Lecture 4. · 2019-09-13 · Lecture 4. dynamics continued 8 Energy budget |In the last lecture we...

Lecture 4: 1

Lecture 4.

Lecture 4: 2

Plan for today

— Summary of the key points of the last lecture.

— Dynamics of fluid motion continued. (Chapter 3 of Arzel’s

notes.)

— Energy equation

— Bernoulli’s equation

— Momentum equation (Euler’s theorem)

— Application of these notions. (Section 3.4 of Arzel’s notes.)

— Using Euler’s theorem to find the force on an array of

rigid bodies

— Using Bernoulli’s equation and Euler’s theorem, Betz’s

law

summary of lecture 3 3

Key points of last lecture

— The Eulerian description uses continuous functions

ui(x, y, z, t) that describe the fluid velocity at each location

(x, y, z, t) in the fluid. In our succinct notation we can write

this ui(xj , t). (I had to use a second index, j, because i is

already taken for the component of the velocity. )

— Eulerian and material time derivatives.


Table 1 – Two different time derivatives.

math physical interpretation standard name

∂∂tθ(xi, t) time rate of change of θ at

point xi

Eulerian time deriva-

tive

DDtθ(xj , t) time rate of change follo-

wing fluid parcel

Material (or substan-

tive or total) time de-

rivative

where the material derivative involves two terms :

D

Dtθ(xj , t) ≡

d

dtθ(xj(t), t) =

∂

∂tθ(xj , t) + uj

∂

∂xjθ(xj , t).

(1)

— Conservation of mass applied to a continuous fluid leads to


the continuity equation

Dρ

Dt= −ρ∂ui

∂xi(2)

— Incompressible flow has negligible divergence ∂ui

∂xi' 0 (or in

vector notation ∇ · ~u ' 0). Using the continuity equation and

the result from Lecture 2 that ∆ρ/ρ = O(M2/2)� 1 for

flows much slower than the sound speed, U � cs, we showed

that M � 1 is the condition for ∂ui

∂xi= 0. This is true for all

oceanic flows and all but the most extreme atmospheric

flows (M = 0.4 is the most extreme tornado ever measured.)

Recall we found the individual terms in ∂ui

∂xi= 0 can be (and

generally are) much larger than their sum. The individual

terms are of order U/L, the inverse of the advective time

scale, but the three terms tend to cancel each other.

— We applied Newton’s second law to a continuous fluid and


obtained the fundamental result :

∂ui∂t

+ uj∂ui∂xj

= Fi +1

ρ

∂σij∂xj

, (3)

where Fi is the i-th component of the body force per unit

mass.

— One obtains the Navier-Stokes equations by restricting the

above to a fluid with Newtonian viscosity and assuming

gravity is the only body force :

∂ui∂t

+ uj∂ui∂xj

= Fi +1

ρ

∂

∂xj(−pδij + 2µeij −

2

3µekkδij),

= −gδiz −1

ρ

∂p

∂xi+ ν

(∂2ui∂x2j

+1

3

∂

∂xi

∂uj∂xj

).

(4)

Lecture 4. dynamics and applications 7

Lecture 4. Dynamics of fluids continued

and some applications

Lecture 4. dynamics continued 8

Energy budget

— In the last lecture we derived the fundamental equation

governing fluid motions,

ρDuiDt

= ρFi +∂σij∂xj

, (5)

were Fi is the body force per unit mass. This was derived

from Newton’s second law, and represents a balance between

the rate of change of momentum of a fluid parcel with the

sum of body and surface forces acting on the fluid parcel.

— To obtain an energy equation, we take the inner product

with velocity

uiρDuiDt

= ui

(ρFi +

∂σij∂xj

), energy equation (6)


— Developing the LHS of the energy equation, Eq(6),

uiρDuiDt

= ρD(12uiui

)Dt

(7)

the rate of change of kinetic energy per unit mass times the

mass per unit volume.

