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# Solved Problems: Design of Energy Storing Elements

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A helical spring is made from a wire of 6 mm diameter and has outside diameter of 75 mm. If the permissible shear stress is 350 MPa and modulus of rigidity 84 kN/mm2, find the axial load which the spring can carry and the deflection per active turn. Solution.

Given :

d = 6 mm ; Do = 75 mm ;

= 350 MPa = 350 N/mm2 ;

G = 84 kN/mm2 = 84 × 103 N/mm2

We know that mean diameter of the spring,

= Do – d

=75 – 6

=69 mm

Spring index, C = D/d

= 69 / 6

= 11.5

Let         W   = Axial load, and

δ/ n   = Deflection per active turn.

1. Neglecting the effect of curvature

We know that the shear stress factor,

Ks  = 1 + (1 / 2C)

= 1 + (1/ 2x11.5)

=1.043

Maximum shear stress induced in the wire (τ),

= Ks (8WD / πd3)

=1.043 (8W x 69 / π x 63) W = 350 / 0.848

=412.7 N

We know that deflection of the spring, δ = 8WD3n / Gd4

Deflection per active turn,

δ/n  = 8WD3 / Gd4

=9.96 mm

Considering the effect of curvature

We know that Wahl’s stress factor,

= 4 ( 11 . 5)− 1 + 0. 615 4(11.5)−4 11.5

= 1.123

We also know that the maximum shear stress induced in the wire (τ),

= K x (8WC / πd2) = 0.913 W

W = 350 / 0.913

= 383.4 N

Deflection of the spring,

= 8WD3n / Gd4

Deflection per active turn,

δ/n  = 8WD3 / Gd4

=9.26 mm

Design a helical spring for a spring loaded safety valve (Rams bottom safety valve) for the following conditions. Diameter of valve seat = 65 mm ; Operating pressure = 0.7 N/mm2; Maximum pressure when the valve blows off freely = 0.75 N/mm2; Maximum lift of the valve when the pressure rises from 0.7 to 0.75 N/mm2 = 3.5 mm: Maximum allowable stress = 550 MPa ;Modulus of rigidity = 84 kN/mm2; Spring index = 6.

Solution.

Given :

D1 = 65 mm

p1 = 0.7 N/mm2

p2 = 0.75 N/mm2 ;

= 3.5 mm

δ = 550 MPa = 550 N/mm2

G = 84         C = 6

kN/mm2 = 84

× 103 N/mm2

1. Mean diameter of the spring coil

Let D = Mean diameter of the spring coil, and d = Diameter of the spring wire.

Since the safety valve is a Rams bottom safety valve, therefore the spring will be under tension. We know that initial tensile force acting on the spring (i.e. before the valve lifts),

W1  = (π/4) x (D1)2 x p1

=(π/4) x (65)2 x 0.7

=2323 N

Maximum tensile force acting on the spring (i.e. when the valve blows off freely), W2 = (π/4) x (D1)2 x p2

=(π/4) x (65)2 x 0.75

=2489 N

Force which produces the deflection of 3.5 mm,

= W2 – W1

=2489 – 2323

=166 N

Since the diameter of the spring wire is obtained for the maximum spring load (W2), therefore maximum twisting moment on the spring,

= W2 x D/2

=2489x 6 d / 2

=7467 d

We know that maximum twisting moment (T ), 7467 d = (π/16) x τ x d3

=(π/16) x 550 x d3 d2 = 7467 / 108

=69.14

d = 8.3 mm

From Table 23.2, we shall take a standard wire of size SWG 2/0 having diameter (d)

d = 8.839 mm

Mean diameter of the coil,

= 6 d

=6 × 8.839

=53.034 mm

Outside diameter of the coil,

Do = D + d

=53.034 + 8.839

=61.873 mm

Inside diameter of the coil,

Di         = D – d

=53.034 – 8.839

=44.195 mm

Number of turns of the coil

Let n = Number of active turns of the coil. We know that the deflection of the spring (δ),

3.5= 8WC3n / Gd

=8 x 166 x 63 x n / 84x103 x 8.839 n = 3.5 / 0.386

=9.06 say 10

For a spring having loop on both ends, the total number of turns, n' = n + 1

=10 + 1

=11

Free length of the spring

Taking the least gap between the adjacent coils as 1 mm when the spring is in free state, the free length of the tension spring,

