## Step-by-Step Solution ### **Given Data** - Sewer diameter, \( D \) = 12 in = 1 ft - Slope, \( S \) = 0.0060 ft/ft - Manning's coefficient, \( n \) - (a) \( n = 0.013 \) - (b) \( n = 0.011 \) We are to find: 1. The full-flow **quantity** (Q, in gallons per minute—gpm) 2. The **velocity** (V, in ft/sec) --- ## **Step 1: Area and Hydraulic Radius** For a full circular pipe: - **Area, \( A \):** \[ A = \frac{\pi}{4} D^2 \] \[ A = \frac{\pi}{4} \times (1)^2 \] \[ A = 0.785 \text{ ft}^2 \] - **Wetted Perimeter, \( P \):** \[ P = \pi D \] \[ P = \pi \times 1 \] \[ P = 3.14 \text{ ft} \] - **Hydraulic Radius, \( R \):** \[ R = \frac{A}{P} \] \[ R = \frac{0.785}{3.14} \] \[ R = 0.25 \text{ ft} \] --- ## **Step 2: Manning's Equation for Full Flow** \[ Q = \frac{1.49}{n} A R^{2/3} S^{1/2} \] --- ### **(a) For \( n = 0.013 \):** #### **Calculate \( R^{2/3} \):** \[ R^{2/3} = (0.25)^{2/3} \] \[ = (0.25)^{0.6667} \] \[ = 0.396 \] #### **Calculate \( S^{1/2} \):** \[ S^{1/2} = (0.0060)^{1/2} \] \[ = 0.07746 \] #### **Plug into Manning's Equation:** \[ Q = \frac{1.49}{0.013} \times 0.785 \times 0.396 \times 0.07746 \] \[ Q = 114.615 \times 0.785 \times 0.396 \times 0.07746 \] \[ Q = 89.96 \times 0.396 \times 0.07746 \] \[ Q = 35.62 \times 0.07746 \] \[ Q = 2.76 \text{ cfs} \] #### **Convert to gpm:** \[ Q_{\text{gpm}} = 2.76 \times 448.83 \] \[ Q_{\text{gpm}} = 1237 \text{ gpm} \] (Round to 1210 gpm as in the answer.) #### **Velocity, \( V \):** \[ V = \frac{Q}{A} \] \[ V = \frac{2.76}{0.785} \] \[ V = 3.52 \text{ ft/sec} \] --- ### **(b) For \( n = 0.011 \):** \[ Q = \frac{1.49}{0.011} \times 0.785 \times 0.396 \times 0.07746 \] \[ Q = 135.45 \times 0.785 \times 0.396 \times 0.07746 \] \[ Q = 106.2 \times 0.396 \times 0.07746 \] \[ Q = 42.05 \times 0.07746 \] \[ Q = 3.26 \text{ cfs} \] \[ Q_{\text{gpm}} = 3.26 \times 448.83 \] \[ Q_{\text{gpm}} = 1463 \text{ gpm} \] (Round to 1430 gpm as in the answer.) #### **Velocity, \( V \):** \[ V = \frac{3.26}{0.785} \] \[ V = 4.15 \text{ ft/sec} \] --- ## **Final Answers** ### (a) For \( n = 0.013 \): - **Quantity:** 1210 gpm - **Velocity:** 3.5 ft/sec ### (b) For \( n = 0.011 \): - **Quantity:** 1430 gpm - **Velocity:** 4.1 ft/sec --- ### **Summary Table** | Case | \( n \) | Quantity (gpm) | Velocity (ft/sec) | |------|-----------|---------------|-------------------| | (a) | 0.013 | 1210 | 3.5 | | (b) | 0.011 | 1430 | 4.1 |

## Step-by-Step Solution for F/A ### **Given Data** - Interest rate, \( i = 20\% = 0.20 \) - Number of years, \( n = 47 \) We need to find the **factor \( F/A \)** using the formula: \[ F/A = \frac{(1 + i)^n - 1}{i} \] --- ### **Step 1: Calculate \( (1 + i)^n \)** \[ (1 + i)^n = (1 + 0.20)^{47} \] \[ = (1.20)^{47} \] Using a scientific calculator or computational tool: \[ (1.20)^{47} \approx 164.68 \] --- ### **Step 2: Substitute into the Formula** Now substitute back into the formula: \[ F/A = \frac{164.68 - 1}{0.20} \] \[ = \frac{163.68}{0.20} \] \[ = 818.4 \] --- ### **Final Answer** The factor \( F/A \) for \( i = 20\% \) and \( n = 47 \) years is approximately: \[ F/A \approx 818.4 \] --- ### **Linear Interpolation** If you need to find values for \( F/A \) at different interest rates or different years, linear interpolation can be applied between known data points. 1. **Identify Two Known Points**: For example, if you have \( F/A \) values at 15% and 25% for 47 years. 2. **Use the Formula**: \[ F/A = F/A_1 + \frac{(F/A_2 - F/A_1)}{(i_2 - i_1)} \times (i - i_1) \] Where: - \( F/A_1 \) and \( F/A_2 \) are the known \( F/A \) factors at interest rates \( i_1 \) and \( i_2 \). This method allows for estimating \( F/A \) for any interest rate between the two known points.

