### Diagram: Bent Pipe Flow Analysis#### DescriptionThis diagram illustrates the flow of fluid through a bent pipe, demonstrating various forces and measurements involved in the system.#### Details- **Pipe Geometry**: - The inlet has a diameter of **36 inches**. - The outlet has a diameter of **24 inches** and exits at an angle of **45°** to the horizontal.- **Fluid Flow**: - **Flow Rate (Q₁)**: The fluid enters the pipe at a flow rate of **20 cubic feet per second (cfs)**. - **Inlet Pressure (P₁)**: The pressure at the inlet is **15 psi**.- **Outlet Conditions**: - **Outlet Pressure (P₂)**: The pressure at the outlet is **14.78 psi**.- **Forces**: - **F₁**: Represents the force exerted by the fluid at the inlet. - **F₂**: Represents the force exerted by the fluid at the outlet. - **Fₙ**: The net force exerted at the bend. - **Fₙₓ and Fₙᵧ**: Components of the net force in the x and y directions.- **Coordinate System**: - The diagram uses a coordinate system with x and y axes to show directionality.This diagram is used for analyzing fluid dynamics and forces in a bent pipe scenario, common in civil and mechanical engineering applications. The image contains mathematical notations and diagrams related to fluid dynamics, specifically addressing forces in a bent pipe. Below is a transcription and explanation suitable for an educational website:---### Fluid Dynamics: Calculating Forces in a Bent Pipe**Given Parameters:**- **Flow Rate (Q):** 20 cubic feet per second (cfs)- **Pressure at Point 1 (P₁):** 15 psi- **Pressure at Point 2 (P₂):** 14.78 psi**Pipe Dimensions and Angle:**- **Pipe Length:** 36 inches- **Angle of Bend:** 45°- **Section Diameter:** Not specified**Diagram:**The diagram illustrates a bend in a pipe with horizontal and vertical components of force. The pipe initially extends straight before making a 45° bend. There are forces denoted as **Fₙₓ** and **Fₙᵧ** acting at the bend, with components aligned to the x-axis and y-axis respectively.**Objective:**Neglecting head loss in the bend, calculate the force acting on the bend.**Force Equations:**- **x-component of force:** \[ \rho Q (u₂ - u₁) = \sum E_{\text{ext}, x} = F₁ - F₂ \cos 45° - F_{N,x} \]- **y-component of force:** \[ \rho Q (v₂ - v₁) = \sum E_{\text{ext}, y} = F_{N,y} - F₂ \sin 45° \]**Solution Approach:**To determine the resultant force (**|Fₙ|**) and direction (angle α), solve for both **Fₙₓ** and **Fₙᵧ**:- **Resultant Force:** \[ |F_N| = \sqrt{F_{N,x}^2 + F_{N,y}^2} \]- **Angle (α):** \[ \alpha = \arctan \left(\frac{F_{N,y}}{F_{N,x}}\right) \]These equations and diagram details provide the basis for calculating the forces exerted on a pipe bend, incorporating principles of fluid mechanics and vector components.

Answered: Image /qna-images/answer/078c3d52-80e7-42bd-80a0-32f7747f2c9f.jpg

Answered: Q₁ = 20cfs P₁ = 15 psi LFE 36in k P₂²…

Engineering

Mechanical Engineering

Q₁ = 20cfs P₁ = 15 psi LFE 36in k P₂² FNA •> 24 in 45 0 = 14.78 Psi

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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