Durst F. Fluid Mechanics: An Introduction to the Theory of Fluid Flows
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8.3 Integral Form of the Mechanical Energy Equation 227
of spheres treated in mechanics, for which the known momentum and en-
ergy equations from (8.15) or (8.19) and (8.20) or (8.24) can be derived as
follows:
• The left side of (8.15) yields for ρ = const. for an integration over the
entire sphere volume
#
V
K
ρ
∂U
j
∂t
+ ρU
i
∂U
j
∂x
i
dV =
#
V
K
D
Dt
(ρU
j
)dV =
D
Dt
#
V
K
ρU
j
dV, (8.25)
and thus
D
Dt
#
V
K
ρU
j
dV =
d
dt
(m
K
U
j
). (8.26)
• With (8.25) and (8.26) for spheres 1 and 2, (8.15) can be written:
d
dt
(m
K
U
j
)
1
=(K
j
)
1
and
d
dt
(m
K
U
j
)
2
=(K
j
)
2
, (8.27)
or rewritten, because (K
j
)
1
= −(K
j
)
2
:
d
dt
7
(m
K
U
j
)
1
+(m
K
U
j
)
2
8
=0; (m
K
U
j
)
1
+(m
K
U
j
)
2
= constant
(8.28)
• The left-hand side of (8.20) yields for ρ = constant
#
V
K
3
ρ
∂
∂t
1
2
U
2
j
+ ρU
i
∂
∂x
i
1
2
U
j
2
4
dV =
#
V
K
D
Dt
ρ
1
2
U
2
j
=
D
Dt
#
V
K
ρ
1
2
U
2
j
dV (8.29)
and thus
d
dt
m
K
1
2
U
2
j
=
D
Dt
#
V
K
ρ
1
2
U
2
j
dV. (8.30)
• With (8.29) and (8.30), (8.20) yields for spheres
d
dt
m
K
1
2
U
2
j
1
=
˙
E
1
and
d
dt
m
K
·
1
2
U
2
j
2
=
˙
E
2
, (8.31)
or rewritten because
˙
E
1
+
˙
E
2
=0:
d
dt
m
K
1
2
U
2
j
1
+
m
K
1
2
U
2
j
2
=
˙
E
1
+
˙
E
2
=0, (8.32)
m
K
1
2
U
j
1
+
m
K
1
2
U
j
2
= constant (8.33)
228 8 Integral Forms of the Basic Equations
Fig. 8.1 Possible motions of spheres following an elastic collision
The insights gained on the elastic collision by employing (8.28) and (8.33)
are sketched in Fig. 8.1. The representations are stated for different mass ra-
tios of the spheres. From the integral forms of the basic equations of flow
mechanics result the collision laws for spheres which are known from appli-
cations of mechanics in physics. This makes clear the general applicability of
the integral form of the mechanical energy equation stated in (8.24).
8.4 Integral Form of the Thermal Energy Equation
In Sect. 5.6, the thermal energy equation was derived and stated for an ideal
gas in (5.77) as follows:
ρc
v
DT
Dt
= λ
∂
2
T
∂x
2
i
− P
∂U
i
∂x
i
− τ
ij
∂U
j
∂x
i
, (8.34)
For an ideal liquid with ρ = const. it was stated with (5.78):
ρc
v
DT
Dt
= λ
∂
2
T
∂x
2
i
− τ
ij
∂U
j
∂x
i
. (8.35)
When one chooses (8.34) for further considerations, this equation can also be
written:
ρc
v
∂T
∂t
+ U
i
∂T
∂x
i
= −
∂ ˙q
i
∂x
i
− P
∂U
i
∂x
i
− τ
ij
∂U
j
∂x
i
. (8.36)
Adding to (8.36) the continuity equation multiplied by c
v
T :
c
v
T
∂ρ
∂t
+ c
v
T
∂(ρU
i
)
∂x
i
=0,
8.4 Integral Form of the Thermal Energy Equation 229
one obtains the initial equation for the derivation of the integral form of the
thermal energy equation:
∂(ρc
v
T )
∂t
+
∂(ρc
v
TU
i
)
∂x
i
= −
∂ ˙q
i
∂x
i
− P
∂U
i
∂x
i
− τ
ij
∂U
j
∂x
i
. (8.37)
With c
v
T = e (inner energy), one obtains
∂(ρe)
∂t
+
∂(ρeU
i
)
∂x
i
= −
∂ ˙q
i
∂x
i
− P
∂U
i
∂x
i
− τ
ij
∂U
j
∂x
i
. (8.38)
The integration of (8.38) over a control volume yields
#
V
K
∂(ρe)
∂t
dV +
#
V
K
∂(ρeU
i
)
∂x
i
dV = −
#
V
K
∂ ˙q
1
∂x
i
dV −
#
V
K
P
∂U
i
∂x
i
dV −
#
V
K
τ
ij
∂U
j
∂x
i
dV
+
(
˙
Q +
˙
E). (8.39)
Rewriting (8.39) with consideration of Gauss’s integral theorem and the
reversibility of the sequence of integration and differentiation, one obtains
∂
∂t
⎛
⎝
#
V
K
ρe dV
⎞
⎠
I
+
#
O
K
ρeU
i
dA
i
II
= −
#
O
K
˙q
i
dA
i
III
−
#
V
K
P ·
∂U
i
∂x
i
dV
IV
−
#
V
K
τ
ij
ρU
j
∂x
i
dV
V
+
(
˙
Q +
˙
E)
VI
. (8.40)
The terms of the resulting equation can be interpreted as follows:
I: Temporal change of the inner energy of the fluid within the control
volume V
K
.
