Torque N-mTurning Moment (Crank Effort) Diagram for a 4-stroke I C engine Excess Energy (Shaded area) T max Expansion T mean 0 Suction Crank Angle Compression P R Venkatesh, Mech Dept, RVCE Exhaust Turning Moment (Or Crank Effort) Diagram (TMD) 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. Uses of TMD 1. The area under the TMD gives the work done per cycle. 2. The work done per cycle when divided by the crank angle per cycle gives the mean torque Tm. P R Venkatesh, Mech Dept, RVCE Uses of TMD 3. The mean torque Tm multiplied by the angular velocity of the crank gives the power consumed by the machine or developed by an engine. 4. The area of the TMD above the mean torque line represents the excess energy that may be stored by the flywheel, which helps to design the dimensions & mass of the flywheel. P R Venkatesh, Mech Dept, RVCE FLYWHEEL Flywheel is a device used to store energy when available in excess & release the same when there is a shortage. Flywheels are used in IC engines, Pumps, Compressors & in machines performing intermittent operations such as punching, shearing, riveting, etc. A Flywheel may be of Disk type or Rim Type Flywheels help in smoothening out the P R Venkatesh, Mech Dept, RVCE D DISK TYPE FLYWHEEL DISK TYPE FLYWHEEL P R Venkatesh, Mech Dept, RVCE X t D X b RIM TYPE FLYWHEEL P R Venkatesh, Mech Dept, RVCE Section X-X Comparision between Disk Type & Rim Type Flywheel : Flywheels posess inertia due to its heavy mass. Mass moment of inertia of a flywheel is given by I = mk 2 , where m=Mass of the flywheel. k=Radius of gyration of the flywheel. D For rim type, k= where D=Mean diameter of the flyheel 2 D For Disk type, k= where D=Outer diameter of the flywheel 2 2 D 2 D 2 Hence I=m Rim and I=m Disk 4 8 Hence for a given diameter & inertia, the mass of the rim type flywheel is half the mass of a disk type flywheel P R Venkatesh, Mech Dept, RVCE Important Definitions (a) Maximum fluctuation of speed : It is the difference between the maximum & minimum speeds in a cycle. (=n1 n2 ) (b) Coefficient of fluctuation of speed : (C s or K s ) It is the ratio of maximum fluctuation of speed to the mean speed. It is often expressed as a % of mean speed. 1 2 2 n where =Angular velocity= 60 n1 n2 Cs (or K s ) n P R Venkatesh, Mech Dept, RVCE Important Definitions (c) Coefficient of fluctuation of energy : (C e or K e ) It is the ratio of maximum fluctuation of energy to the mean kinetc energy. E1 E2 E e Ce (or K e ) E E E ** It is often expressed as the ratio of excess energy e to the work done per cycle. C e ( or K e ) = W.D / cycle (d) Coefficient of steadiness : It is the reciprocal of coefficent of fluctuation of speed. Coefficient of steadiness= 2 P R Venkatesh, Mech1Dept, RVCE EXPRESSION FOR ENERGY STORED BY A FLYWHEEL Let I be the mass moment of inertia of the flywheel 1 & 2 be the max & min speeds of the flywheel Mean speed of the flywheel m=Mass of the flywheel, k=Radius of gyration of the flywheel Cs =Coefficient of fluctuation of speed The max fluctuation of energy (to be stored by the flywheel) 1 2 1 2 1 e E1 E2 I 1 I 2 I 12 22 2 2 2 1 e I 1 2 (1 2 ) 2 P R Venkatesh, Mech Dept, RVCE EXPRESSION FOR ENERGY STORED BY A FLYWHEEL 1 Putting the mean agular speed = 1 2 , 2 We get e = Iω(ω1 - ω2 ) Multiplying & dividing by , (ω1 - ω2 ) (ω1 - ω2 ) Also Cs , the coefficient of fluctuation of speed Hence e = Iω2C s e = Iω2 2 2 Putting I=mk 2 , we get e = mkω C s 1 2 Note: 1.Alternatively, if Mean kinetic energy E= I , 2 I 2 2 E , e=2EC s C e e 2Cs But Ce OR e = 2 E E Cs P R Venkatesh, Mech Dept, RVCE EXPRESSION FOR ENERGY STORED