A tube 12 cm in diameter is carrying engine oil a distance of 4 meters from...

A tube 12 cm in diameter is carrying engine oil a distance of 4 meters from barrel to outlet. If the pressure difference between the ends of the tube is 1064.28 atm , what is the oil flow rate?

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Expert Solution

The pressure difference between the ends of the tube dP= 1064.28 atm

dP = 1064.28 x1.01x10 ⁵ Pa

Diameter d = 12 cm

Radius r = d/ 2= 6 cm = 0.06 m

Area of crosssection A = r ²

= 11.309 x10 ^-3 m ²

We know dP = (1/2) v ²

From this velocity of the oil v = [2xdP/ ]

Where = density of engine oil = 888 kg / m3

Substitutue values you get v = [2(1064.28 x1.01x10 ⁵ ) /888]

= 492 m/s

The oil flow rate = A v

= 11.309 x10 ^-3 x 492

= 5.564 m³ /s

genius_generous answered 11 months ago

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