NBS-M016 Contemporary Issues in Climate Change and Energy 2010

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 Revision Session Some worked examples 1 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 Building uses only electricity. What can you deduce? • Balance Temperature for no heating cooling ~ 16o C • Separate into two groups • <= 16o C and > 16o 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 Cooling Functional Energy Use Heating Intercept of two lines @ 400000 kWh and 16.5o C. Annual functional energy use = 12 * 400000 = 4800000 kWh = 50% of total use Heating is (3810000-6*400000)/9600000 = 14.6%)

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 4

Predicting Output from Wind Turbines – Worked Example – part 2 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 CO2 = 2409.8 * 0.52 = 1253 tonnes 5

Variant of previous example – wind speed data given as percentage 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

Question 4 from 2008 Exam A large hotel in India has a total window area of 6000 m2 which are single glazed with a U-Value of 5 W m-2oC-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.

Question 4 from 2008 Exam • 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 oC-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 oC-1 • As COP of air-conditioner is 2.75 heat gain rate in summer is • 37.5 * 2.75 = 103.125 kWoC-1 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

Question 4 from 2008 Exam 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 5 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 oC-1 and have a neutral internal temperature of 15.5oC. • 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

Question 5 from 2008 Exam 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 • 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.

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

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 kWoC-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: 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 CO2 savings if emission factors are 0.186 tonnes per MWh for gas and 0.544 tonnes per MWh for electricity

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 oC 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

CHP worked example

CHP worked example 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. • 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] 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 • 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] 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 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 • 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 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]

Assumes number of days in each month is 30

CHP worked example % heat supplied by CHP = 8.71/11.84 = 73.6% Carbon Emissions – situation before installation

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

CHP worked example

NBS-M016 Contemporary Issues in Climate Change and Energy 2010