Chapter 10. Relief Sizing

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Chapter 10. Relief Sizing

The learning objectives for this chapter are to understand relief sizing for:

Liquid and vapor service.
Two-phase flow.
Fire exposure.
Liquid thermal expansion.

Relief sizing calculations are performed to determine the vent area of the relief device. Relief sizing is a complex subject with many important subtleties too numerous to present here–experts in this area must be consulted. This chapter presents only a few simple cases. Also, relief sizing technology continues to evolve with time, particularly with respect to two-phase relief sizing.

Relief devices are considered preventive safeguards since they protect process equipment from the effects of high pressure. The pressure relief device might never be used, but it must operate properly every time it is required. Plant personnel must have confidence that all pressure relief systems are sized properly.

The relief sizing calculation uses an appropriate equation based on fundamental hydrodynamic principles to determine the relief device vent area. The relief vent area calculation depends on the specific application, the type of flow (liquid, vapor, or two-phase), and the type of relief device selected. See Section 9-4 and Table 9-2 for more details on the complete relief design procedure.

Figure 9-1 shows that the relief device begins to open at the relief set pressure, with the pressure then continuing to rise past the set pressure. A relief designed to maintain the pressure at the set pressure would require a very large and impractical relief area. The relief overpressure is a design feature of the device and must be specified prior to relief sizing.

Figure 10-1 shows that the relief vent area is reduced substantially as the overpressure increases. As mentioned in Chapter 9, the main purpose of a relief is to remove energy from the process. As the overpressure increases, the pressure driving force through the relief increases and the energy density of the fluid being discharged increases. Collectively, these changes result in a smaller vent area.

A graph of vent area versus overpressure for two-phase flow is shown. — **Figure 10-1** Required vent area as a function of overpressure for two-phase flow. The vent area is decreased appreciably as the overpressure increases. Data from J. C. Leung. “Simplified Vent Sizing Equations for Emergency Relief Requirements in Reactors and Storage Vessels.” *AICHE Journal* 32, no. 10 (1986): 1622.

A graph depicts the required vent area as a function of overpressure for two-phase flow. The horizontal axis, representing percent overpressure, ranges from 0 to 60, in increments of 10. The vertical axis, representing vent area (in meter squared), ranges from 0.00 to 0.12, in increments of 0.02. The vent area begins at (0, 0.15), after which it decreases steadily to reach the overpressure value at 55 and vent area at 0.02.

Spring-operated relief devices typically require a minimum flow to maintain the valve seat in the open position. Lower flows may result in “chattering,” caused by rapid opening and closing of the valve disc (see Section 9-5). This can lead to destruction of the relief device and a hazardous situation. A relief device with an area that is too large for the required flow is likely to chatter. In addition, the cost of the relief device increases dramatically with its size. For these reasons, reliefs must be designed with the proper vent area, neither too small nor too large. It is essential that the relief be sized properly to provide confidence that it will operate correctly when required.

This chapter presents methods for calculating the relief device vent areas for the following configurations:

Reliefs in liquid or vapor–gas service
Two-phase flow during runaway reactor relief
Dust and vapor explosions
Fires external to process vessels
Thermal expansion of process fluids

This chapter does not include relief sizing for atmospheric or low-pressure storage vessels. These reliefs, which are called conservation vents, have a completely different design and sizing basis.

10-1 Set Pressure and Accumulation Limits for Reliefs

Figure 9-3 shows the code requirements for the set pressure and accumulation limits for relief devices. These limits are summarized in Table 10-1. All percentages are with respect to the maximum allowable working pressure (MAWP) of the protected equipment. In the table, the accumulated pressure is the maximum pressure at the inlet to the relief device during the entire relief process expressed as a percentage of the MAWP.

Table 10-1 Set pressure and accumulation limits code requirements for pressure-relieving devices. (Source: Adapted from API RP 520, Recommended Practice for the Sizing, Selection, and Installation of Pressure-Relieving Systems in Refineries, 9th ed. (Washington, DC: American Petroleum Institute, 2014).)

Relief configuration	Single relief device installation		Multiple relief device installation
Relief configuration	Maximum set pressure (% of MAWP)	Maximum accumulated pressure (% of MAWP)	Maximum set pressure (% of MAWP)	Maximum accumulated pressure (% of MAWP)
Nonfire Scenario
Primary relief device	100	110	100	116
Additional devices	—	—	105	116
Fire Scenario
Primary relief device	100	121	100	121
Additional devices	—	—	105	121
Supplemental device^a	—	—	110	121

^aIn addition to a nonfire relief device.

Two relief scenarios are shown in Table 10-1: nonfire and fire. The fire scenario considers reliefs for fires external to equipment (discussed in more detail in Section 10-6). Reliefs for fire scenarios are typically much larger in size to handle the large heat flux from the external fire. The primary relief would be sized for a nonfire scenario while the supplemental fire relief would be sized separately for the fire scenario.

As shown in Figure 9-3 and Table 10-1, the set pressure for a primary relief device must never exceed the MAWP.

Example 10-1

Consider a process vessel with a MAWP of 100 psig. For the nonfire case, determine the maximum accumulated pressure, the allowable overpressure, and the maximum relieving pressure for a set pressure of (a) 100 psig and (b) 90 psig.

Solution

The answers are provided in Table 10-1. The results are shown here.

Set pressure at 100 psig equal to the MAWP.

Maximum accumulated pressure: 110.0 psig (110% of MAWP; see Table 10-1)

Allowable overpressure: 10.0 psi (maximum accumulated pressure – set pressure)

Maximum relieving pressure: 110.0 psig (set pressure + allowable overpressure)
Set pressure at 90 psig, which is below the MAWP.

Maximum accumulated pressure: 110.0 psig (110% of MAWP; see Table 10-1)

Allowable overpressure: 20.0 psi (maximum accumulated pressure – set pressure)

Maximum relieving pressure: 110.0 psig (set pressure + allowable overpressure)

Even though the set pressure is lower in (b), the maximum relieving pressure is the same as in (a) because it is based on the MAWP.

Example 10-2

Rework Example 10-1 (a), but now add another nonfire relief device.

Solution

The answers are provided in Table 10-1. The results are shown here.

First relief device:

Relief set pressure: 100.0 psig (equal to the MAWP)

Maximum accumulated pressure: 116.0 psig (116% of MAWP; see Table 10-1)

Allowable overpressure: 16.0 psig (maximum accumulated pressure – set pressure

Maximum relieving pressure: 116.0 psig (set pressure + allowable overpressure)

Additional nonfire relief device:

Relief set pressure:105.0 psig (105% of MAWP; see Table 10-1)

Maximum accumulated pressure: 116.0 psig (116% of MAWP; see Table 10-1)

Allowable overpressure: 11.0 psi (maximum accumulated pressure – set pressure)

Maximum relieving pressure:116.0 psig (set pressure + allowable overpressure)

A few comments about Examples 10-1 and 10-2:

For primary and additional relief devices, the maximum set pressure for the primary relief is set at the MAWP and the maximum set pressure for the additional relief is increased to 105% of the MAWP. The primary relief will open first.
Adding another relief device increases the maximum accumulated pressure for the primary relief device from 110% to 116% of the MAWP. The maximum accumulated pressure for the additional relief device is also at 116% of the MAWP.
The primary relief device is usually smaller in size to handle more routine relief scenarios, such as a regulator failure. The supplemental device might have a much larger area for a more robust relief scenario, such as a runaway reaction.

10-2 Relief Sizing for Liquid Service

Flow through spring-type reliefs is approximated as flow through an orifice. An equation representing this flow is derived from the mechanical energy balance (Equation 4-1). The result is similar to Equation 4-6, except that the pressure is represented by a pressure difference across the valve seat:

u¯=Co2gcΔPρ−−−−−−√(10-1) $\begin{matrix} \bar{u} = C_{o} \sqrt{\frac{2 g_{c} Δ P}{ρ}} & (10-1) \end{matrix}$

where

ū is the liquid velocity through the spring relief (distance/time),

C_o is the discharge coefficient (unitless),

ΔP is the pressure drop across the valve seat (force/area), and

ρ is the liquid density (mass/volume).

The volumetric flow, Q_v, of the liquid is the product of the velocity times the area, or ūA. Substituting the liquid velocity, u, from Equation 10-1 into the equation for Q_v and solving for the vent area A of the relief, we obtain

A=QvCo2gc−−−√ρΔP−−−−√(10-2) $\begin{matrix} A = \frac{Q_{v}}{C_{o} \sqrt{2 g_{c}}} \sqrt{\frac{ρ}{Δ P}} & (10-2) \end{matrix}$

A working equation with fixed units is derived from Equation 10-2 by (1) replacing the density ρ with the specific gravity (ρ /ρ_ref) and (2) making the appropriate substitutions for the specific unit conversions. The result is

A=[in2(psi)1/238.0 gpm]QvCo(ρ/ρref)ΔP−−−−−−−√(10-3) $\begin{matrix} \begin{matrix} A = [\frac{{in}^{2} {(psi)}^{1 / 2}}{38.0 gpm}] \frac{Q_{v}}{C_{o}} \sqrt{\frac{(ρ / ρ_{ref})}{Δ P}} \end{matrix} & (10-3) \end{matrix}$

where

A is the computed relief area (in²),

Q_v is the volumetric flow through the relief (gpm),

C_o is the discharge coefficient (unitless),

(ρ /ρ_ref) is the specific gravity of the liquid (unitless), and

ΔP is the pressure drop across the spring relief (lb_f /in²).

Equation 10-3 represents flow through an orifice. A true orifice has a constant, fixed area. However, flow through a spring-type relief is actually flow through an annulus. As the pressure increases, the relief spring is compressed, increasing the annulus discharge area and increasing the flow.

Equation 10-3 also does not consider the viscosity of the fluid. Many process fluids have high viscosities; in such a case, the relief vent area must increase as the fluid viscosity increases.

Finally, Equation 10-3 does not consider the special case of a balanced bellows type of relief.

The American Society of Mechanical Engineers (ASME) pressure vessel code¹ now requires all pressure relief valves (PRVs) in liquid service to be flow certified. ASME-certified relief valves are required to reach full rated capacity at 10% or less of overpressure. Relief valves are certified by the National Board of Boiler and Pressure Vessel Inspectors. The ASME code provides valve flow testing methods for code certification.

¹ASME Boiler and Pressure Vessel Code (New York, NY: American Society of Mechanical Engineers, 2017).

For capacity certified valves, the following sizing equation applies:²

²API RP 520, Sizing, Selection, and Installation of Pressure-Relieving Devices in Refineries, 6th ed. (Washington, DC: American Petroleum Institute, 2017).

A=[in2(psi)1/238.0 gpm]QvCoKbKcKv(ρ/ρref)P1−P2−−−−−−−√ (10-4) $\begin{matrix} \begin{matrix} A = [\frac{{in}^{2} {(psi)}^{1 / 2}}{38.0 gpm}] \frac{Q_{v}}{C_{o} K_{b} K_{c} K_{v}} \sqrt{\frac{(ρ / ρ_{ref})}{P_{1} - P_{2}}} \end{matrix} & (10-4) \end{matrix}$

where

A is the computed relief area (in²),

Q_v is the volumetric flow through the relief (gpm),

C_o is the rated discharge coefficient (unitless),

K_b is the backpressure correction (unitless),

K_c is the combination correction factor (unitless),

K_v is the viscosity correction (unitless),

(ρ /ρ_ref) is the specific gravity of the liquid (unitless),

P₁ is the gauge upstream relieving pressure (lb_f /in²), which is the set pressure plus the allowable overpressure, and

P₂ is the total gauge backpressure (lb_f /in²).

