MD2 Poster
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Transcript of MD2 Poster
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Diagram of Fly wheel
∆ E=I ksω2
k 2=D o
2+(Do−2h) ²8
for rim flywheel
≅D ²4
when h is small
k 2=D o
2
8 for disc flywheel
v=πDN
For CI flywheel
v < 1500 m/min for HP<100 < 2000 m/min for HP>100for cast steel flywheelv = 3000 4000 m/min
w =πDbhgb/h = 0.65-2
For belt pulley flywheel b should be 3 to 5cm wider than the belt
Stresses in the rim:
Tensile stress due to centrifugal force,
σ t=γv ²g
Bending stress,
σ b=π ² v ²Dγn ²gh
σ total=34σ t+
14σ t
≤ 350 kgf/cm2 for CI σ t< 65 kgf/cm2 for CI < 280 kgf/cm2 for cast steel
∆E - excess energyI - mass moment of inertia of the flywheelk s - coefficient of fluctuation of speed k s= maximum, minimum and mean angular velocities respectivelyk – radius of gyration of flywheelDo, D outside and mean diameters of the rimH – thickness of the rimv – mean rim speedw – weight of the rim of the flywheelb – width of the flywheelg – specific weightn – number of arms 6,8 or 10Mt- transmitted torque T1, T2 belt tensions
Stresses in the arm (at the hub end): BendingStress
σ b1=M t (D−d)n ZwD
For CIσ b1< 130 kgf/cm < 65 kgf/cm2, for severe load
Section of arm:
α=3√ 64Z yyπ
with C = α/2
Government Engineering College, Dahod Mechanical Engineering Department Title: Design of rope sheave and drum Year : 2015 PREPARED BY: 120183119014 : Gadhiya Mehul 120183119015 : Badi Iliyas 120183119016 : Patel Sandip 120183119018 : Bhatt Dhruvin 120183119019 : Patel Udit Guided by: R.I.Patel