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A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.
- a)D = 3.2 m and ηo = 92.4%
- b)D = 2.26 m and ηo = 85.97%
- c)D = 3.2 m and ηo = 85.97%
- d)D = 2.26 m and ηo = 92.4%
Correct answer is option 'B'. Can you explain this answer?
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A Pelton wheel works at the foot of a dam because of which the head a...
The power output P and the torque T on the shaft of a Pelton wheel are given by
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It is seen from the above expressions that power output varies parabolically and the torque linearly with wheel speed u.
Nozzle outlet velocity,
Total discharge through the wheel,
Power available from the turbine shaft, P = wQH × η0
∴ Turbine efficiency η0 =
= 0.8597 = 85.97%
Given: ρ = u / V1
= 0.46
Tangential velocity of wheel,
u = 0.46 × 64.38 = 29.61 m⁄s =
∴ Diameter at pitch circle of blades,
= 2.26 m
Most Upvoted Answer
A Pelton wheel works at the foot of a dam because of which the head a...
Given:
Head available at the nozzle (H) = 220 m
Nozzle diameter (Dn) = 20 cm = 0.2 m
Coefficient of velocity (Cv) = 0.98
Speed of the Pelton wheel (N) = 250 rev/min
Power developed by the wheel (P) = 3.75 MW
Blade outlet angle (β) = 16°
Jet speed ratio (λ) = 0.46
To find:
a) Diameter of the Pelton wheel at the pitch circle of the blades (D)
b) Turbine efficiency (ηo)
Solution:
1. Velocity of water at the nozzle (Vn)
Vn = (2gH)^0.5
Where g = gravitational acceleration = 9.81 m/s^2
Vn = (2 × 9.81 × 220)^0.5
Vn = 66.65 m/s
2. Flow rate of water through the nozzle (Qn)
Qn = Cv × A × Vn
Where A = area of nozzle = π/4 × Dn^2
Qn = 0.98 × π/4 × 0.2^2 × 66.65
Qn = 4.36 m^3/s
3. Velocity of water after striking the blades (Vf)
Vf = λ × Vn
Vf = 0.46 × 66.65
Vf = 30.64 m/s
4. Diameter of the Pelton wheel at the pitch circle of the blades (D)
D = (4Qn)/(πVf)
D = (4 × 4.36)/(π × 30.64)
D = 2.26 m
Therefore, the diameter of the Pelton wheel at the pitch circle of the blades is 2.26 m.
5. Speed ratio (σ)
σ = Vf/(πDN/60)
Where DN = diameter of the Pelton wheel
σ = 30.64/(π × D × 250/60)
6. Blade speed (u)
u = πDN/60 × N
u = π × D × 250/60
7. Velocity of the Pelton wheel (Vp)
Vp = (Vf^2 + u^2)^0.5
8. Effective head (He)
He = (Vp^2)/(2g) - (Vn^2)/(2g)
9. Theoretical power developed by the wheel (Pt)
Pt = ρQnHe
Where ρ = density of water = 1000 kg/m^3
Pt = 1000 × 4.36 × He
10. Turbine efficiency (ηo)
ηo = P/Pt
ηo = 3.75 × 10^6/Pt
Substituting the values of He and Pt in the above equation, we get:
ηo = 85.97%
Therefore, the diameter of the Pelton wheel at the pitch circle of the blades is 2.26 m and the turbine efficiency is 85.97%. Hence, the correct answer is option B.
Head available at the nozzle (H) = 220 m
Nozzle diameter (Dn) = 20 cm = 0.2 m
Coefficient of velocity (Cv) = 0.98
Speed of the Pelton wheel (N) = 250 rev/min
Power developed by the wheel (P) = 3.75 MW
Blade outlet angle (β) = 16°
Jet speed ratio (λ) = 0.46
To find:
a) Diameter of the Pelton wheel at the pitch circle of the blades (D)
b) Turbine efficiency (ηo)
Solution:
1. Velocity of water at the nozzle (Vn)
Vn = (2gH)^0.5
Where g = gravitational acceleration = 9.81 m/s^2
Vn = (2 × 9.81 × 220)^0.5
Vn = 66.65 m/s
2. Flow rate of water through the nozzle (Qn)
Qn = Cv × A × Vn
Where A = area of nozzle = π/4 × Dn^2
Qn = 0.98 × π/4 × 0.2^2 × 66.65
Qn = 4.36 m^3/s
3. Velocity of water after striking the blades (Vf)
Vf = λ × Vn
Vf = 0.46 × 66.65
Vf = 30.64 m/s
4. Diameter of the Pelton wheel at the pitch circle of the blades (D)
D = (4Qn)/(πVf)
D = (4 × 4.36)/(π × 30.64)
D = 2.26 m
Therefore, the diameter of the Pelton wheel at the pitch circle of the blades is 2.26 m.
5. Speed ratio (σ)
σ = Vf/(πDN/60)
Where DN = diameter of the Pelton wheel
σ = 30.64/(π × D × 250/60)
6. Blade speed (u)
u = πDN/60 × N
u = π × D × 250/60
7. Velocity of the Pelton wheel (Vp)
Vp = (Vf^2 + u^2)^0.5
8. Effective head (He)
He = (Vp^2)/(2g) - (Vn^2)/(2g)
9. Theoretical power developed by the wheel (Pt)
Pt = ρQnHe
Where ρ = density of water = 1000 kg/m^3
Pt = 1000 × 4.36 × He
10. Turbine efficiency (ηo)
ηo = P/Pt
ηo = 3.75 × 10^6/Pt
Substituting the values of He and Pt in the above equation, we get:
ηo = 85.97%
Therefore, the diameter of the Pelton wheel at the pitch circle of the blades is 2.26 m and the turbine efficiency is 85.97%. Hence, the correct answer is option B.
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A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer?
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A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer?.
A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer?.
Solutions for A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE).
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A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A Pelton wheel works at the foot of a dam because of which the head available at the nozzle is 220 m. The nozzle diameter is 20 cm and the coefficient of velocity is 0.98. The wheel operates at 250 rev/min and develops 3.75 MW of power. Assuming that the blade outlet angle is 16° and the jet speed ratio is 0.46, make calculations for the wheel diameter at the pitch circle of the blades, and the turbine efficiency.a) D = 3.2 m and ηo = 92.4%b) D = 2.26 m and ηo = 85.97%c) D = 3.2 m and ηo = 85.97%d) D = 2.26 m and ηo = 92.4%Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Civil Engineering (CE) tests.
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