#$&* course PHY-242 007. `query 6
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Given Solution: ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . ** Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: query video experiment terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The greater the velocity the more push (or drag) force we had. Too little velocity caused the sphere not to move. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Since the altitude does not change, we can use Pressure_1 + (Density_1 * Velocity_1^2)/2 = Pressure_2 + (Density_2 * Velocity_2^2)/2 Since we know the Pressure drop, this means we know Pressure_1 and Pressure_2. Since the liquid does not change composition or temperature, then density_1 and density_2 should be the same. Since we know the speed of pipe with greater diameter, this means we know Velocity_1. The only variable we do not know is the velocity_2. We would solve for velocity_2 Then, we use d_1 * A_1 * V_1 = d_2 * A_2 * V_2 Since the density is the same, A_1 * V_1 = A_2 * V_2 Using this equation, we would find the area of the pipe As we know, the area of a circle is pi * r^2 After we find the radius, we can use the formula Diameter = 2 * Radius to find the diameter of the pipe confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok ********************************************* Question: query univ phy problem 12.93 / 14.91 11th edition14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: To solve this problem, we would need to use the Bernoulli’s equation. P1 + 0.5 * rho * V1^2 + rho * g * y1 = P2 + 0.5 * rho * v2^2 + rho * g * y2 However, since I am not given any specific values, but only variables, I can not give a specific answer. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 Rating: 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: Since the altitude does not change, we can use Pressure_1 + (Density_1 * Velocity_1^2)/2 = Pressure_2 + (Density_2 * Velocity_2^2)/2 Since we know the Pressure drop, this means we know Pressure_1 and Pressure_2. Since the liquid does not change composition or temperature, then density_1 and density_2 should be the same. Since we know the speed of pipe with greater diameter, this means we know Velocity_1. The only variable we do not know is the velocity_2. We would solve for velocity_2 Then, we use d_1 * A_1 * V_1 = d_2 * A_2 * V_2 Since the density is the same, A_1 * V_1 = A_2 * V_2 Using this equation, we would find the area of the pipe As we know, the area of a circle is pi * r^2 After we find the radius, we can use the formula Diameter = 2 * Radius to find the diameter of the pipe If a U tube containing mercury articulates with the pipe at the two points, then one point will have a greater pressure and the other point will have a lower pressure. Difference in pressure from both sides will cause mercury to move either into one direction or the other direction. I would calculate the difference in the pressures (Pressure_2 - Pressure_1). Then, I would use the static pressure formula Pressure = density * gravity * height The density of mercury is 13.534 g/cm^3 and gravity is 9.8 m/s^2. So my equation would be: Pressure = 13.534 g/cm^3 * 9.8 m/s^2 * height Then I would solve for the height and determine by how much the mercury rises in the u-tube confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok 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: Since the altitude does not change, we can use Pressure_1 + (Density_1 * Velocity_1^2)/2 = Pressure_2 + (Density_2 * Velocity_2^2)/2 Since we know the Pressure drop, this means we know Pressure_1 and Pressure_2. Since the liquid does not change composition or temperature, then density_1 and density_2 should be the same. Since we know the speed of pipe with greater diameter, this means we know Velocity_1. The only variable we do not know is the velocity_2. We would solve for velocity_2 Then, we use d_1 * A_1 * V_1 = d_2 * A_2 * V_2 Since the density is the same, A_1 * V_1 = A_2 * V_2 Using this equation, we would find the area of the pipe As we know, the area of a circle is pi * r^2 After we find the radius, we can use the formula Diameter = 2 * Radius to find the diameter of the pipe If a U tube containing mercury articulates with the pipe at the two points, then one point will have a greater pressure and the other point will have a lower pressure. Difference in pressure from both sides will cause mercury to move either into one direction or the other direction. I would calculate the difference in the pressures (Pressure_2 - Pressure_1). Then, I would use the static pressure formula Pressure = density * gravity * height The density of mercury is 13.534 g/cm^3 and gravity is 9.8 m/s^2. So my equation would be: Pressure = 13.534 g/cm^3 * 9.8 m/s^2 * height Then I would solve for the height and determine by how much the mercury rises in the u-tube confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok #*&!