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Feyzin Oil Refinery Disaster

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1 Feyzin Oil Refinery Disaster
Feyzin Oil Refinery (near Lyons), France 4th January   A large storage tank at an oil refinery holding liquefied propane exploded and killed a number of fire fighters. The incident was an important lesson for the hydrocarbon industries Thanks to Ann-Marie McSweeney, John Barrett & Jacinta Sheehan Ware Department of Process Engineering, UCC

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A fire developed in a tank farm at an oil refinery. No person was in apparent danger. The fire service was called out  but as the fire had already taken hold they decided to simply monitor it until it safely burnt itself out. However the firemen appeared unaware of the significance of the enormous radiant heat flux from the fire that was impinging on adjacent pressurized storage tanks and spheres. This was raising their temperature and the temperature of the products within them. More importantly it was weakening the integrity (tensile strength) of the wall material (material tensile strength falls with higher temperature). When the membrane stress in the storage vessels due to the raised internal pressure exceeded the reduced tensile strength of the wall material, the vessels burst open. The contents of the vessels then ignited in a fireball and killed the fire fighters.

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A very good description of the incident is given in the book SAFETY & LOSS PREVENTION AUTHOR: FRANK LEES (In the UCC Library under Classification ) The course notes for PE 3005 should also be consulted especially the material dealing with the temperature dependence of material strength. Similarly the notes of PE 2003 dealing with pressure vessel analysis.

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Refinery Storage Vessels – Pressurized Spheres

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PRODUCT DESCRIPTION The material in the storage tanks was Propane C3H8 Third member in the saturated hydrocarbon group known as the alkanes. Commonly used as a fuel.

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Thermodynamic Properties Boiling Point is – 42 °C at Patm. Colourless Gas at Room Temperature and Atmospheric Pressure Molecular Weight M = 44 Gas Constant R = 189 J/kgK Calorific Value = 47 MJ/kg Flammability Limits (in air) 2.5 % to 9.5 %

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Thermodynamic Properties Liquid Density = 588 kg/m3 (at 1bar) Vapour Density = kg/m3 (at 1bar) It can be seen that the liquid is lighter than water and the vapour is heavier than air. Specific Heat Cp= 1679 J/kgK Ratio of Specific Heats  = Cp/Cv = 1.126 Latent heat of evaporation  = 428 kJ/kg

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Vapour Pressure Curve (actually that of Propene) From this chart, knowing the temperature of the propane, its vapour pressure (i.e. tank internal pressure) can be found.

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CONTAINMENT DESCRIPTION The vessel in question was a large outdoor spherical vessel resting on vertical legs designed for the bulk storage of liquefied propane. The vessel was amongst other similar storage vessels. There was a pressure relief valve (safety valve) at the top on a pipeline leading to a flare. In emergencies this valve would open and the escaping vapour flared off. No information on the diameter of the safety valve orifice (subsequently assume it is 100 mm). Vessel Geometry Vessel Diameter D = 14 m Vessel Radius, R = 7 m Total Volume Vtotal = 4/3πR3 = m3 Ullage (i.e. free space) set at 20 % Working Volume Vworking = 0.8 x = 1150 m3

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Illustration of storage sphere showing vessel supports and pressure relief system Safety Valve Flare Support Legs

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Vessel Geometry Material of construction is structural steel with a density s = 7800 kg/m3 Wall thickness, t = 45 mm ·        Mass of tank wall = 4πR2 t s = 216 tonnes ·        Surface area of tank = 4πR2 = 616 m2 ·        Projected area of tank = πR2 = 154 m2

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Vessel Pressure Stress Analysis Tensile Strength (maximum strength) of structural steel TS = 620 MN/m2 Rupture Pressure can be estimated from knowledge of the membrane stress in a spherical vessel PR = 80 bar Under normal conditions, the vessel would not be expected to rupture until the internal (propane) pressure reached 80 bar. Vessel was un-insulated Propane temperature = Ambient outside temperature.

