#$&* course Phy 232 1230pm 7/9 007. `query 6
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Given Solution: ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: query video experiment terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Evidence from this experiment that the drag force increases with velocity is the fact that at a certain point of velocity, the added weights showed less and less of a change on the velocity. So, with increasing velocity comes increasing drag force. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** When weights were repetitively added the velocity of the sphere repetitively 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. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: `q001. If you know the pressure drop of a moving liquid between two points in a narrowing round pipe, with both points at the same altitude, and you know the speed and pipe diameter in the section of pipe with the greater diameter, how could you determine the pipe diameter at the other point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: With the information given, we can find the final velocity through Bernoulli's equation. Since the height is equal, we can take out the (density)(gravity)(height) part of the equation. (1/2)(density)(v1)^2 + P1 = (1/2)(density)(v2)^2 + P2 P2 - P1 = (1/2)(density)(v1)^2 - (1/2)(density)(v2)^2 Since we are given the pressure drop, and the first velocity, we can solve for the final velocity. v2 = sqrt([(1/2)(density)(v1)^2 - change in pressure]/[(1/2)(density)]) With the seconds velocity, we can use the Continuity Equation: A1 v1 = A2 v2 to solve for the second area. We can get the first area using the diameter from the first hole, so A1 = pi*(d/2)^2 Then, we can find the second area, A2 = (A1*V1)/(V2) Once we have the value for A2, we can set up: A2 = pi*(d/2)^2 and then solve for d to get the second diameter. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: OK ********************************************* Question: query univ phy problem 12.93 / 14.91 11th edition14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I understand that Bernoulli's equation will be used to solve this, but I have no thoughts on how to start the problem or which steps to take. At the moment the facts that I know are that the water moves into the narrowed tube, and that water from tank F that is connected to the narrowed tube will rise up to a height of h2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 and the pressure in this part of the tube is 1 atmosphere. 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. However the fact that the widened end of the tube isn't full is not consistent with the assumption made by the text. So let's assume that it is somehow full, though that would require either an expandable fluid (which would make the density rho variable) or a non-ideal situation with friction losses. We will consider a number of points: point 0, at the highest level of the fluid in the top tank; point 1, in the narrowed tube; point 2 at the point where the fluid exits; point 3 at the top of the fluid in the vertical tube; and point 4 at the level of the fluid surface in the lower container. At point 2 the pressure is atmospheric so the previous analysis holds and velocity is vExit such that .5 rho vExit^2 = rho g h1. Thus v_2 = vExit = sqrt(2 g h1). At point 1, where the cross-sectional area of the tube is half the area at point 2, the fluid velocity is double that at point 1, so v_1 = 2 v_2 = 2 sqrt( 2 g h1 ). Comparing points 1 and 2, there is no difference in altitude so the rho g y term of Bernoulli's equation doesn't change. It follows that P_1 + 1/2 rho v_1^2 = P_2 + 1/2 rho v_2^2, so that P_1 = 1 atmosphere + 1/2 rho (v_2^2 - v_1^2) = 1 atmosphere + 1/2 rho ( 2 g h1 - 8 g h1) = 1 atmosphere - 3 rho g h1. There is no fluid between point 1 and point 3, so the pressure at point 3 is the same as that at point 1, and the fluid velocity is zero. There is continuous fluid between point 3 and point 4, so Bernoulli's Equation holds. Comparing point 3 with point 4 (where fluid velocity is also zero, but where the pressure is 1 atmosphere) we have P_3 + rho g y_3 = P_4 + rho g y_4 where y_3 - y_4 = h_2, so that h_2 = y_3 - y_4 = (P_4 - P_3) / (rho g) = (1 atmosphere - (1 atmosphere - 3 rho g h1) ) / (rho g) = 3 h1. Self-Critique: So each part of the system acts a different way, and understanding this is key to understanding the problem. Because the tube narrows and then gets bigger, it doesn't make sense that water would fill the tube back up again so the change in pressure exiting the tube would be 1 atm. Since pressure is 1 atm, it is assumed that the VExit = sqrt(2*g*h1), since (1/2)(density)(Vexit)^2 = (density)(gravity)(h1) Since the narrow tube is half the area, the narrow tube would therefore have twice the velocity (2sqrt(2*g*h1)) based on A1*V1 = A2*V2 Because there is no fluid between the top of the verical tube and the narrow tube, the velocity would be zero and the pressure would be whatever pressure in in the narrow tube. The pressure of the narrow tube can be found by using Bernoulli's equation. Since it is level, the equation would be P1 + (1/2)(density)(v1)^2 = P2 + (1/2)(density)(v2)^2. Since P2 is open to the atmosphere, it would be 1 atm. So, p1 = 1 atm + (1/2)(density)(v2^2 - v1^2) = 1 atm + (1/2)(density)[(2*g*h1) - (8*g*h1)] = 1 atm - 3(density)*(gravity)*(h1) There is water between the bottom tank, and the vertical tube. The velocity would still be zero, but the pressure would now be 1 atm. From this, we get: P1 + (density)(gravity)(h_a) = P2 + (density)(gravity)(h_b), with h_a -h_b = h2 So h_a - h_b = (P2 - P1)/ (density x g) Since h_a - h_b is equal to h2, and the pressure P1 would be equal to the pressure in the narrow tube, h2 = (1 atm - (1 atm - 3(density)*(gravity)*(h1))/(density*gravity) = 3*h1 ------------------------------------------------ Self-Critique Rating: 3 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, or with the implicit assumption that rho remains constant. However note that I am often (though not always) wrong when I disagree with the textbook's solution. ** ********************************************* Question: `q002. If you know the pressure drop of a moving liquid between two points in a narrowing round pipe, with both points at the same altitude, and you know the speed and pipe diameter in the section of pipe with the greater diameter, how could you determine the pipe diameter at the other point? If a U tube containing mercury articulates with the pipe at the two points, how can you find the difference between the mercury levels in the two sides of the pipe? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: You can use Bernoulli's equation, without the (density)(gravity)(height) part since the altitude does not change. (1/2)(density)(v1)^2 + P1 = (1/2)(density)(v2)^2 + P2 P2 - P1 = (1/2)(density)(v1)^2 - (1/2)(density)(v2)^2 Then solve for v2 since that is the only thing not given. v2 = sqrt([(1/2)(density)(v1)^2 - change in pressure]/[(1/2)(density)]) With the second velocity, we can use the Continuity Equation: A1 v1 = A2 v2 to solve for the second area. (the first area can be found using the first diameter, in the equation A = pi*(d/2)^2) A2 = (A1*V1)/(V2) Then set A2 = pi*(d/2)^2 and solve for d to get the second diameter. To find the difference between the mercury levels in the two sides of the pipe, you can use Bernoulli's equation: (1/2)(density)(v1)^2 + (density)(gravity)(height1) + P1 = (1/2)(density)(v2)^2 + (density)(gravity)(height2) + P2 (1/2)(density)(v1)^2 - (1/2)(density)(v2)^2 = (density)(gravity)(height2) - (density)(gravity)(height1) + P2 - P1 (1/2)(density)(v1)^2 - (1/2)(density)(v2)^2 = (density)(gravity)(height2-height1) + P2 - P1 change in altitude = [(1/2)(density)(v1)^2 - (1/2)(density)(v2)^2 - change in pressure ] / [(density)(gravity)] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!