Environmental Physics
Chapter 5:
Home Energy Conservation and HeatTransfer Control
Copyright © 2012 by DBS
Introduction
~20% of all the energy
in US is used for
heating and cooling
buildings
Residential sector uses
50% for space heating
Energy (conservation)
efficiency should be
first step in dealing
with environmental
impacts…
Figure 4.1: U.S. household energy consumption by
end use. 1 Quad = 1015 Btu.
Introduction
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Amount of our total energy that is used for heating and cooling of buildings is massive (20%)
Heat is transferred from hot objects to cold objects by conduction, convection and radiation
Important to learn how to control the exchange of heat with our surroundings
– New home designs
– Retrofitting
Building Materials
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Conductors and insulators
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Good conductors of electricity are usually good conductors of heat energy
e.g. copper in electrical wires and water pipes
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Poor conductors of electricity are usually poor conductors of heat energy
e.g. Styrofoam, fiberglass (or any other material containing trapped air)
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Insulating materials are said to have a high ‘resistance’ to heat transfer
Aerogel is composed of 99.8% air
and is chemically similar to
ordinary glass. Being the world's
lightest known solid, it weighs only
three times that of air.
Building Materials
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Amount of heat radiated by an object depends on its temperature (SB law)
Color of material also influences heat transfer via radiation
– Black objects are excellent emitters as well as excellent absorbers
– White objects reflect and does not absorb/emit as much
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Kirchoff’s law: Objects that are good emitters of radiation are also good absorbers
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Starting at the same temperature, a hot black object will radiate energy faster than a similar hot
light-colored object
Demo
Question
How is heat lost from a cup of coffee?
I have just poured myself a cup of coffee but not yet added the milk, when there is a knock at the
door and I have to go and see someone. I’ll be gone a few minutes – should I add the cream before
I go or after I get back in order to have the coffee at the maximum temperature when I drink it?
Building Materials
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Before you go!
– Cream makes the coffee lighter in color = poorer radiator
– Temperature will drop when cream is added so the rate of
heat transfer by conduction and radiation will decrease as a
result of the smaller ΔT
Figure 5.3: For coffee to be the hottest when you are ready to drink it at a later time, you should
add the cream initially, not just before drinking, as these data for the two cases illustrate.
Building Materials
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Thermos – reduces heat transfer
A bottle within a bottle separated by a vacuum
Vacuum prevents conduction and convection
Radiation is controlled by silvered surfaces
Figure 5.2: Thermos bottle (cutaway view).
House Insulation and Heating Calculations
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Insulation of houses is one of the easiest and most cost-effective means of reducing energy
consumption
House Insulation and Heating Calculations
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Rate of heat flow via conduction is given by:
Where Q = heat (J) transferred in time t (s), k = thermal conductivity (W m-1 K-1), A = surface area,
δ = thickness, T1 and T2 are temperatures on each side
House Insulation and Heating Calculations
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“R-value” is a measure of the resistance of a material to heat flow, it is a function of the type of
material and its thickness:
R=δ/k
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High R-value equals better insulating properties
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Substituting this equation into the conductivity equation:
Qc = 1 x A x ΔT
t
R
House Insulation and Heating Calculations
R-values of some common materials
e.g. 1” expanded polystyrene is rated R-4
1” Brick is rated R-0.20
polystyrene is 20x as insulating
would need a 2 ft wide brick wall
House Insulation and Heating Calculations
Show why the units are ft2.h.°F/Btu
Qc = 1 x A x ΔT
t
R
House Insulation and Heating Calculations
Figure 5.5: To obtain an R-value of 22 ft2-h-°F/Btu, you would have to use the
indicated thicknesses of various materials.
Question
Calculate the total heat transfer for 12 hours through an insulated window that measures 4ft x 7 ft,
when the outside temperature is 5 °F and the inside temperature is 65 °F
Calculate the total heat transfer for 3” of expanded polystyrene foam
R-value = 1.54 ft2.h.°F/Btu
A = 28 ft2
ΔT = 60 °F
 Qc = 13,100 Btu
Qc(foam) = 1680 Btu
House Insulation and Heating Calculations
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Homes are made up of many different materials
To find the total thermal resistance of a composite structure add the R-values
Rtotal = R1 + R2 + R3 + …
Figure 5.6: Example of the calculation of thermal resistance value for a composite wall.
House Insulation and Heating Calculations
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R-values change depending on wind speed
Tabulated values calculated based on 10 mph
End
• Review
House Insulation and Heating Calculations
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Heat loss by convection on the inside of a window
– Increases if air is allowed to move
– Can be decreased using curtains/drapes that touch the floor
Layer of still air
next to window
Figure 5.7: (a) Thermal drapes mounted in this arrangement will contribute to heat losses
rather than stop them. (b) This arrangement is better for reducing convective losses.
