NBS-M016 Contemporary Issues in
Climate Change and Energy 2010
10.
11.
12.
13.
Introduction to Thermodynamics
Combined Cycle Gas Turbines
Combined Heat and Power
Heat Pumps
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 Innovation
Lecture 1
Lecture 2
Lecture 3
1
10. Elementary Thermodynamics - History.
Problem:
Cylinder continually is
cooled and heated.
1) Boil Water > Steam
2) Open steam valve
pushes piston up
(and pumping rod down)
3) At end of stroke, close steam
value open injection valve
4) Water sprays in condenses steam
in cylinder creating a vacuum and
sucks piston down - and pumping
rod up
Newcomen Engine
2
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
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 Course
of limited relevance for this
4
10. Elementary Thermodynamics.
Carnot’s
reasoning
Water at top has
potential energy
Water at bottom has lost
potential energy but gained
kinetic energy
5
10. Elementary Thermodynamics.
Carnot’s
reasoning
Water looses potential
energy
H1
Part converted into
rotational energy of
wheel
Potential Energy = mgh
H2
• Theoretical Energy Available = m g (H1 - H2)
• Practically we can achieve 85 - 90% of this
6
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
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
Over a complete cycle
Q =
W
Heat in +ve
Q - heat
Heat out -ve
Work in -ve
Work out +ve
FIRST LAW: “You can’t get something for nothing”
8
10. Elementary Thermodynamics.
First Law:
Heat In
Q1
W = Q 1 - Q2
so efficiency
Heat Engine
Work Out W

Heat Out Q2
Work Out
Heat In

Q1  Q 2
Q1
But Carnot saw that
Schematic Representation
of a Power Unit
Heat
• What do we mean by temperature?
Celcius,
Fahrenheit, Reamur,

Temperature
T1  T2

T1
Rankine,
Kelvin?
• Which should we use?
9
10. Elementary Thermodynamics.
T1  T2

T1
Is this a sensible definition of efficiency?
If T1 = 527oC ( = 527 + 273 = 800K)
and T2 =
27oC ( = 300K)
800  300

 62.5%
800
Note: This is a theoretical MAXIMUM efficiency
Work Out  Heat Out W  Q 2
why not  

Heat In
Q1
Q1  Q 2  Q 2

 100%
Q1
10
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%
800  360

 55.0%
800
11
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:-
isen
actual work out

work predicted from Carnot Efficiency
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
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
10. Elementary Thermodynamics.
(566 + 273) - (30 + 273)
Carnot efficiency = ------------------------------ = 63.9%
566 + 273
so overall efficiency in power station:-
=
0.9
|
x
0.639
|
x
0.75
|
x
0.95
x
|
combustion Carnot
Isentropic Generator
loss
efficiency efficiency efficiency
0.94
= 0.385
|
Station
use
14
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
10. Elementary Thermodynamics.
T1  T2

T1
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
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:
T1  T2

T1
17
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
11. Applications of Thermodynamics.
Combined Cycle Gas Turbines
Combined Cycle Gas Turbine
Waste Heat
Practical Efficiencies:Gas Turbine alone
20 - 25%
Steam Turbine alone
35 - 38%
CCGT 47 - 52%
19
11. Applications of Thermodynamics.
Combined Cycle Gas Turbines: Multiple Shaft Example
0.23
1.0
Gas Turbine
Generator
0.77
Waste Heat
Boiler
0.62
0.22
Electricity
0.01
0.15
0.02
0.30
Generator
Steam Turbine
0.28
Electricity
0.32
Condenser
Gas turbine
T1 = 950oC = 1223 K
T2 = 500oC = 823K
1223 823

* 0.8  23%
1223
Isentropic efficiency ~ 80%
Steam turbine
T1 = 500oC = 773 K
T2 = 30oC = 303K
773 303

