Let's solve the problem **step-by-step**: --- ## **Given Data** - Stack height (\( h_s \)) = 80 m - SO₂ concentration in stack gas = 4 g/m³ - Volumetric flow rate (\( Q \)) = 25 m³/s - Stack diameter (\( d \)) = 2 m - Wind speed at stack (\( u \)) = 4 m/s - Exit gas temperature (\( T_s \)) = 120°C - Ambient temperature (\( T_a \)) = 30°C - Pressure (\( P \)) = 750 mmHg = 100,000 Pa (approx, for calculations) - Distance for part (b) (\( x \)) = 1 km = 100 m --- ## **(a) Effective Stack Height (\( H \))** The **effective stack height** is: \[ H = h_s + \Delta h \] where \( \Delta h \) is the **plume rise**. For a **momentum-dominated plume** (assume since exit velocity is high): ### **Momentum Plume Rise Formula (Briggs, for rural, neutral conditions):** \[ \Delta h = 1.6 \left( \dfrac{u_s d}{u} \right) \] Where: - \( u_s \) = exit velocity of stack gas - \( d \) = stack diameter - \( u \) = wind speed at stack height #### **Calculate exit velocity \( u_s \):** \[ u_s = \frac{Q}{A} = \frac{Q}{\frac{\pi d^2}{4}} \] \[ A = \frac{\pi (2)^2}{4} = \pi \text{ m}^2 \] \[ u_s = \frac{25}{\pi} = 7.96 \text{ m/s} \] #### **Calculate plume rise \( \Delta h \):** \[ \Delta h = 1.6 \left( \frac{7.96 \times 2}{4} \right) = 1.6 \times 3.98 = 6.37 \text{ m} \] #### **Effective stack height:** \[ H = 80 + 6.37 = \boxed{86.4~\text{m}} \] --- ## **(b) Maximum Ground Level Concentration at 1 km** ### **Gaussian Plume Equation (at ground, centerline, \( y=, z= \)):** \[ C(x,,) = \frac{Q_{SO_2}}{2\pi u \sigma_y \sigma_z} \exp\left(-\frac{H^2}{2\sigma_z^2}\right) \] #### **Calculate \( Q_{SO_2} \):** \[ Q_{SO_2} = (\text{Concentration}) \times (\text{Flow rate}) = 4~\text{g/m}^3 \times 25~\text{m}^3/\text{s} = 100~\text{g/s} \] #### **Stability Class** Sunny summer day: **Class B** (slightly unstable) #### **For \( x = 100~\text{m} \), from Pasquill-Gifford curves:** \[ \sigma_y = .16x (1 + .0004x)^{-.5} \] \[ \sigma_z = .12x (1 + .0003x)^{-.5} \] \[ \sigma_y = .16 \times 100 (1 + .0004 \times 100)^{-.5} = 160 (1.4)^{-.5} = 160 / 1.183 = 135.25~\text{m} \] \[ \sigma_z = .12 \times 100 (1 + .0003 \times 100)^{-.5} = 120 (1.3)^{-.5} = 120 / 1.140 = 105.3~\text{m} \] #### **Plug into Gaussian equation:** \[ C = \frac{100}{2\pi \times 4 \times 135.25 \times 105.3} \exp\left(-\frac{(86.4)^2}{2 \times (105.3)^2}\right) \] Calculate exponent: \[ \frac{(86.4)^2}{2 \times (105.3)^2} = \frac{7465}{22180} = .336 \] \[ \exp(-.336) = .714 \] Calculate denominator: \[ 2\pi \times 4 \times 135.25 \times 105.3 \approx 2 \times 3.1416 \times 4 \times 135.25 \times 105.3 \approx 6.2832 \times 4 \times 135.25 \times 105.3 \] \[ = 25.1328 \times 135.25 = 3401.52 \] \[ 3401.52 \times 105.3 = 358204 \] So, \[ C = \frac{100}{358204} \times .714 = .000199 \times .714 = .000142~\text{g/m}^3 \] \[ C = \boxed{1.42 \times 10^{-4}~\text{g/m}^3} \] --- ## **(c) Distance for Maximum Ground Concentration** For ground level, centerline, max occurs when: \[ \frac{H}{\sigma_z} = 1.6 \implies \sigma_z = \frac{H}{1.6} \] \[ \sigma_z = \frac{86.4}{1.6} = 54~\text{m} \] From earlier: \[ \sigma_z = .12x(1+.0003x)^{-.5} \] Set equal and solve for \( x \): Let’s try iteratively or estimate \( x \): \[ 54 = .12x (1+.0003x)^{-.5} \] Let \( y = (1+.0003x)^{.5} \) \[ 54y = .12x \implies x = \frac{54y}{.12} = 450y \] But \( y = (1+.0003x)^{.5} \), so: \[ y = (1+.0003 \times 450y)^{.5} \implies y^2 = 1 + .135y \implies y^2 - .135y - 1 = \] Quadratic formula: \[ y = \frac{.135 \pm \sqrt{.135^2 + 4 \times 1}}{2} = \frac{.135 \pm \sqrt{.0182 + 4}}{2} = \frac{.135 \pm 2.006}{2} \] Take positive root: \[ y = \frac{.135 + 2.006}{2} = 1.07 \] \[ x = 450 \times 1.07 = 480~\text{m} \] \[ \boxed{480~\text{m}} \] --- ## **(d) Concentration at \( x = 480~\text{m} \)** Calculate \( \sigma_y \) and \( \sigma_z \): \[ \sigma_y = .16 \times 480 \times (1+.0004 \times 480)^{-.5} = 76.8 \times (1.192)^{-.5} = 76.8 / 1.092 = 70.4~\text{m} \] \[ \sigma_z = .12 \times 480 \times (1+.0003 \times 480)^{-.5} = 57.6 \times (1.144)^{-.5} = 57.6 / 1.07 = 53.8~\text{m} \] \[ C = \frac{100}{2\pi \times 4 \times 70.4 \times 53.8} \exp\left(-\frac{(86.4)^2}{2 \times (53.8)^2}\right) \] Exponent: \[ \frac{7465}{2 \times 2897} = \frac{7465}{5794} = 1.29 \] \[ \exp(-1.29) = .275 \] Denominator: \[ 2\pi \times 4 \times 70.4 \times 53.8 = 25.1328 \times 70.4 \times 53.8 = 1768.9 \times 53.8 = 95199 \] So, \[ C = \frac{100}{95199} \times .275 = .00105 \times .275 = .000289~\text{g/m}^3 \] \[ C = \boxed{2.89 \times 10^{-4}~\text{g/m}^3} \] --- ## **Summary Table** | Part | Answer | |------|------------------------------------| | (a) | Effective stack height: **86.4 m** | | (b) | Max ground conc. at 1 km: **1.42 × 10⁻⁴ g/m³** | | (c) | Distance of max ground conc.: **480 m** | | (d) | Concentration at (c): **2.89 × 10⁻⁴ g/m³** | --- **If you need the units in mg/m³, just multiply by 100!** Let me know if you'd like further explanation for any step.