House Insulation and Heating Calculations
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Heat loss through insulated glass windows
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Wider gaps in double glazed windows not much higher R-values
Smaller gaps have greater heat loss via conduction, larger gaps allow
more convection leading to higher convective heat loss
Variations:
(i) evacuated or filled with inert gas (e.g. Ar) which has a lower
thermal conductivity than air,
(ii) low-e coating, thin metal coating reflects heat
House Insulation and Heating Calculations
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Windows: can be a large source of heat loss or gain
~35% of the energy requirements or a typical home are a result of heat loss through windows
Double pane windows required in most parts
Storm windows (inside or outside) cut down heat losses considerably (colder climates)
In most areas energy savings
not significant enough to
warrant installation
House Insulation and Heating Calculations
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Air infiltration: major source of heat losses
Cold air from outside leaks into the house from cracks, doors, windows, basement
Can be reduced by sealing the gaps
Figure 5.10: Cold air infiltration can account for 50% of the
energy needs of a house. Wherever there is a way for the air to
get in, it will. Winds can significantly increase infiltration rates.
House Insulation and Heating Calculations
Figure 5.11: Types of weather stripping. Caulking all cracks and weather stripping all
windows will reduce air infiltration. Such work is easy to do and very cost-effective
(with payback times of one heating season or less).
House Insulation and Heating Calculations
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Infiltration heat losses can be calculated using the “air-exchange” method
Qinfil = 0.018 x V x K x ΔT
t
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Where V = volume (ft3), ΔT = inside/outside temperature difference, 0.018 = volume heat capacity
of air (Btu/ft3-°F), K = no. of air changes per hour
K is measured using a ‘blower door’ and measuring the rate of decrease of a
tracer gas usually 0.5 – 1.5 h-1
House Insulation and Heating Calculations
5 °C is the ‘design temp.’ or the
lowest expected for the locality
House Insulation and Heating Calculations
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When calculating the heating load for a house efficiency must also be taken into account
e.g. natural gas furnace at 70 % efficiency, we would need:
36,000/0.70 = 51,000 Btu/h
House Insulation and Heating Calculations
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To calculate heat losses for an entire year you could sum A x ΔT/Rtotal for each day x 24 h/d
Easier to use “degree-day”
Degree Day (DD) = 65 °F – Tavg
(-ve DD are taken as zero)
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Total no. of DD for a year is the sum of the individual DD
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Annual total heating needs for conductive loses:
Q total = Σ (A/R) x (24 h/day) x (no. annual DD)
Figure 5.12: Annual heating degree-days (DD).
House Insulation and Heating Calculations
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If the insulated house was located in Buffalo, NY, with 7000 DD per yr,
Q total
Q total
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= Σ (A/R) x (24 h/day) x (no. annual DD)
= 1/14 Btu/ft2-h-°F x 4600 ft2 x 24 h/d x 7000 DD
= 55.3 x 106 Btu
Add to infiltration losses (calculated using blower door):
Qinfil
= 0.018 Btu/ft3- ° F x 15,000 ft3 x 1/h x 24 h/d x 7000 DD = 45.4 x 106 Btu
Total load = Q total + Qinfil = 101 x 106 Btu = 101 MBtu
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Assume cost electricity Is 0.07 cents per kWh
$0.07/kWh x 1 kWh / 3413 Btu x 106 Btu/Mbtu = $20.51 / Mbtu
Cost = $20.51 /Mbtu x 101 Mbtu = $2071 (does not take into account efficiency
Site Selection
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Wind increases convective heat loss Increase from 10 – 20 mph increases
infiltration by 100-200 %
Figure 5.13: Planting trees on the leeward side of a hill can substantially reduce
the wind velocity over the site. Vegetation or walls can block or deflect natural air
flow patterns and so reduce convective heat loss.
Impact of Energy Conservation Measures
Impact of Energy Conservation Measures
Figure 5.14: Summary of the effects of energy conservation measures on
space heating requirements for a typical 1500-ft2 house in three climates.
