Detailed AC system design

From Open Source Solar Project
Jump to navigation Jump to search

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 the number of parallel battery circuits.

  • Photo of the building.

  • Location of the building.

Contents

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 150 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

  • 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 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

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

DC system voltage = 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:[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 ÷ DC system voltage × 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

Specifications:
  • Battery type: FLA
  • Nominal voltage: 6 V
  • C/20 rated capacity: 390 Ah

Step 5: Calculate number of batteries in series

Batteries in series = DC system voltage ÷ 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 battery 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 DC 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 PV source circuit 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 PV source circuit configuration of 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 DC 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) ÷ DC 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 DC 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.

2 × Victron MPPT 250/60 (250 volt maximum input voltage, 60 amps maximum output current) SmartSolar charge controller will be used for this design as it was determined to be the ideal configuration. This design process was attempted with a 100 A charge controller - as it has sufficient capacity to handle the final PV source size, but it resulted that the terminals on the charge controller were too small to accept the size wire specified by this design process. The next best option was to use 2 × 250/60 Smart solar charge controllers (the smallest 250 V charge controller that Victron makes). This is a good example of how a proper design is often the result of many different attempts at hypothetical designs. Specifications sheet

Specifications:
  • Nominal output voltage: 48V
  • Maximum rated output current: 60 A
  • Maximum input voltage: 250 V
  • Maximum rated PV power source size (W) at 48 V = 3440 W
Charge controller maximum input voltage = From charge controller specifications sheet
= 250 V
Charge controller current rating = From charge controller specifications sheet
= 60 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 DC system voltage 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.[2]

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 PV source configuration

It is necessary to test various different configurations of PV modules to find the best configuration of modules connected in series per PV soruce circuit. The proposed number of PV modules per PV source circuit must be less than the calculated maximum number of PV modules in series (Step 5) and greater than the calculated minimum number of PV modules in series (Step 6).

If the minimum number of PV modules (Step 2) required by the design exceeds the maximum number of modules that the charge controller can handle in series (Step 6), then multiple PV source circuits will be required. All of the PV source circuits must have the same number of PV modules if there is a single charge controller or else it will not function properly (the number of modules in series therefore must divide evenly into the minimum number of PV modules required). If there are multiple charge controllers, then the number of modules connected in series per PV source circuit should be the same for each one.

As long as the voltage doesn't exceed the rating of the charge controller(s), more PV modules per PV source circuit is generally preferrable as it permits smaller sized wires and minimizes voltage drop.

If there is not a configuration that is meets these criteria then a different PV module or charge controller should be used in the design.

The design calls for a minimum of 12 PV modules. There are two charge controllers. The maximum number of PV modules connected in series is 6 and the minimum is 3. The possible configurations are:

  • 4 PV source circuits (2 per charge controller) of 3 PV modules in series
  • 3 PV source circuits (1 for one charge controller, 2 for the other) of 4 PV modules in series
  • 2 PV source circuits (1 per charge controller) of 6 PV modules in series

The option with the highest voltage - 6 PV modules in series - is the best option. The modules would be evenly split between the two charge controllers - 6 modules per charge controller.

Proposed PV source power rating = PV module power rating (Step 1) × Number of PV modules in series × Number of PV source circuits
= 285 W × 6 × 2 = 3420 W

Step 10: 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 9) × 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 ) ÷ DC system voltage
= 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 PV source configuration

The design calls for a minimum of 12 PV modules, but this array size was undersized. The array will be increased 25% with the hope of finding a suitable configuration.

This design calls for a minimum of 16 PV modules. There are two charge controllers. The maximum number of PV modules connected in series is 6 and the minimum is 3. The possible configurations are:

  • 4 PV source circuits (2 per charge controller) of 4 PV modules in series

There is only one possible configuration. The modules will be evenly split between the two charge controllers - 6 modules per charge controller.

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

Step 10: 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 9) × 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 ) ÷ DC system voltage
= 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 11: 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.[3] AGM batteries can should be charged with a current that is between .05-.2 (5-20%) of their C/20 rating.[3] 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 9) ÷ 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 PV source configuration is within the acceptable range. The design is okay. ÷

Step 12: 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 PV source circuits
= 285 W × 4 modules in series × 4 PV source circuits = 4560 W

Step 13: Verify PV source and charge controller compatability

MPPT charge controllers have a maximum PV source power rating in watts that limits the size of the PV source. Verify that the maximum PV source power rating is greater than the final PV source power rating. If it is not, the charge controller size needs to be increased.

