Chapter 3

Thermal-electrical analogy: thermal network

3.1 Expressions for resistances

Recall from circuit theory that resistance 𝑅"#"$ across an element is defined as the ratio of electric potential difference Δ𝑉 across that element, to electric current I traveling through that element, according to Ohm’s law,

𝑅"#"$ =𝑉𝐼 (3.1)

Within the context of heat transfer, the respective analogues of electric potential and current are temperature difference Δ𝑇 and heat rate q, respectively. Thus we can establish “thermal circuits” if we similarly establish thermal resistances R according to

𝑅 =Δ𝑇  𝑞 (3.2)

Figure 3.1: System geometries: planar wall (left), cylindrical wall (right)

Medium:    λ, cross-­‐sectional  area  A  

 Distance  x  

Temperature  T

 

 

L  

 Distance  r  

Temperature  T

   

 

 

Medium:    λ, length  L    

(into  the  page)    

 

 

Planar wall conductive resistance: Referring to Figure 3.1 left, we see that thermal resistance may be obtained according to

𝑅,  $-./ =T1,3 − T1,5  

𝑞,=

𝐿𝜆𝐴 (3.3)

The resistance increases with length L (it is harder for heat to flow), decreases with area A (there is more area for the heat to flow through) and decreases with conductivity 𝜆. Materials like styrofoam have high resistance to heat flow (they make good thermal insulators) while metals tend to have high low resistance to heat flow (they make poor insulators, but transmit heat well).

Cylindrical wall conductive resistance: Referring to Figure 3.1 right, we have conductive resistance through a cylindrical wall according to

𝑅9  $-./ =T1,3 − T1,5  

𝑞9=ln 𝑟5 𝑟32𝜋𝜆𝐿 (3.4)

Convective resistance: The form of Newton’s law of cooling lends itself to a direct form of convective resistance, valid for either geometry.

𝑅,  $-.? =T1 − T@  𝑞,

=𝐿ℎ𝐴 (3.5)

3.2 Series and parallel thermal networks

With expressions for calculating thermal resistances in hand, we move on to the important task of choosing appropriate models for thermal networks. The utility of thermal resistances exists in the ease with which otherwise complicated thermal systems are modeled. This section considers series and parallel thermal networks, drawing analogies to circuit theory’s rules for equivalent resistance. For simplicity, only Cartesian geometries are considered.

Series networks: Recall from circuit theory that resistors in series produce an equivalent resistance between input-output terminals that is the sum of individual resistances, owing to the fact that each individual resistor has the same current flowing through it.

𝑅1"9B"1  "C = 𝑅BB

(3.6)

Likewise, systems in which multiple elements are intercepted by a single heat flowline are modeled serially.

Figure 3.2: Layered planar wall.

Figure 3.2 illustrates an example series model and circuit schematic for a double-exposed, layered window (note the presence of both conduction and convection transfer modes). In this example, the equivalent thermal resistance would be the sum of each resistance shown, and would relate overall temperature difference across the network according to

𝑅"C =T@,3 − T@,D  

𝑞,=1𝐴

1ℎ3+𝐿G𝜆G+𝐿H𝜆H+𝐿I𝜆I+1ℎ5

(3.7)

Parallel networks: Recall that electrical resistors in parallel produce equivalent resistance

𝑅JK9K##"#  "CL3 = 𝑅BL3B

(3.8)

Such a model exists when multiple current paths exist between two nodes. A similar situation occurs in thermal networks when two paths (through different media) exist between two points of the same temperature. An example system is shown in Figure 3.3, with two equivalent schematics.

Figure 3.3: Parallel conduction network.

The equivalent resistance between the left and right faces of material F (or G) is thus

𝑅"CL3 =𝐿M

𝜆M 𝐴 2

L3

+𝐿N

𝜆N 𝐴 2

L3

(3.9)

Note that this equivalent resistance will be less than either individual component. Note that this circuit model is valid only when heat flow is assumed approximately one-dimensional (if significant heat flow occurred vertically between materials F and G, the resistors-in-parallel model would be invalidated).

