Problem Set Topic 1 Solutions PDF

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Lecture 2 – Electricity Supply and Demand In the UK, the electricity grid operates at a frequency of 50 Hz. What can happen to infrastructure and equipment when there is deviation from this frequency? SolutionLarge frequency drops damage transformers and induction motors due to the high magnetising ...

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Lecture 2 – Electricity Supply and Demand 1. In the UK, the electricity grid operates at a frequency of 50 Hz. What can happen to infrastructure and equipment when there is deviation from this frequency? Solution Large frequency drops damage transformers and induction motors due to the high magnetising currents required for maintaining the flux. These devices are widely used in transmission and distribution networks as well as in consumer appliances. Components, such as turbine blades, are designed to operate in a narrow band of frequencies to avoid mechanical vibrations of blades at their natural frequencies. The maximum operating time of a steam turbine at full load with is 100 minutes with a 2% change in frequency, 10 minutes with a 3% change in frequency, 1 minute with a 4% change in frequency, etc. When frequency drops, the air flow in generators and turbines will be reduced thus reducing cooling. Furthermore, the reduction in frequency causes generator control systems to increase their input power to maintain the generation and demand balance. The reduction in cooling and increase of power may result in an increase of the internal temperature of the turbine and generator windings. As the internal temperature increases, the protection devices will trip the generator thus increasing the imbalance between demand and generation. 2. *The table below shows the number of power plants online on a given day. If no frequency response was available, how long could the grid continue to meet the increased power demand if all the renewable generation tripped simultaneously? Why would this not work in practice? Power plant

Hi (s)

Si (MW)

Number

Gas CCGT

7

2000

9

Coal

8

2000

2

Nuclear

9

2000

2

Solar array

0

500

12

Wind farm

0

1000

14

Solution ∑ 𝐸 =∑ 𝐻𝑖 𝑆𝑖 = 9(7 ∙ 2000) + 2(8 ∙ 2000) + 2(9 ∙ 2000) = 194 GWs Power lost = (12 ∙ 500) + (14 ∙ 1000) = 20 GW 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑛𝑒𝑟𝑡𝑖𝑎

𝑝𝑜𝑤𝑒𝑟 𝑖𝑚𝑏𝑎𝑙𝑎𝑛𝑐𝑒

=

194 𝐺𝑊𝑠 20 𝐺𝑊

= 9.7 s

This wouldn’t work in practice as the ROCOF is too high and LFDD would occur in > 1 second 𝑑𝑓 𝑑𝑡

=

∆𝑃∙ 𝑓𝑟𝑒𝑓 2 ∙ ∑𝐸

=

20 𝐺𝑊∙ 50 𝐻𝑧 2 ∙ 194 𝐺𝑊𝑠

= 2.58 𝐻𝑧/𝑠

ES4E0 Topic 1 Problem Set

3. Describe how the increasing use of renewables in the United Kingdom can affect ‘grid stability’, which increases the risk of power cuts. In your answer, define what is meant by ‘grid stability’ and explain what effect a loss of frequency can have on the grid. Also describe the different methods currently in place to counteract any loss of generation. The below answer is a lengthy example – full marks could b be e given for a much shorter answer. Solution When there is a mismatch between electricity generation and demand (either from a loss of generation or a sudden increase in demand), National Grid will increase the amount of energy generated. This may be through ramping up gas turbines or turning on additional generators, or through pumped hydro. The ability of the grid to deal with planned (e.g. afternoon peak, big sporting events) or unplanned demand changes (generators tripping) is known as grid stability. However, increasing generation still takes several seconds, and in the meantime there is only one place for the extra energy to come from – the kinetic energy stored as inertia. The more inertia present in the grid, the slower the rate of change of frequency. ‘Traditional’ renewables that operate without a synchronous generator (Wind, Solar, etc) do not have mechanical inertia, but do provide synthetic inertia (i.e. they can deliver below their maximum power to make up for shortfalls): • Wind turbines do not provide system inertia, as the turbines rotating masses are de-coupled from the grid by AC to DC convertors. They do still store kinetic energy, as there are rotating parts. They can therefore provide ‘synthetic inertia’, sometimes called Fast Frequency Response “FFR”, via their power electronics. They rely on a mixture of power stored in the flywheel energy of the wind turbine and any ability to increase output power (for example if the turbine is running below its rated power). • Solar PV installations have no rotating parts and therefore do not provide natural inertia. However, by varying their output, it is possible for them to provide some degree of synthetic inertia. • HVDC links also have no natural inertia, however they can be operated at below their rated power to provide additional output when needed. This requires coordination between both National Grid and its foreign counterparts, and is only useful if the interconnector is not being used to full capacity into the UK. Large frequency drops damage transformers and induction motors due to the high magnetising currents required for maintaining the flux. These devices are widely used in transmission and distribution networks as well as in consumer appliances. Very large frequency fluctuation cause power generators to trip (to purposely disconnect) to prevent a phenomenon known as power islanding. National Grid hold various responses in reserve. Primary Response (within 10 seconds) and Enhanced Response (within 1 second) are used to stop frequency fluctuations and trips, and National Grid hold enough primary reserve to cover a ‘largest loss’ scenario – approximately 1 GW. Primary Response includes: • Mandatory Frequency Response (e.g. more steam diverted to a turbine in a CCGT plant to increase output). MFR is mandatory – under normal conditions thermal power stations (except nuclear) must generate a lower than maximum power with some reserve. ES4E0 Topic 1 Problem Set

