Engineering Article

Sizing Relief Systems for Heat Exchanger Tube Rupture Scenarios

A comprehensive step-by-step methodology to evaluate API 521 exemption criteria and accurately size pressure relief valves for both liquid and gas tube failure cases.

Last Updated: April 12, 2026
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In shell-and-tube heat exchangers, a significant pressure differential often exists between the tube-side and shell-side fluids. A sudden, complete rupture of a high-pressure tube can instantly subject the lower-pressure side to overpressure. Designing a pressure relief system for this scenario requires a clear understanding of API Standard 521 and API Standard 520.

Mitigating Heat Exchanger Tube Rupture Scenarios: Sizing Relief Systems

Figure 1: Typical shell and tube rupture dynamics. Comparison of unmitigated hazard vs engineered relief solution.

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1. The 10/13 Rule: Is Relief Actually Required?

Before performing complex calculations, engineers must determine if a relief device is legally and technically required for the tube rupture scenario. According to API 521, a relief device may not be required if the design pressure of the low-pressure (LP) side is sufficiently high to withstand the incident without catastrophic failure.

For equipment designed to ASME Section VIII Div. 1, the hydrotest pressure is 1.3 times the design pressure. Because a tube rupture is considered an unlikely and highly localized shock, API allows the LP side to experience pressures up to its test pressure. Thus, exemption is granted if:

$$P_{LP, design} \ge \frac{10}{13} P_{HP, design}$$

If this ratio (which equals $\approx 0.77$) is met, the LP side is structurally robust enough to withstand a tube rupture without a PSV. If the ratio is less than 0.77, you must size a relief valve.

2. Determining the Rupture Area ($A_r$)

When a tube breaks, fluid escapes from both sides of the cleanly severed tube (the long length and the short stub at the tubesheet). Therefore, the effective flow area for the rupture is conventionally taken as twice the cross-sectional internal area of a single tube.

$$A_r = 2 \times \frac{\pi}{4} \cdot (d_{internal})^2$$
Discharge Coefficient Warning: When calculating flow through the broken tube, the rupture acts as a sharp-edged orifice. We use a rupture discharge coefficient ($C_{d,rup}$) of typically 0.60. Do not confuse this with the PSV valve discharge coefficient ($K_d$), which is 0.65 for liquid valves and 0.975 for gas valves!

3. Case 1: Non-Flashing Liquid Calculation

If the high-pressure fluid is a liquid that does not flash into vapor upon dropping to the relief pressure, the calculation is straight-forward fluid dynamics.

Step A: Rupture Flow Rate ($q$)

The volumetric flow rate across the ruptured tube is calculated using the standard orifice equation driven by the differential pressure ($\Delta P = P_{HP,op} - P_{relieving}$):

$$q = 38 \cdot A_r \cdot C_{d,rup} \sqrt{\frac{P_{HP,op} - P_{relieving}}{G}} \quad \text{(US Customary)}$$

Step B: Relief Valve Area ($A$)

Once $q$ is known, the PSV orifice area is calculated per API 520 Part I:

$$A = \frac{q}{38 \cdot K_d \cdot K_w \cdot K_v \cdot K_c} \sqrt{\frac{G}{P_1 - P_2}} \quad \text{(US Customary)}$$
Correction Factors Legend:
  • $K_d$: 0.65 for liquid preliminary sizing.
  • $K_w$: Backpressure correction factor (typically 1.0 for conventional valves discharging to atmosphere).
  • $K_v$: Viscosity correction factor (iterated based on Reynolds number).
  • $K_c$: 0.9 if a rupture disk is installed upstream of the valve, otherwise 1.0.

Example 1: Liquid Water Rupture

Scenario: Water flows in the tubes at 550 psig. The shell side is protected by a PSV set at 150 psig with 10% overpressure and discharging to atmosphere (0 psig). The tube is 1.0 in OD with a 0.109 in wall thickness.

