Potential Flow
We will cover following topics
PYQs
Potential Flow
1) For a two-dimensional potential flow, the velocity potential is given by ϕ=x2y=xy2+13(x3−y3). Determine the velocity components along the direction x and y. Also, determine the stream function ψ and check whether ϕ represents a possible case of flow or not.
[2018, 15M]
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.
2) If the velocity of an incompressible fluid at the point (x,y,z) is given by (3xzr5,3yzr5⋅3z2−r2r5), r2=x2+y2+z2 then prove that the liquid motion is possible and that the velocity potential is zr3. Further, determine the streamlines.
[2017, 15M]
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
∂u∂x=3z[1r5+(−5x)r−6∂r∂x]=3zr5−15x2zr7Since 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
dϕ=∂ϕ∂xdx+∂ϕ∂ydy+∂ϕ∂zdz=−(udx+vdy+wdz), by definition of ϕ=−[3xr5dx+3yzr5dy+3z2−r2r5dz]=r2dz−3z(xdx+ydy+zdz)r5 Thus, dϕ=r3dz−3r2zdr(r3)2=d(zr3), using (2)Integrating,
ϕ=z/r3[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
dxu=dyv=dzw i.e., dx3xz/r5=dy3yz/r5=dz(3z2−r2)/r5So,
dx3xz=dy3yz=dz3z2−r2…(4)Taking the first two members of (4) and simplifying, we get
dx/x=dy/y ⟹dx/x−dy/y=0Integrating, 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
=xdx+ydy+zdz3z(x2+y2+z2)−z(x2+y2+z2)=xdx+ydy+zdz2z(x2+y2+z2), by (2)…(6)Taking the first member in (4) and (6), we get
dx3xz=xdx+ydy+zdz2z(x2+y2+z2) or 23dxx=122xdx+2ydy+2zdzx2+y2+z2Integrating, (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).
3) A simple source of strength m is fixed at the origin O in a uniform stream of incompressible fluid moving with velocity U→i, show that the velocity potential ϕ at any point P of the stream is mr−Urcosθ, where OP=r and θ is the angle which OP makes with the direction i. Find the deferential equation of the streamlines and show that the lie on the surfaces Ur2sin2θ−2mcosθ= constant.
[2016, 15M]
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
⇒−∂ϕ∂x=U+mxr2 ⇒dϕ=(−U−mxr2)dxIntegrating both sides, we get f=−Ux−mlogr+f(y) where, f(y) is arbitrary function of y Also, −∂ϕ∂y=v=myr2…(i)
⇒dϕ=−myr2dy ⟹ϕ=−mlogr+g(x)…(ii)where g(x) is an arbitrary function of x. From (i) and (ii) ϕ=−Ux−mlogr
⇒ϕ=−Urcosθ−mlogrNow, equation of streamlines is given by:
dxu=dyv ⇒dxu+mxr2=dymyr2⇒dydx=myr2(U+mxr2), which is the differential equation of streamlines.
4) The space between two concentric spherical shells of radii a, b(a<b) is filled with a liquid of density ρ. If the shells are set in motion, the inner one with velocity U in the x−direction and the outer one with velocity V in the y−direction, then show has the initial motion of the liquid is given by velocity potential ϕ={a3U(1+12b3r−3)x−b3V(1+12a3r−3)y}(b3−a3), where r2=x2+y2+z2, the coordinate being rectangular. Evaluate the velocity at any point of the liquid.
[2016, 20M]
Since the motion is irrotational, consequently, the corresponding velocity potential ϕ exists such that:
∇2ϕ=0…(1)The boundary conditions for ϕ are:
(−∂ϕ∂r)r=a=Ucosθ…(2) (−∂ϕ∂r)r=b=Vsinθ…(3)
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,
⟹u=−a3u(1+b3r−3)(b3−a3)+3a3b3x2Ur−52(b3−a3)−3a3b3xyVr−52(−b3−a3)…(9) v=−∂ϕ∂y=−a3Ux(−32b2r−4yr)b3−a3+b3V(1+12a3r−3)b3−a3+b3Vy(−32a3r−4yr)b3−a3 ⟹v=−3a3b3xyUr−52(b3−a3)+b3v(1+12a3r−3)b3−a3−3a3b3y2Vr−52(b3−a3)…(10)∴ velocity is given by:
→q=uˆi+ˆj, where u and v are given by (9) and (10).
5) Given the velocity potential ϕ=12log[(x+a)2+y2(x−a)2+y2], determine the streamlines.
[2014, 20M]
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),
∂ψ∂x=−∂ϕ∂y=−y(x+a)2+y2+y(x−a)2+y2By (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.
6) If n rectilinear vortices of the same strength K are symmetrically arranged as generators of a circle cylinder of radius a in an infinite liquid, prove that the vortices will move round the cylinder uniformly in time 8π2a2(n−1)K. Find the velocity at any point of the liquid.
