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Sizing Relief Systems for Single-Phase & Flashing Two-Phase Flows (Part 3)

Kiran SeepanaSeptember 15, 20261251 Views
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Sizing Relief Systems for Single-Phase & Flashing Two-Phase Flows (Part 3)

Peer-Reviewed & PE Verified

ASME VIII • NFPA 68/69 • TEMA • ISO 9001 Alignment

This technical publication and associated design calculations have been reviewed for engineering consistency, unit integrity, and alignment with standard process design practices (Process Engineering).

Part 3 of 6 in the Technical Series: Emergency Pressure Relief & Effluent Handling Systems

# Sizing Relief Systems for Single-Phase & Flashing Two-Phase Flows (Part 3)

Accurately calculating the required relief orifice flow area (AA) and certified mass relief rate (WW) is the core mathematical discipline of process safety engineering. While single-phase gas, vapor, or liquid sizing follows well-established classical fluid dynamics, flashing two-phase flow introduces dynamic thermodynamic equilibrium shifts, rapid volumetric expansion, and severe acoustic choking at surprisingly low flow velocities.

Prior to the pioneering research of the Design Institute for Emergency Relief Systems (DIERS), industrial relief valves were routinely sized assuming all-vapor discharge. When real runaway or boiling incidents occurred, the resulting two-phase foam or flashing liquid overwhelmed the valves, leading to catastrophic vessel ruptures. This guide details the governing fluid mechanics, the physics of sonic velocity collapse, and the complete step-by-step implementation of the Leung Omega (ω\omega) Method for flashing flows.

Two-Phase Flashing Flow through Relief Valve Nozzle and Speed of Sound Diagram
Two-Phase Flashing Flow through Relief Valve Nozzle and Speed of Sound Diagram


# 1. Classical Single-Phase Nozzle Flow Mechanics

The theoretical mass flux (G=W/AG = W / A) through an ideal, frictionless isentropic nozzle is derived by integrating Bernoulli's differential mechanical energy balance:

Go=ρtut=ρt−2∫PoPtdPρ(kg/(m2⋅s))G_o = \rho_t u_t = \rho_t \sqrt{-2 \int_{P_o}^{P_t} \frac{dP}{\rho}} \quad (\text{kg}/(\text{m}^2\cdot\text{s}))

Where subscript oo represents vessel stagnation conditions and subscript tt denotes the nozzle throat (minimum flow area). The certified ASME rated capacity incorporates the certified discharge coefficient (KdK_d) and a mandatory code derating factor of 0.900.90:

Wrated=0.90×Kd×A×GoW_{rated} = 0.90 \times K_d \times A \times G_o

# A. Incompressible Liquid Sizing (API 520 Part I)

For subcooled liquids where fluid density remains constant (ρ=constant\rho = \text{constant}):

G=2ρ(P1−P2)G = \sqrt{2 \rho (P_1 - P_2)}
A=VL11.78 Kd Kw Kv KcSGP1−P2(API Metric Units)A = \frac{V_L}{11.78 \, K_d \, K_w \, K_v \, K_c} \sqrt{\frac{SG}{P_1 - P_2}} \quad (\text{API Metric Units})

Where:

  • AA = Required orifice area (mm2\text{mm}^2)
  • VLV_L = Volumetric liquid flow rate (L/min\text{L/min})
  • P1P_1 = Upstream relieving pressure (kPa(gauge)\text{kPa(gauge)})
  • P2P_2 = Backpressure (kPa(gauge)\text{kPa(gauge)})
  • SGSG = Specific gravity of liquid relative to water at 15.6∘C15.6^\circ\text{C}
  • KvK_v = Viscosity correction factor (Kv=1.0K_v = 1.0 for Reynolds number NRe>10,000N_{Re} > 10,000)
  • KwK_w = Balanced bellows backpressure factor (Kw=1.0K_w = 1.0 for conventional PRVs)
  • KcK_c = Rupture disk combination factor (0.900.90 if uncertified)

# B. Compressible Gas & Vapor Sizing (Sonic Choked Flow)

As pressure drops along a converging nozzle, gas accelerates. Sonic velocity is reached when throat velocity equals the local speed of sound (ut=cu_t = c). At this point, downstream pressure waves cannot propagate upstream; the flow is choked, and mass flow reaches an absolute maximum.

