1 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 19. Energy Conservation in Buildings: The Basics 20. Heat Loss Calculations 21. Energy.

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1 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 19. Energy Conservation in Buildings: The Basics 20. Heat Loss Calculations 21. Energy Management N.K. Tovey ( ) M.A, PhD, CEng, MICE, CEnv .. .., - Energy Science Director CRed Project HSBC Director of Low Carbon Innovation 1 Lecture 1 Lecture 2 Lecture 3 Slide 2 2 19. Energy Conservation in Buildings: The Basics Intrinsic Energy Use provision of comfortable thermal environment Functional Energy Use energy 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 percentage In a well insulated building, functional energy can be the dominant use representing over 50% in ZICER. Slide 3 3 19. Energy Conservation in Buildings: The Basics Heat Loss / Heat Gain Three forms of heat transfer Conduction Radiation Convection For buildings - conductive losses are main issue to address in heat loss/heat gain. Q d A T1T1 T2T2 T1T1 T2T2 Temperature Profile Heat Flow is proportional to temperature difference Slide 4 4 Construction of Typical UK Walls: Solid Walls most houses built pre-war. 19. Energy Conservation in Buildings: The Basics Brick plaster In addition to resistance of brick and plaster there is: Internal surface resistance External surface resistance Slide 5 5 Construction of Typical UK Walls: Solid Walls most houses built pre-war. 19. Energy Conservation in Buildings: The Basics Brick U-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 = r brick + r plaster + r int + r ext But resistance = as A = 1.0 k for brick ~ 1.0 W m -1 o C -1 r brick = 0.22 / 1.0 = 0.22 m 2 o C W -1 For plaster k = 0.7 W m -1 o C -1 so r plaster = 0.013/0.7 = 0.02 m 2 o C W -1 Total resistance = 0.22 + 0.02 + 0.123 + 0.055 = 0.418 m2 oC W-1 So U- value = 1 / 0.418 = 2.39 W m -2 o C -1 13 mm 220 mm Slide 6 6 Construction of Typical UK Walls: post war 19. Energy Conservation in Buildings: The Basics Brick cavity plaster Cavities provide an extra air-space and hence extra resistance to heat flow. Slide 7 7 Construction of Typical UK Walls: post war 19. Energy Conservation in Buildings: The Basics Brick cavity plaster 110 mm Components of resistance Internal surface Plaster Brick Cavity Brick External Surface r internal = 0.123 r plaster = 0.013 / 0.7 = 0. 02 r brick = 0.11 / 1 = 0.11 r cavity = 0. 18 r brick = 0.11 / 1 = 0.11 r external = 0.055 Total resistance = 0.123 + +0.02 + 0.11 + 0.18 + 0.11 + 0.055 = 0.598 m 2 o C W -1 U value = 1 / 0.598 = 1.67 W m -2 o C -1. Or 70% of solid wall. Slide 8 8 Construction of Typical UK Walls: post ~ 1960 19. Energy Conservation in Buildings: The Basics Brick cavity 110 mm Components of resistance Internal surface Plaster Block Cavity Brick External Surface r brick = 0.11 / 1 = 0.11 r block = 0.11 / 0.14 = 0.76 r plaster = 0.013 / 0.7 = 0. 02 r external = 0.055 r internal = 0.123 r cavity = 0. 18 Total resistance = 0.123 + +0.02 + 0.76 + 0.18 + 0.11 + 0.055 = 1.248 m 2 o C W -1 U value = 1 / 1.248 = 0.8 W m -2 o C -1. Or 50% of brick / cavity / brick wall. Block plaster Slide 9 9 Construction of Typical UK Walls: post ~ 1960 19. Energy Conservation in Buildings: The Basics Brick Cavity insulation 110 mm Components of resistance Internal surface Plaster Block Cavity insulation Brick External Surface r brick = 0.11 / 1 = 0.11 r block = 0.11 / 0.14 = 0.76 r plaster = 0.013 / 0.7 = 0. 