fluid_mechanics

212 Chapter 5 ■ Finite Control Volume Analysis
When the gate is closed, the horizontal forces acting on the
contents of the control volume are identified in Fig. E5.16c. Application
of Eq. 5.22 to the contents of this control volume yields
0 1no flow2
(1)
Note that the hydrostatic pressure force, gH 2 b2, is used. From
Eq. 1, the force exerted on the water by the gate 1which is equal to
the force necessary to hold the gate stationary2 is
R x 1 2 gH 2 b
which is equal in magnitude to the hydrostatic force exerted on
the gate by the water.
When the gate is open, the horizontal forces acting on the contents
of the control volume are shown in Fig. E5.16d. Application
of Eq. 5.22 to the contents of this control volume leads to
cs
urV nˆ dA 1 2 gH 2 b R x 1 2 gh 2 b F f
Note that because the water at sections (1) and (2) is flowing
along straight, horizontal streamlines, the pressure distribution at
those locations is hydrostatic, varying from zero at the free surface
to g times the water depth at the bottom of the channel (see
Chapter 3, Section 3.4). Thus, the pressure forces at sections (1)
and (2) (given by the pressure at the centroid times the area) are
gH 2 b2 and gh 2 b2, respectively. Also, the frictional force between
the channel bottom and the water is specified as F f . The
surface integral in Eq. 3 is nonzero only where there is flow
across the control surface. With the assumption of uniform velocity
distributions,
cs
urV nˆ dA 1u 1 2r1u 1 2Hb 1u 2 2r1u 2 2hb
Thus, Eqs. 3 and 4 combine to form
ru 2 1Hb ru 2 2hb 1 2 gH 2 b R x 1 2 gh 2 b F f
(2)
(3)
(4)
(5)
cs
urV nˆ dA 1 2 gH 2 b R x
If the upstream velocity, is much less than so that the
Closed sluice
gate
Control volume
H
Control volume
H
Open sluice
gate
h
z
x
u
(a)
Control volume
Water only
F I G U R E E5.16
Control volume
H h,
u 1 ,
contribution of the incoming momentum flow to the control surface
integral can be neglected and from Eq. 5 we obtain
R x 1 2 gH 2 b 1 2 gh 2 b F f ru 2 2hb
By using the continuity equation, m # bHu 1 bhu 2 , Eq. (6)
can be rewritten as
R x 1 2 H 2 b 1 2 h 2 b F f m # 1u 2 u 1 2
Hence, since u 2 7 u 1 , by comparing the expressions for R x (Eqs.
2 and 7) we conclude that the reaction force between the gate and
the water (and therefore the anchoring force required to hold the
gate in place) is smaller when the gate is open than when it is
closed.
(Ans)
(b)
Water only
u 1
R x
1_
( 2 γ H )
Hb
Section (2)
u 2
1_
( 2 γ H )
Hb 1_
Section F f ( )
(c)
R x
(1)
(d)
2 γ h hb
u 2
(6)
(7)
The linear momentum
equation can
be written for a
moving control
volume.
V5.9 Jelly fish
All of the linear momentum examples considered thus far have involved stationary and nondeforming
control volumes which are thus inertial because there is no acceleration. A nondeforming
control volume translating in a straight line at constant speed is also inertial because there is
no acceleration. For a system and an inertial, moving, nondeforming control volume that are both
coincident at an instant of time, the Reynolds transport theorem 1Eq. 4.232 leads to
D
Dt sys
Vr dV 0 0t cv
Vr dV cs
VrW nˆ dA
When we combine Eq. 5.23 with Eqs. 5.19 and 5.20, we get
0
0t cv
Vr dV cs
VrW nˆ dA a F contents of the
control volume
(5.23)
(5.24)
When the equation relating absolute, relative, and control volume velocities 1Eq. 5.142 is used with
Eq. 5.24, the result is
0
0t cv
1W V cv 2r dV cs
1W V cv 2rW nˆ dA a F contents of the
control volume
(5.25)