course Phy 122 dᩀx|assignment #016
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23:01:13 query introset change in pressure from diameter change given original vel and diameter
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RESPONSE --> I am assuming this question means to explain the change in pressure from the diameter change if you are given the initial velocity and the diameter? Velocity ratio is inversely proportional to area ratio and area ratio is directly proportional to the square of the diameter ratio v2 / v1 = (A1 / A2) = (d1 / d2)^2 so So we know that v2 = (d1/d2)^2 * v1 We are assuming that the height is constant that way the pgy goes to nothing in Bernoulli's equation. So we have P1 + .5* rho * v1^2 = P2 + .5 rho * v2^2 So you can fill in the (d1/d2)^2 * v1 for v2^2.
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23:01:24 ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . **
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RESPONSE --> Okay
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23:17:53 query video experiment 4 terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity?
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RESPONSE --> When we added more weight to the sphere, the velocity increased. Once we added a lot of paper clips (about six or seven paper clips) the velocity stayed about the same. As the velocity increased so did the drag force from the water.
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23:17:59 ** When weights were repetitively added the velocity of the sphere repetively increased. As the velocities started to aproach 0.1254 m/sec the added weights had less and less effect on increasing the velocity. We conclude that as the velocity increased so did the drag force of the water. **
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RESPONSE --> ok
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23:18:15 query univ phy problem 14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises.
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RESPONSE --> Phy 122 Not assigned.
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23:18:18 ** The fluid exits the narrowed part of the tube at atmospheric pressure. The widened part at the end of the tube is irrelevant--it won't be filled with fluid. So Bernoulli's Equation will tell you that the fluid velocity in this part is vExit such that .5 rho vExit^2 = rho g h1. Now compare three points: point 1, in the narrowed tube; point 2 at the top of the fluid in the lower tube; and point 3 at the level of the fluid surface in the lower container. At point 1 the pressure is atmospheric and velocity is vExit. Pressure is atmospheric because there is no pressure loss within the tube, since friction and viscosity are both assumed negligible. At point 2 fluid velocity is zero and since there is no fluid between the narrowed tube and this point there is no net rho g h contribution to Bernoulli's equation. So we have .5 rho v1^2 + P1 = .5 rho v2^2 + P2. P1 is atmospheric and v1 is vExit from above, while v2 = 0 so P2 = atmospheric pressure + .5 rho vExit^2 = atmospheric pressure + rho g h1. Now comparing point 2 with point 3 we see that there is a difference h in the fluid altitude, with velocity 0 at both points and atmospheric pressure at point 3. Thus P2 + rho g h2 = P3 + rho g h3, or (atmospheric pressure + rho g h1) + rho g h2 = atmospheric pressure + rho g h3. Thus rho g (h3 - h2) = rho g h1 and h3 - h2, which is the height of the fluid in the lower tube, is just equal to h1. If we assume that somehow the fluid manages to expand on escaping the narrowed tube so that it fills the once-again-widened tube, and exits with vExit as above, then the velocity in the narrowed tube will be 2 * vExit. This leads to the conclusion that pressure change from small to large tube is .5 rho (2 vExit)^2 - .5 rho vExit^2 = .5 rho (3 vExit^2). Since pressure is atmospheric in the large tube, pressure in the small tube is atmospheric pressure + 3 ( .5 rho vExit^2). If we use this pressure for point 1 and follow the steps given above we conclude that h3 - h2, the height of the fluid column in the lower tube, is 3 h1. This is the book's answer. Again I don't have the problem in front of me and I might have missed something, but the idea of the fluid expanding to refill the larger pipe doesn't seem consistent with the behavior of even ideal liquids, which are pretty much incompressible at ordinary pressures. However note that I am sometimes wrong when I disagree with the textbook's solution. **
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RESPONSE -->
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23:18:25 query univ phy problem 14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises.
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RESPONSE --> Phy 122 Not assigned
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23:18:27 ** The fluid exits the narrowed part of the tube at atmospheric pressure. The widened part at the end of the tube is irrelevant--it won't be filled with fluid. So Bernoulli's Equation will tell you that the fluid velocity in this part is vExit such that .5 rho vExit^2 = rho g h1. Now compare three points: point 1, in the narrowed tube; point 2 at the top of the fluid in the lower tube; and point 3 at the level of the fluid surface in the lower container. At point 1 the pressure is atmospheric and velocity is vExit. Pressure is atmospheric because there is no pressure loss within the tube, since friction and viscosity are both assumed negligible. At point 2 fluid velocity is zero and since there is no fluid between the narrowed tube and this point there is no net rho g h contribution to Bernoulli's equation. So we have .5 rho v1^2 + P1 = .5 rho v2^2 + P2. P1 is atmospheric and v1 is vExit from above, while v2 = 0 so P2 = atmospheric pressure + .5 rho vExit^2 = atmospheric pressure + rho g h1. Now comparing point 2 with point 3 we see that there is a difference h in the fluid altitude, with velocity 0 at both points and atmospheric pressure at point 3. Thus P2 + rho g h2 = P3 + rho g h3, or (atmospheric pressure + rho g h1) + rho g h2 = atmospheric pressure + rho g h3. Thus rho g (h3 - h2) = rho g h1 and h3 - h2, which is the height of the fluid in the lower tube, is just equal to h1. If we assume that somehow the fluid manages to expand on escaping the narrowed tube so that it fills the once-again-widened tube, and exits with vExit as above, then the velocity in the narrowed tube will be 2 * vExit. This leads to the conclusion that pressure change from small to large tube is .5 rho (2 vExit)^2 - .5 rho vExit^2 = .5 rho (3 vExit^2). Since pressure is atmospheric in the large tube, pressure in the small tube is atmospheric pressure + 3 ( .5 rho vExit^2). If we use this pressure for point 1 and follow the steps given above we conclude that h3 - h2, the height of the fluid column in the lower tube, is 3 h1. This is the book's answer. Again I don't have the problem in front of me and I might have missed something, but the idea of the fluid expanding to refill the larger pipe doesn't seem consistent with the behavior of liquids, which are pretty much incompressible at ordinary pressures. However note that I am sometimes wrong when I disagree with the textbook's solution. **
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RESPONSE -->
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