Chapter 2 Source term

Chapter 2

Source term 1 INTRODUCTION

Major accidents always start with a loss of containment. A material contained inside a piece of equipment (for example a tank, a distillation column or a pipe) exits to the atmosphere through an opening such as a hole, a crack or an open valve. The origin may be corrosion, a mechanical impact or a human error. The loss of containment itself can also be an accident, as in the case of the explosion of a pressurized tank. Once the loss of containment has started, the evolution of the event will depend on a series of circumstances such as the condition of the material (gas, liquid or a mixture of the two), its properties, the meteorological conditions and the measures taken to mitigate the leak. In order to predict the effects and consequences of a given accident, we must calculate (or, better, estimate) the velocity at which the material will be released, the size of the liquid pool that will form and the velocity at which the liquid will evaporate. This information is required in order to apply mathematical models of the various dangerous phenomena that can occur (fire, explosion, toxic cloud, etc.) and thus predict the physical effects of the accident (concentration, thermal radiation, blast, etc.) as a function of time and distance. As mentioned in Chapter 1, a number of accidents can occur in the event of a loss of containment of a hazardous material. Fig. 2-1 summarizes the various possibilities. The released material is often a fluid (a gas or a liquid). The loss of containment can be continuous over a certain time or instantaneous. Continuous releases can take place through a hole in a tank, a broken pipe or a safety valve. In these cases, we must calculate the mass flow rate (sometimes as a function of time), the total amount released or the time during which the release takes place. If the released material is a liquid, a pool will probably form (or the liquid will be retained in a dike) and evaporate at a certain velocity. Vapour can also be released if a liquid is depressurized in the loss of containment and flashes. The loss of containment can also be instantaneous, e.g. if a storage tank breaks or a pressurized tank bursts. A set of equations is used to perform these calculations. Often, when too many variables are unknown, the calculation is not direct and an iterative procedure must be applied. Furthermore, to predict the effects of a hypothetical accident, some assumptions must be made: the size of the hole, its position (if it is located at the bottom of a tank, the release is likely to be a liquid, whereas if it is located at the top, the release will probably be a gas), the time during which the release will take place, the ground temperature, etc. Here, the experience of the person performing the calculations is very important. For a very specific case or unit, a few probable sources can usually be identified. For risk analysis of industrial installations, however, some general rules are often applied (see the last section of this chapter).

19

g Instantaneous release of the complete inventory (gas)

~

~

~

~

Instantaneous release of the complete inventory (liquid or liquified pressurized gas)

Continuous two-phase flow through a hole, a broken pipe, a PSV, etc.

Continuous flow of liquid through a hole, a broken pipe, etc.

Continuous flow of gas through a hole, a broken pipe, etc.

Calculate the mass flow rate (as a function of time?), the total amount released, the release time

Calculate the mass flow rate (as a function of time?), the total amount released, the release time

Estimate the liquid/vapour flash at ambient pressure and calculate the amount of vapour

Liquid pool: calculate the evaporation rate

Fig. 2-L Simplified scheme of the various source terms that can occur when there is a loss of containment in a plant or during transportation.

This chapter discusses the source term models for the most common loss-of-containment events: Flow of liquid through a hole in a tank. Flow of liquid through a pipe.

20

Flow of gas or vapour through a hole. Flow of gas through a pipe. Two-phase flow. Evaporation of a liquid from a pool. Some illustrative examples have also been included. For a more extensive treatment, additional literature is recommended [1, 2, 3, 4]. 2 LIQUID RELEASE

Liquids are common in industrial installations, where they are stored in tanks and flow through pipes and equipment. If a liquid is released through a hole or a broken pipe, its subsequent behaviour will depend on the conditions prior to the loss of containment (pressure and temperature): it can flash, form a pool and then evaporate. In order to foresee what will happen, it is necessary to estimate the liquid flow rate. The following sections analyse the most common situations. For a flowing fluid, the mechanical energy balance can be written as follows [1]:

(2-1) where g is the acceleration of gravity (m s· 2) z is the height above an arbitrary level (m) u is the velocity of the fluid (m s· 1) P is the pressure (Pa) Ws is the shaft work (kJ kg" 1) F1is the friction loss (kJ kg" 1), and pis the density of the fluid (kg m- 3). For incompressible fluids, the density is constant and (2-2) Furthermore, if there is no pump or turbine in the line (see Fig. 2-2), Ws = 0. Then, the mechanical energy can be simplified to:

(2-3)

2.1 Flow of liquid through a hole in a tank Consider a tank containing a liquid up to a certain level, and with an absolute pressure Pcont above the liquid. This pressure can be kept constant (as, for example, when the tank is padded with nitrogen to avoid the formation of a flammable atmosphere) at a certain value or simply at atmospheric pressure. In the latter case, if the tank is emptied, a gas entry must be provided to prevent a vacuum from forming, which could lead to the collapse of the vessel.

21

Finally, if the tank contains partly liquefied gas, the vapour-liquid equilibrium will maintain constant the pressure above the liquid.

'1:ont

p

Fig. 2-2. Flow ofliquid from a hole in a tank.

If there is a hole in the wall of the tank (Fig. 2-2), there will be a liquid leak. The mass flow rate can be calculated with the following expression, which is based on the mechanical energy balance and a discharge coefficient CD, in order to take into account the frictional losses in the hole:

(2-4) where m is the liquid mass flow rate (kg s- 1) Aor is the cross-sectional area of the orifice (m2) CD is a discharge coefficient (-) Pcont is the pressure above the liquid (Pa) Po is the outside pressure (usually the atmospheric pressure) (Pa), and h1 is the height ofliquid above the leak (m). The discharge coefficient CD is a complicated function of the leak geometry, the ratio d0 /dpipe and the Reynolds number in the hole. For turbulent flow, the following values can be used: sharp-edged orifices, CD= 0.62; straight orifices, CD= 0.82; and rounded orifices, CD= 0.97. If, due to depressurization, the liquid undergoes a flash vaporization, it can be assumed that this phenomenon occurs downstream from the hole, so it is not taken into account for the calculation of m. As the release of liquid proceeds, the liquid level in the tank decreases, and as a consequence the flow rate through the hole decreases as well. The mass discharge rate at any time t can be calculated with the following expression [2]:

22

(2-5)

2

where A1 is the cross-sectional area of the tank (m ) and tis the time from the onset of the leak (s). However, in risk analysis, the conservative assumption of constant discharge rate-at the initial flow rate-is sometimes applied until all of the liquid above the level of the hole has been released. Finally, the time required for the vessel to empty to the level of the leak is, for a constant tank cross section:

t

e

_1 (~J [ 2( A

=C

Dg

pcont -

or

Po +

P1

h

g ~

J_

(2-6)

For a non-constant cross-sectional area of a tank (sphere, horizontal cylinder), see [4, 5]. Example 2-1 A cylindrical tank 10 m tall and 5 m in diameter contains toluene at 20 °C. The pressure above the liquid surface is kept at essentially atmospheric pressure with nitrogen. The tank is filled to 85%. A collision creates a hole in the tank wall (dar= 50 mm) 1 m above the bottom. The leak is repaired and stopped 30 min after the onset. Calculate: a) the initial flow rate through the hole; b) the amount of toluene spilled, and c) the time during which the toluene would have been spilled if the leak had not been repaired. 3 (ptoluene = 867 kg m- ). Solution a) Initial height ofliquid in the tank: 0.85 · 10 = 8.5 m Initial height ofliquid above the leak: 8.5 -1 = 7.5 m The initial (maximum) flow rate is calculated with Eq. (2-4):

m

052

= ;r 0 ·

4

867 · 0.62

~2 · 9.81 · 7.5 = 12.8 kg s- 1

b) As toluene is spilled, the height of liquid in the tank decreases and, consequently, the leak flow rate decreases as well. Therefore, the release must be divided into several discrete time segments, assuming constant liquid head and leak flow rate over each segment. Ten segments are taken for this case. For the first time segment (from t = 0 s tot= 180 s): m = 12.8 kg s- 1, ht= 7.5 m.

During the first 180 s, the amount of toluene spilled is: 12.8 · 180 = 2304 kg=> 2.66 m

3

After the first time segment, the height of the liquid above the leak is:

23

2.66 h1 = 7.5---2- = 7.365 m Jrd 4

The following table summarizes the calculation results for the various time segments. t, s

m, kg s- 1

ht.m

12.80 12.68 12.57 12.45 12.34 12.23 12.11 11.99 11.88 11.76

7.500 7.365 7.231 7.099 6.968 6.832 6.704 6.577 6.451 6.324 6.201

0-180 180-360 360-540 540-720 720-900 900-1,080 1,080-1,260 I ,260-1,440 1,440-1,620 I ,620-1,800 1,800

Accumulated spilled mass, kg 2,304 4,586 6,850 9,092 11,313 13,514 15,694 17,853 19,991 22,107

The amount spilled is 22, I 07 kg. c) The time is calculated with Eq. (2-6): 2

/[5 1 4 te= 0.62. 9.81 [ /[ 0.05 2 4

j

~2-9.81·7.5=19,944s::::>5.54h.

2.2 Flow of liquid through a pipe When a fluid flows through a pipe, there is friction between the fluid and the pipe wall (depending on the roughness of the wall) and the mechanical energy of the fluid is partially converted into thermal energy. For a given flow rate, the fluid experiences a certain pressure drop, and the pressure upstream must be high enough to overcome it. Even though the pressure changes, for incompressible fluids (liquids) the density is constant along the pipe. In risk analysis, a common problem is to calculate the liquid flow through a hole in a pipe or through a broken pipe (Fig. 2-3).

