A uniform flow of velocity 6 m/s is flowing along x axis over a source & sink which are situated along x axis. The strength of source and sink is 15 m2/s and they are at a distance of 1.5m apart. Determine: - i. Location of stagnation points - ii. Lengths & width of the Rankine oval - iii. Equation of profile of the Rankine body -

Given:

$U=6 m/s \text{(Velocity of Uniform flow)} \\ k=15 m^2/s \text{(Strength of Source and Sink)} \\ 2s=1.5 m \text{(Distance between source and sink)}$

To find:

1. Location of stagnation points $(S_1 and S_2)$

2. Lengths & width of the Rankine oval (L and W)

3. Equation of profile of the Rankine body

Sol:

We know the Stream function for the Uniform Stream, Source and Sink combination is given by,

$$ψ=Uy-\dfrac{q}{2π} \tan^{-1}\bigg(\dfrac{2ys}{x^2+y^2-s^2}\bigg)$$

For the stagnation point S(x,y), we put u=0 at y=0

Now,

$\begin{aligned} u &=\dfrac{∂ψ}{∂y} \\ \therefore u&=0=\bigg[\dfrac{∂}{∂y}\bigg(Uy-\dfrac{q}{2π}tan^{-1} \bigg(\dfrac{2ys}{x^2+y^2-s^2}\bigg)\bigg)\bigg]_{y=0} \\ \therefore 0&=\bigg[U-\dfrac{q}{2π}\dfrac{1}{\bigg(1+\bigg(\dfrac{2ys}{x^2+y^2-s^2} \bigg)^2\bigg)} \bigg(\dfrac{(x^2+y^2-s^2 )2s-4sy^2}{(x^2+y^2-s^2 )}\bigg)\bigg]_{y=0} \\ \therefore 0&=\bigg[U-\dfrac{q}{2π} \bigg(\dfrac{2s}{x^2-s^2} \bigg)\bigg] \\ \therefore x&=±\sqrt{\dfrac{qs}{πU}+s^2} \\ \therefore x&=±\sqrt{\dfrac{qs}{πU}+s^2} \hspace{3cm} (\therefore q=2πk=94.2478 m^2/s and s=0.75 m) \\ \therefore x&=±\sqrt{\dfrac{94.2478(0.75)}{6π+0.75^2}} \\ \therefore x&=±2.0767 m \\ &\text{The location of stagnation points are} S_1 (-2.0767,0)and S_2 (2.0767,0) \end{aligned}$

$\begin{aligned} \therefore &\text{Length of the Rankine oval} = L=2x=4.1534 m \\ &\text{For the width of the Rankine oval, we put} ψ=0 at x=0 \\ \therefore ψ &=0=\bigg[Uy-\dfrac{q}{2π} \tan^{-1}\bigg(\dfrac{2ys}{x^2+y^2-s^2}\bigg) \bigg]_{x=0} \\ \therefore 0 &=\bigg[Uy-\dfrac{q}{2π} \tan^{-1}\bigg(\dfrac{2ys}{y^2-s^2}\bigg) \bigg] \\ \therefore Uy &=\dfrac{q}{2π} \tan^{-1}\bigg(\dfrac{2ys}{y^2-s^2}\bigg) \\ \therefore \bigg(\dfrac{2ys}{y^2-s^2}\bigg)&=\tan\bigg(\dfrac{2 \pi yU}{q}\bigg) \end{aligned}$

Solving by trial and error,

$\begin{aligned} ∴y&≈1.885 m \\ ∴& \text{Width of the Rankine oval }=W=2y=3.77 m \\ &\text{We know that equation of Profile for the Rankine Body is,} \\ ψ&=0 \\ ∴&Uy-\dfrac{q}{2π} \tan^{-1}\bigg(\dfrac{2ys}{x^2+y^2-s^2}\bigg)=0 \\ ∴y&=\dfrac{q}{2πU} \tan^{-1}\bigg(\dfrac{2ys}{x^2+y^2-s^2}\bigg) \\ &\text{…this is the required equation.} \end{aligned}$