TURNING MOMENT DIAGRAM
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1) A turning moment diagram (TMD) graphically represents torque over crank angle and is used to calculate work done per cycle and mean torque. 2) A flywheel stores excess energy from the engine during the power stroke and releases it during other strokes to reduce fluctuations in speed. There are disc and rim types of flywheels. 3) To calculate the energy stored in a flywheel, formulas use mass, radius of gyration, angular velocity, and the coefficient of fluctuation of speed. The problem gives values for an engine to calculate the required flywheel's mass moment of inertia.
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Here are the steps to solve this problem:
a) Power of the engine
- Power stroke area = 6000 mm2 = 6000 x 10 Nm = 60,000 Nm
- Resisting torque is uniform, so work done = Area x Resisting torque
- Assuming resisting torque is 10 Nm, work done per cycle = 60,000 Nm
- Work done per cycle x rpm = Power
- Given: Work done per cycle is 60,000 Nm
- Assuming rpm is N
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b) Energy to be stored by flywheel
- Total area under the diagram = Power stroke + Exhaust + Suction + Compression
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1. The document describes a turning moment diagram (TMD) for a 4-stroke IC engine, showing torque versus crank angle.
2. The TMD can be used to determine the work done per cycle, mean torque, power developed, and excess energy that may be stored in the flywheel.
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Here are the steps to solve this problem:
a) Power of the engine
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- Resisting torque is uniform, so work done = Area x Resisting torque
- Assuming resisting torque is 10 Nm, work done per cycle = 60,000 Nm
- Work done per cycle x rpm = Power
- Given: Work done per cycle is 60,000 Nm
- Assuming rpm is N
- Then, Power = 60,000 x N W
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- Total area under the diagram = Power stroke + Exhaust + Suction + Compression
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2) Flywheels absorb energy when the engine produces excess torque and release energy when engine torque is insufficient, helping maintain constant rotational speed despite cyclic torque variations. Their moment of inertia determines how effectively they can reduce speed fluctuations.
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3. The document also examines the gyroscopic couple and centrifugal couple in vehicles like cars and motorcycles taking turns, and how this affects their stability.
Equivalent
The document discusses equivalent dynamical systems used to model the motion of rigid bodies acted on by external forces. It defines an equivalent dynamical system as one where: 1) the sum of the masses equals the total body mass, 2) the center of gravity of the masses matches the body's, and 3) the sum of the mass moments of inertia about the center of gravity equals the body's. It then gives an example of calculating the equivalent dynamical system for a connecting rod suspended and oscillating, determining the radius of gyration and placing one mass at the small end center and the other a distance l2 from the center of gravity.
flywheel
A punching press punches 38mm holes in 32mm thick plates, requiring 7 N-m of energy per square mm of sheared area. It punches one hole every 10 seconds. The mean flywheel speed is 25 m/s.
To calculate the motor power required, the energy per hole is calculated based on the sheared area of the hole. The mass of the flywheel required to limit speed fluctuations to 3% of the mean is then calculated using the energy variation and flywheel properties.
Unit 3 free vibration
This document covers free and damped vibrations. It defines key terms like natural frequency, damping, and damping ratio. It describes the equations of motion for an undamped single degree of freedom system and how to calculate the natural frequency. It also covers calculating the natural frequency of damped systems and defines types of damping like overdamped, underdamped, and critically damped systems. Formulas are provided for damped vibration frequency, logarithmic decrement, and damping ratio. Examples are given on calculating natural frequency, damping coefficient, and damping ratio from data provided on an oscillating system.
Transverse vibrations
This document provides instructions on how to calculate the natural frequency of vibration of a shaft that is carrying multiple loads at different points. It describes calculating: 1) the moment of inertia of the shaft, 2) the static deflection caused by each point load individually, 3) the total static deflection by summing the individual deflections, and 4) the natural frequency of transverse vibration of the shaft based on the total deflection. It then gives two example problems involving shafts of different sizes, materials, and load configurations to calculate the natural frequency.
