1 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 Revision Session Some worked examples 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 1

Format of Exam Duration 2 hours Two sections of equal value One Section Three questions – choose one Entirely descriptive - Covering all aspects of course Do read the question - they are frequently specific – so do not merely write everything you know on say “SOLAR” for an answer – the question is likely to be more specific Second Section Two questions – choose one Part descriptive (30+%), Part Numeric (up to 70%) Do not forget the descriptive part!!!!!!!!! NOTE: Two sections may be ordered so that descriptive section is first or second – this is to minimise pagination problems

Energy Management Mean Temperature Electricity Consumption (kWh) January February March April May June July August September October November December Mean Temperature Electricity Consumption (kWh) Total Building uses only electricity. What can you deduce? Balance Temperature for no heating cooling ~ 16 o C Separate into two groups 16 o C Plot points against Mean External Temperature Check there is no problem with balance temperature, and if there is move data points between two categories Intercept of two kWh and 16.5 o C. Annual functional energy use = 12 * = kWh = 50% of total use Heating is ( *400000)/ = 14.6%) Functional Energy Use Heating Cooling

Range of Wind Speed (m/s) daysmean wind speed (m/s) output (kW) - < > Predicting Output from Wind Turbines – Worked Example – part 1 Step 1: Work out mean wind speed Read of Graph for output at each mean wind speed

5 Predicting Output from Wind Turbines – Worked Example – part 2 Range of Wind Speed (m/s)days mean wind speed (m/s) Output (kW) Generated in period (MWh) (1)(2) (3)(4) from graph (5) = (2)*(4)*24/ < > Total Output = MWh per annum – Maximum Possible = 1* 8760 = 8760 MWh So Load Factor = / 8760 = 27.5% If carbon factor = 0.52, saving in CO 2 = * 0.52 = 1253 tonnes

Variant of previous example – wind speed data given as percentage Wind speedoutputWind speed Frequency Power * Frequency [m/s][kW]% cutout00.80 Summation180 Output at 100% load factor = 600 * Load Factor =180/ % Electricity generated in year = 180*8760 = kWh Under Feed in Tariff Revenue is 9.7p per kWh So annual income is * = £

Question 4 from 2008 Exam A large hotel in India has a total window area of 6000 m 2 which are single glazed with a U-Value of 5 W m -2 o C -1 and is cooled by an electrically driven air conditioner having an average coefficient of performance of Data for total electricity consumption at selected mean average external temperatures during the summer months are shown in Table 1. Mean daily external temperatureMean Electricity Consumption (kW)

b)Comment on the relationship between electricity consumption and temperature. [10%] c) Estimate the annual carbon emissions associated with cooling if the mean cooing degree days in India are 3120 and the overall carbon emission factor for electricity in India is g / kWh. [25%] d) The windows are replaced by double glazing units with a U-value of 2.5 W m -2 o C -1. Estimate the annual savings in carbon emissions. [25%]. Question 4 from 2008 Exam

Plot consumption against consumption. Two parts to curve – winter – no cooling, summer cooling Gradient of line is 37.5 kW o C -1 As COP of air-conditioner is 2.75 heat gain rate in summer is 37.5 * 2.75 = kW o C -1 Question 4 from 2008 Exam The carbon emissions associated with cooling = 37.5 * 3120 * 24 /1000 = 2808 MWh | | | Gradient of line degree days hours in a day As the carbon factor for India is g/kWh, the total carbon emissions will be 2808 * = 2750 tonnes

The change in the heat loss rate from installing double glazing will be: ( 5 – 2.5) * 6000 = 15 kW oC-1 So the saving in carbon emissions will be: 15 * 3120 * 24 * / 2.75 Remember to include COP! = 400 tonnes Question 4 from 2008 Exam

As a senior manager in a small office firm which is constructing a new building you are asked to make recommendations on the mode of heating that should be employed and have been given guidance that you should use a discount rate of 5%. The building is designed to have a heat loss rate of 10 kW o C -1 and have a neutral internal temperature of 15.5 o C. You have two options to consider, an oil condensing boiler system with an efficiency of 90% or a heat pump system, the coefficient of performance of which is shown in Table 2. The oil boiler system costs £25000 to install while the heat pump installation would cost £ Both systems have an expected life of 10 years and both have similar annual maintenance charges. Which option would you recommend? Data relating to the climate data are given in Table 3 while other relevant data are given in Table 4. You may assume that the energy tariffs do not change in real terms over the period. Question 5 from 2008 Exam

Table 2. Coefficient of Performance of Heat Pump Table 3. Mean External Temperature at location of Building External Temperature ( o C) Coefficient of Performance Months Mean Temperature ( o C) 01.7January - March April - June July - September October - December Question 5 from 2008 Exam Calorific Value of oil37 MJ/litre Cost of oil42.67 per litre Cost of electricity4.75 p per kWh

