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Republic of the Philippines Nueva Ecija University of Science and Technology Cabanatuan City

Problems and Solutions In ME 414D (Flywheel)

Submitted By: Jefferson Dancel Archie Molina Jun Ceazar Soriano

1. Flywheel turns 550 rev/min (RPM). Determine the magnitude of the normal acceleration of the flywheel point which are at a distance of 15 cm from the rotation axis. Solution: 𝑔=

15 550 2 (2𝜋 ) = 331.73 𝑚⁄𝑠 2 100 60

2. A cast iron flywheel with a mean diameter of 24 inches changes speed from 550 rpm to 530 while it gives up 5800 ft-lb of energy. What is the coefficient of fluctuation. Solution: Cf = V1 – V2 / Vave = N1 – N2 / Nave = 550 – 530 / (550+530/2) = 0.0370 answer

3. A mechanical press is used to punch 6 holes per minute is 25 mm diameter and the plates has an ultimate strength in shear of 420 Mpa. The normal operating speeed 200 rpm. And it slows down to 180 rpm during the process of punching. The flywheel has a mean diameter of one meter and the rim width is 3 times the thickness. Assume that the hub and arm account for 5% of the rim weight concentrated at the mean diameter and the density of cast iron is 7200 kg per cubic meter. Find the power in KW required to drive the press. Solution: E = ½ Ftp but : Ss = Sus F/A = 420 Mpa F/(𝜋Dtp) = 420 F = 420[𝜋Dtp] Then: E = ½(420)(𝜋)(D)(tp)2 = ½(420)(𝜋)(25)(25)2 = 10 308 351 N-mm = 10 308.351 Nm 1 𝑚𝑖𝑛 60 𝑠

t = 6 ℎ𝑜𝑙𝑒𝑠[𝑚𝑖𝑛] = 10 s/holes P = E/t =

10 308 10

= 1031 W power required

answer

4. Find the weight of the flywheel needed by a machine to punch 20.5 mm holes in 15.87 mm thick steel plate. The machine is to make 30 strokes per minute and a hole be punched every stroke, the hole is to be formed during 30 degrees rotation of the puncher crankshaft. A gear train with a ratio of 12 to 1 is to correct the flywheel shaft to

the crankshaft. Let mean diameter of a flywheel rim to the 91.44 cm the minimum flywheel speed is to be 90% of the maximum and assume mechanical efficiency of the machinr to be 80%. Assume an ultimate stress of 49000 psi. Solution: Solving for the energy needed by the process: E = ½ Ftp Where: F/A = 49000 psi 0.101325 F/(𝜋Dtp) = 49000[ ] 14.7 F = 337.75(𝜋Dtp) Then: E = ½(337.75)(𝜋)(D)(tp)2 E = ½(337.75)(𝜋)(20.5)(15.87)2 E = 2739.195 Nm Solving for the maximum speed of the flywheel: N1 = 30

𝑠𝑡𝑟𝑜𝑘𝑒𝑠 1 ℎ𝑜𝑙𝑒 𝑚𝑖𝑛

[

𝑠𝑡𝑟𝑜𝑘𝑒

360

][

30

][

1 𝑟𝑒𝑣 ℎ𝑜𝑙𝑒

] = 360 rpm

Solving for the minimum speed of the flywheel: N2 = 0.90(N1) = 0.90(360) = 324 rpm Solving for the weight of the flywheel: (based on energy equation) ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

ΔKE = -2739.195 N-m Since the flywheel will supply / release this energu, then: -2739.195 = Wf / 2(9.81) [( 𝜋 x 0.9144 x 324/60)2 – ( 𝜋 x 0.9144 x 360/60)2] Wf = 952.11 N or 97.05 kgf

answer

5. It is found that the shearing machine requires 205 joules ofenergy to shear a specific gauge of sheet metal. The mean diameter of the flywheel is to be 76.2 cm, the normal operating speed is 200 rpm and slow down to 180 rpm during shearing process. The rim width is 30.48 cm and the weigth of ast iron is 7,196.6 kg/m3. Find the thicknes of the

rim, assuming that the hub and arm account for 10% of the rim weight concentrated on the mean diameter. Solution: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-205 = Wf / 2(9.81) [( 𝜋 x 0.762 x 180/60)2 – ( 𝜋 x 0.762 x 180/60)2] Wf = 33.89 kg weight of the flywheel Solving for the weight of the rim: Wf = Wrim + WHA 33.89 = 1.10(Wrim)

where: WHA = 0.10 Wrim Wrim = 30.81 kg

Solving for the thickness of the rim, Wrim = ƴ(Vrim) = ƴ(𝜋Dmbt)

where : = 7196.6 kg/m3

30.81 = 7196.6 [𝜋(0.762)(0.3048)(t)] t = 5.867 mm

answer

6. A sheet metal working company purchase a shearing machine from a surplus dealer without a flywheel. It is calculated that the machine will use 2380 joules of energy to shear a 1.2 mm thick sheet metal. The flywheel to be used will have a mean diameter of 91.44 cm with a width of 25.4 cm. The normal operating speed is 180 rpm and slows down to 160 rpm during the shearing process. Assuming that the arms and the hub will account for 12% of the rim weight concentrated at the mean diameter and that the material density is 0.26 lb/cu. In. Compute for the weight of the flywheel. Solution: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1 V2 = 𝜋Dmn2 2380 = Wf / 2(9.81) [( 𝜋 x 0.9144 x 160/60)2 – ( 𝜋 x 0.9144 x 160/60)2]

Wf = 2996 N = 305 kgf answer

7. Calculate the moment of inertia, mass, for the circular flywheel of which the section. The material of the flywheel is steel, the density of which is 7800 kg.m 3. Solution: We can divide the area into three parts: I1, I2, and I3. I1 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – R14)

2 7800 𝑥 0.05 𝑥 𝜋

= x (0.1754 – 0.154) 2 = 0.264 kg.m2 I2 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – R14)

2 7800 𝑥 0.03 𝑥 𝜋

= x (0.154 – 0.1354) 2 = 0.064 kg.m2 I3 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – 0)

2 7800 𝑥 0.015 𝑥 𝜋

= x (0.1354 – 0) 2 = 0.061 kg.m2 The total moment of inertia is ITotal = 0.264 + 0.064 + 0.061 = 0.389 kg.m2 8. A solid cast-iron flywheel 12 in. in diameter, rotating at 2500 rpm and having a weight of 50 lbs., is used for a punch press. Calculate the mean torque obtained, if the flywheel's energy is restored during (1/4)th of a revolution. Solution: In a punch press energy is stored in the flywheel for a small portion of the cycle in the form of kinetic energy = (1/2)Iω2 where I = (1/2)MR2(For a solid circular disk.) = (1/2)(W / g)R2 lbs.-in.sec2. Where; W = weight and R = radius of flywheel. Now, this kinetic energy is restored in the form of work done by the flywheel during the punching operation. ∴Work done = Tmɸ Where; ɸ= Portion of flywheel's rotation during which the work is done = (π / 2) radians And, Tm = Mean torque obtained from flywheel rotation. Equating kinetic energy and work done by the flywheel

Tmɸ = (1/2)Iω2 Or, Tm = (1/2)[(Iω2) / ɸ]

(1)

Now, I = (1/2)(W/g)R2 g = 386 in./sec.2 = (1/2)[50 / (386)](6)2 = 2.33 lbs.-in.-sec2. And, ω(angular velocity) = [(2πN) / 60][(rads) / (sec)] or, ω = [{2π(2500)} / 60] = 262 rads/sec. Substituting values of I, ω and ɸ in equation (1) Tm = (1/2)[{2.33 × (262)2} / (π/2)] Tm = 50,910 in.-lb. The same torque is required to apply the brake to stop the flywheel during (1/4)th of a revolution

9. A flywheel used in a punch-press with a total energy requirement of 7000 lbs.-ft. is powered by an electric motor. The torque supplied by the motor is smooth and the torque required by the flywheel is rapidly varying. Both torques are shown in the loadtorque diagram, Figure 1. The motor rpm varies from 1100 to 1400 and the motor supplies (1/4) of its rpm to the flywheel. Calculate the mass-moment of inertia and the outside diameter of the flywheel if the rest of the flywheel's dimensions are in suitable proportions as shown in Figure 2. Assume density of the flywheel material, p = 0.289 lb./in.3 Solution:

MOMENT OF INERTIA: The moment of inertia of the flywheel If can be found by using the relation for change in rotational kinetic energy K.E. ROT, i.e., δK.E.ROT = (1/2) If(ωmax2 – ωmin2) From the load-torque diagram, the total energy required for punching = 7000 lbs.-ft.

(1)

Also, energy supplied by the motor for the punching stroke = 1910 × (π / 4) = 1500 lbs.-ft. Change in K.E. of the flywheel = Energy provided by the flywheel for punching operation = Total energy for punching – Energy provided by the motor = 7000 – 1500 = 5500 lbs.-ft. or δK.E. can also be found from the load-torque diagram as δK.E. = (1/4) × Area of approximate ΔPST = (1/4) × 14,000 × (π / 4) = 5500 lbs.-ft. Also maximum angular velocity of the flywheel ωmax = (1/4) × max. angular velocity of the motor = (1/4) × [(2πN) / 60] = (1/40 × [{2π(1400)} / 60] = 37 rads./sec. Minimum angular velocity of the flywheel = (1/4) × [{2π(1100)} / 60] = 29 rads./sec. Substituting all the above values back into eq.(l), 5500 = (1/2) If[(37)2 – (29)2] from which If = 21 lbs.-ft.-sec.2 OUTSIDE DIAMETER OF FLYWHEEL'Do’: Since polar moment of inertia of the flywheel is Jf = [{π(Do4 – Di4)} / 32] [for a hollow disk]

Also, mass moment of inertia of the flywheel If and polar moment of inertia are related together as, If = ebJf = eb[{π(Do4 – Di4)} / 32] (2) 3 Since, ρ = 0.289 lbs./in . (given) = 499 lbs./ft?.3 And, b = 0.15Do (from Fig. 2) Di = 0.75Do Further, If = 21 lbs.-ft.-sec.2 Substituting these in eq.(2) gives, 21 = (π / 32)(0.15Do)(499)[Do4 – (0.75Do)4] ⇒ Do5 = 4.169 or Do = 1.35 ft. 10. A 26 in. mean diameter cast-iron flywheel is used in a shearing machine. After performing the shearing operation at 160 rpm, the flywheel comes back to its normal speed of 190 rpm. The flywheel provides 1200 lbs.-ft. of energy for shearing a sheet

metal. Calculate the thickness of the rim if the width of the rim is 10 in. and arms and hub of the flywheel constitute 12% of the mass of the rim. Solution:

The thickness of the rim tr = (Vr / Ar)

Where,

Vr = Volume of rim Ar = Area of X-section of rim. (Fig. 1) To find Vr, the weight of the rim Wr should be known, Since the arm and the hub have 0.12Wr weight, ∴ Total weight of the flywheel = Wr + 0.12Wr = 1.12Wr Or, 1.12Wr = [{2g(δK.E.)} / (VN2 – VS2)] (2) Where, δK.E.= Loss of kinetic energy of the flywheel used in shearing operation = 1200 lbs.-ft. (given) VN = Normal working speed VS = Speed during shearing operation, VN = [(πDmN) / (12 × 60)] = 0.0044 × 26 × 190 = 22 ft./sec. VS = [(πDmN) / (12 × 60)] = 0.0044 × 26 × 160 = 18 ft./sec. Putting these parameters back into eq.(2) 1.12Wr = [(2 × 32 × 1200) / {(22)2 – (18)2}] ∴ Wr = 428.6 lbs. Specific weight of cast-iron = 0.26 [lb. / (in.3)] ∴ Vr = [Wr / (0.26)] = [(428.6) / (0.26)] = 1648.5 in.3 Also from Figure 1 Ar = w × L = w × (πDm) = 10 × (π × 26)

(1)

= 260π in.2 Putting the values of Vr and Ar in eq.(1) tr = [(1648.5) / (260π)] = 2 in. 11. A cast iron flywheel with a mean diameter of 75 inches changes speed from 550 rpm to 430 while it gives up 24000 ft-lb of energy. What is the coefficient of fluctuation.

