Mathematical Models of Thermal Systems

Example: Two thermal resistances in series

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_isand 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.

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 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.

 

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 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). 

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 R_1a, between room 2 and ambient is R_2a 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 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, θ₁and θ₂, 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 R1a and the heat through 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 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.

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

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 Rmt.		In this case we have heat in from the fluid flow and from the metal block. We have heat out to ambient through Rta.