Presentation on theme: "1 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 19. Energy Conservation in Buildings: The Basics 20. Heat Loss Calculations 21. Energy."— Presentation transcript:
1 NBS-M016 Contemporary Issues in Climate Change and Energy Energy Conservation in Buildings: The Basics 20. Heat Loss Calculations 21. Energy Management N.K. Tovey ( ) 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 1 Lecture 1 Lecture 2 Lecture 3
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 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
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
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 = = m2 oC W-1 So U- value = 1 / = 2.39 W m -2 o C mm 220 mm
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.
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 = r plaster = / 0.7 = r brick = 0.11 / 1 = 0.11 r cavity = r brick = 0.11 / 1 = 0.11 r external = Total resistance = = m 2 o C W -1 U – value = 1 / = 1.67 W m -2 o C -1. Or 70% of solid wall.
8 Construction of Typical UK Walls: post ~ 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 = / 0.7 = r external = r internal = r cavity = Total resistance = = m 2 o C W -1 U – value = 1 / = 0.8 W m -2 o C -1. Or 50% of brick / cavity / brick wall. Block plaster
9 Construction of Typical UK Walls: post ~ 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 = / 0.7 = r external = r internal = r cavity insulation = 0.05/0.04 = 1.25 Total resistance = = m 2 o C W -1 U – value = 1 / = 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
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 mm W m -2 o C mm W m -2 o C mm W m -2 o C mm W m -2 o C mm W m -2 o C mm W m -2 o C 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.
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
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 = * ach * volume 19. Energy Conservation in Buildings: The Basics
13 NBS-M016 Contemporary Issues in Climate Change and Energy Heat Loss Calculations 21. Energy Management N.K. Tovey ( ) 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
14 Five components to heat loss / gain parameter Losses through Floor Roof Windows Walls Ventilation 20: Heat Loss / Heat Gain Calculations Ventilation: Volume * * ach Fabric Components: Area * U - value 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)
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
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
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 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
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 = 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 20: Heat Loss / Heat Gain: Degree Days
19 Annual Degree Days 20 year average 1959 – year average 1979 – year average : Heat Loss / Heat Gain: Degree Days 20 year averageJanFebMarAprMayJunJulAugSepOctNovDec 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 * ( ) * = GJ Or 450 * ( ) * 24 / 1000 = 9850 kWh is seconds in a day 24 is hours in a day
20 20: Heat Loss / Heat Gain: Dynamic Considerations Hours Boiler Output for a house during early January Temperature o C Boiler Output (kW)
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
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 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? 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
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: * ( ) = 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 = tonnes 20: Heat Loss / Heat Gain: Worked Example 1
26 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 * 1000 * = £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 20: Heat Loss / Heat Gain: Worked Example 1 Data for India
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 Capital costs: Oil Boiler £2000, Heat Pump £4000
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 - Mar Apr - Jun Jul - Sept NoHeating needed Oct - Dec Total energy requirement Boiler efficiency90% Energy input boiler option as oil 9456 Total effective electrical input via heat pump 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 = MJ So / 37 = 920 litres are needed Cost of oil = 920 * 0.45 = £414 Cost of electricity for heat pump = * = £ and an annual saving of £ Analysis is best done in tabular form Heat Loss Rate for house
20: Heat Loss / Heat Gain: Worked Example 2 Annual saving in energy costs = £ 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.
30 NBS-M016 Contemporary Issues in Climate Change and Energy Energy Management N.K. Tovey ( ) 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
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 ~ £ £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
CRed carbon reduction Local Authority Offices NorwichGreat Yarmouth Kings Lynn Naturally ventilated Air- conditioned Good Practice Typical85178 Electricity kWh/m 2 kWh/employee Great Yarmouth Kings Lynn Norwich HSBC Sustainability Audit Report Norwich Branch and Office
CRed carbon reduction ElectricityCarbon Dioxide kWh/sqm kWh/ employee kg/sqm tonnes/ employee Great Yarmouth Kings Lynn Norwich 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: & GJ Corresponding degree days: 1100 and 500 Total Energy consumed = W + H * degree days* * H * W = (1) 500 * H * W = (2) Simultaneous equations (subtract 2 from 1) H = (31.76 – 18.80) * 10 9 = 250 Watts ( )*86400 Substitute for H in either equation to get W W = * * 250 * = 8 * 10 9 = 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 3 rd quarter = 400 –Consumption predicted to be –400 * 250 * * 10 9 = GJ H W If actual consumption is 17.5 GJ then energy has been wasted
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
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
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 Two parts to graph Heating part represented by sloping line Base load for cooking/hot water by horizontal line.
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)
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) * 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
Cumulative deviation method Time Saving Excess 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
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 59.8 kW o C -1 heating and 72.4 kW o C -1 cooling =43.05/720 = 52.12/720
Neutral Temperature o C 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. Baseline consumption Analysis of Energy Data - Fudan University – Twin Towers