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HW09-solutions

# HW09-solutions - baum(lmb2768 HW09 distler(56295 This...

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baum (lmb2768) – HW09 – distler – (56295) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. For the siphon problem, note that the pres- sure in the hose must be everywhere non- negative 001 10.0 points The picture below shows water forced out of a fire extinguisher by compressed air. The water jets out at speed 11 m / s from a nozzle located at height 1 . 78 m above the water level inside the extinguisher. h v Calculate the gauge pressure of the com- pressed air inside the extinguisher; that is, the pressure di ff erence P gauge air = P inside air - P outside air . The acceleration of gravity is 9 . 8 m / s 2 . Water density is 1000 kg / m 3 and the extinguisher’s body is much wider than the water jet. Correct answer: 77 . 944 kPa. Explanation: Let : v jet = 11 m / s , h = 1 . 78 m , g = 9 . 8 m / s 2 , and ρ = 1000 kg / m 3 . This problem is a straightforward application of Bernoulli’s equation P + ρ g y + ρ 2 v 2 = const . The flow begins at the water level inside the extinguisher and continues all the way to the jet emerging from the nozzle. At the inside water level, the pressure of the water equals the pressure of the compressed air inside the extinguisher. In the jet emerging from the nozzle, however, the water pressure equals the atmospheric pressure of the inside air. Thus, P inside air + ρ 2 v 2 init = P outside air + ρ 2 v 2 jet + ρ g h so that P gauge air = P inside air - P outside air = ρ g h + ρ 2 ( v 2 jet - v 2 init ) . Since the extinguisher’s body is much wider than the jet, the initial speed of the water flow inside the extinguisher is much slower than the jet ( v init v jet ) so v 2 jet - v 2 init v 2 jet , and consequently P gauge air ρ g h + ρ 2 v 2 jet = (1000 kg / m 3 ) (9 . 8 m / s 2 ) (1 . 78 m) + 1000 kg / m 3 2 (11 m / s) 2 = 77944 Pa . 002 10.0 points A large water tank is emptying by having water spurt out two holes at the bottom of the tank, where hole 1 is twice the diameter of hole 2. Which of the following statements is cor- rect? 1. The water speed out of hole 1 is twice that of the water speed out of hole 2, while the mass flow rate is twice that of hole 2. 2. The water speed out of hole 1 is equal to the water speed out of hole 2, as is the mass flow rate out of hole 2. 3. The water speed out of hole 1 is equal to that of the water speed out of hole 2, while

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baum (lmb2768) – HW09 – distler – (56295) 2 the mass flow rate is four times that of hole 2. correct 4. The water speed out of hole 1 is equal to the water speed out of hole 2, while the mass flow rate is twice that of hole 2. 5. The water speed out of hole 1 is twice that of the water speed out of hole 2, while the mass flow rate is four times that of hole 2. Explanation: The openings at the top of the tank and at both holes are at atmospheric pressure, p a . If we approximate the water speed at the top, y = y t , of the large tank to be v t = 0, and take y 1 = y 2 = 0 at the holes, then Bernoulli’s equation for flow between the top of the tank and hole 1 is p a + ρ g y t = p a + 1 2 ρ v 2 1 , and between the top of the tank and hole 2 is p a + ρ g y t = p a + 1 2
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HW09-solutions - baum(lmb2768 HW09 distler(56295 This...

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