Environmental Physics Chapter 5: Home Energy Conservation and Heat-Transfer Control Copyright © 2008 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 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 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 http://www.boingboing.net/2006/02/06/chunks-of-aerogel-fo.html 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 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 http://www.boingboing.net/2006/02/06/chunks-of-aerogel-fo.html

Demo http://www.boingboing.net/2006/02/06/chunks-of-aerogel-fo.html

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 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.

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).

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

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

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

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 5.3: 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 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.

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

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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.

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

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) http://www.doortechllc.com/leader_window.asp In most areas energy savings not significant enough to warrant installation

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.

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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).

House Insulation and Heating Calculations Infiltration heat losses can be calculated using the “air-exchange” method Qinfil = 0.018 x V x K x ΔT t 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 http://www.southface.org/web/resources&services/publications/journal/sfjv106/106art/richard_shields_blowerdoor.jpg

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

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

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).

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 = 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 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

Site Selection Wind increases convective heat loss Increase from 10 – 20 mph increases infiltration by 100-200 % 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.

Impact of Energy Conservation Measures

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.

Question Problem 4 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

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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

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. 10-20 °F Coolth tubes Building skin features – insulation, thermal mass, glazing Opposing windows, skylights and high windows – act as thermal chimneys

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 http://www.ohioenergyrater.com/services http://www.thurmancustomhomes.com/features.htm Figure 5.12: Radiant barriers can reduce heat gain in the attic of a house.

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

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.

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 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.

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

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

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