Turning Moment (Crank Effort) Diagram for a 4-stroke I C engine
m N e T u max q r o T
Excess Excess Energy Energ y (Shaded area)
Expansion T mean
0 Crank Angle
Turnin urning g 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 T m.
Turnin urning g 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 T m.
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.
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 fluctuations of the torque on the crankshaft &
DISK TYPE FLYWHEEL
D
X
X
D t
b
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 gyratio n of the flywheel. For rim type, k=
D
2
For Disk type, k=
where D=Mean diameter of the flyheel D
2 2
D 2 Hence I=m Rim 4
where D=Outer diameter of the flywheel
D2 and I=m Disk 8
Hence for a given diameter & inertia, the mass of the
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. n1 n2 1 2 C s (or K s ) n 2 n where =Angular velocity= 60
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 E 2 E e C e (or K e ) E E E * * It is often expressed as the ratio of excess energy to the work done per cycle. C e ( or K = )e
e W.D / cycle
(d) Coefficient of steadiness :
It is the reciprocal of coefficent of fluctuation of speed.
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=Rad ius of gyration of the flywheel Cs =Coefficient of fluctuation of speed The max fluctuation of energy (to be stored by the flywheel) e E1 E2
e
1 2
1 2
2 I 1
1 2
2 I 2
I 1 2 (1 2 )
1 2
2 2 I 1 2
EXPRESSION FOR ENERGY STORED BY A FLYWHEEL Putting the mean agular speed =
1
1 2 ,
2 We get e = I ω(ω1 - ω2 ) Multiplying & dividing by ,
e = I ω Also
2
( ω1 - ω2 )
( ω1 - ω2 )
C s , the coefficient of fluctuation o f speed
Hence e = I ω2C s Putting I=mk 2 , we get e = m k 2 ω2C s 1 Note: 1.Alternatively, if Mean kinetic energy E= I 2 , 2
I 2 2 E , e
e=2EC s e
EXPRESSION FOR ENERGY STORED BY A FLYWHEEL 2
e= I Cs , Putting mean Kinetic energy E=
1 2
2
I
and expressing C s as a percentage, e
2EC s
100 e 0.02 E C s Note: 2. Alternatively, if Mean kinetic energy E=
(k ) v 2
e mv 2cs
2
, E=
1 2
mv
2
1 2
2
2
mk ,
MASS OF FLYWHEEL IN TERMS OF DENSITY & CROSSECTION AREA
We know that mass m=Density Volume For Disk type flywheel, Volume =
D 2
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: (i)Velocity of the flywheel v=
Dn
m / sec
60 (ii) Hoop Stress (Centrifugal stress) in the flywheel
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.
TURNING MOMENT DIAGRAM
m N e T u max q r o T
Excess Energy
Expansion T mean
x
0 Suction
Exhaust Crank Angle Compression
Data :
Power P=18.4 KW=18.4 103 W, Mean speed n=300 rpm Work done during expansion W E 3 Work done during compression C s
2% 0.02
Given 4-stroke cycle engine
Crank angle per
cycle=
Solution :
Angular Velocity of flywheel = i.e.
2 300 60
2 n 60
31.416 rad /sec
Also power P=Tm 18.4 103 18 4 103
Tm 31.416
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=(W E WC ) Given W E
3W C
Or WC
W E
3
, we can write
W E 2 W E W E 11040 N-m 7360= W E 3 3
This work represents the area under triangle for expansion stroke i e 11040
1
T
Excess energy stored by the flywheel The shaded area represents the excess energy. i.e.excess energy stored by flywheel e=
1
x (Tmax Tmean )
2 where x is the base of shaded triangle, given by
x
(Tmax T mean )
x
T max
(Tmax
Hence e=
T mean )
T max
1 2
(7028.3 585.7) 7028.3
2.88rad
2.88 (7028.3 585.7) 9276.67 N-m
We know that excess energy is given by 2
e=I Cs
9276.64 I (31.416) 0.02 2
Hence mass moment of inertia of flywheel I=470 Kg-m
2
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.
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= Density =7.2 Mg/m3
7200 Kg/m3 , Hoop stress =5.5 MPa
Solution :
Angular Velocity of flywheel = i.e.
2 360 60
2 n 60
37.7 rad/sec
Also power P=Tm 75 103 75 10
3
Tm 37.7
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 Diameter of the flywheel : Hoop stress = v 2
5.5 106
=7200 (v 2 )
Hence, velocityof flywheel v = 27.64m / sec Also v =
Dn 60
27.64 =
D 360 60
The energy stored by the flywheel is given by 2
2
s
.
