Fluid Mechanics Mechanics
Deeptanshu
Fluid Kinematics Normal force per unit area - normal stress/pressure Tangential force per unit area - shearing stress , µ - coeffiecient of viscosity Newton’s law of viscosity- Shear stress τ = µ du dy For 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 valued and 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 ∂ (∂ u(x,y,z = 0, the transformation can be inverted. In ,u ,u ) such a case, ( u1 , u2 , u3 ) are the curvilinear coordinates of P .If at every point P (x,y,z ), the coordinate axes are mutually perpendicular, u1 , u2 , u3 form orthogonal curvilinear coordinates of P . Let e1 , e2 , e3 be the right-handed system of unit vectors extending in the direction of increasing u1 , u2 , u3 respectively. Scalar Factors or Material Coefficients h 1 , h2 , h3 are defined as 1
2
3
∂ r hi = ; ∂u i
where r = r (u1 , u2 , u3 )
∂ r = h i ei ∂u i
(scalar factors)
Quadratic Differential Form
dr =
hi dui ei
So, the differential of an arc length, ds satisfies (ds)2 = d r · dr =
h2i (dui )2
Differential Operators dψ = ∇ψ.d r
∇ui =
∇ψ =
i
∂ψ ∇ui = ∂u i
1
i
ei hi
ei ∂ψ hi ∂u i
(Gradient)
Let F (u1 , u2 , u3 ) = F 1 e1 + F 2 e2 + F 3 e3 ∇ · F =
1 h1 h2 h3
∂ (F 1 h2 h3 ) ∂ (F 2 h3 h1 ) ∂ (F 3 h1 h2 ) + + ∂u 1 ∂u 2 ∂u 3
∇ × F =
2
∇ ψ = ∇ · (∇ψ ) = ∇ ·
1 h1 h2 h3
1 ∂ψ h1 ∂u 1
h1 e1
h 2 e2
h3 e3
h1 F 1
h2 F 2
h3 F 3
∂ ∂u 1
e1 +
∂ ∂u 2
1 ∂ψ h2 ∂u 2
e2 +
∂ ∂u 3
1 ∂ψ h3 ∂u 3
(Divergence)
(Curl)
e3
(Laplacian)
The differential operators for Cartesian, Cylindrical and Spherical coordinate systems may be easily determined by the application of these general formula to the specific coordinate systems.
Kinematics There are two methods for studying fluid motion. • Lagrangian Method: Study the individual time-rate of change i.e. given initial position of a particleP 0 (x0 , y0 , z0 ) at t = t 0 , find out P (x,y,z ) at any time t. However, the fundamental equations are non-linear and difficult to solve in most cases. x = f 1 (x0 , y0 , z0 , t)
y = f 2 (x0 , y0 , z0 , t)
z = f 3 (x0 , y0 , z0 , t)
• Eulerian Method: Study the local time-rate of change i.e. select any fixed point in space and study the fluid as it passes through that point. u = F 1 (x,y,z,t)
v = F 2 (x,y,z,t )
z = F 3 (x,y,z,t)
, v = * To convert from Eulerian to Lagrangian system, use u = dx dt d D = dt to simplify in case none of the relations is an ODE.
dy . dt
Use
Derivatives , v = dy ,w = Let q = (u,v,w) be the velocity of the particle where u = dx dt dt Also, let f (x,y,z,t) be any scalar parameter associated with the fluid flow. δf =
dz . dt
∂f ∂f ∂f ∂f δx + δy + δz + δt ∂x ∂y ∂z ∂t
Dividing by δt and taking limit δt → 0, we get Df ∂f ∂f ∂f ∂f = +u +v +w Dt ∂t ∂x ∂y ∂z D Dt
Material Derivative
=
∂ ∂t
+
(q · ∇)
Convective Derivative
Local Derivative
2
(Derivatives)
The acceleration of a fluid particle of fixed identity is given by a =
Dq ∂ q = + (q · ∇)q Dt ∂t
(Acceleration of a Particle)
Note that the q · ∇ operator is on the whole velocity vector including the unit vectors.
