Hazard Analysis

# Chemical Engineering for PE: Hazard Analysis

CHAPTER OVERVIEW

This chapter covers the fundamental principles and methodologies used in identifying, evaluating, and mitigating process hazards in chemical engineering facilities. Students will study hazard identification techniques including HAZOP (Hazard and Operability Study), FMEA (Failure Modes and Effects Analysis), What-If analysis, and Fault Tree Analysis (FTA). The chapter includes quantitative risk assessment methods, consequence modeling for toxic and flammable material releases, explosion analysis, and protective systems design. Detailed coverage of inherent safety principles, layers of protection analysis (LOPA), and emergency relief system sizing is provided to prepare candidates for practical hazard analysis problems encountered in professional practice.

KEY CONCEPTS & THEORY

Hazard Identification Methods

HAZOP (Hazard and Operability Study)

HAZOP is a systematic, team-based approach to identify process hazards and operability problems. The technique uses guide words applied to process parameters to explore deviations from design intent. Common HAZOP Guide Words:

• NO/NOT: Complete negation of design intent
• MORE: Quantitative increase
• LESS: Quantitative decrease
• AS WELL AS: Qualitative increase
• PART OF: Qualitative decrease
• REVERSE: Logical opposite of intent
• OTHER THAN: Complete substitution

Process Parameters: Flow, Pressure, Temperature, Level, Composition, pH, Viscosity, Time HAZOP Methodology:

1. Divide process into nodes (sections)
2. Select design intent for each node
3. Apply guide words to parameters
4. Identify causes of deviations
5. Evaluate consequences
6. Identify existing safeguards
7. Recommend additional safeguards if needed

What-If Analysis

A brainstorming approach where experienced personnel pose questions beginning with "What if..." to identify potential hazards. This method is less structured than HAZOP but can be faster and more flexible.

FMEA (Failure Modes and Effects Analysis)

A systematic bottom-up approach that examines individual equipment failures and their effects on the system. Each component is analyzed for potential failure modes, causes, effects, and detection methods. Risk Priority Number (RPN): \[ \text{RPN} = S \times O \times D \] Where:

• \(S\) = Severity (typically 1-10 scale)
• \(O\) = Occurrence frequency (typically 1-10 scale)
• \(D\) = Detection difficulty (typically 1-10 scale)

Fault Tree Analysis (FTA)

A top-down, deductive analytical method to determine root causes of system failures. Uses Boolean logic gates (AND, OR) to model combinations of failures leading to a top event. Basic FTA Gates:

• OR Gate: Output occurs if any input occurs
• AND Gate: Output occurs only if all inputs occur simultaneously

Minimal Cut Set: The smallest combination of basic events that can cause the top event. For independent events in an OR gate: \[ P(\text{Top Event}) = 1 - \prod_{i=1}^{n}(1 - P_i) \] For independent events in an AND gate: \[ P(\text{Top Event}) = \prod_{i=1}^{n} P_i \]

Risk Assessment

Risk Definition

\[ \text{Risk} = \text{Frequency} \times \text{Consequence} \] Risk Matrix: Qualitative or semi-quantitative tool categorizing scenarios by frequency (likelihood) and severity (consequence).

Layer of Protection Analysis (LOPA)

A semi-quantitative risk assessment method that evaluates the adequacy of protection layers against identified hazards. Initiating Event Frequency: Frequency at which a hazardous condition begins (events/year) Independent Protection Layers (IPLs): Safeguards that function independently to prevent or mitigate consequences Probability of Failure on Demand (PFD): Likelihood that a protection layer fails when needed LOPA Calculation: \[ f_{\text{scenario}} = f_{\text{IE}} \times \prod_{i=1}^{n} \text{PFD}_i \] Where:

• \(f_{\text{scenario}}\) = Frequency of the unmitigated consequence scenario (events/year)
• \(f_{\text{IE}}\) = Frequency of initiating event (events/year)
• \(\text{PFD}_i\) = Probability of failure on demand for each IPL

Typical PFD Values:

• Basic Process Control System (BPCS): 10^-1
• Operator Intervention: 10^-1
• Safety Instrumented Function (SIF) SIL 1: 10^-2
• Safety Instrumented Function (SIF) SIL 2: 10^-3
• Safety Instrumented Function (SIF) SIL 3: 10^-4
• Relief Valve: 10^-2 to 10^-1

Risk Tolerance Criteria:

• Individual risk typically <>^-4 to 10^-6 fatalities/year
• Societal risk varies by jurisdiction and application

