Chapter 3 x Integral Relations for a Control Volume
The integral term
³
wV
dV
ds | 1 h is very small and will be neglected, and p1 p2.. Then
wt
dt
12
ª 2gh º
V1 | «
¬ D 1 »¼
h
³
ho
dh
h1/2
293
, where D
(D/d)4; but also V1
³ dt,
or: h
o
(b) the tank is empty when []
dh
, separate and integrate:
dt
2
1/2 t
ª 2g º
«
¬ D 1 »¼
1/2
ª 1/2 ­ g
½ º
«ho ®
¾ t» , D
¯ 2(D 1) ¿ ¼»
¬«
0 in (a) above, or tfinal
3.180 The large tank of incompressible
liquid in Fig. P3.180 is at rest when, at t 0,
the valve is opened to the atmosphere.
Assuming h | constant (negligible velocities
and accelerations in the tank), use the
unsteady frictionless Bernoulli equation to
derive and solve a differential equation for
V(t) in the pipe.
§D·
¨ ¸
©d¹
4
Ans. (a)
[2(D 1)g/ho]1/2. Ans. (b)
Fig. P3.180
Solution: Write unsteady Bernoulli from 1 to 2:
2
³
1
wV
V22
V12
gz 2 |
gz1 , where p1
ds wt
2
2
p2 , V1 | 0, z 2 | 0, and z1
dV
dV
L, so the diff. eqn. is 2L
V2
dt
dt
This first-order ordinary differential equation has an exact solution for V
The integral approximately equals
V
§V t·
Vfinal tanh ¨ final ¸ , where Vfinal
© 2L ¹
2gh
h
const
2gh
0 at t
0:
Ans.
3.181 Modify Prob. 3.180 as follows. Let the top of the tank be enclosed and under
constant gage pressure po. Repeat the analysis to find V(t) in the pipe.