Thermal Mass -- under construction

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Basic Characteristics

The basic properties that dictate the thermal behavior of materials are the density, specific heat and conductivity. An effective thermal mass mass material should have a high heat capacity, a moderate conductance, a moderate density and a high emissivity.

Specific Heat (Cp)

It takes 100 calories of energy to change the temperature of water from 0 deg C to 100 deg C. We know that 100 deg C is the boiling point of water. However, if all you did was raise the temperature of the water to 100 degrees, it won't boil. You need to add an additional 539 calories to get it to boil. When you add the first 100 calories, that energy goes to increasing the kinetic energy of the molecules. The heat energy that was added is absorbed by the individual water molecules and causes them to be more energetic but it has not had any effect on the bonds between water molecules.

Once a temperature of 100 degrees is obtained, the bonds in between water molecules have to be broken in order for a particular water molecule to break loose from the liquid lattic and escape. It takes 539 calories of heat energy to accomplish this. While putting in this 539 calories of heat energy, the temperature of the water will not change 1 bit from the 100 degrees that it started with. This energy required to break the cohesive bonds between water molecules is called the latent heat of vaporization. A similar thing happens when freezing water. The temperature of the water may be down to 0 degrees C but the water has not crystallized into ice. In order to cause the bonds to arrange themselves in a crystal, the energy state of the system has to be reduced. Heat has to be taken out so the molecules don't have enough energy to move around any more. This is called the latent heat of fusion.

There are 2 different heat concepts involved in getting water to boil. The first concept is the specific heat. The specific heat of a material is the amount of heat energy it takes, per unit mass to raise the temperature of the material one degree celcius. Per unit mass is important. The unit mass means 1 unit of mass of a specific measure. In this case its grams. Specifically, it takes 1 calorie to raise the temperature of 1 gram of water 1 degree Celcius. But, that only works up to 100 degrees. This characteristic is also known as the Specific Heat Capacity. Heat capacity in general terms is a measured physical quantity for a material that characterizes how much heat is required to change the temperature of the material a specific amount.

Once the water is boiling, you have to keep putting heat in to keep it boiling. As each water molecule escapes the surface tension of water and becomes a water vapor molecule, heat is lost from the system. That heat is carried away with the escaped water vapor molecule. Evaporation is a process of cooling, not heating. The amount of cooling or evaporation of water vapor molecules out of the liquid water will remain in balance with the amount of heat energy added to the system. No matter how much heat you put into the system, the water will continue to boil and its temperature will not change from the 100 degrees Celcius. That is the boiling point of water on the Celcius scale. Every substance has a boiling point. When that boiling point is achieved, it does not change its temperature past the boiling point, it just continues to boil provided the correct amount of heat or energy is added to the system.

The relationship between heat and temperature change of a material is expressed by the specific heat capacity formula: Q = m*c*(delta T).

Q is the amount of heat energy added to the system. It can be expressed in calories. M is the mass of the material, in grams. Delta T is the temperature difference you are trying to achieve, in degrees C. The specific heat capacity (c) of a material is a measure of how many calories it takes to get 1 gram of the material to raise its temperature 1 degree C. In the case of water, the specific heat is 1 calorie per gram of water, or 1 Cal/gm, up to 100 degrees C. After that, we cannot use the specific heat, rather we must use the latent heat of vaporization.

In SI units, the specific heat capacity has units of Joules / kg -Kelvins

To convert to US imperial units, the following conversion factors apply:

1 joule = 0.000947 Btu

1 Btu = 252.1644 calories

1 kg = 2.20 lbs

1 kelvin = -457.87 degrees Fahrenheit, but, to do the conversion: Degrees F = (Kelvins * 9/5) - 459.67. In terms of the Celcius scale, Degrees C = Kelvins - 273.15

For Example:

Water at 25 deg C in the liquid phase has a specific heat capacity of 4.1813 J/g-K

To convert that to calories, grams and degrees celcius, we use the following conversion factors:

4.1813 J/g-K * 0.000947 Btu/1 Joule * 252.1644 Calories/Btu

= 0.9984 calories/g - K

A kelvin unit is the same as a celcius unit. Its just a different starting point on the scale. There is no need to try to convert K to degrees Celcius. The result is 1 calorie/g-degrees C, rounded.

The only difference between the heat capacity of a material and the specific heat capacity of a material is that the latter applies to a specific mass of the material. The volumetric heat capacity is the heat capacity per unit volume instead of unit mass. This is important because the ability of a material to store heat is an extensive property. The more of it you have in one place the better. One cup of water will not be as effective as a heat storage device as 100 gallons.

