Detailed AC/DC system design
Location: Puerto Arturo, Madre de Dios, Peru
GPS coordinates: -12.48694444, -69.21305556
Altitude: 3378m
Description: A community building with lighting and AC power needs. The system is used all year long, but it is typically only used three to four times a week by community members for meetings, parties, or training sessions. Load usage is typically during the day. The community does not intend on adding any major appliances in the near future.
The system will use DC for lighting and AC for powering loads. DC is used for lighting so that the system continually provides light regardless of whether the inverter is turned on. As the building is used intermittently, the inverter can be turned off to reduce wear and to lessen the liklihood of an accident or damage from lightning.
Photo of the building.
Location of the building.
Contents
Load evaluation
Although the system is used only one day a week, inputting 1 day a week of usage for the loads will lead to an undersized array and a poor system design. We will input 4 days a week to ensure that the PV source is still of a reasonable size.
DC load evaluation
Step 1: Fill out DC load chart
| April - September | October - March | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| # | Load | Quantity | Watts | Total watts | Duty cycle | Hours per day | Days per week | Average daily DC watt-hours | Hours per day | Days per week | Average daily DC watt-hours |
| 1 | LED light | 8 | 5 W | 40 W | 1 | 3 hours | 4 days | 69 Wh | 3 hours | 4 days | 69 Wh |
| 2 | Inverter | 1 | 7 W | 7 W | 1 | 3 hours | 4 days | 12 Wh | 5 hours | 4 days | 12 Wh |
- Load: The make and model or type of load.
- Quantity: The number of the particular load.
- Watts: The power rating in watts of the load.
- Total watts = Quantity × Watts
- Duty cycle = Rated or estimated duty cycle for the load. If the load has no duty cycle a value of 1 should be used. A load with a duty cycle of 20% would be inputted as .2
- Hours per day: The maximum number of hours the load(s) will be operated per day. If the load has a duty cycle 24 hours should be used.
- Days per week: The maximum number of days the load(s) will be operated per week.
- Average daily DC watt-hours = Total watts × Duty cycle × Hours per day × Days per week ÷ 7 days
Step 2: Determine DC energy demand
| Total average daily DC watt-hours (April - September) | = sum of Average daily DC watt-hours for all loads for April - September |
|---|---|
| = 81 Wh |
| Total average daily DC watt-hours (October - March) | = sum of Average daily DC watt-hours for all loads for October - March |
|---|---|
| = 81 Wh |
AC load evaluation
Step 1: Determine inverter efficiency
A conservative inverter efficiency value of .85 is going to be used.
| Inverter efficiency | .85 |
|---|
Step 2: Fill out AC load chart
| April - October | March - September | ||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| # | Load | Quantity | Watts | Total watts | Duty cycle | Surge factor | Surge watts | Power factor | Volt-amperes (VA) | Hours per day | Days per week | Average daily AC watt-hours | Hours per day | Days per week | Average daily AC watt-hours |
| 1 | Projector | 1 | 300 W | 300 W | 1 | 0 | 0 | .9 | 333 VA | 3 hours | 4 days | 605 Wh | 3 hours | 4 days | 605 Wh |
| 2 | Stereo | 1 | 30 W | 30 W | 1 | 0 | 0 | .9 | 33 VA | 3 hours | 4 days | 61 Wh | 3 hours | 4 days | 61 Wh |
| 3 | Cell phone | 5 | 5 W | 25 W | 1 | 0 | 0 | .9 | 28 VA | 1 hour | 4 days | 17 Wh | 1 hour | 4 days | 17 Wh |
- Load: The make and model or type of load.
- Quantity: The number of the particular load.
- Watts: The power rating in watts for the load.
- Total watts = Quantity × Watts
- Duty cycle = Rated or estimated duty cycle for the load. If the load has no duty cycle a value of 1 should be used. A load with a duty cycle of 20% would be inputted as .2
- Surge factor = Rated or estimated duty cycle for the load. Common values are between 3-5. If the load does not have a surge requirement a value of 0 should be used.
- Power factor = Rated or estimated power factor for the load.
- Volt-amperes (VA) = Total watts ÷ Power factor
- Hours per day: The maximum number of hours the load(s) will be operated per day. If the load has a duty cycle 24 hours should be used.
- Days per week: The maximum number of days the load(s) will be operated per week.
- Average daily AC Watt-hours = Total watts × Duty cycle ÷ Inverter efficiency (Step 1) × Hours per day × Days per week ÷ 7 days
Step 3: Deteremine AC energy demand
| Total average daily AC watt-hours (April - September) | = sum of Average daily AC watt-hours for all loads for April - September |
|---|---|
| 682 Wh |
| Total average daily AC watt-hours (October - March) | = sum of Average daily AC watt-hours for all loads for October - March |
|---|---|
| 682 Wh |
Step 4: Determine AC power demand
| Total VA | = sum of volt-amperes (VA) |
|---|---|
| 394 VA |
| Total VA with surge watts | = sum of Surge watts for all loads + Total VA |
|---|---|
| 394 VA |
Total average daily energy demand
The total energy demand for the system is the added Average daily DC-watt hours and Average daily AC watt-hours for each time period.
| Average daily watt-hours required (April - September) | = Total average daily DC watt-hours (April - October) + Total average daily AC watt-hours (April - September) |
|---|---|
| = 81 Wh + 682 Wh = 763 Wh |
| Average daily watt-hours required (April - September) | = Total average daily DC watt-hours (October - March) + Total average daily AC watt-hours (October - March) |
|---|---|
| = 81 Wh + 682 Wh = 763 Wh |
Weather and solar resource evaluation
Maximum ambient temperature = 35°C
Minimum ambient temperature = 15°C
Maximum indoor temperature = 30°C
Minimum indoor temperature = 20°C
Retrieving PVGIS monthly insolation data.
