Fluid Mechanics

Deeptanshu

Fluid Kinematics

Normal force per unit area - normal stress/pressureTangential force per unit area - shearing stressNewton’s law of viscosity- Shear stress τ = µdudy , µ - coeffiecient of viscosityFor an ideal fluid, τ = 0 i.e. ideal fluids are inviscid.

Orthogonal Curvilinear Coordinates

Suppose P (x, y, z) can be expressed in terms of 3 independent, single valuedand continuously differentiable scalar point functions u1, u2, u3 as

x = x(u1, u2, u3) y = y(u1, u2, u3) z = z(u1, u2, u3)

If the Jacobian ∂(x,y,z)∂(u1,u2,u3)

6= 0, the transformation can be inverted. In

such a case, (u1, u2, u3) are the curvilinear coordinates of P .If at every pointP (x, y, z), the coordinate axes are mutually perpendicular, u1, u2, u3 form or-thogonal curvilinear coordinates of P .

Let e1, e2, e3 be the right-handed system of unit vectors extending in thedirection of increasing u1, u2, u3 respectively.

Scalar Factors or Material Coefficients h1, h2, h3 are defined as

hi =

∣∣∣∣ ∂r

∂ui

∣∣∣∣ ; ∂r

∂ui= hiei (scalar factors)

where r = r(u1, u2, u3)

Quadratic Differential Form

dr =∑

hiduiei

So, the differential of an arc length, ds satisfies

(ds)2 = dr · dr =∑

h2i (dui)2

Differential Operators

dψ = ∇ψ.dr ∇ui =ei

hi

∇ψ =∑i

∂ψ

∂ui∇ui =

∑i

ei

hi

∂ψ

∂ui(Gradient)

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Let F(u1, u2, u3) = F1e1 + F2e2 + F3e3

∇ · F =1

h1h2h3

[∂(F1h2h3)

∂u1+∂(F2h3h1)

∂u2+∂(F3h1h2)

∂u3

](Divergence)

∇× F =1

h1h2h3

∣∣∣∣∣∣h1e1 h2e2 h3e3∂∂u1

∂∂u2

∂∂u3

h1F1 h2F2 h3F3

∣∣∣∣∣∣ (Curl)

∇2ψ = ∇ · (∇ψ) = ∇ ·(

1

h1

∂ψ

∂u1e1 +

1

h2

∂ψ

∂u2e2 +

1

h3

∂ψ

∂u3e3

)(Laplacian)

The differential operators for Cartesian, Cylindrical and Spherical coordinatesystems may be easily determined by the application of these general formulato the specific coordinate systems.

Kinematics

There are two methods for studying fluid motion.

• Lagrangian Method: Study the individual time-rate of change i.e. giveninitial position of a particleP0(x0, y0, z0) at t = t0, find out P (x, y, z) at anytime t. However, the fundamental equations are non-linear and difficultto solve in most cases.

x = f1(x0, y0, z0, t) y = f2(x0, y0, z0, t) z = f3(x0, y0, z0, t)

• Eulerian Method: Study the local time-rate of change i.e. select anyfixed point in space and study the fluid as it passes through that point.

u = F1(x, y, z, t) v = F2(x, y, z, t) z = F3(x, y, z, t)

* To convert from Eulerian to Lagrangian system, use u = dxdt , v = dy

dt . Use

D = ddt to simplify in case none of the relations is an ODE.

Derivatives

Let q = (u, v, w) be the velocity of the particle where u = dxdt , v = dy

dt , w = dzdt .

Also, let f(x, y, z, t) be any scalar parameter associated with the fluid flow.