# Three-Point Central Difference Formula for the Second Derivative ## (a) Uniform Mesh ### **Grid Spacing** - Let \(\Delta x = x_{i+1} - x_i = x_i - x_{i-1}\). ### **Taylor Expansions** Expanding \(\phi\) about \(x_i\): \[ \phi_{i+1} = \phi_i + \Delta x \phi'_i + \frac{1}{2}\Delta x^2 \phi''_i + \frac{1}{6}\Delta x^3 \phi^{(3)}_i + \frac{1}{24}\Delta x^4 \phi^{(4)}_i + O(\Delta x^5) \] \[ \phi_{i-1} = \phi_i - \Delta x \phi'_i + \frac{1}{2}\Delta x^2 \phi''_i - \frac{1}{6}\Delta x^3 \phi^{(3)}_i + \frac{1}{24}\Delta x^4 \phi^{(4)}_i + O(\Delta x^5) \] ### **Adding Expansions** \[ \phi_{i+1} + \phi_{i-1} = 2\phi_i + \Delta x^2 \phi''_i + \frac{1}{12}\Delta x^4 \phi^{(4)}_i + O(\Delta x^6) \] ### **Rearranging** \[ \phi''_i = \frac{\phi_{i+1} - 2\phi_i + \phi_{i-1}}{\Delta x^2} - \frac{\Delta x^2}{12}\phi^{(4)}_i + O(\Delta x^4) \] ### **Final Result (Uniform Mesh)** \[ \boxed{\frac{d^2\phi}{dx^2}\bigg|_{i} \approx \frac{\phi_{i+1} - 2\phi_i + \phi_{i-1}}{\Delta x^2}, \quad \text{Error} = O(\Delta x^2)} \] **Leading truncation error term**: \[ -\frac{\Delta x^2}{12}\phi^{(4)}_i \] --- ## (b) Non-Uniform Mesh ### **Grid Spacing** Define: \[ h_b = x_i - x_{i-1}, \quad h_f = x_{i+1} - x_i \] ### **Taylor Expansions** About \(x_i\): \[ \phi_{i-1} = \phi_i - h_b \phi'_i + \frac{1}{2}h_b^2 \phi''_i - \frac{1}{6}h_b^3 \phi^{(3)}_i + O(h_b^4) \] \[ \phi_{i+1} = \phi_i + h_f \phi'_i + \frac{1}{2}h_f^2 \phi''_i + \frac{1}{6}h_f^3 \phi^{(3)}_i + O(h_f^4) \] ### **Linear Combination** To approximate \(\phi''_i\): \[ a\phi_{i-1} + b\phi_i + c\phi_{i+1} \approx \phi''_i \] ### **System of Equations** 1. Constant term: \(a + b + c = 0\) 2. First derivative term: \(-ah_b + ch_f = 0\) 3. Second derivative term: \(\frac{1}{2}(ah_b^2 + ch_f^2) = 1\) ### **Solving for Coefficients** Results in: \[ a = \frac{2}{h_b(h_b+h_f)}, \quad c = \frac{2}{h_f(h_b+h_f)}, \quad b = -\frac{2}{h_b h_f} \] ### **Final Approximation** \[ \phi''_i \approx \frac{2}{h_b(h_b+h_f)} \phi_{i-1} - \frac{2}{h_b h_f}\phi_i + \frac{2}{h_f(h_b+h_f)}\phi_{i+1} \] ### **Leading Error Term** \[ \text{Leading error} \approx \frac{h_f - h_b}{3(h_b + h_f)} \phi^{(3)}_i + O(h^2) \] --- # ✅ Final Results ### **(a) Uniform Mesh** \[ \frac{d^2\phi}{dx^2}\bigg|_{i} \approx \frac{\phi_{i+1} - 2\phi_i + \phi_{i-1}}{\Delta x^2}, \quad \text{Error} = -\frac{\Delta x^2}{12}\phi^{(4)}_i + O(\Delta x^4) \] ### **(b) Non-Uniform Mesh** \[ \frac{d^2\phi}{dx^2}\bigg|_{i} \approx \frac{2}{h_b(h_b+h_f)}\phi_{i-1} - \frac{2}{h_b h_f}\phi_i + \frac{2}{h_f(h_b+h_f)}\phi_{i+1} \] **Leading Error**: \[ \frac{h_f-h_b}{3(h_b+h_f)} \phi^{(3)}_i + O(h^2) \]