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

The differential equation for the height h of the water at time t is d(h)/dt = -A * sqrt(2g/(πr^2)) * sqrt(h), where A is the area of the hole, r is the radius of the tank, and g is the acceleration due to gravity.

Step-by-step explanation:

To determine the differential equation for the height h of the water in the tank at time t, we can use Torricelli's theorem, which states that the speed at which water flows out of a hole in a tank is equal to the speed it would reach if it fell a distance h from the surface of the reservoir. In this case, the speed can be expressed as sqrt(2gh), where g is the acceleration due to gravity. The volume of water flowing out of the tank per unit time can be expressed as A * sqrt(2gh), where A is the area of the hole. The area of the hole is π * (3/4)^2. The change in volume of water in the tank per unit time is equal to the negative of the volume of water flowing out of the tank per unit time, since the water level is decreasing. Therefore, the differential equation for the height h of the water is dV/dt = -A * sqrt(2gh), where V is the volume of water in the tank. Since the tank is in the shape of a right-circular cylinder, the volume V can be expressed as πr^2h, where r is the radius of the tank. Substituting this expression into the differential equation, we get d(πr^2h)/dt = -A * sqrt(2gh). Simplifying, we have d(h)/dt = -A * sqrt(2g/(πr^2)) * sqrt(h). Therefore, the differential equation for the height h of the water at time t is d(h)/dt = -A * sqrt(2g/(πr^2)) * sqrt(h).

Answer 2

`(dh)/(dt) = -3.48 * 10^(-3)√(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

`(3)/(4) = 0.75 inches = 0.75 * 0.083` = 0.0625 feet.

Now the surface area,

=
`\pi * 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.

`(1)/(2)mv^2 =mgh`

`v^2 = (2mgh)/(m)`

`v^2 = 2gh`

`v= √(2gh)`

The rate of water flowing through the hole is =
`a* v`

= >
`a * √(2gh)`

At any time t

`V(t) = A * h(t)`

`(dV)/(dt) = -a√(2gh)`

`(d(Ah(t)))/(dt) = -a√(2gh)`

`A (d(h(t)))/(d(t)) = -a√(2gh)`

`(dh)/(dt) = -(a)/(A) √(2gh)`

On substituting the values, we get

`(dh)/(dt) = -(0.0625\pi)/(4\pi)\sqrt{2* 32 * 10`

`(dh)/(dt) = -3.48 * 10^(-3)√(h)` feet per second