Pipe Radius: Calculate Diameter of Middle Part

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The discussion revolves around calculating the diameter of the middle section of a pipe based on water flow and travel time. The lengths of the pipe sections A and B are each 30m with a diameter of 2cm, while the middle section is 110m long. An engineer observed that a coin took 88.8 seconds to travel from A to B, leading to calculations that suggest the middle section's diameter is approximately 2.43cm. However, the model answer claims the diameter is 3.60cm, raising questions about its accuracy. The conversation concludes with a clarification that 2cm refers to the radius, not the diameter.
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Homework Statement


The following image shows the parts A and B of a pipe where the middle part goes under a hill the length of both parts A and B is 30m and the diameter of them each is 2cm where as the middle part has a length of 110m to determine the middle part's diameter one engineer decided to push water in the pipe with a speed of 2.50 m\s in part A he dropped a small coin in the flow and noticed that it took 88.8 seconds for it to start from A and exit from B.
1)calculate the diameter of the middle pipe

Homework Equations



v=d\s
A1*v1=A2*v2. (Continuity equation)
A=pi*r^2[/B]

The Attempt at a Solution


Finding the area of a and b:
0.02^2*pi=0.00125m^2
Finding the time it would take to travel through A and B (not including mid pipe)
2.50=30/s
s=12
So for a and b it would take 24 seconds
For mid part it would take 88.8-24=64.8
Mid part speed
110/64.8=1.697m\s
So
0.00125*2.50=1.697*A2
Area=0.001841
0.0018141=pi*r^2
R=0.024m
Answer sheet says that the diameter is 3.60cm
help!
 

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Finding the area of a and b:
0.02^2*pi=0.00125m^2
... the diameter of the pipe is 2cm.

Note: you do not have to compute the area ...
You can express: ##A_1v_1 = A_2v_2## in terms of the two diameters ##d_1## and ##d_2## using ##A=\pi d^2/4##.
 
Now I've had a chance to check.
Assuming the coin travels at the same (average) speed as the water - it travels 60m at u=2.5m/s, this takes 60/2.5=24s
Total travel time is 88.8s to time through the entire length, so the time spent in the middle section is 88.8-24=64.8s
This gives an average speed through that section as v=110/64.8 = 1.6975m/s = 169.75cm/s
Volume flow rate is a constant for incompressible flow: we want to find hidden diameter d, given revealed diameter, u, and v
##\pi (d^2/4)v = \pi (2^2/4)u \implies d = 2\sqrt{2.5/1.6975} = 2.43##cm (2dp)
... in agreement with your answer. (notice I did not need to convert the units of diameter to meters.)

The model answer says d=3.60cm.
This suggests the model answer is incorrect, or uses a different model.
You can work out what the time through the entire pipe would be if the middle section were 3.6cm in diameter.
 
Simon Bridge said:
Now I've had a chance to check.
Assuming the coin travels at the same (average) speed as the water - it travels 60m at u=2.5m/s, this takes 60/2.5=24s
Total travel time is 88.8s to time through the entire length, so the time spent in the middle section is 88.8-24=64.8s
This gives an average speed through that section as v=110/64.8 = 1.6975m/s = 169.75cm/s
Volume flow rate is a constant for incompressible flow: we want to find hidden diameter d, given revealed diameter, u, and v
##\pi (d^2/4)v = \pi (2^2/4)u \implies d = 2\sqrt{2.5/1.6975} = 2.43##cm (2dp)
... in agreement with your answer. (notice I did not need to convert the units of diameter to meters.)

The model answer says d=3.60cm.
This suggests the model answer is incorrect, or uses a different model.
You can work out what the time through the entire pipe would be if the middle section were 3.6cm in diameter.

Thank you very much but I made a mistake 2cm is the radius not the diameter thanks!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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