Physics 100, Section L1F

Group members are:

Natalie Leung
Paloma Stanar
Jonathan Ng

Comparing the Relative Heating Costs to Warm Different Homes

Introduction

People live in many types and styles of homes, but not all of them require the same amount of energy to maintain a warm interior temperature throughout the winter.

Project Statement

Using four different kinds of homes, this project will determine which one requires the least amount of energy to stay a certain temperature indoors throughout the winter, and will therefore be the most economically and environmentally efficient. We will use the total surface area that is losing heat per home every month, and with it, determine the total amount of heat lost. The four different kinds of homes we will compare are stated below.

An apartment
A townhouse
A house
A penthouse

Once heating energies are known, we will compare all four and determine the most energy efficient mode of living in a cold area.

Goals

Determine the relative heating costs to maintain a warm interior temperature of four different homes.
Using the cost of energy to heat a home and the amount of energy needed per home per month, come up with a monthly bill for each case and compare relative costs and efficiencies.

Relevant Physics

Surface Area of a wall $m^{2}\,$ : Length of wall $m\,$ x width of wall $m\,$

Surface Area of the ceiling $m^{2}\,$ : Length of ceiling $m\,$ x width of ceiling $m\,$

Total surface area $m^{2}\,$ : Number of walls x Surface area of a wall $m^{2}\,$ + Number of ceiling x Surface area of ceiling $m^{2}\,$

Cost $(\$)\,$ : Price per kilowatt hour $(\$)/Kw*h\,$ x Usage $Kw*h\,$

Heat loss through convection:

Heat does get lost through the windows by convection, but we will not include this formula in our calculations.

Heat loss/gain through radiation:

Heat is gained naturally by radiation from the sun, through the windows. Using the emissivity of a surface, $e\,$ , as well as the Stefan-Boltmann constant, δ, $T\,$ , and $A\,$ , rate of heat transfer by radiation can be found by the formula:

${\big .}{\frac {Q}{\Delta t}}=\epsilon \cdot \sigma \cdot A\cdot T^{4}$

Heat loss through conduction:

${\big .}{\frac {\Delta Q}{\Delta t}}=-kA{\frac {\Delta T}{\Delta x}}$

${\Delta Q}=-kA{\frac {\Delta T}{\Delta x}}*{\Delta t}$

where

A is the cross-sectional surface area,

\Delta T

is the temperature difference between the ends,

\Delta x

is the distance between the ends.

k\,

is a quantity that characterizes whether a material is a good conductor or not, and is called the thermal conductivity.

Q is thermal energy while

\Delta Q

is the difference in thermal energy

Such that

The difference in Thermal energy $J\,$ = Total surface area $m^{2}\,$ x Thermal conductivity $(W/K*M)\,$ x Difference in temperature $k\,$ x Difference in time $sec\,$ / Thickness $m\,$

Data

Number of walls/ceilings/windows in which heat escapes:

House: 4 walls/1 ceiling/4 windows (on each walls)

Townhouse: 2 walls/1 ceiling/2 windows

Apartment: 1 wall/0 ceiling/1 window

Penthouse: 4 wall/0 ceiling/4 windows

Wall size 4 meters wide x 3 meters tall

Ceiling size 16 square meters

Window size 1 square meter

Clear Models

Assumptions

Assumptions:

Walls of all homes (except wooden) are of the same size, shape, thickness and are made from the same material (concrete).
Wooden home is of the same size and shape as the other homes, and walls have the same thickness. The walls will be made from wood.
Assume there is absolutely no space in which air could get out/in.
Assume the interior of the house is only made of 6 surfaces.
Windows are all single pane, made from the same material and have the same thickness
Heating is from the same source and, thus, have the same cost per kilowatt hour. The internal temperature will also be identical.
The exterior temperature for all homes are also identical.
The temperature of the ground is the same as inside the homes. (no heat loss through ground)
The temperature of adjacent homes to the townhouse is the same as inside the townhouse. (no heat loss through two walls in townhouse)
The temperature of adjacent homes/spaces (below, above, and 3 sides) to the apartment is the same as inside the apartment. (no heat loss through three walls, the floor and the ceiling)
The temperature of the adjacent homes/spaces (below and above) to the penthouse is the same as inside the penthouse. (no heat loss through floor and ceiling)
No heat gain by sunlight due to the winter months receiving little sunlight.
No heat gain by radiation from other objects outside of homes.

Conclusion

Our final answers serve not as a set cost for heating during winter, but as a reference to the difference in heating costs between the four homes. This is mainly due to the assumption of no heat gain by sunlight, which should generally have an effect on the change in heat energy. As a reference, these results tells us that...

References

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatra.html

http://www.staygarnethill.com/photos.htm

http://en.wikipedia.org/w/index.php?title=Thermal_radiation&action=edit&section=5

http://en.wikipedia.org/wiki/Conduction_%28heat%29