Detailed AC/DC system design

Physical evaluation

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.

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  Photo of the building.

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  Location of the building.

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

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  Retrieving PVGIS monthly insolation data.

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  Retrieving PVGIS monthly insolation data.

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  Retrieving PVGIS weather data.

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  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 20°C. An AGM battery will be used to avoid regular maintenance.

• Battery temperature correction factor = 1.03

Correction factors for various battery types:

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)
	= 763 Wh ÷ 24 V × 1.03 × 2 days ÷ .5 = 131 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
	= 24 V ÷ 12 V
	= 2 × 12 V battery

Step 6: Calculate number of batteries in parallel
A Trojan 12 V 135 Ah AGM battery will be used with this design. Specifications sheet

Batteries in parallel	= Total Ah required (step 4) ÷ Chosen battery Ah rating
	= 131 Ah ÷135 Ah = .97
	= Round up to 1 × 135 Ah battery.

Step 7: Calculate final Ah capacity

Final Ah capacity	= Number of batteries in parallel (Step 7) × Chosen battery Ah rating
	= 2 batteries in parallel × 135 Ah = 135 Ah

Minimum PV source size

Step 1: Deteremine PV source loss parameters

The PV module(s) will be mounted on a pole mount system.

• Module degradation parameter = .94
• Shading loss parameter = .96
• Soiling loss parameter = .97
• Wiring loss parameter = .96
• Module mismatch parameter = 1
• Mounting system temperature adder = 20°C for a pole mount
• PV source temperature loss parameter = -.41%/°C

PV source temperature loss parameter	= 1 + (Maximum ambient temperature + Mounting system temperature adder - 25°C) x Temperature coefficient of max power %/°C ÷ 100
	= 1 + (35°C + 20°C - 25°C) x -.41%/°C ÷ 100 = .88

Total PV source loss parameter	= Module degradation parameter × Shading loss parameter × Soiling loss parameter × Wiring loss parameter × Module mismatch parameter × PV source temperature loss parameter
	= .94 × .96 × .97 × .96 × 1 × .88 = .74

Step 2: Charge controller efficiency parameter

All charge controllers lose a certain percentage of all energy that is produced as heat as it is transferred to the energy storage system and loads. For both PWM and MPPT charge controllers a value of .98 (98% efficient) can be used.

Step 3: Energy storage efficiency parameter

The system will use an AGM battery, which is a VRLA battery.

• Valve-regulated lead acid (VRLA) battery efficiency = .85 (85% efficient)

Step 4: Deteremine minimum size of the PV source

Minimum PV source size	= Design daily watt-hours required ÷ Design daily insolation ÷ Total PV source loss parameter (Step 1) ÷ Charge controller efficiency parameter (Step 2) ÷ Energy storage efficiency parameter (Step 3)
	= 763 Wh ÷ 3.7 kWh/m² ÷ .74 ÷ .98 ÷ .85 = 335 W

Step 5: Determine charge controller type

A PWM charge controller is a reliable, low-cost option for a small system like this.

PWM charge controller sizing and selection

Step 1: Determine PV module power rating

The chosen system voltage limits the choices of modules and configurations that are possible with a PWM charge controller. A 24 volt system requires 1 × 72-cell module per string. The minimum PV source size was determined to be 335 W.
A Canadian Solar 340W 72-cell polycrystalline PV module will be used for the design. Specifications sheet

PV module power rating	= 340 W

Number of modules in series	= 1 module

Step 2: Determine proposed module configuration

This calculation will give a minimum number of PV modules - the result should always be rounded up. Different modules sizes and configurations can be explored to find the optimal design.

Minimum number of PV modules	= Minimum PV source size ÷ PV module power rating (Step 1)
	= 335 W ÷ 340 W = .99
	= 1 × 340 W module

Minimum number of modules in parallel	= Minimum number of PV modules ÷ Number of modules in series (Step 1)
	= 1 ÷ 1 = 1

Step 3: Verify excess production

This system is not used everyday, therefore it will generate sufficient energy on the days when it is not used to recharge the energy storage system.

Step 4: Verify charging current

This system will use an AGM battery, so its charge current should be between .05 (5%) and .2 (20%) of its C/20 rating.

Minimum required charge current	= Final Ah capacity × .05
	= 135 Ah × .05 = 6.75 A

It is necessary to check the minimum required charge current against the available charge current from the proposed PV source power rating.

Available charging current	= Maximum power current (Imp) × Minimum number of PV modules in parallel
	= 9.05 A × 1 = 9.05 A

Percentage of C/20 rate	= Available charging current ÷ Final Ah capacity
	= 9.05 A ÷ 135 Ah = .07

The PV source can supply .07 (7%) of the C/20 current rating of the energy storage system, which is more than .05 (5%) and less than .2 (20%). The PV source configuration is okay.

Step 5: Determine final number of PV modules

Determine a final number of modules and a series/parallel configuration that can meet the requirements of Step 1, Step 2, Step 3, and Step 4.

Final number of PV modules	= 1

Final number of PV modules in series	= 1

Final number of PV modules in parallel	= 1

Step 6: Total PV source current

This calculation will give a minimum current rating to use as a basis for selecting the charge controller. The Isc rating of the PV module can be found on its specifications sheet.

Total PV source current	= Final number of PV modules (Step 5) × Isc rating of chosen module (Step 1) × Irradiance safety parameter
	= 1 × 9.62 A × 1.25 = 12 A

Step 7: Select a charge controller

The final chosen charge controller should:

1. Function at the system voltage.
2. The charge controller(s) should have a total current rating that is larger than the minimum current rating (Step 6).

A Victron BlueSolar PWM-Pro 20 A charge controller will be used. Specifications sheet

The result of the following equation should always be rounded up.

Number of charge controllers	= Total PV source current (Step 6) ÷ Chosen charge controller current rating
	= 12 A ÷ 20 A = .6 = 1 (round up to 1 charge controller)

Step 8: Determine final PV source power rating

PV source power rating	= PV module power rating (Step 1) × Final number of PV modules in parallel (Step 5)
	= 340 W × 1 = 340 W