Assignment 07

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course PHY 232

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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:

I would first apply A1v1 = A2v2, or (v1/v2) = (A2/A1). I also know that the proportion (A2/A1) = (d2/d1)^2, so I ultimately have the proportion:

(v1/v2) = (d2/d1)^2

Using these ratios, I would be able to find the final velocity of the water. I would then use Bernoulli’s equation to find the final pressure.

P1 + (1/2)*’rho*v1^2 +’rho*g*h1 = P2 + (1/2)*’rho*v2^2 + ‘rho*g*h2

Assuming h is constant, I am left with:

P1 + (1/2)*’rho*v1^2 = P2 + (1/2)*’rho*v2^2

Since the question asks for the change in pressure, I can rearrange the equation further:

P2 - P1 = (1/2)*’rho*v1^2 - (1/2)*’rho*v2^2

‘delta P = - (1/2)*’rho* ‘delta v

In other words, since h is constant, the change in pressure is equal to the negative change in .5*’rho*v^2

confidence rating #$&*:

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confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

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3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

• 2 indicates that you believe you addressed most of the discrepancies between the given solution and your solution but are unsure of some aspects of the situation; you would at this point consider including a question or a statement of what you're not sure you understand

• 1 indicates that you believe you understand the overall idea of the solution but have not been able to address the specifics of the discrepancies between your solution and the given solution; in this case you would normally include a question or a statement of what you're not sure you understand

• 0 indicates that you don't understand the given solution, and/or can't make a reasonable judgement about whether or not your solution is correct; in this case you would be expected to address the given solution phrase-by-phrase and state what you do and do not understand about each phrase)

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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:

Adding the weights increased the velocity of the sphere. However, as time progressed, the added weight had less effect on increasing the velocity.

confidence rating #$&*:

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confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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:

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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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?

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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 #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Self-Critique Rating:

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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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.

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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 #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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.

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Self-Critique Rating:

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Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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. **

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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?

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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.

(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 #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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Self-Critique Rating:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

------------------------------------------------

Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

Self-critique (if necessary):

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Self-critique rating:

#*&!

Assignment 07

&#Your work looks good. Let me know if you have any questions. &#