— Developing the RHS of the energy equation, Eq(6),

ui

(ρFi +

∂σij∂xj

)= uiρFi − ui

∂δijp

∂xj+ ui

∂dij∂xj

, (8)

where because the stress tensor, σij , contains terms of often

very different magnitude, we separate it as we did before into

the (mechanical) pressure part, denoted simply as −δijp, and

the viscose part, dij , the so-called deviatoric stress tensor.

— We consider and interpret the 3 terms on the RHS of the

energy equation, Eq(6), separately :

1. uiρFi is the rate of work done by the body force per unit


volume of the fluid. In vector notation, ρ~u · ~F . Only the

part of the force in the direction of the movement does

work on the fluid. When the body force is gravity, this

work term results in the conversation of kinetic energy to

gravitational potential energy.

2. The second term involves the pressure

ui∂ − δijp∂xj

= −ui∂p

∂xi, summed over j (9)

This is the rate of work by the pressure gradient force.

3. The final term involves the viscose terms, which for a

Newtonian fluid can be written

ui∂dij∂xj

= −uiµ

(∂2ui∂x2j

+1

3

∂

∂xi

∂uj∂xj

), summed over j

(10)

The viscose terms have the local effect of diffusing (or


smearing out) the kinetic energy. But globally the viscose

terms provide a sink for fluid kinetic energy. Where does

the energy go ? When lost from the kinetic energy of the

flow the energy is converted to internal energy via

heating the fluid. You can think of this as a change in

scale from macroscopic length scales of order L to

microscopic scales associated with molecular motions, of

order d� ` (refer back to the continuum hypothesis of

Lecture 1).


Bernoulli’s equation

— Suppose we have a stationary flow (also called a “steady

flow”). This means the Eulerian time derivative is zero

everywhere. There are non-zero velocities but the velocity at

a fixed point in space does not change with time.

— Suppose furthermore the fluid is homogeneous

(ρ =constant), and inviscid (µ = 0).

— Suppose finally that the body force is derived from a

potential, for example gravity, Fi = − ∂ψ∂xi

, or in vector

notation ~F = −∇ψ.

— In this special case a theorem applies analogous to the

particle mechanics law of conservation of kinetic plus

potential energy.

— For an inviscid fluid, the mechanical pressure, for simplicity


we note with p, is the only contribution to the stress tensor

σij = −pδij .— The material derivative contains only the advective term for

a steady flow :

D

Dtui = uj

∂ui∂xj

. Eulerian derivatives vanish for steady flow

(11)

— Putting these simplifications into the Navier-Stokes

equations we have

ρuj∂ui∂xj

= − ∂p

∂xi− ρ ∂ψ

∂xi(12)

— The LHS (left-hand side) of the above equation contains the

velocity adjection, which in vector notation (recall exercise

from Lecture 3) is written as (~u · ∇) ~u.


— We now use a vector calculus identity to re-express the LHS :

(~u · ∇) ~u = ~ω × ~u+1

2∇(~u · ~u), (13)

where ~ω is an important dynamical quantity called the

vorticity, defined by

~ω ≡ ∇× ~u, ωi = εijk∂

∂xjuk. (14)

Eq(13) expresses the vector calculus identity well-known in

fluid mechanics. It is expressed in tensor notation

uj∂uk∂xj

= εi`m∂um∂x`

ujεijk +1

2

∂unun∂xk

(15)

But this is one of the rare occasions where vector notation is

more convenient, so we continue with vector notation.


— Using Eq(13) in the simplified Eq(12) gives

ρ

(~ω × ~u+

1

2∇(~u · ~u)

)= −∇p− ρ∇ψ. (16)

— We define a new scalar field H defined by the sum of terms

with units specific energy

H ≡ ~u · ~u2

+ ψ +p

ρ. (17)

and taking its gradient we find

∇H = ~u× ~ω. (18)

— This equation tells us that the vector ∇H is orthogonal to

both ~u and ~ω, i.e.