BY A FLYWHEEL 1 2 e= I Cs , Putting mean Kinetic energy E= I 2 and expressing C s as a percentage, 2 2EC s e 100 e 0.02 ECs 1 Note: 2. Alternatively, if Mean kinetic energy E= mk 2 2 , 2 1 2 2 2 (k ) v , E= mv 2 e mv 2cs P R Venkatesh, Mech Dept, RVCE MASS OF FLYWHEEL IN TERMS OF DENSITY & CROSSECTION AREA We know that mass m=Density Volume D2 For Disk type flywheel, Volume = t 4 For Rim type flywheel, Volume= D( A) where A= Cross section of the rim =b t b= width of rim & t= thickness of the rim Note: Dn (i)Velocity of the flywheel v= m / sec 60 (ii) Hoop Stress (Centrifugal stress) in the flywheel = v 2 where = density of flywheel material P R Venkatesh, Mech Dept, RVCE Problem 1 A single cylinder 4 stroke gas engine develops 18.4 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. P R Venkatesh, Mech Dept, RVCE Torque N-m TURNING MOMENT DIAGRAM Excess Energy T max Expansion T mean x 0 Suction Exhaust Crank Angle Compression P R Venkatesh, Mech Dept, RVCE Data : Power P=18.4 KW=18.4 103 W, Mean speed n=300 rpm Work done during expansion WE 3 Work done during compression Cs 2% 0.02 Given 4-stroke cycle engine Crank angle per cycle=4π radians( Q 2 rev of crank shaft) Solution : 2 n Angular Velocity of flywheel = 60 2 300 i.e. 31.416 rad/sec 60 Also power P=Tm 18.4 103 Tm 31.416 18.4 103 Mean torque Tm 585.7 N-m 31.41 6 P R Venkatesh, Mech Dept, RVCE Work done per cycle Work done per cycle=Tm Crank angle per cycle i.e. W.D/Cycle =Tm 4 585.7 31.416 W.D/Cycle 7360 N-m W.D/Cycle W.D during expansion W.D during compression (As the W.D during suction & compression are neglected) 7360=(WE WC ) WE Given WE 3WC Or WC , we can write 3 WE 2 7360= WE WE WE 11040 N-m 3 3 This work represents the area under triangle for expansion stroke 1 i.e. 11040 Tmax 2 R Venkatesh, Mech Dept, RVCE Max torque Tmax 7P028. 3 N-m Excess energy stored by the flywheel The shaded area represents the excess energy. 1 i.e.excess energy stored by flywheel e= x (Tmax Tmean ) 2 where x is the base of shaded triangle, given by x (Tmax Tmean ) Tmax (Tmax Tmean ) (7028.3 585.7) x 2.88rad Tmax 7028.3 1 Hence e= 2.88 (7028.3 585.7) 9276.67 N-m 2 P R Venkatesh, Mech Dept, RVCE We know that excess energy is given by e=I Cs 9276.64 I (31.416) 0.02 2 2 Hence mass moment of inertia of flywheel I=470 Kg-m 2 P R Venkatesh, Mech Dept, RVCE Problem 2 A single cylinder internal combustion engine working on 4-stroke cycle develops 75 KW at 360 rpm. The fluctuation of energy can be assumed to be 0.9 times the energy developed per cycle. If the fluctuation of speed is not to exceed 1% and the maximum centrifugal stress in the flywheel is to be 5.5 MN/m2, estimate the diameter and the cross sectional area of the rim. The material of the rim has a density 7.2 Mg/m3. P R Venkatesh, Mech Dept, RVCE Data : Power P=75 KW=75 103 W, Mean speed n=360 rpm Fluctuation of energy e =0.9 W.D/cycle 4 stroke cycle Crank angle per cycle=4π radians Density =7.2 Mg/m 3 7200 Kg/m 3 , Hoop stress =5.5 MPa Solution : 2 n Angular Velocity of flywheel = 60 2 360 i.e. 37.7 rad/sec 60 Also power P=Tm 75 103 Tm 37.7 75 103 Mean torque Tm 1989.4 N-m P 37.7 R Venkatesh, Mech Dept, RVCE Work done per cycle : Work done per cycle=Tm Crank angle per cycle i.e. W.D/Cycle =Tm 4 1989.4 4 W.D/Cycle 25000 N-m Also given e =0.9×W.D / cycle = 22500 N - m Diameter of the flywheel : Hoop stress = v 2 5.5 106 =7200 (v 2 ) Hence, velocityof flywheel v = 27.64m / sec Dn D 360 Also v = 27.64 = 60 60 Diameter of the flywheel = 1.466 m P R Venkatesh, Mech Dept, RVCE The energy stored by the flywheel is given by e = mk 2 2C s . D 1.466 For