11.2 Transient Purging and Air Distribution Geometry

Transient Purging and Air Distribution Geometry

Steady-state equations answer "how much air do I need to hold this concentration?" Transient equations answer the questions that come up in practice: how long until the space is safe to enter, and how long after a release before the concentration falls below the limit.

1. Transient Purging and Non-Steady-State Concentration Decay

When a chemical spill occurs, a batch process finishes, or an unventilated storage room must be cleared prior to worker entry, contaminant generation ceases (G = 0), and ventilation air purges the space over time. Concentration reduction follows first-order exponential decay kinetics.

   +-------------------------------------------------------------------------+
   |                  TRANSIENT EXPONENTIAL PURGE DECAY                      |
   +-------------------------------------------------------------------------+
   |                                                                         |
   |  Concentration (C)                                                      |
   |    ^                                                                    |
   |  C1|---*                                                                |
   |    |    \                                                               |
   |    |     \                                                              |
   |    |      \                                                             |
   |    |       '.                                                           |
   |    |         \_                                                         |
   |  C2|-----------'-._________                                             |
   |    +------------------------------------> Time (t)                      |
   |    0           Purge Time (Δt)                                          |
   |                                                                         |
   |  Master Equation:  ln(C1 / C2) = (Q × Δt × K') / V                      |
   |  Time to Purge:    Δt = (V / [Q × K']) × ln(C1 / C2)                    |
   +-------------------------------------------------------------------------+

Derivation of the Exponential Purge Equation

The instantaneous rate of change of contaminant mass in a room of volume V (extft³) with zero ongoing generation (G = 0) is governed by the differential mass balance:

$V \frac{dC}{dt} = -Q' \cdot C(t) = - (Q \cdot K') \cdot C(t)$

Where:

• V = Room volume (extft³)
• C(t) = Contaminant concentration at time t
• Q = Total volumetric airflow rate (extcfm)
• K' = Fractional mixing factor (K' = 1/K, where 0 < K' ≤ 1.0; K' = 1.0 represents ideal mixing, while K' = 0.2 corresponds to K = 5)
• Q' = Q · K' = Effective dilution airflow rate (extcfm)

Separating variables and integrating from initial concentration C1 at t = 0 to final concentration C2 at time Δ t:

$\int_{C_1}^{C_2} \frac{dC}{C} = -\frac{Q \cdot K'}{V} \int_{0}^{\Delta t} dt$

$\ln\left(\frac{C_2}{C_1}\right) = -\frac{Q \cdot K'}{V} \Delta t \quad \Longleftrightarrow \quad \mathbf{\ln\left(\frac{C_1}{C_2}\right) = \frac{Q \cdot K'}{V} \Delta t = \frac{Q \cdot \Delta t}{K \cdot V}}$

Master Equation for Purge Time (Δ t)

Solving explicitly for the time required (Δ t, in minutes) to purge a room from concentration C1 down to target concentration C2:

$\mathbf{\Delta t = \frac{V}{Q \cdot K'} \ln\left(\frac{C_1}{C_2}\right) = \frac{K \cdot V}{Q} \ln\left(\frac{C_1}{C_2}\right)}$

Air Changes per Hour (ACH) Relationship

The nominal room air change rate (N, in hr⁻¹) is defined as:

$N = \frac{Q \times 60}{V} \quad \Longleftrightarrow \quad \frac{Q}{V} = \frac{N}{60}$

Substituting N into the purge equation yields the purge time in hours (thr) or minutes (Δ t):

$\Delta t = \frac{60}{N \cdot K'} \ln\left(\frac{C_1}{C_2}\right) = \frac{60 \cdot K}{N} \ln\left(\frac{C_1}{C_2}\right) \quad [\text{minutes}]$

Non-Steady-State Concentration Build-Up (Continuous Generation)

If a continuous source is introduced into an initially clean room (C0 = 0) with constant generation G, the concentration build-up over time t approaches steady state exponentially:

$C(t) = \frac{G \times 10^6}{Q \cdot K'} \left(1 - e^{-\frac{Q \cdot K'}{V} t}\right) = C_{\text{steady-state}} \left(1 - e^{-\frac{Q \cdot K'}{V} t}\right)$

• At t = V/(Q · K') (1 effective time constant), C(t) = 63.2% of steady state.
• At t = 3 × V/(Q · K'), C(t) = 95.0% of steady state.
• At t = 4.6 × V/(Q · K'), C(t) = 99.0% of steady state.

