nbs-m016 contemporary issues in climate change and energy 2010

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 Innovation1

Format of ExamDuration 2 hoursTwo 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 3.75 910000

February 9 700000

March 13 540000

April 16 420000

May 23 790000

June 27 1030000

July 31.5 1300000

August 29 1150000

September 25 910000

October 20 610000

November 14 500000

December 8 740000

Mean Temperature

Electricity Consumption

(kWh)

3.75 910000

9 700000

13 540000

16 420000

23 790000

27 1030000

31.5 1300000

29 1150000

25 910000

20 610000

14 500000

8 740000

Total 3810000 5790000

9600000

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

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%)

0200000400000600000800000

100000012000001400000

0 5 10 15 20 25 30 35kW

h/mon

th)

Mean Temperature

Functional Energy Use

HeatingCooling

Range of Wind Speed

(m/s)

days mean wind speed (m/s)

output (kW)

- <1 10 0 01 - 3 30 2 03 - 5 50 4 05 - 7 101 6 977 - 9 84 8 281

9 - 11 40 10 55711 - 13 20 12 84513 - 15 10 14 98315 - 17 8 16 100017 - 19 5 18 100019 - 21 3 20 100021 - 23 2 22 1000 - >23 2 23 0

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

Power Curve

0

200

400

600

800

1000

0 10 20 30

wind speed (m/s)

kW

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

- <1 10 0 0 01 - 3 30 2 0 03 - 5 50 4 0 05 - 7 101 6 97 235.17 - 9 84 8 281 566.5

9 - 11 40 10 557 534.711 - 13 20 12 845 405.613 - 15 10 14 983 235.915 - 17 8 16 1000 19217 - 19 5 18 1000 12019 - 21 3 20 1000 7221 - 23 2 22 1000 48 - >23 2 23 0

Total 2409.8Output = 2409.8 MWh per annum – Maximum Possible = 1* 8760 = 8760 MWhSo Load Factor = 2409.8 / 8760 = 27.5%

If carbon factor = 0.52, saving in CO2 = 2409.8 * 0.52 = 1253 tonnes

Variant of previous example – wind speed data given as percentage

Wind speed output Wind speed Frequency

Power * Frequency

[m/s] [kW] % 4.5 0 0 05 30 21.8 6.547 130 18.6 24.189 280 11.1 31.0811 434 9.5 41.2313 529 6.5 34.38515 585 4.1 23.98517 600 2.1 12.619 600 1 6

cutout 0 0.8 0 Summation 180Output 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

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-2 oC-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 temperature Mean Electricity Consumption (kW)

15 15920 16025 350

30 535

35 725

40 912

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 oC-1. Estimate the annual savings in carbon emissions.

[25%].


• 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


0

100

200

300

400

500

600

700

800

900

1000

0 10 20 30 40 50

External Temperature (oC)

Ele

ctri

city

Co

nsu

mp

tio

n (

kW)

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

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


• 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.


Table 2. Coefficient of Performance of Heat Pump

Table 3. Mean External Temperature at location of Building

External Temperature

(oC)Coefficient of

Performance

Months Mean Temperature

(oC)0 1.7 January - March 6.5

4 2.5 April - June 13.58 3.3 July - September 16

12 3.88October -

December 7.516 4.2


Calorific Value of oil 37 MJ/litre

Cost of oil 42.67 per litre

Cost of electricity 4.75 p per kWh


0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 2 4 6 8 10 12 14 16 18

External Temperature (oC)

Coef

ficie

nc o

f Pef

orm

ance


External Temperature

(oC)

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 6.5 3 90 9 194400 64800Apr - Jun 13.5 4 91 2 43680 10920Jul - Sept 16 4.2 92

Oct - Dec 7.5 3.2 92 8 176640 55200

Total energy requirement 414720

Boiler efficiency 90%

Energy input boiler option 460800

Total effective input via heat pump 130920

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 = 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.


0

1

2

3

4

5

6

7

8

9

0 1 2 3 4 5 6 7 8 9 10 11 12

3 units

7.6 units

Gradient of Line = 3/7.6 = 0.40

Working out the gradient of a line

CHP worked exampleA 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:

Month Temp (oC) Electricity (kW)

Jan 1.9 1600Feb 4.5 1450Mar 9 1300Apr 12 1200May 14 1000June 16 1000July 17 1000Aug 16 1000Sep 13 1200Oct 11 1350Nov 9 1450Dec 4.1 1600

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 are0.186 tonnes per MWh for gas and 0.544 tonnes per MWh for electricity

Month Temp (oC) Space Heat Demand (kW)

Total Heat Demand (kW)

Electricity (kW) given info

[1] [2] [3] [4] [5]