rim type, radius of gyration k= 0.733m 2 2 2 2 22500 = m( 0.733) ( 37.7) 0.01 Hence, Mass of the flywheel m = 2946.4 Kg But , for rim type, mass m= DA (where A=cross section area of the rim) 2946.4 1.466 A 7200 A=0.09m 2 P R Venkatesh, Mech Dept, RVCE Note : If it is given that the rectangular cross section of the rim has width (b)=3 thickness ( t), Then A=b t=3t t=3t 2 0.09 3t t=0.1732m 173 mm b=3t=520 mm 2 P R Venkatesh, Mech Dept, RVCE Problem 3 The crank effort diagram for a 4-stroke cycle gas engine may be assumed to for simplicity of four rectangles, areas of which from line of zero pressure are power stroke =6000 mm2, exhaust stroke =500 mm2, Suction stroke=300 mm2, compression stroke = 1500 mm2. Each Sq mm represents 10 Nm. Assuming the resisting torque to be uniform, find a)Power of the engine b)Energy to be stored by the flywheel c)Mass of a flywheel rim of 1m radius to limit the total fluctuation of speed to ±2% P R Venkatesh, of Mech 150 Dept, RVCE of the mean speed rpm. Torque N-m Excess energy (Shaded area) T max Expansion T mean 0 Suction Crank Angle Compression P R Venkatesh, Mech Dept, RVCE Exhaust Data : 4 stroke cycle Crank angle per cycle=4π radians Radius of gyration k 1 meter, Mean speed n=150 rpm C s 2% 4% 0.04 (Q Total fluctuation=2 Fluctuation on either side) Solution : Angular Velocity of flywheel = 2 n 60 2 150 i.e. 15.71 rad/sec 60 WD/cycle=W.D during Expansion-(W.D during other strokes) W.D/cycle= 6000-(300 1500 500) 3700mm 2 W.D/cycle 3700 scale of diagram=3700 10=37000 N-m W.D/cycle 37000 Mean Torque Tm 2944.4 N-m Crank angle/cycle 4 P R Venkatesh, Mech Dept, RVCE (ii) Energy stored by flywheel : e =Shaded area =π( Tmax -Tm ) But W.D during expansion =Tmax 6000 10 Tmax Tmax = 19098.6N - m Torque N-m (i) Power developed by engine : P Tm 2944.4 15.71 46.256 KW Excess energy (Shaded area) T max Expansion T mean 0 Suction Compression Substituting for Tmax , e (Tmax Tm ) (19098.6 2944.6) e =50749.27 N - m P R Venkatesh, Mech Dept, RVCE Crank Angle Exhaust (iii) Mass of flywheel We know that energy stored by flywheel e mk Cs 2 2 50749.27 m (1) (15.71) 0.04 Mass of flywheel m = 5140.64 Kg 2 2 P R Venkatesh, Mech Dept, RVCE Problem 4 A multi cylinder engine is to run at a speed of 600 rpm. On drawing the TMD to a scale of 1mm=250 Nm & 1mm=30, the areas above & below the mean torque line are +160, -172, +168, -191, +197, -162 mm 2 respectively. The speed is to be kept within ±1% of the mean speed. Density of Cast iron flywheel=7250 kg/m3 and hoop stress is 6 MPa. Assuming that the rim contributes to 92% of the flywheel effect, determine the dimensions of the rectangular cross section of the rim assuming width to be twice the thickness. P R Venkatesh, Mech Dept, RVCE Turning Moment 168 160 1 197 2 172 3 5 4 191 6 162 7 Mean Torque line Crank angle Let the energy at 1=E Energy at 2=(E+160) Energy at 3=(E+160)-172=(E-12) Energy at 4=(E-12)+168=(E+156) Energy at 5=(E+156)-191=(E-35) Energy at 6=(E-35)+197=(E+162) Energy at 7=(E+162)-162=E= Energy at 1 Hence, Maximum fluctuation of energy (in terms of area) = (E+162)-(E-35)=197 Sq mm P R Venkatesh, Mech Dept, RVCE Energy stored by the flywheel : 2 N 2 600 Angular velocity = 62.84 rad / sec, 60 60 Cs 1% 2% 0.02 Scale of the diagram is 3 1mm 250 13.1Nm 180 Max Fluctuation of energy e (Max.K.E-Min K.E) 2 e=(E+162)-(E-35)=197 mm 2 e =197 ×13.1 = 2581Nm i.e.2581=I 2Cs I (62.84) 2 0.02 2 Mass moment of inertia I =32.7Kgm P R Venkatesh, Mech Dept, RVCE Diameter of the flywheel : Using = v 2 ; 6 106 7250 v 2 Velocity v=28.8 m/sec DN D 600 Also v= 28.8= 60 60 Mean dia of flywheel D=0.92 