---

2. Air Distribution Geometry and Supply/Exhaust Placement

The physical location of supply air diffusers and exhaust air grilles determines the flow streamlines and actual mixing efficiency across the workspace.

   +-------------------------------------------------------------------------+
   |                  SUPPLY AND EXHAUST AIR DISTRIBUTION                   |
   +-------------------------------------------------------------------------+
   |                                                                         |
   |  A. CORRECT DILUTION GEOMETRY (Sweeping Fresh Air):                     |
   |     Fresh supply air enters behind the worker, sweeps across the        |
   |     breathing zone, carries contaminant away from worker into exhaust.  |
   |                                                                         |
   |     [Supply Diffuser] ===> ( Worker ) ===> [ Source ] ===> [ Exhaust ]  |
   |          Fresh Air ----------> Clean ----------> Vapor --------> Waste  |
   |                                                                         |
   |  B. INCORRECT DILUTION GEOMETRY (Contaminant Pulled Over Worker):       |
   |     Fresh air enters opposite, pulling concentrated vapors directly     |
   |     through the worker's breathing zone before reaching the exhaust!    |
   |                                                                         |
   |     [Supply Diffuser] ===> [ Source ] ===> ( Worker ) ===> [ Exhaust ]  |
   |          Fresh Air ----------> Vapor --------> Exposure! -------> Waste |
   |                                                                         |
   |  C. SHORT-CIRCUITING FAULT:                                             |
   |     Supply and exhaust grilles placed adjacent on ceiling; fresh air    |
   |     bypasses room volume, leaving stagnant contaminant pools below.     |
   |                                                                         |
   |     [Supply Diffuser] ===== (Direct Bypass) =====> [Exhaust Grille]     |
   |     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~     |
   |                     [ Stagnant Contaminant Zone ]                       |
   +-------------------------------------------------------------------------+

Key Aerodynamic Distribution Rules

1. Sweeping Airflow Trajectory: Supply air must be introduced so that it passes through the worker's breathing zone before it reaches the contamination source, carrying vapors toward the exhaust register.
2. Exhaust Placement Relative to Vapor Density:
   • While solvent vapors diffuse in air as individual molecules (molecular weight effects are largely dominated by turbulent convection at low concentrations), heavy solvent evaporation pools near the floor benefit from low-level exhaust grilles (6--12 inches above the floor) to capture high-density boundary layers.
   • Buoyant thermal plumes (warm gases, steam) require high ceiling exhausts.
3. Elimination of Short-Circuiting: Supply inlets and exhaust registers must be separated across the room footprint to prevent clean supply air from immediately entering the exhaust register without mixing throughout the occupied zone.
4. Cross-Draft Minimization: Supply jet discharge velocities must be engineered to prevent high-velocity drafts (> 50 fpm) from impinging upon nearby local exhaust hoods or disrupting delicate chemical fume hood sashes.

---

3. Worked Step-by-Step Calculation Examples

Worked Example 10.1.1: Sizing Steady-State Dilution Airflow for Solvent Degreasing

Problem: A manual parts cleaning operation in an aircraft maintenance hangar evaporates 1.5 pints per hour of methyl ethyl ketone (MEK; 2-butanone).

• Molecular weight (MW) = 72.11 g/mol
• Specific gravity (SG) = 0.805
• Target airborne concentration (Ctarget) = 100 ppm (OSHA PEL is 200 ppm, but company safety policy enforces a 50% action level)
• Because worker proximity is moderate and room air distribution is average, the industrial hygienist specifies an empirical mixing factor K = 4.0.

Calculate the required steady-state dilution ventilation airflow rate (Q) in cubic feet per minute (extcfm).

