Battery and Motor Sizing for a Solar Cart
A team is building a small solar-assisted cart for a school garden. The cart must carry 80 kg of soil up a 12 m ramp with a 10% gradient (rise 1.2 m) at a steady 0.5 m/s, then return empty. The drive is a 24 V DC gear motor through a chain reduction. Rolling resistance is estimated at 40 N loaded and 20 N empty. The team has a 24 V, 20 Ah lithium battery and a 250 W motor with a rated efficiency of 75%. A 100 W solar panel is mounted on the cart. The event runs for 4 hours, and the cart makes 15 loaded trips and 15 empty returns. The team wants to know whether the battery will last the event without mains charging.
1. Calculate the energy required per loaded trip and per empty return, and the total energy for the event.
Read the worked answer
Loaded trip (up the ramp):
- Gravity component along slope: for a 10% gradient, sin(theta) ≈ 0.0995. Force = m·g·sin(theta) = 80 × 9.81 × 0.0995 ≈ 78.1 N.
- Rolling resistance loaded = 40 N.
- Total tractive force ≈ 118.1 N.
- Distance along ramp = sqrt(12^2 + 1.2^2) ≈ 12.06 m. Work = 118.1 × 12.06 ≈ 1424 J.
- At 0.5 m/s, time = 12.06/0.5 ≈ 24.1 s, power ≈ 1424/24.1 ≈ 59 W mechanical.
- Electrical energy at 75% efficiency = 1424/0.75 ≈ 1899 J ≈ 0.53 Wh per loaded trip.
Empty return (down the ramp):
- Gravity assists: 78.1 N downhill; rolling resistance empty = 20 N opposes motion.
- Net force downhill ≈ 58.1 N, so the motor may need to brake rather than drive. Assume regenerative braking is not available, so energy is dissipated in brakes; motor electrical draw ≈ 0 for the descent. To be conservative, assume the motor still drives at low load: tractive force = 20 N, work = 20 × 12.06 ≈ 241 J, electrical ≈ 321 J ≈ 0.09 Wh.
Totals:
- 15 loaded trips: 15 × 0.53 = 7.95 Wh.
- 15 empty returns: 15 × 0.09 = 1.35 Wh.
- Total traction ≈ 9.3 Wh.
This is tiny compared with the battery capacity, so traction is not the limiting factor.
2. Assess whether the battery will last the 4-hour event and identify the real limiting factors.
Read the worked answer
Battery capacity: 24 V × 20 Ah = 480 Wh nominal. Usable capacity for lithium is typically 80-90% to preserve life, so about 384-432 Wh usable. Traction demand of ~9.3 Wh is negligible - under 3% of usable capacity.
Solar input: 100 W panel over 4 hours gives up to 400 Wh in ideal sun, but real output is far lower: panel derating, angle, shade from the cart and trees, and the fact the cart moves. A realistic harvest might be 150-250 Wh over the event. Even zero solar harvest leaves the battery ample for traction.
So the battery will easily last the event for traction. The real limiting factors are elsewhere:
- Peak current and motor rating. The 250 W motor at 24 V draws about 10.4 A at rated load. Starting on the ramp, stall current could be several times rated and may exceed the battery's continuous discharge rating or trip a BMS. The team must check the battery's continuous current spec and add a soft-start or current limit.
- Voltage sag under load. A 20 Ah pack under 30 A peak will sag; the motor controller may cut out on low voltage. Wiring gauge and connector rating matter.
- Mechanical holding. On a 10% ramp with 80 kg, the cart needs a brake or the motor must hold torque; a gear motor may back-drive. A fail-safe brake is a safety requirement, not an energy one.
- Thermal. Repeated trips heat the motor; 15 trips with short rest may push winding temperature.
Conclusion: energy is not the constraint - the team should redirect effort to current limiting, braking and thermal checks, and verify the battery's continuous discharge rating before the event.