2.2.14 Cable sizing
Cable sizing must be done carefully to minimize voltage losses due to resistance. Cable size is usually given
in terms of the cross sectional area of the cable in mm
2 .
The thicker the cable, the lower the resistance; and the
higher the cost. Generally, the recommended maximum voltage drop is about 5% of the system voltage
and is used as a guide for cable sizing. The cable crosssectional area is related to the voltage drop, length, and
current flowing through the cable by: A = 2ρLI/Vd,
where A is the conductor cross-sectional area (mm
2 ),
ρ is the resistivity of the conductor material in
mm
2 /m (for copper it is 0.0183), L is the conductor length one-way (no return) (m), I is the current
flowing (A), and Vd is the voltage drop along the
conductor (0.05 × system voltage) (Michael Boxwell,
2019).This calculation can also be used to calculate the
maximum cable length to give a voltage drop within
the limit of 5% for a particular cable size. If the cable
size obtained by calculation is not in the market, then
choose the next larger cable size.
2.3 Method 2: The Energy Output Method (EOM)
The energy output method calculates the energy output
of the system components taking into account their
inefficiencies. The starting point is the load, DED, then
compute backwards towards the module as detailed
below. This enables the calculation of the total power of
all the modules we need for our solar system and then
proceed on to size the other components. The method
is demonstrated in Figure 1 below and the explanation
beneath it.
Figure 1. A block layout of the energy flow and component
interconnection for a PV system using alternating current
(AC).
2.3.1 Calculation of the inverter output, Inv out
The energy that should be given out by the inverter is
the energy that should go to our loads and so the DED
is equal to our inverter output.
2.3.2 Calculation of inverter input Inv in /battery
output, Batt out
The energy input to the inverter is what has been
given out by the battery. So the inverter input is equal
to the battery output. So the question is: how much
energy is needed for the inverter input to get the output
equivalent of DED? The question is answered by dividing the DED by the inverter efficiency, as a decimal
0.85, i.e., (DED/0.85).
2.3.3 Determination of the battery input,
Batt in /module output, P out
The battery input is the same as the module output
since the battery gets its energy from the module.
Again the question is: how much energy is needed
at the battery input to get the battery output we have
obtained?. The question is answered by dividing the
Batt out by the battery efficiency, as a decimal 0.80, i.e.
(Batt out /0.80).
2.3.4 Determination of module input, Pin
The Module input is computed by dividing the module output divided by the module efficiency i.e.
Pin = (Pout/0.8).
2.3.5 Lastly, calculate the module capacity, (Pw)
Divide the module input with the product of solar
resource (PSH) and module efficiency, 0.8, i.e.
Pw = (Pin/(PSH × 0.8)). A suitable module combination is then chosen from what is in the market.
2.3.6 Number of module parallel strings
The total system power in watts is divided by the rating
of one module.
2.3.7 Number of modules connected in series per
string
Divide the system voltage by the nominal voltage of
one module.
2.3.8 Total number of modules
Multiply the numbers of parallel strings by the number
connected in series in each parallel string.
2.3.9 Sizing of charge controller
Multiply Isc for one module and the number of modules strings. Then multiply this total by 1.25 (margin
of 25%) to get final charge controller current rating.
Lastly choose the charge controller with this current
rating or higher and same system voltage.
2.3.10 Battery bank capacity
First multiply the battery output Batt out , by 100 and
divide by the DOD in %. Secondly, divide the resulting
value by the system voltage to get the battery capacity
in AHrs. Thirdly, choose a battery from what is in the
market, and divide the battery capacity in AHrs by
the rating in AHrs of one battery to get the number
of parallel strings. Fourthly, calculate the number of
series connected batteries per string by dividing the
system voltage by the voltage of one battery. Fifthly,
multiply the number of parallel strings by the number
connected in series per string to get the total number of
batteries needed. When days of autonomy are factored,
it increases the number of batteries.