m Given 92% of the flywheel effect is provided by the rim, erim 0.92 2581 2375 Nm erim mk 2 2 cs m(k ) 2 cs mv 2cs 2375 m (28.8) 2 0.02 Mass of rim m = 143 kg. P R Venkatesh, Mech Dept, RVCE Dimensions of the crossection of the rim : We know that mass of the flywheel rim m=Volume of rim density=( DA) 143 ( 0.92 A) 7250 A=0.006824m 2 As cross section of rim is rectangular with b=2t, A=(b t)=2t 2 0.006824 2t 2 Hence t = 58.4 mm, b = 2t = 116.8 mm. P R Venkatesh, Mech Dept, RVCE Problem 5 Torque –output diagram shown in fig is a single cylinder engine at 3000 rpm. Determine the weight of a steel disk type flywheel required to limit the crank speed to 10 rpm above and 10 rpm below the average speed of 3000 rpm. The outside diameter of the flywheel is 250 mm. Determine also the weight of a rim type flywheel of 250 mm mean diameter for the same allowable fluctuation of speed. P R Venkatesh, Mech Dept, RVCE 100 T 75 N-m 50 25 0 -25 90 180 360 450 (Degrees) -50 -75 -100 P R Venkatesh, Mech Dept, RVCE 540 630 720 Data : Crank angle per cycle=7200 = 4π radians Mean speed n=3000 rpm, 250 =125 mm =0.125 m (For rim type) 2 0.25 Radius of gyration k= =0.0884 m (For disk type) 2 2 10 20 Cs 0.00667 3000 3000 2 n Solution : Angular Velocity of flywheel = 60 2 3000 i.e. 314.16 rad/sec 60 Radius of gyration k= P R Venkatesh, Mech Dept, RVCE WD/cycle=Net area under TMD =75 50 100 75 50 2 2 2 W.D / cycle =87.5π N - m W.D per cycle Mean torque Tm Crank angle per cycle Tm = 21.875N - m 100 75 2 2 2 87.5 4 P R Venkatesh, Mech Dept, RVCE 100 T 75 N-m 50 25 1 0 -25 Tmean 5 2 90 3 180 4 360 6 450 540 (Degrees) -50 -75 -100 P R Venkatesh, Mech Dept, RVCE 7 8 630 720 Excess energy stored by flywheel Let the energy at 1=E E 26.5625 2 Energy at 3=(E 26.5625 ) - (71.875) E 9.735 2 Energy at 4=(E 9.735 )+(100-21.875) E 68.75 Energy at 2=E+(75-21.875) E 20.3125 2 Energy at 6=(E 20.3125 ) (50-21.875) E 34.375 2 Energy at 5=(E 68.75 )-(96.875) Energy at 7=(E 34.375 )-(121.875) E 26.5625 2 Energy at 8=(E 26.5625 )+(75-21.875) E 2 P R Venkatesh, Mech Dept, RVCE Excess energy e = Max Energy-Min energy e=(E+68.75 ) (E-26.5625 ) 299.43Nm Mass of flywheel We know that energy stored by Rim type flywheel e mk Cs 2 2 299.43 m (0.125) 2 (314.16) 2 0.00667 Mass of flywheel m = 29.11 Kg Fordisk type, k = 0.0884 m Mass of flywheel m = 58.221 Kg P R Venkatesh, Mech Dept, RVCE Problem 6 The torque required for a machine is shown in fig. The motor driving the machine has a mean speed of 1500 rpm and develop constant torque. The flywheel on the motor shaft is of rim type with mean diameter of 40 cm and mass 25 kg. Determine; (i)Power of motor (ii)% variation in motor speed per cycle. P R Venkatesh, Mech Dept, RVCE Torque 2000 N-m 400 N-m Crank angle P R Venkatesh, Mech Dept, RVCE Data : Crank angle per cycle= 2π radians Mean speed n=1500 rpm, 40 Radius of gyration k= =20 cm =0.2 m (For rim type) 2 m=25 kg 2 n Solution : Angular Velocity of flywheel = 60 2 1500 i.e. 157.08 rad/sec 60 P R Venkatesh, Mech Dept, RVCE (i) Power developed by the engine : W.D/cycle area 1+area 2+area 3 1 =400 2 (2000 400) (2000 400) 4 2 2 W.D / cycle =1600π N - m W.D per cycle 1600 Mean torque Tm Crank angle per cycle 2 Tm =800 N - m Power developed by the engine P=Tm 800 157.08 125.664 KW P R Venkatesh, Mech Dept, RVCE Excess energy e (shaded area) Torque 2000 N-m 800 N-m mean x 2 400 N-m 3 1 Crank angle P R Venkatesh, Mech Dept, RVCE (ii) Coefficent of fluctuation of speed C s x 1200 From the similar triangles, x 1.178rad 1600 2 Energy stored by flywheel e = Shaded area 1 e= 2000 800 1.178 2000 800 4 2 e =1649.28Nm