Detailed AC system design

Physical evaluation

Location: Colonia Martires, Alta Verapaz, Guatemala
GPS coordinates: 15.37083333, -90.46555556
Altitude: 1450 m
Grid voltage and frequency: 120 V at 60 Hz
Description:
A school with dozens of students that have significant lighting and AC power needs. Classes are in session from the end of January until the middle of October. The building is used heavily from Monday to Friday with load usage typically occuring during the day. There is also occasional additional usage during the weekends. The community would like a system that offers extra energy and power for the future.

As thes school and power demands are relatively large, the system will rely on AC for all lighting and loads. With a large system of this type a 48 V energy storage system will be the best design to keep wire sizes low and reduce parallel circuits of batteries.

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  Photo of the building.

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  Location of the building.

Load evaluation

As the system is used for 5 consecutive days of the week plus additional usage during the weekend, the system should be designed assuming that the loads will be used 7 days a week.

DC load evaluation

Step 1: Fill out DC load chart

The only DC load is the inverter. The inverter has a regular standby consumption of 30 W. The inverter also offers power saving modes, but we will assume that the inverter will be continuously operating in its regular mode.

						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	Inverter	1	25 W	25 W	1	24 hours	7 days	600 Wh	24 hours	7 days	600 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
	= 600 Wh

Total average daily DC watt-hours (October - March)	= sum of Average daily DC watt-hours for all loads for October - March
	= 600 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	7 days	1059 Wh	3 hours	7 days	1059 Wh
2	Stereo	1	30 W	30 W	1	0	0	.9	33 VA	3 hours	7 days	106 Wh	3 hours	7 days	106 Wh
3	Cell phone	10	5 W	50 W	1	0	0	.9	56 VA	1 hour	7 days	59 Wh	1 hour	7 days	59 Wh
4	Laptop	15	20 W	300 W	1	0	0	.5	600 VA	3 hours	7 days	1059 Wh	3 hours	7 days	1059 Wh
5	LED television	2	30 W	60 W	1	0	0	.75	80 VA	2 hours	7 days	141 Wh	2 hours	7 days	141 Wh
6	Printer	2	70 W	140 W	1	0	0	.65	215 VA	.25 hour	7 days	41 Wh	.25 hour	7 days	41 Wh
7	LED lights	30	5 W	5 W	1	0	0	.75	200 VA	8 hours	7 days	1412 Wh	8 hours	7 days	1412 Wh
8	Refrigerator	1	50 W	50 W	.5	0	0	.5	100 VA	24 hours	7 days	706 Wh	24 hours	7 days	706 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
	= 4582 Wh

Total average daily AC watt-hours (October - March)	= sum of Average daily AC watt-hours for all loads for October - March
	= 4582 Wh

Step 4: Determine AC power demand

Total VA	= sum of volt-amperes (VA)
	= 1618 VA

Total VA with surge watts	= sum of Surge watts for all loads + Total VA
	= 1768 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)
	= 600 Wh + 4582 Wh = 5182 Wh

Average daily watt-hours required (April - September)	= Total average daily DC watt-hours (October - March) + Total average daily AC watt-hours (October - March)
	= 600 Wh + 4582 Wh = 5182 Wh

Weather and solar resource evaluation

Maximum ambient temperature = 31°C
Minimum ambient temperature = 7°C
Maximum indoor temperature = 30°C
Minimum indoor temperature = 12°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	108.4 kWh/m²	5182 Wh	47.83
February	131.6 kWh/m²	5182 Wh	39.39
March	145.2 kWh/m²	5182 Wh	32.32
April	143.5 kWh/m²	5182 Wh	30.29
May	123.2 kWh/m²	5182 Wh	36.33
June	132.8 kWh/m²	5182 Wh	41.55
July	143.3 kWh/m²	5182 Wh	37.64
August	167.6 kWh/m²	5182 Wh	39.65
September	112.4 kWh/m²	5182 Wh	55.72
October	147.0 kWh/m²	5182 Wh	53.15
November	139.0 kWh/m²	5182 Wh	53.75
December	134.3 kWh/m²	5182 Wh	45.70

• 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
	= 93.0 kWh/m² ÷ 30 = 3.1 kWh/m²

Design daily watt-hours required	= Total average daily energy demand from month with the highest ratio
	= 5182 Wh

Design parameters

System voltage parameter = 48 V 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 = 46 V

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 = 3

Step 3: Determine battery temperature correction factor
The minimum indoor temperature was determined to be 12°C. Flooded lead acid batteries for longevity and low-price.

• Battery temperature correction factor = 1.19

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)
	= 5182 Wh ÷ 48 V × 1.19 × 3 days ÷ .5 = 771 Ah

Two series strings of Rolls S6-L16 6 V, 390 Ah flooded lead acid batteries will provide 780 Ah of storage capacity. Specifications sheet

Step 5: Calculate number of batteries in series

Batteries in series	= System voltage parameter ÷ Chosen battery voltage
	= 48 V ÷ 6 V
	= 8 × 6 V batteries

Step 6: Calculate number of parallel battery circuits

Number of parallel battery circuits	= Total Ah required (Step 4) ÷ Chosen battery Ah rating
	= 770 Ah ÷ 390 Ah = 1.97
	= Round up to 2 parallel circuits

Step 7: Calculate final Ah capacity

Final Ah capacity	= Number of parallel battery circuits (Step 6) × Chosen battery Ah rating
	= 2 parallel circuits × 390 Ah = 780 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 = .98
• Mounting system temperature adder = 30°C for a roof mount
• Temperature coefficient of max power %/°C = -.39%/°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 + (31°C + 30°C - 25°C) x -.39%/°C ÷ 100 = .86

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 × .98 × .86 = .71

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 a flooded lead acid battery.

• Flooded lead acid battery battery(FLA) efficiency = .75 (75% 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)
	= 5182 Wh ÷ 3.1 kWh/m² ÷ .71 ÷ .98 ÷ .75 = 3206 W

Step 5: Determine charge controller type

An MPPT is a good option for a system of this size as the additional cost is relatively low compared the total system cost and it will significantly increase system performance.

MPPT charge controller sizing and selection

A MPPT charge controller is rated to operate at a particular system voltage, maximum current and maximum voltage. MPPT charge controllers can charge the battery bank with any series and parallel configuration of modules that doesn't exceed the maximum voltage and maximum current or drop below the required charging voltage of the energy storage system. Exceeding the voltage rating of an MPPT due to cold temperatures can damage it. Many charge controllers allow the current rating to be exceeded to a certain point without damage, just lost energy - it depends on the charge controller. There are several important calculations that must be performed to properly size an MPPT charge controller:

• Should be sized to work with a series and parallel configuration of the PV source that will not damage the charge controller due to high voltages resulting from low temperatures at the project location.
• Should be sized to work with a series and parallel configuration of PV source that will still be able to properly charge the energy storage system under high temperatures and as PV modules age at the project location.

There are various tools that can greatly simplify this process:

• Open Source Solar Project Design Tool
• Many manufacturers provide tools that are specific for their products on their websites.

