Presentation on theme: "NBS-M016 Contemporary Issues in Climate Change and Energy 2010"— Presentation transcript:
1 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 Energy Conservation in Buildings: The BasicsHeat Loss CalculationsEnergy ManagementN.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnvН.К.Тови М.А., д-р технических наукEnergy Science Director CRed ProjectHSBC Director of Low Carbon InnovationLecture 1Lecture 3Lecture 21
2 19. Energy Conservation in Buildings: The Basics Intrinsic Energy Useprovision of comfortable thermal environmentFunctional Energy Useenergy use associated with specific activities in building at time.Intrinsic Energy Use (mostly associated with heating) will vary little with different uses (apart from specifics e.g. warehouses, sports complexes etc).Functional Energy Use will vary depending on use – e.g. office (high computer use), laboratory (equipment), supermarket, hotel etc.In a poorly insulated building, functional energy use over life time will be low as a percentageIn a well insulated building, functional energy can be the dominant use – representing over 50% in ZICER.
3 19. Energy Conservation in Buildings: The Basics Heat Loss / Heat GainThree forms of heat transferConductionRadiationConvectionFor buildings - conductive losses are main issue to address in heat loss/heat gain.dHeat Flow is proportional to temperature differenceQAT1T2T1T2Temperature Profile
4 19. Energy Conservation in Buildings: The Basics Construction of Typical UK Walls: Solid Walls – most houses built pre-war.plasterBrickIn addition to resistance of brick and plaster there is:Internal surface resistanceExternal surface resistance
5 19. Energy Conservation in Buildings: The Basics Construction of Typical UK Walls: Solid Walls – most houses built pre-war.13 mmBrickU-value is amount of heat transferred per sqm for a unit temperature difference between inside and out.It is the reciprocal of aggregate resistance.R = rbrick + rplaster + rint + rextBut resistance = as A = 1.0k for brick ~ 1.0 W m-1 oC-1rbrick = / 1.0 = m2 oC W-1220 mmFor plaster k = 0.7 W m-1 oC-1 so rplaster = / = m2 oC W-1Total resistance = = m2 oC W-1So U- value = 1 / = W m-2 oC-1
6 19. Energy Conservation in Buildings: The Basics Construction of Typical UK Walls: post warcavityplasterBrickBrickCavities provide an extra air-space and hence extra resistance to heat flow.
7 19. Energy Conservation in Buildings: The Basics Construction of Typical UK Walls: post warComponents of resistanceInternal surfacePlasterBrickCavityExternal Surfacerinternal = 0.123rplaster = / 0.7 = 0. 02rbrick = / 1 = 0.11rcavity =rexternal = 0.055cavityplaster110 mmBrickBrickTotal resistance = = m2 oC W-1U – value = 1 / = 1.67 W m-2 oC-1.Or 70% of solid wall.
8 19. Energy Conservation in Buildings: The Basics Construction of Typical UK Walls: post ~ 1960Components of resistanceInternal surfacePlasterBlockCavityBrickExternal Surfacerbrick = / 1 = 0.11rblock = / = 0.76rplaster = / 0.7 = 0. 02rexternal = 0.055rinternal = 0.123rcavity =cavityplaster110 mmBlockBrickTotal resistance = = m2 oC W-1U – value = 1 / = 0.8 W m-2 oC-1.Or 50% of brick / cavity / brick wall.
