Continuum Mechanics - Bernouilli's Equation in a Cylindrical Tank

[lemonkey]

Homework Statement

A large cylindrical tank of radius R is full of water to a height h(t). This drains under gravity
out of the bottom of the tank through a small hole of radius r. The acceleration due to gravity
is g. The pressure of the air can be assumed to be the same at the top and at the bottom of the tank.

Show that if the uniform velocity of the fluid at the top of the tank is U and the uniform
velocity of the fluid leaving through the small hole is u then:
$$u = U\;\frac{R^2}{r^2}$$

By applying Bernoulli’s Theorem on a streamline from the top of the fluid to the bottom,
show that:
$$\frac{dh}{dt} = -\sqrt{2gh}\;(\frac{R^4}{r^4} - 1)^{-\frac{1}{2}}$$

Homework Equations

We've been given Bernoulli's Theorem as:
$$\frac{1}{2}u\bullet u + \frac{P}{\rho} + V = constant$$, where P is pressure, rho is density and V is potential.

The Attempt at a Solution

I tried using Bernoulli's Theorem, but wasn't sure what V was. I'm really not sure what to do. :(

 

[mighty2000]

Thank you for posting your question. Let me help you with the solution.

First, let's define the variables:
- R: radius of cylindrical tank
- h(t): height of water in the tank at time t
- r: radius of the small hole at the bottom of the tank
- g: acceleration due to gravity
- U: uniform velocity of fluid at the top of the tank
- u: uniform velocity of fluid leaving through the small hole

To solve this problem, we will use Bernoulli's Theorem, which states that the total energy of a fluid particle remains constant along a streamline. This can be written as:

\frac{1}{2}\rho u^2 + P + \rho gh = constant

Where \rho is the density of the fluid and P is the pressure.

Now, let's apply this theorem at the top and bottom of the tank. At the top of the tank, the velocity of the fluid is U and the pressure is atmospheric pressure, which we can assume to be the same as the pressure at the bottom of the tank. Therefore, we can write the equation as:

\frac{1}{2}\rho U^2 + Patm + \rho gh = constant

At the bottom of the tank, the velocity of the fluid is u and the pressure is also atmospheric pressure. However, the height of the fluid is now h(t) - r, as the fluid level has decreased by r due to the draining. Therefore, the equation becomes:

\frac{1}{2}\rho u^2 + Patm + \rho g(h(t)-r) = constant

Since the total energy of the fluid particle remains constant, we can equate these two equations and solve for u:

\frac{1}{2}\rho U^2 + Patm + \rho gh = \frac{1}{2}\rho u^2 + Patm + \rho g(h(t)-r)

\frac{1}{2}\rho U^2 + \rho gh = \frac{1}{2}\rho u^2 + \rho g(h(t)-r)

\frac{1}{2}\rho U^2 + \rho gh = \frac{1}{2}\rho u^2 + \rho gh - \rho gr

\frac{1}{2}\rho U^2 = \frac{1}{2}\rho u^2