Step 1: Determine PV module power rating

A Yingli YGE 60 Cell, Series 2 polycrystaline module with a power rating of 285 watts will be used. Specifications sheet

Module specifications:

• Power = 285 W
• Open circuit voltage (Voc) = 38.2 V
• Short circuit current (Isc) = 9.45 A
• Max power voltage (Vmp) = 31.5 V
• Max power current (Imp) = 8.95 A
• Temperature coefficient of open circuit voltage (TkVoc) = -.30 %/°C
• Temperature coefficient of max power (TkPmp) = -.39 %/°C

Step 2: Determine minimum number of PV modules

This calculation will give a minimum number of modules. The final array size should always be larger than this value, thus if the result of the calculation is a decimal, it should 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)
	= 3206 W ÷ 285 W = 11.25 modules
	= 12 modules (rounded up)

Step 3: Minimum PV source power rating

This calculation will give a power rating of the PV source based upon the chosen module size and the number of modules required.

Minimum PV source power rating	= Minimum number of PV modules (Step 2) × PV module power rating (Step 1)
	= 12 modules × 285 W = 3420 W

Step 4: Determine the minimum PV source current

An MPPT charge controller is capable of of accepting varying voltages from the array and converting them into current at the proper charging voltage for the energy storage system. The maximum current of the PV source can be calculated by dividing the power rating of the array by the system voltage. If the charge controller manufacturer explicitly permits it, the PV source may be oversized somewhat (typically 110-125%). Larger systems often require multiple charge controllers operating in parallel.

Minimum PV source current	= Minimum PV source power rating (Step 3) ÷ System voltage
	= 3420 W ÷ 48 V = 71.25 A

Step 5: Select a charge controller

The final chosen charge controller should:

1. Function at the system voltage.
2. Have a current rating that is larger than the minimum PV source current rating (Step 4) or multiple charge controllers will have to be used. A single charge controller is the simplest and most cost-effective option.

A Victron MPPT 250/100 SmartSolar charge controller will be used for this design. (It would be possible to use the 75 A version here, but the array size will be increased in later steps, so the largest charge controller will be used here to avoid creating confusion). Specifications sheet

Charge controller maximum input voltage	= From charge controller specifications sheet
	= 250 V

Charge controller current rating	= From charge controller specifications sheet
	= 100 A

Step 6: Determine maximum number of PV modules in series

PV module cell temperatures below 25°C will increase the voltage a PV module beyond its rating. In locations that experience low temperatures, it is necessary to determine the maximum number of modules in series that will be possible given the minimum temperature at the project location. PV module manufacturers provide a temperature coefficient for voltage that can be used to calculate increases or decreases in power based upon the environmental conditions. This coefficient is referred to as temperature coefficient of open circuit voltage (Voc) and can typically be found on module specifications sheets in -%/°C.

The maximum voltage of the module under standard test conditions - open circuit voltage (Voc) - will be used for this calculation.

% change in Voc at minimum temperature	= (Minimum ambient temperature - 25°C) × Temperature coefficient of open circuit voltage (Voc)
	= (12°C - 25°C) * -.30%/°C = 3.9%

Voc at minimum temperature	= PV module open circuit voltage (Voc) × ((% change in Voc at minimum temperature ÷ 100) + 1)
	= 35.5 V × ((3.9 ÷ 100) + 1) = 38.3 V

Maximum number of PV modules in series	= Maximum Voc rating of charge controller ÷ Voc at minimum temperature
	= 250 V ÷ 38.3 V = 6.5
	= 6 modules (round down)

This value must be rounded down to the next whole number (there are no partial modules).

Step 7: Determine maximum charging voltage parameter

The minimum number of PV modules in series must be calculated based upon the maximum required charging voltage for the energy storage system or the minimum charging voltage provided by the MPPT charge controller manufacturer in the specifications sheet. The maximum system charging voltage parameter is the value for the maximum voltage at which the energy storage system will be charged. This value depends upon the system voltage parameter and the energy storage system type. The specifications sheet or user manual for the battery that is used in the system should be consulted.

The maximum charging voltage for Rolls S6-L16 6 V, 390 Ah flooded lead acid batteries is 60 volts. Specifications sheet

Step 8: Determine minimum number of PV modules in series

PV module cell temperatures above 25°C will decrease the voltage a PV module beyond its rating. PV module voltage will also decrease as the module ages. It is therefore important to make sure that the PV source is adequately sized to ensure that at high temperatures and with the passage of time that the array will still be able to provide sufficient voltage to charge the energy storage system. PV module manufacturers provide a temperature coefficient for power that can be used to calculate increases or decreases in power based upon the environmental conditions. This coefficient is referred to as temperature coefficient of max power and can typically be found on module specifications sheets in -%/°C. The value from the specifications sheet of a module can be used in these calculations if a module has been chosen, but a standard average value of (-.48%/°C) will work for both poly and monocrystalline modules.

The operating voltage of the module under standard test conditions - maximum power voltage (Vmp) - will be used for this calculation.

The mounting system will also affect the ability of the PV source to cool itself. A mounting system temperature adder should be added to the maximum temperature that is used to calculate the decrease in Voc:

• 20°C for pole mount
• 25°C for ground mount
• 30°C for roof mount

This system is a roof mounted system, so it will use a value of 30°C.

% change in Vmp at maximum temperature	= (Maximum ambient temperature + Array temperature adder - 25°C) × Temperature coefficient of max power %/°C
	= (31°C + 30°C - 25°C) × -.39%/°C = -14.0%

Vmp at maximum temperature	= Maximum power voltage (Vmp) × ((% change in Vmp at maximum temperature ÷ 100) + 1) × Module degradation parameter
	= 31.5 V × ((-14% ÷ 100) + 1) × .94 = 25.5 V

Minimum number of PV modules in series	= Maximum charging voltage parameter (Step 7) ÷ Vmp at maximum temperature
	= 60 V ÷ 25.5 V = 2.35 modules
	= 3 modules (round up)

This value must be rounded up to the next whole number (there are no partial modules).

Attempt 1:

The configuration of the PV source and MPPT charge controller must meet all of the considerations in the following steps. It often necessary to perform this process various times to find the optimal design.

Step 9: Determine the number of PV modules in series per string

The proposed number of PV modules per series string must be less than the maximum number of PV modules in series (Step 5) and greater than the minimum number of PV modules in series (Step 6). If there is not a configuration that is meets these criteria then another module or charge controller should be used in the design. As long as the voltage doesn't exceed the rating of the charge controller, more PV modules per string is generally preferrable to use a higher operating voltage and lower current.

The design calls for a minimum of 12 PV modules. A maximum of six modules per string is possible. It is an even number that will work well.

Step 10: Determine the number of strings of PV modules in parallel

The proposed number of parallel PV circuits will be determined by dividing the minimum array size by the final number of PV modules per series string. This calculation should be rounded up. The total short circuit current (Isc) of the parallel PV circuits should not exceed the current rating of the charge controller unless permitted by the manufacturer. A larger charge controller should be chosen if an appropriate series/parallel configuration cannot be found.

The 6 modules per string would mean that there would have to be 2 parallel circuits to have 12 modules.

Minimum number of parallel PV circuits	= Minimum number of PV modules (Step 2) ÷ Minimum number of PV modules per series string (Step 8)
	= 12 modules ÷ 6 modules in series = 2 parallel PV circuits

Step 11: Determine proposed PV source power rating

The total power rating of the PV source can be calculated by multiplying the power rating of the chosen PV module by the final number of PV modules in series (Step 7) and the number of parallel PV circuits (Step 8).

