Pressure Rise In Vessel Due To Blocked Discharge (Fire Case)

Hello,

I am attempting to do a hand calculation of the following scenario.

Scenario:

Fire impinging on an unwetted vessel and in this case the PSV and other outlets are blocked. I would like to calculate the pressure rise of the fluid in the vessel. Assuming ideal gas law, I will first need the temperature rise of the fluid in the vessel during the fire duration.

So far, I have the flame properties from API 521. With this information, I am able to get the net heat gain by the vessel; i.e.

Q total = Q convection + Q radiation

And so the temperature vessel temperature rise can be computed with: ΔT = [Δt(QR + QC)]/CSρSWS

Having this new vessel surface temperature, how do I proceed to calculate the heat transfer to the gas in the vessel to get the new temperature?

I've tried Q = hc (T surface - T in)

But my Tin seems to be very high which does not make sense.

Thermal diffusivity of the base material plays a major role when fire impinges on metal. Consider the affect of a hand held cutting torch that can melt a hole in a metal vessel wall by rapidly increasing the temperature in a local area above the base material melting point. How is your scenario any different? At some point the temperature of the gas inside becomes irrelevant when the vessel no longer retains pressure due to a new hole. Flame temperature is very high but metal temperature rise will be localized during impingement. The entire mass of the vessel is not affected and the vessel wall temperature is not uniform.

Fire case + PSV blocked + other outlets blocked.

So, at least 4 simultaneous, independent failures, or 2 if blockage has a common cause failure mechanism (like unmonitored vapor phase polymerization).  Which is it?

Either way, it would be a very rare event in a properly run PHA.

Alright. Re-imagining a scenario with a standalone vessel with no PSV exposed to a pool fire engulfing the vessel. Is it possible to calculate the temperature rise of the air/gas present in the vessel?

a standalone vessel with no PSV exposed to a pool fire engulfing the vessel. Is it possible to calculate the temperature rise of the air/gas present in the vessel?

Sure.  Using the vessel volume and initial T and P calculate the mass of air/gas.  n = PV/RT.  Calculate the fire heat input to the vessel.  Consult API 520/521.  Calculate T vs time.  Q = MCpDT.

Since air/gas does not "temper" the temperature rise of a vessel, it will over-temperature and fail very quickly.  A PSV is not much protection for a vessel with air/gas in a fire.  Consider other protection measures, like fire resistant insulation and/or a fixed fire water spray system (NFPA 15).

 

a standalone vessel with no PSV exposed to a pool fire engulfing the vessel. Is it possible to calculate the temperature rise of the air/gas present in the vessel?

 

Sure.  Using the vessel volume and initial T and P calculate the mass of air/gas.  n = PV/RT.  Calculate the fire heat input to the vessel.  Consult API 520/521.  Calculate T vs time.  Q = MCpDT.

 

Since air/gas does not "temper" the temperature rise of a vessel, it will over-temperature and fail very quickly.  A PSV is not much protection for a vessel with air/gas in a fire.  Consider other protection measures, like fire resistant insulation and/or a fixed fire water spray system (NFPA 15).

 

That's very helpful! Thank you!

 

a standalone vessel with no PSV exposed to a pool fire engulfing the vessel. Is it possible to calculate the temperature rise of the air/gas present in the vessel?

 

Sure.  Using the vessel volume and initial T and P calculate the mass of air/gas.  n = PV/RT.  Calculate the fire heat input to the vessel.  Consult API 520/521.  Calculate T vs time.  Q = MCpDT.

 

Since air/gas does not "temper" the temperature rise of a vessel, it will over-temperature and fail very quickly.  A PSV is not much protection for a vessel with air/gas in a fire.  Consider other protection measures, like fire resistant insulation and/or a fixed fire water spray system (NFPA 15).

 

Hi Latexman.

Sorry to bother but I just want to re-confirm my calculations here.

Step 1: Getting Total Heat Input from Fire to Content in Vessel

Radiative Heat from Jet Fire @ T1= 187.71 kW/m2

Vessel Surface Area = 34.4 m2

Total Radiative Heat Transfer to the Gas in the Vessel = 6462.35 kW

Step 2: Getting Temperature Rise of Gas In Vessel

Gaseous Mass in Vessel = 1647 kg hydrocarbon

Heat Capacity of Gas = 2.712 kJ/kg.K

Applying: Q = m x Cp x Del T

Delta T = 1.45 oC

New Temperature of Gas after T1 = 20 oC + 1.45 oC = 21.45 oC

Step 3: Applying P1/T1 = P2/T2 to get New Pressure

P1 = 85 bar

T1 = 20 oC

T2 = 21.45 oC

Hence, P2 = 85.42 barg

I iterated this calculations for two more time steps (T2 and T3). It just seems baffling that a vessel being exposed to jet fire, the gas temperature in the vessel only rise by 3 oC after 3 minutes of exposure.

Edited by kc45, 20 June 2022 - 09:03 PM.

Hi,

I hope you are using Kelvin !

Breizh

Hi,

I hope you are using Kelvin !

Breizh

Hi Breizh,

Yes, just cross checked. I'm using Kelvin

And, absolute pressure!

I see nothing wrong with your logic and math.

Do you really have a jet fire big enough to impinge on the entire 34.4 m² of the vessel?  That's at least (4*34.4/3.14)^1/2 = 6.6 meter diameter jet.  That's huge!  The nozzle diameter on one RL 10 engine in the propulsion system of a Delta IV rocket is 2.1 meters.  You should build a fire wall between the pressurized source and the vessel.

The dynamics of the situation will follow Pilesar's post.  Metal temperature will rise fast, and average gas temperature will rise slower.  The metal will melt and a hole will form pretty quickly.

And, absolute pressure!

 

I see nothing wrong with your logic and math.

 

Do you really have a jet fire big enough to impinge on the entire 34.4 m² of the vessel?  That's at least (4*34.4/3.14)^1/2 = 6.6 meter diameter jet.  That's huge!  The nozzle diameter on one RL 10 engine in the propulsion system of a Delta IV rocket is 2.1 meters.  You should build a fire wall between the pressurized source and the vessel.

 

The dynamics of the situation will follow Pilesar's post.  Metal temperature will rise fast, and average gas temperature will rise slower.  The metal will melt and a hole will form pretty quickly.

Just assuming worst case scenario where I took the entire area of the vessel.

If I were to take the impinged area, I suppose the total temperature rise and pressure rise will be even lesser!

Anyway, thank you for looking through my calculation process.