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The koyna hydroelectric project is equipped with four units of vertical shaft pelton turbines to be coupled with 7000kvA, 3 phase, 50 hertz generators. the generator are provided with 10 pairs of poles. the gross design head is 505m and the transmission efficiency of head race tunnel and penstock together is 94%. the four units together provide for a power of 260MW at efficiency of 91%. the nozzle efficiency is 98% and each turbine has four jets. take speed ratio to be 0.48 and flow coefficient to be 0.98 and nozzle diameter 25% bigger than jet diameter, estimate the; a)discharge for the turbine b)jet diameter c)nozzle tip diameter d)pitch circle diameter of the wheel e)specific speed f)number of buckets on the wheel

1 Answer

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Answer:

a) Discharge for the turbine: 0.053 m³/s

b) Jet diameter: 5.53 meters

c) Nozzle tip diameter: 6.91 meters

d) Pitch circle diameter of the wheel: 11.48 meters

e) Specific speed: 26.76

f) Number of buckets on the wheel: 43

Step-by-step explanation:

To estimate the required values for the Koyna hydroelectric project, we'll use the given information and apply relevant formulas. Let's calculate each value step by step:

a) Discharge for the turbine:

The power output of the four units combined is given as 260 MW. Since the efficiency is 91%, we can calculate the actual power output:

Power output = Efficiency * Total power output

Power output = 0.91 * 260 MW

The power output is related to the discharge (Q) and gross head (H) by the following formula:

Power output = Q * H * ρ * g / 1000

Where:

ρ = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

From this equation, we can solve for Q:

Q = (Power output * 1000) / (H * ρ * g)

Substituting the given values:

Q = (260 MW * 1000) / (505 m * 1000 kg/m³ * 9.81 m/s²)

Q = 0.053 m³/s

Therefore, the discharge for the turbine is 0.053 m³/s.

b) Jet diameter:

The flow coefficient (φ) is given as 0.98. The jet diameter (D) can be calculated using the following formula:

φ = (π * D² * Q * √(2 * g * H)) / (4 * A * √(2 * g * H))

Where:

A = Number of jets

Rearranging the formula, we get:

D² = (4 * A * φ * A * √(2 * g * H)) / (π * Q)

Substituting the given values:

D² = (4 * 4 * 0.98 * 4 * √(2 * 9.81 m/s² * 505 m)) / (π * 0.053 m³/s)

D² ≈ 30.66

Taking the square root:

D ≈ √30.66

D ≈ 5.53 m

Therefore, the jet diameter is approximately 5.53 meters.

c) Nozzle tip diameter:

The nozzle tip diameter (d) is given to be 25% larger than the jet diameter (D):

d = D + 0.25 * D

d = 5.53 m + 0.25 * 5.53 m

d ≈ 6.91 m

Therefore, the nozzle tip diameter is approximately 6.91 meters.

d) Pitch circle diameter of the wheel:

The speed ratio (λ) is given as 0.48. The pitch circle diameter (D_p) is related to the jet diameter (D) by the following formula:

D_p = D / λ

Substituting the given values:

D_p = 5.53 m / 0.48

D_p ≈ 11.48 m

Therefore, the pitch circle diameter of the wheel is approximately 11.48 meters.

e) Specific speed:

The specific speed (N_s) can be calculated using the formula:

N_s = (n * √Q) / (√H^(3/4))

Where:

n = Rotational speed of the turbine (rpm)

The rotational speed (n) can be calculated using the formula:

n = (120 * f) / p

Where:

f = Frequency (Hz)

p = Number of poles

Substituting the given values:

n = (120 * 50 Hz) / 10

n = 600 rpm

Substituting the calculated values into the specific speed formula:

N_s = (600 rpm * √0.053 m³/s) / (√505 m)^(3/4)

N_s ≈ 26.76

Therefore, the specific speed is approximately 26.76.

f) Number of buckets on the wheel:

The number of buckets (B) on the wheel is related to the specific speed (N_s) by the formula:

N_s = (n * B) / (√H^(5/4))

Solving for B:

B = (N_s * √H^(5/4)) / n

Substituting the given values:

B = (26.76 * √505 m^(5/4)) / 600 rpm

B ≈ 43.09

Therefore, the number of buckets on the wheel is approximately 43.

answered
User Amir Khorsandi
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