2.2.1 Liquid flow rate The relationship between pressure drop and fluid velocity for an incompressible liquid flowing through a piping system can be obtained from the Fanning equation:

(2-7-a)

24

or

(2-7-b)

U=

where iJP is the pressure drop over the pipe (Pa) L is the pipe length (m) u is the fluid velocity (m s- 1) dp is the diameter of the pipe (m), and fF is the Fanning friction factor(-). The Fanning friction factor is the ratio between the mechanical energy dissipated by friction and the kinetic energy of the flowing fluid. The mass flow rate in the pipe can then be calculated by:

(2-8)

where Ap is the cross-sectional area of the pipe (m2).

~ont

ff L, dp,

?~ u

Fig. 2-3. Flow ofliquid from a pipe.

The Fanning factor is a function of the Reynolds number, which depends on the velocity of the liquid in the pipe; therefore, the mass flow rate in the pipe must be calculated by iteration [3]. The following procedure can be followed: 1. Guess a value for the Reynolds number. 2. Calculate the value of the Fanning friction factor. 3. Calculate the liquid velocity in the pipe. 4. Calculate a new Reynolds number.

25

5. Compare the two Reynolds numbers. If they are not equal, correct the value of Re and repeat the procedure. The pressure drop in the piping system is not only due to friction with the pipe itself. Pipe fittings, elbows, valves, contractions, etc., also play a role. This contribution is usually expressed as an equivalent length of straight pipe; thus: L = L straight pipe + I, L equivalent

(2-9)

Table 2-1 shows the equivalent length of various pipe fittings for turbulent flow. Table 2-1 Equivalent length of pipe fittings_{turbulent flow on_!y2_* Pipe fitting Globe valve, wide open Angle valve, wide open Gate valve, wide open %open Y2 open V..open 90° elbow, standard long radius 45° elbow, standard Tee, used as elbow, entering the stem Tee, used as elbow, entering one of two side arms Tee, straight through 180° close return bend Ordinary entrance (pipe flush with wall of vessel) Borda entrance (pipe protruding into vessel) Rounded entrance, union, coupling Sudden enlargement from dp to D

""'300 ""' 170 :::::7 ""'40 ""'200 ""'900 30 20 15 90

60 20 75 16 30 Negligible

~;H~JJ

Laminar flow in dp

~J,.,H~Jr

Turbulent flow in dP Sudden contraction from D to dp; all conditions except high-speed gas flow where P 1/P 2 :::>: 2 Laminar flow in dP

~[1.25-[d~ ]] 160 D 2

Turbulent flow in dP

*Taken from 0. Levensp1el, Engineering Flow and Heat Exchange, p. 25, Plenum Press, New York (1984), with permission of Springer Science and Business Media.

26

2.2.2 Friction factor The Fanning friction factor ifF) can be calculated as follows [1]. For laminar flow:

f F =_!i Re

(2-1 0)

In the transition regime, the value of /F is uncertain. For turbulent flow, the Colebrook equation can be used:

_1_ =

fj;

_ 4 log[-1-~+ 3.7 dP

1.255] Re fj;

(2-11)

where sis the roughness of the pipe(-) (Table 2-2). Table 2-2 Roughness of clean pipes* Pipe material Riveted steel Concrete Wood stave Cast iron Galvanized iron Asphalted cast iron Commercial steel or wrought iron Drawn tubing Glass Plastic (PVC, ABS, polyethylene)

e,mm 1 - 10

0.3-3 0.2- 1

0.25-0.26 0.15 0.12

0.043 - 0.046 0.0015 0 0

*Taken from 0. Levenspiel, Engineering Flow and Heat Exchange. Plenum Press, New York, 1984.

For fully developed turbulent flow in rough pipes,JF can be calculated with:

(2-12)

For smooth pipes andRe< 100,000,/F can be calculated with the Blasius equation:

fF = 0.079 Re -o.2s

(2-13)

And for smooth pipes andRe> 100,000, the following expression can be applied [6]:

JF = 0.0232 Re-o 1507

(2-14)

The value of the Fanning factor can also be obtained from the classical plot of fF as a function of dd andRe (see Fig. 2-4).

27

0.025

,___ ~~m/n~~ flo~ ~- f ~ 1§_ ~ F Re 1

0.02

r-...

/;:

Complete turbulence rough pipe

1--.

0.05

/, /;:

0.015

0.04

I"'

~

!"-

~

\

0.03 0.025

~

8::

t-

--r-

0.02

\ ;; ~ 'r-

~

0.01

1I ~

0.009

\

0.008

I

"

ll~ I"'

I%:~

0.007 0.006

0.015

t---.......

Uncertainregion

0.01 0.008

t---

,~

0.006

&

0.004 0.003

--

0.002 0.0015

1:11

0.005

t-

......

0.001 0.0008 0.0006

........

~['-,

0.004

I~ ~

!'-,

r--

~

t-

~ ~

...... Smooth pipe (E = 0)

.s::

Ol :J

e Q)

>

~

Qj

0::

0.0001

~

0.0025

Q)

c

0.0002

]'... ~

0.003

vi

(/)

0.0004

~

I~

"'ll:l

I'.

Jt' ,......

5x10 -s 3x1 0 -5 2x10 -s 5x10 -6 1x10 -5

"'23468 2 3 4 68 2 3 4 68 2 34 6 8 0.002 6 8 23468 7 6 5 4 3 10 8 10 10 10 10 10 Re

Fig. 2-4. The Fanning factor as a function of the Reynolds number and pipe roughness. Taken from 0. Levenspiel, Engineering Flow and Heat Exchange, p.20, Plenum Press, New York (1984), with permission of Springer Science and Business Media.

Example 2-2 A cylindrical tank 10 m tall and 5 m in diameter contains toluene at 20 °C. The pressure above the liquid surface is kept at essentially atmospheric pressure with nitrogen. The tank is filled to 85%. Estimate the initial liquid outflow through a commercial steel pipe (with an inside diameter of 100 mm) connected to the bottom of the tank. The pipe has been broken 65 m from the tank; it is horizontal and has a gate valve (open) and two 90° elbows. 4 1 3 Data: f.ltoluene. 20 "C = 5.9 . 10- kg m·l s· ; Ptoluene, 20 "C = 867 kg m· ; Ccomrnerciol steel= 0.046 mm.

28

Solution For the commercial steel pipe:

0 046 _!_ = · = 0.00046 dp 100 The pressure drop through the piping system will be the difference between the pressure at the pipe inlet and the atmospheric pressure:

M = p 1 g h1 = 867 · 9.81· (0.85 ·10)= 72,295 Pa The equivalent length of the pipe, including the fittings, is: L=65+0.1(7+2·30+16+32)=76.5 m Guessed Re = 25,000 u=

,u Re = (5.9 ·10-4 )25,000 = O.l 7 m s-1 dp P1 0.1· 867

From Fig. 2-4,/F = 0.0064. By applying Eq. (2-7-b ):

U=

72,295 · 0.1 = 2 _92 m s·I 2. 0.0064. 867.76.5

Therefore, Re must be corrected. The following table summarizes the results of the trial-anderror procedure. -[

Re

u,ms

25,000 70,000 300,000 400.000 500,000 520,000 530,000

0.170 0.544 2.041 2.722 3.402 3.539 3.631

fF

u (Eq. (2-7b)), m s· 1

0.00640 0.00520 0.00450 0.00440 0.00435 0.00433 0.00430

2.920 3.240 3.480 3.520 3.540 3.544 3.560

Finally, u = 3.547 m s- 1• Thus, the mass flow rate is: 2

ff 0.1 -1 m = 3.547 · 867 · - - = 24.1 kg s .

4

29

3 GASN APOUR RELEASE

When a gas or a vapour is released from a given piece of equipment (pipe, tank, etc.), the pressure energy contained in the gas is converted into kinetic energy as the gas leaves and expands through the exit. The density, pressure and temperature of the gas change during the loss of containment. In practice, if the density change of the gas is small (p/p 2 < 2, P /P 2 < 2) and the velocity is relatively low (u < 0.3 times the velocity of sound in the gas), then the flow can still be considered incompressible. However, at high pressure changes and high flow velocities, kinetic energy and compressibility effects become dominant in the mechanical energy balance and the flow is considered compressible. Therefore, the accurate analysis of such systems involves four equations: the equation of state and those of continuity, momentum and energy. This makes the analysis rather complicated. To simplify matters, it is usually assumed that the flow is reversible and adiabatic, which implies isentropic flow. Furthermore, it is often assumed that the fluid is an ideal gas with constant specific heat (average value). The models for estimating the release flow rate of a gas can also be applied to a vapour, as long as no condensation occurs. Therefore, in this chapter, both categories (gas and vapour) will be referred to as "gas". 3.1 Flow of gas/vapour through a hole For liquids and gases with low pressure changes and low velocities (P/P2 < 2, Ma < 0.3) (Ma is the Mach number), the flow can be considered incompressible and the expressions presented in the previous section can be applied. However, with a gas flow, if the pressure change is significant and the velocity is high, then kinetic and compressibility effects play an important role [I] and the pressure, temperature and density change significantly when the gas flows through an opening. The flow is considered compressible and different expressions must be applied to calculate it.

