1 Design of flywheel

Design of flywheel

1. DESIGN OF FLYWHEEL

2. Flywheel Application

3. A Flywheel is used to maintain constant angular velocity of the crankshaft in a reciprocating engine. In this case, the flywheel—which is mounted on the crankshaft— stores energy when torque is exerted on it by a firing piston and it releases energy to its mechanical loads when no piston is exerting torque on it.

4. INTRODUCTION FLYWHEEL A flywheel is an inertial energy-storage device. It absorbs mechanical energy and serves as a reservoir, storing energy during the period when the supply of energy is more than the requirement and releases it during the period when the requirement of energy is more than the supply.

5. FLYWHEELS-FUNCTION NEED AND OPERATION The main function of a fly wheel is to smoothen out variations in the speed of a shaft caused by torque fluctuations. If the source of the driving torque or load torque is fluctuating in nature, then a flywheel is usually called for. Many machines have load patterns that cause the torque time function to vary over the cycle. Internal combustion engines with one or two cylinders are a typical example. Piston compressors, punch presses, rock crushers etc. are the other systems that have fly wheel. Flywheel absorbs mechanical energy by increasing its angular velocity and delivers the stored energy by decreasing its velocity.

6. FLYWHEEL MATERIALS Traditionally, flywheel are made of cast iron. From design consideration, cast iron flywheel offer the following advantages:- Cast iron flywheels are the cheapest. Cast iron flywheel can be given any complex shape without involving machining operations. Cast iron flywheel has excellent ability to damp vibrations. however, cast iron has poor tensile strength compare to steel. The failure of cast iron is sudden and total. The machinability of cast iron flywheel is poor compared to steel flywheel.

7. Applications Providing continuous energy when the energy source is discontinuous. For example, flywheels are used in reciprocating engines because the energy source, torque from the engine, is intermittent. Delivering energy at rates beyond the ability of a continuous energy source. This is achieved by collecting energy in the flywheel over time and then releasing the energy quickly, at rates that exceed the abilities of the energy source. Dynamic balancing of rotating elements. Energy storage in small scale electricity generator sets

8. Advance and Modern Flywheel Flywheels have also been proposed as a power booster for electric vehicles. Speeds of 100,000 rpm have been used to achieve very high power densities. Modern high energy flywheels use composite rotors made with carbon-fibre materials. The rotors have a very high strength-to- density ratio, and rotate at speeds up to 100,000 rpm. in a vacuum chamber to minimize aerodynamic losses.

9. Benefits in Aerospace Fly w heels are preferr e d over co n ven t io n al b at t eries in m any aerospace applications because of the following benefits: 5 to 10+ times greater specific energy Lower mass / kW output Long life. Unaffected by number of charge / discharge cycles 85-95% round trip efficiency Fewer regulators / controls needed Greater peak load capability Reduced maintenance / life cycle costs

10. Disadvantages There are safety concerns associated with flywheels due to their high speed rotor and the possibility of it breaking loose & releasing all of it's energy in an uncontrolled manner. Its Bulkier, adds more weight to the vehicle.

11. Energy stored in a flywheel Rotational Kinetic Energy, E = ½ Iω 2 where, I - moment of inertia of the flywheel (ability of an object to resist changes in its rotational velocity) ω - rotational velocity (Rad / sec) The moment of inertia, I = kMr 2 where, M - mass of the flywheel r - radius of flywheel k - inertial constant.

12. k depends on the shape of the rotating object. Shape-factor K for different planar stress geometries So for a solid disk ; I = Mr 2 /2

13. Co efficient of fluctuation of speed ( Cs ) The difference between the max and min speeds during a cycle is called the max fluctuation of speed. The ratio of the max fluctuation of speed to the mean speed is called coefficient of fluctuation of speed. Cs = (N 1 - N 2 )/N = 2( N 1 -N 2 ) / N 1 + N 2 w h ere N 1 = max speed in r.p.m. N 2 = min speed in r.p.m. N = mean speed in r.p.m. = (N 1 + N 2 ) / 2

14. Permissible values for C S S.NO Types of machines Coefficient of fluctuation of speed (C S ) 1 Engines with belt transmission 0.030 2 Gear wheel transmission 0.020 3 Crushing machines 0.200 4 Electrical machines 0.003 5 Hammering machines 0.200 6 Pumping machines 0.03-0.05 7 Machine tools 0.030

15. Stresses in a flywheel rim A flywheel consists of a rim at which the major portion of the mass or weight of flywheel is concentrated, a boss or hub for fixing the flywheel on to shaft and a number of arms for supporting the rim on the hub. The following stresses are induced in the rim. Tensile stress due to centrifugal force. Tensile bending stress caused by the restraint of the arms.

16. Stresses in flywheel arms The f o ll o wing st r e s ses are i n d u c e d in t he a r m s o f t h e flywheel. Tensile stresses due to centrifugal force acting on the rim Bending stress due to the torque transmitted from the rim to the shaft or from the shaft to the rim.

