Bernoulli equation (self made diagram)

[Kaxa2000]

An object is accidentally dropped into a water pipe. Pressure applied to pipe is 10 atm. The depth of lake is 10m. What will be speed of object when it exits the pipe and first enters the lake? Assume that the object is carried along with the surrounding water and does not affect flow of water in any way. Neglect friction.

Diagram is in attachment

I know you have to use Bernoullis equation for this problem. The examples I've looked at usually have a different area at each end. This problem has apparently the same area at each end. Also I don't understand how to factor in the height included in the problem.

 

[Kaxa2000]

.5*density *(0)² + 0 + 10atm(convert) = constant

(.5)(density)(vf)² + density(g)(y height) + 2atm

Would I set these equal and solve for vf??

What about the height of the lake and the height of the pipe?? Do those add more pressure??

 

[kerimek]

I would first start by breaking down the Bernoulli equation and understanding each component. The Bernoulli equation states that the total energy of a fluid remains constant along a streamline. In this case, the fluid is water and the streamline is the pipe.

The first component of the equation is the static pressure, which is the pressure applied to the pipe. In this case, it is given as 10 atm.

The second component is the dynamic pressure, which is the kinetic energy of the fluid. In this case, the object is dropped into the pipe, so the fluid is moving and therefore has kinetic energy.

The third component is the potential energy, which is determined by the elevation or height of the fluid. In this case, the pipe is connected to a lake, which is 10m deep. This means that the potential energy of the fluid is affected by the depth of the lake.

Now, to solve for the speed of the object when it exits the pipe and first enters the lake, we need to use the Bernoulli equation in the form of:

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2

Where P1 is the initial pressure (10 atm), v1 is the initial velocity (unknown), h1 is the initial height (unknown), P2 is the final pressure (atmospheric pressure), v2 is the final velocity (unknown), and h2 is the final height (10m).

Since we are neglecting friction, we can also assume that the pressure remains constant throughout the pipe, so P1 = P2.

Plugging in the known values and solving for v2, we get:

10 atm + 0 + ρgh1 = 1 atm + 1/2ρv2^2 + ρgh2

Since the object is dropped into the pipe, we can assume that its initial height (h1) is 0. Also, since the object is carried along with the surrounding water, we can assume that its final height (h2) is also 0. This simplifies the equation to:

10 atm = 1 atm + 1/2ρv2^2

Solving for v2, we get:

v2 = √(2(10 atm - 1 atm)/ρ)

To determine the density (ρ) of