Question 5 from 2008 Exam

External Temperature ( o C) COP from graph Number of days Difference from balance temperature Heat Requirement (kWh) Heat Requirement after allowing for COP (kWh) (1)(2)(3)(4)(5)(6)(7) Jan - Mar Apr - Jun Jul - Sept Oct - Dec Total energy requirement Boiler efficiency90% Energy input boiler option Total effective input via heat pump Col (2) from Table 2 Col (3) from graph Col (4) number of days in period – i.e Jan (31) Feb (28) Mar (31) = 90 days Col (5) balance temperature is 15.5, so col(5) = 15.5 – col(2) Col (6) heat loss rate (10 kW oC-1 ) * col (5) * col (4) * 24 (hours in a day) Col (7) col(6)/col(3)

Question 5 from 2008 Exam External Temperature ( o C) COP from graph Number of days Difference from balance temperature Heat Requirement (kWh) Heat Requirement after allowing for COP (kWh) (1)(2)(3)(4)(5)(6)(7) Jan - Mar Apr - Jun Jul - Sept Oct - Dec Total energy requirement Boiler efficiency90% Energy input boiler option Total effective input via heat pump Col (2) from Table 2 Col (3) from graph Col (4) number of days in period – i.e Jan (31) Feb (28) Mar (31) = 90 days Col (5) balance temperature is 15.5, so col(5) = 15.5 – col(2) Col (6) heat loss rate (10 kW oC-1 ) * col (5) * col (4) * 24 (hours in a day) Col (7) col(6)/col(3)

Question 5 from 2008 Exam External Temperature ( o C) COP from graph Number of days Difference from balance temperature Heat Requirement (kWh) Heat Requirement after allowing for COP (kWh) (1)(2)(3)(4)(5)(6)(7) Jan - Mar Apr - Jun Jul - Sept Oct - Dec Total energy requirement Boiler efficiency90% Energy input boiler option Total effective input via heat pump Col (2) from Table 2 Col (3) from graph Col (4) number of days in period – i.e Jan (31) Feb (28) Mar (31) = 90 days Col (5) balance temperature is 15.5, so col(5) = 15.5 – col(2) Col (6) heat loss rate (10 kW oC-1 ) * col (5) * col (4) * 24 (hours in a day) Col (7) col(6)/col(3)

Question 5 from 2008 Exam External Temperature ( o C) COP from graph Number of days Difference from balance temperature Heat Requirement (kWh) Heat Requirement after allowing for COP (kWh) (1)(2)(3)(4)(5)(6)(7) Jan - Mar Apr - Jun Jul - Sept Oct - Dec Total energy requirement Boiler efficiency90% Energy input boiler option Total effective input via heat pump Col (2) from Table 2 Col (3) from graph Col (4) number of days in period – i.e Jan (31) Feb (28) Mar (31) = 90 days Col (5) balance temperature is 15.5, so col(5) = 15.5 – col(2) Col (6) heat loss rate (10 kW oC-1 ) * col (5) * col (4) * 24 (hours in a day) Col (7) col(6)/col(3)

Question 5 from 2008 Exam External Temperature ( o C) COP from graph Number of days Difference from balance temperature Heat Requirement (kWh) Heat Requirement after allowing for COP (kWh) (1)(2)(3)(4)(5)(6)(7) Jan - Mar Apr - Jun Jul - Sept Oct - Dec Total energy requirement Boiler efficiency90% Energy input boiler option Total effective input via heat pump Col (2) from Table 2 Col (3) from graph Col (4) number of days in period – i.e Jan (31) Feb (28) Mar (31) = 90 days Col (5) balance temperature is 15.5, so col(5) = 15.5 – col(2) Col (6) heat loss rate (10 kW oC-1 ) * col (5) * col (4) * 24 (hours in a day) Col (7) col(6)/col(3)

Question 5 from 2008 Exam External Temperature ( o C) COP from graph Number of days Difference from balance temperature Heat Requirement (kWh) Heat Requirement after allowing for COP (kWh) (1)(2)(3)(4)(5)(6)(7) Jan - Mar Apr - Jun Jul - Sept Oct - Dec Total energy requirement Boiler efficiency90% Energy input boiler option Total effective input via heat pump Col (2) from Table 2 Col (3) from graph Col (4) number of days in period – i.e Jan (31) Feb (28) Mar (31) = 90 days Col (5) balance temperature is 15.5, so col(5) = 15.5 – col(2) Col (6) heat loss rate (10 kW oC-1 ) * col (5) * col (4) * 24 (hours in a day) Col (7) col(6)/col(3)

