Fire case - Relieving temperature 1

[sheiko]

Hello,

Could you please confirm that the below formula for calculating relieving temperature is always correct for fire case: T1=T0*(P0/P1)with T in K?

eg:
T0=92°C
P0=1.79 bara
P1=5.41 bara
Then, T1=278°C

I ask this because we receive from subcontractor a value of 155°C for this example...

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[sheiko]

then T1=831°C sorry

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[sheiko]

And given that:
P0,T0: normal operating conditions
P1, T1: Relieving conditions

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[MortenA]

The formula that you state is a simplified version on the perfect gas law - but i think you switched P1/P0? Or else T1 will be lower than T0:

P1*V=nRT1
and
P0*V=nRT0

<=>

P1/P0=T1/T0

<=>

P1*T0/P0=T1

So the validity of said formula will depend on you P/T and gas.

At the stated pressure/temperature its propably not too far off.

Best regards Morten

 

[sheiko]

Indeed i switched P1/P0

Thanks

"We don't believe things because they are true, things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[sheiko]

Subcontractor answere: "Relieving pressure is 5.41 bara. I assume the water will evaporata --> T sat for water @ 5.41 bara = 154.9°C" Is it valid?


"We don't believe things because they are true, things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[MortenA]

Well - the formula YOU mentioned is for a dry vessel.

If the vessel is filled with water then the temperature will increase to the boiling point at the RP+21% (assuming API sizing). If this pressure is 5.41 bara then my steam table says 148 deg C - but thats close enough

Best regards

Morten

 

[JoeWong88]

sheiko,

When we are dealing with fire contingency, first thing is to establish the scenario & condition of the fluid in vessel :

i) completely fill with liquid ==> Hydraulic expansion
The temperature would be properly close to initial temperature

ii) partially fill with liquid ==> Boiling liquid
The temperature would be at the boiling temperature

iii) completely fill with vapor, gas or supercritical fluid ==> gas expansion
The temperature would be very high. The temperature rise would follow the path between isothermal and adiabatic. Nevertheless, API has considered isothermal...

You may see that the fluid temperature for each case would varies significantly.

After scenario is properly defined, then only think of the way to estimate the temperature.

 

[sheiko]

in the case iii) completely filled with gas/vapor/supercritical fluid, how can the temperature rise under ISOTHERMAL (constant temperature => Pressure/Density=Constant) condition? Please explain

"We don't believe things because they are true, things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[JoeWong88]

Sheiko,

I have mixed gas expansion in vessel and valve. My apologize.

The last statement
"The temperature would be very high. The temperature rise would follow the path between isothermal and adiabatic. Nevertheless, API has considered isothermal..."

should be reworded as
" The temperature would be very high. The temperature rise would follow the isochoric process (isovolumetric). "



JoeWong
Chemical & Process Technology

 

[sheiko]

In the case i) completely fill with liquid you sait that the temperature would be properly close to initial temperature. Why? Please explain because i would rather say that the temperature is, as for case ii), the boiling temperature...

"We don't believe things because they are true, things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[JoeWong88]

Liquid is incompressible...A small temperature rise lead to liquid expansion and large pressure rise...

JoeWong
Chemical & Process Technology

 

[sheiko]

Thanks

"We don't believe things because they are true, things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[sheiko]

Additionnal question if you allow me:
Do we still calculate the latent heat in the case of full liquid discharging at initial temp (in above mentionned case i)? If yes i assume at the relieving pressure.Correct?

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"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[Bill3752]

Sheiko, might be good to start back at beginning. As Joe et al indicated you first need to consider the scenario.

Generally either wetted surface (liquid present) or unwetted (gas filled).

Usually wetted relief temp will be vapor pressure at Prel. When tank is very full (say over 80-90%) you need to consider two phase flow. You could rigorously evaluate the time-dependency of the type and amount of relief, which I have found to be quite consuming. I now spend quite a bit of time confirming whether I can base my design on vapor flow only before going the rigorous route. Yes, if any liquid is in the tank you will be using the Hvap in your calcs. There is another thread today you may want to read.

Unwetted, you are using ideal gas law to estimate the temperature (Tr = Tn x Pr/Pn). API 521 does a good job of explaining how to evaluate the unwetted case.