Energy Conversion Tecnologies

Energy Conversion Technologies
(Part 3)

Source: General Electric

Brunel University 2013, Dr. Z. Dehouche

1

CONTENTS – Part III
1. Conversion of thermal to mechanical energy
2. Conversion
electricity

of

thermal

and

chemical

to

3. Nuclear energy and its operation
4. Environmental impact of power plant operation
Reading List:
1.

Stephen R. Turns, Thermodynamics - Concepts And Applications, Cambridge University Press, 2006

2.

A.W. Culp, Principles of Energy Conversion, McGraw-Hill Series in Mechanical Engineering, 2nd Edition, 1991

3.

T. D. Eastop and A. McConkey, Applied Thermodynamics for Engineering Technologists - Prentice Hall, fifth
edition (1996)

4.

M.J. Moran and H.N. Shapiro, Fundamentals of Engineering Thermodynamics, John Wiley & Sons, 4th
edition (2000)

5.

R. OHayre, W. Colella, SukWon Cha and F.B. Prinz, Fuel Cell Fundamentals, John Wiley & Sons; 2nd edition
(2009)
2

Learning Objectives
• Understand the principles of different thermodynamic power
cycles
• Be able to calculate the thermal properties at state points
under some conditions
• Be able to assess the overall performance of power cycles
• Be aware of the principle of the direct conversion of thermal
to electricity
• Know the basic principle of fuel cells; their types and
applications
• Be aware of the basic concepts of nuclear reactors and their
operations
• Be aware of the pollutions caused by the energy conversion
systems
3

Introduction
Basic Concepts and Definitions
• The first law of thermodynamics defines the
relationship between

the various forms of energy present in a system
(kinetic and potential):
 the work which the system performs and the
transfer of heat
• The 1st Law of Thermodynamics is a statement of
the Conservation of Energy Principle:

in terms of heat, work, and internal energy
4

Introduction

Energy is neither created nor destroyed during a
process; it can only change forms

Energy of the universe (system + surroundings) is
constant

The heat Q transferred to the control volume is
equal to the shaft work W and the change in the
internal energy ∆U

Universe=System+ Surroundings
5

Second Law - Heat Engines
• The second law of thermodynamics asserts that energy has
quality as well as quantity, and
 that processes occur in the direction of decreasing quality of
energy

It is impossible to extract an amount of heat QH from a hot
reservoir and use it all to do work W

Some amount of heat QC must be exhausted to a cold reservoir

This precludes a perfect heat engine

6

Chapter 1 - Conversion
Mechanical Energy

of

Thermal

to

 The

devices or systems used to produce a net power
output are often called engines, and
 the thermodynamic cycles they operate on are
called power cycles

Thermodynamic cycles often use quasistatic (ideal)
processes to model the workings of actual devices
 A thermodynamic cycle is a series of thermodynamic
processes transferring heat and work, while
 varying pressure, temperature, and
other state variables, eventually returning a
system to its initial state
7

 Thermodynamic cycles are categorized as gas cycles and
vapour cycles, depending on the phase of the working
fluid:
 In gas cycles, the working fluid remains in the
gaseous phase throughout the entire cycle, whereas
 In vapour cycles the working fluid exists in the
vapour phase during one part of the cycle and
in the liquid phase during another part
 Thermodynamic cycles are also categorized as closed
and open cycles:
 In closed cycles, the working fluid is returned to the
initial state at the end of the cycle and is recirculated
 In open cycles, the working fluid is renewed at the
end of each cycle instead of being recirculated
8

 Heat

engines are
combustion engines,

categorized

as

internal

and

external

depending on how the heat is supplied to the working fluid
 In external combustion engines (such as steam power
plants), heat is supplied to the working fluid from an external
source:


furnace, geothermal well, nuclear reactor, or solar
radiation

Geothermal power plant

9

In internal combustion
(such
as
engines
automobile engines), this
is done by burning the
fuel within the system
boundaries
10

 Heat engines are designed
for
the
purpose
of
converting thermal energy
to work:
The thermal efficiency
ηth is the ratio of the net
work produced by the
engine to the total heat
input:
η th =

net work output (kJ )
added heat (kJ )

Wnet
=
Qa
11

 The working fluid (such as an ideal gas or water)

moves through many thermodynamic states in a
never-ending cyclic process
The detailed structure of the heat engine contains
the following four steps:
1. Isothermal absorption of heat from the hightemperature reservoir
2. Adiabatic (isentropic) production of work
3. Isothermal rejection of heat to the lowtemperature reservoir
4. Adiabatic work done on the fluid to return it
to the state at the start of step 1
12

