Sizing a Solar Street-Light Pole Against a Frozen Budget
A government polytechnic in Kanpur is replacing 12 damaged street lights along its hostel lane with LED fittings. Each fitting draws 30 W and is switched on for 10 hours a night, all year. For each pole the supplier has quoted a 100 Wp polycrystalline panel, a 12 V, 100 Ah tubular battery and a PWM charge controller. The site receives an average of 4.5 peak sun hours per day. The college's specification, written after an earlier failure, demands two things: at least two nights of backup with no sun at all, and calculation of panel output after a 20 per cent derating for dust, heat and wiring loss. The battery must never be taken beyond 60 per cent depth of discharge. The quotation is the lowest received, the budget is frozen, and the electrical workshop in-charge must approve or reject it on the numbers before the file goes to the principal.
1. For one pole, find the nightly energy demand and the usable energy in the quoted battery. Does the quotation satisfy the two-night backup clause?
Read the worked answer
Nightly demand = 30 W × 10 h = 300 Wh.
Usable battery energy = 12 V × 100 Ah × 0.60 = 720 Wh.
Backup = 720 / 300 = 2.4 nights.
The clause is technically met, but with only 0.4 of a night (about 120 Wh) in hand, and that assumes a new battery at about 25 °C. A two-year-old tubular battery at 80 per cent of rated capacity gives 12 × 100 × 0.8 × 0.6 = 576 Wh, i.e. 1.9 nights, which fails. So the battery passes on paper and fails in practice within a couple of years. The honest verdict is 'conditionally acceptable, minimum size only' - the purchase file should record that this is not a comfortable margin, so that nobody later claims the design promised more than it does.
2. Check whether the 100 Wp panel can recharge the battery after two consecutive sunless nights while still running the lamp every night. Which component fails first?
Read the worked answer
Daily panel yield = 100 Wp × 4.5 h = 450 Wh ideal; after the 20 per cent derating = 360 Wh per day.
The lamp alone consumes 300 Wh per day, so the surplus available for recharging is only 60 Wh per day.
Two sunless nights drain 2 × 300 = 600 Wh. At 60 Wh per day, restoring that deficit takes 600 / 60 = 10 days of perfect sunshine, and any further cloudy day restarts the clock.
The panel, not the battery, is the weak link: the battery stores 2.4 nights but the panel can never refill it within a realistic weather window. Winter makes it worse, since 4.5 peak sun hours is an annual average and December delivers far less. Verdict: reject the panel rating, keep the battery.
3. Recommend the single change with the best value and justify it with numbers.
Read the worked answer
Upgrade only the panel, from 100 Wp to 200 Wp.
New yield = 200 × 4.5 × 0.80 = 720 Wh per day. Surplus after the 300 Wh nightly load = 420 Wh per day, so a 600 Wh deficit from two cloudy nights is cleared in under 1.5 days and normal operation ends every day with the battery full.
Keeping the 100 Ah battery is right: its 720 Wh usable capacity already meets the two-night autonomy clause, and enlarging it adds cost without adding energy. Enlarging the battery instead is the classic wrong answer - more storage that the panel still cannot refill.
One check before ordering: a 200 Wp module at 12 V delivers roughly 11–12 A, so a 10 A PWM controller in the quotation must be replaced by a 20 A unit, or the extra generation will simply be clipped.