Let's solve the problem step by step: First, note the given data: - Emission rate (Q): .05 lb/day of benzene - Stack height (H): 75 ft - Cancer risk target (LCP): 1 × 10⁻⁶ - The risk must be kept below the LCP at the fenceline by choosing the appropriate distance (x). **Step 1: Convert emission rate to grams per second (g/s)** 1 lb = 453.6 g, 1 day = 86400 s So, \( Q = .05\, \text{lb/day} \times \frac{453.6\, \text{g}}{1\, \text{lb}} \times \frac{1}{86400\, \text{s}} = .000263\, \text{g/s} \) **Step 2: Use the Unit Risk Factor (URF) for benzene** Typical URF for benzene (inhalation) = \( 2.2 \times 10^{-6} \text{ per } \mu\text{g/m}^3 \) The risk is: \[ \text{Risk} = \text{Concentration} \times \text{URF} \] Set risk = \( 1 \times 10^{-6} \): \[ \text{Concentration} = \frac{1 \times 10^{-6}}{2.2 \times 10^{-6}} = .4545\, \mu\text{g/m}^3 \] **Step 3: Estimate ground-level concentration at distance x from a stack** For a continuous point source stack, ground-level centerline concentration: \[ C(x) = \frac{Q}{\pi \sigma_y \sigma_z u} \] Where: - \( Q \) = emission rate (g/s) - \( \sigma_y, \sigma_z \) = dispersion coefficients (m) - \( u \) = wind speed (m/s) Assume a typical wind speed \( u = 2 \, \text{m/s} \). Assume neutral atmosphere, use Pasquill-Gifford equations for dispersion coefficients: \[ \sigma_y = .08x (1 + .0002x)^{-.5} \] \[ \sigma_z = .06x (1 + .0015x)^{-.5} \] Stack height H = 75 ft = 22.86 m. For maximum ground-level concentration, the downwind distance where the plume touches the ground is approximately: \[ x = (H / .06)^{.625} \] But for a conservative estimate, let's solve numerically. Let C(x) = .4545 μg/m³ = .0004545 mg/m³ = .0004545 g/m³ So, \[ .0004545 = \frac{.000263}{\pi \sigma_y \sigma_z \cdot 2} \] \[ \pi \sigma_y \sigma_z \cdot 2 = \frac{.000263}{.0004545} \] \[ \pi \sigma_y \sigma_z \cdot 2 = .5786 \] \[ \sigma_y \sigma_z = \frac{.5786}{2\pi} = .0921 \] Assume \( \sigma_y \approx .08x \) and \( \sigma_z \approx .06x \) for small x. So, \[ \sigma_y \sigma_z \approx .08x \cdot .06x = .0048x^2 \] Set \( .0048x^2 = .0921 \) \[ x^2 = \frac{.0921}{.0048} = 19.19 \] \[ x = \sqrt{19.19} = 4.38 \, \text{m} \] However, this value is *very* small and unrealistic due to simplifications. Usually, with a 75-ft stack, the required distance is in hundreds of meters. Let's check with a more typical estimate. Using EPA's screening formula for ground-level concentration: \[ C = \frac{Q}{uH^2} \] So, \[ C = \frac{.000263}{2 \times (22.86)^2} = \frac{.000263}{2 \times 522.9} = \frac{.000263}{1045.8} = 2.52 \times 10^{-7} \text{g/m}^3 \] Convert to μg/m³: \( 2.52 \times 10^{-7} \times 10^{6} = .252 \mu\text{g/m}^3 \) This value is already below the .4545 μg/m³ needed for 1 × 10⁻⁶ risk, which means that at the minimum downwind distance close to the stack (fenceline), risk is already below the threshold. **Final Answer:** Given the provided emission rate and stack height, the cancer risk at the fenceline is already below 1 × 10⁻⁶ for typical meteorological conditions, even at relatively short distances from the stack. Therefore, the fenceline can be located as close as about 23 meters (75 ft) from the stack and still meet the risk goal. If you want a more conservative buffer, increasing this distance to 50–100 meters would further reduce risk, but with the given emission rate and stack height, the requirement is already satisfied at the minimum distance.