NBS-M016 Contemporary Issues in Climate Change and Energy 2010

1. N.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnv Н.К.Тови М.А., д-р технических наук Energy Science DirectorCRedProject HSBC Director of Low Carbon Innovation Lecture 1 Lecture 3 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 • Energy Conservation in Buildings: The Basics • Heat Loss Calculations • Energy Management Lecture 2 1

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.

3. 19. Energy Conservation in Buildings: The Basics d Q T1 T2 Temperature Profile • 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. Heat Flow is proportional to temperature difference A T1 T2

4. 19. Energy Conservation in Buildings: The Basics plaster Brick • Construction of Typical UK Walls: Solid Walls – most houses built pre-war. In addition to resistance of brick and plaster there is: Internal surface resistance External surface resistance

5. 19. Energy Conservation in Buildings: The Basics 13 mm Brick 220 mm • Construction of Typical UK Walls: Solid Walls – most houses built pre-war. 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 = rbrick + rplaster + rint + rext But resistance = as A = 1.0 k for brick ~ 1.0 W m-1 oC-1 rbrick = 0.22 / 1.0 = 0.22 m2 oC W-1 For plaster k = 0.7 W m-1 oC-1 so rplaster = 0.013/0.7 = 0.02 m2oC 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 oC-1

6. 19. Energy Conservation in Buildings: The Basics cavity plaster Brick Brick • Construction of Typical UK Walls: post war Cavities provide an extra air-space and hence extra resistance to heat flow.

7. 19. Energy Conservation in Buildings: The Basics cavity plaster 110 mm Brick Brick • Construction of Typical UK Walls: post war • Components of resistance • Internal surface • Plaster • Brick • Cavity • Brick • External Surface • rinternal = 0.123 • rplaster = 0.013 / 0.7 = 0. 02 • rbrick = 0.11 / 1 = 0.11 • rcavity = 0. 18 • rbrick = 0.11 / 1 = 0.11 • rexternal = 0.055 Total resistance = 0.123 + +0.02 + 0.11 + 0.18 + 0.11 + 0.055 = 0.598 m2oC W-1 U – value = 1 / 0.598 = 1.67 W m-2 oC-1. Or 70% of solid wall.

8. 19. Energy Conservation in Buildings: The Basics cavity plaster 110 mm Block Brick • Construction of Typical UK Walls: post ~ 1960 • Components of resistance • Internal surface • Plaster • Block • Cavity • Brick • External Surface • rbrick = 0.11 / 1 = 0.11 • rblock = 0.11 / 0.14 = 0.76 • rplaster = 0.013 / 0.7 = 0. 02 • rexternal = 0.055 • rinternal = 0.123 • rcavity = 0. 18 Total resistance = 0.123 + +0.02 + 0.76 + 0.18 + 0.11 + 0.055 = 1.248 m2oC W-1 U – value = 1 / 1.248 = 0.8 W m-2 oC-1. Or 50% of brick / cavity / brick wall.

9. 19. Energy Conservation in Buildings: The Basics Cavity insulation plaster 110 mm Block 50 mm Brick • Construction of Typical UK Walls: post ~ 1960 • Components of resistance • Internal surface • Plaster • Block • Cavity insulation • Brick • External Surface • rbrick = 0.11 / 1 = 0.11 • rblock = 0.11 / 0.14 = 0.76 • rplaster = 0.013 / 0.7 = 0. 02 • rexternal = 0.055 • rinternal = 0.123 • rcavity insulation = 0.05/0.04 = 1.25 Total resistance = 0.123 + +0.02 + 0.76 + 1.25 + 0.11 + 0.055 = 2.318 m2oC W-1 U – value = 1 / 2.318 = 0.43 W m-2 oC-1. Or 50% of uninsulated brick / cavity / block wall. Brick / cavity / brick wall with insulation has U – Value = 0.59 W m-2oC-1

10. 19. Energy Conservation in Buildings: The Basics • 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 oC-1 • U – value double glazing ~ 2.5 – 2.86 W m-2 oC-1 • Floors – typically 1.0 unless there is insulation. • Roofs – depends on thickness of insulation • Uninsulated post war ~ 2.0 W m-2 oC-1 • 25 mm - 0.89 W m-2 oC-1 • 50 mm - 0.57 W m-2 oC-1 • 100 mm - 0.34 W m-2 oC-1 • 150 mm - 0.25 W m-2 oC-1 • 200 mm - 0.18 W m-2 oC-1 • 250 mm - 0.15 W m-2 oC-1 There 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 • 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

12. 19. Energy Conservation in Buildings: The Basics • Ventilation: equivalent parameter to U-value • i.e. Proportional to temperature difference Volume * ach * specific heat of air / 3600 W m-2 0C-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

13. N.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnv Н.К.Тови М.А., д-р технических наук Energy Science DirectorCRedProject HSBC Director of Low Carbon Innovation NBS-M016 Contemporary Issues in Climate Change and Energy 2010 • Heat Loss Calculations • Energy Management 13

14. 20: Heat Loss / Heat Gain Calculations Fabric Components: Area * U - value • Five components to heat loss / gain parameter • Losses through • Floor • Roof • Windows • Walls • Ventilation Ventilation: Volume * 0.361 * ach Total Heat Loss / Heat Gain Rate (H) H = ΣArea * U–value 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)

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 –1oC In more extreme parts -30C 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.

