Let's solve the problem step by step: --- ## **Given Data** - Stack height, \( H_s = 80) m - SO₂ emission, \( C_s = 4 \) g/m - Volumetric flow rate, \( Q = 25 \) m³/s - Stack diameter, \( d = 2 \) m - Wind speed at stack height, \( u = 4 \) m/s - Temperature of exit gas, \( T_s = 120^\circ C = 393 \) K - Ambient temperature, \( T_a = 30^\circ C = 303 \) K - Pressure, \( P = 750 \) mmHg (not directly needed for calculations) - Downwind distance (for part b), \( x = 1 \) km \( = 100 \) m --- ## **(a) Determine the Effective Stack Height** ### **Effective Stack Height Formula** The effective stack height is: \[ H = H_s + \Delta h \] Where: - \( H_s \): Physical stack height - \( \Delta h \): Plume rise For a **momentum-dominated plume** (Briggs equation, urban or rural, neutral conditions): \[ \Delta h = \frac{1.6 \, u_s \, d}{u} \] Where: - \( u_s \): Exit velocity of stack gas - \( d \): Stack diameter - \( u \): Wind speed at stack height #### **Calculate Exit Velocity (\( u_s \)):** \[ u_s = \frac{Q}{A} = \frac{Q}{\pi d^2/4} \] \[ A = \frac{\pi}{4} \times (2)^2 = \pi\, \text{m}^2 \] \[ u_s = \frac{25}{\pi} \approx 7.96 \text{ m/s} \] #### **Calculate Plume Rise (\( \Delta h \))** \[ \Delta h = \frac{1.6 \times 7.96 \times 2}{4} = \frac{25.472}{4} = 6.368 \text{ m} \] #### **Effective Stack Height** \[ H = 80 + 6.37 = 86.37 \text{ m} \] --- ## **(b) Maximum Ground Level Concentration at 1 km Downwind** ### **Step 1: Calculate SO₂ Emission Rate** \[ Q_{SO_2} = C_s \times Q = 4 \text{ g/m}^3 \times 25 \text{ m}^3/\text{s} = 100 \text{ g/s} \] ### **Step 2: Gaussian Plume Equation (for ground level, centerline, \( y = , z = \)):** \[ C(x, , ) = \frac{Q_{SO_2}}{2\pi u \sigma_y \sigma_z} \exp\left( -\frac{H^2}{2\sigma_z^2} \right) \] ### **Step 3: Calculate Dispersion Coefficients for \( x = 100 \) m** On a sunny summer day (unstable, "Class B"): - \( \sigma_y = .16x [1 + .0004x]^{-.5} \) - \( \sigma_z = .12x [1 + .0002x]^{-.5} \) \[ \sigma_y = .16 \times 100 \times [1 + .0004 \times 100]^{-.5} = 160 \times [1.4]^{-.5} \approx 160 \times .845 = 135.2 \text{ m} \] \[ \sigma_z = .12 \times 100 \times [1 + .0002 \times 100]^{-.5} = 120 \times [1.2]^{-.5} \approx 120 \times .9129 = 109.6 \text{ m} \] ### **Step 4: Plug Values into Gaussian Equation** \[ C(100, , ) = \frac{100}{2\pi \times 4 \times 135.2 \times 109.6} \exp\left( -\frac{(86.37)^2}{2 \times (109.6)^2} \right) \] First, calculate the denominator: \[ 2\pi \times 4 \times 135.2 \times 109.6 \approx 2 \times 3.1416 \times 4 \times 135.2 \times 109.6 \approx 6.2832 \times 4 \times 135.2 \times 109.6 \] \[ = 25.1328 \times 135.2 \times 109.6 = 3398.5 \times 109.6 = 372,529.2 \] Now, the exponent: \[ \left( \frac{86.37^2}{2 \times 109.6^2} \right) = \frac{746.5}{2 \times 12,017.2} = \frac{746.5}{24,034.4} = .3104 \] \[ \exp(-.3104) = .7334 \] Now, the full equation: \[ C(100,,) = \frac{100}{372,529.2} \times .7334 = .000266 \text{ g/m}^3 \] --- ## **(c) Distance for Maximum Ground Level Concentration** For ground-level centerline, the maximum occurs where \( \frac{H}{\sigma_z} \approx 1.6 \): \[ \sigma_z = \frac{H}{1.6} = \frac{86.37}{1.6} = 53.98 \text{ m} \] From the Class B formula: \[ \sigma_z = .12x[1 + .0002x]^{-.5} \] Set \( \sigma_z = 53.98 \): \[ 53.98 = .12x[1 + .0002x]^{-.5} \] \[ 53.98[1 + .0002x]^{.5} = .12x \] \[ \frac{53.98}{.12} [1 + .0002x]^{.5} = x \] \[ 449.83 [1 + .0002x]^{.5} = x \] Let \( y = [1 + .0002x]^{.5} \), so \( x = 449.83y \), and \( y^2 = 1 + .0002x = 1 + .0002 \times 449.83y \): \[ y^2 = 1 + .08997y \] \[ y^2 - .08997y - 1 = \] Quadratic formula: \( y = \frac{.08997 \pm \sqrt{(.08997)^2 + 4 \times 1}}{2} \) \[ = \frac{.08997 \pm \sqrt{.0081 + 4}}{2} = \frac{.08997 \pm \sqrt{4.0081}}{2} = \frac{.08997 \pm 2.002}{2} \] Choose positive root: \[ y = \frac{.08997 + 2.002}{2} = 1.04598 \] \[ x = 449.83 \times 1.04598 \approx 470 \text{ m} \] --- ## **(d) Concentration at Location from (c)** ### **Calculate \( \sigma_y \) and \( \sigma_z \) at \( x = 470 \) m** \[ \sigma_y = .16 \times 470 \times [1 + .0004 \times 470]^{-.5} \] \[ = 75.2 \times [1.188]^{-.5} = 75.2 \times .917 = 68.97 \text{ m} \] \[ \sigma_z = .12 \times 470 \times [1 + .0002 \times 470]^{-.5} \] \[ = 56.4 \times [1.094]^{-.5} = 56.4 \times .956 = 53.92 \text{ m} \] ### **Plug into Gaussian Equation** \[ C(470,,) = \frac{100}{2\pi \times 4 \times 68.97 \times 53.92} \exp\left( -\frac{(86.37)^2}{2 \times (53.92)^2} \right) \] Calculate denominator: \[ 2\pi \times 4 \times 68.97 \times 53.92 \approx 6.2832 \times 4 \times 68.97 \times 53.92 = 25.1328 \times 68.97 = 1,733.3 \times 53.92 = 93,456.3 \] Exponent: \[ \left( \frac{86.37^2}{2 \times 53.92^2} \right) = \frac{746.5}{2 \times 2,910.7} = \frac{746.5}{5,821.4} = 1.282 \] \[ \exp(-1.282) = .2773 \] Final value: \[ C(470,,) = \frac{100}{93,456.3} \times .2773 = .000297 \text{ g/m}^3 \] --- # **Summary of Answers** ### **(a) Effective stack height:** \[ \boxed{86.4 \text{ m}} \] ### **(b) Maximum ground level concentration at 1 km:** \[ \boxed{.00027 \text{ g/m}^3} \] ### **(c) Distance where maximum ground-level concentration occurs:** \[ \boxed{470 \text{ m}} \] ### **(d) Concentration at location in (c):** \[ \boxed{.00030 \text{ g/m}^3} \] --- **Let me know if you want to see the calculations in more detail, or if you need the answers in different units!**