Torsional vibrations
This document discusses torsional vibrations in shafts. It provides equations to calculate the natural frequency of torsional vibrations based on the shaft's torsional stiffness, mass moment of inertia, and material properties. As an example, it calculates the natural frequency of a flywheel mounted on a vertical shaft. It then discusses multi-rotor shaft systems and how to determine the location of nodes. Finally, it provides methods to calculate the natural frequency and node locations of stepped shafts with varying diameters connecting multiple flywheels.
Free vibrations
The document provides information on vibrations including definitions of key terms like natural frequency, forced vibrations, and damped vibrations. It also discusses single degree of freedom systems and how to calculate the natural frequency of longitudinal and transverse vibrations in beams and shafts using different methods. An example is provided to show how to calculate the natural frequencies of longitudinal and transverse vibrations for a cantilever shaft with a mass at the free end. Formulas are given for stiffness, static deflection, and natural frequency of beams and shafts.
Damped vibrations
The document discusses different types of damping in vibrating systems, including viscous, Coulomb, and critical damping. It provides equations to calculate damping coefficient, logarithmic decrement, damping ratio, natural frequency, and damped vibration frequency. Examples are given to show how to determine damping coefficient, critical damping coefficient, damping factor, logarithmic decrement, and ratio of damped to undamped frequencies based on given mass, spring constant, amplitude decay between cycles.
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TURNING MOMENT DIAGRAM
- 2. Session-06
- 4. 4 stroke engine – Turning moment diagram
- 5. 4 stroke engine – Turning moment diagram
- 6. Purpose •Turning moment diagram is a graphical representation of turning moment or torque (along Y-axis) versus crank angle (X-axis) for various positions of crank. •The area under the TMD gives the work don per cycle. •The work done per cycle when divided by the crank angle per cycle gives the mean torque Tm. •The mean torque Tm multiplied by the angular velocity of the crank gives the power consumed by the machine or developed by an engine.
- 7. Fly wheels The Fly wheel stores the excessive mechanical energy from the engine. It stores the energy during power stroke and releases it during the other strokes. Two types Disk type Rim type
- 8. Types of flywheels • Disc type • I = mD2/8 • Rim type • I = mD2/4
- 9. Important Terms 1. Maximum fluctuation of speed ◦ It is the difference between the maximum & minimum speeds in a cycle. N1 – N2 2. Coefficient of fluctuation of speed ◦ It is the ratio of maximum fluctuation of speed to the mean speed ◦ It is expressed as percentage of mean speed
- 10. Energy stored in flywheel E = mk2w2Cs Or Iw2Cs Where, m = mass (kg) k = radius of gyration (m) [ k2 = D/4 for rim type k2 = D/8 for disc type] w = angular velocity of flywheel (rad/s) Cs = coefficient of fluctuation of speed (no unit) E = excessive energy (J) I = mass Moment of inertia (kg-m2)
- 11. Session-07
- 12. Energy in fly wheels problems Formulae •Excess energy, stored in flywheel, E = mk2w2Cs or Iw2Cs •Work done /cycle = area under the curve •Work done /cycle = Tmean x (angle for one cycle) • i.e. Net Area = Tmean x 4p (for 4 stroke engine)
- 15. Coefficient of Fluctuation of Speed
- 16. Energy Stored in a Flywheel
- 17. A single cylinder 4 stroke gas engine develops 20 KW at 300 rpm with work done by the gases during the expansion being 3 times the work done on the gases during compression.The work done during the suction & exhaust strokes is negligible. The total fluctuation of speed is 2% of the mean. The TMD may be assumed to be triangular in shape. Find the mass moment of inertia of the flywheel.
- 18. Given: Power = 20 kw N = 300rpm WE= 3xWC Cs = 0.04 4 stroke engine i.e q = 4p (2revs) I = ?
- 21. Since the area above the mean turning moment line represents the maximum fluctuation of energy, therefore maximum fluctuation of energy,
- 23. Energy in multi cylinder engine