9 19. Energy Conservation in Buildings: The Basics Construction of Typical UK Walls: post ~ 1960Components of resistanceInternal surfacePlasterBlockCavity insulationBrickExternal Surfacerbrick = / 1 = 0.11rblock = / = 0.76rplaster = / 0.7 = 0. 02rexternal = 0.055rinternal = 0.123rcavity insulation = 0.05/0.04 = 1.25Cavity insulationplaster110 mmBrickBlock50 mmTotal resistance = = m2 oC W-1U – value = 1 / = W m-2 oC-1.Or 50% of uninsulated brick / cavity / block wall.Brick / cavity / brick wall with insulation has U – Value = W m-2 oC-1
10 19. Energy Conservation in Buildings: The Basics U – values for non-standard constructions can be estimated in a similar wayU – values are tabulated for standard componentsU – value single glazing ~ – W m-2 oC-1U – value double glazing ~ – W m-2 oC-1Floors – typically 1.0 unless there is insulation.Roofs – depends on thickness of insulationUninsulated post war ~ 2.0 W m-2 oC-125 mm W m-2 oC-150 mm W m-2 oC-1100 mm W m-2 oC-1150 mm W m-2 oC-1200 mm W m-2 oC-1250 mm W m-2 oC-1There are diminishing returns after first ~ 100mm and other conservation strategies become more sensible both economically and in carbon savings.
11 19. Energy Conservation in Buildings: The Basics Ventilationin poorly insulated buildings may be only 25 – 30% of lossesIn well insulated buildings may be > 80% of total heat losses.>> ventilation heat recovery – e.g. ZICER.Ventilation occursThrough door/window openingThrough crack around windows / doors / floorsThrough fabric itselfThrough vents, chimneys etc.Adequate ventilation is required for healthCovered by specifying a particular number of air-changes per hour (ach) i.e. whole volume is changed in an hour.In a typical house 1 – 1.5 achIn a crowded lecture room may need 3 – 4 ach
12 19. Energy Conservation in Buildings: The Basics Ventilation: equivalent parameter to U-valuei.e. Proportional to temperature differenceVolume * ach * specific heat of air / W m-2 0C-1Specific heat: quantity of energy required to raise temperature of unit mass (volume) of material by 1 degree.For air, specific heat ~ J m-3Ventilation heat loss rate = volume * ach * 1300/3600= * ach * volume
13 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 Heat Loss CalculationsEnergy ManagementN.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnvН.К.Тови М.А., д-р технических наукEnergy Science Director CRed ProjectHSBC Director of Low Carbon Innovation13
14 20: Heat Loss / Heat Gain Calculations Five components to heat loss / gain parameterLosses throughFloorRoofWindowsWallsVentilationFabric Components: Area * U - valueVentilation: Volume * * achTotal Heat Loss / Heat Gain Rate (H)H = ΣArea * U–value of fabric components + Volume * * achHeat lost from a building in a cool climate (or heat gained in warm climate) must be replaced (removed) by the heating (cooling) appliance – e..g boiler (air-conditioner)Heat to be replaced (removed) = H * temperature difference (inside – outside)
15 20: Heat Loss / Heat Gain Calculations Design considerations:Heating/Cooling Capacity depends on internal and external temperatures.What should design temperature be?Internally – comfortable temperature – thermostat setting.Externally ?In heating mode if set too high, heating will not be supplied in extreme conditions.But extra cost is often implied.Design External temperature in UK for heating –1oCIn more extreme parts -30C is sometimes selectedHeavy weight buildings do store heat to allow for some carry over to colder conditions.Heating appliances usually come in standard sizes – size to next size above requirement.
16 20: Heat Loss / Heat Gain Calculations Design considerations:Hot water heating is often provided by same sourceProvides an extra buffer for peak heating demand.Heating / Cooling must be designed to cope with peak design demand.Annual Energy ConsumptionIncidental gains arise fromBody heatLightingHot water useAppliance useSolar gainDecrease / Increase overall annual heating (cooling) energy consumption – typically by several degrees.
