Question

A motor is being self-started against a resisting torque of 60 N-m and at each start, the engine is cranked at 75 r.p.m. for 8 seconds. For each start, energy is drawn from a lead acid battery. If the battery has the capacity of 100 Wh, calculate the number of starts that can be made with such a battery. Assume an overall efficiency of the motor and gears as 25%.

Answer 1

Angular speed ω = 2π N/60 rad/s = 2π × 75/60 = 7.85 rad/s 

Power required for rotating the engine at this angular speed is 

P = torque × angular speed = ωT watt = 60 × 7.85 = 471 W

Energy required per start is = power × time per start = 471 × 8 = 3,768 watt-s = 3,768 J 

= 3,768/3600 = 1.047 Wh 

Energy drawn from the battery taking into consideration the efficiency of the motor and gearing 

= 1.047/0.25 = 4.188 Wh 

No. of start possible with a fully-charged battery = 100/4.188 = 24 (approx.)