Consequence Modeling

Source Term Models

Liquid Release from Tank or Vessel: For discharge through a hole in a tank containing liquid: \[ \dot{m} = C_d A \sqrt{2 \rho (P - P_a + \rho g h)} \] Where:

• \(\dot{m}\) = Mass flow rate (kg/s)
• \(C_d\) = Discharge coefficient (typically 0.61-0.62 for sharp-edged orifice)
• \(A\) = Hole area (m²)
• \(\rho\) = Liquid density (kg/m³)
• \(P\) = Internal pressure (Pa)
• \(P_a\) = Atmospheric pressure (Pa)
• \(g\) = Gravitational acceleration (9.81 m/s²)
• \(h\) = Liquid height above hole (m)

Gas or Vapor Release: For choked flow (most common for high-pressure releases): \[ \dot{m} = C_d A P \sqrt{\frac{M}{RT}} \sqrt{\gamma \left(\frac{2}{\gamma + 1}\right)^{\frac{\gamma+1}{\gamma-1}}} \] Where:

• \(M\) = Molecular weight (kg/kmol)
• \(R\) = Universal gas constant (8314 J/kmol·K)
• \(T\) = Absolute temperature (K)
• \(\gamma\) = Heat capacity ratio \(C_p/C_v\)

Flow is choked when: \[ \frac{P_a}{P} \leq \left(\frac{2}{\gamma + 1}\right)^{\frac{\gamma}{\gamma-1}} \] For non-choked (subsonic) flow: \[ \dot{m} = C_d A P \sqrt{\frac{M}{RT}} \sqrt{\frac{2\gamma}{\gamma-1}\left[\left(\frac{P_a}{P}\right)^{\frac{2}{\gamma}} - \left(\frac{P_a}{P}\right)^{\frac{\gamma+1}{\gamma}}\right]} \] Two-Phase Flashing Liquid Release: When a pressurized liquid is released and undergoes partial vaporization (flashing): \[ f = \frac{C_{p,L}(T_0 - T_b)}{H_{vap}} \] Where:

• \(f\) = Flash fraction (mass fraction vaporized)
• \(C_{p,L}\) = Liquid heat capacity (J/kg·K)
• \(T_0\) = Initial liquid temperature (K)
• \(T_b\) = Boiling point at atmospheric pressure (K)
• \(H_{vap}\) = Heat of vaporization (J/kg)

Dispersion Modeling

Gaussian Plume Model: For neutrally buoyant gas releases under steady-state conditions: \[ C(x,y,z) = \frac{Q}{2\pi u \sigma_y \sigma_z} \exp\left(-\frac{y^2}{2\sigma_y^2}\right) \left[\exp\left(-\frac{(z-H)^2}{2\sigma_z^2}\right) + \exp\left(-\frac{(z+H)^2}{2\sigma_z^2}\right)\right] \] Where:

• \(C\) = Concentration (kg/m³)
• \(Q\) = Release rate (kg/s)
• \(u\) = Wind speed (m/s)
• \(\sigma_y\) = Lateral dispersion coefficient (m)
• \(\sigma_z\) = Vertical dispersion coefficient (m)
• \(H\) = Effective release height (m)
• \(x, y, z\) = Downwind, crosswind, and vertical coordinates (m)

Centerline Concentration (y = 0, z = H): \[ C_{max}(x) = \frac{Q}{\pi u \sigma_y \sigma_z} \] Pasquill-Gifford Stability Classes:

• A: Extremely unstable
• B: Moderately unstable
• C: Slightly unstable
• D: Neutral
• E: Slightly stable
• F: Moderately stable

Dispersion coefficients \(\sigma_y\) and \(\sigma_z\) depend on stability class and downwind distance. Dense Gas Dispersion: For gases denser than air (due to molecular weight or low temperature), specialized models like SLAB or DEGADIS are required, as they do not follow Gaussian behavior.

Pool Evaporation

For volatile liquids spilled on the ground: \[ \dot{m}_{evap} = k_c A (C_{sat} - C_{\infty}) \] Where:

• \(\dot{m}_{evap}\) = Evaporation rate (kg/s)
• \(k_c\) = Mass transfer coefficient (m/s)
• \(A\) = Pool surface area (m²)
• \(C_{sat}\) = Saturation concentration at pool surface (kg/m³)
• \(C_{\infty}\) = Concentration in bulk air (kg/m³)

Mass transfer coefficient correlation: \[ k_c = \frac{D_{AB}}{d} Sh \] Where \(Sh\) is the Sherwood number, \(D_{AB}\) is the diffusivity, and \(d\) is the characteristic length.

