SSC JE Mechanical Engineering 2018 Set 4 Paper

The most economical section of a rectangular channel for maximum discharge is obtained when its depth is equal to half the breadth.
Most economical rectangular section
• Let consider a rectangular channel as shown in figure in which width of channel is b and depth of flow is h.
Cross - sectional area of flow = A = bh
Wetted Perimeter = P = b + 2h = A/h + 2h
• For most economical section, perimeter should be minimum, i.e.,
$\frac{\mathit{d}\mathit{P}}{\mathit{d}\mathit{h}}=0$
$\frac{\mathit{d}\mathit{P}}{\mathit{d}\mathit{h}}=\frac{\mathit{d}}{\mathit{d}\mathit{h}}\left(\mathit{b}+2\mathit{h}\right)=\frac{\mathit{d}}{\mathit{d}\mathit{h}}\left(\frac{\mathit{A}}{\mathit{h}}+2\mathit{h}\right)=0$
$-\frac{\mathit{A}}{{\mathit{h}}^2}+2=0\Rightarrow \mathit{A}=2{\mathit{h}}^2$
$\mathit{b}\mathit{h}=2{\mathit{h}}^2$
$\mathit{b}=2\mathit{h}\mathrm{or}\mathit{h}=\frac{\mathit{b}}{2}$
Hence for most economical rectangular section, width is twice the depth of channel.