* 0.8  48.6%
773
• 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
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.
21
12. Applications of Thermodynamics.
Combined Heat and Power (2)
• Heat is normally rejected at ~ 30oC
• Too low a temperature for useful space heating
(566  273)  ( 30  273)

 63.9%
(566273)
• Reject heat at 100oC
(566  273)  (100  273)
466


55.5%
(566273)
(566273)
• 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
22
12. Applications of Thermodynamics.
Combined Heat and Power (1)
• 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
W Q1 Q2
• and efficiency =
 
Q1
Q1
•
• If we could utilise all of rejected heat
W  Q 2 Q1  Q 2  Q 2


1
Q1
Q1
• The 1947 Act stated Electricity must be generated as
efficiently as possible
• i.e. Work/Electricity (not energy) was King
23
12. Applications of Thermodynamics.
Combined Heat and Power (3)
Boiler
Boiler
Heat Exchanger
Heat Exchanger
To District Heat
Main ~ 90oC
Back Pressure Steam Turbine
To District Heat
Main ~ 90oC
ITOC or Pass out Steam Turbine
GT
Heat Exchanger
Gas in
Normal
Condenser
To District Heat
Main ~ 90oC
Gas Turbine with CHP also
Diesel/gas engine with CHP
Problem:
For most CHP plant, electrical output will
be limited if there is no requirement for
heat.
ITOC provides greater flexibility
24
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
25
12. Applications of Thermodynamics.
CCGT with CHP (1)
Heat Lost
24 MW
Fuel in
239 MW
GT Temp
1127oC
Steam Turbine
Generator
Gas Turbine
Electricity
55 MW
Generator
Electricity
62 MW
Useful Heat
98 MW
26
9. Applications of Thermodynamics.
Combined Heat and Power
Engine
Generator
27
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
20000
18000
16000
14000
12000
10000
8000
6000
4000
2000
0
S upplementary Heat
Dec
Nov
Oct
Sep
Aug
July
June
May
April
Mar
Feb
Heat from CHP
Jan
kW
Heat Supply
28
12. Applications of Thermodynamics.
Example of Small Scale Scheme (5)
Electricity generation in summer is restricted and import
is highest when demand is least
Electricity Supply
9000
7000
CHP electricity
6000
Imported Electricity
5000
4000
3000
2000
1000
Dec
Nov
Oct
Sep
Aug
July
June
May
April
Mar
Feb
0
Jan
kW
8000
29
12. Applications of Thermodynamics.
Example of Small Scale CHP Scheme 6000 kWe (1)
1
Mean Temperature
(oC)
1.9
mean Electricity
Demand (kW)
7800
Hot water and process heat
demand is constant over the
year at 4000 kW
2
4.5
7200
3
9
6800
Heat loss rate for buildings is
1000 kW oC-1
4
12
6250
5
14
5800
6
16
5200
7
17
4800
Mean space heat demand in
January
8
16
4800
= (15.5 – 1.9) * 1000
9
13
5200
10
11
6200
11
9
6800
12
4.1
7800
Month
Existing Heating provided by
gas (80% efficiency).
= 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
30
12. Applications of Thermodynamics.
Example of Small Scale Scheme (2)
Column
is electricity
Month [5]
Temp
Space Heat
demand from
Sheet
(oC)Previous
Demand
(kW)
Column [6] indicates the
[1] amount
[2] of heat
[3]
potential
Jan [7]would
1.9be available.
13600of
Col
is actual
amount
which
Feb that can
11000
heat