• Fast Frequency Response (A commercial provider turning things on/off – including engines, generators and large plant) Enhanced Frequency Response must deliver within 1 second, and are batteries. Secondary Response is used more for predicted changes in demand and includes: • Spinning reserve, where thermal power stations keep generators spinning but not connected (response within 2 minutes). This also includes Dinowig pumped storage. • STOR (Short Term Operating Reserve) – thermal power stations with generators not spinning – response within 20 minutes.

National Grid only hold enough Primary and Enhanced reserve to cover a ‘largest loss scenario’. When this isn’t enough to cover freak loss of generation events, the last resort is Low Frequency Demand Disconnection (LFDD) at 48.8 Hz. This is more commonly known as a power-cut, and is a deliberate disconnection of customers with no warning. This is rare, but when it does happen is causes widespread disruption. The increase in renewables means that the UK grid frequency is dropping below the statutory minimum of 0.5 Hz (1%) more regularly, and closer to LFDD.

Note: the wording of Question 4 has been changed slightly. 4. *The table below shows the inertia constant, Hi, and maximum rated power, Si, for three types of power plant. Power plant Gas CCGT Solar array Wind farm

Hi (s) 7 0 0

Si (MW) 2000 500 1000

Assume that on a given day, the United Kingdom requires 30 GW of electricity, which can be provided by any number of the above power plants only. In the worst-case scenario there is a 2 GW loss of power and so there must always be enough primary frequency response (PFR) present to counteract this loss of power. What is the maximum percentage of power that can be generated by renewables to ensure that the rate of change of frequency (ROCOF) remains below a limit of 0.5 Hz/s. Assumptions: • •

Before the tripping event, power generation and load are equal to 30 GW. The largest loss of generation power possible is 2 GW.

• Each synchronous generator is operating at 70% of its rated power, with the remaining 30% held in reserve for PFR. All non-synchronous generators are operating at 100% of their rated power.

ES4E0 Topic 1 Problem Set

Solution Loss of generation power is equal to a loss of 2 GW. Reserve capacity of each synchronous generator (in this case gas power stations) is 30% x 2000 MW = 600 MW, therefore a minimum of four are required to cover a 2 GW loss. The RoCoF must not exceed 0.5 Hz/s, and is defined as: 𝑑𝑓 ∆𝑃 ∙ 𝑓𝑟𝑒𝑓 = 𝑑𝑡 2 ∙ ∑𝐸

∴ ∑𝐸 =

∆𝑃 ∙ 𝑓𝑟𝑒𝑓 −2000 𝑀𝑊 ∙ 50 𝐻𝑧 = 2 ∙ −0.5 Hz/s 𝑑𝑓 2 ∙ 𝑑𝑡

= 100000 MWs = total inertia required to stop RoCoF exceeding 0.5 Hz/s ∑ 𝐸 =∑ 𝐻𝑖 𝑆𝑖

