Vacuum Purging

Vacuum purging is the most commonly used inerting procedure for vessels. This procedure is not used for large storage vessels, however, because those containers are usually not rated for vacuum. Most can withstand only a pressure of a few inches of water gauge.

Many process vessels, including reactors, are often designed for full vacuum—that is, –760 mm Hg gauge or 0.0 mm Hg absolute. Consequently, vacuum purging is a common procedure employed with reactors. The steps in the vacuum purging process include (1) drawing a vacuum on the vessel until the target pressure is reached, (2) relieving the vacuum with an inert gas, and (3) repeating steps 1 and 2 until the target oxygen concentration is obtained.

The process for vacuum purging using the stepwise procedure is shown in Figure 7-1. A vessel of known size is vacuum-purged from an initial oxygen concentration y_o to a final target oxygen concentration y_j. The vessel is initially at pressure P_H and is vacuum-purged using a vacuum at pressure P_L. The objective is to determine the number of cycles required to achieve the target oxygen concentration. Note that the high and low pressures selected for the cycles shown in Figure 7-1 depend on the design pressure of the vessel and the low pressure capability of the facility’s vacuum system.

The oxygen concentration at the end of each vacuum cycle is the same as the concentration at the beginning of the vacuum cycle: Evacuation of the gas from the vessel does not change the concentration. For each pressurization cycle, if the inert does not contain any oxygen, then the number of moles of oxygen remains constant.

Assuming a high pressure P_H and a low pressure or vacuum P_L for each cycle and assuming ideal gas behavior, the total moles of gas at each pressure are

nH=PHVRgT(7-1)

nL=PLVRgT(7-2)

where n_H and n_L are the total moles at the high and low pressures, respectively.

The number of moles of oxidant for the low-pressure P_L and high-pressure P_H are computed using Dalton’s law:

(noxy)1L=y0nL(7-3)

(noxy)1H=y0nH(7-4)

where 1L and 1H are the first vacuum and first atmospheric states, respectively.

When the vacuum is relieved with pure inert gas, the moles of oxidant remains the same as in the vacuum state and the moles of nitrogen increases. The new (lower) oxidant concentration is

y1=(noxy)1LnH(7-5)

where y₁ is the oxygen concentration after the first purge with nitrogen. Substituting Equation 7-3 into Equation 7-5 gives

y1=(noxy)1LnH=y0(nLnH)

If the vacuum and inert relief process is repeated, the concentration after the second purge is

y2=(noxy)2LnH=y1nLnH=y0(nLnH)2

This process is repeated as often as required to decrease the oxidant concentration to a desired level. The concentration after j purge cycles—that is, vacuum and relief—is given by the following general equation:

yj=y0(nLnH)j=y0(PLPH)j(7-6)

This equation assumes that the pressure limits P_H and P_L are identical for each cycle.

The total moles of inert gas added for each cycle is constant. For j cycles, it is given by

ΔnN2= j(PH−PL)VRgT(7-7)

Equations 7-1 through 7-7 assume the gas is well mixed in the vessel, the ideal gas law applies, and the temperature is constant.

Pressure Purging

Vessels can be pressure-purged by adding inert gas under pressure. After the inert gas is added and mixes throughout the vessel, it is vented to the atmosphere, usually down to atmospheric pressure. More than one pressure cycle may be necessary to reduce the oxidant content to the desired concentration. For pressure purging, the high pressure is specified by the inert gas pressure and the pressure rating of the vessel.

The cycles required to reduce the oxygen concentration to a target value are shown in Figure 7-2. In this case, the vessel is initially at P_L and is pressurized using a source of pure inert at P_H. The objective is to determine the number of pressure purge cycles required to reach the desired concentration.

Because the vessel is pressurized with pure inert gas, the number of moles of oxygen remains constant during pressurization, and the mole fraction of oxygen decreases. During depressurization, the composition of the gas within the vessel remains constant, but the total number of moles is reduced. Thus, the oxygen mole fraction remains unchanged.

The equation used for this pressure purging is identical to Equation 7-6, where n_L is now the total moles at atmospheric pressure (low pressure) and n_H is the total moles under pressure (high pressure). In this case, however, the initial concentration of oxidant in the vessel (y_o) is computed after the vessel is pressurized (the first pressurized state). The number of moles for this pressurized state is n_H and the number of moles for the atmospheric case is n_L.

One practical advantage of pressure purging versus vacuum purging is the potential for cycle time reductions. The pressurization process is much more rapid compared to the relatively slow process of pulling a vacuum. Also, the capacity of vacuum systems decreases significantly as the absolute vacuum is decreased. Pressure purging, however, uses more inert gas. Therefore, the best purging process is selected based on cost and performance.

Combined Pressure–Vacuum Purging

In some cases, both pressure and vacuum capabilities are available and are used simultaneously to inert a vessel. The computational procedure depends on whether the vessel is first evacuated or pressurized.

The purging cycles for a pressure-first purge are shown in Figure 7-3. In this case, the beginning of the cycle is defined as the end of the initial pressurization. If the initial oxygen mole fraction is 0.21, the oxygen mole fraction at the end of this initial pressurization is given by

y0=0.21(PoPH)(7-8)

At this point, the remaining cycles are identical to those for pressure purging and Equation 7-6 applies. However, the number of cycles j is the number of cycles after the initial pressurization.

The purging cycles for an evacuate-first purge are shown in Figure 7-4. In this case, the beginning of the cycle is defined as the end of the initial evacuation. The oxygen mole fraction at this point is the same as the initial mole fraction. Furthermore, the remaining cycles are identical to those in the vacuum purge operation and Equation 7-6 is directly applicable. However, the number of cycles j is the number of cycles after the initial evacuation.

Vacuum and Pressure Purging with Impure Nitrogen

The equations developed for vacuum and pressure purging apply to processes using pure inert gases only. In reality, many of the nitrogen separation processes available today do not provide pure nitrogen; instead, they typically provide nitrogen in the 98% range, with the remainder of the gas being usually oxygen.