Verify PV source and charge controller size = Final PV source power rating must be less than the maximum PV source power rating of the charge controller
= 4560 W ÷ 2 charge controllers = 2580 W per charge controller
= 2280 W per charge controller is less than the 3440 W maximum PV source power rating of each charge controller

The PV source size and charge controller size is okay.

Inverter sizing and selection

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

  1. The voltage must match the DC 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 DC 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. This inverter allows a generator to be incorporated into the system, which is very useful in larger systems. 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) ÷ DC system voltage ÷ 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 wet/dry 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 .94 because it is a 90°C wet/dry 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).
    4. Total wire correction parameter = Smaller of (Ambient temperature correction factor × Conduit fill correction factor) or .8
    = Smaller of (.94 × 1) or .8 = .8
  4. Determine minimum wire ampacity. Divide the maximum circuit current (Phase 1) by the total wire correction factor (Step 3)
    5. 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 25 A 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 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
    2. Minimum OCPD size = Maximum circuit current (Phase 1) × 1.25
    = 11.81 A × 1.25 = 14.8 A
  2. A standard 15 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. 15 A is also the largest fuse size permitted by the module manufacturer.
  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 OCPD size is used from the list of standard OCPD sizes. If the current rating of the chosen OCPD size is larger than the maximum OCPD size under conditions of use, then the wire size must be increased until it passes this test.
    Maximum current under conditions of use = Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
    = 25 A × .8 = 20 A

    Determine the maximum OCPD size under conditions of use.

    Maximum OCPD size under conditions of use = Can be equal to the maximum current under conditions of use. If between standard OCPD sizes, the next largest OCPD is used.
    = 20 A OCPD
    Verify OCPD under conditions of use = The current rating of the chosen OCPD (Step 2) must be less than or equal to the maximum OCPD size under conditions of use
    = 15 A OCPD is less than the maximum 20 A OCPD size

The OCPD and wire size are 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 3%.

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.

PV output circuit

There are 2 × PV output circuits. Each one will carry the current of 2 × PV source circuits.

Phase 1: Maximum circuit current

Maximum circuit current = PV module Isc × Number of PV source circuits × Irradiance safety parameter
= 9.45 A × 2 × 1.25 = 23.63 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum ambient temperature 35°C. 90°C dry/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 .94 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 be 4 × current-carrying conductors in the conduit, so the conduit correction fill factor is (.8).
  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).
    4. Total wire correction parameter = Smaller of (Ambient temperature correction factor × Conduit fill correction factor) or .8
    = Smaller of (.96 × .8) or .8 = .75
  4. Determine minimum wire ampacity. Divide the maximum circuit current (Phase 1) by the total wire correction factor (Step 3)
    5. Minimum wire ampacity = Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
    = 23.63 ÷ .75 = 31.5 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 10 mm² 90°C wet/dry rated wire with an ampacity rating of 50 A will be used. 50 A is larger than the required 31.5 A ampacity.

Phase 3: Overcurrent protection and disconnects

The appropriate overcurrent protection device size can be determined by:

  1. Determine the minimum size
    2. Minimum OCPD size = Maximum circuit current (Phase 1) × 1.25
    = 31.5 A × 1.25 = 39.4 A
  2. A standard 40 A DC breaker will be used. This is breaker is larger than the required minimum OCPD size, but smaller than the rated 50 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 OCPD size is used from the list of standard OCPD sizes. If the current rating of the chosen OCPD size is larger than the maximum OCPD size under conditions of use, then the wire size must be increased until it passes this test.
    Maximum current under conditions of use = Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
    = 50 A × .8 = 40 A

    Determine the maximum OCPD size under conditions of use.

    Maximum OCPD size under conditions of use = Can be equal to the maximum current under conditions of use. If between standard OCPD sizes, the next largest OCPD is used.
    = 40 A OCPD
    Verify OCPD under conditions of use = The current rating of the chosen OCPD (Step 2) must be less than or equal to the maximum OCPD size under conditions of use
    = 40 A OCPD is equal to the maximum 40 A OCPD size

The OCPD and wire size are 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 3%.

PV source circuit current = Max power current of the PV module × Number of PV source 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.

Charge controller output circuit

There are 2 × 250 volt, 60 amp charge controllers. They 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
= 60 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum indoor temperature 30°C. 75° wet / 90°C dry 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 75° wet / 90°C dry 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).
    4. 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)
    5. Minimum wire ampacity = Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
    = 60 A ÷ .8 = 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 35 mm² 75° wet / 90°C dry rated wire with an ampacity rating of 115 A will be used for this circuit as this is a commonly used wire type for a circuit of this type. 115 A is larger than the required 75 A ampacity. This is the largest size wire that the terminals on the charge controller will accept.

