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Chapter 4: l%e Momentum Balances 173 
I, v, p (v cos a) dA + 2 Sv v, p dV = I; Fx (4.1-12) 
~ v , p ( v c o s a ) d A + $ S V v , p d V = LCF, (4.1-13) 
(4.1 - 14) 
In general, the convective terms as well as the sum of the forces may have a 
non-zero component in any direction, although, if possible, we usually try to 
choose a coordinate system which minimizes the number of non-zero 
components. Also notice that the forces exerted by the surroundings on the 
control volume (the forces in the momentum balance) are equal and opposite 
in sign to the forces exerted by the control volume on the surroundings. 
le 4.1-1 . . m e n t m flux of fLrud rn lgmmw -flow UL 
circular p - 
The (axially symmetric) velocity profile for fully developed laminar flow in 
a smooth tube of constant circular cross-section is 
v = v,, [l -(;)'I 
(4.1-15) 
Calculate the momentum flux. 
Choose the flow direction to be the x-direction. 
Solution 
a) The momentum flux across any arbitrary pipe cross-section is 
Momentum flux = V I , ( v . 4 
= ~(v-[l-(~)2](i)}(v~[l-(~)2](i*n)}27crLh 0 
(4.1 - 16) 
Noting that for this choice of coordinates, the outward normal is in the x- 
direction 
174 Chapter 4: The Momentum Balances 
( i - n ) = 1 
Then 
(4.1- 17) 
Momentum flux = 2 ~t vkx i 
(4.1-18) 
= 2 n v L i l ( r - ~ + $ ) c i r (4.1-19) 
= ~ ~ V ~ ~ [ C - L + Y - ] I R 
2 2R2 6R4 (4.1-20) 
x R2 vLx 
(4.1-21) Momentum flux = i 
4.1.1 Types of forces 
There are two types of forces with which we must contend: 
body forces 
surface forces. 
Body forces arise from the same types of fields that we considered when 
discussing potential energy - gravitational, electromagnetic, etc. The change in 
energy from position changes in a potential field can be accomplished by a force 
acting on a body during its displacement. Such a force - for example, gravity - 
does not necessarily disappear when the body is at rest. All potential fields do 
not produce what we call body forces, but all body forces are produced by 
potential fields. Body forces are proportional to the amount of mass and act 
throughout the interior of the system or control volume. 
The second type of force is the surface force. Surface forces are exerted on 
the system by the surroundings via components of the traction on the surface 
of the system. Traction (force) acting at a point on the surface is a vector since 
it has both magnitude and direction, and it can therefore be resolved into two 
Chapter 4: The Momentum Balances I75 
components: one parallel to the surface (shear), and one component normal to 
the surface. 
The normal component in many problems is constituted of only the 
negative of the pressure multiplied by the area on which it acts, although non- 
Newtonian fluids can also have viscous forces acting in the normal direction. 
At a point in a fluid at rest, the normal force is the Same in all directions 
and is called the hydrostatic pressure. Hydrostatic pressure is the same as the 
pressure used in a thermodynamic equation of state. There is no shear in static 
classical fluid systems because by definition such a fluid would deform 
continuously under shear. (This classic definition of a fluid becomes somewhat 
fuzzy when we later discuss non-Newtonian fluids, such a Bingham plastics, that 
exhibit characteristics of both fluids and solids.) 
When a general fluid is flowing, the normal stresses are not always equal. If 
the fluid is incompressible, however, the pressure can be defined satisfactorily as 
the mean of the normal stresses in the three coordinate directions. 
The pressure field in either a static or a moving fluid exhibits a gradient 
caused by gravity. This gradient can often be neglected except in special 
circumstances; for example, where buoyancy forces enter the problem. 
4.1.2 
entire surface of irregular objects 
Influence of uniform pressure over 
To obtain the net force resulting from pressure on a control volume, we 
must integrate over the entire outside surface 
net force produced by 
on surface of control volume (4.1.2-1) 
For irregularly shaped volumes, this can lead to cumbersome integrals. We 
often can avoid integrating the force produced by pressure over the non-flow 
surface of irregular volumes by performing the integration in gauge pressure. 
This artifice removes a constant pressure over the entire surface of the 
control volume from the integral. Adding or subtracting a constant normal stress 
over the entire outside surface of a control volume, no matter how irregularly 
shaped, does not give a net contribution to the external force term. 
176 Chapter 4: The Momentum Balances 
This can be seen by dividing an irregular control volume into elementary 
prisms as shown in Figure 4.1.2- 1, adding the forces on the left- and right-hand 
faces, and taking the limit as the face areas of the prisms approach zero to obtain 
the area integral. 
Xk 
xi 
Figure 4.1.2-1 Approximation of solid by prisms 
(7" . n) dA 
+ normal stress x area 
)left facc I 1 rime faa (normal stress x area 
(4.1.2-2) 
Consider the typical rectangular prism shown in Figure 4.1.2-2 of area dXj 
by dxk (the coordinate xk is normal to the page) on its left face, a, and whose 
right face, b, is sliced at some angle p as indicated. The angle y is a right angle, 
and the outward normal to b makes an angle a with the horizontal. The normal 
stress is assumed to arise only from the pressure. 
Notice that if the magnitude of the normal stress 
(4.1-2-3) 
Chapter 4: The Momentum Balances I77 
Figure 4.1.2-2 Detail of prism 
is the same on each face, one obtains no net force in the xi-direction, because the 
increase in area over which the normal force acts on the right face is always just 
compensated by the decrease in the component of the force acting in that 
direction 
Fxi on left face 
component of normal stress in area on which it acts] 
(4.1-24) 
In this case the nonnal stress is just the pressure, which acts normal to the 
surface and in a direction opposed to that of the outward normal. 
Fxi on left face = p dx, dx, (4.1-2-5) 
Considering the right face and noting that the angle a and the angle p are equal 
because their sides are perpendicular 
Fxi on right face 
1 = component of normal stress in area on which it acts I L 
dxj dx, 
= -pdxjdxk (4.1-26) 
The same conclusion would be reached if the left-hand face were sliced 
at some different angle. 
I78 Chapter 4: The Momentum Balances 
Therefore, the imposition of a uniform pressure (normal force) over the 
outer surface of an irregularly shaped object contributes no net force tbe surface 
of the object. 
4.1.3 Averages and the momentum equation 
If we examine the fifst term in the macroscopic momentum balance in 
cylindrical coordinates as applied to constant-density steady axial flow in a 
uniform-diameter pipe between upstream cross-section 1 and down-stream cross- 
section 2, with the positive z-coordinate in the direction of flow, we see that U = 
0, that the magnitude of the z-component of v is the same as the magnitude of 
v itself, and that the cosine is -1 at cross-section 1 and +1 at cross-section 2, so 
the term reduces to 
/Av,pvcosadA = l j A 2 v2 M - j A l v2 
= P [ (4) - (41 A 
We often make the assumption in evaluating such a term that 
(v\u2019) = (v)2 
(4.1.3-1) 
(4.1.3-2) 
In the following Sections we examine this assumption for a particular turbulent 
flow profile and for laminar flow.5 
Momentum balance approximation - turbulent flow 
The velocity distribution for turbulent flow in pipes is fairly well 
represented by the following empirical profile: 
v = v,(l-*)+ (4.1.3-3) 
In using the momentum balance for such flows, we often make the further 
approximation