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N.K. Tovey ( 杜伟贤 ) M.A, PhD, CEng, MICE, CEnv Н.К.Тови М.А., д-р технических наук Energy Science Director C Red Project HSBC Director of Low Carbon Innovation. NBS-M016 Contemporary Issues in Climate Change and Energy 2010. Introduction to Thermodynamics Combined Cycle Gas Turbines

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slide1

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

  • Introduction to Thermodynamics
  • Combined Cycle Gas Turbines
  • Combined Heat and Power
  • Heat Pumps

Lecture 2

Lecture 3

Lecture 1

1

slide2

10. Elementary Thermodynamics - History.

1) Boil Water> Steam

Problem:

Cylinder continually is cooled and heated.

2) Open steam valve

pushes piston up

(and pumping rod down)

3) At end of stroke, close steam valueopen injection valve

4) Water sprays in condenses steam in cylinder creating a vacuum and sucks piston down - and pumping rod up

Newcomen Engine

2

slide3

10. Elementary Thermodynamics - Watt Engine.

1) Cylinder is always warm

2) cold water is injected into condenser

3) vacuum is maintained in condenser so “suck” out exhaust steam.

4) steam pushes piston down pulling up pumping rod.

Higher pressure steam used in pumping part of cycle.

  • Watt Engine

3

slide4

10. Elementary Thermodynamics.

  •  Thermodynamics is a subject involving logical reasoning.
    • Much of it was developed by intuitive reasoning.
  • 1825 - 2nd Law of Thermodynamics - Carnot
  • 1849 - 1st Law of Thermodynamics - Joule
  • Zeroth Law - more fundamental - a statement
  • about measurement of temperature
  • Third Law - of limited relevance for this Course

4

slide5

10. Elementary Thermodynamics.

Carnot’s reasoning

Water at top has potential energy

Water at bottom has lost potential energy but gained kinetic energy

5

slide6

H1

H2

10. Elementary Thermodynamics.

Carnot’s reasoning

Water looses potential energy

Part converted into rotational energy of wheel

Potential Energy = mgh

  • Theoretical Energy Available = m g (H1 - H2)
  • Practically we can achieve 85 - 90% of this

6

slide7

10. Elementary Thermodynamics.

Carnot’s reasoning

  • Temperature was analogous to Head of Water
    • Energy  Temperature Difference
    • Energy  (T1 - T2)
    • T1 is inlet temperature
    • T2 is outlet temperature
  • Just as amount of water flowing in = water flowing out.
  • Heat flowing in = heat flowing out.
  • In this respect Carnot was wrong
  • However, in his day the difference was < 1%

7

slide8

10. Elementary Thermodynamics.

  • Joule 1849
  • Identified that “Lost” Heat = energy out as Work
  • Use a paddle wheel to stir water - the water will heat up
  • Mechanical Equivalent of Heat
    • Berlin Demonstration
  • Symbols
      • W - work Q - heat
  • Over a complete cycle
      • Q = W
      • Heat in +ve Heat out -ve
      • Work in -ve Work out +ve
  • FIRST LAW: “You can’t get something for nothing”

8

slide9

Heat In Q1

Work Out W

Heat Out Q2

Schematic Representation

of a Power Unit

10. Elementary Thermodynamics.

First Law:

W = Q1 - Q2

so efficiency

Heat Engine

But Carnot saw that

Heat  Temperature

  • What do we mean by temperature?

Rankine,

Kelvin?

Reamur,

Fahrenheit,

Celcius,

  • Which should we use?

9

slide10

10. Elementary Thermodynamics.

Is this a sensible definition of efficiency?

If T1 = 527oC ( = 527 + 273 = 800K)

and T2 = 27oC ( = 300K)

Note: This is a theoretical MAXIMUM efficiency

10

slide11

10. Elementary Thermodynamics.

  • Second Law is more restrictive than First
    • “It is impossible to construct a device operating in a cycle which exchanges heat with a SINGLE reservoir and does an equal amount of work on the surrounds”
  • This means Heat must always be rejected
  • Second Law cannot be proved
  • - fail to disprove the Law
  • If heat is rejected at 87oC (360K)

By keeping T2 at a potentially useful temperature, efficiency has fallen from 62.5%

11

slide12

10. Elementary Thermodynamics.

The Practical efficiency will always be less than the Theoretical Carnot Efficiency.