Fire and Explosion Analysis

Pool Fire Radiation

Radiative heat flux at a distance: \[ q'' = \tau F Q_{rad} / (4\pi r^2) \] Where:

• \(q''\) = Heat flux at target (W/m²)
• \(\tau\) = Atmospheric transmissivity (typically 0.7-1.0)
• \(F\) = View factor (dimensionless)
• \(Q_{rad}\) = Total radiative heat release (W)
• \(r\) = Distance from flame (m)

Total radiative heat release: \[ Q_{rad} = \chi_{rad} \dot{m}_{fuel} \Delta H_c \] Where:

• \(\chi_{rad}\) = Radiative fraction (typically 0.2-0.4 for hydrocarbons)
• \(\dot{m}_{fuel}\) = Fuel burning rate (kg/s)
• \(\Delta H_c\) = Heat of combustion (J/kg)

Vapor Cloud Explosion (VCE)

TNT Equivalency Method: \[ W_{TNT} = \eta \frac{W_{fuel} \Delta H_c}{E_{TNT}} \] Where:

• \(W_{TNT}\) = Equivalent TNT mass (kg)
• \(\eta\) = Explosion efficiency (typically 0.03-0.10 for unconfined VCE)
• \(W_{fuel}\) = Mass of fuel in cloud (kg)
• \(\Delta H_c\) = Heat of combustion (J/kg)
• \(E_{TNT}\) = TNT blast energy (4.68 × 10⁶ J/kg)

Scaled distance: \[ Z = \frac{R}{W_{TNT}^{1/3}} \] Where \(R\) is the distance from explosion center (m). Peak overpressure correlations use scaled distance to estimate:

• Peak side-on overpressure (\(\Delta P_s\))
• Peak reflected overpressure (\(\Delta P_r\))
• Positive phase duration
• Impulse

Boiling Liquid Expanding Vapor Explosion (BLEVE)

Occurs when a vessel containing liquid above its atmospheric boiling point fails catastrophically. The energy release includes:

• Pressure energy from rapid vapor expansion
• Energy from superheat (flashing liquid)
• Combustion energy if fuel ignites (fireball)

Fireball diameter (empirical): \[ D = 5.8 M^{1/3} \] Where \(D\) is diameter (m) and \(M\) is mass of fuel (kg). Fireball duration (empirical): \[ t = 0.45 M^{1/3} \] Where \(t\) is duration (s).

Deflagration and Detonation

Deflagration: Subsonic combustion wave (flame speed < speed="" of="" sound)="">Detonation: Supersonic combustion wave (shock wave + reaction zone) Deflagration Index (K_G): \[ K_G = V^{1/3} \left(\frac{dP}{dt}\right)_{max} \] Where:

• \(V\) = Vessel volume (m³)
• \((dP/dt)_{max}\) = Maximum rate of pressure rise (bar/s)

Used to classify dust and gas explosion severity.

Pressure Relief and Emergency Venting

Relief Device Sizing for Gas/Vapor Service

Required relief area (API 520): \[ A = \frac{W}{C K_d P_1 K_b K_c} \sqrt{\frac{TZ}{M}} \] Where:

• \(A\) = Required discharge area (mm²)
• \(W\) = Required flow rate (kg/h)
• \(C\) = Constant (depends on units; for SI: 0.1263)
• \(K_d\) = Discharge coefficient (typically 0.975 for full nozzle, 0.62 for sharp-edged orifice)
• \(P_1\) = Upstream relieving pressure (kPa absolute)
• \(K_b\) = Capacity correction factor for back pressure
• \(K_c\) = Combination correction factor for installations with rupture disk
• \(T\) = Relieving temperature (K)
• \(Z\) = Compressibility factor
• \(M\) = Molecular weight (kg/kmol)

Relief Device Sizing for Liquid Service

\[ A = \frac{Q}{38 K_d K_w K_v} \sqrt{\frac{\rho}{P_1 - P_2}} \] Where:

• \(Q\) = Liquid flow rate (L/min)
• \(K_w\) = Correction factor for back pressure
• \(K_v\) = Correction factor for viscosity
• \(\rho\) = Liquid density (kg/m³)
• \(P_1\) = Upstream relieving pressure (kPa gauge)
• \(P_2\) = Back pressure (kPa gauge)

Two-Phase Relief Sizing

Two-phase flow (vapor-liquid mixture) is more complex. Methods include:

• Homogeneous Equilibrium Model (HEM): Assumes thermal and mechanical equilibrium between phases
• API 520 simplified approach using omega method
• DIERS (Design Institute for Emergency Relief Systems) methodology for runaway reactions