Solution: Cf = V1 – V2 / Vave = N1 – N2 / Nave = 550 – 430 / (550+430/2) = 0.245

answer

12. A 38 in diameter spoked steel flywheel having a 12 in wide x 10 in deep rim rotates at 200 rpm. How long a cut (in inches) can be stamped in one inch thick aluminum plate if ultimate shearing strength of the aluminum is 40,000 lb/in2. During stamping, the force exerted by the stamo varies from a maximum F lb at the point of contact to zero lb when the stamp emerges from the metal. Neglect the weight of the flywheel and spokes and use 0.28 lb/in3 density for flywheel material. Solution: Solving for the mean diameter, Dm = D – t = 48 – 10 = 38 in Solving for the weight of the flywheel: Wf = Wrim + WHA Wf = ƴ(Vrim) = ƴ(𝜋Dmbt)

but : WHA = 0 where : ƴ= o.28 lb/in3

Wf =4011.19 lb Solving for the energy needed by the process: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1 Assuming that all the energy of the flywheel is released during the process, we have V2 = 0 4011.19 38 200 ΔKE = 2(32.2) [0 − (𝜋 𝑥 12 𝑥 60 )2] = -68493.41 ftlb Solving for the force,

E = ½ Ftp 68493.41 = ½ F(1/12) F 1 643 842 lb Solving for the length of the aluminum plate Ss = Sus F/A = 40 000 psi F/Lt = 40 000 1 643 842/L(1) = 40 000 L = 41.10 in answer

13. What would be the weight of a flywheel in kg if the weight of the rim is 3 times the sum of the weight of the hub and arms. Given the outside diameter and inside diameter to be 24 in and 18 in respectively and the rim width is 4.5 in. (assume steel flywheel) Solution: where : ƴ= 0.284 lb/in3 for steel

Wf = ƴ(Vrim) = ƴ(𝜋Dmbt) Dm = ½ (D + d) t = ½ (D – d) then: Wrim = ƴ[𝜋(

𝐷+𝑑 2

)(𝑏)(

= 0.284[𝜋(

24+18 2

𝐷−𝑑 2

)]

)(𝑏)(

24−18 2

)]

= 252.94 lb = 114.71 kgf Solving for the weight of the flywheel, Wf = Wrim + WHA

but : Wrim = 3 WHA or WHA = 1/3 (Wrim)

Wf = Wrim + 1/3 (Wrim) = 4/3 (114.71) = 152.95 kgf

answer

14. Find the rim thickness for a cast iron flywheel width of 200 mm, a mean diameter of 1.2 m, a normal operating speed of 300 rpm, a coefficient fluctuation of 0.05 and which is capable of hanging 3000 N-m of kinetic energy. Assume that the hub and arms represent 10% of therim weight and the specific weight of cast iron is 7200 kg/m 3.

Solution: Solve for the minimum speed: Cf = N1 – N2 / Nave = N1 – N2 / (N1 – N2 /2) 0.05 = 300 – N2 / (300 – N2 /2) N2 = 285 rpm The weigth of the flywheel: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

3000 = Wf / 2(9.81) [( 𝜋 x 1.2 x 285/60)2 – ( 𝜋 x 1.2 x 300/60)2] Wf = 1740.47 N = 177.42 kgf Weight of the rim: Wf = Wrim + WHA

but : WHA = 0.10(Wrim)

Wf = Wrim + 0.10(Wrim) Wf = 1.10(Wrim) 177.42 = 1.10(Wrim) Wrim = 161.29 kgf Solving for the rim thickness, Wrim = ƴ(V) = ƴ(𝜋Dmbt)

where : ƴ= 7200 kg/m3 for cart iron

161.29 = 7200[𝜋(1.2)(0.2)(t)] = 0.02971m t = 29.71 mm

15. It is found that the shearing machine requires 305 joules ofenergy to shear a specific gauge of sheet metal. The mean diameter of the flywheel is to be 86.2 cm, the normal operating speed is 300 rpm and slow down to 280 rpm during shearing process. The rim width is 40.48 cm and the weigth of ast iron is 8,196.6 kg/m3. Find the thicknes of the rim, assuming that the hub and arm account for 10% of the rim weight concentrated on the mean diameter. Solution:

ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-305 = Wf / 2(9.81) [( 𝜋 x 0.862 x 280/60)2 – ( 𝜋 x 0.862 x 300/60)2] Wf = 25.81 kg weight of the flywheel Solving for the weight of the rim: Wf = Wrim + WHA 25.81 = 1.10(Wrim)

where: WHA = 0.10 Wrim Wrim = 23.46 kg

Solving for the thickness of the rim, Wrim = ƴ(Vrim) = ƴ(𝜋Dmbt)

where : = 8196.6 kg/m3

23.46 = 8196.6 [𝜋(0.862)(0.4048)(t)] t = 2.611 mm

16. A flywheel has a mean diameter of 4 ft and is required to handle 2200 ft-lb of kinetic energy. The flywheel has a width of 8 in. Normal operating speed is 300 rpm and the coefficient of fluctuation is to be 0.05. find the weight of the rim assuming that the arms and hub are equivalent is 10% of the specific weight. Soluttion: Solving for the weight of the flywheel: (Note : N2 = 285 rpm) ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-2200 = Wf / 2(32.2) [( 𝜋 x 4 x 285/60)2 – ( 𝜋 x 4 x 300/60)2] Wf = 368.08 lb Solving for the weight of the rim, Wf = Wrim + WHA Wf = 1.10(Wrim) 368.08 = 1.10(Wrim)

= Wrim + 0.10(Wrim) where : WHA = 0.10(Wrim)

Wrim = 334.62 lb

answer

17. The mass of a flywheel is 175 kg and its radius of gyration is 380 mm. Find the torque required to attain a speed of 500 rpm from the rest in 30 seconds. Solution: Solving for the mass moment of inertia: Im = mk2 = 175(0.380)2 = 25.27 kgm2 Solving for the angular acceleration: α = Wf – Wo / t =

( 2𝜋 𝑥

500 )−0 60

30

= 1.745 red/s2 Solving for the torque: T = Im α = 25.27(1.75) = 44.10 N-m

answer

18. A cast iron flywheel is rotated at a speed of 1200 rpm and having a mean rim radius of 1 foot. If the weight of the rim is 30 lbs. What is the centrifugal force? Use factor C=41. Solution: V = 2 𝜋RN 1200 = 2 𝜋(1) 60 = 125.66 fps Solving for the centrifugal force: Fc = WV2/gR (30)(125.66) Fc = 32.2(1) 2 Fc = 14712 lb

19. A flywheel rim which weights 800 pounds has a mean diameter of 48 inches. The speed is to be maintained between 100 and 120 rpm. Considering that the effect of the arms and hub accounts for 12% of the rim weight, determine the capacity of the flywheel. Express your answer in ft-lb. Solution: V1 = (𝜋)DN

=

(𝜋)(48)(120) 12(60)

= 25.13 fps (𝜋)(48)(100) V2 = 12(60) = 20.94 fps ΔKE = 1.12 W [ V22 – V12]/ 2g = 1.12 (800) [ (25.13)2 – (20.94)2]/ 2(32.2) ΔKE = 2,685.68 ft-lbs

20. The moment of inertia of a flywheel is to be calculated for use in a horizontal pressing machine. The machine is operated by a belt driven pulley B of 42 in. diameter, at 300 rpm and a crankshaft S, requiring a work of 7500 lbs.-ft per cycle is driven by a gear G, which in turn is driven by a pinion P with speed reduction ratio of 7 between P and G.The mechanical efficiency of the crankshaft drive is 78% and the flywheel supplies energy for (1/3)rd of crankshaft revolution.

Solution:

The moment of inertia I of the flywheel is found by the equation, I = [{32.2(ΔU)} / (Crω2)] Where, ΔU= Change in energy of the flywheel required for pressing-operation ω = Mean angular speed Cr = Coefficient of speed regulation, = 0.25 (given).

(1)

Now, Where,

ΔU = ES – EB ES = Work required per crankshaft revolution EB = Work done by the belt during operating part of the cycle.

To find EB: Since the speed of pinion P = NP = 300 rpm Speed reduction ratio = r = 7 ∴ Speed of gear G = NG = (NP / 7) = [(300) / 7] = 42.86 rpm. Therefore the crankshaft S has the same rpm, ∴ NS = 42.86 rpm. Now, the horsepower required by the crankshaft to operate the press is given by H.P. = [(TNS) / (33,000)] ; where; T= Torque, lb.-ft. NS= rpm. = [(7500 × 42.86) / (33,000)] = 9.74 h.p. ∵ Mechanical friction ηm = 0.78 ∴ Actual horsepower required = [(9.74) / (0.78)] = 12.5 h.p. Now, the net tension (force) in the belt ’P' is found by the equation, H.P. = [(P × V) / (33,000)] Where, P = Force, lb. V = Speed, ft./min. The belt speed, V = [(πDN) / 12] = [(π × 42 × 300) / 12] = 3298.7 ft./min. ∴ P = 33,000 × [(H.P.) / V] = 33,000 × [(12.5) / (3298.7)] = 125 lbs. Now the distance, the belt traverses around the pulley during the pressing-operation period of the shaft revolution is, L = [{0.33 × 3298.7(ft./min.)} / {42.86(rev./min.)}] = 25.4 ft./rev. ∴ EB = P × L = 125 × 25.4 = 3175 lbs.-ft.

To find ES: The actual work required by crankshaft during pressing- operation is ES = [(7500) / (0.78)] = 9615.4 lbs.-ft. Putting the values of ES and EB in eq.(2)

(2)

ΔU = 9615.4 – 3175 = 6440.4 lbs.-ft. Now, ω = [(2πN) / 60] = [(2 × π × 300) / 60] N = Flywheel's rpm = 31.4 rads./sec. Therefore from equation (1), I = [(32.2 × 6440.4) / {0.25 × (31.4)2}] = 841.3 lbs.-ft.2

21. A 2.2 kw, 960 rpm motor powers the cam driven ram of a press through a gearing of 6:1 ratio. The rated capacity of the press is 20 kN and has a stroke of 200 mm. Assuming that the cam driven ram is capable of delivering the rated load at a constant velocity during the last 15% of a constant velocity stroke. Design a suitable flywheel that can maintain a coefficient of Speed fluctuation of 0.02. Assume that the maximum diameter of the flywheel is not to exceed 0.6m.

