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    Calculates locally in your browser · API 520 relief valve sizing
    Process / relief systems

    Pressure relief valve sizing

    A relief valve is the last barrier on a vessel, so the sizing is done to a standard rather than to a rule of thumb. The gas equation is a choked nozzle whose capacity depends on upstream pressure and a gas constant that is a function of k alone; the liquid equation is a square root of differential pressure. Both produce an effective area, which is then rounded up to the next API 526 orifice letter.

    P1179.7 psiaC347Kd0.975
    Required area
    1.274 in²
    gas and vapour service
    Orifice
    J — 1.287 in²
    next API 526 size up
    Capacity
    20,207 lb/hr
    99% of the rated flow
    Flow regime
    Choked
    capacity depends on upstream pressure only
    The sizing calculation
    A choked nozzle: capacity depends on upstream pressure, temperature and the gas constant C.
    J orifice
    The sizing calculation
    TermValueNote
    Set pressure150 psigas entered
    Relieving pressure P1179.7 psia110% of set plus atmosphere
    Back pressure14.7 psia0 psig
    Required flow20,000 lb/hrfrom the relieving case
    Gas constant C347.0k = 1.30
    Critical ratio0.546below this the flow is choked
    Temperature660 °Rabsolute
    Molecular weight44.0lb/lbmol
    Discharge coefficient Kd0.975gas and vapour
    Required area1.274 in²effective orifice
    Selected orificeJ — 1.287 in²API 526 letter
    Rated capacity20,207 lb/hrat the selected area

    A J orifice (1.287 in²) passes 20207 lb/hr at these conditions, 99% of its rated capacity. The flow is choked, so the capacity depends on upstream pressure only.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Required area against set pressure
    Higher set pressure means a smaller orifice for the same flow — the driving force is the pressure itself.
    Required area against set pressure0.0000.7361.472.212.943.6850 psig50 psig100 psig100 psig150 psig150 psig200 psig200 psig300 psig300 psig400 psig400 psigin²
    Required against selected
    The API 526 letters are a discrete set, so there is always some over-capacity.
    Required against selected0.0000.3040.6070.9111.211.521.27RequiredRequired1.29SelectedSelectedin²
    Canonical equations
    The exact relationships implemented by this engine.

    A = W √(TZ/M) / (C Kd P1 Kb Kc) · API 520 Part I, Eq. 1.

    C = 520 √(k (2/(k+1))^((k+1)/(k−1))) · A function of k only.

    A = Q √(G/ΔP) / (38 Kd Kw Kc Kv) · API 520 Part I, Eq. 5.

    next API 526 letter ≥ A · D through T.

    engineering methodology

    A clear path from inputs to results.

    A relief valve is the last barrier on a vessel, so the sizing is done to a standard rather than to a rule of thumb. The gas equation is a choked nozzle whose capacity depends on upstream pressure and a gas constant that is a function of k alone; the liquid equation is a square root of differential pressure.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Choose gas or liquid service.
    2. Set the set pressure, back pressure and required flow.
    3. Add the gas properties, or the liquid specific gravity.
    4. Read the required area and the API 526 orifice that covers it.

    The formula

    A=WCKdP1TZMA = \frac{W}{C K_d P_1}\sqrt{\frac{TZ}{M}}

    Reference: API 520 Part I, Sizing, Selection and Installation of Pressure-Relieving Devices; orifice letters from API 526.

    Worked example

    20,000 lb/hr of gas at a 150 psig set pressure

    Inputs
    • 150 psig set, atmospheric back pressure.
    • 20,000 lb/hr, molecular weight 44, k = 1.3.
    • 200 °F relieving temperature, Z = 1.
    Calculation path
    1. 1

      The relieving pressure is 110% of set plus atmosphere: 179.7 psia.

    2. 2

      The gas constant C is 347.0 for k = 1.3.

    3. 3

      The required area is (20,000 / (347 × 0.975 × 179.7)) × √(659.7/44) = 1.274 in².

    4. 4

      The next API 526 orifice is J at 1.287 in².

    5. 5

      At atmospheric back pressure the flow is choked, so capacity depends on upstream pressure only.

    Result

    1.274 in² required · J orifice · choked flow.

    Interpreting the result

    Choked flow is the normal case for a gas relief valve discharging to atmosphere, and it means the back pressure has no effect on capacity. Once the back pressure rises above about 10% of set, a conventional valve’s capacity starts to fall and a balanced bellows valve is needed — which is why the back-pressure correction is the term most often mishandled.

    Assumptions & limitations

    • A conventional valve with 10% overpressure, Kb = 1 and Kc = 1.
    • Gas service uses Kd = 0.975 and liquid service 0.65.
    • The relief case is the governing one; fire and thermal relief are separate cases.
    • Two-phase and flashing flow are not modelled.

    Frequently asked questions

    What is the gas constant C?

    A function of the specific-heat ratio only: C = 520 √(k(2/(k+1))^((k+1)/(k−1))). It is about 356 for air and 347 for k = 1.3.

    Why is Kd different for gas and liquid?

    Because the two are calibrated against different test fluids and the discharge behaviour differs. Gas and vapour use 0.975; liquid uses 0.65.

    How do I pick between a conventional and a balanced valve?

    By back pressure. Above about 10% of set, a conventional valve’s capacity falls and a balanced bellows valve is required to keep it predictable.

    What does choked mean?

    The flow has reached the speed of sound at the orifice throat, so lowering the downstream pressure further does not increase the flow. Capacity then depends only on upstream conditions.

    What is the 10% overpressure?

    The allowance above set pressure at which the valve is rated to reach full lift. It is 10% for most cases and 21% for fire exposure.

    Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.