Heating and cooling

Heating and cooling are among the most energy-gobbling, CO2 – belching things we do. Refrigerators, freezers, and air conditioners keep things cold. Central heating, ovens, and kilns keep things hot. For all these, it is the use phase of life that contributes most to energy consumption and emissions. Some, like refrigeration, central heating, or cooling, aim to hold tempera­tures constant over long periods of time—the fridge or building is heated or cooled once and then held like that. Others, like ovens and kilns, heat up and cool down every time they are used, zigzagging up and down in tem­perature over the span of a few hours. The best choice of material to mini­mize heat loss depends on what the use cycle looks like.

Refrigerators. To get into the topic, take a look at refrigerators. The func­tion of a fridge is to provide a cold space. A fridge is an energy-using product (an "EuP"), and like most EuPs, it is the use phase of life that dominates energy consumption and emission release. Thus a measure of eco-excel­lence for a fridge is the energy per year per cubic meter of cold space, H*, the * signifying "per cubic meter." Minimizing this is the objective.

But suppose that the fridges that are good by this criterion are expensive and the ones that are not so good are cheap. Then economically minded consumers will perceive a second measure of excellence in choosing a fridge: the initial cost per cubic meter of cold space, C* . Minimizing this becomes an objective, one that, almost certainly, conflicts with the first. Resolving the conflict needs the tradeoff methods of Chapter 8.

Figure 9.8 plots the two measures of excellence for 95 contemporary (2008) fridges. The tradeoff line is sketched (remember that it is just the convex-down envelope of the occupied space). The fridges that lie on or near it are the best choices; several are identified. They are "nondominated solutions"—offering lower energy for the same price or lower price for the same energy than any of the others.

That still leaves many, and they differ a lot. So wheel out penalty functions. The easiest unit of penalty is that of cost, in whatever currency you choose (here US$). We want to minimize life cost, which we take to be the sum of the initial cost and the cost of the energy used over the prod­uct’s life. So, define the penalty function

Z* = C* +aeH*t (9Л0)

where ae, the exchange constant, is the cost of energy per kWhr and t is the service life of the fridge in years, making Z* the life cost of the fridge per cubic meter of cold space. The grand objective is to minimize Z*.

Take the service life to be 10 years and the cost of electrical power to be US$0.2 per kW. hr. Then the penalty function becomes

Z* = C* +2H* (9Л1)

or, solving for H * :

H* = 1Z* – 1C* (9.12)

f 2 2 f

The axes of Figure 9.8 are H* and C*, so this equation describes a family of straight lines with a slope of -1/2, one for any given value of pen­alty Z*. Five are shown for Z* values between $2000 and $6000. The best choices are the fridges with the lowest value of Z*—the ones where the Z* contour is tangent to the tradeoff line.

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0 1000 2000 3000 4000 5000 6000

Initial cost per m3, C* ($/m3)

A tradeoff plot for fridges with contours of the penalty function Z*. Data from 2008


If by some miracle the cost of energy drops by a factor of 10, the Z* con­tours get 10 times steeper—almost vertical—and the best choice becomes the cheapest fridge, regardless of power consumption. If, more probably, it rises by a factor of 10, the contours become almost flat and the best choice shifts to those that use least energy, regardless of initial cost.

You could argue that this purely economic view of selection is mis­guided. The environment is more important than that; reducing use energy and emissions has a greater value than $0.2 per kW. hr. Fine. Then you must define what you believe it to be worth and use that for ae instead, bas­ing your selection on the Z* contours that result. But if you want the rest of the world to follow your example, you must persuade them—or if you are in government, make them—use the same value. There are no absolutes in this game. Everything has to be assigned a value.

Figure 9.8 is a tradeoff plot with a sharp nose at the lower left. When the plots are like this, the optimal choice is not very sensitive to the value of the exchange constant ae. But when the nose is more rounded, its value influ­ences choice more strongly. We will see an example in a later case study.

So fridges use energy. Where does it go? And what choice of material would minimize it? To answer that we need a little modeling.