A Phase Change Material (PCM) as a thermal mass depends on the same concept as the latent heat of vaporization of water. A good PCM material takes some energy to get its temperature up to a certain point. After that, it has the capacity to store a lot of heat per unit mass without changing its temperature. Good PCM materials for solar design do this within a reasonable temperature range more in alignment with interior thermostat setpoints. It would be extremely difficult to raise the temperature of water with solar energy to 212 degrees Fahrenheit.

With respect to non PCM materials to act as a thermal mass, the first characteristics we look for are a high specific heat capacity and a high density. The higher the specific heat capacity the more energy it will take to raise the temperature of the material 1 degree.

Density (p)

To find out how much heat we can store in each building element, we take the specific heat capacity of that element and mulitiply it by the density of that element. The specific heat of most masonry type materials or stone is very similar. They are all in the range of about 0.2 to 0.25 W-h/Kg-C. .

If we go to Wikipedia and look under Heat Capacity, at the bottom of the article we will find a table with heat capacities of some common building materials.

Concrete has a specific heat capacity of 0.88 J/g-K. We know from converting the specific heat of water above that we have a conversion factor to go from Joules/g-K to Btu/g -K. which is 0.000947 Btu/1 Joule. One gram is 0.0022 lbs. The heat capacity of concrete in US imperial units is 0.88 J/g-K * 0.000947 Btu/1 Joule * 1 gram/0.0022 lbs, yielding a specific heat capacity of 0.3788 Btu/lbs-C. Remember that to go between degrees C and K does not require a conversion. That means that 1 deg C is equal to 1 degree K, they just start at different points on the scale. Finally, we want to convert the degrees C to degrees F. The conversion is Degrees F = (9/5 * degrees C) + 32. The Celcius scale freezes water at 0 degrees and boils it at 100. The Fahrenheit scale freezes water at 32 degrees and boils it at 212 degrees. Every celcius degree is worth almost 2 fahrenheit degrees (1.8 to be more exact). The reason you add the 32 degrees on is because of the starting point of the scale itself. If you wanted to know the equivalent temperature in Degrees C and Degrees F, then you would add on the 32 degrees. In this case, we want to know what the degree change is in F if we add a certain amount of heat.

We complete the conversion of our concrete. 0.3788 Btu/lbs- C * 1 deg C/ (5/9 deg F) to obtain 0.210 Btu/lbs - deg F.

It takes roughly 5x as much concrete, by weight, as water to store the same amount of energy as the same weight of water. 1 lb of concrete can store 0.21 Btu's of energy before it changes its temperature 1 degree F. One lb of water can store 1 Btu of energy before it changes it temperature 1 degree F.

To get an idea of how much mass is involved, we use the density of the materials. Many times densities will be given in terms of Kg/m^3 or kilograms per cubic meter. In the US, we are more used to thinking in terms of lbs/cf or pounds per cubic foot. The density of water is 1,000 kg/cm. To convert that to lbs/cf, we use 2 conversion factors. 1 kg = 2.2 lbs and 1 cubic meter = 35.314 cubic feet. The math is: 1,000 kg/cm * 2.2 lbs/kg * 1 cm/35.314 cf, to obtain 62.29 lbs of water per cubic foot. Incidently, 1 cubic foot is 7.481 gallons of water.

The density of portland cement with some gravel mixed in is on the order of about 150 lbs per cubic foot of material. A slab of 1" thickness and 1 SF in surface area will weigh about 12.5 lbs. Just divide the cubic feet by 12 for 12". That amount of concrete has the capacity to hold (0.0.21 Btu/lbs-F * 12.5 lbs) about 2.62 Btu of heat energy before changing its temperature 1 degree F.

One cubic foot of water is 7.481 gallons. A container of water 1" deep and 1 foot in surface area is 0.623 gallons. This can be visualized with a 1 gallon liquid container filled just shy of 2/3 full. This amount of water has the capacity to absorb (62.29 lbs/cf * 1 cf/7.481 gallons * 0.623 gallons * 1 btu/lbs - F) 5.18 Btu of heat energy before changing its temperature 1 degree F. You can alternatively take the lbs per CF of water at 62.29 and divide by 12 to obtain 5.19 lbs of water at 1" thick and 1 foot square. Since the thermal heat capacity of water is 1, you need not do anything else. From a volume point of view, you need roughly twice as much volume of cement to hold the same heat as 1 of the same volume of water.

The typical first floor is designed to hold about 50 PSF (pounds per square foot). Planning for thermal mass frequently involves structural planning. Higher floors are designed to hold less weight than the first floor.

