Transcript 投影片 1

THERMODYNAMIC
an introduction
Closed and open systems
Forms of energy
macroscopic
microscopic
Properties of a system
Intensive properties
Extensive properties
State and equilibrium
Zeroth Law
1. Theromodynamics, an engineering
approach, 2nd ed., by Yunus A.
Çengel & Michael A. Boles,
McGraw-Hill, Inc., 1994
2. http://www.wikipremed.com/image
_archive.php?code=010304
SYSTEMS AND CONTROL VOLUMES

System:
System: the
thematerial
materialinin the
theportion
portionof
of space
space to
tobe
beanalyzed
analyzed (closed
(closed
or
oropen)
open)

Boundary:
Boundary: AAseparator,
separator,real
realor
orimaginary,
imaginary,between
betweensystem
systemand
and
surroundings
surroundings (can
(canbe
befixed
fixedor
ormovable.)
movable.)

Surroundings:
Surroundings:exterior
exteriorenvironment
environment
U
 Q, W
Mass in
Energy in




An open system (a control
volume) with one inlet and
one exit.

Mass out
Energy out
Closed system (Control mass): A fixed amount
of mass, and no mass can cross its boundary.
Open system (control volume): A properly
selected region in space.
It usually encloses a device that involves mass
flow such as a compressor, turbine, or nozzle.
Both mass and energy can cross the boundary
of a control volume.
Control surface: The boundaries of a control
volume. It can be real or imaginary.
Forms of energy
Fluid Mechanics
Combustion
Heat Transfer: Conduction,
Convection, Radiation
Mass Transfer
Thermodynamics: The science of energy.
The name thermodynamics stems from
the Greek words therme (heat) and
dynamis (power).
First law of thermodynamics—
Conservation of energy principle:
During an interaction, energy can change from one
form to another but the total amount of energy
remains constant.
Energy cannot be created or destroyed

In thermodynamic analysis, it is often helpful to consider the
various forms of energy that make up the total energy of a
system in two groups:
Macroscopic and Microscopic energy
The macroscopic forms of
energy , are those a system
possesses as a whole with
respect to some outside
reference frame , such as kinetic
energy (K.E.)and potential
energy (P.E.)
Internal energy is defined above as
the sum of all the microscopic forms of
energy of a system such as :
K.E. of the molecules  sensible energy,
Phase changed  latent energy
(inter-molecular forces)
Bonds in a molecule  chemical (or bond) energy
(combustion, catalytic electrochemical
reaction)
Electronic energy
Bonds within the nucleus of the atom itself  nuclear energy
Temperature of
the system
Fuel
Injector
Diffuser
Compressor
Turbine
Combustion
Chamber
Hot
exhaust
Nozzle
Combustion
(Fuel+Air)
Qin→  H↑
Compressor do
Work on air
Win→  H↑
Vair, P↑
KE→  H
 H →Wout
Backwork
ratio
 H → KE
PROPERTIES OF A SYSTEM






Property: Any characteristic of a
system.
Some familiar properties are
pressure P, temperature T, volume
V, and mass m.
Properties are considered to be
either intensive or extensive.
Intensive properties: Those that
are independent of the mass of a
system, such as temperature,
pressure, and density.
Extensive properties: Those
whose values depend on the size—
or extent—of the system.
Specific properties: Extensive
properties per unit mass.
Criterion to differentiate
intensive and extensive
properties.
EQUILIBRIUM







Zeroth Law of
thermodynamic~!!
Thermodynamics deals with
equilibrium states.
Equilibrium: A state of balance.
In an equilibrium state there are no
unbalanced potentials (or driving
forces) within the system.
Thermal equilibrium: If the
temperature is the same throughout
the entire system.
Mechanical equilibrium: If there is
no change in pressure at any point
of the system with time.
Phase equilibrium: If a system
involves two phases and when the
mass of each phase reaches an
equilibrium level and stays there.
Chemical equilibrium: If the
chemical composition of a system
does not change with time, that is,
no chemical reactions occur.
A system at two different states.
A closed system reaching thermal
equilibrium.
TEMPERATURE AND THE ZEROTH LAW OF
THERMODYNAMICS


The zeroth law of thermodynamics: If two bodies are in thermal
equilibrium with a third body, they are also in thermal equilibrium with
each other.
By replacing the third body with a thermometer, the zeroth law can
be restated as two bodies are in thermal equilibrium if both have the
same temperature reading even if they are not in contact.
Two bodies reaching
thermal equilibrium
after being brought
into contact in an
isolated enclosure.
Any question ?
THERMODYNAMIC
The first law
The 1st Law
States and Processes
Work done during volume changed
Path between states
Cyclic Processes
The State Postulate


