Answer:
The velocity at R/2 (midway between the wall surface and the centerline) is given by (3/4)(Vmax) provided that Vmax is the maximum velocity in the tube.
Explanation:
Starting from the shell momentum balance equation, it can be proved that the velocity profile for fully developedblaminar low in a circular pipe of internal radius R and a radial axis starting from the centre of the pipe at r=0 to r=R is given as
v = (ΔPR²/4μL) [1 - (r²/R²)]
where v = fluid velocity at any point in the radial direction
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
But the maximum velocity of the fluid occurs at the middle of the pipe when r=0
Hence, maximum veloxity is
v(max) = (ΔPR²/4μL)
So, velocity at any point in the radial direction is
v = v(max) [1 - (r²/R²)]
At the point r = (R/2)
r² = (R²/4)
(r²/R²) = r² ÷ R² = (R²/4) ÷ (R²) = (1/4)
So,
1 - (r²/R²) = 1 - (1/4) = (3/4)
Hence, v at r = (R/2) is given as
v = v(max) × (3/4)
Hope this Helps!!!