3.3 Elements of thermal network

There are two fundamental physical elements that make up thermal networks, thermal resistances and thermal capacitance. There are also three sources of heat, a power source, a temperature source, and fluid flow.

A note on temperature

In practice temperature when we discuss temperature we will use degrees Celsius (°C), while SI unit for temperature is to use Kelvins (0°K = -273.15°C). However, we will generally be interested in temperature differences, not absolute temperatures (much as electrical circuits deal with voltage differences). Therefore, we will generally take a reference temperature (which we will label T1), and measure all temperatures relative to this reference. We will also assume that the reference temperature is constant. Thus, if T1 is =25°C, and the temperature of interest is Ti=32°C, we will say that Ti=7° above reference. Note: this is consistent with electrical systems in which we assign one voltage to be ground (and assume that it is constant) and assign it the value of zero volts. We then measure all voltages relative to ground.

Thermal resistance

Consider the situation in which there is a wall, one side of which is at a temperature T1, with the other side at temperature T2. The wall has a thermal resistance of R12. This is depicted in Figure 3.4 in three different, but equivalent, diagrams.

Figure 3.4: Thermal-electrical analogy of conduction heat transfer.

Thermal capacitance

In addition to thermal resistance, objects can also have thermal capacitance (also called thermal mass). The thermal capacitance of an object is a measure of how much heat it can store. If an object has thermal capacitance its temperature will rise as heat flows into the object, and the temperature will lower as heat flows out. To understand this, envision a rock in the sun. During the day heat goes in to the rock from the sunlight, and the temperature of the rock increases as energy is stored in the rock as an increased temperature. At night energy is released, and the rock cools down. We represent a thermal capacitance in isolation in diagrams (and equations) as shown in Figure 3.5 (in the drawing at the left the coil represents a power source and the stippled object is the thermal capacitance). In the thermal analogy, one end of the capacitor is always connected to the constant ambient temperature.

Figure 3.5: Thermal-electrical analogy of thermal capacitance.

The electrical model will always have one side of the capacitance connected to ground, or

reference. Also, we could write the equation as 𝐶 / PQLPR/S

= 𝑞 but since T1 is constant, it can be

removed from the derivative. The thermal capacitance of an object is determined by its mass and specific heat

𝐶 = 𝑚𝑐J (3.10) where C is the thermal capacitance, m is the mass in kilograms, and cp is the specific heat in J/(kg-°K). It is always assumed that the capacitor is at a single uniform temperature, though this is obviously a simplification in many cases.

Example

The specific heat of water is 4.2 kJ/kg-°C.

-   What is the thermal capacitance of 5 liters of water?

-   If the water starts at θc=25°C, how hot will it be if it is heated with a 1 kW heater for 1 minute?

Solution.

a) C = m·cp, and 5 liters of water has a mass of 5 kg. So C = 5·(4.2·103) = 21 kJ/°C.

b) First, calculate the rate of increase of temperature

𝑑𝑇$𝑑𝑡 =

𝑞𝐶 =

1𝑘𝑊

21𝑘𝐽 °𝐶=

1𝑘𝐽 𝑠21 𝑘𝐽 °𝐶

=121°𝐶 𝑠

then find the total increase:

∆𝑇$ =𝑑𝑇$𝑑𝑡 ∆𝑇 =

121°𝐶 𝑠 ∙ 60𝑠 = 2.9°𝐶

so the final temperature is

25 + 2.9 = 27.9°𝐶

Power source (or heat source)

A common part of a thermal model is a controlled power source that generates a predetermined amount of power, or heat, in a system. This power can either be constant or a function of time. In the electrical analogy, the power source is represented by a current source. An example of a power source is the quantity q in the diagrams for the thermal capacitance, above. In practice a power source is often an electrical heating element comprised of a coil of wire that is heated by a current flowing through it. Therefore, we use a diagram of a coil of wire to represent the power source. An ideal power source generates power that is independent of temperature.

