Temperature Simulation in Residential Building

By Use of Electrical Circuits

Steven Levine levine.steven@wustl.edu 914-263-8281 B.S. in Systems Science and Engineering, 2015

Abstract The most optimal, in terms of accuracy and computational complexity, method to model

and predict temperatures in a building is still being explored. My report will explore the

potential of electrical circuits to achieve this objective. I will first develop the general

method for converting a thermal system model and its parameters into an electrical circuit

and respective parameters. I will then use this methodology to simulate the temperature

profile of a two-room structure. I will use the simulation results to determine the best

method to integrate windows and doors into the model. These results will then be applied

to the development of an electrical circuit to represent the thermal system of a real

apartment. The simulation of the temperature profile of this apartment will demonstrate

the ability of this type of model to successfully fulfill its objective and serve as a tool to

analyze the physical heat flows around a building.

I. Introduction Reducing energy usage through increased energy efficiency is a very important

objective. One very large source of demand of energy comes from buildings, which

account for 34% of total energy usage within the United States [1]. Furthermore, HVAC,

i.e., heating, ventilation, and air conditioning, systems account for half of energy usage

within buildings [1]. Therefore, increasing HVAC systems’ efficiency, largely by

minimizing their energy usage, is necessary to meet the goal of increasing overall energy

efficiency. In order to accomplish this goal, the true temperature and climate profile of a

room must be known. However, the current room climate model that commercial

thermostats are based on is very simple. Another type of model, learned about in

conversations with my advisor and research, that holds promise for simulating room

climate profiles is an equivalent electrical circuit model [1], [2], [3]. Electrical circuits are

generally much more ubiquitous and easier for people to conceptualize and model than

thermal systems. More importantly, this model has the potential to accurately predict

room temperature and have low enough computational cost to be used commercially.

This is in contrast to room temperature models based on partial differential equations.

While these can very accurately predict temperature, their computational complexity is

too high to be used commercially.

II. Problem Statement and Objectives In my project, I will expand on the research done using equivalent electrical

circuit models and analyze their ability to be used in room climate control. I will first

explain how a thermal system in the inside of a building, with heat inputs from the

outside temperature and a HVAC system, can be converted into an equivalent circuit.

Then, using these principles, I will solve for the conversion of the thermal system of an

example two-room system into an electrical circuit. In solving for this conversion, my

goal is to design a circuit that can respond to varying changes in room and outside

conditions in a way that makes physical sense. The third part of the problem will be

expanding these principles with a real building. I will model a one-floor apartment in an

apartment building at 749 Westgate Avenue as an electrical circuit. When designing the

circuit, I will have the same objective as when designing the circuit of the two-room

system. Additionally, another goal is to design for a circuit that accounts for the

differences between rooms, in terms of size and other factors, when determining their

climate profiles.

The overall objective of my project is to demonstrate the ability that equivalent

electrical circuits have in being the model for an indoor thermal system. This ability will

be demonstrated by looking at the performance of the model in simulations, focusing on

two characteristics of the temperature performance, the time constant and steady-state

value. Additionally, I will demonstrate the ability to easily adjust parameters in this

model to account for possible real-life changes in room and outside conditions. This

flexibility would be very beneficial in applying the model to real-life situations.

III. General Conversion of Thermal into Electrical Model Converting different types of systems and models into electrical circuits is an idea

that has been used for a long time, due to the ubiquitous nature of electrical circuits, as

previously mentioned. Mechanical systems, due to their typically linear nature, are one

class where these transformations have been successfully used [4]. While many thermal

models have been transformed into electrical equivalents, there has been little research on

the relation between physical parameters of a building and their electrical parameter

equivalents. The procedure for performing the thermal to electrical transformation is

explained below.  

The basic components and respective parameters of a thermal system are the

following, with the symbols for the parameters in reference to the thermal system

diagram above:

• Heat source, quantified by heat source u1.

• Thermal resistor, quantified by resistance R1.

• Thermal capacitor, quantified by thermal capacitance C1 and temperature y.

The basic components and respective parameters of an electrical circuit are the

following, with the symbols for the parameters in reference to the circuit pictured above:

• Voltage or current source, quantified by source u.

• Electrical inductor, quantified by inductance L.

• Electrical resistor, quantified by electrical resistance R.

• Electrical capacitor, quantified by electrical capacitance C.

The basic components of a thermal system are converted to the basic components

of an electrical circuit as follows [3]:

• A heat source becomes a current source, with a heat rate quantity, measured in

Watts, equivalent to a current quantity, measured in Amperes.

Basic  Thermal  System   Basic  Electrical  Circuit  

L  

R  

C  

u  

• An outside temperature, measured in degrees Celsius, becomes a voltage source,

measured in Volts.

• A thermal resistor and thermal resistance quantity, measured in degrees Celsius

per Watt, becomes an electrical resistor and electrical resistance quantity,

measured in Ohms, respectively.

• A thermal capacitor and thermal capacitance quantity, measured in Joules per

degree Celsius, becomes an electrical capacitor and electrical capacitance

quantity, measured in Farads, respectively.

• There is no component and respective parameter in the thermal system diagram

that corresponds to an inductor and the inductance, respectively. As will be seen,

this will not present a challenge to achieving the objectives of using an equivalent

electrical circuit model.

Next, formulas for the quantities of an electrical circuit, namely the electrical capacitance

and resistance, that are functions of the properties of the thermal system components

must be devised. The formulas are as follows:

•                  𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙  𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒  𝐶 =  𝑡ℎ𝑒𝑟𝑚𝑎𝑙  𝑚𝑎𝑠𝑠 = 𝑣 ∗ 𝑑 ∗ 𝐶! [5]

𝑣 = 𝑣𝑜𝑙𝑢𝑚𝑒,  𝑚!

