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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.
summary of lecture 3 4
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
summary of lecture 3 5
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
summary of lecture 3 6
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)
Lecture 4. dynamics continued 9
— 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
Lecture 4. dynamics continued 10
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
Lecture 4. dynamics continued 11
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).
Lecture 4. dynamics continued 12
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
Lecture 4. dynamics continued 13
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.
Lecture 4. dynamics continued 14
— 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.
Lecture 4. dynamics continued 15
— 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)
Lecture 4. dynamics continued 16
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
Lecture 4. dynamics continued 17
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)
Lecture 4. dynamics continued 18
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
= ρFi +∂σij∂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)
Lecture 4. dynamics continued 19
— Add the two equations,
ρ∂ui∂t
+ ui∂ρ
∂t+ ui
∂
∂xj(ρuj) + ρuj
∂ui∂xj
= ρFi +∂σij∂xj
,
∂ρui∂t
+∂ρujui∂xj
= ρFi +∂σij∂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
= ρFi +∂σij∂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
Lecture 4. dynamics continued 20
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
Lecture 4. dynamics continued 21
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)
Lecture 4. dynamics continued 22
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
Lecture 4. applications 24
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
Lecture 4. applications 25
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.
Lecture 4. applications 26
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
Lecture 4. applications 27
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
Lecture 4. applications 28
the array. We can infer Farray from Eq(31)
Farray + pdownδA− pupδA = 0,
Farray = δA(pup − pdown). (32)
Lecture 4. applications 29
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
Lecture 4. applications 30
locations. The stream-tube broadens in the downstream
direction because the fluid slows down while respecting (as it
must) conservation of mass.
Lecture 4. applications 31
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).
Lecture 4. applications 32
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)
Lecture 4. applications 33
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
Lecture 4. applications 34
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)
Lecture 4. applications 35
— 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
Lecture 4. applications 36
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)
Lecture 4. applications 37
— 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.
Lecture 4. applications 38
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.