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**NBS-M016 Contemporary Issues in Climate Change and Energy 2010**

Energy Conservation in Buildings: The Basics Heat Loss Calculations Energy Management N.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnv Н.К.Тови М.А., д-р технических наук Energy Science Director CRed Project HSBC Director of Low Carbon Innovation Lecture 1 Lecture 3 Lecture 2 1

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**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.

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**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. d Heat Flow is proportional to temperature difference Q A T1 T2 T1 T2 Temperature Profile

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**19. Energy Conservation in Buildings: The Basics**

Construction of Typical UK Walls: Solid Walls – most houses built pre-war. plaster Brick In addition to resistance of brick and plaster there is: Internal surface resistance External surface resistance

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**19. Energy Conservation in Buildings: The Basics**

Construction of Typical UK Walls: Solid Walls – most houses built pre-war. 13 mm 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 = rbrick + rplaster + rint + rext But resistance = as A = 1.0 k for brick ~ 1.0 W m-1 oC-1 rbrick = / 1.0 = m2 oC W-1 220 mm For plaster k = 0.7 W m-1 oC-1 so rplaster = / = m2 oC W-1 Total resistance = = m2 oC W-1 So U- value = 1 / = W m-2 oC-1

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**19. Energy Conservation in Buildings: The Basics**

Construction of Typical UK Walls: post war cavity plaster Brick Brick Cavities provide an extra air-space and hence extra resistance to heat flow.

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**19. Energy Conservation in Buildings: The Basics**

Construction of Typical UK Walls: post war Components of resistance Internal surface Plaster Brick Cavity External Surface rinternal = 0.123 rplaster = / 0.7 = 0. 02 rbrick = / 1 = 0.11 rcavity = rexternal = 0.055 cavity plaster 110 mm Brick Brick Total resistance = = m2 oC W-1 U – value = 1 / = 1.67 W m-2 oC-1. Or 70% of solid wall.

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**19. Energy Conservation in Buildings: The Basics**

Construction of Typical UK Walls: post ~ 1960 Components of resistance Internal surface Plaster Block Cavity Brick External Surface rbrick = / 1 = 0.11 rblock = / = 0.76 rplaster = / 0.7 = 0. 02 rexternal = 0.055 rinternal = 0.123 rcavity = cavity plaster 110 mm Block Brick Total resistance = = m2 oC W-1 U – value = 1 / = 0.8 W m-2 oC-1. Or 50% of brick / cavity / brick wall.

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**19. Energy Conservation in Buildings: The Basics**

Construction of Typical UK Walls: post ~ 1960 Components of resistance Internal surface Plaster Block Cavity insulation Brick External Surface rbrick = / 1 = 0.11 rblock = / = 0.76 rplaster = / 0.7 = 0. 02 rexternal = 0.055 rinternal = 0.123 rcavity insulation = 0.05/0.04 = 1.25 Cavity insulation plaster 110 mm Brick Block 50 mm Total resistance = = m2 oC W-1 U – 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

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**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 ~ – W m-2 oC-1 U – value double glazing ~ – 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 W m-2 oC-1 50 mm W m-2 oC-1 100 mm W m-2 oC-1 150 mm W m-2 oC-1 200 mm W m-2 oC-1 250 mm 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.

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**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

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**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 / 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 ~ J m-3 Ventilation heat loss rate = volume * ach * 1300/3600 = * ach * volume

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**NBS-M016 Contemporary Issues in Climate Change and Energy 2010**

Heat Loss Calculations Energy Management N.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnv Н.К.Тови М.А., д-р технических наук Energy Science Director CRed Project HSBC Director of Low Carbon Innovation 13

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**20: Heat Loss / Heat Gain Calculations**

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

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**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.

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**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 annual heating (cooling) energy consumption – typically by several degrees.

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**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 = 4.5oC 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

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**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 = 5.5 For -1oC we add – (-1) = 16.5 For -10oC we add – (-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

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**20: Heat Loss / Heat Gain: Degree Days**

Annual Degree Days 20 year average – 20 year average – 20 year average 20 year average Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 374 336 291 228 145 72 36 69 157 266 341 337 303 272 212 128 65 33 31 62 143 258 338 Example: 1 Heat Loss Rate is WoC-1 What is estimated energy consumption for heating in January to March based on latest 20 year data = * ( ) * = GJ Or 450 * ( ) * 24 / = kWh 86400 is seconds in a day 24 is hours in a day

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**20: Heat Loss / Heat Gain: Dynamic Considerations 1**

Boiler Output for a house during early January 1985. 20 15 10 5 -5 -10 10 9 8 7 6 5 4 3 2 1 Temperature oC Boiler Output (kW) Hours

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**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

