# Calculate flywheel mass effect's influence

p>As energy contractors, we are required to perform a basic heat loss or heat gain calculation in order to properly size the heating or cooling plant necessary to provide comfort to the end user. As I have noted in previous columns, these heat loss/gain calculations always assume worst-case scenarios. They only look at the theoretical side of the equation,assuming in the case of a heat load, no contributions

p>As energy contractors, we are required to perform a basic heat loss or heat gain calculation in order to properly size the heating or cooling plant necessary to provide comfort to the end user. As I have noted in previous columns, these heat loss/gain calculations always assume worst-case scenarios. They only look at the theoretical side of the equation,assuming in the case of a heat load, no contributions from any outside or inside sources. This methodology results in a substantial over-sizing of the equipment for the majority of the equipment's operating lifecycle.

In this column, we will look into the flywheel mass effect's influence on a given home's energy demands. Once again, I will use my own home for the purposes of demonstration.

My home has 950 sq. ft. of living space per floor, with two floors, one of which is above grade, and 75% of the basement is below grade. In order to simplify matters, we will only look at the upper floor construction.

The home was built in the early 1950s using conventional stick frame construction. The framing was then covered with 1/2 in. Sheetrock. The floors on the main level are finished red oak tongue and groove.

The outside dimensions of the exterior walls are 38 ft. by 25 ft. for a total of 126 linear running ft. of wall. Of this upper wall space, windows and doors occupy 205 sq. ft.. There are 82 linear ft. of wall with 1/2 in. of sheet rock on each side, and ceilings that are 8-ft. tall. The ceiling itself is 950 sq. ft. of the same Sheetrock material.

Hence, I have 3,065 sq. ft. of 1/2-in. thick Sheetrock inside my home. This equates to 95 sheets of 1/2-in. thick Sheetrock. Each sheet weighs 56 lb. Multiply 95 x 56 = 5,320 lb. of gypsum. At a specific heat capacity of 0.259 Btu per lb., the Sheetrock represents 1,378 "free" Btu per degree Fahrenheit difference. With a typical end of summer temperature of 76°F, and a minimum room setting of 70°F, this represents a total potential of 8,267 Btu.

### Heat loss/gain calculations always assume worst-case scenarios.

As for wooden framing, based on typical 2-by-4 construction with studs at 16 in. on center, I have 69 cu. ft. of Douglas fir pine. At an average density of 33 lb./cu. ft., this represents 2,277 lb. of Douglas fir pine. At a specific heat capacity of 0.29 Btu/lb./°F difference, the framing represents 660 Btu of "free" energy per degree Fahrenheit difference. With a typical end of summer temperature of 76°F, and a minimum room setting of 70°F, this represents a total potential of 3,961 Btu.

As for hardwood flooring, I have 59 cu. ft. of red oak hardwood. At an average density of 44 lb./cu. ft., this represents 2,596 lb. of hardwood. With a specific heat of 0.39 Btu/lb./°F, this represents 1,012 Btu of energy per degree Fahrenheit difference. With a typical end of summer temperature of 76°F, and a minimum room setting of 70°F, this represents a total potential of 6,074 Btu.

The hardwood floor is laid upon 59 cu. ft. of 1-by-6 white pine. At an average density of 38 lb./cu. ft., this represents 2,265 lb. of pine. With a specific heat of 0.39 Btu/lb./°F, this represents 883 Btu per degree difference. With a typical end of summer temperature of 76°F, and a minimum room setting of 70°F, this represents a total potential of 5,301 Btu.

In total, excluding plumbing fixtures, glass, ceiling and floor framing components (assumed to be outside the heated envelope) and other components, this mass of construction for this scenario represents 23,603 Btu of "free" energy that is not taken into consideration in the typical heat-loss calculations. This is a considerable amount of energy that helps buffer the temperature swings within the house during temperature changes.

In reality, my home is relatively low in internal mass compared with some of the older brick homes built around the turn of the century. These homes may have as much as three to four times the internal mass that my home has, which means it has three to four times the flywheel mass effect to help buffer temperature swings and reduce real-time energy consumption.

This information should be taken into consideration when evaluating real-time energy consumption because it can either work for you or against you.

Tune in next month as we look at the state-of-the-art solar homes undergoing testing at the second annual Solar Decathalon. Until then, Happy Massive Hydronicing!

TAGS: Hydronics