qz10sol_3706Hs11 - p s = 150 kN/m 2 . a) For the conditions...

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TA: Tomoyuki Nakayama Tuesday, April 12, 2011 PHY 2048: Physic 1, Discussion Section 3706H Quiz 10 (Homework Set #12) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ Fresh water flows horizontally from pipe section 1 of cross-sectional area A 1 into pipe section 2 of cross-sectional area A 2 . The figure below gives a plot of the pressure difference p 2 - p 1 versus the inverse area squared A 1 -2 that would be expected for a volume flow rate of a certain value if the water flow were laminar under all circumstances. The scale on the vertical axis is set by
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Unformatted text preview: p s = 150 kN/m 2 . a) For the conditions of the figure, what is the value of A 2 (in square meters)? Applying Bernoullis equation at section 1 and section 2, we have p 1 + (1/2) v 1 2 = p 2 + (1/2) v 2 2 p = (1/2) (v 1 2 v 2 2 ) The continuity equation yields A 1 v 1 = A 2 v 2 v 1 = (A 2 /A 1 )v 2 Combining the two equation leads to the mathematical expression for the graph: p = (1/2) [(A 2 /A 1 ) 2 1]v 2 2 According to this expression, p = 0 when A 1 = A 2 . Therefore, A 2-2 = 16 A 2 = 0.25 m 2 b) What is the value of the volume flow rate? When A 1-2 = 32 m-4 , (A 2 /A 1 ) 2 = (32/16) = 2 and p = 150 kN/m 2 . The expression obtained in part a) yields p = (1/2) [2 1]v 2 2 v 2 = (2 p/ ) = 17.3 m/s The volume flow rate is A 1 v 1 = 4.33 m 3 /s...
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