hw12b' - HW#12b Note: numbers used in solution steps are...

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HW#12b Note: numbers used in solution steps are different from your WebAssign values. Page 1 of 3
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HW#12b Note: numbers used in solution steps are different from your WebAssign values. Page 2 of 3 First, The thinner the pipe the higher the speed. A 1 v 1 =A 2 v 2 , due to the continuity, the total flow rate is the same. When pipe becomes wide at a location, flow speed reduces. When pipe becomes narrower flow speed is high. Pipe area is proportional to diameter square. A=3.14*D 2 /4, here D is diameter. A 1 v 1 =A 2 v 2 ; So, A 1 v 1 /A 2 = v 2 , So, D 1 2 v 1 / D 2 2 = v 2 So, at narrower location: v 2 = (0.02 2 /0.018 2 )*0.85 =1.05 m/s The change of the speed is due to the change of pipe diameter. Once we know the speed and height at the new location we can use the Bernoulli’s equation P 1 +1/2 ρ v 1 2 + ρ g y 1 = P 2 +1/2 ρ v 2 2 + ρ g y 2 To find out the pressure P 2 at the new location. 165*10
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This note was uploaded on 08/25/2010 for the course PHYS 111 taught by Professor Jamesm.lockhart during the Spring '09 term at S.F. State.

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hw12b' - HW#12b Note: numbers used in solution steps are...

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