Assume that the nitrogen used as an inerting gas also contains oxygen at a constant mole fraction of y_oxy. For a pressure purging procedure. the total moles of oxygen present at the end of the first pressurization is given by the moles initially present plus the moles included with the nitrogen. This amount is

noxy=y0(PLVRgT)+yoxy(PH−PL)VRgT(7-9)

The total moles in the vessel at the end of the first pressurization is given by Equation 7-1. Thus, the mole fraction of oxygen at the end of this cycle is

y1=noxyntot=y0(PLPH)+yoxy(1−PLPH)(7-10)

This result is generalized into a recursive equation (Equation 7-11) and a generalized equation (Equation 7-13), for the oxygen concentration at the end of the jth pressure cycle:

yj=yj−1(PLPH)+yoxy(1−PLPH)(7-11)

Rearrange Equation 7-11 to give:

yj−yoxy=(yj−1−yoxy)(PLPH)

Use the preceding equation to determine the concentration after one purge:

y1−yoxy=(y0−yoxy)(PLPH)(7-12)

Repeat this process for the second purge:

y2−yoxy=(y1−yoxy)(PLPH)

Replace (y1−yoxy) with the (y1−yoxy) as shown earlier, resulting in:

y2−yoxy=(y0−yoxy)(PLPH)(PLPH)=(y0−yoxy)(PLPH)2

Repeat this process to give the following general equation:

yj−yoxy=(y0−yoxy)(PLPH)j(7-13)

Equation 7-13 is used in place of Equation 7-6 for both pressure and vacuum purging with impure inert gases.

Pressure purging is faster because the pressure differentials are greater but it uses more inert gas than vacuum purging. Vacuum purging uses less inert gas because the oxygen concentration is reduced primarily by vacuum on the very first cycle. When combining vacuum and pressure purging, less nitrogen is used to if the initial cycle is a vacuum cycle.

Sweep-Through Purging

The sweep-through purging process adds the purge gas into a vessel at one opening and withdraws the mixed gas from the vessel to the atmosphere (or scrubber) from another opening. This purging process is commonly used when the vessel or equipment is not rated for pressure or vacuum; the purge gas is added and withdrawn at a fixed pressure, normally atmospheric.

Purging results are defined by assuming perfect mixing within the vessel, constant temperature, and constant pressure. Under these conditions, the mass or volumetric flow rate for the exit stream is equal to the inlet stream. The material balance around the vessel is

VdCdt=C0Qv−CQv(7-14)

where

V is the vessel volume,

C is the concentration of oxidant within the vessel (mass or volumetric units),

C₀ is the inlet oxidant concentration (mass or volumetric units),

Q_v is the volumetric flow rate (volume/time), and

t is time.

The mass or volumetric flow rate of oxidant into the vessel is C₀Q_v, and the flow rate of oxidant exiting is CQ_v. Equation 7-14 is rearranged and integrated:

Qv∫0tdt=V∫C1C2dC(C0−C)(7-15)

The volumetric quantity of inert gas required to reduce the oxidant concentration from C₁ to C₂ is Q_vt, and it is determined using the following equation:

Qvt=Vln(C1−C0C2−C0)(7-16)

If pure inert gas is used for the sweep purge, then C₀ = 0.

Siphon Purging

The sweep-through process requires large quantities of nitrogen, as shown by Equation 7-16. This could be expensive when purging large storage vessels. In such a case, siphon purging is used to minimize this purging expense.

The siphon purging process starts by filling the vessel with liquid water or any liquid compatible with the material to be stored in the vessel. The purge gas is subsequently added to the vapor space of the vessel as the liquid is drained from the vessel. The volume of purge gas is equal to the volume of the vessel, and the rate of purging is equivalent to the volumetric rate of liquid discharge.

It is typically not possible to fill the vessel entirely with liquid. Thus, when using the siphon purging process, it may be desirable to first fill the vessel as much as possible with liquid and then use the sweep-through purge process to remove oxygen from the residual head space. This method decreases the oxygen concentration to low concentrations, but requires only a small extra expense for the additional sweep-through purging.

Using the Flammability Diagram to Avoid Flammable Atmospheres

The flammability diagram introduced in Chapter 6 is important for determining whether a flammable mixture exists and for providing target concentrations for inerting and purging procedures. The objective is to avoid flammable mixtures.

The procedure for taking a vessel out of service is illustrated in Figure 7-5. The vessel is initially at point A and contains pure fuel gas at atmospheric pressure. If air is used to purge the vessel, the composition follows line AR, which goes through the flammability zone. If nitrogen is first pumped into the vessel, the gas composition follows along line AS. One approach is to continue the nitrogen flow until the vessel contains pure nitrogen. However, this requires a large amount of nitrogen and is costly. A more efficient procedure is to inert with nitrogen until point S is reached. Then air can be introduced, and the gas composition follows along the line SR. In this case, the flammability zone is avoided and a safe vessel preparation procedure is ensured.

One might suggest an even more optimized procedure, which involves first pumping air into the vessel until a point is reached on the air line above the UFL. This is followed by pumping nitrogen into the vessel, followed by pumping in air. This approach avoids the nose of the flammability zone and minimizes the consumption of nitrogen. Unfortunately, the air forms a flammable mixture at the entry point as the pure air mixes with the fuel-rich gas mixture in the vessel. The flammability diagram reflects only the average gas composition within the vessel. Using nitrogen first avoids this problem.

When using the nitrogen purge process, one must determine the location of point S in Figure 7-5. The approach is shown in Figure 7-6. Point S is approximated by a line starting at the pure air point R and connecting through a point M at the intersection of the LFL with the stoichiometric combustion line. Because the gas compositions at points R and M are known, the composition at point S is determined graphically or by using the following equation:

OSFC=LFL1−z(LFL21)(7-17)

where OSFC is the out-of-service fuel concentration—that is, the fuel concentration at point S in Figure 7-6. The LFL is the volume percent of fuel in air at the lower flammability limit, and z is the stoichiometric oxygen coefficient from the combustion reaction given in Chapter 6. The derivation of Equation 7-17 is provided on the website for this textbook.

