54 CIVIL WORKS GUIDELINES FOR MICRO-HYDROPOWER IN NEPAL
B = 0.405 x 1.236
B = 0.50 m
Calculate the top width up to the design water level, T:
T = B + (2HN)
T = 0.50 + (2 x 0.405 x 0.5)
T = 0.905 m
Check if V< 0.8 Vc
Vc = Ag/T = 0.285x9.8 / 0.905
Vc = 1.76m/s
0.8 Vc = 1.41 m/s > V = 1.0 m/s OK
Calculate the wetted perimeter, P:
P = B+2xH (1+N2)
P = 0.5 + 2 x 0.405 (l+0.52)
P = 1.406m
Calculate the hydraulic radius, R:
R = A/P = 0285/1.406 = 0.230 m
Figure 4.7 Proposed internal canal dimensions for Example 4.1
Calculate the required canal bed slope, S:
( )S = nV
2
R0.667
S = (0.035x1/0.2030.667)2
S = 0.0103 or 1:97 (i.e. 1 m of drop in 97 m of horizontal canal length)
Finally allow 300 mm of freeboard. The canal dimensions can be seen in Figure 4.7.
Check the flow depth for maximum flood flow in the canal.
(BH+NH2)5/3 S
Q = n[(B + 2H (l+N2)]2/3
(0.5H+0.52)5/3 x 0.0103
0.480 = 0035[(0.5+2H (l+0.52)]2/3
By trial and error method, the above equation is balanced when H = 0.55 m. Therefore, the flood flow occupies 50% of the freeboard
(the maximum allowed, as discussed earlier) and the head on the spillway (hovertop) will be 100 mm.
Check the size of particle that will settle in the canal at a velocity of 1.0 m/s.
D = 11RS
= 11x0.203x0.0103
= 23 mm
i.e., particles larger than 23 mm would settle in this headrace canal. Therefore, to avoid deposition upstream of the settling basin, the
gravel trap must be designed to remove all particles greater than 23 mm.