Solution: Work done by the press= U = 20 x 103 x 0.2 x 0.15 = 600 Nm Energy absorbed= work done= 600 Nm Mean torque on the shaft: 2.2 𝑥 103 = 21.88 𝑁𝑚 960 2 𝑥 𝜋 𝑥 60

Energy supplied= work don per cycle = 2𝜋 x 21.88 x 6 = 825 𝑁𝑚 Thus the mechanical efficiency of the system is = 𝜂=

600 825

Therefore the fluctuation in energy is =

= 0.727 = 72%

𝐸𝑘 = 𝐸𝑛𝑒𝑟𝑔𝑦 𝑎𝑏𝑠𝑜𝑟𝑏 − 𝐸𝑛𝑒𝑟𝑔𝑦 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 600 − 825 𝑥 0.075(21.88 𝑥 6 𝑥 𝜋 𝑥 0.15) 538.125 𝑁𝑚 𝐼=

𝐸𝑘 538.125 = 2 960 𝐶𝑓 (𝜔𝑎𝑣𝑔 ) 0.02(2𝜋 𝑥 60 )2 = 2.6622 𝑘𝑔 𝑚2 𝐼=

Assuming;

𝑟𝑖 𝑟𝑜

𝜋 𝑟 2 𝑥 (𝑟 − 𝑟𝑖 2 )𝑡 2 𝑔 𝑜

= 0.8 2.6622 =

𝜋 78500 (0.304 − 0.244 )𝑡 𝑥 2 9.86 = 59.805 𝑡

∴

𝑡=

2.6622 59.805

= 0.0445

𝑜𝑟 = 45 𝑚𝑚 𝜎𝑡 = 𝜎𝑡 =

𝑟 2 3+𝑦 1 + 3𝑦 2 𝜔 ( )(𝑟𝑖 2 + 𝑟𝑜 2 − 𝑟 ) 𝑔 8 3+𝑦

78500 2 3 + 0.3 1.9 𝜔 ( )(0.242 + 0.32 − 0.242 ) 9.81 8 3.3 𝜎𝑡 = 0.543(2𝜋 𝑥

960 2 ) 60

= 55667 𝑁/𝑚2 = 0.556 𝑀𝑃𝑎 Or if 𝜎𝑡 = 150 𝑀𝑃𝑎 150𝑥106 = 7961.4𝜔2 (0.4125)(0.0376)(0.090)(0.0331) = 0.548𝜔2 𝜔 = 16544 𝑟𝑎𝑑/𝑠𝑒𝑐 2

𝑁𝑂𝑆 =

𝜔𝑦𝑖𝑒𝑙𝑑 16544 = 𝜔 32𝜋

= 164.65 22. A flywheel used for energy storage. Find (a) the rotational kinetic energy stored in the flywheel, and (b) the energy storage capacity of the flywheel (W-hr/lbm).

Solution: (a) The rotational kinetic energy of the flywheel is, 𝐼𝜔2 (𝐾𝐸)𝑟𝑜𝑡 = 2𝑔𝑡 The mass moment of inertia of a solid circular cylinder rotating about its center, 𝑚𝑟 2 1 27 𝑖𝑛 2 1 𝑓𝑡 2 𝐼= = ( )(100 𝑙𝑏𝑚)( ) │( ) = 63.28 𝑓𝑡 2 − 𝑙𝑏𝑚 2 2 2 12 𝑖𝑛 Then,

(𝐾𝐸)𝑟𝑜𝑡 =

𝑟𝑒𝑣 2𝜋 𝑟𝑎𝑑 𝑚𝑖𝑛 (63.28 𝑓𝑡 2 − 𝑙𝑏𝑚)(40,000 𝑚𝑖𝑛)2 │( 𝑟𝑒𝑣 )2 (60 𝑠)2 𝑙𝑏𝑚 − 𝑓𝑡 2 (32.174 ) 𝑙𝑏𝑓 − 𝑠 2

= 1.73𝑥107 𝑓𝑡 − 𝑙𝑏𝑓

Converting to W-hr, (𝐾𝐸)𝑟𝑜𝑡 = (1.73𝑥107 𝑓𝑡 − 𝑙𝑏𝑓)│ (

1055.0𝐽 778.16𝑓𝑡−𝑙𝑏𝑓

)(

ℎ𝑟 3600𝑠

)(

𝑊−𝑠 𝐽

) = 6498 𝑊 − ℎ𝑟

(b) The energy storage capacity of the flywheel is the energy stored per unit mass, 𝑘𝑒𝑟𝑜𝑡 =

(𝐾𝐸)𝑟𝑜𝑡 6498𝑊 − ℎ𝑟 𝑊 − ℎ𝑟 = = 6.498 𝑚 100𝑙𝑏𝑚 𝑙𝑏𝑚

23. The punch press shown is driven by a motor which rotates the flywheel in the counter clockwise direction at 20 revolutions per minute. During the punch operation, the clutch engages, disconnecting the motor, allowing the flywheel to run the press through its large inertial properties. The total punch operation takes t1 = 1.5 seconds. After the punch operation, the flywheel rotates through a 1/2 revolution with a constant angular acceleration and then the motor engages accelerating the flywheel at 0.1 rad/s2 until it reaches its original angular speed of 20 rpm. The angular acceleration of the punch cycle is shown in the figure provided. Find the total punch cycle time t3.

Solution: Angular speed For any calculations involving angular speed or acceleration, it is always a good idea to convert their units to radians per second or radians per second squared, respectively. 𝜔0 = 20

𝑟𝑒𝑣 2𝜋𝑟𝑎𝑑 𝑚𝑖𝑛 2𝜋 𝑟𝑎𝑑 = 𝑚𝑖𝑛 𝑟𝑒𝑣 60𝑠 3 𝑠

Using the information provided regarding the angular acceleration, we can determine the angular speed at the end of the punch operation. 𝜔 =∝0 (𝑡 − 𝑡0 ) + 𝜔0 For this portion of the cycle, ∝𝑜 =∝0−1, 𝑡 − 𝑡𝑜 = 𝑡1, 𝑎𝑛𝑑 𝜔𝑜 = 𝜔0

𝜔1 = −0.2(1.5) +

2𝜋 𝑟𝑎𝑑 = 1.794 3 𝑠

Angular displacement After the punch operation the flywheel rotates through 0.5 revolutions. We will use this information plus that fact that it rotates at a constant angular acceleration to determine the time and angular speed at end of this stage. 1

𝜃 = ∝𝑜 (𝑡 − 𝑡𝑜 )2 + 𝜔𝑜 (𝑡 − 𝑡𝑜 ) + 𝜃𝑜

𝜔 =∝𝑜 (𝑡 − 𝑡𝑜 ) + 𝜔𝑜

2

For this portion of the cycle, ∝𝑜 =∝1−2, 𝑡 − 𝑡𝑜 = 𝑡2 − 𝑡1, 𝑎𝑛𝑑 𝜔𝑜 = 𝜔1 1

1

𝜃1−2 = 2 ∝1−2 (𝑡2 − 𝑡1 )2 + 𝜔1 (𝑡2 − 𝑡1 ) 0 = 𝑡22 − 38.88𝑡2 + 118.8

𝜋 = 2 (−0.1)(𝑡2 − 1.5)2 + 1.794(𝑡2 − 1.5) 𝑡2 =

38.88±√38.882 −4(118.8) 2

𝜔2 = 𝜔1−2 (𝑡2 − 𝑡1 ) + 𝜔1 = −0.1(3.35 − 1.5) + 1.794 = 1.61

= 3.35𝑠

𝑟𝑎𝑑 𝑠

Punch cycle time To determine the punch cycle time t3, we can again apply the constant acceleration equation. 𝜔 =∝𝑜 (𝑡 − 𝑡𝑜 ) + 𝜔 𝑜 For this portion of the cycle, ∝𝑜 =∝2−3, 𝑡 − 𝑡𝑜 = 𝑡2 − 𝑡3, 𝜔𝑜 = 𝜔2, 𝑎𝑛𝑑 𝜔 = 𝜔3 𝜔3 =∝2−3 (𝑡3 − 𝑡2 ) + 𝜔2

2𝜋 3

= 0.1(𝑡3 − 3.35) + 1.61

𝑡3 = 8.19𝑠

24. A flywheel rim which weights 800 pounds has a mean diameter of 48 inches. The speed is to be maintained between 100 and 120 rpm. Considering that the effect of the arms and hub accounts for 12% of the rim weight, determine the capacity of the flywheel. Express your answer in ft-lb. Solution: V1 = (𝜋)DN =

(𝜋)(48)(120) 12(60)

= 25.13 fps

V2 =

(𝜋)(48)(100) 12(60)

= 20.94 fps ΔKE = 1.12 W [ V22 – V12]/ 2g = 1.12 (800) [ (25.13)2 – (20.94)2]/ 2(32.2) ΔKE = 2,685.68 ft-lbs

25. A shearing machine requires 2000 ft-lbs of kinetic energy to operate. The mean diameter of its flywheel is 36 in. The rated speed is 66 fps at the mean diameter. The coefficient of fluctuation is 0.20. Compute the weight of the rim is needed if the effects of the flywheel arms and hub are to be neglected. Solution: ΔE = (Wf)(Cf)(V)2/g Wf = (ΔE)(g)/(cf)(v)2 (2000)(32.2) = 0.20(66) 2 = 73.92 lbs

answer

26. A cast iron flywheel with a mean diameter of 36 inches changes speed from 300 rpm to 280 while it gives up 8000 ft-lb of energy. What is the coefficient of fluctuation. Solution: Cf = V1 – V2 / Vave = N1 – N2 / Nave = 300 – 280 / (300+280/2) = 0.069 answer

27. The mass of the flywheel of a shearing machine is 3 tons and its radius of gyration is 0.55 m. At the beginning of its cutting stroke the speed of the flywheel is 110 rev/min. If 12 kJ of work is done in shearing a plate in 3 seconds, calculate the rotation speed of the flywheel at the end cutting stroke in rev/min. Solution: ΔKE = W ½ [3000(0.552)(11.522 – W22)] = 12000 W2 = 10.31 rad/s = 98.45 rpm answer

28. A flywheel weighing 457 kg has a radius of 375 mm. Compute how much energy, in N-m, does the flywheel loss from 3 rps to 2.8 rps? Solution: V1 = 2𝜋(0.375)(3) = 7.069 m/s V2 = 2𝜋(0.375)(2.8) = 6.597 m/s ΔKE = ½ [m(V1 – V2)] = 457[(7.069)2 – (6.597)2]/2 = 1473.91 Nm answer

29. A flywheel if manufactured with R2 = 0.2m and R1 = 0.15m, W = 30mm. If it is manufactured from aluminum, calculate its inertia. [density for aluminum is 2720 kg m3]. Solution: 𝜋 𝑥 2720 𝑥 0.03 If = (0.24 – 0.154) 2 = 0.14 kg.m2 answer

30. Repeat example 17, if the flywheel was: (a) Completely solid. What do you notice? (b) Made of carbon-steel. Solution: (a) Using the same formula, we can substitute a value of zero for R1: 𝜋 𝑥 2720 𝑥 0.03

If = (0.24 – 04) 2 = 0.205 kg.m2 (b) The density of carbon steel is 7850 kg/m3. Substituting in the formula gives: 𝜋 𝑥 7850 𝑥 0.03

If = (0.24 – 0.154) 2 = 0.404 kg.m2

31. Calculate the moment of inertia, mass, for the circular flywheel of which the section. The material of the flywheel is steel, the density of which is 7800 kg.m 3.

