Refer to Fig.1-4. A fluid has absolute viscosity 0.0010 lb-sec/ft2 and specific gravity 0.913. Calculate the velocity gradient and the intensity of shear stress at the boundary and at points 1 in, 2 in, and 3 in from the boundary, assuming (a) a straight-line velocity distribution and (b) a parabol

Question

Refer to Fig.1-4. A fluid has absolute viscosity 0.0010 lb-sec/ft[latex]^{2}[/latex] and specific gravity 0.913. Calculate the velocity gradient and the intensity of shear stress at the boundary and at points 1 in, 2 in, and 3 in from the boundary, assuming (a) a straight-line velocity distribution and (b) a parabolic velocity distribution. The parabola in the sketch has its vertex at A. Origin is at B.

Accepted Answer

(a) For the straight-line assumption, the relation between velocity and distance y is V = 15y. Then dV = 15 dy, or the velocity gradient is dV/dy = 15.

For y = 0, V = 0, dV/dy = 15 sec[latex]^{-1}[/latex] and

τ = μ(dV/dy) = 0.0010 x 15 = 0.015 1b/ft[latex]^{2}[/latex]

Similarly, for other values of y we also obtain τ = 0.015 lb/ft[latex]^{2}[/latex].

(b) The equation of the parabola must satisfy the condition that the velocity is zero at the boundary B. The equation of the parabola is V = 45 — 5(3 — y)[latex]^{2}[/latex]. Then dV/dy = 10(3 — y), and tabulation of results yields the following:

y	V	dV/dy	τ = 0.0010(dV/dy)
0	0	30	0.030 lb/ft[latex]^{2}[/latex]
1	25	20	0.020 lb/ft[latex]^{2}[/latex]
2	40	10	0.010 lb/ft[latex]^{2}[/latex]
3	45	0	0

It will be observed that where the velocity gradient is zero (which occurs at the centerline of a pipe flowing under pressure, as will be seen later) the shear stress is also zero.

Note that the units of velocity gradient are sec[latex]^{-1}[/latex], and therefore the product μ(dV/dy) = (1b-sec/ft[latex]^{2}[/latex])(sec[latex]^{-1}[/latex]) = lb/ft[latex]^{2}[/latex], the correct dimensions of shear stress τ.

Question 1.11: Refer to Fig.1-4. A fluid has absolute viscosity 0.0010 lb-s...

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