For rim type, radius of gyration k=
2
D
2
1.466 2
37.7) 2 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
0.733m
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 0.09 3t 2
2
t=0.1732m 173 mm
b=3t=520 mm
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 mm 2, exhaust stroke =500 mm2, Suction stroke=300 mm2, compression stroke = 1500 mm 2. 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% of the mean
m N e u Tmax q r o T
Excess energy (Shaded area)
Expansion T mean
0 Suction Compression
Crank Angle
Exhaust
Data :
4 stroke cycle Crank angle per cycle= Radius of gyration k 1 meter, Mean speed n=150 rpm Cs
2% 4% 0.04 (
Total fluctuation=2 Fluctuation on either side)
Solution : Angular Velocity of flywheel = i.e.
2 150
2 n 60
15.71 rad/sec
60 WD/cycle=W.D during Expansion-(W.D during other strok es)
W.D/cycle= 6000-(300 1500 500) 3700mm 2
W.D/cycle 3700 scale of diagram=3700 10=37000 N-m
Mean Torque Tm
W.D/cycle Crank angle/cycle
37000 4
2944.4 N-m
(i) Power developed by engine : P T m 2944.4 15.71 46.256 K W
(ii) Energy stored by flywheel :
But W.D during expansion =Tmax
6000 10 Tmax T max = 19098.6N - m Substituting for Tmax , e (Tmax T m ) (19098.6 2944.6)
m N e u Tmax q r o T
Excess energy (Shaded area)
Expansion T mean
0 Suction Compression
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) 2
Mass of
2
0.04
flywheel m = 5140.64 Kg
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/mm3 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
168 160 t n e m o 1 2 3 M g n i n r u 172 T
197
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
Energy stored by the fl ywheel : Angular velocity =
2 N
60 C s 1% 2% 0.02
2 600 60
62.84rad / sec,
Scale of t he dia gra m is 3 1mm 250 13.1 Nm 180 Max Fluctuation of energy e (Max.K.E-Min K.E) e=(E+162)-(E-35)=197 mm2 2 2 i.e.2581=I C s I (62.84) 0.02 2
Dia meter of the flywheel : Using = v 2 ; 6 106 7250 v 2
Velocity v=28.8 m /sec Also v=
DN
28.8=
D 600
60 60 Mean dia of flywheel D=0.92 m Given 92% of the flywheel effect is provi ded by the rim, erim 0.92 2581 2375 Nm erim mk 2 2cs m(k ) 2 cs mv 2cs
2375 m (28.8)2 0.02 Mass of rim m = 143 kg.
Dimensions of the crossection of the rim : We know that mass of the flywheel rim
density=( DA) 143 ( 0.92 A) 7250 2 A=0.006824m m=Volume of rim
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.
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.
100 T 75 N-m 50 25 0 -25 -50 -75 -100
90
180
360
(Degrees)
450
540
630
720
Data :
Crank angle per cycle=720 cycle=7200 Mean speed n=3000 rpm, Radius of gyration k= Radius of gyration k= Cs
10 3000
20 3000
250 2 0.25 2 2
=125 mm =0.125 m (For rim type) =0.0884 m (For disk type)
0.00667
Solut So lutio ion n : Angular Velocity of flywheel = i.e.
2 3000 60
314.16 rad rad/sec
2 n 60
WD/cycle=Net area under TMD =75 50 100 75 50 100 75 2 2 2 2 2 2
Mean torque Tm
W.D per cycle Crank angle per cycle ycle
87.5 4
100 T 75 N-m 50 25 1
0 -25 -50 -75 -100
Tmean
5 2
90
3
180
4
360 (Degrees)
6
450
540
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 5=( E 68.75 )-(96.875) E 20.3125 2 Energy at 6=( E 20.3125 ) (50-21.875) E 34.375 2 Energy at 7=( E 34.375 )-(121.875) E 26.5625 2 E t 8 ( E 26 5625 )+ )+((75 21 875) E Energy at 2=E+(75-21.875)
Excess energy e = Max Energy-Min energy e=(E+68.75 ) (E-26.5625 ) Mass of flywheel
We know that energy stored by Rim type flywheel e mk Cs 2
2
299.43 m (0.125) (314.16) 2
Mass of
2
0.00667
flywheel m = 29.11 Kg
Fordisk type, k = 0.0884 m
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.
2000 N-m
e u q r o T
400 N-m
Crank angle
Data :
Crank angle per cycle= Mean speed n=1500 rpm, Radius of gyration k=
40 2
=20 cm =0.2 m (For rim type)
m=25 kg Solution : Angular Velocity of flywheel = i.e.
