THERMODYNAMICan introduction

Closed and open systems

Forms of energymacroscopicmicroscopic

Properties of a systemIntensive propertiesExtensive 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: 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 inEnergy in

Mass outEnergy 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, RadiationMass Transfer

Fluid MechanicsCombustion

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 ofenergy 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 energyBonds within the nucleus of the atom itself nuclear energy

Temperature of the system

DiffuserFuel Injector

TurbineHot exhaust

Compressor

Combustion Chamber Nozzle

Vair, P↑KE→ H

Compressor doWork on airWin→ H↑

Combustion (Fuel+Air)

Qin→ H↑

H →Wout

D 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 ProcessesWork done during volume changedPath between states

The 1st Law

Cyclic Processes

The State PostulateThe 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.

ProcessProcess: 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

Either heating or stirring can raise T of the water.

Joule’s apparatus

1st Law of Thermodynamics :Increase in internal energy = Heat added Work done

U Q W

Thermodynamic state variable = variable independent of history.

e.g., U, T, P, V,

…

Not Q, W, …

dU dQ dW

dt dt dt

PE of falling weight

KE of paddle

Heat in water

Another example of energy transformaitonQin

Win

The 1st Law of Thermodynamics

1st Law of Thermodynamics :

( . . . .)

(flow work)

m h p e k e Q W

h u pv

Path between states :Isothermal Processes

Isothermal process : T = constant.

2

1

V

VW p dV 2

1

V

V

m R TdV

V 2

1ln V

Vm R T V

2

1

lnV

W m R TV

3

2U N k T 0U Q W

2

1

lnV

Q W m R TV

Isothermal processes on ideal gas

3

2m R TFor

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.PV n R T2

1

lnV

W n R TV

1 1 2 2P V P VT = const

0.94 J

ln 3.5W n R T

2 1

1 2

V P

V P

3.5

1

atm

atm 3.5

1 1 ln 3.5p V 34350,000 0.008 ln 3.5

3Pa m

Constant-Volume Processes & Specific Heat (Cv)

Constant-volume process ( isometric, isochoric, isovolumic ) : V = constant

0V 0W p V

U Q

VU Q mC T

CV = molar specific heat at constant volume

Ideal gas: U = U(T)

ideal gas VU mC T for all processes

isometric processes

VQ mC T only for const-vol processes

1V

V

dQC

m dT

Isobaric Processes & Specific Heat (Cp)

Isobaric Process : constant P

2 1W p V V p V

Q U W U p V

isobaric processesPQ mC T

CP = molar specific heat at constant pressure

P VmC T mC T p V Ideal gas, isobaric :

VmC T m R T

P VC C R Ideal gas

Isotherms

1P

P

dQC

m dT

Adiabatic Processes Adiabatic process: Q = 0 (Compression is always a adiabatic process if it is fast enough)

U W

pV const adiabat, ideal gas

1P

V

C

C

1T V const

2 2 1 1

1

p V p VW

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.

mm

mR

mR

Implies reversible process no friction and equilibrium processReversibl work!!

Diesel PowerFuel 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 500C, 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.

1T V const

1 / 1.4 1273 500

273 20

K K

K K

1 / 1

max min

min max

V T

V T

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 , V =const, U .

(b) adiabatic ; T , p , V , U .

(c) isobaric ; T , p =const, V , U .

Cyclic ProcessesCyclic 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?1A A B B

AB

p V p VW

741 J

AB (adiabatic):

0BCW BC (isometric):

ln ACA

C

VW n R T

VCA

(isothermal):

1.4 1100 4.0 1 4

1.4 1

kPa L

AB A

B

Vp p

V

1

11

A A AAB

B

p V VW

V

ln 4A Ap V 555 J

work done by gas:ABCA AB BC CAW W W W 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

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 Carnot1796-1832

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 2nd 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….

0S

Any question ?

CyclesA 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: dsdQ/T -> Q=integral of Tds!

If you can convert some of the heat to work, you have an engine!

Cycle TypesPremixed Charge – Otto Cycle,

gasoline, spark-ignition engineNon-premixed charge or stratified

charge engine (compression ignition or Deisel cycles)

Gas Turbines – Brayton CycleOther cycles: Rankine, …

Where to start: Air (ideal gas) cycles

Assume no changes in gas properties (cp, MW, g, …) 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 AssumptionMass in cylinder

Other info

1 -> 2 Intake P = const Inc.

1. Intake valve open1. Exhaust valve closed2. intake valve closed3. spark fires5. 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

35

v

v

s

s

Expand

Burn:Consta

nt Volume

Compress

Blowdown

Simplify  

Otto Cycle

T

S1,2,6

4

35

v

v

s

sCompression

Expansion

Heat Added

Heat Rejected

P

V1

4

2, 6

3

5v

v

s

s

Thermal efficiency

hth=what you get/what you pay for

in Heatin Workout Work th

3T4T)2T3T4T5T-

)3T4(Tvc)2T3(Tvc)4T5(Tvc-

th

Adiabatic reversible compression/ expansion)1(

2V

3V

2T3T

)1(

5V

4V

5T4T

and

cr1

2V3V

5V4V

thus where rc: compression ratio

Thermal efficiency

independent of heat inputefficiency increases as rc increases

◦ why not go to rc -> why not?

◦ geometrical limitations, heat loss, irreversibilities

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

cr11ottoth,

• After some algebra:

Thermal efficiency

• Example: Auto engine: rc~8; g~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)

cr11th

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 toburn than in the premixed case. This is bec. fuel must mix, vaporize, than burn. Takes time.

- To model this, combustion processassumed to occur at increasing volume, constant pressure

Combustion

Compression

Expansion

New ratio V3/V2 introduced

Diesel Cycle

2V3V

βDefine: depends on the heat input

1β1β

cr

11dieselth,can show:

>1 for b>1

ottoth,dieselth, thus:

ottoth,dieselth, and: when b=1

Ideal Brayton Cycle (Gas Turbines)

P

V

1

32

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.

WnetWc1

2 3

4

4

Ideal Brayton Cycle

T

s1

3

24

v

v

s

sCompression

Expansion

Heat Adde

d

Heat Rejecte

d

P

V

1

32

4

Ideal Brayton Cycle

1

T2

TpcmcW

4

T3

TpcmtW

2

T3

TpcminQ

1

T4

TpcmoutQ

Wnet = Wt – Wc =

1

T2

T4

T3

Tpcm

/)(

PR11

inQnet

W

th

4T3T

1T2T

1P2

P

1

where PR=

1P2

P

Note:

Quake-damaged Japanese nuclear power plant—Fukushima Daiichi plant

What do we learn from this catastrophe?