Question

The power transmitted through a pipeline is maximum when the head lost due to friction in the pipe is equal to

• A. the total supply head
• B. half of the total supply head
• C. one-third of the total supply head
• D. one-fourth of the total supply head

Answer 1

The correct option is C one-third of the total supply head

If H is the total supply head and $h_f$ is the loss of the head due to firction, then head available at the outlet of pipe (neglecting the minor losses) is $H-h_f$. If Q is the discharge through the pipe, V is the velocity of flow and L and D are the length and diameter of the pipe respecting, then from dercy-Weisbach equation we have

$h_f=\frac{fL{V}^2}{2gD}andQ=\frac{\pi D^2}{4}\times V$

$\Rightarrow P=\gamma \times \frac{\pi D^2}{4}\times V\times (H-\frac{fL{V}^2}{2gD})=0$

For a given pipe and head at the entrance the condition for the maximum power transmitted through the pipe line may be obtained by differentiating P w.r.t. V and equating it to zero

$\therefore \frac{dP}{dV}=\gamma \times \frac{\pi D^2}{4}\times (H-\frac{3fL{V}^2}{2gD})=0$

$\Rightarrow H-\frac{3fL{V}^2}{2gD}=0$

$\Rightarrow H-3h_f=0$

$\Rightarrow h_f=\frac{H}{3}$