usefully
Typically
it4.5
isbe
around
1.4
used.
i.e electricity
if9col [6] 6500
is
times
Mar the
greater
demand 3500
then the
generation
Apr thanso
12
useful amount = demand
May[6] = 1.4
14 * col [5]
1500
Col
subject
maximum 0
June to a16
electricity
generation
of
July
17
0 6000
kW
Aug
16
0
i.e.
demand
Sepwhen electrical
13
2500 >
6000kW, only 6000 * 1.4 =
Oct
11
4500
8400 kW will be available
Novheat. 9
6500
for
Dec
4.1
11400
Total Heat Electricity CHP Heat Useful CHP
Demand
(kW)
available
Heat (kW)
(kW)
(kW)
Column
3 values
[4]
[5]
[6]
[7]
Column
[4] values
17600
7800
8400
8400
15000
8400
8400
= (15.5 –7200
col [2])* 1000
10500
6800
8400
8400
= col[3] 8400
+ 4000 7500
7500
6250
5500
5800
oC is the
15.5
balance or8120
neutral 5500
4000
5200
7280
4000
temperature
at
which
no
heating
is
The
4000 is 6720
hot water and
4000
4800
4000
required. process
Incidentalheat
gainsrequirement.
from appliance
4000
6720 temperature
4000 to
heat
and body4800
heat increase
6500 level.5200
7280
6500
comfort
8500
10500
15400
6200
6800
7800
8400
8400
8400
8400
8400
8400
Maximum Electricity generation = 6000 kW electrical 8400kW heat
31
12. Applications of Thermodynamics.
Example of Small Scale Scheme (3)
Month
Total
Electricity Useful
Suppleactual
Supplementary
Column [9]
is actual
Heat
CHP mentary
electricity
Electricity
electricity that
can be (kW)
Column
[10]
is additional
Demand
Heat Heat (kW)
generated
Needed
generated.
electricity needed.
(kW)
(kW)
If[1]
the heat demand
is [5]
[4] import
[7]
[8]
[9]
[10]
Note:
highest
occurs
greater than
8400, then
Column
[8] is supplementary
Jan
17600
7800
8400
9200
6000
1800
when
electricity
demand
is
units
can
be
run
at
full
heat6000
required from back
Feb
15000
7200
8400
6600
1200 up
least.
output – i.e.
6000 kW.6800
boilers
Mar
10500
8400
2100
6000
800
April
7500
7500
5357***
893
If heat requirement
is6250
less
May
5500 then output
5800
5500
3929***
1871
than 8400kW,
Col
[8] = col [4] – col
[7]
June
4000will be 5200
4000
2857***
2343
of generators
July
4800of
4000
2857***
1943
restricted 4000
to a maximum
Aug
4000
4800
4000
2857***
1943
Col
[7]
/
1.4
Sep
6500
5200
6500
4643***
557
Oct
8500
6200
8400
100
6000
200
Nov
10500
6800
8400
2100
6000
800
Dec
15400
7800
8400
7000
6000
1800
GWh
GWh
GWh
GWh
GWh
GWh
TOTAL 78.48
53.75
58.97
19.51
42.12
11.63
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
32
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
20000
18000
16000
14000
12000
10000
8000
6000
4000
2000
0
S upplementary Heat
Dec
Nov
Oct
Sep
Aug
July
June
May
April
Mar
Feb
Heat from CHP
Jan
kW
Heat Supply
33
12. Applications of Thermodynamics.
Example of Small Scale Scheme (5)
Electricity generation in summer is restricted and import
is highest when demand is least
Electricity Supply
9000
7000
CHP electricity
6000
Imported Electricity
5000
4000
3000
2000
1000
Dec
Nov
Oct
Sep
Aug
July
June
May
April
Mar
Feb
0
Jan
kW
8000
34
12. Applications of Thermodynamics.
CCGT with CHP (1) large scale
1.0
Gas Turbine
Generator
Electricity
Out
Irrecoverable
Losses
