- Differentiate this equation with respect to L to obtain the rate
of change of D; substitute numerical values
Differentiating gives
dD sb - se~a
=
i -No*)(31a)or in accordance with Eq. 28,dD sb - se
-E- -Tw? (31*)Substituting yields QlKgD^3 ) = 282 /(32.2 x 3.2^3 ) = 0.743; dDldL = (0.0010 - 0.0004)/(1 -
0.743) = 0.00233 ft/ft (0.00233 m/m). The depth is increasing in the downstream direc-
tion, and the water is therefore being decelerated.
As Eq. 3IZ? reveals, the relationship between the actual depth at a given section and the
critical depth serves as a criterion in ascertaining whether the depth is increasing or de-
creasing.
DISCHARGE BETWEEN COMMUNICATING
VESSELS
In Fig. 13, liquid is flowing from tank A to tank B through an orifice near the bottom. The
area of the liquid surface is 200 ft
2
(18.58 m
2
) in A and 150 ft
2
(13.93 m
2
) in B. Initially,
the difference in water levels is 1.4 ft (4.3 m), and the discharge is 2 ft
3
/s (56.6 L/s). As-
suming that the discharge coefficient remains constant, compute the time required for the
water level in tank A to drop 1.8 ft (0.54 m).
Calculation Procedure:- By expressing the change in h during an elemental time
interval, develop the time-interval equation
Let Aa and Ab denote the area of the liquid surface in tanks A and B, respectively;
let subscripts 1 and 2 refer to the beginning and end, respectively, of a time interval t.
Then
24A(Jt 1 -[MJ
0
'
5
) ,
fii(^ + A)- Find the value of h when ya
diminishes by 1.8 ft (0.54 m)
Thus, Ayb = (-AJAMAyJ = -<200/l50)(-1.8) = 2.4 ft
(0.73 m); Ah = Ay 0 -Ayb = -l.Z- 2.4 = -4.2 ft (-1.28
FIGURE 13 m); hi = 14 ft (4.3 m); A 2 = 14 - 4.2 = 9.8 ft (2.99 m).