Question

EG: For the main spindle of shaper, In one working circulation, the resisting moment is M, = 600 Nm , and the driving moment M, is constant. The average velocity of the main spindle is n=60r/min and the non-uniform coefficient is 8=0.1. Calculate the proper rotary inertia for flywheel (ignoring the inertia of other components). AM M.=600 N·m TT TE 416 TT (27)

Answer 1

Given 2 Data: 0 Resisting moment = m = 600 N-m (constant) of main spindje = n = 60 ppm ftuctuation of speed = $=0.1 Droit ving moment = md لي Average Coefficient of velocity 4 Solution : M Average Angular speed of 600 Nm C F G vo the Mp 211n main Spindle = w= 60 W= 25x60 60 w= 2lrad/s 125 + N-m A we ma к L MIN Spinde Angie T TYL > К% %, 2T and the Area of this mo' foro one This during the due to know that, Area under moment dicogram (M-03 diagram) repraesents Energy. of the spindle. diagram working cycle is the Energy pero cycle fluctuates energy cycle Variation in the resisting varrying shown in figure Driving moment (md) is also represented in the Energy pero cycle = = E= summation of all areas "M-o' diagram. . E=E Areas of 'm-o diagroom E = Area (IKBCL) + ACOMFGN) E = (600 x 14) + (600x146) moment (ie, va doad) as figure so, in the

(+) Moment (md) on the spindle. .. E= 600 T 6 - 600 x 10 24 E = 250 M Joule this eno the total de which should 1 Now, Energy is required energy in one the by driving Ez Driving moment X cycle to Angle E = md X 25 2 equating eqns eans ¢ 2 ř 2500= md X25 Now, manimum Speed of I = can be Calculated ag 1 Ima = 12 N-m Let, 'Wi' be the the frywheel and "Wh be the minimum speed of the flywheel Rotaroy Inertia of the flycoheel. Hence, the Rotational Kinetic energy of the figoheel (marimum Energy I Emax = 1 / I W} Energy) manimum fluctuation of the energy i AE = Emax - Emin = 1 / I w? - 1/2 1 w? = 1/2 I (w? - Z) I (wit W₂) (Ws - Wh) & Emin 1 1/ I want (minimum is, ald ald AE 2

AE = I w Wyby) where W = Wit Ave verage Angular are speed fluctuation in the speed 102 = maximum LE I ? wp-W₂ where Wi-W2 - 8 = coefficient of fluctuation in the Speed non-uniforom coefficient 3 ing cycle ome work DE= I. w? d Now, manimum fluctuation the Energy during can be fan calculated from "m-o' diagram By observation of im diagroom fluctuation of energy is the Area of Rectangle "ABCD! AE = Anea ( ABCD) the maximum A B XAD ! (Mr-md) x (600- 125) x 17 / 4. 475 T 4 : Equating ating egne equs 3 ¢ © 3 4759 I. w?. f 475T 4 - I X (270)? x0.1 1 475XT 4x4 T² xo.1 I = = 94-4982 kg- m2
area under resisting moment-spindle angle diagram for one cycle represents total energy required per cycle, this energy should be equivalent to the energy available in one cycle which is the product of constant driving moment and total cycle angle, from this the value of mean constant driving moment is calculated and mean moment line is represented in the moment-angle diagram. Then the maximum area above or below the mean moment line represents maximum fluctuation of energy in one cycle so by equating this maximum fluctuation of energy from diagram with the value calculated from the formula and putting all the known values, the unknown value of rotary inertia of the flywheel is calculated as

I= 94.4982 kg-m2