sem4 termo slide nota dr wan fulll

Week-1

Thermodynamics(Classical Thermodynamics)

“Therme” meaning heat

“Dynamics” meaning strength

Thermodynamics

- the science concerned with the causes and effects arising from energy transformations,(e.g heat & work) and the physical properties of substances that are involved in energy transformations.

Thermodynamics is based on definitions, concepts, and physical laws.

Thermodynamics

Statistical

(Microscopic)Classical

(Macroscopic)

Microscopic viewpoints- A collection of atoms within a container, each with a unique velocity

Macroscopic viewpoints (classical)-the properties of system are

assigned to the system as a whole.(does not require knowledge of the behaviour of individual particle)

The energy in both cases is the same, E

In the macroscopic description,atomistic concepts are disregarded.

Thermodynamics, is fields of physics that describes and correlates the physical properties of macroscopic systems of matter and energy

Thermodynamics

Processes

properties principles

Thermodynamics Triangle

Definitions:System : the part of the universe that has been chosen to study.

Surroundings: the rest of the universe

Boundary: the surface (can be real or imaginary) dividing the system from the surroundings.

boundarysystem

surroundings

System can be:Closed: energy can transfer between the system and the surroundings, but not the mass.Open: Mass and energy can transfer between the system and the surrounding.Isolated: Neither mass nor energy can transfer between the system and the surrounding

(OPEN SYSTEM) or control volume.e.g: pumps, compressors, turbines, valves, heat exchangers.

(CLOSED SYSTEM)or control mass.Mass = constante.g: sealed tanks, piston & cylinder devices.Boundary can move.

(ISOLATED SYSTEM)

Mass=constant

System

Surr 1

Surr 4

Surr 3Mass

Surr 2

Isolated system Boundary

Heat=0

Work=0

Mass=0

Across

Isolated

boundary

Property of a System

– Any characteristic of a system in

equilibrium that can be calculated or

measured. The property is independent of

the path used to arrive at the system

condition. Also known as variables of the

system.

Some thermodynamic properties are

pressure P, temperature T, volume V, and

mass m.

Two kinds of properties:

1) Intensive properties – independent of the

size of the system, e.g. T, P, r, any mass

independent property.

2) Extensive properties – proportional to the

quantity of material, or those that vary directly

with size of the system. e.g. m,V, E(total), H

(mass dependent property)

system

m, V, T

½ m, ½ V,

T, P, r

½ m, ½ V,

T, P, r

System 1

System 2

pseudointensive properties:

- extensive properties expressed per unit quantity of material.Or extensive properties per unit massEg:specific volume : v = V/m

(m3/kg)

[molar properties : expressed on a unit mole basis]

Basic thermodynamics properties

1.Density : mass per unit volume

)/( 3mkgV

volume

Low densiy

Density : mass per unit volume

High density

specific gravity or relative density:

the ratio of the density of a substance to the

density of water at a specific temperature.

(usually 4oC, for which rH2O = 1000 kg/m3)

rs = r/rH2O

1 2 3 ………..12 ……………….6.022 x 1023

Amount of substance: how much is there?

Avogadro’s

NumberDozen

Impact

Weight

2. Pressure (P) : force per unit area

3. Temperature (T) (thermal potential) :

-a measure of relative hotness or coldness of a material

4. Volume (V) (mechanical displacement) :

the quantity of space possessed by a material.

5. Entropy (S) (thermal displacement) :

the quantity of disorder possessed by a material.

6. Internal energy (U) :

the energy of a material which is due to the kinetic and potential energies of its constituent parts (atoms and molecules, usually).

Secondary thermodynamic properties

Enthalpy [H] - internal energy plus the pressure-volume product.

Heat capacity [Cp or Cv] (specific heatcp or cv) –-the amount of energy required to increase the temperature of one unit quantity of material by one degree, under specific conditions.

State (of System)

Consider a system that is not undergoing any change. The properties can be measured or calculated throughout the entire system. This gives us a set of properties that completely describe the condition or state of the system. At a given state all of the properties are known and have fixed values. If the value of even one properties changes, the state will change to another state.

Thermodynamic state, or macrostate,

is specified by stating a complete set of state variables, minimum any two of the three STATE VARIABLES (P, V or T).

Eg: Water

T = 40oC

v= 0.87 m3/kg

The state of the water is fixed by two properties

The system can change states only through an exchange of internal energy with the environment (unless it's a vacuum).

WaterT = 40oCV = 0.87 m3/kg

(State 1)

WaterT = 90oCV = 1.7 m3/kg

(state 2)

State FunctionAny physical quantity having a unique value for a given state of the system.

Examples: Temperature, entropy, enthalpy, pressure, volume,

N.B.: "work" and "heat", although thermodynamic variables, are not state functions.

Equation of State :-any equation that relates properties of a system at equilibrium

Equilibrium : implies a state of balance

Thermodynamic equilibrium: is achieved when all the

relevant types of equilibrium are satisfied.

iii. Phase equilibrium: the mass of each phase

reaches an equilibrium level and stays there.

iv. Chemical equilibrium: chemical composition

does not change with time.

ii. Mechanical equilibrium: no change in

pressure at any point of the system with time.

i.Thermal equilibrium: temperature is the same

throughout the entire system

A Simple Compressible System:

- a system without the effects of electrical, magnetic, gravitational, motion and surface tension.

The State Postulate- The state of a simple compressible system is completely specified by two independent, intensive properties.

e.g. T and v are independent properties.T and P : are independent properties for single phase systemT and P : are dependent properties for multiphase system.

Thermodynamic Processes and Cycles.

Process - any succession of events; or any change

from one state to another.

States of a thermodynamic system can be changed by

interacting with its surrounding through work and heat.

When this change occurs in a system, it is said that the

system is undergoing a process.

thermodynamic process - is the specific way in which internal energy is exchanged.

Path of the process:a series of states through which a system passes

during a process

State 2

State 1 Process path

Property B

Process path

The points (P1,V1) and (P2,V2) characterize the initial and

final equilibrium states.

The thermodynamic process can be indicated by a line

passing through the intermediate equilibrium states the

system passed through.

Quasi-static process: when a process proceeds

in such a manner that the system remains

infinitesimally close to an equilibrium state at all

time ………slow-process

Prefix iso- : to designate a process for which a particular

property remains constant

Types of Thermodynamic Processes

Reversible: Can happen slowly in either direction.

Irreversible: Involves net increase in entropy (can’t go

backwards).

for a sample of ideal gas: Isothermal: Constant temperature T = 0

State 1 State 2

Isobaric: Constant pressure P or P = 0

State 1 State 2

Isochoric: Constant volume V or V= 0

and W = 0

State 1

State 2

Adiabatic: No heat added or release or Q = 0

Cycle:is a sequence of different processes that begins and ends

at the same thermodynamic state.

Initial state

final state

initial state

Forces and Potentials

Solid matter is comprised of interacting atoms and

molecules – how might we model those

interactions?

xU = ½(kx)

Potential energy curve for SHM

Morse curve – approximation to potential energy variation for two neutral atoms.