Proposed PV source power rating	= PV module power rating (Step 1) × Number of PV modules in series (Step 8) × Number of parallel PV circuits (Step 9).
	= 3420 W

Step 12: Verify excess production

During periods of poor weather or low solar resource, an off-grid PV system is designed to discharge the battery to a certain depth of discharge which can leave the energy storage system depleted. It is important that the energy storage system is brought back up to a full state of charge in short period of time or the cycle life of the batteries will be reduced. The PV array therefore must be sized to generate sufficient excess energy, while continuing to meet all of the power needs from the load evaluation. It is recommended that the array be sufficiently sized to reach a full state of charge within 7 days or that the system incorporate a generator to ensure adequate charging.

If the system is not used heavily everyday, then the number of days to reach full state of charge can be more than 7 as the system will have extra energy on days when it is not used or used lightly to charge the energy storage system.

Proposed PV source low insolation production	= Proposed PV source power rating (Step 11) × Total PV source loss parameter × Design daily insolation × Charge controller efficiency parameter × Energy storage efficiency parameter
	= 3420 W × .71 × 3.1 kWh/m² × .98 × .75 = 5532 Wh

Daily excess production in Ah	= (Proposed PV source low insolation production - Design daily watt-hours required ) ÷ System voltage parameter
	= 5532 Wh - 5182 Wh ÷ 48 V = 7.3 Ah

Ah used at full depth of discharge	= Final Ah capacity × Depth of discharge parameter
	= 780 Ah × .5 = 390 Ah

Time to reach full state of charge	= Ah used at full depth of discharge ÷ Daily excess production in Ah
	= 390 Ah ÷ 7.3 Ah = 53.4 days

Array is undersized. 53.4 days is far too long.

Attempt 2:

The configuration of the PV source and MPPT charge controller must meet all of the considerations in the following steps. It often necessary to perform this process various times to find the optimal design.

Step 9: Determine the number of PV modules in series per string

The proposed number of PV modules per series string must be less than the maximum number of PV modules in series (Step 5) and greater than the minimum number of PV modules in series (Step 6). If there is not a configuration that is meets these criteria then another module or charge controller should be used in the design. As long as the voltage doesn't exceed the rating of the charge controller, more PV modules per string is generally preferrable to use a higher operating voltage and lower current.

The design calls for a minimum of 12 PV modules, but this array size was undersized. This attempt will call for 16 PV modules.

Step 10: Determine the number of strings of PV modules in parallel

The proposed number of parallel PV circuits will be determined by dividing the minimum array size by the final number of PV modules per series string. This calculation should be rounded up. The total short circuit current (Isc) of the parallel circuits should not exceed the current rating of the charge controller unless permitted by the manufacturer. A larger charge controller should be chosen if an appropriate series/parallel configuration cannot be found.

If there need to be 16 PV modules, then the only number of series modules per string that divides evenly into 16 is 4 (must be 3-6 modules per string).

Minimum number of parallel PV circuits	= Minimum number of PV modules (Step 2) ÷ Minimum number of PV modules per series string (Step 8)
	= 16 modules ÷ 4 modules in series = 4 parallel circuits

Step 11: Determine proposed PV source power rating

The total power rating of the PV source can be calculated by multiplying the power rating of the chosen PV module by the final number of PV modules in series (Step 7) and the number of parallel PV circuits (Step 8).

Proposed PV source power rating	= PV module power rating (Step 1) × Number of PV modules in series (Step 8) × Number of parallel PV circuits (Step 9).
	= 4560 W

Step 12: Verify excess production

During periods of poor weather or low solar resource, an off-grid PV system is designed to discharge the battery to a certain depth of discharge which can leave the energy storage system depleted. It is important that the energy storage system is brought back up to a full state of charge in short period of time or the cycle life of the batteries will be reduced. The PV array therefore must be sized to generate sufficient excess energy, while continuing to meet all of the power needs from the load evaluation. It is recommended that the array be sufficiently sized to reach a full state of charge within 7 days or that the system incorporate a generator to ensure adequate charging.

If the system is not used heavily everyday, then the number of days to reach full state of charge can be more than 7 as the system will have extra energy on days when it is not used or used lightly to charge the energy storage system.

Proposed PV source low insolation production	= Proposed PV source power rating (Step 11) × Total PV source loss parameter × Design daily insolation × Charge controller efficiency parameter × Energy storage efficiency parameter
	= 4560 W × .71 × 3.1 kWh/m² × .98 × .75 = 7377 Wh

Daily excess production in Ah	= (Proposed PV source low insolation production - Design daily watt-hours required ) ÷ System voltage parameter
	= 7377 Wh - 5182 Wh ÷ 48 V = 45.73 Ah

Ah used at full depth of discharge	= Final Ah capacity × Depth of discharge parameter
	= 780 Ah × .5 = 390 Ah

Time to reach full state of charge	= Ah used at full depth of discharge ÷ Daily excess production in Ah
	= 390 Ah ÷ 45.73 = 8.53 days

The battery will be able to reach full state of charge while using loads in 8.53 days, which is acceptable as the system will have the weekend to recharge. The design is okay.

Step 13: Verify charging current

Lead acid batteries last longer and perform better when they are regularly recharged with a current in a certain range that depends upon battery type. Flooded lead acid and gel batteries should be charged with current that is between .05-.13 (5-13%) of their C/20 rating. AGM batteries can should be charged with a current that is between .05-.2 (5-20%) of their C/20 rating. If a system uses many loads during the day, this will limit the available charging current for the energy storage system and should be taken into account by increasing the PV source size. Most designs should have a charge rate between 5-10% - closer to 10% if the system is used heavily during the day. It is necessary to consult the manual or manufacturer for recommended maximum and minimum charging currents.

These calculations are performed with the Ah rating of the total energy storage system.

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	= Proposed PV source power rating (Step 11) ÷ Maximum charging voltage parameter (Step 7)
	= 4560 W ÷ 60 V = 76 A

Percentage of C/20 rate	= Available charging current ÷ Final Ah capacity
	= 76 ÷ 780 = .097 (9.7%)

The array configuration is within the acceptable range. The design is okay.

Step 14: Determine final PV source power rating

Final PV source power rating	= PV module power rating (Step 1) × Final number of PV modules in series × Final number of parallel PV circuits
	= 285 W × 4 modules in series × 4 parallel PV circuits = 4560 W

Inverter sizing and selection

The basic considerations for sizing and selecting an inverter are the following:

1. The voltage must match the system voltage.
2. The inverter should be able to meet the continuous power demand for all loads that will operate at the same time.
3. The inverter should be able to meet the continuous power demand for all loads and the surge power demand for all loads that will operate at the same time.

Step 1: Determine the inverter continuous output rating

Minimum inverter continuous duty rating	= Total VA
	1618 VA

Step 2: Determine minimum inverter surge rating

Minimum inverter surge rating	= Total VA with surge watts
	1768 VA

Step 3: Determine final inverter continuous duty rating

The chosen inverter should:

1. Match the system voltage.
2. Have a larger continuous output rating than determined in Step 1.
3. Have a larger surge rating than the determined in Step 2.

A Victron Quattro 48V, 3000VA 120V Inverter/Charger Phoenix will be used. Specifications sheet

Specifications:

• Nominal input voltage = 48 V
• Nominal output voltage = 120 V
• Nominal frequency = 60 Hz
• Continuous power rating (W) at 25°C = 2400 W
• Continuous power rating (VA) at 25°C = 3000 VA
• Peak/surge power (W) = 6000 W

Step 4: Verify that maximum continuous current of inverter is acceptable

The energy storage system must be large enough to be able to supply the continuous current that the inverter requires. Using an inverter that is oversized relative to the energy storage system can damage the batteries and can cause the system to function improperly. It is recommended that the continous current required by the inverter not exceed .13 (13%) of the Final Ah capacity of the energy storage system for lead acid batteries and .2 (20%) for AGM batteries. If this value is exceeded, then the size of the energy storage system should be increased.