The metric equivalent to Equation 10-4 is

A=[(bar)1/2s14.16 m]QvCoKbKcKv(ρ/ρref)P1−P2−−−−−−−√ (10-5) $\begin{matrix} A = [\frac{{(bar)}^{1 / 2} s}{14.16 m}] \frac{Q_{v}}{C_{o} K_{b} K_{c} K_{v}} \sqrt{\frac{(ρ / ρ_{ref})}{P_{1} - P_{2}}} & (10-5) \end{matrix}$

where the variables have the same meaning as Equation 10-4 but with the following units:

A: m2Qv: m3/ sCo, Kb, Kc, Kv: unitless(ρ/ρref): unitlessP1, P2: bar $\begin{array}{l} A : m^{2} \\ Q_{v} : m^{3} / s \\ C_{o}, K_{b}, K_{c}, K_{v} : unitless \\ (ρ / ρ_{ref}) : unitless \\ P_{1}, P_{2} : bar \end{array}$

The capacity correction factors in Equations 10-4 and 10-5 are defined as follows:

C_o is the discharge coefficient and should be obtained from the valve manufacturer. For preliminary sizing, an effective discharge coefficient can be selected as follows: 0.65 for a spring-operated relief installed with or without a rupture disc in combination and 0.62 for a rupture disc.
K_b is the backpressure correction. For conventional and pilot-operated valves, use a value of 1.0. If the backpressure is atmospheric, use a value of 1.0. Balanced bellows valves require a correction factor determined from Figure 10-2.

Figure 10-2 Backpressure correction K_b for 25% overpressure on balanced bellows reliefs in liquid service. For metric units, barg can be used for psig. This is drawn using the equation K_b = 1.165 − 0.01P_G, derived from data from API RP 520, Recommended Practice for the Sizing, Selection, and Installation of Pressure-Relieving Systems in Refineries, 9th ed. (Washington, DC: American Petroleum Institute, 2014).

The graph shows backpressure correction (k subscript p equals capacity with variable backpressure over rated capacity based on the square root of 1.25P minus p subscript b) plotted against the gauge backpressure (p subscript g equals backpressure, pounds per square inch over set pressure, pounds per square inch times 100). The horizontal axis, representing gauge backpressure, ranges from 0 to 50, in increments of 10. The vertical axis, representing backpressure correction (k subscript p), ranges from 0.50 to 1.00, in increments of 0.05. The plot remains fairly constant from (0, 1.00) to (15, 16), which then decreases steeply through the end of the plot at (50, 0.66).
K_c is the combination correction factor; it corrects for installations with a rupture disc upstream of the PRV. Use a value of 1.0 if a rupture disc is not installed. Use a value of 0.9 if a rupture disc is installed with the PRV.
K_v is the viscosity correction; it corrects for the additional frictional losses resulting from flow of higher-viscosity material through the valve. This correction is given in Figure 10-3. The required relief vent area becomes larger as the viscosity of the liquid increases (lower Reynolds numbers). Since the Reynolds number is required to determine the viscosity correction and because the vent area is required to calculate the Reynolds number, the procedure is iterative. For most reliefs, the Reynolds number is greater than 16,000 and the correction is near 1. This assumption is frequently used as an initial estimate to begin the calculations.

Figure 10-3 Viscosity correction factor K_v for conventional and balanced bellows reliefs in liquid service. This is drawn using the equation In K_v = 0.08547 − 0.9541/ln R − 35.571/R using data from API RP 520, Recommended Practice for the Sizing, Selection, and Installation of Pressure-Relieving Systems in Refineries, 9th ed. (Washington, DC: American Petroleum Institute, 2014).

The graph shows the viscosity correction factor (K subscript v) is plotted against the Reynolds number. The horizontal axis, representing Reynolds number, ranges from 10 to 100000, in increments of 10. The vertical axis, representing viscosity correction factor (K subscript v), ranges from 0.3 to 1.0, in increments of 0.1. There was a steady rise in the viscosity correction factor and Reynolds number to reach 70,000 Reynolds number at 1.0 viscosity correction factor from (35, 0.3).

Darby and Molavi developed an equation to represent the viscosity correction factor shown in Figure 10-3.³ This equation applies only to Reynolds numbers greater than 100:

³R. Darby and K. Molavi. “Viscosity Correction Factor for Safety Relief Valves.” Process Safety Progress 16, no. 2 (2004), pp. 80-82.

Kv=0.9751170Re +0.98−−−−−−−−−−⎷ (10-6) $\begin{matrix} K_{v} = 0.975 \sqrt{\frac{1}{\frac{170}{R e} + 0.98}} & (10-6) \end{matrix}$

where

K_v is the viscosity correction factor (unitless) and

Re is the Reynolds number (unitless).

Prior to ASME requiring capacity certification, a modified form of Equation 10-4 was used. An overpressure correction factor, K_p, is required for relieving pressures other than 25% overpressure. This results in the following equation:

A=[in2(psi)1/238.0 gpm]QvCoKbKcKvKp(ρ/ρref)1.25 Ps−P2−−−−−−−−−−√ (10-7) $\begin{matrix} \begin{matrix} A = [\frac{{in}^{2} {(psi)}^{1 / 2}}{38.0 gpm}] \frac{Q_{v}}{C_{o} K_{b} K_{c} K_{v} K_{p}} \sqrt{\frac{(ρ / ρ_{ref})}{1.25 P_{s} - P_{2}}} \end{matrix} & (10-7) \end{matrix}$

where

A is the computed relief area (in²),

Q_v is the volumetric flow through the relief (gpm),

C_o is the discharge coefficient (unitless),

K_b is the backpressure correction (unitless),

K_c is the combination correction factor (unitless),

K_v is the viscosity correction (unitless),

K_p is the overpressure correction (unitless),

(ρ /ρ_ref) is the specific gravity of the liquid at standard conditions (unitless),

P_s is the gauge set pressure (lb_f/in²), and

P₂ is the total gauge backpressure (lb_f/in²).

Equation 10-7 was the only liquid sizing equation used in previous editions of this textbook.

The metric equivalent to Equation 10-7 is

A=[(bar)1/2s14.16 m]QvCoKbKcKvKp(ρ/ρref)1.25Ps−P2−−−−−−−−−−√ (10-8) $\begin{matrix} A = [\frac{{(bar)}^{1 / 2} s}{14.16 m}] \frac{Q_{v}}{C_{o} K_{b} K_{c} K_{v} K_{p}} \sqrt{\frac{(ρ / ρ_{ref})}{1.25 P_{s} - P_{2}}} & (10-8) \end{matrix}$

where the variables have the same meaning as Equation 10-3 but with the following units:

A: m2Qv: m3/ sCo, Kb, Kc, Kv, Kp: unitless(ρ/ρref): unitlessPs, P2: bar $\begin{array}{l} A : m^{2} \\ Q_{v} : m^{3} / s \\ C_{o}, K_{b}, K_{c}, K_{v}, K_{p} : unitless \\ (ρ / ρ_{ref}) : unitless \\ P_{s}, P_{2} : bar \end{array}$

The capacity correction factors in Equations 10-7 and 10-8 are defined as follows:

C_o is the discharge coefficient and should be obtained from the valve manufacturer. For preliminary sizing, an effective discharge coefficient of 0.62 is used.
K_b is the backpressure correction factor. For conventional spring-operated and pilot-operated valves, use a value of 1.0. If the backpressure is atmospheric, use a value of 1.0. Balanced bellows valves require a correction factor determined from Figure 10-2.
K_c is the combination correction factor; it corrects for installations with a rupture disc upstream of the PRV. Use a value of 1.0 if a rupture disc is not installed. If a rupture disc is installed, see API RP 510 for details.
K_v is the viscosity correction; it corrects for the additional frictional losses resulting from flow of high-viscosity material through the valve. See note 4 below Equation 10-5. This correction is given by Figure 10-3 or Equation 10-6.
K_p is the overpressure correction and has a value of 1.0 at 25% overpressure. For overpressures other than 25%, use Figure 10-4.

A graph of overpressure correction (K subscript p) versus overpressure (p0) is shown. — **Figure 10-4** Overpressure correction K_p for spring-operated reliefs (conventional and balanced bellows) in uncertified liquid service. This is drawn using the equations shown, derived from data from *API RP 520, Recommended Practice for the Sizing, Selection, and Installation of Pressure-Relieving Systems in Refineries*, 9th ed. (Washington, DC: American Petroleum Institute, 2014).

The graph shows overpressure correction (K subscript p) plotted against overpressure (p0). The horizontal axis, representing overpressure (p0), ranges from 10 to 50, in increments of 5. The vertical axis, representing overpressure correction (K subscript p), ranges from 0.5 to 1.1, in increments of 0.1. The overpressure correction starts at (10, 0.62), reaches a peak of (28, 1.0), after which it increases slightly throughout the end of the graph. The equation for the overpressure range 10 to 17.5 is k subscript p equals 0.2533 plus 0.03632 p0; the equation for the overpressure range 17.5 to 27.5 is k subscript p equals 4.207 minus 0.1217 p0 plus 0.001595 p squared 0 minus 29.35 over p0 and the equation for the overpressure range 27.5 to 50 is k subscript p equals 0.9034 plus 0.003522 p0. These values are tabulated on the graph.

Example 10-3

A positive displacement pump is rated for 200 gpm at a pressure of 200 psig. Because a deadheaded positive displacement pump can be easily damaged, compute the area for a capacity-certified spring-operated relief required to protect the pump. The rated pressure for the pump is 250 psig. The superimposed backpressure is 20 psig and constant. Determine the required relief area assuming an overpressure of 25 psig.

Solution

Equation 10-4 applies for capacity-certified liquid relief.

The set pressure is 250 psig. P₁ is the set pressure plus the overpressure,

P₁ = 250 psig + 25 psig = 275 psig.

P₂ is the total gauge backpressure, 20 psig.

For water, (ρ/ρ_ref = 1.0.

The quantity of material relieved is the capacity of the pump, so Q_v = 200 gpm.

The discharge coefficient C_o is not specified. However, for preliminary sizing purposes, a value of 0.65 is used.

Since this is a conventional spring-operated relief, the backpressure correction is K_b = 1.0.

Pumps do not normally have a spring relief coupled with a rupture disc combination. Thus, the combination correction factor, K_c, has a value of 1.0.

The Reynolds number through the relief device is not known since the relief area is to be determined. However, at 200 gpm, the Reynolds number is assumed to be greater than 16,000. Thus, the viscosity correction is K_v = 1.0.

These numbers are substituted into Equation 10-4:

A=[in2(psi)1/238.0 gpm]QvCoKbKcKv(ρ/ρref)P1−P2−−−−−−−√=[in2(psi)1/238.0 gpm]200 gpm(0.65)(1.0)(1.0)(1.0)1.0(275−20) psig−−−−−−−−−−−−−√=0.507 in2(10−4) $\begin{matrix} A & = [\frac{{in}^{2} {(psi)}^{1 / 2}}{38.0 gpm}] \frac{Q_{v}}{C_{o} K_{b} K_{c} K_{v}} \sqrt{\frac{(ρ / ρ_{ref})}{P_{1} - P_{2}}} \\ = [\frac{{in}^{2} {(psi)}^{1 / 2}}{38.0 gpm}] \frac{200 gpm}{(0.65) (1.0) (1.0) (1.0)} \sqrt{\frac{1.0}{(275 - 20) psig}} \\ = 0.507 {in}^{2} \end{matrix} \begin{matrix} (10 - 4) \end{matrix}$

The diameter of the relief is

d=4Aπ−−−√=4(0.507 in2)3.14−−−−−−−−−−−√=0.80 in $d = \sqrt{\frac{4 A}{π}} = \sqrt{\frac{4 (0.507 {in}^{2})}{3.14}} = 0.80 in$

Manufacturers do not provide relief devices to the nearest 0.01 in. Thus, a selection must be made depending on the relief device sizes available commercially. The valve will open only enough to deliver the 200 gpm flow. The next largest available size–with a higher capacity–is therefore selected.