Retrieving PVGIS monthly insolation data.
Retrieving PVGIS weather data.
Retrieving PVGIS weather data.
Load and solar resource comparison
Step 1: Determine monthly ratio of energy demand to solar resource
| Month | Average monthly insolation | Total average daily energy demand | Ratio |
|---|---|---|---|
| January | 131.2 kWh/m² | 763 Wh | 5.81 |
| February | 110.5 kWh/m² | 763 Wh | 6.90 |
| March | 145.2 kWh/m² | 763 Wh | 5.25 |
| April | 143.5 kWh/m² | 763 Wh | 5.32 |
| May | 123.2 kWh/m² | 763 Wh | 6.19 |
| June | 132.8 kWh/m² | 763 Wh | 5.75 |
| July | 143.3 kWh/m² | 763 Wh | 5.32 |
| August | 167.6 kWh/m² | 763 Wh | 4.55 |
| September | 112.4 kWh/m² | 763 Wh | 6.79 |
| October | 147.0 kWh/m² | 763 Wh | 5.19 |
| November | 139.0 kWh/m² | 763 Wh | 5.49 |
| December | 134.3 kWh/m² | 763 Wh | 5.68 |
- Month: The month of the year.
- Average monthly insolation: Solar resource data obtained for the location from Weather and solar resource evaluation.
- Total average daily energy demand for the month from the load evaluation.
- Ratio = Total average daily energy demand ÷ Average monthly insolation
Step 2: Determine design values
| Design daily insolation | = Average monthly insolation from month with the highest ratio ÷ 30 |
|---|---|
| = 110.5 kWh/m² ÷ 30 = 3.7 kWh/m² |
| Design daily watt-hours required | = Total average daily energy demand from month with the highest ratio |
|---|---|
| = 763 Wh |
Design parameters
System voltage parameter = 24 V
- This system will be built with a 24 volt nominal voltage in order to be able to use a 72-cell module. It could also easily be built as a 12 volt system.
Irradiance safety parameter = 1.25
- The irradiance safety parameter is always the same.
Continuous duty safety parameter = 1.25
- The continuous duty safety parameter is always the same.
Low voltage disconnect parameter = 22.2 V
- A simple charge controller with a pre-programmed low voltage disconnect will be used.
Energy storage sizing and selection
Step 1: Determine depth of discharge parameter
For this project a depth of discharge of .5 (50%) is a good compromise.
- Depth of discharge = .5
Step 2: Determine days of autonomy parameter
The building is used intermittently, but as the building serves an important need for the community it should still have at least 2 days of autonomy. The energy storage system size has already been reduced as all loads in the load evaluation were put in as only being used 4 days per week, but this calculation only works if the usage is spread out throughout the week. In this case, all of the usage will occurr during one day.
- Days of autonomy = 2
Step 3: Determine battery temperature correction factor
The minimum indoor temperature was determined to be 10°C. An AGM battery will be used to avoid regular maintenance.
- Battery temperature correction factor = 1.08
Correction factors for various battery types:[1]
| Temperature | FLA | AGM | Gel |
|---|---|---|---|
| 25°C | 1.00 | 1.00 | 1.00 |
| 20°C | 1.06 | 1.03 | 1.04 |
| 15°C | 1.13 | 1.05 | 1.07 |
| 10°C | 1.19 | 1.08 | 1.11 |
| 5°C | 1.29 | 1.14 | 1.18 |
| 0°C | 1.39 | 1.20 | 1.25 |
| -5°C | 1.55 | 1.28 | 1.34 |
| -10°C | 1.70 | 1.35 | 1.42 |
Step 4: Calculate total Ah required
| Total Ah required | = Average daily Watt-hours required ÷ System voltage parameter × Battery temperature correction factor (Step 3) × Days of autonomy parameter (Step 2) ÷ Depth of discharge parameter (Step 1) |
|---|---|
| = 140 Wh ÷ 12 V × 1.08 × 2 days ÷ .4 = 43 Ah |
Step 5: Calculate number of batteries in series
A 12 V battery is ideal for a system of this size.
| Batteries in series | = System voltage parameter ÷ Chosen battery voltage |
|---|---|
| = 12 V ÷ 12 V | |
| = 1 × 12 V battery is sufficient |
Step 6: Calculate number of batteries in parallel
An MK Deka 12V 55Ah AGM battery will be used with this design. Specifications sheet
| Batteries in parallel | = Total Ah required (step 4) ÷ Chosen battery Ah rating |
|---|---|
| = 43 Ah ÷ 55 Ah = .78 | |
| = Round up to 1 × 55 Ah battery. |
Step 7: Calculate final Ah capacity
| Final Ah capacity | = Number of batteries in parallel (Step 7) × Chosen battery Ah rating |
|---|---|
| = 1 battery in parallel × 55 Ah = 55Ah |
- ↑ Trojan Battery Company - Battery Sizing Guidelines https://www.trojanbattery.com/pdf/TRJN0168_BattSizeGuideFL.pdf
