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1 NBS-M016 Contemporary Issues in Climate Change and Energy 2010 Revision Session Some worked examples N.K. Tovey ( ) M.A, PhD, CEng, MICE, CEnv .. .., - Energy Science Director CRed Project HSBC Director of Low Carbon Innovation 1 Slide 2 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 Slide 3 Energy Management Mean Temperature Electricity Consumption (kWh) January3.75910000 February9700000 March13540000 April16420000 May23790000 June271030000 July31.51300000 August291150000 September25910000 October20610000 November14500000 December8740000 Mean Temperature Electricity Consumption (kWh) 3.75910000 9700000 13540000 16420000 23790000 271030000 31.51300000 291150000 25910000 20610000 14500000 8740000 Total38100005790000 9600000 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 lines @ 400000 kWh and 16.5 o C. Annual functional energy use = 12 * 400000 = 4800000 kWh = 50% of total use Heating is (3810000-6*400000)/9600000 = 14.6%) Functional Energy Use Heating Cooling Slide 4 Range of Wind Speed (m/s) daysmean wind speed (m/s) output (kW) - 232 230 4 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 Slide 5 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)*24/1000 - 232230 Total2409.8 Output = 2409.8 MWh per annum Maximum Possible = 1* 8760 = 8760 MWh So Load Factor = 2409.8 / 8760 = 27.5% If carbon factor = 0.52, saving in CO 2 = 2409.8 * 0.52 = 1253 tonnes Slide 6 Variant of previous example wind speed data given as percentage Wind speedoutputWind speed Frequency Power * Frequency [m/s][kW]% 4.50 0 0 53021.86.54 713018.624.18 928011.131.08 114349.541.23 135296.534.385 155854.123.985 176002.112.6 1960016 cutout00.80 Summation180 Output at 100% load factor = 600 * 1 600 Load Factor =180/600 30.00% Electricity generated in year = 180*8760 = 1576800 kWh Under Feed in Tariff Revenue is 9.7p per kWh So annual income is 1576800 * 0.097 = 152949.6 Slide 7 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 2.75. 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) 15159 20160 25350 30535 35725 40912 Slide 8 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 979.4 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 Slide 9 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 = 103.125 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 979.4 g/kWh, the total carbon emissions will be 2808 * 979.4 = 2750 tonnes Slide 10 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 * 979.4 / 2.75 Remember to include COP! = 400 tonnes Question 4 from 2008 Exam Slide 11 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 105 000. 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 Slide 12 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 - March6.5 42.5April - June13.5 83.3July - September16 123.88 October - December7.5 164.2 Question 5 from 2008 Exam Calorific Value of oil37 MJ/litre Cost of oil42.67 per litre Cost of electricity4.75 p per kWh Slide 13 Question 5 from 2008 Exam Slide 14 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 - Mar6.5390919440064800 Apr - Jun13.549124368010920 Jul - Sept164.292 Oct - Dec7.53.292817664055200 Total energy requirement414720 Boiler efficiency90% Energy input boiler option460800 Total effective input via heat pump130920 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) Slide 15 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 - Mar6.5390919440064800 Apr - Jun13.549124368010920 Jul - Sept164.292 Oct - Dec7.53.292817664055200 Total energy requirement414720 Boiler efficiency90% Energy input boiler option460800 Total effective input via heat pump130920 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) Slide 16 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 - Mar6.5390919440064800 Apr - Jun13.549124368010920 Jul - Sept164.292 Oct - Dec7.53.292817664055200 Total energy requirement414720 Boiler efficiency90% Energy input boiler option460800 Total effective input via heat pump130920 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) Slide 17 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 - Mar6.5390919440064800 Apr - Jun13.549124368010920 Jul - Sept164.292 Oct - Dec7.53.292817664055200 Total energy requirement414720 Boiler efficiency90% Energy input boiler option460800 Total effective input via heat pump130920 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) Slide 18 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 - Mar6.5390919440064800 Apr - Jun13.549124368010920 Jul - Sept164.292 Oct - Dec7.53.292817664055200 Total energy requirement414720 Boiler efficiency90% Energy input boiler option460800 Total effective input via heat pump130920 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) Slide 19 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 - Mar6.5390919440064800 Apr - Jun13.549124368010920 Jul - Sept164.292 Oct - Dec7.53.292817664055200 Total energy requirement414720 Boiler efficiency90% Energy input boiler option460800 Total effective input via heat pump130920 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) Slide 20 From above table input energy = 460800 kWh = 1658880 MJ But calorific value of oil is 37 MJ/litre so number of litres required = 460800/37 = 44834.59 litres Annual running costs with oil = 44834.59* 42.67 /100 = 19130.92 Annual running costs of heat pump = 130920 * 4.75/100 = 6218.70 Annual saving in running costs = 19130.92 - 6218.70 = 12912.22 From discount tables the cumulative discount factor is 8.721735 So the discounted savings over life of project = 8.721735 * 12912.22 = 112617 Net present Value = 25000 105000 + 112617 = 32617 i.e Heat Pump scheme is best option to choose. Question 5 from 2008 Exam Slide 