course Phy 202 üFˆöÙ´É’Õy‡Ý‹šœÛÏŠfñJ€ÆÐÞH±assignment #016
......!!!!!!!!...................................
09:09:29 query introset change in pressure from diameter change given original vel and diameter
......!!!!!!!!...................................
RESPONSE --> First the change in velocity must be found using the statement of proportion: A1v1 = A2v2 After this is found Bernoulli's equation can be used, removing the statements involving y since the difference is negligible in this problem. Left with P1 + .5 'rho v1^2 = P2 + ..5 'rho v2^2, this equation can be set to solve for P2 and the numbers we have can be plugged in. Alternatively, the way it was solved in the intro set solution, the two sides of the equation can separately be set to 0 and the difference between the two Pressure values can be found. This value is subtracted from the given P1 since we are aware that P has to decrease to compensate for v's increase.
.................................................
......!!!!!!!!...................................
09:11:47 ** 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) . **
......!!!!!!!!...................................
RESPONSE --> v2 / v1 = A1 / A2 = (d1 / d2)^2 v2 = (d1/d2)^2 * v1 P2 - P1 = .5 * rho (v1^2 - v2^2) I think I have a decent understanding of this scenario.
.................................................
......!!!!!!!!...................................
09:12:24 query video experiment 4 terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity?
......!!!!!!!!...................................
RESPONSE --> I managed to break my DVDs, so I am unable to answer this.
.................................................
......!!!!!!!!...................................
09:12:53 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Drag force of water increases as velocity increases.
.................................................
......!!!!!!!!...................................
09:12:56 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:12:59 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:13:02 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
09:13:05 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................