# Chapter 1:Fundamentals of Electric Drives¶

## Example no:1.1,Page no:10¶

In :

import math

#Variable declaration
J=12 #journey per hour
Load=5500 #[Load]  Kg
t_up=1.5 #[time]  min
W_cage=500 #[Weight of cage]  Kg
t_down=1.25 #[time]  min
h=50 #[height]  m
Wb=3000 #Kg(Balance weight)
Eff_hoist=0.8     #Effeciency of hoist
Eff_motor=0.85    #Efficiency of motor
g=9.81 #gravity constant

#Calculation
E_upward=(Load+W_cage-Wb)*g*h #J
E_downward=(Wb-W_cage)*g*h #J
Edj=E_upward+E_downward #J(Energy used in double journey)
Ein=Edj/Eff_hoist/Eff_motor #J or W-s
Ein=Ein/1000/3600 #KWh
Ein_hour=Ein*J #KWh

Rating=E_upward/Eff_hoist/t_up/60 #W

#Result
print"\n\n(a).Electrical energy used in upward journey :",E_upward,"J"
print"    Electrical energy used in downward journey :",E_downward,"J"
print"    Electrical energy used in each double journey :",Edj,"J"
print"    Energy input in each double journey=",round(Ein,3),"kWh\n"
print"(b).Electrical energy consumption in one hour : ",round(Ein_hour,3),"kWh\n"
print"(c) kW Rating of motor: ",Rating/1000,"kW"


(a).Electrical energy used in upward journey : 1471500.0 J
Electrical energy used in downward journey : 1226250.0 J
Electrical energy used in each double journey : 2697750.0 J
Energy input in each double journey= 1.102 kWh

(b).Electrical energy consumption in one hour :  13.224 kWh

(c) kW Rating of motor:  20.4375 kW