10-3 Relief Sizing for Vapor and Gas Service

For most vapor or gas discharges through reliefs, the flow is critical. However, the downstream pressure must be checked to ensure that it is less than the choked pressure computed using Equation 4-49. Thus, for an ideal gas Equation 4-50 is valid:

(Qm)choked=CoAPγgcMRgT(2γ+1)(γ+1)/(γ−1)−−−−−−−−−−−−−−−−−−−−√(4−50) $\begin{matrix} {(Q_{m})}_{choked} = C_{o} A P \sqrt{\frac{γ g_{c} M}{R_{g} T} {(\frac{2}{γ + 1})}^{(γ +1) / (γ - 1)}} & (4 - 50) \end{matrix}$

where

(Q_m)_choked is the discharge mass flow (mass/time),

C_o is the discharge coefficient (unitless),

A is the area of the discharge (area),

P is the absolute upstream relieving pressure, equal to the set pressure plus the overpressure (force/area),

γ is the heat capacity ratio for the gas (unitless),

g_c is the gravitational constant (length mass/force time²)

M is the molecular weight of the gas (mass/mole)

R_g is the ideal gas constant, and

T is the absolute temperature of the discharge (degree).

Equation 4-50 is solved for the area of the relief vent given a specified mass flow rate Q_m:

A=QmCoPT/MγgcRg(2γ+1)(γ+1)/(γ−1)−−−−−−−−−−−−−−−−−−−⎷(10-9) $\begin{matrix} A = \frac{Q_{m}}{C_{o} P} \sqrt{\frac{T / M}{\frac{γ g_{c}}{R_{g}} {(\frac{2}{γ + 1})}^{(γ +1) / (γ - 1)}}} & (10-9) \end{matrix}$

Equation 10-9 is simplified by defining a function χ:

χ=γgcRg(2γ+1)(γ+1)/(γ−1)−−−−−−−−−−−−−−−−−−√(10-10) $\begin{matrix} χ = \sqrt{\frac{γ g_{c}}{R_{g}} {(\frac{2}{γ + 1})}^{(γ +1) / (γ - 1)}} & (10-10) \end{matrix}$

Then the required relief vent area for an ideal gas is computed using a simplified form of Equation 10-9:

A=QmCoχPTM−−−√(10-11) $\begin{matrix} A = \frac{Q_{m}}{C_{o} χ P} \sqrt{\frac{T}{M}} & (10-11) \end{matrix}$

Equation 10-11 is modified by (1) including the compressibility factor z to represent a nonideal gas; (2) including a backpressure correction K_b; and (3) including a combination correction factor K_c to account for the presence of a rupture disc below the spring relief. The result is⁴

⁴API RP 520, Sizing, Selection, and Installation of Pressure-Relieving Devices in Refineries, 6th ed. (Washington, DC: American Petroleum Institute, 2017).

A=QmχCoKbKcPTzM−−−√(10-12) $\begin{matrix} \begin{matrix} A = \frac{Q_{m}}{χ C_{o} K_{b} K_{c} P} \sqrt{\frac{T z}{M}} \end{matrix} & (10-12) \end{matrix}$

where

A is the area of the relief vent (area),

Q_m is the discharge mass flow (mass/time),

C_o is the effective discharge coefficient (unitless),

K_b is the backpressure correction (unitless),

K_c is the combination correction factor (unitless),

P is the upstream relieving pressure, equal to the set pressure plus the allowable overpressure (force/area, absolute),

T is the absolute temperature (degrees),

z is the compressibility factor (unitless), and

M is the average molecular weight of the discharge material (mass/mole).

Note that Equation 10-12 does not include a downstream pressure since critical flow through an orifice is independent of the downstream pressure.

The constant χ is calculated using Equation 10-10. It is conveniently calculated using the following fixed-unit expressions:

χ=[519.5(lb-mole lbmoR)1/2lbfhr]γ(2γ+1)(γ+1)/(γ−1)−−−−−−−−−−−−−−−−√(10-13) $\begin{matrix} χ = [519.5 \frac{{({lb-mole lb}_{m}^{o} R)}^{1 / 2}}{{lb}_{f} hr}] \sqrt{γ {(\frac{2}{γ + 1})}^{(γ +1) / (γ - 1)}} & (10-13) \end{matrix}$

If Equation 10-13 is used with Equation 10-12, then the following fixed units must be used: Q_m in lb_m/hr, P in psia, T in °R, and M in lb_m/lb-mole. The area computed is in in².

The metric equivalent to Equation 10-13 is:

χ=[3.468×104(mol kg K)1/2bar s m2]γ(2γ+1)(γ+1)/(γ−1)−−−−−−−−−−−−−−−−√(10-14) $\begin{matrix} χ = [3.468 \times 10^{4} \frac{{(mol kg K)}^{1 / 2}}{{bar s m}^{2}}] \sqrt{γ {(\frac{2}{γ + 1})}^{(γ +1) / (γ - 1)}} & (10-14) \end{matrix}$

If Equation 10-14 is used with Equation 10-12, then the following fixed units must be used: Q_m in kg/s, P in bar, T in K, and M in kg/mol. The area computed is in m².

The capacity correction factors in Equation 10-12 are defined as follows:

C_o is the discharge coefficient and should be provided by the manufacturer. For preliminary sizing, the following values are used:
- When the PRV is installed with or without a rupture disc in combination: 0.975
- When sizing for a rupture disc alone: 0.62
K_b is the capacity correction factor for backpressure and should be obtained from the manufacturer. For preliminary sizing, use Figure 10-5 for balanced bellows valves only. For conventional and pilot-operated valves, use a value of 1.0.

Figure 10-5 Backpressure correction K_b for balanced-bellows reliefs in vapor or gas service. For metric units, barg can be used for psig. This is drawn using the equations provided. Data from API RP 520, Recommended Practice for the Sizing, Selection, and Installation of Pressure-Relieving Systems in Refineries, 9th ed. (Washington, DC: American Petroleum Institute, 2014).

A graph shows the backpressure (k subscript b equals capacity with back pressure over rated capacity without back pressure) drawn against the gauge pressure (G equals back pressure, pound per square inch over set pressure, pound per square inch). The horizontal axis, representing gauge pressure (G in percentage), ranges from 0 to 50, in increments of 5. The vertical axis, representing backpressure (k subscript b), ranges from 0.5 to 1.0, in increments of 0.1. The graph plots two lines that remain unchanged from (0, 1.0) to (30, 1.0), lying on the top of the other lines. The line representing 20 percentage overpressure drops from (30, 1.0) and steadily decreases for the rest of the values. The line representing 10 percentage overpressure experiences a steep fall to reach the value at (50, 0.68). The values of a, b, and range for the overpressure is derived from the equation k subscript b equals 1 over (a plus b G cubed). The values of overpressure, a, b, and range are as follows: 10 percentage; 0.8707; 4.724 times 10 power negative 6; 30 to 50 and 20 percentage; 0.9760; 8.360 times 10 power negative 7; 30 to 50.
K_c is the combination correction factor for a rupture disc upstream of the valve. If no rupture disc is upstream, use a value of 1.0. Use 0.9 when a rupture disc is installed in combination with the PRV and the combination does not have a certified value; see API RP 520 for more details.

Subcritical Vapor/Gas Flow

Critical flow of vapors/gases through relief devices is the usual case, but subcritical flow is also possible. See Sections 4-5 and 4-6 to determine whether subcritical or critical flow occurs in your particular case. For subcritical flows through relief devices, consult API RP 520 (2014), page 66.

Steam Flow Relief Sizing

Steam is a specific case of vapor flow and the physical properties of steam are well known. API RP 520 has a dedicated relief sizing procedure for steam (page 72 of the standard). Consult this reference for steam relief applications.

Rupture Disc Sizing

Rupture discs may be used as the primary relief device for gas, vapor, liquid, or multiphase flow. There are two accepted methods for rupture disc sizing, according to API RP 520.

The first method uses the equations already presented here for liquid and vapor service. Use Equation 10-4 or Equation 10-7 for liquid service and Equation 10-12 for vapor service. A coefficient of discharge, C_o, of 0.62 must be used. This method may only be used in the following scenarios:

When the rupture disc discharges directly to the atmosphere
When the rupture disc is installed within eight pipe diameters from the vessel entry nozzle
When the rupture disc has a length of discharge no greater than five pipe diameters
When the inlet and outlet discharge piping is equal to or greater than the diameter of the rupture disc

Consult API RP 520, page 81, for additional details.

The second method for rupture disc sizing is to use the flow resistance method outlined in Section 4-4 with the velocity head method. This approach must include the rupture disc device, piping and other piping components, entrance and exit losses, elbows, tees, reducers, valves, and other piping components. The calculated relieving capacity should be multiplied by 0.90 or less to allow for uncertainties in this method. Consult API RP 520, page 81, for more details.

Pilot-Operated Relief Sizing

For pilot-operated valves, neither the set pressure nor the capacity is affected by backpressure since the valve lift is not affected by backpressure.

Buckling Pin Relief Sizing

See UG-127(b) and UG-138 of the ASME Code for sizing, application, and minimum requirements for buckling pin devices.

Example 10-4

One high pressure scenario identified for a reactor vessel is the failure of a nitrogen regulator, allowing high-pressure nitrogen to enter the reactor through a 10-cm-diameter supply pipe. The source of the nitrogen is at 25°C and 10 barg. The reactor vessel MAWP is 3.0 barg. Determine the diameter of a balanced bellows spring-type vapor relief required to protect the reactor from this incident. Assume a constant superimposed backpressure of 0.5 bar.

Solution

From Table 10-1, the maximun set pressure is equal to the vessel MAWP at 3.0 barg. The maximum accumulated pressure is 1.10 × 3.0 barg = 3.3 barg = 4.3 bara.

