It is hoped this will be a useful resource for your revision. I intend to add queries and answers as and when these are asked, so please consult this page before contacting me and check the answer(s) do not already exist on this page.

• General Questions on format of exam and also procedures for efficient answering.
• General Definitions/Units including query on Specific Heat from Kathryn
• The Pool, New Electricity Trading Arrangements, NFFO, Renewable Obligation
• 2003 Numeric Example of New Electricity Trading Arrangements
• Wind Energy Includes 2 questions from 2003 class
• Tidal Energy A question about basin heights from 2003 class
• Questions on Biomass/ Chp / Cofiring and Specific heat of water asked by Kathryn

Queries about Specific Questions in Examples Handouts

• A factor of 1000 often appears in questions in examples booklet - these are almost always assoicated with the density of water which is 1000 kg per cubic metre in the SI system of units. Such conversions will occur in hydro examples, cases where storages is involved and also may occur in some types of geothermal question. Since water is the basis behind the whole definition of SI units, it is assumed that this is known. Only where the density differs - e.g. for sea water for tidal/wave questions would a spcific value be given.
• Why does a factor of 10 often occur - should this not be 9.81 representing the acceleration due to gravity - yes it is the acceleration due to gravity, but 10 is close enough - remember the questions will almost always ask "Estimate" and not "Calculate" which allows approximation for ease of use - but you should always specify that this is your assumption.
• How do you convert from Watts to Joules - these are two completely different units and YOU CANNOT CONVERT between them. Watts is Joules per second so if you have something running at X watts for say an hour then the energy consumed in that period will be 3600 X where 3600 is number of seconds in an hour. In other words, the quantity JOULES requires a time period. If talking about a powere generation system e.g. nuclear or renewable etc., then there will be a load factor which must be also used in the multiplication. e.g. if the load factor in the above example is 30% the total energy consumed in one hour would be 0.3 * 3600 X.
• What is "Load Factor" - the load factor is an indication of how much a system is used. If it is used at its maximum or rated output continuously, then it would have a load factor of 100%. If it is running at less than full output then the laod factor will be the average overall output as a proportion of the maximum potential output. Thus the laod factor allows for maintenance, breakdowns, variations in availability of wind etc.
• What is "Load Following" - Load following is a measure of how well a generation plant can follow changes in demand. Thus if the demand falls, then it would be helpful if the output from a generator could reduce in response. Clearly, renewable resources cannot load follow at all, while fossil fuels are probably the best. Selected nuclear stations are better than others, thus PWR's can load follow while Magnox reactors cannot. The issue here is one of xenon poisoning which occurs in all reactors, but is more significant in Magnox reactors as they use unenriched uranium (at 0.71% compared to 4% in PWRs).

• Is it true that the Pool ceased operation in March 2001? - Yes the Pool ceased operation on 27th March 2001, and electricity is now supplied under the New Electricity Trading Arrangement. However, it is important to understand the changes which have taken place and the interim situation the POOL had for 10 years following Privatisation. A list of key WEB links is accessible from the main Energy WEB links page - for the latest information consult the OFGEM WEB site linked from the above Energy WEB Link. While a numeric example of the operation of the POOL appeared some years ago, such a question would now be irrelevant. Equally, except in very simple terms, it is unlikely that any numeric question would be set on NETA as its operation is still under review.
• I am confused over the term "SUPPLIER" in the context of NETA - A SUPPLIER does not GENERATE electricity. GENERATORS do the generation but are NOT IN GENERAL SUPPLIERS. In the context of NETA, the word SUPPLIER has a very special meaning. In essence it refers to a TRADER of ELECTRICIY who SUPPLIES electricity to a CUSTOMER. It has nothing to do with generation.
• What are the main differences between the POOL and NETA - this is lengthy to answer - consult full lecture notes. However, the most important poitns are -
  • in the pool only generators bid and there was little involvement of suppliers /consumers into the prices set.
  • in NETA most electricity is traded by bilateral agreements and it is only the balancing to ensure there is security of supply that come into the NETA trading arrangements. The balancing mechanism is operated by the National Grid. Currently GENERATORS and SUPPLIERS must notify their agreed traing position by 3.5 hours before the start of the contract period. After that time balancing is in operation in which generators can be asked to increase or decrease output and conversely suppliers can be asked to decrease or increase demand. Note that generators increase supply (+VE) or decrease supply (-VE) whereas the reverse sign convention applies to suppliers (i.e. increase in demand (more negative), decrease in demand (less negative).

    In July 2002, the Gate Closure Period was reduced to 1 hour before real time.