[2013, 20M]
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:
w=ik2π[log(z−z0)+log(z−z1)+⋯+log(z−zn−1)]=ik2πlog(z−z0)(z−z1)⋯(z−zn−1)=ik2πlog(zn−an)…(1)For the point A0, z=a, so that r=a and θ=0.
If w′ is the complex potential at A0, then
w′=w−ik2πlog(z−a)=ik2π[log(zn−an)−log(z−a)] ϕ′+iψ′=ik2π[log(rneinθ−an)−log(reiθ−a)] ψ′=k4π[log(r2n+a2n−2rnancosnθ)−log(r2+a2−2racosθ)] ∂ψ′∂r=k4π[2nr2n−1−2nrn−1ancosnθr2n+a2n−2rnancosnθ−2r−2acosθr2+a2−2racosθ] ∂ψ′∂θ=k4π[2nrnansinnθr2n−2rnancosnθ+a2n−2racosθr2+a2−2racosθ] (∂ψ′∂r)r=a=k4πa[n(1−cosnθ1−cosuθ)−(1−cosθ1−cosθ)]=k4πa(n−1) (∂ψ′∂θ)r=a=k4π[nsinnθ1−cosnθ−sinθ1−cosθ]=k4π[n2cosnθnsinnθ−cotθsinθ]asθ→0 (using L'Hospital's Rule)=k4π[−n3sinnθn2cosnθ−(−sinθ)cosθ] as θ→0 (using L'Hospital's Rule)=k4π[0+0]=0Finally, ∂ψ′∂θ=k4πa(n−1)=0, as r→a,θ→0.
Consequently, the velocity q0 of the vertex A0 is given by:
q0=[(∂ψ′∂r)2+1r(∂ψ′∂θ)2]1/2=k(n−1)4πa7) Show that ϕ=xf(r) is a possible form for the velocity potential for an incompressible fluid motion. If the fluid velocity ˙q→0 as r→∞, find the surfaces of constant speed.
[2012, 30M]
It is given that velocity potential, ϕ=xf(r).
For motion of fluid to be possible, ∇2ϕ=0
⇒∂2ϕ∂x2+∂2ϕ∂y2+∂2ϕ∂z2=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)
Now, →q=∇ϕ=f(r)ˆi+xrf′(r)ˆr
⟹ˉq=(−c13r3+c2)ˆı+(xc1r5)ˆr [using(1) and (2)] Since ˉq→0 as λ→∞⇒c2=0⇒ˉq=c13r3(ˆı−3xr2ˆr)Surfaces of constant speed are given by q2= Constant ⇒→q.→q=constant=(c13r3)2[(ˆi−3xr2ˆr)⋅(ˆi−3xr2ˆr)]=c219r6(1+3x2r2)=c219r8(r2+3x2)
∴ Surfaces of constant speed are given by (r2+3x2)r8=Constant
8) A rigid sphere of radius a is placed in a stream of fluid whose velocity in the undisturbed state is V. Determine the velocity of the fluid at any point of the disturbed stream.
[2012, 12M]
Let ϕ velocity potential of sphere and U be velocity of fluid along z−axis. Then ϕ satifies Laplace’s equation, thus ∂2ϕ∂r2+2r∂ϕ∂r+1r2∂2ϕ∂θ2+cotθr2∂ϕ∂θ=0
⇒ϕ is of the form f(r)cosθ Let ϕ=(Ar+Br2)cosθ
At boundary of sphere, i.e, r=a, normal velocity of sphere=velocity of fluid at that point.
⇒−∂ϕ∂r=0⇒−(A−2Ba3)cosθ=0⇒A=2Ba3…(1)⟹ϕ=[r(2Ba3)+Br3]cosθ[using (1)] −∂ϕ∂r=−[2Ba3−2Br3]cosθ As r→∞, velocity =Ucosθ(=∂ϕ∂x)⇒Ucosθ=−[2Ba3−2Br3]cosθ⇒B=−a3U2⇒A=−U[using (1)] ⇒ϕ=(−U)(r+a32r2)cosθ Now, velocity components are given by:
qr=−∂ϕ∂r=U(1−a3r3)cosθ qθ=−1r∂ϕdθ=(−u)(1+a32r3)9) If the velocity potential of a fluid is ϕ=1r3ztan1(yz), r2=x2+z2+z2 then show that the stream lines lie on the surfaces x2+y2+z2=c(x2+y2)2/3, c being a constant.
[2008, 12M]
10) Show that the velocity potential ϕ=12a(x2+y2−2z2) satisfies the Laplace equation, and determine the stream lines.
[2002, 12M]
11) If the velocity distribution of an incompressible fluid at the point (x,y,z) is given by (3xzr5,3yzr5,kz2−r2r5), then determine the parameter k such that it is a possible motion. Hence find its velocity potential.
[2001, 12M]