Temperature source

Another common source used in thermal systems is a controlled temperature source that maintains a constant temperature. An ideal temperature source maintains a given temperature independent of the amount of power required. A refrigerator is an example of such a source. Another such source is the ambient surroundings. We will assume that the temperature of the ambient surroundings is constant regardless of the heat flow in or out (we will also take ambient temperature to be our reference temperature).

Mass transfer (fluid flow)

If fluid with specific heat cp J/(kg-°K) flows into a system with a flow rate of G kg/sec and a temperature of Tin °C above reference, and flows out at a temperature of Tout °C below reference then the rate of heat flow into the system is given by

𝑞B. = 𝐺𝑘𝑔𝑠𝑒𝑐 ∙ 𝑐J

𝐽𝑘𝑔  𝐾 ∙ 𝑇B. − 𝑇-iS °𝐶 = 𝐺𝑐J 𝑇B. − 𝑇-iS 𝑊

(3.11)

We can cancel the K and °C since a temperature difference (Tin-Tout) is the same in Kelvin or Celsius.

If you carefully observe this equation, it makes sense intuitively. Heat into a system goes up with mass flow rate into the system (increased mass flow, yields increased heat flow). Heat into a system also goes up with the specific heat of the mass (higher specific heat indicates increased capacity to store heat). Finally, heat into a system increases with an increased inflow temperature, or a decreased outflow temperature (if the temperature difference between inflow and outflow increases, more heat is being taken from the fluid). Note: the mass flow rate at the input and output must be equal or the mass (and thermal capacitance) of the system would be changing. This is not allowed for the systems being studied (time-invariant systems).

The energy balance

To develop a mathematical model of a thermal system we use the concept of an energy balance. The energy balance equation simply states that at any given location, or node, in a system, the

heat into that node is equal to the heat out of the node plus any heat that is stored (heat is stored as increased temperature in thermal capacitances).

Heat in = Heat out + Heat stored

To better understand how this works in practice it is useful to consider several examples.

3.4 Examples involving only thermal resistance and capacitance

Example: Two thermal resistances in series

Consider a situation in which we have an internal temperature, Ti, and an ambient temperature, Ta with two resistances between them. We will call the resistance between the internal temperature and the separation temperature Ris, and the temperature between separation and ambient Rsa.

Solve for the temperature Ts if Ti = 37°C, Ta = 9°C, Ris = 0.75 °C/W and Rsa= 2.25 °C/W.

Solution.

Temperatures are drawn as voltage sources. Ambient temperature is taken to be zero (i.e., a ground temperature), all other temperatures are measured with respect to this temperature).

c) There is only one unknown temperature (at Ts), so we need only one energy balance (and, since there is no capacitance, we don't need the heat stored term).

𝐻𝑒𝑎𝑡  𝑖𝑛 = 𝐻𝑒𝑎𝑡  𝑜𝑢𝑡 + 𝐻𝑒𝑎𝑡  𝑠𝑡𝑜𝑟𝑒𝑑

𝑇B − 𝑇1𝑅B1

=𝑇1 − 𝑇K𝑅1K

=𝑇1𝑅1K

Note: the first equation included Ta, but the second does not, since Ta is our reference temperature and is taken to be zero.

Solving for Ts

𝑇B − 𝑇1𝑅B1

=𝑇1𝑅1K

𝑇B − 𝑇1 𝑅1K = 𝑇1𝑅B1

𝑇1 = 𝑇B𝑅1K

𝑅B1 + 𝑅1K

Note: you may recognize this result as the voltage divider equation from electrical circuits.

We can now solve numerically (we use 28°C for the internal temperature since it is 28°C above ambient (37°-9°=28°)

𝑇1 = 𝑇B𝑅1K

𝑅B1 + 𝑅1K= 28°

2.250.75 + 2.25 = 28 0.75 = 21°

This says that Ts is 21°C above ambient. Since the ambient temperature is 9°C, the actual temperature is 30°C.