3.1.1 Critical velocity When a gas or vapour exits through a hole, there are two possible situations: sonic velocity and subsonic velocity. This is discussed below.

y

~ont u=o

~ont ~ ~~J y-:___ 2

I

~

~

U hole

~nt

~

<

!y+1 j Z

_J_ Y·l

~ subsontc veloctly

Fig. 2-5. Flow of gas or vapour through a hole.

30

somcvelocrty

Let us assume that gas is flowing from a tank at a certain pressure (Pcont) through a hole in the wall (Fig. 2-5). If the pressure downstream from the hole (Pout) decreases, the velocity of the gas through the hole increases. This velocity will increase until, at a certain value of Pout, it reaches the velocity of sound in that gas (at that temperature). Further decrease of Pout will not cause any increase in fluid velocity: the velocity of sound at Pehoked, Tehoked, is the maximum velocity at which the gas can flow through the orifice (to reach supersonic speed, specially designed converging-diverging nozzles would be required). The pressure at the hole outlet will be Pehoked, even though Pout decreases further. Pehoked is called choked or critical pressure, and the velocity at the hole in these conditions is called choked or critical velocity. Assuming isentropic expansion, the relationship between the choked pressure and the pressure inside the tank can be expressed as: r pchoked pcont

2 ] r-I = y+l

(2-15)

[

where

Peont is the pressure inside the container or the pipe (Pa), and yis the ratio of heat capacities, c/cv (-). The choked velocity is the maximum possible velocity in an accidental release. It is found in most accidental gas releases. Since Pout is usually the atmospheric pressure (essentially constant), the same conditions are reached if the pressure inside the container (a tank, a pipe) increases up to a certain value: further increases in Peont will not produce any further increase in the gas exit velocity. Therefore, critical velocity will be reached if the following condition is fulfilled:

(2-16) In fact, r is the isentropic coefficient of the gas or vapour at the relieving conditions. However, for gases with properties similar to those of an ideal gas, r is the ratio of heat capacities. r is always greater than unity. For most gases, it ranges from 1.1 to 1.4; therefore, sonic velocity will be reached when Pcon/Po :2: 1.9. For air (y= 1.4), for example, sonic velocity is reached when Pcon/Po :2: 1.893, i.e. when the downstream absolute pressure is 52.8% of the upstream absolute pressure. The speed of sound in an ideal gas at a temperature T can be calculated with the following expressiOn:

(2-17) is expressed in m s- 1 R is the ideal gas constant (8.314 kJ kmole- 1 K- 1) Mv is the molecular weight of the gas (kg kmole- 1) Table 2-3 shows the ratio of heat capacities, y= cJcv, for various gases. The density of a gas increases with pressure. Therefore, once the critical velocity has been reached, if Peont is further increased, the release velocity will still be the speed of sound, but

where

Us

31

the density of the gas will be higher. Therefore, the mass flow rate will increase with Pcant· It is therefore clear that what becomes critical or choked is the velocity (m/s) of the gas rather than the flow. Thus, critical velocity is a better term than critical flow. Critical flow -i.e. both choked gas velocity and mass flow rate- can be reached when there is a given pressure upstream from the hole and vacuum conditions downstream from the hole so that the sonic velocity is reached. In this case, the inlet gas density is constant and therefore the mass flow rate is also choked. The temperature of the gas in the jet at the orifice is:

p

Tchoked -- Tcon!

(~) p coni

(r;I)

2 - Tcon! ( -1)

(2-18)

y+

where Tcant is the temperature in the container or pipe (K). Table 2-3 Molecular weight, heat capacity ratio and sonic velocity for various gases and vapours at 298 K and I 01.3 kPa. Calculated from [7, 8] Gas Molecular weight y = c/cv Acetylene 26.0 1.247 345 Acrylonitrile 53.1 1.149 232 Air 29.0 1.400 346 Ammonia 17.0 1.311 437 Benzene 78.1 1.112 188 Butane 58.1 1.091 216 Carbon dioxide 44.0 1.301 271 Carbon monoxide 28.0 1.400 352 Chlorine 216 70.9 1.330 Cyclohexane 84.2 1.085 179 Ethane 30.1 1.188 313 Ethylene 28.0 1.253 333 261 Ethylene oxide 44.0 1.215 Helium 4.0 1.660 1014 Hexane 1.062 175 86.2 Hydrogen chloride 36.5 1.399 308 Hydrogen 2.0 1314 1.405 Hydrogen sulphide 34.1 1.326 310 Methane 16.0 1.304 449 Natural gas• 18.1 1.280 419 Nitrogen 28.0 1.406 352 Oxygen 32.0 1.395 329 Propane 44.1 1.146 253 Propylene 42.1 1.148 260 Sulphur dioxide 64.1 1.264 221 Toluene 92.1 171 1.087 a

86.15% CH4, 12.68% C2H6, 0.09% C4Hw, 0.68% N2

32

3.1.2 Mass flow rate The mass flow rate of gas through an orifice can be calculated with the following expression, obtained from the mechanical energy balance by assuming isentropic expansion and introducing a discharge coefficient: y+l

m

-A C p

hole-

or

contl.f/

D

(2-I9)

(-2-Jr-1 Mv r +I ZT R 103

r

cont

where m is the mass flow rate (kg s· 1) cd is a dimensionless discharge coefficient (-) Aor is the cross-sectional area of the orifice (m2), and Z is the gas compressibility factor at Pcont, Tcont (-)(for ideal gas behaviour, Z = I). l.j/is a dimensionless factor that depends on the velocity of the gas. For sonic gas velocity: l.j/=I

(2-20)

and for subsonic gas velocity:

l.f/ 2

=~(L:!J);:: (~J~ [~-(~Jr;l r I

2

?,ont

(2-2I)

Pcont

The value of If/ has been plotted as a function of Pr/Pcont, for various heat capacity ratios, in Fig. 2-6. The length of a free jet of a gas can be estimated [9] with the following expression:

(2-22) where

is the velocity of the jet at the source (m s· 1) dar is the diameter of the source, and 1 1 Uw is the average ambient wind speed (m s· ) (default value: 5 m s· ).

Uj

3.1.3 Discharge coefficient CD is a coefficient that takes into account the fact that the process is not isentropic. Its value is CD = 1.0 for a full-bore rupture in a pipe. For sharp-edged orifices in accidental releases (high Reynolds number), some authors recommend CD = 0.62 and others recommend a conservative value of 1.0. Example 2-3 Due to an incorrect manoeuvre, an impact creates a hole with an approximate diameter of 2 em in the top of a tank containing propane at 25 oc and 10 bar. The level of the liquid is low, so gas is released through the hole. Calculate the mass flow rate.

33

Solution For propane, y= 1.15. Therefore: r

-y+l]r-1

[ 2

l.l5 [1.15+1]1.15-1 = --

2

= 1.74

Since 10 bar = _ 9 87 1.013 bar

Pcont

P0

the propane velocity at the orifice is critical. Therefore, by applying Eq. (2-19):

=

ff

1.15+1 2 0.02 CD (10 ·l 05 ) l l 15 (-2-) l.l5-l 44.1 = {0.525 kg/s if CD = 0.62 0.847kg/s ifCD=l.O 4 . 1.15+1 1(273+25)8.314-10 3

The velocity at the hole can now be calculated. The conditions at the choked jet are: l.IS

Pchoked

-)u

2 -10 (l.l - +l 5

51 - -- 5.744 bar

l.IS-1 Tchoked

Pchoked

5.74)l:i5

= 298 ( lo

= 277.2 K

(5.744-10 5 )44.1 (8.314 ·10 3 ) 277.2

II kg m- 3

The velocities at the hole corresponding to the two mass flow rates are: 0 525 ·

u

2

( ff

0.~2 ) ll

= 152 m s- 1

34

The speed of sound in propane at choked conditions is:

