EXAMPLE #01

Data:

1. Triangular channel with side slopes having ratio of 1:1.5 2. Flowrate is 0.68 ft3/s 3. Channel type – clean, excavated earth

Calculate:

1. Critical depth 2. Minimum specific energy 3. Plot specific energy curve 4. Determine energy for 0.45 ft and alternate depth 5. Velocity of flow and Froude number 6. Calculate require slope for given flow

SOLUTION FOR EXAMPLE #01 Used Excel. z = 1.5 Q = 0.68 ft3/s n = 0.022 y = independence number Area, ( )( ) 22

21 zyyzyA ==

Flow velocity, AQV =

Free surface width, zyT 2=

Hydraulic depth,

TAyh =

We cannot solve this question directly because the dimension of the triangular weir is not given. Trial and error method need to be applied. First step, create excel worksheet and calculate all the important parameter.

(1) Critical depth Critical depth occur at minimum energy, or at Froude number, Fr = 1.0 Find in the excel worksheet. Critical depth is 0.53 ft

(2) Minimum specific energy, Emin Minimum specific energy occur at critical depth. Emin = 0.66275 ft

(c) Plot the graph y-axis is E (energy) and x-axis is depth (y)

Specific energy

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.5 1 1.5 2

Energy (ft)

Dep

th (f

t)

(d) depth 0.45 ft, Energy = 0.705438 ft Alternate depth = 0.645 ft (which has same specific energy)

(e) Velocity for y = 0.450 ft, V = 2.24 ft/m, Fr = 1.51, super-critical flow for y = 0.645 ft, V = 1.09 ft/m, Fr = 0.61, sub-critical flow (f) slope ? Take y = 0.45 ft For triangle, Cross section area, A = 1

22zy( ) y( ) = zy2 = 0.30375

Wetted perimeter, A = 2y 1+ z2 =1.6225

Hydraulic radius, R = A

WP= 0.30375

1.6225= 0.1872

Flowrate, Q = 1.49

nAR

23S

12

So, slope, S = Qn

1.49( )AR23

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2

= 0.01

Or 1 foot drop for every 100 feets.