Sizing solar for Te Whare Hui: read the load profile before the invoice
Te Whare Hui marae trustees have $18,000 and a $6,400 annual power bill. The electrician's load schedule for a typical winter day is:
- two heat pumps drawing an average of 0.9 kW each, for 6 h;
- a 3 kW hot-water cylinder element, on for 4 h;
- a chest freezer averaging 0.15 kW for 24 h;
- 18 LED lights at 12 W for 5 h;
- a 2 kW urn, used for 30 min per day.
An installer quotes 6.0 kW of panels on the north roof plus a 10 kWh battery for $18,000. For the marae's region, December averages 5.6 peak-sun hours per day and June averages 2.9; the quotation's yield figures use 4.2 h/day. A pōhutukawa shades the north roof from about 3:00 pm in winter. Most hui and tangi happen in winter, when the kitchen and hot water run for longer than the schedule shows.
1. Calculate the typical winter daily demand from the schedule, and the realistic winter daily output of the 6.0 kW array assuming 15% system losses and a further 10% winter shading loss. Show your working.
Read the worked answer
Winter daily demand
| Load | Working | kWh/day |
|---|---|---|
| Heat pumps | 2 × 0.9 kW × 6 h | 10.8 |
| Hot-water cylinder | 3 kW × 4 h | 12.0 |
| Freezer | 0.15 kW × 24 h | 3.6 |
| Lights | 18 × 0.012 kW × 5 h | 1.08 |
| Urn | 2 kW × 0.5 h | 1.0 |
| Total | ≈ 28.5 kWh |
The hot-water cylinder alone is 42% of the load - that is where any efficiency work should start.
Realistic winter output
- Ideal: 6.0 kW × 2.9 h = 17.4 kWh/day
- After 15% system losses: 17.4 × 0.85 = 14.8 kWh
- After 10% shading loss: 14.8 × 0.90 = 13.3 kWh/day
Comparison: 13.3 / 28.5 = 47% of winter demand. The array covers less than half, leaving a shortfall of about 15 kWh/day drawn from the grid. Note also that the pōhutukawa shading starts at 3:00 pm, exactly when winter generation is already falling, so the 10% figure is an assumption that should be checked with a site measurement rather than accepted.
2. The installer's figures use 4.2 h/day. Explain what difference that makes, and why the choice of that number matters to the trustees' decision.
Read the worked answer
The arithmetic
- At 4.2 h/day: 6.0 × 4.2 = 25.2 kWh, × 0.85 losses = 21.4 kWh/day.
- At the June value: 13.3 kWh/day (with shading).
- That is about 8 kWh/day more - roughly 60% higher - than the realistic winter figure.
Where 4.2 h/day comes from It is close to the average of the December and June values ((5.6 + 2.9)/2 = 4.25). So the number is probably a genuine annual average, not an error of fact. The problem is that it is being applied to a load that peaks in winter, when output is at its lowest. Averaging hides exactly the month the marae most needs the panels.
Why it matters to the decision
- At 4.2 h/day, the array appears to cover about 75% of the 28.5 kWh load; in June it covers under half. A trustee could reasonably vote for the system on a coverage figure that will not hold in the months the marae is fullest.
- Overstated yield also overstates savings and shortens the apparent payback. At about $0.28/kWh, 8 kWh/day for a year is roughly 2,900 kWh, around $800/year of savings that may not appear.
- It changes the design question: with honest winter figures, the trustees should ask about reducing the load (cylinder timer or heat-pump hot water, insulation, LED) and about load shifting, not just about adding panels.
- The right question to put to the installer is: what is the modelled output for June, and what assumption about shading and losses is behind it?
3. A trustee argues: 'With a 10 kWh battery we will be self-sufficient.' Evaluate this claim using the numbers you have calculated.
Read the worked answer
The claim is not supported.
- Self-sufficiency requires generation to match demand over the period. In winter, generation is about 13.3 kWh/day and demand about 28.5 kWh/day. A battery moves energy in time; it does not create it. The 10 kWh battery can at most shift the daytime share of 13.3 kWh into the evening, and after a round-trip loss of roughly 10% it delivers about 12 kWh. It cannot touch the missing 15 kWh/day.
- In summer the position reverses: December output (6.0 × 5.6 × 0.85 ≈ 28.6 kWh, less shading) is close to demand, so the battery will often fill early and the surplus will be exported at a low export rate rather than stored.
What the battery does buy It shifts generation into the evening peak, when lights, cooking and hot water coincide, and it provides outage resilience - a real benefit on a marae that may host people during a civil defence event.
What the trustees should weigh instead
- Attack the 12 kWh/day hot-water load - a cylinder timer or heat-pump water heater changes the economics more than a battery of the same cost.
- Check the quotation's winter figure before signing.
- Ask whether the $18,000 is better split between a smaller battery and demand reduction, since the marae is winter-peaking and the array is winter-troughing.
Verdict: the battery improves self-consumption and resilience, but 'self-sufficient' is false in the season that matters most.