Introduction to Fluid Mechanics Tien-Tsan Shieh

April 16, 2009

What is a Fluid? The key distinction between a fluid and a solid lies in the mode of resistance to change of shape. The fluid, unlike the solid, cannot sustain a finite deformation under the action of a shear force. I

Hooke’s law σ = E holds for solids up to the proportion limit of strain. σ: strain and : normal stress.

I

For most fluids, Newton noted that τ ∝ ˙ where τ is a shear force and ˙ is the time rate of change of a fluid element’s deformation.

Consider a three-dimensional element. I

Hooke’s law of shear: pyx = G yx

I

du Newton’s law of viscosity: pyx = 2µ˙ = µ dy

Classification of Fluid Flows

I

Gases versus Liquid

I

Continuum versus Discrete Fluid

I

Perfect versus Real Fluids

I

Newtonian and Non-Newtonian Fluids

I

Compressible and Incompressible Fluids

I

Steady and Unsteady Fluid Flows

I

One, Two, Three-Dimensional Flows

I

Rotational versus Irrotational Flow

Properties of Fluids I I I I

I I

I

I I I

Mass: M Density: ρ Specific Weight: γ weight per unit volume Specific Gravity: S = ρg the ratio of density to the density of pure water Pressure: p the normal stress Bulk Modulus of Elasticity: K = − Vdp dV T a measure of the compressibility of liquids Absolute or Dynamic Viscosity: µ The viscosity of a gas increases with an increase of temperature. The viscosity of a liquid decreases with an increase of temperature. Kinematic Viscosity: ν = µρ Surface Tension: σ Capillary Rise or Depression: h

Aerohydrostatics For static state, the sum of all external forces acting on the fluid control colume is zero, so is the sum of all momnets of these forces. Consider a static fluid on earth. we will have ρ1 ∇p = g. Examples: I

Hydrostatics is the science of the static equilibrium of incompressible fliuds. p2 − p1 = ∆p = γh

I

Aerostatics differs from hydrostatics in the specific wight γ and/or density is no longer considered constant. I

Hally’s law: p = p0 exp

−gz RT0

by assuming the eq of state of air ρ = and isothermal T = T0 . I

αz T0

g Rα

p RT

(the perfect gas law)

Logrithmic Law: p = p0 1 − by assuming the eq of state and T = T0 − αz. (The typical value of α is 6.5 C/km)

Lagrange Description

The Lagrangian decription describes the history of the particle exaclty. But it is rarely used in fluid mechanics because of its significant mathematical complexities and experimental limitations. The Lagrangian description is often used to describe the dynamic behaviour of solids. u = x

= x(x0 , y0 , z0 , t)

y

= y (x0 , y0 , z0 , t)

z

= z(x0 , y0 , z0 , t)

v

=

w

=

∂x ∂t ∂y ∂t ∂z ∂t

ax

=

ay

=

aw

=

∂u ∂2x = 2 ∂t ∂t ∂v ∂2y = 2 ∂t ∂t ∂w ∂2z = 2 ∂t ∂t

Euler Description The Eulerian description is used to describe what is happening at a given spatial location P(x, y , z) in the flow field at a given instant of time. D : the Stoke deriv. Substantive Derivative Dt D ∂ ∂ ∂ ∂ u = f1 (x, y , z, t) ≡ +u +v +w Dt ∂t ∂x ∂y ∂z v = f2 (x, y , z, t) ∂ w = f3 (x, y , z, t) = +V·∇ ∂t The acceration in the Eulerian Description: Du Dt Dv ay = Dt Dw az = Dt ax =

= = =

∂u ∂u ∂u ∂u +u +v +w ∂t ∂x ∂y ∂z ∂v ∂v ∂v ∂v +u +v +w ∂t ∂x ∂y ∂z ∂w ∂w ∂w ∂w +u +v +w ∂t ∂x ∂y ∂z

Differential equations of fluid behaviour

6 unkown variable: three scalar velocity components, the temperature, the pressure and the density of the fluid. Here, we use Eulerian description. I

The equation of state (1)

I

The equation of continuity (1)

I

The equation of conservation of fluid momnetum (3)