17 20: Heat Loss / Heat Gain Annual Energy Requirements If incidental gains from all sources amount to 2250 watts, and the heat loss rate is 500 W C-1.Free temperature rise from incidental gains = / 500 = 4.5oCIf thermostat is set at 20 oC, No heating is needed until internal temperature falls below 20 – 4.5 = 15.5 oC.15.5oC is the neutral/base/or balance temperature.In UK and USA and used internationally the balance temperature for heating is on average 15.5oC (60oF).Each building is different and for accurate analysis, corrections must be applied.To allow rapid assessment of annual energy consumptionHeating Degree Days (HDD)……Cooling Degree Days (CDD)There appears to be no standard for the base temperature for Cooling Degree Days but UKCIP02 uses 22oC
18 20: Heat Loss / Heat Gain: Degree Days Degree Days are an indirect measure of how cold or how warm a given period is.Used for estimating annual energy consumption.Heating Degree DaysFor every 1oC MEAN temperature on a particular day is below base temperature we add 1.For 10oC we add = 5.5For -1oC we add – (-1) = 16.5For -10oC we add – (-10) = 26.5For days when MEAN temperatures above base temperature we do not add anything.Total Degree Days over a period is sum of all individual daysGives approximate estimate – see shaded box in hand out for more accurate method.Monthly Degree Days are published at
19 20: Heat Loss / Heat Gain: Degree Days Annual Degree Days20 year average –20 year average –20 year average20 year averageJanFebMarAprMayJunJulAugSepOctNovDec37433629122814572366915726634133730327221212865333162143258338Example: 1Heat Loss Rate is WoC-1What is estimated energy consumption for heating in January to March based on latest 20 year data= * ( ) * = GJOr 450 * ( ) * 24 / = kWh86400 is seconds in a day24 is hours in a day
20 20: Heat Loss / Heat Gain: Dynamic Considerations 1 Boiler Output for a house during early January 1985.2015105-5-1010987654321Temperature oCBoiler Output (kW)Hours
21 20: Heat Loss / Heat Gain: Dynamic Considerations 2 If no heating is provided and mean external temperature is 20oCInternal temperature has a much lower amplitude and lags by several hoursCan be used in effective management
22 20: Heat Loss / Heat Gain: Dynamic Considerations 3 In morning period, boiler is full on during period, but throttles back during evening period
23 20: Heat Loss / Heat Gain: Dynamic Considerations 4 With time switching – larger boiler is required to get temperature to acceptable levels
24 20: Heat Loss / Heat Gain: Worked Example 1 Large building in tropical country has sqm of single glazingElectricity consumption is as shownIf Cooling Degree Days are 3000, and coefficient of performance of air-conditioner is 2.5, what is annual energy consumption?Gradient of cooling line is 75 kW /oCAnnual consumption is 75 * 3000 * 24= 5400 MWhIf carbon factor is 800 kg /MWhCarbon emitted= 5400 * 800 / 1000= 4320 tonnesAppliance / Base Load demandCooling demand
25 20: Heat Loss / Heat Gain: Worked Example 1 Gradient of line = 75 kWoC-1actual heat gain rate = 75 *2.5 = 225 kW oC-1must allow for COP of air-conditionerInstalling double glazing reduces heat gain rate by:12000 * ( ) = 30 kW oC-1U – values before and after double glazingSaving in electricity with be 30 /2.5 = 12 kWoC-1Saving in electricity consumed= 12 *3000 * 24 = MWhcarbon saving= 864*800 / 1000 = tonnes