II: Convective outflow and inflow of inner energy per unit time over the
surface O
K
of the control volume.
III: Molecular heat flow per unit time, i.e. the sum of the outflow and inflow,
over the surface O
K
of the control volume.
IV: The work carried out during expansion by the total volume per unit time.
V: The mechanical energy dissipated per unit time in the entire control
volume.
VI: External heat and energy supply per unit time which is added to the
entire control volume.
The above equation holds likewise for an ideal liquid, but for this term IV
is equal to zero, as no work can be done during expansion because of ρ =
constant.
230 8 Integral Forms of the Basic Equations
8.5 Applications of the Integral Form of the Basic
Equations
The importance of the integral forms of the basic equations of fluid mechan-
ics becomes clear from applications that are listed below. Many books on the
basics of fluid mechanics treat flow problems of this kind, so that the consid-
erations carried out in the following sections can be brief. Typical examples
are treated that make it clear that the derived integral forms of the basic
equations represent the basis for a variety of engineering problem solutions.
However, attention has to be paid to the fact that solutions often can be
derived only by employing simplifications to the general form of the equa-
tions. Reference is made to these simplifications for each of the treated flow
problems and their implications for the obtained solutions in the framework
of the derivations.
In order to introduce the reader to the methodically of the correct han-
dling of the integral form of the equations, each of the problems treated below
is solved by starting from the employed basic equations. Starting with the
general form of the integral form of the equations, terms are deleted which
are equal to zero for the treated flow problem. In addition, by introducing
simplifications, terms in the equations are also removed which are small and
therefore have very little influence on the treated flow problem, so that easily
comprehensible solutions are obtained. Below only examples for the applica-
tions of the integral forms of the basic equations are given. More examples
are found in refs. [8.1] to [8.5].
8.5.1 Outflow from Containers
In Fig. 8.2, a simple container is sketched, having a diameter D,whichis
partly filled with a fluid and is assumed to be closed at the top. Between the
fluid surface and the container lid there is a gas having a constant pressure
P
H
. The fluid height is H and at the bottom of the container there is an
opening with diameter d. Sought is the outflow velocity from the container,
i.e. the velocity U
d
.
From Fig. 8.2, it can be seen that the water surface is moving downwards
with a velocity U
D
, because of the fluid flowing out, which exits with U
d
from
Fig. 8.2 Diagram for the treatment of outflows
from containers
8.5 Applications of the Integral Form of the Basic Equations 231
the container opening. Through the integral form of the continuity equation
one obtains
˜ρ
˜
UF = constant ; ρU
D
π
4
D
2
= ρU
d
π
4
d
2
; U
D
=
d
2
D
2
U
d
(8.41)
By employing the Bernoulli equation between the points (A) and (B), one
obtains
1
2
U
2
D
+
P
H
ρ
+ gH =
1
2
U
2
d
+
P
0
ρ
. (8.42)
Hence
1
2
U
2
d
=
1
2
U
2
D
+ gH +
1
ρ
(P
H
− P
0
) , (8.43)
or, after insertion of (8.41)
1
2
U
2
d
=
1
2
d
4
D
4
U
2
d
+ gH +
1
ρ
(P
H
− P
0
) , (8.44)
the following relationship for U
d
results:
U
d
=
9
:
:
;
2gH +
2
ρ
(P
H
− P
0
)
1 −
d
4
D
4
. (8.45)
For P
H
= P
0
and d<<D, the well known equation U
d
=
√
2gH results.