The number of properties
required to fix the state of a
system is given by the state
postulate:
◦ The state of a simple
compressible system is
completely specified by
two independent, intensive
properties (P ,T , v ).
Simple compressible
system: If a system involves
no electrical, magnetic,
gravitational, motion, and
surface tension effects.
The state of nitrogen is
fixed by two independent,
intensive properties.
Process
Process: Any change that a system undergoes from one equilibrium state to
another.
Path: The series of states through which a system passes during a process.
To describe a process completely, one should specify the initial and final states,
as well as the path it follows, and the interactions with the surroundings.
Quasistatic or quasi-equilibrium process: When a process proceeds in such a
manner that the system remains infinitesimally close to an equilibrium state at all
times.
Work done during volume changed
Work done during volume changed
Path between states
m
m
(isometric, isovolumic)
Path between states
The 1st Law of Thermodynamics
Another example of energy
transformaiton
Qin
PE of falling weight
 KE of paddle
Win
 Heat in water
Either heating or stirring can raise T of the water.
1st Law of Thermodynamics :
Increase in internal energy = Heat added  Work done
U  Q  W
dU dQ dW


dt
dt
dt
Thermodynamic state variable
= variable independent of history.
e.g., U, T, P, V, …
Not Q, W, …
Joule’s apparatus
The 1st Law of Thermodynamics
1st Law of Thermodynamics :

m(h  p.e.  k .e.)  Q  W
h  u  pv(flow work)
Path between states :Isothermal
Processes
Isothermal process : T = constant.
V2
V2
V1
V1
W   p dV  
W  m R T ln
m RT
dV  m R T ln V
V
V2
V1
V2
V1
3
3
For
U  N k T  m RT 
monoatomic gas
2
2
e.g. He
Q  W  m R T ln
V2
V1
U  0  Q  W
Isothermal processes
on ideal gas
Example : Bubbles !
A scuba diver is 25 m down, where the pressure is 3.5 atm
( 350 kPa ).
The air she exhales forms bubbles 8.0 mm in radius.
How much work does each bubble do as it arises to the surface,
assuming the bubbles remain at 300 K.
W  n R T ln
V2
V1
PV  n RT
T = const 
P1 V1  P2 V2
V2 P1  3.5 atm 


 3.5
V1 P2
1
atm


3
 4
W  n R T ln 3.5  p1 V1 ln 3.5   350,000 Pa  
 0.008 m   ln 3.5
 3

 0.94 J
Constant-Volume Processes &
Specific Heat (Cv)
Constant-volume process ( isometric, isochoric, isovolumic ) :
V = constant
V  0
W  p V  0

U  Q
CV = molar specific heat at constant volume

U  Q  m CV T
Ideal gas: U = U(T)

CV 
1  dQ 


m  dT V
isometric processes
U ideal gas  m CV T
Q  mCV T
for all processes
only for const-vol processes
Isobaric Processes & Specific Heat
(Cp)
Isobaric Process : constant P
Isotherms
W  p  V2 V1

 p V
Q  U  W  U  p V
CP = molar specific heat at constant pressure
CP 
Q  mCP T
isobaric processes
m CP T  m CV T  p V
Ideal gas, isobaric :
 m CV T  m R T

CP  CV  R
1  dQ 


m  dT  P
Ideal gas
Adiabatic Processes
No heat lost
Q=0
Adiabatic process: Q = 0
(Compression is always a adiabatic process if it is fast enough)
Think it in a common sence:
Pumping the handle results inU  W
what?
p V   const
if there is no heat lose (Q=0)
T V  1  const
adiabat,
ideal gas

1. gas pressure increased
2. gas temperature increased
p V  p1 V1
W 2 2
 1
Adiabatic: larger p
CP
1
CV
Summary:
Q/A
m
m
The ideal gas law says p V = n R T,
but the adiabatic equation says p V  = const.
Which is true,
(a) the ideal gas law ,
(b) the adiabatic equation, or
(c) both?
Explain.
mR
Implies reversible process
no friction and equilibrium
process
Reversibl work!!
mR
Diesel Power
Fuel ignites in a diesel engine from the heat of compression (no spark plug needed).
Compression is fast enough to be adiabatic.
If the ignite temperature is 500C, what compression ratio Vmax / Vmin is needed?？
Given : Air’s specific heat ratio is  = 1.4, & before the compression the air is at 20 C.
T V  1  const
1 /   1
Vmax  Tmin 


