A garden hose with an internal diameter of $\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{cm}$is connected to a (stationary) lawn sprinkler that consists merely of a container with$\mathbf{24}$holes, each $\mathbf{0}\mathbf{.}\mathbf{13}\mathbf{}\mathbf{cm}$in diameter. If the water in the hose has a speed of $0.9\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$, at what speed does it leave the sprinkler holes?

i) The diameter of a garden hose is${\mathrm{d}}_1=1.9\mathrm{cm}$

ii) The number of holes of the lawn sprinkler is$\mathrm{N}=24$

iii) The diameter of each hole of the sprinkler is${\mathrm{d}}_2=0.13\mathrm{cm}$

iv) The speed of water in the hose is$\mathrm{v}=0.91\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

An ideal fluid is incompressible and lacks viscosity, and its flow is steady and irrotational. A streamline is a path followed by an individual fluid particle. A tube of flow is a bundle of streamlines. The flow within any tube of flow obeys the equation of continuity:

${\mathbf{R}}_{\mathbf{v}}\mathbf{=}\mathbf{Av}\mathbf{=}\mathbf{}\mathbf{Constant}\mathbf{}$

Which ${\mathbf{R}}_{\mathbf{v}}$is the volume flow rate, $\mathrm{A}$is the cross-sectional area of the tube of flow at any point, and $\mathbf{v}$is the speed of the fluid at that point.

Write the continuity equation in terms of the diameter of each hole and the diameter of a garden hose. Then inserting the given values in it, find the speed of water through the sprinkler holes.

Formulae are as follows:

${\mathrm{A}}_1{\mathrm{v}}_1={\mathrm{A}}_2{\mathrm{v}}_2$

where $\mathrm{A}$is area and $\mathrm{v}$is velocity.

Let, ${\mathrm{A}}_2$be the area of a single hole of the sprinkler. Then, the equation of continuity becomes,

${\mathrm{A}}_1{\mathrm{V}}_1={\mathrm{NA}}_2{\mathrm{V}}_2$

${\mathrm{V}}_2-\frac{{\mathrm{A}}_1{\mathrm{V}}_1}{{\mathrm{NA}}_2}-\frac{{\mathrm{V}}_1}{\mathrm{N}}\times \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}$

So,

$\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{{\mathrm{r}}_1^2}{{\mathrm{r}}_2^2}$

$=\frac{{\mathrm{d}}_1^2}{{\mathrm{d}}_2^2}$

Substituting this in the above equation,

${\mathrm{V}}_2=\frac{{\mathrm{V}}_1}{\mathrm{N}}\times \frac{{\mathrm{d}}_1^2}{{\mathrm{d}}^2}$

$=\frac{0.91\mathrm{m}/\mathrm{s}}{24}\times \frac{{\left(1.9\times {10}^{-2}\mathrm{m}\right)}^2}{{\left(0.13\times {10}^{-2}\mathrm{m}\right)}^2}$

$=8.1\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

Therefore, the speed of water through the sprinkler holes is ${\mathrm{v}}_2=8.1\mathrm{m}/\mathrm{s}$.