16. 20: Heat Loss / Heat Gain Calculations 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 annualheating (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 = 2250 / 500 = 4.5oC • If 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 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 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 Days • For every 1oC MEAN temperature on a particular day is below base temperature we add 1. For 10oC we add 15.5 -10 = 5.5 For -1oC we add 15.5 – (-1) = 16.5 For -10oC 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

19. 20: Heat Loss / Heat Gain: Degree Days • Annual Degree Days • 20 year average 1959 – 1978 - 2430 • 20 year average 1979 – 1988 - 2351 • 20 year average 1988 - 2007 - 2182 Example: 1 Heat Loss Rate is 450 WoC-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

20. 20: Heat Loss / Heat Gain: Dynamic Considerations 1 Boiler Output for a house during early January 1985. 10 9 8 7 6 5 4 3 2 1 0 20 15 10 5 0 -5 -10 Temperature oC Boiler Output (kW) 120 132 144 156 168 180 192 204 216 228 240 Hours

21. 20: Heat Loss / Heat Gain: Dynamic Considerations 2 If no heating is provided and mean external temperature is 20oC Internal temperature has a much lower amplitude and lags by several hours Can 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 Appliance / Base Load demand Cooling demand • 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? Gradient of cooling line is 75 kW /oC Annual consumption is 75 * 3000 * 24 = 5400 MWh If carbon factor is 800 kg /MWh Carbon emitted = 5400 * 800 / 1000 = 4320 tonnes

25. 20: Heat Loss / Heat Gain: Worked Example 1 • Gradient of line = 75 kWoC-1 actual heat gain rate = 75 *2.5 = 225 kW oC-1 must allow for COP of air-conditioner • Installing double glazing reduces heat gain rate by: 12000 * ( 5 - 2.5) = 30 kW oC-1 U – values before and after double glazing • Saving in electricity with be 30 /2.5 = 12 kWoC-1 • Saving in electricity consumed = 12 *3000 * 24 = 864 MWh • carbon saving = 864*800 / 1000 = 691.2 tonnes

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 26

27. 20: Heat Loss / Heat Gain: Worked Example 2 • New house designed with heat loss rate of 0.2 kW oC-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. Capital costs: Oil Boiler £2000, Heat Pump £4000

28. 20: Heat Loss / Heat Gain: Worked Example 2 Analysis is best done in tabular form 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 Heat Loss Rate for house

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.

30. N.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnv Н.К.Тови М.А., д-р технических наук Energy Science DirectorCRedProject HSBC Director of Low Carbon Innovation NBS-M016 Contemporary Issues in Climate Change and Energy 2010 21. Energy Management 30

31. CRed carbon reduction HSBC Sustainability Audit ReportNorwich Branch and Office Nearly double early 2005 level 33% higher than historic level Cost increase ~ £10000 - £12000 pa CO2 increase ~ 100 tonne pa Appears to be associated with malfunction of air conditioner Electricity Consumption improves in 2004 Implementation of conservation measures - Low Energy Lighting phased over autumn Sudden jump in consumption

32. CRed carbon reduction HSBC Sustainability Audit ReportNorwich Branch and Office

33. CRed carbon reduction HSBC Sustainability Audit ReportNorwich Branch and Office Annual Household consumption of Electricity in Norwich 3720 kWh

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).

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

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

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) * 109 = 250 Watts (1100-500)*86400 Substitute for H in either equation to get W W = 31.76 * 109 - 1100 * 250 * 86400 = 8 * 109 = 8GJ H - heat loss W - hot water

38. Degree day method Once H & W have been calculated • Performance for subsequent quarters can be estimated • If degree days for 3rd quarter = 400 • Consumption predicted to be • 400 * 250 * 86400 + 8 * 109 = 16.64 GJ H W • If actual consumption is 17.5 GJ then energy has been wasted

39. CRed carbon reduction HSBC Sustainability Audit ReportNorwich Branch and Office Gas Consumption is relatively low Extensive use of fixed and free standing electrical heaters Double carbon dioxide emission when heating with electricity

40. Analysis of lighting (non-electrically heated building) Lighting Appliances and Refrigeration • 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 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

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

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. 42

43. Analysis of lighting (non-electrically heated building) 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.

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

45. No energy conservation – horizontal line Winter following improved insulation Summer – no savings – heat conservation only Winter – parallel to 2 Summer - improved management of hot water Should be (4) + (5) but gradient is in fact less - energy conservation performance has got worse Cumulative deviation method + Time + + + + + + + Cumulative Saving + + + + + + + + + + + + + + + + + + + + + + + + 4 6 3 2 1 5 Excess Saving

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 kWoC-1 heating and 72.4 kWoC-1 cooling =43.05/720 = 52.12/720

47. Analysis of Energy Data - Fudan University – Twin Towers Baseline consumption Neutral Temperature - 17.5oC 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.