Analogy Between Thermal and Electrical Conduction
In the development of his law for electrical circuits, Georg Ohm performed experiments that modeled Fourier’s law of heat conduction. Consequently, an analogy between heat and electrical conduction can be observed. Symbolically Ohm’s law can be expressed as
where I is the current flowing through an element, ΔV is the voltage across the element, and Re is the electrical resistance across the element. Similarly Fourier’s law can be written as
where q is the rate of heat conduction, ΔT is the temperature difference between the surfaces of a slab, and Rt is the thermal resistance between the surfaces defined as
where k is the thermal conductivity and A is the effective area of the resistance. By using the convective heat transfer coefficient h, the convective thermal resistance from a surface to a fluid can also be calculated by the equation
Schematic Representation of Thermal-Electrical Analogy
Since heat is understood to transfer through lattice vibrations between adjacent atoms, and electricity is conducted by free valence electrons of atoms, it may seem intuitive that heat and electrical conduction are analogous. Although this is most often true, some non metallic materials like diamond exhibit dissimilar heat and electrical conductivities.
Uses of the Thermal-Electrical Analogy
As found in the referenced literature, the analogy is useful in the analysis of several steady heat transfer problems from property measurement to modeling. In modeling, a complicated heat transfer analysis can be made much simpler by creating an “electric circuit” like model of the problem. This can be seen in the illustrative examples below. In the first example, a simple comparison between single and triple paned windows is modeled. In the second example an entire house is modeled using a thermal network. Also evident from the illustrative examples is the fact that one should take additional care in determining if system components are acting in series or parallel.
Limits of the Thermal-Electrical Analogy
Although generally extremely useful, the analogy is not without its own limitations. These include the non-linearities that occur between voltage and current at extremely high and low values. The literature survey would also lead to special cases where the analogy doesn’t quite fit the actual physics of a problem. Recall the constitutive models of heat transfer are based on the continuum concept. These special cases are at extremely large or small values of current and voltage as well as heat transfer problems at the nano-scale level. Despite this, modifications can be made to still find the relationship helpful in the analysis.
Consider the steady heat loss out of a residential glass window. The first case is a single pane window, and the second is a triple paned window.
Area of windows: 1 m2
Thickness of single paned window L: 1 cm
Thickness of panes in triple paned window l: 0.3 cm
Thickness of air gap between panes: 0.5 cm
Temperature in house T∞ in : 20°C
Temperature outside T∞ out : -10°C
Constant thermal conductivity of glass k1: 0.78 W/(m*°C)
Constant thermal conductivity of air k2: 0.026 W/(m*°C)
Convective heat transfer coefficient outside hout: 40 W/(m*°C)
Convective heat transfer coefficient inside hin: 10 W/(m*°C)
Radiative heat transfer resistance Rrad: 0.03 °C/W
Instead of doing a heat conduction analysis of each layer starting from first principles, a simplified thermal resistance network can be used.
Calculate thermal resistances:
Rrad = given
Rconv1=1/(hout×A) = 0.025 °C/W
Rconv2 = 1/(hin×A) = 0.1 °C/W
Rglass=L/(k1×A) = 0.0128 °C/W
Calculate thermal resistances:
Rrad = given = 0.03 °C/W
Rconv1=1/(hout×A) = 0.025 °C/W
Rconv2 = 1/(hin×A) = 0.1 °C/W
Rglass = l/(k1×A) = 0.00385 °C/W
Rair=(L/2)/(k2×A) = 0.1923 °C/W
Comparison: The rate of heat loss in Case 2 with triple paned windows was much less than Case 1 with a single pane. Note the high resistance of the air gaps. Temperatures on the inside and outside surfaces as well as in the gaps can now be calculated as well.
Energy prices are increasing. This is likely due to depleting resources, environmental concerns about its production, and diplomatic conflicts in traditional energy rich nations. Consequently, the need for energy conservation is becoming an important economic, political, and environmental issue. A major consumer of energy is heating systems in residential homes. Heating and cooling systems frequently use over 60% of the energy consumed in residential buildings and even more in office and educational buildings . Therefore, to conserve energy, the heating system in residential buildings is a logical place to start. In addition to making heating furnaces more efficient, it is essential to retain the heat supplied by the furnace. This is done by insulating a home during its construction. Insulating materials in residential home construction are often rated using R-values. This is similar to the thermal resistance value discussed in the preceding sections, except that thermal resistances are the R-values divided by the effective area. To further describe the thermal electrical analogy, a simple thermal resistance example is discussed using a typical residential home in the Midwest United States.
Building used in Example:
To illustrate the
thermal resistance concept, a house constructed in 1985 in the southwest
Calculating the Overall Thermal Resistance
As opposed to performing a one-dimensional heat transfer analysis on each contributing layer of construction, the thermal-electrical analogy was used and an overall thermal resistance network or circuit was used. This was done by taking into account all the building materials between the inside and outside environment acting in either series or parallel. A schematic illustration of this circuit can be seen in figure 3.
Figure 3. Thermal Resistance Network of House
Next, the individual thermal resistance of each contributing material was determined. This was done by using the dimensional specifications of the building in combination with either published thermal conductivities or R values of the various building materials in the house. These values are tabulated in Table 2. Using the thermal circuit model, a total thermal resistance for the house can be calculated by the equation
By substitution the equation for the total thermal resistance Rtotal of the house can now be written as
and by substituting the values in table 1, the total thermal resistance is
From the initial temperature a heat loss rate could be calculated using the total thermal resistance of the house calculated above using equation (5) and listed in table 1 resulting in 591.41 W. Using $1.87 per gallon as published on the DOE website , this converts to about 4.1 cents per hour.
Table 1. Calculated and Measured Parameters
Table 2. Component Resistance Values
Confirmation of Calculated Thermal Resistance
To check the calculated thermal resistance, a simple experiment was conducted by taking temperatures with the furnace turned off for four hours during the middle of the day. During this period there was negligible wind and temperature variation outside of the house. Temperatures were taken before shut off and after 4 hours. One way heat loss can be measured is by using the equation
where ρair is the mass density of the indoor air, Vair is the volume of the indoor air, Cp air is the specific heat of the indoor air, Tinitial is the initial indoor temperature and Tfinal is the final indoor temperature after the 4 hour shut-off. The values of these parameters are given in table 2. However a more accurate manner to measure the heat loss is a psychrometric chart. This takes into account the relative humidity of the air inside the home. Assuming a constant relative humidity of 50%, the heat loss during the four hour shut-off is also in table 1.
From table 1 we can see a substantial discrepancy in the rate of heat loss calculated using the thermal resistance and the rate of heat loss using the psycrometric chart. Before immediately abandoning the analysis of thermal resistances, several factors contributing to heat gain, other than the furnace, need to be considered. For one, light. The house is lit with incandescent light bulbs which give off considerable amounts of heat due to their inefficiencies. For instance, assume during the 4 hour shut-off period there were 5- 60 W incandescent light bulbs on and running at a typical 10% efficiency. That would contribute 270 W to the house. In addition to light, consider there was one adult man sitting in the house during the shut-off. The average adult man gives off body heat at a rate of 108 W when seated in a room . These factors alone when subtracted from the heat losses calculated using the thermal resistances result in a much smaller discrepancy of 7.1%.
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Special thanks to the Lindberg household for letting me freeze them for 2 days.