Conductivity (k)

The thermal conductivity of a material is a measure of its ability to conduct heat. It is usually measured in Watts per K per meter. If you take the thermal conductivity of a material and multiply it by a temperature difference in K, and an area in meters, then divide by the thickness of the material (in meters) the result will be the energy loss in Watts (W) through the piece of material.

If you know the area and thickness of a material and the difference in temperature on opposite faces of it, then you can compute the quantity of heat that passes through it in some unit time. The amount of heat that passes through it is given by the following formula:

W/K = [thermal conductivity (k) * A (surface area of material)] / L (thickness of material). The inverse of thermal conductivity is thermal resistance. The thermal resistance is also known as the R-Value of a material. Generally, the thicker the insulation material, the lower the R value is per inch.

The heat transfer coefficient of a material is very similar. It is expressed in units of Watts/K-m^2. It is simply the ratio of the thermal conductivity over the thickness (k/L). The inverse of the thermal heat transfer coefficient is the thermal insolance.

Heat transfer coefficients do not have units. They are a proportion. The numerator of the ratio has the amount of heat that was transferred given a specific set of conditions. The denominator of the ratio is the product of the driving force for the heat transfer (delta T) and the area over which the heat transfer took place.

There is yet another term. The thermal transmittance. This incorporates the thermal conductance of a material (heat moving by conduction heat transfer) as well as convection and radiation heat transfer. It is measured in the same units as the thermal conductance. This is also called the U value. Therefore, thermal transmittance, thermal conductivity and U value are all synonyms.

The R value of red brick, 4" thick is 0.8 Btu/SF-deg F-hr. In the US a brick is 8x4x2.25". Converting that to cubic feet, (8" * 1 ft/12") * (4" * 1'/12") * (2.25" * 1'/12") we obtain 0.04 cubic feet. The storage capacity of red brick is 24 Btu / CF before it increases in temperature 1 degree F. This amount of brick has a storage capacity of about 1 Btu (0.96). If the temperature difference between one face of the brick and the other is 1 degree F, then the heat transfer across it will be given by Q = (delta T) / R * A. We can find the area A, because the R value was given for the thickness of 4". Therefore, the area exposed to the temperature difference is 0.125 SF. The heat transfer then is Q = (1 deg F/ 0.8 Btu/SF-deg F-hr) * 0.125 = 0.156 Btu/hr. It will take 7.68 hours for that amount of heat to make it from one end of the brick to the other before the brick starts to change in temperature. This is likely what is happening to a brick that is not in the path of direct sunlight. If the temperature difference between the 2 sides of the brick was 5 degress F, then Q = (5/0.8) * 0.125 = 0.78 Btu. It would take 1.23 hours for the brick to gain enough energy before it raised its temperature 1 degree F. You can see that the temperature difference is the driving force for heat transfer. The higher the temperature difference, the faster the heat transfer from the warm side to the cool side takes place.

Compare the brick to a 2x4 which has similar thickness at 3.5". The R value is 4.38. The R value is 5.47 times bigger so the same heat transfer experiment would require 5.47 times the amount of time to get the same amount of heat through a 2x4. A cubic foot of softwood has a storage capacity somewhere in the vicinity of almost 20 btu before it changes its temperature 1 deg F. The storage capacity of 1 SF and 1" thickness is about 1.66 Btu. If we multiply that by the 3.5" thickness, we obtain 5.81 Btu. But, a 2x4 is not 12"x12" in face. We adjust by dividing by 12 then multiplying by 1.5". In the end, a 1' long piece of 2x4 has the capacity to hold 0.72 Btu of energy. The heat transfer with a 1 degree temperature difference is Q = (1/4.38) * 0.125 SF = 0.0285 Btu/hr. It will take 36 hours for that piece of 2x4 to gain enough heat to raise its temperature 1 degree F with only a 1 degree temperature change. This is what is likely to happen to framing members that are not in the direct path of sunlight. In the course of the 36 hours, the temperature may not be maintained at a constant 1 degree difference between the ambient air and the framing member.

A typical solar collection day in the middle of winter is on the order of 6 hours. A good thermal mass material for a heating dominated condition will display the ability to absorb the heat it is subject to in a reasonable amount of time.

If you put the brick in direct sunlight and it receives 200 Btu of energy per hour, then we can calculate the temperature change of the brick after 1 hour. 200 Btu = (delta T / 0.8 Btu/SF-deg F-hr) * 0.125 SF . Solving for delta T, we multiply 200 * the R value and divide the result by the area. We obtain 1280 degrees F. We know if we leave a red brick out in direct sunlight it could get hot. However, we have no reason to believe it is going to achieve a temperature of 1280 deg F, if it started at 0 deg F. At some point there is an equilibrium that is reached.

