A nozzle is fitted to a pipe 120 mm in diameter and 250 m long, with coefficient of friction as 0.01. if the available head at the nozzle is 100 m find the diameter of the nozzle and the maximum power transmitted by a jet of water discharging freely out of a nozzle.

Given :

Diameter of pipe D = 120 mm = 0.12 m

Length of pipe L = 250 m

Co-efficient of friction F = 0.01

Head available at nozzle = 100 m

For maximum power transmission through nozzle, the diameter at the outlet of nozzle (d),

$d = (\frac{D^5}{8fL})^1/4 = (\frac{0.12^5}{8 \times 0.01 \times 250})^1/4$

$\therefore$ Diameter of Nozzle is d = 0.0334 m

Now, Area at the nozzle,

$a = \frac{\pi }{4} \times d^2 = \frac{\pi }{4} \times 0.0334^2 = 8.76 \times 10^-4 m^2$

The nozzle at outlet, discharges water into atmosphere and hence the total head available at the nozzle is converted into kinetic head,

$\therefore$ Head available at outlet = $V^2/2g$

i.e. $100 = V^2 / 2 \times 9.81$

$\therefore$ v = 44.29 m/s

$\therefore$ Discharge through nozzle, $Q = a \times v = 8.76 \times 10^-4 \times 44.29 = 0.0388 m^3/s$

Now, maximum power transmitted = $\frac{ \rho g \times Q \times \text{Head at outlet of nozzle}}{1000}$

Max. power transmitted = $\frac{1000 \times 9.81 \times 0.0388 \times 100}{1000} = 38.064 KW$