How to calculate how many solar modules you need (Solar Energy Course 2020 Part 8 of 12)

How many solar modules are needed to produce
a certain amount of power? How would you calculate this? What is the effect of module efficiency on
your result? Find out in this video. This video is part of iPolytek’s online course
on solar energy. iPolytek, Professional Development Courses
for Engineers. Sizing a PV System involves a number of steps. The first is to determine the required PV
power capacity based on – the energy consumption of the load and
– the number of peak sun hours per day (at 1000 W/m2) at the location in question Let’s look at an example: Let’s say the annual energy consumption of
the load is 9125 kWh/year. This gives us an average daily energy consumption
of 25 kWh/day Let’s say 50% of energy consumption will be
provided by PV. So, the PV system energy output will equal
12.5 kWh/day. Now determine the number of peak sun hours
available per day at the installation location. There are two ways of doing this. In the first method, you look up the number
of peak sun hours available per day on a map or chart such as the ones shown below. Remember to use the value of the worst month
of the year for critical applications. Here we see a World Insolation Map. This map
shows the amount of energy in hours, received each day on an optimally inclined surface
during the worst month of the year. Or you can use tables such as this one which
shows the Sun Hours Available Per Day in Canada. This table shows the number of sun hours at
peak solar irradiance (1000 W/m2) during the summer, winter and the average over the year. Remember, that it is more difficult to produce
electricity during the winter because of shorter days, increased cloudiness and the sun’s lower
position in the sky. So, if you are using your system primarily in the summer, use the
summer value. If you are using your system year-round, especially for a critical application,
use the winter value. If you are using the system most of the year (spring, summer and
fall) or the application is not critical, use the average value. Here is a similar table for the USA. In the second method, you calculate the number
of peak sun hours available per day based on the Global Horizontal Irradiation map.
However, note that this method will give you the yearly average value. Let’s look at an example. Here is the GLOBAL HORIZONTAL IRRADIATION
(GHI) map. Remember that this map shows us the energy available at a solar irradiation of
1000 W/m2 both on a daily basis and on an annual basis. So, taking Montreal Canada as an example we
see that the daily GHI is 3.5 kWh/m2*day. Converting this value to watts and dividing
by 1000 W/m2 we obtain 3.5 peak sun hours/day. NOTE:
Referring back to tables we can see that we calculated the same value as the average value
listed in this table using the Global Horizontal Irradiation map. Continuing with our example, let’s use the
number of sun hours available per day in Montreal during the winter as indicated in the table. Recall that the PV system energy output required
was 12.5 kWh / day And note that the # of hours of peak sunshine
/ day (Montreal, winter) is 2.3 h We can calculate the PV Power required by
dividing the PV energy output by the number of peak sun hours in the winter. This gives us 5.4 kW Now we must take into account the losses
found in the overall system, including those due to the conversion of direct current
into alternating current. We do this by taking into account a number
of derating factors. To do so we multiply the Pmax of the solar module by these factors.
These are shown in the table below along with the standard value used for each one and the
typical range used for each factor. On the far right you have a more detailed explanation
for each factor which you may find in the notes you downloaded at the beginning of this course. The factors are: PV module nameplate DC rating=0.95 Tilt Factor / Orientation Adjustment=1.00 Inverter and Transformer=0.95 Mismatch=0.98 Diodes and connections=0.99 DC wiring=0.98 AC wiring=0.99 Soiling=0.95 System availabilty=0.98 Shading=0.95 Sun-tracking=1.00 Age=0.99 Multiplying all of these derating factors
together we obtain an Overall Pmax Multiplier of 0.74 This means that a module having a rated capacity
of 100W, for example, would in reality produce 74W
because of the losses found in the system. Retrurning now to our example we can calculate
the PV installed capacity by diving the Power
required by the overall Derating Factor. This gives us 7.3 kW. The third step in sizing a PV system is to choose a module and correct its Pmax the actual
operating temperature. Here is a Sample Commercial Module Specification
sheet for modules manufactured by Canadian Solar. Their Nominal Max Power (Pmax) is 285W, measured
at 25 degrees Celsius. If we look under the TEMPERATURE CHARACTERISTICS
of the modules we find the Temperature Coefficient (Pmax)
and the Nominal Module Operating Temperature (NMOT) These coefficients allow us to calculate Pmax
under the operating conditions, that is when the Module Temperature equals
the NMOT (or the Nominal Temperature) The power output of a solar module can be
estimated using the following equation where delta T=the Temperature increase compared
to standard conditions of 25 degrees C and A equals the Temperature coefficient Pmax
(%/degree C) In this case, the Temperature coefficient (Pmax)
equals -0.39 % / degree C and the Nominal Temperature (NMOT) equals
43C Using the equation we obtain a Pmax at 43C
of 265 W per module. Let’s take a moment to discuss the NMOT. In reality, the modules operate at higher
temperatures and slightly lower solar irradiation conditions
than those used in standard tests. (Namely, 25C and 1000 W/m2) And as we have just seen, the NMOT is used
to determine the power output of the solar module. The NMOT is measured under the following conditions, Solar irradiance=800 W/m2
Air temperature=20C Wind speed=1 m/s
Mounting open back side From this definition we can see that the the
cell temperature varies according to: • solar irradiation,
• ambient temperature, and • module cooling The graph on the right shows the difference
between the cell temperature and the ambient temperature (see y-axis) at different irradiation levels for a well
cooled module, a poorly cooled module and a typically cooled module. The data presented is for a wind speed of
1 m/s. The temperature of the module will be lower than shown in the figure if the wind
speed is greater than 1 m/s and higher under calmer conditions. The following formula can be used to estimate
the temperature of the cell NOCT=Normal Operating Cell Temperature in
degrees Celcuis, S=irradiation in mW/cm2 Here’s a closer look at the graph The Red line represents a poorly cooled module The Black line shows typical data for an Average
Module and the Blue Line is for a well cooled module Now back to our example… Now we are ready to calculate the number
of PV modules required To do so, we divide the PV Installed Capacity
of 7300 W / by Pmax (corrected for the Actual Operating temperature or the NMOT), in this
case 265W and we obtain 28 modules. Recall that efficiency is Pmax divided by
the product of the standard solar irradiation and the collector area. So, if the efficiency increases thanks to
a higher Pmax, fewer modules would be required. If the efficiency increases thanks to a smaller
collector area, the installation will become smaller. Researchers are constantly working to improve
the efficiency of solar cells. Each new wave of solar cells gives rise to
a new generation of photovoltaic devices. Today, there exists three generations. We’ll
learn more about these in this course. Now that we’ve seen some basic calculations
and understand efficiency We think the time has come to ask the question What different types of cells exist today
and how efficient are they? Find out in our next video! Thanks for watching and see you soon.

Leave a Reply

Your email address will not be published. Required fields are marked *