2 1500 60
157.08 rad/sec
2 n 60
(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
Mean torque Tm
W.D per cycle Crank angle per cycle
Power developed by the en gine P=Tm 800 157.08 125.664 KW
1600 2
Excess energy e (shaded area)
2000 N-m
e u q r o T 800 N-m 400 N-m
mean
x
2
3 1
Crank angle
(ii) Coefficent of fluctuation of speed C s
From the similar triangles,
x
2
1200 1600
x 1.178 rad
Energy stored by flywheel e = Shaded area e= 2000 800
4
1
1.178 2000 800 2
We know that energy st ored by the flywheel 2 2 e mk Cs
1649.28 25 (0.2) 2 (157.08) 2 C s
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.
80N-m T (N-m)
0 60
120 degrees
180
240
300
360
Data :
Crank angle per cycle= Mean speed n=600 rpm, Radius of gyration k=8cm =0.08 m m=10 kg Solution : Angular Velocity of flywheel = i.e.
2 600 60
62.83 rad/sec
2 n 60
(i) Mean to rque Tm : W.D/cycle area of 3 triangles 1 =3 80 377 2 Mean torque Tm
N -m
W.D per cycle Crank angle per cycle
377 2
As the maxim um torque (T max ) is 80 Nm, and T mean = 60 Nm, the minimum torque (T min ) will be = 40 N - m. Hence the modified TMD
80 N-m T (N-m)
60 Nm
40 Nm
0 60
120
180
240
300
degrees
Modified TMD for 3 Cylinder engine
360
(i) Power developed : P T mean 60 62.83 x From the similar triangles, 2
20 40
x 3 rad
3 Due to symmetry,the energy stored by flywheel =Area of any one traingle (Shaded portion) 1 e= 3 80 60 2
(ii) Coefficeint of fluctuation of speed : We know that energy stored by the flywheel 2 2 e mk Cs
10.47 10 (0.08) 2 (62.83) 2 C s
Coefficient of
fluctuation of speed = 0.0414 = 4.1 4%
(iii) Maximum angular accelera tion of flywheel We know that T=I ,where
T=Max fluctuation of torque=(Tmax T mean ) I=mk 2 , the mass moment of inertia of flywheel = Max angular acceleration, rad/sec
2
Problem 8 A torque delivered by a two stroke is represented by T=(1000+300 sin 2 cos 2 m 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 mm . Determine (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.
Data :
As the torque is a function of 2 , equate 2
360 180 0
The crank angle
0
per cycle = 180
0
Mean speed n=250 rpm, Radius of gyration k=400mm =0.4 m m=400 kg Solution : Angular Velocity of flywheel =
2 250
2 n 60
(i) Power developed by engine : Mean torque Tm
1
1
W.D per cycle Crank angle per cycle
Td (1000 300 sin 2 500 cos 2 ) d 0
0
Mean torque Power developed by engine P = P 1000 26.18
Sl No
Angle
Torque 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
180
(ii) Coefficient of fluctuation of speed : The excess energy stored by the flywheel is given by integrating
between the limits 1
& 2 where 1 & 2 correspond to points where T=Tmean Or (T-Tmean ) =
=0
(As the torque curve intercepts the mean torque line at these points)
500 cos 2 ) 0 500 1.667 Hence tan2 = 300 i.e. (300 sin 2
(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 perf ormed using calcul ator by keeping in radian mode and sub sti tuting the limits of integration in radia ns)
e = 583.1 N - m 2 2 2 2 Also e=mk Cs 583.1 400 (0.4) (26.18) C s
0
(iii) Angular acceleration at 60 crank position Acceleration (or retardation) is caused by excess (or deficit) torque measured from mean torque at any instant. i.e T.
300 sin(2 60) 50 0 cos(2 60) 509.8 N m I 0
At =60 ,
Now,
509.8 400 (0.4) 2 Hence Angular acceleration at 60 0crank positi on
(iv) Maximum angular acceleration :
Maximum acceleration (or retardation) is caused by maximum fluctuation of torque from mean, i.e
max
(To find ΔTmax first find the crank positio ns at which
ΔT is ma ximum & then substitute those values in the equation of ΔT.)
