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6. windmill gear assembly

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windmill gear assembly-DESING OF MACHINE ELEMENTS

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6. windmill gear assembly

  1. 1. Group Members: PC Mubashar Sharif (2009-462) NS Abdullah Bin Masood (2009-204) NS Haider Iqbal (2009- ) Dated: 20 January 2014 Windmill Gear Assembly Design of Machine Elements Sir Raja Amer Azeem
  2. 2. Design of Machine elements 1 Wind Mill Actually the idea of our project is to design different components used in the wind mill pump which includes shaft on which the fan is mounted, bearings in which shaft is mounted and after that gears and the process which is usually used for joining of two components on a single point/place. A brief introduction: Wind energy is the actually the conversion of wind energy into a some kind of useful form of energy, such as using wind turbines to make electricity, wind mills for mechanical power, wind pumps for pumping water or drainage, or sails to propel ships as described in the applications. It works when wind hits the windmills blade it spins the rotor and the gearbox and the mechanic box activate the rotor and the wires inside take all the wind and make it into energy and the electricity poles carry the electricity and powers and gives 1000 houses in an hour or so. Here we are describing about the windmill pump. Windmill for pumping the water: Actually windmill pump is a device that pumps the water from different surfaces. Water is pumped to the surface when the pump rod raises the piston. When piston check valve closes and holds the water above the piston. As the piston rises, water is moved up the pipe towards the surface. Water is also drawn into the lower section of the pump cylinder through a screen and the lower check valve. When the pump rod reverses and begins to descend, the lower check valve closes and the piston check valve opens allowing the water in the cylinder to pass through the piston check valve and become trapped above the piston when the check valve closes. This cycle is constantly repeated as the wind wheel turns, operating the reciprocating mechanism in the gearbox, which operates the pump rod and pump
  3. 3. Design of Machine elements 2 The gear assembly we are trying to design and fabricate is an assembly which has 2 gears of different diameters and number of teeth, fixed mounted on the two different shafts of same geometry. The two gears are meshed in such a way with a certain input on one shaft you get a different speed (rpm) and torque on the other. For this purpose, the two shafts have been mounted on two same bearing (since both of the shafts are same) and fixed in a support. In our project, following are the major components that need to be designed: - Gears - Shafts - Bearings - Welding We are a group of three members and have divided the tasks such that each member is responsible for one component. The name of member and his respective job/task is mentioned below: Abdullah Bin Masood: Designing and calculations of Gears Mubashar Sharif: Designing and calculations of Shafts Haider Iqbal: Calculations and selection of Bearings First of all the calculations for the shaft have been made and finalized and then before fabrication of shaft we went to buy the bearings of the appropriate size that could fit mount on our shaft but after moving in the market we found out that the bearings are not available in every, every single size we design. There are in fact, some certain specific sizes of the bearings. Same happened to our case. Our shaft was basically much smaller than the bearings size available so we had to revise our design and with everything same but the Factor of safety increased significantly due to over-designing of the shaft. However, in order to keep the report brief and to the point, only finalized calculations have been included.