Maximum recommended discharge current	= Final Ah capacity × (.13 for FLA and gel batteries, .2 for AGM batteries)
	= 780 Ah × .13 = 101.4 A

Inverter continuous duty input current	= Final inverter continuous duty rating (Step 3) ÷ System voltage parameter ÷ Inverter efficiency parameter
	= 3000 W ÷ 48 V ÷ .85 = 73.5 A

The inverter at maximum capacity will draw less than .13 (14%) of the C/20 rate of the energy storage system. The inverter size is okay.

Wire, overcurrent protection, and disconnect sizing and selection

The chosen wire for a circuit must meet the requirements set out in each phase of this process. The wire size must be increased if it fails to meet any of these phases and then the process must be performed again with the new wire size.

PV source circuit

Phase 1: Maximum circuit current

Maximum circuit current	= PV module Isc × Irradiance safety parameter
	= 9.45 A × 1.25 = 11.81 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum ambient temperature 35°C. 90°C rated PV wire will be used for this circuit.

The minimum wire size for a circuit can be using the following steps:

1. Determine the ambient temperature correction factor based upon the maximum ambient temperature using the ambient temperature correction factor table. The ambient temperature correction factor for this circuit is .96 because it is a 90°C rated wire with a maximum ambient temperature of 31°C.
2. Determine the conduit fill correction factor based upon the number of number of conductors in the conduit using the conduit fill correction factor table There will only be two current-carrying conductors in the conduit, so the conduit correction fill factor is 1.
3. Determine the total wire correction parameter based upon the smaller of: the Ambient temperature correction multiplied by the Conduit fill correction factor or .8 (a safety factor from the US National Electrical code).

Total wire correction parameter	= Smaller of (Ambient temperature correction factor × Conduit fill correction factor) or .8
	= Smaller of (.96 × 1) or .8 = .8

4. Determine minimum wire ampacity. Divide the maximum circuit current (Phase 1) by the total wire correction factor (Step 3)

Minimum wire ampacity	= Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
	= 11.81 A ÷ .8 = 14.76 A

5. Select a wire size with a maximum rated ampacity equal to or above the minimum wire ampacity calculated in the previous step using the allowable wire ampacity table. A 4 mm² PV wire with an ampacity rating of 30 A at 90°C will be used for this circuit as this is a commonly used wire type for a circuit of this type. 30 A is larger than the required 14.76 A ampacity.

Phase 3: Overcurrent protection and disconnects

Each PV source circuit - there are 4 - will require overcurrent protection.

The appropriate overcurrent protection device size can be determined by:

1. Determine the minimum size

Minimum OCPD size	= Maximum circuit current (Phase 1) × 1.25
	= 11.81 A × 1.25 = 14.76 A

2. A standard 16 A DC fuse will be used. This is fuse is larger than the required minimum OCPD size, but smaller than the rated 30 A ampacity of the wire that it will protect.
3. Verify that the chosen OCPD size from Step 2 will protect the wire size chosen in Phase 2 from excessive current under the conditions of use. The current rating of the chosen OCPD size (Step 2) must be less than the calculated maximum current under conditions of use unless the calculated maximum current under conditions of use is between standard OCPD values, in this case the next largest size can be chosen.

   Maximum current under conditions of use	= Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
   	= 30 A × .8 = 24 A

   Verify OCPD under conditions of use	= Maximum current under conditions of use must be greater than or equal to the current rating of the chosen OCPD (Step 2)
   	= 24 A is greater than 16 A.

The OCPD size is okay.

Phase 4: Voltage drop

If the voltage drop for the wire chosen in Phase 2 for a particular circuit is not within the recommended values, then using a larger sized wire should be considered. Increasing the wire size will not affect any other part of this process; the calculated OCPD size can remain the same.

This circuit will be 4 meters long one-way. It is 4 mm² wire with a resistance value in (Ω/Km) of 6.73 Ω. It is recommended that the voltage drop between the PV source and the charge controller be kept below 2%.

PV source circuit current	= Max power current of the PV module
	= 8.96 A

PV source circuit nominal voltage	= Number of PV modules in series × Max power voltage of the PV module
	= 31.5 V × 4 = 126 V

Voltage drop	= 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
	= 2 × 8.96 A × 4m × 6.73 Ω ÷ 1000 = .48 V

Percentage voltage drop	= Voltage drop ÷ Nominal circuit voltage x 100
	= .73 V ÷ 126 V × 100 = .38%

.38% voltage drop for this circuit is acceptable. The wire size is okay.

Phase 5: Terminal temperature ratings

The PV wire is rated at 90°C, but the terminals on for the fuse holder are rated at 75°C. It must be verified that the OCPD will still protect these terminals at their lower temperature rating. If the wire is not protected by the OCPD at the temperature rating of the terminals in the circuit, then the wire size must be increased until the circuit passes this test. No other calculations need to be performed again. Increasing the wire size will have the added benefit of decreasing voltage drop.

Step 1: Maximum ampacity based upon terminal temperature rating

Maximum ampacity based upon terminal temperature rating	= Wire size (Phase 2) ampacity at terminal temperature rating from allowable wire ampacity table × Total wire correction parameter (Phase 2)
	= 25 A × .8 = 20 A

Step 2: Verify OCPD protection of terminals

Verify OCPD protection of terminals	= Maximum current considering terminal temperature rating (Step 1) must be greater than or equal to the current rating of the chosen OCPD (Phase 3)
	= 20 A is greater than the 16 A current rating of the fuse

Wire and OCPD size are okay.

PV output circuit

Phase 1: Maximum circuit current

Maximum circuit current	= PV module Isc × Number of parallel PV circuits × Irradiance safety parameter
	= 9.45 A × 4 × 1.25 = 47.25 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum ambient temperature 35°C. 90°C rated PV wire will be used for this circuit.

The minimum wire size for a circuit can be using the following steps:

1. Determine the ambient temperature correction factor based upon the maximum ambient temperature using the ambient temperature correction factor table. The ambient temperature correction factor for this circuit is .96 because it is a 90°C rated wire with a maximum ambient temperature of 31°C.
2. Determine the conduit fill correction factor based upon the number of number of conductors in the conduit using the conduit fill correction factor table There will only be two current-carrying conductors in the conduit, so the conduit correction fill factor is 1.
3. Determine the total wire correction parameter based upon the smaller of: the Ambient temperature correction multiplied by the Conduit fill correction factor or .8 (a safety factor from the US National Electrical code).

Total wire correction parameter	= Smaller of (Ambient temperature correction factor × Conduit fill correction factor) or .8
	= Smaller of (.96 × 1) or .8 = .8

4. Determine minimum wire ampacity. Divide the maximum circuit current (Phase 1) by the total wire correction factor (Step 3)

Minimum wire ampacity	= Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
	= 47.25 ÷ .8 = 59.1 A

5. Select a wire size with a maximum rated ampacity equal to or above the minimum wire ampacity calculated in the previous step using the allowable wire ampacity table. A 25 mm² 90°C dry / 75° rated wire with an ampacity rating of 85 A at 75°C will be used for this circuit as it will be exposed to water. 85 A is larger than the required 59.1 A ampacity.

