Question

A garden hose with a small puncture is stretched horizontally along the ground. The hose is attached to an open water faucet at one end and a closed nozzle at the other end. As you might suspect, water pressure builds up and water squirts vertically out of the puncture to a height of 0.73 m. Determine the pressure inside the hose. (Enter your answer to the nearest 1000 Pa.) 7161.3 If you use the subscripts "1" and "2" respectively to represent a point in the hose before the leak and the site of the leak, how does h compare to h2? How does v1 compare to v2? What is the value of the pressure at the site of the puncture?

Answer 1

Answer:

The pressure inside the hose 7000 Pa to the nearest 1000 Pa.

$h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}$

$V_{1} =V_{2}$

The pressure at the site of the puncture is $P₂ =7161.3 Pa$

Explanation:

According to Poiseuille's law, $P_{1} - P_{2} = \frac{128μLQ}{πD^{4} }$

Where $P_{1}$ is the pressure at a point $1$ before the leak, $P_{2}$ is the pressure at the point of  the leak $2$, μ = dynamic viscosity, L = the distance between points $1$ and $2$, Q = flow rate, D = the diameter of the garden hose.

Also,  from the equation $P =ρgh$, the equations $h₁ = \frac{P₁} {ρg}$ and $h₂ = \frac{P₂} {ρg}$ can be derived.

Combining Poseuille's law with the above, we get $h₁ - ρgh₂ = \frac{128μLQ}{πD^{4} }$

$h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}$

$V =\frac{Q}{A}$

Since the hose has a uniform diameter, the nozzle at the end is closed and neither point $1$ nor $2$ lie after the puncture,

$V_{1} =V_{2}$

The pressure at the site of the puncture $P₂ =ρgh₂$

$P₂ =1kgm⁻³ ×9.81ms⁻¹×0.73m$

$P₂ =7161.3 Pa$