  4. 4. Design of Machine elements 3 Design Layout/Modeling:
  5. 5. Design of Machine elements 4 Wind Power Calculation: Shaft is basically a mechanical component that transmits power. We know that from wind the available energy to the pump is 𝑷 π’‚π’—π’‚π’Šπ’π’‚π’—π’π’†= 𝟏 𝟐 ƿɛ𝑨𝑽 πŸ‘ 𝑉 𝑀𝑖𝑛𝑑=20π‘š/𝑠𝑒𝑐 Diameter of shaft=1cm 𝜌 = 1.225π‘˜π‘”/π‘š3 πœ€ = 0.59 By putting values in above equation we get π‘ƒπ‘Žπ‘£π‘Žπ‘–π‘™π‘Žπ‘£π‘™π‘’= 190.9562 π‘€π‘Žπ‘‘π‘‘π‘  Assumptions Blade tip speed = 1/√2 Γ—wind speed V=rΟ‰ by putting values we get Ο‰=1308.917 rev/min N=108 (for infinite cycles assumption) Calculations of Shafts: Shaft is basically a mechanical component that transmits power. In our assembly two shafts are being used. Since both are same therefore we will make calculations for one shaft only. Assuming the following constraints/given condition for our system: Power to be transmitted: π‘ƒπ‘Žπ‘£π‘Žπ‘–π‘™π‘Žπ‘£π‘™π‘’= 190.95 π‘€π‘Žπ‘‘π‘‘π‘  Angular velocity: Ο‰=1308.917 rev/min Ο‰=136.97 rad/sec
  6. 6. Design of Machine elements 5 Designing for the infinite life: N=108 (since shafts will be continuously subject to fatigue in our case) Diameter of shaft= 0.7 cm The material chosen is CD 1020 steel Following the iterations approach: 1 β†’ P=TΟ‰ Torque=T=190.95/136.97 T = 1.39 Nm 2 β†’ Shear stress=S=16T/πœ‹π‘‘3 After putting values S =20.65 Mpa Now since, 3 →𝑺 𝒆=𝑲𝒂×𝑲𝒃×𝑲𝒄×𝑲𝒅×𝑲𝒆×𝑲𝒇×𝑺𝒆′ 𝑆 𝑒𝑑 = 470π‘€π‘π‘Ž 𝑆 𝑦=390Mpa 𝑆𝑒′ = 0.5𝑆 𝑒𝑑 = 0.5Γ— 470 = 235π‘€π‘π‘Ž Ka=a𝑆 𝑒𝑑 𝑏 =4.51Γ— 470βˆ’0.265 = 0.8832 Kb=1.24(1 Γ— 10βˆ’2 )βˆ’0.1.07 =0.969 Kc=0.59 for torsion Kd=Ke=Kf=1 Se=0.8832Γ—0.969Γ—0.59Γ—235 Se=118.66Mpa 4β†’a= [𝒇 Γ— (𝑺 𝒖𝒕)]Λ†πŸ 𝑺 𝒆 f=0.9 b= βˆ’1 3 Γ— log(𝑓𝑆𝑒𝑑/𝑆𝑒)
  7. 7. Design of Machine elements 6 After putting values we get a=1507.91Mpa b=-0.189655 𝑆𝑓=a𝑁 𝑏 Put values in above where N=108 cycles We get 𝑆𝑓= 47.53 Mpa This is the fatigue strength for the rotating shaft. Now from the stress concentration factor chart using our fillet radius and diameter of shaft we calculated Ks after which we calculated Kfs: Where, Notch radius=0.508 mm Notch sensitivity=q(s)=0.64 Kt=2.2 Kfs=1+q (Kt-1) β†’Kfs=1.76 Now β†’Ζ¬max=20.06 Γ— 1.76 = 35.30 Mpa Ζ¬a=35.30/2=17.65Mpa Ζ¬m=17.65Mpa From soderberg criteria:
  8. 8. Design of Machine elements 7 Ζ¬a +Ζ¬m = 0.577𝑆𝑦 𝑛 Putting values we get; n= 0.577Γ—390 14.88 = 6 So, Factor of safety = n= 6 Comment: This factor of safety is reasonable and this design can easily sustain under given conditions. Selection of Bearing:- Here we have chosen deep groove ball bearing and not the roller bearing because roller bearing is used for high speed applications but here we are dealing with small velocities. As our machineβ€˜s working service is 24hrs. So using Shigley’s Chap 11 (table 11-4)
  9. 9. Design of Machine elements 8 Since, LD=55kh; We have selected our bearing for 1 million cycles. So, Solution: We have selected our bearing for a lifespan of 1 million cycles. L10 = LRnR60 = 106 Cycles F 𝐷 = mg Desired life = ld= 55000 hrs desired radial load = FD = 1.47N = 0.00147kN L10 = LRnR60 = 106 Cycles = 394.04N = 0.394kN
  10. 10. Design of Machine elements 9 Ο‰ = nD = 500 rev min⁄ C10 = FD( LDnD60 LRnR60 ) 1 a⁄ C10 = 0.00147 Γ— ( 55000 Γ— 500 Γ— 60 106 ) = 2.42kN So using catalog provided by Timken catalog given in the book. Here, we have got the rating for the bearing i.e. 2.42kN. Now if we see in the catalog, we get the nearest possible result i.e. C10rated = 5.07kN For this, we have selected our bearing on the basis of our rated value from catalog Deep Groove Single Row 02-Series. Bore = 10mm Outer Diameter= = 30mm Width = 9mm Fillet radius = 0.6mm ds = 12.5mm dH = 27mm However, the bearings installed/used in our assembly are of the following characteristic value which was nearest to the calculated one: C10rated = 6.89kN For this, we have selected our bearing on the basis of our rated value from catalog Deep Groove Single Row 02-Series. Bore = 12mm Outer Diameter = 32mm Width = 10mm Fillet radius = 0.6mm ds = 14.5mm dH = 28mm