200 ))](exp(1[ rrEU

Bottom of interatomic potential curve may be approximated by SHM potential.

Temperature

•a concept originates with our sense perceptions.

•a term used to quantify the difference between

warm and cold level of internal energy of a

substance.

•a measure of the average translational kinetic

energy associated with the disordered microscopic

motion of atoms and molecules.

•a physical quantity which gives us an idea of how

hot or cold an object is.

temperature

Molecular kinetic energy

Temperature scale

Standard temperature

MAY BE DEFINED IN TERM OF

And experessed in term of

Based on

The temperature of an object depends on

how fast the atoms and molecules which

make up the object can shake, or

oscillate.

As an object is cooled, the oscillations of its atoms

and molecules slow down.

With increasing temperature the inter-atomic bond

length increases (centre of oscillations shifts to

larger r).

Hence object expands at higher temperatures.

Thermometric Properties : properties

of materials that change with temperature. (may be used to define a temperature)

volumes of most liquids increase with temperature.

length of a metal rod increases as the temperature increases.

pressure of a constant volume of gas

increases with temperature.

volume of a gas at constant pressure

increases with temperature.

electrical resistance of a piece of wire

changes with temperature.

Thermometer : a device for measuring

temperature.

0 K = - 273.15 oCkelvin or K is absolute temperature scale

pressure

-273.15 temperature

0K T (K)

Heat is internal energy that is

exchanged between two objects due to their difference in temperature.

Heat always transfers, or flows, from a body at higher temperature to one at lower temperature.

Thermal Equilibrium

HOT COLD

Internal energy of A is greater than internal energy of B

UA > UB

UAAUBA

UA: decreasing UB : increasing

to UAA to UBA

UAA ≡ UBA

No energy transfer

Thermal equilibrium is the subject of the Zeroth Law of thermodynamics.

hot cold

Heat is transferred from hot object to cold object

Both objects are at the same temperature –

THERMAL EQUILIBRIUM - Zeroth Law of Thermodynamics

Th TcTh > Tc

The "zeroth law" states that if two systems are at the same

time in thermal equilibrium with a third system, they are in thermal

equilibrium with each other.

A is in thermal equilibrium with B; B is in thermal equilibrium with CTherefore A and C are in thermal equilibrium

No heat transferA C

The Zeroth Law implies existence of some universal property of systems in thermal equilibrium.

Temperature is that property of thermal equilibrium that is the same for any two systems in thermal equilibrium.

Temperature being a property of a body and heat being an energy flow to or from a body by virtue of a temperature difference.

The Zeroth law is used to define temperature

Let TA is the temperature of ATB is the temperature of BTC is the temperature of C

If TA = TB and TB = TC Then TA = TC

Any two objects in thermal equilibrium have the same temperature

So temperature determines if two objects are in thermal equilibrium.

Any one thermodynamic system can be used as a “thermometer” to define a temperature scale.

Internal energy

Energy is often defined as the ability to do work

Two types of energy:Kinetic Energy is energy of motion Potential Energy is energy of position

(technically position in a gradient "field")

Potential energy is often thought of as "stored" kinetic energy, meaning that bodies remain stationary in a potential field while held in place by some force, and upon change in this force (breaking the bond between two atoms), potential energy is converted to kinetic form.

Many forms of energy:

Nuclear Energy :The potential energy required

to bind nucleons in the nucleus.

Light Energy: The potential energy possessed

by the oscillating electric and magnetic fields that make up electromagnetic radiation.

Chemical Energy: The potential energy stored

in the electrostatic bonding relationships among atoms in a molecule.

Electrical Energy : The potential energy

involved in initiating and maintaining electron flow.

Mechanical Energy :The energy generated

(or stored) by machines which induces (or results

from) concerted motion processes in a system.

Heat Energy: The kinetic energy associated

with random motion of matter including the

vibratory and rotatory action of molecules

Internal energy is defined as the energy

associated with the random, disordered motion of molecules.

Molecular attractive forces are associated with potential energy

Molecular kinetic energy is a part of internal energy

Internal energy involves energy on the microscopic scale.

Kinetic : energy due to velocity, Ek = ½ mv2

Potential : energy due to position in a force field.

Internal : Sum of kinetic and potential energies

acting on all molecules in a system.

Specific internal energy : internal energy per unit

mass [Joules/kg]

latent energy : a portion of internal energy

associated with the phase of the system (binding forces between the molecules

chemical energy : is a potential energynuclear energy: is a potential energy

Internal energy is the sum of sensible energy +

latent energy + chemical energy + nuclear energy

sensible energy : a portion of internal energy

associated with kinetic energies of the molecules

Internal energy for the system with the same temperature

Solid or fluidMolecular gasMono- atomic gas

Potential energy

Vib. and rot. K.E

Trans. K.E

U: internal energy

Rotation about x-axis(vertical)

Rotation about y-axis (horizontal)

Vibrational motion along the bond

Molecular translation Electron

translation Nuclear spinElectron spin

Total energy of the system

static : energy stored in a system

dynamic or energy interaction:-able to cross the system boundry

2 forms of energy interaction:

i. heat transfer (or heat): the driving force is

a temperature difference.

ii. work: other than heat interaction.

organized form of energy (work):orderly motion of all molecules in one direction.

disorganized form of energy (heat):random motion of molecules

Random motion

Oderly motion

Major area of thermodynamics – conversion of

disorganized energy (heat) into organized energy (work).

THE END

WEEK 2

PURE SUBSTANCES

matter

Homogenous?

yesHeterogenous

mixture

Separated by

physical means ?yes

Homogenous

mixturePure substance

Decomposed by

chemical processes?

compoundelement

Pure substances have an invariable

composition and are composed of either

elements or compounds.

Elements - Substances which cannot be

decomposed into simpler substances by

chemical means.

Compounds can be decomposed into two or more elements.

In Thermodynamics we’ll define it as

something that has a fixed chemical

composition throughout.

Mixtures vs Pure Substances

Matter

Mixtures Pure Substances

Mechanical

mixtures

one can

separate

mixtures by

physical means.

Solutions

Completely

mixed.

Suspensions

Example:

homogenized milk.

Particles slightly

clumpled together

but remain in

suspension

Elements

Substances

made of only

1 type of atom

Compounds

Substances that have

more than 1 element

in them (chemically

united)

Eg: NaCl

Physical Versus Chemical Change

Physical Changes Chemical Changes

Water freezing

Ice melting

Sugar dissolving in water

Detergent removing

grease from dirty pot

Piece of wood burns in a

A bomb explodes

burning of fuel inside the

engine

A pure substance may exist in more than one phase.

Different phases of a substance have the same chemical composition but different physical structures, such as solid, liquid, and gas.

e.g : water; water + steam

Air : a mixture of nitrogen,oxygen and argon, therefore is not a pure substance. However since air is nonreactive mixture and has a constant chemical composition on macroscopic level, air can be approximated as pure substance.

Molecules are

(a) closely bound, therefore relatively dense

(b) arranged in rigid three-dimensional patterns so that they do not easily deform.