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**20: Heat Loss / Heat Gain: Dynamic Considerations 3**

In morning period, boiler is full on during period, but throttles back during evening period

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**20: Heat Loss / Heat Gain: Dynamic Considerations 4**

With time switching – larger boiler is required to get temperature to acceptable levels

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**20: Heat Loss / Heat Gain: Worked Example 1**

Large building in tropical country has 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 Appliance / Base Load demand Cooling demand

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**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 * ( ) = 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 = MWh carbon saving = 864*800 / 1000 = tonnes

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**20: Heat Loss / Heat Gain: Worked Example 1**

Annual electricity saved = 864 MWh - Annual carbon saved = tonnes Marginal cost is 740 Paise/Unit p per unit at Exchange Rate on 07April 2008 Total saving in monetary terms would be 864 * * = £80,594 per year With a life time of 30 years say, this represents a saving of £2.4 million A total of MWh saved and tonnes of carbon dioxide. If ‘K’ glass (low emissivity glass were installed) savings would be around 50% larger Data for India 26

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**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 Temperature COP 6.5 3 12.5 4.0 16 4.2 8 3.2

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**20: Heat Loss / Heat Gain: Worked Example 2**

Analysis is best done in tabular form External Temperature (oC) 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 - Mar 6.5 3 90 9 3888 1296 Apr - Jun 12.5 4 91 1310.4 327.6 Jul - Sept 16 4.2 92 No Heating needed Oct - Dec 8 3.2 7.5 3312 1035 Total energy requirement 8510.4 Boiler efficiency 90% 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 kWh = MJ So / 37 = 920 litres are needed Cost of oil = 920 * 0.45 = £414 Heat Loss Rate for house Cost of electricity for heat pump = * 0.045 = £ and an annual saving of £294.36

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**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 So 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.

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**NBS-M016 Contemporary Issues in Climate Change and Energy 2010**

21. Energy Management N.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnv Н.К.Тови М.А., д-р технических наук Energy Science Director CRed Project HSBC Director of Low Carbon Innovation 30

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**HSBC Sustainability Audit Report Norwich Branch and Office**

Nearly double early 2005 level 33% higher than historic level Cost increase ~ £ £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 CRed carbon reduction

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**HSBC Sustainability Audit Report Norwich Branch and Office**

Local Authority Offices Norwich Great Yarmouth Kings Lynn Naturally ventilated Air- conditioned Good Practice 54 97 289 125 140 Typical 85 178 Electricity kWh/m2 kWh/employee Great Yarmouth 125.3 4695 Kings Lynn 140.0 3226 Norwich 289.4 3817 CRed carbon reduction

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**HSBC Sustainability Audit Report Norwich Branch and Office**

Electricity Carbon Dioxide kWh/sqm kWh/ employee kg/sqm tonnes/ Great Yarmouth 125.3 4695 89.2 3.34 Kings Lynn 140.0 3226 105 2.43 Norwich 289.4 38171 188.2 2.48 Annual Household consumption of Electricity in Norwich kWh CRed carbon reduction

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**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).

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**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

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**Total Energy = W + H*degree days*86400**

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

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**Degree day method - example**

Energy consumption 2 successive quarters: & GJ Corresponding degree days: and 500 Total Energy consumed = W + H * degree days*86400 1100 * H * W = (1) 500 * H * W = (2) Simultaneous equations (subtract 2 from 1) H = (31.76 – 18.80) * 109 = 250 Watts ( )*86400 Substitute for H in either equation to get W W = * * 250 * 86400 = 8 * 109 = 8GJ H - heat loss W - hot water

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**Once H & W have been calculated **

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 * * 109 = GJ H W If actual consumption is 17.5 GJ then energy has been wasted

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**HSBC Sustainability Audit Report Norwich 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 CRed carbon reduction

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**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 Lighting Appliances and Refrigeration 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

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**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

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**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 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 Two parts to graph Heating part represented by sloping line Base load for cooking/hot water by horizontal line. 42

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**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.

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**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) * 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

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**Cumulative deviation method**

Excess 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 Time + 1 + 2 Cumulative Saving + + 3 4 Saving + 5 6 +

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**Actual data from large residential building in Shanghai in 2006**

Fudan University – Twin Tower Gradients of lines MWh per deg C per month heating MWh per deg C per month cooling Assume 720 hours in a month kWoC-1 heating and 72.4 kWoC-1 cooling =43.05/ = 52.12/720

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**Analysis of Energy Data - Fudan University – Twin Towers**

Baseline consumption Neutral Temperature oC Baseline consumption MWh/month Annual Heating Demand MWh Annual Cooling Demand MWh Annual Baseline (Functional) Demand MWh Functional Energy Use is 35.5% of total energy use.

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