Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 10

Q. 10.14.3

A fluid motion for which the Reynolds number is small (so that nonlinear terms in velocity are negligible) is known as a creeping flow or Stokes’s flow. For a steady creeping flow of an incompressible viscous fluid under zero body force, show that p is a harmonic function. Deduce that ψ defined by (10.14.19) is a biharmonic function in this case.

\textbf{v}=\psi_{,2}\textbf{e}_{1}-\psi_{,1}\textbf{e}_{2}        (10.14.19)

Step-by-Step

Verified Solution

For creeping flow,

\frac{Dv}{Dt}=\frac{\partial \textbf{v}}{\partial t}+(\textbf{v}.\triangledown)\textbf{v}\approx \frac{\partial \textbf{v}}{\partial t}                  (10.14.27)

Then the Navier-Stokes equation (10.14.6) becomes

v\triangledown^2\textbf{v}-\frac{1}{\rho}\triangledown p+\textbf{b}=\frac{D\textbf{v}}{Dt}          (10.14.6)

v\triangledown^2\textbf{v}-\frac{1}{\rho}\triangledown p+\textbf{b}=\frac{\partial\textbf{v}}{\partial t}                      (10.14.28)

For steady flow with zero body force, this equation reduces to

v\triangledown^2\textbf{v}=\frac{1}{\rho}\triangledown p                      (10.14.29)

Taking the divergence of this equation and using the equation of continuity div v = 0, we get

\triangledown^2 p=0                    (10.14.30)

Thus, p is a harmonic function.
For ψ defined through the relation (10.14.19), we have

v_{1}=ψ_{,2},\ \ v_{2}=-ψ_{,1},\ \ v_{3}=0

Using these in equation (10.14.29), we get

p_{,1}=\rho v\triangledown^2(\psi_{,2});\ \ p_{,2}=-\rho v\triangledown^2(\psi_{,1})                  (10.14.31)

From these, we obtain, respectively,

p_{,12}=\rho v\triangledown^2(\psi_{,22});\ \ p_{,21}=-\rho v\triangledown^2(\psi_{,11})                    (10.14.32)

so that

v\rho\triangledown^2(\psi_{,11}+\psi_{,22})=-(p_{,21}-p_{,12})=0\ or\ \triangledown^2\psi=0

Thus, ψ is a biharmonic function.