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
    2. Minimum OCPD size = Maximum circuit current (Phase 1) × 1.25
    = 60 A × 1.25 = 75 A
  2. A standard 80 A DC breaker will be used. This is breaker is larger than the required minimum OCPD size, but smaller than the 115 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 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 OCPD size is used from the list of standard OCPD sizes. If the current rating of the chosen OCPD size is larger than the maximum OCPD size under conditions of use, then the wire size must be increased until it passes this test.
    Maximum current under conditions of use = Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
    = 115 A × .8 = 92 A

    Determine the maximum OCPD size under conditions of use.

    Maximum OCPD size under conditions of use = Can be equal to the maximum current under conditions of use. If between standard OCPD sizes, the next largest OCPD is used.
    = 100 A OCPD
    Verify OCPD under conditions of use = The current rating of the chosen OCPD (Step 2) must be less than or equal to the maximum OCPD size under conditions of use
    = 80 A OCPD is less than the maximum 100 A OCPD size

The OCPD and wire size are 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 ÷ DC system voltage) or Charge controller current rating
= Smaller of (2280 W ÷ 48 V = 47.5 A) or (60 A) = 47.5 A
Charge controller circuit nominal voltage = DC system voltage
= 48 V
Voltage drop = 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
= 2 × 47.5 A × .5m × .661 Ω ÷ 1000 = .03 V
Percentage voltage drop = Voltage drop ÷ Nominal circuit voltage voltage x 100
= .03 V ÷ 48 V × 100 = .06%

.06% voltage drop for this circuit is acceptable. The wire size is 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
= 3000 VA ÷ 46 V ÷ .85 = 76.7 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum indoor temperature 30°C. 75° wet/dry 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 75° wet/dry 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).
    4. 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)
    5. Minimum wire ampacity = Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
    = 76.7 A ÷ .8 = 95.9 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 50 mm² 75°C dry/wet rated wire with an ampacity rating of 130 A at 75°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 95.9 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
    2. Minimum OCPD size = Maximum circuit current (Phase 1) × 1.25
    = 77 A × 1.25 = 96.3 A
  2. A standard 100 A DC breaker will be used. This is breaker is larger than the required minimum OCPD size and 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 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 OCPD size is used from the list of standard OCPD sizes. If the current rating of the chosen OCPD size is larger than the maximum OCPD size under conditions of use, then the wire size must be increased until it passes this test.
    Maximum current under conditions of use = Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
    = 130 A × .8 = 104 A

    Determine the maximum OCPD size under conditions of use.

    Maximum OCPD size under conditions of use = Can be equal to the maximum current under conditions of use. If between standard OCPD sizes, the next largest OCPD is used.
    = 125 A OCPD
    Verify OCPD under conditions of use = The current rating of the chosen OCPD (Step 2) must be less than or equal to the maximum OCPD size under conditions of use
    = 100 A OCPD is less than the maximum 125 A OCPD size

The OCPD and wire size are 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 1.5 meters long one-way as it only needs to reach the DC busbars where the battery circuit is connected. It is 50 mm² wire with a resistance value in (Ω/Km) of 0.524 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 DC system voltage: 48 V.

Voltage drop = 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
= 2 × 97 A × 1.5 m × 0.524 Ω ÷ 1000 = .15 V
Percentage voltage drop = Voltage drop ÷ Nominal circuit voltage voltage x 100
= .06 V ÷ 48 V × 100 = .32%

.32% voltage drop for this circuit is acceptable. The wire size is 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
= 3000 VA ÷ 120 V = 25 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum indoor temperature 30°C. 75° wet / 90°C dry 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 75° wet / 90°C dry 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).
    4. 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)
    5. Minimum wire ampacity = Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
    = 25 A ÷ .8 = 31.3 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 10 mm² 75° wet / 90°C dry rated wire with an ampacity rating of 50 A will be used for this circuit as this is a commonly used wire type for a circuit of this type. 50 A is larger than the required 31.3 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
    2. Minimum OCPD size = Maximum circuit current (Phase 1) × 1.25
    = 25 A × 1.25 = 31.3 A
  2. A standard 32 A AC residual current device will be used. This is breaker is larger than the required minimum OCPD size, but smaller than the 55 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 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 OCPD size is used from the list of standard OCPD sizes. If the current rating of the chosen OCPD size is larger than the maximum OCPD size under conditions of use, then the wire size must be increased until it passes this test.
    Maximum current under conditions of use = Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
    = 50 A × .8 = 40 A

    Determine the maximum OCPD size under conditions of use.