To obtain the "real" efficiency we define the term Isentropic Efficiency as follows:-

Thus "real" efficiency = carnot x isen

Typical values of isen are in range 75 - 80%

Hence in a normal turbine, actual efficiency = 48%

A power station involves several energy conversions. The overall efficiency is obtained from the product of the efficiencies of the respective stages.

12

slide13

10. Elementary Thermodynamics.

  • EXAMPLE:
  • In a large coal fired power station like DRAX (4000MW), the steam inlet temperature is 566oC and the exhaust temperature to the condenser is around 30oC.
  • The combustion efficiency is around 90%, while the generator efficiency is 95% and the isentropic efficiency is 75%.
  • If 6% of the electricity generated is used on the station itself, and transmission losses amount to 5% and the primary energy ratio is 1.02, how much primary energy must be extracted to deliver 1 unit of electricity to the consumer?

13

slide14

10. Elementary Thermodynamics.

  • (566 + 273) - (30 + 273)
  • Carnot efficiency = ------------------------------ = 63.9%
  • 566 + 273
  • so overall efficiency in power station:-

= 0.9 x

|

combustion

loss

0.639 x

|

Carnot

efficiency

0.75 x

|

Isentropic

efficiency

0.94

|

Station

use

0.95 x

|

Generator

efficiency

= 0.385

14

slide15

10. Elementary Thermodynamics.

Transmission Loss ~ 91.5% efficient

Primary Energy Ratio for Coal ~ 1.02

Overall efficiency

1 x 0.385 x 0.915

= -------------------------- = 0.345 units of delivered energy

1.02

i.e. 1 / 0.345 = 2.90 units of primary energy are needed to deliver 1 unit of electricity.

15

slide16

10. Elementary Thermodynamics.

  • How can we improve Carnot Efficiency?
  • Increase T1 or decrease T2
  • If T2 ~ 0 the efficiency approaches 100%
    • T2 cannot be lower than around 0 - 30oC i.e. 273 - 300 K
    • T1 can be increased, but properties of steam limit maximum temperature to around 600oC, (873K)
  • We can improve matters by the use of combined cycle gas turbine stations CCGTs.

16

slide17

10. Elementary Thermodynamics.

In this part of the lecture we shall explore ways to improve efficiency

Most require us to ensure we work with thermodynamics rather than against it

The most important equation:

slide18

11. Applications of Thermodynamics. Other modes of Electricity Generation: Open Circuit Gas turbinesc

Efficiency is low - exhaust temperature is high --- (T1 - T2)/T1

- similar to an aircraft engine

18

slide19

11. Applications of Thermodynamics.

Combined Cycle Gas Turbines

Waste Heat

Combined Cycle Gas Turbine

Practical Efficiencies:-

Gas Turbine alone

20 - 25%

Steam Turbine alone

35 - 38%

CCGT 47 - 52%

19

slide20

11. Applications of Thermodynamics.

Combined Cycle Gas Turbines: Multiple Shaft Example

0.22

0.23

1.0

Generator

Gas Turbine

0.77

0.01

0.15

Waste Heat Boiler

0.02

0.28

0.62

0.30

Generator

Steam Turbine

0.32

Condenser

Gas turbine

T1 = 950oC = 1223 K

T2 = 500oC = 823K

Electricity

Steam turbine

T1 = 500oC = 773 K

T2 = 30oC = 303K

Electricity

Isentropic efficiency ~ 80%

  • Output from Gas Turbine: 0.23 units of power to generator and 0.77 units to WHB
  • Generator is ~ 95% efficient so output ~ 0.22 units
  • Waste Heat boiler is ~ 80% efficient so there will be ~ 0.15 units lost with 0.8*0.77=0.62
  • units effective for raising steam.
  • Shaft power from Steam turbine = 0.62 * 0.486 = 0.30 units with 0.32 units to condenser
  • Total electrical output = 0.22 + 0.28 = 0.50units of which 0.03 units are used on station
  • Overall efficiency = 47%

