How to Size a Rupture Disc for Gas Service (API 520)

When this method applies

API 520 Part I offers two ways to size a rupture disc. The simpler one, the coefficient-of-discharge method, treats the burst disc as a nozzle with a fixed discharge coefficient K_d = 0.62. It may be used only when all of these are true:

the disc discharges directly to atmosphere;
it is installed within 8 pipe diameters of the vessel nozzle;
the discharge pipe after it is no longer than 5 pipe diameters;
the inlet and outlet piping are at least the nominal size of the disc.

If the disc feeds a long discharge line or a header, the piping restricts the flow more than the disc does, and the resistance-to-flow method (using the disc's certified K_R and a full pressure-drop calculation of the line) must be used instead.

The steps

Find the relieving pressure P₁. This is the pressure in the vessel while the disc is relieving, in absolute units. With the disc's burst pressure at the vessel's maximum allowable working pressure:
P₁ = burst pressure × (1 + overpressure) + atmospheric pressure
The overpressure is 10 % for a single device, 16 % where several devices share the duty, and 21 % for an external fire case.
Check that the flow is critical. Gas flow through the disc is sonic (critical) when the back pressure is low enough:
P₂ / P₁ ≤ (2 / (k + 1))^{k / (k − 1)}
For air (k = 1.40) the limit is 0.528, so a disc venting to atmosphere is in critical flow whenever P₁ is above about 1.9 bar absolute.
Calculate the gas constant C from the ratio of specific heats k (US customary units):
C = 520 · √( k · (2 / (k + 1))^{(k + 1) / (k − 1)} )
Calculate the required area.
A = W / (C · K_d · P₁) · √(T · Z / M)
with A in in², W in lb/h, P₁ in psia, T in °R (°F + 459.67), Z the compressibility factor and M the molecular weight.
Choose the disc. Take the next standard size whose net flow area is at least A, then check its capacity.

Worked example

Nitrogen (M = 28.01, k = 1.40, Z = 1) must be relieved at 2000 kg/h from a vessel protected by a disc with a burst pressure of 5 bar g. The gas is at 100 °C, the disc vents to atmosphere and it is the only relief device.

Step	Calculation	Result
Relieving pressure	5 × 14.504 × 1.10 + 14.696	94.47 psia
Critical flow?	14.696 / 94.47 = 0.156, below 0.528	Yes
Temperature	100 × 1.8 + 32 + 459.67	671.67 °R
Mass flow	2000 × 2.2046	4409 lb/h
Gas constant C	520 · √(1.40 × 0.8333⁶)	356.1
Required area	4409 / (356.1 × 0.62 × 94.47) × √(671.67 / 28.01)	1.035 in² (668 mm²)
Disc size	1″ gives 0.785 in², too small; 1½″ gives 1.767 in²	1½″
Capacity of the 1½″ disc	0.62 × 356.1 × 1.767 × 94.47 / √(671.67 / 28.01)	7525 lb/h (3413 kg/h)

You can reproduce these figures with the quick estimate on the rupture disc sizing page.

Flow areas used for sizing

Nominal size	Flow area (in²)	Flow area (mm²)
1″	0.785	506
1½″	1.767	1140
2″	3.142	2027
3″	7.068	4560
4″	12.566	8107
6″	28.274	18241
8″	50.265	32429
10″	78.540	50671

Check the actual disc. These are full-bore areas. The certified minimum net flow area of a particular disc model and holder can be smaller, so confirm it on the manufacturer's datasheet before finalising the size.

Mistakes to avoid

Gauge instead of absolute pressure. P₁ must include atmospheric pressure.
Celsius or Fahrenheit instead of absolute temperature.
Ignoring the manufacturing range and burst tolerance. The marked burst pressure can differ from the pressure you asked for; the vessel must still be protected at the upper end.
Using K_d = 0.62 with long discharge piping. That case needs the resistance-to-flow method.
Forgetting temperature. A disc's burst pressure changes with temperature, so specify it at the real operating temperature.