δf =∂f

∂xδx+

∂f

∂yδy +

∂f

∂zδz +

∂f

∂tδt

Dividing by δt and taking limit δt→ 0, we get

Df

Dt=∂f

∂t+ u

∂f

∂x+ v

∂f

∂y+ w

∂f

∂z

D

Dt︸︷︷︸Material Derivative

=∂

∂t︸︷︷︸Local Derivative

+ (q · ∇)︸ ︷︷ ︸Convective Derivative

(Derivatives)

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The acceleration of a fluid particle of fixed identity is given by

a =Dq

Dt=∂q

∂t+ (q · ∇)q (Acceleration of a Particle)

Note that the q · ∇ operator is on the whole velocity vector including theunit vectors.

Equation of Continuity

∂ρ

∂t+∇ · (ρq) = 0 (Eulerian Form)

Since ∇ · (ρq) = ∇ρ · q + ρ∇ · q

D(log ρ)

Dt+∇ · q = 0

It is derived from the Law of Conservation of Mass.The equation can be applied to any orthogonal coordinate system by using

the general definition of ∇ · F

∂ρ

∂t+

1

h1h2h3

[∂(ρq1h2h3)

∂u1+∂(ρq2h3h1)

∂u2+∂(ρq3h1h2)

∂u3

]= 0 (General Form)

Coordinate System Coordinates (h1, h2, h3)Cartesian (x, y, z) (1, 1, 1)

Cylindrical (r, θ, z) (1, r, 1)Spherical (r, θ, φ) (1, r, r sin θ)

Working Rule

For any general point P , construct a parallelopiped with edge lengths λ1δα1, λ2δα2, λ3δα3.Let the velocity components of fluid in these 3 directions be u, v, w respectively.

Total excess flow-in through the first face = −λ1δα1∂

λ1∂α1(ρuλ2δα2λ3δα3)

So, conservation of mass gives

∂

∂t(ρλ1δα1λ2δα2λ3δα3) = −

∑λ1δα1

∂

λ1∂α1(ρuλ2δα2λ3δα3)

Condition for Boundary Surface

A fluid and the surface with which contact is preserved must havezero relative velocity along the normal.

For a surface F (r, t) = 0 to be a boundary surface of a fluid

∂F

∂t+ q · ∇F = 0 (Boundary Surface Condition)

The normal at boundary, n is the unit vector along ∇F

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Streamlines

Streamline is a curve such that the tangent at any point is in the direction ofthe velocity of fluid at that point. By definition, for a streamline, q× dr = 0.

For Cartesian Coordinates, this reduces to

dx

u=dy

v=dz

w

Velocity Potential

φ is the velocity potential of a fluid flow if it satisfies

q = −∇φ (Velocity Potential)

The − sign is to ensure that flow takes place from higher to lower potential.The necessary and sufficient condition for velocity potential to exist is that

∇× q = 0. If such φ exists, the flow is known as irrotational.The surfaces φ = constant are called equipotentials and they intersect the

streamlines orthogonally.For an incompressible, irrotational fluid, the equation of continuity demands

that φ is a harmonic function

∇2φ = 0 (Incompressible Irrotational Fluid)

Vorticity

Ω = ∇× q (Vorticity Vector)

Like stream lines, a vortex line is a curve such that the tangent to it at anypoint is in the direction of the vorticity vetor i.e Ω× dr = 0.

A flow where Ω is not zero everywhere is said to be rotational or vortexmotion.

The rotation or angular velocity vector of a fluid element, ω satisfies

Ω = 2ω (Rotation)

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Inviscid Flow

Euler’s Equation of Motion

Dq

Dt= F− 1

ρ∇p (Euler’s Dynamical Equation)

Here, F is the external force acting per unit mass.

