Structural Analysis
Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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### Problem Statement A two-dimensional square channel redirects a water flow from horizontal uniform to vertical nonuniform. The pipe width is \( h = 75.5 \, \text{mm} \). The pressure is \( 185 \, \text{kPa (abs)} \) at the inlet and at atmospheric pressure at the exit with \( v_{\text{max}} = 2v_{\text{min}} \). The mass of the pipe is \( M = 2.05 \, \text{kg} \). The internal volume of the pipe is \( V = 0.00355 \, \text{m}^3 \). a. Find \( v_{\text{min}} \) if \( U = 7.5 \, \text{m/s} \). b. Calculate the force exerted by the pipe assembly on the supply. ### Diagram Explanation The diagram depicts a two-dimensional flow channel where water enters horizontally at velocity \( U \) and exits vertically. The flow transitions from a uniform horizontal flow to a nonuniform, vertical flow. The channel consists of: - **Horizontal Section:** Water enters with a velocity \( U \) and width \( h \). - **90-degree Bend:** Water is redirected upwards. - **Vertical Section:** Water exits with velocities \( v_{\text{max}} \) and \( v_{\text{min}} \) at different points across the pipe. ### Calculations 1. **Find \( v_{\text{min}} \)** - Assume continuity and/or Bernoulli's principle may be used to solve this part. 2. **Calculate the Force** - Consider momentum conservation and force due to the pressure difference and weight of the pipe water system to determine the force exerted by the pipe assembly. This context will enhance understanding for students on topics such as fluid dynamics, pressure-flow relationships, and forces in fluid systems.
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Transcribed Image Text:### Problem Statement A two-dimensional square channel redirects a water flow from horizontal uniform to vertical nonuniform. The pipe width is \( h = 75.5 \, \text{mm} \). The pressure is \( 185 \, \text{kPa (abs)} \) at the inlet and at atmospheric pressure at the exit with \( v_{\text{max}} = 2v_{\text{min}} \). The mass of the pipe is \( M = 2.05 \, \text{kg} \). The internal volume of the pipe is \( V = 0.00355 \, \text{m}^3 \). a. Find \( v_{\text{min}} \) if \( U = 7.5 \, \text{m/s} \). b. Calculate the force exerted by the pipe assembly on the supply. ### Diagram Explanation The diagram depicts a two-dimensional flow channel where water enters horizontally at velocity \( U \) and exits vertically. The flow transitions from a uniform horizontal flow to a nonuniform, vertical flow. The channel consists of: - **Horizontal Section:** Water enters with a velocity \( U \) and width \( h \). - **90-degree Bend:** Water is redirected upwards. - **Vertical Section:** Water exits with velocities \( v_{\text{max}} \) and \( v_{\text{min}} \) at different points across the pipe. ### Calculations 1. **Find \( v_{\text{min}} \)** - Assume continuity and/or Bernoulli's principle may be used to solve this part. 2. **Calculate the Force** - Consider momentum conservation and force due to the pressure difference and weight of the pipe water system to determine the force exerted by the pipe assembly. This context will enhance understanding for students on topics such as fluid dynamics, pressure-flow relationships, and forces in fluid systems.
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Can you please sketch the diagram to see where the forces are acting, as well as can you exlain why the Pressure at the atmosphere is negative 

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where did you -Patm from? and what is that supposed to be in the calculations 

 

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for the second question when solving for fx, why did you negelct density?

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for the second problem why did you do pressure x area - force(x)? where did the pressure x area come from in that equation 

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how did you get 1.5 for the first part 

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Can you please sketch the diagram to see where the forces are acting, as well as can you exlain why the Pressure at the atmosphere is negative 

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Follow-up Question

where did you -Patm from? and what is that supposed to be in the calculations 

 

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Follow-up Question

for the second question when solving for fx, why did you negelct density?

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for the second problem why did you do pressure x area - force(x)? where did the pressure x area come from in that equation 

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how did you get 1.5 for the first part 

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