𝐸𝑖,𝑔𝑎𝑠 = 7 s ∙ (0.7 ∙ 2000 MW) = 9800 MWs ∴ number of power stations needed is 100000 MWs / 9800 MWs = 10.2 ∴ 11 are needed to provide sufficient inertia Power from 11 gas power stations = 11 ∙ 1400 MW = 15400 MW = 15.4 GW ∴ Power from renewables = 30 GW – 15.4 GW = 14.6 GW Percentage power from renewables =

𝑝𝑜𝑤𝑒𝑟 𝑓𝑟𝑜𝑚 𝑟𝑒𝑛𝑒𝑤𝑎𝑏𝑙𝑒𝑠 𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑤𝑒𝑟

x 100% =

ES4E0 Topic 1 Problem Set

14.6 𝐺𝑊 30 𝐺𝑊

x 100% = 48.7%

Lecture 3 – Bioenergy and Waste

1. France has a population of 67 million and consumes roughly 3000 TWh of energy each year. If the average power density from biomass is 0.5 W/m2, can France sustain itself on biomass alone given that is has a land area of 640,000 km² of which 50% is considered suitable for agricultural use? Solution France annual energy consumption = 3000 TWh ∴ average daily energy use per person = 3000 x 109 kWh/365/67000000 = 122.7 kWh. Power density of biomass: 0.5 W/m2. Maximum area of France that could be covered: 50% (= 320,000 km2). Area available for each person = 320,000 km2/67,000,000 = 0.00478km2 = 4780 m2. Power available per person = 0.5 W/m2 x 4780 m2 = 2388 W. Daily energy available per person = 2388 W x 86400 seconds = 206000 kJ = 57 kWh. ∴ No, as there is only half the available energy available with the above assumptions.

ES4E0 Topic 1 Problem Set

2. Consider a biomass power plant that operates as a Carnot heat engine between the temperature limits of 10°C and 300°C. The power plant bums black locust wood as fuel. Assuming combustion and generator efficiencies of 0.90 and 0.95, find the electrical output power for a unit mass per second of fuel burned. Solution

3. In a direct-fuelled power plant, hardwood from a forest is burned at a rate of 9.6 kg/s at 360°C. The combustion efficiency is 0.85, and the turbine output power is 65 MW. If the electrical generator efficiency is 0.94, find the thermal efficiency, overall efficiency, electrical output power, rate of heat injection to a 20°C atmosphere and Carnot efficiency. If the plant capacity factor is 0.3, how much electrical energy does on plant generate in one year? ES4E0 Topic 1 Problem Set

Solution:

ES4E0 Topic 1 Problem Set

4. *A company proposes a biomass power plant that bums fuel at a temperature of 275°C in a 20°C environment. They claim that the combustion and generator efficiencies are 0.85 and 0.90, respectively. They also claims that the overall efficiency of the power plant is 0.45. Evaluate their claims. Solution: The Carnot efficiency is:

The company claims overall efficiency is 0.45. If this were the case, the thermal efficiency would be:

This exceeds the Carnot efficiency, so the company’s claims violate the second law of thermodynamics. You could also calculate the overall efficiency based on the Carnot efficiency:

The company’s claimed overall efficiency exceed this value for overall efficiency, again violating the second law of thermodynamics.

ES4E0 Topic 1 Problem Set

5. Show, using a diagram, what happens to the carbon in waste when it is sent to Energy-fromWaste plants and landfill? Solution: The diagrams below are taken ffrom rom Lecture 3, and use the approximate percentages used in L3 and the coursework. A diagram using di different fferent percentages and taking a different form, as long as justified, could also receive full ma marks. rks.

ES4E0 Topic 1 Problem Set

6. Describe the differences between ‘sequestered carbon’, ‘fossil-carbon’ and ‘biogenic-carbon’. Solution: The below answer is a length lengthy y example – full marks could b be e given for a much shorter answer. Biogenic carbon is carbon found in biologically-based materials other than fossil fuels. This includes all plants, organisms and animal waste – often referred to as biomass. Biogenic carbon dioxide emissions are defined as emissions directly resulting from the combustion or decomposition of biogenic carbon. Fossil carbon is the carbon present in fossil fuels. Fossil fuels, such as coal or gas, formed in the geological past from the remains of living organisms. Burning fossil fuels releases carbon that has been locked up in the ground for millions of years, while burning biomass emits carbon that is part of the biogenic carbon cycle. In other words, fossil fuel use increases the total amount of carbon in the biosphere-atmosphere system while bioenergy systems operates within this system; biomass combustion simply returns to the atmosphere the carbon that was absorbed as the plants grew Carbon sequestration describes long-term storage of carbon dioxide or other forms of carbon to either mitigate or defer global warming and avoid dangerous climate change. It has been proposed as a way to slow the atmospheric and marine accumulation of greenhouse gases, which are released by burning fossil fuels. Burying fossil-carbon in waste is one such was to sequester it.