Vmin  Tmax 
1 / 1.4 1
 273 K  500 K 


273
K

20
K


 11
Q/A :
Name the basic thermodynamic process involved when each of the following is
done to a piston-cylinder system containing ideal gas,
tell also whether T, p, V, & U increase or decrease.
(a) the piston is lock in place & a flame is applied to the bottom of the cylinder,
(b) the cylinder is completely insulated & the piston is pushed downward,
(c) the piston is exposed to atmospheric pressure & is free to move, while the
cylinder is cooled by placing it on a block of ice.
(a) isometric;
T , p ,
(b) adiabatic ; T , p ,
V =const, U .
V,
(c) isobaric ; T , p =const, V  ,
U .
U .
Cyclic Processes
Cyclic Process : system returns to same thermodynamic
state periodically.
。
A four-process cycle
Example : Finding the Work done in a cycle
An ideal gas with  = 1.4 occupies 4.0 L at 300 K & 100
kPa pressure.
It’s compressed adiabatically to ¼ of original volume,
then cooled at constant V back to 300 K,
& finally allowed to expand isothermally to its original V.
How much work is done on the gas?
AB (adiabatic):
WAB 
WAB
BC (isometric):
CA (isothermal):
work done by gas:
p A VA  pB VB
 1
V 
pB  p A  A 
 VB 

 1




pA VA
VA
100 kPa  4.0 L  1  41.41 

1     

 741 J
  1   VB  
1.4  1


WBC  0
WCA  n R T ln
VA
VC
 pA VA ln 4  555 J
WABCA  WAB  WBC  WCA  186 J
From 1st law of thermodynamic, We know that:
“You cannot build a perpetual motion ! Since sou
cannot get more energy out than you put
in(conservation of energy).”
But……
About the efficiency:
Can we know how much work done at least we can get after
putting energy into the machine?
About the direction of heat:
When you’re holding a cup of coffee , Why doesn’t your hand get
colder as the coffee become hotter and hotter , It does not
against with the 1st law!
The 1st law is not enough to explain both questions~!
We are going to the world of 2nd law
Any question ?
THERMODYNAMIC
The second law
The 2nd Law
Clasusius statements
Kelvin-Planck statements
Limits on performance
Irreversible
Carnot cycle
Entropy statement
We’ll miss you,
Qc …
(Clausius statement) no process is possible
where the sole result is the removal of heat
from a low-temp reservoir and the absorption
of an equal amount of heat by a high temp
reservoir
(Kelvin-Planck) no process is possible in which
heat is removed from a single reservoir w/
equiv amount of work produced
Rudolf Clausius
(1828-1888)
Lord Kelvin
(1824-1907)
Max Planck
(1858-1947)
Heat Engine Efficiency
Limits on performance
An irreversible processes normally include one or more
of the following processes :
1.
2.
3.
4.
5.
6.
7.
8.
Heat transfer through a finite temperature difference
Unrestrained expansion of a gas or liquid to a lower pressure
Spontaneous chemical reactions
Spontaneous mixing of matter at different compositions or states
Friction-sliding friction as well as friction in the flowing fluids
Electric current flow through a resistance
Magnetization or polarization with hystersis
Inelastic deformation
Limits on performance
Reversible cycle  Carnot Cycle
Nicolas Léonard
Sadi Carnot
1796-1832
A Carnot Cycle consists of four steps:
 Isothermal expansion
(in contact with the heat reservoir)
 Adiabatic expansion
(after the heat reservoir is removed)
 Isothermal compression
(in contact with the cold reservoir)
 Adiabatic compression
(after the cold reservoir is removed)
Every processes in the cycle are reversible! How about its efficiency ~!
Efficiency of a Carnot cycle
Since no one can create a 0 k cold reservoir or a ∞ k
heat reservoir . Carnot efficiency is a theoretical
maximum and it can’t reach 100%
Entropy
T v.s. S diagram of Carnot cycle
The 2nd law of thermodynamic
S  0

If a process occurs in an isolated (closed
and adiabatic) system the entropy of the
system increases for irreversible process
and remains constant for reversible
processes. IT NEVER DECREASES….
Any question ?
Cycles

A diagram can be drawn with any pair of properties
◦ P-T
◦ P-V (allows the net work of a cycle to be determined:
W=integral of pdV
◦ T-S (gives the net heat of a cycle; recall 2nd law which
states: dsdQ/T -> Q=integral of Tds!