~u · ∇H = 0,

~ω · ∇H = 0, (19)


This means that along a streamline, and separately along a

line of vorcity, this scalar H is constant.

— Gravity is the most common body force, ψ = −gz, for which

Eqs(17) and (19) become (when multiplied by the constant

density ρ) we have the famous Bernoulli’s Equation

ρH = ρ~u · ~u

2+ ρgz + p = const. along a streamline (20)

This equation looks similar to the particle mechanics law of

conservation of energy applied to a fluid parcel of unit

volume, with an additional term involving pressure. Because

the flow field is steady, the pressure field is static and

provides a reservoir of potential energy p per unit volume

that augments the gravitational potential energy per unit

volume ρgz. The kinetic energy ρ~u·~u2 per unit volume of a

fluid parcel will increase (to keep ρH constant) as the fluid


parcel drops to lower height (smaller z) and to a region of

lower pressure (smaller p). Conversely, as the fluid parcel

rises to greater z or a region of higher pressure the velocity

slows down (smaller ~u · ~u).

— The vector quantity ~ω is a measure of the rate of rotation of

an infinitesimal fluid parcel about its centre. There are

special flows, called irrotational flow, for which the vorticity

vanishes everywhere, ~ω = ~0. In this case, from Eq(18), we

have ∇H = 0 so that

ρH = ρ~u · ~u

2+ ρgz + p = const. throughout domain (21)


Euler’s equation

— We return to the general momentum equation (Lecture 3,

Eq(34)) expressing Newton’s second law from which we

derived the Navier-Stokes equations :

∂ui∂t

+ uj∂ui∂xj

= Fi +1

ρ

∂σij∂xj

, (Lecture 3, Eq(34))

ρ∂ui∂t

+ ρuj∂ui∂xj


. multiplied by ρ (22)

— Furthermore, recall the continuity equation

∂ρ

∂t+

∂

∂xj(ρuj) = 0, (Lecture 3, Eq(20))

ui

(∂ρ

∂t+

∂

∂xj(ρuj)

)= 0. multiplied by ui (23)


— Add the two equations,

ρ∂ui∂t

+ ui∂ρ

∂t+ ui

∂

∂xj(ρuj) + ρuj

∂ui∂xj


,

∂ρui∂t

+∂ρujui∂xj


. (24)

This is another form of the general momentum equation, like

Eq(22), that is valid for a general continuum fluid.

— We restrict attention to a steady flow (stationary in time) so

the Eulerian time derivative vanishes in Eq(24)

∂ρujui∂xj


. (25)

— Now we consider a so-called control volume V bounded by a

non-moving surface A (where this surface could be a real

surface like the walls of a vessel, or completely imaginary, or

both like the walls of a pipe with imaginary cross sections


through the pipe) that is completely in the fluid. We

integrate Eq(25) over V∫V

∂ρujui∂xj

dV =

∫V

ρFidV +

∫V

∂σij∂xj

dV,∫A

ρujuinj dA =

∫V

ρFidV +

∫A

σijnj dA, used Gauss’ theorem

(26)

— Now consider three more restrictions : (1) the body force per

unit mass is conservative, Fi = −∂ψ/∂xi, (2) the fluid is

homogeneous ρ =constant and (3) inviscid so σij = −pmecδij

or for simplicity we drop the “mec” subscript and write

simply p for pressure. Introducing these three restrictions

into the integral momentum balance Eq(26). Let’s consider


the terms one by one. For the body force we can write :∫V

ρFidV = −ρ∫V

∂ψ

∂xidV,

= −ρ∫A

ψnidA = −∫A

ρψnidA. (27)

For the pressure term∫A

σijnj dA = −∫A

pδijnj dA,

= −∫A

pnidA. summed over j (28)

— Combining the two results above and inserting in the

integral momentum balance Eq(26) we arrive at the

momentum theorem (or Euler’s theorem) :∫A

ρujuinj + ρψni + pni dA = 0. (29)


The physical meaning of this result is as follows. The net

momentum flux across the control surface A by the fluid is

balanced by the sum of the net pressure force and

conservative force exerted on the fluid inside the control

surface.