Solution: We can divide the area into three parts: I1, I2, and I3. I1 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – R14)

2 7800 𝑥 0.05 𝑥 𝜋

= x (0.1754 – 0.154) 2 = 0.264 kg.m2 I2 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – R14)

2 7800 𝑥 0.03 𝑥 𝜋

= x (0.154 – 0.1354) 2 = 0.064 kg.m2 I3 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – 0)

2 7800 𝑥 0.015 𝑥 𝜋

= x (0.1354 – 0) 2 = 0.061 kg.m2 The total moment of inertia is ITotal = 0.264 + 0.064 + 0.061 = 0.389 kg.m2 answer

32. A cast iron flywheel with a mean diameter of 36 inches changes speed from 300 rpm to 280 while it gives up 8000 ft-lb of energy. What is the coefficient of fluctuation. Solution: Cf = V1 – V2 / Vave = N1 – N2 / Nave = 300 – 280 / (300+280/2) = 0.0513 answer

33. A mechanical press is used to punch 8 holes per minute is 35 mm diameter and the plates has an ultimate strength in shear of 520 Mpa. The normal operating speed 200 rpm. And it slows down to 180 rpm during the process of punching. The flywheel has a mean diameter of one meter and the rim width is 3 times the thickness. Assume that the hub and arm account for 5% of the rim weight concentrated at the mean diameter and the density of cast iron is 7200 kg per cubic meter. Find the power in KW required to drive the press. Solution: E = ½ Ftp Then:

but : Ss = Sus F/A = 520 Mpa F/(𝜋Dtp) = 520 F = 520[𝜋Dtp]

E = ½(520)(𝜋)(D)(tp)2 = ½(520)(𝜋)(35)(25)2 = 17 867 808 N-mm = 17 867.808 Nm 1 𝑚𝑖𝑛 60 𝑠

t = 8 ℎ𝑜𝑙𝑒𝑠[𝑚𝑖𝑛] = 7.5 s/holes 17 868

P = E/t =

7.5

= 2382 W power required

answer

34. Find the weight of the flywheel needed by a machine to punch 30.5 mm holes in 25.87 mm thick steel plate. The machine is to make 40 strokes per minute and a hole be punched every stroke, the hole is to be formed during 30 degrees rotation of the puncher crankshaft. A gear train with a ratio of 12 to 1 is to correct the flywheel shaft to the crankshaft. Let mean diameter of a flywheel rim to the 91.44 cm the minimum flywheel speed is to be 90% of the maximum and assume mechanical efficiency of the machinr to be 80%. Assume an ultimate stress of 59000 psi. Solution: Solving for the energy needed by the process: E = ½ Ftp Where: F/A = 59000 psi 0.101325 F/(𝜋Dtp) = 59000[ 14.7 ] F = 406.68(𝜋Dtp) Then: E = ½(406.68)(𝜋)(D)(tp)2 E = ½(406.68)(𝜋)(30.5)(25.87)2 E = 1303.963 Nm Solving for the maximum speed of the flywheel: N1 = 40

𝑠𝑡𝑟𝑜𝑘𝑒𝑠 1 ℎ𝑜𝑙𝑒 𝑚𝑖𝑛

[

𝑠𝑡𝑟𝑜𝑘𝑒

360

][

30

][

1 𝑟𝑒𝑣 ℎ𝑜𝑙𝑒

] = 480 rpm

Solving for the minimum speed of the flywheel: N2 = 0.90(N1) = 0.90(480) = 432 rpm Solving for the weight of the flywheel: (based on energy equation) ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

ΔKE = -1303.963 N-m Since the flywheel will supply / release this energu, then: -1303.963 = Wf / 2(9.81) [( 𝜋 x 0.9144 x 432/60)2 – ( 𝜋 x 0.9144 x 480 /60)2] Wf = 254.95 N answer

35. A shearing machine requires 5000 ft-lbs of kinetic energy to operate. The mean diameter of its flywheel is 38 in. The rated speed is 58 fps at the mean diameter. The coefficient of fluctuation is 0.40. Compute the weight of the rim is needed if the effects of the flywheel arms and hub are to be neglected.

Solution: ΔE = (Wf)(Cf)(V)2/g Wf = (ΔE)(g)/(cf)(v)2 =

(5000)(32.2) 0.30(56)2

= 171.13 lbs

answer

36. It is found that the shearing machine requires 305 joules ofenergy to shear a specific gauge of sheet metal. The mean diameter of the flywheel is to be 86.2 cm, the normal operating speed is 300 rpm and slow down to 280 rpm during shearing process. The rim width is 40.48 cm and the weigth of ast iron is 8,196.6 kg/m3. Find the thicknes of the rim, assuming that the hub and arm account for 10% of the rim weight concentrated on the mean diameter. Solution: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-305 = Wf / 2(9.81) [( 𝜋 x 0.862 x 280/60)2 – ( 𝜋 x 0.862 x 300/60)2] Wf = 25.81 kg weight of the flywheel Solving for the weight of the rim:

Wf = Wrim + WHA 25.81 = 1.10(Wrim)

where: WHA = 0.10 Wrim Wrim = 23.46 kg

Solving for the thickness of the rim, Wrim = ƴ(Vrim) = ƴ(𝜋Dmbt)

where : = 8196.6 kg/m3

23.46 = 8196.6 [𝜋(0.862)(0.4048)(t)] t = 2.611 mm

answer

37. A sheet metal working company purchase a shearing machine from a surplus dealer without a flywheel. It is calculated that the machine will use 1380 joules of energy to shear a 1.2 mm thick sheet metal. The flywheel to be used will have a mean diameter of 11.44 cm with a width of 25.4 cm. The normal operating speed is 80 rpm and slows down to 60 rpm during the shearing process. Assuming that the arms and the hub will account for 12% of the rim weight concentrated at the mean diameter and that the material density is 0.26 lb/cu. In. Compute for the weight of the flywheel. Solution: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1 V2 = 𝜋Dmn2 1380 = Wf / 2(9.81) [( 𝜋 x 0.1144 x 60/60)2 – ( 𝜋 x 0.1144 x 80/60)2] Wf = 2695 N

answer

38. A 38 in diameter spoked steel flywheel having a 12 in wide x 10 in deep rim rotates at 200 rpm. How long a cut (in inches) can be stamped in one inch thick aluminum plate if ultimate shearing strength of the aluminum is 20,000 lb/in2. During stamping, the force exerted by the stamo varies from a maximum F lb at the point of contact to zero lb when the stamp emerges from the metal. Neglect the weight of the flywheel and spokes and use 0.28 lb/in3 density for flywheel material. Solution: Solving for the mean diameter, Dm = D – t = 48 – 10 = 38 in Solving for the weight of the flywheel:

Wf = Wrim + WHA Wf = ƴ(Vrim) = ƴ(𝜋Dmbt)

but : WHA = 0 where : ƴ= o.28 lb/in3

Wf =4011.19 lb Solving for the energy needed by the process: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1 Assuming that all the energy of the flywheel is released during the process, we have V2 = 0 4011.19 38 200 ΔKE = 2(32.2) [0 − (𝜋 𝑥 12 𝑥 60 )2] = -68493.41 ftlb Solving for the force, E = ½ Ftp 68493.41 = ½ F(1/12) F 1 643 842 lb Solving for the length of the aluminum plate Ss = Sus F/A = 20 000 psi F/Lt = 20 000 1 643 842/L(1) = 20 000 L = 82.192 in answer

39. The flywheel of a machine having weight of 4500 N and radius of gyration of 2 m has cyclic fluctuation of speed from 125 rpm to 120 rpm. Assuming g = 10m/s 2, the maximum fluctuation of energy is? Solution: Mass of flywheel = weight of flywheel/Acceleration due to gravity = 4500/10kg Moment of Inertia = mk2 = 1800 kgm2 ω1 = 2π/60 x 125rad/sec ω2 = 2π/60 x 120rad/sec Emax = 1/2 I(?)2 = 12087.2 N-m = 12100 Nm

40. A circular solid disc of uniform thickness 20 mm, radius 200 mm and mass 20 kg, is used as a flywheel. If it rotates at 600 rpm, the kinetic energy of the flywheel, in Joules is?

Solution: For flywheel K.E = 1/2Iω2 ω = 2πN/60 = 62.83 rad/s I (for solid circular disk) = 1/2mR2 = 0.4 kg m2 Hence, K.E = 790 Joules. 41. The speed of an engine varies from 210 rad/s to 190 rad/s. During the cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in kg/m2 is? Solution: Given ω1 = 210 rad/ sec, ω2 = 190 rad/ sec, ΔE= 400 Nm As the speed of flywheel changes from ω1 to ω2, the maximum fluctuation of energy, ΔE = 1/2I [(ω1)2 (ω2)2] I = 0.10 kgm2 42. For a certain engine having an average speed of 1200 rpm, a flywheel approximated as a solid disc, is required for keeping the fluctuation of speed within 2% about the average speed. The fluctuation of kinetic energy per cycle is found to be 2 kJ. What is the least possible mass of the flywheel if its diameter is not to exceed 1 m? Solution: Given N = 1200 rpm, ΔE = 2kJ = 2000 J, D = 1m, Cs = 0.02 Mean angular speed of engine, ω = 2πN/60 = 125.66 rad/ sec Fluctuation of energy of the flywheel is given by, ΔE = Iω2Cs = 1/2mR2ω2Cs For solid disc I = mR2/2 m = 51 kg 43. Table 16–6 lists values of the torque used to plot Fig. 16–28. The nominal speed of the engine is to be 250 rad/s. (a) Integrate the torque-displacement function for one cycle and find the energy that can be delivered to a load during the cycle. (b) Determine the mean torque Tm (see Fig. 16–28). (c) The greatest energy fluctuation is approximately between θ = 15◦ and θ = 150◦ on the torque diagram; see Fig. 16–28 and note that To = −Tm. Using a coefficient of speed fluctuation Cs = 0.1, find a suitable value for the flywheel inertia. (d) Find ω2 and ω1. Solution: (a) Using n = 48 intervals of ∆θ = 4π/48, numerical integration of the data of Table 16–6 yields E = 3368 in · lbf. This is the energy that can be delivered to the load.

