## Solar Thermal Math (air heating)

**Efficiency & Calculating BTU's**

Believe it or not, too high of a temperature is bad for efficiency. Technically speaking, the heat generated by a solar air heater is a relatively "low-grade" heat. That's not a disparaging term, it simply refers to the relatively low temperatures (80 to 140°F) that air heaters operate. If the airflow rate were reduced, a collector would heat air above this range, but in terms of Btu's delivered (*see note below), there is a lot more heat in a strong blast of 80°F air than there is in a tiny trickle of 160°F+ air from the same collector. Why? If the airflow rate is too low, then that will increase the exhaust temperature but also cause heat to build up in the collector which will subsequently be lost outside through the collector's components (sides, glazing, trim, etc) and negatively affect efficiency. A higher air flow resulting in a lower temperature within the ideal range (80-140°F) means that heat is being distributed to more air, more quickly, so there is far less heat being lost outside through the collector's components and the exhaust temperature will remain comfortable. A general rule for ensuring a reasonable balance between temperature and airflow without an intelligent fan speed controller is to size the fan so that it has a high static pressure rating (0.5"H2O+) and is capable of moving 2 to 3 cfm of air for every square foot of collector; for example if the collector has an area of 10 square feet then the fan should produce 20-30 cfm, no more or less. A collector capable of producing high temperatures is just an indication of energy potential, really, and is only half of the equation - actually transferring the heat to where it's needed without compromising efficiency or the exhaust temperature to the point that it feels more like a cool draft, is what makes the system work properly. This can be demonstrated using the following equation for calculating btu per hour production from a solar heater: (*NOTE: Btu is an abbreviation for British Thermal Units, a traditional unit of energy equal to about 1055 joules. It is the amount of energy needed to cool or heat one pound of water by one degree Fahrenheit.)

**Btu*h = [V x Oa x Ad x (To - Ti) x Sa] x 60**where:

**V**= air flow velocity from the heater/collector in linear feet per minute (measured with an anemometer)

**Oa**= hot air exhaust outlet area in square feet

**Ad**= air density at sea level and 60°F = 0.065lb/cu.ft

**To**= temperature of hot air exhaust in degrees Fahrenheit

**Ti**= temperature of cold air intake in degrees Fahrenheit

**Sa**= specific heat of air; the amount of energy required to heat 1lb of air 1 degree fahreneit = 0.24btu/lb-F

**60**= number of minutes in an hour

*Example 1:*

*(Given values: exhaust temp = 200F, intake temp = 70F, velocity = 200Lft/m, outlet area = 0.088ft2 )*

Btu*h = [200Lfm x 0.088ft2 x 0.065lbft3 x (200F - 70F) x 0.24btulb-F] x 60

= 200 x 0.088 x 0.065 x 130 x 0.24 x 60

=

__2,141.6 btu*h__

*Example 2: (Given values: exhaust temp = 140F, intake temp = 70F, velocity = 500Lft/m, outlet area = 0.088ft2 )*

Btu*h = [500Lfm x 0.088ft2 x 0.065lbft3 x (135F - 70F) x 0.24 btulb-F] x 60

= 500 x 0.088 x 0.065 x 70 x 0.24 x 60

=

__2,882.9 btu*h__Notice that with a lower temperature difference but higher flow rate, we in turn generate more btu*h in the second example than we did in the first because with the high flow rate, heat has less opportunity to be absorbed into the collectors shell/glazing and radiate back outside - the system is instead heating and circulating more air, more quickly.

**Collector Sizing**

For systems without thermal storage, the collector area shouldn't exceed %20 of the floor area of the space to be heated. Systems that do incorporate storage should have a collector surface of no less than 200ft2 . Referring to

__NRC's solar resource map__, you can see that New Brunswick receives ~3 kwh/m2 of solar energy per day on average during the winter months, which equates to about 10,340 btuh/m2 , given that 1btu = 0.29 watts, and a solar heating system will harness anywhere from %50-%95 of that energy depending on it's design and application. This is important to know because it will determine how large your collector should be to do the job it's intended to do. Below is a basic equation for determining how many btu's is required to heat a space in ideal conditions:

*BtuR = [C x (Ti - To)] / 55*where:

