```Chapter 3 x Integral Relations for a Control Volume
symmetrically along the barrier.
195
Calculate the horizontal force F needed, per unit
thickness into the paper, to hold the barrier stationary.
Solution:
For water take U = 998 kg/m. The control volume (see figure) cuts through all
four jets, which are numbered. The velocity of all jets follows from the weight flow at (1):
V1, 2,3,4
m 1
w 1
UgA1
V1
w 1
g
1960 N / s
(9.81m / s 2 )(998kg / m 3 )(0.04m)(1m)
1960 N / s m
9.81N / s
2
200
kg
; m 2
sm
0.3m 1
60
5.0
kg
; m 3
sm
m
s
m 4
70
kg
sm
Then the x-momentum relation for this control volume yields
6Fx
m 2 u 2 m 3u3 m 4 u 4 m 1u1
F
F
(60)(5.0) (70)(5.0 cos 55D ) (70)(5.0 cos 55D ) 200(5.0) , or :
F
1000 201 201 300 |
1100 N per meter of width
3.31 A bellows may be modeled as a
deforming wedge-shaped volume as in
Fig. P3.31. The check valve on the left
(pleated) end is closed during the stroke.
If b is the bellows width into the paper,
derive an expression for outlet mass flow
m o as a function of stroke T(t).
Solution: For a control volume enclosing
the bellows and the outlet flow, we obtain
d
out
( UX ) m
dt
0, where X
o
since L is constant, solve for m
Fig. P3.31
bhL
bL2 tan T
d
dT
( U bL2 tan T ) U bL2 sec 2T
dt
dt
Ans.
Ans.
```