From above table input energy = kWh = MJ But calorific value of oil is 37 MJ/litre – so number of litres required = /37 = litres Annual running costs with oil = * /100 = £ Annual running costs of heat pump = * 4.75/100 = £ Annual saving in running costs = £ £ = £ From discount tables the cumulative discount factor is So the discounted savings over life of project = * = £ Net present Value = £25000 – = £32617 i.e Heat Pump scheme is best option to choose. Question 5 from 2008 Exam

3 units 7.6 units Gradient of Line = 3/7.6 = 0.40 Working out the gradient of a line

CHP worked example A firm is considering installing a CHP scheme to replace the existing gas boiler (which has an efficiency of 80%) It is proposed to have three 400 kW CHP power plants to provide electricity and heat the buildings. The buildings have a heat loss rate of 250 kW o C -1 and there is a requirement of 100 kW for hot water. Assuming that there are 30 days in each month estimate the saving in energy compared to the existing system if the external temperature and electricity demand data are as follows: MonthTemp (oC)Electricity (kW) Jan Feb Mar91300 Apr May June July Aug Sep Oct Nov91450 Dec In the proposed scheme 1.4 units of heat are rejected for each unit of electricity generated. Overall efficiency of CHP plant is 80% What proportion of heat is supplied by CHP? What are CO 2 savings if emission factors are tonnes per MWh for gas and tonnes per MWh for electricity

MonthTemp ( o C)Space Heat Demand (kW) Total Heat Demand (kW) Electricity (kW) given info [1][2][3][4][5] Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec CHP worked example Heat demand = Heat loss rate * (temp. diff) = 250 * (15.5 – ext temp) e.g. for January = 250 * (15.5 – 1.9) = 3400 kW 15.5 o C is the balance or neutral temperature and is the temperature at which no heating is required. Total Heat demand = Space Heat Demand + Hot Water/Process Demand

MonthTemp ( o C)Space Heat Demand (kW) Total Heat Demand (kW) Electricity (kW) given info [1][2][3][4][5] Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec CHP worked example

MonthTotal Heat Demand (kW) Electricity demand (kW) Potential CHP Heat available (kW) Useful CHP Heat (kW) Supple- mentary Heat (kW) Actual Electricity Generated (kW) Supple- mentary Electricity Needed (kW) [1][4][5][6][7][8][9][10] Jan Feb Mar Apr ***504 May ***661 June ***929 July ***929 Aug ***929 Sep ***682 Oct Nov Dec When there is insufficient heat provided by CHP units, supplementary heat must be provided by existing boilers. Thus supplementary heat (col[8]) = Col [4] – col 7] However, not all CHP heat may be needed. Need to compare with actual demand. If CHP heat available is less than demand we can use all the heat. col [7] = col [6] in Jan – Mar and Nov – Dec If CHP heat is greater than demand we can only utilise actual heat demand. col [7] = col [4] Heat available from CHP units will be 1.4 times electrical output. But maximum electrical output = 3 *400 = 1200 kW If electrical demand is greater than 1200 kW, the maximum heat available from CHP Units will be 1200 *1.4 = 1680 kW If electrical demand is less than 1200, then heat available will be actual electricity demand * 1.4 e.g. for May – August heat available = 1000 * 1.4 = 1400 kW Unless there is export of electricity. Start by assuming no export. CHP worked example

MonthTotal Heat Demand (kW) Electricity demand (kW) Potential CHP Heat available (kW) Useful CHP Heat (kW) Supple- mentary Heat (kW) Actual Electricity Generated (kW) Supple- mentary Electricity Needed (kW) [1][4][5][6][7][8][9][10] Jan Feb Mar Apr ***504 May ***661 June ***929 July ***929 Aug ***929 Sep ***682 Oct Nov Dec When there is insufficient heat provided by CHP units, supplementary heat must be provided by existing boilers. Thus supplementary heat (col[8]) = Col [4] – col 7] However, not all CHP heat may be needed. Need to compare with actual demand. If CHP heat available is less than demand we can use all the heat. col [7] = col [6] in Jan – Mar and Nov – Dec If CHP heat is greater than demand we can only utilise actual heat demand. col [7] = col [4] Heat available from CHP units will be 1.4 times electrical output. But maximum electrical output = 3 *400 = 1200 kW If electrical demand is greater than 1200 kW, the maximum heat available from CHP Units will be 1200 *1.4 = 1680 kW If electrical demand is less than 1200, then heat available will be actual electricity demand * 1.4 e.g. for May – August heat available = 1000 * 1.4 = 1400 kW Unless there is export of electricity. Start by assuming no export. CHP worked example