**C**= total capacity of space to be heated in cubic feet (length x width x height)

**Ti**= desired temperature to be achieved inside the space in degrees Fahrenheit

**To**= lowest outside temperature

**55**= 55ft3 ; the amount of air that one btu of energy will raise one degree Fahrenheit

*Small home example:*

BtuR = [(46ft long x 28ft wide x 8 ft high) x (70F - -4F)] / 55ft3

= [10304ft3 x 74F] / 55

= 762,496 / 55

=

__13,863.6 btu__

Btu requirements depend on more than just the volume to be heated and the temperature range. Other major factors include the rate of heat loss through roof, windows, walls, and doors. Air leaks and poor insulation increase btu requirements, as does normal ventilation. A good approach would be to add 20-30% to your total from the equation above to compensate for these losses. Below is a link to an on-line calculator that also factors in these variables and determines minimum/maximum btu requirements for your space and climate conditions:

__http://www.hearth.com/calc/btucalc.html__

Using the small home spec's from the above example, inputting them into the on-line calculator then establishing the average between the minimum and maximum, we can see that approximately 30,000 btu's is needed to heat a small, well insulated home in a cold climate. To determine how much solar collector surface area is needed to meet that requirement, simply divide the total btu requirement by the available average btu/m2 of solar energy for your area during the winter months as follows:

*Ca = R / Bm*where:

**R**= total Btu requirement

**Bm**= total available heat energy per square meter of collector surface in Btu/m2

But we need to do a few things first. From the solar resource map we can see that New Brunswick receives about 3 kwh/m2 of solar energy per day. Because kilowatt hours is a measurement of energy used or produced over a period of time - in this case an average winter day - we need to break it down to represent energy produced per hour by dividing the total 3 kwh per day by the average number of daylight hours per day during winter, which in New Brunswick is about 4 hours:

*Kilowatts per hour per square meter = total kilowatt hours / total daylight hours per day*= 3kw / 4hr

=

__0.75Kw per hour per sq.m , or 750w per hour per sq.m__

Next, we need to convert 750w/sq.m to it's equivalent in btu/sq.m , (remember that 1 btu is equal to 0.29 watts):

*Btu per hour per sq.m = watts per hour per sq.m / watts per btu*= 750w per hour per sq.m / 0.29w per btu

=

__2586.2 btu per hour per sq.m__

Now we can calculate the collector area needed to generate 30, 000btu's:

Ca = 30,000btu / 2586.2btu-m 2

=

__11.6 sq.m , or 124 sq.ft__

Keep in mind that we're only going to be able to extract and use a certain percentage of the available energy, from between 50-95%, so a good medium to use to help make the final calculation would be about 75%. With that said, we're going to need to add %25 more collector area to compensate for potential losses in the absorption process, so:

*TCa = (Ca x 0.25) + Ca*= (11.6m2 x 0.25) + 11.6m2 = 2.9 + 11.6

=

__14.5 sq.m , or 156 sq.ft__

So to create 30,000 btu's, we'll need a collector surface area of about 14.5 sq.m , or 156 sq.ft.

**Storage Sizing**

A general rule on storage sizing calls for 50 - 60 lbs of rock per square foot of collector area. It's recommended that storage systems be incorporated with collectors that have a minimum area of 200 sq.ft , so that equates to about 11,000 lbs of rock. Working with btu's, the 'specific heat' of rock is such that one cubic foot stores 20 btu's for every 1 degree Fahrenheit that it rises in temperature. In the case of a 70°F rise, a cubic foot would store:

20 btu per cu.ft x 70°F

=

__1400 btu per cu.ft__

Borrowing from our previous 'small home' examples, we can calculate that our 156ft2 30,000btu collector working for an average of 4 hours/day will generate 120,000btuh's:

30,000btu x 4hr

=

__120,000 btu per hr__

In order to store that much heat (at a 70°F design temperature rise):

120,000 btu per day / 1400 btu per cu.ft of rock

=

__85.7 cu.ft__

Now we've determined that for a 156 sq.ft collector, we'll need at least 85.7 cu.ft of rock for storing heat to reach a 70°F heat gain. To decrease losses and increase efficiency, the rock bin should be located as close to the space to be heated as possible. Separate stand alone systems are possible, but the less efficient. The best place for storage is always in or under the space, that way any heat loss goes directly where it's needed. Generally, a 6-8 ft tall bin is sufficient. So (again - borrowing from previous examples and given an average 7 ft height), to determine the other two dimensions of the bin needed to store 85.7 cu.ft of rock:

√(85.7cu.ft / 7ft)

= √12.24

=

__3.5ft__

So, we've determined that we need 30,000 btu's, or 120,000 btu per hour per day to heat our small home, and to store enough thermal mass for one day's use we need a bin that is at least 7ft high x 3.5ft x 3.5ft to hold 85.7 cu.ft of rock. To increase the storage to suit a specific number of days, simply multiply the amount of rock needed for one day's worth of storage by the number of days worth of storage desired:

85.7cu.ft x 3 days

=

__257 cu.ft__