Another approach is to estimate the fuel concentration at point S by extending the line from point R through the intersection of the minimum oxygen concentration (M ) and the stoichiometric combustion line. The analytical result is

OSFC=LOCz(1−LOC21)(7-18)

where

LOC is the limiting oxygen concentration (also called the minimum oxygen concentration) in volume percent oxygen,

z is the stoichiometric coefficient for oxygen given in Equation 6-9, and

LFL is the fuel concentration at the lower flammability limit, in volume percent of fuel in air.

The derivation of Equation 7-18 is provided on the website for this textbook.

Equations 7-17 and 7-18 are approximations of the fuel concentration at point S. Fortunately, these equations are mostly (but not always) conservative—that is, less than the experimentally determined OSFC value. For instance, for methane the LFL is = 5.3% and z is 2. Thus Equation 7-17 predicts an OSFC of 10.7% fuel. This is compared to an experimentally determined OSFC of 14.5% (from Table 7-2). By using an experimental LOC of 12%, an OSFC value of 14% is determined. This is closer to the experimental value but still conservative. For ethylene, 1,3-butadiene, and hydrogen, Equation 7-18 predicts a higher OSFC than the experimentally determined value. For all other species in Table 7-2, Equation 7-17 estimates an OSFC that is less than the experimental value.

Table 7-2 Experimental In-Service Oxygen Concentrations (ISOCs) and Out-Of-Service Fuel Concentrations (OSFCs)

Source: C. V. Mashuga and D. A. Crowl. “Application of the Flammability Diagram for Evaluation of Fire and Explosion Hazards of Flammable Vapors.” Process Safety Progress 17, no. 3 (1998): 176.

Figure 7-7 shows the procedure for placing a vessel into service. The vessel begins with air, shown as point A. Nitrogen is pumped into the vessel until point S is reached. Then fuel is pumped in following line SR until point R is reached. The problem is to determine the oxygen (or nitrogen) concentration at point S. The in-service oxygen concentration (ISOC) represents the maximum oxygen concentration at point S in Figure 7-7 that just avoids the flammability zone, with a small margin of safety.

If a detailed flammability diagram is lacking, then the ISOC is estimated. One approach is to use the intersection of the LFL with the stoichiometric combustion line. A line is drawn from the top apex of the triangle (R) through this intersection to the nitrogen axis, as shown in Figure 7-8. The composition at S is determined graphically or with

ISOC=zLFL1−(LFL100)(7-19)

Where

ISOC is the in-service oxygen concentration in volume percent oxygen,

z is the stoichiometric coefficient for oxygen given in Equation 6-9, and

LFL is the fuel concentration at the lower flammability limit, in volume percent of fuel in air.

Equation 7-19 is derived on the website for this textbook. The nitrogen concentration at point S is equal to 100 – ISOC.

An expression to estimate the ISOC using the intersection of the minimum oxygen concentration and the stoichiometric line is also found using a similar procedure. The analytical result is

ISOC=zLOCz−LOC100(7-20)

where LOC is the minimum oxygen concentration in volume percent oxygen.

A comparison of the estimates using Equations 7-19 and 7-20 with the experimental values in Table 7-2 shows that Equation 7-19 predicts a lower oxygen value than the experimental values for all species except for methyl formate. Equation 7-20 predicts a lower oxygen concentration than the experimental value for all species in Table 7-2 except for butane, methyl-1-butene, 1,3-butadiene, isobutyl formate, and acetone. The calculated values are deliberately not shown in Table 7-2. Obtaining direct and reliable experimental data under conditions as close as possible to process conditions is recommended.

Fundamentals of Static Charge

Static charge buildup is a result of physically separating a poor conductor from a good conductor or another poor conductor. When different materials touch each other, the electrons move across the interface from one surface to the other. Upon separation, more of the electrons remain on one surface than on the other—that is, one material becomes positively charged and the other negatively charged. In such a case, the charges are equal but opposite in polarity.

If both the materials are good conductors, the charge buildup due to separation is small because the electrons are able to quickly scurry between the surfaces. Conversely, if one or both of the materials are insulators or poor conductors, electrons are not as mobile and may become trapped on one of the surfaces, resulting in charge accumulation.

Household examples that result in static charge accumulation include walking across a rug, placing different materials in a tumble dryer, removing a sweater, and combing hair. The clinging fabrics and sometimes audible sparks are the result of the buildup of static charge.

Common industrial examples include pumping a nonconductive liquid through a pipe, mixing immiscible liquids, pneumatically conveying solids, and leaking steam that contacts an ungrounded conductor. The static charges in these examples accumulate to develop large voltages. Subsequent grounding produces large and energetic sparks.

For industrial operations where flammable vapors may be present, any charge accumulation exceeding 0.1 mJ is considered dangerous. Static charges of this magnitude are easy to generate. For example, the static buildup created just by walking across a carpet averages about 20 mJ and exceeds several thousand volts.

Basic electrostatic relationships are used to understand and investigate these situations. These relationships may include field strengths produced by static charges, electrostatic potential, capacitance, relaxation times, currents, and potentials in flow systems.

An electrostatic discharge occurs when two materials at different potentials or polarities come close enough together to generate a charge transfer. In an explosive environment, this sudden transfer of charges may be energetic enough to act as an ignition source. To prevent ignitions created in this way, one must understand (1) how charges accumulate on objects, (2) how charges discharge by means of charge transfer, and (3) how to estimate the resulting energy discharged in relationship to the minimum ignition energy (MIE) of the material present.

Charge Accumulation

Four charge accumulation mechanisms are relevant to dangerous electrostatic discharges in a chemical plant:²

²J. A. Cross. Electrostatics: Principles, Problems, and Applications (Bristol, UK: Adam Higler, 1987).