Waste Heat
Boiler
Generator
Steam Turbine
Station Electricity Use
Condenser
Heat Out
Useful
Heat
Heat Losses
from Mains
35
12. Applications of Thermodynamics.
CCGT with CHP (2) large scale
1.0
Gas Turbine
0.25
Electricity
0.0125
0.75
Waste Heat
Boiler
0.2375
Generator
0.125
Irrecoverable
Losses
0.625
Inlet temperature to gas turbine
Exhaust temperature from gas turbine
Combustion Losses from stack, generator
and WHB
Isentropic efficiency of both turbines
Generator efficiencies
Temperature
Temperature (K)
1127 oC
1400
660 oC
933
12.50%
Note Error in
Gas turbine efficiency (1400)(933)*0.750.25
(1400)
Electricity generated: 0.25 * 0.95 = 0.2375
75.0%
95.0%
Handout
Energy to Steam Turbine
= 0.75 – 0.125 = 0.625
36
12. Applications of Thermodynamics.
CCGT with CHP (3) large scale
1.0
Gas Turbine
0.25
0.0125
0.75
Waste Heat
Boiler
0.2375
Generator
0.125
0.0133
0.625
0.2656
Steam Turbine
Generator
0.3594
Condenser
(850)( 368)
*0.750.425
steam turbine efficiency
(850)
Inlet temperature to steam turbine
Condenser temperature
Irrecoverable
Losses
0.2523
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
Temperature Temperature (K)
577 oC
850
95 oC
368
37
12. Applications of Thermodynamics.
CCGT with CHP (4) large scale
1.0
0.25
Gas Turbine
Generator
0.0125
0.75
0.125
Waste Heat
Boiler
Irrecoverable
Losses
0.0133
0.625
0.2656
Steam Turbine
0.3594
0.2375
Generator
Electricity
Out
0.470
0.2523
Station Electricity Use
Heat Out
Condenser
0.0196
Useful Heat
0.3594
Heat Losses
from Mains
0.0546
Station use of electricity
4.0%
Distribution losses on heating mains 15.2%
0.3048
Station use of electricity
= (0.2375 + 0.25230) * 0.04 = 0.196
Useful Heat
= 0.3594 * (1 – 0.152) = 0.3048
38
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%
39
NBS-M016 Contemporary Issues in
Climate Change and Energy
2010
13. Heat Recovery : Heat Pumps
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 Innovation
40
13. Heat Recovery Systems and Heat Pumps
• Parallel Plate Heat Exchanger
Hot Fluid
In
Cold Fluid In
41
13. Heat Recovery Systems and Heat Pumps
Parallel Flow Shell and Tube Exchanger
Hot Fluid In
Temperature
Cold
Fluid
In
Hot Fluid
Cold Fluid
Distance
Inefficient: maximum temperature achieved is ~ 50% of temperature difference
42
13. Heat Recovery Systems and Heat Pumps
Contra Flow Shell and Tube Exchanger
Hot Fluid In
Temperature
Cold
Fluid
In
Hot Fluid
Cold Fluid
Distance
Inefficient: maximum temperature achieved is ~ 50% of temperature difference
43
Operation of Regenerative
Heat Exchangers
Fresh
Air
B
A
Stale air passes through
Exchanger A and heats it up
before exhausting to atmosphere
Stale
Air
After ~ 90 seconds the flaps switch
over
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
Fresh Air is heated by exchanger
B before going into building
Fresh
Air
B
A
Stale
Air
44
13. Applications of Thermodynamics.
Heat Pumps
Heat
Out Q1
Heat
Pump
Work IN
W
Heat In
Q2
Schematic Representation
of a Heat Pump
A Heat Pump is a reversed
Heat Engine: NOT a
reversed Refrigerator
COP 
Heat Out
Work In
COP 

