Brewster H.D. Fluid Mechanics
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212
Integral Forms
of
the Basic Equations
2 2
-2
-pUAH
A
-pUAH
B
+pU
(HA
+HB
)=(P
1
-P
2
)(H
A
+HB)·
When one now inserts the above expression for
[]
and solves the equation
for
P 2 one obtains:
( )
2
HAHB
P
2
=P
1
+PU
A
-U
B
2·
(HA
+H
B
)
The pressure thus increases as a consequence
of
mixing the two flows
from position (1) to position (2).
Force
on
a Turbine
Blade
in
a
Viscosity-Free
Fluid
In the flow machines blade wheels are used to exploit the momentum
of
fluid
jets
for propulsion purposes. A
jet
from a rectangular nozzle (flat
jet
having the width
II,
and the depth b )hits a stationary blade which deflects the
jet
symmetrically to two sides around the angle 180 ±
/1
The sizes
of
the
inflowing and out flowing fluid flows are given. The pressure along the surface
of
the diagrammed control volume is equal all over to the ambient pressure,
so that the integral form
of
the momentum equation
Fig. Diagram for Explaining the Force on a Turbine Blade
~
J
pU
·dV + J pU.U
·dF:
=-
J
PdF.
at)
1)1
)
V
K
OK
OK
-J tijdF; + f
pgi
dV
+
LK
j
OK
V
K
can be simplified in the following way, the gravitation being negligible.
a
a
J pU
jdV
= 0, Stationary flow problem,
t V
K
Integral Forms
o/the
Basic Equations
J P dF
j
=
0,
as P =
Po
on all surfaces
of
the control volume,
OK
J
tijdE';
=
0,
as/! = 0 was set to zero,
OK
f pgjdV =
0,
as gravitation is negligible.
v
K
Thus it results:
J pUjU
j
dE';
=
LK
j
.
OK
+
213
As the problem is symmetrical, only the horizontal componentj = 1 has
to be considered, i.e.
inj
= 2 no resultant force appears so that the conservation
of
momentum can be written as follows:
-pHbU
2
+
2!pHbU
(-U . cos
13)
=K
1
•
2
The resultant force on the blade thus results in
Integral Forms
of
the Basic Equations
Ks =
-Kl
= pHbU2
(1
+ cos
~).
The formula makes clear that by deflecting the jets
in
direction
of
the
incoming flow, an increase
of
the force
KI
acting on the turbine blade can be
obtained.
Force
on
a Periodical
Blade
Grid
Assumptions: g
= 0,
Jl
= 0
Two-dimensional blade grids are used in flow machines
in
order to exploit
forces caused by flows for propulsion purposes, e.g. in turbines for driving
mechanical aggregates.
U
A
= {(U1)A,
(U
2
)A,
O}
or
U 3 - component
of
the flow field is zero. Because
of
the deflection
of
the flow by the blade grid the flow-o velocity differs from the attack velocity,
so that it holds:
The statements for
U A and U B ' for which always holds
(U
3
)A
=
(U
3
)B
= 0
make clear that the flow field remains "two-dimensional" and also has
two components. Thus for the attack-flow/inflow and off-flow the following
214
Integral Forms
of
the Basic Equations
equation can be derived from the continuity equation, where B is the width
of
the blade grid in
x3
-direction:
P(UI)A
tB
= P(UI)B
tB
.
~
7T
o t P
B
P.t
I
/,~/)bT
~
b'
//@-i
t
cxB
'
~-)l
(UZ)B U
I
K~
B
/ I C
/
IJ
//
%2
(Ut)A
a
(/
recht zur Zeichen-
/ K/ Breite B senk-
\
It
ebene
cxA
v~
Schaufel
(U
2
)A , d
U
A
Fig. Diagram Explaining the Effect
of
Blade Grids
For the attack-flow/inflow and off-flow thus the same velocity components
in
xl
-direction result. The purpose
of
blade grids is therefore to change the
velocity component
(U
2
)A
of
the attack-flow.
For the computation
of
the force on a grid blade a control volume
(a,
b,
c, d). is chosen. In
x2
- direction two flow lines are chosen as boundary
of
the
control volume which are positioned along
(a-
b) and
(d-c).
with the distance
of
the blade grid
t.
Thus the inflow and outflow to the chosen control volume
take place only over the areas
(d
- a)B. and (c - d)B. When one neglects the
gravitational effects, the Bernoulli equation yields the following connection:
P
A
1
[(
2 )2] P
B
1
[{)2
)2
]
-+2
UI)A
+(U
2
A
=-+
2 U
I
B
+(U
2
B .