• Why is large scale HYDRO not exempt from the Climatic Change Levy? - This is Government Policy - it does not seem logical - there are many inconsistencies about the Renewable Obligation - e.g.:-
  • The premable says the driving force is the need to reduce Carbon Dioxide levels. However, the proposed 10% renewable component will do absolutely nothing towards this. Since the whole of the new renewables will be taken up meeting the increase in the demand for electricity. In addition, if nuclear stations are closed and not replaced, we shall see an increase in carbon dioxide - despite what Prescott and Meacher may say.
  • the emphasis is solely on grid connected electricity - nothing is proposed that would help heat generation - e..g solar water heaters. - see the the main Energy WEB links page and in particular the link to the Royal Commission on Environmental Pollution.

• question about cut in and cut out speeds
• question about wind energy rating curve and determination of power at a given windspeed
• area required for wind turbines - section 1.5 of notes

• What is the cut in speed, rated output and cut out speed when referred to Wind Energy - The output from a Wind Turbine is proportional to the cube of the wind velocity. This means that doubling wind speed the output goes up by 8 times. For various technical reasons, it is in appropriate to have generation below a given threshold - typically 3.5 - 5 m/s. In any case the actual output below these speeds will be very small. In Question 7 of 1999, the cut in speed would be around 4 m/s. The rated output in the example in question 7 (1999) is 330 kw and this is achieved once the wind spead reaches 12 m/s. Above this the blades are partially feathered such that the output is constant at the rated output speed. Above 21 m/s in this example the wind is so strong that damage may occur and the machine owuld be shut down and no output would be possible. Thus speed is the cut out speed.

• I have a query re q1 1999 How do you calculate the output when the wind speed is below 12m/s? Energy notes page 91 states that at 10m/s the output is 76% ie 228 kw, but I can't get that since 330 x .76 = 250.8 kw, when checking against the first example answer of the 1999 paper it shows 280 kw at a mean wind speed of 10m/s.
  

  
  Fig. 16.3 Turbine rating Curve.

  In the example shown in Fig. 16.3 (Page 91) there is a turbine rating curve for a 300 kW machine. Each machine will have a different rating curve, and to get the output at any windspeed you must read of the relev ant power for that windspeed. In the example shown in Fig. 16.3, the power at 10 m/s is 228 kW which happens to be 76% of the rated output of 300 kW. However, the rating curves vary, and this percentage figure is only relevant for that machine.

  In question 1 (1999), there is a different rating curve as shown below, and in this case reading off from the graph gives a figure of 280 kW. It is necessary to read off the power at each windspeed when constructing the table.

  

  I am confused over some of the area calculations with regards to Wind Turbines

  I agree, there is a little confusion which arose because I changed the turbine size this year from the older 300 kW to the 1.5 MW size, and one or two lines remain which refer to the older version. The overall figures are correct. The following summarises the differences which are higlighted where there are changes. In one or two cases redundant lines have been deleted.

  Hopefully the following will clear up the position

  1.5.1. Nuclear Lobby calculation

  Sizewell generates 1188 MW at a load factor of 75% i.e. equivalent to 891 MW (891000 kW) continuous output

  The Swaffham Turbine is rated at 1.5 MW and has a blade diameter of 67m

  However, Wind Turbines have a load factor of 30% (extensive data now available from Llandinam), and an availability of 95%

  Mean output per turbine is 1.5*0.3 *0.95 = 427.5kW

  So total number of turbines needed = 891000/427.5

  = 2084 turbines - this final value is correct despite above line

  for a square array there will be one turbine on each area 670 x 670 m or 0.4489 sq km [ or 0.389 sq km for a triangular array].

  So area for turbines is 935 sq km for a square array. - but 810 sq km for a triangular arrayP>

  1.5.3 Compare like with like

  Each Wind Turbine needs an area of 67 m dia around each turbine on which the ground has been stabilised to take weight of 20+ tonne cranes. Also tracks for access to each turbine are needed.

  and an area of track 2.5m wide (and 670m long for each turbine) would be 1675 sqm = (2.5 * 670)

  area around each turbine ~ 3525 sqm [ p D²/4 = p 67 x 67 /4 ]

  So the total area of land unavailable for other purposes per turbine is around 5200 sq m or for all turbines is 10.8 sq km

---

Section 7c - 1991 Worked Examples

Questions appear in itallic - answers in red

NO QUERIES AT PRESENT

---

Section 7d - 1993 Worked Examples

Questions appear in itallic - answers in red

• Question 5: does the 300 in this line refer to the output power or height:
  5 x 3600 x 5 x 300 x 10^6 = 2.7 x 10^13 J
  - here we are calculating the energy generated in 5 hours from a knowledge of the rated output of the turbines so it refers to 300 MW
• Question 5: does the 300 in this line refer to the output power or height: 0.9 x 10 x 300 x10^3
  - in this case it is the height as we are trying to get the volume of water - note the denominator is or the form ....m g h and the 10^3 is to allow for the density of water ..1000 kg per cu metre
• Question 5: why does 0.9 appear twice as the denominator on this line: 0.9 x 0.9
  - this is because we must allow for the 90% efficiency pumping the water uphill and also the 90% efficiency during the generation phase.