20

slide21

11. Applications of Thermodynamics.

Combined Cycle Gas Turbines:

  • Early CCGTs had multiple shafts with separate generators attached to gas turbines
  • Some had two or more gas turbines providing heat to waste heat boilers which powered a single steam turbine
  • Modern CCGT’s tend to have a common shaft with a gas turbine and steam turbine turning a single generator.
  • Advantages of single shaft machines:
    • tend to have lower capital cost
    • Tend to have higher overall efficiencies up to 55/56% - e.g. Great Yarmouth
  • Disavantages:
    • No option to run gas turbine by itself
    • Gas Turbines can reach full output in a matter of minutes.
    • Steam turbines take 12 hours or more
  • Gas Turbines tend to have higher NOx emissions and special provision is needed to reduce these levels – e.g. injecting steam into gas turbine.
slide22

12. Applications of Thermodynamics.

Combined Heat and Power (2)

  • Heat is normally rejected at ~ 30oC
  • Too low a temperature for useful space heating
  • Reject heat at 100oC
  • i.e. Less electricity is generated, but heat is now useful
  • Typically there are boiler and other losses before steam is raised
    • Assume only 80% of energy available in coal is available.
    • And technical (isentropic efficiency) is 75%
  • Then for 1 unit of coal - electricity generated
  • case 1 = 0.8*0.75*0.639 = 0.38 units
  • case 2 = 0.8*0.75*0.555 = 0.33 units + up to 0.47 units of heat
  • or up to 0.8 units in total. Typically 10% of heat is lost so 0.73 units available
slide23
The first Law of Thermodynamics states that we can neither create or destroy energy

ie Work out = Heat in – Heat Out

  Second Law states we must always reject Heat

and efficiency =

If we could utilise all of rejected heat

The 1947 Act stated Electricity must be generated as efficiently as possible

i.e. Work/Electricity (not energy) was King

12. Applications of Thermodynamics.

Combined Heat and Power (1)

slide24

12. Applications of Thermodynamics.

Combined Heat and Power (3)

Boiler

Heat Exchanger

To District Heat Main ~ 90oC

GT

Heat Exchanger

Gas in

To District Heat Main ~ 90oC

Boiler

Heat Exchanger

Normal Condenser

To District Heat Main ~ 90oC

Back Pressure Steam Turbine

ITOC or Pass out Steam Turbine

Problem:

For most CHP plant, electrical output will be limited if there is no requirement for heat.

ITOC provides greater flexibility

Gas Turbine with CHP also Diesel/gas engine with CHP

slide25

12. Applications of Thermodynamics.

Combined Heat and Power (4)

Process

Integrated Electricity Generation, Process Heat, Space Heat and Air compression at ICI Wallerscote Plant in late 1970s

slide26

12. Applications of Thermodynamics.

CCGT with CHP (1)

Heat Lost 24 MW

Fuel in 239 MW

Steam Turbine

GT Temp

1127oC

Generator

Generator

Electricity 62 MW

Electricity 55 MW

Gas Turbine

Useful Heat 98 MW

slide27

Engine

Generator

9. Applications of Thermodynamics.

Combined Heat and Power

slide28

12. Applications of Thermodynamics.

Example of Small Scale Scheme (4)

In most cases, CHP plant is based on an approximate summer time heat load with supplementary heating provided by normal boilers in coldest months of year

slide29
Electricity generation in summer is restricted and import is highest when demand is least

12. Applications of Thermodynamics.

Example of Small Scale Scheme (5)

slide30

12. Applications of Thermodynamics.

Example of Small Scale CHP Scheme 6000 kWe (1)

Hot water and process heat demand is constant over the year at 4000 kW

Heat loss rate for buildings is 1000 kW oC-1

Existing Heating provided by gas (80% efficiency).

Mean space heat demand in January

= (15.5 – 1.9) * 1000

= 13 600 kW

This is the balance temperature – we shall discuss this in 2 weeks time.

