Tools › Fluids and Heat Transfer › Compressible Flow › Pressure Relief Valve and Rupture Disk Sizing › Example
Air relief valve at 150 psig set, 10 % overpressure, 5,000 lb/h: required area by the nozzle equation, API 526 orifice G, inlet 3 % check, built-up back pressure and reaction force
| Tool | Pressure Relief Valve and Rupture Disk Sizing |
| Source | American Petroleum Institute (2020). Sizing, Selection, and Installation of Pressure-Relieving Devices, Part I: Sizing and Selection. API Standard 520, 10th ed., Washington, DC (the gas equation and its constants, cited; the equation is derived from the isentropic nozzle on the method page); Bhowmik, P. K., and Sabharwall, P. (2024). "Sizing and Selection of Pressure Relief Valves for High-Pressure Thermal–Hydraulic Systems." Processes, Vol. 12, No. 1, Article 21 (Table 6 orifice areas, Table 7 worked case). |
| Level | Undergraduate |
Problem
A vessel holding air at 100 °F needs a relief valve set at 150 psig with 10 % overpressure for a required relieving rate of 5,000 lb/h (M = 28.97, k = 1.4, Z = 1) venting to atmosphere through a 3-in Schedule 40 line 20 ft long; the inlet is 3 ft of 2-in Schedule 40 with an entrance and an elbow. Find the relieving pressure, the choked-flow coefficient C, the required effective area with K_d = 0.975, the API 526 orifice letter and its rated capacity, the inlet pressure loss against the 3 % rule, the built-up back pressure, and the discharge reaction force.
Inputs to Enter
- Gas or vapour case, certified relief valve: set pressure 150 psig, overpressure 10 %, atmospheric 14.696 psia, back pressure 0; K_d 0.975, K_c 1.0, K_b computed (conventional valve)
- Air: W 5,000 lb/h, T 100 °F, M 28.97, k 1.4, Z 1.0, viscosity 0.019 cP
- Piping: inlet 2.067 in ID, 3 ft, ΣK 1.4; outlet 3.068 in ID, 20 ft, ΣK 1.0; roughness 1.8 mil; limits 3 % inlet, 10 % outlet; reaction force on
Answers
- P_1 = 150 (1.10) + 14.696 = 179.70 psia; C = 519.45 √(1.4 (2/2.4)^6) = 355.7 (API table 356); critical ratio r_c = 0.5283, P_crit = 94.93 psia, choked at zero back pressure
- Ideal nozzle flux G = 14,542 lb/(h in²); A = W √(T Z/M)/(C K_d P_1 K_b K_c) = 5,000 √(559.67/28.97)/(355.7 · 0.975 · 179.70) = 0.3527 in² (0.3524 in² with C = 356)
- Orifice G (0.503 in² = 324.5 mm²) is the smallest letter covering 0.3527 in²; rated capacity of G at the relieving conditions 7,132 lb/h (ratio 1.43 to the requirement)
- Inlet at the rated capacity: V = 98.1 ft/s, Re 1.15 × 10^6, f 0.0193, ΔP = 0.302 + 1.260 = 1.56 psi = 1.04 % of the set pressure, under the 3 % rule; outlet (isothermal compressible flow, 3-in line): built-up back pressure 5.80 psi = 3.87 % of the set pressure, under 10 %
- Reaction force at the rated capacity through the 3-in exit: subsonic exit at atmospheric pressure (M_e = 0.46), F = W u_e = 32.2 lbf; the API 520 Part II form, which assumes a sonic exit, gives 65.4 lbf
Notes
The tool's default inputs are this valve; press Submit. The case is illustrative (stated on the page); every value is a closed form checked by test_relief_valve_sizing.php (330 checks) against an independent numpy / scipy script (References/Relief-Valve/checks/), and the gas equation is checked on the published case of Bhowmik and Sabharwall (2024), Table 7 (93.17 cm² for 50,000 kg/h at 4.725 bar, k = 1.27) to the printed digits.