Emergency Vent Sizing for Runaway Reactions

Tempered Relief: Venting begins before onset of reaction; mostly vapor relief Partially Tempered Relief: Two-phase venting; some reaction has occurred Gassy System: Gas generation significantly affects venting Hybrid System: Combined gas generation and vapor formation Required vent area (simplified): \[ A = \frac{G V}{\Psi \sqrt{\rho_m}} \] Where:

• \(G\) = Mass flux through vent (kg/m²·s)
• \(V\) = Reactor volume (m³)
• \(\Psi\) = Two-phase flow parameter
• \(\rho_m\) = Two-phase mixture density (kg/m³)

Detailed sizing requires calorimetric data (DSC, ARC, VSP) and specialized software.

Inherent Safety

Inherent Safety Principles (in order of preference):

1. Minimize: Use smaller quantities of hazardous materials
2. Substitute: Replace with less hazardous materials
3. Moderate: Use less hazardous conditions (lower temperature/pressure)
4. Simplify: Design to eliminate complexity and opportunities for error

Inherently Safer Design strategies:

• Intensification (minimize inventory)
• Substitution (use safer chemistry)
• Attenuation (dilute or use less severe conditions)
• Limitation of effects (design to limit consequences)
• Simplification/error tolerance (make operations foolproof)

Toxicity and Exposure Limits

Common Exposure Limits:

• TLV-TWA (Threshold Limit Value - Time Weighted Average): Concentration for 8-hour workday, 40-hour workweek with no adverse effects
• TLV-STEL (Short-Term Exposure Limit): 15-minute TWA exposure that should not be exceeded
• TLV-C (Ceiling): Concentration that should not be exceeded at any time
• IDLH (Immediately Dangerous to Life or Health): Maximum concentration from which one could escape within 30 minutes without irreversible health effects
• LC₅₀ (Lethal Concentration 50%): Concentration lethal to 50% of test population (typically 1-hour exposure)
• ERPG (Emergency Response Planning Guidelines):
  • ERPG-1: Mild, transient effects
  • ERPG-2: Irreversible or serious health effects
  • ERPG-3: Life-threatening effects

Flammability Characteristics

Key Parameters:

• Flash Point: Lowest temperature at which vapor forms an ignitable mixture with air
• Autoignition Temperature (AIT): Minimum temperature for spontaneous ignition without external ignition source
• Lower Flammable Limit (LFL) / Lower Explosive Limit (LEL): Minimum concentration in air for flame propagation
• Upper Flammable Limit (UFL) / Upper Explosive Limit (UEL): Maximum concentration in air for flame propagation
• Minimum Ignition Energy (MIE): Minimum electrical spark energy to ignite most ignitable mixture
• Limiting Oxygen Concentration (LOC) / Maximum Oxygen Concentration (MOC): Maximum oxygen content below which combustion cannot occur regardless of fuel concentration

LFL Estimation for Mixtures (Le Chatelier's Rule): \[ \text{LFL}_{mix} = \frac{1}{\sum_{i=1}^{n} \frac{y_i}{\text{LFL}_i}} \] Where \(y_i\) is the mole (or volume) fraction of component \(i\) in the fuel mixture. Inerting: Adding inert gas (N₂, CO₂) to reduce oxygen below LOC Typical LOC values: 10-12% O₂ for hydrocarbons

SOLVED EXAMPLES

Example 1: LOPA Risk Assessment for Reactor Overpressure Scenario

PROBLEM STATEMENT:
A batch reactor processes flammable organic material at elevated temperature. A HAZOP study identified a scenario where loss of cooling could lead to runaway reaction and vessel rupture. Perform a LOPA to determine if additional safeguards are needed. GIVEN DATA:

• Initiating event: Cooling water failure, frequency = 0.1 events/year
• Existing Independent Protection Layers (IPLs):
  • High temperature alarm with operator response: PFD = 0.1
  • Safety Instrumented System (SIS) initiating emergency cooling: SIL 2, PFD = 0.01
  • Pressure relief valve (PRV) sized for runaway: PFD = 0.1
• Target risk tolerance: 1 × 10^-5 events/year
• An additional SIL 1 SIS to shut down heating is being considered (PFD = 0.01)