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Vessel Pressure versus Ambient Temperature Storage pressure (i.e. propane vapour pressure) varies with ambient (i.e. propane) temperature.  For this location: °C < TAMB < 40 °C Hence can tabulate the normal pressures that might exist within the gas storage spheres.

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Vessel Pressure Analysis Thus normal tank internal pressure is well below the tank failure pressure. To prevent internal pressure for whatever reason rising to and reaching the rupture pressure, the safety valve was set to lift (i.e. open) at 20 bar; corresponding to a propane temperature of about 60 °C. Thus the maximum membrane stress that could be developed in the tank wall would be when internal pressure was 20 bar.

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Reduction in Steel Tensile Strength with Temperature The mechanical strength of metals depends on their temperature; as the temperature rises, the strength falls off. Unless otherwise stated, any quoted mechanical strength value is the value that exists at ambient temperatures. sTS MN/m2 100 T °C 200 700 600 500 400 300

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Significance of the Previous Chart At ‘cold’ i.e. ambient temperatures, the vessel can contain internal pressures of up to 80 bar because the tensile strength TS = 620 MN/m2 (note it could even be higher!). However, if the vessel wall temperature rises to 700 °C, in which case the steel tensile strength, TS falls to 150 MN/m2, then the vessel will rupture even with the safety valve open. PSET = 20 bar => σ = 155 MN/m2 σ > σTS RUPTURE!

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INCIDENT DESCRIPTION What follows is a simplified account of what actually happened. At the date in question, there were 400 tonnes of propane in the tank. Volume occupied V = 680 m3 Given the working volume VW = 1150 m3 Tank was approximately 60 % full.

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INCIDENT DESCRIPTION An adjacent hydrocarbon storage tank at the depot caught fire and burnt fiercely (the actual incident was a good deal more involved). => Propane storage sphere exposed to intense radiant heat. => Temperature of propane (and steel wall) will rise. => Vapour pressure of propane rises. When the vapour pressure reaches relief valve pressure setting, PSET of 20 bar, the safety valve lifted and propane vapour was expelled from the vessel and sent to the flare.

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INCIDENT DESCRIPTION Assuming the pressure inside the vessel henceforth remains at 20 bar We have a kind of controlled equilibrium; boiling off propane vapour at a constant pressure of 20 bar. The fire fighting strategy was to let the vessel empty itself over time much like a kettle boiling itself dry; at the end all that would be left would be a burnt out empty vessel. The fire fighters that had been called to the scene remained in proximity to the fire.

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INCIDENT DESCRIPTION What the firemen didn’t know! The lower part of the tank wall in contact with the boiling liquid propane will remain at something close to the liquid propane temperature (60 °C at 20 bar) due to the very high heat transfer coefficient (say 10,000 W/m2K) between a boiling liquid and metal wall. However the upper part of the tank wall in contact with the vapour receives no such cooling (H.T.C. of 100 W/m2K) and it will rise towards the of the radiant flame temperature. This would be an upper theoretical limit of about 1300 °C.

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INCIDENT DESCRIPTION  A race is on! If the wall of the vessel reaches 700 °C before the vessel has emptied itself, the wall will rupture and the remaining propane will go up in a fireball! Two times must be calculated 1] Time for upper surface of tank wall to reach 700 ºC due to radiant heat transfer from adjacent fire. 2] How much propane will have left the sphere through the open safety valve in this time.