17. 1. Tensile stress due to the centrifugal force. The tensile stress in the rim due to the centrifugal force, assuming that the rim is unstrained by the arms, is determined in the similar way as the thin cylinder subjected to internal pressure. f t = ρ.R 2 .ω 2 = ρ.v 2 ( v = R.ω ) When ρ is in kg/m 3 , v is in m/sec, f t will be in N/m 2 where ρ = density of the flywheel material ω = angular speed of the flywheel R = mean radius of the flywheel v = linear velocity of the flywheel

18. 2.Tensile bending stress caused by restraint of arms . The tensile bending stress in the rim due to the restraint of arms is based on the assumption that each portion of the rim between a pair of arms behaves like a beam fixed at both ends and uniformly loaded, such that length between fixed ends, L = π.D/n = 2.π.R / n where n - number of arms

19. The max bending moment, M = w.l 2 /12 = b.t. ρ . ω 2 .R/12(2. π .R/n) Section modulus, Z = 1/6 (b.t 2 ) So bending stress f b = M/Z = b.t. ρ . ω 2 .R/12 (2. π .R/n) * 6 / (b.t 2 ) Total stress in the rim f = f t + f b

20. Construction of Flywheel  Flywheels are typic a l l y m ade of st e e l a n d rot a te on conve n tion a l be a r i ng s ; the s e are gene r a lly l i m ited t o a revolution rate of a few thousand RPM  The flywheel of smaller size( upto 600 mm dia )are casted in one piece. The rim and the hub are joined together by means of web.

21. Construction If flywheel is of larger size (upto 2-5 meters diameter ), then it is made of arms. The number of arms depends upon the size of the flywheel and its speed of rotation. But the flywheels above 2-5 meters are usually casted in two pieces. Such a flywheel is known as “ split flywheel “. A split flywheel has the advantage of relieving the shrinkage stresses in the arms due to unequal rates of cooling of casting.

22. DESIGN APPROACH There are two stages to the design of a flywheel. First, the amount of energy required for the desired degree of smoothening must be found and the (mass) moment of inertia needed to absorb that energy determined. Then flywheel geometry must be defined that caters the required moment of inertia in a reasonably sized package and is safe against failure at the designed speeds of operation. Design Parameters:- It depend upon acceptable changes in the speed. Speed fluctuation:- The change in the shaft speed during a cycle is called the speed fluctuation and it is given by Fl =ω max −ω min

23. DESIGN OF FLYWHEEL Design Equation:- S I = 𝑬 𝒌 𝑪 𝒇 ∗ 𝝎 𝒂𝒗𝒈 𝟐 where “ C f ”is the co-efficient of speed fluctuation and “ E k ”is the kinetic energy and “ 𝝎 avg ” is the average rotational motion. Torque Variation and Energy:- The required change in kinetic energy E k is obtained from the known torque time relation or curve by integrating it for one cycle and it is given by 𝜃 @ 𝜔 𝜃 @ 𝜔 𝑚 𝑎𝑥 (𝑇 1 − 𝑇𝑎 𝑣𝑔 ) d 𝜃 =E k

24. GEOMETRY OF FLYWHEEL It can be a solid cylindrical disc. It can be like conventional wheel design. But energy requirements and size of the flywheel increases the geometry changes to disc of central hub and peripheral rim connected by webs and to hollow wheels with multiple arms.

25. GEOMETRY OF FLYWHEEL

26. GEOMETRY OF FLYWHEEL For a solid disc geometry with inside radius r i and out side radius r o , the mass moment of inertia I is The mass of a hollow circular disc of constant thickness t is Combing the two equations we can write Where is material’s weight density 

27. STRESSES IN FLYWHEEL Flywheel being a rotating disc, centrifugal stresses acts upon its distributed mass and attempts to pull it apart. Its effect is similar to those caused by an internally pressurized cylinder.  = material weight density, ω= angular velocity in rad/sec. ν= Poisson’s ratio, is the radius to a point of interest, r i and r o are inside and outside radii of the solid disc flywheel.

28. ADVANCE AND MODERN FLYWHEEL Advanced flywheels are also now used for protecting against interruptions to the national electricity grid. – The flywheel provides power during period between the loss of utility supplied power and either the return of utility power or the start of a sufficient back-up power system Flywheels have also been proposed as a power booster for electric vehicles. Speeds of 100,000 rpm have been used to achieve very high power densities. Modern high energy flywheels use composite rotors made with carbon- fibre materials. The rotors have a very high strength-to-density ratio, and rotate at speeds up to 100,000 rpm. in a vacuum chamber to minimize aerodynamic losses.

29. BENEFITS IN AEROSPACE Flywheels are preferred over conventional batteries in many aerospace applications because of the following benefits: 5 to 10+ times greater specific energy Lower mass / kW output Long life. Unaffected by number of charge / discharge cycles 85-95% round trip efficiency Fewer regulators / controls needed Greater peak load capability Reduced maintenance / life cycle costs

30. APPLIC A TIONS

31. Example 22.1. The turning moment diagram for a petrol engine is drawn to the following scales: Turning moment, 1 mm = 5 N-m; Crank angle, 1 mm = 1º. The turning moment diagram repeats itself at every half revolution of the engine and the areas above and below the mean turning moment line, taken in order are 295, 685, 40, 340, 960, 270 mm2. Determine the mass of 300 mm diameter flywheel rim when the coefficient of fluctuation of speed is 0.3% and the engine runs at 1800 r.p.m . Also determine the cross-section of the rim when the width of the rim is twice of thickness. Assume density of rim material as 7250 kg / m3.