PYQs

Let →q=uˆi+vˆȷ
Then,
u=−∂ϕ∂x
⟹u=−(2xy−y2+x2)
⟹u=y2−x2−2xy

Also,
v=−∂ϕ∂y
⇒v=−(x2−2xy−y2)
⟹v=y2−x2−2xy
We know that ϕ+iψ is an analytic function and salisfies Cauchy Riemann equations.

So, ∂ψ∂y=u and ∂ψ∂x=v
⇒∂ψ∂x=y2−x2+2xy
Integrating wrt x:
ψ=xy2−x33+x2y+f(y)

Now, ∂ψ∂y=2xy+x2+f′(y), and
∂ψ∂y=−u

⇒x2+2xy+f′(y)=x2+2xy−y2
⇒f′(y)=−y2
dfdy=−y2
⇒f=−y33+c

So,
ψ=xy2−(x3+y3)3+x2y+c
ψ=0 at origin ⟹c=0

So, ψ=xy2+x2y−(x3+y3)3

Checking possible flow:
Since∂u∂x+∂y∂y=−2x−2y+2y+2x=0
⟹ Equation of continuity is satisfied,
⟹ It is a possible flow.

Here u=3xzr5, v=3yzr5, w=3z2−r2r5=3z2r5−1r3… (1) where
r2=x2+y2+z2… (2)

From (2),∂r/∂x=x/r,∂r/∂y=y/r,∂r/∂z=z/r From (1), ( 2 ) and (3), we have

Since the equation of continuity is satisfied by the given values of u,v and w, the motion is possible. Let ϕ be the required velocity potential. Then

Integrating,

[Omitting constant of integration, for it has no significance in ϕ]

In spherical polar coordinates (r,θ,ϕ), we know that z=rcosθ. Hence the required potential is given by ϕ=(rcosθ)/r3=(cosθ)/r2. We now obtain the streamlines. The equations of streamlines are given by

So,

Taking the first two members of (4) and simplifying, we get

Integrating, logx−logy=logc1 l.e. x/y=c1,c1 being a constant …(5)

Now, each member in (4)=xdx+ydy+zdz3x2z+3y2z+3z3−r2z=xdx+ydy+zdz3z(x2+y2+z2)−r2z

Taking the first member in (4) and (6), we get

Integrating, (2/3)×logx=(1/2)×log(x2+y2+z2)+logc2 or x2/3=c2(x2+y2+z2)1/2,c2 being an arbitrary constant. …(7).

The required streamlines are the curves of intersction of (5) and (7).

Given, strength of source = m.

⟹ The complex potential at any point z is given by, w=−mlogz

∴dwdz=−mz=−mreiθ=−mre−iθ⟹dwdz=−mr(cosθ−isinθ)=−mcosθr+imsinθr ⟹dwdz=−mxr2+imyr2=−um+ivm

where um and vm are velocity components due to source at m. ∴um=mxr2,vm=myr2

Also, fluid is moving with velocity Uˆi. ∴ Total velocity. →q=(U+mxr2)ˆi+(myr2)ˆj=uˆi+ˆjv Let ϕ be the velocity potential. ∴u=−∂ϕ∂x

Integrating both sides, we get f=−Ux−mlogr+f(y) where, f(y) is arbitrary function of y Also, −∂ϕ∂y=v=myr2…(i)

where g(x) is an arbitrary function of x. From (i) and (ii) ϕ=−Ux−mlogr

Now, equation of streamlines is given by:

⇒dydx=myr2(U+mxr2), which is the differential equation of streamlines.