The interior of a car left out in the hot sun can easily rise 50 degrees F in an hour but at some point the amount of heat lost is equal to the amount of heat gained. This is why when we return to our vehicles after leaving them in the hot sun for hours, we don't find everything inside melted.

The same thing happens to the brick. At some temperature the amount of heat it gains is in balance with the amount of heat it is losing. What about the 2x4? We leave the 1' length of 2x4 out in the sun where it receives 200 Btu of energy in 1 hour. 200 Btu = (Delta T /4.38) * 0.125 SF. We divide both sides by 0.125 SF and multiply both sides by the R value, 4.38. We obtain 7008 deg F. We know if we leave a 2x4 out in the sun, right next to the brick, that the 2x4 will likely feel a lot cooler than the brick after 1 hour and it will also likely be very handleable.

Diffusivity (alpha)

The diffusivity of a material is a thermodynamic property that measures how quickly a material adjusts its own temperature to the temperature of the surround.

The thermal diffusivity is the thermal conductivity (k) in W/m-k / [(the density(p) in kg/m^3) * The specific heat capacity (Cp) in J/kg - K)] . The denominator is actually the volumetric heat capacity of the material. The thermal diffusivity of a material is a rate of heat loss. It is usually given in square meters/second. The higher the thermal diffusivity of a material, the more quickly it loses its heat to the surround.

The thermal effusivity is like it. It is the square root of the product, rather than the quotient of the above terms. This is a measure of a material's ability to exchange heat with its surround. The thermal effusivity is also called the thermal inertia. The units are Joules/m^2-K-s. It tell you how many Joules of heat are lost per square meter of the material per second. It can be computed by taking the square root of the (thermal conductivity * density * specific heat capacity).

The higher the volumetric heat capacity of a material, the longer it takes for the system to reach equilibrium with its surround.

The thermal inertia of water is 0.038, Sandy soil is 0.024, Basalt is 0.053 and Stainless Steel is 0.168.

For the brick, take the square root of the following products ( 1.25 sf-F/ btu-hr *119.61 lbs/cf * 0.2 Btu/lbs-F), or the square root of (29.9) to obtain 5.46 Btu-hr/deg F. The brick can lose 5.46 Btu's per hour for every degree F it is increased in temperature above ambient conditions. The standard red brick can store about 2 Btu of energy before its temperature goes up 1 degree F. If it is subjected to 200 Btu of solar radiation per hour, its temperature can rise 100 degrees F in one hour, if it wasn't also losing energy. It will achieve a temperature increase of 36 degrees in approximately 18 minutes. From there, it can lose what it gains. If the brick started off at 70 degrees F, then it may not get any hotter than 106 degrees F.

If the brick is in direct sunlight and it has managed to gain 36 deg F in temperature increase, then it can lose 200 Btu/hr. If the solar insolation upon it is 200 Btu/SF-hr, then we can see how it can lose just as much as it gains. As soon as it reaches some critical temperature, it will not increase past that.

Diurnal Heat Capacity

References: Diurnal Heat Storage in Direct Gain Passive Solar Buildings by D. Balcomb and D. Neeper, Los Alamos NL, http://library.lanl.gov/cgi-bin/getfile?00248851.pdf

and http://www.nmsea.org/Curriculum/Courses/Passive_Solar_Design/IV.htm

The diurnal heat capacity is a measure of the effective amount of heat stored during a sunny day and then released at night in a typical 24 hour diurnal cycle. This enables the prediction of maximum temperature swings in a 24 hour period.

The Diurnal Heat Capacity (DHC) of a material is the amount of heat a given mass will store or release per unit area of exposed surface per degree temperature difference between it and its surround. The units are Btu/ (square foot-degree F). It is the exposed surface that participates in energy gained or lost, not the volume or the density. A mass with very high volume but little exposed surface area is not likely to be as efficient as a mass with very high surface area.

The DHC can be divided into 2 parts. Direct DHC is energy gained from direct solar radiation and Indirect DHC is energy from other sources such as people, light bulbs, appliances and other objects in its line of sight that are warmer than it.

Q stored and returned after 6 pm = A*Cd*Delta T. The values for Cd, the diurnal heat capacity must either be for a specified thickness or for an infinite thickness. The material must be thick enough to approximate infinite thickness.