For max value of T,
d
(300sin 2
d d
( T) 0
500 cos 2 ) 0
d i.e 600 cos 2 +1000 sin2 0
Hence tan2 =-0.6
2 =-310 &
(iv) Maximum angular acceleration (contd) :
At 2 =-310 , T 583.1N-m (causes retardation) At 2 =149 , T 583.1 (Causes acceleration) 0
Max angular acceleration of flywheel T max 583.1 max 2 I
400(0.4)
Max angular retardation of flywheel T min 583.1 min 2 I
400(0.4)
(-ve sign indicates retardation)
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
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
T
m
00
30 0
60 0
90 0
Crank Angle
T
E
= Engine torque
0
120
m
Data :
As the torque is a function of 3 , equate 3
360 120 0
The crank angle
0
per cycle = 120
0
2 3
Mean speed n=400 rpm, m=350 kg Radius of gyration k=220mm =0.22 m Solution : Angular Velocity of flywheel =
2 400
2 n 60
(i) Power developed by engi ne :
Mean torque Tm 2
1 2
W.D per cycle Crank angle per cycle 2
3
Td
3 0 Mean torque
1 2
3
3
(800 180sin 3 )d
0
Power developed by engine P = P 800 41.89
(ii) Coefficient of fluctuation of speed : (a) When the resisting torque is constant The excess energy stored by the flywheel is given by integrating
between the limits 1
& 2 , where To find 1 & 2 are crank positions at which T=Tmean Or (T-Tmean
0
(As the torque curve intercepts the mean torque line at these points)
i.e. (800 180 sin 3 800) 0
180 sin 3 0 3 0 & 3 180 0
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 substitutin g the limits of integration in radians)
e = 120 N - m Also e=mk 2 2Cs Coefficie
120 350 (0.22)2 (41.89)2 C s
of fluctuatio of
ed C
0 00404
0 404%
(b) When the resisting torque is (800 + 80sin ) T
Excess Energy
E
T (N-m)
T
M
00
30 0
60 0
90 0
Crank Angle
T
E
= Engine torque
0
120
0
180
(ii) Coefficient of fluctuatio n of speed : (b) When the resisting torque is (800 + 80sin ) The excess energy stored by the flywheel is given by integrating
between the limits 1
& 2 , where To find 1 & 2 are crank positions at which T E =TM Or (TE -TM
0
(As the engine torque curve intercepts the machine torque curve at these points)
i.e. (800 180 sin 3 ) (800 80 sin )
180(3sin 4sin 3 ) 80 sin =0 2 0 sin 0.799 (460-720sin
0
(ii) Coeff icient of f luctuation 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 substi tuting 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 C s Coefficie
of fluctuati
of
eed C
0 007
0.7%
Problem 10 A certain machine requires a torque of N-m to drive it, where is the (500+50sin angle of rotation of the shaft. The machine is directly coupled to an engine which produces a m. The flywheel and torque of (500+60 sin2 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
T
M
T
E Excess Energy
T (N-m)
00
0
Crank Angle
T
E
= Engine torque
180
Data :
As the torque is a function of 2 , equate 2
360 180 0
The crank angle
0
per cycle = 180
0
Mean speed n=150 rpm, Radius of gyration k=400mm =0.4 m m=500 kg Solution : Angular Velocity of flywheel =
2 150
2 n 60
(i) Excess energy stored by flywheel : The excess energy stored by the flywheel is between the limits 1
given by integrating
& 2 , where To find 1 & 2 are crank positions at which T E =TM Or (TE -TM
0
i.e. (500 60 sin 2 ) (500 50 sin )
0
(60 sin 2 50 sin )=0 sin 12 cos
5 0.
Either sin 0 00 ,1800
or 12 cos 5 0 cos 5
65.370 12
Considering max difference between consecutive crank positions, 0
θ = 65 37 & θ = 180
0
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 radian s)
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 C s
Coefficient o f fluctuation of
ed C
0 0061
(iii) Maximum & Minimum angular acceleration : Maximum acceleration (or retardation) is caused by maximum fluctuation of torque from mean, i.e
max
(To find ΔTmax first find the crank positio ns at which
ΔT is maximum & then substitute those values in the equation of ΔT.)
For max value of T,
d d
(60 sin 2
d d
(T) 0
50sin ) 0
i.e.12 cos 2 - 5cos 0
(2 cos 2 1), we get 12(2 cos 2 1) - 5cos 0 Put cos 2
At
0
& 0
0
:
T=60 sin(2 35)-50sin(35)= 27.7 Maximum acceleration max = At
0
N - m
27.7 80
0.346 ra d
:
T=60 sin(2 127.6)-50sin(127.6)=-97.62 N - m 97.62
/ sec 2
Flywheel for Punch press Flywheel
Crank shaft
Crank
connecting rod
Punching tool t
Plate d Die
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 •
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•
If the energy or work done /sheared area is given, the work done per hole =W.D/mm2 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
1
2
)
1 2
(E2 = Energy supplied per sec× Actual time of punching)
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.
Data :
Mean speed of flywheel v=20m/sec C s
3% 0.03,
Diameter of hole d=40 mm
Thickness of plate t=35 mm, Energy/mm2 =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 = 4398.23 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
As the punch travels 95 × 2 = 190 mm in 10 secs ⇒ actual time taken to punch one hole
T actual = T actual =
Thickness of plate 2×stroke length 35
Time taken per cycle
10=1.842 Secs
190 Excess energy supplied by flyw heel
e=Energy required/hole Energy supplied during actual punching e=
(
2
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.
Data :
Maximum speed of flywheel n1 =240 rpm
1
2 240 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
Number of rivets closed per hour will be=
2500 3600 J 10000