1. Contact and frictional charging: When two materials, one or both of which are insulators, are brought into contact, a charge separation occurs at the interface. If the two objects are then separated, some of the charges remain separated, giving the two materials opposite but equal charges.

2. Double-layer charging: Charge separation occurs on a microscopic scale in a liquid at any interface (solid–liquid, gas–liquid, or liquid–liquid). As the liquid flows, it carries a charge and it leaves a charge of opposite sign on the other surface—for example, a pipe wall.

3. Induction charging: This phenomenon is applicable only to materials that are electrically conductive. A person with insulated shoes, for example, may approach an overhead vessel that is positively charged (previously filled with positively charged solids). Electrons in the person’s body (head, shoulders, and arms) migrate toward the positive charge of the vessel, thereby causing an equal quantity of positive charges to accumulate on the opposite side of the body. This leaves a part of the body positively charged by induction. When a metal object is touched, a transfer of the electrons occurs, creating a spark.

4. Charging by transport: When charged liquid droplets or solid particles settle on an isolated object, the object becomes charged. The transferred charge is a function of the object’s capacitance and of the conductivities of the particles.

Some nonconductive hydrocarbon liquids have such low conductivities that once they receive a charge, they retain it for a relatively long time. This can result in the potential hazard of a hydrocarbon liquid storage tank. This critical factor is described in a CSB video on static electricity.³

³Static Sparks Explosion in Kansas. https://www.csb.gov/videos/static-sparks-explosion-in-kansas/.

Electrostatic Discharges⁴

⁴T. B. Jones and J. L. King. Powder Handling and Electrostatics (Chelsea, MI: Lewis Publishers, 1991).

A charged object can be discharged to a ground or to an oppositely charged object when the field intensity exceeds 3 MV/m (breakdown voltage of air) (where 1 MV/m = 10⁶ V/m = 1000 V/mm) or when the surface reaches a maximum charge density of 7 × 10^-5 coulombs/m². This discharge can occur by six methods: (1) spark, (2) propagating brush, (3) conical pile (sometimes known as Maurer discharge), (4) brush, (5) lightning-like, and (6) corona discharges.

A spark discharge (Figure 7-9) is a discharge between two metallic objects. Because both objects are conductive, the electrons move to exit at a single point of the charged object, and they enter the second object at a single point. This will result in an energetic spark that can ignite a flammable dust or gas.

A propagating brush discharge (Figures 7-9 and 7-10) is a discharge from a grounded conductor when it approaches a charged insulator that is backed by a conductor. These discharges are energetic and can ignite flammable gases and dusts. Data show that propagating brush discharges are not possible if the breakdown voltage of the insulator is 4 kV or less.⁵

⁵B. Maurer. “Discharges Due to Electrostatic Charging of Particles in Large Storage Silos.” German Chemical Engineering 3 (1979): 189–195.

A conical pile discharge (Figure 7-9) is a form of a brush-type discharge that occurs at the conical surface of a pile of powder.⁶ The necessary conditions for this discharge are (1) a powder with a high resistivity (greater than 10¹⁰ ohm m), (2) a powder with coarse particles 0.1 mm in diameter, (3) a powder with a high charge-to-mass ratio (for example, charged by pneumatic transport), and (4) a filling rate greater than approximately 0.5 kg/s. Conical pile discharges are relatively intense discharges, with energies up to several hundred millijoules, and can ignite flammable gases and dusts. To ignite dusts, the coarse particles need a fraction of fines to give an explosive atmosphere.

⁶M. Glor and B. Maurer. “Ignition Tests with Discharges from Bulked Polymeric Granules in Silos (Cone Discharge).” Journal of Electrostatics 30 (1993): 123–134.

A brush discharge (Figure 7-9) is a discharge between a relatively sharp-pointed conductor (radius of 0.1 to 100 mm) and either another conductor or a charged insulated surface. This discharge radiates from the conductor in a brush-like configuration. Such a discharge is less intense compared to a point-to-point spark discharge, and it is unlikely to ignite dusts. However, brush discharges can ignite flammable gases.

Lightning-like discharges (Figure 7-9) are discharges from a cloud in the air over the powder. Experiments have shown that lightning-like discharges do not occur in vessels with volumes less than 60 m³ or in silos with diameters less than 3 m.⁷ There is currently no physical evidence indicating that lightning-like discharges have resulted in industrial deflagrations.

⁷P. Boschung, W. Hilgner, G. Luttgens, B. Maurer, and A. Wider. “An Experimental Contribution to the Question of the Existence of Lightning-like Discharges in Dust Clouds.” Journal of Electrostatics 3 (1977): 303–310.

A corona discharge (Figure 7-11) is similar to a brush discharge. The electrode conductor has a sharp point. The discharge from such an electrode has sufficient energy to ignite only the most sensitive gases (for example, hydrogen).

Energy from Electrostatic Discharges

Figure 7-12 illustrates the energy generated in electrostatic discharges versus the minimum ignition energies of gases and vapors and dusts. In general, the results illustrate that most flammable gases and vapors can be ignited by spark, brush, conical pile, and propagating brush discharges, and that most flammable dusts can be ignited only by sparks, propagating brush, and conical pile discharges. The regions enclosed by the dotted lines in Figure 7-12 indicate regions of uncertainty.

Energy of Electrostatic Ignition Sources

A spark is generated between two conductors when the distance between the conductors is small compared to the diameter of the conductors and when the electric field intensity between the conductors is approximately 3 MV/m. A brush discharge is generated if the distance between the conductors is large compared to the radius of curvature of the conductor.

The energy of a spark discharge is a function of the accumulated charge (Q in coulombs) on the object, the capacitance of the object (C in farads), and the potential, or voltage (V, in volts), of the object. These three variables are related by the expression C = Q/V. The actual energy (expressed in joules) associated with the discharge process is given by

J=Q22C(7-21)

Equation 7-21 assumes a capacitance-type discharge (that is, a spark); however, capacitance and voltage are not defined in nonconductive systems. As a consequence, Equation 7-21 is valid only for capacitive sparks, but it is used qualitatively for the other discharges.

