Question

The efficiency of the motor of an electric vehicle is 80% both in motor and generator mode. At velocities up to 50km/h the friction force of 500 N has to be overcome. The mass of the vehicle equals one ton. The velocity diagram below describes the motion pattern of the vehicle on a rectilinear path. Calculate the energy recovered during braking phase an the net energy consumed.

Answer 1

The given data states while braking the speed of vehicle is 50 kmph or 13.88 m/s and eventually stops in 615 s with braking beginning from 600 s.

t0=600s, t1=615 s, m=1000 kg, v=50kmph or 13.88 m/s, Friction force F= 500 N, Efficiency of the generator= 80 %

From 600 -615 s the velocity dependency on time is as: v= 0.2777(50 – (10/3)*(t-600)) m/s

Hence the energy recovered is the net energy considering the kinetic energy overcoming the frictional and air drag force.

E_recovered=0.5m.V² -
SF.vdt
=0.5*(1000)*(13.88)² -
615 500.vdt

= 96.327 kJ – 500*0.277()

S605 50dt - (10/3) S65 (t - 600)dt

=96.327 kJ – 52.083 kJ =44.244 kJ

Net E_recovered = 0.8*44.244 kJ = 35.3952 kJ