Question 7.P.22: Water is drawn from a reservoir through a pipe of diameter D......

Consider a length dx at location x, using the equation, the drop dh over length dx is

h_{f}={\frac{f L u^{2}}{2g D}}        ∴        d h={\frac{f d x}{D}}{\frac{u^{2}}{2g}}           (1)

At this location, the flow rate Q can be obtained as

Q = K(L – x), as total flow is KL and draw off upto x is Kx.

u=\frac{4Q}{\pi D^{2}},\,u^{2}=\frac{16Q^{2}}{\pi^{2}D^{4}}=\frac{16K^{2}~(L-x)^{2}}{\pi^{2}D^{4}}

Substituting in eqn. (1)

d h={\frac{f d x}{2g}}\,{\frac{16K^{2}\,(L-x)^{2}}{\pi^{2}D^{5}}}={\frac{8fK^{2}}{g\pi^{2}\,D^{5}}}\,\,(L-x)^{2}\,d x

Integrating from x = 0 to L

h_{\mathrm{2}}-h_{\mathrm{1}}=h_{\mathrm{f}}={\frac{8\,f K^{2}}{g\pi^{2}\,D^{5}}} \int_{0}^{L}\left[L^{2}-2L x+x^{2}\right]d x =\frac{8\ f K^{2}}{g\pi^{2}D^{5}}\left[L^{3}-L^{3}+\frac{L^{3}}{3}\right]

∴             h_{f}={\frac{8\,f L^{3}\,K^{2}}{\,3g\pi^{2}\,D^{5}}}( Note: K has a unit m³/sm)

for the following data,

f=0.024,\,K=7.5\,l/h r/m=2.085\times10^{-6}\,\mathrm{m^{3}/s/m}, \\ D=0.1~{\mathrm{m}},L=4.8\times10^{3}~{\mathrm{m}}, \\ \mathbf{h}_{\mathrm{f}}={\frac{8\times0.024\times(4.8\times10^{3})^{3}\, (2.085\times10^{-6})^{2}}{3\times9.81\times\pi^{2}\times0.1^{5}}}=31.73\ \mathrm{m}

The head drop between lengths L_{1}\operatorname{and}L_{2} can be determined by difference i.e., (h_{f_{2}}-h_{f_{1}})