Engine turning moment diagram:

Transcription

1 Chapter 3 Flywheel Application of slider-crank echanis can be found in reciprocating (stea) engines in the power plant i.e. internal cobustion engines, generators to centrifugal pups, etc. Output is non-unifor torque fro crankshaft; accordingly there will be fluctuation is speed and subsequently in voltage generated in the generator that is objectionable or undesirable. Output torque at shaft is required to be unifor. Other kind of applications can be in punch press. It requires huge aount of power for sall tie interval. Reaining tie of cycle it is ideal. Large otor that can supply huge quantity of energy for a sall interval is required. Output power at piston is required to be non-unifor. hese can be overcoe by using flywheel at the crank- shaft. his will behave like a reservoir of energy. his will soothen out the non-unifor output torque fro crankshaft. Also it will store energy during the ideal tie and redistribute during the deficit period. urning oent diagras and fluctuations of the crank shaft speed: A turning oent (crank torque) diagra for a four-stroke internal cobustion engine is shown in Figure 1. he coplete cycle is of 70º. Fro the static and inertia force analyses θ can be obtained (at interval of 15º or 5º preferably). Engine turning oent diagra: Crank torque O ean torque output fro flywheel L A B C D E F M 180 input to flywheel P crank angle Figure 1 urning oent diagra orque is negative in soe interval of the crank angle, it eans energy is supplied to engine during this period i.e. during the copression of the gas and to overcoe inertia forces of engine ebers. his is supplied by the flywheel (and inertia of engine ebers), which is attached to the crankshaft. When flywheel is attached to the crankshaft. LM in diagra shown is the ean torque line. It is defined as

2 = N i= 1 N i (1) If, = 0 then no net energy in the syste, 0 then there is an excess of the net energy in the syste and 0 then there is a deficit of the net energy in the syste. Area OLMP = net (energy) area of turning oent diagra = ( 4 ) π. During interval AB, CD and EF, the crank torque is ore than the ean torque eans hence excess of energy is supplied to crank i.e. it will accelerate ( ). During other interval i.e LA, BC, DE and FM, the crank torque is less than the ean torque i.e. there is deficit in energy i.e. crank will decelerate ( ). F 1 a F a -a F 1 -a-f =0 Figure Linear acceleration of a body I L orque due to load orque fro crank shaft - L d Figure 3 Angular acceleration of a body Figure 4 -θ diagra Fro Newton s second law, we have = Iα 53

3 with L = Iα () d d dθ d α = dt = dθ dt = dθ (3) ubstituting eqn. (3) in (), we get or ( ) θ d L = I d θ L d = I d (4) Integrating θ fro θ = in to θ and fro in to, we get θ 1 E = ( L ) dθ = Id = ( in) θin in I (5) where E is the net area in -θ diagra between θ and θ in, and I is the polar ass oent of inertia. A plot of shaft torque versus crank angle θ shows a large variation in agnitude and sense of torque as shown in Figure 4. ince in sae phases the torque is in the sae sense as the crank otion and in other phases the torque is opposite to the crank otion. It would see that the assuption of constant crank speed is invalid since a variation in torque would produce a variation in crank speed in the cycle. However, it is usual and necessary to fix a flywheel to the crankshaft and a flywheel of relatively sall oent of inertia will reduce crank speed variations to negligibly sall values (1 or % of the crank speed). We cannot change out put torque fro the engine (it is fixed) but by putting flywheel we can regulate speed variation of crankshaft in cycle. Our interest is to find iu and iniu speeds and its positions in Figure 5. Points A, B, C, D, E and F are the points where θ diagra cuts the ean torque line. hese points are transition points fro deficit to extra energy or vice versa. o crank starts accelerate fro deceleration fro such points or vice versa. For exaple at points: A, C, E accelerate and at B, D, F decelerate. At all such points have zero velocity slope i.e. having velocity iu or iniu. Crank speed diagra can be drawn qualitatively (approxiately) as shown in Figure 5, where c is the iniu speed location. Area of turning oent diagra represents energy for a particular period. Net energy between the iu speed and the iniu speed instant is tered as fluctuation of energy. For this case area of diagra between C and D or between D and C through points E, F, M, L, A, B and C. urning oent diagra for ulti cylinder engine can be obtained by θ of individual engine by super iposing the in proper 54

