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1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.?
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1) A Pelton wheel is required to develop 6 MW under a head of 300 m. I...
Solution:
Given:
Power to be developed(P) = 6 MW
Head(H) = 300 m
Speed of the wheel(N) = 550 rpm
Jet ratio = 10
Overall efficiency(η) = 85%
Cv = 0.98
Ku = 0.46
i) Diameter of wheel(D):
The power developed by the Pelton wheel is given by the formula:
P = ρQgHη
where, ρ = density of water, Q = quantity of water, g = acceleration due to gravity
We know, Q = (Cv)(A)(N)
where, A = area of the jet
Jet ratio = D/d = 10
where, D = diameter of the wheel, d = diameter of the jet
So, D = 10d
We also know, Ku = (π/4)(d^2)/(π/4)(D^2) = d^2/D^2
So, D = d/√Ku
Substituting these values in the formula for power, we get:
P = (ρCvπ/4)(d^2)(N)(g)(H)(η)/Ku
6 × 10^6 = (ρCvπ/4)(d^2)(550/60)(9.81)(300)(0.85)/0.46√Ku
On solving, we get:
D = 2.7 m
ii) Quantity of water required(Q):
From the equation Q = (Cv)(A)(N), we can find the area of the jet:
A = Q/(CvN)
We know, A = (π/4)(d^2)
Substituting these values, we get:
d = √(4Q/(CvNπ))
We know, D = 10d
Substituting these values in the expression for power, we get:
P = (ρCvπ/4)(d^2)(N)(g)(H)(η)/Ku
On substituting the given values, we get:
6 × 10^6 = (ρ × 0.98 × π/4) (4Q/(CvNπ)) (550/60) (9.81) (300) (0.85) / 0.46 (√Ku)
On solving, we get:
Q = 115.7 m^3/s
iii) Number of jets:
We know, Q = (Cv)(A)(N)
Substituting the given values, we get:
115.7 = 0.98 (π/4) (d^2) (550/60)
On solving, we get:
d = 0.377 m
We know, D = 10d
Substituting the value of D, we get:
D = 3.77 m
Area of the jet, A = (π/4)(d^2) = 0.111 m^2
Area of the water passage = πD^2/4 = 11.24 m^2
Number of jets = Q/(CvA) = 115.7/(0.98 × 0.111) = 1058
Therefore, the diameter of the wheel is 2.7 m, the quantity of
Given:
Power to be developed(P) = 6 MW
Head(H) = 300 m
Speed of the wheel(N) = 550 rpm
Jet ratio = 10
Overall efficiency(η) = 85%
Cv = 0.98
Ku = 0.46
i) Diameter of wheel(D):
The power developed by the Pelton wheel is given by the formula:
P = ρQgHη
where, ρ = density of water, Q = quantity of water, g = acceleration due to gravity
We know, Q = (Cv)(A)(N)
where, A = area of the jet
Jet ratio = D/d = 10
where, D = diameter of the wheel, d = diameter of the jet
So, D = 10d
We also know, Ku = (π/4)(d^2)/(π/4)(D^2) = d^2/D^2
So, D = d/√Ku
Substituting these values in the formula for power, we get:
P = (ρCvπ/4)(d^2)(N)(g)(H)(η)/Ku
6 × 10^6 = (ρCvπ/4)(d^2)(550/60)(9.81)(300)(0.85)/0.46√Ku
On solving, we get:
D = 2.7 m
ii) Quantity of water required(Q):
From the equation Q = (Cv)(A)(N), we can find the area of the jet:
A = Q/(CvN)
We know, A = (π/4)(d^2)
Substituting these values, we get:
d = √(4Q/(CvNπ))
We know, D = 10d
Substituting these values in the expression for power, we get:
P = (ρCvπ/4)(d^2)(N)(g)(H)(η)/Ku
On substituting the given values, we get:
6 × 10^6 = (ρ × 0.98 × π/4) (4Q/(CvNπ)) (550/60) (9.81) (300) (0.85) / 0.46 (√Ku)
On solving, we get:
Q = 115.7 m^3/s
iii) Number of jets:
We know, Q = (Cv)(A)(N)
Substituting the given values, we get:
115.7 = 0.98 (π/4) (d^2) (550/60)
On solving, we get:
d = 0.377 m
We know, D = 10d
Substituting the value of D, we get:
D = 3.77 m
Area of the jet, A = (π/4)(d^2) = 0.111 m^2
Area of the water passage = πD^2/4 = 11.24 m^2
Number of jets = Q/(CvA) = 115.7/(0.98 × 0.111) = 1058
Therefore, the diameter of the wheel is 2.7 m, the quantity of
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1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.?
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1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about 1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.?.
1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about 1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.?.
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1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.?, a detailed solution for 1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.? has been provided alongside types of 1) A Pelton wheel is required to develop 6 MW under a head of 300 m. If rotates with a speed of 550 rpm. Assuming Jet ratio as 10 and overall efficiency as 85%. Take: Cv=0.98 and Ku=0.46. Determine :i) Diameter of wheel ii) Quantity of water required.iii) Number of jets.? theory, EduRev gives you an
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