Guide on How to Develop a Small Hydropower Plant
ESHA 2004
The main Pelton dimensions would be D1 = 0.962 m, B2 = 0.274 m and De = 0.108 m which are quite reasonable. If we now select a Francis turbine, the maximum value of nQE would be 0.33 (table 6.2). Using equation 6.5, the corresponding speed would be n = 76.43 t/s or 4'765.8 rpm, which is far from a realistic synchronous rotational speed. For this reason, we will choose the maximum usual value, which is 1,500 rpm. According to 6.5, the corresponding nQE would be:
n
QE
=
n⋅ Q E
3 4
=
25⋅ 1.5
(9.81⋅200)
3 4
= 0.104
[-]
The main Francis runner dimensions according to 6.20, 6.21 and 6.22 would be:
D
3
= 84.5 ⋅ (0.31 + 2.488 ⋅ n QE) ⋅
)⋅D D = (0.4 + 0.0950 n 1
3
QE
H
n
60 ⋅ n
= 84.5 ⋅ (0.31 + 2.488 ⋅ 0.104) ⋅
= (0.4 + 0.0950 ) ⋅ 0.453 = 0.595 0.104
200 = 0.453 60 ⋅ 25
[m]
[m]
As nQE < 0.164, we can consider than D2 = D1 = 0.595 m. According to equation 6.28, the cavitation coefficient would be: 1.41
σ = 1.2715 ⋅ n QE +
v
2
2⋅g⋅H n
= 1.2715 ⋅ 0.1041.41 +
2
2 = 0.0533 2⋅9.81⋅200
[-]
According to equation 6.27, the setting would be: 2
−Pv + v − σ⋅H = 90'250−880 + 2 − 0.0533 ⋅ 200 = - 1.35 Hs = Patm n 1'000 ⋅9.81 2⋅9.81 ρ⋅g 2⋅g 2
[m]
A setting that requires important excavation. If we have selected a Francis running at 1000 rpm we would have had: nQE = 0.069, D3 = 0.576 m, D1 = 1.02 m, σ = 0.0305 and Hs = 3.21 m which does not need excavation. The final choice will be economical. If the flow strongly varies, a 4-nozzle Pelton could be a good choice. If it is not the case, a 1000-rpm Francis that does not need any excavation could be the best alternative.
1
By Vincent Denis (MHyLab), Jean-Pierre Corbet (SCPTH), Jochen Bard (ISET), Jacques Fonkenell (SCPTH) and Celso Penche (ESHA)
2
L. Vivier, "Turbines hydrauliques et leur régulation", Albin Michel, Paris, 1966
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