Lecture 4. applications 23

Applications of Bernoulli’s equation and

Euler’s Theorem


Using Euler’s theorem to find the force

on an array of rigid bodies

— Consider an incompressible flow of steady uniform velocity

U far upstream, impinging on an array of similar rigid

bodies distributed regularly over a plane normal to the

stream, for instance a wire mesh used to filter the water.

The flow is bounded laterally by walls, for instance flow in a

pipe of uniform diameter with axes in the x-direction. The

flow my become turbulent in the wake of the array but far

enough downstream the flow returns to a steady uniform

flow of the same velocity U , required by continuity.

— The fluid exerts a force on the array, which is directed

downstream (as required by symmetry considerations). We

wish to determine this force per unit area using only


measured properties far upstream and downstream of the

array.

— We choose a control surface shown by the broken lines in

Fig. 1, adapted from (Batchelor , 2000, Fig. 5.15.1). The

surface A1 surrounds the elements of the array within

volume bounded by the outer boundary A2. The outer

boundary A2 could be in the form of a cylinder with axis

along the pipe and is chosen long enough to be in the steady

regimes far up and downstream of the array, and large

enough to contain a large sample of the array.


A2

A1dA,U, pup

dA,U, pdown

Y

X

Figure 1 – Use of Euler’s theorem to determine the force on a regular

array of rigid bodies such as a mesh filtering water flowing in a pipe

of uniform diameter, adapted from (Batchelor , 2000, Fig. 5.15.1).

— We apply Eq(29). We assume the pipe diameter is

sufficiently small and the pressure sufficiently large that we

can ignore the gravitational body force ρψ � p.

— Note that because the velocity is the same at the upstream

entrance and downstream exit of the control volume, the net


momentum flux vanishes,∫A

ρujuinjdA = 0. (30)

— Thus Eq(29) simplifies in this arrangement to∫A

pnidA = 0. (31)

Consider the x component, i = 1 or i = x. On the surface A2

only the cross-section normal to the flow contributes δA,

with the upstream outward normal −nx where pressure is

measured to be a steady pup and the downstream outward

normal nx where pressure is measured to be a steady pdown.

— Furthermore we have the unmeasured pressure distribution

on the array, that by symmetry provides a net force

Farray/δA in the horizontal (i.e. x-direction) transmitted to


the array. We can infer Farray from Eq(31)

Farray + pdownδA− pupδA = 0,

Farray = δA(pup − pdown). (32)


Using Bernoulli’s equation and Euler’s

theorem, Betz’s law

— Betz’s law expresses the maximum theoretical power that

can be extracted from an ideal wind turbine.

— Fluid assumptions : Homogeneous, incompressible and

frictionless fluid in steady flow. Furthermore, the far

upstream and downstream pressures are taken to be the

ambient pressure, pamb.

— Turbine assumptions : an infinite number of turbine blades,

modeled as an effective “disc actuator”; uniform thrust over

the disc ; non-rotating wake.

— We take as a control surface, the stream-tube shown in

Fig. 2 terminated by the two cross-sections indicated by the

vertical dashed lines at the upstream and downstream


locations. The stream-tube broadens in the downstream

direction because the fluid slows down while respecting (as it

must) conservation of mass.


dAup pambUup

dAa pbeforeUa

dAa pafterUa

stream-tube

actuatordisk

dAdown pambUdown

Figure 2 – An ideal turbine. Incompressible and frictionless fluid

approaches the actuator disk with steady uniform flow, the velocity

far upstream being Uup and ambient pressure pamb. The flow reaches

the actuator at velocity Ua and remains essentially unchanged exiting

the actuator, while the pressure drops across the actuator from pbefore

to pafter. Far downstream from the actuator, the pressure returns to

the ambient pressure but the flow is Udown ; adapted from (Manwell

et al., 2009, Fig. 3.1).


