Make Sure That this is correct, because the answer in mannings equation is wrong. Detailed Solution for Open Channel Design (Rectangular Section) We will design a rectangular open channel with the given specifications and provide detailed calculations for discharge, cost estimate, and structural analysis. Given Data: ⚫Discharge (Q: 18 m³/s • Bed Slope (5): S- (bed slope) Manning's Roughness Coefficient (n): -0.015 Channel Shape: Rectangular section ⚫Efficiency Goal: Minimize friction loss by designing a hydraulically efficient section Now, substitute this into Manning's equation: 9-() Substitute values: 2/3 8/2 18 (2d) *-*)()()" 0.015 Now, let's simplify this expression step by step. Step 2: Simplifying and Solving for d 1500 Step 1: Hydraulic Design of the Rectangular Channel Manning's Equation for Open Channel Flow: Manning's equation describes the flow in an open channel 1 0.015 First, simplify the term () Now, substitute this into the equation: -(0)-()" (金) 1500 2/3 18 0.015 x 2 x 22/3 x 1500 1500, V1500 1 38.72983 -0.0258 Where: ⚫ Q = Discharge (m³/s) ⚫n Manning's roughness coefficient ⚫ A = Cross-sectional area of flow (m²) ⚫R Hydraulic radius (m) ⚫ S = Slope of the bed ⚫ A is calculated as A =Bxd, where B is the bottom width and d is the depth of the channel. Since we are designing the most efficient rectangular section, we can assume the bottom width B - 2d, where d is the depth. Hydraulic Radius R: The hydraulic radius R is defined as: Calculate the constant ()/2 Now the equation becomes: Simplify further: where: R- ⚫P is the wetted perimeter (m). For a rectangular channel, the wetted perimeter is: Thus: 18- x2dx 0.015 f/3 x 0.0258 22/3 18-> 2d²x²/3 0.015 x 22/3 x 0.0258 To simplify this equation, we'll use the value 22/3 1.587, so: Now, compute the constants: 20/3 0.015 x 1.587 x 0.0258 0.015 x 1.587 0.023805 P B+2d 2d +2d-4d 0.023805 x 0.0258 0.0006137 Thus, the equation becomes: 2d² d P 4d 18- 2/3 0.0006137 Solve for d Now, solve for d 18 x 0.0006137 2 -0.0055344 d=(0.0055344) 0.4775m So, the flow depth d is approximately 0.48 meters. Width of the Channel B: Since B-2d, the bottom width is B-2x 0.48 0.96 m Step 3: Cross-sectional Area and Hydraulic Radius Now we can calculate the cross-sectional area (A) and hydraulic radius (R): A-Bxd-0.96 x 0.48 -0.4608 m² d 0.48 R- 2 -0.24 m Step 4: Cost Estimate To estimate the cost of the channel, we need to consider the following: 1. Excavation Cost: The volume of soil removed to create the channel is approximately A x Length of the Channel. For simplicity, we'll assume a unit length of 1 meter. Volume to excavate: 0.4608 m2 x 1m 0.4608 m³ Cost per cubic meter of excavation (assuming 50 USD/m³ Excavation cost -0.4608 x 50 - 23.04 USD 2. Concrete Lining Cost (if needed): Assuming the channel is lined with concrete and the cost is 150 USD/m². Surface area of the channel: P x Length - 4 x 0.48 - 1.92 m² • Concrete lining cost Lining cost 1.92 x 150-288 USD Step 5: Structural Analysis 1. Shear Stress at the Bed: The shear stress at the bed r can be calculated using: T-pgSR where: • p = density of water 1000 kg/m³ ⚫ g = acceleration due to gravity 9.81 m/s² ⚫ S = slope 1500 ⚫ R = hydraulic radius 0.24 m 1 T1000 × 9.81 × x 0.24 1.57 Pa 1500 This shear stress is within acceptable limits for a concrete-lined channel.