A criterion that is commonly used to estimate the potential hazard of a discharge is to compare the MIE of the fuel–air mixture to the equivalent energy of the discharge. A precise experimental determination of the MIE is often required under the specific conditions of the system. MIEs are shown in Table 6-4 for a number of flammable gases and in Table 6-7 for dusts.

The static discharge energy is a function of the accumulated charge. In an industrial setting, this accumulated charge is usually the result of either contact or friction charging for flowing solids and double-layer charging for flowing liquids. In each case, the charge (electrons) is transported with the material. The measure of this flow of electrons is a streaming current and is expressed in coulombs per second or amps.

Streaming Current

A streaming current I_S is the flow of electricity produced by transferring electrons from one surface to another by a flowing fluid or solid. When a liquid or solid flows through a pipe (metal or glass), an electrostatic charge develops due to the streaming material. This current is analogous to a current in an electrical circuit. The relation between a liquid streaming current and the pipe diameter, pipe length, fluid velocity, and fluid properties is given by the following equation:⁸

⁸L. G. Britton. Avoiding Static Ignition Hazards in Chemical Operations (New York: American Institute of Chemical Engineers, 1999).

Is=[10×10−6amp(m/s)2(m)2](ud)2[1−exp(−Luτ)](7-22)

where

I_s is the streaming current (amps),

u is the velocity (m/s),

d is the pipe diameter (m),

L is the pipe length (m), and

τ is the liquid relaxation time (seconds).

The relaxation time of the liquid is the time for the charge to fall by approximately 37% of its initial value, assuming no charges are added—that is, the time required for a charge to dissipate by leakage. It is determined using

τ=εrε0γc(7-23)

where

τ is the relaxation time (seconds),

ε_r is the relative dielectric constant (unitless),

ε₀ is the permittivity constant, with a value of

8.85×10−12coulomb2Nm2=2.7×10−12coulombvolt ft=8.85×10−12coulombvolt m8.85×10−14coulombvolt cm=8.85×10−14sohm cm=8.85×10−14mhos cm, and

γ_c is the specific conductivity (mho/cm).

Specific conductivities and relative dielectric constants are listed in Table 7-3. Notice that resistance = 1/conductivity = 1/(mho/cm) = ohm cm.

Table 7-3 Properties for Electrostatic Calculations

^aResistance = 1/conductivity = 1/(mho/cm) = ohm cm.

Source: J. H. Perry. Chemical Engineers’ Handbook, 3rd ed. (New York, NY: McGraw-Hill, 1950), p. 1734.

Charges also accumulate when solids are transported. The buildup results from the separation of solid particle surfaces. Because solid geometries are almost always ill defined, electrostatic calculations for solids are handled empirically.

The streaming current that is generated while transporting solids is a function of the solids processing method (see Table 7-4) and the flow rate, as shown by

Is=(coulombskg)(kgs)(7-24)

where

I_S is coulombs/second or amps,

coulombs/kg is given in Table 7-4, and

kg/s is the solids flow rate.

Table 7-4 Charge Buildup for Various Operations

Source: R. A. Mancini. “The Use (and Misuse) of Bonding for Control of Static Ignition Hazards.” Plant/Operations Progress 7, no. 1 (1988): 24.

Some generally accepted guidelines for electrostatic calculations are shown in Table 7-5.

Table 7-5 Accepted Electrostatic Values for Calculations

Source: F. G. Eichel. “Electrostatics.” Chemical Engineering (March 13, 1967): 163.

Electrostatic Voltage Drops

Figure 7-13 illustrates a tank with a feed line. Fluid flows through the feed line and drops into the tank. The streaming current builds up a charge and voltage in the feed line to the vessel and in the vessel itself. The voltage from the electrical ground in the metal line to the end of the glass pipe is calculated using

V=IsR(7-25)

The resistance R (in ohms) is computed using the conductivity of the fluid γ_c (in mho/cm), the length of the conductor L (in cm), and the area A of the conductor (in cm²):

R=LγcA(7-26)

This relationship shows that as the area of the conductor increases, the resistance decreases; if the conductor length increases, the resistance increases.

Energy of Charged Capacitors

A capacitor is used to store electrostatic energy, with generally two electrical conductors (plates) separated by a dielectric (i.e., insulator). The amount of work required to increase the charge on a capacitor from Q to Q + dQ is dJ = VdQ, where V is the potential difference and the charge is Q. Because V = Q/C, the integration gives Equation 7-21, and substitutions give

J=CV22(7-27)

J=QV2(7-28)

The units used in Equations 7-27 and 7-28 are usually C in farads, V in volts, Q in coulombs, and J in joules.

Capacitances of various materials used in the chemical industry are given in Table 7-6.

Table 7-6 Capacitance of Various Objects

Source: R. A. Mancini. “The Use (and Misuse) of Bonding for Control of Static Ignition Hazards.” Plant/Operations Progress 7, no. 1 (1988): 24.

Charges can accumulate as a result of a streaming current dQ/dt = I. Assuming a constant streaming current,

Q=Ist(7-29)

where I_S is in amps and t is in seconds. Equation 7-29 assumes that the system starts with no accumulation of charge, only one constant source of charge I_S, and no current or charge loss terms.

Example 7-4

Determine the voltage developed between a charging nozzle and a grounded tank, as shown in Figure 7-14. Also, compute the energy stored in the nozzle and the energy accumulated in the liquid. Explain the potential hazards in this process for a flow rate of

1. 1 gpm

2. 150 gpm

The data are

• Hose length: 20 ft

• Hose diameter: 2 in.

