Calculating Fluid Eflux: The Impact of a Cylindrical Tank on a Platform

Click For Summary
SUMMARY

The discussion focuses on calculating the fluid eflux from a cylindrical tank with a diameter of 90 cm and a height of 6 m. The tank is filled with water to a depth of 3 m, and a plug with an area of 3 cm² is removed from the orifice at the bottom. The speed of the water striking the ground is derived using the equation H=0.5gt², leading to a time calculation of approximately 1.4 seconds. Additionally, the rate of change of volume over time is expressed as D(vol)/Dt = A*(2gh)½, confirming the relationship between velocity and height in fluid dynamics.

PREREQUISITES
  • Understanding of basic physics principles, specifically fluid dynamics
  • Familiarity with kinematic equations, including H=0.5gt²
  • Knowledge of calculus, particularly in relation to rates of change
  • Experience with geometric calculations involving cylinders
NEXT STEPS
  • Study fluid dynamics principles, focusing on Bernoulli's equation
  • Learn about the application of kinematic equations in real-world scenarios
  • Explore calculus applications in fluid mechanics, specifically differential equations
  • Investigate the effects of orifice size on fluid eflux rates
USEFUL FOR

Engineers, physicists, and students studying fluid dynamics or related fields will benefit from this discussion, particularly those interested in practical applications of fluid flow calculations.

alamin
Messages
8
Reaction score
0
A cylindrical tank of diameter 90 cm rests on top of a platform 6m
high. Initially the tank is filled with water to a depth of 3m. A plug
whose are is 3 cm^2 is removed from the orifice on the side of the
tank at the bottom.

1) at what speed will water strike the ground?
11) how long will it take for the tank to be empty?
 
Physics news on Phys.org
What are your thoughts on this? Anything you got so far?
 
Yeah, for part one:
H=0.5gt^2
t^2 = 12/9.8
t= (12/9.8)^.5

v=u +at
v= 0 + (12*9.8)^0.5
this we could achieve

But for part 2:
Velocity of eflux changes with time. thus
D(vol)/Dt = A*(2gh)^.5 = ((Pi*d^2)/4)*dh/dt
Is this correct?
 

Similar threads

  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
50
Views
7K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 24 ·
Replies
24
Views
4K
Replies
8
Views
3K
  • · Replies 49 ·
2
Replies
49
Views
5K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K