# hmwk09 - Su (ycs73) HW09 Tsoi (58020) 1 This print-out...

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Unformatted text preview: Su (ycs73) HW09 Tsoi (58020) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points The horizontal constricted pipe illustrated in the figure (a venturi tube), can be used to measure flow velocities in an incompressible fluid. 1 2 P v P 1 v 2 Which equation best represents Bernoullis equation for the given flow configuration? 1. P 1 + v 2 1 = P 2 + v 2 2 2. P 1 + 1 2 v 2 2 = P 2 + 1 2 v 2 1 3. P 1 + 1 2 v 2 1 = P 2 + 1 2 v 2 2 correct Explanation: Bernoullis equation is P + 1 2 v 2 + g y = constant . In Bernoullis equation for horizontal flow ( y constant), there is no change in the grav- itational potential energy, so P + 1 2 v 2 = constant . This is true for all parts of the tube, so P 1 + 1 2 v 2 1 = P 2 + 1 2 v 2 2 . 002 (part 2 of 3) 10.0 points Which equation best represents the conti- nuity equation for the given flow configura- tion? 1. A 1 v 2 A 2 v 1 = 0 2. A 1 v 2 + A 2 v 1 = 0 3. A 1 v 1 A 2 v 2 = 0 correct 4. A 1 v 1 + A 2 v 2 = 0 Explanation: The continuity equation for an incompress- ible fluid (which means that the density of the fluid is a constant) is A 1 v 1 = A 2 v 2 = constant . The continuity equation for incompressible flow is given by d m 1 dt = d m 2 dt or d m 1 dt = A 1 d x 1 dt = A 1 v 1 = d m 2 dt = A 2 d x 2 dt = A 2 v 2 = A 1 v 1 = A 2 v 2 . 003 (part 3 of 3) 10.0 points The ratio for the cross section areas of the tube is A 2 A 1 = 0 . 54, the difference in the pres- sures is P 1 P 2 = dP = 11 . 5 Pa , and the density of the fluid is 1 . 64 kg / m 3 . Find the speed of the fluid near the right hand end of the tube ( i.e. , find v 2 ). Correct answer: 4 . 44941 m / s. Explanation: Let : A 2 A 1 = 0 . 54 , dP = 11 . 5 Pa , and = 1 . 64 kg / m 3 . From the continuity equation v 1 = A 2 A 1 v 2 . We substitute this value for v 1 into Bernoullis equation to obtain P 1 + 1 2 parenleftbigg A 2 A 1 parenrightbigg 2 v 2 2 = P 2 + 1 2 v 2 2 . Su (ycs73) HW09 Tsoi (58020) 2 Solving for v 2 gives v 2 = radicaltp radicalvertex radicalvertex radicalvertex radicalvertex radicalbt 2 ( P 1 P 2 ) bracketleftBigg 1 parenleftbigg A 2 A 1 parenrightbigg 2 bracketrightBigg = radicalBigg 2 (11 . 5 Pa) (1 . 64 kg / m 3 ) [1 (0 . 54) 2 ] = 4 . 44941 m / s . 004 10.0 points Consider an object that floats in water but sinks in oil. When the object floats in a glass of water, half of the object is submerged. We now slowly pour oil into the glass so it completely covers the object. Water Oil Figure: The second figure does not necessarily show the correct changes, only the experimental setup....
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## This note was uploaded on 04/15/2009 for the course PHY 58020 taught by Professor Tsoi during the Spring '09 term at University of Texas at Austin.

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hmwk09 - Su (ycs73) HW09 Tsoi (58020) 1 This print-out...

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