1. Environmental Physics
Chapter 5:
Home Energy Conservation and Heat-Transfer Control
Copyright © 2008 by DBS

2. 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.

3. Introduction
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

4. Building Materials Conductors and insulators
Good conductors of electricity are usually good conductors of heat energy
e.g. copper in electrical wires and water pipes
Poor conductors of electricity are usually poor conductors of heat energy
e.g. Styrofoam, fiberglass (or any other material containing trapped air)
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.

5. Building Materials
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
Kirchoff’s law: Objects that are good emitters of radiation are also good absorbers
Starting at the same temperature, a hot black object will radiate energy faster than a similar hot light-colored object

6. Demo

7. 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?

8. Building Materials 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 5.1: 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.

9. Building Materials Thermos – reduces heat transfer
A bottle within a bottle separated by a vacuum
Vacuum prevents conduction and convection
Radiation is controlled by silvered surfaces
Thermos.com
Figure 5.2: Thermos bottle (cutaway view).

10. House Insulation and Heating Calculations
Insulation of houses is one of the easiest and most cost-effective means of reducing energy consumption

11. House Insulation and Heating Calculations
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

12. House Insulation and Heating Calculations
“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
High R-value equals better insulating properties
Substituting this equation into the conductivity equation:
Qc = 1 x A x ΔT
t R

13. 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

14. House Insulation and Heating Calculations
Show why the units are ft2.h.°F/Btu Qc = 1 x A x ΔT t R

15. House Insulation and Heating Calculations
Figure 5.3: To obtain an R-value of 22 ft2-h-°F/Btu, you would have to use the indicated thicknesses of various materials.

16. 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

17. House Insulation and Heating Calculations
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 5.4: Example of the calculation of thermal resistance value for a composite wall.

18. House Insulation and Heating Calculations
R-values change depending on wind speed
Tabulated values calculated based on 10 mph

19. End
Review

20. House Insulation and Heating Calculations
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 5.5: (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.

21. House Insulation and Heating Calculations
Heat loss through insulated glass windows
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

22. House Insulation and Heating Calculations
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

23. House Insulation and Heating Calculations
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 5.6: 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.

25. House Insulation and Heating Calculations
Figure 5.7: 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).

26. House Insulation and Heating Calculations
Infiltration heat losses can be calculated using the “air-exchange” method
Qinfil = x V x K x ΔT t
Where V = volume (ft3), ΔT = inside/outside temperature difference, = 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

27. House Insulation and Heating Calculations
5 °C is the ‘design temp.’ or the lowest expected for the locality

28. House Insulation and Heating Calculations
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

29. House Insulation and Heating Calculations
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)
Total no. of DD for a year is the sum of the individual DD
Annual total heating needs for conductive loses:
Q total = Σ (A/R) x (24 h/day) x (no. annual DD)
Figure 5.8: Annual heating degree-days (DD).

30. House Insulation and Heating Calculations
If the insulated house was located in Buffalo, NY, with 7000 DD per yr,
Q total = Σ (A/R) x (24 h/day) x (no. annual DD)
Q total = 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:
Qinfil = 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
Assume cost elecricity 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

31. Site Selection
Wind increases convective heat loss Increase from 10 – 20 mph increases infiltration by %
Figure 5.9: 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.

32. Impact of Energy Conservation Measures

33. Impact of Energy Conservation Measures
Figure 5.10: Summary of the effects of energy conservation measures on space heating requirements for a typical 1500-ft2 house in three climates.

34. Question Problem 4 a.) 0.5 + 1.0 + 0.8 = 2.3 b.) 19 + 2.3 = 21.3,
/ 21.3 = 89%
c.) Qtotal = (1/2.3) x 6500 x 1 ft2 x 24 h/d = 67,826 Btu / ft2.yr
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 MBtu / ft2.yr x $10/MBtu = $0.60 / ft2.yr
Since fiberglass is 40 cents a ft2, payback is less than a season

35. End
Review

36. Cooling Degree-day (CDD) = Tavg - 75 °F
Use same equations with cooling degree days (average of days high and low temp. minus 75 °F)
Cooling Degree-day (CDD) = Tavg - 75 °F
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
Demand for air conditioning:
66 % of US households
83 % commercial space
6 % of household energy use

37. Cooling Passive cooling (far less expensive)
Site selection – location, orientation, vegetation
Natural ventilation - evaporative cooling of human body (perspiration)
Evaporative coolers (swamp coolers in SE USA)
Desiccants
Architectural features – surface-to-volume ratio, overhangs, window sizes, shades
Shading can reduce indoor temps °F
Coolth tubes
Building skin features – insulation, thermal mass, glazing
Opposing windows, skylights and high windows – act as thermal chimneys

38. Cooling Insulation using ‘radiant barrier’ materials
Restrict passage of low wavelength IR radiation, e.g. aluminum foil backed fiberglass insulation
May be painted on, chips or film
Figure 5.12: Radiant barriers can reduce heat gain in the attic of a house.

39. Air Conditioners and Heat Pumps
Heat from cooler indoors is transferred to the warmer outdoors
Not a violation of 2nd law! Why?

40. Air Conditioners and Heat Pumps
Requires and energy source (electricity)
CONDENSER working fluid temperature > outside air so condenses → liquid
EVAPORATOR working fluid evaporates → gas
COMPRESOR Raises temp. and pressure of gas
Figure 5.13: Air conditioner operation.

41. 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)

42. Air Conditioners and Heat Pumps
Refrigerant liquid absorbs heat from outside air and evaporates
Resulting gas is compressed
Hot gas transfers heat to room and is condensed
Cold refrigerant liquid absorbs heat from warmer room
Resulting gas is compressed
Transfers heat to outside air
A residential heat pump.

43. Air Conditioners and Heat Pumps
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)
Heat pump is 2.75 x cheaper to operate than a electrical resistance system
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

44. Figure 5. 16: Relative costs of different types of heating fuels
Figure 5.16: 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).
Figure 5.16: 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).
Fig. 5-16, p. 153

45. Summary
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
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)
Infiltration losses must be added to this to find the total heating load for the house