4. While traveling on a dirt road, the bottom of a car hits a sharp rock and a small hole develops at the bottom of its gas tank. If the height of the gasoline in the tank is 30 cm, determine the initial velocity of the gasoline at the hole. Discuss how the velocity will change with time and how the flow will be affected if the lid of the tank is closed tightly.

Initial velocity at hole = 2.426 m/s

Upon closure of lid, Velocity will change to: v2 = √[ (2gz1) - 2 (P2 - P1) / ρ]

Explanation:

This is a Bernoulli equation problem.

Now, let's assume that the outer and inner pressure of the gas tank are the same. Also, that the velocity of the fluid inside the tank v1 = 0 and the outside height z2 = 0 which will be the reference height.

The Bernoulli equation is;

P1 + ρ (v1) ²/2 + ρgz1 = P2 + ρ (v2) ²/2 + ρgz2

Putting v1 and z2 as 0, and P1 and P2 will cancel out as they are the same value, we have;

ρgz1 = ρ (v2) ²/2

ρ will cancel out to give;

g•z1 = (v2) ²/2

v2 = √2gz1

In the question, we are told that the height of the gasoline in the tank is 30 cm. So, z1 = 30cm = 0.3m

So,

v2 = √2 * 9.81 * 0.3

v2 = √5.886

v2 = 2.426 m/s

From the equation of v2 gotten earlier, we can see that the exit velocity depends on the height of the gas in the tank and that as the gas leaks out, the speed at which it is leaking out will decrease.

Now, if we close the lid of the tank, the inner and outer pressure would be not be the same as earlier assumed and so the velocity equation gotten earlier would change to;

v2 = √[ (2gz1) - 2 (P2 - P1) / ρ]