# Simple Fluid mechanics (bernoulli's?)

• mechEstudent
In summary, the pressure inside the can of Coca-Cola with a small pinhole leak spraying vertically into the air at a height of 0.5 m can be found using Bernoulli's equation with the assumption of atmospheric pressure at the top of the can and no initial velocity. The equation simplifies to p1 = (g*z2)/ρ, where z2 is the height of the spray and ρ is the density of Coca-Cola. This is because v1 = 0 everywhere inside the can and the pressure term is exchanged for potential energy in the Bernoulli equation.
mechEstudent

## Homework Statement

A can of Coca‐Cola has a small pinhole leak in it. The Coke sprays vertically
into the air at a height of 0.5 m. What is the pressure inside the can?
ρ = 1.11 g/mL

## Homework Equations

Bernoulli: p1/ρ + (v1^2)/2 +g*z1 = p2/ρ + (v2^2)/2 +g*z2

## The Attempt at a Solution

I've assumed:
z1=0, z2= 0.5
p2= atmospheric = 0 (gage)
v2=0

This simplifies bernoulli's to p1 = (g*z2-(v1^2)/2)/ρ
However v1 is still unknown and i have no way to solve for it

Last edited:
mechEstudent said:

## Homework Statement

A can of Coca‐Cola has a small pinhole leak in it. The Coke sprays vertically
into the air at a height of 0.5 m. What is the pressure inside the can?
ρ = 1.11 g/mL

## Homework Equations

Bernoulli: p1/ρ + (v1^2)/2 +g*z1 = p2/ρ + (v2^2)/2 +g*z2

## The Attempt at a Solution

I've assumed:
z1=0, z2= 0.5
p2= atmospheric = 0 (gage)
v2=0

This simplifies bernoulli's to p1 = (g*z2-(v1^2)/2)/ρ
However v1 is still unknown and i have no way to solve for it

Pretend you're a tiny bug just under the top surface of the can (where the hole is). How fast would you be moving? Or if you were at the bottom of the can?

Try to find $v_1$ from the simple equations of motion, in fact we have here the case of a vertical throw upwards.

ah, i see now that v1=0 as well, since point 1 is to be taken inside the can. I was mistakenly assuming v1 to be an exit velocity. i knew it was something simple. Thank you

mechEstudent said:
ah, i see now that v1=0 as well, since point 1 is to be taken inside the can. I was mistakenly assuming v1 to be an exit velocity. i knew it was something simple. Thank you

Correct! v1 = 0 everywhere inside the can. Inside the can, the pressure term in Bernoulli is exchanged for potential energy as we go from the bottom of the can to the top. But v1 = 0 everywhere inside.

## What is Bernoulli's principle?

Bernoulli's principle states that in a fluid flow, an increase in the speed of the fluid will result in a decrease in pressure, and vice versa.

## How is Bernoulli's principle applied in everyday life?

Bernoulli's principle can be observed in everyday life in situations such as the lift generated by an airplane wing, the flow of water through a garden hose, and the creation of sound by wind instruments.

## What is the equation for Bernoulli's principle?

The equation for Bernoulli's principle is P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2, where P is pressure, ρ is density, v is velocity, g is the acceleration due to gravity, and h is height.

## What are some limitations of Bernoulli's principle?

Bernoulli's principle assumes that the fluid is incompressible, inviscid, and irrotational, which may not always be the case in real-world situations. It also does not take into account external forces, such as friction, that may affect the flow of the fluid.

## How is Bernoulli's principle related to the conservation of energy?

Bernoulli's principle is a manifestation of the conservation of energy in fluid mechanics. It states that the total energy of the fluid, including its pressure, velocity, and potential energy, will remain constant as long as there are no external forces acting on it.

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