(a) Tm =

3368 4𝜋

= 268 lbf · in

(b) The largest positive loop on the torque-displacement diagram occurs between θ = 0◦ and θ = 180◦. We select this loop as yielding the largest speed change. Subtracting 268 lbf · in from the values in Table 16–6 for this loop gives, respectively, −268, 2532, 1822, 2162, 1892, 1572, 1322, 942, 798, 535, 264, −84, and −268 lbf · in. Numerically integrating T − Tm with respect to θ yields E2 − E1 = 3531 lbf · in. We now solve Eq. (16–64) for I. This gives

I=

E2 − E1 Csω2

3531

= 0.1(250)2 = 0.565 lbf · s 2 in

(c) Equations (16–62) and (16–63) can be solved simultaneously for ω2 and ω1. Substituting appropriate values in these two equations yields ω2 =

ω 2

(2 + Cs) =

250 2

(2 + 0.1) = 262.5 rad/s

ω1 = 2ω − ω2 = 2(250) − 262.5 = 237.5 rad/s These two speeds occur at θ = 180◦ and θ = 0◦, respectively.

44. What would be the weight of a flywheel in kg if the weight of the rim is 6 times the sum of the weight of the hub and arms. Given the outside diameter and inside diameter to be 24 in and 18 in respectively and the rim width is 4.5 in. (assume steel flywheel) Solution:

where : ƴ= 0.284 lb/in3 for steel

Wf = ƴ(Vrim) = ƴ(𝜋Dmbt) Dm = ½ (D + d) t = ½ (D – d) then: Wrim = ƴ[𝜋(

𝐷+𝑑 2

)(𝑏)(

= 0.284[𝜋(

24+18 2

𝐷−𝑑 2

)]

)(𝑏)(

24−18 2

)]

= 252.94 lb = 114.71 kgf Solving for the weight of the flywheel, Wf = Wrim + WHA

but : Wrim = 6 WHA or WHA = 1/6 (Wrim)

Wf = Wrim + 1/6 (Wrim) = 7/6 (114.71) = 133.82 kgf

answer

45. Find the rim thickness for a cast iron flywheel width of 200 mm, a mean diameter of 1.2 m, a normal operating speed of 600 rpm, a coefficient fluctuation of 0.05 and which is capable of hanging 6000 N-m of kinetic energy. Assume that the hub and arms represent 10% of therim weight and the specific weight of cast iron is 5200 kg/m3. Solution: Solve for the minimum speed: Cf = N1 – N2 / Nave = N1 – N2 / (N1 – N2 /2) 0.05 = 300 – N2 / (300 – N2 /2) N2 = 285 rpm The weigth of the flywheel: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

6000 = Wf / 2(9.81) [( 𝜋 x 1.2 x 285/60)2 – ( 𝜋 x 1.2 x 600/60)2] Wf = -106.96 kgf

Weight of the rim: Wf = Wrim + WHA

but : WHA = 0.10(Wrim)

Wf = Wrim + 0.10(Wrim) Wf = 1.10(Wrim) 106.96 = 1.10(Wrim) Wrim = -97.23 kgf Solving for the rim thickness, where : ƴ= 5200 kg/m3 for cart iron

Wrim = ƴ(V) = ƴ(𝜋Dmbt)

-97.23 = 5200[𝜋(1.2)(0.2)(t)] = 0.02479m t = 24.79 mm

46. A flywheel has a mean diameter of 8 ft and is required to handle 1100 ft-lb of kinetic energy. The flywheel has a width of 8 in. Normal operating speed is 500 rpm and the coefficient of fluctuation is to be 0.05. find the weight of the rim assuming that the arms and hub are equivalent is 10% of the specific weight. Soluttion: Solving for the weight of the flywheel: (Note : N2 = 285 rpm) ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-1100 = Wf / 2(32.2) [( 𝜋 x 8 x 285/60)2 – ( 𝜋 x 8 x 5 00/60)2] Wf = 2.39 lb Solving for the weight of the rim, Wf = Wrim + WHA

= Wrim + 0.10(Wrim) where : WHA = 0.10(Wrim)

Wf = 1.10(Wrim) 2.39 = 1.10(Wrim) Wrim = 2.17 lb

answer

47. The mass of a flywheel is 115 kg and its radius of gyration is 280 mm. Find the torque required to attain a speed of 450 rpm from the rest in 25 seconds. Solution: Solving for the mass moment of inertia: Im = mk2 = 115(0.280)2 = 90.16 kgm2 Solving for the angular acceleration: α = Wf – Wo / t =

( 2𝜋 𝑥

450 )−0 60

25

= 1.885 red/s2 Solving for the torque: T = Im α = 90.16(1.885) = 169.95 N-m answer

48. A cast iron flywheel is rotated at a speed of 2400 rpm and having a mean rim radius of 3 foot. If the weight of the rim is 55 lbs. What is the centrifugal force? Use factor C=41. Solution: V = 2 𝜋RN 1200 = 2 𝜋(3) 60 = 753.98 fps Solving for the centrifugal force: Fc = WV2/gR (55)(753.98) Fc = 32.2(3) 2 Fc = 323672 lb

49. A flywheel rim which weights 450 pounds has a mean diameter of 14 inches. The speed is to be maintained between 150 and 170 rpm. Considering that the effect of the arms and hub accounts for 12% of the rim weight, determine the capacity of the flywheel. Express your answer in ft-lb. Solution: V1 = (𝜋)DN (𝜋)(14)(170) = 12(60)

= 10.38 fps (𝜋)(14)(150) V2 = 12(60) = 9.163 fps ΔKE = 1.12 W [ V22 – V12]/ 2g = 1.12 (450) [ (10.38)2 – (9.163)2]/ 2(32.2) ΔKE = 186.134 ft-lbs

50. A shearing machine requires 4000 ft-lbs of kinetic energy to operate. The mean diameter of its flywheel is 36 in. The rated speed is 56 fps at the mean diameter. The coefficient of fluctuation is 0.30. Compute the weight of the rim is needed if the effects of the flywheel arms and hub are to be neglected. Solution: ΔE = (Wf)(Cf)(V)2/g Wf = (ΔE)(g)/(cf)(v)2 (4000)(32.2) = 0.30(56) 2 = 136.90 lbs

answer

51. A cast iron flywheel with a mean diameter of 72 inches changes speed from 800 rpm to 680 while it gives up 16000 ft-lb of energy. What is the coefficient of fluctuation. Solution: Cf = V1 – V2 / Vave = N1 – N2 / Nave = 800 – 680 / (800+680/2) = 0.1621 answer

52. The mass of the flywheel of a shearing machine is 3 tons and its radius of gyration is 0.55 m. At the beginning of its cutting stroke the speed of the flywheel is 110 rev/min. If 12 kJ of work is done in shearing a plate in 3 seconds, calculate the rotation speed of the flywheel at the end cutting stroke in rev/min. Solution: ΔKE = W ½ [3000(0.552)(11.522 – W22)] = 12000 W2 = 10.31 rad/s = 98.45 rpm answer

53. A flywheel weighing 642 kg has a radius of 275 mm. Compute how much energy, in N-m, does the flywheel loss from 5 rps to 4.8 rps? Solution: V1 = 2𝜋(0.275)(5) = 8.64 m/s V2 = 2𝜋(0.275)(4.8) = 8.29 m/s ΔKE = ½ [m(V1 – V2)] = 642[(8.64)2 – (8.29)2]/2 = 1902.09 Nm answer

54. A flywheel if manufactured with R2 = 0.45 m and R1 = 0.17m, W = 30mm. If it is manufactured from aluminum, calculate its inertia. [density for aluminum is 3420 kg m3]. Solution: 𝜋 𝑥 3420 𝑥 0.03 If = (0.454 – 0.174) 2 = 6.47 kg.m2 answer

55. Repeat example 17, if the flywheel was: (c) Completely solid. What do you notice? (d) Made of carbon-steel. Solution: (c) Using the same formula, we can substitute a value of zero for R1: 𝜋 𝑥 3420 𝑥 0.03

If = (0.454 – 04) 2 = 6.609 kg.m2 (d) The density of carbon steel is 7850 kg/m3. Substituting in the formula gives: 𝜋 𝑥 7850 𝑥 0.03

If = (0.454 – 0.174) 2 = -30896284 kg.m2

56. Calculate the moment of inertia, mass, for the circular flywheel of which the section. The material of the flywheel is steel, the density of which is 7800 kg.m3. Solution:

We can divide the area into three parts: I1, I2, and I3. I1 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – R14)

2 7800 𝑥 0.05 𝑥 𝜋

= x (0.1754 – 0.154) 2 = 0.264 kg.m2 I2 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – R14)

2 7800 𝑥 0.03 𝑥 𝜋

= x (0.154 – 0.1354) 2 = 0.064 kg.m2 I3 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – 0)

2 7800 𝑥 0.015 𝑥 𝜋

= x (0.1354 – 0) 2 = 0.061 kg.m2 The total moment of inertia is ITotal = 0.264 + 0.064 + 0.061 = 0.389 kg.m2 answer

57. A cast iron flywheel with a mean diameter of 74 inches changes speed from 600 rpm to 480 while it gives up 24000 ft-lb of energy. What is the coefficient of fluctuation. Solution: Cf = V1 – V2 / Vave = N1 – N2 / Nave = 600 – 480 / (600+480/2) = 0.222 answer 58. A mechanical press is used to punch 12 holes per minute is 25 mm diameter and the plates has an ultimate strength in shear of 537 Mpa. The normal operating speeed 200 rpm. And it slows down to 180 rpm during the process of punching. The flywheel has a mean diameter of one meter and the rim width is 3 times the thickness. Assume that the hub and arm account for 5% of the rim weight concentrated at the mean diameter and the density of cast iron is 7200 kg per cubic meter. Find the power in KW required to drive the press. Solution: E = ½ Ftp but : Ss = Sus F/A = 537 Mpa F/(𝜋Dtp) = 537 F = 537[𝜋Dtp] Then: E = ½(537)(𝜋)(D)(tp)2 = ½(537)(𝜋)(25)(25)2

= 13 179 962 N-mm = 13 179.962 Nm 1 𝑚𝑖𝑛

60 𝑠

t = 12 ℎ𝑜𝑙𝑒𝑠[𝑚𝑖𝑛] = 5 s/holes P = E/t =

13 180 5

= 2636 W power required

answer

59. A typical 26-inch bicycle wheel rim has a diameter of 559 mm (22.0") and an outside tire diameter of about 26.2" (665 mm). For our calculation we approximate the radius - r - of the wheel to r = ((665 mm) + (559 mm) / 2) / 2 = 306 mm = 0.306 m The weight of the wheel with the tire is 2.3 kg and the inertial constant is k = 1. The Moment of Inertia for the wheel can be calculated I = (1) (2.3 kg) (0.306 m)2 = 0.22 kg m2 The speed of the bicycle is 25 km/h (6.94 m/s). The wheel circular velocity (rps, revolutions/s) - nrps - can be calculated as nrps = (6.94 m/s) / (2 π (0.665 m) / 2) = 3.32 revolutions/s The angular velocity of the wheel can be calculated as ω = (3.32 revolutions/s) (2 π rad/revolution) = 20.9 rad/s The kinetic energy of the rotating bicycle wheel can then be calculated to Ef = 0.5 (0.22 kg m2) (20.9 rad/s)2 = 47.9 J