233
Cable sizing must be done carefully to minimize voltage losses due to resistance. Cable size is usually given
in terms of the cross sectional area of the cable in mm
2 .
The thicker the cable, the lower the resistance; and the
higher the cost. Generally, the recommended maximum voltage drop is about 5% of the system voltage
and is used as a guide for cable sizing. The cable crosssectional area is related to the voltage drop, length, and
current flowing through the cable by: A = 2ρLI/Vd,
where A is the conductor cross-sectional area (mm
2 ),
ρ is the resistivity of the conductor material in
mm
2 /m (for copper it is 0.0183), L is the conductor length one-way (no return) (m), I is the current
flowing (A), and Vd is the voltage drop along the
conductor (0.05 × system voltage) (Michael Boxwell,
2019).This calculation can also be used to calculate the
maximum cable length to give a voltage drop within
the limit of 5% for a particular cable size. If the cable
size obtained by calculation is not in the market, then
choose the next larger cable size.
2.3 Method 2: The Energy Output Method (EOM)
The energy output method calculates the energy output
of the system components taking into account their
inefficiencies. The starting point is the load, DED, then
compute backwards towards the module as detailed
below. This enables the calculation of the total power of
all the modules we need for our solar system and then
proceed on to size the other components. The method
is demonstrated in Figure 1 below and the explanation
beneath it.
Figure 1. A block layout of the energy flow and component
interconnection for a PV system using alternating current
(AC).
2.3.1 Calculation of the inverter output, Inv out
The energy that should be given out by the inverter is
the energy that should go to our loads and so the DED
is equal to our inverter output.
2.3.2 Calculation of inverter input Inv in /battery
output, Batt out
The energy input to the inverter is what has been
given out by the battery. So the inverter input is equal
to the battery output. So the question is: how much
energy is needed for the inverter input to get the output
equivalent of DED? The question is answered by dividing the DED by the inverter efficiency, as a decimal
0.85, i.e., (DED/0.85).
2.3.3 Determination of the battery input,
Batt in /module output, P out
The battery input is the same as the module output
since the battery gets its energy from the module.
Again the question is: how much energy is needed
at the battery input to get the battery output we have
obtained?. The question is answered by dividing the
Batt out by the battery efficiency, as a decimal 0.80, i.e.
(Batt out /0.80).
2.3.4 Determination of module input, Pin
The Module input is computed by dividing the module output divided by the module efficiency i.e.
Pin = (Pout/0.8).
2.3.5 Lastly, calculate the module capacity, (Pw)
Divide the module input with the product of solar
resource (PSH) and module efficiency, 0.8, i.e.
Pw = (Pin/(PSH × 0.8)). A suitable module combination is then chosen from what is in the market.
2.3.6 Number of module parallel strings
The total system power in watts is divided by the rating
of one module.
2.3.7 Number of modules connected in series per
string
Divide the system voltage by the nominal voltage of
one module.
2.3.8 Total number of modules
Multiply the numbers of parallel strings by the number
connected in series in each parallel string.
2.3.9 Sizing of charge controller
Multiply Isc for one module and the number of modules strings. Then multiply this total by 1.25 (margin
of 25%) to get final charge controller current rating.
Lastly choose the charge controller with this current
rating or higher and same system voltage.
2.3.10 Battery bank capacity
First multiply the battery output Batt out , by 100 and
divide by the DOD in %. Secondly, divide the resulting
value by the system voltage to get the battery capacity
in AHrs. Thirdly, choose a battery from what is in the
market, and divide the battery capacity in AHrs by
the rating in AHrs of one battery to get the number
of parallel strings. Fourthly, calculate the number of
series connected batteries per string by dividing the
system voltage by the voltage of one battery. Fifthly,
multiply the number of parallel strings by the number
connected in series per string to get the total number of
batteries needed. When days of autonomy are factored,
it increases the number of batteries.
233