SSC JE Civil Engineering 2018 Set 5 Solved Paper

Major loss in long pipes is due to friction, Mainly given by Darcy weisbach equation ${\mathit{h}}_{\mathrm{f}}=\frac{{\mathrm{fLV}}^2}{2\mathit{g}\mathit{D}}$
f → Friction coefficient
L → Length of pipe
V → Velocity of flow
g → Gravitational acceleration = 9.81 m/sec${}^2$
D → Diameter of pipe
Minor losses in pipe:-
→ Losses due to sudden expansion
$\mathit{h}\mathit{e}=\frac{{\left({\mathit{V}}_1-{\mathit{V}}_2\right)}^2}{2\mathit{g}}$
→ Losses due to sudden contraction
$\mathit{h}\mathit{e}=\frac{{\mathit{V}}_2^2}{2\mathit{g}}{\left(\frac{1}{{\mathit{C}}_{\mathit{c}}}-1\right)}^2$
$\mathit{h}\mathit{e}=0.5\frac{{\mathit{V}}_2^2}{2\mathit{g}}$
${\mathit{C}}_{\mathit{c}}=$ Coefficient of contraction ${\mathit{C}}_{\mathit{c}}=\frac{{\mathit{V}}_2}{{\mathit{V}}_{\mathit{c}}}$
→ Losses at exit of pipe
$\mathit{h}\mathit{e}=\frac{{\mathit{V}}^2}{2\mathit{g}}$
→ Losses at entrance of pipe
$\mathit{h}\mathit{e}=\frac{0.5{\mathit{V}}^2}{2\mathit{g}}$
→ Losses due to bends
$\mathit{h}\mathit{e}=\frac{\mathit{K}{\mathit{V}}^2}{2\mathit{g}}$
K = Bend coefficient
≈ 1.2 for 90° bend
≈ 0.4 for 45° bend.