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A hydraulic turbine has a discharge of 5 m3/s, when operating under a head of 20 m with a speed of 500 rpm. If it is to operate under a head of 15 m, for the same discharge, the rotational speed in rpm will approximately be
A. 433
B. 403
C. 627
D. 388
B. 403
C. 627
D. 388
Correct answer is '433'. Can you explain this answer?
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Verified Answer
A hydraulic turbine has a discharge of 5 m3/s, when operating under a ...
Hydraulic turbine has a discharge of 5 m3/s, when operating under a head of 20 m with a speed of 500 rpm. If it is to operate under a head of 15 m, for the same discharge, the rotational speed in rpm will approximately be.
Most Upvoted Answer
A hydraulic turbine has a discharge of 5 m3/s, when operating under a ...
Given Data:
Discharge Q = 5 m3/s
Head H1 = 20 m
Speed N1 = 500 rpm
Head H2 = 15 m
Formula used:
Theoretical Power developed by a turbine P = ρQgH, where
ρ = Density of water
g = Acceleration due to gravity
H = Head
Q = Discharge
Specific speed Ns = (N√Q)/(H)^(3/4), where
N = Speed of turbine in rpm
Q = Discharge
H = Head
Calculation:
From the given data, theoretical power developed by the turbine at H1 is
P1 = ρQgH1
Let us assume that the efficiency of the turbine remains the same at H2. Then, the theoretical power developed at H2 is
P2 = ρQgH2
Since the discharge remains constant, the power developed by the turbine is directly proportional to the head. Therefore, the ratio of powers is given as
P2/P1 = H2/H1
Substituting the values, we get
P2/P1 = 15/20
P2 = (15/20)P1 = (3/4)P1
Now, let us find the specific speed of the turbine at H1 using the given formula.
Ns1 = (N1√Q)/(H1)^(3/4)
Substituting the values, we get
Ns1 = (500√5)/(20)^(3/4) ≈ 47.5
Since the specific speed is constant for a given turbine, we can use the same value of Ns to find the speed of the turbine at H2.
Using the formula for specific speed, we get
Ns2 = (N2√Q)/(H2)^(3/4)
Substituting the values, we get
47.5 = (N2√5)/(15)^(3/4)
Solving for N2, we get
N2 ≈ 433 rpm
Therefore, the rotational speed of the turbine at H2, for the same discharge, will be approximately 433 rpm.
Answer: Option A. 433
Discharge Q = 5 m3/s
Head H1 = 20 m
Speed N1 = 500 rpm
Head H2 = 15 m
Formula used:
Theoretical Power developed by a turbine P = ρQgH, where
ρ = Density of water
g = Acceleration due to gravity
H = Head
Q = Discharge
Specific speed Ns = (N√Q)/(H)^(3/4), where
N = Speed of turbine in rpm
Q = Discharge
H = Head
Calculation:
From the given data, theoretical power developed by the turbine at H1 is
P1 = ρQgH1
Let us assume that the efficiency of the turbine remains the same at H2. Then, the theoretical power developed at H2 is
P2 = ρQgH2
Since the discharge remains constant, the power developed by the turbine is directly proportional to the head. Therefore, the ratio of powers is given as
P2/P1 = H2/H1
Substituting the values, we get
P2/P1 = 15/20
P2 = (15/20)P1 = (3/4)P1
Now, let us find the specific speed of the turbine at H1 using the given formula.
Ns1 = (N1√Q)/(H1)^(3/4)
Substituting the values, we get
Ns1 = (500√5)/(20)^(3/4) ≈ 47.5
Since the specific speed is constant for a given turbine, we can use the same value of Ns to find the speed of the turbine at H2.
Using the formula for specific speed, we get
Ns2 = (N2√Q)/(H2)^(3/4)
Substituting the values, we get
47.5 = (N2√5)/(15)^(3/4)
Solving for N2, we get
N2 ≈ 433 rpm
Therefore, the rotational speed of the turbine at H2, for the same discharge, will be approximately 433 rpm.
Answer: Option A. 433
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Community Answer
A hydraulic turbine has a discharge of 5 m3/s, when operating under a ...
For constant discharge, model law
H/N2D2 = CONSTANT
20/(500^2*D^2)= 15/(N^2*D^2) from here N =433 rpm
H/N2D2 = CONSTANT
20/(500^2*D^2)= 15/(N^2*D^2) from here N =433 rpm
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A hydraulic turbine has a discharge of 5 m3/s, when operating under a head of 20 m with a speed of 500 rpm. If it is to operate under a head of 15 m, for the same discharge, the rotational speed in rpm will approximately beA. 433B. 403C. 627D. 388Correct answer is '433'. Can you explain this answer?
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A hydraulic turbine has a discharge of 5 m3/s, when operating under a head of 20 m with a speed of 500 rpm. If it is to operate under a head of 15 m, for the same discharge, the rotational speed in rpm will approximately beA. 433B. 403C. 627D. 388Correct answer is '433'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A hydraulic turbine has a discharge of 5 m3/s, when operating under a head of 20 m with a speed of 500 rpm. If it is to operate under a head of 15 m, for the same discharge, the rotational speed in rpm will approximately beA. 433B. 403C. 627D. 388Correct answer is '433'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydraulic turbine has a discharge of 5 m3/s, when operating under a head of 20 m with a speed of 500 rpm. If it is to operate under a head of 15 m, for the same discharge, the rotational speed in rpm will approximately beA. 433B. 403C. 627D. 388Correct answer is '433'. Can you explain this answer?.
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