(c) having large attractive forces between atoms or molecules

(d) in constant motion – they vibrate in place

(e) vibrating more at higher temperature

Liquid

Molecules are

(a) closely bound, therefore also relatively

dense and unable to expand to fill a space.

(b) no longer rigidly structured so that they are

free to move within a fixed volume.

(c) look a lot like a solid.

Molecules have

(a) virtually no attraction one to another.

(b) not arranged in a fixed pattern. They are far apart.

(c ) high kinetic energy.

heating of water at constant pressure

Saturation Pressure is the

pressure at which the liquid and vapour phases are in equilibrium at a given temperature (point 2 &

Compressed or Subcooled

liquid: (Between states 1

& 2) NOT about to vaporize

(Sub-cooed liquid) e.g., water

at 20oC and 1 atmosphere

Saturation Temperature is

the temperature at which the liquid and vapour phases are in equilibrium at a given pressure (point 2 & 4).

Saturated liquid:

(State 2) about to vapourize e.g., water at 100oC and 1

atmosphere

Saturated Liquid-Vapour

Mixture or Quality

region: (Between states 2 &

4) Liquid and vapour exist together in a mixture.

Saturated Vapour:

(State 4) about to condense

e.g., water vapour (steam) at

100oC and 1 atm.

Superheated Vapour:

(Between States 4 & 5)

NOT about to condense

e.g., water vapour (steam) at

>100oC and 1 atm

changing the pressures

CRITICAL POINT:

the saturated liquid and saturated vapour states are identical.No saturated mixture exists - the substance changes

directly from the liquid to vapour states.

Above the critical point there is no sharp difference

between liquid and gas!!

single phase liquid region

single phase vapour region

supercritical fluid region

T – V diagram

P-v diagram

P-v Diagram of a Substance that Contracts on Freezing

P-v Diagram of a Substance that Expands on Freezing

P-T Diagram

At any point in the areas separated by the curves, the pressure and temperature allow only one phase (solid, liquid, or gas) to exist, and changes in temperature and pressure, up to the points on the curves, will not alter this phase.

At any point on the curves, the temperature and pressure allow two phases to exist in equilibrium: solid and liquid, solid and vapor, or liquid and vapor.

Triple point: all three phases of a substance

will exist in equilibrium

P-v-T surfacePutting all the three properties on the same diagram

Contracts on Freezing

P-v-T surfacePutting all the three properties on the same diagram

Expands on Freezing

ENTHALPY, H

another property of pure substances (like internal energy, U)

U is a function ONLY of the temperature of the substance

H = U + PV ; U : internal energy ; PV : work done

(note: PV (N/m2)(m3) = N-m = Joules )

on a per unit mass basis: h = u + Pv (specific enthalpy)

H is a function of BOTH temperature and pressure.

Property Tables

The steam table are taken as an example.

Data on the table are;

P pressure

T temperature

v specific volume

u specific internal energy

h specific enthalpy

s specific entropy

subscript fg : difference between the saturated vapour and saturated liquid values

vf : specific volume of saturated liquid

vg: specific volume of saturated vapour

vfg = vg - vf

example of the table

using the steam tables:

given P and T find the saturation pressure Psat corresponding to T

for P > Psat the point corresponds to compressed liquid

for P < Psat the point corresponds to superheated vapour

for P = Psat the point is in saturation region

but P and T are not independent, needs additional properties to determine state

given P and v or T and v first find vf and vg

if vg < v the state is in superheated vapour

if vf v vg the state is in saturation (SLVM)

if vf > v the state is in compressed liquid

same procedure for

P and u, T and u, etc.

example:

Find the volume, the internal energy, and the enthalpy of 2 kg water at the following condition;

a. T = 300oC, P=2 MPa

b. T = 300oC, P= 20 MPa

c. T = 300oC, P= 8.59 MPa

a. T = 300oC, P=2 MPa, mass = 2 kg, to find V, H, U

See table A4: at T=300oC,

Psat = 8.588 Mpa.

So P < Psat, this implies

that the point corresponds

to Superheated water

(vapour), so see Table A6

See Pressure P=2.00MPa

See at Temperature

T=300oC

Read: v = 0.12551 m3/kg

v= V∕m or V = mv

V= 2 kg (0.12551)m3/kg

V= 0.25102 m3

V≈0.25 m3.

Read: u = 2773.2 kJ/kg

u= U∕m or U = mu

V= 2 kg (2773.2)kJ/kg

V= 5546.4 kJ

Read h=3024.2 kJ/kg

(H=6,048.4 kJ)

b. T = 300oC, P= 20 MPa

Psat = 8.588 Mpa.

So P › Psat, this implies

to Compressed Liquid, so

see Table A7

See Pressure P=20MPa

See at Temperature

T=300oC

Read: v = 0.0013611

v= V∕m or V = mv

V= 2 kg (0.0013611)

V= 0.0027222 m3

V≈0.003 m3. Read: u = 1307.2 kJ/kg

u= U∕m or U = mu

V= 2 kg (1307.2)kJ/kg

V= 2,614.4 kJ

Read h=1334.4 kJ/kg

(H=2668.8 kJ)

c. T = 300oC, P= 8.581 MPa

Psat = 8.588 Mpa.

So P = Psat, this implies

to SLVM, so see Table A4

See Pressure P=8.581MPa

See at Temperature

T=300oC

Have to consider X:

Quality Factor

Saturated Liquid – Vapour Mixture(SLVM)

U kJ/kg

QUALITY ( symbol x ) : the ratio of mass of vapour to the total mass of the mixture.

m msaturated vapour

X = mg /mt

since mt = mf + mg

mt : total mass of the mixture.X = 0 : the mixture is all saturated liquid

X = 1 : the mixture is all saturated vapour

Properties at point P (in the diagram above)

-a mixture of liquid & vapour

Vapour

liquid

Let say x = 0.4

40% Vapour

60% liquid

mf : mass of sat.

liquid

mg : mass of sat.

vapour

mt : total mass

mt = mf + mg

total volume V = Vf + Vg

but V = mvso mtvav = mfvf + mgvg

mtvav = (mt – mg )vf + mgvg

vav = [(mt/mt – (mg/mt) ]vf + (mg/mt)vg

vav = [1 – X ]vf + X vg

or vav = vf – x vf + x vg

vav = vf + x [vg - vf ]

** vav = vf + Xvfg

specific properties of saturated mixture

uav = uf + x·ufg

vav = vf + x·vfg

hav = hf + x·hfg

sav = sf + x·sfg

x = 0 : uav = uf : all liquid

x = 1 : uav = uf + ufg = uf + (ug – uf) =ug

: all vapour

0 < x < 1 : liquid + vapour48

i) At the given temperature, determine the sat. pressure (from the steam table A4)

T = 90oC , P = 70.18 kPa

Properties of Saturated Water -Temperature Table (SI)

Vapour

Temperature of the mixture is 90oC.

i) pressure in the tank (look at table)

ii) quality of the mixture (X)

iii) volume of the tank (V=mt vav

Vav=vf+Xvfg )