• The mathematical equation for an ideal fluid undergoing a
reversible (i.e., no entropy generation) adiabatic process:

k

PV = C
where P is pressure, V is volume, and k the specific heat ratio
defined as:

k=

cp
cv

CP being the specific heat for constant pressure and CV being
the specific heat for constant volume
For a monatomic ideal gas, k = 1.66, and for a diatomic gas
(such as nitrogen and oxygen, the main components of air)
k =1.4
http://buphy.bu.edu/~duffy/semester1/c27_process_adiabatic_sim.html

13

For reversible adiabatic processes
transferred), it is also true that
k −1

P T

−k

=C

;

V

(no

heat

k −1

T =C

where T is an absolute temperature
Using the ideal gas law, this can also be written as:

P1  T1 
=  
P2  T2 

k
k −1

;

V2  T1 
=  
V1  T2 

http://www.youtube.com/watch?v=t6cGw1scLvc

1
k −1

;

P2  V1 
=  
P1  V2 

k

14

The Ideal Power Cycle― Carnot Heat Engine Cycle
• Two reversible isothermal processes alternated with two
reversible adiabatic (isentropic) processes
Thermal efficiency:

ηth ,max

TL
= 1−
TH

P-v and T-s diagrams of a Carnot cycle

15

EXAMPLE 1
The turbine in a power station extracts kinetic energy from
steam at a temperature of 800 K. The steam emerging from
the turbine has a temperature of 370 K. What is the Carnot
maximum efficiency of the turbine?
The Carnot efficiency of the heat engine is given by:

ηth ,max

TL
370
= 1−
= 1−
= 0.5375 or 53.75%
TH
800

This will be the maximum efficiency of the power station as a
whole

16

1.2 Internal combustion engine cycles
Principle of the reciprocating internal combustion
engine
 The
reciprocating
engine (basically a
piston–cylinder device)
is the powerhouse of
the vast majority of:

automobiles,
trucks, light aircraft,
ships, and electric
power generators
Reciprocating engine

17

 The piston reciprocates in the cylinder
between two fixed positions called:


The top dead center (TDC): the position
of the piston when it forms the smallest
volume in the cylinder (vmin), and

The bottom dead center (BDC): the
position of the piston when it forms the
largest volume (vmax) in the cylinder
• The distance between the TDC and the BDC
is the largest distance that the piston can
travel in one direction, and it is called the
stroke of the engine
 The diameter of the piston is called the bore

Nomenclature
reciprocating engines

for

 The air–fuel mixture is drawn into the cylinder through the intake

valve, and
 the combustion products are expelled from the cylinder through the
exhaust valve
18

 The minimum volume formed

in the cylinder when the piston
is at TDC is called the
clearance volume
 The volume displaced by the
piston as it moves between
TDC and BDC is called the
displacement volume
 The ratio of the maximum
volume formed in the cylinder
to the minimum (clearance)
volume
is
called
the
compression ratio rv of the
engine:

Displacement and clearance volumes of
a reciprocating engine

vmax
rv =
vmin

https://www.youtube.com/watch?v=Nplacsumrfw

19

 Another term frequently used in

conjunction
with
reciprocating
engines is the mean effective
pressure (MEP), given by:

Wnet
MEP =
vmax − vmin

The net work output of a cycle is
equivalent to the product of the
mean effective pressure and the
displacement volume:

(kPa )

 MEP

is the theoretical pressure
exerted on the top of the piston
during the power stroke
The engine with a larger value of MEP delivers more
net work per cycle (the power output of the engine)
20

 Reciprocating

engines are classified as:

spark-ignition (SI) engines or
 compression-ignition (CI) engines,
 depending on how the combustion
process in the cylinder is initiated:

In SI engines, the combustion of the air–fuel
mixture is initiated by a spark plug

In CI engines, the air–fuel mixture is selfignited
 as a result of compressing the mixture
above its self-ignition temperature
21

Otto cycle: The ideal cycle for spark-ignition engines
 The Otto cycle is the ideal cycle for spark-ignition reciprocating engines
 In most spark-ignition engines, the piston executes four complete
strokes within the cylinder for each thermodynamic cycle

Actual and ideal cycles in spark-ignition engines and their P-v diagrams

22

 Initially, both the intake and the exhaust valves are closed, and the
piston is at its lowest position (BDC):

During the compression stroke, the piston moves upward, compressing
the air–fuel mixture:
 Shortly before the piston reaches its highest position (TDC), the
spark plug fires and the mixture ignites, increasing the pressure and
temperature of the system

The high-pressure gases force the piston down,
 which in turn forces the crank-shaft to rotate, producing a useful
work output during the expansion or power stroke

At the end of this stroke, the piston is at its lowest position (BDC), and
the cylinder is filled with combustion products