LF  = n.d + (n – 1) 1

=10 × 8.839 + (10 – 1) 1

=97.39 mm

Pitch of the coil

We know that

Pitch of the coil       = Free length / n – 1

=97.39 / 10 – 1

=10.82 mm

The areas of the turning moment diagram for one revolution of a multi-cylinder engine with reference to the mean turning moment, below and above the line, are – 32, + 408, – 267, + 333, – 310, + 226, – 374, + 260 and – 244 mm2. The scale for abscissa and ordinate are: 1 mm = 2.4° and 1 mm = 650 N-m respectively. The mean speed is 300 r.p.m. with a percentage speed fluctuation of ± 1.5%. If the hoop stress in the material of the rim is not to exceed 5.6 MPa, determine the suitable diameter and cross-section for the flywheel, assuming that the width is equal to 4 times the thickness. The density of the material may be taken as 7200 kg / m3. Neglect the effect of the boss and arms.

Solution.

Given :

N = 300 r.p.m.

= 2 ×π x 300/60 = 31.42 rad/s ;

σt = 5.6 MPa = 5.6 × 106 N/m2 ; ρ = 7200 kg/m3

Diameter of the flywheel

Let D = Diameter of the flywheel in metres.

We know that peripheral velocity of the flywheel,

v = π D N / 60

= π D ×300 / 60

= 15.71 D m/s

We also know that hoop stress (σt ),

× 106  = ρ × v2

=7200 (15.71 D)2

=1.8 × 106 D2

D2 = 5.6 × 106 / 1.8 × 106 = 3.11

D  = 1.764 m

Cross-section of the flywheel

Let     t = Thickness of the flywheel rim in metres, and

b = Width of the flywheel rim in metres     = 4 t’

Cross-sectional area of the rim, A     = b × t

= 4 t × t

Now let us find the maximum fluctuation of energy

Since the scale of crank angle is 1 mm        = 2.4º

= 2.4 × (π / 180)

The scale of the turning moment is 1 mm   = 650 N-m,

1 mm2 on the turning moment diagram     = 650 × 0.042

= 27.3 N-m

Let the total Energy at A = E. Therefore from Fig. 22.13, we find that

Energy at B = E – 32

Energy at C = E – 32 + 408 = E + 376

Energy at D = E + 376 – 267 = E + 109

Energy at E = E + 109 + 333 = E + 442

Energy at F = E + 442 – 310 = E + 132

Energy at G = E + 132 + 226 = E + 358

Energy at H = E + 358 – 374 = E – 16

Energy at I = E – 16 + 260 = E + 244

Energy at J = E + 244 – 244 = E = Energy at A

From above, we see that the energy is maximum at E and minimum at B.

Maximum energy            = E + 442

and     minimum energy  = E – 32

We know that maximum fluctuation of energy,

∆ E = Maximum energy – Minimum energy = (E + 442) – (E – 32)

=474 mm2

=474 × 27.3

=12 940 N-m

Since the fluctuation of speed is ± 1.5% of the mean speed, therefore total fluctuation of speed, ω1 – ω2 = 3% of mean speed

= 0.03 ω

coefficient of fluctuation of speed,

CS = (ω1 − ω2) / ω

= 0.03

Let m = Mass of the flywheel rim.

We know that maximum fluctuation of energy (∆E), 12940 = m.R22.CS

m x (1.764/2)2 x (31.42)2 x 0.03

=23 m

m = 12 940 / 23

= 563 kg

We also know that mass of the flywheel rim (m),

= A × π D × ρ

=4 t2 × π × 1.764 × 7200

=159624 t2

t2  = 563 / 159624

=0.00353 t = 0.0594 m

=59.4 say 60 mm

= 4 t

=4 × 60

=240 mm

A punching machine makes 25 working strokes per minute and is capable of punching 25 mm diameter holes in 18 mm thick steel plates having an ultimate shear strength of 300 MPa. The punching operation takes place during 1/10 th of a revolution of the crank shaft. Estimate the power needed for the driving motor, assuming a mechanical efficiency of 95 percent. Determine suitable dimensions for the rim cross-section of the flywheel, which is to revolve at 9 times the speed of the crank shaft. The permissible coefficient of fluctuation of speed is 0.1. The flywheel is to be made of cast iron having a working stress (tensile) of 6 MPa and density of 7250 kg / m3. The diameter of the flywheel must not exceed 1.4 m owing to space restrictions. The hub and the spokes may be assumed to provide 5% of the rotational inertia of the wheel. Check for the centrifugal stress induced in the rim.

Solution.

