# Mathematical Models of Thermal Systems

## Introduction

While the previous page (System Elements) introduced the fundamental elements of thermal systems, as well as their mathematical models, no systems were discussed.  This page discusses how the system elements can be included in larger systems, and how a system model can be developed.   The actual solution of such models is discussed elsewhere.

## 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.

## Examples Involving only Thermal Resistance and Capacitance

##### Example: Two thermal resistances in series

Consider a situation in which we have an internal temperature, θi, and an ambient temperature, θa with two resistances between them.  An example of such a situation is your body.  There is a (nearly) constant internal temperature, there is a thermal resistance between your core and your skin (at θs), and there is a thermal resistance between the skin and ambient.  We will call the resistance between the internal temperature and the skin temperature Ris, and the temperature between skin and ambient Rsa.
a) Draw a thermal model of the system showing all relevant quantities.
b) Draw an electrical equivalent
c) Develop a mathematical model (i.e., an energy balance).
d) Solve for the temperature of the skin if  θi, =37°C, θa =9°C, Ris=0.75°/W; for a patch of skin and Rsa= 2.25°/W for that same patch.

Solution:

a) In this case there are no thermal capacitances or heat sources, just two know temperatures ( θi, and θa), one unknown temperature (θs), and two resistances ( Risand Rsa.) b) 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 θs), so we need only one energy balance (and, since there is no capacitance, we don't need the heat stored term). Note: the first equation included θa, but the second does not, since θa is our reference temperature and is taken to be zero.

d) Solving for θs 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°) This says that θs is 21°C above ambient.  Since the ambient temperature is 9°C, the actual skin temperature is 30°C.

Note: If Rsa is lowered, for example by the wind blowing, the skin gets cooler, and it feels like it is colder.  This is the mechanism responsible for the "wind chill" effect.

##### Example: Heating a Building with One Room

Consider a building with a single room.   The resistance of the walls between the room and the ambient is Rra, and the thermal capacitance of the room is Cr, the heat into the room is qi, the temperature of the room is θr, and the external temperature is a constant, θa.

a) Draw a thermal model of the system showing all relevant quantities.
b) Draw an electrical equivalent
c) Develop a mathematical model (i.e., a differential equation).

Solution:

a)  We draw a thermal capacitance to represent the room (and note its temperarature).  We also draw a resistance between the capacitance and ambient. b) To draw the electrical system we need a circuit with a node for the ambient temperature, and a node for the temperature of the room.  Heat (a current source) goes into the room.  Energy is stored (as an increased temperature) in the thermal capacitance, and heat flows from the room to ambient through the resistor. c)  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 qi, heat leaves the room through a resistor and energy is stored (as increased temperature) in the capacitor. by convention we take the ambient temperature to be zero, so we end up with a first order differential equation for this system. ##### Example: Heating a Building with One Room, but with Variable External Temperature.

Consider the room from the previous example.  Repeat parts a, b, and c if the temperature outside is no longer constant but varies.  Call the external temperature θe(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.

a)  The image is as before with the external temperature replaced by θe(t). b) To draw the electrical system we need a circuit with a node for the external temperature and a node for the temperature of the room.  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 room to the external temperature through the resistor. c)  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 room is qi, heat leaves the room through a resistor and energy is stored (as increased temperature) in the capacitor. (the ambient temperature is taken to be zero in this equation).  In this case we end up with a system with two inputs (qi and θe).

##### Example: Heating a Building with Two Rooms

Consider a building that consists of two adjacent rooms, labeled 1 and 2.   The resistance of the walls room 1 and ambient is R1a, between room 2 and ambient is R2a and between room 1 and room 2 is R12. The capacitance of rooms 1 and 2 are C1 and C2, with temperatures θ1 and θ2,  respectively.  A heater in in room 1 generates a heat qin. The temperaturexternal temperature is a constant, θa.

a) Draw a thermal model of the system showing all relevant quantities. b) Draw an electrical equivalent c) Develop a mathematical model (i.e., a differential equation).

In this case there are two unknown temperatures, θ1and θ2, so we need two energy balance equations.  In both cases we will take θa to be zero, so it will not arise in the equations.

 Room 1: Heat in = Heat out + Heat Stored Room 2: Heat in = Heat out + Heat Stored  In this case there are two parts to the "Heat Out"term, the heat flowing through R1a and the heatthrough R12. In this case we take heat flow through R12 to (from 1 to 2) to be an input. We could also take this energy balance to have no heat in, and write the heat flow from2 to 1 as a second "Heat out" term.  (note thechange of subscripts in the subtracted terms) The two first order energy balance equations (for room 1 and room 2) could be combined into a single second order differential equation and solved.  Details about developing the second order equation are here.

## Examples Involving Fluid Flow

So far we have not considered fluid flow in any of the examples; let us do so now.

##### Example: Cooling a Block of Metal in a Tank with Fluid Flow.

Consider a block of metal (capacitance=Cm, temperature=θm).  It is placed in a well mixed tank (at termperature θt, with capacitance Ct).  Fluid flows into the tank at temperature θin 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.

 Metal Block: Heat in = Heat out + Heat Stored Tank: Heat in = Heat out + Heat Stored  In this case there is not heat in, and heat outis to the tank through Rmt. In this case we have heat in from the fluid flowand from the metal block.We have heat out to ambient through Rta.
##### Aside: Modeling a Fluid Flow with and Electrical Analog

To model this system with an electrical analog, we can represent the fluid flow as a voltage source at θin, with a resistance equal to 1/(Gin·cp).  If you sum currents at the nodes θt and θm you can show that this circuit is equivalent to the thermal system above. ## Solving the Model

Thus far we have only developed the differential equations that represent a system.  To solve the system, the model must be put into a more useful mathematical representation such as transfer function or state space.   Details about developing the mathematical representation are here.

References