    Maximum OCPD size under conditions of use = Can be equal to the maximum current under conditions of use. If between standard OCPD sizes, the next largest OCPD is used.
    = 40 A OCPD
    Verify OCPD under conditions of use = The current rating of the chosen OCPD (Step 2) must be less than or equal to the maximum OCPD size under conditions of use
    = 32 A OCPD is less than the maximum 40 A OCPD size

The RCD/OCPD and wire size are 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 1.5 meters long one-way. It is 10 mm² wire with a resistance value in (Ω/Km) of 4.226 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: 120 V.

Inverter output circuit current = Maximum circuit current (Phase 1)
= 25 A
Voltage drop = 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
= 2 × 25 A × 1.5 m × 4.226 Ω ÷ 1000 = .32 V
Percentage voltage drop = Voltage drop ÷ Nominal circuit voltage x 100
= .01 V ÷ 120 V × 100 = .27%

.27% voltage drop for this circuit is acceptable. The wire size is okay.

AC branch circuit example

All AC branch circuits can be calculated in the same way as this example. This circuit will have an that will power:

Load Quantity Watts Total watts
Projector 1 300 W 300 W
Stereo 1 30 W 30 W
Cell phone 10 5 W 50 W
Laptop 2 20 W 40 W
LED television 1 30 W 30 W
Printer 1 70 W 70 W
LED lights 20 5 W 100 W

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
= 620 W ÷ 120 V = 5.2 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum indoor temperature 30°C. 75° wet / 90°C dry 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 75° wet / 90°C dry 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 be 8 current-carrying conductors in the conduit, so the conduit correction fill factor is .7
  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).
    4. Total wire correction parameter = Smaller of (Ambient temperature correction factor × Conduit fill correction factor) or .8
    = Smaller of (1.00 × .7) or .8 = .7
  4. Determine minimum wire ampacity. Divide the maximum circuit current (Phase 1) by the total wire correction factor (Step 3)
    5. Minimum wire ampacity = Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
    = 5.2 A ÷ .7 = 7.4 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 75° wet / 90°C dry rated wire with an ampacity rating of 20 A will be used for this circuit as this is a commonly used wire type for a circuit of this type. 20 A is larger than the required 7.4 A ampacity.

Phase 3: Overcurrent protection and disconnects

This circuit will use an OCPD to protect it from excessive currents from the inverter.

The appropriate overcurrent protection device size can be determined by:

  1. Determine the minimum size
    2. Minimum OCPD size = Maximum circuit current (Phase 1) × 1.25
    = 7.4 A × 1.25 = 9.25 A
  2. A standard 10 A AC breaker will be used. This is breaker is larger than the required minimum OCPD size and 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 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 OCPD size is used from the list of standard OCPD sizes. If the current rating of the chosen OCPD size is larger than the maximum OCPD size under conditions of use, then the wire size must be increased until it passes this test.
    Maximum current under conditions of use = Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
    = 20 A × .8 = 16 A

    Determine the maximum OCPD size under conditions of use.

    Maximum OCPD size under conditions of use = Can be equal to the maximum current under conditions of use. If between standard OCPD sizes, the next largest OCPD is used.
    = 16 A OCPD
    Verify OCPD under conditions of use = The current rating of the chosen OCPD (Step 2) must be less than or equal to the maximum OCPD size under conditions of use
    = 10 A OCPD is less than the maximum 16 A OCPD size

The OCPD and wire size are 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 15 m 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%. 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 × 7.4 A × 15 m × 10.7 Ω ÷ 1000 = 2.37 V
Percentage voltage drop = Voltage drop ÷ Circuit operating voltage x 100
= 2.37 V ÷ 120 V × 100 = 1.99%

1.99% voltage drop for this circuit is less than the recommended maximum of 2%. The wire size is okay.

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 76.7 A or (2 × 60 A) = 120 A

Phase 2: Wire ampacity

There will only be two current-carrying conductors in the conduit. The maximum indoor temperature 30°C. 75° wet/dry 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 75° wet/dry 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).
    4. 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)
    5. Minimum wire ampacity = Maximum circuit current (Phase 1) ÷ Total wire correction parameter (Step 3)
    = 120 A ÷ .8 = 150 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 95 mm² 75°C dry/wet rated wire with an ampacity rating of 200 A will be used for this circuit as this is a commonly used wire type for a circuit of this type. 165 A is larger than the required 150 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
    2. Minimum OCPD size = Maximum circuit current (Phase 1) × 1.25
    = 120 A × 1.25 = 150 A
  2. A 150 A DC breaker will be used. This is breaker is larger than the required minimum OCPD size and smaller than the 165 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 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 OCPD size is used from the list of standard OCPD sizes. If the current rating of the chosen OCPD size is larger than the maximum OCPD size under conditions of use, then the wire size must be increased until it passes this test.
    Maximum current under conditions of use = Wire ampacity from allowable wire ampacity table × Total wire correction parameter (Phase 2).
    = 165 A × .8 = 132 A

    Determine the maximum OCPD size under conditions of use.