  11. 11. Design of Machine elements 10 Conclusion: We have selected our bearing using standard catalog provided by Timken, but unfortunately this type of bearing was not available so we have to change our bearing according to the availability. So, in our project single side shielded bearing No. Z-0009 has been used. Gear designing: Assumptions: The assumptions taken for our gears are as under: Failure acts only the yielding point. We have used here spur gear because they are for low loads only. Actually Spur Gears are the most common type of gears. They are mostly used where the application is of rotary type while maintaining constant torque and speed. Material selection: The main problem while designing was that either we will get our required material or not. Here the material used for the manufacturing of gears was mainly dependent on working conditions like wear, noise etc. So, here we have used annealed steel for our working. The Gear: Pressure angle = ΞΈ = 20 Number of teeth =N= 70 Face width = b = 8 π‘šπ‘š Power = 𝑃 = 192.95 watt = 0.258 hp n = 1308 rev/min=136.97 rad /sec T = P/n =190.95/136.97=1.39 N-m Pitch diameter= Dp = 70 mm=0.07 m Diameteral pitch = Pd = number of teeth/pitch dia Pd=70/63.5
  12. 12. Design of Machine elements 11 Pd = 1.102 teeth mm Module=m= Dp number of teeth =63.5/70=0.90 mm Circular pitch=p= (Ο€*Dp)/N= (Ο€*m)=2.84 mm Now there will be many terms to obtain as shown in following figures
  13. 13. Design of Machine elements 12 Transmitted force = Ft = 2T Dp = 43.77 N Normal force = Fn = Ft tan ΞΈ = 15.93 N Resultant force = Fr = Ft cos ΞΈ = 46.58 N Surface speed = Vm = Ο€Dpn 12 =2.27 m/sec Now , we use Lewis equation Fs = SnYb Pd Where Fs = Allowable dynamic bending force Y = Lewis form factor The material used is 4140 annealed steel Sut = 655 MPa Sn = 0.5 Sut = 327.5 MPa Y = 0.429 (from table for 70 teeths) Fs = SnYb Pd = 1.019 kN Dynamic force = Fd = 600 + Vm 600 Ft = 46.76 N Design Method: Strength of gear tooth should be greater than the dynamic force Fs β‰₯ Fd Including factor of safety Nsf: Fs Nsf β‰₯ Fd So , we have
  14. 14. Design of Machine elements 13 Fs Fd β‰₯ Nsf π‚π¨π¦π¦πžπ§π­ 𝐨𝐧 𝐝𝐞𝐬𝐒𝐠𝐧 21 is the factor of safety for designing of big dia gear. This shows that our gear is over designed. This is a huge value and logically we should review our design but in our case neither the cost nor the size matter that much. Pinion: Pressure angle = ΞΈ = 20 Number of teeth = 18 Face width = b = 7.6 mm Power = 192.95 watt = 0.258 hp N1/N2 = n2/n1 N2=number of teeth of gear N1=number of teeth of pinion n2=speed of gear So, n1=speed of pinion=5199 rev/min=544.43 rad/sec n = 544.43 rad/sec N1/N2 = T2/T1 N2=number of teeth of gear N1=number of teeth of pinion T2=torque of gear So, T1=torque of pinion=0.18 N-m Pitch diameter= Dp = 18 mm
  15. 15. Design of Machine elements 14 Diameteral pitch = Pd = number of teeth/pitch dia Pd = 18 18 = 1 teeth mm Module=m= Dp number of teeth =18/18=1 mm Circular pitch=p= (Ο€*Dp)/N= (Ο€*m)=3.14 mm Transmitted force = Ft = 2T Dp = 20 N Normal force = Fn = Ft tan ΞΈ = 7.27 N Resultant force = Fr = Ft cos ΞΈ = 21.2 N Surface speed = Vm = Ο€Dpn 12 = 2.56 m sec Now , we use Lewis equation Fs = SnYb Pd Where Fs = Allowable dynamic bending force Y = Lewis form factor The material used is 4140 annealed steel Sut = 655 MPa Sn = 0.5 Sut = 327.5 MPa Y = 0.309 (from table for 18 teeths) Fs = SnYb Pd = 769.1 N Dynamic force = Fd = 600 + Vm 600 Ft = 21.29 N Design Method: Strength of gear tooth should be greater than the dynamic force Fs β‰₯ Fd