02 r external = 0.055 r internal = 0.123 r cavity insulation = 0.05/0.04 = 1.25 Total resistance = 0.123 + +0.02 + 0.76 + 1.25 + 0.11 + 0.055 = 2.318 m 2 o C W -1 U value = 1 / 2.318 = 0.43 W m -2 o C -1. Or 50% of uninsulated brick / cavity / block wall. Brick / cavity / brick wall with insulation has U Value = 0.59 W m -2 o C -1 Block plaster 50 mm Slide 10 10 U values for non-standard constructions can be estimated in a similar way U values are tabulated for standard components U value single glazing ~ 5.0 5.7 W m -2 o C -1 U value double glazing ~ 2.5 2.86 W m -2 o C -1 Floors typically 1.0 unless there is insulation. Roofs depends on thickness of insulation Uninsulated post war ~ 2.0 W m -2 o C -1 25 mm - 0.89 W m -2 o C -1 50 mm - 0.57 W m -2 o C -1 100 mm - 0.34 W m -2 o C -1 150 mm - 0.25 W m -2 o C -1 200 mm - 0.18 W m -2 o C -1 250 mm - 0.15 W m -2 o C -1 19. Energy Conservation in Buildings: The Basics There are diminishing returns after first ~ 100mm and other conservation strategies become more sensible both economically and in carbon savings. Slide 11 11 Ventilation in poorly insulated buildings may be only 25 30% of losses In well insulated buildings may be > 80% of total heat losses. >> ventilation heat recovery e.g. ZICER. Ventilation occurs Through door/window opening Through crack around windows / doors / floors Through fabric itself Through vents, chimneys etc. Adequate ventilation is required for health Covered 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 ach In a crowded lecture room may need 3 4 ach 19. Energy Conservation in Buildings: The Basics Slide 12 12 Ventilation: equivalent parameter to U-value i.e. Proportional to temperature difference Volume * ach * specific heat of air / 3600 W m -2 0 C -1 Specific heat: quantity of energy required to raise temperature of unit mass (volume) of material by 1 degree. For air, specific heat ~ 1300 J m -3 Ventilation heat loss rate = volume * ach * 1300/3600 = 0.361 * ach * volume 19. Energy Conservation in Buildings: The Basics Slide 13 13 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 20. Heat Loss Calculations 21. Energy Management N.K. Tovey ( ) M.A, PhD, CEng, MICE, CEnv .. .., - Energy Science Director CRed Project HSBC Director of Low Carbon Innovation 13 Slide 14 14 Five components to heat loss / gain parameter Losses through Floor Roof Windows Walls Ventilation 20: Heat Loss / Heat Gain Calculations Ventilation: Volume * 0.361 * ach Fabric Components: Area * U - value Total Heat Loss / Heat Gain Rate (H) H = Area * Uvalue of fabric components + Volume * 0.361 * ach Heat 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) Slide 15 15 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 1 o C In more extreme parts -3 0 C is sometimes selected Heavy 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. 20: Heat Loss / Heat Gain Calculations Slide 16 16 Design considerations: Hot water heating is often provided by same source Provides an extra buffer for peak heating demand. Heating / Cooling must be designed to cope with peak design demand. Annual Energy Consumption Incidental gains arise from Body heat Lighting Hot water use Appliance use Solar gain Decrease / Increase overall annual heating (cooling) energy consumption typically by several degrees. 20: Heat Loss / Heat Gain Calculations Slide 17 17 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 = 2250 / 500 = 4.5 o C If thermostat is set at 20 o C, No heating is needed until internal temperature falls below 20 4.5 = 15.5 o C. 15.5 o C is the neutral/base/or balance temperature. In UK and USA and used internationally the balance temperature for heating is on average 15.5 o C (60 o F). Each building is different and for accurate analysis, corrections must be applied. To allow rapid assessment of annual energy consumption Heating Degree Days (HDD)Cooling Degree Days (CDD) There appears to be no standard for the base temperature for Cooling Degree Days but UKCIP02 uses 22 o C 20: Heat Loss / Heat Gain Annual Energy Requirements Slide 18 18 Degree Days are an indirect measure of how cold or how warm a given period is. Used for estimating annual energy consumption. Heating Degree Days For every 1 o C MEAN temperature