Cylindrical Coordinates

Dq

Dt=

(DqrDt− q2θ

r,DqθDt

+qrqθr,DqzDt

)∇ ≡

(∂

∂r,

1

r

∂

∂θ,∂

∂z

)

Spherical Coordinates

Dq

Dt=

(DqrDt−q2θ + q2φ

r,DqθDt

+qrqθr−q2φ cot θ

r,DqφDt

+qθqφ cot θ

r

)

∇ ≡(∂

∂r,

1

r

∂

∂θ,

1

r sin θ

∂

∂φ

)Euler’s equation can be written as

∂q

∂t+∇

(q2

2

)+ Ω× q = F− 1

ρ∇p (Lamb’s Hydrodynamical Equation)

If an ideal fluid satisfies the following conditions:

• Density ρ is a function of pressure p only

• The motion is in steady state

• The external forces are conservative

There must exist a function P such that ∇P = 1ρ∇p.

n = ∇(P + V +q2

2) = q×Ω

n is normal to both q and Ω. So, the family of surfaces V +P + q2

2 = C containsboth the streamlines and vortex lines.

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Working Rule

• Write down the equation of continuity as f(r)v = f(r0)v0 = F (t) wherer0, v0 are parameters for a known point.

• For spherical symmetry, it is r2v = F (t). For cylindrical symmetry, it isrv = F (t).

• Differentiate it to obtain ∂v∂t = F ′(t)

f(r) . Do not differentiate f(r) as we are

looking at a fixed point.

• Plug in the value of ∂v∂t in the Euler’s equation of motion and integrate

with respect to r

• Use v ∂v∂r = 12∂(v2)∂r if required.

• Determine the constant of integration using given conditions.

Impulsive Action

Generated due to sudden velocity changes at the boundary or impulsive forcesbeing made to act on the interior. The impulsive pressure is same in everydirection and propagates instantaneously through the fluid.

If the velocity changes instantaneously from q1 to q2 under the influence ofimpulsive pressure ω and impulsive body force per unit mass I

q2 − q1 = I− ∇ωρ

(Impulsive forces)

For an incompressible fluid, ∇ · q1 = ∇ · q2. So, if I = 0

∇2ω = 0 (Incompressible fluid, I = 0)

If q1 = 0, I = 0, q = −∇(ωρ

). So, the flow is irrotational and φ = ω

ρ is the

potential function.

Energy Equation

The rate of change of total energy(kinetic,potential,intrinsic) of anyportion of a compressible inviscid fluid is equal to the rate at whichwork is being done by the pressure on the boundary, provided thethe potential due to extraneous forces is time-invariant.

T =

∫V

1

2ρq2dV (Kinetic Energy)

W =

∫V

ρσdV (Potential Energy)

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I =

∫V

ρEdV (Intrinsic Energy)

E is the intrinsic energy per unit mass defined as E =∫ V0

VpdV .

dI

dt= −

∫V

p∇ · qdV

Let n be the inward normal at the surface. Then, the net rate at which workis being done by the fluid pressure

R =

∫S

pq · ndS

.The Law of Conservation of Energy,also known as the Volume Integral form

of Bernoulli’s Equation is

d

dt(T +W ) = R− dI

dt=

∫S

pq · ndS +

∫V

p∇ · qdV (Energy Equation)

d

dt(T +W ) = R =

∫S

pq · ndS (Incompressible Fluids)

Bernoulli’s Equation

The general form, obtained by integrating Euler’s equation of motion, is

−d(∂φ

∂t

)+

1

2dq2 + dV +

dp

ρ= 0

where φ is the velocity potential and V is the force potential.If ρ is a function of p only,

−∂φ∂t

+1

2q2 + V +

∫dp

ρ= F (t)

If ρ is constant, then

−∂φ∂t

+1

2q2 + V +

p

ρ= F (t)

If the motion is steady,

1

2q2 + V +

∫dp

ρ= C (Steady Motion)

If the velocity potential does not exit, 12q

2 + V +∫dpρ is constant along a

streamline.

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2D Motion

A fluid is said to have 2-dimensional motion if, at any given instant, the flowpattern in a certain plane is the same as that in all other parallel planes withinthe fluid.