---

Section 7e - 1995 Worked Examples

Questions appear in itallic - answers in red

NO QUERIES AT PRESENT

---

Section 7f - 1997 Worked Examples - [see part 1 document]

Questions appear in itallic - answers in red

• Question 1: Should not the geothermal energy production be modified to allow for the distribution losses? -
  NO - we have included the distribution losses in the calculation already. The demand for heat is 20 MW, but allowing for distrubution losses we need to supply 22.22 MW. Initially the boielrs will have to supply all of this, but after including geothermal component we will reduce the heat needed from coal by 6MW, and hence this is the figure used to evaluate saving. Please note that the 10 (cubed) in the following line arises solely from the need to convert a volume of water into a mass - i.e. 1000 kg per cubic metre

  = 71.65 * (80 - 60) * 4.1868 * 10^3 = 6.00 MW

---

Part 2 Section 2 - 1999 Worked Examples

Questions appear in itallic - answers in red

NO QUERIES AT PRESENT

---

Part 2 Section 4 - 2001 Worked Examples

Questions appear in itallic - answers in red

NO QUERIES AT PRESENT

---

Part 2 Section 5 - 2003 Worked Examples

Note: Handout shows Section 4 it should be Section 5

Questions appear in itallic - answers in red

NO QUERIES AT PRESENT

---

---

Questions asked by Kathryn - Sunday 1st May 2005. Questions appear in itallic - answers in red

Hi Keith,

Just a few quick qu I was hoping you could help me with.

• An answer to a past qu said that Biomass used for electricity only was 30% efficient, used for direct combustion was 70%, but this could be further improved by CHP. What exactly did it mean by direct combustion? (Heat/elec?):

  The efficiency of generation of electricity for biomass, just as will coal, oil, nuclear ,gas does depend on the temperatures invovlved. You will remember we went through an example just like this last Thursday (28th April).

  The actual overall efficiency for biomass can be a little more variable depending on the feed stock and a figure in the range of 30+% for overall electricity generation would be typical, but it could be as low as 25% in some cases.

  Modern direct combustion boilers can achieve up to 85 - 90%, but rather lower if using biomass because the quality of the fuel is usually not as precise as say for oil. Indeed, overall efficiencies using CHP might achieve up to 80%, but this depends on many things inlcuding the losses in the heating main which in some cases can be significant. The best way to sort out what is going on is to draw a flow diagram and follow the steps we did in the worked example last Thursday

• I couldn't find the specific heat capacity of water in the env data book. Is this 4.1868kJ/kg/K?

  The specific heat of water is indeed 4.1868 kj/kg/degC and it is in the Data Book on page 113 of the 2004 edition. The page number may be slightly different for the 2005 edition available in the exams, but it will be the first page of the Energy Section.

• Co-firing- this is where biomass is burned in combination with another fuel (coal?) in a coal fired power station isn't it? Or can it apply to just biomass being burnt in a coal fired power st (without the coal)?

  Co-firing can be used when more than one fuel is used in a pwoer station - some stations can burn a combination of coal and oil (or gas). Most coal fired power stations require some oil firing or gas firing to start - this owuld not normally be classified as co-firing. Co-firing would be the deliberate mixing of fuels. This can be beneficial if one fuel is in short supply - e.g. in the miners strike in 1985.

  More recently it has been used in conection of burning biomass in fossil fuel power stations. The rational behind this is that until there is a market, farmers won't grow, and unless there is a fuel chain, investors will not build biomass stations. The idea was thus to tgry to promote a biomass growing industry. Biomass used in co-firing qualifies for ROCs, but will be phased out progressively by 2016 - see last section of handout 5.

  For fossil fuel generators cofiring has an added advantage in that it will reduce carbon emissions and thsu they will not have to have so many carbon emission credits (under EU-ETS which came into force on 1st January 2005).

  Co firing implies two or more fuels. In reality, much of the cofiring to date has not actually helped growers - Ironbridge - for example has been importing Malaysian Palm Oil Nuts for co-firing.

  If a station were converted solely to biomass (no coal) then that would not be classified as co-firing, but all our coal fired stations are at least 500 MW in capacity - way beyond the scope of any biomass plant (remember the practical - 20 MW reuqired around 85 sq km).
  

  ---

  ---

  This page is maintained by Keith Tovey (e-mail: k.tovey@uea.ac.uk).