Given :

= 25 ;

d1    = 25 mm ;

t1    = 18 mm ;

τu    = 300 MPa = 300 N/mm2 ;

ηm = 95% = 0.95 ; CS = 0.1 ;

σt = 6 MPa = 6 N/mm2 ; ρ = 7250 kg/m3 ;

D     = 1.4 m

= 0.7m

Power needed for the driving motor

We know that the area of plate sheared,

AS  = d1 × t1

=π × 25 × 18

=1414 mm2

Maximum shearing force required for punching,

FS  = AS × τu

=1414 × 300

=424200 N

Energy required per stroke = Average shear force × Thickness of plate

=½ x FS × t1

=½ × 424200 × 18

=3817.8 × 103 N-mm

Energy required per min = Energy / stroke × No. of working strokes / min

=3817.8 × 103 × 25

=95.45 × 106 N-mm

=95 450 N-m

We know that

The power needed for the driving motor = Energy required per min / (60 × ηm)

=95450 / (60 × 0.95)

=1675 W

Dimensions for the rim cross-section

Considering the cross-section of the rim as rectangular and assuming the width of rim equal to twice the thickness of rim.

Let    t = Thickness of rim in metres, and

b = Width of rim in metres = 2 t.

Cross-sectional area of rim,

= b × t

=2 t × t

=2 t2

Since the punching operation takes place (i.e. energy is consumed) during 1/10 th of a revolution of the crank shaft, therefore during 9/10th of the revolution of a crank shaft, the energy is stored in the flywheel.

Maximum fluctuation of energy, ∆ E = 9/10 × Energy/stroke

=9/10 × 3817.8 × 103

=3436 × 103 N-mm

=436 N-m

Let m = Mass of the flywheel.

Since the hub and the spokes provide 5% of the rotational inertia of the wheel, therefore the maximum fluctuation of energy provided by the flywheel rim will be 95%.

Maximum fluctuation of energy provided by the rim, (∆ E)rim = 0.95 × ∆ E

=0.95 × 3436

=3264 N-m

Since the flywheel is to revolve at 9 times the speed of the crankshaft and there are 25 working strokes per minute, therefore mean speed of the flywheel,

= 9 × 25

= 225 r.p.m.

Mean angular speed, ω = 2 π × 225 / 60

We know that maximum fluctuation of energy (∆ E), 3264 = m.R22.CS

=m (0.7)2 (23.56)2 x 0.1

=27.2 m

= 3264 / 27.2

= 120 kg

We also know that mass of the flywheel (m),

= A × π D × ρ

=2 t2 × π × 1.4 × 7250

=63 782 t2

t2 = 120 / 63 782 = 0.001 88 or

t  = 0.044 m = 44 mm

= 2 t

= 2 × 44 = 88 mm

Check for centrifugal stress

We know that peripheral velocity of the rim,

= (π D N) / 60

=π ×1.4 ×225 / 60

=16.5 m/s

Centrifugal stress induced in the rim, σt = ρ.v2

=7250 (16.5)2

=1.97 × 106 N/m2

=1.97 MPa

Since the centrifugal stress induced in the rim is less than the permissible value (i.e. 6 MPa), therefore it is safe

A split type flywheel has outside diameter of the rim 1.80 m, inside diameter 1.35 m and the width 300 mm. the two halves of the wheel are connected by four bolts through the hub and near the rim joining the split arms and also by four shrink links on the rim. The speed is 250 r.p.m. and a turning moment of 15 kN-m is to be transmitted by the rim. Determine:

The diameter of the bolts at the hub and near the rim, σtb = 35 MPa.

The cross-sectional dimensions of the rectangular shrink links at the rim, σtl = 40MPa; w = 1.25 h.

The cross-sectional dimensions of the elliptical arms at the hub and rim if the wheel has six arms, σta = 15 MPa, minor axis being 0.5 times the major axis and the diameter of shaft being 150 mm. Assume density of the material of the flywheel as 7200 kg / m3.

Solution

Given :

D0 = 1.8 m Di = 1.35 m

b = 300 mm = 0.3 m N = 250 r.p.m.

T = 15 kN-m = 15 000 N-m σtb = 35 MPa = 35 N/mm2 σtl = 40 MPa = 40 N/mm2 w = 1.25 h

n = 6

b1 = 0.5 a1

σta = 15 MPa = 15 N / mm2 d1 = 150 mm

= 7200 kg / m3.

Diameter of the bolts at the hub and near the rim

Diameter of the bolts at the hub and near the rim

Let dc = Core diameter of the bolts in mm.

