Inert Gas Lel Rules

5/17/2018 Inert Gas Lel Rules

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19SaletyEstimating LEL and Flash 273Tank Blanketing ................................................ 273Equipment Purging 275Static Charge from Fluid Flow 276Mixture Flammability 279Relief Manifolds 282Natural Ventilation ................................................... 288

272


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Safety 273

E stim a tin g L EL a nd F la shThe lower explosive limit (LEL) is the rrnrumum

concentration of a vapor in air that will support a flamewhen ignited. The flash point is the lowest temperature ofa liquid that produces sufficient vapor for an open flameto ignite in air.Gooding provides ways to estimate these two impor-

tant safety-related properties. The methods make use ofthe following observed rules:

I.The LEL occurs at about 50% of the stoichiometricoxidation concentration at ambient temperature andpressure.

2. The flash point occurs at about the temperature atwhich the liquid has a vapor pressure equal to theLEL partial pressure.

3. It follows then, that knowing the stoichiometry andhaving a vapor pressure chart. one can determine theLEL and flash point. Also if either the LEL or flashpoint is known, a vapor pressure chart can be usedto estimate the other.

Example:Estimate the LEL and flash point for ethanol.

The oxidation (combustion) equation is:

For I mol of ethanol we need:3mols of O~or 3/0.21 = 14.28 mols of airThe stoichiometric concentration of ethanol in air is

thus 1/15.28 = 0.0654mol fraction. The LEL is 50~ ofthis or 0.0327mol fraction. This matches the reportedvalue of 3.3% by volume.The partial pressure of LEL ethanol is 0.0327 atm. The

temperature that produces a vapor pressure of 0.0327 atrnis IIC, which is our predicted flash point. This is closeto the reported 13C,The Gooding article presents graphs that show highaccuracy for these methods.

S o u r c eGooding, Charles H., "Estimating Flash Point and LowerExplosive Limit," Chemical Engineering. December12,1983.

T a n k B l an k e tin gInert gas is used to blanket certain fixed-roof tanks for

safety. Here is how to determine the inert gas require-ments. Inert gas is lost in two ways: breathing losses fromday/night temperature differential. and working losses todisplace changes in active level.

B r e a th in g L o s s e sI. Determine the vapor volume, V,

Vo =7tD~/4(avg.outage)where

D = Tank diameter, ftavg. outage = Average vapor space, tt

Vo = Vapor volume, scf

2. Calculate daily breathing loss (DBL)DBL = Vo{[(460 +T, +Tdc. 2)/ (460 +T, + tl- Tdc 2) 1

-l.0}whereT, = Storage temperature, O FTJ" = Daily temperature change, O FDBL = Daily breathing loss, set'~=Adjustment for the differential between blanket-ing and pressure-relief settings (normally 2-4F)

3. See Figures 1 and 2 for T, and T,k


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274 Rules of Thumb for Chemical Engineers

Figure 1. Average storage temperatures for the U.S. used to estimate breathing losses.

I~ : .' J,,~- --"'-j I:t3.

f .J_J23 2c.\Ie

< '6

JO

- , 7.,0 i__, -, 1221.

21.

Figure 2. Average daily temperature changes for the U.S. used to estimate breathing losses.


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W o rk in g L o ss esUse the following displacement equivalents of inert gas

to tank liquid:1gal x 0.1337 = 1scf inert gasIbbl x 5.615 = 1scf inert gasExample: A fixed-roof lank

D = 128ftHeight = 36 ft

Avg. outage = 12ftAnnual throughput = 300,000 bbl

Location = New OrleansDetermine monthly inert gas usageSolution:

T, = 75F (Figure I)Telc= 15F (Figure 2)l'. = 2P (Assumed)

Safety 275

Vapor volume = rr(l28)2/4(l2) = 155,000ft3DBL = IS5,000{[(460 + 75 + 15/2)/(460 + 75

+2 - 15/2)] - I.O} scfld= 3,805 scfld or 3,805(30)= 114,000 scflmo

Monthlyworking loss = 300,000112(5.615) = 140,000scfmo

Total inert gas usage 254,000scf/mo

S o u r c eBlakey, P. and Orlando, G., "Using Inert Gases ForPurging, Blanketing and Transfer." Chemical Engi-neering, May 28, 1984.

E q u ip m e n t P u rg in gBlakey and Orlando give useful methods for deter-

mining inert gas purging requirements.

D ilu tio n P u rg in g

The inert gas simply flows through the vessel andreduces the concentration of unwanted component. It is used for tanks, reactors, and other vessels. Use Figure 1 to determine the quantity of inert purgegas required.

Example: A tank full of air (21% O~) needs to bepurged to 1% O2, From Figure I, 3.1 vessel volumes ofinert gas are required.