In such schemes approximately 1.4 kW heat is rejected for every 1 kW electricity generated. In this case 8400kW

slide31

12. Applications of Thermodynamics.

Example of Small Scale Scheme (2)

Col [7] is actual amount of heat that can be usefully used. i.e if col [6] is greater than demand then the useful amount = demand

Column 3 values

= (15.5 – col [2])* 1000

15.5oC is the balance or neutral temperature at which no heating is required. Incidental gains from appliance heat and body heat increase temperature to comfort level.

Column [5] is electricity demand from Previous Sheet

Column [6] indicates the potential amount of heat which would be available. Typically it is around 1.4 times the electricity generation so

Col [6] = 1.4 * col [5] subject to a maximum electricity generation of 6000 kW

i.e. when electrical demand > 6000kW, only 6000 * 1.4 = 8400 kW will be available for heat.

Column [4] values

= col[3] + 4000

The 4000 is hot water and process heat requirement.

Maximum Electricity generation = 6000 kW electrical 8400kW heat

slide32

12. Applications of Thermodynamics.

Example of Small Scale Scheme (3)

Column [9] is actual electricity that can be generated.

If the heat demand is greater than 8400, then units can be run at full output – i.e. 6000 kW.

If heat requirement is less than 8400kW, then output of generators will be restricted to a maximum of

Col [7] / 1.4

Column [8] is supplementary heat required from back up boilers

Col [8] = col [4] – col [7]

Column [10] is additional electricity needed.

Note: highest import occurs when electricity demand is least.

The totals represent the total amount of heat or electricity generated or required over the year. Using 30 day months the totals in each column will be: mean values * 24 * 30

slide33

12. Applications of Thermodynamics.

Example of Small Scale Scheme (4)

In most cases, CHP plant is based on an approximate summer time heat load with supplementary heating provided by normal boilers in coldest months of year

33

slide34
Electricity generation in summer is restricted and import is highest when demand is least

12. Applications of Thermodynamics.

Example of Small Scale Scheme (5)

34

slide35

12. Applications of Thermodynamics.

CCGT with CHP (1) large scale

1.0

Gas Turbine

Waste Heat Boiler

Generator

Steam Turbine

Station Electricity Use

Heat Out

Condenser

Heat Losses from Mains

Generator

Electricity Out

Irrecoverable Losses

Useful Heat

slide36

12. Applications of Thermodynamics.

CCGT with CHP (2) large scale

Gas Turbine

0.2375

Electricity

1.0

0.0125

Waste Heat Boiler

Generator

0.25

0.75

0.125

Irrecoverable Losses

0.625

Gas turbine efficiency

Electricity generated: 0.25 * 0.95 = 0.2375

Energy to Steam Turbine

= 0.75 – 0.125 = 0.625

slide37

12. Applications of Thermodynamics.

CCGT with CHP (3) large scale

Gas Turbine

0.2375

0.25

1.0

0.0125

0.75

Waste Heat Boiler

0.125

Irrecoverable Losses

0.0133

0.625

0.2523

0.2656

Generator

Steam Turbine

Condenser

Generator

0.3594

Mechanical power to generator

= 0.425 * 0.625 = 0.2656

Heat to Condenser

= 0.625 – 0.2656 = 0.3594

Electricity out

= 0.95 * 0.2656 = 0.2523

steam turbine efficiency

slide38

12. Applications of Thermodynamics.

CCGT with CHP (4) large scale

Gas Turbine

0.2375

0.25

1.0

Electricity Out

0.0125

0.75

Waste Heat Boiler

0.125

0.470

Irrecoverable Losses

0.0133

0.625

0.2523

0.2656

Generator

Steam Turbine

Station Electricity Use

0.3594

Heat Out

0.0196

Useful Heat

0.3594

Condenser

0.3048

Heat Losses from Mains

0.0546

Generator

Station use of electricity

= (0.2375 + 0.25230) * 0.04 = 0.196

Useful Heat

= 0.3594 * (1 – 0.152) = 0.3048

slide39

12. Applications of Thermodynamics.

CCGT with CHP (5)