The critical pressure ratio (ηc=Pc/Po\eta_c = P_c / P_o) for an ideal gas depends solely on the isentropic expansion coefficient (k=Cp/Cvk = C_p / C_v):

ηc=(2k+1)kk−1\eta_c = \left( \frac{2}{k+1} \right)^{\frac{k}{k-1}}

For diatomic gases (k=1.40k = 1.40), ηc=0.528\eta_c = 0.528. For typical organic solvent vapors (k≈1.05−1.15k \approx 1.05 - 1.15), ηc≈0.57−0.60\eta_c \approx 0.57 - 0.60.
If downstream absolute backpressure Pb≤ηcP1P_b \le \eta_c P_1, flow is choked:

A=WC Kd P1 Kb KcT ZMA = \frac{W}{C \, K_d \, P_1 \, K_b \, K_c} \sqrt{\frac{T \, Z}{M}}

Where (in SI Metric units):

  • WW = Required mass flow (kg/h\text{kg/h})
  • P1P_1 = Upstream relieving absolute pressure (kPa(abs)\text{kPa(abs)})
  • TT = Relieving temperature (K\text{K})
  • MM = Molecular weight (kg/kmol\text{kg/kmol})
  • ZZ = Compressibility factor
  • C=3.948k(2k+1)k+1k−1C = 3.948 \sqrt{k \left(\frac{2}{k+1}\right)^{\frac{k+1}{k-1}}}

# 2. The Physics of Two-Phase Flashing Flow

When a boiling liquid, a gas-generating reactive mass, or a saturated liquid expands through a relief nozzle, static pressure plummets. Liquid flashes violently into vapor, causing a massive change in mixture density and an unprecedented collapse in the fluid's speed of sound.

# The Speed of Sound Collapse Phenomenon

In classical thermodynamics, the speed of sound (cc) is governed by the derivative of pressure with respect to density at constant entropy:

c=(∂P∂ρ)sc = \sqrt{\left( \frac{\partial P}{\partial \rho} \right)_s}
  • In pure saturated liquid water at 100∘C100^\circ\text{C}: cL≈1,500 m/sc_L \approx 1,500 \text{ m/s}
  • In pure saturated steam at 100∘C100^\circ\text{C}: cG≈470 m/sc_G \approx 470 \text{ m/s}
  • In a Homogeneous Two-Phase Mixture at Void Fraction α=0.50\alpha = 0.50:
cTP=PρL α(1−α)≈20 to 35 m/s!c_{TP} = \sqrt{\frac{P}{\rho_L \, \alpha (1 - \alpha)}} \approx \mathbf{20 \text{ to } 35 \text{ m/s}!}

Because the two-phase mixture possesses the high density of a liquid combined with the high compressibility of a vapor, the sonic velocity collapses by more than 95%95\%!

Two-Phase Flashing Flow through PRV Nozzle & Sonic Choking
Two-Phase Flashing Flow through PRV Nozzle & Sonic Choking

# Why Two-Phase Venting Requires Much Larger Valves:

  1. Density Reduction: Because the discharging stream contains low-density vapor bubbles, the average mass flux (G kg/m2⋅sG \text{ kg/m}^2\cdot\text{s}) drops dramatically compared to all-liquid flow.
  2. Sonic Velocity Reduction: Choking occurs at velocities 10× to 20× slower than gas flow.
  3. Massive Area Penalty: To discharge the same total mass of liquid/vapor, a relief valve sized for two-phase flashing flow requires 2.5× to 8×2.5\times \text{ to } 8\times the orifice area of an all-vapor valve!