Phase 3: Overcurrent protection and disconnects

The appropriate overcurrent protection device size can be determined by:

1. Determine the minimum size

Minimum OCPD size	= Maximum circuit current (Phase 1) × 1.25
	= 47.25 A × 1.25 = 59.1 A

2. A standard 63 A DC breaker will be used. This is breaker is larger than the required minimum OCPD size, but smaller than the rated 30 A ampacity of the wire that it will protect.
3. Verify that the chosen OCPD size from Step 2 will protect the wire size chosen in Phase 2 from excessive current under the conditions of use. The current rating of the chosen OCPD size (Step 2) must be less than the calculated maximum current under conditions of use unless the calculated maximum current under conditions of use is between standard OCPD values, in this case the next largest size can be chosen.

   Maximum current under conditions of use	= Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
   	= 85 A × .8 = 68 A

   Verify OCPD under conditions of use	= Maximum current under conditions of use must be greater than or equal to the current rating of the chosen OCPD (Step 2)
   	= 68 A is greater than 63 A.

The OCPD size is okay.

Phase 4: Voltage drop

If the voltage drop for the wire chosen in Phase 2 for a particular circuit is not within the recommended values, then using a larger sized wire should be considered. Increasing the wire size will not affect any other part of this process; the calculated OCPD size can remain the same.

This circuit will be 10 meters long one-way. It is 25 mm² wire with a resistance value in (Ω/Km) of 1.053 Ω. It is recommended that the voltage drop between the PV source and the charge controller be kept below 2%.

PV source circuit current	= Max power current of the PV module × Number of parallel PV circuits
	= 8.96 A × 4 = 35.84 A

PV source circuit nominal voltage	= Number of PV modules in series × Max power voltage of the PV module
	= 31.5 V × 4 = 126 V

Voltage drop	= 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
	= 2 × 35.84 A × 10m × 1.053 Ω ÷ 1000 = .75 V

Percentage voltage drop	= Voltage drop ÷ Nominal circuit voltage x 100
	= .75 V ÷ 126 V × 100 = .60%

.60% voltage drop for this circuit is acceptable. The wire size is okay.

Phase 5: Terminal temperature ratings

The terminals of the charge controller and breaker are rated for 75°C, but these calculations do not need to be performed as all previous calculations were already performed using 75°C rated ampacity values.

Wire and OCPD size are okay.

Charge controller output circuit

The charge controller output circuit will be connected to common DC busbars (along with the inverter) that will be connected to the energy storage system by the energy storage circuit.

Phase 1: Maximum circuit current

Maximum circuit current	= Current rating of the charge controller
	= 100 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum indoor temperature 30°C. 90°C dry / 75° wet rated wire will be used for this circuit.

The minimum wire size for a circuit can be using the following steps:

1. Determine the ambient temperature correction factor based upon the maximum ambient temperature using the ambient temperature correction factor table. The ambient temperature correction factor for this circuit is 1.00 because it is a 90°C dry / 75° wet rated wire with a maximum indoor temperature of 30°C.
2. Determine the conduit fill correction factor based upon the number of number of conductors in the conduit using the conduit fill correction factor table There will only be two current-carrying conductors in the conduit, so the conduit correction fill factor is 1.
3. Determine the total wire correction parameter based upon the smaller of: the Ambient temperature correction multiplied by the Conduit fill correction factor or .8 (a safety factor from the US National Electrical code).

Total wire correction parameter	= Smaller of (Ambient temperature correction factor × Conduit fill correction factor) or .8
	= Smaller of (1.00 × 1) or .8 = .8

4. Determine minimum wire ampacity. Divide the maximum circuit current (Phase 1) by the total wire correction factor (Step 3)

Minimum wire ampacity	= Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
	= 100 A ÷ .8 = 125 A

5. Select a wire size with a maximum rated ampacity equal to or above the minimum wire ampacity calculated in the previous step using the allowable wire ampacity table. A 35 mm² 90°C dry / 75° wet rated wire with an ampacity rating of 130 A at 90°C will be used for this circuit as this is a commonly used wire type for a circuit of this type. 130 A is larger than the required 125 A ampacity. It would be better to use 50 mm² wire, but the largest wire that the terminals on the charge controller will accept is 35 mm² wire, so we will have to use that size in the design.

Phase 3: Overcurrent protection and disconnects

This circuit will use an OCPD to protect it from excessive currents from the energy storage system and as an equipment disconnect, so that it can be disconnected from the energy storage system as needed.

The appropriate overcurrent protection device size can be determined by:

1. Determine the minimum size

Minimum OCPD size	= Maximum circuit current (Phase 1) × 1.25
	= 100 A × 1.25 = 125 A

2. A standard 125 A DC breaker will be used. This is breaker is larger than the required minimum OCPD size, but smaller than the 130 A rated ampacity of the wire that it will protect.
3. Verify that the chosen OCPD size from Step 2 will protect the wire size chosen in Phase 2 from excessive current under the conditions of use. The current rating of the chosen breaker size (Step 2) must be less than the calculated maximum current under conditions of use unless the calculated maximum current under conditions of use is between standard OCPD values, in this case the next largest breaker can be chosen.

   Maximum current under conditions of use	= Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
   	= 130 A × .8 = 104 A

   Verify OCPD under conditions of use	= Maximum current under conditions of use must be greater than or equal to the current rating of the chosen OCPD (Step 2)
   	= 104 A is smaller than 125 A.

A 125 A OCPD is too large to properly protect the wire.

The OCPD size will be reduced to a standard 100 A breaker. This will properly protect the wire under the conditions of use, but could potentially disconnect the circuit during periods of sustained high current due to excess heat as it is smaller than the recommended size.

A 100 A OCPD is okay.

Phase 4: Voltage drop

If the voltage drop for the wire chosen in Phase 2 for a particular circuit is not within the recommended values, then using a larger sized wire should be considered. Increasing the wire size will not affect any other part of this process; the calculated OCPD size can remain the same.

This circuit will be .5 meters long one-way as it only needs to reach the DC busbars where the battery circuit is connected. It is 35 mm² wire with a resistance value in (Ω/Km) of 0.661 Ω. It is recommended that the voltage drop between the charge controller and energy storage system be less than 1.5%. The current of the circuit is the

Charge controller circuit output current	= Smaller of (Total PV source power rating ÷ System voltage parameter) or Charge controller current rating
	= Smaller of (5440 W ÷ 48 = 113 A) or (100 A) = 100 A

Charge controller circuit nominal voltage	= System voltage parameter
	= 48 V

Voltage drop	= 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
	= 2 × 100 A × .5m × .661 Ω ÷ 1000 = .07 V

Percentage voltage drop	= Voltage drop ÷ Nominal circuit voltage voltage x 100
	= .07 V ÷ 48 V × 100 = .15%

.15% voltage drop for this circuit is acceptable. The wire size is okay.

Phase 5: Terminal temperature ratings

The wire is rated at 90°C dry / 75° wet, but the terminals on the charge controller and breaker are rated at 75°C. It must be verified that the OCPD will still protect these terminals at their lower temperature rating. If the wire is not protected by the OCPD at the temperature rating of the terminals in the circuit, then the wire size must be increased until the circuit passes this test. No other calculations need to be performed again. Increasing the wire size will have the added benefit of decreasing voltage drop.

