Question

To appreciate how viscous drag are being computed if we have the velocity field, let us consider the following velocity field for a flow of Newtonian viscous fluid u = x + 2y v = x 2 + y w = 0 (4.158) The fluid has viscosity of 10−2 kgm−1 s −1 (a) Determine the viscous stresses on the reference fluid element.(b) Determine the viscous stress vector acting on a surface aligned at an angle of 45 degrees (counterclockwise) from the x−axis.(c) Determine the viscous force acting on the same surface if surface spans from x = 2 to x = 5 and its length in the z-direction being 2 m.

Answer 1

Firstly let's find see the relation of the stress components in terms of velocity field

Now

u = x + 2y and v = 2x + y and w = 0

mu = 0.01

So we have

So we have a constant stress in the flow field , no rotating the stress tensor by 45 deg with x-axis to have required stress tensor

  

  

  

Now finding general force vector field on such a plane

  

  

  

Integrating from x=2 to x=5

Since we have constant argument for integration so we need to find only total area

From triangle corresponding a and b we have

so we get

A = 2 X (5-2) / cos(45) = 8.4853

So

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