Oil flows in a pipe 100 mm bore diameter with a Reynolds’ Number of 5...
Since Re is less than 2000 flow is laminar so Poiseuille’s equation applie
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Oil flows in a pipe 100 mm bore diameter with a Reynolds’ Number of 5...
Given data:
Bore diameter (D) = 100 mm
Reynolds’ Number (Re) = 500
Density (ρ) = 800 kg/m3
Viscosity (µ) = 0.08 Ns/m2
Radius (r) = 40 mm
We know that Reynolds’ Number (Re) is given by the formula:
Re = (ρvd)/µ
where,
v = velocity of the fluid
d = diameter of the pipe
Let’s rearrange the formula to find the velocity (v):
v = (Re µ)/ρd
We can find the diameter (d) from the bore diameter of the pipe:
d = 100/1000 = 0.1 m
Substituting the given values, we get:
v = (500 × 0.08)/(800 × 0.1) = 0.04 m/s
Now, we can use the formula for the velocity profile of a laminar flow in a pipe:
v(r) = (vmax/2) (1 - (r/R)^2)
where,
vmax = maximum velocity at the centerline of the pipe
r = radial distance from the centerline of the pipe
R = radius of the pipe
At the centerline of the pipe, r = 0, so vmax = v.
Substituting the given values, we get:
v(r) = (0.04/2) (1 - (0.04/0.1)^2) = 0.36 m/s
Therefore, the velocity of a streamline at a radius of 40 mm is 0.36 m/s. Hence, option (c) is the correct answer.