  1. Geometry: Tube ID = 0.782 in. The total rupture area ($A_r$) for 2 ends = 0.9606 in².
  2. Pressures: Relieving pressure ($P_1$) = Set + 10% = 165 psig. The differential across the break is 385 psi. (Note: We assume discharge to atmosphere where $P_2 = 0$. If tying into a flare header, the superimposed gauge backpressure must be accounted for).
  3. Flow Rate ($q$): Using $C_{d,rup} = 0.6$ and $G = 1.0$, the flow escaping the tube is 429.6 gpm.
  4. PSV Sizing ($A$): Using API preliminary liquid coefficients ($K_d = 0.65$, $K_w = 1.0$, $K_v = 1.0$), the required PSV orifice area is 1.354 in². (Select an API "K" orifice).
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4. Case 2: Gas or Vapor Calculation

When the high-pressure fluid is a gas, the calculation must account for compressibility and thermodynamic expansion. We must first determine if the flow across the broken tube is critical (sonic velocity) or subcritical.

Step A: Check Critical Flow

Calculate the critical pressure ($P_{cf}$) using the HP operating pressure and the specific heat ratio ($k$):

$$P_{cf} = P_{HP,abs} \times \left( \frac{2}{k+1} \right)^{\frac{k}{k-1}}$$

If the LP relieving pressure ($P_1$) is less than $P_{cf}$, the flow is choked (critical). Otherwise, it is subcritical.

Step B: Gas Mass Flow Rate ($W$)

For critical flow, the mass flow escaping the tube is calculated using the API gas formula, substituting the valve area with our rupture area ($A_r$) and rupture coefficient ($C_{d,rup}$):

$$W = A_r \cdot C_{d,rup} \cdot P_{HP,abs} \cdot C \sqrt{\frac{MW}{Z \cdot T_{HP,abs}}}$$

Where $C$ is a function of $k$, generally ranging between 315 and 377.

Step C: Relief Valve Area ($A$)

Using the mass flow $W$, the required area of the PSV is calculated based on the relieving conditions. Note that the PSV flow state (critical vs subcritical) is evaluated independently based on $P_1$ and atmospheric backpressure $P_2$.

$$A = \frac{W}{C \cdot K_d \cdot P_{1,abs} \cdot K_b \cdot K_c} \sqrt{\frac{T_{rel,abs} \cdot Z}{MW}} \quad \text{(Critical Flow)}$$
Correction Factors Legend:
  • $K_d$: 0.975 for gas preliminary sizing.
  • $K_b$: Backpressure capacity correction factor for gas (often 1.0 depending on bellows presence and backpressure % limits).

Example 2: Methane Gas Rupture

Scenario: Methane ($MW=16.04$, $k=1.31$, $Z=1.0$) flows at 550 psig and 150 °F. The shell side relieves at 165 psig. Tube geometry is identical to Example 1 ($A_r = $ 0.9606 in²).

  1. Critical Check: $P_{hp,abs}$ = 564.7 psia. The critical pressure $P_{cf}$ evaluates to 307.0 psia. Since relieving pressure (179.7 psia) is less than $P_{cf}$, flow across the tube rupture is critical.
  2. Mass Flow ($W$): The gas constant $C$ for $k=1.31$ evaluates to 347.4. Calculating critical flow through the rupture area yields $W = $ 18362 lb/hr.
  3. PSV Sizing ($A$): Checking the PSV, discharge to atmosphere is also critical. Using $K_d = 0.975$ for gas valves, the required relief area calculates to 1.86 in². (Select an API "L" orifice).

5. Special Case: Flashing Liquids

Two-Phase Flow Consideration: The methodologies described above assume pure, non-flashing states. However, if a high-pressure, high-temperature liquid ruptures into a lower-pressure shell and vaporizes (flashes) during the pressure drop, the fluid becomes a two-phase mixture.

Sizing for two-phase flow requires advanced methodologies like the API 520 Omega Method or detailed numerical integration (HEM Model), as the drastic increase in specific volume greatly limits the capacity of the relief valve.
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Conclusion

Tube ruptures dictate some of the largest relief valve sizes in process facilities due to the immense driving force of the high-pressure fluid. By accurately applying the API 10/13th rule and distinguishing between rupture flow dynamics and valve sizing dynamics, engineers can ensure absolute safety and compliance.

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