I

The equation of conservation of fluid energy (1)

The conservation of mass I

I

The general property of balance: if φ is an intensive continuum qunatitiy of the fluid. Z Z Z D ∂ φ dx = φ dx + φV · dA Dt Ω ∂t Ω ∂Ω Dφ ∂φ = + ∇ · (φV) ∂t ∂t The equation of the conservation of mass ∂ρ + ∇ · (ρV) = 0 ∂t

I

If the fluid is incompressible (ρ =constant), the continuity equation is expressed as ∇·V =0

Decomposition of the motion of particles(I)

Express the velocity u, v , w of a particle at Q(x, y , z) near P(x0 , y0 , z0 ) in Taylor’s series form: ∂u ∂u ∂u x − x0 u u0 ∂x ∂y ∂z ∂v ∂v ∂v v = v0 + ∂x ∂y ∂z y − y0 +O(high order) ∂w ∂w ∂w z − z0 w w0 ∂x ∂y ∂z 0

V = V0 + A(r − r0 ) + B(r − r0 ) + O(high order) T the anti-symetric part A = 21 DV − (DV) the symetric part B = 12 DV + (DV)T

Decomposition of the motion of particles(II) The velocity can be expressed as V = V0 − (r − r0 ) × ω + (r − r0 ) · S˙ where the angular rotation is 1 ω = ∇×V 2 the strain rate dyadic is ˙S =

1 2 1 2

∂u ∂x ∂u ∂y + ∂u ∂z +

1 2

∂v ∂x ∂w ∂x

∂u ∂y + ∂v ∂y 1 ∂v 2 ∂z +

∂v ∂x

∂w ∂y

1 ∂u 2 ∂z + 1 ∂v 2 ∂z + ∂w ∂z

∂w ∂x ∂w ∂y

0

The strain rate dyadic S˙

˙xx ˙S = ∂ ij = ˙yx ∂t ˙zx =

1 2 1 2

∂u ∂x ∂u ∂y + ∂u ∂z +

˙xz ˙yz ˙zz

˙xy ˙yy ˙zy

∂v ∂x ∂w ∂x

∂u ∂y + ∂v ∂y 1 ∂v 2 ∂z + 1 2

∂v ∂x

∂w ∂y

1 ∂u 2 ∂z + 1 ∂v 2 ∂z + ∂w ∂z

∂w ∂x ∂w ∂y

The strain rate dyadic S˙ involves the dilatation and shearing strain of th fluid particle at P. The dilatation D is defined as D = ˙xx + ˙yy + ˙zz = ∇ · V

The stress dyadic P

Consider the most general form of a linear relation between a stress and a rate of strain. P = aS˙ + bI where tensor a contains 36 constants ”a“ and tensor b contains 3 constant ”b”. It is called the constitutive equation of fluid dynamics.

The stress dyadic P

Consider an isotropic fluid (no preferred direction), I

Incompressible fluid: P = 2µS˙ − pI

I

Compressible fluid: 2 P = 2µS˙ − (p + ∇ · V )I 3

where the pressure is p = − 13 (pxx + pyy + pzz ).

Newton’s viscosity potulates

Consider the isotropic incompressible fluid. Express the stress tensor pij as ( ∂uj i + µ ∂u ∂xj ∂xi , j 6= i pij = i −p + 2µ ∂u j =i ∂xi , Comparing the stresses with the strain rate tensor S, we see pxy

= 2µ˙xy

pxz

= 2µ˙xz

pyz

= 2µ˙yz

These relations are called Newton’s viscosity potulates.

Surface forces Fs and Vorticity ξ I

Surface forces: Normal part: Z Fσ =

P · dA,

i =j

P · dA,

i 6= j

A

Tangential part: Z Fτ = A I

Vorticity is defined by ξ =∇×V Note that ξ = 2ω.