Jan 1.9 3400 3500 1600

Feb 4.5 2750 2850 1450

Mar 9 1625 1725 1300

Apr 12 875 975 1200

May 14 375 475 1000

June 16 0 100 1000

July 17 0 100 1000

Aug 16 0 100 1000

Sep 13 625 725 1200

Oct 11 1125 1225 1350

Nov 9 1625 1725 1450

Dec 4.1 2850 2950 1600

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

Month Temp (oC) Space Heat Demand (kW)

Total Heat Demand (kW)

Electricity (kW) given info

[1] [2] [3] [4] [5]

Jan 1.9 3400 3500 1600

Feb 4.5 2750 2850 1450

Mar 9 1625 1725 1300

Apr 12 875 975 1200

May 14 375 475 1000

June 16 0 100 1000

July 17 0 100 1000

Aug 16 0 100 1000

Sep 13 625 725 1200

Oct 11 1125 1225 1350

Nov 9 1625 1725 1450

Dec 4.1 2850 2950 1600

CHP worked example

Month Total Heat

Demand (kW)

Electricitydemand

(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 3500 1600 1680 1680 1820 1200 400

Feb 2850 1450 1680 1680 1170 1200 250

Mar 1725 1300 1680 1680 45 1200 100

Apr 975 1200 1680 975 696*** 504

May 475 1000 1400 475 339*** 661

June 100 1000 1400 100 71*** 929

July 100 1000 1400 100 71*** 929

Aug 100 1000 1400 100 71*** 929

Sep 725 1200 1680 725 518*** 682

Oct 1225 1350 1680 1225 875 475

Nov 1725 1450 1680 1680 45 1200 250

Dec 2950 1600 1680 1680 1270 1200 400

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

Month Total Heat

Demand (kW)

Electricitydemand

(kW)


(kW)

Useful CHP Heat

(kW)

Supple-mentary

Heat (kW)


(kW)

Supple-mentary


[1] [4] [5] [6] [7] [8] [9] [10]

Jan 3500 1600 1680 1680 1820 1200 400

Feb 2850 1450 1680 1680 1170 1200 250

Mar 1725 1300 1680 1680 45 1200 100

Apr 975 1200 1680 975 696*** 504

May 475 1000 1400 475 339*** 661

June 100 1000 1400 100 71*** 929

July 100 1000 1400 100 71*** 929

Aug 100 1000 1400 100 71*** 929

Sep 725 1200 1680 725 518*** 682

Oct 1225 1350 1680 1225 875 475

Nov 1725 1450 1680 1680 45 1200 250

Dec 2950 1600 1680 1680 1270 1200 400



= 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

Month Total Heat

Demand (kW)

Electricitydemand

(kW)


(kW)

Useful CHP Heat

(kW)

Supple-mentary

Heat (kW)


(kW)

Supple-mentary


[1] [4] [5] [6] [7] [8] [9] [10]

Jan 3500 1600 1680 1680 1820 1200 400

Feb 2850 1450 1680 1680 1170 1200 250

Mar 1725 1300 1680 1680 45 1200 100

Apr 975 1200 1680 975 696*** 504

May 475 1000 1400 475 339*** 661

June 100 1000 1400 100 71*** 929

July 100 1000 1400 100 71*** 929

Aug 100 1000 1400 100 71*** 929

Sep 725 1200 1680 725 518*** 682

Oct 1225 1350 1680 1225 875 475

Nov 1725 1450 1680 1680 45 1200 250

Dec 2950 1600 1680 1680 1270 1200 400



= Col [4] – col 7]

CHP worked example

Month Total Heat

Demand (kW)

Electricity demand

(kW)


(kW)

Useful CHP Heat

(kW)

Supple-mentary

Heat (kW)


(kW)

Supple-mentary


[1] [4] [5] [6] [7] [8] [9] [10]

Jan 3500 1600 1680 1680 1820 1200 400

Feb 2850 1450 1680 1680 1170 1200 250

Mar 1725 1300 1680 1680 45 1200 100

Apr 975 1200 1680 975 696*** 504

May 475 1000 1400 475 339*** 661

June 100 1000 1400 100 71*** 929

July 100 1000 1400 100 71*** 929

Aug 100 1000 1400 100 71*** 929

Sep 725 1200 1680 725 518*** 682

Oct 1225 1350 1680 1225 875 475

Nov 1725 1450 1680 1680 45 1200 250

Dec 2950 1600 1680 1680 1270 1200 400

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

Month Total Heat

Demand (kW)

Electricity demand

(kW)


(kW)

Useful CHP Heat

(kW)

Supple-mentary

Heat (kW)


(kW)

Supple-mentary


[1] [4] [5] [6] [7] [8] [9] [10]