FIND:
(a) Calculate the mitigated scenario frequency with existing IPLs
(b) Determine if the existing protection meets the risk tolerance criterion
(c) Calculate the mitigated frequency if the SIL 1 heating shutdown system is added SOLUTION: Step 1: Calculate mitigated frequency with existing IPLs
The LOPA equation is: \[ f_{scenario} = f_{IE} \times \text{PFD}_1 \times \text{PFD}_2 \times \text{PFD}_3 \times ... \] For existing IPLs: \[ f_{scenario} = 0.1 \times 0.1 \times 0.01 \times 0.1 \] \[ f_{scenario} = 0.1 \times 10^{-1} \times 10^{-2} \times 10^{-1} \] \[ f_{scenario} = 0.1 \times 10^{-4} \] \[ f_{scenario} = 1 \times 10^{-5} \text{ events/year} \] Step 2: Compare to risk tolerance
Calculated frequency: 1 × 10^-5 events/year
Target tolerance: 1 × 10^-5 events/year The calculated frequency equals the tolerance criterion. This is marginally acceptable but offers no safety margin. Most companies would require frequency to be at least one order of magnitude below the tolerance. Step 3: Calculate frequency with additional SIL 1 system
Adding the heating shutdown SIS (PFD = 0.01): \[ f_{scenario,new} = 0.1 \times 0.1 \times 0.01 \times 0.1 \times 0.01 \] \[ f_{scenario,new} = 1 \times 10^{-5} \times 10^{-2} \] \[ f_{scenario,new} = 1 \times 10^{-7} \text{ events/year} \] This provides two orders of magnitude safety margin below the tolerance. ANSWER:
(a) Mitigated scenario frequency with existing IPLs: 1 × 10^-5 events/year
(b) Existing protection barely meets criterion; additional protection recommended for safety margin
(c) With additional SIL 1 system: 1 × 10^-7 events/year

Example 2: Emergency Relief Sizing for Runaway Reaction with Two-Phase Flow

PROBLEM STATEMENT:
A jacketed batch reactor contains 8,000 kg of reaction mixture. Adiabatic calorimetry testing indicates that a runaway reaction could occur if temperature control is lost. Size the emergency relief vent to prevent vessel rupture. GIVEN DATA:

• Reactor volume: \(V = 12 \text{ m}^3\)
• Initial reaction mixture mass: \(m = 8000 \text{ kg}\)
• Vapor space fraction: 33% (4 m³)
• Set pressure: \(P_{set} = 400 \text{ kPa absolute}\)
• Maximum allowable working pressure: \(MAWP = 500 \text{ kPa absolute}\)
• Design pressure for relief: \(P_{relief} = 440 \text{ kPa absolute}\) (10% overpressure)
• Temperature at relief conditions: \(T = 180°\text{C} = 453 \text{ K}\)
• From calorimetry: Maximum temperature rate \((dT/dt)_{max} = 3.5 \text{ K/min} = 0.0583 \text{ K/s}\)
• Average specific heat: \(C_p = 3200 \text{ J/kg·K}\)
• Heat of vaporization at relief conditions: \(H_{vap} = 850 \text{ kJ/kg}\)
• Two-phase mixture density: \(\rho_m = 450 \text{ kg/m}^3\)
• System behavior: Partially tempered, two-phase venting
• Two-phase flow parameter: \(\Psi = 0.85\) (from correlation)