𝑑 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑘𝑔/𝑚!

𝐶! = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐  ℎ𝑒𝑎𝑡  𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒, 𝐽/ ℃ ∗ 𝑘𝑔

• Electrical resistance R can be quantified by the equations for resistance to heat

conduction or convection, or a combination of both, depending on the nature of

the thermal system being analyzed.

𝐻𝑒𝑎𝑡  𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒  𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑥/ 𝑘 ∗ 𝐴    [6]

                                                                                                       𝑥 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠  𝑜𝑟  𝑙𝑒𝑛𝑔𝑡ℎ,𝑚

 𝑘 = 𝑡ℎ𝑒𝑟𝑚𝑎𝑙  𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦,  𝑊 𝑚 ∗℃

                                                                                                   𝐴 = 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙  𝑎𝑟𝑒𝑎,𝑚!

                                     𝐻𝑒𝑎𝑡  𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛  𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 1/ ℎ ∗ 𝐴          [3]

ℎ = ℎ𝑒𝑎𝑡  𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟  𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡,𝑊/ 𝑚! ∗℃

                                                                                         𝐴 = 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙  𝑎𝑟𝑒𝑎,𝑚!

IV. Basic Two-Room Modeling Example

1) Example Set-Up and Objective The picture of the set-up of the room is below.  

The length and width of each room is five feet, and the height of each room is two

and a half feet. Each room has thermostats on both the exterior and interior walls, and

there is a door in between the rooms. Additionally, there is a HVAC system, represented

by a heater, which has an outlet into the first room, but not the second room. The

surrounding environment of the rooms is a standard ambient environment. The objective

of this example is to demonstrate that an equivalent electrical circuit model can be

applied to a thermal system. There is no hard data that can be used to compare against the

simulation of this model. Therefore, when looking at the performance of the simulation,

the utility of the model will be judged by the performance parameters not being an exact

value, but by being in a reasonable range of values.

2) Design of the Model The basic design of the equivalent electrical circuit for the thermal system of this

two – room model is below. It was designed using SimScape in Matlab.

The links between the two-room picture and the components of this

electrical model are as follows.

• The two voltage sources both represent the ambient environment, more

specifically the outside temperature.

• The current source represents the heat being generated from the HVAC system.

• The two exterior structures can be thought of as thermal resistors, as they prevent

heat from being transported across its structure. Therefore, they are modeled as

electrical resistors, with the “WallL” resistor, the one all the way to the left, and

the “WallR” resistor, the one all the way to the right, representing room one and

room two, respectively.

• The middle electrical resistor represents the interior wall, also a thermal resistor.

• Once heat is transferred into one of the two rooms, air resists its transport within

the rooms. Therefore, there must be electrical resistors that account for this

resistance of air. These resistors are the two ones adjacent to the wall resistors.

• Each room acts as a thermal capacitor, storing heat, and must be represented by an

electrical capacitor. However, each room contains two thermostats. As

thermostats measure temperature, which is equivalent to voltage, each thermostat

must be represented by an electrical capacitor, as they measure changes in

voltage. Therefore, they are four capacitors in the model, with the two outer

capacitors representing the thermostats on the exterior walls and the two inner

capacitors representing the thermostats on the two sides of the interior wall.

• As the output of a residential thermal system is the temperature on the thermostat,

the output of the electrical model will be its corresponding parameter to

temperature, the voltage across each of the four capacitors, i.e, thermostats. The

four voltage sensors connected to each of the capacitors output the measurements.

Besides linking the components of the thermal system to the components of an

electrical circuit, there were two other important considerations when designing this

model. Most importantly, the resistors are connected in series, rather than in parallel. This

is because of how heat is transported across a building. Once heat enters the building,

either from the HVAC system or from the ambient environment, it must overcome the

resistance of air to be transported within the building. Similarly, it must overcome

resistance from the interior wall to be transported to the adjacent room. Putting the

resistors in any type of parallel sequence would allow the current in the circuit, which

represents heat transfer, to bypass some of the resistors, which is not realistic.

Additionally, there was not a need to have an inductor in this circuit. This is

because there is no change in current throughout the model. As current represents heat

rate, a change in current would correspond to a change in the heat rate. However, there is

a constant heat rate into the thermal system. Therefore, the change in heat rate, and

therefore, change in current, is not a state variable that needs to be measured in this

model, which is what the inductor would be used for. Now, the parameters for the

resistors and capacitors must be calculated.

3) Calculation of Model Parameters

a. Model Parameter Calculation – Capacitance

𝐶! = 5  𝑚 ∗ 5  𝑚 ∗ 2.5  𝑚 ∗ 1.184 !"!! ∗ 1.01 !"

!"∗℃= 74.74 !"

℃  ∀𝑖

Each capacitor represents one of the two rooms in the model. Therefore, the

volume term is the volume of each room, which is the same for both rooms. Then, since

the substance that transports heat in the room is air, the density and specific heat

capacitance in the capacitance equation are the values of those parameters for air at

standard temperature and pressure.

b. Model Parameter Calculation – Resistance

𝑅!"#!$%&$  !"#$%"$#&,!"#$%!&'"# =.!"#$  !

(.!"#$  ! ℃∗!)∗(!".!  !!)=   .25  ℃ 𝑚

The standard thickness of a wall is one foot, which corresponds to .3048 meters.