Stream Function

The stream function or current function ψ for a 2D motion satisfies

u = −∂ψ∂y

v =∂ψ

∂x(Stream Function)

qr =1

r

∂ψ

∂θqθ =

∂ψ

∂r(Polar Coordinates)

It always exists for a 2D flow, even if the velocity potential φ does not.The equation of streamline is given by dψ = 0 or ψ = C.Also, ψ2 − ψ1 = flow across any line joining 1&2.

ξ = 0 η = 0 ζ =1

2∇2ψ (Spin Components/Vorticity)

ζ = 0 =⇒ ∇2ψ = 0 (Irrotational Flow)

w = φ+ iψ (Complex Potential)

dw

dz=∂φ

∂x− i∂φ

∂y= −u+ iv (Complex Velocity)

w = −V e−iαz is the complex potential of a uniform flow with magnitude Vat an angle α to X− axis.

Cauchy-Riemann Equation

∂φ

∂x=∂ψ

∂y

∂φ

∂y= −∂ψ

∂x(Cartesian Form)

∂φ

∂r=

1

r

∂ψ

∂θ

1

r

∂φ

∂y= −∂ψ

∂r(Polar Form)

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Sources and Sinks

The mass m of fluid coming out of a source or going in to the sink per unit timeis known as its strength. They are singularities in the flow field as infinitelymany streamlines meet at a source/sink. So, the velocity vector is not uniqueat the point of a source or sink.

A sink is regarded as a source of strength −m.

Complex Potential

In 2D, the flow across any small curve surrounding the source is 2πm. So, if qris the radial velocity and the source is taken as the origin, 2πrqr = 2πm. Thisgives φ = −m log r, ψ = −mθ

In general, for a source of stength m situated at z = z1 is

w = −m log z − z1 = −m log r︸ ︷︷ ︸φ

+i · (−mθ)︸ ︷︷ ︸ψ

(Complex Potential)

The velocity of fluid at a point P due to a source of strength µ at A is µr

along AP. ????????

Doublet/Dipole

A source and sink of equal strength,m placed very at a very small distance δsapart.

µ = mδs (Strength of Doublet)

The line from −m to +m is the axis of the doublet.For a doublet making an angle α with the X− axis, situated at z = z′

w =µeiα

z − z′(Complex Potential of Doublet)

Images

If in a liquid, a surface S can be drawn across which there is no flow,then any system of sources, sinks and doublets on the opposite sideof the surface are known as the image of the system with regard tothe surface

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If the surface S is treated as a rigid boundary and all the liquid removedfrom one side of it, the motion on the other side will remain unchanged. Sincethere is no flow across it, S must be a streamline.

The image of a source or a doublet with respect to a line(in 2D) is its mirrorimage with respect to the line.

Conformal Mapping

Consider two complex variables, z = x+ iy, ζ = ξ + iη such that there exists amapping of the z−plane into the ζ−plane,ζ = f(z). The necessary condition forexistence of such a mapping is that dζ

dz must exist independent of the directionof δz.

Suppose f maps neighbouring points P, P1, P2 in the z−plane to Q,Q1, Q2

in the ζ−plane. Then, ∠Q1QQ2 = ∠P1PP2 i.e. a conformal mapping preservesangles.

QQ1

PP1=QQ2

PP2= |f ′(z)| =

∣∣∣∣dζdz∣∣∣∣

So, the map scales distances by |f ′(z)|.

δζ

δz=δ(ξ + iη)

δ(x+ iy)=

(∂ξ∂x + i ∂η∂x

)δx+

(∂ξ∂y + i∂η∂y

)δy

δx+ iδy

If this is independent of δxδy ,(

∂ξ

∂y+ i

∂η

∂y

)= i

(∂ξ

∂x+ i

∂η

∂x

)So, ζ satisfies the Cauchy-Riemann Conditions ∂ξ

∂x = ∂η∂y

∂ξ∂y = −∂η∂x

∴dζ

dz=∂ξ

∂x+ i

∂η

∂x=∂η

∂y− i ∂ξ

∂y

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