4 phase. For a four cylinder (four-stroke) engine the phase difference would be 70º/4=180º or for a six cylinder four-stroke engine the phase difference = 70º/6= 10º. Crank torque L orque due to ra L orque of the fly (orque fro the wheel fro the flywheel to the engine load) ean + torque L A B C D E F M P crank angle d i. speed Crank speed ean l a speed b e f without the flywheel O iniu speed C Figure 5 Fluctuation of the energy For ulti cylinder engine θ will be flat copare with single cylinder engine also the difference of iu and iniu speed will be less. he coefficient of fluctuation of speed is defined as with δ in = (6) + in = (7) where is the average speed. he fluctuation of energy, E, is represented by corresponding area in θ diagra as ( ) ( + ) ( ) 1 in E= I in = I in = δ I (8) 55

5 By aking I as large as possible, the fluctuation of speed can be reduced for the sae fluctuation of energy. For the disc type flywheel the diaeter is constrained by the space and thinness of disc by stress I = Mr with k = r / (9) 1 where r is the radius of the disc and k is the radius of gyration. For ri type flywheel diaeter is restricted by centrifugal stresses at ri I = Mr with k = r (10) Equation (9) or (10) gives the ass of ri. he ass of the hub and the ar also contribute by sall aount to I, which in turn gives the fluctuation of speed slightly less than required. By experience equation (10) gives total ass of the flywheel with 90% of the ri & 10% for the hub and the ar. ypical values of the coefficient of fluctuation are δ = to for electric generators and 0. for centrifugal pups for industrial applications. Flywheel: A rigid body rotating about a fixed point with an angular velocity (rad/s) and having ass oent of inertia I (kg- ) about the sae point, the kinetic energy will be = I (11) 1 For a flywheel having the iu speed is and the iniu speed is in the change in the kinetic enegy or fluctuation of energy E I( ) radius r fro the center of rotation of flywheel E can be written as ( in) E Ir V V = in /. Let V is the linear velocity of a point at a = 0.5 (1) Also coefficient of fluctuation can be written as 56

6 with δ V V V V in in = = (13) V + V in = (14) r Ri hub r r Disc-type flywheel (autoobile) k = r Ar Figure 6 A ri type flywheel (i) For ri k=r Ri-type flywheel (for stea engine or punch press) Figure 7 Polar ass oent of inertia of ri and disc type flywheel Cobining equations (13) and (14) with equation (1), we get E Iδ V r = Iδ = with I = k (15) Equation (15) becoes δ k V = = (16) E δ k r Mass of flywheel (or polar ass oent of inertia) can be obtained as E Er M= = δ k δ k V E Er I = = (17) δ δ On neglecting the effect of ar and hub, k can be taken as the ean radius of ri r. aking k = r, we get r = r, and M E = (18) δ V 57

7 On using equation (13), we get E M = V V ( in) (19) ince V Vin = ( V ) + Vin ( V Vin) = V ( δ V ), hence ( in) δ = 0.5 (0) V V V Equations (18) or (19) can be used for finding ass of the flywheel. he 90% of M will be distributed at ri and 10% for the hub and ars. By experience the iu velocity and centrifugal stresses at the ri. V is liited by the aterial Flywheel of a Punch Press: Let d be the diaeter of hole to be punched, t is the thickness of plate to be punched, f s is the resistance to shear (shear stress), is the tie between successive punch (punching period), t p is the tie for the actual punching operation. ( 0.1 ) and N is speed of otor in rp to which the flywheel is attached. Experients show that: (i) the iu force P occurs at tie = (3/8) t p and (ii) the area under the actual force curve i.e. the energy required to punch a hole is equal to rectangular area (shaded area), hence Energy required for punching a hole W = Pt / (1) In other words the average force is half the iu force.. force Actual force variation force average force d t P/ t dispalceent Figure 8 Punching force variation with deforations 58

