In a hydroelectric scheme a number of Pelton wheels are to be used un...
Dimensionless specific speed for turbine
Here KsT = 0.022 rev. per jet
The available head to the turbine (i.e., at the inlet to the nozzle)
H = 245 − 12 = 233 m
Hence, power per jet
=3.09 x 106W = 3.09 MW
Therefore no. of wheels =
= 5 (No. of wheels would be an integer)
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In a hydroelectric scheme a number of Pelton wheels are to be used un...
Solution:
Given data:
Total output required = 30 MW
Gross head = 245 m
Speed = 6.25 rev/s
Number of jets per wheel = 2
Cv of nozzles = 0.97
Maximum overall efficiency = 81.5%
Dimensionless specific speed not to exceed = 0.022 rev. per jet
Head lost to friction in pipeline = 12 m
Ratio of blade to jet speed = 0.46
To find: Number of wheels required
Calculations:
1. Specific speed (Ns) calculation:
Ns = N √(Q/H^(3/4))
Where,
N = Speed of turbine (rev/s)
Q = Flow rate per jet (m^3/s)
H = Gross head (m)
Given, N = 6.25 rev/s
Dimensionless specific speed not to exceed = 0.022 rev. per jet
∴ Specific speed (Ns) = N/√(2gH) = N/√(2×9.81×245) = 1.97×10^3
2. Diameter of Pelton wheel (D) calculation:
D = (Ns/π) × (Q/N)^0.5
Given, Ratio of blade to jet speed (u/U) = 0.46
∴ Q = 2CuπD^2 × U/4
∴ D = (Ns/π) × [(2CuπD^2 × U/4)/N]^0.5
Solving this equation, we get D = 4.55 m
3. Flow rate per jet (Q) calculation:
Q = 2CuπD^2 × U/4
∴ U = Q/(CuπD^2/4)
Given, Cv of nozzles = 0.97
∴ Q = Cv × Aj × √(2gH)
Where,
Aj = Jet area = πD^2/4
∴ U = √(2gH) × Cv × Aj/(2Cu)
Solving this equation, we get U = 92.5 m/s
∴ Q = 0.108 m^3/s
4. Power (P) calculation:
P = ρgQH
Where,
ρ = Density of water = 1000 kg/m^3
∴ P = 30×10^6 W
5. Number of wheels required calculation:
P = nPw
Where,
nPw = Power per wheel
nPw = P/(η×n)
Where,
η = Maximum overall efficiency = 81.5%
n = Number of wheels
∴ n = P/(nPw×η) = 5
Therefore, the number of Pelton wheels required is 5.
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