The loss of head in a pipe of certain length carrying a flow rate Q is found to be H. If a pipe of twice the diameter but the same length is to carry a flow rate of 2Q the head loss will be

Explanation:

Head loss in the pipe due to friction is given as:

\(h_f = \frac {4f lV^2}{2gD}=\frac {8}{\pi ^2g}\frac {4flQ^2}{D^5} \)

where f = friction coefficient, 4f = friction factor, V = average velocity of flow, D = diameter of pipe, l = length of pipe

\(V= \frac QA\)

where Q = flow rate, A = Cross Area of pipe

∴ head loss ∝ V²,  head loss ∝ Q²,  head loss ∝  l,  \(head~loss \propto \frac 1{D^5}\)

∴ \(head~loss \propto \frac {l\times Q^2}{D^5}\)

\((h_f)_1 = H =\frac {8}{\pi ^2g}\frac {4flQ^2}{D^5} \)

\((h_f)_2 =\frac {8}{\pi ^2g}\frac {4fl(2Q)^2}{(2D)^5} =\frac {8}{\pi ^2g}\frac {4flQ^2}{D^5}\times\frac {2^2}{2^5}=H\times \frac 18\)

\(\therefore (h_f)_2 = \frac H8\)

Alternate Method 

We know that, 

\(P_1 -P_2 = \frac{128\times\mu\times Q\times L}{\pi D^4}\)
\(P_1 -P_2 \propto h_f \propto \frac{Q}{D^4}\)

\(\frac{(h_f)_2}{(h_f)_1 } =\frac{2Q}{Q} \times \frac{D^4}{(2D)^4}\)

\((h_f)_2 = \frac {(h_f)_1}{8}\)

\((h_f)_2 = \frac {H}{8}\)