Handbook of Civil Engineering Calculations

3. Substitute numerical values in Eq. 32 and solve for t
   Thus, t = 2(200)(150)[14 - (14 x 9.8)°^5 ]/[2(200 + 15O)] = 196 s = 3.27 min.

VARIATION IN HEAD ON A WEIR WITHOUT

INFLOW TO THE RESERVOIR

Water flows over a weir of 60-ft (18.3-m) length from a reservoir having a surface area of
50 acres (202,350 m^2 ). If the inflow to the reservoir ceases when the head on the weir is 2
ft (0.6 m), what will the head be at the expiration of 1 h? Consider that the instantaneous
discharge is given by Eq. Ila.

Calculation Procedure:

1. Develop the time-interval equation
   Let A = surface area of reservoir and C = numerical constant in discharge equation; and
   subscripts 1 and 2 refer to the beginning and end, respectively, of a time interval t. By ex-
   pressing the change in head during an elemental time interval,

2A

*~ CZ>(l//#^5 -l//*?'^5 ) (33)

2. Substitute numerical values in Eq, 33; solve for H 2
   Thus, A = 50(43,560) = 2,178,000 ft^2 (202,336.2 m^2 ); t = 3600 s; solving gives h 2 = 1.32
   ft (0.402 m).

VARIATION IN HEAD ON A WEIR

WITH INFLOW TO THE RESERVOIR

Water flows over an 80-ft (24.4-m) long weir from a reservoir having a surface area of
6,000,000 ft
2
(557,400.0 m
2
) while the rate of inflow to the reservoir remains constant at
2175 fWs (61,578.9 L/s). How long will it take for the head on the weir to increase from
zero to 95 percent of its maximum value? Consider that the instantaneous rate of flow
over the weir is 3Abh^1 -^5.

Calculation Procedure:

1. Compute the maximum head on the weir by equating
   outflow to inflow
   The water in the reservoir reaches its maximum height when equilibrium is achieved, i.e.,
   when the rate of outflow equals the rate of inflow. Let Q 1 = rate of inflow; Q 0 = rate of
   outflow at a given instant; t = time elapsed since the start of the outflow.
   Equating outflow to inflow yields 3.4(8OAJ^x) = 2175; /zmax = 4.0 ft (1.2 m); 0.95/zmax
   = 3.8 ft (1.16m).