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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
Correct answer is '433'. Can you explain this answer?
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hydraulic turbine has a discharge of 5 m3/s, when operating under a he...
Solution:
Given parameters:
Discharge Q1 = 5 m3/s
Head H1 = 20 m
Speed N1 = 500 rpm
To find:
Speed N2 at head H2 = 15 m
Formula used:
The specific speed of the turbine is given by the formula:
N_s = N * √Q/H^(3/4)
where,
N = rotational speed in rpm,
Q = discharge in m3/s, and
H = head in meters
The specific speed is a dimensionless quantity that is used to compare the performance of hydraulic turbines.
Calculation:
Using the specific speed formula,
N_s1 = N1 * √Q1/H1^(3/4)
N_s1 = 500 * √5/20^(3/4)
N_s1 = 37.53
Now, at head H2 = 15 m, the specific speed remains constant.
Therefore,
N_s2 = N2 * √Q1/H2^(3/4)
N_s2 = 37.53
Substituting the given values,
N2 * √5/15^(3/4) = 37.53
N2 = 37.53 * 15^(3/4)/√5
N2 ≈ 433 rpm
Therefore, the approximate rotational speed of the hydraulic turbine operating under a head of 15 m for the same discharge of 5 m3/s is 433 rpm.
Given parameters:
Discharge Q1 = 5 m3/s
Head H1 = 20 m
Speed N1 = 500 rpm
To find:
Speed N2 at head H2 = 15 m
Formula used:
The specific speed of the turbine is given by the formula:
N_s = N * √Q/H^(3/4)
where,
N = rotational speed in rpm,
Q = discharge in m3/s, and
H = head in meters
The specific speed is a dimensionless quantity that is used to compare the performance of hydraulic turbines.
Calculation:
Using the specific speed formula,
N_s1 = N1 * √Q1/H1^(3/4)
N_s1 = 500 * √5/20^(3/4)
N_s1 = 37.53
Now, at head H2 = 15 m, the specific speed remains constant.
Therefore,
N_s2 = N2 * √Q1/H2^(3/4)
N_s2 = 37.53
Substituting the given values,
N2 * √5/15^(3/4) = 37.53
N2 = 37.53 * 15^(3/4)/√5
N2 ≈ 433 rpm
Therefore, the approximate rotational speed of the hydraulic turbine operating under a head of 15 m for the same discharge of 5 m3/s is 433 rpm.
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Community Answer
hydraulic turbine has a discharge of 5 m3/s, when operating under a he...
N1√(Q1)/H1^3/4=N2√(Q2)/H2^3/4
Q1=Q2=5m^3/s, N1=500rpm H1=20m, H2=15m
N2=N1{H2/H1}^3/4
=500{15/20}^3/4=403
hence answer is 403
Q1=Q2=5m^3/s, N1=500rpm H1=20m, H2=15m
N2=N1{H2/H1}^3/4
=500{15/20}^3/4=403
hence answer is 403
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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 beCorrect answer is '433'. Can you explain this answer?
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