MonthTotal Heat Demand (kW) Electricity demand (kW) Potential CHP Heat available (kW) Useful CHP Heat (kW) Supple- mentary Heat (kW) Actual Electricity Generated (kW) Supple- mentary Electricity Needed (kW) [1][4][5][6][7][8][9][10] Jan Feb Mar Apr ***504 May ***661 June ***929 July ***929 Aug ***929 Sep ***682 Oct Nov Dec When there is insufficient heat provided by CHP units, supplementary heat must be provided by existing boilers. Thus supplementary heat (col[8]) = Col [4] – col 7] However, not all CHP heat may be needed. Need to compare with actual demand. If CHP heat available is less than demand we can use all the heat. col [7] = col [6] in Jan – Mar and Nov – Dec If CHP heat is greater than demand we can only utilise actual heat demand. col [7] = col [4] CHP worked example

MonthTotal Heat Demand (kW) Electricity demand (kW) Potential CHP Heat available (kW) Useful CHP Heat (kW) Supple- mentary Heat (kW) Actual Electricity Generated (kW) Supple- mentary Electricity Needed (kW) [1][4][5][6][7][8][9][10] Jan Feb Mar Apr ***504 May ***661 June ***929 July ***929 Aug ***929 Sep ***682 Oct Nov Dec When there is insufficient heat provided by CHP units, supplementary heat must be provided by existing boilers. Thus supplementary heat (col[8]) = Col [4] – col 7] CHP worked example

MonthTotal Heat Demand (kW) Electricity demand (kW) Potential CHP Heat available (kW) Useful CHP Heat (kW) Supple- mentary Heat (kW) Actual Electricity Generated (kW) Supple- mentary Electricity Needed (kW) [1][4][5][6][7][8][9][10] Jan Feb Mar Apr ***504 May ***661 June ***929 July ***929 Aug ***929 Sep ***682 Oct Nov Dec In Jan – Mar CHP units will run fully with 1200 kW electricity and 1680 kW heat. However, electricity will be restricted if less than 1680 kW of heat is supplied. In April only 975 kW heat is required. 1.4 units of heat are rejected for each unit of electricity. So actual output of electricity will be less than rated output at 975/1.4 = 696 kW. *** indicates electricity is restricted. CHP worked example

MonthTotal Heat Demand (kW) Electricity demand (kW) Potential CHP Heat available (kW) Useful CHP Heat (kW) Supple- mentary Heat (kW) Actual Electricity Generated (kW) Supple- mentary Electricity Needed (kW) [1][4][5][6][7][8][9][10] Jan Feb Mar Apr ***504 May ***661 June ***929 July ***929 Aug ***929 Sep ***682 Oct Nov Dec Col [9] shows the maximum electricity that can be generated. Where the output is 1200 kW, the units will be running at their rated output. Otherwise output is restricted because of limited heat requirements. Supplementary electricity must be imported – i.e. Col[10] = col[5] – col[9] CHP worked example

MonthTemp ( o C) Space Heat Demand (kW) Total Heat Demand (kW) Electricity demand (kW) CHP Heat available (kW) Useful CHP Heat (kW) Supple- mentary Heat (kW) Actual Electricity Generated Supple- mentary Electricity Needed [1][2][3][4][5][6][7][8][9][10] Jan Feb Mar Apr ***504 May ***661 June ***929 July ***929 Aug ***929 Sep ***682 Oct Nov Dec Sum of numbers in column GWh equals numbers in red * 30 * Assumes number of days in each month is 30

MonthTemp ( o C) Space Heat Demand (kW) Total Heat Demand (kW) Electricity (kW) CHP Heat available (kW) Useful CHP Heat (kW) Supple- mentary Heat (kW) Actual Electricity Generated Supple- mentary Electricity Needed [1][2][3][4][5][6][7][8][9][10] Jan Dec Sum of numbers in column GWh equals numbers in red * 30 * % heat supplied by CHP = 8.71/11.84 = 73.6% Carbon Emissions – situation before installation Total Demand EfficiencyCO 2 factor (tonnes/MWh) CO 2 (tonnes) GWh heating col [4] electricity col [5] total CHP worked example

Total Demand EfficiencyCO 2 factor (tonnes/MWh) CO 2 (tonnes) GWh CHP heating [7]8.71 CHP electricity [9]4.95 Total CHP supplementary heating [8] supplementary electricity [10] CHP worked example Carbon emissions before installation 8688 tonnes Carbon emissions after installation: 4776 tonnes Saving = 8688 – 4776 = 3911 tonnes or 45%

3 units 7.3 units Gradient of Line = 3/7.3 = 0.411

MonthTotal Heat Demand (kW) Electricity (kW) Potential CHP Heat available (kW) Useful CHP Heat (kW) Supple- mentary Heat (kW) Actual Electricity Generated (kW) Supple- mentary Electricity Needed (kW) [1][4][5][6][7][8][9][10] Jan Feb Mar Apr ***504 May ***661 June ***929 July ***929 Aug ***929 Sep ***682 Oct Nov Dec CHP worked example