Engr Help

Laminar Flow/Navier-Stokes Equations

Example The pressure drop needed to force water through a horizontal, 1 in. diameter pipe

is 0.60 psi for every 12 ft length of pipe. Find the shear stress on the pipe wall. Also find the

shear stress at distances of 0.3 and 0.5 in. away from the pipe wall.

Obtain the shearing stress at the wall from ∆P = 4 L τ_w / D So

τ_w = ∆P D / 4L = (0.6 psi)(144 in²/ft²) (1 in)(1ft/12 in) ] / 4(12 ft) = 0.15 lb/ft² (result)

Now the shear stress varies linearly with distance from the centerline of the pipe.

τ = 2 τ_w (r/D)

In this example the centerline is 0.5 in from the wall, so r = 0 and τ = 0 psf (result).

At 0.3 in from the wall, r = 0.2 in. So τ = 2 (0.15) (0.2/1) = 0.06 psf (result).

Example Find the distance from the centerline that the actual velocity equals the average velocity

for laminar flow in a round pipe of diameter, D.

Strategy: The fluid velocity for laminar flow is parabolic. u(r) = V_c [ 1 – (2r/D)²]

r=D/2

So the average velocity is: U_ave = ∫ u(r) dA / A = { ∫ V_c [ 1 – (2r/D)²] (2πr dr) } / (πD²/4)

r=0

Evaluation of the integral gives: U_ave = V_c / 2

Now u(r*) = V_c [ 1 – (2r*/D)²] = V_c / 2 so [ 1 – (2r*/D)²] = ½

And (2r*/D)² = ½ r* = D / √8 (result)