    Maximum OCPD size under conditions of use = Can be equal to the maximum current under conditions of use. If between standard OCPD sizes, the next largest OCPD is used.
    = 150 A OCPD
    Verify OCPD under conditions of use = The current rating of the chosen OCPD (Step 2) must be less than or equal to the maximum OCPD size under conditions of use
    = 150 A OCPD is equal to the maximum 150 A OCPD size

The OCPD and wire size are 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 DC system voltage: 24 V.

Voltage drop = 2 x Circuit current x One-way circuit length (m) x Resistance (Ω/Km) ÷ 1000
= 2 × 120 A × 3m × 0.2610 Ω ÷ 1000 = .19 V
Percentage voltage drop = Voltage drop ÷ Nominal circuit voltage voltage x 100
= .19 V ÷ 48 V × 100 = .40%

.40% voltage drop for this circuit is acceptable. The wire size is 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.

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
  • PV output circuit: 63 A OCPD = 10 mm² EGC
  • Charge controller output circuit: 80 A OCPD = 10 mm² EGC
  • Inverter input circuit: 100 A OCPD = 10 mm² EGC
  • Inverter output circuit: 32 A OCPD = 4 mm² EGC
  • AC branch circuit: 10 A OCPD = 2.5 mm² EGC
  • Energy storage circuit: 150 A OCPD = 16 mm² EGC

Design summary

A three-line wiring diagram for this system. Equipment grounding wires are not labeled for simplicity.
  • DC nominal voltage: 48 V
  • AC nominal voltage: 120 V
  • Mounting system: Roof mount
  • Tilt angle: 15°
  • Azimuth: 10°

Components

  • PV module: 16 × Yingli YGE 60 Cell, Series 2 polycrystaline module with a power rating of 285 watts. Specifications sheet
  • Charge controllers: 2 × Victron MPPT 250/60 (250 volt maximum input voltage, 60 amps maximum output current) SmartSolar charge controller Specifications sheet
  • Energy storage system: 16 × Rolls S6-L16 6 V, 390 Ah flooded lead acid batteries. Specifications sheet
  • Inverter: 1 × Victron Quattro 48V, 3000VA 120V Inverter/Charger Phoenix will be used. Specifications sheet

Circuits

  • PV source circuits: 4 mm² 90°C PV wire. 16 A DC fuses.
  • PV output circuit: 25 mm² 75° wet / 90°C dry rated wire. 63 A breaker as a power source disconnect.
  • Charge controller output circuit: 35 mm² 75° wet / 90°C dry rated wire. 80 A breaker as a power source disconnect.
  • Inverter input circuit: 50 mm² 75°C dry/wet rated wire. 100 A breaker.
  • Inverter output circuit: 10 mm² 90°C dry/wet rated wire. 32 A residual current device.
  • AC branch circuit: 2.5 mm² 90°C dry/wet rated wire. 10 A breaker.
  • Energy storage circuit: 95 mm² 75°C dry/wet rated wire. 150 A breaker.

Grounding

  • Grounding electrode: Ground rod
  • Grounding electrode conductor: 16 mm² wire
  • System grounding jumpers: 16 mm² wire
  • Ground fault protection device: 80 A GFPD
  • PV source circuit: 16 A OCPD = 4 mm² EGC
  • PV output circuit: 63 A OCPD = 10 mm² EGC
  • Charge controller output circuit: 80 A OCPD = 10 mm² EGC
  • Inverter input circuit: 100 A OCPD = 10 mm² EGC
  • Inverter output circuit: 32 A OCPD = 4 mm² EGC
  • AC branch circuit: 10 A OCPD = 2.5 mm² EGC
  • Energy storage circuit: 150 A OCPD = 16 mm² EGC

Notes/references

  1. ↑ Trojan Battery Company - Battery Sizing Guidelines https://www.trojanbattery.com/pdf/TRJN0168_BattSizeGuideFL.pdf
  2. ↑ HOMER - PV Temperature Coefficient of Power https://www.homerenergy.com/products/pro/docs/latest/pv_temperature_coefficient_of_power.html
  3. ↑ 3.0 3.1 Trojan Battery Company - User's Guide https://www.trojanbattery.com/pdf/TrojanBattery_UsersGuide.pdf
Retrieved from "http://www.opensourcesolar.org/w/index.php?title=Detailed_AC_system_design&oldid=4475"