8.5.2 Exit Velocity of a Nozzle
In fluid mechanics, it is necessary to calibrate indirectly operating measur-
ing processes (e.g. stagnation-pressure tubes, hot-wire anemometers) in flow
fields in which the flow velocity is known. By letting a fluid flow through a
nozzle, it can be achieved that at the nozzle exit the flow velocity required for
calibration can be adjusted via the pressure in the input pipe of the nozzle
(Fig. 8.3).
With the statements made in Sect. 8.1, the integral form of the continuity
equation holds in the following form:
˙m =˜ρ
˜
UF = constant ; ρU
A
π
4
D
2
= ρU
B
π
4
d
2
, (8.46)
i.e. one can write for U
A
U
A
=
d
2
D
2
U
B
. (8.47)
For the planes (A) and (B) it can be written as a result of the Bernoulli
equation:
232 8 Integral Forms of the Basic Equations
Fig. 8.3 Diagram of a nozzle-calibrating test rig for velocity-measuring sensors
1
2
U
2
A
+
P
A
ρ
=
1
2
U
2
B
+
P
B
ρ
=
1
2
d
4
D
4
· U
2
B
+
P
A
ρ
. (8.48)
From this, the following results:
U
B
=
$
2(P
A
− P
B
)
ρ
!
1 −
d
4
D
4
". (8.49)
When one chooses D d, one obtains for U
B
, to a good approximation:
U
B
=
0
2
ρ
(P
A
− P
B
)=
0
2
ρ
(P
A
− P
0
). (8.50)
By adjusting different P
A
values, the entire velocity regime required for
the calibration of measuring tubes can be set. Hence, measuring P
A
= P
0
and knowing ρ yields U
B
against which velocity sensors can be calibrated.
8.5.3 Momentum on a Plane Vertical Plate
When flowing fluid jets are decelerated, forces appear which are used in many
fields of technology. For the flow problem shown in Fig. 8.4, the question
arises as to what force needs to be applied in the x
1
direction to prevent the
deflection of the plate due to the momentum impact of the plane fluid jet.
The plane jet has a thickness H in the x
2
direction and a width b in the x
3
direction. The jet velocity far away from the plate is known and is U
A
.The
density ρ is known and g
1
=0,asthex
1
direction axis is horizontal.
Employing the integral form of the continuity equation gives
ρU
A
Hb = ρU
C
H
C
b + ρU
B
H
B
b. (8.51)
From the Bernoulli equation, one obtains
1
2
U
2
A
+
P
A
ρ
=
1
2
U
2
B
+
P
B
ρ
and P
A
= P
B
= P
0
, (8.52)
8.5 Applications of the Integral Form of the Basic Equations 233
Fig. 8.4 Diagram for consideration of the momentum impact on a vertical plate
and thus U
A
= U
B
. Analogous considerations yield U
A
= U
C
. Owing to the
symmetry of the problem, one obtains
H =2H
C
=2H
B
; hence H
C
= H
B
.
For the solution of the problem to yield K, the integral form of the momentum
equation can be employed, as it is stated in (8.19):
∂
∂t
#
V
K
ρU
j
dV +
#
O
K
ρU
i
U
j
dA
i
= −
#
O
K
P dA
j
−
#
O
K
τ
ij
dA
i
+
#
V
K
ρg
i
dV +
F
j
.
(8.53)
For the considered flow problem, the following simplifications of the above
universally valid equation hold:
∂
∂t
#
V
K
ρU
j
dV =0, stationary flow problem
#
O
K
P dA
j
=0, as P = P
0
on all surfaces of the chosen control volume
#
O
K
τ
ij
dA
i
=0, absence of viscosity in the fluid
#
˙
V
ρg
j
dV =0, gravitation term, here g
j
= g
1
=0.
(8.54)
Therefore, the following holds for the simplified form of (8.53):
#
O
K
ρU
i
U
j
dA
i
= F
j
, (8.55)
234 8 Integral Forms of the Basic Equations
and thus one obtains by integration for j =1:
F
1
= −ρU
2
A
HB. (8.56)
The result of the above derivations shows that F
1
must act in the negative
x
1
direction, in order to prevent deflection of the plane plate by the incoming
plane fluid jet. This gives an example to make clear the kind of force terms
F
j
that occur in (8.19). All forces need to be included for a particular flow
problem that act on the considered control volume.
8.5.4 Momentum on an Inclined Plane Plate
For a fluid jet hitting an inclined plane plate, the jet behavior is shown in
Fig. 8.5. Because of the inclination of the plate, which encloses the angle α
with the axis of the incoming plane fluid jet, the jet splits into two parts of
unequal thickness. The thicker jet goes upwards and has a height h
B
= εH
A
and the thinner jet goes downwards and has a height h
C
=(1− ε)H
A
.This
results from the continuity equation. Applying this equation, U
A
= U
B
= U
C
results when g = 0 is introduced into the Bernoulli equation.