The nitrogen source is at 10 barg = 11.01 bara. If the regulator fails, the nitrogen will flood the reactor, increasing the pressure and exceeding the MAWP. A relief vent must be installed to vent the nitrogen as fast as it is supplied through the 10-cm pipe. Because no other information on the piping system is provided, the flow from the pipe is assumed to be represented by critical flow through an orifice. Equation 4-50 applies:

(Qm)choked=CoAPγgcMRgT(2γ+1)(γ+1)/(γ−1)−−−−−−−−−−−−−−−−−−−−√(4-50) $\begin{matrix} {(Q_{m})}_{choked} = C_{o} A P \sqrt{\frac{γ g_{c} M}{R_{g} T} {(\frac{2}{γ + 1})}^{(γ +1) / (γ - 1)}} & (4-50) \end{matrix}$

First, however, the choked pressure across the pipe must be determined to ensure critical flow. For diatomic gases, the choked pressure is given as (see Section 4-5)

Pchoked=0.528P=(0.528)(11.01 bara)=5.8 bara $P_{choked} = 0.528 P = (0.528) (11.01 bara) = 5.8 bara$

The maximum accumulated pressure in the reactor vessel during the relief venting is 4.3 bara. Thus, the pressure in the reactor is less than the choked pressure and the flow through the 10-cm line into the reactor vessel will be sonic. The required quantities for Equation 4-50 are

APγTM(2γ+1)(γ+1)/(γ−1)=πd24=(3.14)(0.10 m)24=7.85×10−3 m2=11.01 bara=1.40 for diatomic gases=25oC+273=298 K=0.028 kg/mol=(21.4+1)(2.4/0.4)=0.335 $\begin{matrix} A & = \frac{π d^{2}}{4} = \frac{(3.14) {(0.10 m)}^{2}}{4} = 7.85 \times 10^{- 3} m^{2} \\ P & = 11.01 bara \\ γ & = 1.40 for diatomic gases \\ T & = 25^{o} C + 273 = 298 K \\ M & = 0.0 28 kg/mol \\ {(\frac{2}{γ +1})}^{(γ +1) / (γ - 1)} & = {(\frac{2}{1 . 4+1})}^{(2.4 / 0.4)} = 0.335 \end{matrix}$

The discharge coefficient C_o is assumed to be 1.0 for sonic flow through the pipe.

Substituting into Equation 4-50, we obtain

(Qm)choked=(1.0)(7.85×10−3 m2)(11.01 bara)(105 N / m2bar)×(1.4)(1.0 kg m / N s2)(0.028 kg / mol)(0.335)(8.314×10−5 bar m3/ mol K)(298 K)(105 N / m2bar)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷=(8.64×103 N)5.30×10−6 kg2/ N2s2−−−−−−−−−−−−−−−−−√=19.9 kg /s $\begin{matrix} {(Q_{m})}_{choked} & = (1.0) (7.85 \times 10^{- 3} m^{2}) (11.01 bara) (\frac{10^{5} N / m^{2}}{bar}) \times \sqrt{\frac{(1.4) (1.0 kg m / {N s}^{2}) (0.0 28 kg / mol) (0.335)}{(8.314 \times 10^{- 5} {bar m}^{3} / mol K) (298 K) (10^{5} \frac{N / m^{2}}{bar})}} \\ = (8.64 \times 10^{3} N) \sqrt{5.30 \times 10^{- 6} {kg}^{2} / N^{2} s^{2}} \\ = 19.9 kg /s \end{matrix}$

The area of the relief vent is computed using Equations 10-12 and 10-14 with a backpressure correction K_b determined from Figure 10-5. The backpressure is 0.5 barg. Thus,

(backpressure, bargset pressure, barg)×100=(0.5 barg3.0 barg)×100=16.7% $(\frac{backpressure, barg}{set pressure, barg}) \times 100 = (\frac{0.5 barg}{3.0 barg}) \times 100 = 16.7 %$

From Figure 10-5, K_b = 1.0.

The effective discharge coefficient C_o for the relief is 0.975.

The gas compressibility factor z is 1 at these pressures.

The pressure P is the upstream relieving pressure, which is the set pressure plus the superimposed backpressure or 3.0 barg + 0.5 barg = 3.5 barg = 4.51 bara.

The constant χ is computed from Equation 10-14:

χ=[3.468×104(mol kg K)1/2bar s m2]γ(2γ+1)(γ+1)/(γ−1)−−−−−−−−−−−−−−−−√=[3.468×104(mol kg K)1/2bar s m2](1.4)(0.335)−−−−−−−−−−√=2.37×104(mol kg K)1/2bar s m2 $\begin{matrix} χ & = [3.468 \times 10^{4} \frac{{(mol kg K)}^{1 / 2}}{{bar s m}^{2}}] \sqrt{γ {(\frac{2}{γ + 1})}^{(γ +1) / (γ - 1)}} \\ = [3.468 \times 10^{4} \frac{{(mol kg K)}^{1 / 2}}{{bar s m}^{2}}] \sqrt{(1.4) (0.335)} \\ = 2.37 \times 10^{4} \frac{{(mol kg K)}^{1 / 2}}{{bar s m}^{2}} \end{matrix}$

The required vent area is computed using Equation 10-9:

A=QmCoχKbPTzM−−−√=19.9 kg/s(0.975)[2.37×104(mol kg K)1/2bar s m2](1.0)(4.51 bara)(298 K)(1.0)0.028 kg / mol−−−−−−−−−−−−√=1.97 × 10−2 m2 $\begin{matrix} A & = \frac{Q_{m}}{C_{o} χ K_{b} P} \sqrt{\frac{T z}{M}} \\ = \frac{19.9 kg/s}{(0.975) [2.37 \times 10^{4} \frac{{(mol kg K)}^{1 / 2}}{{bar s m}^{2}}] (1.0) (4.51 bara)} \sqrt{\frac{(298 K) (1.0)}{0.028 kg / mol}} \\ = 1.97 \times 10^{- 2} m^{2} \end{matrix}$

The required vent diameter is

d=4Aπ−−−√=4(1.97 × 10−2 m2)3.14−−−−−−−−−−−−−−−−√=0.16 m=16 cm $d = \sqrt{\frac{4 A}{π}} = \sqrt{\frac{4 (1.97 \times 10^{- 2} m^{2})}{3.14}} = 0.16 m = 16 cm$

The next largest valve size is selected from the manufacturer. If the vapor relief valve is pop-acting, then the valve will be open for a period of time and then closed for a period of time. The average flow over the open/closed intervals equals the required nitrogen flow.

Example 10-5

Compute the rupture disc vent diameter required to protect the vessel described in Example 10-4.

Solution

Since we do not know any information regarding the actual plumbing, we will use the capacity discharge correction method and assume that all the conditions listed are satisfied. The solution is provided by Equation 10-12. The solution is identical to Example 10-4, but with the discharge coefficient C_o = 0.62. The area is therefore

A=(1.97 × 10−2 m2)(0.9750.62)=3.05 × 10−2 m2 $A = (1.97 \times 10^{- 2} m^{2}) (\frac{0.975}{0.62}) = 3.05 \times 10^{- 2} m^{2}$

The rupture disc diameter is

d=4Aπ−−−√=4(3.05 × 10−2 m2)3.14−−−−−−−−−−−−−−−√=0.197 m=19.7 m $d = \sqrt{\frac{4 A}{π}} = \sqrt{\frac{4 (3.05 \times 10^{- 2} m^{2})}{3.14}} = 0.197 m = 19.7 m$

This compares to a spring relief diameter of 16 cm.

10-4 Two-Phase Flow during Runaway Reaction Relief

Two-phase flow is normally expected during the relief process when a runaway reaction occurs within a reactor vessel. A detailed study using one or more of the calorimeters described in Table 8-8 provides the much-needed data for two-phase relief area sizing. Two-phase flow must be confirmed and the complex flow regimes identified using these calorimeters.

Two-phase flow through reliefs is much more complex than the introduction provided here might suggest. Furthermore, the technology is still undergoing substantial development.

There are three dominant methods currently available for two-phase relief sizing:

Leung method
Fauske method
API method

Specific references are available at the end of this chapter for each of these methods. Only the Leung method is presented here.

Figure 10-6 shows the most common type of reactor system, called a tempered reactor. It is called “tempered” because the reactor contains a volatile liquid that vaporizes or flashes during the relieving process. This vaporization removes energy by means of the heat of vaporization and tempers the rate of temperature rise resulting from the exothermic reaction.

A tempered reaction system presents the most important energy terms. — **Figure 10-6** A tempered reaction system showing the important energy terms.

The energy terms of the tempered reaction system are illustrated. A reactor labeled "energy generated by the reaction" shows a narrow opening at the top for two-phase flow. The water level at the top indicates energy loss by vaporization of the liquid. The region below the water level contains the energy accumulated by heating of liquid beyond set temperature. A relief device is installed at the beginning of the narrow opening of the reactor.

The runaway reactor is treated as entirely adiabatic. The energy terms include (1) the energy accumulation resulting from the sensible heat of the reactor fluid as a result of its increased temperature due to overpressure and (2) the energy removal resulting from the vaporization of liquid in the reactor and subsequent discharge through the relief vent.

The first step in the relief sizing calculation for two-phase vents is to determine the mass flux through the relief. This is computed using Equation 4-105, representing choked two-phase flow through a hole:

Qm=ΔHvAvfggcTsCP−−−−−√(4-105) $\begin{matrix} Q_{m} = \frac{Δ H_{v} A}{v_{fg}} \sqrt{\frac{g_{c}}{T_{s} C_{P}}} & (4-105) \end{matrix}$

The heat losses through the reactor walls are assumed to be negligible where, for this case,

Q_m is the mass flow through the relief (mass/time),

ΔH_V is the heat of vaporization of the fluid (energy/mass),

A is the area of the hole (area),

v_fg is the change of specific volume of the flashing liquid (volume/mass),

C_P is the heat capacity of the fluid (energy/mass degrees), and

T_s is the absolute saturation temperature of the fluid at the set pressure (degrees).

The mass flux G_T is given by

GT=QmA=ΔHVvfggcTsCP−−−−−√(10-15) $\begin{matrix} G_{T} = \frac{Q_{m}}{A} = \frac{Δ H_{V}}{v_{fg}} \sqrt{\frac{g_{c}}{T_{s} C_{P}}} & (10-15) \end{matrix}$

Equation 10-15 applies to two-phase relief through a hole. For two-phase flow through pipes, an overall dimensionless discharge coefficient ψ is applied. Equation 10-15 is the so-called equilibrium rate model (ERM) for low-quality choked flow.⁵ Leung showed that Equation 10-15 must be multiplied by a factor of 0.9 to bring the value in line with the classic homogeneous equilibrium model (HEM).⁶ The result should be generally applicable to homogeneous venting of a reactor (low quality, not restricted to just liquid inlet condition):

⁵H. K. Fauske. “Flashing Flows or Some Practical Guidelines for Emergency Releases.” Plant Operations Progress 4, no. 3 (July 1985), pp. 132-134.

⁶J. C. Leung. “Simplified Vent Sizing Equations for Emergency Relief Requirements in Reactors and Storage Vessels.” AICHE Journal 32, no. 10 (1986): 1622.

GT=QmA=0.9ψΔHVvfggcTsCP−−−−−√(10-16) $\begin{matrix} \begin{matrix} G_{T} = \frac{Q_{m}}{A} = 0.9 ψ \frac{Δ H_{V}}{v_{fg}} \sqrt{\frac{g_{c}}{T_{s} C_{P}}} \end{matrix} \end{matrix} \begin{matrix} (10-16) \end{matrix}$

Values for ψ are provided in Figure 10-7. For a pipe of length 0, ψ = 1. As the pipe length increases, the value of ψ decreases.

A graph compares the correction factor (psi) and L over D for two-phase flashing flow through pipes. — **Figure 10-7** Correction factor ψ correcting for two-phase flashing flow through pipes. Data from J. C. Leung and M. A. Grolmes. “The Discharge of Two-Phase Flashing Flow in a Horizontal Duct,” *AICHE Journal* 33, no. 3 (1987): 524–527.

A graph shows the correction factor (psi) plotted against L over D. The horizontal axis, representing L over D, ranges from 0 to 400, in increments of 100. The vertical axis, representing correction factor (psi), ranges from 0.5 to 1.1, in increments of 0.1. The graph begins at (0, 1.0), after which there is a steady downward trend when it falls to the lowest point at (400, 0.56).