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Radiant Heat Transfer Calculation Early morning in January, so initial temperature of Propane tank Ti  0 °C How long will it take for upper tank wall to reach 700 °C? Do a very crude energy balance! Radiant heat flux: QR = ..A.(TFlame4 – TWall4)  Take TFlame = 1300 °C = 1600 K  TWall = ½(0 °C °C) = 350°C = 620 K A is the projected area of the sphere = 154 m2 Stefan-Boltzman Constant  = 5.67 x 10-8 W/m2K Emissivity  (really a fudge factor), Take  = 0.5 QR = x (16004 – 6204) QR = 28 MW i.e. 28 MJ/s

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Temperature Rise in System Calculate the amount of thermal energy needed to produce the corresponding temperature rise of the system so that the upper wall reaches 700 C. Note the total heat in has sensible heat transfer and latent heat transfer components: 1.   Bring 400 tonnes of propane from 0 °C up to 60 °C. 2.   Bring approximately half of tank wall (108 tonnes of steel) from 0 °C up to 60 °C. 3.    Bring other half of tank wall from 0 °C up to 700 °C. 4.    Evaporate off some portion (say half) of the propane. Note the specific heat capacity of steel cp = 450 J/kgK

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Temperature Rise in System Q = 400 x (60 – 0) (1) + 108 x (60 – 0) (2) + 108 x (700 – 0) (3) + 200 x x (4) Q = GJ = 163 GJ Dividing the total heat requirement by the heat flux to obtain time t = 5821 s = 97 minutes So, very roughly we might expect that one hour and a half after the outbreak of the initial fire, the tank wall temperature will reach 700 °C.

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Rate of Vessel Emptying How much vapour has been expelled through the safety valve after an hour and a half (and assuming the safety valve lifts soon after the fire starts)? Require the mass flux through the safety valve; model the process as isentropic expansion of an ideal gas across a nozzle with choked flow at outlet. J Mass flow rate kg/s P Vessel pressure bar A Valve outflow area m2 T Propane vapour temperature (absolute) K

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Rate of Vessel Emptying P = 20 bar, T = 60 °C = 333 K 0.008 m2  = 1.126, R = 189 J/kgK (Propane) J = 38.3 kg/s Total outflow of propane in 5821 s J = 38.3 x 5821 = 223 tonnes Thus after an hour and a half, the amount of propane remaining in the tank is 400 – 223 = 177 tonnes.

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Accident Progression Plotting tank wall temperature and tank internal pressure versus time

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Accident Progression Plotting mass of propane in vessel, wall membrane stress and wall tensile strength versus time.

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CONSEQUENCES An hour and a half or so after the commencement of the fire: ·        Wall temperature reached 700 °C ·        Membrane stress was at 155 MN/m2 ·        Steel tensile strength had fallen to 150 MN/m2 ·        Hence the Storage Sphere ruptured ·        177 tonnes of propane exploded into a fireball ·        17 people killed

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FIRE BALL (BLEVE) CALCULATIONS Model the instantaneous combustion of the escaped vapour. Duration of burning of fire ball is The radiative power of the fire can be calculated from QR Radiative power W HC Calorific Value J/kg td Duration of fire ball s M Mass of fuel in fire ball kg

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FIRE BALL (BLEVE) CALCULATIONS A point source model of the fire gives the radiative heat flux as  Radiative flux W/m2 QR Radiative power of flame W r Distance from source m In turn the thermal radiation dosage can be calculated as L Thermal radiation dosage (kW/m2)1.33s  Intensity of radiation (radiation flux) kW/m2 t Duration of exposure s

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FIRE BALL (BLEVE) CALCULATIONS Note the duration of exposure is equal to the duration of the fire ball. Damage to people exposed to the fire can be quantified with Hence can estimate how close people must have been to the fire to have been killed or injured.

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POSSIBLE ACCIDENT PREVENTION STRATEGIES 1. Install a larger pressure relief valve so if the tank is exposed to fire, its contents can be flared more rapidly. 2. Space the tanks further apart so that an outbreak of fire in one does not impose excessive heat radiation on adjacent units. 3. Place thermal insulation on tank. 4. Spray water over the top of tank to keep it cool and hence maintain its mechanical strength. 5. Choose a different material of construction for the tank; specifically a heat resistant steel that can maintain substantial mechanical strength even at elevated temperatures.

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