective area of 2.853 in², the rated capacity is calculated using
(Eq. 7):
𝑤𝑤!"# =
0.85 3,164 2.853
= 191,800 lb/h 0.04
The results calculated above are compared to results obtained in Aspen HYSYS in Table 6.
Variable
Units
Example Calculation
Aspen HYSYS
Required Relief Load (w)
lb/h
138,900
138,900
psig
176
176.0
122
122.0
3,164
3,114
Relief Pressure (P0)
Relief Temperature (T0)
F
Rated Capacity (wmax)
lb/h
Maximum Mass Flux (G)
Required Area (A)
lb/s-ft
in
2
2
Blue = Calculation input
2.066
2.099
191,800
188,800
Gray = Calculated value
Table 6: Comparison of example calculation and Aspen HYSYS calculation for direct integration
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Noise
Equations
The equations for computing noise level for atmospheric relief valves are given in section 5.8.10.3
of API Standard 521.2 The noise level at 30 m from the point of discharge is calculated using the
following equation below, where w is the mass flow through the valve in kg/h, c is the speed of
sound in the gas at the valve in m/s, and L is the sound level in decibels.
Eq. 16
𝐿𝐿!" = 𝐿𝐿 + 10 log!"
1
𝑤𝑤 𝑐𝑐 ! 7200
The speed of sound may be computed as shown below, where k is the ideal specific heat ratio in the
gas, T is the temperature
in K, and M is the molecular weight of the gas.
Eq. 17
𝑐𝑐 = 91.2
𝑘𝑘 𝑇𝑇
𝑀𝑀
The value of L in (Eq. 16) is read from a figure in the text. It may be approximated using two semilog
curve fits shown below, where X is the pressure ratio across the pressure relief valve.
Eq. 18
𝐿𝐿 =
39.461 ln 𝑋𝑋 + 12.648 ∶ 𝑋𝑋 ≤ 2.866
2.1716 ln 𝑋𝑋 + 51.914 ∶ 𝑋𝑋 > 2.866
Example
The example is based on the following conditions:
Composition
100% air using the Aspen HYSYS PR package
for physical properties
Relief conditions
330 kPaa, 311 K
Required relief load
52,560 kg/h
Setting up a stream in Aspen HYSYS at the relief conditions will yield the following properties:
M = 28.95
k = 1.404
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Figure 7: Noise example calculated
in Aspen HYSYS
The speed of sound may be calculated
using (Eq. 17) as:
𝑐𝑐 = 91.2
1.404 311
= 354.2 m/s 28.95
Based on a pressure ratio of 330⁄101.325=3.26, the contribution due to pressure ratio may be
estimated by (Eq.
18) to be 54.5 dB. Then, the total noise level is calculated using (Eq. 16):
𝐿𝐿!" = 54.5 + 10 log!"
1
52,560 354.2
7200
!
= 114.1 dB The results calculated above are compared to results obtained in Aspen HYSYS in Table 7.
Variable
Units
Example Calculation
Aspen HYSYS
Required Relief Load (w)
kg/h
52,560
52,560
330 kPaa
2.287 barg
311 K
37.85 C
114.1
114.1
Relief Pressure (P0)
Relief Temperature (T0)
Sound Level (L30)
dB
Blue = Calculation input
Gray = Calculated value
Table 7: Comparison of example calculation and Aspen HYSYS calculation for noise
Figure 8: Reaction force example
calculated in Aspen HYSYS
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Reaction Forces
Equations
The equations for computing reaction forces on the outlet piping of atmospheric relief valves are
given in section 4.4.1 of API Recommended Practice 520 Part II.3 The reaction force due to vapor
discharge is calculated using the following equation below, where F is the reaction force in N, w is the
required relief load in kg/h, k is the ideal specific heat ratio of the gas, T is the discharge temperature
in K, M is the molecular weight of the gas, A is the area of the pipe at the point of discharge in mm²,
and p3 is the static pressure at the outlet in barg (that is, 0 for flow that is unchoked at the outlet, or
the choke pressure for flow that is choked at the outlet).
Eq. 19
𝐹𝐹 =
1
𝑤𝑤
27.9
𝑘𝑘 𝑇𝑇
+ 0.1 𝐴𝐴 𝑝𝑝! 𝑘𝑘 + 1 𝑀𝑀
Example
The example is based on the following conditions:
Composition
30% propane, 70% n-butane using the Aspen
HYSYS SRK package for physical properties
Relief conditions
4.4 barg, 92.15 C
Required relief load
36,970 kg/h
Fluid properties at relief may be observed to be:
M = 53.92
k = 1.081
P* = 0.8201 bara
The reaction force may be calculated using (Eq. 19), as shown below. Because the choke pressure is
less than atmospheric pressure, the second term may be neglected.
𝐹𝐹 =
1
37,150
27.9
1.081 92.15 + 273.15
= 2,498 N 2.081 53.92
The results calculated above are compared to results obtained in Aspen HYSYS in Table 8.
Variable
Units
Example Calculation
Aspen HYSYS
Required Relief Load (w)
kg/h
37,150
37,150
C
92.15
92.15
N
2,498
2,497
Relief Temperature (T)
Reaction Force (F)
Blue = Calculation input
Gray = Calculated value
Table 8: Comparison of example calculation and Aspen HYSYS calculation for reaction force
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Conclusion
Safety is of the highest priority to every process, and ensuring accurate, validated calculations is a
key component of this work. To view additional validation papers, access tutorial documents and
videos, and learn more information about the tools AspenTech provides to address process safety
work, please visit the safety page on our company website, today!
References
1.
American Petroleum Institute, API Standard 520 Part I 9th Ed.: Sizing, Selection, and Installation
of Pressure-relieving Devices, Washington, DC: API Publishing Services, 2014.
2. American Petroleum Institute, API Standard 521 6th Ed.: Pressure-relieving and Depressuring
Systems, Washington, DC: API Publishing Services, 2014.
3. American Petroleum Institute, API Recommended Practice 520 Part II 5th Ed., Reaffirmed,
Washington: API Publishing Services, 2011.
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