Modeling Thermal Loss. Creating and maintaining cold or hot space costs energy. The analysis is the same for both (though the choice of material is not). The result depends on how often the space is heated and cooled and how long it is held like that on each cycle. Take, as a generic example, the heated space, oven, or kiln sketched in Figure 9.9; we will refer to it as "the kiln." The design requirements are listed in Table 9.5.

When a kiln is fired, the internal temperature rises from ambient, To, to the operating temperature, T, where it is held for the firing time t. The


energy consumed in one firing has two contributions. The first is the heat absorbed by the kiln wall in raising it to T. Per unit area, it is


where Cp is the specific heat of the wall per unit mass (so Cpp is the specific heat per unit volume) and w is the insulation wall thickness. It is min­imized by choosing a wall material with a low heat capacity Cpp and by making it as thin as possible.

The second contribution is the heat conducted out: at steady state the heat loss by conduction, Q2, per unit area, is given by the first law of heat flow. If held for time t it is

Q1 = – A d^t = A T———— І2І t (9.14)

dx w

It is minimized by choosing a wall material with a low thermal conduct­ivity A and by making the wall as thick as possible.

The total energy consumed per unit area is the sum of these two:

where AT = (T – To). Consider first the limits when the wall thickness w is fixed. When the heating cycle is short, the first term dominates and the best choice of material is that with the lowest volumetric heat capacity, Cpp. When instead the heating cycle is long, the second term dominates and the best choice of material is that with the smallest thermal conductivity, A.

A wall that is too thin loses much energy by conduction but little to heat the wall itself. One that is too thick does the opposite. There is an optimum thickness, which we find by differentiating Equation 9.15 with respect to wall thickness w and equating the result to zero, giving:

/ U/2

= (2at)1/2 (9.16)

where a = X/pCp is the thermal diffusivity. The quantity (2at)1/2 has dimen­sions of length and is a measure of the distance heat can diffuse in time t. Substituting Equation 9.16 back into Equation 9.5 to eliminate w gives:

Q = (ACpP)1/2 AT(2t)1/2 (9.17)

This is minimized by choosing a material with the lowest value of the quantity

M = (ACpp)1/2 = (9.18)

Figure 9.10 shows the A – a chart of Chapter 8, expanded to include more materials that are good thermal insulators. All three of the criteria we have derived—minimizing Cpp, A and (ACpp)1/2—can be plotted on it; the "guidelines" show the slopes. For long heating times it is A we want to minimize, and the best choices are the materials at the bottom of the chart: polymeric foams or, if the temperature T is too high for them, foamed glass, vermiculite, or carbon. But if we are free to adjust the wall thickness to the optimum value of Equation 9.16, the quantity we want to minimize is (ACpp)1/2. A selection line with this slope is plotted in the figure. The best choices are the same as before, but now the performance of vermicu – lite, foamed glass, and foamed carbon are almost as good as that of the best polymer foams. Here the limitation of the hard-copy charts becomes appar­ent: there is not enough room to show a large number of specialized mater­ials such as refractory bricks and concretes. The limitation is overcome by the computer-based methods mentioned in Chapter 8, allowing a search over a much greater number of materials.

Exactly the same analysis works for refrigerators and ovens. For workspace and housing there is an additional complication: humans have to breathe, and that means ventilation, and ventilation means that hot or cold air is pumped out of the "space" and replaced by new air that has to be heated or cooled.

Materials can help here too by acting as heat exchangers, extracting heat or cold from the outgoing air and using it to precondition the air coming in.

Postscript. It is not generally appreciated that, in an efficiently designed kiln, as much energy goes into heating up the kiln itself as is lost by ther­mal conduction to the outside environment. It is a mistake to make kiln walls too thick; a little is saved in reduced conduction loss, but more is lost in the greater heat capacity of the kiln itself.

That is the reason that foams are good; they have a low thermal con­ductivity and a low heat capacity. Centrally heated houses in which the heat is turned off at night suffer a cycle like that of the kiln. Here (because T is lower) the best choice is a polymeric foam, cork, or fiberglass (which has thermal properties like those of foams). But as this case study shows,
turning the heat off at night doesn’t save you as much as you think, because you have to supply the heat capacity of the walls in the morning.

Updated: October 3, 2015 — 10:57 am