26 20: Heat Loss / Heat Gain: Worked Example 1 Annual electricity saved = 864 MWh - Annual carbon saved = tonnesMarginal cost is 740 Paise/Unit p per unit at Exchange Rate on 07April 2008Total saving in monetary terms would be864 * * = £80,594 per yearWith a life time of 30 years say, this represents a saving of £2.4 millionA total of MWh saved and tonnes of carbon dioxide. If ‘K’ glass (low emissivity glass were installed) savings would be around 50% largerData for India26
27 20: Heat Loss / Heat Gain: Worked Example 2 New house designed with heat loss rate of 0.2 kW oC-1Two optionsOil boiler - oil costs 45p/litre: calorific value 37 MJ/litreHeat Pump – electricity costs 4.5 per kWhExamine most cost effective option.Heat pump data as shown in graph.Capital costs:Oil Boiler £2000, Heat Pump £4000TemperatureCOP6.5312.54.0164.283.2
28 20: Heat Loss / Heat Gain: Worked Example 2 Analysis is best done in tabular formExternal Temperature (oC)COP from graphNumber of daysDifference from balance temperatureHeat Requirement (kWh)Requirement after allowing for COP (kWh)(1)(2)(3)(4)(5)(6)(7)Jan - Mar6.5390938881296Apr - Jun12.54911310.4327.6Jul - Sept164.292 NoHeating needed Oct - Dec83.27.533121035Total energy requirement8510.4Boiler efficiency90%Energy input boiler option as oil9456Total effective electrical input via heat pump2658.6Col (5) = 15.5 – col (2)Col (6) = 0.2 * col (5) * col (4) * 24Col (7) = col (6) / col (3)Oil required kWh = MJSo / 37 = 920 litres are neededCost of oil = 920 * 0.45 = £414Heat Loss Rate for houseCost of electricity for heat pump = * 0.045= £ and an annual saving of £294.36
29 20: Heat Loss / Heat Gain: Worked Example 2 Annual saving in energy costs = £294.36At 5% discount rate, cummulative discount factor over 10 years isSo the discounted savings over life of project= * = £This is greater than the capital cost difference of £2000 (i.e (£ £2000), there will be a net saving of £ over the project life and the heat pump scheme is the more attractive financially.
30 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 21. Energy ManagementN.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnvН.К.Тови М.А., д-р технических наукEnergy Science Director CRed ProjectHSBC Director of Low Carbon Innovation30
31 HSBC Sustainability Audit Report Norwich Branch and Office Nearly double early 2005 level33% higher than historic levelCost increase ~ £ £12000 paCO2 increase ~ 100 tonne paAppears to be associated with malfunction of air conditionerElectricity Consumption improves in 2004Implementation of conservation measures - Low Energy Lighting phased over autumnSudden jump in consumptionCRedcarbon reduction
32 HSBC Sustainability Audit Report Norwich Branch and Office Local Authority OfficesNorwichGreatYarmouthKingsLynnNaturallyventilatedAir-conditionedGoodPractice5497289125140Typical85178ElectricitykWh/m2kWh/employeeGreat Yarmouth125.34695Kings Lynn140.03226Norwich289.43817CRedcarbon reduction
33 HSBC Sustainability Audit Report Norwich Branch and Office ElectricityCarbon DioxidekWh/sqmkWh/employeekg/sqmtonnes/Great Yarmouth125.3469589.23.34Kings Lynn140.032261052.43Norwich289.438171188.22.48Annual Household consumption of Electricity in Norwich kWhCRedcarbon reduction
34 Annual energy consumption Basic analysisAim: Assess overall energy performance of buildingNormalise to a standard time periodAssess variation with external temperaturePredictionAim set targets for energy consumption following improvementsIssues to addressConvert all units to GJ or kWhfor GJ multiply heat loss rate by Degree Days and no of seconds in a day (86400).for kWh multiply heat loss rate by Degree Days and no of hours in a day (24).
