import math #variable declaration Jo=0.2 # inertia of the motor in kg-m2 a1=0.1 # reduction gear J1=10 # inertia of the load in kg-m2 Tl1=10 # load torque v=1.5 # speed of the translational load M1=1000 # mass of the translational load N=1420 # speed of the motor n1=.9 # efficiency of the reduction gear n1_=0.85 # efficiency of the translational load and the motor F1=M1*9.81 # force of the translational load #Calculation Wm=N*math.pi/30 #angular speed J=Jo+a1**2*J1+ M1*(v/Wm)**2 # total equivalent moment of inertia Tl= a1*Tl1/n1+F1/n1_*(v/Wm) # total equivalent torque #Result print"\nEquivalent moment of inertia is :",round(J,1),"kg-m2" print"\nEquivalent load torque :",round(Tl,2),"N-m"
Equivalent moment of inertia is: 0.4 kg-m2 Equivalent load torque : 117.53 N-m
import scipy from scipy import integrate import math # variable declaration J=10 #moment of inertia of the drive in kg-m2 print("Passive load torque during steady state is :Tl=0.05*N in N-m") print("And load torque : T=100-0.1*N in N-m ") print("load torque when the direction is reversed T=-100-0.1*N in N-m") #Calculation print("T-Tl=0") print("100-0.1*N-0.05*N=0") N=100/0.15 #Required speed of the motor in rpm during steady state N2=-100/0.15 #During reversal speed is in opposite direction print("\nJdWm/dt=-100-0.1*N-0.05*N during reversing") print("dN/dt=30/(J*pi)*(-100-0.15*N)") print("dN/dt=(-95.49-0.143*N)") N1=N N2=N2*0.95 #for speed reversal x2 = lambda N: 1/(-95.49-0.143*N) t=integrate.quad(x2, round(N1), round(N2)) #result print"\nHence Time of reversal is :",round(t,2),"s"
Passive load torque during steady state is :Tl=0.05*N in N-m And load torque : T=100-0.1*N in N-m load torque when the direction is reversed T=-100-0.1*N in N-m T-Tl=0 100-0.1*N-0.05*N=0 JdWm/dt=-100-0.1*N-0.05*N during reversing dN/dt=30/(J*pi)*(-100-0.15*N) dN/dt=(-95.49-0.143*N) Hence Time of reversal is : 25.51 s
import math from __future__ import division #variable declaration Tlh=1000 # load torque in N-m Tmax=700 # maximum motor torque Tll=200 # light load for the motor to regain its steady state Tmin=Tll # minimum torque t_h=10 # period for which a load torque of 1000 N-m is apllied in sec Jm=10 # moment of inertia of the motor in Kg-m2 No=500 # no load speed in rpm Tr=500 # torque at a given no load speed in N-m #Calculation Wmo=No*2*math.pi/60 # angular no load speed in rad/sec s=0.05 # slip at a torque of 500 N-m Wmr=(1-s)*Wmo # angular speed at a torque of 500 N-m in rad/sec y=math.log((Tlh-Tmin)/(Tlh-Tmax)) x=Tr/(Wmo-Wmr) J=x*t_h/y Jf=J-Jm #Result #answer in the book is wrong print"\n\nMoment of inertia of the flywheel : ", round(Jf,1),"Kg-m2"
Moment of inertia of the flywheel : 1937.2 Kg-m2