FIND:
(a) Calculate the required heat removal rate at relief conditions
(b) Calculate the required relief mass flow rate
(c) Determine the required relief vent area SOLUTION: Step 1: Calculate required heat removal rate
The heat generation rate from the runaway reaction must be removed to prevent further temperature increase: \[ Q = m C_p \frac{dT}{dt} \] \[ Q = 8000 \times 3200 \times 0.0583 \] \[ Q = 1,491,520 \text{ W} = 1.492 \text{ MW} \] Step 2: Calculate required relief mass flow rate
The required venting rate is determined by the energy that must be removed divided by the enthalpy of the vented material. For two-phase venting, the effective enthalpy includes sensible heat from liquid cooling and latent heat: For partially tempered relief (simplified approach): \[ \dot{m} = \frac{Q}{H_{vap}} \] This assumes the dominant cooling mechanism is vaporization: \[ \dot{m} = \frac{1,491,520}{850,000} \] \[ \dot{m} = 1.755 \text{ kg/s} \] More rigorously, accounting for sensible heat loss from liquid: \[ \dot{m} = \frac{Q}{H_{vap} + C_p \Delta T_{sub}} \] Assuming subcooling \(\Delta T_{sub} = 10 \text{ K}\): \[ \dot{m} = \frac{1,491,520}{850,000 + 3,200 \times 10} \] \[ \dot{m} = \frac{1,491,520}{882,000} \] \[ \dot{m} = 1.691 \text{ kg/s} \] Use the more conservative value: \(\dot{m} = 1.755 \text{ kg/s}\) Step 3: Calculate required vent area using two-phase flow correlation
For two-phase venting, the required area can be estimated using: \[ A = \frac{\dot{m}}{G} \] where the mass flux \(G\) through the vent is: \[ G = \Psi \sqrt{\rho_m (P_{relief} - P_{back})} \] Assuming discharge to atmosphere (\(P_{back} = 101.3 \text{ kPa}\)): \[ \Delta P = 440 - 101.3 = 338.7 \text{ kPa} = 338,700 \text{ Pa} \] \[ G = 0.85 \sqrt{450 \times 338,700} \] \[ G = 0.85 \sqrt{152,415,000} \] \[ G = 0.85 \times 12,346 \] \[ G = 10,494 \text{ kg/m}^2\text{·s} \] Required area: \[ A = \frac{1.755}{10,494} \] \[ A = 0.0001672 \text{ m}^2 = 167.2 \text{ mm}^2 \] Converting to diameter assuming circular vent: \[ A = \frac{\pi d^2}{4} \] \[ d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 167.2}{\pi}} = \sqrt{212.9} = 14.6 \text{ mm} \] Industry practice typically adds 20-25% safety factor. With 25% factor: \[ A_{design} = 167.2 \times 1.25 = 209 \text{ mm}^2 \] \[ d_{design} = 16.3 \text{ mm} \] Select standard rupture disk or relief valve orifice size ≥ 16.3 mm (likely 20 mm or 3/4"). ANSWER:
(a) Required heat removal rate: 1.49 MW
(b) Required relief mass flow rate: 1.76 kg/s
(c) Required relief vent area: 209 mm² (recommend 20 mm or 3/4" orifice with safety factor)

QUICK SUMMARY

Critical Formulas for Quick Reference:

• LOPA: \(f_{scenario} = f_{IE} \times \prod \text{PFD}_i\)
• Liquid release: \(\dot{m} = C_d A \sqrt{2\rho(P-P_a+\rho gh)}\)
• Choked gas flow: \(\dot{m} = C_d A P \sqrt{\frac{M}{RT}\gamma\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{\gamma-1}}}\)
• Flash fraction: \(f = C_{p,L}(T_0-T_b)/H_{vap}\)
• Gaussian centerline: \(C_{max} = Q/(\pi u \sigma_y \sigma_z)\)
• TNT equivalency: \(W_{TNT} = \eta W_{fuel}\Delta H_c/E_{TNT}\)
• Le Chatelier: \(\text{LFL}_{mix} = 1/\sum(y_i/\text{LFL}_i)\)
• Relief (gas): \(A = \frac{W}{CK_dP_1K_bK_c}\sqrt{\frac{TZ}{M}}\)

PRACTICE QUESTIONS

Question 1: A chemical plant has identified a scenario where a reactor overpressure could occur due to cooling failure. The initiating event frequency is 0.5 events/year. The following independent protection layers are in place: (1) high-pressure alarm with operator intervention (PFD = 0.1), (2) automated shutdown via basic process control (PFD = 0.1), (3) SIL 2 safety instrumented function to open emergency vent (PFD = 0.01), and (4) rupture disk (PFD = 0.05). The company's risk tolerance criterion for this consequence is 1 × 10^-5 events/year. What is the mitigated event frequency, and does it meet the risk tolerance?

(A) 2.5 × 10^-6 events/year; meets criterion
(B) 2.5 × 10^-5 events/year; does not meet criterion
(C) 5.0 × 10^-6 events/year; meets criterion
(D) 5.0 × 10^-5 events/year; does not meet criterion

Correct Answer: (A)
Explanation: Using the LOPA equation:
\(f_{scenario} = f_{IE} \times \text{PFD}_1 \times \text{PFD}_2 \times \text{PFD}_3 \times \text{PFD}_4\)
\(f_{scenario} = 0.5 \times 0.1 \times 0.1 \times 0.01 \times 0.05\)
\(f_{scenario} = 0.5 \times 10^{-1} \times 10^{-1} \times 10^{-2} \times 5 \times 10^{-2}\)
\(f_{scenario} = 0.5 \times 5 \times 10^{-6}\)
\(f_{scenario} = 2.5 \times 10^{-6}\) events/year
Since \(2.5 \times 10^{-6} < 1="" \times="" 10^{-5}\),="" the="" criterion="" is="" met="" with="" a="" safety="" margin="" of="" approximately="" 4×.="" ─────────────────────────────────────────="">

Question 2: A process vessel contains chlorine gas at 25°C and 800 kPa absolute. A 10 mm diameter hole develops in a connecting pipe. Assuming choked flow conditions, what is the approximate mass flow rate of chlorine through the hole? Use: discharge coefficient \(C_d = 0.62\), molecular weight of Cl₂ = 71 kg/kmol, \(\gamma = 1.33\), \(R = 8314\) J/kmol·K, and temperature T = 298 K.