Standard insulation material has a thermal conductivity between .035 and .16 W/°C*m

[7]. Therefore, I chose the midpoint, .0975 W/°C*m. The effective cross-sectional area

of resistance is the length or width of one of the rooms, 5 feet, multiplied by the height

of the room, 2.5 feet, resulting in an area of 12.5 m2.

𝑅!"#$%!&%  !"##,!"#$%!&'"# =  (.!"#$  !)/!

.!"  ! ℃∗! ∗(!".!  !!)=   .0508  ℃/𝑚

In determining this resistance, I used the resistance of the exterior wall as a

starting basis to make further assumptions. I assumed that an interior wall would be a

third as thick as an exterior wall. Additionally, I assumed that an interior wall would have

poorer insulation than an exterior wall. Therefore, I chose the value of thermal

conductivity of the interior wall to be the highest in the range of thermal conductivity

values for standard insulation. There is no change in the area value.

𝑅!"#,!"#$%&'"# =!

!""  ! ℃∗!! ∗(!".!  !!)= 8 ∗ 10!!  ℃/𝑊

As the transport of heat by air primarily occurs by the movement of air itself, the

term characterizing the resistance of air to heat transfer will use the convection equation.

The area value stays the same from the conduction resistance terms, as the same area is

still being heated. For forced convection of air, which describes the inside air in this

model because of the presence of the HVAC system, the heat transfer coefficient ranges

from 10 to 1000 W/°C*m2 [8]. Therefore, I chose a value of 100 W/°C*m2 for the heat

transfer coefficient.

At this point, all parameters have been calculated and the model can be simulated.

c. Simulation and Discussion The model was simulated with a constant current source of 1 A, equivalent to a

constant heat rate of 1 W, and a constant voltage source of 23 V, equivalent to an ambient

temperature of 23°C. The voltage and current sources are acting as step inputs.

Additionally, the initial condition for each state was zero Volts.

For all states of the model, i.e., the temperatures of all the thermostats, the steady-

state value, the value at which the temperature no longer changed, equated to its

maximum value. The time at which all states reached their steady-state value was

approximately 488550 seconds, or over 5 days. Other performance values are in the

table below. Values are not necessarily exact.

Thermostat Steady-State

Value (V)

Time constant (hr)

*63% of s.s value

Time constant (hr)

*2 V below s.s

Left Exterior 23.1368 10.3 25.4

Left Interior 23.1365 10.3 25.4

Right Interior 23.1135 10.3 25.5

Right Exterior 23.1131 10.3 25.4

Firstly, two different time constants are given because of the range at which

humans can feel differences in temperature. The standard time constant used is in terms

of achieving 63% of the final steady-state value. For 23°C, this corresponds to

approximately 14°C. However, humans can certainly feel and be affected by a

temperature deviation less than 9°C. Therefore, a time constant for when the temperature

reaches 21°C, at which point it is reasonable a human might not feel a difference from

23°C, may be more appropriate in this situation.

As evidenced by the table above, all thermostats, and therefore, both rooms,

exhibit an almost identical temperature performance. Given that each room has the same

dimensions and no other differentiating features, and that the impact of the heater is

minimal, as evidenced by all the steady-state temperatures being very close to the

ambient temperatures, this is expected. More importantly, this is not a realistic

simulation. It is not reasonable that it would take the lengths of time found in this

simulation for an indoor thermal system, even one with minimal influence from a heater,

to reach steady-state and equilibrate. Therefore, changes must be made to the model for it

to be considered an accurate and useful representation of the temperature profile of an

indoor structure.

The main changes that can be made to this model concern the calculation of the

parameters. While the actual design of the model can be changed, this is not advisable

because the steady-state behavior of the thermostats and final values demonstrate that the

model is accurately portraying the general mechanisms of heat transfer. Rather, the

capacitance and resistance values should be changed, to reflect considerations not yet

taken into account.

d. Transformation into two-dimensional model The initial calculation of the model parameters regarded the two rooms as one-

dimensional, as the area used in the resistance equation for the exterior structures was

only the area of one of the walls of each exterior structure. However, as one resistor is

used to represent all exterior walls for each room, it must account for the heat coming

through all the walls. Therefore, the area of each exterior resistor must be the summed

area of all the walls. Additionally, heat transfer through the air in each room was

previously characterized as convection, with the air moving in a horizontal line from the

end of one room to the end of the other room. To account for the movement of air in all

directions, a conduction term is introduced. This term represents an imaginary wall of

air in the middle of the room, with a thickness of one foot and a thermal conductivity

value of .024 W/°C*m for air, that is parallel to the convection term for heat transfer of

air, as heat can be transferred through either method. The new resistance values are

below.

𝑅!"#!$%&$  !"#$%"$#&,!"#$%!&'"# =.!"#$  !

(.!"#$  ! ℃∗!)∗(!∗!".!  !!)=   .08℃ 𝑊

!!!"#  

= !!!,!"#$.

+ !!!,!"#$.

= !.!"#$  !

.!"#  ! ℃∗! ∗ !".!  !!+ !

!∗!"!!  ℃/!  

 ⟹ 𝑅!"# = 8 ∗ 10!!  ℃/𝑊

As can be seen, air transfer of heat is dominated by convection over conduction.

Also, there is no change in the resistance of the interior wall. Additionally, the

capacitance of each capacitor must be reduced. Previously, the volume term for each

capacitor was the volume of the entire room. However, as there are two capacitors in

each room, each capacitor only stores the heat of half the room. Therefore, the value for

each capacitance is reduced by half, to 37370 J/°C. The model is now simulated again.