P re ss ur e- Cy cle P u rg in g

The vessel at I atrn is alternately pressured with inertgas and vented. It is used for vessels that can withstand 30psig ormore, vessels with only one port, or vessels with coilsor baffles inside. It is useful when pressurization isneeded anyway, such as for testing.

The dilution ratio is (liP)"whereP = pressure, atmn = number of cycles

1,000800600400

200c0'c:'" 100~ 80uc 60cc~ 40


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The quantity of inert gas required for each cycle is p,I vessel volumes.

Example: Purge O2 from 21 % to I%. Use pressurecyclepurging to 5 atm.Two cycles will do the job since (1I5).:' = 0.04 and

0.04(21) = 0 . 8 4 o / c O2 For each cycle, purge volume is5 - I = 4 vessel volumes. So total purge volume is 8vessel volumes.

V a c u u m - C y c le P u r g in g The vessel is alternately evacuated and fed with inertgas to 1atm . It can only be used for vessels capable of withstand-ing a vacuum.

Concentration of unwanted component is reducedfrom C to Cl"

whereC = concentration. o /cP = pressure, atmn = number of cyclesThe quantity of inert gas required for each cycle i

P vessel volumes.Example: Purge 0:, from 21 % to 1%. Use vacuu

cycle purging to 0.5 atm.C= 2IO/C;Cl" = IO/Cor 21(0.5)n = 1 ; ( 0.5 )" = 0.048n = 4.4 so use 5 cyclesPurge gas required =5(1 - 0.5) =2.5 vessel volumes

S o u r c eBlakey. P. and Orlando, G., "Using Inert GasesPurging. Blanketing and Transfer," Chemical Eneering, May 28, 1984.

S ta tic C ha rg e fro m F lu id F lo wThe following article written by Adam Zanker for

Hydrocarbon Processing (March 1976) is reprinted herein its entirety.Development of electrostatic charges in tanks and

vessels in which hydrocarbons are pumped or stirred iswidely recognized as a serious hazard. This electricity,which is generated in tanks during these normal opera-tions, occasionally causes a spark in a tank vapor space.As statistics show, during 10 years in one state and one

oil company only, 18 fires have been attributed to staticelectricity, causing damage and product losses of millionsof dollars. IThe order of magnitude of currents and voltages related

to static electricity are of different orders of magnitudefrom those common to us from everyday life.Voltages are extremely high; hundreds of thousands

of volts are common. Currents, however, are usually lessthan one-millionth of an ampere. It is due to the extremelyhigh resistivity of hydrocarbon products. This resistivityvaries from io' to 10150hm-cm.Voltage drop may be calculated from Ohms Law:

V =1 RWhen the resistivity equals l O'


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It is especially important for the IP-4 type prod-ucts since they produce a typical explosive atmos-phere at ambient temperatures above the liquid level.The atmosphere in gasoline tanks is too rich in

hydrocarbons, and in kerosene tanks is too lean inhydrocarbons, so that theoretically, both mixtures ofair and vapor are outside the explosivity range.

When the change of roof is impossible. it is advis-able to exchange the air in the tank free space tocarbon dioxide which will prevent ignition in the caseof sparking.

A widely used practice is the addition of antistaticadditives which strongly reduces the resistivityof hydrocarbons, thus, decreasing the voltagedifference.

An enlarged list of less common methods used to preventthe formation of static electricity or to reduce its danger-ous influence is listed in the excellent work of W. M.Bustin &. Co.1 As has been told. a knowledge of currentgenerated by pumping and stirring of hydrocarbons is ofessential importance.The order of magnitude of pumping currents is in the

range of -100 to - L O O O X 10-10A. When the hydrocar-bon is pumped through filters or discharged in jet-form,these values may be even larger.As experience shows (which, however, is not quite in

agreement with the theory), the strongest current genera-tion occurs when the hydrocarbon resistivity lies between1012to 101~ohm-cm and decreases beyond this range.

P ip e L in e C h ar ge G e ne ra tio n . The problem of electricalcharges produced by flowing hydrocarbon products IIIpipe lines was studied by many investigators.Bustin, Culbertson, and Schleckser' designed a

research unit consisting of:( I ) Source drum and pump section(2) Test line(3) Receiving drum

(Section I)(Section 2)(Section 3)

Hydrocarbons were pumped through this unit and thecurrents-to-ground from each section were measured.On the basis of these measurements, they have

developed a semi-empirical and semi-theoretical formulawhich allows prediction of the current generated in pipeline during hydrocarbons flow.This formula is as follows:

Safety 27

whereI L = Current from Section 2 (pipe line) to ground (i

amperes x I0-1)Is = Current from Section I (source drums and pump)

ground (in amperes xlO-lo )T = Time constant (found experimentally as equal 3sec.)