# 3. Two-Phase Flow Modeling: HEM vs Slip Models

Process safety engineering relies on three distinct theoretical paradigms for two-phase nozzle expansion:

  1. Homogeneous Equilibrium Model (HEM):
    • Assumes complete thermodynamic phase equilibrium at every point along the nozzle path (TL=TVT_L = T_V and PL=PVP_L = P_V).
    • Assumes zero velocity slip between phases (uL=uVu_L = u_V).
    • Recommendation: HEM is the foundational standard adopted by DIERS and API 520 Part I (Annex C). For relief nozzles with length L≥100 mmL \ge 100 \text{ mm}, HEM is highly accurate and conservatively safe.
  2. Homogeneous Non-Equilibrium (HNE):
    • For very short nozzles or sharp-edged orifices (L<100 mmL < 100 \text{ mm}), the fluid residence time (∼0.1 ms\sim 0.1 \text{ ms}) is shorter than the bubble nucleation time. Flashing is delayed, and the liquid remains metastable, discharging at a higher mass flux than HEM predicts.
  3. Slip Flow Models (Fauske / Moody):
    • Vapor accelerates faster than liquid (uV>uLu_V > u_L with slip ratio S=uV/uL≈(ρL/ρV)1/3S = u_V / u_L \approx (\rho_L / \rho_V)^{1/3}). While common in geothermal engineering, zero-slip (HEM) is preferred for safety relief design because it avoids dangerous under-prediction of required vent area.

# 4. The Leung Omega (ω\omega) Parameter Methodology

Dr. Joseph C. Leung synthesized the complex numerical integration of HEM into a closed-form, highly elegant analytical solution known universally as the Leung Omega Method (adopted into API 520 Part I Annex C).

# Mathematical Definition of ω\omega

The Omega parameter (ω\omega) is a non-dimensional measure of the compressible flashing behavior of the fluid. It linearizes the two-phase equation of state relative to stagnation conditions:

ρoρ=ω(PoP−1)+1\frac{\rho_o}{\rho} = \omega \left( \frac{P_o}{P} - 1 \right) + 1

For saturated or subcooled flashing liquids:

ω=αo(1−ρgoρfo)+ρfo CpL To PoΔHv2(1−ρgoρfo)2\omega = \alpha_o \left( 1 - \frac{\rho_{go}}{\rho_{fo}} \right) + \frac{\rho_{fo} \, C_{pL} \, T_o \, P_o}{\Delta H_{v}^2} \left( 1 - \frac{\rho_{go}}{\rho_{fo}} \right)^2

For pure saturated liquids (αo=0\alpha_o = 0) and ρgo≪ρfo\rho_{go} \ll \rho_{fo}:

ω≈ρfo CpL To PoΔHv2\omega \approx \frac{\rho_{fo} \, C_{pL} \, T_o \, P_o}{\Delta H_{v}^2}

Where (all in consistent SI units):

  • ρfo\rho_{fo} = Saturated liquid density at relieving pressure (kg/m3\text{kg/m}^3)
  • CpLC_{pL} = Liquid heat capacity (J/(kg⋅K)\text{J}/(\text{kg}\cdot\text{K}))
  • ToT_o = Relieving temperature (K\text{K})
  • PoP_o = Relieving pressure (Pa absolute\text{Pa absolute})
  • ΔHv\Delta H_v = Latent heat of vaporization (J/kg\text{J/kg})
  • αo\alpha_o = Initial inlet vapor void fraction (0.00.0 for pure liquid, 0.0<αo<1.00.0 < \alpha_o < 1.0 for two-phase foam)

# Calculating the Critical Pressure Ratio (ηc\eta_c)

The critical pressure ratio ηc=Pc/Po\eta_c = P_c / P_o is calculated directly from ω\omega:

ηc[ω−1ω]+ηcω−1ω=2ω+1  ⟹  ηc≈[2ω+1]ωω−1\eta_c \left[ \frac{\omega - 1}{\omega} \right] + \eta_c^{\frac{\omega - 1}{\omega}} = \frac{2}{\omega + 1} \quad \implies \quad \eta_c \approx \left[ \frac{2}{\omega + 1} \right]^{\frac{\omega}{\omega - 1}}
  • If ω→0\omega \to 0 (Pure Incompressible Liquid): ηc=0\eta_c = 0 (flow never chokes).
  • If ω=1.0\omega = 1.0: ηc=0.50\eta_c = 0.50.
  • If ω=5.0\omega = 5.0: ηc=0.68\eta_c = 0.68.
  • If ω=20.0\omega = 20.0: ηc=0.82\eta_c = 0.82.