Now the piston moves upward one more time, purging the exhaust
gases through the exhaust valve (the exhaust stroke),

and down a second time, drawing in fresh air–fuel mixture through the
intake valve (the intake stroke)
23

P-v and T-s diagrams of the ideal Otto cycle

• The ideal Otto cycle consists of four internally reversible
processes:
1-2 Isentropic compression
2-3 Constant-volume heat addition
3-4 Isentropic expansion
4-1 Constant-volume heat rejection
24

• The Otto cycle is executed in a closed system, the overall
energy balance for any of the processes is expressed as:

(Qa − Qr ) + (Wa − Wr ) = ∆U

(kJ )

• No work is involved during the two heat transfer processes
(2-3 and 4-1) since both take place at constant volume
• Therefore, heat transfer to and from the working fluid can be
expressed as

Qa = U 3 − U 2 = mcv (T3 − T2 )
Qr = U 4 − U1 = mcv (T4 − T1 )
Where m is the mass of the working fluid
25

• The thermal efficiency of the ideal Otto cycle

 T4

T1  − 1
T1
Wnet Qa − Qr
T4 − T1


=
=1−
=1−
η th =
Qa
Qa
T3 − T2
 T3

T2  − 1
 T2

Processes 1-2 and 3-4 are isentropic, and v2=v3 and v4=v1 .Thus,

T1  V2 
=  
T2  V1 

k −1

 V3 
=  
 V4 

k −1

T4
=
T3

Substituting these equations into the thermal efficiency relation give

(1− k )

η th = 1 − rv

26

Where rv is the compression ratio :

vmax v1 v4
rv =
= =
vmin v2 v3
and k is the specific heat ratio

k=

cp
cv
27

Diesel cycle: The ideal cycle for compressionignition engines
 The Diesel cycle is the ideal

cycle for
engines

CI

reciprocating

 In spark-ignition engines, the
air–fuel mixture is compressed
to a temperature that is below
the auto-ignition temperature
of the fuel,
 and the combustion process
is initiated by firing a spark
plug

In diesel engines, the spark plug is
replaced by a fuel injector, and only air
is compressed during the compression
process

 In CI engines, the air is compressed to a temperature that is above the
auto-ignition temperature of the fuel,
 and combustion starts on contact as the fuel is injected into this hot air
28

T-s and P-v diagrams for the ideal Diesel cycle

 The ideal Diesel cycle processes are: 1-2 isentropic compression, 2-3
constant-pressure heat-addition, 3-4 isentropic expansion, and 4-1
constant-volume heat rejection
 The thermal efficiency of the ideal Diesel cycle

 T4 
T1  − 1
T1
Wnet
mcv (T4 − T1 )
Qr
T4 − T1


= 1−
= 1−
= 1−
= 1−
η th =
Qa
Qa
mc p (T3 − T2 )
k (T3 − T2 )
 T3 
kT2  − 1
 T2 

29

• We now define a new quantity, the cut-off ratio rcf, as the
ratio of the cylinder volumes after and before the combustion
process:

v3
rcf =
v2

• Utilizing rcf and the isentropic ideal-gas relations for
processes 1-2 and 3-4, the thermal efficiency relation reduces
to

ηth = 1 −

k
cf
( k −1)
v

r −1

kr

(r

cf

− 1)

Where rv is the compression ratio :

vmax v1
rv =
=
vmin v2

30

Take the cv=0.7176 kJ/(kg K) and the specific heat ratio, k=1.4 for the compression
and the expansion processes

Solution: The Otto cycle and given data
following p-v and T-s plots,

are shown in

31

• The temperature and pressure at state 1 are given:

T1=27˚C=300 K, p1=100 kPa
• The process from state 1 to state 2 is isentropic, so
the reversible adiabatic relationships for an ideal
gas can be used, as

V

k −1

T =C

and

V1
= rv = 8
V2

k −1

T2  V1 
=  
T1  V2 
T2 = 300 x8 0.4 = 689 .2 K
32

k

k

PV = C

p2  V1 
=  
p1  V2 
p2 = 100 x81.4 = 1837 .9kPa

• The process from state 2 to state 3 is constant volume, so
the heat transfer is equal to the change of internal energy
according to the first law,

qa = (u3 − u 2 ) = cv (T3 − T2 )
kJ
kJ
(T3 − 689.2 )
1840
= 0.7176
kg
kg .K
P
T3 = 3253 .5 K
=C
T
 T3 
 3253 .5 


p3 = p2   = 1837 .9
 = 8676 .1 kPa
 689.2 
 T2 
33

• The process from state 3 to state 4 is isentropic, so the
following relationships hold
k

