In: Physics
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?
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