91Ó°ÊÓ

The downspouts height $h_1=11m.$

The floor drain height ${\mathrm{h}}_2=1.2\mathrm{m}.$

Radius of pipe $\mathrm{M}$is

$$\begin{gathered} {\mathrm{R}}_{\mathrm{m}}=3\mathrm{cm} \\ =0.03\mathrm{m} \end{gathered}$$

Side width of the house is$\mathrm{w}=30\mathrm{m}.$

Front length of the house is$\mathrm{L}=60\mathrm{m}.$

All water striking the roof goes through pipe$\mathrm{M}.$

Initial speed of the water in a downspout is negligible.

Wind speed is negligible.

By using Bernoulli’s equation, we find the speed ${\mathrm{V}}_{\mathbf{2}}$of water in pipe M and by putting this value of ${\mathbf{V}}_{\mathbf{2}}$in equation of continuity, we can find the rainfall rate $\mathit{c}\mathit{m}\mathbf{/}\mathit{h}\mathit{r}$in as the water from pipe $\mathbf{M}$reaches the height ${\mathbf{h}}_{\mathbf{2}}$ of the floor drain and threatens to flood the basement.

Formula:

Bernoulli's equation,

$\mathrm{P}+\frac{1}{2}{\mathrm{ÒϳÕ}}^2+\mathrm{Òϲ\mu ²â}=\mathrm{constant}$

Equation of continuity

$\mathrm{av}=\mathrm{constant}$

By applying Bernoulli’s equation, we get,

${\mathrm{P}}_1+\frac{1}{2}{{\mathrm{ÒϳÕ}}_1}^2+{\mathrm{Òϲ\mu ³ó}}_1={\mathrm{P}}_2+\frac{1}{2}{{\mathrm{ÒϳÕ}}_2}^2+{\mathrm{Òϲ\mu ³ó}}_2=\mathrm{constant}$

When the water level rises to high ${\mathrm{h}}_2$just on the verge of flooding, the speed $V_2$of the water in pipe $M$is given by

${\mathrm{Òϲ\mu ³ó}}_1=\frac{1}{2}{{\mathrm{ÒϳÕ}}_2}^2+{\mathrm{Òϲ\mu ³ó}}_2$

$\{\xcancel{}{V}_1$is negligible and${\mathrm{P}}_1={\mathrm{P}}_2\}$

$\mathrm{ÒÏ}g(h_1-h_2)=\frac{1}{2}\mathrm{ÒÏ}{V_2}^2$

$V_2=2g(h_1-h_2)$

$V_2=\sqrt{2g(h_1-h_2)}$

$=\sqrt{2x9.8m/s^2(11.0m-1.2m)}$

$=13.8593\mathrm{m}/\mathrm{s}$

Now, using the equation of continuity, we get

$V_1{A}_1=V_2{A}_2$

$V_1=\frac{V_2{A}_2}{A_1}$

Here, areas $A_1$and $A_2$are

${\mathrm{A}}_1=\mathrm{w}\mathrm{x}\mathrm{L}$

$=30mx60m$

$=1800{\mathrm{m}}^2$

${\mathrm{A}}_2={{\mathrm{Ï€¸é}}_{\mathrm{m}}}^2$

$=3.14\mathrm{x}{(0.03\mathrm{m})}^2$

$=0.002826{\mathrm{m}}^2$

Putting these values in the equation of continuity, we get

${\mathrm{V}}_1=\frac{13.8593\mathrm{m}/\mathrm{s}\mathrm{x}0.002826{\mathrm{m}}^2}{1800{\mathrm{m}}^2}$

$=2.176\mathrm{x}{10}^{-5}\mathrm{m}/\mathrm{s}$

But we need the rainfall rate in $\mathrm{cm}/\mathrm{hr},$

Conversion factor is

$1\mathrm{m}/\mathrm{s}=359999.999972\mathrm{cm}/\mathrm{hr}$

We get

${\mathrm{V}}_1\mathrm{in}\mathrm{m}/\mathrm{s}\mathrm{x}359999.999972={\mathrm{V}}_1\mathrm{in}\mathrm{cm}/\mathrm{hr}$

${\mathrm{V}}_1\mathrm{in}\mathrm{cm}/\mathrm{hr}=2.176\mathrm{x}{10}^{-5}\mathrm{m}/\mathrm{s}\mathrm{x}359999.999972$

$=7.8336\mathrm{cm}/\mathrm{hr}$

${\mathrm{V}}_1\mathrm{in}c\mathrm{m}/\mathrm{s}=7.8336cm/hr$

$=7.8\mathrm{cm}/\mathrm{hr}$

The rainfall rate in $\mathrm{cm}/\mathrm{hr}$as the water from pipe $M$reaches the height $h_2$of the floor drain and threatens to flood the basement is$7.8cm/hr.$