Cauchy’s equation of motion P P Applying Newton’s second law: Fs + Fb = Ma where Fs surface forces and Fb body forces If there is only gravitational force acted on the body, we have Z Z Z aρ dx = gρ dx + P · dA Ω

Ω

∂Ω

1 a=g+ ∇·P ρ This is called the Cauchy’s equation of motion. ∂pyx ∂u ∂u ∂u ∂u 1 ∂pxx ∂pzx +u +v +w = gx + + + ∂t ∂x ∂y ∂z ρ ∂x ∂y ∂z ∂pyy ∂pzy ∂v ∂v ∂v 1 ∂pxy ∂v +u +v +w = gy + + + ∂t ∂x ∂y ∂z ρ ∂x ∂y ∂z ∂pyz ∂w ∂w ∂w ∂w 1 ∂pxz ∂pzz +u +v +w = gz + + + ∂t ∂x ∂y ∂z ρ ∂x ∂y ∂z

The Navier-Stokes Equations Consider a compressible fluid with the consitutive equation 2 P = 2µS˙ − (p + D)I 3 Pluging into Cauchy’s eq of motion, we obtain 1 ∂V + (V · ∇)V = g − ∇p + ν∇2 V + ν∇D ∂t ρ Navier-Stokes eq for incompressible flows: ∂V 1 + (V · ∇)V = g − ∇p + ν∇2 V ∂t ρ

Euler’s equation and Stokes flow

Navier-Stoke equation for inviscid fluid flow (ν = 0): 1 ∂V + (V · ∇)V = g − ∇p ∂t ρ It is usually called Euler’s equation.

For the case of very slow fluid motion, the Navier-Sokes equation becomes ∇p = µ∇2 V It is popularly called Stokes flow.

The Gromeka-Lamb form of the Navier-Stokes eq ∂V +ξ×V =g−∇ ∂t

p V2 + ρ 2

− ν(∇ × ξ)

where ξ = ∇ × V is a vorticity vector. I

For a steady and irrotational flow, g = ∇ ∂V ∂t

+ ξ × V = −∇

p ρ

+

V2 2

p ρ

+

V2 2

.

+Ω .

I

For inviscid fluid flow, where g = −∇Ω.

I

For a steady, andincompressible flow, inviscid p V2 V × ξ = ∇ ρ + 2 + Ω Crocco’s or lamb’s eq, this gives Bernoulli’s equation:

I

p ρ

+

V2 2

+ Ω = const.

For an irrotational flow (∇ × V = 0), we can set V = φ. 2 Bernoullli’s eq: pρ + V2 + Ω + ∂φ ∂t = c(t)

Conservation of energy(I) 2

Specific energy: e = i + V2 + gz + e nuclear + e elect + e magn + other In the present discussion, we shall neglect all energies except internal,kinetic and potential. D(ρe) ∂(ρe) Dt = ∂t + ∇ · (eρV) I

The first law of thermodynamics:

I

Conservation of heat:

I

Fourier’s Law: q = −k∇T dw The net power: dw dt = dt mech +

I

dq dt

dq dt

+

dw dt

D(ρe) Dt

=

= −∇ · q dw dt v

I

Lossof power due to viscous stress: dw ˙ ˙ dt v = −2µ∇ · (V · S) + 2µ(S · ∇) · V

I

The power due to the normal stresses:

dw dt p

+

dw dt p

= −∇ · (pV)

Conservation of energy(II)

dw dt

mech

∂(ρe) p ˙ ˙ = +∇· e + ρV − k∇T + 2µV · S −2µ(S·∇)·V ∂t ρ

This equation applies to any Newtonian fluid in a field where the only transfer of heat is by conduction. Examples I

∂(ρe) ∂t

= 0, No heat transfer: 0, Inviscid ∇T = V2 dw flow: µ = 0, we obtain dt mech = ∇ · h + 2 + gz ρV Steady flow:

where specific enthalpy h = i + pρ 2 The solution is wmech = ∆ h + V2 + gz . I

For a fluid at rest or moving with negligible velocity and having no mechanical energy transfer: ∂(ρi) ∂t = ∇ · k∇T

I

In paritcular, if the fluid is a perfect gas, then ∂ Cv ∂t (ρT ) = k∇2 T

Dimensional Analysis

I

The Buckingham π theorem is a key theorem in dimensional analysis. The theorem loosely states that if we have a physically meaningful equation involving a certain number, n, of physical variables, and these variables are expressible in terms of k independent fundamental physical quantities, then the original expression is equivalent to an equation involving a set of p = n − k dimensionless variables constructed from the original variables: it is a scheme for nondimensionalization.