A somewhat more convenient expression is derived by rearranging Equation 4-103 to yield

ΔHVvfg=TsdPdT(10-17) $\begin{matrix} \frac{Δ H_{V}}{v_{fg}} = T_{s} \frac{d P}{d T} & (10-17) \end{matrix}$

Substituting into Equation 10-16, we obtain

GT=0.9ψdPdTgcTsCP−−−−√(10-18) $\begin{matrix} G_{T} = 0.9 ψ \frac{d P}{d T} \sqrt{\frac{g_{c} T_{s}}{C_{P}}} & (10-18) \end{matrix}$

The exact derivative is approximated by a finite-difference derivative to yield

GT≅0.9ψΔPΔTgcTsCP−−−−√(10-19) $\begin{matrix} \begin{matrix} G_{T} ≅ 0.9 ψ \frac{Δ P}{Δ T} \sqrt{\frac{g_{c} T_{s}}{C_{P}}} \end{matrix} & (10-19) \end{matrix}$

where

ΔP is the overpressure (force/area), and

ΔT is temperature rise corresponding to the overpressure (degrees).

The required vent area is computed by solving a particular form of the dynamic energy balance. Details are provided elsewhere.⁷ The result is

⁷J. C. Leung. “Simplified Vent Sizing Equations for Emergency Relief Requirements in Reactors and Storage Vessels.” AICHE Journal 32, no. 10 (1986): 1622.

A=moqGT(VmoΔHVvfg−−−−−−−−√+CVΔT−−−−−−√)2(10-20) $\begin{matrix} \begin{matrix} A = \frac{m_{o} q}{G_{T} {(\sqrt{\frac{V}{m_{o}} \frac{Δ H_{V}}{v_{fg}}} + \sqrt{C_{V} Δ T})}^{2}} \end{matrix} & (10-20) \end{matrix}$

An alternative form is derived by applying Equation 10-17:

A=moqGT(VmoTsdPdT−−−−−−−−√+CvΔT−−−−−√)2(10-21) $\begin{matrix} \begin{matrix} A = \frac{m_{o} q}{G_{T} {(\sqrt{\frac{V}{m_{o}} T_{s} \frac{d P}{d T}} + \sqrt{C_{v} Δ T})}^{2}} \end{matrix} & (10-21) \end{matrix}$

For Equations 10-20 and 10-21, the following additional variables are defined:

m_o is the total mass contained within the reactor vessel before relief (mass),

q is the exothermic heat release rate per unit mass (energy/time),

V is the volume of the vessel (volume), and

C_v is the liquid heat capacity at constant volume (energy/mass degrees).

For both Equations 10-20 and 10-21, the relief area is based on the total heat added to the system (numerator) and the heat removed or absorbed (denominator). The first term in the denominator corresponds to the net heat removed by the liquid and vapor leaving the system; the second term corresponds to the heat absorbed as a result of increasing the temperature of the liquid because of the overpressure.

The heat input q resulting from an exothermic reaction is determined using fundamental kinetic information or from a calorimeter (see Table 8-8). For data obtained using the Vent Sizing Package (VSP), the equation

q=12Cv[(dTdt)s+ (dTdt)m](10-22) $\begin{matrix} q = \frac{1}{2} C_{v} [{(\frac{d T}{d t})}_{s} + {(\frac{d T}{d t})}_{m}] & (10-22) \end{matrix}$

is applied, where the derivative denoted by the subscript “s” corresponds to the heating rate at the set pressure and the derivative denoted by the subscript “m” corresponds to the temperature rise at the maximum turnaround pressure. Both derivatives are determined experimentally using the VSP.

The equations include the following assumptions:

Uniform froth or homogeneous vessel venting occurs.
The mass flux G_T varies little during the relief.
The reaction energy per unit mass q is treated as a constant.
The physical properties C_V, ΔH_V, and v_fg are constant.
The system is a tempered reactor system. This applies to most reaction systems.

Units are a particular problem when using the two-phase equations. The best procedure is to convert all energy units to their mechanical equivalents before solving for the relief area, particularly when English engineering units are used.

Example 10-6

Leung⁸ reported on the data of Huff ⁹ involving a 3500-gal reactor with styrene monomer undergoing adiabatic polymerization after being heated inadvertently to 70°C. The MAWP of the vessel is 5 bar. Given the following data, determine the relief vent diameter required. Assume a set pressure of 4.5 bar and a maximum pressure of 5.4 bar absolute:

⁸J. C. Leung. “Simplified Vent Sizing Equations for Emergency Relief Requirements in Reactors and Storage Vessels.” AICHE Journal 32, no. 10 (1986): 1622.

⁹J. E. Huff. “Emergency Venting Requirements.” Plant/Operations Progress 1, no. 4 (1982): 211.

Data:

Volume (V ): 3500 gal = 13.16 m³

Reaction mass (m_o): 9500 kg

Set temperature (T_s): 209.4°C = 482.5 K

Data from VSP:

Maximum temperature (T_m): 219.5°C = 492.7 K

(dT/dt)_s = 29.6°C/min = 0.493 K/s

(dT/dt)_m = 39.7°C/min = 0.662 K/s

Physical Property Data:

	4.5-Bar set	5.4-Bar peak
v_f (m³/kg)	0.001388	0.001414
v_g (m³/kg)	0.08553	0.07278
C_P (kJ/kg K)	2.470	2.514
ΔH_V (kJ/kg)	310.6	302.3

Solution

The heating rate q is determined using Equation 10-22:

q=12CV[(dTdr)s+(dTdt)m](10-22) $\begin{matrix} q = \frac{1}{2} C_{V} [{(\frac{d T}{d r})}_{s} + {(\frac{d T}{d t})}_{m}] & (10-22) \end{matrix}$

Assuming that C_V = C_P, we have

q=12(2.470 kJ / kg K)(0.493+0.662)(K / s)=1.426 kJ / kg s $\begin{matrix} q & = \frac{1}{2} (2.470 kJ / kg K) (0.493 + 0.662) (K / s) \\ = 1.426 kJ / kg s \end{matrix}$

The mass flux is given by Equation 10-16. Assuming L/D = 0, ψ = 1.0:

GT=0.9ψΔHVνfggcTsCP−−−−−√=(0.9)(1.0)(310,600 J / kg)(1 N m / J)(0.08553−0.001388) m3/ kg×[1 (kg m / s2) / N](2470 J / kg K)(482.5 K)[1 N m / J]−−−−−−−−−−−−−−−−−−−−−−−−−−−√=3043 kg / m2s $\begin{matrix} G_{T} & = 0.9 ψ \frac{Δ H_{V}}{ν_{fg}} \sqrt{\frac{g_{c}}{T_{s} C_{P}}} \\ = (0.9) (1.0) \frac{(310, 600 J / kg) (1 N m / J)}{(0.08553 - 0.001388) m^{3} / kg} \times \sqrt{\frac{[1 (kg m / s^{2}) / N]}{(2470 J / kg K) (482.5 K) [1 N m / J]}} \\ = 3043 kg / m^{2} s \end{matrix}$

The relief vent area is determined from Equation 10-20. The change in temperature ΔT is T_m – T_s = 492.7 – 482.5 = 10.2 K:

$\begin{matrix} A & = \frac{m_{o} q}{G_{T} {(\sqrt{\frac{V}{m_{o}} \frac{Δ H_{V}}{ν_{fg}}} + \sqrt{C_{V} Δ T})}^{2}} \\ = \frac{(9500 kg) (1426 J / kg s) (1 N m/J)}{3043 kg / m^{2} s} \\ \times {(\sqrt{(\frac{13.16 m^{3}}{9500kg}) {\frac{(310, 600 J / kg) (1 N m/J)}{0.084l4 m^{3} / kg}}} + \sqrt{(2470 J / kg K) (10.2 K) (1 N m/J)})}^{- 2} \\ = 0 . {084m}^{2} \end{matrix}$

The required relief diameter is

$d = \sqrt{\frac{4 A}{π}} = \sqrt{\frac{(4) (0.084 m^{2})}{314}} = 0.327 m$

Suppose that all vapor relief was assumed. The size of the required vapor-phase rupture disc is determined by assuming that all the heat energy is absorbed by the vaporization of the liquid. At the set temperature, the heat release rate q is

$= C_{V} {(\frac{d T}{d t})}_{s} = (2.470 kJ / kg K) (0.493 K / s) = 1.218 kJ / kg s$

The vapor mass flow through the relief is then

$\begin{matrix} Q_{m} & = \frac{q m_{o}}{Δ H_{v}} \\ = \frac{(1218 J / kg s) (9500 kg)}{(310, 600 J / kg)} \\ = 37.2 kg / s \end{matrix}$

Equation 10-9 provides the required relief area. The molecular weight of styrene is 104. Assume that γ = 1.32 and C_o = 1.0. Then,

$\begin{matrix} \begin{matrix} \end{matrix} A & = \frac{Q_{m}}{C_{o} P} \sqrt{\frac{R_{g} T}{γ g_{c} M} {(\frac{2}{γ + 1})}^{(γ +1) / (1 - γ)}} \\ = \frac{37.2 kg / s}{(1.0) (4.5 bar) (100, 000 Pa / bar) [1 (N / m^{2}) / Pa]} \\ \times \sqrt{\frac{(8314 Pa m^{3} / kg-mole K) (482.5 K) [1 (N / m^{2}) / Pa)}{(1.32) [1 (kg m / s^{2}) / N] (104 kg / kg-mole)}} \times \sqrt{{(\frac{2}{2.32})}^{232 / (- 032)}} \\ = 0.0242 m^{2} \end{matrix}$

This requires a relief device with a diameter of 0.176 m, a significantly smaller diameter than for two-phase flow. Thus, if the relief were sized assuming all-vapor relief, the result would be physically incorrect and the relief would be inadequate to protect the reactor.

10-5 Deflagration Venting for Dust and Vapor Explosions

An uncontrolled dust or gas/vapor explosion in a warehouse, storage bin, or processing unit can eject high-velocity structural debris over a considerable area, propagating the accident and resulting in increased injuries. Deflagration venting reduces the impact of dust and vapor cloud explosions by controlling the release of the explosion energy. The energy of the explosion is directed away from plant personnel and equipment. Deflagration vents are considered mitigative safeguards since they reduce the magnitude of the incident.

Deflagration venting in buildings and process vessels is achieved by using blowout panels, as shown in Figure 10-8. The blowout panel is designed to have less strength than the walls of the structure. Thus, during an explosion, the blowout panels are preferentially detached and the explosive energy is vented. Damage to the remaining structure and equipment is minimized. For particularly explosive dusts or vapors, it is not unusual for the walls (and perhaps roof) of the entire structure to be constructed of blowout panels.

A figure depicts the deflagration vents for structures and process vessels. — **Figure 10-8** Deflagration vents for structures and process vessels.

Three figures illustrate the deflagration vents for structure and process vessels. The first figure shows a rectangular layout named "structure" and the upper portion is assembled in the form of a triangle. A vent is placed at the right end of the structure. Here, the position of the vent is said to be normal. The second figure depicts the dust or vapor explosion occurrence in the structure. At this stage, the deflagration vent is held in open position to reduce the overpressure and resulting damage. An attachment of vent prevents the projection of vent debris. The third diagram shows a bent pipe, passing through the process vessel which is attached with a vent at its right end.

The actual construction details of blowout panels are beyond the scope of this text. A detached blowout panel moving at high velocity can cause considerable damage. Therefore, a mechanism must be provided to retain the panel during the deflagration process. Furthermore, thermal insulation of panels is required for heating and cooling. Construction details are available in manufacturers’ literature.