35 Analysis of heating requirements Degree day methodQuickerOil & coal heating difficult – general estimates of consumptionMean temperature methodMore accuratePlot mean consumption against mean external temperature
36 Total Energy = W + H*degree days*86400 Degree day methodTwo component partsTemperature relatedIndependent of temperatureHot water & cooking if by gasTotal Energy = W + H*degree days*86400W energy for hot water + cooking (gas)H is heat loss rate for the homeTwo unknowns W & H,Know degree days & energy consumption in two different periods of yearEstimate heat loss & steady energy requirement
37 Degree day method - example Energy consumption 2 successive quarters: & GJCorresponding degree days: and 500Total Energy consumed = W + H * degree days*864001100 * H * W = (1)500 * H * W = (2)Simultaneous equations (subtract 2 from 1)H = (31.76 – 18.80) * 109 = 250 Watts( )*86400Substitute for H in either equation to get WW = * * 250 * 86400= 8 * 109 = 8GJH - heat lossW - hot water
38 Once H & W have been calculated Degree day methodOnce H & W have been calculatedPerformance for subsequent quarters can be estimatedIf degree days for 3rd quarter = 400Consumption predicted to be400 * 250 * * 109 = GJH WIf actual consumption is 17.5 GJ then energy has been wasted
39 HSBC Sustainability Audit Report Norwich Branch and Office Gas Consumption is relatively lowExtensive use of fixed and free standing electrical heatersDouble carbon dioxide emission when heating with electricityCRedcarbon reduction
40 Analysis of lighting (non-electrically heated building) Electricity consumption varies during year.Base load for appliances and refrigerationVariable lighting Load depending on number of hours required for lightingIntercept is base load (A)Gradient is Lighting Load Parameter LLightingAppliances and RefrigerationInstalling Low Energy Lighting will decrease gradient by a factor 5Installing more efficient appliances will reduce base loadInstalling both measures will reduce both L and A
41 Mean temperature method - similar to degree Day Method (non electrical heating) Plot the mean consumption over a specific period against mean external temperatureGenerally more accurate than Monthly Degree Day Method as short term variations can be explored.With Daily readings, variations with day of week can be explorede.g. Weekend –shut down, do Mondays see extra consumptionTwo parts to graphHeating part represented by sloping lineBase load for cooking/hot water by horizontal line.Do not merely do a regression line
42 Efficiencies of all boilers are available on SEDBUK Database Mean temperature method - similar to degree Day Method (non electrical heating)Gradient of line is heat loss rateAdjust for boiler efficiencyMultiply by to get heat loss rate– e.g.70% for non condensing boiler,90% for condensing boiler300% for heat pumpEfficiencies of all boilers are available on SEDBUK DatabaseTwo parts to graphHeating part represented by sloping lineBase load for cooking/hot water by horizontal line.42
43 Analysis of lighting (non-electrically heated building) Data before conservationIntercept = appliance and refrigeration load (A).Gradient is Lighting load (L)Low energy lighting installed – should reduce L by 80%Actual data after installationSuggests that improvement of 80% is not achieved.If actual data are shown as blue line – improvements in energy management have taken place – or replacement of appliances with more energy efficient ones.
44 Analysis of heating & lighting in electrically heated building A – appliance LoadW – water heating LoadH – heat loss parameterL – lighting Load parameterMore complex as both H & L are unknownCombine A & W to give overall appliance + hot water load (A*)E = (degree days * H + lighting hours * L) * A*Where E = energy consumption3 unknowns – H, L & AIf we have data for 3 quartersEstimate values for H, L & A by solving 3 simultaneous equationsIf appliance load is known calculation is easier
45 Cumulative deviation method ExcessNo energy conservation – horizontal lineWinter following improved insulationSummer – no savings – heat conservation onlyWinter – parallel to 2Summer - improved management of hot waterShould be (4) + (5)but gradient is in fact less- energy conservation performance has got worseTime+1+2Cumulative Saving++34Saving+56+
46 Actual data from large residential building in Shanghai in 2006 Fudan University – Twin TowerGradients of lines MWh per deg C per month heatingMWh per deg C per month coolingAssume 720 hours in a month kWoC-1 heating and 72.4 kWoC-1 cooling=43.05/ = 52.12/720
47 Analysis of Energy Data - Fudan University – Twin Towers Baseline consumptionNeutral Temperature oCBaseline consumption MWh/monthAnnual Heating Demand MWhAnnual Cooling Demand MWhAnnual Baseline (Functional) Demand MWhFunctional Energy Use is 35.5% of total energy use.