I

The Rayleigh Method

Dimensionless parameters Dimensionless Navier-Stokes equation: ∂V∗ k 1 ∗2 ∗ + (V∗ · ∇∗ )V∗ = −∇∗ p ∗ − 2 + ∇ V ∂τ Fr RL L: a constant characteristic length U: a constant characteristic velocity I

Reynolds number RL =

I

Froude number Fr =

I

Mach number M =

I

Weber number W =

I

Euler number E = Cp =

∆p 1/2ρU 2

UL µ

√U gL

U c

and Cauch number C =

K ρU 2

U 2 Lρ σ

ρ∞ , ρU 2

and the Pressure coefficient

Reynolds number RL RL =

UL µ

Examples where RL is very large or infinite: I

Turbulent flows

I

Inviscid flows

I

Potential flows

I

Flows far removed from boundary

Examples where RL is very small: I

Creeping flows

I

Laminar flows

I

Stokes flows and lubrication theory

I

Bubble flows

I

Flows very close to a boundary

SOme other parameters I

Reynoolds number can be defined the ratio of the momentum flux to the shearing stress.

I

Fr > 1: tranquil flow or rapid flow

I

For large Mach number M ≥ 0.3, the effect of compressibility must be considered. 0.3 < M < 1: subsonic flow, M > 1: supersonic flow Mach number can be viewed as the ratio of teh intertial force to the compressibility.

I

Cauchy number is the ratio of the compressibility force to the intertial force. M = √1C .

I

Large Weber number W indicates surface tension is relatively unimportant, compared to the inertial force.

Types of flows

I

Viscous Fluid Flows

I

Laminar Pipe Flow

I

Turbulent Pipe Flow

I

Potential Flow

I

Open-Channel Flow

I

Boundary Layer Flows

I

One-dimensional Compressible Flows

References

I

Robert A. Granger, Fluid Mechanics, Dover, 1985.

April 16, 2009

What is a Fluid? The key distinction between a fluid and a solid lies in the mode of resistance to change of shape. The fluid, unlike the solid, cannot sustain a finite deformation under the action of a shear force. I

Hooke’s law σ = E holds for solids up to the proportion limit of strain. σ: strain and : normal stress.

I

For most fluids, Newton noted that τ ∝ ˙ where τ is a shear force and ˙ is the time rate of change of a fluid element’s deformation.

Consider a three-dimensional element. I

Hooke’s law of shear: pyx = G yx

I

du Newton’s law of viscosity: pyx = 2µ˙ = µ dy

Classification of Fluid Flows

I

Gases versus Liquid

I

Continuum versus Discrete Fluid

I

Perfect versus Real Fluids

I

Newtonian and Non-Newtonian Fluids

I

Compressible and Incompressible Fluids

I

Steady and Unsteady Fluid Flows

I

One, Two, Three-Dimensional Flows

I

Rotational versus Irrotational Flow

Properties of Fluids I I I I

I I

I

I I I

Mass: M Density: ρ Specific Weight: γ weight per unit volume Specific Gravity: S = ρg the ratio of density to the density of pure water Pressure: p the normal stress Bulk Modulus of Elasticity: K = − Vdp dV T a measure of the compressibility of liquids Absolute or Dynamic Viscosity: µ The viscosity of a gas increases with an increase of temperature. The viscosity of a liquid decreases with an increase of temperature. Kinematic Viscosity: ν = µρ Surface Tension: σ Capillary Rise or Depression: h

Aerohydrostatics For static state, the sum of all external forces acting on the fluid control colume is zero, so is the sum of all momnets of these forces. Consider a static fluid on earth. we will have ρ1 ∇p = g. Examples: I

Hydrostatics is the science of the static equilibrium of incompressible fliuds. p2 − p1 = ∆p = γh

I

Aerostatics differs from hydrostatics in the specific wight γ and/or density is no longer considered constant. I

