assignment 7

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course phy 232

007. `query 6

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Question: query introset How do we find the change in pressure due to diameter change given the original velocity of the flow and pipe diameter and final diameter?

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Your Solution:

Well A1(V1)=A2(V2)=(d1/d2)^2 so

(P2-P1)=.5*rho (V1^2-V2^2)

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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) . **

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Question: query video experiment terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity?

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Your Solution:

When the weights were added to the sphere the velocity increased. But as the velocity got higher and higher the weights had less and less of an effect on increasing the velocity resulting in the drag force increasing.

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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. **

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Question: query univ phy problem 14.89 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.

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Your Solution:

I have the 12th edition so this is the solution to 14.89

Starting off with Bernoulli's Equation

P1+rho g y1+.5rho v1^2= P2+rho g y2+.5rho v2^2

Plugging in the known variables we get a rate of 70 m^3/second

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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. 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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