When employing the continuity equation in integral form, i.e. when con-
sidering that for the solution of the problem the mass conservation law can
be used, it follows that
ρU
A
H
A
b = ρU
B
h
B
b + ρU
C
h
C
b, (8.57)
or, because U
A
= U
B
= U
C
from the Bernoulli equation:
U
A
H
A
= U
B
h
B
+ U
C
h
C
; H
A
= h
B
+ h
C
. (8.58)
For the two split jets forming on the plate, it can therefore be stated that
h
B
= εH
A
and h
C
=(1− ε) H
A
. (8.59)
Fig. 8.5 Diagram for explain-
ing the independence of fluid
flow considerations on the chosen
coordinate system
F
1
F
2
=
0
because
=
0
F
F
2
b
- Width of plate
in
x
2
-
direction
8.5 Applications of the Integral Form of the Basic Equations 235
When one employs the integral momentum equation:
∂
∂t
#
V
K
ρU
j
dV +
#
O
K
ρU
i
U
j
dA
i
= −
#
O
K
P dA
j
−
#
O
K
τ
ij
dA
i
+
#
V
K
ρg
j
dV +
F
j
.
(8.60)
For the problems sketched in Fig. 8.5, the following simplifications can be
introduced:
∂
∂t
#
V
K
ρU
j
dV =0, stationary problem
#
O
K
P dA
j
=0, as P = P
O
on all surfaces of the chosen control volume
#
O
K
τ
ij
dA
i
=0, viscosity-free fluid
#
V
K
ρg
j
dV =0, insignificant term or g
j
=0.
(8.61)
For the simplified form of the above integral momentum equation it thus
holds that
#
0
K
ρU
i
U
j
dA
i
= F
j
. (8.62)
When one chooses a coordinate system oriented along the plate, then for K
p
,
integration over the planes (A), (B)and(C), yields three contributions:
F
P
= −ρU
2
A
H
A
b sin α + ρU
2
A
εH
A
b − ρU
2
A
(1 − ε)H
A
b. (8.63)
As in the present problem µ = 0 was set, the following results for the force
F
P
= 0, due to the moving fluid along the plate, can be deduced from (8.63):
−1 − sin α +2ε =0, (8.64)
or one obtains for
ε =
1
2
(1 + sin α). (8.65)
For the force acting vertically on the plate, it can be computed that K
S
is
F
S
= ρU
2
A
H
A
b cos α. (8.66)
It is evident that the above considerations have to be independent from the
chosen coordinate system. When one chooses the coordinate system indicated
by x
1
and x
2
in Fig. 8.6, one obtains for K
1
the following contributions,
derived by integration over the planes (A), (B)and(C):
F
1
= −ρU
2
A
H
A
b + ρU
2
A
εH
A
b sin α − ρU
2
A
(1 − ε)H
A
b sin α. (8.67)
236 8 Integral Forms of the Basic Equations
Fig. 8.6 Jet deflection at a knife edge and
its cause
For the force K
2
similar derivations yield:
F
2
= ρU
2
A
εH
A
b cos α − ρU
2
A
(1 − ε)H
A
b cos α. (8.68)
Since, because µ = 0, the total force on the plate resulting from F
1
and
F
2
has to act on the plate K vertically, the following holds:
tan α =
F sin α
−F
1
=
F
2
F cos α
=
2ε cos α −cos α
1 − 2ε sin α +sinα
. (8.69)
From this it can be derived by introduction of 2ε cos α − cos α =sinα −
2ε sin α +sinα or for ε:
ε =
1
2
(1 + sin α), (8.70)
which is the same result as (8.65).
8.5.5 Jet Deflection by an Edge
When a fluid jet (height H,widthb) hits, with part of its cross-sectional area,
a plate standing vertically to the jet, the arriving fluid is partitioned into two
partial jets. One of the two partial jets runs, vertically to the original jet
direction, downwards along the plate, and the other partial jet is deflected
upwards by an angle α with respect to the original jet direction. Neglecting
viscosity forces and gravitational forces and assuming a constant ambient
pressure, it results from the Bernoulli equation that the two partial jets have
the same velocity, which is equal to the velocity of the fluid in the original
jet. Because of the continuity equation, the two partial jets have jet heights
εH and (1 − ε)H. In the integral form of the momentum equation
∂
∂t
#
V
K
ρU
j
dV +
#
O
K
ρU
i
U
j
dA
i
= −
#
O
K
P dA
j
−
#
O
K
τ
ij
dA
i
+
#
V
K
ρg
j
dV +
F
j
,
(8.71)