Solution Steps:

1. Calculate the vapor generation rate (G) in CFM: $G = \frac{403 \cdot SG \cdot ER_{\text{pt/hr}}}{60 \cdot \text{MW}}$ $G = \frac{403 \times 0.805 \times 1.5}{60 \times 72.11} = \frac{486.6225}{4326.6} = 0.11247\text{ cfm of pure MEK vapor}$

2. Calculate required dilution airflow (Q): $Q = \frac{G \times 10^6}{C_{\text{target}}} \times K$ $Q = \frac{0.11247 \times 10^6}{100\text{ ppm}} \times 4.0 = 1124.7 \times 4.0 = 4498.8\text{ cfm} \approx 4500\text{ cfm}$

3. Direct single-step formula verification: $Q = \frac{6.72 \times 10^6 \cdot SG \cdot ER_{\text{pt/hr}} \cdot K}{\text{MW} \cdot C_{\text{target}}} = \frac{6.72 \times 10^6 \times 0.805 \times 1.5 \times 4.0}{72.11 \times 100}$ $Q = \frac{32,457,600}{7211} = 4501.1\text{ cfm} \approx 4500\text{ cfm}$

Result: The required dilution ventilation airflow rate is 4500 cfm.

---

Worked Example 10.1.2: Confined Space Purge Time Calculation

Problem: A sealed industrial storage vault with dimensions 20 ft × 30 ft × 10 ft (Volume V = 6000 ft³) contains an initial airborne toluene concentration of C1 = 800 ppm following the repair of a solvent piping leak. A portable explosion-proof ventilation blower delivering Q = 1200 cfm is connected to purge the space before entry.

• Target entry concentration (C2) = 20 ppm (ACGIH TLV-TWA)
• Due to internal structural racking and equipment baffling, the mixing factor is estimated at K = 3.0 (effective mixing fraction K' = 1/K = 0.3333).

Calculate the purge time (Δ t) in minutes required to safely reduce the airborne toluene concentration from 800 ppm to 20 ppm.

Solution Steps:

1. Identify parameters:

   • Room volume V = 20 × 30 × 10 = 6000 ft³
   • Airflow rate Q = 1200 cfm
   • K = 3.0 → K' = 1 / 3.0 = 0.3333
   • Initial concentration C1 = 800 ppm
   • Final concentration C2 = 20 ppm

2. Calculate the concentration ratio logarithm: $\frac{C_1}{C_2} = \frac{800}{20} = 40.0$ $\ln(40.0) = 3.6889$

3. Calculate purge time (Δ t): $\Delta t = \frac{V \cdot K}{Q} \ln\left(\frac{C_1}{C_2}\right) = \frac{6000 \times 3.0}{1200} \times 3.6889$ $\Delta t = \frac{18000}{1200} \times 3.6889 = 15.0 \times 3.6889 = 55.33\text{ minutes} \approx 55.3\text{ minutes}$

Result: The blower must operate for 55.3 minutes before workers may enter the space without air-supplied respirators.

---

Worked Example 10.1.3: Concentration Build-Up During Non-Steady-State Operation

Problem: A paint mixing room of volume V = 10,000 ft³ has a continuous dilution ventilation rate Q = 2000 cfm with a mixing factor K = 2.0 (K' = 0.50). An open dip tank begins continuously releasing acetone vapor at a generation rate G = 0.20 cfm. Assuming the room was initially clean (C0 = 0 ppm), calculate the acetone concentration in the room after 15 minutes of operation.

Solution Steps:

1. Calculate steady-state concentration (Css): $C_{\text{ss}} = \frac{G \times 10^6}{Q \cdot K'} = \frac{0.20 \times 10^6}{2000 \times 0.50} = \frac{200,000}{1000} = 200\text{ ppm}$

2. Calculate the exponential time constant term ((Q · K')/V t): $\frac{Q \cdot K'}{V} t = \frac{2000 \times 0.50}{10,000} \times 15 = \frac{1000}{10,000} \times 15 = 0.10 \times 15 = 1.50$

3. Calculate concentration at t = 15 min: $C(15) = C_{\text{ss}} \left(1 - e^{-1.50}\right)$ $e^{-1.50} = 0.2231$ $C(15) = 200 \times (1 - 0.2231) = 200 \times 0.7769 = 155.38\text{ ppm} \approx 155\text{ ppm}$

Result: After 15 minutes, the airborne concentration reaches 155 ppm (77.7% of the steady-state maximum of 200 ppm).