We know that energy stored by the flywheel e mk 2 2Cs 1649.28 25 (0.2) (157.08) Cs 2 2 Coefficient of fluctuation of speed = 0.0668 = 6.68% P R Venkatesh, Mech Dept, RVCE Problem 7 A 3 cylinder single acting engine has cranks set equally at 1200 and it runs at 600 rpm. The TMD for each cylinder is a triangle, for the power stroke with a maximum torque of 80 N-m at 600 after dead center of the corresponding crank. The torque on the return stroke is zero. Sketch the TMD & determine the following; (i)Power developed (ii)Coefficient of fluctuation of speed if mass of flywheel is 10 kg and radius of gyration is 8 cm. (iii)Maximum angular acceleration of flywheel. P R Venkatesh, Mech Dept, RVCE 80N-m T (N-m) 0 60 120 180 240 degrees P R Venkatesh, Mech Dept, RVCE 300 360 Data : Crank angle per cycle= 2π radians Mean speed n=600 rpm, Radius of gyration k=8cm =0.08 m m=10 kg 2 n Solution : Angular Velocity of flywheel = 60 2 600 i.e. 62.83 rad/sec 60 P R Venkatesh, Mech Dept, RVCE (i) Mean torque Tm : W.D/cycle area of 3 triangles 1 =3 80 377 N - m 2 W.D per cycle 377 Mean torque Tm Crank angle per cycle 2 Tm =60 N - m As the maxim um torque (Tmax ) is 80 Nm, and Tmean = 60 Nm, the minimum torque (Tmin ) will be = 40 N - m. Hence the modified TMD may be drawn as shown in fig. P R Venkatesh, Mech Dept, RVCE 80 N-m T (N-m) 60 Nm 40 Nm 0 60 120 180 240 300 degrees Modified TMD for 3 Cylinder engine P R Venkatesh, Mech Dept, RVCE 360 (i) Power developed : P Tmean 60 62.83 3.77 Kw x From the similar triangles, 2 20 x rad 3 40 3 Due to symmetry,the energy stored by flywheel =Area of any one traingle (Shaded portion) 1 e= 80 60 e =10.47 N - m 3 2 P R Venkatesh, Mech Dept, RVCE (ii) Coefficeint of fluctuation of speed : We know that energy stored by the flywheel e mk 2 2Cs 10.47 10 (0.08) (62.83) Cs 2 2 Coefficient of fluctuation of speed = 0.0414 = 4.14% (iii) Maximum angular acceleration of flywheel We know that T=I ,where T=Max fluctuation of torque=(Tmax Tmean ) I=mk 2 , the mass moment of inertia of flywheel = Max angular acceleration, rad/sec 2 20 10(0.08) 2 ∴α =312.5 rad / sec 2 P R Venkatesh, Mech Dept, RVCE Problem 8 A torque delivered by a two stroke is represented by T=(1000+300 sin 2 cos 2m where is the angle turned by crank from IDC. The engine speed is 250 rpm. The mass of the flywheel is 400 kg and the radius of gyration is 400 mmDetermine (i)The power developed (ii)Total percentage fluctuation of speed (iii)The angular acceleration and retardation of flywheel when the crank has rotated through an angle of 600 from the IDC (iv)Max & Min angular acceleration & retardation of flywheel. P R Venkatesh, Mech Dept, RVCE Data : As the torque is a function of 2 , equate 2 360 180 0 0 The crank angle per cycle = 180 π radians 0 Mean speed n=250 rpm, Radius of gyration k=400 mm =0.4 m m=400 kg 2 n Solution : Angular Velocity of flywheel = 60 2 250 i.e. 26.18 rad/sec 60 P R Venkatesh, Mech Dept, RVCE (i) Power developed by engine : W.D per cycle Mean torque Tm Crank angle per cycle 1 2 2 1 0 Td 2 2 (1000 300sin 2 500 cos 2 )d 0 Mean torqueTm =1000 N - m Power developed by engine P = Tm P 1000 26.18 26.18 KW P R Venkatesh, Mech Dept, RVCE Sl No Angl e Torqu e T N-m 1 0 500 2 30 1010 3 60 1510 4 90 1500 5 120 990 6 150 490 7 180 500 mean ) T (N-m) Excess Energy T mean =1000 N-m 500 Nm 0 Crank Angle P R Venkatesh, Mech Dept, RVCE 180 (ii) Coefficient of fluctuation of speed : The excess energy stored by the flywheel is given by integrating ΔT between the limits 1 & 2 where 1 & 2 correspond to points where T=Tmean Or (T-Tmean ) =ΔT = 0 (As the torque curve intercepts the mean torque line at these points) i.e. (300sin 2 500 cos 2 ) 0 500 Hence tan2 = 0.6 300 2θ1 =590 & 2θ2 =( 1800 +590 )= 2390 0 Hence θ1 = 29.5 & θ2 =119.5 0 P R Venkatesh, Mech Dept, RVCE (ii) Coefficient of fluctuation of speed (contd.....) 