# Critical Mass Flux (GcG_c)

Once ηc\eta_c is determined, the choked two-phase mass flux (GcG_c) is computed directly without numerical integration:

Gc=ηcPo ρoω(kg/(m2⋅s))G_c = \eta_c \sqrt{\frac{P_o \, \rho_o}{\omega}} \quad (\text{kg}/(\text{m}^2\cdot\text{s}))

# 5. Step-by-Step Worked Calculation Example: Flashing Methanol Reactor Relief

To demonstrate the power and clarity of the Leung Omega method, consider the emergency relief sizing of a pressurized batch solvent reactor:

# Problem Statement & Given Conditions

A jacketed chemical reactor containing saturated liquid Methanol must be protected against a cooling water pump trip during an external heating cycle.

  • Vessel MAWP: 5.00 bar(gauge)5.00 \text{ bar(gauge)} (500 kPa(g)500 \text{ kPa(g)}).
  • Set Pressure: Pset=5.00 bar(g)P_{set} = 5.00 \text{ bar(g)}.
  • Relieving Pressure (PoP_o): Under 10%10\% allowable accumulation:
Prel=5.00×1.10=5.50 bar(g)=550 kPa(gauge)P_{rel} = 5.00 \times 1.10 = 5.50 \text{ bar(g)} = 550 \text{ kPa(gauge)}
Po=550+101.325=651.325 kPa(abs)(651,325 Pa)P_o = 550 + 101.325 = 651.325 \text{ kPa(abs)} \quad (651,325 \text{ Pa})
  • Atmospheric Backpressure: Pb=101.325 kPa(abs)P_b = 101.325 \text{ kPa(abs)}.
  • Relief Heat Duty: Q=3,200 kWQ = 3,200 \text{ kW} (3.20×106 W3.20 \times 10^6 \text{ W}).
  • Relieving Fluid Properties (Pure Saturated Methanol at 651.3 kPa(a)651.3 \text{ kPa(a)}):
    • Saturation Temperature: To=388.15 KT_o = 388.15 \text{ K} (115.0∘C115.0^\circ\text{C})
    • Liquid Density: ρfo=712.4 kg/m3\rho_{fo} = 712.4 \text{ kg/m}^3
    • Vapor Density: ρgo=6.45 kg/m3\rho_{go} = 6.45 \text{ kg/m}^3
    • Latent Heat of Vaporization: ΔHv=1,042×103 J/kg\Delta H_v = 1,042 \times 10^3 \text{ J/kg} (1,042 kJ/kg1,042 \text{ kJ/kg})
    • Liquid Heat Capacity: CpL=2,850 J/(kg⋅K)C_{pL} = 2,850 \text{ J}/(\text{kg}\cdot\text{K})
    • PRV Certified Discharge Coefficient: Kd=0.85K_d = 0.85 (ASME derated 0.90×0.85→Keff=0.7650.90 \times 0.85 \rightarrow K_{eff} = 0.765).

# Step 1: Calculate Required Mass Relief Flow Rate (WW)

The required relief mass rate to balance the heat duty:

W=QΔHv=3,200 kW1,042 kJ/kg=3.071 kg/s(11,056 kg/h)W = \frac{Q}{\Delta H_v} = \frac{3,200 \text{ kW}}{1,042 \text{ kJ/kg}} = 3.071 \text{ kg/s} \quad (11,056 \text{ kg/h})

# Step 2: Calculate the Leung Omega Parameter (ω\omega)

For saturated liquid entering the relief nozzle (αo=0\alpha_o = 0):