Example: Three thermal resistances in series

𝑇B 𝑥 = 0

𝑇3 𝑥 = 𝑥3

𝑇5 𝑥 = 𝑥5

𝑇K 𝑥 = 𝑥r

𝑥3 − 𝑥s = 𝑙B3            , 𝑥5 − 𝑥3 = 𝑙35            , 𝑥r − 𝑥5 = 𝑙5K

𝑅B3 =𝑙B3𝜆B3𝐴

           ,            𝑅35 =𝑙35𝜆35𝐴

           ,            𝑅5K =𝑙5K𝜆5K𝐴

𝑞 = −𝑇3 − 𝑇B𝑅B3

=  −𝑇5 − 𝑇3𝑅35

= −𝑇K − 𝑇5𝑅5K

         𝑇3 − 𝑇B𝑅B3

   =      𝑇5 − 𝑇3𝑅35

   =    𝑇K − 𝑇5𝑅5K

1st 2nd 3th

𝑞 = −𝜆"C𝑇K − 𝑇B

𝑙B3 + 𝑙35 + 𝑙5K

1uv =  2wx  2wx =  3vz    

𝑇3𝑅B3

−𝑇B𝑅B3

−𝑇5𝑅35

+𝑇3𝑅35

= 0

𝑇5𝑅35

−𝑇3𝑅35

−𝑇K𝑅5K

+𝑇5𝑅5K

= 0

1𝑅B3

+1𝑅35

𝑇3 −1𝑅35

𝑇5 =1𝑅B3

𝑇B

−1𝑅35

𝑇3 +1𝑅35

+1𝑅5K

𝑇5 =1𝑅5K

𝑇K

1𝑅B3

+1𝑅35

−1𝑅35

−1𝑅35

1𝑅35

+1𝑅5K

𝑇3𝑇5

=

1𝑅B3

𝑇B1𝑅5K

𝑇K

𝐴 𝑇 𝐵

𝑇 = 𝐴L3  𝐵

If we use the reference temperature 𝑇K = 0 , the calculus is semplified:

1𝑅B3

+1𝑅35

−1𝑅35

−1𝑅35

1𝑅35

+1𝑅5K

𝑇3𝑇5

=1𝑅B3

𝑇B0

Example: Three thermal resistances in parallel

𝑞BK = 𝑞3 + 𝑞5 + 𝑞r

−𝑇B − 𝑇K𝑅"C

= −𝑇B − 𝑇K𝑅3

−𝑇B − 𝑇K𝑅5

−𝑇B − 𝑇K𝑅r

or

−𝑇B𝑅"C

= −𝑇B𝑅3−𝑇B𝑅5−𝑇B𝑅r

1𝑅"C

=1𝑅3+1𝑅5+1𝑅r

Example: Heating of an insulated enclosure

Consider a single insulated enclosure. The resistance of the walls between the enclosure and the ambient is Rra, and the thermal capacitance of the enclosure is Cr, the heat into the enclosure is q, the temperature of the enclosure is Tr, and the external temperature is a constant, Ta (temperature prescribed on the external surface of the insulation).

Solution.

To draw the electrical network, we need a circuit with a node for the ambient temperature, and a node for the temperature of the enclosure. Heat (a current source) goes into the enclosure. Energy is stored (as an increased temperature) in the thermal capacitance, and heat flows from the enclosure to ambient through the resistor.

We only need to develop a single energy balance equation, and that is for the temperature of the thermal capacitance (since there is only one unknown temperature). The heat into the room is q, heat leaves the room through a resistor and energy is stored (as increased temperature) in the capacitor.

Heat in = Heat out + Heat stored

𝑞 =𝑇9 − 𝑇K𝑅9K

+ 𝐶𝑑𝑇9𝑑𝑡

by convention we take the ambient temperature to be zero, so we end up with a first order differential equation for this system.

𝑞 =𝑇9𝑅9K

+ 𝐶𝑑𝑇9𝑑𝑡

Example: Heating a single insulated enclosure, with variable external temperature.

Consider the enclosure from the previous example. Repeat if the temperature outside is no longer constant but varies. Call the external temperature Te(t) (this will be the temperature relative to the ambient temperature). We will also change the name of the resistance of the walls to Rre to denote the fact that the external temperature is no longer the ambient temperature.

Solution.