2 Excess energy e = T .d 1 119.5 e= (300sin 2 500 cos 2 ) d 29.5 (The above integration may be performed using calculator by keeping in radian mode and substituting the limits of integration in radians) e = 583.1 N - m Also e=mk 2 2Cs 583.1 400 (0.4) 2 (26.18) 2 Cs Coefficient of fluctuation of speed C s 0.0133 1.33% P R Venkatesh, Mech Dept, RVCE (iii) Angular acceleration at 60 0 crank position Acceleration (or retardation) is caused by excess (or deficit) torque measured from mean torque at any instant. i.e T. At =600 , ΔT 300sin(2 60) 500 cos(2 60) ΔT 509.8 N m Now, ΔT I 509.8 400 (0.4) 2 0 Hence Angular acceleration at 60 crank position α600 =7.965 rad / sec 2 P R Venkatesh, Mech Dept, RVCE (iv) Maximum angular acceleration : Maximum acceleration (or retardation) is caused by maximum fluctuation of torque from mean, i.e ΔTmax (To find ΔTmax first find the crank positions at which ΔT is maximum & then substitute those values in the equation of ΔT.) d For max value of T, ( T) 0 d d (300sin 2 500 cos 2 ) 0 d i.e 600 cos 2 +1000 sin2 0 Hence tan2 =-0.6 2 =-310 & 0 0 2 (1800 ( 31 ) 149 P R Venkatesh, Mech Dept, RVCE (iv) Maximum angular acceleration (contd) : At 2 =-31 , T 583.1N-m (causes retardation) 0 At 2 =149 , T 583.1 (Causes acceleration) 0 Max angular acceleration of flywheel Tmax 583.1 2 max 9.11 rad / sec 2 I 400(0.4) Max angular retardation of flywheel Tmin 583.1 2 min -9.11 rad / sec 2 I 400(0.4) (-ve sign indicates retardation) P R Venkatesh, Mech Dept, RVCE Problem 9 A machine is coupled to a two stroke engine which produces a torque of (800+180 Sin 3 N-m where is the crank angle. The mean engine speed is 400 rpm. The flywheel and the rotating parts attached to the engine have a mass of 350 kg at a radius of gyration of 220 mm. Calculate; (i)The power developed by the engine (ii)Total percentage fluctuation of speed when, (a) The resisting torque is constant (b) The resisting torque is (800+80 Sin P R Venkatesh, Mech Dept, RVCE Sl No Angle Torque T N m 1 0 800 2 30 980 3 60 800 4 90 620 5 120 800 (a) When the resisting torque is Constant T E T (N-m) Excess Energy Tm m 00 300 600 90 0 Crank Angle T = Engine torque E Tm =mean Torque P R Venkatesh, Mech Dept, RVCE 1200 Data : As the torque is a function of 3 , equate 3 360 120 0 0 2π The crank angle per cycle = 120 radians 3 Mean speed n=400 rpm, m=350 kg 0 Radius of gyration k=220 mm =0.22 m 2 n Solution : Angular Velocity of flywheel = 60 2 400 i.e. 41.89 rad/sec 60 P R Venkatesh, Mech Dept, RVCE (i) Power developed by engine : W.D per cycle Mean torque Tm Crank angle per cycle 1 2 2 3 1 Td 2 2 3 (800 180sin ) d 3 0 3 0 Mean torqueTm = 800 N - m Power developed by engine P = Tm P 800 41.89 33.51 KW P R Venkatesh, Mech Dept, RVCE (ii) Coefficient of fluctuation of speed : (a) When the resisting torque is constant The excess energy stored by the flywheel is given by integrating ΔT between the limits 1 & 2 , where To find 1 & 2 are crank positions at which T=Tmean Or (T-Tmean ) = ΔT = 0 (As the torque curve intercepts the mean torque line at these points) i.e. (800 180sin 3 800) 0 180sin 3 0 3 0 & 3 180 0 θ1 =0 & θ2 =60 0 0 P R Venkatesh, Mech Dept, RVCE 0 (ii) Coefficient of fluctuation of speed (contd.....) 