T(oC) P(kPa) S.Vol., v (m3/kg) Int.Enr, u(kJ/kg) Enthl., h(kJ/kg)

vf vg uf ug hf hg

85 57.868 0.001032 2.8261 355.96 2487.8 356.02 2651.4

90 70.183 0.001036 2.3593 376.97 2494.0 377.04 2659.6

95 84.609 0.001040 1.9808 398.00 2500.1 398.09 2667.6

ii) X = mg / mt = mg / (mf + mg )

= 2 kg / ( 2 + 8 ) kg = 0.2

at 90oC, vf = 0.001036 m3/kg

vg = 2.3593 m3/kg

(from the steam table)

vav = vf + xvfg

= 0.001036 + (0.2)(2.3593-0.001036)

= 0.473 m3/kg

V = mvav = 10 kg (0.473 m3 / kg)

= 4.73 m3

v m3/kg

P=70.18 kPa

T = 90oC , mg = 2 kg

mf = 8 kg

Quality is used to find other properties of

saturated liquid-vapor mixtures (SLVM):

uavg = uf + x ufg and, x = (uavg - uf )/ufg

havg = hf + x hfg and, x = (havg - hf )/hfg

ALWAYS check: hf ≤ havg ≤ hg,

another e.g.

GIVEN: 0.05 kg of water at 25oC in a container of 1.0 m3 volume.

FIND: the phase description, pressure, and quality (if appropriate).

Answer: 1.Find the proper table: Water + given T, use Table, Look up

vf = 0.001003 m3/kg , vg = 43.340 m3/kg

2.Calculate v (= vavg) = V/m = 1.0 m3/0.05 kg = 20 m3/kg

3.Since vf < vavg < vg the phase is saturated liquid-vapor

mixture (SLVM).

4.Since this is a saturated state: P = P sat@25C = 3.1698 kPa

5.Finally, x = (vavg - vf)/(vg - vf) = (20 - 0.001003)/(43.34 -

0.001003) = 0.461 or ( 46.1% )

SUPERHEATED VAPOUR:NOT about to condense.

e.g:Water at P = 0.5 MPa, h = 2890 kJ/kg. Find T.

Answer:

1.Check Table (given P) finding hg = 2748.7

2.since h > hg, this is a superheated vapor !

3.From Table A6 at P = 0.5 MPa, find two rows

that bracket the given h

T= 200 oC h= 2855.8 kJ/kg

T= 250oC h= 2961.0 kJ/kg

h of 2890 kJ/kg is between these values

make interpolation57

COMPRESSED LIQUID:all liquid, NOT about to vaporize

Characteristics:

at a given P, T < TSAT or

at a given T, P > PSAT or

at either a given P or T, v < vf or h < hf or u < uf

approximate compressed liquid behaviour, evaluated at the

given TEMPERATURE (don’t use the pressure)

Note:-given two independent intensive properties and need to find other properties.

• Before the quantitative properties can be found, must find qualitatively where the point is on the phase diagram – saturated, superheated, subcooled or mixture …etc

• Algorithm:

– Given T and P:

• Compare to Tsat or Psat.

Decide whether subcooled, saturated, or

superheated.

– Given T or P and one other property (v, u, h, s)

• Look up vf or uf or hf or sf and vg or ug or hg or sg

from table (sat.)

• If v, u, h, or s < sat values (vf , uf,….): subcooled.

• If v, u, h, or s > sat values (vg, ug,….): superheated.

• If vf or uf or hf or sf <v, u, h, or s < vg or ug or hg or sg

: two-phase (SLVM) -- find the quality x.

Properties of Compressed Liquids

• Generally no tabular data for compressed

liquids

• Since liquids are close to being

incompressible, their properties change

little with pressure.

– We know this is true from our everyday

experience with specific volume (density).

– It also holds for u, h, s …

• To find thermodynamic properties for

compressed liquids, we use the properties

of saturated liquids at the same

temperature.63

Example:1. Complete the following steam table. Locate each point on a PT and a PV diagram.

T oC P, kPa v m3/kg Phase

Description

50 ? 4.16 ?

? 200 ? Saturated

vapour

250 400 ? ?

110 600 ? ?

T oC P, kPa v m

3/kg Phase

Description

50 12.352 4.16 SLVM

120.21 200 0.89 Saturated

vapour

250 400 0.53 Superheated

vapour

110 600 0.001 Compressed

liquid

At Saturated Liquid & saturated Vapour states:

• Latent heat of vapourization

• Enthalpy of vapourization

• represents

• The amount of Energy needed to Vapourize a unit Mass of Saturated Liquid at a given Temperature and Pressure

• At critical point hfg= 0

• h as T , P

2. A piston-cylinder device is used to vaporize 10 kg saturated liquid to saturated vapour at 101 kPa. Determine the volume change and the amount of energy added.[given the Latent heat of vaporization of water at 101

kPa is LV = 2257.0 kJ/kg )

Assumptions:

-The pressure is kept as a constant during the process

Volume Change Vg - Vf

(1)Volume change

The total volume of the saturated liquid

Vf = vfm

The total volume when all the water becomes saturated

vapour.

Vg = vgm

The volume difference is:

Vdifference = vgm - vfm (vg & vf from table)

= 10 (1.673 - 0.001) = 16.72 m3

(2) Amount of energy added

Latent heat of vaporization of water at 101 kPa is LV =

2257.0 kJ/kg. It is the same amount of energy needed to

change 1 kg saturated water to saturated vapour at 100

Q = m Lv = 10kg (2257.0)kJ/kg = 22,570 kJ

THE END

Example:1. Complete the following steam table. Locate each point on a PT and a PV diagram.

1. For a pure subcooled liquidA. the pressure will be lower than the vapour pressure at

the same temperature.

B. the quality will be zero.

C. the temperature will be lower than the boiling

temperature at the same pressure.

2. In the superheated vapour region of a pure substance

A. quality is greater than 1.0

B. the temperature is lower than the boiling temperature

at the same pressure.

C. the pressure is lower than the vapor pressure at the

same temperature.

3. Specific properties

A. apply at a point.

B. include temperature and pressure.

C. are intensive properties.

4. Quality

A. is defined by ML/(ML + MV)

B. is defined by v = xvf + (1-x)vg

C. is 1 for a superheated vapour.

D. is the fraction of mass that is vapour.

E. is only defined in the two phase region.

9. In the piston-cylinder experiment, volume increased in

the two phase region as heat was added at constant

pressure because gases and liquids both expand as they

are heated.

True or false

10. At a given pressure, the melting temperature and the

boiling temperature can not be equal.

True or false

5. The vapour pressure of a pure substance

A. depends on the relative amounts of liquid and vapour

present.

B. is defined between the triple point and critical point.

C. is the pressure exerted by a pure substance in the two

phase (liquid-vapor) region.

6. Which of the following is an extensive property?

A. Temperature

B. Pressure

C. Volume

D. Density

E. None of the above

7. The critical point of a pure substance:

A. represents the highest temperature and pressure for

which liquid and vapour can coexist.

B. is the point at which the saturated liquid and

saturated vapour curves meet.

