Chapter 2, Problem 2.11P

Assuming the velocity field given in Problem 2.6 pertains to an incompressible flow, calculate the stream function and velocity potential. Using your results, show that lines of constant $\varphi$ are perpendicular to lines of constant $\psi$.

To determine

To find:

The stream function and velocity potential for the given velocity functions and to prove it’s perpendicularity.

Answer to Problem 2.11P

Equation of the velocity potential is $\varphi =\frac{c}{2}\left(x^2-y^2\right)$.

Equation of the stream function is $\psi =cxy+d$.

Explanation of Solution

Given:

The horizontal and vertical velocity component of velocity is given as below ( problem 2.6).

$\begin{array}{l}u=cx\mathrm{....................}(1)\\ v=-cy\mathrm{.................}(2)\end{array}$

Stream function and velocity potential has to be found and prove perpendicularity of the stream function and velocity potential.

Stream function:

The equation of the stream function is given as follows:

$\begin{array}{l}u=\frac{\partial \psi }{\partial y}andv=-\frac{\partial \psi }{\partial x}\\ \text{Put the value of the u:}\\ \frac{\partial \psi }{\partial y}=cx\\ \partial \psi =cx\partial y\\ \text{Integrating both sides,}\\ {\displaystyle \int \partial \psi }={\displaystyle \int cx\partial }y\\ \psi =cxy+f(x)\mathrm{...............................}(3)\\ \text{Where f(x) is the integration constant}\text{.}\\ \text{Differentiate equation (3) with respect to x}\text{.}\\ \psi =cxy+f(x)\\ \frac{\partial }{\partial x}\left(\psi \right)=\frac{\partial }{\partial x}\left(cxy\right)+f\text{'}(x)\\ \frac{\partial \psi }{\partial x}=cy+f\text{'}(x)\\ \text{Put the vaue of the }\frac{\partial \psi }{\partial x}=cy.\\ cy=cy+f\text{'}(x)\\ f\text{'}(x)=0\\ f(x)=d\\ \text{Put the value of the f(x) in equation 1}\text{.}\\ \psi =cxy+0\\ \psi =cxy+d\mathrm{.........................}(4)answer\\ \text{Where d is constant}\text{.}\end{array}$

Equation of the stream function is $\psi =cxy+d$

Velocity potential:

The equation of the velocity potential is given as follows:

$\begin{array}{l}u=\frac{\partial \varphi }{\partial x}andv=\frac{\partial \varphi }{\partial y}\\ \text{Put the value of the u}\text{.}\\ \frac{\partial \varphi }{\partial x}=cx\\ \partial \varphi =cx\partial x\\ \text{Integrating both sides,}\\ {\displaystyle \int \partial \varphi }={\displaystyle \int cx\partial }x\\ \varphi =\frac{c{x}^2}{2}+f(y)\mathrm{...............................}(5)\\ \text{Where f(y) is the integration constant}\text{.}\\ \text{Differentiate equation (1) 1 with respect to y}\text{.}\\ \varphi =\frac{c{x}^2}{2}+f(y)\\ \frac{\partial }{\partial y}\left(\varphi \right)=\frac{\partial }{\partial y}\left(\frac{c{x}^2}{2}\right)+f\text{'}(y)\\ \frac{\partial \varphi }{\partial y}=0+f\text{'}(y)\\ \text{Put the vaue of the }\frac{\partial \varphi }{\partial y}=-cy.\\ f\text{'}(y)=-cy\\ f(y)=-\frac{c{y}^2}{2}\\ \text{Put the value of the f(y) in equation 2}\text{.}\\ \varphi =\frac{c{x}^2}{2}-\frac{c{y}^2}{2}\\ \varphi =\frac{c}{2}\left(x^2-y^2\right)\mathrm{.........................}(6)answer\end{array}$

Equation of the stream function is $\varphi =\frac{c}{2}\left(x^2-y^2\right)$.

Proof of the perpendicularity:

Differentiate equation (2) and (3) with respect to x keeping φ and ψ constant.

$\begin{array}{l}\psi =cxy\\ \frac{d}{dx}(\psi )=c\left(x\frac{dy}{dx}+y\right)\\ 0=x\frac{dy}{dx}+y\\ \frac{dy}{dx}=\frac{-y}{x}\mathrm{..............................}(7)\end{array}$

Similarly,

$\begin{array}{l}\frac{d\varphi }{dx}=\frac{c}{2}\left(2x-2y\frac{dy}{dx}\right)\\ 0=\frac{c}{2}\left(2x-2y\frac{dy}{dx}\right)\\ 2x-2y\frac{dy}{dx}=0\\ \frac{dy}{dx}=\frac{x}{y}\mathrm{..............................}(8)\end{array}$

Compare equations 7 and 9.

${\left(\frac{dy}{dx}\right)}_{\text{at constant }\varphi }=\frac{1}{-{\left(\frac{dy}{dx}\right)}_{\text{at constant }\psi }}$

Hence,the stream function and velocity potential are perpendicular to each other.