fluid_mechanics

208 Chapter 5 ■ Finite Control Volume Analysis
Note that the y component of velocity is positive at section 112 but
is negative at section 122. Also, the mass flowrate term is negative
at section 1121flow in2 and is positive at section 1221flow out2. From
the continuity equation 1Eq. 5.122, we get
and thus Eq. 3 can be written as
Solving Eq. 5 for
m # 1v 1 v 2 2 F Ay p 1 A 1 p 2 A 2
F Ay
m # m # 1 m # 2
we obtain
F Ay m # 1v 1 v 2 2 p 1 A 1 p 2 A 2
From the given data we can calculate the mass flowrate, m # , from
Eq. 5.6 as
m # r 1 A 1 v 1 11.94 slugsft 3 210.1 ft 2 2150 fts2
9.70 slugss
For determining the anchoring force, F Ay , the effects of atmospheric
pressure cancel and thus gage pressures for p 1 and p 2 are
appropriate. By substituting numerical values of variables into
Eq. 6, and using the fact that 1 lb 1 slug fts 2 we get
F Ay 19.70 slugss2150 fts 50 fts2
130 psia 14.7 psia21144 in. 2 ft 2 210.1 ft 2 2
124 psia 14.7 psia21144 in. 2 ft 2 210.1 ft 2 2
F Ay 970 lb 220 lb 134 lb 1324 lb (Ans)
F Ay
The negative sign for is interpreted as meaning that the y
component of the anchoring force is actually in the negative y
direction, not the positive y direction as originally indicated in
Fig. E5.12b.
COMMENT As with Example 5.11, the anchoring force for
the pipe bend is independent of the atmospheric pressure. However,
the force that the bend puts on the fluid inside of it, R y ,
(4)
(5)
(6)
depends on the atmospheric pressure. We can see this by using a
control volume which surrounds only the fluid within the bend as
shown in Fig. E5.12c. Application of the momentum equation to
this situation gives
R y m # 1v 1 v 2 2 p 1 A 1 p 2 A 2
where p 1 and p 2 must be in terms of absolute pressure because
the force between the fluid and the pipe wall, R y , is the complete
pressure effect 1i.e., absolute pressure2. We see that forces exerted
on the flowing fluid result in a change in its velocity direction 1a
change in linear momentum2.
Thus, we obtain
R y 19.70 slugss2150 fts 50 fts2
130 psia21144 in. 2 ft 2 210.1 ft 2 2
124 psia21144 in. 2 ft 2 210.1 ft 2 2
1748 lb
We can use the control volume that includes just the pipe
bend 1without the fluid inside it2 as shown in Fig. E5.12d to
determine F Ay , the anchoring force component in the y direction
necessary to hold the bend stationary. The y component of the
momentum equation applied to this control volume gives
F Ay R y p atm 1A 1 A 2 2
(8)
where R y is given by Eq. 7. The p atm 1A 1 A 2 2 term represents the
net pressure force on the outside portion of the control volume.
Recall that the pressure force on the inside of the bend is accounted
for by R y . By combining Eqs. 7 and 8 and using the fact that
p atm 14.7 lbin. 2 1144 in. 2 ft 2 2 2117 lbft 2 , we obtain
F Ay 1748 lb 2117 lbft 2 10.1 ft 2 0.1 ft 2 2
1324 lb
in agreement with the original answer obtained using the control
volume of Fig. E5.12b.
(7)
z
Water in 180° bend
R z
w p 1
A 1
u v y
x
p 2
A 2
R y
Pipe bend only
F Ay
R y p atm
(A 1 + A 2 )
v 2
v 1
(c)
Control volume
R x
Control volume
F I G U R E E5.12 cont.
(d)
V5.8 Fire hose
In Examples 5.10 and 5.12 the force exerted on a flowing fluid resulted in a change in flow
direction only. This force was associated with constraining the flow, with a vane in Example 5.10,
and with a pipe bend in Example 5.12. In Example 5.11 the force exerted on a flowing fluid
resulted in a change in velocity magnitude only. This force was associated with a converging
nozzle. Anchoring forces are required to hold a vane or conduit stationary. They are most easily
estimated with a control volume that contains the vane or conduit and the flowing fluid involved.
Alternately, two separate control volumes can be used, one containing the vane or conduit only
and one containing the flowing fluid only.