• Liquid conductivity: 10^–8 mho/cm

• Dielectric constant ε_r: 25.7

• Density: 0.88 g/cm³

Solution

1. Because the hose and nozzle are not grounded, the voltage generated at the nozzle tip is V = IR. The resistance is computed using Equation 7-25 for the conducting fluid with a resistance length equivalent to the hose length (from the ground near the pump to the nozzle) and a resistance area equivalent to the cross-sectional area of the conducting fluid:

   L=(20 ft)(12 in/ft)(2.54 cm/in)=610 cmA=πr2=(3.14)(1 in)2=3.14 m2=20.3 cm2

   Using Equation 7-26, we obtain

   $$\begin{gathered} R=(1γc)(LA) \\ =(108ohm cm)(610 cm20.3 cm2)=3.00×109ohm \end{gathered}$$

   The streaming current is a function of the velocity and the pipe diameter. The average velocity in the pipe is

   u=(1 gal/min3.14 in2)(ft37.48 gal)(144 in2ft2)(1 min60 s)=0.102 ft/s=3.1×10−2 m/s

   The relaxation time is estimated using Equation 7-23:

   τ=εrε0γc=(25.7)(8.85×10−14mho scm)10−8 mho/cm=22.7×10−5 s

   The streaming current is now determined using Equation 7-22:

   Is=[10×10−6amp(m/s)2(m)2](ud)2[1−exp(−Luτ)]=[10×10−6amp(m/s)2(m)2][(3.1×10−2m/s)(2in12in/ft×3.28 ft/m)]2[1−exp(−20 ft0.102ft/s×2.27×10−4s)]=[10×10−6amp(m/s)2(m)2](1.57×10−3m2/s)2(1−0)=2.48×10−11amp

   Method 1: Compute the energy accumulated in the capacitor formed between the flanges at the nozzle. A spark between the flanges may be an ignition source. The voltage drop down the 20-ft line is the same as the voltage drop from the hose flange to the nozzle flange, assuming that the nozzle is grounded. The voltage is therefore

   V=IR=(2.48×10−11amp)(3.0×109ohm)=0.074volt

   The capacitance between the two 2-in flanges is given in Table 7-5:

   C=20×10−12farads=20×10−12coulomb/volt

   The energy is determined using Equation 7-27:

   J=CV22=[20×10−12coulomb/volt (0.074 volt)22]=5.49×10−14J

   This is significantly lower than the energy required to ignite a flammable gas (0.1 mJ). An electrostatic hazard at the nozzle is not expected.

   Method 2: Compute the energy accumulated in the capacitor formed by the tank of liquid. A brush discharge can jump from this liquid to a metal component, such as a grounded thermocouple. The accumulated charge is computed using Equation 7-29:

   Q=Ist

   with the time equal to the filling time of the vessel:

   t=(300gal/1 gpm)(60 s/min)=18,000 s

   Substitution into Equation 7-29 gives

   Q=Ist=(2.48×10−11 amp)(18,000s)=4.46×10−7coulomb

   The capacitance of the liquid is estimated to be one-tenth of the capacitance of a 2000-gal vessel, shown in Table 7-5; therefore

   C=100×10−12farads=100×10−12coulomb/volt

   and the accumulated energy is determined using Equation 7-21:

   J=Q22C=(4.46×10−7 coulomb)22(100×10−12 farads)=9.9×10−4 J=0.99 mJ

   This exceeds the energy required to ignite a flammable gas (0.1 mJ). In this situation the vessel should be purged with nitrogen to maintain the concentration of the flammable vapor below the LOC.

2. This case is identical to case a except that the flow rate is higher, 150 gpm versus 1 gpm for case (a):

   u=(0.102 ft/s)(150 gpm1 gpm)=4.66 m/s

   The resistance is the same as for case (a)—that is, 3 × 10⁹ ohm:

   τ=22.7×10−5s

   The streaming current is, from Equation 7-21,

   Is=[10×10−6 amp(m/s)2(m)2][(4.66 m/s)(0.051 m)]2[1−exp(−20 ft15.3 ft/s×2.27×10−4 s)]=(5.65×10−7 amp)(1−0)=5.65×10−7amp

   Method 1: Compute the energy accumulated in the capacitor formed between the flanges at the nozzle:

   V=IR=(5.65×10−7amp)(3×109ohm)=1695 volts

   The accumulated energy is again computed using Equation 7-26:

   J=CV22=(20×10−12 coulomb)(1695 volts)22=0.000029 J=0.029 mJ

   This is less than the energy required to ignite a flammable gas (0.1 mJ).

   Method 2: Compute the energy accumulated in the capacitor formed by the tank of fluid:

   t=300 gal150 gpm(60 smin)=160 sQ=ISt=(5.65×10−7 amp)(160 s)=9×10−5coulombsJ=Q22C=(9×10−5 coulombs)22(100×10−12 coulomb/volt)=40 J >> 0.1 mJ

   This energy exceeds 0.1 mJ. This problem illustrates the importance of inerting. It is relatively easy to build up energies that exceed 0.1 mJ.

Capacitance of a Body

The buildup of a charge on one surface relative to another surface produces a capacitor. In the chemical industry, the properties of the developed capacitor are estimated by assuming parallel flat plate or spherical geometries. For example, the capacitance of a tank or a person is estimated by assuming spherical geometries, and the capacitance of a person’s shoe sole or of a noncorrosive tank lining is estimated assuming parallel flat plates. Several examples are shown in Figure 7-15.

The capacitance C of a body is Q/V. For a sphere with radius r, the voltage developed when a charge Q is accumulated is determined using the following equation from physics:

V=14πε0Qεrr(7-30)

Because C = Q/V, the capacitance for a spherical body is

C=4πεrε0r(7-31)

where

ε_r is the relative dielectric constant (unitless),

ε₀ is the permittivity (8.85 × 10^–12 coulomb²/Nm² = 2.7× 10^–12 coulomb/volt ft),

r is the sphere radius, and

C is the capacitance.

For two parallel plates,

V=QLεrε0A(7-32)

Therefore, the capacitance between parallel plates is

C=εrε0AL(7-33)

where

A is the area of the surface and

L is the thickness of the dielectric.

