Solar-Powered Vineyard Weather Station in Winter
A vineyard installs a remote weather station powered by a 12 V lead-acid battery and a 10 W solar panel. The sensor and data logger draw an average of 28 mA continuously. A cellular modem draws 120 mA for 90 seconds every 15 minutes to upload data. The battery is rated 18 Ah, and the installer allows only 50% depth of discharge to protect it. At the site, the worst-month daily solar insolation is 1.5 kWh/m²/day, and the installer assumes 30% system losses. The station must run unattended through cloudy winter spells of up to seven days. The installer claims the panel is 'more than enough' because the panel is rated at 10 W and the loads are 'only milliamps'.
1. Calculate the station's average daily energy use in Wh and in Ah at 12 V. Show the contribution of each load.
Read the worked answer
Convert each load to Ah per day, then to Wh at 12 V.
Logger and sensor: 28 mA = 0.028 A. Continuous for 24 h: 0.028 × 24 = 0.672 Ah/day. At 12 V: 0.672 × 12 = 8.064 Wh/day.
Modem: 90 s = 90/3600 = 0.025 h. Each upload uses 0.120 A × 0.025 h = 0.003 Ah. Every 15 minutes means 96 uploads per day: 96 × 0.003 = 0.288 Ah/day. At 12 V: 0.288 × 12 = 3.456 Wh/day.
Total: 0.672 + 0.288 = 0.960 Ah/day, or 0.960 × 12 = 11.52 Wh/day.
The modem is about 30% of the daily energy even though it is on for only 2.4 minutes per day, because 120 mA is much larger than 28 mA. The installer's 'only milliamps' reasoning ignores duty cycle and the number of transmissions.
2. Using the winter insolation and loss figures, determine whether the 10 W panel keeps the battery charged. Include the usable battery capacity and estimate the days of autonomy.
Read the worked answer
The panel's daily output is estimated from rated power and daily insolation. At 1.0 kW/m², a 10 W panel produces 10 W; 1.5 kWh/m²/day is equivalent to 1.5 hours at full rated output, so ideal yield = 10 W × 1.5 h = 15 Wh/day. With 30% losses, usable yield = 15 × 0.70 = 10.5 Wh/day.
Average daily demand is 11.52 Wh/day. The winter deficit is 11.52 − 10.5 = 1.02 Wh/day, so the panel does not keep up on an average winter day. The battery will slowly discharge even in ordinary winter weather, and much faster in a cloudy spell when output is near zero.
Usable battery capacity is 18 Ah × 0.50 = 9 Ah, or 9 × 12 = 108 Wh. At 11.52 Wh/day with no solar input, autonomy is 108 ÷ 11.52 ≈ 9.4 days. That is just over the seven-day requirement, but it assumes a new, healthy battery and ignores the deficit that accumulates before the cloudy spell. A degraded battery or a colder temperature would reduce usable capacity. The design is marginal, not 'more than enough'.
3. Propose one change that would close the winter energy deficit, and show the arithmetic. Then state one practical check before installing it.
Read the worked answer
Reducing the modem upload interval from every 15 minutes to every hour is the simplest change. At hourly uploads there are 24 per day: 24 × 0.003 Ah = 0.072 Ah/day for the modem. Total demand becomes 0.672 + 0.072 = 0.744 Ah/day, or 8.928 Wh/day. Winter supply is 10.5 Wh/day, giving a margin of about 1.57 Wh/day, so the panel covers the load.
An alternative is a larger panel. To supply 11.52 Wh/day after 30% losses at 1.5 h equivalent, the rated power needed is 11.52 ÷ (1.5 × 0.70) = 11.0 W, so a 15 W panel would give 15 × 1.5 × 0.70 = 15.75 Wh/day and a clear margin. A 10 W panel is just below the requirement.
Before installing, check the actual winter load: the modem's 120 mA figure may be a peak, and cold temperature reduces lead-acid capacity. Measure the current with a logging multimeter over 24 h rather than relying on datasheet nominal values.