on a particular day is below base temperature we add 1. For 10 o C we add 15.5 -10 = 5.5 For -1 o C we add 15.5 (-1) = 16.5 For -10 o C we add 15.5 (-10) = 26.5 For days when MEAN temperatures above base temperature we do not add anything. Total Degree Days over a period is sum of all individual days Gives approximate estimate see shaded box in hand out for more accurate method. Monthly Degree Days are published at www.vesma.com 20: Heat Loss / Heat Gain: Degree Days Slide 19 19 Annual Degree Days 20 year average 1959 1978 - 2430 20 year average 1979 1988 - 2351 20 year average 1988 - 2007 - 2182 20: Heat Loss / Heat Gain: Degree Days 20 year averageJanFebMarAprMayJunJulAugSepOctNovDec 1979 - 19983743362912281457236 69157266341 1988 - 200733730327221212865333162143258338 Example: 1 Heat Loss Rate is 450 W o C -1 What is estimated energy consumption for heating in January to March based on latest 20 year data = 450 * ( 337 + 303 + 272) * 86400 = 35.46 GJ Or 450 * (337 + 303 +272) * 24 / 1000 = 9850 kWh 86400 is seconds in a day 24 is hours in a day Slide 20 20 20: Heat Loss / Heat Gain: Dynamic Considerations 1 120 132 144 156 168 180 192 204 216 228 240 Hours Boiler Output for a house during early January 1985. 20 15 10 5 0 -5 -10 Temperature o C 10 9 8 7 6 5 4 3 2 1 0 Boiler Output (kW) Slide 21 21 20: Heat Loss / Heat Gain: Dynamic Considerations 2 If no heating is provided and mean external temperature is 20 o C Internal temperature has a much lower amplitude and lags by several hours Can be used in effective management Slide 22 22 20: Heat Loss / Heat Gain: Dynamic Considerations 3 In morning period, boiler is full on during period, but throttles back during evening period Slide 23 23 20: Heat Loss / Heat Gain: Dynamic Considerations 4 With time switching larger boiler is required to get temperature to acceptable levels Slide 24 24 Large building in tropical country has 12000 sqm of single glazing Electricity consumption is as shown If Cooling Degree Days are 3000, and coefficient of performance of air- conditioner is 2.5, what is annual energy consumption? 20: Heat Loss / Heat Gain: Worked Example 1 Gradient of cooling line is 75 kW / o C Annual consumption is 75 * 3000 * 24 = 5400 MWh If carbon factor is 800 kg /MWh Carbon emitted = 5400 * 800 / 1000 = 4320 tonnes Cooling demand Appliance / Base Load demand Slide 25 25 Gradient of line = 75 kW o C -1 actual heat gain rate = 75 *2.5 = 225 kW o C -1 must allow for COP of air-conditioner Installing double glazing reduces heat gain rate by: 12000 * ( 5 - 2.5) = 30 kW o C -1 U values before and after double glazing Saving in electricity with be 30 /2.5 = 12 kW o C -1 Saving in electricity consumed = 12 *3000 * 24 = 864 MWh carbon saving = 864*800 / 1000 = 691.2 tonnes 20: Heat Loss / Heat Gain: Worked Example 1 Slide 26 26 Annual electricity saved = 864 MWh - Annual carbon saved = 691.2 tonnes Marginal cost is 740 Paise/Unit - 9.328p per unit at Exchange Rate on 07April 2008 Total saving in monetary terms would be 864 * 1000 * 0.09328 = 80,594 per year With a life time of 30 years say, this represents a saving of 2.4 million A total of 25920 MWh saved and 20700 tonnes of carbon dioxide. If K glass (low emissivity glass were installed) savings would be around 50% larger 20: Heat Loss / Heat Gain: Worked Example 1 Data for India Slide 27 27 New house designed with heat loss rate of 0.2 kW o C -1 Two options Oil boiler - oil costs 45p/litre: calorific value 37 MJ/litre Heat Pump electricity costs 4.5 per kWh Examine most cost effective option. Heat pump data as shown in graph. 