The approximate time to reach steady-state value for all thermostats, which again

are all slightly more than 23°C and the maximum values reached for each thermostat, is

8.0306*104 seconds, or 22.3 hours. The time constants for 63% of final value and 2°C

below the steady-state value are approximately the same for all thermostats and

approximately 2 hours and 4.2 hours, respectively. While these values are not as high as

before, they are still unreasonable and do not demonstrate the viability of this type of

model.

V. Integration of Windows and Doors

1) General Approach Windows and doors are components of essentially all buildings. They can also be

factors that heavily influence the temperature profile of a building. To this point, they

have been ignored in modeling the two rooms. This was in order to be able to build a

basic framework of the model. Now this has accomplished, the way to integrate windows

and doors into this model will be explored.

2) Calculations and Simulations - Windows

As with walls, heat will be transported by windows through conduction.

Therefore, with the inclusion of windows, the overall structure of the exterior resistors

becomes two resistors in parallel, representing heat conduction through both the exterior

insulation and windows. The resulting equation for the resistance is

!!!"#!$%&$  !"#$%"$#&

= !!!"#$%!,!"#$.

+ !!!"#$%&'!(!,!"#$.

⟹ 𝑅!" = 1 !!!+ !

!!

In the model, a window will be added to each exterior wall, meaning there will be

three windows in each room. The window size will be 34 by 46 inches, or .8636 by 1.168

meters, one of the more common window sizes in the U.S. [9]. The thickness will be .125

inches, or 3.175*10^-3 meters, a common window thickness [10]. Window glass has a

thermal conductivity of .96 W/°C*m [7]. Additionally, the area term in the insulation

resistance has changed, as the combined area of all the windows must be subtracted out.

Using the equation above, the new resistance for the two exterior structures is 3.16*10-3

°C/W. The model is now simulated again, with no change in the other parameters and the

outside temperature. However, in this simulation, the value of the heater is increased to

1500 A, equivalent to 1500 W.

The graph of the simulation and performance data is on the following page.

Thermostat Steady-State

Value (V)

Time constant (mins)

*63% of s.s. value

Time constant (min)

*2 V below

Left exterior 27.4 3.6 10.9

Left interior 27.4 4.7 11.5

Right interior 23.3 4.7 10.9

Right Exterior 23.2 3.6 10

As can immediately be seen by the graph, the time it takes to heat the rooms is

greatly reduced with the addition of the windows. Additionally, by increasing the heat

rate, there is now a visible impact of the heater on the temperature profile of the rooms.

Unsurprisingly, since the outtake of the heater is in the left room, the heater has a much

bigger impact on the thermostats of that room. Now, the impact of doors can be added to

this simulation.

3) Calculations and Simulations – Doors The opening of a door essentially causes the replacement of a solid portion of a

wall, the area of the door, with air. As demonstrated, heat transfer through air primarily

occurs by convection. Therefore, to model the door opening, a convection term for the

transfer of heat by air in an area equal to the area of the door can be introduced. The

resulting resistance term for a wall with a door is

!!!"##

= !!!"##,!"#$.    !"  !""#

+ !!!"##,!"#$.

⟹ 𝑅!"## = 1 !!!+ !

!!

In this model, the interior wall contains a door. The door will be 32 inches by 80

inches, or .8128 by 2.03 meters, a standard door size [11]. The heat transfer coefficient

will remain at 100 W/°C*m2. As in the window case, the area value in the conduction

term is smaller, with the area of the door being subtracted out. The resulting resistance

for the interior wall is 5.5*10-3 °C/W. The model is now simulated again, with the

inclusion of windows, and the heater value at 1500 A.

The graph of the simulation is below.

As can be seen by the graph, the opening of a door has little effect on the time

characteristics of the model. Rather, the opening of a door causes a decrease in the

temperature difference between the two rooms. The temperature of the two thermostats in

the left room both decrease by 1°C and the temperature of the two thermostats in the right

room both increase by 1°C, reducing the temperature difference from 4°C to 2°C.

4) Analysis & Discussion The simulations that occur after integrating the windows and door validate the

model. Most importantly, they reduce the time constants to reasonable values. With the

lack of hard data, the order of magnitude of the performance measures, rather than their

exact numbers, are more meaningful. This is also because there were a range of values

that could have been reasonably used for parameters, such as the choice of window size

and the heat transfer coefficient, in the resistance equations. However, the ranges of

values were often within one order of magnitude. Therefore, as confirmed by other

simulations, changing these parameters do not change the order of magnitude of the

performance measures.

Very importantly, these simulations also validate that the model is built on sound

physical reasoning, as it adjusts to changes in ways that would be expected. As would be

expected, the simulations demonstrate that windows, which offer very little resistance to

heat transfers, are effective at causing a quick change in temperature. Additionally, the

opening of a door should cause two different zones of air to mix and for their

temperatures to equilibrate, as happens in the simulation. Other simulations performed

further demonstrate that the temperature profile will respond to changes in model

conditions in a physically logically way, as the below examples demonstrate.

• Removing the convection term from the air resistance causes large increase in only

time constant of interior thermostats because there’s negligible heat transfer through

air by conduction.

• Decreasing the window size and increasing the heat transfer coefficient causes

opening of door to have larger effect on reducing temperature differences because the

heat is being more effectively distributed among the two rooms and less heat is being

lost or gained through the windows.

VI. Application: Westgate 749 & Heat Flow Testing

1) Background and Design of Model The goal of this part is to be able to properly simulate the temperature profile of a

real apartment in the apartment building at 749 Westgate, St. Louis, MO. No real data

could be obtained for this apartment. Therefore, as for the two-room example, the success

of applying this model to the apartment will be determined by logical reasoning. The

schematic of the apartment is below.