Hally’s law: p = p0 exp

−gz RT0

by assuming the eq of state of air ρ = and isothermal T = T0 . I

αz T0

g Rα

p RT

(the perfect gas law)

Logrithmic Law: p = p0 1 − by assuming the eq of state and T = T0 − αz. (The typical value of α is 6.5 C/km)

Lagrange Description

The Lagrangian decription describes the history of the particle exaclty. But it is rarely used in fluid mechanics because of its significant mathematical complexities and experimental limitations. The Lagrangian description is often used to describe the dynamic behaviour of solids. u = x

= x(x0 , y0 , z0 , t)

y

= y (x0 , y0 , z0 , t)

z

= z(x0 , y0 , z0 , t)

v

=

w

=

∂x ∂t ∂y ∂t ∂z ∂t

ax

=

ay

=

aw

=

∂u ∂2x = 2 ∂t ∂t ∂v ∂2y = 2 ∂t ∂t ∂w ∂2z = 2 ∂t ∂t

Euler Description The Eulerian description is used to describe what is happening at a given spatial location P(x, y , z) in the flow field at a given instant of time. D : the Stoke deriv. Substantive Derivative Dt D ∂ ∂ ∂ ∂ u = f1 (x, y , z, t) ≡ +u +v +w Dt ∂t ∂x ∂y ∂z v = f2 (x, y , z, t) ∂ w = f3 (x, y , z, t) = +V·∇ ∂t The acceration in the Eulerian Description: Du Dt Dv ay = Dt Dw az = Dt ax =

= = =

∂u ∂u ∂u ∂u +u +v +w ∂t ∂x ∂y ∂z ∂v ∂v ∂v ∂v +u +v +w ∂t ∂x ∂y ∂z ∂w ∂w ∂w ∂w +u +v +w ∂t ∂x ∂y ∂z

Differential equations of fluid behaviour

6 unkown variable: three scalar velocity components, the temperature, the pressure and the density of the fluid. Here, we use Eulerian description. I

The equation of state (1)

I

The equation of continuity (1)

I

The equation of conservation of fluid momnetum (3)

I

The equation of conservation of fluid energy (1)

The conservation of mass I

I

The general property of balance: if φ is an intensive continuum qunatitiy of the fluid. Z Z Z D ∂ φ dx = φ dx + φV · dA Dt Ω ∂t Ω ∂Ω Dφ ∂φ = + ∇ · (φV) ∂t ∂t The equation of the conservation of mass ∂ρ + ∇ · (ρV) = 0 ∂t

I

If the fluid is incompressible (ρ =constant), the continuity equation is expressed as ∇·V =0

Decomposition of the motion of particles(I)

Express the velocity u, v , w of a particle at Q(x, y , z) near P(x0 , y0 , z0 ) in Taylor’s series form: ∂u ∂u ∂u x − x0 u u0 ∂x ∂y ∂z ∂v ∂v ∂v v = v0 + ∂x ∂y ∂z y − y0 +O(high order) ∂w ∂w ∂w z − z0 w w0 ∂x ∂y ∂z 0

V = V0 + A(r − r0 ) + B(r − r0 ) + O(high order) T the anti-symetric part A = 21 DV − (DV) the symetric part B = 12 DV + (DV)T

Decomposition of the motion of particles(II) The velocity can be expressed as V = V0 − (r − r0 ) × ω + (r − r0 ) · S˙ where the angular rotation is 1 ω = ∇×V 2 the strain rate dyadic is ˙S =

1 2 1 2

∂u ∂x ∂u ∂y + ∂u ∂z +

1 2

∂v ∂x ∂w ∂x

∂u ∂y + ∂v ∂y 1 ∂v 2 ∂z +

∂v ∂x

∂w ∂y

1 ∂u 2 ∂z + 1 ∂v 2 ∂z + ∂w ∂z

∂w ∂x ∂w ∂y

0

The strain rate dyadic S˙

˙xx ˙S = ∂ ij = ˙yx ∂t ˙zx =

1 2 1 2

∂u ∂x ∂u ∂y + ∂u ∂z +

˙xz ˙yz ˙zz

˙xy ˙yy ˙zy

∂v ∂x ∂w ∂x

∂u ∂y + ∂v ∂y 1 ∂v 2 ∂z + 1 2

∂v ∂x

∂w ∂y

1 ∂u 2 ∂z + 1 ∂v 2 ∂z + ∂w ∂z

∂w ∂x ∂w ∂y

The strain rate dyadic S˙ involves the dilatation and shearing strain of th fluid particle at P. The dilatation D is defined as D = ˙xx + ˙yy + ˙zz = ∇ · V