Q1
Q1  Q 2
T1
T1  T2
If T1 = 323K (50oC)
and T2 = 273K (0oC)
323
COP 
 6.46
323  273
Schematic Representation of a Heat Pump.
IT IS NOT A REVERSED REFRIGERATOR.
45
Responding to the Challenge: Technical Solutions
The Heat Pump
Heat supplied
to house
High Temperature
High Pressure
Condenser
Throttle
Valve
Compressor
Evaporator
Heat extracted
from outside
Low Temperature
Low Pressure
Any low grade source of heat may be used
• Typically coils buried in garden
• Bore holes
• Example of roof solar panel (Look East: Tuesday)
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.
46
46
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
47
13. Applications of Thermodynamics.
Heat Pumps and Refrigerators
A heat pump refrigerator consists of four parts:-
Throttle
Valve
Condenser
Evaporator
Compressor
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.
48
The Norwich Heat Pump
Original Paper by
John Sumner
Proc. Institution of Mechanical
Engineers (1947): Vol 156 p 338
49
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.
50
The History of the Site
Evaporator
Compressor
Condenser
• The Norwich Heat Pump - note the shape of the columns
51
The Norwich Heat Pump
Schematic of the Norwich Heat Pump
- from John Sumner’s Book - Heat Pumps
52
The Norwich Heat Pump
53
13.6 Types of Heat Pump
air
Heat
Sink
air
air to air
water
air to
water
solid
air to solid
Heat Source
water
water to
air
water to
water
water to
solid
ground
ground to air
ground to
water
ground to solid
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
54
13.6 Types of Heat Pump
Some Examples
Air to air:Air to water:Air to solid
Water to air
Water to water
Water to solid
ground to air
ground to water
ground to solid
Refrigeration vehicles, many simple heat pumps,
most air-conditioning plants.
Proposed UEA scheme in 1981
? May be relevant in a case where heat recovery from
exhaust air is recovered - ?? A variant of ZICER possible use in Academic Building East??
Ditchingham Primary School
Norwich Electricity Board Heat Pump during War;
Royal Festival Hall. Southampton Geothermal
Scheme.
Proposed Duke Street Refurbishment
? A scheme with cooling in summer and heating in
winter with inter-seasonal heat storage
ENV demonstration scheme. No longer available
John Sumner's Bungalow: Now the preferred route
for heat pumps except where water source is available
55
13. Heat Pumps:
Advantages
Air
Water
Heat Sources
Disadvantages
• 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
• source temperature normally
higher than air or ground in
winter: hence improved COP
• source temperature nearly
constant: hence design can be
optimised
• not readily available
• reasonable availability
• moderate source temperature better than air, worse than
Ground water
• moderate variation in source
temperature: some optimisation
possible
• capital cost is great if retrofitted
56
13. Heat Pumps:
Advantages
Air
• relatively low temperature:
hence good COP
• possibility of heat recovery
using mechanical ventilation.
• Possibility of use with airconditioning and inter-seasonal
heat store if used with ground
source.
• more compact: can be
incorporated with existing
systems in use in UK
Water
• Low temperature if installed as
under-floor heating.
Ground • Possibility of using heat store in
fabric.
Heat Supply
Disadvantages
• can only be fitted into hot air
systems:
• cannot be used with most
current Central Heating
systems in UK.
• higher operating temperature:
hence lower COP
• Difficult to incorporate heat
recovery
• Cannot be fitted retrospectively: must be installed
at time of construction.
57
Winnington – Tovey Heat Pump
Under floor
Heating
Condenser
Solar
Compressor
Air Heating
Waste Water
Ground Loop
Stale Air
58
13. Absorption Heat Pump.
Absorption Heat Pump
Heat from external
source
The Win - Win opportunity
Heat rejected
High Temperature
High Pressure
Desorber
Condenser
Heat
Exchanger
Throttle
Valve
Compressor
W
W in
~0
Evaporator
Absorber
Heat extracted for
cooling
Low Temperature
Low Pressure
• More electricity can be generated in summer
• Less electricity demand in summer
• More income from exported electricity
59
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
60
Example: Using an Air-Conditioner in a Tropical Climate.
A large hotel complex
Mean Electricity (kW)
2000
Cooling Load
1500
1000
500
Base Load
0
0
10
20
30
40
50
Mean External; Temperature
• Separate consumption into components: base load for lighting appliances etc.
• Air-conditioning load:
Gradient of line = 75 kW oC-1
61
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
62
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
63