us=

1,00

y= 1,10 0,95

0,90

1,15 1 20 1,25 1,30 1,35 1,40 1,45

0,85

\f

~

j,

0,80

,, 1,10

~~~ ~

[\

0,75

... 0,70

0,65

0,60

0,55

0,50

0,45

0,40 1,00

0,95

0,90

0,85

0,80

0,75

0,70

0,65

0,60

0,55

0,50

PofPcont

Fig. 2-6.ljl as a function of P r/Pcont and y.

3.2 Flow of gas/vapour through a pipe A typical case in risk analysis is the calculation of a gas flow from a pipe connected to an upstream constant pressure source (usually a vessel) (Fig. 2-7). The gas outflow can take place through a full-bore rupture of the pipe or through a hole in the pipe wall. In both cases, the pressure in the pipe must be estimated at a point just in front of the orifice. This requires knowledge of the gas flow rate which, in tum, depends on the pressure drop between the upstream constant pressure source and the aforementioned point. A trial-and-error procedure is therefore required. A relatively simple method [3] is presented below.

35

Rupture

Surroundings

I P, I I I Hole I ~-m-h-ot_e_ _ _ _ __

Upstream constant pressure source

I

I

·~ I

~ont L

Fig. 2-7. Flow of a gas through a pipe.

The overall pressure drop between the upstream pressure source and the environment is the pressure drop in the pipe plus the pressure drop through the opening (a hole or the fully broken pipe): (2-23) where Pp is the pressure inside the pipe just in front of the opening (Pa), and Po is the ambient pressure (Pa). The mass flow rate through the pipe depends on the pressure drop through the pipe and the mass flow rate through the hole depends on the pressure drop through the hole. According to the law of conservation of mass, the mass flow through the pipe due to the loss of containment (mpipe) must be equal to the mass flow through the hole (mhote): m pipe

=

(2-24)

m hole

The mass flow rate through a pipe depends on the pressure at both ends of the pipe, and can be calculated with the following expression [3]:

mpipe

(2-25)

= Ap

where Ap is the cross-sectional area of the pipe (m2) p (P) is the density of the gas (kg m- 3) /F is the Fanning friction factor (-) Lis the length of the pipe (m), and dp is the diameter of the pipe (m). The following relationships apply:

36

I

p

=constant(~);

(2-26)

(2-27) where Z is the compressibility factor. For ideal gas behaviour:

Z = 1 and(=r The integral in Eq. (2-25) may be solved analytically, assuming a constant compressibility factor and constant specific heat at a constant volume, cv; for Z = I:

(2-28)

This set of equations can be solved by trial and error by guessing the internal pressure inside the pipe just in front of the pipe opening, Pp. The following procedure must be followed: I. Guess the pressure inside the pipe just in front of the pipe opening (Po< Pp < Pcont). 2. Calculate the mass flow rate through the pipe opening with Eq. (2-19) (CD= 0.62 for a hole in the pipe wall; CD= I for full-bore rupture). 3. Calculate the mass flow rate through the pipe with Eqs. (2-25) and (2-28). 4. Compare the two mass flow rates. If they are not equal, correct the value of Pp and repeat the procedure. To calculate the mass flow rate through the pipe opening (Eq. (2-19)), the temperature of the gas in the pipe in front of the opening must be estimated. This requires a trial-and-error procedure. By defining a parameter Y, the following expressions can be applied [1]: r-1

Y =1+--Ma 2 I

2

(2-29)

I

(2-30)

(2-31)

r;Irn[Ma~ Y,ant

Macont YP

]-[-\---1_]+ r(4; LJ=O Macont

2

Map

p

37

(2-32)

For sonic flow at the exit, Map= 1 and Eqs. (2-30) and (2-32) become:

~= 2 Ycont Tcont

(2-33)

r+l

y+lln( J-(-~--IJ+r(4fFLJ=O 2 (r +I2Ycont )Ma;ont Ma;ont d

(2-34)

P

For the case of a hole in the wall of the pipe, to calculate the mass flow rate, the pressure in the pipe at a point on a level with the hole must be known. If the only flow is the one caused by the leak, then a trial-and-error procedure must be applied to establish the value of Pp. If there is, furthermore, a certain flow rate through the pipe, it must be taken into account in order to estimate the value of Pp. Example 2-4 A constant pressure source (5 bar, 288 K) of natural gas (Mv = 17.4; 1.27) is connected to a smooth polyethylene pipe with a diameter of 164 mrn. Calculate the release flow rate a) if the pipe is completely broken 330 m from the source, and b) ifthere is a hole in the pipe wall, at L = 330m, with a diameter of 50 mm.

r=

Solution a) For full-bore rupture, Co= 1.0. For a smooth polyethylene pipe (Fig. 2-4), a value of the Fanning factor /F = 0.002 will be assumed. Guessed pressure: Pp = 4 bar. To estimate the temperature of the gas at the end of the length of pipe, just in front of the opening, Eq. (2-34) is applied:

(1.2 7 -I)M

2(1 1.27+lln 2 [

2

)]

+ 2 acont (1.27+l)Ma;ont

-(-~--IJ+I.2 7 (4·0.002·330)=o Ma;ont

I.e.:

1.135ln(2 + 0.27 ~a;ont 2.27 Macont

J-

_1_2_ + 20.44 = 0 Macont

By trial and error, Macont = 0.2. Therefore, by applying Eq. (2-29): 2 yconi =1 + 1.27 - 1 02 · =1.005 2

From Eq. (2-33):

38

0.164

TP

2 ·1.005

288

1.27 + 1

Tp =255 K

Mass flow rate through the opening (Eq. (2-19)): 1.27+1

m hole

= JrO.l 4

642

127 1 2 17 1.0(4·10 )1 1.27(-- ) .4 =16kgs· 1 1.27 +1 255· 8.314·10 3 5

Mass flow rate through the pipe (Eq. (2-25))

2 [- (5 IO' )3 68 (, Jr0.164 2 mpipe = 4

(pcont

= 3.68 kg m·\

~:.:7 )[m';',~' -1)

l

_ _:_______--;------,-----'- = 4.33 kg s .] 4 . 0.002 (

330 0.164

)

Since two different results have been obtained, a new trial is required. The results of the calculation procedure are summarized in the following table: P 0 , bar

4 3 2.7 2.0 1.75 1.74

kf!. s" 16.00 12.00 10.01 8.00 6.99 6.96

mho/e.

1

moioe.

kJ!, s·'

4.33 5.84 6.36 6.78 6.95 6.96

Therefore, m = 6.96 kg s· 1• b) For the hole in the pipe wall, CD= 0.62. The influence of pressure decrease on gas temperature is now neglected. A trial-and-error procedure is again applied. Due to the change in the flow, a new Fanning friction factor value,JF = 0.003, is now assumed (it will be checked later). Guessed pressure: Pp = 4.5 bar. To estimate the temperature of the gas, by trial and error the value of Macont is found: Macont = 0.17. The temperature of the gas just in front ofthe opening is calculated with Eqs. (2-29) and (2-33): Tp = 254.7 K. The mass flow rates are now:

=Jr0.0

m hole

4

52

062(45·10 5 )1 . .

2-):~~:: 17 .4 127(-. 1.27+1 254.7·8.314·10

39

3

=1.04kgs· 1

m P'Pe . =

Jr0.164 2 4

2 [- (s

w' )3 6&

(t !:~7 J[(

45 ;

f':' -t)]

-~---------:,.---.,-------------'- =

330 4. 0.003 ( ) 0.164

2.56 kg s· 1

The two values are different. By trial and error, the following values are ultimately obtained: Pp = 4.9 bar, m = 1.13 kg s- 1• Although the critical velocity is not reached, the use of ljl= 1 is correct (see Fig. 2-6). With this mass flow rate, the average Reynolds number in the pipe is calculated and the value of fF is found. Since two different results are obtained, a new trial is required. The results of the calculation procedure are summarized in the following table: Pp, bar

m,kgs- 1

Re

rF*

4,90 4,50 4,47 4,46

1,13 1,04 1,03 1,03

891,53 819,78 813,17 812,52

0,01795 0,01952 0,01968 0,01969

*fF calculated wtthEq. (2-10) because there rs 1ammar flow.

Therefore, Pp=4.46 bar and m = 1.03 kg s- 1•

A special, relatively complex case is the calculation of accidental releases from distribution systems (gas networks). For these situations, more powerful models are required (see, for example [6]). 3.3 Time-dependent gas release In the above examples, the proposed expressions estimate the mass flow rate of a gas release to a constant pressure source. If the pressure upstream is not constant, as, for example, in the case of a vessel containing a pressurized gas, then these expressions can only be applied to the first few minutes of the release. As the release proceeds, the pressure inside the vessel decreases. Therefore, in this case, the upstream pressure must be corrected at given time intervals in order to calculate the mass flow rate as a function of time (the assumption of a constant flow rate, at the initial value, throughout the period will lead to a conservative prediction). The release decay is basically a function of two factors: the initial leak rate and the initial mass of gas inside the vessel. The decay of the leak, which follows an exponential trend, can be roughly described [10] by the following expression:

m(t)

= minitial eX p ( - minitial w

t)

(2-35)

where mrtJ is the mass flow rate at the timet after the onset of the leak (kg s· 1)

40

is the mass flow rate at the onset of the leak (kg sW is the initial mass of gas in the vessel (kg), and tis the time after the onset of the leak (s).

minitiat

1 )

Example 2-5 A vessel with a volume of9 m3 contains nitrogen at an absolute pressure of 14 bar and 25 °C. Calculate the mass flow rate through a hole with a diameter of 2.54 em a) at the onset of the release, and b) as a function of time. Solution At the initial pressure, the density of nitrogen is 15.8 kg m- 3 • Therefore, the initial content of the vessel is:

W=9·15.8=142.2 kg Due to the pressure in the vessel, it is evident that critical velocity will be reached at the hole. The initial mass flow rate can be calculated with Eq. (2-19): 1.41+1

m hole

2 0 02542 = ;r · 0.62 (14·10 5 )1 1.41(4 1.41+1

28 -)1:4i::r 298·8.314·10

3

l.015kgs-1

(A value of CD= 0.62 has been assumed here.) The flow as a function of time is calculated with Eq. (2-35). For example, thirty seconds after the onset of the release:

m( 3o)

= 1.015 exp (

-1.015. 30J -1 = 0.82 kg s . 142.2

Fig. 2-8 shows the variation of mass flow rate as a function of time.

1.0

'"'

!I'

0.8

.si ~

~

gl

"'

0.6 0.4

:2

0.2

10

Time, min

Fig. 2-8. Flow of a gas through a hole as a function of time.

