Electrical Analogy to Heat Transfer

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Objectives

• Represent Heat Transfer as a set of electrical circuits
• Solve an electrical circuit
• Represent an electrical circuit as a set of differential equations

If you can already complete the above tasks, continue to the challenge section at the bottom of the page

Thermal to Electrical Comparison

Thermal properties can be directly compared to electrical properties in a circuit. For the rest of this course we will be using the thermal equivalent.

• Temperature - Voltage

• Heat Transfer - Current

• Thermal Resistance - Electrical Resistance (Inverse of Conductance) $R = 1/H$

• Thermal Capacitance - Electrical Capacitance

Heat transfer representation

Calculation

The heat transfer through the above conductance can be calculated as

$Q = \frac{T_{in}-T_{out}}{R_w}$

Q is the rate of heat transfer from T1 to T2

Drawing a Circuit

Let us assume that we have a house with the following properties

• Windows with a resistance of 0.2K/W

• Walls with an overall resistance of 0.05 K/W

  The equivalent circuit representation looks like this. The circuits are in parallel because the windows are physically in parallel with each other

If we were to add some additional insulation to the wall this would be represented with an additional resistor in series

Let's add some additional insulation to the wall which has a total resistance of 0.03 K/W

The resistor is added in series as the heat energy first passes through the wall, and then the added insulation.

Solving the Circuit

Resistances in series are added as so

$R_{wall.total} = R_{wall} + R_{extra} \\ R_{wall.total} = 0.05 + 0.03 = 0.08 K/W$

Resistances in parallel are added like so:

$\frac{1}{R_{total}} =\frac{1}{R_{wall.total}} + \frac{1}{R_{window}} \\ \frac{1}{R_{total}} =\frac{1}{0.08} + \frac{1}{0.2} \\ R_{total} = 0.057 K/W$

Note: All series resistors must be added before parallel resistors can be added

Adding Capacitance

Many thermodynamic systems have a thermal capacitance. In a building, this is caused by the mass of the building holding in heat.

A capcitance is connected to one of the temperature nodes. This thermal node will then have some delay. Let's assume that the building in question has a thermal capacitance of 2000Wh/K

We can simplify the resistances as we did before to get

This is known as a 1R1C building circuit which we will use for our first computation model. Although it is simple, it has proven to be accurate enough for simple tasks See Paper Here

Solving

The equation for the heat disipating from the thermal mass is $Q_C = C_m \frac{dT_{in}}{dt}$

Remembering from earlier for the resistor

$Q_R = \frac{T_{in}-T_{out}}{R_w}$

If we think about the conservation of energy. All the heat leaving the thermal mass and entering the thermal node T_in must also leave the thermal node T_in through the resistor. This is also known as Kirchoffs Current Law

so $Q_C = Q_R$

solving we get the differential equation

$\frac{dT_{in}}{dt} = \frac{T_{in}-T_{out}}{C_m R_w}$

This video explains this for electrical circuits

In the next chapter we will use numerical methods to solve this

Test

1) Draw the electrical circuit analogy for a room with two windows and a wall. The room has an internal thermal capacitance.

2) Derive the differential equation for this circuit