Blowout panels must be sized to provide the proper relief area. The relief sizing depends on many factors that impact the design of these vents. A partial list includes:

The physical state of the material combusting, including flammable dusts, gases/vapors, or a combination (hybrid) mixture.
The combustion behavior of the material combusting:

For gases/vapors, this includes the flammable limits, the maximum pressure during combustion, and the deflagration index, K_G.

For dusts, this includes the particle size and shape, the dust loading, the maximum pressure during combustion, and the deflagration index, K_St.
The geometry of the structure being protected, including the volume, surface area, and structure shape.
The congestion inside the structure due to equipment. Equipment within the structure will result in turbulence augmentation during the combustion, which will increase the robustness of the combustion.
The strength of the structure being protected.
The strength and inertia of the vent panel.

Deflagration vent design is segregated into two categories: (1) gases, vapors, and mists and (2) dusts and hybrid mixtures.

Vents for Gases/Vapors and Mists

Vent design for gases/vapors and mists is based on the Runes (pronounced “Roo-ness”) equation:¹⁰

¹⁰NFPA 68, Guide for Venting of Deflagrations (Quincy, MA: National Fire Protection Association, 2018).

$\begin{matrix} A_{v} = \frac{A_{s} C}{\sqrt{P_{red}}} & (10-23) \end{matrix}$

where

A_v is the required vent area (area),

A_s is the internal surface area of the structure being protected, including the walls and ceiling (area),

C is a vent constant that depends on the nature of the combustible material (pressure)^1/2, and

P_red is the maximum internal pressure developed in the structure during the venting process (pressure). It is based on the pressure rating of the weakest element of the structure (force/area).

The vent constant, C, is estimated using the following equation:

$\begin{matrix} C = \frac{S_{u} ρ_{u} λ}{2 G_{u} C_{d}} [{(\frac{P_{max} + 1}{P_{o} + 1})}^{1 / γ_{b}} - 1] {(P_{o} + 1)}^{1 / 2} & (10-24) \end{matrix}$

where

S_u is the burning parameter, discussed later in this section (m/s),

ρ_u is the mass density of the unburned gas–air mixture (kg/m³),

λ is the turbulent augmentation factor (unitless),

G_u is the unburned gas–air mixture sonic flow mass flux (kg/m²s),

C_d is the vent flow discharge coefficient (unitless),

P_max is the maximum unvented explosion pressure by ignition of the same gas–air mixture (barg),

P_o is the initial pressure in the enclosure prior to ignition (barg), and

γ_b is the heat capacity ratio for the burned gas–air mixture.

More details on how to estimate these parameters are provided in NFPA 68.

Equations 10-23 and 10-24 apply only for P_red ≤ 0.1 barg (1.5 psig) and P_stat ≤ P_red = 0.024 barg, where P_stat is the pressure at which the vent activates. NFPA 68 provides equations for other conditions.

The turbulent augmentation factor, λ, is estimated using an arduous procedure provided in NFPA 68. It accounts for enhancement of the combustion due to induced turbulence by objects within the structure. Such objects may include piping, conduit, structural columns and beams, stairways and railings and any other equipment. Turbulence enhancement will require a much larger vent area. Without turbulence enhancement, the turbulent augmentation factor has a value of 1.

It is always best to use experimentally determined values for C. However, for estimation purposes, for P_max less than 9 barg and a fuel concentration less than 10% of the stoichiometric concentration, the following equation can be used:

$\begin{matrix} C = 0.0223 λ S_{u} & (10-25) \end{matrix}$

where

C is the vent constant ( $\sqrt{bar}$ ),

λ is the turbulent augmentation factor (unitless), and

S_u is the burning parameter (m/s).

Table 10-2 provides burning parameters, S_u for various gases.

Table 10-2 Burning parameters, S_u, for a variety of gases.

Gas	Burning parameter (cm/s)
Acetylene	166
Benzene	48
n-Butane	45
Carbon disulfide	58
Carbon monoxide	46
n-Decane	43
Diethyl ether	47
Ethane	47
Ethylene	80
Gasoline (100 octane)	40
n-Hexane	46
Hydrogen	312
Isopropyl alcohol	41
Methane	40
Methyl alcohol	56
n-Pentane	46
Propane	46
Toluene	41

Selected from NFPA 68, Guide for Venting of Deflagrations (Quincy, MA: National Fire Protection Association, 2018).

Britton proposed the following correlation for estimating the burning parameter, S_u:¹¹

¹¹Britton, L. G. “Using Heats of Oxidation to Evaluate Flammability Hazards.” Process Safety Progress 21, no. 1 (2000): 1–24.

$\begin{matrix} S_{u} = 1666 - 34.23 (\frac{Δ H_{c}}{z}) - 0.180 {(\frac{Δ H_{c}}{z})}^{2} & (10-26) \end{matrix}$

where

S_u is the burning parameter (m/s),

$Δ H_{c}$ is the heat of combustion of the fuel (kcal/mol),and

z is the stoichiometric ratio of oxygen to fuel.

For a fuel with a composition of C_cH_hO_mN_nX_x, the stoichiometric ratio of oxygen to fuel is calculated from

$\begin{matrix} z = c + \frac{h - x - 2 m}{4} & (10-27) \end{matrix}$

Example 10-7

A room is used for dispensing flammable liquids. The liquids are expected to have a burning parameter of 50 cm/s. The room is 9 m long by 6 m wide by 6 m in height. Three of the walls are shared with an adjoining structure. The fourth and larger wall of the room is on the outer surface of the structure. The three inside walls are capable of withstanding a pressure of 0.05 barg. Estimate the vent area required for this operation.

Solution

The vent must be installed on the larger outer wall to vent the combustion away from the adjoining structure. The venting constant for this flammable vapor is calculated using Equation 10-25. Assume no turbulent augmentation, so λ = 1. Then,

$C = 0.0223 λ S_{u} = 0.0223 (1) (0.050 m/s) = 1.12 \times 10^{- 3} {bar}^{1 / 2}$

Equation 10-23 is used to estimate the required vent area. The total surface area of the room (including floor and ceiling) is

$A_{s} = (2) (9 m) (6 m) + (2) (6 m) (6 m) + (2) (6 m) (9 m) = 288 m^{2}$

The required vent area is then

$A_{v} = \frac{A_{s} C}{\sqrt{P_{red}}} = \frac{(288 m^{2}) (1.12 \times 10^{- 3} {bar}^{1 / 2})}{\sqrt{0.05 bar}} =1.4 m^{2}$

The area of the large outer wall has more than adequate space to accommodate this vent.

Vents for Dusts and Hybrid Mixtures

The vent design for dusts and hybrid mixtures is based on the definition of a deflagration index for gases or dusts:

$\begin{matrix} K_{St} = {(\frac{d P}{d t})}_{\max} V^{1 / 3} & (10-28) \end{matrix}$

where

K_St is the deflagration index for the dust (bar m/s),

(dP/dt)_max is the maximum pressure increase, determined experimentally (bar/s), and

V is the volume of the vessel (m³).

Chapter 6 discusses how to measure the deflagration index for dusts and provides a table of typical values.

When the structure initial pressure is between –0.2 barg and 0.2 barg, and the vent opening pressure, P_stat, is less than 0.75 barg, then the following equation is used to estimate the vent area required:

$\begin{matrix} A_{v} = 1 \times 10^{- 4} K_{St} V^{3 / 4} (1 + 1.54 P_{stat}^{4 / 3}) \sqrt{\frac{P_{max}}{P_{red}} - 1} & (10-29) \end{matrix}$

where

A_v is the required vent area (m²),

K_St is the deflagration index for the dust, given by Equation 10-28 (bar m/s),

V is the internal volume of the enclosure (m³),

P_stat is the opening pressure for the vent (barg),

P_max is the maximum pressure of an unvented deflagration initially at atmospheric pressure (barg),

P_red is the maximum internal pressure developed in the structure during the venting process (barg). This is based on the pressure rating of the weakest element of the structure.

Equation 10-29 applies only for the following conditions:

$\begin{matrix} 0.1 m^{3} \leq V \leq {10,000 m}^{3} \\ 10 bar m/s \leq K_{S t} \leq 800 bar m/s & (10-30) \\ 5 barg \leq P_{max} \leq 12 barg \end{matrix}$

NFPA 68 provides equations for other conditions.

Example 10-8

Consider again the room of Example 10-7, but now assume it is handling a dust. In this case, the walls have been reinforced to withstand a pressure of 0.4 barg (P_red). The vent will open at 0.2 barg (P_stat), the maximum pressure of the exploding dust is 10 barg (P_max), and K_St of the dust is 200 bar-m/s. Estimate the vent area required to protect this enclosure.

Solution

The volume of the structure is

$(9 m) (6 m) (6 m) = 324 m^{3}$

All of the conditions of Equations 10-30 apply, and Equation 10-29 is used to solve this problem. Substituting the numbers into the fixed units equation,

$\begin{matrix} A_{v} & = 1 \times 10^{- 4} K_{St} V^{3 / 4} (1 + 1.54 P_{stat}^{4 / 3}) \sqrt{\frac{P_{max}}{P_{red}} - 1} \\ = (1 \times 10^{- 4}) (200 bar m/s) {(324 m^{3})}^{3 / 4} [1 + 1.54 {(0.2 barg)}^{4 / 3}] \sqrt{\frac{10 barg}{0.4 barg} - 1} \\ = 8.83 m^{2} \end{matrix}$

10-6 Venting for Fires External to the Process

Fires external to process vessels can result in heating and boiling of process liquids, as shown in Figure 10-9. Venting is required to prevent explosion of these vessels.

A figure displays the heating of process vessels by means of external fire. — **Figure 10-9** Heating of a process vessel as a result of an external fire. Venting is required to prevent vessel rupture. For most fires, only a fraction of the external vessel is exposed to fire.

A figure shows a storage vessel with a narrow opening at the center portion. The part of the vessel is exposed to an external fire. As a result, foam is developed in the liquid. Hence, pressure occurs in the vessel due to the increased vapor pressure or decomposition products.

Two-phase flow during these reliefs is possible, but not likely. For runaway reactor reliefs, the energy is generated by reaction throughout the entire reactor liquid contents. For heating caused by external fire, the heating occurs only at the surface of the vessel. Thus, for vessels exposed to fire, the liquid boiling will occur only next to the wall; the resulting two-phase foam or froth at the liquid surface will not have a substantial thickness. Two-phase flow during fire relief can be prevented by providing a suitable vapor space above the liquid within the vessel.

Two-phase fire relief equations are available to support conservative designs. Leung provides an equation for the maximum temperature based on an energy balance around the heated vessel.¹² This equation assumes a constant heat input rate Q:

¹²J. C. Leung. “Simplified Vent Sizing Equations for Emergency Relief Requirements in Reactors and Storage Vessels.” AICHE Journal 32, no. 10 (1986): 1622.

$\begin{matrix} T_{m} - T_{s} = \frac{Q}{G_{T} A C_{V}} [\ln (\frac{m_{o}}{V} \frac{Q}{G_{T} A} \frac{v_{fg}}{Δ H_{V}}) - 1] + \frac{V Δ H_{V}}{m_{o} C_{V} v_{fg}} & (10-31) \end{matrix}$

where

T_m is the maximum temperature in the vessel (degrees),

T_s is the set temperature corresponding to the set pressure (degrees),

Q is the constant heat input rate (energy/time),

G_T is the mass flux through the relief (mass/volume-time),

A is the area of the relief (area),

C_V is the heat capacity at constant volume (energy/mass-degrees),

m_o is the liquid mass in the vessel (mass),

V is the volume of the vessel (volume),

v_fg is the specific volume difference between the vapor and liquid phases (volume/mass), and

ΔH_v is the heat of vaporization of the liquid (energy/mass).