Summary of Scheme

For each unit of fuel

  • Electricity available = 0.470 units
  • Heat sent out = 0.3594 units
  • Station efficiency = 0.470 + 0.3594

= 82.9%

  • But heat is lost form mains so only 0.3048 is actually useful
  • Overall system efficiency = 0.47 + 0.3048

= 77.5%

slide40

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

13. Heat Recovery : Heat Pumps

40

slide41

13. Heat Recovery Systems and Heat Pumps

  • Parallel Plate Heat Exchanger

Hot Fluid In

Cold Fluid In

13 heat recovery systems and heat pumps
13. Heat Recovery Systems and Heat Pumps

Parallel Flow Shell and Tube Exchanger

Inefficient: maximum temperature achieved is ~ 50% of temperature difference

Hot Fluid In

Cold Fluid In

Hot Fluid

Temperature

Cold Fluid

Distance

13 heat recovery systems and heat pumps1
13. Heat Recovery Systems and Heat Pumps

Contra Flow Shell and Tube Exchanger

Inefficient: maximum temperature achieved is ~ 50% of temperature difference

Hot Fluid In

Cold Fluid In

Hot Fluid

Temperature

Cold Fluid

Distance

slide44

Operation of Regenerative Heat Exchangers

Fresh Air

Fresh Air

Stale Air

Stale Air

B

Stale air passes through Exchanger A and heats it up before exhausting to atmosphere

Fresh Air is heated by exchanger B before going into building

A

After ~ 90 seconds the flaps switch over

B

Stale air passes through Exchanger B and heats it up before exhausting to atmosphere

Fresh Air is heated by exchanger A before going into building

A

slide45

Work IN W

Heat In Q2

Schematic Representation

of a Heat Pump

13. Applications of Thermodynamics.

Heat Pumps

A Heat Pump is a reversed Heat Engine: NOT a reversed Refrigerator

Heat Out Q1

Heat Pump

If T1 = 323K (50oC)

and T2 = 273K (0oC)

Schematic Representation of a Heat Pump.

IT IS NOT A REVERSED REFRIGERATOR.

slide46

Responding to the Challenge: Technical Solutions

The Heat Pump

Compressor

Heat supplied to house

High Temperature

High Pressure

Condenser

  • Any low grade source of heat may be used
  • Typically coils buried in garden
  • Bore holes
  • Example of roof solar panel (Look East: Tuesday)

Throttle Valve

Evaporator

Low Temperature

Low Pressure

A heat pump delivers 3, 4, or even 5 times as much heat as electricity put in.

We are working with thermodynamics not against it.

Heat extracted from outside

46

slide47

13. Applications of Thermodynamics.

Heat Pumps

  • Performance is measured by Coefficient of Performance
  • (COP)
  • Theoretical Performance of 6.46
  • Practical COP in excess of 3.
  • i.e. Three times as much heat is obtained as work put in.
  • Remaining heat comes from the environment
  • The closer the temperature difference, the better the COP
  • Can be used for efficient heat recovery
  • Can recover the energy lost in electricity generation
  • Will out perform even a gas condensing boiler
  • Working with Thermodynamics - NOT against it
slide48

Condenser

Throttle

Valve

Compressor

Evaporator

13. Applications of Thermodynamics.

Heat Pumps and Refrigerators

A heat pump refrigerator consists of four parts:-

1) an evaporator (operating under low pressure and temperature)

2) a compressor to raise the pressure of the working fluid

3) a condenser (operating under high pressure and temperature)

4) a throttle value to reduce the pressure from high to low.

the norwich heat pump
The Norwich Heat Pump

Original Paper by

John Sumner

Proc. Institution of Mechanical

Engineers (1947): Vol 156 p 338

the history of the site
The History of the Site
  • The building was unique - the very first heat pump in the UK.
  • Installed during in early 1940s during the War.
  • Built from individual components which were not ideal.
  • Compressor was second hand built in early 1920’s ! for Ice making.
  • The evaporator and condenser had to be built specifically on site.
  • Refrigerant choice was limited during War - only sulphur dioxide was possible.
  • A COP of 3.45 was obtained - as measured over 2 years.
  • Even in 1940s, the heat pump was shown to perform as well as, if not better than older coal fired boiler.
the history of the site1
The History of the Site