60. Find the weight of the flywheel needed by a machine to punch 20.5 mm holes in 15.87 mm thick steel plate. The machine is to make 30 strokes per minute and a hole be

punched every stroke, the hole is to be formed during 30 degrees rotation of the puncher crankshaft. A gear train with a ratio of 12 to 1 is to correct the flywheel shaft to the crankshaft. Let mean diameter of a flywheel rim to the 91.44 cm the minimum flywheel speed is to be 90% of the maximum and assume mechanical efficiency of the machinr to be 80%. Assume an ultimate stress of 49000 psi. Solution: Solving for the energy needed by the process: E = ½ Ftp Where: F/A = 49000 psi 0.101325 F/(𝜋Dtp) = 49000[ 14.7 ] F = 337.75(𝜋Dtp) Then: E = ½(337.75)(𝜋)(D)(tp)2 E = ½(337.75)(𝜋)(20.5)(15.87)2 E = 2739.195 Nm Solving for the maximum speed of the flywheel: N1 = 30

𝑠𝑡𝑟𝑜𝑘𝑒𝑠 1 ℎ𝑜𝑙𝑒 𝑚𝑖𝑛

[

𝑠𝑡𝑟𝑜𝑘𝑒

360

][

30

][

1 𝑟𝑒𝑣 ℎ𝑜𝑙𝑒

] = 360 rpm

Solving for the minimum speed of the flywheel: N2 = 0.90(N1) = 0.90(360) = 324 rpm Solving for the weight of the flywheel: (based on energy equation) ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

ΔKE = -2739.195 N-m Since the flywheel will supply / release this energu, then: -2739.195 = Wf / 2(9.81) [( 𝜋 x 0.9144 x 324/60)2 – ( 𝜋 x 0.9144 x 360/60)2] Wf = 952.11 N or 97.05 kgf

answer

61. What would be the weight of a flywheel in kg if the weight of the rim is 3 times the sum of the weight of the hub and arms. Given the outside diameter and inside diameter to be 24 in and 18 in respectively and the rim width is 4.5 in. (assume steel flywheel)

Solution: where : ƴ= 0.284 lb/in3 for steel

Wf = ƴ(Vrim) = ƴ(𝜋Dmbt) Dm = ½ (D + d) t = ½ (D – d) then: Wrim = ƴ[𝜋(

𝐷+𝑑 2

)(𝑏)(

= 0.284[𝜋(

24+18 2

𝐷−𝑑 2

)]

)(𝑏)(

24−18 2

)]

= 252.94 lb = 114.71 kgf Solving for the weight of the flywheel, Wf = Wrim + WHA

but : Wrim = 3 WHA or WHA = 1/3 (Wrim)

Wf = Wrim + 1/3 (Wrim) = 4/3 (114.71) = 152.95 kgf

answer

62. Find the rim thickness for a cast iron flywheel width of 500 mm, a mean diameter of 4.2 m, a normal operating speed of 300 rpm, a coefficient fluctuation of 0.05 and which is capable of hanging 3000 N-m of kinetic energy. Assume that the hub and arms represent 10% of therim weight and the specific weight of cast iron is 3800 kg/m3. Solution: Solve for the minimum speed: Cf = N1 – N2 / Nave = N1 – N2 / (N1 – N2 /2) 0.05 = 300 – N2 / (300 – N2 /2) N2 = 285 rpm The weigth of the flywheel: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

3000 = Wf / 2(9.81) [( 𝜋 x 1.2 x 285/60)2 – ( 𝜋 x 1.2 x 300/60)2]

Wf = 1740.47 N = 177.42 kgf Weight of the rim: Wf = Wrim + WHA

but : WHA = 0.10(Wrim)

Wf = Wrim + 0.10(Wrim) Wf = 1.10(Wrim) 177.42 = 1.10(Wrim) Wrim = 161.29 kgf Solving for the rim thickness, where : ƴ= 3800 kg/m3 for cart iron

Wrim = ƴ(V) = ƴ(𝜋Dmbt)

161.29 = 7200[𝜋(4.2)(0.5)(t)] = 0.006434m t = 6.434 mm

63. It is found that the shearing machine requires 205 joules ofenergy to shear a specific gauge of sheet metal. The mean diameter of the flywheel is to be 76.2 cm, the normal operating speed is 200 rpm and slow down to 180 rpm during shearing process. The rim width is 30.48 cm and the weigth of ast iron is 7,196.6 kg/m3. Find the thicknes of the rim, assuming that the hub and arm account for 10% of the rim weight concentrated on the mean diameter. Solution: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-205 = Wf / 2(9.81) [( 𝜋 x 0.762 x 180/60)2 – ( 𝜋 x 0.762 x 180/60)2] Wf = 33.89 kg weight of the flywheel Solving for the weight of the rim: Wf = Wrim + WHA 33.89 = 1.10(Wrim)

where: WHA = 0.10 Wrim Wrim = 30.81 kg

Solving for the thickness of the rim, where : = 7196.6 kg/m3

Wrim = ƴ(Vrim) = ƴ(𝜋Dmbt) 30.81 = 7196.6 [𝜋(0.762)(0.3048)(t)] t = 5.867 mm

answer

64. A flywheel has a mean diameter of 4 ft and is required to handle 2200 ft-lb of kinetic energy. The flywheel has a width of 8 in. Normal operating speed is 300 rpm and the coefficient of fluctuation is to be 0.05. find the weight of the rim assuming that the arms and hub are equivalent is 10% of the specific weight. Soluttion: Solving for the weight of the flywheel: (Note : N2 = 285 rpm) ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-2200 = Wf / 2(32.2) [( 𝜋 x 4 x 285/60)2 – ( 𝜋 x 4 x 300/60)2] Wf = 368.08 lb Solving for the weight of the rim, Wf = Wrim + WHA

= Wrim + 0.10(Wrim) where : WHA = 0.10(Wrim)

Wf = 1.10(Wrim) 368.08 = 1.10(Wrim) Wrim = 334.62 lb

answer

65. It is found that the shearing machine requires 205 joules ofenergy to shear a specific gauge of sheet metal. The mean diameter of the flywheel is to be 76.2 cm, the normal operating speed is 200 rpm and slow down to 180 rpm during shearing process. The rim width is 30.48 cm and the weigth of ast iron is 7,196.6 kg/m3. Find the thicknes of the rim, assuming that the hub and arm account for 10% of the rim weight concentrated on the mean diameter. Solution: ΔKE = Wf / 2G [ V22 – V12]

Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-205 = Wf / 2(9.81) [( 𝜋 x 0.762 x 180/60)2 – ( 𝜋 x 0.762 x 180/60)2] Wf = 33.89 kg weight of the flywheel Solving for the weight of the rim: Wf = Wrim + WHA 33.89 = 1.10(Wrim)

where: WHA = 0.10 Wrim Wrim = 30.81 kg

Solving for the thickness of the rim, Wrim = ƴ(Vrim) = ƴ(𝜋Dmbt)

where : = 7196.6 kg/m3

30.81 = 7196.6 [𝜋(0.762)(0.3048)(t)] t = 5.867 mm

answer

66. A sheet metal working company purchase a shearing machine from a surplus dealer without a flywheel. It is calculated that the machine will use 2380 joules of energy to shear a 1.2 mm thick sheet metal. The flywheel to be used will have a mean diameter of 91.44 cm with a width of 25.4 cm. The normal operating speed is 180 rpm and slows down to 160 rpm during the shearing process. Assuming that the arms and the hub will account for 12% of the rim weight concentrated at the mean diameter and that the material density is 0.26 lb/cu. In. Compute for the weight of the flywheel. Solution: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1 V2 = 𝜋Dmn2 2380 = Wf / 2(9.81) [( 𝜋 x 0.9144 x 160/60)2 – ( 𝜋 x 0.9144 x 160/60)2] Wf = 2996 N = 305 kgf answer

67. A 38 in diameter spoked steel flywheel having a 12 in wide x 10 in deep rim rotates at 200 rpm. How long a cut (in inches) can be stamped in one inch thick aluminum plate if

ultimate shearing strength of the aluminum is 68,000 lb/in2. During stamping, the force exerted by the stamo varies from a maximum F lb at the point of contact to zero lb when the stamp emerges from the metal. Neglect the weight of the flywheel and spokes and use 0.28 lb/in3 density for flywheel material. Solution: Solving for the mean diameter, Dm = D – t = 48 – 10 = 38 in Solving for the weight of the flywheel: Wf = Wrim + WHA

but : WHA = 0

Wf = ƴ(Vrim) = ƴ(𝜋Dmbt)

where : ƴ= 0.28 lb/in3

Wf =4011.19 lb Solving for the energy needed by the process: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1 Assuming that all the energy of the flywheel is released during the process, we have V2 = 0 4011.19 38 200 ΔKE = 2(32.2) [0 − (𝜋 𝑥 12 𝑥 60 )2] = -68493.41 ftlb Solving for the force, E = ½ Ftp 68493.41 = ½ F(1/12) F 1 643 842 lb Solving for the length of the aluminum plate Ss = Sus F/A = 68 000 psi F/Lt = 68 000 1 643 842/L(1) = 68 000 L = 24.17 in answer 68. A flywheel weighing 639 kg has a radius of 375 mm. Compute how much energy, in N-m, does the flywheel loss from 3 rps to 2.8 rps? Solution: V1 = 2𝜋(0.375)(3) = 7.069 m/s V2 = 2𝜋(0.375)(2.8) = 6.597 m/s ΔKE = ½ [m(V1 – V2)] = 639[(7.069)2 – (6.597)2]/2 = 2060.89 Nm answer

69. A flywheel if manufactured with R2 = 0.135m and R1 = 0.148m, W = 30mm. If it is manufactured from aluminum, calculate its inertia. [density for aluminum is 3420 kg m3]. Solution: 𝜋 𝑥 3420 𝑥 0.03 If = (0.1354 – 0.1484) 2 = -0.0238 kg.m2 answer

70. Repeat example 17, if the flywheel was: (a)Completely solid. What do you notice? (b)Made of carbon-steel. Solution: (a)Using the same formula, we can substitute a value of zero for R1: 𝜋 𝑥 3420 𝑥 0.03

If = (0.1354 – 04) 2 = 0.535 kg.m2 (b)The density of carbon steel is 7850 kg/m3. Substituting in the formula gives: 𝜋 𝑥 7850 𝑥 0.03

If = (0.1354 – 0.1484) 2 = -0.0546 kg.m2

71. Calculate the moment of inertia, mass, for the circular flywheel of which the section. The material of the flywheel is steel, the density of which is 3500 kg.m 3. Solution: We can divide the area into three parts: I1, I2, and I3. I1 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – R14)

2 7800 𝑥 0.05 𝑥 𝜋

= x (0.1754 – 0.154) 2 = 0.119 kg.m2

I2 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – R14)

2 3500 𝑥 0.03 𝑥 𝜋

= x (0.154 – 0.1354) 2 = 0.0479 kg.m2 I3 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – 0)

2 3500 𝑥 0.015 𝑥 𝜋

= x (0.1354 – 0) 2 = 0.0913 kg.m2 The total moment of inertia is ITotal = 0.119 + 0.0479 + 0.0913 = 0.2582 kg.m2 answer

72. Flywheel turns 450 rev/min (RPM). Determine the magnitude of the normal acceleration of the flywheel point which are at a distance of 10 cm from the rotation axis. Solution: 𝑔=

10 450 2 (2𝜋 ) = 222.07 𝑚⁄𝑠 2 100 60

73. The torque (in N-m) exerted on the crank shaft of a two stroke engine can be described as T = 10000 + 1000 sin 2θ – 1200 cos 2θ, where θ is the crank angle as measured from inner dead center position. Assuming the resisting torque to be constant, the power (in kW) developed by the engine at 100 rpm is? Solution: T=10,000+1000sinθ−1200cos2θT=10,000+1000sin⁡θ−1200cos2θ its a function of 2θ2θ ,so 2θ2θ=360, θθ=180 T=o∫π10,000+1000sinθ−1200cos2θT=o∫π10,000+1000sin⁡θ−1200cos2θ =[10,000θ−1000cosθ−12002sin2θ]π0[10,000θ−1000cosθ−12002sin2θ]π0 =(10,000π+1000−0)−(0−1000−0)(10,000π+1000−0)−(0−1000−0) =10,000πN⋅m10,000πN⋅m Now, T=Tmean×πTmean×π so, Tmean=10,000N⋅mTmean=10,000N⋅m w=2πN/60=2π×100/60=10.4719rad/sec.w=2πN/60=2π×100/60=10.4719rad/sec.

