What Volume Flow Rate Keeps Pressures Equal in a Bending Pipe System?

[rechitzy]

Homework Statement

A liquid is flowing through a horizontal pipe whose radius is 0.0215 m. The pipe bends straight upward through a height of 10.1 m. The pipe then bends back to the horizontal direction with a different radius of 0.0399 m. What volume flow rate will keep the pressures in the two horizontal sections of pipe the same?

Homework Equations

p1 + (1/2) ρ v1^2 + ρgh1 = p2 + (1/2) ρ v2^2 + ρ g h2

The Attempt at a Solution

p1 + (1/2) ρ v1^2 + ρgh1 = p2 + (1/2) ρ v2^2 + ρ g h2
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(1/2) v1^2 + 0 = (1/2) v2^2 + g h2
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or v1^2 = v2^2 + 2*9.80*10.1
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v1^2 = v2^2 + 203.84
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Now... the flow rate through each is speed * area, and area is π r2 so...
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flow rate 1 = flow rate 2 or v1 * π r1^2 = v2 * π r2^2
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v1 * 0.02152^2 = v2 * 0.0399^2
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simplify... v1 = 3.43764 v2
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plug this into the first equation...
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(3.43764 v2)^2 = v2^2 + 203.84 11.81736877 v2^2 = v2^2 + 203.84
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11.81736877 v2^2 = 203.84 v2^2 = 17.2491811

v2 = 4.153205763 is the speed in the upper pipe.
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flow rate = speed * area = 4.153205763 * π * 0.0399 = 0.5206024 m3 /sec

please help! thanks

 

[BerryBoy]

Does 2*9.8*10.1 really equal 203.84?? (I can see without using a calculator that this number is not going to exceed 200)

And in your flow rate equation, the radius should be squared. I think your equations are fine, but not the "maths".