Equation of Continuity ∂ρ + ∇ · (ρq) = 0 ∂t
(Eulerian Form)
Since ∇ · (ρq) = ∇ρ · q + ρ∇ · q D (log ρ) + ∇ · q = 0 Dt
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 1 + ∂t h1 h2 h3
∂ (ρq 1 h2 h3 ) ∂ (ρq 2 h3 h1 ) ∂ (ρq 3 h1 h2 ) + + = 0 (General Form) ∂u 1 ∂u 2 ∂u 3
Coordinate System
Cartesian Cylindrical Spherical
Coordinates (x,y,z ) (r,θ,z ) (r,θ,φ)
(h1 , h2 , h3 ) (1, 1, 1) (1, r, 1) (1, r , r sin θ )
Working Rule
For any general point P , construct a parallelopiped with edge lengths λ1 δα1 , λ2 δα2 , λ3 δα3 . Let the velocity components of fluid in these 3 directions be u, v,w respectively. ∂ Total excess flow-in through the first face = −λ1 δα 1 λ ∂α (ρuλ2 δα2 λ3 δα 3 ) So, conservation of mass gives 1
∂ (ρλ1 δα1 λ2 δα 2 λ3 δα 3 ) = − ∂t
λ1 δα1
1
∂ (ρuλ2 δα 2 λ3 δα3 ) λ1 ∂α 1
Condition for Boundary Surface A fluid and the surface with which contact is preserved must have zero relative velocity along the normal. For a surface F (r, t) = 0 to be a boundary surface of a fluid ∂F + q · ∇F = 0 ∂t
(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 of the velocity of fluid at that point. By definition, for a streamline, q × dr = 0. For Cartesian Coordinates, this reduces to dx dy dz = = u v 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 ∇2 φ = 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 any point 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 vortex motion. The rotation or angular velocity vector of a fluid element, ω satisfies Ω = 2ω
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(Rotation)
Inviscid Flow Euler’s Equation of Motion Dq 1 = F − ∇ p Dt ρ
(Euler’s Dynamical Equation)
Here, F is the external force acting per unit mass .
Cylindrical Coordinates Dq = Dt
Dq r q 2 Dq θ q r q θ Dq z + , − θ, Dt r Dt r Dt
∇≡
∂ 1 ∂ ∂ , , ∂r r ∂θ ∂z
Spherical Coordinates Dq = Dt
q θ2 + q φ2 Dq θ q φ2 cot θ Dq φ Dq r q r q θ q θ q φ cot θ , , − + − + Dt r Dt r r Dt r
∇≡
∂ 1 ∂ ∂ 1 , , ∂r r ∂θ r sin θ ∂φ
Euler’s equation can be written as ∂ q +∇ ∂t
q 2
2
1 + Ω × q = F − ∇ p ρ
(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 = ρ1 ∇ p. n = ∇(P + V +
q 2
2
) = q × Ω 2
n is normal to both q and Ω . So, the family of surfaces V + P + q2 = C contains
both 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) where r0 , v0 are parameters for a known point. • For spherical symmetry, it is r2 v = F (t). For cylindrical symmetry, it is rv = F (t). ∂v ∂t
• Differentiate it to obtain looking at a fixed point.
∂v in ∂t
• Plug in the value of with respect to r • Use v ∂v = ∂r
2 1 ∂ (v ) 2 ∂r
F (t) . f (r)
=
Do not differentiate f (r) as we are
the Euler’s equation of motion and integrate
if required.
• Determine the constant of integration using given conditions.
Impulsive Action Generated due to sudden velocity changes at the boundary or impulsive forces being made to act on the interior . The impulsive pressure is same in every direction and propagates instantaneously through the fluid. If the velocity changes instantaneously from q 1 to q 2 under the influence of impulsive pressure ω and impulsive body force per unit mass I
q2 − q1 = I −
∇ω
(Impulsive forces)
ρ
For an incompressible fluid, ∇ · q1 = ∇ · q2 . So, if I = 0 ∇2 ω = 0
If q1 = 0 , I = 0 , q = −∇ potential function.
ω ρ
(Incompressible fluid, I = 0 ) . So, the flow is irrotational and φ =
ω is ρ
the
Energy Equation The rate of change of total energy(kinetic,potential,intrinsic) of any portion of a compressible inviscid fluid is equal to the rate at which work is being done by the pressure on the boundary, provided the the potential due to extraneous forces is time-invariant.
T =
1 2 ρq dV V 2
W =
(Kinetic Energy) (Potential Energy)
ρσdV
V
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I =
(Intrinsic Energy)
ρEdV
V
E is the intrinsic energy per unit mass defined as E = dI =− dt
p∇ · qdV
V0 V
pdV .
