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NeX [460]
3 years ago
10

Oil of density 780 kg/m3 is flowing at a velocity of 20 m/s at the atmospheric pressure in a horizontal cylindrical tube elevate

d at 25 m above the sea level. Please find the tube radius if the rate of total mechanical energy (kinetic plus potential plus flow work) for oil is 57.5 kW (i.e. 57.5 kJ/s).
Engineering
1 answer:
Soloha48 [4]3 years ago
3 0

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

E = \dot m ( \frac{v^2}{2] +  zg + p\nu)

E = \dot m( \frac{v^2}{2] +  zg + p_{atm} \frac{1}{\rho})

57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})

solving for flow rate

\dot m = 99.977we know that [tex]\dot m  = \rho AV

\dot m  = 780 \frac{\pi}{4} D^2\times 16

solving for d

99.97 = 780 \times \frac{\pi}{4} D^2\times 16

d = 0.090 m

so radius = 0.045 m

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