# ³ w D

```Chapter 3 x Integral Relations for a Control Volume
The integral term
&sup3;
wV
dV
ds | 1 h is very small and will be neglected, and p1 p2.. Then
wt
dt
12
&ordf; 2gh &ordm;
V1 | &laquo;
&not; D 1 &raquo;&frac14;
h
&sup3;
ho
dh
h1/2
293
, where D
(D/d)4; but also V1
&sup3; dt,
or: h
o
(b) the tank is empty when []
dh
, separate and integrate:
dt
2
1/2 t
&ordf; 2g &ordm;
&laquo;
&not; D 1 &raquo;&frac14;
1/2
&ordf; 1/2 &shy; g
&frac12; &ordm;
&laquo;ho &reg;
&frac34; t&raquo; , D
&macr; 2(D 1) &iquest; &frac14;&raquo;
&not;&laquo;
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
derive and solve a differential equation for
V(t) in the pipe.
&sect;D&middot;
&uml; &cedil;
&copy;d&sup1;
4
Ans. (a)
[2(D 1)g/ho]1/2. Ans. (b)
Fig. P3.180
Solution: Write unsteady Bernoulli from 1 to 2:
2
&sup3;
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
&sect;V t&middot;
Vfinal tanh &uml; final &cedil; , where Vfinal
&copy; 2L &sup1;
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.
```