C. is where vapor pressure has its largest possible value.

8. The normal boiling temperature of water is lower in

Tibet than in Malaysia.

True or false

9. In the piston-cylinder experiment, volume

increased in the two phase region as heat was

added at constant pressure because gases and

liquids both expand as they are heated.

True or false

10. At a given pressure, the melting temperature

and the boiling temperature can not be equal.

True or false

Week 3EQUATION OF STATE

Any equation that relates the pressure, temperature, and specific

volume of a substance

• The simplest is the equation of state for substance in the gas phase, best-known as the ideal-gas equation of state.

Gas & Vapour

• Vapour : a gas closes to the state of condensation

• Gas : the vapour phase when it is above the critical temperature (always)

% of error involved in assuming steam to be an ideal gas & the region where steam can be treated as an ideal gas with less than 1% error.

A gas will fill its container completely and

does not exhibit a free surface.

A liquid will take the shape of its container

but exhibits a free surface.

• very low density that must be enclosed to keep together.

• no definite shape.

• no definite volume, but they completely fill a container. The volume of the container is the volume of the gas in it.

gasgas

containergas

A gas exerts a pressure on all sides of the container that holds it. Gas can be compressed by pressures greater than the pressure the gas on its container .

• particles (atoms or molecules) are not attached to each other.

• When a gas molecule hits another one, they bounce off each other.

Ideal gas

• An ideal gas is a special case of a pure substance in the vapour phase and follow the ideal gas law, which was originally derived from the experimentally measured Charles’ law and Boyle’s law.

Boyle’s law PV = C at constant T

Charles’ law V/T = C at constant P

Boyle’s law, Charles’ law and Avogadro’s hypothesis give the ideal gas law

P = R (T/v) or Pv = RT• P : pressure (absolute)

• T : temperature (absolute)

• v : specific volume

• R : characteristic gas constant

• R is different for each gas R = RU /M• unit : kJ/kg.K or kPa.m3/kg.K or kJ/kmol.K• M : molar mass (molecular weight) of the gas• The mass of one kmol (or kilogram-mole) of a substance in

kilograms.• e.g : molar mass of nitrogen is 28• -it means : the mass of 1 kmol of nitrogen is 28 kg.• -the mass of the system,

m = M N or N = m/M• where N : the mole number• (e.g : volume per unit mass, v, m3/kg

volume per unit mole, , m3/kmol)

• RU: universal gas constant

(same value for all gases)

RU = 8.314 kJ/kmol.K, or kPa/m3/kmol.K

• Several forms of the ideal gas equation

PV = m R T (V = mv)

PV = NRUT (mR = (MN)R = NRU)

P v = RUT (V = N v )

Note : V = mvmR = (MN)R = N RU

V = NV : volume (m3)

v : specific volume (m3/kg): molar specific volume (m3/kmol)v

what is PV ?

PV = (pressure).(volume)

= 1 (kg/(m.s2))(m3)

= 1 ( kg.m2/s2)

= 1 joule

a measure of energy !!!

Compressibility factor Z (a measure of deviation from ideal gas behaviour)

• there is no real substance that satisfies the ideal gas definition for the entire range of state

• the ideal gas is a concept rather than a reality

• the deviation of a real substance to the ideal gas behaviour is given by compressibility factor, Z

Z = Pv/RT or Pv = ZRT• for ideal gases , Z = 1

The further away Z is from

unity, the more the gas deviates

from ideal gas behaviour.

REAL GAS

1< Z 0R Z < 1

Molar mass and gas constant properties.

Substance Formula Molar

Mass,M

kg/kmol

Gas constant, R

kJ/kg.K

Air - 28.97 0.2870

Argon Ar 39.948 0.2081

Carbon dioxide CO2 28.011 0.1889

Chlorine Cl2 70.906 0.1173

Helium He 4.003 2.0769

Nitrogen N2 28.013 0.2968

Oxygen O2 31.999 0.2598

Water H2O 18.015 0.4615

Example 1:Determine the mass of the air in a room whose dimensions are 4 m x 5 m x 6 m at 100 kPa and 25oC.

Answer:

from the table above :

the gas constant of air R = 0.287 kPa.m3/kg.K

absolute temperature of the room

T = 25 oC + 273 = 198 K

volume of the room , V = 4x5x6 m3 = 120 m3

mass of the air in the room

m =(100 kPa)(120 m3)

(0.287kPa. m3/kg.K)(298 K)= 140.3 kg

PV = mRT ; therefore m = PV/RT

Example 2:

Using the ideal gas equation of state find the specific volume of steam at 400oC and at pressure,i) 0.01 MPa ii) 0.1 MPa iii) 20.0 MPa

Answer:

• For steam R = 0.4615 kJ/kg.K ( from table above)

• i)v = RT/P = 0.4615x(400+273) / 10 = 31.066 m3/kg

• ii)v = RT/P = 0.4615x(400+273) / 100= 3.1066 m3/kg

• ii) v = RT/P = 0.4615x(400+273) / 20000 = 0.015533 m3/kg

The combine gas law formula

• The same amount of the same gas is given at two different sets of conditions [m1=m2].

State 1P1, V1, T1, m1

State 2P2, V2, T2, m2=m1

11 at state 1 :

at state 2 : R

• therefore or

Other Equations Of State

• -The ideal gas equation of state is a simple algebraic equation.

• -the range of applicability is quite limited

• -accurate for low pressures and/or high temperatures

• - for improving the accuracy and range of applicability in describing the behaviour of actual system, many equation have been proposed.

(i) Van der Waals Equation

comparing this equation with the ideal gas equation, we can see

• P is replaced with (P+a/v2) – taking care of attraction forces by other molecules when a molecule strikes the container wall.

• V is replaced with (v-b) – the free volume the molecules can move around is the volume that are not occupied by other molecules.

RTb))(vv

Constants of Van der Waals equation of state

Substance

m6kPa/kmol2b

m3/kmol

Carbon dioxide 365.4 0.04280

Helium 3.46 0.02371

Hydrogen 24.96 0.02668

Nitrogen 232.4 0.03864

Oxygen 138.1 0.03184

Water 552.6 0.03042

(ii) Beattie-Bridgeman Equation

AB)v)(

Specific Heats, internal energy and enthalpy

tell us about energy storage capability of a substance.

20oC 30oC20oC 30oC

4.5 kJ41.8 kJ

Specific Heats, internal energy and enthalpy

• All matter has a temperature, T associated with it

• T -- a direct measure of the motion of the molecules:

• -greater the motion the higher the temperature

• -Motion requires energy--more energy ®higher T

• -this energy is supplied by heat

• QUESTION: by how much will the temperature of an object increase or decrease by the gain or loss of heat energy?

• The greater the heat capacity of an object, the more heat

• energy is required to raise the temperature of the object

Specific Heat Capacity : the energy required to raise the temperature of a unit mass of a substance by one degree.

• (The relationship does not apply if a phase change is encountered)

m= 1 kg, T = 1oC

specific heat =5 kJ/kg.oC

Q = m c DT(heat added) = (mass) x (specific heat) x

(change in temperature)

Two kinds of specific heat

• specific heat at constant volume, cV

-the energy required to

raise the temperature of

the unit mass of a

substance by one degree

as the volume is

maintained constant.