Betz’s law

— We derive two expressions for the x-component of the force

per unit area on the actuator, Fx/δAa.

— Applying Eq(29) to a control surface given by the

stream-tube and the two cross sections immediately before

and after the actuator, we have the same situation as in the

previous problem, where now the force on actuator replaces

the force on the array. So just as in the previous problem,

Eq(29) simplifies in this arrangement as well to∫A

pnidA = 0. (33)

In particular,

FxδAa

= pbefore − pafter. (34)


The unknown pressures pbefore and pafter can be eliminated

using Bernoulli’s equation Eq(20) applied to the flow from

the upstream cross-section to the “before cross-section”.

Ignoring changes in the vertical we have

ρ1

2U2up + pamb = ρ

1

2U2a + pbefore,

ρ1

2U2down + pamb = ρ

1

2U2a + pafter (35)

Taking the difference gives

ρ1

2

(U2up − U2

down

)= pbefore − pafter, (36)

which we substitute into Eq(34), giving

FxδAa

= ρ1

2

(U2up − U2

down

). (37)

— We derive a second expression for Fx/δAa by now applying

Eq(29) to a control surface given by the stream-tube and the


two cross sections far upstream and downstream of the

actuator with a hole around the actuator as indicated by the

dashed lines just before and after the actuator. The

x-component of Eq(29) becomes

ρU2upδAup(−nx) + pbeforenxδAa + pafter(−nx)δAa

+ ρU2downδAdownnx = 0. (38)

We recognize the mass flow rate through the control volume :

ρUupδAup = ρUdownδAdown = m, (39)

so that Eq(38) can be written more simply :

(pbefore − pafter)δAa = m(Uup − Udown),

= ρδAaUa(Uup − Udown),

FxδAa

= (pbefore − pafter) = ρUa(Uup − Udown). (40)


— Now we can equate our two expressions Eq(37) and Eq(40)

for the force/unit area on the actuator

ρUa(Uup − Udown) = ρ1

2

(U2up − U2

down

),

Ua(Uup − Udown) =1

2(Uup − Udown) (Uup + Udown) ,

Ua =1

2(Uup + Udown) , (41)

the velocity at the actuator is the average of the up and

down stream values !

— Engineers define the so-called axial induction factor, a, as

the fractional decrease in the wind velocity between the free

stream Uup and the rotor plane (here the actuator) Ua,

a =Uup − Ua

Uup(42)

see (Manwell et al., 2009), so Ua = Uup(1− a) and


Udown = Uup(1− 2a). Note we require a < 1/2 for otherwise

the downstream velocity vanishes and there is no possibility

for fluid to exit the turbine.

— We want to maximize the power output, P = FxUa, which

is, using Eq(37) again

P = ρ1

2δAa

(U2up − U2

down

)Ua,

= ρ1

2δAaU

3up4a(1− a)2. (43)

— We wish to maximize the power P for a given upstream

wind Uup and actuator area δAa. The maximum is found by

setting to zero the derivative with respect to a,

dP

da= 0 = ρ2δAaU

3up

d

da(a(1− a)2),

0 = ρ2δAaU3up[(1− a)2 − 2a(1− a)],

=⇒ a = 1/3. (44)


— The power coefficient, Cp, is defined as

Cp =P

12ρδAaU3

up

=turbine power

wind power,

= 4a(1− a)2 =16

27≈ 0.5926 (45)

which is called Betz’s coefficient.


References

Batchelor, G. K. (2000), An Introduction to Fluid Dynamics, 615

pp., Cambridge University Press, Cambridge, UK, 615 + xviii pp

+ 24 plates.

Manwell, J., J. McGowan, and A. Rogers (2009), Wind energy

explained : theory, design and application, second ed., xvi+689

pp., John Wiley & Sons Ltd.

Lecture 4. · 2019-09-13 · Lecture 4. dynamics continued 8 Energy budget |In the last lecture we...

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