I would like to independently check a vendor's silo design for proper explosion venting. Worst case, it is essentially cylinder filled with a stoichiometric mixture atomized sugar dust and atmospheric oxygen initially at atmospheric pressure. The roof of the cylinder is designed to detach from the structure with, lets assume, zero force. The roof also has some amount of mass per square inch.

Lots of different ways to go about it, but initially to skip transient calcs I was thinking I could calculate the worst case by assuming 100% of oxygen reacts instantly. Put that energy into the ideal gas law and calculate the final pressure if unvented. Not sure if this will truly be the max pressure or if the transient shock wave pressure would exceed this?

The minimum pressure would be the psi required to lift and accelerate the roof up. The vented roof pressure loading should be pretty easy, again not considering transients.

Any thoughts on how to better roughly calculate wall pressures without cfd? Thanks!


2 Answers 2


You have two pressures which are important here. One pressure (the one you are considering and that @Bart has calculated) is caused by the chemical reaction, i.e. instead of sugar dust and air you have some new gases (carbon oxides and water and probably nitrogen oxides) at elevated temperature. This is called the constant volume equilibrium pressure.

Additionally there is a flame front travelling through your explosive mixture that causes a jump in pressure (and density and temperature). This is the transient explosion pressure, and this pressure is much larger.

If you want to estimate the explosion pressure, for a first order guess you can calculate the Chapman-Jouguet pressure (CJ pressure). There is a good youtube video that explains how to do that. Note that this is a 55 minutes lecture video.

Alternatively, if you do not want to do the calculation by hand, you can use NASA's CEA software (Chemical Equilibrium with Applications). This will probably take a similar amount of time to use.

Or, you just use a number from this (8.5 bar) or this (8.5 bar) or this (9 bar) website, which all seem reasonable.

Note that you will have higher pressures in edges and especially corners due to concentration of shock waves. For this to calculate you need special CFD software.

And yes, 8.5 bar are enough to blow away the roof.

  • $\begingroup$ Excellent! Good find on the dust explosion pressure data! I will use that 8.5 bar and see how much safety factor the cylindrical portion of the silo has while temporarily operating as a pressure vessel. I will also have to watch that lecture when I get time. Thanks! $\endgroup$
    – ericnutsch
    Jan 25, 2018 at 9:16

The pressure in the silo is determined by the temperature rise, caused by the combustion. The end temperature after combustion can be calculated like this:

$$T_{end}= \dfrac{E}{(m_{air}+m_{fuel})C_V}+T_{begin}$$

Where $E$ is the added energy added per space(eg. m3), $m$ is the mass of the fuel and air in kg present in that space, and $C_V$ is the specific heat of the mixture at constant volume (take 724 J/kgK for air), and $T_{begin}$ is temperature of the mixture before igniting (room temperature, 293 K).

Assume we have a temperature rise of 980 degrees, which makes the end temperature 1000°C or 1273 K. Then when begin pressure is 1 atmosphere, the end pressure will be:

$$p_{end}=p_{begin}\cdot\dfrac{T_{end}}{T_{begin}}=1013.25\cdot\dfrac{1273}{293} =4402\text{ hPa/mbar}$$

So that's 4.34 times as high; 4.5 bar or about 64 psi. Obviously there's a lot of assumptions and guesses made here, and it won't fully combust instantaneously, but I would be surprised if the figures differ a lot. I'd bet it is indeed about 4 to 5 bar. It's obviously enough to blow a roof away.

  • $\begingroup$ Thanks! I will hold the best answer for a bit and see if anyone wants to take a stab at estimating the shock wave. $\endgroup$
    – ericnutsch
    Jan 25, 2018 at 2:03
  • $\begingroup$ @ericnutsch Depending on the air/fuel ratio, it will combust quicker or slower, and depending on the mass of the roof, it will allow the silo to build up more pressure before exploding. I think these are the biggest factors in determining the size of the shockwave. $\endgroup$
    – Bart
    Jan 25, 2018 at 10:47

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