2. A two-dimensional square channel redirects a water flow from horizontal uniform to vertical nonuniform. The pipe width h=75.5 mm. The pressure is 185 kPa (abs) at the inlet and at atmospheric pressure at the exit with Vmax=2Vmin. The mass of the pipe is M=2.05 kg. The internal volume of the pipe is V=0.00355 m³. a. Find Vmin if U=7.5 m/s b. Calculate the force exerted by the pipe assembly on the supply Vmax 2 Vmin

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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2. A two-dimensional square channel redirects a water flow from horizontal uniform to
vertical nonuniform. The pipe width h=75.5 mm. The pressure is 185 kPa (abs) at the inlet
and at atmospheric pressure at the exit with Vmax=2Vmin. The mass of the pipe is M=2.05 kg.
The internal volume of the pipe is V=0.00355 m³.
a. Find Vmin if U=7.5 m/s
b. Calculate the force exerted by the pipe assembly on the supply
Vmax
2
Vmin
Transcribed Image Text:2. A two-dimensional square channel redirects a water flow from horizontal uniform to vertical nonuniform. The pipe width h=75.5 mm. The pressure is 185 kPa (abs) at the inlet and at atmospheric pressure at the exit with Vmax=2Vmin. The mass of the pipe is M=2.05 kg. The internal volume of the pipe is V=0.00355 m³. a. Find Vmin if U=7.5 m/s b. Calculate the force exerted by the pipe assembly on the supply Vmax 2 Vmin
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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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