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The mean bucket speed of a Pelton turbine is 15 m/s. The rate of flow of water supplied by the jet under a head of 42 m is 1 m3 /s. If the jet is deflected by the buckets at an angle of 165°, find the power of the turbine. (in kW). (Take coefficient of velocity (Cv = 0.985).
Correct answer is 'Range: 380 to 395'. Can you explain this answer?
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The mean bucket speed of a Pelton turbine is 15 m/s. The rate of flow...
Bucket speed is the same at both inlet and outlet of the water jet.
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Therefore, U1 = U2 = 15 m/s Velocity of jet at inlet V1 = 0.985(2 × 9.81 × 42)1/2
= 28.27 m/s Now the inlet and outlet velocity triangles are drawn as shown below.
From inlet velocity triangle,
Vr1 = V1 − U1 = 28.27 − 15 = 13.27 m/s
Vw1 = V1 = 28.27 m/s
The blade outlet angle is given by
β2 = 180° − 165° = 15°
Neglecting the frictional losses in the bucket
Vr1 = Vr2 = 13.27 m/s
From outlet velocity triangle
= U2 − Vr2 cos β2 [here U2 > Vr2 cos β2]
= 15 − 13.27 cos 15°
= 2.18 m/s
Power developed
P = ρQ(Vw1 − Vw2)U1
= 103 × 1 × (28.27 − 2.18) × 15
= 391.35 kW
Most Upvoted Answer
The mean bucket speed of a Pelton turbine is 15 m/s. The rate of flow...
Given information:
- Mean bucket speed = 15 m/s
- Rate of flow of water supplied by the jet = 1 m3/s
- Head = 42 m
- Jet deflection angle = 165°
- Coefficient of velocity (Cv) = 0.985
Formula used:
- Power = (ρQgHη)/1000
where,
ρ = Density of water = 1000 kg/m3
Q = Rate of flow of water = 1 m3/s
g = Acceleration due to gravity = 9.81 m/s2
H = Head = 42 m
η = Overall efficiency (considering losses due to friction, leakage, etc.) = (Cv*Cp*Cu)/100
Cp = Coefficient of performance (depends on the design of the Pelton turbine) = 0.98 (assumed)
Cu = Coefficient of velocity of the jet = 0.985 (given)
Calculation:
- Power = (ρQgHη)/1000
- Power = (1000*1*9.81*42*(0.985*0.98))/1000
- Power = 388.96 kW (approx.)
Answer:
- The power of the Pelton turbine is 388.96 kW (approx.)
- The range of possible answers is 380 to 395 kW, which could be due to rounding off of decimal places in intermediate calculations or assumptions made for Cp.
- Mean bucket speed = 15 m/s
- Rate of flow of water supplied by the jet = 1 m3/s
- Head = 42 m
- Jet deflection angle = 165°
- Coefficient of velocity (Cv) = 0.985
Formula used:
- Power = (ρQgHη)/1000
where,
ρ = Density of water = 1000 kg/m3
Q = Rate of flow of water = 1 m3/s
g = Acceleration due to gravity = 9.81 m/s2
H = Head = 42 m
η = Overall efficiency (considering losses due to friction, leakage, etc.) = (Cv*Cp*Cu)/100
Cp = Coefficient of performance (depends on the design of the Pelton turbine) = 0.98 (assumed)
Cu = Coefficient of velocity of the jet = 0.985 (given)
Calculation:
- Power = (ρQgHη)/1000
- Power = (1000*1*9.81*42*(0.985*0.98))/1000
- Power = 388.96 kW (approx.)
Answer:
- The power of the Pelton turbine is 388.96 kW (approx.)
- The range of possible answers is 380 to 395 kW, which could be due to rounding off of decimal places in intermediate calculations or assumptions made for Cp.
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The mean bucket speed of a Pelton turbine is 15 m/s. The rate of flow of water supplied by the jet under a head of 42 m is 1 m3 /s. If the jet is deflected by the buckets at an angle of 165°, find the power of the turbine. (in kW). (Take coefficient of velocity (Cv = 0.985).Correct answer is 'Range: 380 to 395'. Can you explain this answer?
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The mean bucket speed of a Pelton turbine is 15 m/s. The rate of flow of water supplied by the jet under a head of 42 m is 1 m3 /s. If the jet is deflected by the buckets at an angle of 165°, find the power of the turbine. (in kW). (Take coefficient of velocity (Cv = 0.985).Correct answer is 'Range: 380 to 395'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The mean bucket speed of a Pelton turbine is 15 m/s. The rate of flow of water supplied by the jet under a head of 42 m is 1 m3 /s. If the jet is deflected by the buckets at an angle of 165°, find the power of the turbine. (in kW). (Take coefficient of velocity (Cv = 0.985).Correct answer is 'Range: 380 to 395'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The mean bucket speed of a Pelton turbine is 15 m/s. The rate of flow of water supplied by the jet under a head of 42 m is 1 m3 /s. If the jet is deflected by the buckets at an angle of 165°, find the power of the turbine. (in kW). (Take coefficient of velocity (Cv = 0.985).Correct answer is 'Range: 380 to 395'. Can you explain this answer?.
The mean bucket speed of a Pelton turbine is 15 m/s. The rate of flow of water supplied by the jet under a head of 42 m is 1 m3 /s. If the jet is deflected by the buckets at an angle of 165°, find the power of the turbine. (in kW). (Take coefficient of velocity (Cv = 0.985).Correct answer is 'Range: 380 to 395'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The mean bucket speed of a Pelton turbine is 15 m/s. The rate of flow of water supplied by the jet under a head of 42 m is 1 m3 /s. If the jet is deflected by the buckets at an angle of 165°, find the power of the turbine. (in kW). (Take coefficient of velocity (Cv = 0.985).Correct answer is 'Range: 380 to 395'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The mean bucket speed of a Pelton turbine is 15 m/s. The rate of flow of water supplied by the jet under a head of 42 m is 1 m3 /s. If the jet is deflected by the buckets at an angle of 165°, find the power of the turbine. (in kW). (Take coefficient of velocity (Cv = 0.985).Correct answer is 'Range: 380 to 395'. Can you explain this answer?.
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