• specific heat at constant pressure, cP

-the energy required to

raise the temperature of

the unit mass of a

substance by one degree

as the pressure is

maintained constant.

• cp > cv

because at constant pressure P, the system is allowed to expand, and the energy for this expansion work must also be supplied to the system.

Suppose that a body absorbs an amount of heat Q and its temperature consequently rises by T.

v constantm=1 kgT = 1oC

cV=3.12 kJ/kg.oC

P constantm= 1 kgT = 1oC

cP=5.2kJ/kg.oC

3.12kJ5.2 kJ

The usual definition of the heat capacity of

the body is

• Let say the body absorbs a very small amount of heat, so that its temperature only rises slightly, and its specific heat remains approximately constant

• For a constant volume process, all the heat absorbed will be converted into internal energy ( since dV = 0, no work done by or onto the system)

• ∂Q = du

• for an ideal gas the internal energy is a function of temperature only.

(T)dTcdu v

• For a constant pressure process, the heat absorbed will increase the internal energy and doing some work.

• From the definition of enthalpy,

• h = u + Pv

• dh = du + Pdv (P is constant)

by the definition of specific heat at constant pressure,

• we can write

• The enthalpy of an ideal gas is a function of temperature only.

(T)dTcdh P

Change in internal energy or enthalpy between state 1 and 2,

• can be simplified by taking average value of specific heat, then we can assume constant specific heat.

V12 (T)dTcuuΔu

P12 (T)dTchhΔh

)T(TcdTcuuΔu 12(av)V

(av)V12

)T(TcdTchhΔh 12(av)P

(av)P12

note : three ways to determine the internal energy and enthalpy changes of ideal gases:

• -by using the tabulated data

• -by using the cV and cP and performing the integration

• -by using average specific heat

Specific-heat Relation

• From the ideal gas law, Pv = RT

• And h = u + Pv

• h = u + RT

• differentiating dh = du + RdT

• replacing dh and du with specific heat relations

• cPdT = cVdT + RdT

• or cP = cV + R•

• The specific heat ratio is given the symbol κ

Summary: Ideal Gas• The gas consists of molecules that are in random

motion and obey the laws of mechanics.

• - The total number of molecules is large, but the volume of the molecules is a negligibly small fraction of the volume occupied by the gas.

• - No appreciable forces act on the molecules except during the collisions.

• An Ideal Gas (perfect gas)is one which obeys Boyle's Law and Charles' Law exactly.

• PV = mRT ; PV =NRUT

• R = 8.314 J K-1 mol-1 if P is in kPa, V is in L, T is in K

What volume is needed to store 0.050 moles of helium gas at 202.6kPa and 400K?

Given universal gas constant is 8.314 J/K.mol

What pressure will be exerted by 20.16g hydrogen gas in a 7.5L cylinder at 20 oC?

Given universal gas constant is 8.314 J/K.mol and molecular mass of hydrogen is 2.016g/mol.

A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 30C. How many moles of argon gas are in the cylinder?

Given universal gas constant is 8.314 J/K.mol

Given molecular mass of He is 4.003g/mol and universal gas constant is 8.314 J/K.mol

• A vessel of volume 0.2 m3 contains nitrogen at 1.013 bar and 15oC. If 0.2 kg of nitrogen is now pumped into the vessel, calculate the new pressure when the vessel has returned to its initial temperature. The molar mass of nitrogen is 28 kg/kmol, and RU = 8.3145 kJ/kmol.K

Solution 1

KkgNmM

RR U ./95.296

5.8314

VPm 237.0

28895.296

2.010013.1 5

RTmP 187

28895.296437.0

From P1v1 = m1RT1 --- initial state

m2 = m1 + 0.2 = 0.237 + 0.2 = 0.437 kg

final state P2v2 = m2RT2

but v2 = v1 & T2 = T1

1 Pa = 1 N/m2

1 atm = 101.325 kPa

= 1.01325 bars

= 760 mm Hg at 0 oC

1 mm Hg = 0.1333 kPa

1 kPa = 103 Pa = 10-3 Mpa

•Thank you!

ENERGY ANALYSIS OF CLOSED SYSTEMS

Week 4

(Heat & Work Interaction)

• Examine the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors.

• Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) systems.

• Develop the general energy balance applied to closed systems.

• Solve energy balance problems for closed (fixed mass) systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances.

Objectives

• Heat, Q, energy caused by DT (temp difference)

• Work W, energy caused by DX (physical motion)

• Systems DO NOT have HEAT (Q) or WORK (W)

HEAT AND WORK INTERACTION

• HEAT and WORK cross the system’s boundaries,

• which changes the energy of the system,

• which then changes the properties (can be measured)

• Heat Interaction: the form of energy that is transferred between two systems (or system and its surroundings) by virtue of a temperature difference.

• Symbol : Q : heat (or heat transfer) (kJ)

• q = Q /m heat per unit mass (kJ/kg)

• : Heat transfer rate (kJ/s = kW)

• An object does not possess "heat"

• Work Interaction: if the energy crossing the boundary of a closed system is not heat, it must be work. Definition: Energy transfer accomplished by a force moving an object some distance.

• Symbol: W unit: kJ

• w = W/m (kJ/kg)

• Power (kJ/s or kW)

** Formal sign convention

• OR by using subscript in and out

• Qin : heat is transferred into the system

• Qout : heat lost by the system

• Win : work done onto the system

• Wout: work done by the system

• If the direction is not known, it can be assumed in or out. After solving – positive result indicates the assumed direction is right, otherwise the other way round.

Similarities1. Both are boundary phenomena.

2. System possesses energy, not heat or work.

3. Both are associated with a process, not a

state.

4. Both are path function (their magnitudes

depend on the path followed during a

process as well as the end states).

Note: the final states are indistinguishable – by doing

work or heating

• -Path functions (heat, work) have inexact differentials –

symbol δ

e.g. δQ, δW

(not DW, DQ)

• -Point functions (properties) have exact differentials –

symbol d

• e.g. dV, dT

• Mechanical forms of work

• W = F s

• The work done by a constant force F on a body displaced a distance s

• Moving boundary work (Wb)

• P: constant

• V: constant

MOVING BOUNDARY WORK

• Moving boundary work

(P dV work): The expansion and compression work in a piston-cylinder device.

A gas does a differential amount of work Wb as it forces the piston to move by a differential amount ds.

• Quasi-equilibrium process: A process during which the system remains nearly in equilibrium at all times.

Wb is positive for expansionWb is negative for compression

The work associated with a moving

boundary is called boundary work.

• The area under the process curve on a P-V diagram represents the boundary work.

• The boundary work done during a process

depends on the path followed as well as the

end states.

• The net work done during a cycle is the difference between the work done by the system and the work done on the system

Moving boundary work (Wb)P: constant

P = f(V)

• Total boundary work,

• Wb = δW1 + δW2 + δW3 +…..+ δWN

• For a large N

• the boundary work is equal in magnitude to the area under process curve.