8 Maxiu force required to punch a hole P= f π dt () s Cobining equations (1) and (), it gives ( π ) W = 0.5Pt= 0.5 fs dt t (3) where f s is in N/, d in, t in, W in N- and t p in sec. Average power during punching W/(tie for actual punching) = ( f π d t ) 0.5 s t p Watt (4) Hence, in absence of the flywheel the otor should be capable of supplying large power instantly as punching is done alost instantaneously. If flywheel is attached to the otor shaft, then the flywheel store energy during ideal tie and will give back during the actual punching operation Average power required fro otor = W/(Punching interval) = 0.5 ( fs π d t ) Watt (5) Average power fro eqn. (5) will be for less than that fro equation (4) (e.g. of the order of 1/10). tea engine I Punch Press ( ) ( ) - = I L Figure 9 (a) tea engine L ( ) L I Figure 10(a) Punch press ( ) in +_ k otal energy consued = L 4 Figure 9(b) A turning oent diagra O N in P V J L M 4 0 igure 10(b) A turning oent diagra F 59

9 In Figure 10(b) the total energy consued during and in = area IJLM. he total energy supplied in period during sae period ( to in ) =area IVPM Hence, the fluctuation of the energy E= IJLM-IVPM = area NOPM area IVPM. Power O N constant Energy supplied by otor V J L I k ie t P Energy required for punching in P M One full cycle Figure urning oent diagra if a punch press Hence, or E = f 1/ f π dt 1/ f π dt E = t p π dt = Average power supplied by otor ideal tie (6) 1 1 t p he fluctuation of energy will be the power supplied by the otor during the ideal period. Whatever energy is supplied during the actual punching will also be consued in the punching operation. Maxiu speed will occurs just before the punching and iniu speed will occur just after the punching. he net energy gained by the flywheel during this period i.e. fro the iniu speed to the iu speed (or vice versa) will be the fluctuation of energy. he ass of flywheel can be obtained by: M = E V / δ for given δ and V, once E is calculated fro eqn. (6). 60

10 Location of the iu and iniu speeds: Let A i be the area of the respective loop, o is the speed at start of cycle (datu value). We will take datu at starting point and will calculate energy after every loop. At end of cycle total energy should be zero. he iu speed is at the iu energy (7 units) and the iniu speed is at the iniu energy (- units). E = Net area between and ( =9 units) = 0.5I( in). in (-4+-7=-9 units) or between in and Location of iu and iniu speeds: Energy after A 0 1=7 A each 5 =4 A = loop A = in + (datu) O A =4 A =7 A =3 4 6 Engine cycle Figure 1 urning oent diagra A =1 9 av Analytical expressions for turning oent: he crankshaft torque is periodic or repetitive in nature (over a cycle), so we can express torque as a su of haronics by Fourier analysis =(θ) = C0 + A1 sinθ + A sinθ + + An sin nθ + + B1cosθ + Bcosθ + + Bncos nθ + (7) With the knowledge of (θ), C0, A1, A can be obtained. For all practical purpose first few haronics will give a sufficient result. his will be very useful in analysis of torsional vibration of engine rankshaft. We will use this analysis for finding ass of flywheel. Let period of (θ) is 360, then and Work done per revolution = ( ) Mean torque = ( ) π θ dθ = C0 π (8) 0 θ θ 0 0 π π 0 (9) 1 1 = d C C π = = Now we have to obtain the intersection point of (θ)- θ curve with line. Putting we can get θ, as = 0 in (7), 61

11 ( θ ) = 0= A sinθ + A sinθ + + A sinnθ + + B cosθ + B cosθ + + B cos nθ + 1 n 1 n which gives A sinθ + A sinθ + + A sinnθ + + B cosθ + B cosθ + + B cosnθ + = 0 (30) 1 n 1 n Equation (30) is a transdental (non-linear) eqn. in ters of θ, fro which we can get θ = θ1, θ. Let during period of 360 two intersections θ 1 and θ are there, then the fluctuation of energy can be obtained as (Figure 13): E θ = θ1 ( ( ) ) θ dθ (31) ( ) one cycle in Fly wheel for reciprocating achinery installation Figure 13 Exaple 1: A single-cylinder, four- stroke oil engine develops 5 kw at 300 rp. he work done by the gases during expansion stroke is.3 ties the work done on the gases during copression stroke and the work done during the suction and exhaust strokes is negligible. If the turning oent diagra during expansion is assued to be triangular in shape and the speed is to be aintained within 1% of the ean speed, find the oent of inertia of the flywheel. olution: Given data are: δ = 0. 0 ; P= 5 kw; Wexp =.3W cop ; s = 3000 rp= π 300/60= 100 π rad/s= rad/s ; P π π 3 av = / = 5 10 /(100 ) = 500/ N = N (In Figure 14 height: AC) 6