P=Tmean×wP=Tmean×w = 104.719 KW 74. A flywheel is a rotating mechanical device used to store mechanical energy. When attached to a combined electric motor-generator, flywheels are a practical way to store excess electrical energy. Solar farms only generate electricity when it's sunny and wind turbines only generate electricity when it's windy. Combining energy sources like solar and wind with flywheel energy storage devices like a flywheel is one way to create a renewable energy system that is load balanced.

characteristic

value

shape

solid cylinder

material

4340 steel

density

7,850 kg/m3

diameter

100 cm

height

60 cm

max. energy

32 kWh

max. power

8 kW

dc voltage

800 V

Energy storage flywheel

a.

What is the mass of the flywheel?

b.

What is the top angular speed of the flywheel?

c. For how long could a fully charged flywheel deliver maximum power before it needed recharging?

d.

What is the average angular acceleration of the flywheel when it is being discharged? Solution: a. m = ρ(πr2h) m = (7,850 kg/m3)π(0.50 m)2(0.60 m) m = 3,700 kg 2

b. 𝜔 = 𝜔=

𝑅

𝐾

√𝑀

2 (32,000𝑊)(3,600𝑠) √ 0.50𝑚 3,700𝑘𝑔

𝜔 = 706

c. ∆𝑡 =

𝑟𝑎𝑑 = 112 𝑟𝑝𝑚 𝑠

∆𝑊 𝑃

32𝑘𝑊ℎ ∆𝑡 = 8𝑊 ∆𝑡 = 4ℎ d. ∝=

∆𝜔 ∆𝑡

𝑟𝑎𝑑 706 𝑠 ∝= 4𝑥3,600𝑠 𝑟𝑎𝑑 ∝= 0.049 2 𝑠

75. A flywheel has a mean diameter of 4 ft and is required to handle 2200 ft-lb of kinetic energy. The flywheel has a width of 8 in. Normal operating speed is 300 rpm and the coefficient of fluctuation is to be 0.05. find the weight of the rim assuming that the arms and hub are equivalent is 10% of the specific weight. Soluttion: Solving for the weight of the flywheel: (Note : N2 = 285 rpm) ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-2200 = Wf / 2(32.2) [( 𝜋 x 4 x 285/60)2 – ( 𝜋 x 4 x 300/60)2] Wf = 368.08 lb

Solving for the weight of the rim, Wf = Wrim + WHA

= Wrim + 0.10(Wrim) where : WHA = 0.10(Wrim)

Wf = 1.10(Wrim) 368.08 = 1.10(Wrim) Wrim = 334.62 lb

answer

76. Flywheel turns 600 rev/min (RPM). Determine the magnitude of the normal acceleration of the flywheel point which are at a distance of 17 cm from the rotation axis. Solution: 𝑔=

17 600 2 (2𝜋 ) = 671.13 𝑚⁄𝑠 2 100 60

77. A disk (flywheel) can rotate with maximum angular speed of 6000 revolutions per minute which can provide kinetic energy that allow a truck to move. When there is an average power needed that allow the truck to operate, the truck will be in motion for limited time as the energy eventually will be reduced with time and that will reduce the power (rate of energy consumed per unit time). Solution:

78. The mass of the flywheel of a shearing machine is 3.5 tons and its radius of gyration is 0.60 m. At the beginning of its cutting stroke the speed of the flywheel is 110 rev/min. If 12.5 kJ of work is done in shearing a plate in 3 seconds, calculate the rotation speed of the flywheel at the end cutting stroke in rev/min. Solution: ΔKE = W ½ [3500(0.602)(11.522 – W22)] = 12500 W2 = 11.52 rad/s = 105.55 rpm

79. The mass of a flywheel is 175 kg and its radius of gyration is 380 mm. Find the tourqe required to attain a speed of 500 rpm from rest in 30 seconds. Solution: Solving for the mass moment of inertia: 𝐼𝑚 = m𝑘 2 = 175(0.380)2 = 25.27 kg𝑚2

Solving for the angular acceleration: a=

𝑊𝑓 − 𝑊𝑜 𝑡

=

500 ) 60

(2π x

30

= 1.745𝑟𝑎𝑑/𝑠 2

Solving for the tourqe: T = 𝐼𝑚 a = 25.27(1.75) = 44.10 N-m

80. A cast iron flywheel is rotated at a speed of 1200 rpm and having a mean rim radius of 1 foot. If the weight of the rim is 30 lbs. What is the centrifugal force?

Solution: V = 2π (1)(1200/60) = 125.66 fps Solving for the centrifugal force: 𝐹𝑐 = 𝐹𝑐

𝑊𝑉 2 𝑔𝑅

=

(30)(125.66)2 32.2(1)

=14712 lb

81. The mean coil diameter of a helical-coil spring is 1 inch and a wire diameter of 1/8 inch. Compute the curve correction factor of the spring.

Solution: 𝐹𝑐 =

𝐾𝑤 𝐾𝑠

Where, 𝐾𝑤 =

4C−1 4C−4

D

+

0.615 C

1

where: C = d = 1/8 = 8 = 1.184

𝐾𝑠 =

2C+1 2C

= 1.0625

1.184

𝐾𝑐 = 1.0625 = 1.1144

82. A flywheel rim which weighs 800 pounds has a mean diameter of 48 inches. The speed is to be maintained between 100 and 120 rpm. Considering that the effect of the arms and hub accounts for 12% of the rim weight, determine the capacity of the flywheel. Express your answer in ft-lb.

Solution: 𝑉1 = (π)DN =

𝑉2 =

(π)(48)(120) 12(60)

(π)(48)(100)

KE =

12(60)

= 25.13 fps

= 20.94 fps

1.12W(𝑉1 )2 − (𝑉2)2 2𝑔

=

1.12(800)(25.13)2 − (20.94)2 2 (32.2)

KE = 2,685.68 ft-lbs

83. A cast iron flywheel with a mean diameter of 36 inches changes speed from 300 rpm to 280 while it gives up 8000 ft-lb of energy. What is the coefficient of fluctuation.

Solution:

𝐶𝑓 =

𝑉1 − 𝑉2 𝑉𝑎𝑣𝑒

=

𝑁1 − 𝑁2 𝑁𝑎𝑣𝑒

300−280

𝐶𝑓 = 300+280/2

𝐶𝑓 = 0.069

84. A 38 diameter spoked steel flywheel having a 12 wide x 10 in deep rim rotates at 200 rpm. How long a cut (in inches) can be stamped in one inch thick aluminium plate if ultimate shearing strength of the aluminium is 40,000 lb/in2 . During stamping, the force exerted by the stamp varies from a maximum F lb at the point of contact to zero lb when the stamp emerges from the metal. Neglect the weight of the flywheel and spokes and use 0.28 lb/in3 denstity for flywheel material. Solution: Solving for the mean diameter, 𝐷𝑚 = D – t = 48 – 10 = 38 in Solving for the weight of the flywheel: 𝑊𝑓 = 𝑊𝑟𝑖𝑚 + 𝑊𝐻𝐴 but: 𝑊𝐻𝐴 = 0 𝑊𝑓 = y [𝑉𝑟𝑖𝑚 ] = y (π𝐷𝑚 bt) where: y= 0.28 𝑙𝑏/𝑖𝑛3

𝑊𝑓 = 0.28[π(38)(12)(10)] = 4011.19 lb Solving for the energy needed by the process: KE =

𝑊𝑓 2g

[V22 – V12] where: V1 = πDmn1

Assuming that all the energy of flywheel is released during the process, we have V2 = 0 KE =

4011.19 2(32.2)

38

[0 – (π x12 x

200 2 )] 60

= -68493.41 ftlb

Solving for the force, 1

1

1

E = 2 ftp  68493.41 = 2 F (12) F = 1643842 lb

85. What would be the weight of a flywheel in kg if the weight of the rim is 3 times the sum of the weight of the hub and arms. Given the outside diameter and inside diameter to be 24 in and 18 in respectively and the rim width is 4.6 in. (assume steel flywheel). Solution: Solving for the weight of the rim, lb

Wrim = y (V) = y(πDmbt) where: y = 0.284 𝑖𝑛3 for steel 1

Dm = 2 (D + d) 1

T = 2 (D – d) Then: D+d D−d 24+18 24−18 Wrim = y [π ( )(b)( )] = 0.28[π ( )(4.5)( )] 2

2

2

2

Wrim = 252.94 lb = 114.71 kgf Solving for the weight of the flywheel, 1

Wf = Wrim + WHA where: Wrim = 3(WHA) or WHA = 3 (Wrim) 1

4

Wf = Wrim + 3 (Wrim) = 3 (114.71) = 152.95kgf

86. Find the rim thickness for a cast iron flywheel with a width of 200 mm, a mean diameter of 1.2 m, a normal operating speed of 300 rpm, a coefficient fluctuation of 0.05 and

which is capable of hanging 3000 N-m of kinetic energy. Assume that the hub and arms represent 10% of the rim weight and the specific weight of cast iron is 7200 kg/m 3 Solution: Solving for the minimum speed: 𝑁 −𝑁 𝑁 −𝑁 Cf = 1𝑁 2 = 𝑁11 + 𝑁22 𝑎𝑣𝑒

0.05 =

300− 𝑁2

2

300 + 𝑁2 2

N2 = 285rpm The weight of the flywheel: KE =

𝑊𝑓 2𝑔

[V22 – V12] where: V1 = 𝜋DmN1 and V2 = 𝜋DmN2

𝑊

𝑓 3000 = 2(9.81) [(𝜋x1.2 x

285 2 ) -(𝜋x1.2 60

x

300 2 )] 60

Wf = 1740.47N = 177.42kgf The weight of the rim: Wf = Wrim + WHA but: WHA = 0.10(Wrim) Wf = Wrim + 0.10 (Wrim) Wf = 1.10(Wrim) 177.42 = 1.10(Wrim) Wrim = 161.29kg, Wrim = y (V) = y (𝜋Dmbt) where: y = 7200 kg/m3 for cart iron 161.29 7200 [𝜋(1.2)(0.2)(t)] = 0.02971m

t = 29.71mm

87. A flywheel rim which weights 900 pounds has a mean diameter of 48 inches. The speed is to be maintained between 100 and 120 rpm. Considering that the effect of the arms and hub accounts for 12% of the rim weight, determine the capacity of the flywheel. Express your answer in ft-lb. Solution: V1 = (𝜋)DN