21 3 units 7.6 units Gradient of Line = 3/7.6 = 0.40 Working out the gradient of a line Slide 22 Slide 23 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) Jan1.91600 Feb4.51450 Mar91300 Apr121200 May141000 June161000 July171000 Aug161000 Sep131200 Oct111350 Nov91450 Dec4.11600 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 0.186 tonnes per MWh for gas and 0.544 tonnes per MWh for electricity Slide 24 MonthTemp ( o C)Space Heat Demand (kW) Total Heat Demand (kW) Electricity (kW) given info [1][2][3][4][5] Jan1.9340035001600 Feb4.5275028501450 Mar9162517251300 Apr128759751200 May143754751000 June1601001000 July1701001000 Aug1601001000 Sep136257251200 Oct11112512251350 Nov9162517251450 Dec4.1285029501600 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 Slide 25 MonthTemp ( o C)Space Heat Demand (kW) Total Heat Demand (kW) Electricity (kW) given info [1][2][3][4][5] Jan1.9340035001600 Feb4.5275028501450 Mar9162517251300 Apr128759751200 May143754751000 June1601001000 July1701001000 Aug1601001000 Sep136257251200 Oct11112512251350 Nov9162517251450 Dec4.1285029501600 CHP worked example Slide 26 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] Jan350016001680 18201200400 Feb285014501680 11701200250 Mar172513001680 451200100 Apr97512001680975 696***504 May47510001400475 339***661 June10010001400100 71***929 July10010001400100 71***929 Aug10010001400100 71***929 Sep72512001680725 518***682 Oct1225135016801225 875475 Nov172514501680 451200250 Dec295016001680 12701200400 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 Slide 27 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] Jan350016001680 18201200400 Feb285014501680 11701200250 Mar172513001680 451200100 Apr97512001680975 696***504 May47510001400475 339***661 June10010001400100 71***929 July10010001400100 71***929 Aug10010001400100 71***929 Sep72512001680725 518***682 Oct1225135016801225 875475 Nov172514501680 451200250 Dec295016001680 12701200400 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 Slide 28 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] Jan350016001680 18201200400 Feb285014501680 11701200250 Mar172513001680 451200100 Apr97512001680975 696***504 May47510001400475 339***661 June10010001400100 71***929 July10010001400100 71***929 Aug10010001400100 71***929 Sep72512001680725 518***682 Oct1225135016801225 875475 Nov172514501680 451200250 Dec295016001680 12701200400 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 Slide 29 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] Jan350016001680 18201200400 Feb285014501680 11701200250 Mar172513001680 451200100 Apr97512001680975 696***504 May47510001400475 339***661 June10010001400100 71***929 July10010001400100 71***929 Aug10010001400100 71***929 Sep72512001680725 518***682 Oct1225135016801225 875475 Nov172514501680 451200250 Dec295016001680 12701200400 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 Slide 30 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] Jan350016001680 18201200400 Feb285014501680 11701200250 Mar172513001680 451200100 Apr97512001680975 696***504 May47510001400475 339***661 June10010001400100 71***929 July10010001400100 71***929 Aug10010001400100 71***929 Sep72512001680725 518***682 Oct1225135016801225 875475 Nov172514501680 451200250 Dec295016001680 12701200400 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 Slide 31 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] Jan350016001680 18201200400 Feb285014501680 11701200250 Mar172513001680 451200100 Apr97512001680975 696***504 May47510001400475 339***661 June10010001400100 71***929 July10010001400100 71***929 Aug10010001400100 71***929 Sep72512001680725 518***682 Oct1225135016801225 875475 Nov172514501680 451200250 Dec295016001680 12701200400 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 Slide 32 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] Jan1.93400350016001680 18201200400 Feb4.52750285014501680 11701200250 Mar91625172513001680 451200100 Apr1287597512001680975 696***504 May1437547510001400475 339***661 June16010010001400100 71***929 July17010010001400100 71***929 Aug16010010001400100 71***929 Sep1362572512001680725 518***682 Oct1111251225135016801225 875475 Nov91625172514501680 451200250 Dec4.12850295016001680 12701200400 Sum of numbers in column16450151501904012100435068756509 GWh equals numbers in red * 30 *24 11.8410.9113.718.713.134.954.69 Assumes number of days in each month is 30 Slide 33 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] Jan1.93400350016001680 18201200400 Dec4.12850295016001680 12701200400 Sum of numbers in column16450151501904012100435068756509 GWh equals numbers in red * 30 *24 11.8410.9113.718.713.134.954.69 % 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]11.840.80.1862753 electricity col [5]10.9110.5445935 total22.75 8688 CHP worked example Slide 34 Total Demand EfficiencyCO 2 factor (tonnes/MWh) CO 2 (tonnes) GWh CHP heating [7]8.71 CHP electricity [9]4.95 Total CHP13.660.80.1863176 supplementary heating [8] 3.130.80.186728 supplementary electricity [10] 4.6910.544872 4776 CHP worked example Carbon emissions before installation 8688 tonnes Carbon emissions after installation: 4776 tonnes Saving = 8688 4776 = 3911 tonnes or 45% Slide 35 Slide 36 3 units 7.3 units Gradient of Line = 3/7.3 = 0.411 Slide 37 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] Jan350016001680 18201200400 Feb285014501680 11701200250 Mar172513001680 451200100 Apr97512001680975 696***504 May47510001400475 339***661 June10010001400100 71***929 July10010001400100 71***929 Aug10010001400100 71***929 Sep72512001680725 518***682 Oct1225135016801225 875475 Nov172514501680 451200250 Dec295016001680 12701200400 CHP worked example