How can I describe the amount of material that can burn in cFast

[Katharina Michaelis]

In the example case, I read a column with Fuel, but there is only the chemical formula like C9H6O2N2. How can I increase or decrease it? Wouldn't it become a different chemical thing if I change the amount of molecules? For example, if I want to have wood or concrete as burning material. How can I describe that as burning fuel? For example, if I have 2kg of it. How can I describe that?

Thank you.

[Kevin McGrattan]

CFAST makes simple assumptions about gases. For example, the molecular weight of the mixture of gases is assumed to be 29 g/mol. So these details are not important. The amount of mass consumed in the fire is

m = int (HRR dt) / HoC

Check the units to make sure you get a mass.

[Katharina Michaelis]

Thank you.  What is int (HRR dt)? And HoC is Heat of combustion? Which units do I need?

I don't understand int and dt.

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[dr_jfloyd]

int = integral. int (HRR dt) is the integral over time of the heat release rate.

HoC is the heat of combustion. 

[Katharina Michaelis]

How can I build an integral with the heat release rate? Is it for example in the case of HRR = 1000 --> 0.5×1000x?

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[Katharina Michaelis]

And if for example I have 0.5*1000x / 5000. What do I put in for x to have a result for the mass?

[Katharina Michaelis]

Which unit do I get as a result?

[Kevin McGrattan]

The integral is the area under the HRR curve, which you can see in the CFAST user interface. The units of this area is the product of the quantities on the axes, s * kW = kJ

[Katharina Michaelis]

Isn't it kg? I would need a mass that burns, not energy...

[Günter Becker]

Hm. You have a result in kJ, and also a specific reaction enthalpy in kJ / kg. Consider dividing your result in kJ by that specific reaction enthalpy. This should yield a result in kg.

Best regards

Günter

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[Katharina Michaelis]

I just recognized: If I divide the integral of the HRR, which has the unit KJ, through KJ, then I don't have a unit anymore?

[Günter Becker]

??? I do not quite understand your argument.  If I perform the aforementioned division, I get:

kJ / (kJ / kg) = kJ * (kg / kJ) = kg

So why is kg not a unit?

Best regards

Günter

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[Katharina Michaelis]

Oh, thank you, I hadn't understood yet that it's not only dividing through kg (kJ/kg), but through kJ/(kg/kJ).

[Katharina Michaelis]

I haven't known about a reaction enthalpy before you mentioned it.

[Katharina Michaelis]

So I guess I am right that the integral of the HRR shows how much material burned during that time? And the enthapy only is a unit in this case. Or do I also have to divide the integral of the HRR through a number? 

[Katharina Michaelis]

As an example, just to see if I got it right: If I have a heat release rate of 100KW, and I want to find out how many kg burned during the first 60 seconds. (so the time is from 0 till 60 seconds) Then the integral is 100*60 - 100*0 = 6000kJ. If I divide it through (kJ/kg), then I get 6000kg which burned. Is this correct? 

[Katharina Michaelis]

And If I had for example a heat release rate of 200kW at 180 seconds, but a heat release rate of 150 kW at 120 seconds. Wouldn't the solution then be 200*180 - 150*120 kg = 18000 kg if I consider the time between 120 and 180 seconds with those two different heat release rates? It's the case in the CFAST user guide example, therefore I am also curious how I would calculate with 2 different heat release rates.

Thank you.

[Katharina Michaelis]

If I calculated like that, a heat release rate of 200kW at 180s and of 150kW at 240s would lead to 0 kg between 180s and 240s? 150*240 - 200*180 kg = 0kg. But the graph of the user guide example shows a space between the y-Axis and the integral function. So there still burns something. What is wrong?

[Katharina Michaelis]

And how can I find out the fire compartment area (Brandabschnittsfläche) with CFAST? 

[Katharina Michaelis]

As the integral has the units kJ: I guess I would have to change that unit kJ to MJ, with kJ/1000, and then multiplying with the considered floor area of the fire? Is that correct?

[Katharina Michaelis]

I just read that kg/kJ is the heat of combustion, so to have the correct kg, I have to divide the integral of the HRR through the heat of combustion?

[Günter Becker]

https://en.wikipedia.org/wiki/Heat_of_combustion

I just notice that in English, we talk about heat of combustion or enthalpy of combustion (sorry, I'm not a native speaker). Wikipedia seems to know it.

Best regards

Günter

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[Katharina Michaelis]

Then in the user guide example case, from 0 - 60 seconds there would burn 0,12kg, and from 60 - 120 seconds 0,24kg. I guess then it's correct? But the time from 180s - 240s would stay 0?

[Günter Becker]

Well, heat of combustion is different from 1 kJ /kg in most cases. So, you do not just divide by a unit, but also by a number. This number depends on what is burnt. wikipedia has a sub page "Heat of combustion tables". It gives values im MJ/kg instead of kJ/Kg, but this should not be a problem, should it?

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[Katharina Michaelis]

Yes, thank you, I guess now it's so much more clear. Only I guess I can't see the result in kg anywhere in CFAST, I only have the kgs from the calculations... But now it's much more clear.