Constant P, Wb = PDV

unit : Wb = kPa-m3 = kJConstant V, Wb = 0

W= (P x dV)

Path (A) WA = W12 + W23 = (0) + ( 4 x (4-2)) = 8 unit of work

Path (B) WB = W14 + W43= (2 x (4 – 2)) + (0) = 4 unit of work

• The work done in changing states is dependent on the way the transformation takes place

• Q and W are "path functions":

• the amount of Q and W depends on the process path (NOT on the initial & final states).

V: constant

P: constant

Polytropic, Isothermal, and Isobaric processes

Polytropic and for ideal gas

Polytropic process:

C, n (polytropic exponent) constants

• When n = 1 (isothermal process)

Constant pressure process.

What is the boundary work for a

constant-volume process?

Schematic and P-V diagram for a polytropic process.

Wb in an IDEAL GAS

Constant Volume process: dV = 0 ; Wb = 0

Constant Pressure process

• Wb = P(V2 – V1) = P (mRT2/P – mRT1/P)

= mR(T2 – T1), for n = 1

Constant Temperature process (isothermal) and more ?

• for n = 1

FLOW WORK

• Mass flow rate :

the amount of mass flowing through a cross section per unit time

• For flow in tube = VmA unit : kg/s

where : density of fluid

Vm : mean fluid velocity normal to A

A : cross-sectional area normal to flow direction

• Volume flow rate: : the amount of the fluid flowing through a cross-section per unit time

=VmA unit : m3/s

Control Volume

F = PA

The work done in pushing the fluid element across the

boundary is called the flow work

Flow work is the energy needed to push a fluid into or out of a control volume, and it is equal to PV.

The total energy (per unit mass),

total energy = flow energy + internal energy + kinetic energy +

potential energy

Note: for non-flow process, Pv = 0.

• Total energy of the flowing fluid,

The rate of energy being transported

ENERGY BALANCE FOR CLOSED SYSTEMS

• Energy balance for any system undergoing any process

• Energy balance in the rate form

-The total quantities are related to the quantities per unit time is

- Energy balance per unit mass basis

- Energy balance in differential form

- Energy balance for a cycle

Energy balance when sign convention is used (i.e., heat

input and work output are positive; heat output and work

input are negative).

For a cycle DE = 0, thus Q = W.

Various forms of the first-law relation for closed systems when sign convention is used.

The first law cannot be proven mathematically, but no

process in nature is known to have violated the first law, and

this should be taken as sufficient proof.

HWU b DD

For a constant-pressure expansion

or compression process:

An example of constant-pressure process

Three ways of calculating Du.

THANK YOU FOR YOUR ATTENTION!!

Energy balance for a constant-pressure expansion or compression process

• General analysis for a closed system undergoing a quasi-equilibrium constant-pressure process.

(Q is to the system and W is from the system.)

For a constant-pressure expansion

or compression process:

HWU b DD

Three ways of calculating Du and Dh

1. By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available.

2. By using the cv or cp relations (Table A-2c) as a function of temperature and performing the integrations. This is very inconvenient for hand calculations but quite desirable for computerized calculations. The results obtained are very accurate.

3. By using average specific heats. This is very simple and certainly very convenient when property tables are not available. The results obtained are reasonably accurate if the temperature interval is not very large.

• An example of constant-pressure process

Week 5

First Law of ThermodynamicsConservation of energy principle:

Energy can be neither created nor destroyed;

it can only change forms.

Conservation of energy the total amount of energy in an isolated system remains constant

The 1st law the application of the conservation of energy principle to heat and thermodynamic processes;

-the kinetic energy of a moving car is converted into

heat energy at the brakes and tire surfaces.

-when chemical energy is released in burning and is

converted into light and heat energy.

-an object can be lifted by heating a rubber band.

Heat is converted into gravitational potential energy

NOTE: Process changes in the properties of that system.

When a very small part of a process occurs in a system, and is accompanied by a correspondingly very small change in the properties of the system

-A large (finite) change in the quantity X DX

-A small (infinitesimal) change in X dX

Q: -denotes heat, it has units of energy (it is an

energy interaction): Joules (J) or kJ (kilo-Joules).

-energy in transit due to the thermal interaction (temperature difference between the system and

its surroundings).

W: -denotes work interaction, it has units of energy (it is also an energy interaction): Joules (J) or

kilo-Joules (kJ).

-energy in transit due to mechanical interaction between the system and its surroundings

E: -Energy contained by the system (stored energy),

property of the system.

How much work can we possibly get out of heat?

E: represents the change in energy experienced by the system. If the system experience a change from state 1 to state 2, then

In many cases in thermodynamic analysis

Energy balance

b. The third process is adiabatic, means Q = 0

DE3 = Q3 –W3 = 32 kJ

= 0 - W3 = 32 kJ

W3 = -DE3 = -32 kJ

Process a: W’ = 0, why? DV = 0

1st law: Q’ – W’ = DU

Q’ – 0 = DU

Q’ = DU

or DU = Q’

Process b: no heat Q” = 0,

work of stirrer = W”s

1st law: Q” – W” = DU

0 – (-W”s) = DU

or DU = W”s

Let say u = f(T,v)

Integrating

T2 = 300K

V2 = 0.05 m3 +0.03 m3

= 0.08 m3

means Q = 0

means T2 = 300 K= T1

Answer

Hence final state: T2 = 300 K, P2 = 112.5 kPa , V2= 0.08 m3

or W = 0

therefore

V2 = 0.05 m3 +0.03 m3

= 0.08 m3

Examples:

Heat exchangers

Throttling Valves

Throttling valves are any kind of flow-

restricting devices that cause a

significant pressure drop in the fluid.

The pressure drop in the fluid is often

accompanied by a large drop in

temperature, and for that reason

throttling devices are commonly used in

refrigeration and air-conditioning

applications.

The temperature of an ideal gas does

not change during a throttling

(h = constant) process since h = h(T).

During a throttling process, the enthalpy

of a fluid remains constant. But internal

and flow energies may be converted to

each other.

Energy

balance

Heat exchangers

Heat exchangers are devices where two

moving fluid streams exchange heat without

mixing. Heat exchangers are widely used in

various industries, and they come in various

designs.A heat exchanger can be as

simple as two concentric pipes.

Mass and energy balances for the

adiabatic heat exchanger in the figure

The heat transfer associated with a heat

exchanger may be zero or nonzero depending

on how the control volume is selected.

Mass and Energy balances for a steady-flow process

A water

heater in

steady

operation.

Mass balance

WEEK 6

THE SECOND LAW OF

THERMODYNAMICS

Objectives• Introduce the second law of thermodynamics.

• Identify valid processes as those that satisfy both the first and second

laws of thermodynamics.

• Discuss thermal energy reservoirs, reversible and irreversible

processes, heat engines, refrigerators, and heat pumps.

• Describe the Kelvin–Planck and Clausius statements of the second law

of thermodynamics.

• Discuss the concepts of perpetual-motion machines.