41

4 TWO-PHASE FLOW In some cases, a mixture of liquid and vapour or gas can occur. Two-phase flow may occur, for example, when a hot, pressurized liquid is significantly depressurized, causing it to boil and suddenly vaporize. If liquid droplets are entrained or foam is formed, a spray will be released through the opening or the relief device. If the mixture of liquid and gas/vapour is ejected into the atmosphere, the spray droplets may be evaporated or may fall to the ground. The existence of two phases has a significant influence on the mass flow rate of the release. Two-phase flow is a complex phenomenon, not yet sufficiently well understood, so conservative design is often applied in order to estimate relief requirements. The following paragraphs explain a simple approach. Furthermore, Section 5.2 discusses a particular case for runaway reactions. For a more accurate prediction, more complex methodologies or computer programs should be used [11]. 4.1 Flashing liquids If there is a leak or a relief to the atmosphere from a vessel containing a hot pressurized liquid, upon depressurization the liquid becomes superheated, its temperature being higher than its boiling temperature at atmospheric pressure. Therefore, it undergoes a sudden, or "flash", vaporization: a mixture of vapour and droplets exits to the atmosphere. This process is so fast that it can be assumed to be adiabatic. The vapour takes the excess energy from the remaining liquid to become vaporized, and the resulting vapour/liquid mixture reaches its atmospheric boiling temperature. Thus, the vaporization of a differential mass of liquid dw1 implies a decrease in the temperature of the remaining liquid dT: (2-36) This expression can be integrated between the initial temperature of the liquid before depressurization, Tcant, and the final temperature of the mixture (the atmospheric boiling point if the mixture is released into the atmosphere), Tb, in order to calculate the mass of vapour:

(2-37)

where

Wv

is the mass of vapour (kg)

wu is the initial mass of liquid (kg) iJHv is the mean latent heat of vaporization between Tcont and Tb (kl kf( 1), and 1 cp1 is the mean heat capacity of the liquid between Tcant and Tb (kl kg· K- ). The ratio between the mass of vapour formed and the initial mass of liquid is usually called the vaporization fraction: cp(Tmm-Th)

f=~=l-e

Ml,.

(2-38)

Wu

42

In practice, a significant amount of droplets will be entrained by the vapour, incorporated into the vapour cloud (thus increasing its density) and later vaporized. Therefore, the "real" value off will be higher than that predicted by Eq. (2-38). In some studies, it has been observed that in fact there was no rainout, and all of the liquid droplets were incorporated into the cloud. Thus, although a value of 2f has been suggested for the vaporization fraction by some authors, a conservative approach is to assume that all of the mass released is entrained in the cloud. For multicomponent mixtures, there is preferential vaporization of the more volatile components and the calculation of/becomes significantly complicated. 4.2 Two-phase discharge When a mixture of liquid and vapour or liquid and gas is discharged, the cross-sectional area required for a given relief is significantly larger than that corresponding to vapour alone. The relationship between the venting area and the discharge rate is complex. An accurate design requires the use of complex procedures. The following paragraphs present a set of equations that allow an approximate calculation. The flashing process approaches equilibrium conditions if the pipe is sufficiently long, the minimum length being approximately 0.1 m or greater than 10 diameters. For a pipe length less than 0.1 m (non-equilibrium regime), the flow rate (kg s· 1 m· 2) increases sharply as length decreases, approaching liquid flow as the length approaches zero. For non-equilibrium regime, flashing flow is choked and can be estimated by [12]:

(2-39) where Gne is the discharge rate (kg m· 2 s· 1) iJHv is the latent heat of vaporization at boiling temperature ofliquid (kJ kg- 1) Tcont is the storage temperature (K) v1v is the change in specific volume brought about by the change from liquid to vapour (m3 kg- 1), and 1 1 cP is the specific heat of the liquid (kJ kg- K" ). 1 N is a non-equilibrium parameter: L

M£2

N= ( ) v 2 2 2 pcont- Po P1 CD Vlv

where

+T;.ont C p 1

(2-40)

Le

is the pressure inside the vessel (Pa) Po is the atmospheric pressure (Pa) Lis the pipe length to opening (ranging from zero to 0.1 m) (m) Pt is the liquid density (kg m· 3) CD is the discharge coefficient (-), and Le = 0.1 m. Eq. (2-40) shows that the degree of non-equilibrium varies directly with the pipe length L. For Lldp = 0, there is no flashing and Eq. (2-39) reduces to the orifice equation for incompressible liquid flow. For L ::: 0.1 m (or greater than 10 diameters), we can assume that equilibrium flashing conditions are reached; the flow rate is a weak function of the Lldp ratio and Eq. (2-39) becomes: Pcont

43

(2-41)

If the physical properties (&fv,

VIv)

are unknown, Eq. (2-41) can be expressed as:

(2-42)

The effect of subcooling on the discharge rate is: (2-43) where Pv is the vapour pressure at the storage temperature (Pa). The mass flow rate for a two-phase discharge of subcooled or saturated liquids can be expressed as [13] [14]: (2-44) is the discharge mass flow rate (kg m- 2 s- 1), and CD is a discharge coefficient(-). As mentioned in Section 3.1, in the discharge of flashing liquids through a hole in a vessel wall, we can assume that the discharge is a liquid, and that flashing occurs downstream of the hole. where

G2p

5 SAFETY RELIEF VALVES A safety valve is a device used to prevent overpressure in a given piece of equipment (vessel or pipe). When a predetermined maximum (set) pressure is reached, the safety valve reduces the excess pressure by releasing a volume of fluid. Safety valves can act in various situations, such as exposure to plant fires, failure of a cooling system and runaway chemical reactions. A safety relief valve has been defined by the ASME as a pressure relief valve characterized by rapid opening or pop action, or by opening in proportion to the increase in pressure over the opening pressure, depending on the application. It may be used either for liquid or compressible fluid. In general, safety relief valves act as safety valves when used in compressible gas systems, but open in proportion to overpressure when used in liquid systems. The basic elements of the design of a safety valve consist of a right-angle pattern valve body with a valve inlet mounted on the pressure-containing system and an outlet connected to a discharge system or venting directly into the atmosphere. Fig. 2-9 shows a typical safety valve design [15]. Under normal operating conditions, a disc is held against the nozzle seat by a spring, which is housed in a bonnet mounted on top of the body. The amount of compression on the spring (which provides the force that closes the disc) is usually adjustable, so the pressure at which the disc is lifted off its seat to allow relief may vary. When the

44

pressure inside the equipment rises above the set pressure, the disc begins to lift off its seat. As the spring starts to compress, the spring force increases and further overpressure is required for any further lift. The additional pressure rise required before the safety valve will discharge at its rated capacity is called the overpressure [15]. The allowed overpressure depends on the standards being followed and on each specific case. For compressible fluids, it normally ranges from 3 to 10%, and for liquids, from 10 to 25%. Once normal operating conditions have been restored or the pressure inside the equipment has dropped below the original set pressure, the valve closes again.

Spring --H/--tttl adjuster

Sprrrrg housing -----+1 {bonnet)

Body

Inlet tract (approach channel)

Fig. 2-9. Typical safety valve design (DIN valve). 5.1 Discharge from a safety relief valve The discharge flow from a safety relief valve depends on the pressure inside the equipment and the cross-sectional flow area of the valve. The relationship between these three variables is established by the existing methods for sizing safety valves. The following paragraphs present the standard AD-Merkblatt A2, DIN 3320, TRD 421. The minimum required orifice area for a safety valve used in air and gas applications can be calculated (for sonic flow) with the following expression:

A = 0.179lmsv sv 'P aw Pcont

~T Z

(2-45)

Mv

and for liquid applications: A =

sv

0.6211msv aw ~ p, (~ant- pback)

(2-46)

45

where Asv is the minimum cross-sectional flow area of the safety valve (mm2) 1 msv is the discharge mass flow rate (kg h- ) Pcant is the absolute relieving pressure (bar) Pback is the absolute backpressure (bar) Tis the inlet temperature (K) Pt is the liquid density (kg m- 3) Mv is the molecular weight (kg kmor 1) Z is the compressibility factor (-)

aw is an outflow coefficient specified by the manufacturer (-), and '¥ is an outflow function (see Fig. 2-1 0). 0.6

r

r.vmar

::.....

1.8 - 0.527 1.6 - · 0. 0.5 i- 1.4- 0.484 r- 1.2 - 0.459

~ -........;

1.0 - 0.429 0.4

;.

~ "

" 0

"'"c 2

0.3

~

~:; 0

0.2

\

'

0.1

0 0.2

0.3

0.4

0.5

0.6

Pressure ratiO (P,..

0.7

0.8

0.9

1.1l

~)

Fig. 2-10. The outflow function as used in AD-Merkb1att A2, DIN 3320, TRD 421. Taken from Spirax Sarco Steam Engineering Tutorials [15], by permission.