The equivalent electrical model of the thermal system of the model is

below.

The same principles for designing the resistors in series as in the two-room

example are followed. There are three constant voltage sources, representing the ambient,

 107  

111    101  

105  

• Room 101 is being considered as the area below the

blue line.

• The left wall of Room 105 is a wall with another

apartment.

• The left wall of Room 101 is a wall to a hallway.

• All other walls, meaning the right walls and the top

and bottom walls are exterior walls to outside.

• The apartment is on the first floor, but heat transfer

from the apartment above or from below the floor

will not be considered.

• Each room has an HVAC outlet, but there is only

one central system.

adjacent apartment, and hallway temperatures. There are four constant current sources,

one for each room, but they all must have the same value.

2) Calculation of Model Parameter Values

As there are 14 resistors in this model, I used a Matlab script, which can be found

in the appendix, to calculate and input the resistance values. Certain values in the

equations still had to be approximated, and they are noted in the appendix. Two changes

in how the values were calculated were made from the two-room model. One was the

calculation of the resistance of the exterior insulation. The R-value of the apartment

insulation, a value used in the construction industry, was given. This value divided by the

area of the relevant structure gives the resistance of the structure. Therefore, this was the

equation used to determine the exterior insulation resistances. Additionally, two different

heat transfer coefficients, both approximate values, were used for the exterior air and

interior air. This was done to account for the forced convection of interior, but not

exterior, air by the HVAC system. The four capacitance values were also calculated in

the same script.

3) Simulations

Simulations were calculated with the door open and closed. In addition to

measuring temperatures, i.e., voltages, these simulations used current sensors in order to

determine the magnitude of the influence of windows and doors on heat transfer. This

was conducted by the individual components of the aggregate resistors that contained

windows or doors i.e., the window or door and solid wall, being put in parallel. The exact

Simulink model that was used in the simulations can be found in the appendix. In both

simulations, the temperature of the ambient environment, adjacent apartment, and

hallway were set at 23°C, 20°C, and 18°C, respectively, and all four heaters were set to a

heat rate of I Watt. In the first simulation, the initial temperature of each room was set at

18°C. The results of the first simulation are shown below.  

 

Room Steady-State

Value (V)

Time to steady-

state (min)

Time constant (mins)

*63% of s.s. value

Time constant (min)

*.5 V below

107 23.002 51.76 3.26 6.72

111 22.998 36.89 2.2 5.07

101 22.9607 41.8 2.88 9.19

105 22.9598 44.5 5.07 9.93

As the initial temperatures are not 0°C, there are changes in the time constant

measured. The standard time constant is now for when the increase in temperature has

been 63% of the difference between 18°C, not 0°C, and the final temperature. Since these

increases ended up being approximately 2°C below the final temperature, the second time

constant was changed. As can be seen from the data table and is evident from the graph,

while the values for the time to steady-state are relatively high, a very large percentage of

the temperature increase comes within the first ten minutes. Also, there is not a very large

variation in performance measures for the different rooms.

The second simulation was conducted with the doors being open. Additionally,

the initial temperature of Room 105 was changed to 20°C. All other conditions were

unchanged. The results of the second simulation are below.  

                 

 

With the doors open, the time characteristics of the temperature profile of the

apartment do not change. However, there is a change in the steady-state temperature

values, though not necessarily as significant as the temperature graph seemingly

indicates. The temperatures range from 22.75°C, for Room 107, to 22.5173°C, for Room

101, a slightly bigger range of values than when the door was closed.

4) Analysis & Discussion

Applying an equivalent electrical circuit model to 749 Westgate produces viable

results that demonstrate the usefulness of this type of model. Again, while there was no

hard data available, the temperature profile in the two simulations is what can be

reasonably expected from this apartment. The largest room, Room 101, takes the longest

time to significantly heat, as defined by its time constants. Additionally, with the heater

exerting minimal influence, it is unsurprising the rooms have very similar steady-state

temperatures.

The use of the current sensors is very valuable, as they provide information about

how the heat is flowing into and within the apartment. The heat rate graph for the first

simulation demonstrates that essentially all the heat from the ambient environment is

transferred through the windows, as the lines that stay around zero represent the exterior

insulation. This highlights the ability of the current sensors to validate and explain the

scientific events that the model is representing.

The heat rate graph for the second simulation fulfills a similar objective. Based on

insights from the two-room model, the door is not functioning properly in this model, as

it decreases the steady-state temperatures and slightly increases the range of temperature

values. However, the door heat rate graph demonstrates why these effects are happening.

The heat rate graph shows that because Room 105 is initially at a higher temperature than

the other rooms, heat is initially transferred from Room 105, as evidenced by the blue and

green lines, before quickly flowing back in the other direction. Additionally, as evidenced

by the bottom red line, with the hallway door in Room 101 open, heat is constantly

flowing from Room 101 to the hallway to try to equilibrate their temperatures. These are

two likely factors that cause the decrease in the steady-state temperatures and Room 101

to have the lowest temperature. Also, as expected, the heat rate graph highlights that

opening doors causes most heat to be transferred by air through the open door space. The

current sensors for the windows and doors both reveal the large influence of these

features on heat transfer, highlighting their importance to energy efficiency goals.

VII. Conclusions & Next Steps

The use of equivalent circuit models for indoor thermal systems has not gained

much attention. First developing the proper techniques to model a thermal system in such

a way by the use of a two-room model, I then demonstrated the ability to successfully

apply the model to a real apartment, 749 Westgate. The simulations showed the ability of

the model to use voltage and current sensors to accurately predict temperatures and gain a

very detailed understanding of the heat flows through a building, respectively.