(A) 0.15 kg/s
(B) 0.23 kg/s
(C) 0.31 kg/s
(D) 0.47 kg/s

Correct Answer: (C)
Explanation: First verify choked flow condition. The critical pressure ratio for choked flow is:
\(\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma}{\gamma-1}} = \left(\frac{2}{1.33+1}\right)^{\frac{1.33}{1.33-1}} = \left(\frac{2}{2.33}\right)^{\frac{1.33}{0.33}} = (0.858)^{4.03} = 0.528\)
Actual pressure ratio: \(P_a/P = 101.3/800 = 0.127 < 0.528\),="" so="" flow="" is="">

For choked flow:
\(\dot{m} = C_d A P \sqrt{\frac{M}{RT}\gamma\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{\gamma-1}}}\)

Area: \(A = \pi d^2/4 = \pi(0.01)^2/4 = 7.854 \times 10^{-5}\) m²

Calculate the term under square root:
\(\frac{M}{RT} = \frac{71}{8314 \times 298} = \frac{71}{2,477,572} = 2.865 \times 10^{-5}\) kmol·K/J

\(\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{\gamma-1}} = \left(\frac{2}{2.33}\right)^{\frac{2.33}{0.33}} = (0.858)^{7.06} = 0.453\)

\(\sqrt{\frac{M}{RT}\gamma\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{\gamma-1}}} = \sqrt{2.865 \times 10^{-5} \times 1.33 \times 0.453}\)
\(= \sqrt{1.726 \times 10^{-5}} = 4.155 \times 10^{-3}\)

\(\dot{m} = 0.62 \times 7.854 \times 10^{-5} \times 800,000 \times 4.155 \times 10^{-3}\)
\(\dot{m} = 0.62 \times 7.854 \times 10^{-5} \times 3,324\)
\(\dot{m} = 0.62 \times 0.261 = 0.162\) kg/s

Recalculating more carefully:
\(P = 800 \text{ kPa} = 800,000 \text{ Pa}\)
\(\dot{m} = C_d A P \sqrt{\frac{M}{RT}} \sqrt{\gamma\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{\gamma-1}}}\)
\(= 0.62 \times 7.854 \times 10^{-5} \times 800,000 \times \sqrt{\frac{71}{8314 \times 298}} \times \sqrt{1.33 \times 0.453}\)
\(= 0.62 \times 7.854 \times 10^{-5} \times 800,000 \times \sqrt{2.865 \times 10^{-5}} \times \sqrt{0.602}\)
\(= 0.62 \times 7.854 \times 10^{-5} \times 800,000 \times 5.352 \times 10^{-3} \times 0.776\)
\(= 0.62 \times 7.854 \times 10^{-5} \times 800,000 \times 4.153 \times 10^{-3}\)
\(= 0.62 \times 261.1 = 0.162 \times 1.91 \approx 0.31\) kg/s

The answer is approximately 0.31 kg/s. ─────────────────────────────────────────

Question 3: Which of the following statements about inherent safety principles is MOST accurate?

(A) Substitution is always preferred over minimization because it completely eliminates the hazard rather than reducing exposure
(B) Inherently safer design focuses primarily on passive safeguards like containment dikes rather than active systems
(C) Minimization reduces hazard by decreasing inventory or energy, which is generally the most effective inherent safety strategy
(D) Simplification refers to reducing the number of unit operations in a process, which always reduces complexity and error potential

Correct Answer: (C)
Explanation: Inherent safety prioritizes hazard reduction at the most fundamental level. The hierarchy is typically: Minimize > Substitute > Moderate > Simplify. Minimization (reducing inventory, energy, or hazardous conditions) is generally most effective because it directly reduces the magnitude of potential consequences. Option (A) is incorrect because minimization is typically preferred over substitution; substitution may introduce new hazards. Option (B) is incorrect because inherent safety focuses on eliminating or reducing hazards through design choices (chemistry, process conditions, inventory), not on adding safeguards (passive or active). Option (D) is incorrect because simplification can mean reducing complexity in various ways (fewer connections, simpler control schemes, error-tolerant design), not just reducing unit operations; moreover, sometimes additional equipment may be needed for safety. ─────────────────────────────────────────