Question
Problem 18
a.) 0.5 + 1.0 + 0.8 = 2.3
b.) 19 + 2.3 = 21.3,
21.3-2.3 / 21.3 = 89%
c.) Qtotal = (1/2.3) x 6500 x 1 ft2 x 24 h/d = 67,826 Btu / ft2.yr
0.067826 MBtu/ft2.yr x $10 / MBtu = $0.68 / ft2.yr
d.) Qnew total = (1 ft2)(24)(6500) / 21.3 = 7324 Btu
savings of 67,826 – 7324 = 60,502 Btu / ft2.yr
Electricity cost = $10 / MBtu
savings of 0.060502 MBtu / ft2.yr x $10/MBtu = $0.60 / ft2.yr
Since fiberglass is 40 cents a ft2, payback is less than a season
End
• Review
Cooling
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Use same equations with cooling degree days (average of days high and low temp. minus 75 °F)
Cooling Degree-day (CDD) = Tavg - 75 °F
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Factors affecting cooling
– Outdoor temperature and intensity of solar radiation are important
– Heat gains from solar heating of walls, roofs and windows must be taken into account
– Heat gains from lights, appliances and people
– thermal storage causes a time delay in indoor temperatures
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Demand for air conditioning:
– 66 % of US households
– 83 % commercial space
–
6 % of household energy use
Cooling
Passive cooling (far less expensive)
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Site selection – location, orientation, vegetation
– Natural ventilation - evaporative cooling of human body (perspiration)
– Evaporative coolers (swamp coolers in SE USA)
– Desiccants
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Architectural features – surface-to-volume ratio, overhangs, window sizes,
shades
– Shading can reduce indoor temps. 10-20 °F
– Coolth tubes
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Building skin features – insulation, thermal mass, glazing
– Opposing windows, skylights and high windows
– act as thermal chimneys
Cooling
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Insulation using ‘radiant barrier’ materials
Restrict passage of low wavelength IR radiation, e.g. aluminum foil backed fiberglass insulation
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May be painted on, chips or film
Figure 5.16: Radiant barriers can reduce
heat gain in the attic of a house.
Air Conditioners and Heat Pumps
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Heat from cooler indoors is transferred to the warmer outdoors
Not a violation of 2nd law! Why?
Air Conditioners and Heat Pumps
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Requires an energy source (electricity)
CONDENSER
working fluid temperature >
outside air so condenses → liquid
Figure 5.17: Air conditioner operation.
EVAPORATOR
working fluid
evaporates → gas
COMPRESOR
Raises temp. and
pressure of gas
Question
A room air conditioner has a capacity of 6000 Btu/h. Would this be sufficient to maintain the
temperature of a small hut at 70 °F when the outside temperature is 95 °F? Assume the hut is 10
ft x 10 ft x 6 ft and the exterior surfaces are made of 1 in softwood
From the table of R-values - R = 1.25
Qc = 1 x A x ΔT
t
R
Qc = (4 x 60 ft2 + 2 x 100 ft2) x 25 °F = 8800 Btu/h
1.25
Answer is NO! (no infiltration included)
Air Conditioners and Heat Pumps
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Heat pumps
(1) Refrigerant liquid absorbs
heat from outside air and
evaporates
(2) Resulting gas is compressed
(3) Hot gas transfers heat to
room and is condensed
A residential heat pump.
(1) Cold refrigerant liquid
absorbs heat from warmer
room
(2) Resulting gas is compressed
(3) Transfers heat to outside air
Air Conditioners and Heat Pumps
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Coefficient of Performance (COP):
COP = heat transferred / electricity input
e.g. a heat pump with COP = 30,000 Bu/h and a compressor rating of 3.2 kW has
COP = 30,000 Btu/h
= 2.75
3.2 kW x (3413 Btu/h/kW)
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Heat pump is 2.75 x cheaper to operate than a electrical resistance heating system
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COP reduces as outside temperature decreases
The lower the outdoor temperature, the less heat there is
e.g. if heat has to be extracted from 10 ° F air, working
fluid must be below 10 °F, compressor must work harder
Figure 5.20: Relative costs of different types of heating fuels. To determine the cost of using a heat pump,
find the unit cost of electricity on the x-axis and move vertically to the appropriate heat pump C.O.P. curve.
Then move horizontally to the y-axis to find the operating cost per million Btu. Find the intersection of this
value with the line for gas or oil, and read off its cost on the x-axis. For example (as shown with dashed
lines), using a heat pump with C.O.P. = 2.0 at an electricity cost of $0.05/kWh is equivalent to using gas at
$0.52/therm or fuel oil at $0.67/gallon (with the furnace efficiencies shown).
Summary
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Household energy conservation has a substantial impact on your energy bill
Heat gains or losses can be reduced by using insulation and infiltration control
Heating load may be calculated using the equation for the rate of heat transfer via conduction
Qc = 1 x A x ΔT
t
R
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To find total heat losses for a season, Qtotal, we use the
Degree-Day concept
Q total = Σ (A/R) x (24 h/day) x (no. annual DD)
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Infiltration losses must be added to this to find the
total heating load for the house