Additionally, physical phenomena, such as heat transfer from other floors, can be

accounted for in the model by simply adding more resistors. Importantly, this technique

can also be employed in order to analyze the energy savings of specific building

management measures.

There are many next steps that can be taken with this model. Simulations that can

be compared with hard data should be conducted in order to refine the model. After this

step, design of control schemes for this model should be the next step. As this is an

entirely linear model, control of this model should be able to be effectively accomplished.

Bibliography

[1] Y. Lin, T. Middelkoop, and P. Barooah. “Issues in identification of control-oriented thermal models of zones in multi-zone buildings,” in Proc. IEEE Conf. Decision and Control, Maui, HI, 2012, pp. 6932-6937. [2] Donghun, Kim et al. "Comparative Evaluation of Model Predictive Control Strategies for a Building HVAC System," in Proc. American Control Conf., Washington, D.C., 2013, pp. 3455-3460. [3] S. Karagol and M. Bikdash. "Generation of Equivalent-Circuit Models From Simulation Data of a Thermal System.” IEEE Trans. Power Electron., vol. 25, no. 4, pp. 820-828, Aprl. 2010. [4] Analogous Electrical and Mechanical Systems [Online]. Available: http://lpsa.swarthmore.edu/Analogs/ElectricalMechanicalAnalogs.html. [5] J. O’Loughlin and D. Loree. “Cooling system transient analysis using an electric circuit program analog,” in Pulsed Power Conf., Dallas, TX, 2003, pp. 767-770. [6] F. Behrens, R.P. Garcia, and R. Ribas. Evaluation of Thermal Devices by Using Equivalent Electrical Circuit Models [Online]. Available: http://www.lsi.usp.br/~dmi/PreviousWorkshops/workshop2000/Articles/Ricardo.pdf. [7] Thermal Conductivity of some common Materials and Gases [Online]. Available: http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html. [8] Convective Heat Transfer [Online]. Available: http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html. [9] Preble, Alec. What Are Standard Window Sizes? [Online]. Available: http://www.ehow.com/facts_7538630_standard-window-sizes.html. [10] Glass Thickness [Online]. Available: http://www.dullesglassandmirror.com/residential/glass_thickness.asp. [11] Deziel, Chris. Standard Indoor Door Sizes [Online]. Available: http://homeguides.sfgate.com/standard-inside-door-sizes-84805.html.

Appendix

I. Model used to generate heat rate graphs

II. Code to generate Westgate 749 parameter values

1) When door is closed % c1 (R105) = 36*7*13*1.184*1.01*10^3*.3048^3 = 1.1093*10^5 % c2 (R107) = 10.5*21*7*1.184*1.01*10^3*.3048^3 = 5.2267e+04 % c3 (R111) = 24*11*7*1.184*1.01*10^3*.3048^3 = 6.2578e+04 % c4 (R101) = 13*11*7*1.184*1.01*10^3*.3048^3 = 3.3896e+04 ftm = .3048; % feet-to-meter conversion htco = 50; % heat transfer coefficient of outdoor air (approximation) htci = 500; % heat transfer coefficient of indoor air (approximation) Tcair = .024; % thermal conductivity of air Thicknessw = 3.175*10^-3; % thickness of window (approximation) Thaw = 3.175*10^-3; % thickness of air gap within window (approximation) Tcw = .96; % thermal conductivity of window glass Thicknesscond = .3048; % thickness of "air walls" in each room (approximation) Rtot = 24.8; % R - value of apartemnt's exterior wall insulation UStoM = 0.176110; % conservation from R value in US units to metric units % Apartment-to-Apartment wall A1 = 36*7*ftm^2 ; R1 = Rtot*UStoM/A1 % R105 Exterior Wall

Ad2 = 7*17*ftm^2/6; Rd2 = 1/(htco*Ad2) Aw2 = (13-17/6-1)*7*ftm^2; Rgw2 = Thicknessw/(Tcw*Aw2) Raw2 = Thaw/(Tcair*Aw2) % conduction resistance in air gap in window Raw2v = 1/(htco*Aw2) % convection resistance in air gap in window Rw2 = Rgw2+1/(1/Raw2+1/Raw2v) % total resistance of window A2 = 13*7*ftm^2 - Aw2 ; Ri2 = Rtot*UStoM/A2 % resistance of exterior insulation R2 = 1/(0/Rw2+0/Rd2+1/Ri2) R19 = Rw2 % R107 Exterior Wall Aw3 = 9*7*ftm^2; Rgw3 = Thicknessw/(Tcw*Aw3) Raw3 = Thaw/(Tcair*Aw3) Raw3v = 1/(htco*Aw3) Rw3 = Rgw3+1/(1/Raw3+1/Raw3v) A3 = (10.5+21)*7*ftm^2 - Aw3 ; Ri3 = Rtot*UStoM/A3 R3 = 1/(0/Rw3+1/Ri3) R16 = Rw3 % R111 Exterior Wall Aw4 = 14.5*7*ftm^2; Rgw4 = Thicknessw/(Tcw*Aw4) Raw4 = Thaw/(Tcair*Aw4) Raw4v = 1/(htco*Aw4) Rw4 = Rgw4+1/(1/Raw4+1/Raw4v) A4 = (24+11)*7*ftm^2-Aw4 ; Ri4 = Rtot*UStoM/A4 R4 = 1/(0/Rw4+1/Ri4) R17 = Rw4 % R101 Exterior Wall Aw5 = 8*7*ftm^2 ; Rgw5 = Thicknessw/(Tcw*Aw5) Raw5 = Thaw/(Tcair*Aw5) Raw5v = 1/(htco*Aw5) Rw5 = Rgw5+1/(1/Raw5+1/Raw5v) A5 = 11*7*ftm^2-Aw5 ; Ri5 = Rtot*UStoM/A5 R5 = 1/(0/Rw5+1/Ri5) R18 = Rw5 % R105/R107 Interior Wall Ad6 = 7*17*ftm^2/6; Rd6 = 1/(htci*Ad6) Thickness6 = 3*ftm/10; Tc6 = .16 ; % thermal conductivity of interior wall (approximation) A6 = 21*7*ftm^2 ; Ri6 = Thickness6/(Tc6*A6)