Sparks resulting from static charge and discharge continue to cause major fires and explosions within the chemical industry. The fundamentals and examples developed in these sections were designed to emphasize the importance of this subject. The emphasis on the fundamentals is intended to assist with understanding to prevent incidents with static electricity.

General Design Methods to Prevent Electrostatic Ignitions

The design objective is to prevent the buildup of charges on a product (liquid or powder) as well as on surrounding objects (equipment or personnel). Also remember that for every charged object, there exists its oppositely charged counterpart. Four methods are used to achieve this objective:

1. Prevent charges from accumulating to dangerous levels by reducing the rate of charge generation and increasing the rate of charge relaxation. This method is generally used when handling liquids.

2. Prevent charges from accumulating to dangerous levels by designing the system to include charge reduction by means of low-energy discharges. This method is generally used when handling powders.

3. When dangerous discharges cannot be eliminated, prevent the possibility of an ignition by maintaining oxidant levels below the combustible levels (inerting) or by maintaining fuel levels below a target concentration, such as the LOC, LFL, or above the UFL.

4. Measures to mitigate the consequences of an explosion are also options for consideration (for example, deflagration venting and explosion suppression).

The special design features for preventing specific electrostatic ignitions are described in the following paragraphs.

Sparks are prevented by grounding and bonding. This procedure prevents two metallic objects (close to each other) from having different potentials. Grounding and bonding are used to avoid the possibility of isolated metal parts or objects. Isolated objects are notorious for building up large potentials and energetic sparks when they are approached by another conductor at a lower potential.

Propagating brush discharges are prevented by keeping the nonconductive surfaces or coatings thin enough or conductive enough to have a breakdown voltage less than 4 kV. These discharges are also prevented by keeping the metallic backings grounded, so as to eliminate the accumulation of a high-density charge on the metallic interface and a countercharge on the nonconductor surface.

Conical pile discharges are prevented by increasing the conductivity (add additives), by decreasing the charge rate to less than 0.5 kg/s, or by using containers with a volume less than 1 m³. The most effective way of preventing ignitions from conical pile discharges is inerting.

Brush discharges are prevented by keeping the nonconductive surfaces thin enough or conductive enough to have a breakdown voltage (ud) of 4 kV. Nonconductive coatings with a thickness greater than 2 mm, however, are capable of brush discharges even with a ud less than 4 kV. To prevent brush discharges, a thickness of less than 2 mm is necessary. This fixes the charges accumulated on the nonconductor, and fixed charges cannot be transferred in a brush discharge. Brush discharges from nonconductive liquids are prevented by increasing the conductivity using conductive additives. The most effective way of preventing ignitions from brush discharges is inerting.

Lightning-like discharges are prevented by keeping the vessel volume to less than 60 m³ or the vessel diameter to less than 3 m. If this condition is not met, then the system needs to be inerted.

Relaxation

When pumping fluids into a vessel through a pipe on top of the vessel, the separation process produces a streaming current I_s, which causes charge buildup. It is possible to substantially reduce this electrostatic hazard by adding an enlarged section of pipe just before entering the tank. This decreases the flow rate and increases the hold time for charge relaxation. The residence time in this relaxation section of pipe should be approximately twice the relaxation time determined from Equation 7-23.

In actual practice, it was found that a hold time equal to or greater than one-half the calculated relaxation time is sufficient to eliminate charge buildup.⁹ The “twice the relaxation time” rule, therefore, provides a safety factor of 4. The American Petroleum Institute recommends a ud, from Equation 7-22, of less than 0.5 m²/s for road tanker filling and 0.8 m²/s for railcar filling.¹⁰

⁹F. G. Eichel. “Electrostatics.” Chemical Engineering (March 13, 1967): 153.

¹⁰API RP 2003, Protection Against Ignitions Arising Out of Static, Lightning, and Stray Currents, 8th ed. (Washington, DC: American Petroleum Institute, 2015).

Bonding and Grounding

The voltage difference between two conductive materials is reduced to zero by bonding the two materials—that is, by bonding one end of a conducting wire to one of the materials and bonding the other end to the second material. When comparing two sets of bonded materials they may have different voltages, however. The voltage difference between these two sets is reduced to zero by bonding each set to ground—that is, by grounding.

Bonding and grounding reduces the voltage of an entire system to ground level or zero voltage. It also eliminates the charge buildup between various parts of a system, eliminating the potential for static sparks. Examples of grounding and bonding are illustrated in Figures 7-16 and 7-17.

Materials inside glass or plastic-lined vessels are grounded using tantalum inserts or metal probes, as illustrated in Figure 7-18. This technique, however, is not effective when handling liquids with low conductivity. In this case, the fill line should extend to the bottom of the vessel (see Figure 7-19), to help eliminate the charge generation (and accumulation) resulting from separation during the filling operation. Also, the inlet velocities should be low enough to minimize the charge generated by streaming current I_S.

Dip Pipes

An extended line, sometimes called a dip leg or dip pipe, reduces the electrical charge that accumulates when liquid is allowed to free fall. However, when using dip pipes, care must be taken to prevent siphoning back when the inlet flow is stopped. One method commonly employed to prevent siphoning is to place a hole in the dip pipe near the top of the vessel. Another technique is to use an angle iron instead of a pipe and to let the liquid flow down the angle iron (see Figure 7-19). These methods are also used when filling drums.

Increasing Conductivity with Additives

The conductivity of nonconducting organic materials can sometimes be increased by using antistatic additives. Examples of antistatic additives include water and polar solvents, such as alcohols. These additives are effective only when they are soluble in the offending liquid, because an insoluble phase gives an additional source of separation and charge buildup.

Handling Solids without Flammable Vapors

Charging solids with a ungrounded and conductive chute can result in a buildup of a charge on the chute. This charge can accumulate and ultimately produce a spark that may ignite a dispersed and flammable dust. Solids are transferred safely by bonding and grounding all conductive parts and/or by using nonconductive parts (drum and chute), as shown in Figure 7-20.