The solution to Equation 10-31 for G_TA is obtained by an iterative or trial-and-error technique. Equation 10-31 is likely to produce multiple roots, in which case the correct solution is the minimum mass flux G_T. For the special case of no overpressure, T_m = T_s, and Equation 10-31 reduces to

$\begin{matrix} A = \frac{Q m_{o} v_{fg}}{G_{T} V Δ H_{v}} & (10-32) \end{matrix}$

Various relationships have been recommended for estimating the heat added to a vessel that is engulfed in fire. For regulated materials, the OSHA 1910.106 criterion is mandatory.¹³ Other standards are also available.¹⁴,¹⁵ Crozier, after analysis of the various standards, recommended the following equations for determining the total heat input Q:¹⁶

¹³OSHA 1910.106, Flammable and Combustible Liquids (Washington, DC: U.S. Department of Labor, 1996).

¹⁴API Standard 2000, Venting Atmospheric and Low-Pressure Storage Tanks (Nonrefrigerated and Refrigerated), 5th ed. (Washington, DC: American Petroleum Institute, 1998).

¹⁵NFPA 30, Flammable and Combustible Liquids Code (Quincy, MA: National Fire Protection Association, 2000).

¹⁶R. A. Crozier. “Sizing Relief Valves for Fire Emergencies.” Chemical Engineering (October 28, 1985), pp. 49-54

$\begin{matrix} \begin{array}{l} \begin{array}{l} Q = 20, 000 A & for 20 < A < 200 \\ Q = 199, 300 A^{0.566} & for 200 < A < 1000 \end{array} \\ \begin{array}{l} Q = 936, 400 A^{0.338} & for 1000 < A < 2800 \\ Q = 21, 000 A^{0.82} & for A > 2800 \end{array} \end{array} & (10-33) \end{matrix}$

where

A is the area absorbing heat (in ft²) for the following geometries:

For spheres, 55% of total exposed area;
For horizontal tanks, 75% of total exposed area;
For vertical tanks, 100% of total exposed area for first 30 ft; and

Q is the total heat input to the vessel (in Btu/hr).

The mass flux G_T is determined using Equation 10-16 or 10-19.

API 520 suggests a slightly different approach to estimate the heat flux to process equipment as a result of a fire.¹⁷ If prompt firefighting is available and if the flammable material is drained away from the vessel, then the heat flux is estimated using the following equation:

¹⁷API RP 520, Sizing, Selection, and Installation of Pressure-Relieving Devices in Refineries, 6th ed. (Washington, DC: American Petroleum Institute, 1993).

$\begin{matrix} Q = 21, 000 F A^{0.82} & (10-34) \end{matrix}$

If adequate firefighting and drainage do not exist, then the following equation is recommended:

$\begin{matrix} Q = 34, 500 F A^{0.82} & (10-35) \end{matrix}$

where

Q is the total heat input through the surface of the vessel (BTU/hr),

F is an environment factor (unitless), and

A is the total wetted surface of the vessel (ft²).

The environment factor F is used to account for vessel protection from insulation. A number of values for various insulation thicknesses are shown in Table 10-3.

Table 10-3 Environment Factors F for Equations 10-34 and 10-35

Insulation thickness (in)	Environment factor F
0	1.0
1	0.30
2	0.15
4	0.075

The surface area A is the area of the vessel wetted by its internal liquid with a height less than 25 ft above the flame source. Wong provided considerably more detail on how to determine this as well as a number of equations for various vessel geometries.¹⁸

¹⁸W. Y. Wong. “Fires, Vessels, and the Pressure Relief Valve.” Chemical Engineering (May 2000) 107: 84.

Example 10-9

Leung reported on the computation of the required relief area for a spherical propane vessel exposed to fire¹⁹. The vessel has a volume of 100 m³ and contains 50,700 kg of propane. A set pressure of 4.5 bars absolute is required. This corresponds to a set temperature, based on the saturation pressure, of 271.5 K. At these conditions, the following physical property data are reported:

¹⁹J. C. Leung. “Simplified Vent Sizing Equations for Emergency Relief Requirements in Reactors and Storage Vessels.” AICHE Journal 32, no. 10 (1986): 1622.

$\begin{matrix} C_{P} & = C_{V} = 2.41 \times 10^{3} J / kg K \\ Δ H_{v} & = 3.74 \times 10^{5} J / kg \\ v_{fg} & = 0 . {1015 m}^{3} / kg \end{matrix}$

The molecular weight of propane is 44.

Solution

The problem is solved by assuming no overpressure during the relief. The relief vent area calculated is larger than the actual area required for a real relief device with overpressure.

The diameter of the sphere is

$d = {(\frac{6 V}{π})}^{1 / 3} = {[\frac{(6) (100 m^{3})}{(3.14)}]}^{1 / 3} = 5.76 m$

The surface area of the sphere is

$π d^{2} = (3.14) {(5.76 m)}^{2} = 104.2 m^{2} = 1121 {ft}^{2}$

The area exposed to heat is given by the geometry factors provided with Equation 10-33:

$A = (0.55) (1121 {ft}^{2}) = 616 {ft}^{2}$

The total heat input is found using Equation 10-33:

$\begin{matrix} Q & = 199, 300 A^{0.566} & = (199, 300) (616 f t^{2})^{0.566} =7.56 \times 10^{6} Btu/hr \\ = 2100 Btu/sec & = 2.22 \times 10^{6} J/s \end{matrix}$

If we use Equation 10-34 and assume that the vessel is full of liquid, then the entire vessel surface area is exposed to the fire. If we also assume that no insulation is present, then F = 1.0. Then,

$Q = 21, 000 F A^{0.82} = (21, 000) (1.0) {(1121 {ft}^{2})}^{0.82} = 6.65 \times 10^{6} Btu / hr$

which is close to the value estimated using Equation 10-33. From Equation 10-16 and assuming ψ = 1.0, we obtain

$\begin{matrix} G_{T} & = \frac{Q_{m}}{A} = 0.9 ψ \frac{Δ H_{V}}{v_{fg}} \sqrt{\frac{g_{c}}{T_{s} C_{P}}} \\ = (0.9) (1.0) (\frac{3.74 \times 10^{5} J / kg}{0 . {1015 m}^{3} / kg}) (\frac{1 N m}{J}) \times \sqrt{\frac{1 (kg m / s^{2}) / N}{(271.5 K) (2.41 \times 10^{3} J / kg K) (1 N m / J)}} \\ = 4.10 \times 10^{3} kg / m^{2} s \end{matrix}$

The required vent area is determined from Equation 10-32:

$\begin{matrix} A & = \frac{Q m_{o} ν_{fg}}{G_{T} V Δ H_{v}} \\ = \frac{(2.22 \times 10^{6} J / s) (50, 700 kg) (0.1015 m^{3} / kg)}{(4.10 \times 10^{3} kg / m^{2} s) (100 m^{3}) (3.74 \times 10^{5} J / kg)} \\ = 0.0745 m^{2} \end{matrix}$

The required relief diameter is

$\begin{matrix} d & = \sqrt{\frac{4 A}{π}} = \sqrt{\frac{(4) (0.0745 m^{2})}{3.14}} \\ = 0.308 m = 12.1 in \end{matrix}$

An alternative way to look at the problem might be to ask the question: What initial fill fraction should be specified in the tank to avoid two-phase flow during a fire exposure incident? No tested correlations are presently available to compute the height of a foam layer above the boiling liquid.

For fire reliefs with single-phase vapor flow, the equations provided in Section 10-3 are used to determine the size of the relief.

As mentioned previously, two-phase flow discharges for fire scenarios are possible, but not likely. To size the relief for fire and a single-vapor phase, use the heat input determined from Equations 10-33 through 10-35, and determine the vapor mass flow rate through the relief by dividing the heat input by the heat of vaporization of the liquid. This assumes that all the heat input from the fire is used to vaporize the liquid. The relief area is then determined using Equations 10-12 through 10-14.

10-7 Reliefs for Thermal Expansion of Process Fluids

Liquids contained within process vessels and piping will normally expand when heated. The expansion will damage pipes and vessels if the pipe or vessel is filled completely with fluid and the liquid is blocked in.

A typical situation is thermal expansion of water in cooling coils in a reactor, shown in Figure 10-10. If the coils are filled with water and are accidentally blocked in, the water will expand when heated by the reactor contents, leading to damage to the cooling coils.

A figure illustrates the damage of cooling coils due to external heating of blockage in cooling fluid. — **Figure 10-10** Damage to cooling coils as a result of external heating of blocked-in cooling fluid.

The damage of cooling coils due to external heating of blockage in cooling fluid is explained. The reactor consists of four cooling coils filled with liquid. The initial coil is attached with a cooling water outlet and the last coil is attached with a cooling water inlet. A hot process liquid inlet (left) and pressure indicator (right) are connected at the top region of the reactor. Both the cooling water valves are closed.

Relief vents are installed in these systems to prevent damage resulting from liquid expansion. Although this may appear to be a minor problem, damage to heat exchange systems can result in (1) contamination of product or intermediate substances, (2) subsequent corrosion problems, (3) substantial plant outages, and (4) large repair expenses. Failure in heat exchange equipment is also difficult to identify, and repairs are time-consuming.

A thermal expansion coefficient for liquids, β, is defined as

$\begin{matrix} β = \frac{1}{V} (\frac{d V}{d T}) & (10-36) \end{matrix}$

where

V is the volume of the fluid (volume), and

T is the temperature (degrees).

Table 10-4 lists thermal expansion coefficients for a number of substances. Water behaves in an unusual fashion. Its thermal expansion coefficient decreases with increasing temperature up to about 4°C, after which the thermal expansion coefficient increases with temperature. Coefficients for water are readily determined from the steam tables.

Table 10-4 Thermal Expansion Coefficients for a Variety of Liquids

Liquid	Density at 20°C (kg/m³)	Thermal expansion coefficient (°C^–1)
Alcohol, ethyl	791	112 × 10^–5
Alcohol, methyl	792	120 × 10^–5
Benzene	877	124 × 10^–5
Carbon tetrachloride	1595	124 × 10^–5
Ether, ethyl	714	166 × 10^–5
Glycerin	1261	51 × 10^–5
Mercury	13,546	18.2 × 10^–5
Turpentine	873	97 × 10^–5

Source: G. Shortley and D. Williams. Elements of Physics, 4th ed. (Englewood Cliffs, NJ: Prentice Hall, 1965), p. 302.

The volumetric expansion rate Q_v through the relief resulting from thermal expansion is

$\begin{matrix} Q_{v} = \frac{d V}{d t} = \frac{d V}{d T} \frac{d T}{d t} & (10-37) \end{matrix}$

By applying the definition of the thermal expansion coefficient, given by Equation 10-36, we obtain

$\begin{matrix} Q_{v} = β V \frac{d T}{d t} & (10-38) \end{matrix}$

For a pipe or process vessel heated externally by a hot fluid, the energy balance of the fluid is given by

$\begin{matrix} m C_{P} \frac{d T}{d t} = U A (T - T_{a}) & (10-39) \end{matrix}$

where

T is the temperature of the fluid (degrees),

C_P is the heat capacity of the liquid (energy/mass-degrees),

UA is an overall heat transfer coefficient (energy/time-degrees), and

T_a is the ambient temperature (degrees).

