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A rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2
Correct answer is '4.92'. Can you explain this answer?
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A rimmed flywheel (mass density = 7000 kg/m3) is used on a punching p...
∆E = 3000 Nm
Rim contributes only 90% of MOI
k = 90%
Cs = 0.2
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A rimmed flywheel (mass density = 7000 kg/m3) is used on a punching p...
To find the area of the rim, we need to follow the given information and solve step by step.
Step 1: Find the moment of inertia of the flywheel
The moment of inertia of the flywheel is given by the equation:
I = m * r^2
where m is the mass of the flywheel and r is the radius of the flywheel.
Given information:
- Mass density of the flywheel = 7000 kg/m^3
- Radius of the flywheel = 0.45 m
- The rim contributes 90% of the required moment of inertia
First, we need to find the mass of the rim. The mass can be calculated using the volume and density of the material:
Density = Mass/Volume
Mass = Density * Volume
Since the cross-section of the rim is square, the volume can be calculated as the product of the area and thickness of the rim:
Volume = Area * Thickness
Step 2: Find the area of the rim
To find the area of the rim, we need to rearrange the equation for volume:
Volume = Mass / Density
Volume = (Density * Area * Thickness) / Density
Volume = Area * Thickness
Now, we can calculate the area of the rim using the given information:
Given information:
- Coefficient of speed fluctuations = 0.2
- Mean speed of the punching press = 250 rpm
- Energy required per punch = 3000 Nm
- Number of punches per revolution = 1/4
Step 3: Calculate the energy required per revolution
The energy required per revolution can be calculated as the product of the energy per punch and the number of punches per revolution:
Energy per revolution = Energy per punch * Number of punches per revolution
Energy per revolution = 3000 Nm * 1/4
Energy per revolution = 750 Nm
Step 4: Calculate the kinetic energy of the flywheel
The kinetic energy of the flywheel can be calculated using the equation:
Kinetic energy = 1/2 * I * w^2
where I is the moment of inertia of the flywheel and w is the angular velocity of the flywheel.
Since the flywheel is running at a mean speed of 250 rpm, the angular velocity can be calculated as:
w = 2 * π * n / 60
w = 2 * 3.1416 * 250 / 60
w = 26.18 rad/s
Now, we can calculate the moment of inertia of the flywheel:
I = (Mass of the rim * Radius^2) + (Mass of the remaining part * Radius^2)
I = (0.9 * Mass of the rim * Radius^2) + (0.1 * Mass of the remaining part * Radius^2)
Since the rim contributes 90% of the required moment of inertia, we can rewrite the equation as:
I = (0.9 * Mass of the rim * Radius^2) + (0.1 * Mass of the rim * Radius^2)
I = Mass of the rim * Radius^2 * (0.9 + 0.1)
I = Mass of the rim * Radius^2
Step 5: Calculate the mass of the rim
To find the mass of the rim, we can rearrange the equation for the moment of inertia:
Mass of the rim =
Step 1: Find the moment of inertia of the flywheel
The moment of inertia of the flywheel is given by the equation:
I = m * r^2
where m is the mass of the flywheel and r is the radius of the flywheel.
Given information:
- Mass density of the flywheel = 7000 kg/m^3
- Radius of the flywheel = 0.45 m
- The rim contributes 90% of the required moment of inertia
First, we need to find the mass of the rim. The mass can be calculated using the volume and density of the material:
Density = Mass/Volume
Mass = Density * Volume
Since the cross-section of the rim is square, the volume can be calculated as the product of the area and thickness of the rim:
Volume = Area * Thickness
Step 2: Find the area of the rim
To find the area of the rim, we need to rearrange the equation for volume:
Volume = Mass / Density
Volume = (Density * Area * Thickness) / Density
Volume = Area * Thickness
Now, we can calculate the area of the rim using the given information:
Given information:
- Coefficient of speed fluctuations = 0.2
- Mean speed of the punching press = 250 rpm
- Energy required per punch = 3000 Nm
- Number of punches per revolution = 1/4
Step 3: Calculate the energy required per revolution
The energy required per revolution can be calculated as the product of the energy per punch and the number of punches per revolution:
Energy per revolution = Energy per punch * Number of punches per revolution
Energy per revolution = 3000 Nm * 1/4
Energy per revolution = 750 Nm
Step 4: Calculate the kinetic energy of the flywheel
The kinetic energy of the flywheel can be calculated using the equation:
Kinetic energy = 1/2 * I * w^2
where I is the moment of inertia of the flywheel and w is the angular velocity of the flywheel.
Since the flywheel is running at a mean speed of 250 rpm, the angular velocity can be calculated as:
w = 2 * π * n / 60
w = 2 * 3.1416 * 250 / 60
w = 26.18 rad/s
Now, we can calculate the moment of inertia of the flywheel:
I = (Mass of the rim * Radius^2) + (Mass of the remaining part * Radius^2)
I = (0.9 * Mass of the rim * Radius^2) + (0.1 * Mass of the remaining part * Radius^2)
Since the rim contributes 90% of the required moment of inertia, we can rewrite the equation as:
I = (0.9 * Mass of the rim * Radius^2) + (0.1 * Mass of the rim * Radius^2)
I = Mass of the rim * Radius^2 * (0.9 + 0.1)
I = Mass of the rim * Radius^2
Step 5: Calculate the mass of the rim
To find the mass of the rim, we can rearrange the equation for the moment of inertia:
Mass of the rim =
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A rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2Correct answer is '4.92'. Can you explain this answer?
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A rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2Correct answer is '4.92'. 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 rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2Correct answer is '4.92'. 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 rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2Correct answer is '4.92'. Can you explain this answer?.
A rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2Correct answer is '4.92'. 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 rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2Correct answer is '4.92'. 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 rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2Correct answer is '4.92'. Can you explain this answer?.
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A rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2Correct answer is '4.92'. Can you explain this answer?, a detailed solution for A rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2Correct answer is '4.92'. Can you explain this answer? has been provided alongside types of A rimmed flywheel (mass density = 7000 kg/m3) is used on a punching press running at a mean speed of 250 rpm. The punching operation consists of one-quarter revolution during which the flywheel is required to supply 3000 Nm of energy. The coefficient of speed fluctuations is limited to 0.2. the rim which contributes 90% of the required moment of inertia has a mean radius of 0.45 m due to space limitation. The cross-section of the rim is square, its area will be ________________ x 10-3m2Correct answer is '4.92'. Can you explain this answer? theory, EduRev gives you an
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