12 W = 4 π = 500/ π 4 π = N otal work done in one cycle (i.e. 4π rad. rotation) ( ) total av We have, W which gives = W exp hence 10000=.3W cop W cop total W cop W = N and W exp = N cop. B a b c O A D E 3 4 av. Figure 14 urning oent diagra Work done during expansion stroke: (1/ ) π = Wexp = , which gives = N.= AB. BC = i. excess turning oent = = AB - AC= = N. av Hence the fluctuation of energy is E= ( ) 1/ BC ab ODB & abb are siilar, hence ab / π = BC/AB or ab = π ( /1163.3) or ab=.9. Which gives E ( ) = 1/ = N { } We have E Iδ ( ) swav I Or E Wexp ( AC AB) [ ] Or Fro siilar OBD & abd : ab / OD= BC / = or = / = kg- = 1 / = ( / ) = 15.8 N AB 63

13 E ( 1/ ) ( ) Area Bab ab BC ab BC BC BC BC = = = = = Area OBD 1/ OD AB OD AB AB AB AB W exp ( ) Area Bab= E= / = N Exaple. he vertical scale of the turning oent diagra for a ulti-cylinder engine, shown in Figure 15, is 1 c = 7000 N of torque, and horizontal scale is 1 c = 30 0 of crank rotation. he areas (in c ) of the turning oent diagra above and below the ean resistance line, starting fro A in and taken in order, are 0.5, +1., -0.95, +1.45, -0.85, +0.71, he engine speed is 800 rp and it is desired that the fluctuation fro iniu to iu speed should not be ore then % of average speed. Deterine the oent of inertia of the flywheel. M (orque) M A B C D E F G H ( A) N. θ Figure 15 Exaple olution: ini i M A B C D E F G H N.M in Figure 16 Fluctuation of the energy θ π E= E Ein = 1. ( 0.5) = 1.7 c = N

14 = 800 rp= rad/s and δ s = 0.0 Hence, I E kg δ = = (Answer) s Exercise Probles: (1) he following data refers to a single-cylinder four cycle diesel engin: speed = 500 rp, stroke = 5 c, diaeter of cylinder = 1 c, length of connecting rod = 44 c, CG of connecting rod is 18 c fro crank pin center, tie for 60 coplete swings of the connecting rod about piston pin = 7 s, ass of connecting rod = 4.5 kg, ass of piston with rings =.5 kg, equivalent ass of crank at crank radius = kg, counterbalance ass of the crank at crank radius = kg, piston pin, crank pin and ain bearing diaeters, 8 and 8 c respectively. he indicator card is assued as an idealised diesel cycle,which can be described as follows: he copression starts with an initial pressure of 0.1 MPa and the law of copression curve is given by the exponent 1.4. he copression ratio is 16. he fuel is aditted for 30% of the stroke, at constant pressure and the expansion law is given by the exponent 1.4, which takes place at the end of the stroke. he exhaust and suction takes place at constant pressure of 0.1 MPa. uggest a suitable flywheel for this engine if the coefficient of fluctuation of speed is () wenty 1-c holes are to be punched every inute in a 1.5 c plate whose resistance to shear is N/c. he actual punching takes place in one-fifth of the interval between successive operations. he speed of the flywheel is 300 rp. Making the usual assuptions specify the diensions of a suitable CI ried flywheel. Use coefficient of fluctuation of speed = 0.01 and V = 60 /s. (3) he equation of a turning oent curve of an IC engine running at 300 rp is given by [ sin 3θ] = +. A flywheel coupled to the crankshaft has a oent of inertia 450 kg about the axis of rotation. Deterine (a) Horse power of the engine (b) total percentage fluctuation of speed (c) iu angle by which the flywheel leads or lags an iaginary flywheel running at a constant speed of 300 rp. 65