R6 = 1/(0/Rd6+1/Ri6) % R105/R111 Interior Wall Ad7 = (2*7+2.5*7)*ftm^2; Rd7 = 1/(htci*Ad7) Thickness7 = 3*ftm/10 ; Tc7 = .16 ; A7 = 11*7*ftm^2 ; Ri7 = Thickness7/(Tc7*A7) R7 = 1/(0/Rd7+1/Ri7) % R105/R101 Interior "Air" Wall A8 = 10*7*ftm^2; Rconv8 = 1/(htci*A8) Rcond8 = Thicknesscond/(Tcair*A8) R8 = 1/(1/Rcond8+1/Rconv8) % R107/R111 Interior Wall Thickness9 = 3*ftm/5 ; Tc9 = .16 ; A9 = 9*7*ftm^2 ; R9 = Thickness9/(Tc9*A9) % R101/R111 Interior Wall Ad10 = 10*7*ftm^2/12; Rd10 = 1/(htci*Ad10) Thickness10 = 3*ftm/10; Tc10 = .16 ; A10 = 13*7*ftm^2 ; Ri10 = Thickness10/(Tc10*A10) R10 = 1/(0/Rd10+1/Ri10) % R107 Air Resistance A11 = 21*7*ftm^2; Rconv11 = 1/(htci*A11) Rcond11 = Thicknesscond/(Tcair*A11) R11 = 1/(1/Rcond11+1/Rconv11) % R111 Air Resistance A12 = 24*7*ftm^2; Rconv12 = 1/(htci*A12) Rcond12 = Thicknesscond/(Tcair*A12) R12 = 1/(1/Rcond12+1/Rconv12) % R101 Air Resistance A13 = 13*7*ftm^2; Rconv13 = 1/(htci*A13) Rcond13 = Thicknesscond/(Tcair*A13) R13 = 1/(1/Rcond13+1/Rconv13) % R105 Air Resistance A14 = 36*7*ftm^2; Rconv14 = 1/(htci*A14) Rcond14 = Thicknesscond/(Tcair*A14) R14 = 1/(1/Rcond14+1/Rconv14)

% Apartment-to-Hallway Wall Ad15 = 7*17*ftm^2/6; Rd15 = 1/(htco*Ad15) A15 = 13*7*ftm^2 ; Ri15 = Rtot*UStoM/A15 R15 = 1/(0/Rd15+1/Ri15)

2) When door is open % c1 (R105) = 36*7*13*1.184*1.01*10^3*.3048^3 % c2 (R107) = 10.5*21*7*1.184*1.01*10^3*.3048^3 % c3 (R111) = 24*11*7*1.184*1.01*10^3*.3048^3 % c4 (R101) = 13*11*7*1.184*1.01*10^3*.3048^3 ftm = .3048; % feet-to-meter conversion htco = 50; % heat transfer coefficient of outdoor air (approximation) htci = 500; % heat transfer coefficient of indoor air (approximation) Tcair = .024; % thermal conductivity of air Thicknessw = 3.175*10^-3; % thickness of window (approximation) Thaw = 3.175*10^-3; % thickness of air gap within window (approximation) Tcw = .96; % thermal conductivity of window glass Thicknesscond = .3048; % thickness of "air walls" in each room (approximation) Rtot = 24.8; % R - value of apartemnt's exterior wall insulation UStoM = 0.176110; % conversion from R value in US units to metric units % Apartment-to-Apartment wall A1 = 36*7*ftm^2 ; R1 = Rtot*UStoM/A1 % R105 Exterior Wall Ad2 = 7*17*ftm^2/6; Rd2 = 1/(htco*Ad2) % convection resistance in area of door - exterior htc Aw2 = (13-17/6-1)*7*ftm^2; Rgw2 = Thicknessw/(Tcw*Aw2) Raw2 = Thaw/(Tcair*Aw2) % conduction resistance in air gap in window Raw2v = 1/(htco*Aw2) % convection resistance in air gap in window Rw2 = Rgw2+1/(1/Raw2+1/Raw2v) % total resistance of window A2 = 13*7*ftm^2 - Aw2 - Ad2 ; Ri2 = Rtot*UStoM/A2 % resistance of exterior insulation R2 = 1/(0/Rw2+0/Rd2+1/Ri2) R19 = Rw2 R20 = Rd2 % R107 Exterior Wall Aw3 = 9*7*ftm^2; Rgw3 = Thicknessw/(Tcw*Aw3)

