Fluid Dynamics - Please check my answers

  • Thread starter stuplato
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  • #1
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I just like someone to help check my answers please o:)
Gil is a perfectly spherical fish bred specifically for use in physics problems. Gil has mass of 33.5 kg. By inflating/deflating his air bladder, Gil has the ability to change his volume while his mass does not change appreciably. Gil sleeps in a small cylinder chamber. The small chamber is connected to a larger chamber by a tapering passageway. The walls of the containers are made of aluminium with a small glass window (diameter = 0.4 m) in the larger chamber. Both chambers are sealed by large (thin) weights that can move up/down as water levels change. Gil is initially 5 m from the bottom of the smaller cylinder.
Part A: Both tanks are filled with water (p=1000 kg/m^3). The small weight has a mass of 10,000 kg and radius of 2 m. The large weight has mass of 2,000,000 kg and radius of 10m. If the water is 15 m from the floor, what is the water level in the larger cylinder? The system is in equilibrium. I get 9.03 m
Part B: Gil adjusts his position in the tank by adjusting his bladder (and thus his volume). What must Gil's radius be to remain stationary in the water tank? I get 0.2 m
Part C: Gil wants to rise to the top of the tank. If he doubles his radius, what will his resultant acceleration be? Ignore frictional effects. I get 68.6 m/s^2
Part D: How long will it take Gil to strike the floating weight at the top of the tank? Ignore friction and turbulence. What will his speed be? I get .274 s and 18.8 m/s
Part E: In order to conduct an experiment, a scientist wishes to force gil into the larger tank. She accomplishes this by adding an additional 63,000 kg of mass to the smaller cylinder. Qualitatively describe the acceleration and velocity. I put: The velocity of the small disk will change until it and the acceleration will equal zero. The acceleration changes due to changes in Net Force by changes in pressure
Part F: Calculate the initial accelerations of both the small and large weights at the moment the 63,000 kg mass is added. Are these values the same? Why or why not? I get 8.49 m/s^2 for small and .309 m/s^2 for large. No they are not same because they are spread over different areas
Part G: What will the new water levels be once the system reaches equilibrium? i get 12.91 for small and 9.45 for large
Part H: Gil is carried into large tank by flow of water. If he enters the large end of the passage (radius 2 m) with velocity of 3.2 m/s, with what velocity will he come out of the other end (radius .5 m)? I get 12.8 m/s
There is more, but I like to know this part is right first before continuing... :smile:
 

Answers and Replies

  • #2
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Okay, I got something wrong cause for a later part, I get an impossible solution, could someone tell me where I went wrong? I just need to know the part.
 
  • #3
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Working through part A again, somehow I got a little over twelve... did I make my mistake there?:uhh:
 
  • #4
Fermat
Homework Helper
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A) I get h = 9.43m

B) correct

C) correct

D) I get t = 0.54s and v = 37 m/s
my time and speed are twice yours.

E) Sounds OK

F) Sorry, don't know how to do this :redface:

G) didn't do this

H) I get v = 51.2 m/s
The volume flow rates should be equal: ->A1*V1 = A2*V2
 
  • #5
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Thanks, I'll be back later... :smile:
 
  • #6
Fermat
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I'll go watch my DVD :smile:
 
  • #7
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For Part I: there is a window of diameter .4 m which is 5 m from the bottom of the large tank, what is the average pressure?

i like to know if the middle point will give me the average pressure? Or how else do I get it?
 
  • #8
Fermat
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The average pressure is the pressure acting on the Centre of Pressure, which in this case, is the centroid of a circle. i.e. its geometric centre.
 
  • #9
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Okay... How did you get 9.43 for A?

I used P= P(initial) + pgh for small cylin and assumed that if I used height of (15-h) and set that equal to Fg of large mass...
 
  • #10
Fermat
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Let p1 be the pressure at the bottom of Cylinder 1 (the small cylinder)
Let p2 be the pressure at the bottom of Cylinder 2 (the large cylinder)

p1 is made up from the weight of m1 (the small disc) and the pressure of h1, the height of water in cyl1.
ditto for p2 in cyl2.

So, p1 = m1g/A1 + pgh1 (p = rho, density)
p2 = m2g/A2 + pgh2

and p1 = p2

for p1 I got,

p1 = 7806.55 + 147,000
p1 = 154,55 N/m²
=============

for p2 I got,

p2 = 62,452.4 + 9,807*h2 N/m²
=======================

Solving for h2 gave h2 = 9.43m
 
  • #11
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Thanks for that... now back to the other parts... :frown:
 

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