For two-phase flow, a conservative approach consists in calculating the area required to discharge the vapour fraction and, separately, that required for the liquid fraction, and then adding them together in order to establish the minimum cross-sectional area. Example 2-6 In a train accident, a derailed wagon tank containing propane is heated by an external fire to 47 oc (Pcant = 16.3 bar). A safety relief valve, now located in the liquid zone, opens and discharges into the atmosphere. Estimate the discharge rate.

46

Data: boiling temperature of the liquid at atmospheric pressure = - 42 oc; average latent heat of vaporization between - 42 oc and 47 oc = 358 kJ ki 1; average cp of the liquid between these two temperatures= 2.54 kJ kg- 1 K" 1; p 1 = 440 kg m· 3; k = 1.15; molecular weight= 44.1; Av = 3.25 cm 2 . Solution Taking into account the liquid temperature in the tank and its boiling temperature at atmospheric pressure, flash vaporization will occur. The vaporization fraction is (Eq. (2-38)):

f=~=l-exp(- 2.54·(320-231))= 358

Wu

0.4 68

Taking into account the pressure values inside and outside the tank, it can be assumed -with a certain degree of uncertainty- that sonic flow will occur in the valve. Because this is a two-phase flow (although there is some uncertainty related to this type of flow at the nozzle), vapour and liquid flow is estimated separately with Eqs. (2-45), (2-46). Vapour flow:

0.468msv

Ag 'I' awPcom 0.1791

(M":: = Ag·0.45·0.7·16.3 ~44.1

vrz

0.1791

320

Liquid flow: _ m =A, aw~Pt (Pcont -PbacJ 0 532 0.6211 SV

A1 0.7)440(16.3-1.013) 0.6211

Furthermore,

By solving these three equations, the discharge mass flow rate is obtained: msv = 6,535 kg h- 1

6 RELIEF DISCHARGES

When a vessel containing a liquid or a gas is heated (by an external fire or by an exothermic chemical reaction, for example), the internal pressure increases. If both temperature and pressure continue to increase, at a certain value, the vessel will burst. To avoid this, a relief device is usually provided to allow the discharge of fluid once a given (set) pressure is reached. If the discharge is controlled and adequately treated (by using a secondary tank, a cold water tank, a scrubber, a flare, etc.), no dangerous release into the atmosphere will occur. However, if the discharge is released directly into the atmosphere, it can lead to an accident. To foresee its eventual effects, the discharge flow rate must be estimated.

47

6.1 Relief flow rate for vessels subject to external fire If a vessel containing a liquid is subject to a certain heat flux, the liquid will evaporate and the pressure will rise. Vapour or gas will have to be released through a pressure relief valve to prevent the pressure from exceeding a given value. However, even with a relief device, an accident (the failure of the tank) can occur if the metal above the wetted surface area (i.e. the metal that is not cooled by the liquid inside the vessel) is excessively heated and loses its strength (Fig. 2-11 ).

Vessel wall unprotected ~

Vessel wall cooled by the liquid

Fig. 2-11. Relief in a vessel exposed to fire.

The following paragraphs describe a method for calculating the gas discharge rate from the relief device (for vessels containing stable liquids) proposed in NFPA 30 [16]. 1. The surface of the vessel exposed to fire is assumed to be: spherical tank: 55% horizontal tank: 75% rectangular tank : 100% (excluding the top surface) vertical tank: 100% (up to a height of9 m). 2. The heat flux to the vessel is estimated as a function of the exposed vessel surface and the design pressure: Aexp < 18.6 m ~ Q=63,080Aexp

(2-45-a)

18.6 m < Aexp < 92.9 m ~ Q = 224,130 A~~~66

(2-45-b)

2 0338 929m < Aexp <260m2 ~ Q=630240 Aexp • '

(2-45-c)

Aexp >260m and Pdesign < 0.07 barg~ Q=4,130,000

(2-45-d)

2

2

2

2

2

Aexp >260m and Pdesign > 0.07 barg~ Q = 43,185 A~~: 2

where

is the exposed vessel surface (m2), and Q is the heat flux (W).

Aexp

48

(2-45-e)

For tanks containing LPG (pressurized tanks), NFPA58 [17] recommends the following expression: 082 Q = 70 ' 945 Aexp

(2-45-f) 2

where Aexp is the entire surface area of the vessel (m ). 3. The gas discharge rate is calculated using the following expression:

FQ

m=--

(2-46)

Mfv

where m is the discharge rate (kg s- 1) F is a reduction factor (-), and 1 &-!vis the latent heat of vaporization at the boiling temperature of the liquid (kJ kg- ). The reduction factor F depends on the protective measures applied: 2 Aexp greater than 18.6 m and drainage with a minimum slope of 1% leading the spill to a remote (at a distance of at least 15m) impounding area: F = 0.5. adequately designed water spray system and drainage: F = 0.3. adequate thermal insulation: F = 0.3. water spray system plus thermal insulation plus drainage: F = 0.15. none of the aforementioned protective measures: F = 1.

Example 2-7 Calculate the mass discharge rate required in a horizontal cylindrical tank containing nhexane that is being exposed to fire. The tank (length = 7 m, diameter = 4 m) is located inside a containing dike and drainage of the spilled liquid is provided. 1 Data: &-!vn-hexane= 334 kJ kg- . Solution The area exposed to fire is.

Taking into account the drainage system, F = 0.5. The heat flux received by the tank is:

Q = 630,240 A~;~38 =630,240·103.7° 338 = 3,025,790 W And the required discharge mass flow rate is: 0.5·3,025,790 4 ,530 kg h-I= 1.2 6 kg s_ 1 334

6.2 Relief flow rate for vessels undergoing a runaway reaction Runaway reaction is the term used to define the uncontrolled development of one or more exothermic chemical reactions. Runaway reactions have been the origin of a number of

49

accidents in chemical plants, including the well-known cases of Seveso (Italy, 1976) and Bhopal (India, 1984). They may involve the loss of control of a desired chemical reaction or the development of an undesired reaction. Highly exothermic chemical reactions are potentially dangerous, and slightly exothermic reactions can cause an increase in temperature that may set off highly exothermic reactions. Uncontrolled exothermic reactions can occur not only in chemical reactors, but also in other units such as distillation columns, storage tanks, etc. If the rate at which the reacting material generates heat is higher than the rate at which the system can dissipate it, the temperature will increase up to a value at which the process is uncontrollable. The essential condition is the existence of a self-accelerated heating process: as the temperature increases, the reaction rate increases exponentially up to very high values. This process can be very slow in its first steps, but very fast in its final step. The formation of gas products or an increase in vapour pressure will raise the pressure inside the vessel. If there is a relief device, gas/vapour of two-phase flow will be released when the set pressure is reached. This will prevent explosion, but if the released stream is not adequately treated, there will be a loss of containment of some dangerous material. The potential danger involved in a runaway reaction is not always taken into account, at least in comparison with other risks involving much smaller amounts of energy. Take, for example, the case of a storage tank containing acrylonitrile [ 18] at 10 °C. If the volume of the tank is 13,000 m3 and it has been filled to 90%, it will contain 725,000 kg. If no inert gas has been used, in the vapour head ( 100 m3) the mixture will contain 11% acrylonitrile and be within flammability limits. The danger associated with this explosive gas mixture is obvious. However, if the energy released in the combustion of the acrylonitrile contained in this gas mixture is compared with the energy released in the exothermic polymerization of the acrylonitrile, the following values are obtained: energy released in the polymerization process: 900,000 MJ. energy released in the combustion of the vapour phase 900 MJ. Therefore, the risk associated with the reactivity of a chemical can be higher than the generally more evident- risk posed by some of its properties (flammability, toxicity, etc.).

E:aled

I

nting

T

Time into venting

Fig. 2-12. Evolution of pressure in a vessel with a runaway reaction. Taken from [ 19], by permission.

50

The evolution of pressure in a vessel in which a runaway reaction takes place will depend on the kinetics of the reaction and on the type of discharge vented. If the vessel is closed, the pressure will increase exponentially. If there is a relief device and it is opened, the pressure will still increase somewhat, reaching a maximum and then decreasing (Fig. 2- 12). The value of the maximum pressure reached during the loss of control will depend on the kinetics of the reaction, the temperature of the mixture when the runaway starts, the initial concentration of the reactants, the mass initially contained in the vessel, the relationship between this mass and the cross-sectional area of the relief device, and the set pressure of the device. The relief mass flow rate can be estimated from the mass and heat balances by applying some simplifying assumptions: the discharge mass flow rate is essentially constant. the heat of reaction per unit mass (q) is practically constant. the physical and thermodynamic properties are constant. With these assumptions, the differential equation obtained from the mass and heat balances can be integrated. Thus, Leung [19] proposed the following expression to estimate the relief mass flow rate for the case of homogeneous-vessel venting, i.e. assummg zero disengagement of liquid and vapour within the vessel:

Wq

m=GA=[~+~l 2

(2-47)

{w~

1