=

(𝜋)(48)(120) 12(60)

= 25.13 fps V2 =

(𝜋)(48)(100) 12(60)

= 20.94 fps ΔKE = 1.12 W [ V22 – V12]/ 2g = 1.12 (900) [ (25.13)2 – (20.94)2]/ 2(32.2) ΔKE = 3,021.40 ft-lbs

88. A cast iron flywheel is rotated at a speed of 3000 rpm and having a mean rim radius of 3.5 foot. If the weight of the rim is 60 lbs. What is the centrifugal force? Use factor C=41. Solution: V = 2 𝜋RN 1500 = 2 𝜋(3.5) 60 = 549.78 fps Solving for the centrifugal force: Fc = WV2/gR Fc =

(60)(549.78) 2 32.2(3.5)

Fc = 85670.74 lb

89. A cast iron flywheel with a mean diameter of 36 inches changes speed from 300 rpm to 260 while it gives up 8000 ft-lb of energy. What is the coefficient of fluctuation. Solution: Cf = V1 – V2 / Vave

= N1 – N2 / Nave = 300 – 280 / (300+260/2) = 0.071 90. The mass of the flywheel of a shearing machine is 3 tons and its radius of gyration is 0.55 m. At the beginning of its cutting stroke the speed of the flywheel is 110 rev/min. If 12 kJ of work is done in shearing a plate in 3 seconds, calculate the rotation speed of the flywheel at the end cutting stroke in rev/min. Solution: ΔKE = W ½ [3000(0.552)(11.522 – W22)] = 12000 W2 = 10.31 rad/s = 98.45 rpm

91. A flywheel weighing 457 kg has a radius of 375 mm. Compute how much energy, in Nm, does the flywheel loss from 3 rps to 2.8 rps? Solution: V1 = 2π(0.375)(3) = 7.069 m/s V2 = 2π(0.375)(2.8) = 6.597 m/s ΔKE = ½ [m(V1 – V2)] = 457[(7.069)2 – (6.597)2]/2 = 1473.91 Nm

92. Calculate the moment of inertia, mass, for the circular flywheel of which the section. The material of the flywheel is steel, the density of which is 7800 kg.m 3. Solution: We can divide the area into three parts: I1, I2, and I3.

I1 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – R14)

2 7800 𝑥 0.05 𝑥 𝜋

= x (0.1754 – 0.154) 2 = 0.264 kg.m2 I2 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – R14)

2 7800 𝑥 0.03 𝑥 𝜋

= x (0.154 – 0.1354) 2 = 0.064 kg.m2 I3 =

𝑝𝑥𝑤𝑥𝜋

x (R24 – 0)

2 7800 𝑥 0.015 𝑥 𝜋

= x (0.1354 – 0) 2 = 0.061 kg.m2 The total moment of inertia is ITotal = 0.264 + 0.064 + 0.061 = 0.389 kg.m2

93. A mechanical press is used to punch 8 holes per minute is 35 mm diameter and the plates has an ultimate strength in shear of 520 Mpa. The normal operating speed 200 rpm. And it slows down to 180 rpm during the process of punching. The flywheel has a mean diameter of one meter and the rim width is 3 times the thickness. Assume that the hub and arm account for 5% of the rim weight concentrated at the mean diameter and the density of cast iron is 7200 kg per cubic meter. Find the power in KW required to drive the press. Solution: E = ½ Ftp Then:

but : Ss = Sus F/A = 520 Mpa F/(𝜋Dtp) = 520 F = 520[𝜋Dtp]

E = ½(520)(𝜋)(D)(tp)2 = ½(520)(𝜋)(35)(25)2 = 17 867 808 N-mm = 17 867.808 Nm 1 𝑚𝑖𝑛 60 𝑠

t = 8 ℎ𝑜𝑙𝑒𝑠[𝑚𝑖𝑛] = 7.5 s/holes

P = E/t =

17 868 7.5

= 2382 W power required

94. Find the weight of the flywheel needed by a machine to punch 30.5 mm holes in 25.87 mm thick steel plate. The machine is to make 40 strokes per minute and a hole be punched every stroke, the hole is to be formed during 30 degrees rotation of the puncher crankshaft. A gear train with a ratio of 12 to 1 is to correct the flywheel shaft to the crankshaft. Let mean diameter of a flywheel rim to the 91.44 cm the minimum flywheel speed is to be 90% of the maximum and assume mechanical efficiency of the machinr to be 80%. Assume an ultimate stress of 59000 psi. Solution: Solving for the energy needed by the process: E = ½ Ftp Where: F/A = 59000 psi 0.101325 F/(𝜋Dtp) = 59000[ 14.7 ] F = 406.68(𝜋Dtp) Then: E = ½(406.68)(𝜋)(D)(tp)2 E = ½(406.68)(𝜋)(30.5)(25.87)2 E = 1303.963 Nm Solving for the maximum speed of the flywheel: N1 = 40

𝑠𝑡𝑟𝑜𝑘𝑒𝑠 1 ℎ𝑜𝑙𝑒 𝑚𝑖𝑛

[

𝑠𝑡𝑟𝑜𝑘𝑒

360

][

30

][

1 𝑟𝑒𝑣 ℎ𝑜𝑙𝑒

] = 480 rpm

Solving for the minimum speed of the flywheel: N2 = 0.90(N1) = 0.90(480) = 432 rpm Solving for the weight of the flywheel: (based on energy equation) ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

ΔKE = -1303.963 N-m Since the flywheel will supply / release this energu, then: -1303.963 = Wf / 2(9.81) [( 𝜋 x 0.9144 x 432/60)2 – ( 𝜋 x 0.9144 x 480 /60)2]

Wf = 254.95 N

95. It is found that the shearing machine requires 305 joules ofenergy to shear a specific gauge of sheet metal. The mean diameter of the flywheel is to be 86.2 cm, the normal operating speed is 300 rpm and slow down to 280 rpm during shearing process. The rim width is 40.48 cm and the weigth of ast iron is 8,196.6 kg/m3. Find the thicknes of the rim, assuming that the hub and arm account for 10% of the rim weight concentrated on the mean diameter. Solution: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-305 = Wf / 2(9.81) [( 𝜋 x 0.862 x 280/60)2 – ( 𝜋 x 0.862 x 300/60)2] Wf = 25.81 kg weight of the flywheel Solving for the weight of the rim: Wf = Wrim + WHA 25.81 = 1.10(Wrim)

where: WHA = 0.10 Wrim Wrim = 23.46 kg

Solving for the thickness of the rim, Wrim = ƴ(Vrim) = ƴ(𝜋Dmbt)

where : = 8196.6 kg/m3

23.46 = 8196.6 [𝜋(0.862)(0.4048)(t)] t = 2.611 mm

96. A 38 in diameter spoked steel flywheel having a 12 in wide x 10 in deep rim rotates at 200 rpm. How long a cut (in inches) can be stamped in one inch thick aluminum plate if ultimate shearing strength of the aluminum is 20,000 lb/in2. During stamping, the force exerted by the stamo varies from a maximum F lb at the point of contact to zero lb when the stamp emerges from the metal. Neglect the weight of the flywheel and spokes and use 0.28 lb/in3 density for flywheel material. Solution: Solving for the mean diameter, Dm = D – t = 48 – 10 = 38 in

Solving for the weight of the flywheel: Wf = Wrim + WHA

but : WHA = 0 where : ƴ= o.28 lb/in3

Wf = ƴ(Vrim) = ƴ(𝜋Dmbt) Wf =4011.19 lb

Solving for the energy needed by the process: ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1 Assuming that all the energy of the flywheel is released during the process, we have V2 = 0 4011.19 38 200 ΔKE = 2(32.2) [0 − (𝜋 𝑥 12 𝑥 60 )2] = -68493.41 ftlb Solving for the force, E = ½ Ftp 68493.41 = ½ F(1/12) F 1 643 842 lb Solving for the length of the aluminum plate Ss = Sus F/A = 20 000 psi F/Lt = 20 000 1 643 842/L(1) = 20 000 L = 82.192 in

97. What would be the weight of a flywheel in kg if the weight of the rim is 6 times the sum of the weight of the hub and arms. Given the outside diameter and inside diameter to be 24 in and 18 in respectively and the rim width is 4.5 in. (assume steel flywheel) Solution: where : ƴ= 0.284 lb/in3 for steel

Wf = ƴ(Vrim) = ƴ(𝜋Dmbt) Dm = ½ (D + d) t = ½ (D – d) then: Wrim = ƴ[𝜋(

𝐷+𝑑 2

)(𝑏)(

= 0.284[𝜋(

24+18 2

𝐷−𝑑 2

)]

)(𝑏)(

24−18 2

)]

= 252.94 lb = 114.71 kgf Solving for the weight of the flywheel, Wf = Wrim + WHA

but : Wrim = 6 WHA or WHA = 1/6 (Wrim)

Wf = Wrim + 1/6 (Wrim) = 7/6 (114.71) = 133.82 kgf

98. A flywheel has a mean diameter of 8 ft and is required to handle 1100 ft-lb of kinetic energy. The flywheel has a width of 8 in. Normal operating speed is 500 rpm and the coefficient of fluctuation is to be 0.05. find the weight of the rim assuming that the arms and hub are equivalent is 10% of the specific weight. Soluttion: Solving for the weight of the flywheel: (Note : N2 = 285 rpm) ΔKE = Wf / 2G [ V22 – V12] Where: V1 = 𝜋Dmn1

V2 = 𝜋Dmn2

-1100 = Wf / 2(32.2) [( 𝜋 x 8 x 285/60)2 – ( 𝜋 x 8 x 5 00/60)2] Wf = 2.39 lb Solving for the weight of the rim, Wf = Wrim + WHA

= Wrim + 0.10(Wrim) where : WHA = 0.10(Wrim)

Wf = 1.10(Wrim) 2.39 = 1.10(Wrim) Wrim = 2.17 lb

99. A cast iron flywheel is rotated at a speed of 2400 rpm and having a mean rim radius of 3 foot. If the weight of the rim is 55 lbs. What is the centrifugal force? Use factor C=41. Solution: V = 2 𝜋RN

1200

= 2 𝜋(3) 60 = 753.98 fps Solving for the centrifugal force: Fc = WV2/gR Fc =

(55)(753.98) 2 32.2(3)

Fc = 323672 lb 100. A cast iron flywheel with a mean diameter of 36 inches changes speed from 320 rpm to 300 while it gives up 8000 ft-lb of energy. What is the coefficient of fluctuation.

Solution:

𝐶𝑓 =

𝑉1 − 𝑉2 𝑉𝑎𝑣𝑒

=

320−300

𝑁1 − 𝑁2

𝐶𝑓 = 320+300/2

𝐶𝑓 = 0.043

𝑁𝑎𝑣𝑒