V = Linear velocity of hydrocarbon in line, fpsL = Length of pipe, feetK = Proportionality factor, determined by pipe diamete

and charging tendency of hydrocarbon pipinterface.

A smooth stainless steel pipe with a diameter of I/~incwas used for these tests.The K value found for this particular case was:

K = 0.632This K factor is approximately proportional to th

roughness of pipe (friction factor) and inversely proportional to the pipe diameter:

fKoc-DThe tested hydrocarbon was aviation fuel IP-4.When taking into account the actual conditions of the

tests (more than 40 runs performed), the authors computed the following equation:

which shows excellent agreement with their test result(the error below 5 percent). The (-) shows that currenin the pipe line has negative polarity while currenmeasured in the drum always had positive polarity. IThe algebraic sum of the three currents measured from

all three sections have to be theoretically always zerohowever, minute differences have been found in thseparate tests.

T he N om o gra ph . The calculation of current generatedin pipe during hydrocarbon flow, according to formula (3is troublesome and time-consuming.Therefore: a nomograph (Figure 1) is presented whic

allows the calculation of the I L value within a couple oseconds using one movement of a ruler only.


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278 Rules of Thumb for ChemicalEngineers

- 9 5 0

80-.i----~7 5

. . . . .wwu..w 7 0o,s;u.. ' [ 7 ]~ 6 5(!Jzw_ .II 6 0~

5 5

5 0

Figure 1. Nomograph for determining current generated in pipe lines from flowing hydrocarbons.

coI0-enwawe,~ "- ~ -,~ " VN:0/~ ! I . . ~"~I . . . . . . ~ " ~ r-,, I" " .0.''" (70.0 I " " ' \ : ~ " " ' " - . .. . . - . .. . .I~~ . . . . . ."~ "...t1- I'IiI; ~0o.,"'""-' ' ' ' 0 ' " ~ " " -N o . . . . " ~" " I\tv ~ ~.r - - " ~~~ - . . . . : I--,l iI . .o.l~ O '"I~.r

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.0.00.006. 0 0:004.003.002

"-'IQ, .001~ .0008Q)c::i'o .00065 .0005c: :Q) .0004>~ .0003'"::

.0002

.0001.00008.00006.00005.00004.00003.00002

.00001.000008.000006.0 00 00 5 1 3 4 5 6 8 1 0 20 30 40 5060 80 100 200 300

P ipe D iam eter, in Inches - d

Safety 28

0106

.0 5

0 4

035

03

0 2 5VI'"c.0 : : :

0 2 0":=;0: : : : :0 18 ~u016 ~

E01~ 0 : ' 3ou . . . .0 1 2

01

0 0 9

008

Figure 3. Relative roughness of pipe materials and friction factors for complete turbulence."


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286 Rules of Thumbfor ChemicalEngineers

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I"I 1.0r : ~ 5~,~ ).p 0.95I I . IH-0.9 )~

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N a t u ra l V e n t il a ti o nAs part of a Process Hazards Analysis (PHA), I was

required to check a naturally ventilated building contain-ing electrical equipment and a fuel gas supply, for ade-quate air flow due to thermal forces (stack effect). APIRP 500' has a method that they recommend for buildingsof 1,000 fr' or less. The building in question was muchlarger. I used the RP 500 method for investigationbecause:I. The building had exhaust fans so the calculation wasonly to check for natural ventilation when the fanswere off.

2. Perhaps the calculation would show far more naturalventilation than required.

First, it was verified that the building ventilationgeometry was to API Standards (no significant internalresistance and inlet and outlet openings vertically sepa-rated and on opposite walls). The building had manywindows on both sides and 3-ft wide louvers at the roofpeak. So the inlets were on both sides and the outlet atthe top middle. This arrangement was judged adequate.The calculations involve first finding:

where h = Height from the center of the lower openingto the Neutral Pressure Level (NPL), in feet.The NPL is the point on the vertical surfaceof a building where the interior and exteriorpressures are equal.

H = Vertical distance (center to center) betweA, and A2, ft. This was weight averagedvarious windows and open doors.A, = Free area of lower opening, ft"

A2 = Free area of upper opening, ft"Then the required opening area is determined

provide 12 changes per hour:

where A = Free area of inlet (or outlet) opings. fr', to give 12 air changeshour (includes a 50% effectivenefactor)V = VoLumeof building to be ventilatfr'

T, = Temperature of indoor air, "RTo = Temperature of outdoor air, "R

(T, - To) = Absolute temperature difference,will always be positive

Fortunately, our calculations indicated that wemore than twice the free area required.

S o u r c eAPI Recommended Practice 500 (RP 500), "Recomended Practice for Classification of LocationsElectrical Installations at Petroleum Facilities," 19American Petroleum Institute.

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Inert Gas Lel Rules