fluid_mechanics

5.2 Newton’s Second Law—The Linear Momentum and Moment-of-Momentum Equations 217
z
^e z
Control volume
θ
^e r
^
r
e θ
U 2
W 2
ω
Control volume
Section (2)
V5.10 Rotating
lawn sprinkler
Flow out
ω
T shaft
Section (1)
Flow out
Section (2)
V 2
W 2
Section (1)
r 2 V θ 2
T shaft
U 2 = r 2 ω
(a)
Flow in
(b)
Control volume
Section (1)
F I G U R E 5.4 (a) Rotary water
sprinkler. (b) Rotary water sprinkler, plane view.
(c) Rotary water sprinkler, side view.
Flow
(c)
Change in moment
of fluid velocity
around an axis can
result in torque and
rotation around
that same axis.
2. We confine ourselves to steady or steady-in-the-mean cyclical flows. Thus,
0
0t cv
1r V2r dV0
at any instant of time for steady flows or on a time-average basis for cyclical unsteady
flows.
3. We work only with the component of Eq. 5.42 resolved along the axis of rotation.
Consider the rotating sprinkler sketched in Fig. 5.4. Because the direction and magnitude of the flow
through the sprinkler from the inlet [section 112] to the outlet [section 122] of the arm changes, the
water exerts a torque on the sprinkler head causing it to tend to rotate or to actually rotate in the direction
shown, much like a turbine rotor. In applying the moment-of-momentum equation 1Eq. 5.422
to this flow situation, we elect to use the fixed and nondeforming control volume shown in Fig. 5.4.
This disk-shaped control volume contains within its boundaries the spinning or stationary sprinkler
head and the portion of the water flowing through the sprinkler contained in the control volume at
an instant. The control surface cuts through the sprinkler head’s solid material so that the shaft torque
that resists motion can be clearly identified. When the sprinkler is rotating, the flow field in the stationary
control volume is cyclical and unsteady, but steady in the mean. We proceed to use the axial
component of the moment-of-momentum equation 1Eq. 5.422 to analyze this flow.
The integrand of the moment-of-momentum flow term in Eq. 5.42,
(1)
z
r
r × V
V
1r V2rV nˆ dA
cs
can be nonzero only where fluid is crossing the control surface. Everywhere else on the control
surface this term will be zero because V nˆ 0. Water enters the control volume axially through
the hollow stem of the sprinkler at section 112. At this portion of the control surface, the component
of r V resolved along the axis of rotation is zero because as illustrated by the figure in the
margin, r V lies in the plane of section (1), perpendicular to the axis of rotation. Thus, there is
no axial moment-of-momentum flow in at section 112. Water leaves the control volume through
each of the two nozzle openings at section 122. For the exiting flow, the magnitude of the axial
component of r V is r 2 V u2 , where r 2 is the radius from the axis of rotation to the nozzle centerline
and V u2 is the value of the tangential component of the velocity of the flow exiting each nozzle as