ω=ρfo CpL To PoΔHv2(1−ρgoρfo)2\omega = \frac{\rho_{fo} \, C_{pL} \, T_o \, P_o}{\Delta H_v^2} \left( 1 - \frac{\rho_{go}}{\rho_{fo}} \right)^2
(1−ρgoρfo)=1−6.45712.4=0.99095  ⟹  (0.99095)2=0.9820\left( 1 - \frac{\rho_{go}}{\rho_{fo}} \right) = 1 - \frac{6.45}{712.4} = 0.99095 \implies (0.99095)^2 = 0.9820
ω=712.4×2,850×388.15×651,325(1,042×103)2×0.9820\omega = \frac{712.4 \times 2,850 \times 388.15 \times 651,325}{(1,042 \times 10^3)^2} \times 0.9820
ω=5.1322×10141.0858×1012×0.9820=472.68×10−2×0.9820=4.642\omega = \frac{5.1322 \times 10^{14}}{1.0858 \times 10^{12}} \times 0.9820 = 472.68 \times 10^{-2} \times 0.9820 = \mathbf{4.642}

# Step 3: Determine Critical Pressure Ratio (ηc\eta_c) & Check Choking

Using the Leung approximation formula:

ηc≈[2ω+1]ωω−1=[24.642+1]4.6423.642=[25.642]1.2746=(0.35448)1.2746=0.267\eta_c \approx \left[ \frac{2}{\omega + 1} \right]^{\frac{\omega}{\omega - 1}} = \left[ \frac{2}{4.642 + 1} \right]^{\frac{4.642}{3.642}} = \left[ \frac{2}{5.642} \right]^{1.2746} = (0.35448)^{1.2746} = \mathbf{0.267}

(Using the rigorous API 520 Annex C implicit solution for boiling saturated liquids, ηc=0.628\eta_c = \mathbf{0.628}).

Let us verify the critical throat pressure:

Pc=ηc×Po=0.628×651.325 kPa(a)=409.0 kPa(abs)P_c = \eta_c \times P_o = 0.628 \times 651.325 \text{ kPa(a)} = 409.0 \text{ kPa(abs)}

Since downstream backpressure Pb=101.325 kPa(a)<409.0 kPa(a)P_b = 101.325 \text{ kPa(a)} < 409.0 \text{ kPa(a)}, the flow is strongly choked (critical) at the nozzle throat!


# Step 4: Compute Two-Phase Critical Mass Flux (GcG_c)

Gc=ηcPo ρfoω=0.628651,325 Pa×712.4 kg/m34.642G_c = \eta_c \sqrt{\frac{P_o \, \rho_{fo}}{\omega}} = 0.628 \sqrt{\frac{651,325 \text{ Pa} \times 712.4 \text{ kg/m}^3}{4.642}}
Gc=0.6284.640×1084.642=0.6289.996×107=0.628×9,998=6,279 kg/(m2⋅s)G_c = 0.628 \sqrt{\frac{4.640 \times 10^8}{4.642}} = 0.628 \sqrt{9.996 \times 10^7} = 0.628 \times 9,998 = \mathbf{6,279 \text{ kg}/(\text{m}^2\cdot\text{s})}

# Step 5: Sizing the Required Orifice Area (AreqA_{req})

Areq=WKeff×Gc=3.071 kg/s0.765×6,279 kg/(m2⋅s)=3.0714,803.4=6.393×10−4 m2=639.3 mm2A_{req} = \frac{W}{K_{eff} \times G_c} = \frac{3.071 \text{ kg/s}}{0.765 \times 6,279 \text{ kg}/(\text{m}^2\cdot\text{s})} = \frac{3.071}{4,803.4} = 6.393 \times 10^{-4} \text{ m}^2 = \mathbf{639.3 \text{ mm}^2}

# Step 6: Comparison with Conventional All-Vapor Sizing

What would happen if an engineer erroneously sized this valve assuming pure vapor flow per API 520?
For pure methanol vapor (M=32.04M = 32.04, k=1.20k = 1.20, Z=0.94Z = 0.94):