2 Excess energy e = T .d 1 60 e = (180sin 3 ) d 0 (The above integration may be performed using calculator by keeping in radian mode and substituting the limits of integration in radians) e = 120 N - m Also e=mk 2 2Cs 120 350 (0.22) 2 (41.89) 2 Cs Coefficient of fluctuation of speed C s 0.00404 0.404% P R Venkatesh, Mech Dept, RVCE (b) When the resisting torque is (800 + 80sin ) T Excess Energy E T (N-m) 00 300 600 90 0 Crank Angle T 1200 T = Engine torque E TPM =Machine Torque R Venkatesh, Mech Dept, RVCE M 0 180 (ii) Coefficient of fluctuation of speed : (b) When the resisting torque is (800 + 80sin ) The excess energy stored by the flywheel is given by integrating ΔT between the limits 1 & 2 , where To find 1 & 2 are crank positions at which TE =TM Or (TE -TM )= ΔT = 0 (As the engine torque curve intercepts the machine torque curve at these points) i.e. (800 180sin 3 ) (800 80sin ) 0 180(3sin 4sin 3 ) 80 sin =0 (460-720sin 0 sin 0.799 2 θ1 =530 & θ2 =( 180 - 53)=127 0 P R Venkatesh, Mech Dept, RVCE (ii) Coefficient of fluctuation of speed (contd.....) 2 Excess energy e = T .d 1 127 e= (180sin 3 80sin )d 53 (The above integration may be performed using calculator by keeping in radian m ode and substituting the limits of integration in radians) e = -208.3 N - m (Take absolute value) Also e=mk 2 2Cs 208.3 350 (0.22) 2 (41.89) 2 Cs Coefficient of fluctuation of speed C s 0.007 0.7% P R Venkatesh, Mech Dept, RVCE Problem 10 A certain machine requires a torque of (500+50sin N-m to drive it, where is the angle of rotation of the shaft. The machine is directly coupled to an engine which produces a torque of (500+60 sin2 m. The flywheel and the other rotating parts attached to the engine have a mass of 500 kg at a radius of 400 mm. If the mean speed is 150 rpm. Find; (a)The maximum fluctuation of energy (b)Total % fluctuation of speed (c)Max & Min angular acceleration of the flywheel & the corresponding shaft positions. P R Venkatesh, Mech Dept, RVCE T M TE Excess Energy T (N-m) 00 0 Crank Angle TE = Engine torque T M =Machine Torque P R Venkatesh, Mech Dept, RVCE 180 Data : As the torque is a function of 2 , equate 2 360 180 0 0 The crank angle per cycle = 180 π radians 0 Mean speed n=150 rpm, Radius of gyration k=400 mm =0.4 m m=500 kg 2 n Solution : Angular Velocity of flywheel = 60 2 150 i.e. 15.71 rad/sec 60 P R Venkatesh, Mech Dept, RVCE (i) Excess energy stored by flywheel : The excess energy stored by the flywheel is given by integrating ΔT between the limits 1 & 2 , where To find 1 & 2 are crank positions at which TE =TM Or (TE -TM ) = ΔT = 0 i.e. (500 60sin 2 ) (500 50sin ) 0 (60sin 2 50sin )=0 Put sin2θ = 2sinθcosθ sin 12 cos 5 0. Either sin 0 00 ,1800 or 12 cos 5 0 cos 5 65.37 0 12 Considering max difference between consecutive crank positions, θ1 = 65.37 0 &θ2 = 180 0 P R Venkatesh, Mech Dept, RVCE 2 Excess energy e = T .d 1 180 e= (60sin 2 50sin ) d 65.37 (The above integration may be performed using calculator by keeping in radian mode and substituting the limits of integration in radians) e = -120.42 N - m (Take absolute value) (ii) Coefficient of fluctuation of speed : Also e=mk 2 2Cs 120.42 500 (0.4) 2 (15.71) 2 Cs Coefficient of fluctuation of speed C s 0.0061 0.61% P R Venkatesh, Mech Dept, RVCE (iii) Maximum & Minimum angular acceleration : Maximum acceleration (or retardation) is caused by maximum fluctuation of torque from mean, i.e ΔTmax (To find ΔTmax first find the crank positions at which ΔT is maximum & then substitute those values in the equation of ΔT.) d For max value of T, ( T) 0 d d (60sin 2 50sin ) 0 d i.e.12 cos 2 - 5cos 0 P R Venkatesh, Mech Dept, RVCE i.e.12 cos 2 - 5cos 0 Put cos 2 (2 cos 2 1), we get 12(2 cos 2 1) - 5cos 0 2 