Handling Solids with Flammable Vapors

A safe design for handling solids that contain flammable vapors includes handling of the solids and liquids in a closed and inerted atmosphere (see Figure 7-21). For solvent-free solids, the use of nonconductive containers is permitted. For solids containing flammable solvents, only conductive and grounded containers are recommended.¹¹

¹¹Expert Commission for Safety in the Swiss Chemical Industry. “Static Electricity: Rules for Plant Safety,” Plant/Operations Progress 7, no. 1 (1988): 1.

Explosion-Proof Housings

In an XP area, the electrical equipment and some instrumentation must have special explosion-proof housings. These housings are not intended to prevent flammable vapors and gases from entering, but rather are designed to withstand an internal explosion and prevent the combustion from spreading beyond the inside of the enclosure. A motor starter, for example, is enclosed in a heavy cast walled box with the strength needed to withstand explosive pressures.

The explosion-proof design includes the use of conduit with special sealed connections around all junction boxes. Figure 7-22 shows features of electrical equipment rated for an XP area. The on/off switch housing is especially large and bulky. The thickness of the metal is approximately ¾ inch. The flanges are large so that if flammable vapors enter the box and are ignited, the burning gas is quenched as it flows out through the flange faces. This bulkiness is necessary to remove the heat of a burning gas and to extinguish the flame before it exits into a potentially flammable environment.

The sealing pipe fitting below the switch in Figure 7-22 is called an EYS. It is approximately 2½ inches long and 1½ inches in diameter. The procedure to seal this fitting is as follows:

1. The wires are separated within the fitting.

2. A fiber dam is created at the lower part of the fixture.

3. An epoxy liquid is poured into the fixture, completing the sealing of the fixture once the epoxy hardens.

A reasonably complex chemical plant may include hundreds of these fixtures, all of which must be sealed properly. Also, any scratches or dirt on the flanges will defeat the gas quenching and result in the combustion propagating to the external environment.

Area and Material Classification

The design of electrical equipment and instrumentation is based on the nature of the process hazards or specific process classifications. The classification method is defined in the NEC; it is a function of the nature and degree of the process hazards within a particular area. The rating method includes Classes I, II, and III; Groups A–G; and Divisions 1 and 2.

The classes are related to the nature of the flammable material:

The groups designate the presence of specific chemical types. Chemicals that are grouped have equivalent hazards:

Division designations are categorized in relationship to the probability of the material being within the flammable or explosive regions:

Design of an XP Rated Area

When designing an XP rated area, all electrical equipment and instrumentation must be specified for the class, group, and division. All equipment and instrumentation within an area must be appropriately specified and installed. The overall classification of an area is only as good as the piece of equipment with the lowest classification.

Open-Air Plants

Open-air plants are recommended because the average wind velocities are high enough to safely dilute volatile chemical leaks that may exist within a plant. Although safety precautions are always practiced to minimize leaks, accidental releases from pump seals and other potential release points may occur.

Plants Inside Buildings

Frequently, processes cannot be constructed outside. In this case, local and dilution ventilation systems are required. These ventilation systems were discussed in detail in Chapter 3, Section 3-3.

Local ventilation is the most effective method of controlling flammable gas releases. Dilution ventilation, however, is also used because the potential points of release are usually numerous and it may be mechanically or economically impossible to cover every potential release point with only local ventilation.

Empirically determined design criteria for ventilation systems for flammable materials inside storage and process areas are identified in Table 7-7. The NFPA ventilation requirements and classification of flammables are given in Table 7-7. There are more details regarding ventilation and classifications in the referenced NFPA document.

Table 7-7 NFPA Ventilation Requirements and Classifications of Flammable Materials

Source: Data taken from National Fire Protection Association, Flammables and Combustible Liquids Code (Quincy, MA: National Fire Protection Association, 2018).

The effectiveness of a ventilation system is determined using material balance equations, described in Chapter 2 and illustrated in Example 7-9.

Practices

Many incidents are caused by the failure to use RAGAGEP. This acronym incorporates five practices: (1) adherence to government regulations; (2) adherence to American Petroleum Institute (API) standards; (3) adherence to AICHE’s Center for Chemical Process Safety (CCPS) guidelines; (4) adherence to National Fire Protection Association (NFPA) fire codes and standards; and (5) industrial experience acquired by sharing information within industry and at professional meetings.

Passive and Active Systems

Passive and active safety systems include:

• Elimination or reduction of fire hazards

• Detection and control to minimize effects after a fire has started

• Mitigation or reduction of the severity of fire consequences

• Emergency response

Passive systems are responses to incidents without manual or control actions. They include (1) fireproofing (coating structures with fire-resistant cements); (2) drainage (slope surfaces away from process units to drain the flammable material to a safer location); (3) containment (increase the strength of vessels to contain fires and explosions); (4) separation (spacing between equipment); and (5) fire walls to minimize the spread of the fire.

Active systems are responses that require manual or control actions. They include (1) sprinklers to extinguish fires activated by heat detection; (2) deluge systems to spray a large quantity of water on process equipment to extinguish fires and prevent BLEVEs; (3) foam application to suppress the magnitude of the fire or release; and (4) fire brigade emergency responders using both plant and local fire departments.

Plant Fire Protection Infrastructure

The plant layout must be designed to ensure that equipment is accessible to maintenance workers as well as to emergency response teams and vehicles. The plant infrastructure also requires water for sprinklers and fire hydrants. A well-designed system includes a reliable source of water such as a municipal supply, well, lake, river, or water tower. Also, the design should include a good distribution system including pumps and backup pumps. The distribution and pump systems need redundancy to ensure that a common cause of failures—such as lightning or power failure—does not interrupt the source of water.

Documentation of Fire and Explosion Protection Strategy

The plant’s fire protection strategy must be documented and audited frequently. The purpose of the audit is to identify changes in equipment, procedures, or technology.