If you can convert some of the heat to work, you have
an engine!
Cycle Types
Premixed Charge – Otto Cycle, gasoline,
spark-ignition engine
 Non-premixed charge or stratified charge
engine
(compression ignition or Deisel cycles)
 Gas Turbines – Brayton Cycle
 Other cycles: Rankine, …

Where to start:
Air (ideal gas) cycles

Assume no changes in gas properties (cp,
MW, , …) due to changes in composition,
temp., …called the IDEAL air cycle!
•REAL cycles must consider fuel-air mixture
which is compressed, burned, expanded,…
with accompanying changes in thermodynamic
properties
Premixed Charge – Otto Cycle
How can we take that into calculation? We need
to simplify it !
Premixed Charge – Otto Cycle
Expand
4
P
s
v
Burn:
Constant
Volume
Simplify
5
3
s
1
2, 6
Compress
Process
v
Blowdown
V (cylinder volume)
Description Assumption
Mass in
Other info
cylinder
1 -> 2
Intake
P = const
Inc.
2 -> 3
Compress
s = const
Const
3 -> 4
Burn
v = const
Const
4 -> 5
Expand
s = const
Const
5 -> 6
Blowdown
v = const
Dec.
6 -> 1
Exhaust
P = const
Dec.
1. Intake valve open
1. Exhaust valve closed
2. intake valve closed
3. spark fires
5. exhaust valve opens –
pressure “blows down”
Otto Cycle
P
Heat
Added
4
s
v
T
5
4
v
Expansion
s
3
s
1
v
2, 6
3
Compression
V
1,2,6
s
5
v
Heat
Rejected
S
Thermal efficiency
hth=what you get/what you pay for
hth  Work out  Work in
Heat in
hth 
- cv (T5  T4)  cv (T3  T2) - T5  T4  T3  T2)

cv (T4  T3)
T4  T3
Adiabatic reversible compression/ expansion







T3 V3 

T2 V 
2
 ( 1)
thus
and






T4 V4

T5 V
5
V4 V3 1


V5 V2 rc






 ( 1)
where rc: compression ratio
Thermal efficiency
• After some algebra:
h
1 1
th,otto
rc  
independent of heat input
 efficiency increases as rc increases

◦ why not go to rc -> 

why not?
◦ geometrical limitations, heat loss, irreversibilities
(high compression -> high T -> high heat loss), knock
Thermal efficiency
h 1 1
th
rc  
• Example: Auto engine: rc~8; ~1.3
hth~0.46 (theoretical); hth~0.30 at best (expt)
Differences:
•
•
•
•
•
Heat Loss to valves, cylinder walls
Incomplete combustion
Friction
Blow by, valves leak
Throttling (Pexhaust Pintake)
Diesel Cycle
Combustion
2
3
P
Compression
6
Stratified charge engine
- fuel injected after air compressed
Expansion - heat release doesn’t occur instantly
since fuel will take more time to
burn than in the premixed case.
4
This is bec. fuel must mix,
vaporize, than burn. Takes time.
- To model this, combustion process
1,5
assumed to occur at increasing
V
volume, constant pressure
New ratio V3/V2 introduced
Diesel Cycle
V
V2
Define: β  3 depends on the heat input
can show:
h
1
th,diesel


























1
β 1
  β 1
rc
>1 for b>1
thus:
h
h
th,diesel th,otto
and:
h
h
th,diesel th,otto
when b=1
Ideal Brayton Cycle
(Gas Turbines)
1.
2.
3.
4.
Isentropic Compression (1->2)
Constant pressure heat addition (2->3)
Isentropic expansion (3->4)
Constant pressure heat rejection (4->1)
P
2
Combustor
.
m
2
3
4
1
Wc
Compressor
3
Wnet
4
Turbine
1
V
Ideal Brayton Cycle
Heat
Added
P
2
3
T
3
Expansion
s
v
2
4
1
Compression
1
V
s
4
v
Heat
Rejected
s
Ideal Brayton Cycle
 cp T T 
Wc  m
1
 2
 cp T T 
Wt  m
4
 3



 cp T T
Qin  m
3 2



 cp T T 
Qout  m
1
 4


( ) / 
W
 1 


h  net  1  
where PR=
 PR 
th Q
in
Note:








 cp T  T  T  T 
m
4
2 1
 3
Wnet = Wt – Wc =
 1
 
P 
2
P 
1
T2 T3
 
T1 T4
P
2
P
1

Tsunami &Quake-induced nuclear power
plant crisis—six nuclear reactors at
Fukushima Daiichi

What do we learn from this catastrophe?