24cosθ - 5cosθ -12 =0 =35 0 & =127.6 0 At =35 0 : T=60 sin(2 35)-50sin(35)= 27.7 N - m 27.7 Maximum acceleration max = 0.346 rad / sec 2 80 At =127.6 0 : T=60 sin(2 127.6)-50sin(127.6)=-97.62 N - m 97.62 Maximum retardation max = 1.22 rad / sec 2 0 P R Venkatesh, Mech Dept,8RVCE Flywheel for Punch press Flywheel Crank shaft Crank connecting rod Punching tool t Plate d Die Flywheel for Punch Press P R Venkatesh, Mech Dept, RVCE Flywheel for Punch press • If ‘d’ is the diameter of the hole to be punched in a metal plate of thickness ‘t’ , the shearing area A= d t mm2 • If the energy or work done /sheared area is given, the work done per hole =W.D/mm 2 x Sheared area per hole. • As one hole is punched in every revolution, WD/min=WD/hole x No of holes punched /min • Power of motor required P=WD per min/60 P R Venkatesh, Mech Dept, RVCE Excess energy Stored by Flywheel : e =Energy required per hole ×( 1- x) where x = fraction of crank rotation during which actual punching operation take place. Actual time of punching x= Time required per revolution P R Venkatesh, Mech Dept, RVCE Problem 11 A punching machine carries out 6 holes per min. Each hole of 40 mm diameter in 35 mm thick plate requires 8 N-m of energy/mm2 of the sheared area. The punch has a stroke of 95 mm. Find the power of the motor required if the mean speed of the flywheel is 20 m/sec. If the total fluctuation of speed is not to exceed 3% of the mean speed, determine the mass of the flywheel. P R Venkatesh, Mech Dept, RVCE Data : Mean speed of flywheel v=20m/sec Cs 3% 0.03, Diameter of hole d=40 mm Thickness of plate t=35 mm, Energy/mm 2 =8 N-m Stroke length =95 mm, No of holes/min=6 Speed of crank=6rpm Time required to punch one hole= 10 secs Solution : Sheared area per hole =πdt = π× 40 × 35 = 4398.2 3 mm 2 W.D/hole= 4398.23 8 = 35186 N - m W.D/hole holes/min Power of motor= KW 3 60 10 35185.4 6 P 3.5186KW 3 60 10 P R Venkatesh, Mech Dept, RVCE As the punch travels 95 ×2 =190 mm in10 secs ⇒ actual time taken to punch one hole Thickness of plate T actual = Time taken per cycle 2×stroke length 35 T actual = 10= 1.842 Secs 190 Excess energy supplied by flywheel Tactual e=Energy required/hole(1-x) where x= Time per revolution 1.842 e=35186 1 = 28705 N - m 10 Also e = mv 2C s 28705 m( 20)2 0.03 m = 2392 Kg P R Venkatesh, Mech Dept, RVCE Problem 12 A constant torque 2.5 KW motor drives a riveting machine. The mass of the moving parts including the flywheel is 125 kg at 700 mm radius. One riveting operation absorbs 10000 J of energy and takes one second. Speed of the flywheel is 240 rpm before riveting. Determine; (i)The number of rivets closed per hour (ii)The reduction in speed after riveting operation. P R Venkatesh, Mech Dept, RVCE Data : Maximum speed of flywheel n1 =240 rpm 1 2 240 25.133 rad / sec 60 Energy required per rivet =10000 J Time taken to close one rivet =1 sec Energy supplied by motor=Power of motor =2.5KW=2500 J/sec Mass of flywheel m=125 kg, Rad. of gyration k=700 mm Mass M.O.I of flywheel I=125 (0.7) 2 61.25 Kg m 2 (i) Number of rivets closed per hour : Energy supplied by motor =2.5KW=2500 J/sec Energy supplied per hour =2500 3600 J Energy required per rivet =10000 J 2500 3600 J Number of rivets closed per hour will be= 900 rivets / hr 10000 P R Venkatesh, Mech Dept, RVCE ( ii) Excess energy supplied by flywheel : e=Energy required/rivet-Energy supplied by motor e=(10000 2500) 7500 J 1 2 2 Also e = I 1 2 2 1 2 2 7500 61.25 (25.13) 2 2 19.66 60 2 19.66rad / sec n2 188rpm. 2 Reduction in speed after riveting=(n 1 n2 ) (240 - 188) = 52 rpm P R Venkatesh, Mech Dept, RVCE