We know that mean diameter of the rim,

D = (Do + Di)/2

=(1.8 + 1.35) / 2

=1.575 m

Thickness of the rim,

= (Do - Di)/2

=(1.8 - 1.35) / 2

=0.225 m

Peripheral speed of the flywheel,

= (π D N) / 60

=π ×1.575 ×250 / 60

=20.6 m / s

We know that centrifugal stress (or tensile stress) at the rim, σt = ρ × v2

=7200 (20.6)2

=3.1 × 106 N/m2

=3.1 N/mm2

Cross-sectional area of the rim,

= b × t

=0.3 × 0.225

=0.0675 m2

Maximum tensile force acting on the rim

=σt × A

=3.1 × 106 × 0.0675

=209250N

……………………………………. (i)

We know that tensile strength of the four bolts

=(π/4) (dc)2 σtb× No. of bolts

=(π/4) (dc)2 x 35 x 4

=110(dc)2

………………………………………(ii)

Since the bolts are made as strong as the rim joint, therefore from equations (i) and (ii), we have (dc)2 = 209250 / 110

= 1903

dc = 43.6 mm

The standard size of the bolt is M 56 with dc = 48.65 mm.

2. Cross-sectional dimensions of rectangular shrink links at the rim

Let    h = Depth of the link in mm, and

w = Width of the link in mm = 1.25 h Cross-sectional area of each link,

Al = w × h

= 1.25 h2 mm2

We know that

Maximum tensile force on half the rim = 2 × σt for rim × Cross-sectional area of rim = 2 × 3.1 × 106 × 0.0675

= 418 500 N        ……….……………………..(iii)

Tensile strength of the four shrink links

=σtl × Al × 4

=40 × 1.25 h2 × 4

=200h2

………..……………………………..(iv)

From equations (iii) and (iv), we have

h2 = 418500 / 200 = 2092.5

h = 45.7 say 46 mm

and w  = 1.25 h

=1.25 × 46

=57.5 say 58 mm

Cross-sectional dimensions of the elliptical arms

Let    a1 = Major axis,

b1 = Minor axis = 0.5 a1 n = Number of arms = 6

Since the diameter of shaft (d1) is 150 mm and the diameter of hub (d) is taken equal to twice the diameter of shaft, therefore

d = 2 d1

=2 × 150

=300 mm = 0.3 m

We know that maximum bending moment on arms at the hub end,

=(T/Dn) (D-d)

=(15000 / 1.575 × 6) (1.575 – 0.3)

=2024 N-m

=2024 × 103 N-mm

Section modulus, Z     = (π / 32) × b1 (a1)2

=(π / 32)× 0.5 a1 (a1)2

=0.05 (a1)3

We know that bending stress for arms (σta),

= M / Z

=2024 x 103 / 0.05 (a1)3 (a1)3 = 40.5 × 106 / 15

=2.7 × 106

a1         = 139.3 say 140 mm

and b1  = 0.5 a1

0.5 × 140 =70 mm

A punching press pierces 35 holes per minute in a plate using 10 kN-m of energy per hole during each revolution. Each piercing takes 40 per cent of the time needed to make one revolution. The punch receives power through a gear reduction unit which in turn is fed by a motor driven belt pulley 800 mm diameter and turning at 210 r.p.m. Find the power of the electric motor if overall efficiency of the transmission unit is 80 per cent. Design a cast iron flywheel to be used with the punching machine for a coefficient of steadiness of 5, if the space considerations limit the maximum diameter to 1.3 m.

Allowable shear stress in the shaft material = 50 MPa Allowable tensile stress for cast iron = 4 MPa Density of cast iron = 7200 kg / m3

Solution.

Given :

No. of holes = 35 per min ;

Energy per hole = 10 kN-m = 10 000 N-m ; d = 800 mm = 0.8 m ;

N = 210 r.p.m. ; η = 80% = 0.8 ;

1/CS = 5 or CS = 1/5 = 0.2 ; Dmax = 1.3 m ;

τ = 50 MPa = 50 N/mm2 ;

σt = 4 MPa = 4 N/mm2 ; ρ = 7200 kg / m3

Power of the electric motor

We know that energy used for piercing holes per minute

=No. of holes pierced × Energy used per hole

=35 × 10 000

=350 000 N-m / min

Power needed for the electric motor, P       = Energy used per minute / (60 × η)

=350000 / (60 × 0.8)

=7292 W = 7.292 kW

Design of cast iron flywheel

First of all, let us find the maximum fluctuation of energy.