Step 1: Maximum ampacity based upon terminal temperature rating

Maximum ampacity based upon terminal temperature rating	= Wire size (Phase 2) ampacity at terminal temperature rating from allowable wire ampacity table × Total wire correction parameter (Phase 2)
	= 115 A × .8 = 92 A

Step 2: Verify OCPD protection of terminals

Verify OCPD protection of terminals	= Maximum current considering terminal temperature rating (Step 1) must be greater than or equal to the current rating of the chosen OCPD (Phase 3)
	= 92 A is less than the 100 A current rating of the breaker

A 100 A OCPD is too large to properly protect the wire.

The wire size of the charge controller is limited by the manufacturer to 35 mm² and therefore cannot be increased in size. This wire/size OCPD combination does not meet the design specifications outlined here, but we will rely on the design of the manufacturer and continue with 35 mm² wire and a 100 A breaker.

Wire and OCPD size are okay.

Inverter input circuit

The charge inverter input circuit will be connected to common DC busbars (along with the charge controller) that will be connected to the energy storage system by the energy storage circuit.

Phase 1: Maximum circuit current

Maximum circuit current	= Final inverter continuous duty rating ÷ Low voltage disconnect parameter ÷ Inverter efficiency parameter
	= 500 W ÷ 22.2 V ÷ .85 = 27 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum indoor temperature 30°C. 90°C dry / 75° wet rated wire will be used for this circuit.

The minimum wire size for a circuit can be using the following steps:

1. Determine the ambient temperature correction factor based upon the maximum ambient temperature using the ambient temperature correction factor table. The ambient temperature correction factor for this circuit is 1.00 because it is a 90°C dry / 75° wet rated wire with a maximum indoor temperature of 30°C.
2. Determine the conduit fill correction factor based upon the number of number of conductors in the conduit using the conduit fill correction factor table There will only be two current-carrying conductors in the conduit, so the conduit correction fill factor is 1.
3. Determine the total wire correction parameter based upon the smaller of: the Ambient temperature correction multiplied by the Conduit fill correction factor or .8 (a safety factor from the US National Electrical code).

Total wire correction parameter	= Smaller of (Ambient temperature correction factor × Conduit fill correction factor) or .8
	= Smaller of (1.00 × 1) or .8 = .8

4. Determine minimum wire ampacity. Divide the maximum circuit current (Phase 1) by the total wire correction factor (Step 3)

Minimum wire ampacity	= Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
	= 27 A ÷ .8 = 33.75 A

5. Select a wire size with a maximum rated ampacity equal to or above the minimum wire ampacity calculated in the previous step using the allowable wire ampacity table. A 16 mm² 75°C dry/wet rated wire with an ampacity rating of 65 A at 75°C will be used for this circuit as this is a commonly used wire type for a circuit of this type. 65 A is larger than the required 33.75 A ampacity.

Phase 3: Overcurrent protection and disconnects

This circuit will use an OCPD to protect it from excessive currents from the energy storage system and as an equipment disconnect, so that it can be disconnected from the energy storage system as needed.

The appropriate overcurrent protection device size can be determined by:

1. Determine the minimum size

Minimum OCPD size	= Maximum circuit current (Phase 1) × 1.25
	= 33.75 A × 1.25 = 42 A

2. A standard 50 A DC breaker will be used. This is breaker is larger than the required minimum OCPD size and smaller than the 65 A rated ampacity of the wire that it will protect.
3. Verify that the chosen OCPD size from Step 2 will protect the wire size chosen in Phase 2 from excessive current under the conditions of use. The current rating of the chosen breaker size (Step 2) must be less than the calculated maximum current under conditions of use unless the calculated maximum current under conditions of use is between standard OCPD values, in this case the next largest breaker can be chosen.

   Maximum current under conditions of use	= Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
   	= 65 A × .8 = 52 A

   Verify OCPD under conditions of use	= Maximum current under conditions of use must be greater than or equal to the current rating of the chosen OCPD (Step 2)
   	= 52 A is greater than 50 A.

The OCPD size is okay.

Phase 4: Voltage drop

If the voltage drop for the wire chosen in Phase 2 for a particular circuit is not within the recommended values, then using a larger sized wire should be considered. Increasing the wire size will not affect any other part of this process; the calculated OCPD size can remain the same.

This circuit will be .5 meters long one-way as it only needs to reach the DC busbars where the battery circuit is connected. It is 16 mm² wire with a resistance value in (Ω/Km) of 1.671 Ω. It is recommended that the voltage drop between the charge controller and energy storage system be less than 1.5%. The nominal circuit voltage is the system voltage: 24 V.

Voltage drop	= 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
	= 2 × 33.75 A × .5m × 1.671 Ω ÷ 1000 = .06 V

Percentage voltage drop	= Voltage drop ÷ Nominal circuit voltage voltage x 100
	= .06 V ÷ 24 V × 100 = .25%

.25% voltage drop for this circuit is acceptable. The wire size is okay.

Phase 5: Terminal temperature ratings

This wire is rated at 75°C, which is the same temperature rating as the terminals on the inverter and breaker. A calculation was already performed in Phase 3 to ensure that the wire, which has a 75°C rating, would be protected by the breaker against excessive current. There is no need to perform the calculation again.

Wire and OCPD size are okay.

Inverter output circuit

The inverter will limit the current that it can output. It is recommended that a residual current device (RCD) be installed on this circuit of the individual AC branch circuits. To minimize costs only one RCD will be installed on the inverter output circuit.

Phase 1: Maximum circuit current

Maximum circuit current	= = Final inverter continuous duty rating ÷ Inverter AC voltage
	= 500 W ÷ 220 V = 2.3 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum indoor temperature 30°C. 90°C dry / 75° wet rated wire will be used for this circuit.

The minimum wire size for a circuit can be using the following steps:

1. Determine the ambient temperature correction factor based upon the maximum ambient temperature using the ambient temperature correction factor table. The ambient temperature correction factor for this circuit is 1.00 because it is a 90°C dry / 75° wet rated wire with a maximum indoor temperature of 30°C.
2. Determine the conduit fill correction factor based upon the number of number of conductors in the conduit using the conduit fill correction factor table There will only be two current-carrying conductors in the conduit, so the conduit correction fill factor is 1.
3. Determine the total wire correction parameter based upon the smaller of: the Ambient temperature correction multiplied by the Conduit fill correction factor or .8 (a safety factor from the US National Electrical code).

Total wire correction parameter	= Smaller of (Ambient temperature correction factor × Conduit fill correction factor) or .8
	= Smaller of (1.00 × 1) or .8 = .8

4. Determine minimum wire ampacity. Divide the maximum circuit current (Phase 1) by the total wire correction factor (Step 3)

Minimum wire ampacity	= Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
	= 2.3 A ÷ .8 = 2.9 A

5. Select a wire size with a maximum rated ampacity equal to or above the minimum wire ampacity calculated in the previous step using the allowable wire ampacity table. A 2.5 mm² 90°C dry / 75° wet rated wire with an ampacity rating of 25 A at 90°C will be used for this circuit as this is a commonly used wire type for a circuit of this type. 25 A is larger than the required 2.9 A ampacity.

Phase 3: Overcurrent protection and disconnects

This circuit could function without an additional OCPD because it is current limited by the inverter, but a residual current device will be sized and added for safety.

