Question

A tank in the form of a right-circular cylinder of radius 2 feet and height 10 feet is standing on end. If the tank is initially full of water, and water leaks from a circular hole of radius 3 4 inch at its bottom, determine a differential equation for the height h of the water at time t. Ignore friction and contraction of water at the hole. (Assume the acceleration due to gravity g is 32.)

Answer 1

Answer:

$\frac{dh}{dt} = -3.48 \times 10^{-3}\sqrt{h}$ feet per second is the differential equation

Step-by-step explanation:

Given:

The radius of the cylindrical tank= 2 feet

The height of the cylindrical tank = 10 feet

The radius of the circular hole = 3/4 inches

To Find:

The differential equation for the height h of the water at time t.

Solution:

Finding the surface area(A) of the tank

Surface area  = $\pi R^2$

On substituting the values

Surface area =$\pi(2)^2$

= $4\pi$square feet

Finding the surface area(a) of the hole

The radius is given in  inches, so converting into feet we have

1 inch =  0.083 foot

similarly

$\frac{3}{4} = 0.75 inches = 0.75 \times 0.083$ =  0.0625 feet.

Now the surface area,

= $\pi \times 0.0625$

= $0.0625 \pi$ square feet

Now let the velocity of water through the hole is v

According law of conservation of energy, the penitential energy due to the height h of the  water gets converted into kinetic energy.

$\frac{1}{2}mv^2 =mgh$

$v^2 = \frac{2mgh}{m}$

$v^2 = 2gh$

$v= \sqrt{2gh}$

The rate of water flowing through the hole is  = $a\times v$

= >$a \times \sqrt{2gh}$

At any time t

$V(t) = A \times h(t)$

$\frac{dV}{dt} = -a\sqrt{2gh}$

$\frac{d(Ah(t))}{dt} = -a\sqrt{2gh}$

$A \frac{d(h(t))}{d(t)} = -a\sqrt{2gh}$

$\frac{dh}{dt} = -\frac{a}{A} \sqrt{2gh}$

On substituting the values, we get

$\frac{dh}{dt} = -\frac{0.0625\pi}{4\pi}\sqrt{2\times 32 \times 10$

$\frac{dh}{dt} = -3.48 \times 10^{-3}\sqrt{h}$  feet per second