Since the motion is irrotational, consequently, the corresponding velocity potential ϕ exists such that:

The boundary conditions for ϕ are:

The above equations suggest that ϕ must involve terms containing sinθ and cosθ. So, we take ϕ as:

ϕ=(Ar+Br2)cosθ+(Cr+Dr2)sinθ…(3) ⟹−∂ϕ∂r=(−A+2Br3)cosθ+(−C+2Dr3)sinθ…(4)

Using boundary condiitons for (1) and (2) in (4), we get:

(−A+2Ba3)cosθ+(−C+2Da3)sinθ=Ucosθ…(5) (−A+2Bb3)cosθ+(−C+2Db3)sinθ=Vsinθ…(6)

Comparing coefficients of cosθ and sinθ in (5) and (6), we get:

−A+2Ba3=U,−C+2Da3=0…(7), and
−A+2Bb3=0,−C+2Db3=V…(8)

Solving (7) and (8), we get:

A=Ua3b3−a3, B=Ua3b32(b3−a3), C=−Ub3b3−a3, D=−Ua3b32(b3−a3)

Putting these values in (3), we get:

ϕ=(Ua3rb3−a3+ua3b32(b3−a3)r2)cosθ+(−Ub3b3−a3⋅r−Ua3b32(b3−a3)r2)sinθ
⟹ϕ={Ua3(1+b32r3)rcosθ−Vb2(1+a32r3)rsinθ}(b3−a3) ⟹ϕ={a3U(1+b3r−32)x−b3V(1+a3r−32)y}b3−a3

Now, let velocity at any point= →q=uˆi+vˆj where, u=−∂ϕ∂x and v=−∂ϕ∂y

Therefore,

u=−∂ϕ∂x=−a3U(1+b3r−32)b3−a3−a3Ux(−32b3r−4xr)(b3−a3) + b3Vy(−32a3r−4,xr)b3−a3 ⟹u=−a3U(1+b3r−32)b3−a3+32a3b3x2Ur−5(b3−a3)−32a3b3xyvr−5b3−a3

Also,

∴ velocity is given by:

→q=uˆi+ˆj, where u and v are given by (9) and (10).

Given, ϕ=12log{(x+a)2+y2]−12log((x−a)2+y2)…(1)

To determine stream lines.

−∂ϕ∂x=u=−∂ψ∂y,−∂ϕ∂y=v=∂ψ∂x Hence ∂ϕ∂x=∂ψ∂y,∂ϕ∂y=−∂ψ∂x

Now, ∂ϕ∂y=x+a(x+a)2+y2−x−a(x−a)2+y2

Integrating wrt y, ψ=tan−1yx+a−tan−1yx−a+F(x)…(2)

where F(x) is constant of integration. To determineF(x),

By (4), ∂ψ∂x=−y(x+a)2+y2+y(x−a)2+y2+F′(x) Equating (5) to (6), F′(x)=0. Integrating this F(x)= absolute const and hence neglected. Since it has no effect on the fluid motion.

Now (2) becomes

ψ=tan−1yx+a−tan−1yx−a =tan−1−2ayx2−a2+y2 ∴ Stream lines are given by ψ= const. i.e., tan−1[−2ayx2−a2+y2]=const.or,yx2−a2+y2=const. If we take const. =0, then we get y=0, i.e., x -axis. If we take conts.=∞, then we get circle x2−a2+y2=0.

From the fig, the n vortices are at A0, A1, A2…, An−1 such that ∠A0OA1=∠A1OA2=…=∠An−1OA1=2πn The coordinates of the points Ar are given by: z=zr=ae(2π/n)ir where r=0,1,2⋯n−1. These are n roots of the equation zn−an=0

Now, [zn−an=0⇒zn=ane2πri] Hence, zn−an=(z−z0)(z−z1)−⋯(z−zn−1) The complex potential due to n vortices at P is given by:

For the point A0, z=a, so that r=a and θ=0.

If w′ is the complex potential at A0, then

Finally, ∂ψ′∂θ=k4πa(n−1)=0, as r→a,θ→0.

Consequently, the velocity q0 of the vertex A0 is given by:

It is given that velocity potential, ϕ=xf(r).

For motion of fluid to be possible, ∇2ϕ=0

[xrf′(r)+x3r2f′′(r)+2xrf′(r)−x3r3f′(r)]+[xrf′(r)−xy2r3f′(r)+xy2r2f′′(r)] +[xrf′(r)−xz2r3f′(r)−x3r2f′′(r)]=0

⇒xf′′(r)+4xf′(r)r=0 f′′(r)f′(r)=−4r logf′(r)=−4logr+c1 f′(x)=c1r4…(1) f=−c13r3+C2…(2)