Practical PV System Designs SS1 Solar Photovoltaic Installation & Maintenance Lesson Note

The Case Study – Meeting the Yusuf Family

The Yusuf family lives in a three-bedroom bungalow. They are tired of the constant noise from their “I-pass-my-neighbor” generator and the high cost of petrol. They want a solar system that can handle their basic needs at night.

The Yusuf Family’s Needs (The Load):

• 6 LED Bulbs: To keep the house bright (10W each).
• 2 Standing Fans: To keep the rooms cool at night (60W each).
• 1 LED TV: For news and football (100W).
• 1 Decoder: For the TV (20W).
• Phones: Charging 3 phones (10W each).

The Goal: They want these items to run for 8 hours every night after the sun goes down.

Step 1 – Calculating the Daily Energy Demand

First, we need to find out how much “energy” (Watt-hours) the Yusuf family consumes in one night.

Total Nightly Need: 2,160Wh (or roughly 2.2kWh).

Step 2 – Sizing the “Translator” (Inverter)

The total power used at the exact same time is 330 Watts. However, we must remember that fans have small motors that “kick” when they start.

• Total Running Watts: 330W.
• Safety Margin: We multiply by 1.25 (to give the inverter “breathing room”).
• 330W×1.25=412.5W.

The Decision: A 1KVA (800W) Inverter is perfect for this family. It can easily handle the 330W load and has enough extra strength to start the fans.

Step 3 – Sizing the “Energy Tank” (Batteries)

The Yusufs are using Tubular (Lead-Acid) batteries, so we must remember the 50% Depth of Discharge rule. We don’t want to drain them to zero!

• Energy Needed: 2,160Wh.
• To stay safe (50% DoD): We need a battery capacity of 2,160×2=4,320Wh.
• In Amp-Hours (Ah): Since it’s a 12V system: 4,320Wh÷12V=360Ah.

The Decision: We will buy two 200Ah batteries and connect them in Parallel. This gives them 400Ah total, which is plenty of “water in the tank” to last the whole night.

Shutterstock

Step 4 – Sizing the “Harvest” (Solar Panels)

To fill up a 2,160Wh “hole” left in the batteries every day, the panels must produce that much power in about 5 Peak Sun Hours.

• Energy Needed: 2,160Wh.
• Efficiency Loss (30% extra): 2,160×1.3=2,808Wh.
• Panel Watts Needed: 2,808Wh÷5 hours=561.6 Watts.

The Decision: We will buy two 300W Solar Panels. 300W×2=600W total. This is more than the 561W we calculated, so the batteries will get full even if the day is a bit cloudy.

Getty Images

Final System Summary for the Yusuf Family:

• Panels: 2 x 300W (Monocrystalline)
• Controller: 60A MPPT
• Inverter: 1KVA Pure Sine Wave
• Batteries: 2 x 200Ah Tubular

Class Discussion:

1. If the Yusuf family decides to buy a refrigerator next month, which part of this system will fail first?
2. Why did we use a 60A controller instead of a 30A one? (Hint: Think about the 600W panels).
3. What happens to this design if the family leaves the TV on for 12 hours instead of 5?