The appropriate overcurrent protection device size can be determined by:

1. Determine the minimum size

Minimum OCPD size	= Maximum circuit current (Phase 1) × 1.25
	= 2.3 A × 1.25 = 2.9 A

2. A standard 4 A AC residual current device will be used. This is breaker is larger than the required minimum OCPD size, but smaller than the 25 A rated ampacity of the wire that it will protect.
3. Verify that the chosen OCPD size from Step 2 will protect the wire size chosen in Phase 2 from excessive current under the conditions of use. The current rating of the chosen breaker size (Step 2) must be less than the calculated maximum current under conditions of use unless the calculated maximum current under conditions of use is between standard OCPD values, in this case the next largest breaker can be chosen.

   Maximum current under conditions of use	= Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
   	= 25 A × .8 = 20 A

   Verify OCPD under conditions of use	= Maximum current under conditions of use must be greater than or equal to the current rating of the chosen OCPD (Step 2)
   	= 20 A is greater than 4 A.

The residual current device size is okay.

Phase 4: Voltage drop

If the voltage drop for the wire chosen in Phase 2 for a particular circuit is not within the recommended values, then using a larger sized wire should be considered. Increasing the wire size will not affect any other part of this process; the calculated OCPD size can remain the same.

This circuit will be .25 meters long one-way. It is 2.5 mm² wire with a resistance value in (Ω/Km) of 10.7 Ω. It is recommended that the voltage drop between the inverter and any loads be kept below 2%. The nominal circuit voltage is the Inverter AC voltage: 220 V.

Inverter output circuit current	= Maximum circuit current (Phase 1)
	= 2.3 A

Voltage drop	= 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
	= 2 × 2.3 A × .25m × 10.7 Ω ÷ 1000 = .01 V

Percentage voltage drop	= Voltage drop ÷ Nominal circuit voltage x 100
	= .01 V ÷ 220 V × 100 = .005%

.005% voltage drop for this circuit is acceptable. The wire size is okay.

Phase 5: Terminal temperature ratings

It is not necessary to perform this calculation as the inverter cannot supply sufficient current to exceed the amapacity rating of the wire under any condition.

Wire and OCPD size are okay.

AC branch circuit example

All AC branch circuits can be calculated in the same way as this example. This circuit will have an outlet that must be capable of powering all of the AC loads at the same time.

Phase 1: Maximum circuit current

Maximum circuit current	= Power rating of all AC loads on the circuit from the AC load evaluation ÷ Inverter AC voltage
	= 394 W ÷ 220 V = 1.8 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum indoor temperature 30°C. 90°C dry / 75° wet rated wire will be used for this circuit.

The minimum wire size for a circuit can be using the following steps:

1. Determine the ambient temperature correction factor based upon the maximum ambient temperature using the ambient temperature correction factor table. The ambient temperature correction factor for this circuit is 1.00 because it is a 90°C dry / 75° wet rated wire with a maximum indoor temperature of 30°C.
2. Determine the conduit fill correction factor based upon the number of number of conductors in the conduit using the conduit fill correction factor table There will only be two current-carrying conductors in the conduit, so the conduit correction fill factor is 1.
3. Determine the total wire correction parameter based upon the smaller of: the Ambient temperature correction multiplied by the Conduit fill correction factor or .8 (a safety factor from the US National Electrical code).

Total wire correction parameter	= Smaller of (Ambient temperature correction factor × Conduit fill correction factor) or .8
	= Smaller of (1.08 × 1) or .8 = .8

4. Determine minimum wire ampacity. Divide the maximum circuit current (Phase 1) by the total wire correction factor (Step 3)

Minimum wire ampacity	= Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
	= 1.8 A ÷ .8 = 2.25 A

5. Select a wire size with a maximum rated ampacity equal to or above the minimum wire ampacity calculated in the previous step using the allowable wire ampacity table. A 2.5 mm² 90°C dry / 75° wet rated wire with an ampacity rating of 25 A at 90°C will be used for this circuit as this is a commonly used wire type for a circuit of this type. 25 A is larger than the required 2.25 A ampacity.

Phase 3: Overcurrent protection and disconnects

This circuit does not require an OCPD if the available current is limited by another OCPD or the inverter - in this design the wire is protected by the 4 A OCPD and the inverter output circuit maximum current of 2.3 A - to a level that is below the ampacity rating of the wire. It is necessary to check to make sure that the terminals and wire will be protected by the other OCPD under the conditions of use - the calculation from Phase 5 can be used here.

Step 1: Maximum ampacity based upon terminal temperature rating

Maximum ampacity based on terminal temperature rating	= Wire size (Phase 2) ampacity at terminal temperature rating from allowable wire ampacity table × Total wire correction parameter (Phase 2)
	= 20 A × .8 = 16 A

Step 2: Verify OCPD protection of terminals

Verify OCPD protection of terminals	= Maximum current considering terminal temperature rating (Step 1) must be greater than or equal to the current rating of the chosen OCPD (Phase 3 or the maximum charge controller output)
	= 16 A is greater than the 4 A current rating on the residual current device

No OCPD required

Phase 4: Voltage drop

If the voltage drop for the wire chosen in Phase 2 for a particular circuit is not within the recommended values, then using a larger sized wire should be considered. Increasing the wire size will not affect any other part of this process; the calculated OCPD size can remain the same.

This circuit will be 12m meters long one-way. It is 2.5 mm² wire with a resistance value in (Ω/Km) of 10.7 Ω. It is recommended that the voltage drop between the inverter and any loads be kept below 1.5%. Circuit current is total current required by all of the loads on the circuit (same as calculated in Phase 1). Operating voltage is Inverter AC voltage: 220 V.

Voltage drop	= 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
	= 2 × 1.8 A × 12 m × 10.7 Ω ÷ 1000 = .46 V

Percentage voltage drop	= Voltage drop ÷ Circuit operating voltage x 100
	= .46 V ÷ 220 V × 100 = .21%

.21% voltage drop for this circuit is acceptable. The combined voltage drop of this circuit and the charge controller load circuit voltage drop is .215%, which is less than the recommended maximum of 2%. The wire size is okay.

Phase 5: Terminal temperature ratings

This check was performed in Phase 3 to verify that no OCPD is required.

Energy storage circuit

The energy storage circuit will be connected to common DC busbars (along with the charge controller and inverter. The wire must be sized to carry the current of both the charge controller and the inverter, although not their combined current as there would not be a situation in which the batteries would be charging and discharging at the same time.

The same wire size calculated here should be used for the series connection between the batteries.

Phase 1: Maximum circuit current

Maximum circuit current	= Larger of Inverter input circuit current or Charge controller charging circuit current
	= Larger of 27 A or 20 A = 27 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum indoor temperature 30°C. 90°C dry / 75° wet rated wire will be used for this circuit.

The minimum wire size for a circuit can be using the following steps:

1. Determine the ambient temperature correction factor based upon the maximum ambient temperature using the ambient temperature correction factor table. The ambient temperature correction factor for this circuit is 1.00 because it is a 90°C dry / 75° wet rated wire with a maximum indoor temperature of 30°C.
2. Determine the conduit fill correction factor based upon the number of number of conductors in the conduit using the conduit fill correction factor table There will only be two current-carrying conductors in the conduit, so the conduit correction fill factor is 1.
3. Determine the total wire correction parameter based upon the smaller of: the Ambient temperature correction multiplied by the Conduit fill correction factor or .8 (a safety factor from the US National Electrical code).