Since the overall efficiency of the transmission unit is 80%, therefore total energy to be supplied during each revolution,

ET = 10000/0.8

= 12 500 N-m

We know that velocity of the belt,

= π d.N

= π × 0.8 × 210

= 528 m/min

Net tension or pull acting on the belt = (Px 60) / v

=(7292 x 60) / 528

=828.6 N

Since each piercing takes 40 per cent of the time needed to make one revolution, therefore time required to punch a hole

=0.4 / 35

=0.0114 min

and the distance moved by the belt during punching a hole

=Velocity of the belt × Time required to punch a hole

=528 × 0.0114

=6.03 m

Energy supplied by the belt during punching a hole,

EB = Net tension × Distance travelled by belt

=828.6 × 6.03

=4996 N-m

Thus energy to be supplied by the flywheel for punching during each revolution or maximum fluctuation of energy,

E  = ET – EB

=12 500 – 4996

=7504 N-m

1. Mass of the flywheel

Let m = Mass of the flywheel rim.

Since space considerations limit the maximum diameter of the flywheel as 1.3 m ; therefore let us take the mean diameter of the flywheel,

D = 1.2 m or R = 0.6 m

We know that angular velocity

= 2 π N / 60

=(2 x π x 210) / 60

We also know that the maximum fluctuation of energy ( E), 7504 = m.R2. ω2.CS

=m x (0.6)2 x (22)2 x 0.2

=34.85 m

= 7504 / 34.85

= 215.3 kg

Cross-sectional dimensions of the flywheel rim

Let    t = Thickness of the flywheel rim in metres, and

b = Width of the flywheel rim in metres = 2 t

Cross-sectional area of the rim,

= b × t

=2 t × t

=2 t2

We know that mass of the flywheel rim (m),

= A × πD × ρ

=2 t2 × π × 1.2 × 7200

=54.3 × 103 t2

t2 = 215.3 / 54.3 × 103 = 0.00396

t  = 0.063 say 0.065 m = 65 mm

= 2 t

=2 × 65

=130 mm

Diameter and length of hub

Let      d = Diameter of the hub,

d1 = Diameter of the shaft, and l = Length of the hub.

First of all, let us find the diameter of the shaft (d1). We know that the mean torque transmitted by the shaft,

Tmean        = (P x 60) / 2 π N

=(7292 x 60) / 2π x 210

=331.5 N-m

Assuming that the maximum torque transmitted by the shaft is twice the mean torque, therefore maximum torque transmitted by the shaft,

Tmax  = 2 × Tmean

=2 × 331.5

=663 N-m

=663 × 103 N-mm

We know that maximum torque transmitted by the shaft (Tmax), 663 × 103

=(π/16) × τ x (d1)3

=(π/16) × 50 (d1)3

=9.82 (d1)3

(d1)3 = 663 × 103 / 9.82 = 67.5 × 103

d1 = 40.7 say 45 mm

The diameter of the hub (d ) is made equal to twice the diameter of the shaft (d1) and length of hub ( l ) is equal to the width of the rim (b).

= 2 d1

=2 × 45

90 mm = 0.09 m l = b = 130 mm

Cross-sectional dimensions of the elliptical cast iron arms

Let      a1 = Major axis,

b1 = Minor axis = 0.5 a1 n = Number of arms = 6

We know that the maximum bending moment in the arm at the hub end, which is assumed as cantilever is given by

= (T / R n) (R – r)

=(T / D n) (D – r)

=(663 / 1.2 x 6 ) (1.2 – 0.09) N-m

=102.2 N-m

=102200 N-mm

Section modulus for the cross-section of the arms,

= (π / 32) × b1 (a1)2

=(π / 32) × 0.5 a1 (a1)2

=0.05 (a1)3

We know that bending stress (σt),

= M / Z

=102200 / 0.05 (a1)3 (a1)3 = 2044 × 103 / 4

=511 × 103

a1 = 80 mm

and         b1  = 0.5 a1

= 0.5 × 80

= 40 mm

5. Dimensions of key

The standard dimensions of rectangular sunk key for a shaft of diameter 45 mm are as follows:

Width of key, w = 16 mm

Thickness of key t = 10 mm

The length of key (L) is obtained by considering the failure of key in shearing.

