Answer:
radius = 0.045 m
Explanation:
Given data:
density of oil = 780 kg/m^3
velocity = 20 m/s
height = 25 m
Total energy is = 57.5 kW
we have now
E = kinetic energy+ potential energy + flow work
[tex]E = \dot m ( \frac{v^2}{2] + zg + p\nu)[/tex]
[tex]E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})[/tex]
[tex]57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})[/tex]
solving for flow rate
[tex]\dot m = 99.977
we know that
[tex]\dot m = \rho AV[/tex]
[tex]\dot m = 780 \frac{\pi}{4} D^2\times 16[/tex]
solving for d
[tex]99.97 = 780 \times \frac{\pi}{4} D^2\times 16[/tex]
d = 0.090 m
so radius = 0.045 m
