2Introduction~20% of all the energy in US is used for heating and cooling buildingsResidential sector uses 50% for space heatingEnergy (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.
3IntroductionAmount 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 radiationImportant to learn how to control the exchange of heat with our surroundingsNew home designsRetrofitting
4Building Materials Conductors and insulators Good conductors of electricity are usually good conductors of heat energye.g. copper in electrical wires and water pipesPoor conductors of electricity are usually poor conductors of heat energye.g. Styrofoam, fiberglass (or any other material containing trapped air)Insulating materials are said to have a high ‘resistance’ to heat transferAerogel 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.
5Building MaterialsAmount of heat radiated by an object depends on its temperature (SB law)Color of material also influences heat transfer via radiationBlack objects are excellent emitters as well as excellent absorbersWhite objects reflect and does not absorb/emit as muchKirchoff’s law: Objects that are good emitters of radiation are also good absorbersStarting at the same temperature, a hot black object will radiate energy faster than a similar hot light-colored object
7QuestionHow 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?
8Building Materials Before you go! Cream makes the coffee lighter in color = poorer radiatorTemperature will drop when cream is added so the rate of heat transfer by conduction and radiation will decrease as a result of the smaller ΔTFigure 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.
9Building Materials Thermos – reduces heat transfer A bottle within a bottle separated by a vacuumVacuum prevents conduction and convectionRadiation is controlled by silvered surfacesThermos.comFigure 5.2: Thermos bottle (cutaway view).
10House Insulation and Heating Calculations Insulation of houses is one of the easiest and most cost-effective means of reducing energy consumption
11House 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
12House 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 / kHigh R-value equals better insulating propertiesSubstituting this equation into the conductivity equation:Qc = 1 x A x ΔTt R
13House Insulation and Heating Calculations R-values of some common materialse.g. 1” expanded polystyrene is rated R-41” Brick is rated R-0.20polystyrene is 20x as insulatingwould need a 2 ft wide brick wall
14House Insulation and Heating Calculations Show why the units are ft2.h.°F/Btu Qc = 1 x A x ΔT t R
15House 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.
16QuestionCalculate 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 foamR-value = 1.54 ft2.h.°F/BtuA = 28 ft2ΔT = 60 °F Qc = 13,100 BtuQc(foam) = 1680 Btu
17House Insulation and Heating Calculations Homes are made up of many different materialsTo find the total thermal resistance of a composite structure add the R-valuesRtotal = R1 + R2 + R3 + …Figure 5.4: Example of the calculation of thermal resistance value for a composite wall.
18House Insulation and Heating Calculations R-values change depending on wind speedTabulated values calculated based on 10 mph
20House Insulation and Heating Calculations Heat loss by convection on the inside of a windowIncreases if air is allowed to moveCan be decreased using curtains/drapes that touch the floorLayer of still air next to windowFigure 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.
21House Insulation and Heating Calculations Heat loss through insulated glass windowsWider gaps in double glazed windows not much higher R-valuesSmaller gaps have greater heat loss via conduction, larger gaps allow more convection leading to higher convective heat lossVariations: (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
22House 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 windowsDouble pane windows required in most partsStorm windows (inside or outside) cut down heat losses considerably (colder climates)In most areas energy savings not significant enough to warrant installation
23House Insulation and Heating Calculations Air infiltration: major source of heat lossesCold air from outside leaks into the house from cracks, doors, windows, basementCan be reduced by sealing the gapsFigure 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.
25House 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).
26House Insulation and Heating Calculations Infiltration heat losses can be calculated using the “air-exchange” methodQinfil = x V x K x ΔTtWhere V = volume (ft3), ΔT = inside/outside temperature difference, = volume heat capacity of air (Btu/ft3-°F), K = no. of air changes per hourK is measured using a ‘blower door’ and measuring the rate of decrease of a tracer gas usually 0.5 – 1.5 h-1
27House Insulation and Heating Calculations 5 °C is the ‘design temp.’ or the lowest expected for the locality
28House Insulation and Heating Calculations When calculating the heating load for a house efficiency must also be taken into accounte.g. natural gas furnace at 70 % efficiency, we would need:36,000/0.70 = 51,000 Btu/h
29House 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/dEasier 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 DDAnnual 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).
30House 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 BtuAdd 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 BtuTotal load = Q total + Qinfil = 101 x 106 Btu = 101 MBtuAssume cost elecricity Is 0.07 cents per kWh$0.07/kWh x 1 kWh / 3413 Btu x 106 Btu/Mbtu = $20.51 / MbtuCost = $20.51 /Mbtu x 101 Mbtu = $2071 (does not take into account efficiency
31Site SelectionWind 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.
36Cooling 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 °FFactors affecting coolingOutdoor temperature and intensity of solar radiation are importantHeat gains from solar heating of walls, roofs and windows must be taken into accountHeat gains from lights, appliances and peoplethermal storage causes a time delay in indoor temperaturesDemand for air conditioning:66 % of US households83 % commercial space6 % of household energy use
37Cooling Passive cooling (far less expensive) Site selection – location, orientation, vegetationNatural ventilation - evaporative cooling of human body (perspiration)Evaporative coolers (swamp coolers in SE USA)DesiccantsArchitectural features – surface-to-volume ratio, overhangs, window sizes, shadesShading can reduce indoor temps °FCoolth tubesBuilding skin features – insulation, thermal mass, glazingOpposing windows, skylights and high windows – act as thermal chimneys
38Cooling Insulation using ‘radiant barrier’ materials Restrict passage of low wavelength IR radiation, e.g. aluminum foil backed fiberglass insulationMay be painted on, chips or filmFigure 5.12: Radiant barriers can reduce heat gain in the attic of a house.
39Air Conditioners and Heat Pumps Heat from cooler indoors is transferred to the warmer outdoorsNot a violation of 2nd law! Why?
40Air Conditioners and Heat Pumps Requires and energy source (electricity)CONDENSER working fluid temperature > outside air so condenses → liquidEVAPORATOR working fluid evaporates → gasCOMPRESOR Raises temp. and pressure of gasFigure 5.13: Air conditioner operation.
41QuestionA 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 softwoodFrom the table of R-values - R = 1.25Qc = 1 x A x ΔTt RQc = (4 x 60 ft2 + 2 x 100 ft2) x 25 °F = 8800 Btu/h1.25Answer is NO! (no infiltration included)
42Air Conditioners and Heat Pumps Refrigerant liquid absorbs heat from outside air and evaporatesResulting gas is compressedHot gas transfers heat to room and is condensedCold refrigerant liquid absorbs heat from warmer roomResulting gas is compressedTransfers heat to outside airA residential heat pump.
43Air Conditioners and Heat Pumps Coefficient of Performance (COP):COP = heat transferred / electricity inpute.g. a heat pump with COP = 30,000 Bu/h and a compressor rating of 3.2 kW hasCOP = 30,000 Btu/h = 2.753.2 kW x (3413 Btu/h/kW)Heat pump is 2.75 x cheaper to operate than a electrical resistance systemCOP reduces as outside temperature decreasesThe lower the outdoor temperature, the less heat there ise.g. if heat has to be extracted from 10 ° F air, working fluid must be below 10 °F, compressor must work harder
44Figure 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
45SummaryHousehold energy conservation has a substantial impact on your energy billHeat gains or losses can be reduced by using insulation and infiltration controlHeating load may be calculated using the equation for the rate of heat transfer via conductionQc = 1 x A x ΔTt RTo find total heat losses for a season, Qtotal, we use the Degree-Day conceptQ 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