### A heating system for the garage

In the previous post I presented the results from two blower door tests. The blower door results from the garage testing can be used to determing the heat load required by the building. We can think of the heat loss as a sum of two components: Heat lost through conduction and heat lost due to infiltration. This is a highly simplified method and only partially describes the physical behaviour of heat loss. For example it neglects to account for thermal mass effects, treats conduction as a one dimensional phenomena, ignores convection currents, ignores the thickness of the walls, etc. However, this simplified method typically overestimates the size of the required heat source so its good enough for this purpose.

Calculating yearly demand is a matter of breaking the structure up into individual components. First I calculated the area of each component using my construction drawings. The product of area, R-value, and 24 hours/day gives the heat lost per degree day. The website www.degreedays.net can be used to generate degree days using your own local weather data. You need to know the base temperature of the building in order to generate the degree days. The base temperature of a building is difficult to choose. It is really dependent on many things like internal gains, solar gains, etc. I expect internal gains to be very little unless I am there using it. The building is oriented towards the south so there will be some solar gains but who knows how it will contribute to the overall base temperature. It is likely that I will keep the workshop cool until I decide to use it so to keep things simple I decided to use a base temperature of 65 F. Degreedays.net generated HDD of 8582 F.day. This number can be used to determine the heat loss through the envelope above the ground. For conduction through the slab and into the ground, I assumed that the ground temperature below the slab is 47 F throughout the heating season. So the temperature difference between the ground and the slab would be 18 F. I assumed that the heating season for the garage was the same as was calculated in the WUFI report for the house (303 days). Based on our weather data, it seemed like a reasonable number. The heating degree days for the slab was then estimated to be the product of 303 days x 18 F = 5454 F.days. This is a very simplified method and it ignores a lot of physics going on with the heat transfer. This being said it probably over estimates the heat loss.

Calculating demand (total energy used to maintain constant temperature of 65 F) is a matter of multiplying the BTU/F.day by the degree days and then adding up all the individually calculated heat losses. Likewise, as explored in a previous post, the infiltration heat loss can also be estimated. Figure 1. below illustrates the results of the calculation. The resultant total demand is about 5500 kWh.

For determining the size of the heating equipment, it was necessary to determine our winter temperatures. Temperatures rarely go below 5 F so the rate of heat loss was calculated based on that temperature. The rate of heat loss is calculated using the product of 1/R x Area x Delta T. Once again, I assumed that heat loss through the floor to the ground below was based on the temperature of the ground being 47 F. The resultant calculations, including rate of heat loss due to infiltration, are presented in Figure 2 below. A total load of about 4700 BTU/h gives me confidence that 9000 BTU heat pump is sufficient to provide the heat required to keep the building warm. Even at a design temperature of 68 F, the total load is just below 5000 BTU/h.

Figure 1. Heat Load Calculation Results. Calculations show a heat source of 1.4 kW is all that's required to maintain constant temperature.

I explored several options for heating the garage. Using the walltherm inside the house to feed a heater loop in the garage was one of them. However, hydronics in that space would require that you always keep the building warm to prevent pipes from freezing. I axed this thought pretty quickly. A pellet stove was an option but once I explored the idea further, the price of pellets is on par with electricity and there are no bulk discounts. Shipping for the local manufacturer is also an issue and would required a 800 km trek. Since the stoves power source is electricity, the operational cost is more than using baseboard heaters. A heat pump was the next best option. After talking with a few subcontractors about it they suggested talking with Reliable Heat Pump Services Ltd. The owner/operator, James Parrel-Gough, convinced me to look at the Halcyon line of Fujitsu mini-splits. They are impressive. The 9RLS3H operates down to -15 F which is more than sufficient for our climate and it is able to modulate heating from 3100-12000 BTU/hr. At 5 F the COP is about 2 so I expect that electricity could be as high as 2700 kWh/year based on the calculated yearly heating demand. This being said, our average winter temperature hovers around 23 F so in all likelihood, the electricity usage will be less. Really world performance will tell the truth.

Calculating yearly demand is a matter of breaking the structure up into individual components. First I calculated the area of each component using my construction drawings. The product of area, R-value, and 24 hours/day gives the heat lost per degree day. The website www.degreedays.net can be used to generate degree days using your own local weather data. You need to know the base temperature of the building in order to generate the degree days. The base temperature of a building is difficult to choose. It is really dependent on many things like internal gains, solar gains, etc. I expect internal gains to be very little unless I am there using it. The building is oriented towards the south so there will be some solar gains but who knows how it will contribute to the overall base temperature. It is likely that I will keep the workshop cool until I decide to use it so to keep things simple I decided to use a base temperature of 65 F. Degreedays.net generated HDD of 8582 F.day. This number can be used to determine the heat loss through the envelope above the ground. For conduction through the slab and into the ground, I assumed that the ground temperature below the slab is 47 F throughout the heating season. So the temperature difference between the ground and the slab would be 18 F. I assumed that the heating season for the garage was the same as was calculated in the WUFI report for the house (303 days). Based on our weather data, it seemed like a reasonable number. The heating degree days for the slab was then estimated to be the product of 303 days x 18 F = 5454 F.days. This is a very simplified method and it ignores a lot of physics going on with the heat transfer. This being said it probably over estimates the heat loss.

Calculating demand (total energy used to maintain constant temperature of 65 F) is a matter of multiplying the BTU/F.day by the degree days and then adding up all the individually calculated heat losses. Likewise, as explored in a previous post, the infiltration heat loss can also be estimated. Figure 1. below illustrates the results of the calculation. The resultant total demand is about 5500 kWh.

Figure 1. Heat Demand Calculation Results. Calculations show a yearly demand of about 5500 kWh

For determining the size of the heating equipment, it was necessary to determine our winter temperatures. Temperatures rarely go below 5 F so the rate of heat loss was calculated based on that temperature. The rate of heat loss is calculated using the product of 1/R x Area x Delta T. Once again, I assumed that heat loss through the floor to the ground below was based on the temperature of the ground being 47 F. The resultant calculations, including rate of heat loss due to infiltration, are presented in Figure 2 below. A total load of about 4700 BTU/h gives me confidence that 9000 BTU heat pump is sufficient to provide the heat required to keep the building warm. Even at a design temperature of 68 F, the total load is just below 5000 BTU/h.

Figure 1. Heat Load Calculation Results. Calculations show a heat source of 1.4 kW is all that's required to maintain constant temperature.

I explored several options for heating the garage. Using the walltherm inside the house to feed a heater loop in the garage was one of them. However, hydronics in that space would require that you always keep the building warm to prevent pipes from freezing. I axed this thought pretty quickly. A pellet stove was an option but once I explored the idea further, the price of pellets is on par with electricity and there are no bulk discounts. Shipping for the local manufacturer is also an issue and would required a 800 km trek. Since the stoves power source is electricity, the operational cost is more than using baseboard heaters. A heat pump was the next best option. After talking with a few subcontractors about it they suggested talking with Reliable Heat Pump Services Ltd. The owner/operator, James Parrel-Gough, convinced me to look at the Halcyon line of Fujitsu mini-splits. They are impressive. The 9RLS3H operates down to -15 F which is more than sufficient for our climate and it is able to modulate heating from 3100-12000 BTU/hr. At 5 F the COP is about 2 so I expect that electricity could be as high as 2700 kWh/year based on the calculated yearly heating demand. This being said, our average winter temperature hovers around 23 F so in all likelihood, the electricity usage will be less. Really world performance will tell the truth.

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