OpenSpecEngineering

Finds the minimum effective orifice area a pressure relief valve needs to pass a gas or vapor relief load under critical (choked) flow. It then picks the smallest standard API 526 orifice letter that covers it. Process engineers use it for preliminary sizing before asking a valve maker for a rated selection.

Equations

G = P₁·√(k·M·C_k/(Z·R·T))
C_k = (2/(k + 1))^{(k+1)/(k−1)}
A = W/(K_d·K_b·K_c·G)

G is the mass flux through a choked nozzle for an ideal gas, in kg/(s·m²) [1], [2]. P₁ is the absolute relieving pressure: the set pressure plus the allowed overpressure plus atmospheric pressure. k is the specific heat ratio, M the molecular weight, Z the compressibility factor, T the relieving temperature in kelvin and R = 8314 J/(kmol·K). W is the relieving flow and A the required effective area. K_d is the effective discharge coefficient, K_b the backpressure correction and K_c the combination correction for a rupture disk upstream, as API 520 Part I defines them [3]. API 520 writes the same equation with a coefficient C.

Inputs

• Gas fills M and the ideal-gas k at 27 °C (300 K) [5]. API 520 calls for k at the relieving temperature, so type that value for a hot or cold relief.
• Device / Configuration sets the API 520 preliminary-sizing coefficients: K_d = 0.975 for a relief valve, K_c = 0.9 with a rupture disk upstream and K_d = 0.62 for a rupture disk alone. A balanced-bellows valve’s K_b comes from its maker or the API 520 curve.

Limits and common mistakes

• The equation holds only for critical flow, when the backpressure is below the critical pressure ratio, about 0.53·P₁ for air. Subcritical flow, liquids, two-phase flow and steam need other API 520 methods.
• Entering the set pressure, or a gauge pressure, as P₁. That undersizes the valve.
• API 526 effective areas go with K_d = 0.975. With a rated K_d, the page doesn’t pick a letter; compare against the maker’s actual area [4].
• This is preliminary sizing only. Final selection, inlet and outlet piping and installation must follow API 520 and 526 and be checked by a qualified engineer.

Worked example

A plant air receiver set at 1,000 kPa gauge must relieve the full compressor output of 3,000 kg/h. With 10 % overpressure, P₁ = 1.1 × 1,000 + 101.3 = 1,201.3 kPa absolute. The air relieves at about 310 K.

1. Enter Required Relieving Flow (W) 3000 kg/hr.
2. Choose Gas Air. k fills in 1.4 and M 28.97.
3. Enter Relieving Temperature (T) 310 K, leave Compressibility Factor (Z) at 1, and enter Relieving Pressure, Absolute (P₁) 1201.3 kPa.
4. Leave Device / Configuration at Conventional or pilot-operated PRV.
5. Press Solve. The Solution shows: Mass Flux at Critical Flow (G): 2,757.78 kg/(s·m²)
   Required Orifice Area: 309.923 mm²
   Required Orifice Area: 0.480382 in²
   Next API 526 standard orifice: G — 0.503 in² (325 mm²)

Check by hand with API 520’s SI form [3]: C = 0.03948·√(1.4 × 0.3349) = 0.02703. Then A = 3000/(0.02703 × 0.975 × 1201.3) × √(310/28.97) = 309.9 mm². The G orifice is the next size up, with about 5 % margin.