Problem $6$

The sketch shows one trough of an air humidifier system. The vertical cross$-$section of the trough is triangular with depth $D$ and half$-$angle $45,$ while the horizontal cross$-$section is rectangular with length $L=30cm.$ Initially, the trough is empty. At time $t=0,$ the valve is opened, and water enters at a constant flowrate of $Q_{in}=1\mathrm{.}5\frac{L}{h}.$ Water evaporates from the liquid surface into the air at a rate $Q_{\text{e}\text{v}\text{a}\text{p}}=0\mathrm{.}0015A_s,$ where $A_s$ is the liquid surface area $(wL)$ in $c{m}^2$ and $Q_{\text{e}\text{v}\text{a}\text{p}}$ is in $c\frac{m^3}{s}.$ The aim is to predict the depth $h$ of water in the trough versus time $t,$ and the minimum depth $D$ required for the trough to prevent overflow.

Your analysis should be carried out as follows:

$($i$)$ derive the differential equation for $h(t).$

$($ii$)$ solve the differential equation and the appropriate initial conditions to find $h(t).$

$($iii$)$ calculate the time required for the trough to fill to a depth $h=2cm.$

$($iv$)$ find $D.$