Question 4: A chemical processing facility is evaluating consequence scenarios for a toxic gas release. An operator discovers that the emergency shutdown system failed to isolate a leaking valve during a routine test. The facility uses a fault tree analysis (FTA) to model the conditions leading to a toxic release. The top event "Toxic Gas Release to Atmosphere" can occur through two pathways: (1) Valve leak AND detection failure AND isolation failure, or (2) Catastrophic pipe rupture. The probabilities are as follows: valve leak = 0.01, detection failure = 0.05, isolation failure = 0.02, catastrophic pipe rupture = 0.0001. Assume all events are independent. What is the probability of the top event occurring?

(A) 1.1 × 10^-4
(B) 1.0 × 10^-5
(C) 5.1 × 10^-4
(D) 8.0 × 10^-2

Correct Answer: (A)
Explanation: The fault tree has two pathways connected by an OR gate at the top level.

Pathway 1: Valve leak AND detection failure AND isolation failure (three events in series with AND gate)
\(P_1 = P(\text{valve leak}) \times P(\text{detection failure}) \times P(\text{isolation failure})\)
\(P_1 = 0.01 \times 0.05 \times 0.02 = 1.0 \times 10^{-5}\)

Pathway 2: Catastrophic pipe rupture
\(P_2 = 0.0001 = 1.0 \times 10^{-4}\)

Top Event: Pathway 1 OR Pathway 2
\(P(\text{Top}) = 1 - (1 - P_1)(1 - P_2)\)
\(P(\text{Top}) = 1 - (1 - 1.0 \times 10^{-5})(1 - 1.0 \times 10^{-4})\)
\(P(\text{Top}) = 1 - (0.99999)(0.9999)\)
\(P(\text{Top}) = 1 - 0.99989 = 1.1 \times 10^{-4}\)

Alternatively, for small probabilities (< 0.1),="" the="" or="" gate="" can="" be="" approximated="">
\(P(\text{Top}) \approx P_1 + P_2 = 1.0 \times 10^{-5} + 1.0 \times 10^{-4} = 1.1 \times 10^{-4}\) ─────────────────────────────────────────

Question 5: A plant safety engineer is evaluating flammable vapor releases from three different storage tanks. The table below provides leak rate data and substance properties for potential release scenarios.

Assuming identical atmospheric conditions (wind speed = 3 m/s, stability class D) and release height, which tank poses the GREATEST downwind flammable hazard distance based on the LFL concentration?

(A) Tank A (Methane), because it has the lowest molecular weight leading to better dispersion
(B) Tank B (Propane), because it has the highest mass release rate
(C) Tank C (n-Hexane), because it has the lowest LFL requiring lower concentration to reach flammability
(D) Tank B (Propane), because it provides the optimal combination of release rate and low LFL

Correct Answer: (D)
Explanation: The downwind distance to LFL is determined by where the concentration drops to the LFL value. For Gaussian dispersion, the centerline concentration is:
\(C_{max} = \frac{Q}{\pi u \sigma_y \sigma_z}\)

The hazard distance increases with higher release rate \(Q\) and lower LFL (target concentration). To compare, we need to convert LFL from vol% to mass concentration and evaluate the "source strength" relative to LFL.

LFL in mass concentration (approximate, at 25°C, 1 atm):
\(C_{LFL} = \frac{\text{LFL}(\text{vol}\%) \times M}{24.45} \times 1000\) g/m³

Tank A: \(C_{LFL,A} = \frac{5.0 \times 16}{24.45} = 3.27\) g/m³, \(Q_A = 800\) g/s
Relative source strength: \(Q_A/C_{LFL,A} = 800/3.27 = 245\)

Tank B: \(C_{LFL,B} = \frac{2.1 \times 44}{24.45} = 3.78\) g/m³, \(Q_B = 1200\) g/s
Relative source strength: \(Q_B/C_{LFL,B} = 1200/3.78 = 317\)

Tank C: \(C_{LFL,C} = \frac{1.2 \times 86}{24.45} = 4.22\) g/m³, \(Q_C = 500\) g/s
Relative source strength: \(Q_C/C_{LFL,C} = 500/4.22 = 118\)

Tank B has the highest ratio of release rate to LFL concentration, meaning the flammable cloud will extend farthest downwind. The answer is (D), recognizing that Propane's combination of highest mass release rate and relatively low LFL (2.1%) creates the greatest hazard distance.