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In a hydroelectric scheme a number of Pelton wheels are to be used under the following conditions: Total output required 30 MW; gross head 245 m; speed 6.25 rev/s; 2 jets per wheel; Cv of nozzles 0.97; maximum overall efficiency (based on conditions immediately before the nozzles) 81.5%; dimensionless specific speed not to exceed 0.022 rev. per jet; head lost to friction in pipeline is 12 m. Ratio of blade to jet speed is 0.46.
Calculate the number of wheels required.
  • a)
    1
  • b)
    3
  • c)
    4
  • d)
    5
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
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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Most Upvoted Answer
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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In a hydroelectric scheme a number of Pelton wheels are to be used under the following conditions: Total output required 30 MW; gross head 245 m; speed 6.25 rev/s; 2 jets per wheel; Cv of nozzles 0.97; maximum overall efficiency (based on conditions immediately before the nozzles) 81.5%; dimensionless specific speed not to exceed 0.022 rev. per jet; head lost to friction in pipeline is 12 m. Ratio of blade to jet speed is 0.46.Calculate the number of wheels required.a) 1b) 3c) 4d) 5Correct answer is option 'D'. Can you explain this answer?
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