Page 399 - Civil Engineering Formulas

326 CHAPTER TWELVE

and the critical depth is
2
D c 3 H m (12.99)
Then the discharge per foot (meter) of width is given by
2 3/2 3/2
q g ( 3 ) H m (12.100)
With g 32.16, Eq. (12.100) becomes
3/2
q 3.087H m (12.101)

Triangular Channels
In a triangular channel (Fig. 12.20), the maximum depth D and the mean depth
c
1
D equal 2 D . Then,
c
m
V c gD c (12.102)
B 2
2
2V c
and D c (12.103)
g
As shown in Fig. 12.20, z is the slope of the channel sides, expressed as a
ratio of horizontal to vertical; for symmetrical sections, z e/D . The area, a
c
2
zD c . Then,
g
Q zD c 5/2 (12.104)
B2
With g 32.16,
5/2
Q 4.01zD c (12.105)
5 2Q 2
and D c (12.106)
B gz 2

T
e e

D c e
z =
D c
FIGURE 12.20 Triangular open channel.