The idea
A dust explosion in a closed vessel reaches 7 to 10 bar g within a fraction of a second. Most process vessels fail at well under 1 bar g. An explosion vent is a deliberately weak panel that opens at a low pressure, Pstat, and releases the burning cloud so that the pressure peaks at a lower value, Pred, which the vessel can survive. Sizing a vent means finding the area that keeps Pred at or below what the vessel was built for.
What you need to know first
| Symbol | Meaning | Where it comes from |
|---|---|---|
| V | Volume that can fill with dust cloud (m³) | Vessel dimensions |
| L/D | Length-to-diameter ratio of the vessel | Vessel dimensions |
| KSt | Dust deflagration index (bar·m/s) | Test of the dust |
| Pmax | Maximum unvented explosion pressure (bar g) | Test of the dust |
| Pstat | Pressure at which the vent opens (bar g) | Vent panel chosen, typically 0.1 bar g |
| Pred | Highest pressure during the vented explosion (bar g) | Vessel strength; no more than two-thirds of its design strength |
The steps
- Base vent area for a compact vessel (L/D of 2 or less):
Av0 = 1×10⁻⁴ · (1 + 1.54 · Pstat4/3) · KSt · V3/4 · √(Pmax / Pred − 1)with pressures in bar g, V in m³ and the area in m².
- Tall or long vessels. When L/D is above 2 the flame accelerates along the vessel and more area is needed:
Av1 = Av0 · [ 1 + 0.6 · (L/D − 2)0.75 · exp(−0.95 · Pred²) ]The correction is validated up to L/D = 6. Beyond that, use several vents along the length or another protection method.
- Partly filled vessels. If the dust cloud can only ever occupy a fraction Xr of the volume, as in some dryers, the area may be reduced. With Π = Pred / Pmax:
Av2 = Av1 · Xr−1/3 · √( (Xr − Π) / (1 − Π) )Use this only when the fill fraction is certain. If in doubt, take the vessel as full (Xr = 1).
- Vent ducts. A duct that carries the discharge outside the building adds resistance and raises the pressure in the vessel, so the vent must be larger:
Av3 = Av2 · [ 1 + 17.3 · (Av2 / V0.753)1.6 · Lduct ]Keep ducts short and straight. The correction is validated for ducts up to 6 m.
- Panel efficiency. Divide by the venting efficiency of the panel if it is below 1.0, for example a heavy or hinged door. Lightweight vent panels are normally taken as 1.0.
- Choose the panels. Select standard vents whose vent areas add up to at least the required area, and check that they fit on the vessel wall or roof.
Worked example
A 25 m³ vessel, 4.2 m long with an equivalent diameter of 2.8 m, handles a dust with KSt = 150 bar·m/s and Pmax = 9 bar g. The vessel can withstand 0.5 bar g and the vents open at 0.1 bar g. It vents directly to a safe outdoor area.
| Step | Calculation | Result |
|---|---|---|
| L/D | 4.2 / 2.8 | 1.5, so no correction |
| Pstat term | 1 + 1.54 × 0.14/3 | 1.0715 |
| Volume term | 250.75 | 11.18 |
| Pressure term | √(9 / 0.5 − 1) | 4.123 |
| Vent area | 1×10⁻⁴ × 1.0715 × 150 × 11.18 × 4.123 | 0.741 m² |
| Panel | One 800 × 1000 mm panel, vent area 0.80 m² | 1 panel |
Limits of the equation
The equation was derived from tests and should be used within their range:
- Pmax between 5 and 12 bar g, and KSt between 10 and 800 bar·m/s;
- volume between 0.1 and 10 000 m³;
- Pstat no higher than 0.75 bar g, and Pred no higher than 2 bar g;
- Pred of at least 0.1 bar g. Weaker enclosures such as buildings and rooms use a different method.
Area is only part of the job. The flame from a vent can reach many metres and the vessel is pushed back by the discharge. The vent must face a safe area, the supports must take the reaction force, and connected equipment needs explosion isolation so that the flame cannot travel down the ducting. Typical dust values are listed in KSt and Pmax of common dusts.