The solution is much like that for the previous example. Exceptions are noted below.

To draw the electrical system, we need a circuit with a node for the external surface temperature and a node for the temperature of the enclosure. Though perhaps not obvious at first we still need a node for the ambient temperature since all of our temperatures are measured relative to this, and our capacitors must always have one node connected to this reference temperature. Heat flows from the enclosure to the external temperature through the resistor.

We still only need to develop a single energy balance equation, and that is for the temperature of the thermal capacitance (since there is only one unknown temperature). The heat into the enclosure is q, heat leaves the enclosure through a resistor and energy is stored (as increased temperature) in the capacitor.

Heat in = Heat out + Heat stored

𝑞 =𝑇9 − 𝑇"𝑅9"

+ 𝐶𝑑𝑇9𝑑𝑡

(the ambient temperature is taken to be zero in this equation). In this case we end up with a system with two inputs (q and Te).

Example: Heating a double enclosure

Consider a system that consists of two adjacent insulated enclosures, labelled 1 and 2. The resistance of the walls enclosure 1 and ambient is R1a, between enclosure 2 and ambient is R2a and between enclosure 1 and enclosure 2 is R12. The capacitance of enclosures 1 and 2 are C1 and C2, with temperatures T1 and T2, respectively. A heater is in enclosure 1 only, generates a heat qin. The external temperature is a constant, Ta.

The electrical equivalent network is:

In this case there are two unknown temperatures, T1 and T2, so we need two energy balance equations. In both cases we will take Ta to be zero, so it will not arise in the equations.

The two first order energy balance equations (for enclosure 1 and enclosure 2) could be combined into a single second order differential equation and solved.

Example: Cooling a block of metal in a tank with fluid flow.

Consider a block of metal (capacitance=Cm, temperature=Tm). It is placed in a well mixed tank (at termperature Tt, with capacitance Ct). Fluid flows into the tank at temperature Tin with mass flow rate Gin, and specific heat cp. The fluid flows out at the same rate. There is a thermal resistance to between the metal block and the fluid of the tank, Rmt, and between the tank and the ambient Rta. Write an energy balance for this system.

Note: the resistance between the tank and the metal block, Rmt, is not explicitly shown.

Solution.

Since there are two unknown temperatures, we need two energy balance equations.

To model this system with an electrical analogy, we can represent the fluid flow as a voltage source at Tin, with a resistance equal to 1/(Gin·cp). If you sum currents at the nodes Tt and Tm you can show that this circuit is equivalent to the thermal system above.

Example

Consider a thermal system as in next figure.

Inputs:

𝑞|3      ,      𝑞|3      , 𝑇H      

In the thermal system, the block #1 is heated with the heat source 𝑞|3. 𝑇K is the ambient temperature. While the block #1 is heating depending on their thermal capacitance, at the same time there is a heat transfer from the block #1 to the next block #B.

In the equivalent circuit, the current supply 𝑞|3can supply the line with the resistance 𝑅3K, capacitor 𝐶3 and the line with the resistor 𝑅3H.

𝑅3K corresponds to the thermal resistance between the block #1 and ambient. 𝑅3Hcorresponds to the thermal resistance between the block #1 and block B. 𝐶3 in the circuit corresponds to the thermal capacitance of the block #1.

The block B has a heat source which the temperature is controlled. The value of the temperature in the block B is 𝑇H. There is a heat transfer from the block B to the block #2 or vice versa.

So, a temperature source 𝑇H is placed into the circuit in the thermal system. This current source 𝑞3H can supply the lines with the resistance 𝑅HK and the resistance 𝑅H5

The block #2 is heated with the heat source 𝑞|5. There is a heat transfer between the block #2 and the ambient.

In the equivalent circuit, the current supply 𝑞|5 can supply the line with the resistance 𝑅5G, capacitor 𝐶5 and the line with the resistor 𝑅H5. 𝐶5 in the circuit corresponds to the thermal capacitance of the block #2. The current sources 𝑞|3, 𝑞|5 and the controlled temperature 𝑇5 are the inputs of the thermal system.