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.
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.
Consider a situation in which we have an internal temperature, θ_{i},
and an ambient temperature, θ_{a} 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 θ_{s}), and there is a thermal
resistance between the skin and ambient. We will call the resistance
between the internal temperature and the skin temperature R_{is}, and
the temperature between skin and ambient R_{sa}.
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 θ_{i},
=37°C, θ_{a} =9°C, R_{is}=0.75°/W; for a
patch of skin and R_{sa}= 2.25°/W for that same patch.
Solution:
a) In this case there are no thermal capacitances or heat sources, just two know temperatures ( θ_{i}, and θ_{a}), one unknown temperature (θ_{s}), and two resistances ( R_{is}and R_{sa}.)
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 θ_{a}, but the second does not, since θ_{a} is our reference temperature and is taken to be zero.
d) Solving for θ_{s}
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 θ_{s} is 21°C above ambient. Since the ambient temperature is 9°C, the actual skin temperature is 30°C.
Note: If R_{sa} 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.
Consider a building with a single room. The resistance of the walls between the room and the ambient is R_{ra}, and the thermal capacitance of the room is C_{r}, the heat into the room is q_{i}, the temperature of the room is θ_{r}, and the external temperature is a constant, θ_{a}.
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 q_{i}, 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.
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 θ_{e}(t) (this will be the temperature relative to the ambient temperature). We will also change the name of the resistance of the walls to R_{re} 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 θ_{e}(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 q_{i}, 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 (q_{i} and θ_{e}).
Consider a building that consists of two adjacent rooms, labeled 1 and 2. The resistance of the walls room 1 and ambient is R_{1a}, between room 2 and ambient is R_{2a} and between room 1 and room 2 is R_{12}. The capacitance of rooms 1 and 2 are C_{1} and C_{2}, with temperatures θ_{1} and θ_{2}, respectively. A heater in in room 1 generates a heat q_{in}. The temperaturexternal temperature is a constant, θ_{a}.
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, θ_{1}and θ_{2}, so we need two energy balance equations. In both cases we will take θ_{a} 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 R_{1a }and the heat through R_{12}. 
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 from 2 to 1 as a second "Heat out" term. (note the change 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.
So far we have not considered fluid flow in any of the examples; let us do so now.
Consider a block of metal (capacitance=C_{m}, temperature=θ_{m}). It is placed in a well mixed tank (at termperature θ_{t}, with capacitance C_{t}). Fluid flows into the tank at temperature θ_{in} with mass flow rate G_{in}, and specific heat c_{p}. The fluid flows out at the same rate There is a thermal resistance to between the metal block and the fluid of the tank, R_{mt}, and between the tank and the ambient R_{ta}. Write an energy balance for this system.
Note: the resistance between the tank and the metal
block, R_{mt}, 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 out is to the tank through R_{mt}. 
In this case we have heat in from the fluid flow and from the metal block. We have heat out to ambient through R_{ta}. 
To model this system with an electrical analog, we can represent the fluid flow as a voltage source at θ_{in}, with a resistance equal to 1/(G_{in}·c_{p}). If you sum currents at the nodes θ_{t} and θ_{m} you can show that this circuit is equivalent to the thermal system above.
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.
© Copyright 2005 to 2015 Erik Cheever This page may be freely used for educational purposes.
Erik Cheever Department of Engineering Swarthmore College