The stress dyadic P

Consider the most general form of a linear relation between a stress and a rate of strain. P = aS˙ + bI where tensor a contains 36 constants ”a“ and tensor b contains 3 constant ”b”. It is called the constitutive equation of fluid dynamics.

The stress dyadic P

Consider an isotropic fluid (no preferred direction), I

Incompressible fluid: P = 2µS˙ − pI

I

Compressible fluid: 2 P = 2µS˙ − (p + ∇ · V )I 3

where the pressure is p = − 13 (pxx + pyy + pzz ).

Newton’s viscosity potulates

Consider the isotropic incompressible fluid. Express the stress tensor pij as ( ∂uj i + µ ∂u ∂xj ∂xi , j 6= i pij = i −p + 2µ ∂u j =i ∂xi , Comparing the stresses with the strain rate tensor S, we see pxy

= 2µ˙xy

pxz

= 2µ˙xz

pyz

= 2µ˙yz

These relations are called Newton’s viscosity potulates.

Surface forces Fs and Vorticity ξ I

Surface forces: Normal part: Z Fσ =

P · dA,

i =j

P · dA,

i 6= j

A

Tangential part: Z Fτ = A I

Vorticity is defined by ξ =∇×V Note that ξ = 2ω.

Cauchy’s equation of motion P P Applying Newton’s second law: Fs + Fb = Ma where Fs surface forces and Fb body forces If there is only gravitational force acted on the body, we have Z Z Z aρ dx = gρ dx + P · dA Ω

Ω

∂Ω

1 a=g+ ∇·P ρ This is called the Cauchy’s equation of motion. ∂pyx ∂u ∂u ∂u ∂u 1 ∂pxx ∂pzx +u +v +w = gx + + + ∂t ∂x ∂y ∂z ρ ∂x ∂y ∂z ∂pyy ∂pzy ∂v ∂v ∂v 1 ∂pxy ∂v +u +v +w = gy + + + ∂t ∂x ∂y ∂z ρ ∂x ∂y ∂z ∂pyz ∂w ∂w ∂w ∂w 1 ∂pxz ∂pzz +u +v +w = gz + + + ∂t ∂x ∂y ∂z ρ ∂x ∂y ∂z

The Navier-Stokes Equations Consider a compressible fluid with the consitutive equation 2 P = 2µS˙ − (p + D)I 3 Pluging into Cauchy’s eq of motion, we obtain 1 ∂V + (V · ∇)V = g − ∇p + ν∇2 V + ν∇D ∂t ρ Navier-Stokes eq for incompressible flows: ∂V 1 + (V · ∇)V = g − ∇p + ν∇2 V ∂t ρ

Euler’s equation and Stokes flow

Navier-Stoke equation for inviscid fluid flow (ν = 0): 1 ∂V + (V · ∇)V = g − ∇p ∂t ρ It is usually called Euler’s equation.

For the case of very slow fluid motion, the Navier-Sokes equation becomes ∇p = µ∇2 V It is popularly called Stokes flow.

The Gromeka-Lamb form of the Navier-Stokes eq ∂V +ξ×V =g−∇ ∂t

p V2 + ρ 2

− ν(∇ × ξ)

where ξ = ∇ × V is a vorticity vector. I

For a steady and irrotational flow, g = ∇ ∂V ∂t

+ ξ × V = −∇

p ρ

+

V2 2

p ρ

+

V2 2

.

+Ω .

I

For inviscid fluid flow, where g = −∇Ω.

I

For a steady, andincompressible flow, inviscid p V2 V × ξ = ∇ ρ + 2 + Ω Crocco’s or lamb’s eq, this gives Bernoulli’s equation:

I

p ρ

+

V2 2

+ Ω = const.