where m is the discharge rate (kg s· ) G is the discharge rate per unit cross-sectional area of venting (kg m- 2 s- 1) W is the initial mass contained in the vessel (kg) q is the heat flow released by the chemical reaction (kW kg- 1) 3 Vis the volume of the vessel (m ) 1 1 Cv is the specific heat at constant volume (kJ kg- K- ) iJHv is the latent heat of vaporization of the liquid (kJ kg- 1) v1v is the change in specific volume when changing from liquid to vapour (m3 kg- 1) L1T is the "overtemperature", the increase in temperature corresponding to the overpressure, (K). The heat flow released by the chemical reaction can be calculated as the arithmetic mean of the values corresponding to the temperature rise rate at the moment at which the relief starts and the moment at which the maximum temperature is reached:

1 [(dT)

q=-c 2

v

dt

(dT) ] dt

+set

(2-48)

max

The ratio (iJH/viv) can be obtained from P-T data by applying the following relationship obtained from the Clapeyron relation:

Miv =T dP V/v

set

dT

(2-49)

51

This expression is very accurate for single-component fluids and is a good approximation for multicomponent mixtures in which composition change is minimal [19] The mass flux or mass flow rate per unit area can be calculated as follows:

(2-50)

where iJHv must be expressed in J ki 1 and cp in J ki 1 K- 1• The Leung method is quick and simple and, thus, widely used for approximate calculations. A classic example proposed by Huff [20] and also applied by Leung [ 19] is used to illustrate the application of this method. Example 2-8 Estimate the discharge rate and the vent area for a tank of styrene monomer undergoing adiabatic polymerization after being heated to 70 °C. The maximum allowable working pressure of the tank is 5 bar. System parameters: vessel volume= 13.2 m 3; W = 9,500 kg; Ps = 1 4.5 bar; Ts = 482.5 K; (dT/dt)set = 0.493 K s- (sealed system); Pmax = 5.4 bar; Tm = 492.7 K; 1 (dT/dt)max = 0.662 K s- . Material properties: Ps =4.5 bar 0.001388 0.08553 2.470

Specific volume, liquid, m 3 kg- 1 Specific volume, gas, m 3 kg- 1 Cp,, kJ kg- 1 K- 1 iJHV, kJ kg-]

Pm = 5.4 bar 0.001414 0.07278 2.514 302.3

310.6

Solution From Eq. (2-48), and assuming cv "'cp for an incompressible fluid:

q = _!_ 2.47 [0.662 +0.493 ]= 1.426 kJ kg- 1 s- 1 2 The mass discharge rate and the mass flux can be calculated from Eqs. (2-47) and (2-50): 9,500·1.426 132

310 6 · 9,500 (0.08553-0.001388)

+~2.47·(492.7-482.5)]

2

= 255 kg s_ 1

0.5

G=0.9

310,600 1 (0.08553-0.001388) ( 2,470·482.5 )

= 3040k m-2s-I g

And the corresponding vent area is: 255 2 = 0.084 m A= 3040

52

7 EVAPORATION OF A LIQUID FROM A POOL 7.1 Evaporation of liquids On land or on water, a spilled liquid forms a pool. On water, the pool is unconfined, but on land its size and shape are often established by the existence of a retention dike. Immediately after the spill, the liquid starts to receive heat from the surroundings (the ground, the atmosphere, solar thermal radiation; see Fig. 2-13) and is vaporized. Initially, the vaporization process is usually controlled by heat transfer from the ground, especially in the case of boiling liquids. These two cases, boiling and non-boiling liquid pools, should be considered separately.

(/

Convection from air

/

~"""'""

/

\

from the sun

Heat flow from ground

Fig. 2-13. Evaporation of a liquid from a pool. 7.2 Pool size 7.2.1 Pool on ground If the spill creates a pool on the ground and there is (as is often the case in industrial facilities) a containment barrier, the pool diameter is fixed. If the dike is right-angled, the equivalent diameter must be used: D=

4

surface area of the pool

(2-51)

J(

If there is no dike, the shape and size of the pool must be determined by the features of the release (continuous or instantaneous) and of the ground (slope). An equilibrium diameter can be reached as a function of the vaporization rate (see [9] for more detailed information). 7.2.2 Pool on water For a hydrocarbon spill on water (usually seawater), the expressions applied to pool fires (Chapter 3) (prior to ignition) can be applied. 7.3 Evaporation of boiling liquids When there is a spill of, for example, a pressurized liquefied gas, the liquid will be at its boiling temperature and at a temperature lower than that of the surroundings. Therefore, heat

53

will be transferred from the atmosphere by solar thermal radiation and, mostly in the first stage, from the ground. By analysing the heat conduction from the ground to the liquid, the following expression can be obtained [2, 21]:

(2-52)

where

is the heat flux reaching the pool from the ground (W m- 2) k, is the thermal conductivity of the soil (see Table 2-4) (W m- 1 K" 1) Ts is the temperature of the soil (K) Tpooi is the temperature of the liquid pool (K) a, is the thermal diffusivity of the soil (m2 s- 1) (see Table 2-4), and tis the time after the spill (s).

Qpool

Table 2-4 Values of thermal conductivity and thermal diffusivity of various solid media [18, 19] 2 .] Substance as, m s 4.3. 10-7 Average ground 0.9 2.0. 10-7 Sandy ground (dry) 0.3 3.3. 10-7 Sandy ground (8% water) 0.6 4.5. 10-7 Wood 0.2 11 . 10-7 Gravel 2.5 127. 10-7 Carbon steel 45.0 10. 10-7 Concrete 1.1 If the entire heat flux is devoted to evaporating liquid, then the mass boiling rate is: Qpool Apool

mpool

(2-53)

/IJ{ v

is the mass boiling rate (kg s- 1) A is the area of the pool (m2 ), and &fv is the latent heat of vaporization (expressed in J kg·\ From Eq. (2-52) it is clear that, as time passes and the ground becomes cooler, the heat flux to the pool decreases. After a certain time, the atmospheric and solar heat fluxes become more important and control the process [3].

where

mpool

7.4 Evaporation of non-boiling liquids If the spilled liquid has a boiling temperature higher than the ambient temperature and is stored at a temperature lower than its boiling point (often at or near the ambient temperature), then the pool will not boil. Evaporation will occur essentially by vapour diffusion, the driving force being the difference between the vapour pressure of the liquid and the partial pressure of the liquid in the atmosphere (Pv - Pamb). The mass transfer process will be significantly influenced by the movement of air above the pool. The evaporation rate can be estimated with the following expression [22]:

54

G pool

= 2 . 10 -3 uo.1s r-o.11 Mv Po ln( 1+ Pv -Pamb) w

RT

P-P 0

(2-54)

v

is the evaporation rate (kg m· 2 s- 1) 1 Uw is the wind speed (m s- ) r is the radius of the circular pool (m) Mv is the molecular weight of the liquid (kg kmor 1) 1 1 R is the ideal gas constant (expressed in J kmor K ) Tis the temperature (K) Pv is the vapour pressure of the liquid at its temperature (Pa) Pamb is the partial pressure of the liquid in the atmosphere (Pa), and Po is the atmospheric pressure (Pa). 4 For Pv < 2·10 Pa, Eq. (2-54) may be simplified to:

where

G pool

Gpool

=2·10-3uo7s r-0.11 Mv (P -P w

RT

v

amb

)

(2-55)

If the pool is not circular but rectangular, r must be substituted in Eqs. (2-54) and (2-55) by L, the length of the side of the pool parallel to the wind (m). Example 2-9 A broken pipe creates a pool of n-hexane with a diameter of 22 m. The ambient temperature is and the wind speed is 3 m s- 1. Calculate the evaporation rate. 20 Data: Mv = 86; Pvn-hexane, 20 'C = 121 mrn Hg.

oc

Solution The boiling temperature of n-hexane at atmospheric pressure is 68.7 °C. Therefore, the pool will not boil. The partial pressure of n-hexane in the atmosphere is, of course, negligible. By applying Eq. (2-54):

=2·10-3 ·3o7s ·11-011

G pool

86·(1.0132·105) ln(1+ 16,130-0 l=0.00224k m-2 s-1 8.314·10 3 (273+20) 1.0132·10 5 -16,130) g

And, for the whole pool: m = 0.00224 kg m- 2 s- 1 ·380m2 = 0.851 kg s- 1

8 GENERAL OUTFLOW GUIDELINES FOR QUANTITATIVE RISK ANALYSIS To predict the loss of containment for a given case and a given accidental scenario, the specific conditions corresponding to that situation and equipment must be taken into account and the appropriate hypothesis concerning the outflow -a hole, a broken pipe, etc.- must be considered. However, if a generic quantitative risk analysis must be performed over a given

55

installation, a set of hypotheses concerning the various loss-of-containment events is usually applied systematically in order to cover, with a conservative approach, the most common events that can significantly influence risk estimation. The "Purple Book" [9] offers a very interesting compilation of general guidelines covering such hypotheses. The following paragraphs contain a summary with the essential aspects of these guidelines. In quantitative risk analysis, the maximum duration of a release is usually 30 min. Usually, loss-of-containment events are included only if their frequency of occurrence IS equal to or greater than I o-8 per year and lethal damage occurs outside the establishment.

8.1 Loss-of-containment events in pressurized tanks and vessels The following loss-of-containment events (LOCs) are usually considered m pressure, process and reactor vessels: Instantaneous release of the complete inventory. Continuous release of the complete inventory in I 0 min at a constant rate of release. Continuous release from a hole with an effective diameter of I 0 mm. If the discharge is from the liquid section of the vessel, pure liquid is released. Flashing in the hole is not modelled (flashing takes place outside the vessel). 8.2 Loss-of-containment events in atmospheric tanks In atmospheric storage tanks, pressure is atmospheric or slightly higher. The following LOCs are usually considered: Instantaneous release of the complete inventory: Directly into the atmosphere. From the tank into an unimpaired secondary container. Continuous release of the complete inventory in I 0 min at a constant release rate: Directly into the atmosphere. From the tank into an unimpaired secondary container. Continuous release from a hole with a diameter of I 0 mm: Directly into the atmosphere. From the tank into an unimpaired secondary container. 8.3 Loss-of-containment events in pipes Full-bore rupture (outflow from both sides). A leak with a diameter of IO% of the nominal diameter (with a maximum of 50 mm). For a full-bore rupture in a pipe, CD= 1.0. In other cases, CD= 0.62. Assume that the pipe has no bends and a wall roughness of approximately 45 f.tm. 8.4 Loss-of-containment events in pumps Full-bore rupture of the largest connecting pipeline. A leak with a diameter of 10% of the nominal diameter of the largest connecting pipe (with a maximum of 50 mm). If no pump specifications are available, assume a release rate of I.5 times the nominal pumping rate. 8.5 Loss-of-containment events in relief devices Discharge at the maximum discharge rate.