We know that maximum torque transmitted by the shaft (Tmax),

663 × 103   = L × w × τ × (d1/2)

= L × 16 × 50 × (45 / 2)

= 18 × 103 L

L       = 663 × 103/18 × 103

= 36.8 say 38 mm

L   = 663 × 103/18 × 103

= 36.8 say 38 mm

Let us now check the total stress in the rim which should not be greater than 4 MPa. We know that the velocity of the rim,

= π d.N / 60

=π 1.2 x 210 / 60

=13.2 m/s

Total stress in the rim, =1 .25 × 106 (0.75 + 1.26)

=2 .5 × 106 N/m2

=2 .5 MPa

Since it is less than 4 MPa, there fore the design is safe.

A shaft fitted with a flywhe el rotates at 250 r.p.m. and drives a machine . The torque of machine varies in a cyclic anner over a period of 3 revolutions. The torque rises from 750 N-m to 3000 N-m uniformly during 1 / 2 revolution and remains constant for the following revolution. It the n falls uniformly to 750 N-m during the next 1 / 2 revolution and remains constant for o ne revolution, the cycle being repeated there after. Determine the power required to drive the machine. If the total fluctuation of speed is not to exceed 3% of the mean speed, determine a suitable diameter and cross-sectio n of the flywheel rim. The width of the rim i s to be 4 times the thickness and the safe cen rifugal stress is 6 MPa. The material density may be assumed as 7200 kg / m3.

Given :

N = 250 r.p.m.

ω = 2 π × 250 / 60 = 26.2 rad/s ; ω1 – ω2 = 3% ω o r

=(ω1 – ω2) / ω = CS = 3% = 0.03 ; σt = 6 MPa = 6 × 106 N/m2 ;

ρ = 7200 kg / m3 We know that the torque required for one complete cycle

Area of figure OABCDEF

Area OA EF + Area ABG + Area BCHG + Area CDH

OF × O A + ½ × AG × BG + GH × CH + ½ × HD × CH

6 π × 750 + ½ × π (3000 – 750) + 2π (3000 – 750) + ½ × (3000 – 750)

=4500 π + 1125 π + 4500 π + 1125 π

= 11 250 π N-m

…………………… ………...(i)

If Tmean is the mean torque in N-m, then

Torque required for one complete cycle = Tmean × 6 π N-m  …………… ………....(ii)

From equations (i) and (ii),

Tmean  = 11250 π / 6 π

= 1875 N-m

We know that power required to drive the machine,

= Tmean × ω

=1875 × 26.2

=49125 W

=49.125 kW

Diameter of the flywheel

Let    D = Diameter of the flywheel in metres, and

v = Peripheral velocity of the flywheel in m/s.

We know that the centrifugal stress (σt), 6 ×106 = ρ × v2

= 7200 × v2

v2  = 6 × 106 / 7200

= 833.3

v   = 28.87 m/s

We also know that peripheral velocity of the flywheel (v),

= π D N / 60

=π D × 250 / 60

=13.1 D

D  = 28.87 / 13.1

= 2.2 m Cross-section of the flywheel rim

Let t = Thickness of the flywheel rim in metres, and b = Width of the flywheel rim in metres = 4 t

Cross-sectional area of the flywheel rim,

= b × t

4 t × t

4 t2 m2

First of all, let us find the maximum fluctuation of energy (∆ E) and mass of the flywheel rim (m). In order to find ∆E, we shall calculate the values of LM and NP. From similar triangles ABG and BLM,

LM / AG     = BM / BG

LM / π         = (3000 – 1857) / (3000 – 750)

= 0.5

LM    = 0.5 π

Now from similar triangles CHD and CNP,

NP / HD      = CN / CH

NP / π         = = (3000 – 1857) / (3000 – 750)

= 0.5

NP     = 0.5 π

BM = CN = 3000 – 1875

= 1125 N-m

Since the area above the mean torque line represents the maximum fluctuation of energy, therefore maximum fluctuation of energy,

E  = Area LBCP = Area LBM + Area MBCN + Area PNC

½ × LM × BM + MN × BM + ½ × NP × CN

½ × 0.5 π × 1125 + 2 π × 1125 + ½ × 0.5 π × 1125

=8837 N-m

We know that maximum fluctuation of energy (∆E), 8837 = m.R22.CS

=m (26.2)2 0.03

=24.9 m

= 8837 / 24.9

=355 kg

We also know that mass of the flywheel rim (m),

= A × π D × ρ

=4 t2 × π × 2.2 × 7200

=199 077 t2

t2  = 355 / 199077

=0.00178 or t = 0.042 m

=42 say 45 mm

And                   b   = 4 t

= 4 × 45 = 180 mm

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