V
Let n be the inward normal at the surface. Then, the net rate at which work is being done by the fluid pressure R =
pq · ndS
S
.
The Law of Conservation of Energy,also known as the Volume Integral form of Bernoulli’s Equation is d dI (T + W ) = R − = dt dt
pq · ndS +
S
d (T + W ) = R = dt
p∇ · qdV
(Energy Equation)
V
pq · ndS
(Incompressible Fluids)
S
Bernoulli’s Equation The general form, obtained by integrating Euler’s equation of motion, is −d
1 ∂φ dp + dq 2 + dV + =0 ∂t ρ 2
where φ is the velocity potential and V is the force potential. If ρ is a function of p only, ∂φ 1 2 − + q + V + ∂t 2
If ρ is constant, then −
dp = F (t) ρ
∂φ 1 2 p + q + V + = F (t) ∂t ρ 2
If the motion is steady, 1 2 q + V + 2
dp = C ρ
If the velocity potential does not exit, streamline.
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(Steady Motion) 1 2 2 q
+ V +
dp is ρ
constant along a
2D Motion A fluid is said to have 2-dimensional motion if, at any given instant, the flow pattern in a certain plane is the same as that in all other parallel planes within the fluid.
Stream Function The stream function or current function ψ for a 2D motion satisfies u = − q r =
∂ψ ∂y
1 ∂ψ r ∂θ
v =
∂ψ ∂x
(Stream Function)
q θ =
∂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, ψ 2 − ψ1 = flow across any line joining 1&2. ξ = 0
η = 0
ζ =
1 2 ∇ ψ 2
(Spin Components/Vorticity)
ζ = 0 =⇒ ∇2 ψ = 0
(Irrotational Flow)
w = φ + iψ
(Complex Potential)
dw ∂φ ∂φ = −i = −u + iv dz ∂x ∂y
(Complex Velocity)
w = −V e iα z is the complex potential of a uniform flow with magnitude V at an angle α to X − axis. −
Cauchy-Riemann Equation ∂φ ∂ψ = ∂x ∂y ∂φ 1 ∂ψ = ∂r r ∂θ
∂φ ∂ψ =− ∂y ∂x
1 ∂φ r ∂y
=−
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(Cartesian Form) ∂ψ ∂r
(Polar Form)
Sources and Sinks The mass m of fluid coming out of a source or going in to the sink per unit time is known as its strength. They are singularities in the flow field as infinitely many streamlines meet at a source/sink. So, the velocity vector is not unique at 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 2 πm . So, if q r is the radial velocity and the source is taken as the origin, 2πrq r = 2πm . This gives φ = −m log r, ψ = −mθ In general, for a source of stength m situated at z = z 1 is w = −m log z − z1 = −m log r +i · (−mθ)
φ
(Complex Potential)
ψ
The velocity of fluid at a point P due to a source of strength µ at A is along AP. ????????
µ r
Doublet/Dipole A source and sink of equal strength, m placed very at a very small distance δs apart. 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 side of the surface are known as the image of the system with regard to the surface
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If the surface S is treated as a rigid boundary and all the liquid removed from one side of it, the motion on the other side will remain unchanged. Since there 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 mirror image with respect to the line .
Conformal Mapping Consider two complex variables, z = x + iy,ζ = ξ + iη such that there exists a mapping of the z −plane into the ζ −plane,ζ = f (z ). The necessary condition for existence of such a mapping is that dζ must exist independent of the direction dz of δz . Suppose f maps neighbouring points P, P 1 , P 2 in the z −plane to Q, Q1 , Q2 in the ζ −plane. Then, ∠Q1 QQ2 = ∠ P 1 P P 2 i.e. a conformal mapping preserves angles.
QQ1 QQ2 dζ = = |f (z )| = P P 1 P P 2 dz
So, the map scales distances by | f (z )|. δζ δ (ξ + iη ) = = δz δ (x + iy )
∂ξ + ∂x
i ∂η δx + ∂x
∂ξ ∂y
δy + i ∂η ∂y
δx + iδy
If this is independent of δx , δy
∂ξ ∂η +i ∂y ∂y
= i
∂ξ ∂η + i ∂x ∂x
So, ζ satisfies the Cauchy-Riemann Conditions ∴
∂ξ ∂x
=
∂η ∂y
dζ ∂ξ ∂η ∂η ∂ξ = +i = −i dz ∂x ∂x ∂y ∂y
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∂ξ ∂y
= − ∂η ∂x