Total wire correction parameter	= Smaller of (Ambient temperature correction factor × Conduit fill correction factor) or .8
	= Smaller of (1.00 × 1) or .8 = .8

4. Determine minimum wire ampacity. Divide the maximum circuit current (Phase 1) by the total wire correction factor (Step 3)

Minimum wire ampacity	= Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
	= 27 A ÷ .8 = 33.75 A

5. Select a wire size with a maximum rated ampacity equal to or above the minimum wire ampacity calculated in the previous step using the allowable wire ampacity table. A 25 mm² 75°C dry/wet rated wire with an ampacity rating of 85 A at 75°C will be used for this circuit as this is a commonly used wire type for a circuit of this type. 85 A is larger than the required 33.75 A ampacity.

Phase 3: Overcurrent protection and disconnects

An OCPD and power source disconnect is required on this circuit. Energy storage systems are constantly ready to supply high amounts of current as needed.

The appropriate overcurrent protection device size can be determined by:

1. Determine the minimum size

Minimum OCPD size	= Maximum circuit current (Phase 1) × 1.25
	= 33.75 A × 1.25 = 42 A

2. A standard 50 A DC breaker will be used. This is breaker is larger than the required minimum OCPD size and smaller than the 85 A rated ampacity of the wire that it will protect.
3. Verify that the chosen OCPD size from Step 2 will protect the wire size chosen in Phase 2 from excessive current under the conditions of use. The current rating of the chosen breaker size (Step 2) must be less than the calculated maximum current under conditions of use unless the calculated maximum current under conditions of use is between standard OCPD values, in this case the next largest breaker can be chosen.

   Maximum current under conditions of use	= Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
   	= 85 A × .8 = 68 A

   Verify OCPD under conditions of use	= Maximum current under conditions of use must be greater than or equal to the current rating of the chosen OCPD (Step 2)
   	= 68 A is greater than 50 A.

The OCPD size is okay.

Phase 4: Voltage drop

If the voltage drop for the wire chosen in Phase 2 for a particular circuit is not within the recommended values, then using a larger sized wire should be considered. Increasing the wire size will not affect any other part of this process; the calculated OCPD size can remain the same.

This circuit will be 2 meters long one-way from the batteries to the DC busbars. It is 25 mm² wire with a resistance value in (Ω/Km) of 1.053 Ω. It is recommended that the voltage drop between the energy storage system and the charge controller or inverter be less than 1.5%. The nominal circuit voltage is the system voltage: 24 V.

Voltage drop	= 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
	= 2 × 33.75 A × 2m × 1.053 Ω ÷ 1000 = .14 V

Percentage voltage drop	= Voltage drop ÷ Nominal circuit voltage voltage x 100
	= .14 V ÷ 24 V × 100 = .58%

.58% voltage drop for this circuit is acceptable. The wire size is okay.

Phase 5: Terminal temperature ratings

This wire is rated at 75°C, which is the same temperature rating as the wire terminals and breaker. A calculation was already performed in Phase 3 to ensure that the wire, which has a 75°C rating, would be protected by the breaker against excessive current. There is no need to perform the calculation again.

Wire and OCPD size are okay.

Grounding system sizing and selection

Step 1: Determine the type of grounding electrode

A ground rod will be used with this installation as it is a simple to install and low-cost option.

Step 2: Determine the size of the grounding electrode conductor (GEC)

The grounding electrode conductor (GEC) is chosen on the basis of the type of grounding electrode. A 16 mm² GEC will be used with the ground rod.

Grounding electrode	GEC size
Rod/plate	6 AWG/16 mm²
Building rebar (Ufer)	4 AWG/25 mm²
Ground ring	2 AWG/35 mm²

Step 3: Determine size of the AC and DC system grounding jumpers

A 16 mm² wire will be used for the AC and DC sysem grounding jumpers.

Step 4: Determine the size of the equipment grounding conductors (EGCs)

Each grounding electrode conductor must be sized to the size of the smallest overcurrent protection device that protects the circuit.

• PV source circuit: 16 A OCPD = 4 mm² EGC
• Charge controller output circuit: 25 A OCPD = 6 mm² EGC
• Charge controller load circuit: 16 A OCPD = 4 mm² EGC.
• DC branch circuit: 16 A OCPD = 4 mm² EGC
• Inverter input circuit: 50 A OCPD = 6 mm² EGC
• Inverter output circuit: 4 A OCPD = 2.5 mm² EGC
• AC branch circuit: 4 A OCPD = 2.5 mm² EGC
• Energy storage circuit: 50 A OCPD = 2.5 mm² EGC

Largest OCPD size	Minimum AWG size	Minimum mm² size
15	14	2.5 mm²
20	12	4 mm²
60	10	6 mm²
100	8	10 mm²
200	6	16 mm²
300	4	25 mm²

Design summary

A three-line wiring diagram for this system. Wire sizes related to the grounding system are not labeled for simplicity.

• DC nominal voltage: 24 V
• AC nominal voltage: 220 V
• Mounting system: Pole mount
• Tilt angle: 12°
• Azimuth: 0°

Components

• PV module: 1 × Canadian Solar Maxpower CS6U 340W monocrystaline| Specifications sheet
• Charge controller: 1 × Victron BlueSolar PWM-Pro 20 A, 12 V/ 24 V charge controller Specifications sheet
• Energy storage system: 2 × Trojan Solar 12 V, 135 Ah AGM battery Specifications sheet
• DC-DC converter: 1 × Samlex 12 A, 24 V-12 V step down converter Specifications sheet
• Inverter: 1 × Victron Phoenix 500 W, 24 V pure sine wave inverter Specifications sheet

Circuits

• PV source circuit: 4 mm² 90°C PV wire. 16 A breaker as a power source disconnect.
• Charge controller output circuit: 6 mm² 90°C dry / 75° wet rated wire. 25 A breaker as a power source disconnect.
• Charge controller load circuit: 4 mm² 90°C dry / 75° wet rated wire. 16 A breaker.
• DC branch circuit: 2.5 mm² 90°C dry / 75° wet rated wire. No overcurrent protection device required.
• Inverter input circuit: 16 mm² 75°C dry/wet rated wire. 50 A breaker.
• Inverter output circuit: 2.5 mm² 90°C dry/wet rated wire. 4 A residual current device.
• AC branch circuit: 2.5 mm² 90°C dry/wet rated wire. No overcurrent protection devic erequired.
• Energy storage circuit: 25 mm² 75°C dry/wet rated wire. 50 A breaker.

Grounding

• Grounding electrode: Ground rod
• Grounding electrode conductor: 16 mm² wire
• System grounding jumpers: 16 mm² wire
• PV source circuit: 16 A OCPD = 4 mm² EGC
• Charge controller output circuit: 25 A OCPD = 6 mm² EGC
• Charge controller load circuit: 16 A OCPD = 4 mm² EGC
• DC branch circuit: 16 A OCPD = 4 mm² EGC
• Inverter input circuit: 50 A OCPD = 6 mm² EGC
• Inverter output circuit: 4 A OCPD = 2.5 mm² EGC
• AC branch circuit: 4 A OCPD = 2.5 mm² EGC
• Energy storage circuit: 50 A OCPD = 2.5 mm² EGC

Trojan Battery Company - Battery Sizing Guidelines https://www.trojanbattery.com/pdf/TRJN0168_BattSizeGuideFL.pdf

HOMER - PV Temperature Coefficient of Power https://www.homerenergy.com/products/pro/docs/latest/pv_temperature_coefficient_of_power.html

Trojan Battery Company - User's Guide https://www.trojanbattery.com/pdf/TrojanBattery_UsersGuide.pdf