7. *How does sending one tonne of waste to landfill or EfW differ in terms of CO2e emissions, when compared to sending it to a CCGT plant? Use the data in the table below, that has simplified waste to 70% food and 30% plastic, by mass. CCGT emits 0.373 tCO2/MWh. Mr Carbon = 12, Mr CO2 = 44. The energy generated from methane in a high efficiency landfill gas engine can be approximated to 2.84 MWh per tonne, or an average efficiency landfill gas engine with (2.84 MWh x 0.5)/tonne. tCO2e = tmethane x 28. Waste

Proportion in 1 tonne

Food 0.7 Plastic 0.3

Calorific Value (MWh/tonne) 1.45 7.10

Proportion of biogeniccarbon 0.14 0

Proportion fossilcarbon 0 0.52

Proportion decomposable carbon 0.07 0

Solution: Part 1: EfW Food: Energy potential per tonne waste = 0.7 x 1.45 MWh/tonne = 1.015 MWh/tonne Mass of biogenic carbon per tonne waste = 0.14 x 0.7 = 0.098 tonnes Mass of biogenic CO2 released per tonne waste = 44/12 x 0.098 = 0.359 tonnes Mass of fossil carbon released per tonne waste = 0 x 0.7 = 0 tonnes Mass of fossil CO2 released per tonne waste = 44/12 x 0 = 0 tonnes ES4E0 Topic 1 Problem Set

Mass of CO2 from CCGT per MWh generated = 1.015 MWh x 0.373 tCO2/MWh = 0.379 tonnes Net fossil CO2 from EfW = Mass fossil CO2 released – Mass of CO2 from CCGT offset = 0 tonnes – 0.379 tonnes = -0.379 tonnes Plastics Energy potential per tonne waste = 0.3 x 7.10 MWh/tonne = 2.130 MWh/tonne Mass of biogenic carbon per tonne waste = 0. x 0.3 = 0 tonnes Mass of biogenic CO2 released per tonne waste = 44/12 x 0 = 0 tonnes Mass of fossil carbon released per tonne waste = 0.52 x 0.3 = 0.156 tonnes Mass of fossil CO2 released per tonne waste = 44/12 x 0.156 tonnes = 0.572 tonnes Mass of CO2 from CCGT per MWh generated = 2.130 MWh x 0.373 tCO2/MWh = 0.794 tonnes Net fossil CO2 from EfW = Mass fossil CO2 released – Mass of CO2 from CCGT offset = 0.572 t – 0.794 t = -0.222 tonnes Total Net Fossil CO2 = (-0.379)+(-0.222) = -0.601 tonnes of CO2 per tonne of waste

Part 2: Landfill Food: Mass of decomposable carbon per tonne = 0.7 x 0.07 = 0.049 tonnes Mass of CH4 = 16/12 x 0.5* x 0.049 tonnes = 0.033 tonnes *[0.5 is because half of landfill gas, where the decomposable carbon goes, becomes CO2 and half becomes CH4] Mass of captured methane = 0.75 x 0.033 tonnes = 0.0248 tonnes Energy from methane burned = 2.84 MWh/tonne x 0.5 x 0.0248 tonnes = 0.0352 MWh Mass of CO2 from CCGT per MWh generated = 0.0352 MWh x 0.373 tCO2/MWh = 0.0131 tCO2 Mass of methane released = Mass of CH4 produced x (1 – 0.75 – (0.25 x 0.1)) = 0.033 tonnes x (1 – 0.75 – (0.25 x 0.1)) = 0.033 tonnes x 0.225 = 0.00743 tonnes CO2e of methane released = 0.00743 tonnes x 28 = 0.208 tCO2e/tonne of waste Net fossil CO2 from landfill = CO2e of methane released – Mass of CO2 from CCGT offset = 0.208 t – 0.0131 t = 0.1949 tonnes of CO2 per tonne of waste Plastics: No calculations needed as all carbon in plastics is fossil-carbon and is sequestered.