For an irrotational flow (∇ × V = 0), we can set V = φ. 2 Bernoullli’s eq: pρ + V2 + Ω + ∂φ ∂t = c(t)

Conservation of energy(I) 2

Specific energy: e = i + V2 + gz + e nuclear + e elect + e magn + other In the present discussion, we shall neglect all energies except internal,kinetic and potential. D(ρe) ∂(ρe) Dt = ∂t + ∇ · (eρV) I

The first law of thermodynamics:

I

Conservation of heat:

I

Fourier’s Law: q = −k∇T dw The net power: dw dt = dt mech +

I

dq dt

dq dt

+

dw dt

D(ρe) Dt

=

= −∇ · q dw dt v

I

Lossof power due to viscous stress: dw ˙ ˙ dt v = −2µ∇ · (V · S) + 2µ(S · ∇) · V

I

The power due to the normal stresses:

dw dt p

+

dw dt p

= −∇ · (pV)

Conservation of energy(II)

dw dt

mech

∂(ρe) p ˙ ˙ = +∇· e + ρV − k∇T + 2µV · S −2µ(S·∇)·V ∂t ρ

This equation applies to any Newtonian fluid in a field where the only transfer of heat is by conduction. Examples I

∂(ρe) ∂t

= 0, No heat transfer: 0, Inviscid ∇T = V2 dw flow: µ = 0, we obtain dt mech = ∇ · h + 2 + gz ρV Steady flow:

where specific enthalpy h = i + pρ 2 The solution is wmech = ∆ h + V2 + gz . I

For a fluid at rest or moving with negligible velocity and having no mechanical energy transfer: ∂(ρi) ∂t = ∇ · k∇T

I

In paritcular, if the fluid is a perfect gas, then ∂ Cv ∂t (ρT ) = k∇2 T

Dimensional Analysis

I

The Buckingham π theorem is a key theorem in dimensional analysis. The theorem loosely states that if we have a physically meaningful equation involving a certain number, n, of physical variables, and these variables are expressible in terms of k independent fundamental physical quantities, then the original expression is equivalent to an equation involving a set of p = n − k dimensionless variables constructed from the original variables: it is a scheme for nondimensionalization.

I

The Rayleigh Method

Dimensionless parameters Dimensionless Navier-Stokes equation: ∂V∗ k 1 ∗2 ∗ + (V∗ · ∇∗ )V∗ = −∇∗ p ∗ − 2 + ∇ V ∂τ Fr RL L: a constant characteristic length U: a constant characteristic velocity I

Reynolds number RL =

I

Froude number Fr =

I

Mach number M =

I

Weber number W =

I

Euler number E = Cp =

∆p 1/2ρU 2

UL µ

√U gL

U c

and Cauch number C =

K ρU 2

U 2 Lρ σ

ρ∞ , ρU 2

and the Pressure coefficient

Reynolds number RL RL =

UL µ

Examples where RL is very large or infinite: I

Turbulent flows

I

Inviscid flows

I

Potential flows

I

Flows far removed from boundary

Examples where RL is very small: I

Creeping flows

I

Laminar flows

I

Stokes flows and lubrication theory

I

Bubble flows

I

Flows very close to a boundary

SOme other parameters I

Reynoolds number can be defined the ratio of the momentum flux to the shearing stress.

I

Fr > 1: tranquil flow or rapid flow

I

For large Mach number M ≥ 0.3, the effect of compressibility must be considered. 0.3 < M < 1: subsonic flow, M > 1: supersonic flow Mach number can be viewed as the ratio of teh intertial force to the compressibility.

I

Cauchy number is the ratio of the compressibility force to the intertial force. M = √1C .

I

Large Weber number W indicates surface tension is relatively unimportant, compared to the inertial force.

Types of flows

I

Viscous Fluid Flows

I

Laminar Pipe Flow

I

Turbulent Pipe Flow

I

Potential Flow

I

Open-Channel Flow

I

Boundary Layer Flows

I

One-dimensional Compressible Flows

References

I

Robert A. Granger, Fluid Mechanics, Dover, 1985.