Diurnal Heat Capacity of Common Building Materials

The DHC is given in Btu/ deg F - SF and the thickness in inches

Material DHC Characteristic Thickness

Granite 11.57 5.88

Concrete 10.71 6.05

Limestone 8.33 4.2

Common Brick 6.49 4.17

Paver Brick 9.69 5.07

Adobe 5.52 3.90

Sand 3.62 3.40

Softwood 1.39 3.11

Hardwood 2.18 2.74

Back to Selecting a Thermal Mass Material

In addition to a high heat capacity, a moderate conductance and a moderate density, an effective thermal mass material must also have a high emissivity.

The process of acting like a thermal mass has 4 steps. In the first step, energy is acquired usually through radiant heat energy from the sun or from objects in the vicinity of the thermal mass.

This energy input causes the temperature of the surface of the mass to increase. The second step involves moving that heat energy more interior to the thermal mass. If a material has a poor thermal conductivity, then it is a better insulator than a thermal conductor and it will not readily store the energy upon it, in its interior. When the mass has achieved a temperature higher than its surround, the third step is to radiate that heat energy to objects cooler than it. The forth step involves driving energy from the internal storage of the mass toward the surface so it can be radiated outward.

Cement is not particularly useful above 4" in thickness as the heat energy does not penetrate further into the cement.

Wood is not an optimal choice for thermal mass because it has a low heat storage potential and wood is not a good conductor of heat. Its a better insulator. As a poor conductor, the heat it gains is not conducted to the inside of the wooden member, rather it is readily lost from the surface as radiant energy when the surface temperature of the wood exceeds the temperature in its local environment.

If we look at the table of R values for common building materials from http://www.coloradoenergy.org/procorner/stuff/r-values.htm we can see that the brick has an R value of 0.8 per 4" of thickness and structural framing lumber has an R value on the order of about R = 1.25 per inch of thickness. Using that information, we can guess at what materials are good conductors and which are good insulators.

Steel may seem like a good candidate for thermal mass but its low emissivity indicates that incoming solar radiation mostly reflected rather than absorbed and its high conductivity indicates it can rapidly move heat from its own internal storage to the surface and lose it quickly to the surround. In a diurnal time frame, steel would have a cycle more on the order of minutes rather than hours.

Glass also seems to have a high potential for heat storage. Its transparent to short wave infrared so energy in this frequency series will pass right through it. With respect to wavelengths that correspond to room temperatures or long wave IR, it is fairly reflective.

With glass and water if you add pigments, the pigments behave differently than the parent material and are capable of absorbing energy in other frequencies.

Concrete and masonary products make a good thermal mass. They have a fairly high capacity for heat storage and moderate conductances which allows the heat energy to be stored within the material. They also have high emissivities which means they absorb the heat energy upon them instead of reflecting it off. They won't transmit the solar energy right through them like a transparent or even a translucent material.

For direct solar gain, water is an excellent thermal mass store. It is capable of storing a lot of energy per unit volume, whether that is measured in cubic feet or in gallons. One of the problems with water is getting the solar energy into it. Like glass, radiant solar energy will pass right through it to a certain depth. It is also not very good at absorbing solar energy in the native wavelengths. However, it will take a lot less water per unit volume to achieve the same thermal mass store as a brick or concrete like material.

The Earth As Thermal Mass, Thermo-Coupling

The Earth is a large thermal mass store. The soil near the surface is capable of storing a lot of energy. Deeper, things are more thermally stable. Dirt may not be an ideal thermal mass store but the amount of it is sufficient to cause delays in temperature behavior. The hottest months of the year are in July and August but the summer solstice is in June. Daytime temperatures tend to be highest several hours after solar noon.

At about 2.5 feet below the surface of the crust temperatures tend to be more constant. Temperatures at these depths may fluctuate seasonally but the span of temperatures is much smaller than at the surface. The lag time between the annual high and low for soil at these depths is on the order of 3 months. The amount of temperature fluctuation depends on soil type, its diffusivity, the depth below the surface, ground water conditions and climatic conditions.

Traditionally thermal mass has been materials such as water, rock, earth, brick, concrete and ceramic. The type of things you would find in a natural cave. Phase change materials (PCM) store energy while maintaining constant temperatures. These are materials such as Glauber's salt, and paraffin wax (there are many kinds of paraffin and not all behave this way). Depending on the PCM storage of 5 - 15 times more heat per volume can be stored than traditional materials. At least one gypsum manufacturer has a gypsum board that incorporates paraffin for this purpose.

For detailed thermal information on wood products, http://www.fpl.fs.fed.us/documnts/pdf1988/tenwo88a.pdf. To access this document Click Here

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