Thermal Systems

• User to model heat transfer
  • For example in a house
  • Or in electronic components
• Determine efficiency of elements
• Determine thermal operating ranges for components

Variables

• Rate of heat flow $q(t)$ in watts ($J{s}^{-1}$)

• Temperature, $\theta (t)$ in Kelvins (K)

• Analogous to current and voltage in electrical systems

Elements

Thermal Capacitor

• Stores heat energy in a reversible way

Elemental equation: $$q_c(t)=C\frac{d\theta }{dt}$$

Where $q_c(t)$ is the net heat flowing in, ie $q_{in}(t)-q_{out}(t)$.

Thermal Resistor

• Dissapates heat
  • Non-reversible

Any object that restricts heat flow when heat flows from on medium to another can be modelled as a resistor. Elemental equation: $$q(t)=\frac{1}{R}{\theta }_{12}(t)$$

Where $q_t$ is the flow of heat from the temperature ${\theta }_1$ on one side of the resistor to the temperature ${\theta }_2$ on the other.

Interconnection Laws

Compatibility Law:

• Temperatures are identical where elements touch,
• ${\theta }_1={\theta }_2=...={\theta }_n$

Equilibrium Law:

• Elemental heat flow rates sum to zero at connection points
• $q_1+q_2+...+q_n=0$

Examples

Develop a thermal model for someone doing winter sports. Assume:

• Ambient temperature ${\theta }_a$
• Body temperature $\theta$
• Thermal resistance between body and ambient (the person is wearing a coat) $R$
• Heat generated by body $q_{in}$

The rate of heat flow out is the difference in ambient and body temperature accross the resistor: $$q_{out}=\frac{\theta -{\theta }_a}{R}$$

In the thermal capacitor, the net input heat is proportional to the rate of change of temperature: $$q_{in}-q_{out}=C\frac{d\theta }{dt}$$

Combining the two equations gives: $$RC\frac{d}{dt}\theta +\theta =R{q}_{in}+{\theta }_a$$