56

8.6 Loss-of-containment events for storage in warehouses Solids: dispersion of a fraction of the packaging unit inventory as respirable powder. Solids: spill of the complete packaging unit inventory. Emission of unburned toxics and toxics produced in the fire. 8. 7 Loss-of-containment events in transport units in an establishment LOCs for road tankers and tank wagons in an establishment: Instantaneous release of the complete inventory. Continuous release from a hole (with the size of the largest connection). If the tank is partly filled with liquid, a liquid phase release from the largest liquid connection is assumed. Full-bore rupture of the loading/unloading hose (outflow from both sides). Leak in the loading/unloading hose (effective diameter of 10% of the nominal diameter, with a maximum of 50 mm). Full-bore rupture of the loading/unloading arm (outflow from both sides). Leak in the loading/unloading arm (effective diameter of 10% of the nominal diameter, with a maximum of 50 mm). Fire under the tank (release from the connections under the tank followed by ignition or fire in the surroundings of the tank): instantaneous release of the complete inventory of the tank. Ships in an establishment: Full-bore rupture of the loading/unloading arm (outflow from both sides). Leak in the loading/unloading arm (effective diameter of 10% of the nominal diameter, with a maximum of 50 mm). External impact, large spill: o Gas tanker: continuous release of 180 m3 in 1800 s. o Semi-gas tanker (refrigerated): continuous release of 126m3 in 1800 s. o Single-walled liquid tanker: continuous release of75 m3 in 1800 s. o Double-walled liquid tanker: continuous release of75 m3 in 1800 s. External impact, small spill: o Gas tanker: continuous release of90 m3 in 1800 s. o Semi-gas tanker (refrigerated): continuous release of 32m3 in 1800 s. o Single-walled liquid tanker: continuous release of30 m3 in 1800 s. o Double-walled liquid tanker: continuous release of20 m3 in 1800 s. 8.8 Pool evaporation If the liquid spill is contained in a bund, the spreading of the liquid will be limited: the maximum pool size will be the size of the bund and the pool size and shape will be defined. The effective pool radius can then be calculated from the bund area as follows: r

pool

==~Abund

(2-56)

ff

If there is no bund, the released liquid will spread onto the soil or the water surface; the pool diameter will increase up to the moment in which the evaporation rate from the pool equals the release rate; at this time, the pool will reach its maximum diameter. The thickness of the pool will be a function of the surface roughness. A minimum value of 5 mm can be assumed.

57

As for the maximum surface for unconfined pools, the following values have been proposed: I ,500 m 2 (D = 44 m) for pools on land and I 0,000 m 2 (D = 113 m) for pools on water.

8.9 Outflow and atmospheric dispersion If the release mass flow rate changes with time, this variation should be taken into account in applying the atmospheric dispersion model. Therefore, the release must be divided into several discrete time segments, with constant outflow over each segment. A division into five segments is suitable for most cases. The following general rules can be applied [6]: - Calculate the mass released in the first 30 min (Mref). - Calculate the mass released in each time segment, Msegment = Mrellnumber of segments. - Calculate the duration of the first time segment, tsegment = time required to release Msegment· - Calculate the release rate in the first time segment = Msegmen/tsegment· - Apply the same procedure to the other time segments. As for the cloud dispersion, each time segment is treated as an independent steady-state release neglecting the dispersion downwind until the total release can be considered instantaneous. NOMENCLATURE Abund Aexp

Aor Ap Asv

At

CD Cp CP!

Cv

dp dar F

f FJ

/F G2p

Gsub

hi &fv

L Lj

Ma Mv m msv

N

bund area (m 2 ) vessel surface exposed to external fire (m2) cross-sectional area of the orifice (m2) cross-sectional area of the pipe (m2) minimum cross-sectional flow area of a safety valve (mm2) cross-sectional area of the tank (m2) discharge coefficient (-) specific heat at constant pressure (kJ kg- 1 K- 1) liquid specific heat at constant pressure (kJ kg- 1 K 1) specific heat at constant volume (kJ kg- 1 K- 1) pipe diameter (m) diameter of the orifice (m) reduction factor (-) vaporization fraction (-) friction loss term (J kg- 1) Fanning friction factor (-) discharge mass flow rate (kg m- 2 s- 1) term accounting for subcooling (kg m- 2 s- 1) height ofliquid above leak (m) 1 latent heat of vaporization (kJ ki ) equivalent length of a pipe plus pipe fittings (m) length of a free jet of gas (m) Mach number(= ulus)(-) molecular weight (kg kmole- 1) mass flow rate (kg s- 1) mass flow rate discharged from a safety valve (kg h- 1) dimensionless parameter (-)

58

P Pback Pchoked Pcont P design

Pmax Po Pp Pout Ps

Pv LJP

Q R

Re rpoot

T Tchoked Tcont

Tj Tp

u Uj Us

Uw Vtv

W

wu w1 Wv

Z

aw s

s

r p Pt

'¥ If/

pressure (Pa) safety valve absolute backpressure (bar) choked or critical pressure (Pa) pressure inside the vessel or the pipe (Pa) vessel design pressure (Pa) maximum pressure reached during venting (Pa) atmospheric pressure (Pa) pressure inside the pipe at a given distance from the source (Pa) pressure downstream from the hole (Pa) set pressure (Pa) vapour pressure at the storage temperature (Pa) pressure drop (Pa) heat flux (W) ideal gas constant (8.314 kJ kmole· 1 K" 1) Reynolds number (-) effective pool radius (m) temperature (K) temperature of the gas at choked conditions (K) temperature in the vessel or the pipe (K) temperature of the gas in the jet (K) temperature of the gas in the pipe, at a given distance from the source (K) velocity ofliquid in the pipe (m s· 1) velocity of the jet at the gas outlet (m s· 1) speed of sound in a gas (m s· 1) wind speed (m s· 1) change in specific volume, liquid to vapour (m3 kg- 1) initial mass of gas in the vessel (kg) initial mass ofliquid (gk) mass ofliquid immediately after flash (kg) mass of vapour immediately after the flash (kg) gas compressibility factor (-) safety valve outflow coefficient (-) pipe roughness (m) dimensionless factor (-) ratio of heat capacities, cp/cv (-) density of the gas (kg m· 3 ) density of the liquid (kg m· 3 ) safety valve outflow function (-) dimensionless factor (-)

REFERENCES [I] 0. Levenspiel, Engineering Flow and Heat Exchange. Plenum Press, New York, 1984. [2] D. A. Crowl, J. F. Louvar. Chemical Process Safety. Fundamentals with Applications, 2nd ed. Prentice Hall PTR, Upper Saddle River, 2002.

[3] Committee for the Prevention of Disasters. Methods for the Calculation of Physical Effects (the "Yellow Book"), 3rd ed. The Hague, SDU, 1997.

59

D. A. Crowl. J. Loss Prev. Process Ind. 5 (1992) 73. T. C. Foster. Chern. Eng. May 4 (1981) 105. H. Montiel, J. A. Vilchez, J. Casal, J. Arnaldos. J. Hazard. Mater., 59 (1998) 211. M. Chase. Nist-Janaf Thermochemical Tables. 41h edition. Journal of Physical and Chemical Reference Data. Monograph No.9, 1998. [8] R. C. Reid, J. M. Prausnitz, B. E. Poling. The Properties of Gases & Liquids. McGrawHill, New York, 1987. [9] Committee for the Prevention of Disasters. Guidelines for Quantitative Risk Analysis (the "Purple Book"). The Hague, SDU, 1999. [10] M. Andreassen, B. Bakken, U. Danielsen, H. Haanes, K. D. Oldshausen, G. Solum, J.P. Stensass, H. Thon, R. Wighus. Handbook for Fire Calculations and Fire Risk Assessment in the Process Industry. Scandpower A/S. Sintef-Nbl, Lillestrom, 1992. [II] H. G. Fisher, H. S. Forrest, S. S. Grosse!, J. E. Huff, A. R. Muller, J. A. Noronha, D. A. Shaw, B. J. Tilley. Emergency Relief System Design Using DIERS Technology. DIERSAIChE, New York, 1992. [12] H. K. Fauske. Plant/Operations Progr. 4 (1985) 132. [13] CCPS. Guidelines for Chemical Process Quantitative Risk Analysis. AIChE, New York, 2000. [14] H. K. Fauske, M. Epstein. Source Term Considerations in Connection with Chemical Accidents and Vapor Cloud Modeling. Proc. of the International Conference on Vapor Cloud Modeling. Cambridge, MA. AIChE, New York, 1987. [15] Spirax Sarco. Steam Engineering Tutorials. Learning Modules, Module 9.1, Introduction to Safety Valves. URL: http://www.spiraxsarco.com/learn/, consulted February 2007. [16] NFPA. Flammable and Combustible Liquids Code. NFPA 30. National Fire Protection Association. Quincy, MA. 1987. [17] NFPA. Standard for the Storage and Use of Liquefied Petroleum Gases. NFPA 58. National Fire Protection Association. Quincy, MA. 1987. [18] J. Bond. Loss Prev. Bulletin, no. 65 (1985) 20. [19] J. C. Leung. AIChE Journal, 32 (1986) 1622. [20] J. E. Huff. Plant/Operations Prog., 1 ( 1982) 211. [21] J. M. Santamaria, P. A. Braiia. Amilisis y reducci6n de riesgos en la industria quimica. Editorial MAPFRE. Madrid, 1994. [22] Committee for the Prevention of Disasters. Methods for the Calculation of Physical Effects (the "Yellow Book"), 1st ed. The Hague, SDU, 1979. [4] [5] [6] [7]

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