How Does Water Flow Dynamics Work in a Tank System?

[av8tor]

Homework Statement

Link to picture: http://session.masteringphysics.com/problemAsset/1011222/12/yf_Figure_14_41.jpg

Water flows steadily from an open tank as shown in the figure. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 4.80×10−2 m^2; at point 3, where the water is discharged, it is 1.60×10−2 m^2. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.

Homework Equations

Delta V = (A)(v)( Delta t)

A1(v)1=A2(v)2

The Attempt at a Solution

I can't figure this out.

To know the velocity of at point 3, don't you have to know the area of the main tank, to get the velocity at point 2, and then point 3?

The Area of point 2 and 3 is given, but no velocity, so what am I missing??

 

[av8tor]

So it ask:

Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 {\rm s}. In other words, find the discharge rate \Delta V/\Delta t.

 

[nlightner]

I would first clarify any uncertainties or missing information in the given problem. In this case, it seems that the velocity at points 2 and 3 is not provided, which is necessary to solve the problem. Additionally, the specific units for the cross-sectional areas are not given, which could affect the calculations. I would also suggest checking the given values and equations to ensure they are consistent and appropriate for the problem.

Assuming that the given values are accurate and the equation for continuity (A1v1 = A2v2) is applicable, we can solve for the velocity at point 3 by setting the cross-sectional areas and velocities at points 1 and 2 equal to each other. This will give us the velocity at point 2, which can then be used to solve for the velocity at point 3.

Alternatively, if the tank is open to the atmosphere and the water is flowing out of it, we can use the equation for Bernoulli's principle (P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2) to solve for the velocity at point 3. This equation takes into account the pressure and elevation differences between points 1 and 2, along with the velocity of the water at each point.

In either case, it is important to carefully consider the given information and use appropriate equations to solve for the unknown variables.