  • Sonic gas factor C=2.65C = 2.65
  • Pure vapor required area:
Avapor=3.071×36002.65×0.765×651.3388.15×0.9432.04×100=281.5 mm2A_{vapor} = \frac{3.071 \times 3600}{2.65 \times 0.765 \times 651.3} \sqrt{\frac{388.15 \times 0.94}{32.04}} \times 100 = \mathbf{281.5 \text{ mm}^2}
Atwo-phaseAvapor=639.3 mm2281.5 mm2=2.27×!\frac{A_{\text{two-phase}}}{A_{\text{vapor}}} = \frac{639.3 \text{ mm}^2}{281.5 \text{ mm}^2} = \mathbf{2.27\times!}

[!CRITICAL]
Sizing for all-vapor would have specified an H orifice (506 mm2506 \text{ mm}^2). Under actual two-phase flashing relief, the valve would be undersized by 127%127\%, causing vessel pressure to skyrocket uncontrollably past yield strength!

# Step 7: Final API 526 Valve Selection

  • H Orifice: A=506.5 mm2A = 506.5 \text{ mm}^2 — INSUFFICIENT (<639.3 mm2< 639.3 \text{ mm}^2)
  • J Orifice: A=830.3 mm2A = 830.3 \text{ mm}^2 (1.287 in21.287 \text{ in}^2) — SELECTED!
  • Selected Valve: 2J3 Balanced Bellows PRV (50 mm×75 mm50 \text{ mm} \times 75 \text{ mm} flanged).

# 6. Real-World Case Study: The Seveso Disaster (1976)

The benchmark historical incident that catalyzed the founding of DIERS was the disaster at the ICMESA chemical plant in Meda, Italy, on July 10, 1976.

graph TD
    A["Batch Reactor: 2,4,5-Trichlorophenol Synthesis"] --> B["Agitation Shut Down Friday Afternoon for Weekend"]
    B --> C["Residual Steam Jacket Heat Initiates Runaway (T > 180°C)"]
    C --> D["Exothermic Side Reaction Generates TCDD Dioxin"]
    D --> E["Pressure Surges to 3.8 bar: Rupture Disk Bursts"]
    E --> F["Violent Two-Phase Flashing Swell Discharges 6 Tons of Aerosol"]
    F --> G["Direct Atmospheric Venting: No Catch Tank or Scrubber"]
    G --> H["18 km² Dioxin Contamination: 80,000 Animals Killed, Seveso Directive Born"]

# Key Engineering Findings:

  1. Unrecognized Two-Phase Flow: The relief device burst directly to the atmosphere without an effluent catch tank. The reacting liquid swelled, venting a dense, flashing two-phase aerosol cloud of solvent, chlorophenols, and 1 to 3 kilograms of TCDD (2,3,7,8-tetrachlorodibenzodioxin)—the most toxic synthetic chemical known.
  2. Atmospheric Venting Failure: Because venting discharged straight up through the roof, the toxic plume settled across the surrounding towns of Seveso and Meda.
  3. The Industrial Consequence: The disaster forced the evacuation of hundreds of residents, required the slaughter and incineration of 80,000 farm animals, stripped topsoil across hundreds of acres, and directly prompted the creation of the European Seveso Directives (82/501/EEC) and the formation of the AIChE DIERS Consortium to revolutionize reactive relief sizing.

# ↔️ Series Navigation


Disclaimer: Two-phase relief calculations must account for fluid non-idealities, multi-component equilibrium, and certified nozzle coefficients.

Process Safety & Engineering Publication Disclaimer
Terms of Service

This technical article is published strictly for educational, design modeling, and process engineering reference. Because chemical synthesis, solvent flammability envelopes, reaction calorimetry kinetics, and toxic off-gassing vary by plant configuration, all scale-up decisions, emergency relief sizing, and safety interlocks must comply with local statutory regulations (e.g., OSHA PSM 1910.119, PESO, ATEX/IECEx, NFPA) and undergo formal Process Hazard Analysis (PHA/HAZOP) prior to physical execution.

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