P , I P , I
Entire
kineti~
enery in
(A)
Entire
kineti~
enery
in
(B)
For computation
of
the forces on the blade the integral form
of
the
momentum equation is at disposal:
! ( f
pUjdV
J+
J
pU;UjdFj
=
V
K
OK
-f P
dF
j
-
f
tij
dFj
+ f pg
jdV
+
IK
j
OK
OK V
K
With,the assumptions one obtains
from:
Integral Forms
of
the Basic Equations
! f pU
jdV
p =
0,
as there is a stationary flow,
V
K
f
tij
dF;
=
0,
as the is a viscosity-free fluid,
OK
f
pg
jdV
= 0, as the gravitational forces are negligible,
v
K
the momentum equation results in the simplified relation:
f
pU·U ·dF
=-
f P d£. +
"K·
I J I J
L....
J Oder
OK
OK
K j = f p U
j
•
U j
dE';
+ f P dF
j
OK OK
Thus for the forces in the directions j = 1 and j = 2 results:
Kl = m
[(U
1
)B
-(Ud
A
]+(PB
-PA)B
·t
=(P
B
-PA)B
·t,
,
,.
#
=0
K2
=ni[(U
2
)B
-(U
2
)A
J.
From the Bernoulli equation one obtains:
(P
B
-PA
)=~[(U2)~
-(U
2
)!
J.
Thus for
Kl
results:
Kl
=fBt[(U2)~
-(U
2
)!
J,
2 .
215
or expressed with the attack-flow angles and the flow-off/off -flow) angles:
K
p B
[U
2
•
2 U
2
•
2 ]
1 = - t A
SIn
a A - B
SIn
aB
,
2
and
K2
=-p.Bt[U
A
sinaA
-UB
sinaB].
Here
Kl
and
K2
are the forces acting on the control volume. The forces
acting on the blades are:
(K
1
)s
=-K
1
=-~Bt[ul
sin
2
aA
-U~
sin
2
aB]'
(K
2
)s
='-K
2
=-pBt[U
A
sinaA
-UB
sinaB].
The blade thus experiences a force
(K
1
)s
in the negative
xl
- direction
and a force
(K
B
)2
in the positive x
2
-
direction.
216
Integral Forms
o/the
Basic Equations
Euler's
Turbine
Equation
The considerations carried out related to a blade grid arranged in a plane,
which is located in a plane flow. When disposing the blade grids radially on a
rotating running wheel as diagrammed, one obtains the basic arrangement
of
a radial turbine. The first part
of
the term radial turbine, i.e. the word "radial"
designates the main flow direction,
in
which the flow through the turbine blades
takes place, namely radially from the inside to the outside. When employing
the continuity equation, one obtains:
P(Ur)A
(21tr
A
)
B =
P(Ur)B
(21tr
B
)B
or otherwise stated, the demand for mass conservation results:
(Ur)B
=rA
(Ur)A'
rB
From the BenouIIi equation results:
PA,
+![
(Ur)~
+(Ut)~
] = P
B
+
Integral Forms
of
the Basic Equations
217
For the moment imposed on the control volume by the running wheel
results:
M
t
=
ni
[rB
(Vt)B -rA (Vt)A 1
The mechanical power output connected with the turbine amounts to:
P
Turb
=-Mtro=-niro[rB (Vt)B -rA (Vt)A
J.
Equation as well as equation are called Euler.
's
turbine equation in the
literature. Here for
r
A
=
re
=.
the "inflow radius"
of
the blade rim is assumed
and for
r
B
=
ra
=.
the "outflow radius".
--
Fig. Diagram
of
the course
of
the Flow through an Axial Turbine
Radialpumpe
@)
Fig. Schematic Representation
of
Radial and Axial Pumps
The resulting equation does not only hold for turbines, but generally for
flow machines, like compressors, air blowers (ventilators), pumps etc. Here
pumps and turbines differ in the considerations carried out only with regard
to the sign
of
the energy exchange between running wheel and flowing fluid.
218
Integral Forms
of
the Basic Equations
In
a turbine energy
is
extracted from the fluid flow, so that one can collect a
usable moment at the shaft
of
this power engine, or the corresponding energy
is at disposal, respectively. In a pump on the other hand energy is supplied to
the fluid flow via the
"running wheel, i.e. a torque for driving the machine
is
exerted at the shaft. Thus the energy necessary for pumping
is
supplied.