Condenser

Compressor

Evaporator

  • The Norwich Heat Pump - note the shape of the columns
slide52

The Norwich Heat Pump

Schematic of the Norwich Heat Pump

- from John Sumner’s Book - Heat Pumps

slide54

13.6Types of Heat Pump

For Space Heating Purposes: The heat source with water and the ground will involve laying coils of pipes in the relevant medium passing water, with anti-freeze to the heat exchanger. In air-source heat pumps, air can be passed directly through the heat exchanger.

For Process Heat Schemes: the source may be a heat exchanger in the effluent of one process

slide56
Readily Available

Noise on external fans

Source temperature low when most heat needed: hence performance inferior at times of greatest need

Source temperature varies greatly:- hence cannot optimise design

13. Heat Pumps: Heat Sources

Disadvantages

Advantages

Air

  • not readily available
  • source temperature normally higher than air or ground in winter: hence improved COP
  • source temperature nearly constant: hence design can be optimised

Water

  • capital cost is great if retro-fitted
  • reasonable availability
  • moderate source temperature - better than air, worse than water
  • moderate variation in source temperature: some optimisation possible

Ground

slide57
relatively low temperature: hence good COP

possibility of heat recovery using mechanical ventilation.

Possibility of use with air-conditioning and inter-seasonal heat store if used with ground source.

can only be fitted into hot air systems:

cannot be used with most current Central Heating systems in UK.

13. Heat Pumps: Heat Supply

Disadvantages

Advantages

Air

  • higher operating temperature: hence lower COP
  • Difficult to incorporate heat recovery
  • more compact: can be incorporated with existing systems in use in UK

Water

  • Cannot be fitted retro-spectively: must be installed at time of construction.
  • Low temperature if installed as under-floor heating.
  • Possibility of using heat store in fabric.

Ground

slide58

Winnington – Tovey Heat Pump

Under floor Heating

Solar

Waste Water

Condenser

Compressor

Air Heating

Ground Loop

Stale Air

slide59

13. Absorption Heat Pump.

Desorber

Heat from external source

Heat Exchanger

W ~ 0

Compressor

Win

High Temperature

High Pressure

Heat rejected

Absorber

Condenser

Throttle Valve

Evaporator

Low Temperature

Low Pressure

Heat extracted for cooling

Absorption Heat Pump

The Win - Win opportunity

  • More electricity can be generated in summer
  • Less electricity demand in summer
  • More income from exported electricity
slide60

13. Other Types of Heat Pump

Other Types of Heat Pump

Diesel or Gas driven Heat Pump

Additional Heat can be obtained from exhaust gases

slide61

Example: Using an Air-Conditioner in a Tropical Climate.

A large hotel complex

Cooling Load

Base Load

  • Separate consumption into components: base load for lighting appliances etc.
  • Air-conditioning load:

Gradient of line = 75 kW oC-1

slide62

Example: Using an Air-Conditioner in a Tropical Climate.

  • Gradient of line = 75 kW oC-1
  • But coefficient of performance is say 2.5
          • actual cooling load is 2.5 * 75 = 225 kWoC-1
  • What is energy consumption for cooling over a period?
  • Degree-days are a measure of cooling (or heating) requirements
  • Heating and Cooling Degree-Day data are often available.
      • CDD: Base temperature is say 20.
      • external temperature …. 30 CDD = 10 for that day
      • External temperature …. 40 CDD = 20
  • Cooling degree days is sum of CDD over relevant period.
  • If CDD over period is 3000 (a typical value for some tropical countries)
  • Total demand of electricity

= 75 * 3000 * 24 = 54000000 kWh = 5 400 MWh

If Carbon factor is 800 kg / MWh,

Total Carbon emissions = 5400 * 800 = 4320 tonnes

slide63

13. Applications of Thermodynamics - Conclusions.

650 m

Our Wasteful Society

We behave as though we call in the RAF

The Heat Pump is the analogy with the crane

21 m

273 m