1
$\begingroup$

I am considering the potential flow where I have a uniform flow past wing and two line vortices, one at the origin and one at a position $(x_1,y_1)$ relative to a wing of chord $D$.

I'm using the following potentials $$\phi_f(x,y)=Ux+Vy$$ $$\phi_{v1}(x,y)=\frac{\Gamma_1}{2\pi}\tan^{-1}\left(\frac{y}{x}\right)=\frac{\Gamma_1}{2\pi}\theta$$ $$\phi_{v2}(x,y)=\frac{\Gamma_2}{2\pi}\tan^{-1}\left(\frac{y-y_1}{x-x_1}\right)=\frac{\Gamma_2}{2\pi}\theta_1$$

I've tried to calculate the velocity potential $W(z)=\phi+i\psi$ where $\phi_x=\psi_y$, and $\phi_y=-\psi_x$ and I'm getting stuck when I try to simplify the equations.

Just considering the 1st vortex for now my text uses an example where and it says $\theta-i\ln(r)=\ln(z)$ where $z=x+iy$. Basically, I've been able to show that $$Z=\frac{y}{x}=-i\frac{z-z^*}{z+z^*}$$ where $z^*$ is the complex conjugate. I was able to rewrite the arc tan part of the equation $$\tan^{-1}(Z)=\frac{i}{2}\ln\left(\frac{1-iZ}{1+iZ}\right)$$ and $$\ln(r)=\frac{i}{2}\ln\left(zz^*\right)$$ so that $$\theta-i\ln(r)=\frac{i}{2}\ln\left(\frac{1-iZ}{1+iZ}\right)-\frac{1}{2}\ln\left(zz^*\right).$$

Could I get some help figuring this out? I haven't been able to get any further.

$\endgroup$
0
$\begingroup$

Playing around with the algebra with complex numbers in polar form led to this solution.

Define the complex number $z=x+iy$ in polar form, along with its conjugate. $$z= re^{i\theta},\; z^*=re^{-i\theta}$$ The product and quotient of the complex number and its conjugate respectively are. $$\therefore zz^*=r^2,\; \ln\left(\frac{z}{z^*}\right)=2i\theta$$

The complex potential for a line vortex with the potential and streamfunction is $$w(z)=\phi+i\psi$$ $$\phi_v=\frac{\Gamma}{2\pi}\theta,\; \psi_v=-i\frac{\Gamma}{2\pi}\ln\left(r\right)$$

$$w_v(z)=\frac{\Gamma}{2\pi}\left(\theta-i\ln(r)\right)$$ Follow through with some algebra $$w_v(z)=\frac{\Gamma}{2\pi}\left(\frac{2i}{2i}\theta-\frac{2}{2}i\ln(r)\right)$$ $$w_v(z)=\frac{\Gamma}{2\pi}\left(\frac{1}{2i}\ln\left(\frac{z}{z^*}\right)-\frac{1}{2}i\ln(zz^*)\right)$$ $$w_v(z)=\frac{\Gamma}{4\pi i}\left(\ln(z)-\ln(z^*)+\ln(z)+\ln(z^*)\right)$$ $$w_v(z)=\frac{\Gamma}{4\pi i}(2\ln(z))=\frac{\Gamma}{2\pi i}\ln(z)$$

The methodology shown here is valid even when the vortex is at $(x_1,y_1)$, or represented as $(r_1,\theta_1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.