Jan 3500 1600 1680 1680 1820 1200 400

Feb 2850 1450 1680 1680 1170 1200 250

Mar 1725 1300 1680 1680 45 1200 100

Apr 975 1200 1680 975 696*** 504

May 475 1000 1400 475 339*** 661

June 100 1000 1400 100 71*** 929

July 100 1000 1400 100 71*** 929

Aug 100 1000 1400 100 71*** 929

Sep 725 1200 1680 725 518*** 682

Oct 1225 1350 1680 1225 875 475

Nov 1725 1450 1680 1680 45 1200 250

Dec 2950 1600 1680 1680 1270 1200 400

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

Month Temp (oC)

Space Heat Demand

(kW)

Total Heat

Demand (kW)

Electricitydemand

(kW)

CHP Heat

available (kW)

Useful CHP Heat (kW)

Supple-mentary

Heat (kW)


Supple-mentary

Electricity Needed

[1] [2] [3] [4] [5] [6] [7] [8] [9] [10]Jan 1.9 3400 3500 1600 1680 1680 1820 1200 400Feb 4.5 2750 2850 1450 1680 1680 1170 1200 250Mar 9 1625 1725 1300 1680 1680 45 1200 100Apr 12 875 975 1200 1680 975 696*** 504May 14 375 475 1000 1400 475 339*** 661June 16 0 100 1000 1400 100 71*** 929July 17 0 100 1000 1400 100 71*** 929Aug 16 0 100 1000 1400 100 71*** 929Sep 13 625 725 1200 1680 725 518*** 682Oct 11 1125 1225 1350 1680 1225 875 475Nov 9 1625 1725 1450 1680 1680 45 1200 250Dec 4.1 2850 2950 1600 1680 1680 1270 1200 400Sum of numbers in column 16450 15150 19040 12100 4350 6875 6509

GWh GWh GWh GWh GWh GWh GWhequals numbers in red

* 30 *2411.84 10.91 13.71 8.71 3.13 4.95 4.69

Assumes number of days in each month is 30

Month Temp (oC)

Space Heat Demand

(kW)

Total Heat

Demand (kW)

Electricity (kW)

CHP Heat

available (kW)

Useful CHP Heat (kW)

Supple-mentary

Heat (kW)


Supple-mentary

Electricity Needed

[1] [2] [3] [4] [5] [6] [7] [8] [9] [10]Jan 1.9 3400 3500 1600 1680 1680 1820 1200 400Dec 4.1 2850 2950 1600 1680 1680 1270 1200 400Sum of numbers in column 16450 15150 19040 12100 4350 6875 6509

GWh GWh GWh GWh GWh GWh GWhequals numbers in red

* 30 *2411.84 10.91 13.71 8.71 3.13 4.95 4.69

% heat supplied by CHP = 8.71/11.84 = 73.6%

Carbon Emissions – situation before installation

Total Demand

Efficiency CO2 factor

(tonnes/MWh)

CO2 (tonnes)

GWh

heating col [4] 11.84 0.8 0.186 2753

electricity col [5] 10.91 1 0.544 5935

total 22.75 8688

CHP worked example

Total Demand

Efficiency CO2 factor

(tonnes/MWh)

CO2

(tonnes)

GWh GWh

CHP heating [7] 8.71

CHP electricity [9] 4.95

Total CHP 13.66 0.8 0.186 3176

supplementary heating [8]

3.13 0.8 0.186 728

supplementary electricity [10]

4.69 1 0.544 872

4776

CHP worked exampleCarbon emissions before installation 8688 tonnes

Carbon emissions after installation: 4776 tonnes

Saving = 8688 – 4776 = 3911 tonnes or 45%

0

1

2

3

4

5

6

7

8

9

0 1 2 3 4 5 6 7 8 9 10 11 12

3 units

7.3 units

Gradient of Line = 3/7.3 = 0.411

Month Total Heat

Demand (kW)

Electricity (kW)


(kW)

Useful CHP Heat

(kW)

Supple-mentary

Heat (kW)


(kW)

Supple-mentary


[1] [4] [5] [6] [7] [8] [9] [10]

Jan 3500 1600 1680 1680 1820 1200 400

Feb 2850 1450 1680 1680 1170 1200 250

Mar 1725 1300 1680 1680 45 1200 100

Apr 975 1200 1680 975 696*** 504

May 475 1000 1400 475 339*** 661

June 100 1000 1400 100 71*** 929

July 100 1000 1400 100 71*** 929

Aug 100 1000 1400 100 71*** 929

Sep 725 1200 1680 725 518*** 682

Oct 1225 1350 1680 1225 875 475

Nov 1725 1450 1680 1680 45 1200 250

Dec 2950 1600 1680 1680 1270 1200 400

CHP worked example

nbs-m016 contemporary issues in climate change and energy 2010

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