CIVIL WORKS GUIDELINES FOR MICRO-HYDROPOWER IN NEPAL
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Example 4.1 Headrace sizing for the Galkot MHP
The existing irrigation canal at Galkot needs to be modified as a headrace canal for a power output of 50 kw. The existing
irrigation canal's command area is 20 hectares. The community has requested that the canal be sized such that it would
be possible to irrigate the fields and produce 50 sectional areas should be assumed (i.e. trial and error) kW simultaneously.
The following information was collected through site investigation and detailed survey:
Gross head (h) = 22 m (forebay to powerhouse)
Intake to 130 m downstream: 1:50 slope (S) with one drop structure.
131 m to 231 m downstream: 1:92 slope.
232 m to 405 m downstream: 1:365 slope.
406 m to 730 m downstream: 1:975 slope with 3 crossings.
731 m to 1119 m downstream (forebay): 1:400 slope with one crossing.
Site conditions dictate that the entire headrace alignment be constructed out of stone masonry in cement mortar.
Note that 1 m gap has been provided at change of slopes. This is the transitional length that connects the two different
slopes.
In this example, the required design flow will be calculated and the headrace canal from chainage 406 m to 730 m will be
sized.
Design flow calculations:
Assume 55% overall efficiency (eo = 0.55)
P = Q g h eo (power equation)
Q = P/(gheo)
= 50/(9.8 x 22 x 0.55) = 0.421 m3/s
Therefore flow required for power generation is 4211/s
Assume an irrigation requirement of 1.5 l/s/ha for existing irrigated land.
Irrigation demand (Q) = 1.5 l/s/ha x 20 ha
= 30 l/s
Total design flow for the headrace canal = 421 l/s + 30 l/s = 451 l/s. Therefore use a design flow of 455 l/s to size the
headrace canal.
Canal sizing
Canal type: stone masonry in cement mortar
n = 0.020 for dressed stone masonry (from Table 4.1)
Q = 0.455 m3/s
s = 1/975
From Table 4.2 choose N = 0.5 (lh/2v)
Set the bottom width (B) = 0.450 m which is the size of the existing irrigation canal. This minimises excavation works.
Now use the following form of the Manning’s equation where only the water depth, H, is unknown:
(BH+NH2)5/3 s
Q = n[(B + 2H (l+N2)]2/3
(0.450H+0.5H2)5/3 + 1/975
0.455 = 0.020[0.450+2H (1+0.52) ]2/3
By trial and error method, the above equation is balanced when H = 0.768 m for a flow of 4551/s.
Therefore, the water depth will be about 768 mm. Now check that the velocity is less than the maximum allowable velocity
of 2.0 m/s from Table 4.1.
V = Q/A
or V = 0.455 / (BH+NH2)
or V = 0.455 / (0.450x0.768+0.5x0.7682)
or V = 0.7 m/s < 2.0 m/s OK.
The drawing and dimensions for this canal section can be seen in Drawing 420/04/2A01 (Canal type B) of Appendix C. Note that
the original design was based on an assumed value of 0.017 for Manning’s n, giving a water depth of 705 mm, therefore actual
freeboard may be less than recommended. The other canal sections of this scheme can be verified by similar calculations.