20: Heat Loss / Heat Gain: Worked Example 2 TemperatureCOP 6.53 12.54.0 164.2 83.2 Capital costs: Oil Boiler 2000, Heat Pump 4000 Slide 28 20: Heat Loss / Heat Gain: Worked Example 2 External Temperature ( o C) COP from graph Number of days Difference from balance temperature Heat Requirement (kWh) Requirement after allowing for COP (kWh) (1)(2)(3)(4)(5)(6)(7) Jan - Mar6.5390938881296 Apr - Jun12.549131310.4327.6 Jul - Sept164.292 NoHeating needed Oct - Dec83.2927.533121035 Total energy requirement8510.4 Boiler efficiency90% Energy input boiler option as oil 9456 Total effective electrical input via heat pump 2658.6 Col (5) = 15.5 col (2) Col (6) = 0.2 * col (5) * col (4) * 24 Col (7) = col (6) / col (3) Oil required 9456 kWh = 34042 MJ So 34042 / 37 = 920 litres are needed Cost of oil = 920 * 0.45 = 414 Cost of electricity for heat pump = 2658.5 * 0.045 = 119.64 and an annual saving of 294.36 Analysis is best done in tabular form Heat Loss Rate for house Slide 29 20: Heat Loss / Heat Gain: Worked Example 2 Annual saving in energy costs = 294.36 At 5% discount rate, cummulative discount factor over 10 years is 8.721735 So the discounted savings over life of project = 8.721735 * 294.36 = 2567.36 This is greater than the capital cost difference of 2000 (i.e (4000 - 2000), there will be a net saving of 567.36 over the project life and the heat pump scheme is the more attractive financially. Slide 30 30 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 21. Energy Management N.K. Tovey ( ) M.A, PhD, CEng, MICE, CEnv .. .., - Energy Science Director CRed Project HSBC Director of Low Carbon Innovation 30 Slide 31 CRed carbon reduction Electricity Consumption improves in 2004 Implementation of conservation measures - Low Energy Lighting phased over autumn Sudden jump in consumption Nearly double early 2005 level 33% higher than historic level Cost increase ~ 10000 - 12000 pa CO 2 increase ~ 100 tonne pa Appears to be associated with malfunction of air conditioner HSBC Sustainability Audit Report Norwich Branch and Office Slide 32 CRed carbon reduction Local Authority Offices NorwichGreat Yarmouth Kings Lynn Naturally ventilated Air- conditioned Good Practice 5497 289125140 Typical85178 Electricity kWh/m 2 kWh/employee Great Yarmouth125.34695 Kings Lynn140.03226 Norwich289.43817 HSBC Sustainability Audit Report Norwich Branch and Office Slide 33 CRed carbon reduction ElectricityCarbon Dioxide kWh/sqm kWh/ employee kg/sqm tonnes/ employee Great Yarmouth 125.3469589.23.34 Kings Lynn140.032261052.43 Norwich289.438171188.22.48 Annual Household consumption of Electricity in Norwich 3720 kWh Slide 34 34 Annual energy consumption Basic analysis Aim: Assess overall energy performance of building Normalise to a standard time period Assess variation with external temperature Prediction Aim set targets for energy consumption following improvements Issues to address Convert all units to GJ or kWh for 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). Slide 35 35 Analysis of heating requirements Degree day method Quicker Oil & coal heating difficult general estimates of consumption Mean temperature method More accurate Plot mean consumption against mean external temperature Slide 36 36 Degree day method Two component parts Temperature related Independent of temperature Hot water & cooking if by gas Total Energy = W + H*degree days*86400 W energy for hot water + cooking (gas) H is heat loss rate for the home Two unknowns W & H, Know degree days & energy consumption in two different periods of year Estimate heat loss & steady energy requirement Slide 37 37 Degree day method - example Energy consumption 2 successive quarters: 31.76 & 18.80 GJ Corresponding degree days: 1100 and 500 Total Energy consumed = W + H * degree days*86400 1100 * H * 86400 + W = 31.76 (1) 500 * H * 86400 + W = 18.80 (2) Simultaneous equations (subtract 2 from 1) H = (31.76 18.80) * 10 9 = 250 Watts (1100-500)*86400 Substitute for H in either equation to get W W = 31.76 * 10 9 - 1100 * 250 * 86400 = 8 * 10 9 = 8GJ H - heat loss W - hot water Slide 38 38 Degree day method Once H & W have been calculated Performance