fluid_mechanics

206 Chapter 5 ■ Finite Control Volume Analysis
^
n
^
n
V
V
^
Vr Vn > 0
Control
volume
y
x
^
Vr Vn < 0
^
n
^
n
V5.7 Running on
water
A control volume
diagram is similar
to a free-body
diagram.
V
V
Several important generalities about the application of the linear momentum equation 1Eq.
5.222 are apparent in the example just considered.
1. When the flow is uniformly distributed over a section of the control surface where flow into
or out of the control volume occurs, the integral operations are simplified. Thus, onedimensional
flows are easier to work with than flows involving nonuniform velocity distributions.
2. Linear momentum is directional; it can have components in as many as three orthogonal
coordinate directions. Furthermore, along any one coordinate, the linear momentum of a
fluid particle can be in the positive or negative direction and thus be considered as a positive
or a negative quantity. In Example 5.11, only the linear momentum in the z direction
was considered 1all of it was in the negative z direction and was hence treated as being
negative2.
3. The flow of positive or negative linear momentum into a control volume involves a negative
V nˆ product. Momentum flow out of the control volume involves a positive V nˆ
product. The correct algebraic sign 1 or 2 to assign to momentum flow 1VrV nˆ dA2
will depend on the sense of the velocity 1
in positive coordinate direction, in negative
coordinate direction2 and the V nˆ product 1
for flow out of the control volume, for
flow into the control volume2. This is shown in the figure in the margin. In Example 5.11,
the momentum flow into the control volume past section 112 was a positive 12 quantity
while the momentum flow out of the control volume at section 122 was a negative 12quantity.
4. The time rate of change of the linear momentum of the contents of a nondeforming control
volume 1i.e., 00t cv
Vr dV2 is zero for steady flow. The momentum problems considered in
this text all involve steady flow.
5. If the control surface is selected so that it is perpendicular to the flow where fluid enters or
leaves the control volume, the surface force exerted at these locations by fluid outside the
control volume on fluid inside will be due to pressure. Furthermore, when subsonic flow exits
from a control volume into the atmosphere, atmospheric pressure prevails at the exit cross
section. In Example 5.11, the flow was subsonic and so we set the exit flow pressure at the
atmospheric level. The continuity equation 1Eq. 5.122 allowed us to evaluate the fluid flow
velocities w 1 and w 2 at sections 112 and 122.
6. The forces due to atmospheric pressure acting on the control surface may need consideration
as indicated by Eq. 13 in Example 5.11 for the reaction force between the nozzle and the fluid.
When calculating the anchoring force, F A , the forces due to atmospheric pressure on the control
surface cancel each other 1for example, after combining Eqs. 12 and 13 the atmospheric
pressure forces are no longer involved2 and gage pressures may be used.
7. The external forces have an algebraic sign, positive if the force is in the assigned positive
coordinate direction and negative otherwise.
8. Only external forces acting on the contents of the control volume are considered in the linear
momentum equation 1Eq. 5.222. If the fluid alone is included in a control volume, reaction
forces between the fluid and the surface or surfaces in contact with the fluid [wetted
surface1s2] will need to be in Eq. 5.22. If the fluid and the wetted surface or surfaces are
within the control volume, the reaction forces between fluid and wetted surface1s2 do not appear
in the linear momentum equation 1Eq. 5.222 because they are internal, not external forces.
The anchoring force that holds the wetted surface1s2 in place is an external force, however,
and must therefore be in Eq. 5.22.
9. The force required to anchor an object will generally exist in response to surface pressure
andor shear forces acting on the control surface, to a change in linear momentum flow
through the control volume containing the object, and to the weight of the object and the
fluid contained in the control volume. In Example 5.11 the nozzle anchoring force was required
mainly because of pressure forces and partly because of a change in linear momentum
flow associated with accelerating the fluid in the nozzle. The weight of the water and
the nozzle contained in the control volume influenced the size of the anchoring force only
slightly.