It follows that

$\begin{matrix} \frac{d T}{d t} = \frac{U A}{m C_{P}} (T - T_{a}) & (10-40) \end{matrix}$

Substituting into Equation 10-41, we obtain

$\begin{matrix} Q_{v} = \frac{β V}{m C_{P}} U A (T - T_{a}) & (10-41) \end{matrix}$

and, with the definition of the liquid density ρ,

$\begin{matrix} \begin{matrix} Q_{v} = \frac{β}{ρ C_{P}} U A (T - T_{a}) \end{matrix} & (10-42) \end{matrix}$

Equation 10-42 describes the fluid expansion only at the beginning of heat transfer, when the fluid is initially exposed to the external temperature T_a. The heat transfer will increase the temperature of the liquid, changing the value of T. However, it is apparent that Equation 10-42 provides the maximum thermal expansion rate, sufficient for sizing a relief device.

The volumetric expansion rate Q_v is subsequently used in an appropriate equation to determine the relief vent size.

Example 10-10

The cooling coil in a reactor has a surface area of 10,000 ft². Under the most severe conditions, such coils can contain water at 32°F and can be exposed to superheated steam at 400°F. Given a heat transfer coefficient of 50 BTU/hr-ft²-°F, estimate the volumetric expansion rate of the water in the cooling coils in gpm.

Solution

The expansion coefficient β for water at 32°F should be used. This is estimated using liquid volumetric data from the steam tables over a short range of temperatures around 32°F. However, the steam tables do not provide liquid-water–specific volume data at temperatures below 32°F. A value between 32°F and some appropriate higher temperature will suffice. From the steam tables:

Temperature (°F)	Specific volume (ft³/lb_m)
32	0.01602
50	0.01603

The expansion coefficient is computed using Equation 10-36:

$\begin{matrix} β & = \frac{1}{v} \frac{d v}{d T} = \frac{1}{0.016025 {ft}^{3} {/1b}_{m}} (\frac{0.01603 - 0.01602}{50 - 32}) (\frac{{ft}^{3} {/1b}_{m}}{° F}) \\ = 3.47 \times 10^{- 5}^{o} F^{- 1} \end{matrix}$

The volumetric expansion rate is given by Equation 10-42:

$\begin{matrix} Q_{v} & = \frac{β}{ρ C_{P}} U A (T - T_{a}) \\ = \frac{(3.47 \times 10^{- 5} / ° F) (50 BTU / {hr-ft}^{2}^{o} F) (10, 000 {ft}^{2}) (400 - 32) ° F}{(62 . {4 1b}_{m} / {ft}^{3}) (1 Btu / {1b}_{m} ° F)} \\ {=102 ft}^{3} / hr = 12.7 gpm \end{matrix}$

The relief device vent area must be designed to accommodate this volumetric flow.

Problems

10-1. Fill in the blanks in the table.

Relief configuration	MAWP	Set pressure
a. Single relief, nonfire	100 psig	100 psig
b. Single relief, nonfire	100 psig	90 psig
c. Primary relief, nonfire, with secondary relief	100 psig	100 psig
d. Secondary relief, nonfire	100 psig	100 psig
e. Primary relief, fire	100 psig	90 psig
f. Secondary relief, fire	100 psig	100 psig
g. Supplemental relief	100 psig	90 psig
h. Single relief, nonfire	10 barg	barg
i. Single relief, nonfire	10 barg	9 barg
j. Primary relief, nonfire with secondary relief	10 barg	10 barg
k. Secondary relief, nonfire	10 barg	10 barg
l. Primary relief, fire	10 barg	9 barg
m. Secondary relief, fire	10 barg	10 barg
n. Supplemental relief	10 barg	9 barg

10-2. Calculate the diameter of a certified capacity spring-type liquid relief for the following conditions:

Pump capacity at Δp	Set pressure	Overpressure	Backpressure	Valve type	(ρ /ρ_ref)
a. 100 gpm	50 psig	5 psig	5 psig	Conventional	1.0
b. 200 gpm	100 psig	10 psig	10 psig	Balanced-bellows	1.3
c. 5 m³/s	5 barg	0.5 barg	0.5 barg	Balanced-bellows	1.2
d. 7 m³/s	10 barg	1 barg	1 barg	Conventional	1.0

10-3. Calculate the diameter of a spring-type vapor relief for the following conditions. Assume for each case a heat capacity ratio γ = 1.3 and a compressibility z = 1.0.

Set pressure	Temperature	Molecular weight	Mass flow	Overpressure	Backpressure
a. 100 psig	100°F	28	50 lb_m/hr	10 psig	10 psig
b. 150 psig	200°F	44	100 lb_m/hr	30 psig	15 psig
c. 200 psig	100°F	28	150 lb_m/hr	10 psig	20 psig
d. 8 barg	300 K	44	10 kg/s	1 barg	1 barg
e. 10 barg	400 K	28	20 kg/s	2 barg	2 barg
f. 20 barg	400 K	28	30 kg/s	2 barg	1 barg

10-4. Calculate the required diameter for certified-capacity liquid rupture discs for the following conditions. Assume a liquid specific gravity of 1.2 for all cases.

Liquid flow	Set pressure	Overpressure	Backpressure
a. 500 gpm	100 psig	10 psig	5 psig
b. 100 gpm	50 psig	5 psig	2 psig
c. 5 m³/s	10 barg	1 barg	0.5 barg
d. 10 m³/s	20 barg	2 barg	1 barg

10-5. Calculate the diameter for rupture discs in vapor service for the following conditions. Assume that nitrogen is the vent gas.

Gas flow	Temperature	Set pressure	Overpressure	Backpressure
a. 100 lb_m/hr	100^oF	100 psig	10 psig	5 psi
b. 200 lb_m/hr	150^oF	150 psig	15 psig	10 psi
c. 10 kg/s	400 K	10 barg	1 barg	0.5 barg
d. 20 kg/s	500 K	20 barg	1 barg	0.5 barg

10-6. Determine the relief diameter for the following two-phase flow conditions. Assume in all cases that L/D = 0.0.

	a	b	c	d
Reaction mass	10,000 lb_m	10,000 lb_m	4000 kg	4000 kg
Volume	200 ft³	500 ft³	15 m³	500 m³
Set pressure	100 psia	100 psia	7 bara	7 bara
Set temperature	500°F	500°F	500 K	500 K
(dT/dt)_s	0.5°F/s	0.5°F/s	2.0 K/s	2.0 K/s
Maximum pressure	120 psia	120 psia	8 bara	9 bara
Maximum temperature	520°F	520°F	540 K	540 K
(dT/dt)_m	0.66°F/s	0.66°F/s	2.4 K/s	2.6 K/s
Liquid-specific volume	0.02	0.2	0.02	0.02
Vapor-specific volume	1.4	1.4	1.4	1.4
Heat capacity	1.1 Btu/lb_m°F	1.1 Btu/lb_m°F	5.0 kJ/kg K	5.0 kJ/kg K
Heat of vaporization	130 Btu/lb_m	130 Btu/lb_m	300 kJ/kg	130 kJ/kg

10-7. Determine the deflagration vent size for the following structures:

Vapors	a	b	c	d
Internal area of structure	1000 ft³	1000 ft³	300 m³	300 m³
Turbulent augmentation factor, λ	1.0	1.5	1.0	1.5
Max. internal pressure, P_red	0.05 bar	0.10 bar	0.05 bar	0.10 bar
Gas	Methane	Hydrogen	Methane	Hydrogen

Dusts	e	f	g	h
Volume of structure	1000 ft³	1000 ft³	30 m³	30 m³
Deflagration index, K_St, bar m/s	200	300	200	300
Opening pressure of vent, P_stat	3 psig	6 psig	0.2 barg	0.4 barg
Max. pressure of unvented, P_max	150 psig	200 psig	10 barg	15 barg
Max. internal pressure, P_red	6 psig	8 psig	0.4 barg	0.6 barg

10-8. A cooling coil contains ethyl alcohol. The heat capacity of the alcohol is 0.58 kcal/kg°C, and its density is 791 kg/m³. Determine the relief volumetric flow required. Assume a cooling coil area of 300 m² and a heat transfer coefficient of 1000 J/s m² K. The ambient temperature is 300 K and the maximum temperature to which the alcohol is exposed is 550 K.

10-9. Home hot-water heaters contain relief devices to provide protection in the event that the heater controls fail and the water is heated to a high temperature. A typical water heater contains 40 gal of water and has a heat input of 42,000 Btu/hr. If the heater is equipped with a 150-psia capacity-certified spring relief device, compute the area required for the relief. Single-phase flow is expected. Assume no overpressure.

Also compute the relief vent size assuming all vapor relief. Assume 20% backpressure.

10-10. A cylindrical tank, 4 ft in diameter and 10 ft long, is completely filled with water and blocked in. Estimate the thermal expansion rate of the water if the water is at 50°F and the steel shell of the tank is suddenly heated to 100°F by the sun. Assume a heat transfer coefficient of 50 Btu/hr ft² °F and that only the top half of the tank is heated.

If the tank is exposed to fire, what is the required relief area? Assume no overpressure.

The tank MAWP is 200 psig.

10-11. A 10-ft-wide by 10-ft-long by 10-ft-high shed is used to store tanks of methane. What deflagration vent area is required? Assume a maximum internal overpressure of 0.1 psig.

10-12. A horizontal vessel, 3 m long and 1 m in diameter, contains water. What relief size is required to protect the vessel from fire exposure? Assume the following: vapor relief only, MAWP of 14 barg, conventional capacity-certified spring-operated relief.

Additional homework problems are available in the Pearson Instructor Resource Center.

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Table of Contents for
Chapter 10. Relief Sizing

Chapter 10. Relief Sizing

10-1 Set Pressure and Accumulation Limits for Reliefs

10-2 Relief Sizing for Liquid Service

10-3 Relief Sizing for Vapor and Gas Service

Subcritical Vapor/Gas Flow

Steam Flow Relief Sizing

Rupture Disc Sizing

Pilot-Operated Relief Sizing

Buckling Pin Relief Sizing

10-4 Two-Phase Flow during Runaway Reaction Relief

10-5 Deflagration Venting for Dust and Vapor Explosions

Vents for Gases/Vapors and Mists

Vents for Dusts and Hybrid Mixtures

10-6 Venting for Fires External to the Process

10-7 Reliefs for Thermal Expansion of Process Fluids

Suggested Reading

Deflagration Vents

Relief Codes and Standards

Two-Phase Flow

Problems

Table of Contents for Chapter 10. Relief Sizing

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Chapter 10. Relief Sizing

10-1 Set Pressure and Accumulation Limits for Reliefs

10-2 Relief Sizing for Liquid Service

10-3 Relief Sizing for Vapor and Gas Service

Subcritical Vapor/Gas Flow

Steam Flow Relief Sizing

Rupture Disc Sizing

Pilot-Operated Relief Sizing

Buckling Pin Relief Sizing

10-4 Two-Phase Flow during Runaway Reaction Relief

10-5 Deflagration Venting for Dust and Vapor Explosions

Vents for Gases/Vapors and Mists

Vents for Dusts and Hybrid Mixtures

10-6 Venting for Fires External to the Process

10-7 Reliefs for Thermal Expansion of Process Fluids

Suggested Reading

Deflagration Vents

Relief Codes and Standards

Two-Phase Flow

Problems

Table of Contents for
Chapter 10. Relief Sizing