Finally it
is
mentioned that Euler's turbine formula can also be applied
to flow machines through which flows pass axially.
Power
of
Flow Machines
A typical application
of
a flow machine,
in
the chosen case the application
of
a pump,
is
shown as an example. It sucks on the intake side a certain quantity
of
water
m in order to transport it upward
at
the discharge side. Here,
differences exist
in
the pipe diameter between the intake side and the discharge
side
of
the pump.
Pumpe
Fig.
Diagram
for
the
Computation
of
the
Pumping
Capacity
The suction
of
the pumped fluid takes place from a container (A) and the
transport into a container (B), as diagrammed in the opposite figure. All
quantities located at the suction side
of
the considered problem are designated
by index A, the quantities on the discharge side
of
the pump by index
B.
From the integral form
of
the continuity equation results for the considered
problem:
Integral Forms
of
the Basic Equations
219
2-
2-
m .
DAU
A
=DBU
B
=-=V
P
For the computation
of
the pump capacity it
is
recommended to employ
the integral form
of
the mechanical energy equation, as stated in equation:
~
J.!pU~dV
+ J
pU·
(.!U
2
)dF
=-
J
PU
dF· + J P
aU
j
dV
at 2 ) I 2 } I } } ax .
V
K
OK OK
V
K
}
J
J
aUj
J . " .
- 'tijUjdF
j
+
'tij
-a-
dV
+
pgX
dx j + L.JE
X·
OK
V
K
I v
K
When reducing this equation by assumptions that apply to the present
problem:
! f
~PUJdV
= 0 stationary pumping conditions,
v
K
f
au·
P
-a
}
dV
= 0 no work done during expansion, as r = const and thus
v
Xj
K
aU
j
fu.=
0,
}
and neglecting the
tij
-terms in the above integral energy equation, then it holds:
f
pUi
(~UJ
)dF;
=-
f
PUjdF
j + f
pgjUjdV
+
LE
OK
OK
V
K
or
-m[±u
j
+±U~
]=VAP
A
-VBP
B
+pgVHc
+Pm'
where
Pm
represents the power introduced by the pump into the fluid.
For the capacity
of
the pump thus holds:
Pm
=V
{(~U~
+P
B
+pgHc
)-(~Ul
+P
A
)}.
When one considers now P Band P A more in detail, it holds:
P
B
=
Po
sowie P
A
=P
o
+pgH
A
-~ul,
and U
A
~O
and UA =
~
V
sowie U
B
=
FB
'
So that one obtains the following relation:
220
Integral Forms
of
the Basic Equations
.
{p
(v
2 V 2 J P
(V
)2
}
P =V -
---
-pqH
A
+-
-
+pgHe
,
m 2
F2
F2
2 F
B A A
or
summarized:
p(V
3
J .
Pm
="2
p;
+pq(He
-HA)V.
For the electrical capacity P e
of
the pump results, with
II
as efficiency
factor:
P
e
=.!..Pm
=.!..[~~:
+pqV
(He
-H
A
)]
" " B .
For the pumping capacity results that the stated electrical power is required
to supply the kinetic energy
ofthe
fluid leaving the pipe per time unit, plusthe
power required per time unit for surmounting the hydrostatic pressure level.
Chapter 8
Stream
Tube
Theory
General
Considerations
Stromlinie
Stromfaden
Stromrohre
Fig: Stream Line, Stream Filament and Stream Pipe
The preceding considerations on the derivation
of
the integral form
of
the basic equations can also be used advantageously to derive simplified equa-
tions applicable to so-called stream filaments and to put them to use for the
solution
of
flow problems. Here one starts the considerations from flow lines
that are introduced as lines
of
a flow area and which at a certain point in time
possess/control the direction
of
the flow at each point
of
the flow field. One
can imagine a stream filament built up from a bundle
of
such flow lines and
one can make a subdivision
of
the entire flow field into a multitude
of
stream
filaments. Furthermore, it is possible to bundle stream filaments to obtain
stream pipes.
Here, attention has to be given to that considerations on properties
of
flows applied to stream filaments can only be employed advantageously when
the stream quantities assigned to each area
of
the stream filament can be
considered to be constant over the cross-section
of
the stream filament. This
makes it occasionally necessary to choose the cross-sectional area
of
a stream
filament su ciently small, so that for the considered problem the assumption
of
uniform state and flow quantities over the cross-sectional area
of
the stream
filament can be fulfilled su ciently precisely.