for subsequent quarters can be estimated If degree days for 3 rd quarter = 400 Consumption predicted to be 400 * 250 * 86400 + 8 * 10 9 = 16.64 GJ H W If actual consumption is 17.5 GJ then energy has been wasted Slide 39 Gas Consumption is relatively low Extensive use of fixed and free standing electrical heaters Double carbon dioxide emission when heating with electricity CRed carbon reduction HSBC Sustainability Audit Report Norwich Branch and Office Slide 40 40 Analysis of lighting (non-electrically heated building) Electricity consumption varies during year. Base load for appliances and refrigeration Variable lighting Load depending on number of hours required for lighting Intercept is base load (A) Gradient is Lighting Load Parameter L Appliances and Refrigeration Lighting Installing Low Energy Lighting will decrease gradient by a factor 5 Installing more efficient appliances will reduce base load Installing both measures will reduce both L and A Slide 41 41 Mean temperature method - similar to degree Day Method (non electrical heating ) Plot the mean consumption over a specific period against mean external temperature Generally more accurate than Monthly Degree Day Method as short term variations can be explored. With Daily readings, variations with day of week can be explored e.g. Weekend shut down, do Mondays see extra consumption Two parts to graph Heating part represented by sloping line Base load for cooking/hot water by horizontal line. Do not merely do a regression line Slide 42 42 Mean temperature method - similar to degree Day Method (non electrical heating ) Gradient of line is heat loss rate Adjust for boiler efficiency Multiply by to get heat loss rate e.g. 70% for non condensing boiler, 90% for condensing boiler 300% for heat pump Efficiencies of all boilers are available on SEDBUK Database www.sedbuk.com/index.htm Two parts to graph Heating part represented by sloping line Base load for cooking/hot water by horizontal line. Slide 43 43 Data before conservation Intercept = appliance and refrigeration load (A). Gradient is Lighting load (L) Low energy lighting installed should reduce L by 80% Actual data after installation Suggests 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. Analysis of lighting (non-electrically heated building) Slide 44 44 Analysis of heating & lighting in electrically heated building A appliance Load W water heating Load H heat loss parameter L lighting Load parameter More complex as both H & L are unknown Combine A & W to give overall appliance + hot water load (A*) E = (degree days * H + lighting hours * L) * 86400 + A* Where E = energy consumption 3 unknowns H, L & A If we have data for 3 quarters Estimate values for H, L & A by solving 3 simultaneous equations If appliance load is known calculation is easier Slide 45 45 6 5 4 1 2 3 Cumulative deviation method + + + + + + + + + + + + + + + + + Time Saving Excess + + + + + + + + + + + + + + + 1.No energy conservation horizontal line 2.Winter following improved insulation 3.Summer no savings heat conservation only 4.Winter parallel to 2 5.Summer - improved management of hot water 6.Should be (4) + (5) but gradient is in fact less - energy conservation performance has got worse Cumulative Saving Slide 46 Actual data from large residential building in Shanghai in 2006 Fudan University Twin Tower Gradients of lines 43.05 MWh per deg C per month heating 52.12 MWh per deg C per month cooling Assume 720 hours in a month 59.8 kW o C -1 heating and 72.4 kW o C -1 cooling =43.05/720 = 52.12/720 Slide 47 Neutral Temperature - 17.5 o C Baseline consumption - 192.0 MWh/month Annual Heating Demand - 1675 MWh Annual Cooling Demand - 2515 MWh Annual Baseline (Functional) Demand - 2304 MWh Functional Energy Use is 35.5% of total energy use. Baseline consumption Analysis of Energy Data - Fudan University Twin Towers