PV = C
T4  V3 
=  
T3  V4 

V3 1
=
V4 rv

and
k −1

1
T4 = 3253 .5 
8
k
p4  V3 
=  
p3  V4 

0.4

= 1416 .2 K

1.4

1
p4 = 8676 .1 
8

= 472 .1kPa
34

• Thermal efficiency

η th = 1 −

1
( k −1)

rv

1
= 1 − 0.4 = 0.565 or 56.5%
8

35

I.2 - An ideal Otto cycle has a compression ratio of 8
(see Fig. I.2). At the beginning of the compression
process, air is at 100 kPa and 17°C, and 800 kJ/kg of
heat is transferred to air during the constant-volume
heat-addition process. Take the cv=0.72 kJ/(kg K)
and the specific heat ratio, k=1.4 for the
compression and the expansion processes.
Determine:
1)The temperature and pressure at the end of each
process of the cycle.
2)The heat rejected qout, in kJ/kg
3)The net work output, in kJ/kg
4)The thermal efficiency
36

37

I.3 - An engine operates on the air-standard diesel cycle(see
Fig. next slide). The conditions at the start of compression
are 27°C and 100 kPa. The heat supplied is 1840 kJ/kg, and
compression ratio is 16. Take the cp=1.0047 kJ/(kg K),
cv=0.7176 kJ/(kg K) and the specific heat ratio, k=1.4 for the
compression and the expansion processes. Determine:
a)The maximum temperature and pressure of the diesel cycle
b)The thermal efficiency
c)The mean effective pressure (MEP) in (kPa)
d) The power output at 2000 rpm for an engine displacement

of 5000 cm3
Assumptions
1.The air in the piston-cylinder is a closed system
2.The air in the system is an ideal gas with constant specific heats
3.Kinetic and potential energy changes are negligible
38

Solution: The ideal Diesel cycle and given data are shown in
following p-v and T-s plots,

a)The temperature and pressure values at the end of each
process can be determined by utilizing the ideal-gas
isentropic relations for processes 1-2 and 3-4
But first we determine the volume at state 1 for ideal-gas,
where, T1=300K, p1=100 kPa, and
39

RT1 0.287 x 300
3
v1 =
=
= 0.861 m / kg
p1
100
Process 1-2 (isentropic compression of an ideal gas,
constant specific heats), hence:
k

Pv = C and

v1
v1 0.861
= rv = 16 ; v2 =
=
= 0.0538 m 3 / kg
v2
16
16

v k −1T = C
k −1

T2  v1 
=  
T1  v2 
T2 = 300 x16 0.4 = 909 .4 K
40

k

p2  v1 
=  
p1  v2 
1.4
p2 = 100 x16 = 4850 .3kPa
Process 2-3 (constant-pressure heat addition to an ideal gas):

qa = (h3 − h2 ) = c p (T3 − T2 )
1840 = 1.0047 (T3 − 909 .4 )
T3 = Tmax = 2740 .8 K
p3 = pmax = p2 = 4850 .3 kPa
RTmax 0.287 x 2740 .8
v3 =
=
= 0.1622 m 3 / kg
pmax
4850 .3
41

Process 3-4 (isentropic expansion of an ideal gas, constant
specific heats), and v4=v1 , hence:
k −1

k

Pv = C ; v T = C ; and

v3 0.1622
=
= 0.188
v4
0.861

k −1

T4  v3 
=  
T3  v4 
0.4
T4 = 2740 .8x0.188 = 1405 .7 K
k

p4  v3 
=  
p3  v 4 
1.4
p4 = 4850 .3x0.188 = 468 .6kPa
42

Process 4-1 is a constant-volume heat-rejection process (it
involves no work interactions), and the amount of heat
rejected is

qr = (u1 − u 4 ) = cv (T1 − T4 )
qr = 0.7176 (300 − 1405 .7 )
qr = −793 .4 kJ / kg
Thus, the net work output of a cycle is

wnet = (qa + qr ) = (1840 − 793.4 ) = 1046.6 kJ / kg
b) Then, the thermal efficiency of this Diesel cycle is
determined from

wnet 1046 .6
η th =
=
= 0.569 or 56.9%
qa
1840
43

c) The mean effective pressure is determined from its
definition

wnet
wnet
1046 .6
MEP =
=
=
= 1296 .6 kPa
vmax − vmin v1 − v2 0.861 − 0.0538
d) The power output is calculated as
 2000 s −1 
 rpm 
-6
3 
 = 108.05 kW
Power = (MEP ) piston displacement 
 = 1296.6 kPa x 5000x10 m 

 2 x 60 
 2 x 60 

(

)

(

)

(

)

44