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.