ES4E0 Topic 1 Problem Set

Lecture 4 – Marine Energy

1. What is the power per unit width of a deep-water wave of wavelength 120 m and amplitude 2.5 m? Solution:

𝑔𝑇 2 𝜆 = 𝑣𝑇, 𝜆 = 2𝜋 ∴T= √

2𝜋𝜆 𝑔

= √

2𝜋(120)

= 8.77 𝑠

9.81

h = 2.5 m ∴ H = 2h = 5 m Using 𝐽 = 𝐽=

𝜌𝑔2

32𝜋

(1025)(9.812 ) (8.77)(3)2 32𝜋

OR

𝜆 = 𝑣𝑇, 𝜆 = 2𝜋𝜆

∴T= √ 𝑣= 1

𝑇𝐻2

𝑔𝑇

2𝜋

𝑔

=

= √

𝑔𝑇 2 2𝜋

2𝜋(120) 9.81

(9.81)(8.77) 2𝜋

= 215 kW

= 8.77 𝑠

= 13.7 ms-1

1

3. 𝑃𝑡𝑜𝑡𝑎𝑙 = 4 𝜌𝑔ℎ2 𝑣

= 4 (1025)(9.81)(2.5)2 (13.7) = 215 kW (or 215 kW/m)

ES4E0 Topic 1 Problem Set

2. Water is pumped rapidly from the ocean at high tide to give an increase in water level in a tidal power basin of 3.0 m. If the tidal range is 6.0 m and if the pump/generator system is only 40% efficient, how much extra energy is gained through pumping?

Solution: Energy required to pump extra water is: 𝐸𝑝𝑢𝑚𝑝

𝐿 (𝜌𝐴𝐿)(𝑔)( ) 𝑚𝑔𝑧 2 = = 𝜂 𝜂

At low tide, this raised water falls a distance (R + L/2): 𝐿 𝐸𝑔𝑒𝑛 = (𝜌𝐴𝐿)(𝑔)(𝑅 + )𝜂 2 The proportionate gain in energy is 𝐸𝑔𝑒𝑛 −𝐸𝑝𝑢𝑚𝑝 𝐸𝑝𝑢𝑚𝑝

= =

3

= 𝐿

𝐸𝑔𝑒𝑛

𝐸𝑝𝑢𝑚𝑝

(𝑅+2)𝜂2 𝐿 2

(6+ 2)(0.4)2 1 2

–1

–1

- 1 = 1.4

∴ Energy gained is 40% more than that of pumping

3. What are the advantages and disadvantages of tidal stream generators? Solution: Advantages: • • • •

Installation is easier than with a tidal barrage/lagoon They can be consolidated onto existing structures (e.g. docks or scaffolds) Wildlife steer clear of underwater turbines due to acoustics, which is not the case with wind turbines They are the cheapest to construct of all tidal technologies (barrages, lagoons and TSGs)

Disadvantages: • •

TSG cannot produce as much energy as barrage systems They are susceptible to saltwater corrosion, and must be engineered to minimise the impact this has

ES4E0 Topic 1 Problem Set

4. *The required electrical output power of a tidal barrage is 45 MW. The tidal range is 4.6 m, and the tidal period is 10.4 hours. If the barrage has ten turbines, each with a combined efficiency of 0.30, find the required surface area of the basin.

Solution: Divide power by 10 so that power is 4.5 MW for each turbine. Then use:

Rearrange for area:

= 5.28 x 106 m2

ES4E0 Topic 1 Problem Set

5. The surface area of a tidal basin is 275 acres (1.11 km2). The barrage incorporates six turbines, each with a combined efficiency of 0.33. If the tidal period is 10.5 hours, what is the required tidal range for an electrical output power of 5.7 MW? Solution: Divide power by 6 so that power is 0.95 MW for each turbine. Then use:

Rearranging for tidal range:

6. The diameter of the turbine blades of a tidal stream generator is 4.75 m, and the efficiency of the system is 0.45. If the system consists of 18 turbines, find the electrical output power for an ocean current vel...