# Hydromechanics - Flow from a tank with one outflow

1. Dec 5, 2012

### Maag

1. The problem statement, all variables and given/known data
I'm currently working on a project which has to do with flow from a tank with one outflow. I have to set up a mathematical equation which describes the flow. Furthermore I have to explain the physics behind the model.

The image shows the container and the outflow.
In order to describe the flow I have to assume that the fluid is incompressible (that the density is the same throughout the fluid), and that the flow is laminar. Also, the fluid is under the influence of a pressure p, the gravitation g and the fluid has the density ρ, by this Bernoullis equation can be used:

p1 + ρgy1 + 0,5ρv12 = p2 + ρgy2 + 0,5ρv22

(they're supposed to be 1 and 2 on the different sides, not 22 and 12)
Where y is the height and v is the velocity of the fluid. From this the formula for a free falling body in the gravitational field can be derived:

v = √(2gy)

Here v describes the velocity of the fluid exiting the container.

Now that the formula for a free falling body has been derived I will take a closer look at the mathematical formula that I have to derive:

With starting point in the formula derived from Bernoullis equation we can see a connection between v and y, which means an equation with two unknown quantities, therefore I have to create another equation where v and y is included. One assumes that the amount of water dVcontainer leaving the container during the time dt is the same amount dVoutflow which leaves the outflow, this is the sentence of mass conservation. To set up the equation I will use the height of the the fluid in the container as a function of the time t, h(t). The cross section of the outflow A, and the cross section of the container B. Both B and A are constants.

First I will see how far the fluid flows during dt:

s = v*dt

Afterwards the equation derived from bernoulli will be inserted:

dVoutflow = -√(2g*h(t)) * A * dt

The minus is included because the amount of fluid is reduced. Also the amount of water leaving is dependant on the cross section of the outflow. Now the volume of the container can be set up:

dVcontainer = B*dh(t)

It was assumed that the amount of water leaving the container is the amount of water leaving the outflow, therefore:

dVoutflow = dVcontainer

-√(2g*h(t)) * A * dt = B*dh(t) <=> (dh(t))/dt) = -(A/B)*√(2g*h(t))

This differentialequation can then be solved with the method "separation of variables", which then ends up with:

∫(1/√(h(t))) * dh(t) = ∫-(A/B) *√(2g) * dt

Which then is the solution to the differential equation.

My question is then: Which physical phenomenons and forces should I include in my paper when arguing for the physics behind the equation? I'm already including gravitation and turbulence (Turbulence is created due to high speeds and viscosity, viscosity is internal friction, friction = force)

-Maag

2. Dec 5, 2012

### SteamKing

Staff Emeritus
You have already stated in your assumptions that the fluid is incompressible and the flow is laminar. Therefore, the flow is not turbulent by definition.

3. Dec 5, 2012

### Maag

That was an assumption I had to make in order to make it simple. If I would have to bring in turbulence the mathematics would be incredibly difficult, since turbulence is very hard to define and work with. However, I still have to describe it because turbulence WILL occur in my experiment which I will set up later on.