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A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m3 /s at 450 rpm. Take the water density and the acceleration due to gravity as 103 kg/m3 and 10 m/s2 , respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full sized turbine while working under a head of 30 m is __________ (correct to two decimal places).
Correct answer is '29.05'. Can you explain this answer?
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Given data:
- Scale model of a Francis turbine
- Head of the model turbine: 2 m
- Volumetric flow rate of the model turbine: 1 m3/s
- Speed of the model turbine: 450 rpm
- Water density: 103 kg/m3
- Acceleration due to gravity: 10 m/s2
Assumptions:
- No losses in both the model and prototype turbines
Objective:
To determine the power output of the full-sized turbine when operating under a head of 30 m.
Approach:
To find the power of the full-sized turbine, we need to use the concept of similarity between the model and prototype turbines. The similarity laws, specifically the law of similarity, can be used to relate the performance of the model turbine to that of the prototype turbine.
Geometric Similarity:
The scale model and the prototype turbine are geometrically similar. This means that the corresponding linear dimensions of the model and the prototype turbine are in the same ratio. Therefore, the ratio of the head and the volumetric flow rate is also the same for both the model and the prototype turbines.
Head Ratio:
The head ratio (HR) is given by the formula:
HR = (Head of the prototype turbine) / (Head of the model turbine)
HR = 30 m / 2 m
HR = 15
Volumetric Flow Rate Ratio:
The volumetric flow rate ratio (Qr) is given by the formula:
Qr = (Volumetric flow rate of the prototype turbine) / (Volumetric flow rate of the model turbine)
Qr = (Qr * Qm) / (Qm * Qr)
Qr = 1
Speed Ratio:
The speed ratio (Nr) is given by the formula:
Nr = (Speed of the prototype turbine) / (Speed of the model turbine)
Nr = (Nr * Nm) / (Nm * Nr)
Nr = 1
Power Ratio:
The power ratio (Pr) is given by the formula:
Pr = (Power of the prototype turbine) / (Power of the model turbine)
Pr = (Pr * Pm) / (Pm * Pr)
Pr = 1
Calculating the Power of the Full-sized Turbine:
Given that the power of the model turbine is not provided, but we know that it is operating under a head of 2 m and a volumetric flow rate of 1 m3/s.
The power can be calculated using the formula:
Power = (Density * g * Head * Volumetric flow rate) / 1000000
Substituting the given values:
Powerm = (103 kg/m3 * 10 m/s2 * 2 m * 1 m3/s) / 1000000
Powerm = 0.0206 MW
Since the power ratio is 1, the power of the full-sized turbine is also 0.0206 MW.
Conclusion:
The power of the full-sized turbine, when operating under a head of 30 m, is 0.0206 MW, as the power ratio between the model and the prototype turbines is 1.
- Scale model of a Francis turbine
- Head of the model turbine: 2 m
- Volumetric flow rate of the model turbine: 1 m3/s
- Speed of the model turbine: 450 rpm
- Water density: 103 kg/m3
- Acceleration due to gravity: 10 m/s2
Assumptions:
- No losses in both the model and prototype turbines
Objective:
To determine the power output of the full-sized turbine when operating under a head of 30 m.
Approach:
To find the power of the full-sized turbine, we need to use the concept of similarity between the model and prototype turbines. The similarity laws, specifically the law of similarity, can be used to relate the performance of the model turbine to that of the prototype turbine.
Geometric Similarity:
The scale model and the prototype turbine are geometrically similar. This means that the corresponding linear dimensions of the model and the prototype turbine are in the same ratio. Therefore, the ratio of the head and the volumetric flow rate is also the same for both the model and the prototype turbines.
Head Ratio:
The head ratio (HR) is given by the formula:
HR = (Head of the prototype turbine) / (Head of the model turbine)
HR = 30 m / 2 m
HR = 15
Volumetric Flow Rate Ratio:
The volumetric flow rate ratio (Qr) is given by the formula:
Qr = (Volumetric flow rate of the prototype turbine) / (Volumetric flow rate of the model turbine)
Qr = (Qr * Qm) / (Qm * Qr)
Qr = 1
Speed Ratio:
The speed ratio (Nr) is given by the formula:
Nr = (Speed of the prototype turbine) / (Speed of the model turbine)
Nr = (Nr * Nm) / (Nm * Nr)
Nr = 1
Power Ratio:
The power ratio (Pr) is given by the formula:
Pr = (Power of the prototype turbine) / (Power of the model turbine)
Pr = (Pr * Pm) / (Pm * Pr)
Pr = 1
Calculating the Power of the Full-sized Turbine:
Given that the power of the model turbine is not provided, but we know that it is operating under a head of 2 m and a volumetric flow rate of 1 m3/s.
The power can be calculated using the formula:
Power = (Density * g * Head * Volumetric flow rate) / 1000000
Substituting the given values:
Powerm = (103 kg/m3 * 10 m/s2 * 2 m * 1 m3/s) / 1000000
Powerm = 0.0206 MW
Since the power ratio is 1, the power of the full-sized turbine is also 0.0206 MW.
Conclusion:
The power of the full-sized turbine, when operating under a head of 30 m, is 0.0206 MW, as the power ratio between the model and the prototype turbines is 1.
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A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m3 /s at 450 rpm. Take the water density and the acceleration due to gravity as 103 kg/m3 and 10 m/s2 , respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full sized turbine while working under a head of 30 m is __________ (correct to two decimal places).Correct answer is '29.05'. Can you explain this answer?
Question Description
A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m3 /s at 450 rpm. Take the water density and the acceleration due to gravity as 103 kg/m3 and 10 m/s2 , respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full sized turbine while working under a head of 30 m is __________ (correct to two decimal places).Correct answer is '29.05'. Can you explain this answer? 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 A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m3 /s at 450 rpm. Take the water density and the acceleration due to gravity as 103 kg/m3 and 10 m/s2 , respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full sized turbine while working under a head of 30 m is __________ (correct to two decimal places).Correct answer is '29.05'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m3 /s at 450 rpm. Take the water density and the acceleration due to gravity as 103 kg/m3 and 10 m/s2 , respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full sized turbine while working under a head of 30 m is __________ (correct to two decimal places).Correct answer is '29.05'. Can you explain this answer?.
A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m3 /s at 450 rpm. Take the water density and the acceleration due to gravity as 103 kg/m3 and 10 m/s2 , respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full sized turbine while working under a head of 30 m is __________ (correct to two decimal places).Correct answer is '29.05'. Can you explain this answer? 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 A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m3 /s at 450 rpm. Take the water density and the acceleration due to gravity as 103 kg/m3 and 10 m/s2 , respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full sized turbine while working under a head of 30 m is __________ (correct to two decimal places).Correct answer is '29.05'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m3 /s at 450 rpm. Take the water density and the acceleration due to gravity as 103 kg/m3 and 10 m/s2 , respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full sized turbine while working under a head of 30 m is __________ (correct to two decimal places).Correct answer is '29.05'. Can you explain this answer?.
Solutions for A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m3 /s at 450 rpm. Take the water density and the acceleration due to gravity as 103 kg/m3 and 10 m/s2 , respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full sized turbine while working under a head of 30 m is __________ (correct to two decimal places).Correct answer is '29.05'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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