Solar Energy in a Temperate Climate.

Printer-friendly versionPrinter-friendly version Share this

Does it pay to cover your roof with solar panels in Vancouver?

Big Ideas: 
  • Electrical energy from home solar panels - how to calculate if this is suitable for you.

We all have seen solar panels. They are mounted on traffic signs, light houses (see title photograph), and in our calculators. They are great green sources of energy. So why isn't every roof in [Vancouver] covered with them. This would reduce our reliance on fossil fuels.

Let’s learn something about the solar panels and do some simple calculations.

To create electric potential energy (or colloquially speaking electricity) we have to separate positive and negative charges. In batteries, the work required to separate positive and negative charges is done by an electrochemical reaction. In silicon-based solar cells, the work is done by the incoming solar radiation in a quantum process: a visible photon from the sun has enough energy to separate an electron from a silicon atom leaving behind a site that is positively charged (called “hole”). (Remember the duality of light, we can treat light as electromagnetic wave or as a stream of particles called photons.) Electrons are attracted to the positive site and holes to the negative site, so once we connect a solar panel to some load (for example a light bulb, a motor or a heater) we have a current that can do work on such external load.

Let’s calculate how much power we can obtain from a solar panel in Vancouver. The solar constant (the amount of solar radiation per m2 at the top of the atmosphere) is about 1400 W/ m2. We have on average 12 hours of sunlight a day, which should give us 1400 W/ m2 times 12 hours = 16.8 kWh/ m2 per day. But according to Natural Resources Canada1 we can only expect on average 5.2 kWh/ m2 on a surface perpendicular to the direction from the Sun. This is due to atmospheric absorption and cloud cover.

There is another problem. As the Sun’s position changes during the day (differently in each day of the year) it is expensive to keep the panel perpendicular to the direction from the Sun. To do this requires directional alignment in two axes so the panels cannot be fixed to the roof.

As illustrated Figure 1  the power of radiation delivered to a given surface depends on the incident angle. Notice that the power contained in a 1 m2 column of radiation impacts 1 m2 of surface if it falls vertically but 2 m2 of surface if it is at an angle of 30 degrees. So at 30 degrees we are getting only ½ of power per m2! The angular dependence of the power delivered is:

$ P_{\theta} =  P_{90} \sin(\theta) $

where $  P_{90} $ is the power per unit area of radiation impacting the surface from the direction perpendicular to the surface, $ \theta $ is the angle of the surface from the horizontal and $ P_{\theta} $ is the power per unit area of the same radiation impacting the surface from the direction at the angle $ \theta $ from the horizontal.

Figure 1. The radiation of the same intensity illuminating a surface at different angles 

Notice that the angle is measured from the direction perpendicular to the direction of incoming radiation as described in1. So if we mount our solar panel at the angle equal to our latitude on a south facing roof we will only get on average 3.7 kWh/ m2 of energy per day impacting our panel.

Why should we tilt the panel at the angle equal to our latitude? Lets look at Figure 2. Twice a year at solstice the axis of rotation of the Earth is perpendicular to the direction of the incoming solar radiation (as shown). The panel tilted from the horizontal plane at the angle equal to the latitude of it's position is at noon perpendicular to the direction of the incoming solar radiation.  Over the year the angle between the axis of the Earth and the direction of the incoming solar radiation changes between -23.5° and +23.5°. Also the angle between the solar panel and the direction of the incoming solar radiation will change during the day. But such a position of the panel gives us a best yearly average of the power of the the solar radiation impacting the panel. How much do we loose if the tilt angle of the panel is different.  Between the tilt angle equal to latitude and tilt angle equal to latitude -15° almost nothing.  At  tilt angle equal to latitude +15° about 10%.

Figure 2. An illustration explaining why we set up solar panels at an angle from horizontal roughly equal to latitude.

A two person family living in a small detached house uses about 40kWh per day for cooking, hot water, heating, lights, TV, laundry, computers, microwave and so on at a cost of about $850 a year. It would seem that a 10m2 solar panel should cover their needs. Unfortunately not! Only about 10% of energy hitting the panel is converted to AC electrical energy. This is because of the solar panel efficiency (15-25%), DC/AC conversion efficiency and resistive loses. So from a nominally 1kW panel (the panel which gives us 1 kW electrical power at noon in full sunshine, when oriented perpendicular to incoming solar radiation) we can expect on average only about 3kWh per day (1 gives us so called Photovoltaic potential in kWh/kW, which is how many kWh per month or year can we expect at a particular location from a 1kW panel). And such a panel has an area of 7-10 m2 depending on the technology! At the moment the complete 1kW solar system costs about $7500. On a typical house and garage there is usually about 50 m2 of a south facing roof space to be covered by solar panels. So we could install 5-7 kW panels and get 15 – 20 kWh per day, almost half of the expected consumption. This installation would cost $40-50K and would recoup the initial cost in 50-60 years; this is unfortunately not a good investment.

And there is an other problem: energy storage. Solar panels give us a peak production on a midsummer day. But peak consumption occurs in midwinter evenings. If the house is connected to the grid we can sell energy at peak production and buy it at peak consumption and all our previous analysis is valid. But if our house is located on a small island we have to think about energy storage. We would have to store about ¼ year’s energy = about 4 000 kWh! A big car battery (mass 55 kg) stores about 200Ah at 12 V = 2.4 kWh. We would need about 1700 such batteries!

On the other hand a solar panel and a modest battery bank would be a perfect solution for a summer cottage used only on weekends. It would be a good exercise to try to calculate the expected energy consumption, the necessary size of solar panels, and the capacity of the batteries.

It sounds pessimistic at the moment but the technology is improving. Printed “roll on the roof” solar cells are coming 2. In few years we might be able to save some money and more importantly the enviroment by covering our roofs with solar cells.


I am sure there is new

I am sure there is new information, we do not follow the market prices. The best way to follow up would be the contact one of the multiple companies delivering and installing the panels.

This article was posted in

This article was posted in 2010, just wounding it there is an updated calculation/information with current cost of installing the roof top solar panels?

This was very

This was very informative!
Thank you very much!

Post new comment

Please note that these comments are moderated and reviewed before publishing.

The content of this field is kept private and will not be shown publicly.
By submitting this form, you accept the Mollom privacy policy.

a place of mind, The University of British Columbia

C21: Physics Teaching for the 21st Century
UBC Department of Physics & Astronomy
6224 Agricultural Road
Vancouver, BC V6T 1Z1
Tel 604.822.3675
Fax 604.822.5324

Emergency Procedures | Accessibility | Contact UBC | © Copyright The University of British Columbia