Water is moving through a pipe. The velocity profile at some section is shown in Fig. and is given mathematically as v = (β/4µ) (d 2 /4 – r 2 ), where v = velocity of water at any position r, β = a constant, µ = viscosity of water, d = pipe diameter, and r = radial distance from centerline. What is the shear stress at the wall of the pipe due to the water? What is the shear stress at a position r = d/4? If the given profile persists a distance L along the pipe, what drag is induced on the pipe by the water in the direction of flow over this distance
Text Solution
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Sol. v = ( β /4 μ ) (d 2 /4 – r 2 ) dv/dr = ( β /4 μ ) (– r 2 ) = – 2 β r/4 μ
τ = μ (dv/dr) = μ (–2 β r/4 μ ) = –2 β r/4
At the wall, r = d/2. Hence,
τ wall = 
= – 
τ r=d/4 = 
= 
Drag = ( τ wall ) (area) = ( τ wall ) ( π dL) = ( β d/4) ( π dL) = β d 2 π L/4.
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