Raw3 = Thaw/(Tcair*Aw3) Raw3v = 1/(htco*Aw3) Rw3 = Rgw3+1/(1/Raw3+1/Raw3v) A3 = (10.5+21)*7*ftm^2 - Aw3 ; Ri3 = Rtot*UStoM/A3 R3 = 1/(0/Rw3+1/Ri3) R16 = Rw3 % R111 Exterior Wall Aw4 = 14.5*7*ftm^2; Rgw4 = Thicknessw/(Tcw*Aw4) Raw4 = Thaw/(Tcair*Aw4) Raw4v = 1/(htco*Aw4) Rw4 = Rgw4+1/(1/Raw4+1/Raw4v) A4 = (24+11)*7*ftm^2-Aw4 ; Ri4 = Rtot*UStoM/A4 R4 = 1/(0/Rw4+1/Ri4) R17 = Rw4 % R101 Exterior Wall Aw5 = 8*7*ftm^2 ; Rgw5 = Thicknessw/(Tcw*Aw5) Raw5 = Thaw/(Tcair*Aw5) Raw5v = 1/(htco*Aw5) Rw5 = Rgw5+1/(1/Raw5+1/Raw5v) A5 = 11*7*ftm^2-Aw5 ; Ri5 = Rtot*UStoM/A5 R5 = 1/(0/Rw5+1/Ri5) R18= Rw5 % R105/R107 Interior Wall Ad6 = 7*17*ftm^2/6; Rd6 = 1/(htci*Ad6) % convection resistance in area of door - interior htc Thickness6 = 3*ftm/10; Tc6 = .16 ; % thermal conductivity of interior wall (approximation) A6 = 21*7*ftm^2 - Ad6 ; Ri6 = Thickness6/(Tc6*A6) R6 = 1/(0/Rd6+1/Ri6) R21 = Rd6; % R105/R111 Interior Wall Ad7 = (2*7+2.5*7)*ftm^2; Rd7 = 1/(htci*Ad7) Thickness7 = 3*ftm/10 ; Tc7 = .16 ; A7 = 11*7*ftm^2 - Ad7 ; Ri7 = Thickness7/(Tc7*A7) R7 = 1/(0/Rd7+1/Ri7) R22 = Rd7; % R105/R101 Interior "Air" Wall A8 = 10*7*ftm^2; Rconv8 = 1/(htci*A8) Rcond8 = Thicknesscond/(Tcair*A8)

R8 = 1/(1/Rcond8+1/Rconv8) % R107/R111 Interior Wall Thickness9 = 3*ftm/5 ; Tc9 = .16 ; A9 = 9*7*ftm^2 ; R9 = Thickness9/(Tc9*A9) % R101/R111 Interior Wall Ad10 = 10*7*ftm^2/12; Rd10 = 1/(htci*Ad10) Thickness10 = 3*ftm/10; Tc10 = .16 ; A10 = 13*7*ftm^2 - Ad10 ; Ri10 = Thickness10/(Tc10*A10) R10 = 1/(0/Rd10+1/Ri10) R23 = Rd10; % R107 Air Resistance A11 = 21*7*ftm^2; Rconv11 = 1/(htci*A11) Rcond11 = Thicknesscond/(Tcair*A11) R11 = 1/(1/Rcond11+1/Rconv11) % R111 Air Resistance A12 = 24*7*ftm^2; Rconv12 = 1/(htci*A12) Rcond12 = Thicknesscond/(Tcair*A12) R12 = 1/(1/Rcond12+1/Rconv12) % R101 Air Resistance A13 = 13*7*ftm^2; Rconv13 = 1/(htci*A13) Rcond13 = Thicknesscond/(Tcair*A13) R13 = 1/(1/Rcond13+1/Rconv13) % R105 Air Resistance A14 = 36*7*ftm^2; Rconv14 = 1/(htci*A14) Rcond14 = Thicknesscond/(Tcair*A14) R14 = 1/(1/Rcond14+1/Rconv14) % Apartment-to-Hallway Wall Ad15 = 7*17*ftm^2/6; Rd15 = 1/(htco*Ad15) A15 = 13*7*ftm^2 - Ad15 ; Ri15 = Rtot*UStoM/A15 R15 = 1/(0/Rd15+1/Ri15) R24 = Rd15;

III. Additional Simulations

1) Basic Two-Room Example

• All simulations are with conditions from simulation of two-rooms with

windows except for changes noted.

 

Figure 1. No convection term for air resistance and

door closed; Rex = 3.16*10-3 °C/W, Rair = 1.016

°C/W, Rinw = .0508 °C/W, heater rate = 1 W, Toutside

= 23°C

 

Figure 2. Window size increased to 20% of

exterior wall area and heat transfer coefficient

reduced to 10 W/m2*°C; Rex = 1.3*10-3 °C/W, Rair

= 7.9*10-3 °C/W, Rinw = .0508 °C/W, heater rate

= 750 W, Toutside = 23°C. Same figure when door

opened; Rinw = .0298 °C/W

Figure 3. Window size decreased to .6096 m by

.8636 and htc increased to 750 W/m2*°C. Door

closed; Rex = 5.9*10-3 °C/W, Rair = 1.067*10-4

°C/W, Rinw = .0508 °C/W, heater rate = 1500 W

Figure 4. Same conditions as Figure 3 but door

opened; Rinw = 7.945*10-4 °C/W

2) Westgate 749 Model

• Same Matlab script used to generate resistance and capacitance values.

Figure 5. Doors closed. Heater Rate = 1 W. Tin =

20°C for Room 105 and Tin = 18°C for all other

rooms

Figure 6. Doors closed. Heater Rate = 1 W. Tin =

20°C for Room 105 and Tin = 18°C for all other

rooms

Figure 7. Doors open. Heater Rate = 1 W. Tin =

18°C for all rooms.

Figure 8. Doors open. Heater Rate = 1 W. Tin =

18°C for all rooms.