  16. 16. Design of Machine elements 15 Including factor of safety Nsf Fs Nsf β‰₯ Fd So , we have Fs Fd β‰₯ Nsf This comes out to be 36 So, the factor of safety is 36 π‚π¨π¦π¦πžπ§π­ 𝐨𝐧 𝐝𝐞𝐬𝐒𝐠𝐧: 36 is the factor of safety for designing of big dia gear. This shows that our gear is over designed. This is a huge value and logically we should review our design but in our case neither the cost nor the size matter that much. Bending and Wear of Gear and Pinion: And (Dp)P = 18 mm=0.018m (Dp)G = 70 mm=0.07 m Vm = Ο€Dpn 12 = 2.56 m sec Wt= 74 N as A = 50 + 56(1 βˆ’ B) =60.08 B = 0.25(12 βˆ’ Qv)2/3 = 0.82 Qv=6
  17. 17. Design of Machine elements 16 So Kv=1.29 m/sec From following table Lewis factor= Y=0.309 (for pinion ) Y=0.439( for gear ) From this equation (Ks)P= 1.03 (Ks)G= 1.05 Cmc = 1 (for uncrowned teeth )
  18. 18. Design of Machine elements 17 Cpf = 0.019 Cpm=1 Cma=0.15 Ce=1 Now as Km= 1.23 Assuming constant gear thickness Kb=1
  19. 19. Design of Machine elements 18 Speed ratio =Ng/Np Mg=70/18= 3.88 Pinion cycles =10^8 Gear cycles =10^8/Mg=10^8/3.88 (YN)P=0.977 (YN)G=0.996 For grade 1 steel HBP = 240 and HBG = 200 From following figure (St)P = 216 MPa (St)G =190 MPa Pinion tooth Bending: [(Οƒ)P]B=23.7 MPa
  20. 20. Design of Machine elements 19 [(SF)P ]B= 2.5 Gear tooth bending: [(Οƒ)G]B=23 .7 Mpa [(SF)G]B= 3.6 Pinion tooth Wear: [(Οƒ)P]W=205 MPa [(SF)P]W=1.4 Gear tooth Wear: [(Οƒ)G}W=207 Mpa [(SF)G]W=1.36 Result of analysis : For the pinion, we compare [(SF)P ]B with [(SF)P]W ^2, or 2.5 with =1.96, so the threat in the pinion is from wear. For the gear, we compare [(SF)G]B with [[(SF)G]W]^2 , or 3.6 with 1.85 , so the threat in the gear is also from wear.
  21. 21. Design of Machine elements 20 Final Results: Above calculations show that our design is over designed, therefore, our designed gears assembly can withstand much larger conditions than the given constraints. Although it will be costly and heavier in weight and size as well, but in our case, these are not the parameters of design success. Applications: Windmills are generally used to harness power of wind to perform mechanical work as well as generation of electricity. Applications of windmills are as follows – o Large Scale Electric Power Generation- Huge numbers of windmills are installed on a single site called as wind farms to produce large scale electric power from wind. Small Scale Electric Power Generation- Single windmill can be installed in remote locations to produce electricity, where connection to the general electric network is not available o .Pumping Water for Irrigation- Windmill is used for pumping of water from the well for agricultural purposes. o Milling Grain- Windmills are used to grind grain, cut wood and supply water in some locations where electricity is not available. o Windmills in Warfare- Initially, windmills were used as watchtowers because they were constructed on the tops of hills or mountains. o The energy produced by windmill does not generate any pollution, which is an infinite source of energy. o As windmill generates energy from wind it does not require fuel. o It does not create greenhouse gases. o It is also used for driving all sorts of mechanical devices for manufacturing. o Windmill is generally used to convert wind energy into rotational energy with the help of blades. o Energy produce by windmill is cheap. o It is a renewable energy source. o When wind energy is produced it does not create any ill effect on environment. o Wind power is the cheapest source of energy that is available in current arena. o The operating cost of windmill is very low. o Wind energy provides jobs in production, maintenance, research and installations of windmills, as windmills are becoming popular. o Windmills produce electricity at 5 cents kilowatt per hour. o Current windmills with 2 or 3 thin blades connected to a long pole type base add a more modern aesthetic to land.

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