A train has schedule speed of 30 km pH over a level track,distance between station being 1 km station stopping time is 20 second .assuming braking retardation of 3 km pH and maximum speed 25 % greater the then average speed .calculate the acceleration frequency run with service ?
Answers
Step-by-step explanation:
#A 250-tonne motor coach driven by four motors takes 20 seconds to attain a speed of 42 km/h,starting from rest on an ascending
#gradient of 1 in 80.The gear ratio is 3.5,gear efficiency 92%,wheel diameter 92cm train resistance 40N/t and rotational inertia
#10 percent of the dead weight.Find the torque developed by each motor.
#Given
M = 250.0 #tonne (mass of motor)
Me = 1.1*250.0 #tonne (mass of rotating motor)
Vm = 42.0 #km/h (speed)
t1 = 20.0 #s (time)
G = 1/80.0*100 # (% gradient)
r = 40.0 #N/tonne(train reistance)
D = 0.92 #m (wheel diameter)
gratio = 3.5 # (gear ratio)
geff = 0.92 # (gear efficiency)
a = Vm/t1 # (acceleration)
#Now,tractive force is given by
Ft = 277.8*Me*a + 98*M*G + M*r #N
#Now Ft = 2*gratio*geff*T/D.Therefore torque 'T' is
T = Ft*D/(2*gratio*geff) #N-m
#There are motors.so,torque by each motor is
torque = T/4 #N-m
print "Torque developed by each motor = ",round(torque),"N-m."
Torque developed by each motor = 7181.0 N-m.