# Thermodynamics

Mechanics

Oct 24, 2013 (4 years and 8 months ago)

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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, 2
nd

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
: the material in the portion of space to be analyzed (
closed

or
open
)

Boundary:

A separator, real or imaginary, between system and
surroundings (
can be
fixed
or
movable
.
)

Surroundings
: exterior environment

System
:

the material in the portion of space to be analyzed (
closed

or
open
)

Boundary:

A separator, real or imaginary, between system and
surroundings (
can be
fixed
or
movable
.
)

Surroundings
:
exterior environment

Q, W

U

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.

An open system (a control
volume) with one inlet and
one exit.

Mass in

Energy in

Mass out

Energy out

Forms of energy

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

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

Heat Transfer: Conduction,

Convection, Radiation

Mass Transfer

Fluid Mechanics

Combustion

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

Diffuser

Fuel
Injector

Turbine

Hot
exhaust

Compressor

Combustion
Chamber

Nozzle

Vair

, P

KE

H

Compressor do

Work on air

W
in

H

Combustion
(
Fuel+Air
)

Q
in

H

H

W
out

H

KE

Backwork

ratio

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

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 closed system reaching thermal
equilibrium.

A system at two different states.

Zeroth

Law of
thermodynamic~!!

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

States and Processes

Work done during volume changed

Path between states

The 1
st

Law

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 1
st

Law of Thermodynamics

Either heating or stirring can raise
T

of the water
.

Joule’s
apparatus

1
st

Law of Thermodynamics
:

Increase in internal energy = Heat

added

Work done

Thermodynamic state variable

= variable independent of history
.

e.g.,
U
,
T
,
P
,
V
, …

Not
Q
,
W
, …

PE

of falling weight

KE

of paddle

Heat

in water

Another example of energy
transformaiton

Qin

W
in

The 1
st

Law of Thermodynamics

1
st

Law of Thermodynamics
:

Path between states :Isothermal
Processes

Isothermal process
:
T

= constant.

Isothermal processes
on ideal
gas

For
monoatomic gas
e.g. He

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
.

T

= const

Constant
-
Volume Processes &

Specific Heat (
Cv
)

Constant
-
volume process
( isometric, isochoric,
isovolumic

)

:

V

= constant

C
V

= molar specific heat at constant volume

Ideal gas:
U = U
(
T
)

for all
processes

isometric
processes

only for const
-
vol

processes

Isobaric Processes & Specific Heat
(Cp)

Isobaric Process
: constant
P

isobaric
processes

C
P

= molar specific heat at constant pressure

Ideal gas, isobaric
:

Ideal
gas

Isotherms

Adiabatic Processes

Adiabatic process
:
Q

=
0

(Compression is always a adiabatic

process if it is fast enough)

adiabat
,
ideal
gas

Adiabatic: larger

p

No heat lost

Q=0

Think it in a common
sence
:

Pumping the handle results in
what?

if there is no heat lose (Q=0)

1. gas pressure increased

2. gas temperature increased

Summary:

Q/A

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.

m

m

m
R

m
R

Implies reversible process

no friction and equilibrium

process

Reversibl

work!!

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
V
max

/
V
min

is needed
?

Given : Air’s
specific heat ratio is

= 1.4, & before the compression the air is at 20

C
.

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

,
V

=const,
U

.

(b)

adiabatic
;
T

,

p

,
V

,
U

.

(c)

isobaric
;
T

,

p

=const,
V

,
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
):

BC (isometric
):

CA (
isothermal
):

work done by gas
:

From 1
st

law of thermodynamic, We know that:

“You cannot build a perpetual motion ! Since
s
ou

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 1
st

law!

The 1
st

law is not enough to explain both questions~!

We are going to the world of 2
nd

law

Any question ?

THERMODYNAMIC

The

second law

The 2
nd

Law

C
lasusius

statements

Kelvin
-
P
lanck 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

Lord Kelvin

(1824
-
1907
)

Max
Planck

(1858
-
1947)

Rudolf
Clausius

(1828
-
1888)

Heat Engine Efficiency

An
irreversible

processes normally include one or more
of the following processes :

1.
Heat transfer through a finite temperature difference

2.
Unrestrained expansion of a gas or liquid to a lower pressure

3.
Spontaneous chemical reactions

4.
Spontaneous mixing of matter at different compositions or states

5.
Friction
-
sliding friction as well as friction in the flowing fluids

6.
Electric current flow through a resistance

7.
Magnetization or polarization with
hystersis

8.
Inelastic deformation

Limits on performance

Limits on performance

Reversible cycle

Carnot Cycle

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 ~!

Nicolas Léonard
Sadi
Carnot

1796
-
1832

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 2
nd

law of thermodynamic

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 2
nd

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

Process

Description

Assumption

Mass in
cylinder

Other info

1
-
> 2

Intake

P = const

Inc.

1. Intake valve open

1. Exhaust valve closed

2. intake valve closed

3. spark fires

5. exhaust valve opens

pressure “blows down”

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.

P

V (cylinder volume)

1

4

2, 6

3

5

v

v

s

s

Expand

Burn:

Constant
Volume

Compress

Blowdown

Simplify

Otto Cycle

T

S

1,2,6

4

3

5

v

v

s

s

Compression

Expansion

Heat
Added

Heat
Rejected

P

V

1

4

2, 6

3

5

v

v

s

s

Thermal efficiency

h
th
=what you get/what you pay for

Adiabatic reversible compression/ expansion

and

thus

where r
c
: compression ratio

Thermal efficiency

independent of heat input

efficiency increases as
r
c

increases

why not go to
r
c

-
>

why not?

geometrical limitations, heat loss,
irreversibilities

(high compression
-
> high T
-
> high heat loss), knock

After some algebra:

Thermal efficiency

Example: Auto engine: r
c
~8;

~1.3

h
th
~0.46 (theoretical);
h
th
~0.30 at best (expt)

Differences:

Heat Loss to valves, cylinder walls

Incomplete combustion

Friction

Blow by, valves leak

Throttling (Pexhaust

Pintake)

Diesel Cycle

P

V

6

3

1,5

2

4

Stratified charge engine

-

fuel injected after air compressed

-

heat release doesn’t occur instantly

since fuel will take more time to

burn than in the premixed case.

This is
bec
. fuel must mix,

vaporize, than burn. Takes time.

-
To model this, combustion process

assumed to occur
at increasing

volume,
constant pressure

Combustion

Compression

Expansion

New ratio V
3
/V
2

introduced

Diesel Cycle

Define: depends on the heat input

can show:

>1 for
b
>1

thus:

and:

when
b
=1

Ideal
Brayton

Cycle

(Gas Turbines)

P

V

1

3

2

1.
Isentropic Compression (1
-
>2)

2.
Constant pressure heat addition (2
-
>3)

3.
Isentropic expansion (3
-
>4)

4.
Constant pressure heat rejection (4
-
>1)

Combustor

Compressor

Turbine

m

.

W
net

W
c

1

2

3

4

4

Ideal
Brayton

Cycle

T

s

1

3

2

4

v

v

s

s

Compression

Expansion

Heat
Added

Heat
Rejected

P

V

1

3

2

4

Ideal
Brayton

Cycle

W
net
= W
t

W
c

=

where
PR=

Note
:

Tsunami &Quake
-
induced nuclear power
plant crisis

six nuclear
reactors at
Fukushima Daiichi

What do we learn from this catastrophe?