fluid_mechanics

202 Chapter 5 ■ Finite Control Volume Analysis
E XAMPLE 5.10
Linear Momentum—Change in Flow Direction
FIND Determine the anchoring force needed to hold the vane
GIVEN As shown in Fig. E5.10a, a horizontal jet of water exits
0
in the positive directions at both the inlet and the outlet.
0t u r dV cs
u r V nˆ dA a F x
cv The water enters and leaves the control volume as a free jet at
a nozzle with a uniform speed of V 1 10 ft/s, strikes a vane, stationary if gravity and viscous effects are negligible.
and is turned through an angle .
SOLUTION
We select a control volume that includes the vane and a portion of and
the water (see Figs. E5.10b, c) and apply the linear momentum
equation to this fixed control volume. The only portions of the
w 2 A 2 V 2 w 1 A 1 V 1 gF z
(2)
control surface across which fluid flows are section (1) (the entrance)
where V uî wkˆ, and F x and F z are the net x and z compo-
and section (2) (the exit). Hence, the x and z components nents of force acting on the contents of the control volume. De-
of Eq. 5.22 become
pending on the particular flow situation being considered and the
0 1flow is steady2
coordinate system chosen, the x and z components of velocity, u
and w, can be positive, negative, or zero. In this example the flow is
and
atmospheric pressure. Hence, there is atmospheric pressure surrounding
the entire control volume, and the net pressure force on
0 1flow is steady2
the control volume surface is zero. If we neglect the weight of the
water and vane, the only forces applied to the control volume contents
are the horizontal and vertical components of the anchoring
0
0t w r dV cs
w r V nˆ dA a F z
cv force, F Ax and F Az , respectively.
or
With negligible gravity and viscous effects, and since p 1 p 2 ,
the speed of the fluid remains constant so that V 1 V 2 10 ft/s
u 2 A 2 V 2 u 1 A 1 V 1 gF x
(1) (see the Bernoulli equation, Eq. 3.7). Hence, at section (1),
u 1 V 1 , w 1 0, and at section (2), u 2 V 1 cos , w 2 V 1 sin .
By using this information, Eqs. 1 and 2 can be written as
Nozzle
Nozzle
z
A 1 = 0.06 ft 2
V 1
Control
volume
V 1
F I G U R E E5.10
(a)
(b)
θ
Vane
V 2
θ
(2)
V 1
x
(1)
F Ax
(c)
F Az
and
Equations 3 and 4 can be simplified by using conservation of
mass, which states that for this incompressible flow A 1 V 1
A 2 V 2 , or A 1 A 2 since V 1 V 2 . Thus
and
With the given data we obtain
and
V 1 cos A 2 V 1 V 1 A 1 V 1 F Ax
V 1 sin A 2 V 1 0 A 1 V 1 F Az
F Ax A 1 V 2 1 A 1 V 2 1 cos A 1 V 2 1 11 cos 2
F Az A 1 V 2 1 sin
F Ax 11.94 slugs/ft 3 210.06 ft 2 2110 ft/s2 2 11 cos u2
11.6411 cos u2 slugs ft/s 2
11.6411 cos u2 lb
F Az 11.94 slugs/ft 3 210.06 ft 2 2110 ft/s2 2 sin u
11.64 sin u lb
(3)
(4)
(5)
(6)
(Ans)
(Ans)
COMMENTS The values of F Ax and F Az as a function of are
shown in Fig. E5.10d. Note that if 0 (i.e., the vane does not
turn the water), the anchoring force is zero. The inviscid fluid
merely slides along the vane without putting any force on it. If
90°, then F Ax 11.64 lb and F Az 11.64 lb. It is necessary
to push on the vane (and, hence, for the vane to push on the water)