Conserving Fuel in Seldom-Used Buildings

The problem of minimal heating appears when buildings are seldom used but not disused. While a disused building can be winterized by draining pipes and filling water closets with antifreeze, this treatment is inapplicable in a building that is used occasionally. A sometimes overlooked consideration is that heat consumption is not linear with the temperature setpoint, in part since on days when the temperature is above the setpoint the furnace shuts off completely. (For Bradford, PA reducing the setpoint from 65 to 50 degrees reduces the number of degree days by about 50% rather than 23%, what would be expected were heat consumption linear.) The minimal heating problem becomes one of determining a temperature setpoint that will allow adequate time for intervention before pipes begin to freeze in the event of failure of the heating system. In order to calculate the time to the onset of possible freezing, a thermal model of the building is necessary.
A strategy for thermal modeling is to first find the area of the building that is most at risk. This will be an area that has plumbing and cools most quickly. The onset of freezing can be taken as 32 degrees F since removal of latent heat introduces a time lag for freezing to take place, but local cold spots can offset the benefit of latent heat. My purpose here is to present two simple models both of which I have used.

The first model is what I call a first-order model. It is very simple and applies when the area of the building selected obeys Newton's Law of Cooling. This case can be diagramed with the electrical schematic shown at left. There is a single thermal mass and a single thermal resistance. The product of the measures of these two is called the thermal time constant. When the first-order model is obeyed, the temperature (versus time) after the heating system is shut down decays exponentially. This type of relation is given by , where a is the time constant, the time it takes for the building to cool to 37% of its initial temperature difference between the inside and outside temperatures. To is here the outdoor temperature and Ti is the initial inside temperature. The usual way of setting up a first-order model is to plot temperature difference versus time on a semilog scale. If the building obeys Newton's Law of cooling, the plot will be a straight line. The time constant can be found from the 37% point. Many plotters will fit an exponential curve. In that case the plotter will give you the equation directly. Setting T to 32 degrees and solving for t gives the time to freezing.

Sometimes a building does not obey Newton's Law of cooling. Then the semilog plot will not be a straight line.
 
The diagram at right illustrates this case. The natural logarithm of the temperature difference is plotted on vertical axis. The intial slope and the final slope are quite different. This case applies to the building shown at the top of this page. This building is built into a hill. The foundation of the building stays at a cooler but much steadier temperature than does the building itself. When heating is interrupted, the building at first cools rapidly. Then heat from the foundation begins to sustain the temperature, and the building cools more slowly. This case is described by a second-order model having two thermal masses, one for the building superstructure and one for its foundation. The simplest form of the second-order model has three thermal resistances. The crossover temperature, constructed as shown on the graph, is approximately the foundation temperature. In this case, a radio-transmitting thermometer (available for $9 at Wal-Mart) was inserted into the foundation underground crawl space and monitored for weeks under different circumstances. The value obtained from crossover on the chart, 47.5 degrees was close to the measured value of the foundation temperature.
The simplest second-order model is shown diagramed at the right. Simplicity is necessary because the more elements in the model, the more constants need fitting. If the temperature of the foundation can be measured directly, there are only three constants that need fitting. Even so, the equation is relatively complicated. It is derived by LaPlace transformation of the two simultaneous differential equations that describe the network shown.
 
 
 

 
 
 
 
 
 

 
where r is given by the equation shown,To is now the initial indoor temperature, Ta is the outdoor temperature, Tf is the intial foundation temperature, f₁₁ is the reciprocal of the time constant for the superstructure with the foundation removed, f₂₂ is the time constant for the foundation with the superstructure held at the outdoor temperature and f₁₂ is a coupling constant. It turns out that f₂₂ is very close to the (positive) coefficient of time in the negative exponential obtained for the slope for the right side of the curve as in the diagram above. f₁₁ is somewhat less than the (positive) coefficient of time in the negative exponential obtained for the slope for the left side (t=0) of the curve as in the diagram above. f₁₂ seems typically on the order of f₁₁. As f₁₂ increases, the elbow becomes more pronounced. The constants can be fitted to three points of the curve either by trial and error or Newton-Raphson with numerically estimated partial derivatives. The time to freezing is easily found by plotting temperature versus time.