• Apply the second law of thermodynamics to cycles and cyclic devices.

• Apply the second law to develop the absolute thermodynamic

temperature scale.

• Describe the Carnot cycle.

• Examine the Carnot principles, idealized Carnot heat engines,

refrigerators, and heat pumps.

• Determine the expressions for the thermal efficiencies and coefficients

of performance for reversible heat engines, heat pumps, and

refrigerators.

INTRODUCTION TO THE SECOND LAW

A cup of hot coffee

does not get hotter in

a cooler room.

Transferring

heat to a wire

will not

generate

electricity.

Transferring

heat to a

paddle wheel

will not cause

it to rotate.

These processes

cannot occur

even though they

are not in violation

of the first law.

Processes occur in a

certain direction, and not

in the reverse direction.

A process must satisfy both

the first and second laws of

thermodynamics to proceed.

MAJOR USES OF THE SECOND LAW

1. The second law may be used to identify the direction of processes.

2. The second law also asserts that energy has quality as well as quantity.

The first law is concerned with the quantity of energy and the

transformations of energy from one form to another with no regard to its

quality. The second law provides the necessary means to determine the

quality as well as the degree of degradation of energy during a process.

3. The second law of thermodynamics is also used in determining the

theoretical limits for the performance of commonly used engineering

systems, such as heat engines and refrigerators, as well as predicting the

degree of completion of chemical reactions.

THERMAL ENERGY RESERVOIRS

Bodies with relatively large thermal

masses can be modeled as thermal

energy reservoirs.

A source

supplies

energy in the

form of heat,

and a sink

absorbs it.

• A hypothetical body with a relatively large thermal energy capacity (mass x

specific heat) that can supply or absorb finite amounts of heat without

undergoing any change in temperature is called a thermal energy reservoir,

or just a reservoir.

• In practice, large bodies of water such as oceans, lakes, and rivers as well as

the atmospheric air can be modeled accurately as thermal energy reservoirs

because of their large thermal energy storage capabilities or thermal masses.

HEAT ENGINES

Work can always

be converted to

heat directly and

completely, but the

reverse is not true.

Part of the heat

received by a heat

engine is

converted to work,

while the rest is

rejected to a sink.

The devices that convert heat to

1. They receive heat from a high-

temperature source (solar energy,

oil furnace, nuclear reactor, etc.).

2. They convert part of this heat to

work (usually in the form of a

rotating shaft.)

3. They reject the remaining waste

heat to a low-temperature sink

(the atmosphere, rivers, etc.).

4. They operate on a cycle.

Heat engines and other cyclic

devices usually involve a fluid to

and from which heat is

transferred while undergoing a

cycle. This fluid is called the

working fluid.

A steam power plant

A portion of the work output

of a heat engine is consumed

internally to maintain

continuous operation.

Thermal efficiency

Some heat engines perform better

than others (convert more of the

heat they receive to work).

Schematic of

a heat engine.

Even the most

efficient heat

engines reject

almost one-half

of the energy

they receive as

waste heat.

Can we save Qout?

A heat-engine cycle cannot be completed without

rejecting some heat to a low-temperature sink.

In a steam power plant,

the condenser is the

device where large

quantities of waste

heat is rejected to

rivers, lakes, or the

atmosphere.

Can we not just take the

condenser out of the

plant and save all that

waste energy?

The answer is,

unfortunately, a firm

no for the simple

reason that without a

heat rejection process

in a condenser, the

cycle cannot be

completed.

Every heat engine must waste

some energy by transferring it to a

low-temperature reservoir in order

to complete the cycle, even under

idealized conditions.

The Second Law of

Thermodynamics:

Kelvin–Planck Statement

A heat engine that violates the

Kelvin–Planck statement of the

second law.

It is impossible for any device

that operates on a cycle to

receive heat from a single

reservoir and produce a net

amount of work.

No heat engine can have a thermal

efficiency of 100 percent, or as for a

power plant to operate, the working fluid

must exchange heat with the

environment as well as the furnace.

The impossibility of having a 100%

efficient heat engine is not due to

friction or other dissipative effects. It is a

limitation that applies to both the

idealized and the actual heat engines.

REFRIGERATORS AND HEAT PUMPS• The transfer of heat from a low-

temperature medium to a high-

temperature one requires special

devices called refrigerators.

• Refrigerators, like heat engines,

are cyclic devices.

• The working fluid used in the

refrigeration cycle is called a

refrigerant.

• The most frequently used

refrigeration cycle is the vapor-

compression refrigeration cycle.

Basic components of a

refrigeration system and

typical operating conditions.

In a household refrigerator, the freezer compartment

where heat is absorbed by the refrigerant serves as

the evaporator, and the coils usually behind the

refrigerator where heat is dissipated to the kitchen

air serve as the condenser.

Coefficient of Performance

The objective of a refrigerator is to

remove QL from the cooled space.

The efficiency of a refrigerator is expressed

in terms of the coefficient of performance

(COP).

The objective of a refrigerator is to remove

heat (QL) from the refrigerated space.

Can the value of COPR be

greater than unity?

The objective

of a heat

pump is to

supply heat

QH into the

warmer

space. The work

supplied to a

heat pump is

used to extract

energy from the

cold outdoors

and carry it into

the warm

indoors.

for fixed values of QL and QH

Can the value of COPHP

be lower than unity?

What does COPHP=1

represent?

When installed backward,

an air conditioner

functions as a heat pump.

• Most heat pumps in operation today have a

seasonally averaged COP of 2 to 3.

• Most existing heat pumps use the cold outside air

as the heat source in winter (air-source HP).

• In cold climates their efficiency drops considerably

when temperatures are below the freezing point.

• In such cases, geothermal (ground-source) HP

that use the ground as the heat source can be

• Such heat pumps are more expensive to install,

but they are also more efficient.

• Air conditioners are basically refrigerators whose

refrigerated space is a room or a building instead

of the food compartment.

• The COP of a refrigerator decreases with

decreasing refrigeration temperature.

• Therefore, it is not economical to refrigerate to a

lower temperature than needed.

Energy efficiency rating (EER): The amount of heat removed from the

cooled space in Btu’s for 1 Wh (watthour) of electricity consumed.

The Second Law of Thermodynamics:

Clasius Statement

It is impossible to construct a device that

operates in a cycle and produces no effect

other than the transfer of heat from a lower-

temperature body to a higher-temperature

It states that a refrigerator cannot operate unless

its compressor is driven by an external power

source, such as an electric motor.

This way, the net effect on the surroundings

involves the consumption of some energy in the

form of work, in addition to the transfer of heat

from a colder body to a warmer one.

To date, no experiment has been conducted that

contradicts the second law, and this should be

taken as sufficient proof of its validity.

A refrigerator that

violates the Clausius

statement of the second

Equivalence of the Two Statements

Proof that the

violation of the

Kelvin–Planck

statement leads

to the violation

of the Clausius

statement.

The Kelvin–Planck and the Clausius statements are equivalent in their

consequences, and either statement can be used as the expression of the

second law of thermodynamics.

Any device that violates the Kelvin–Planck statement also violates the

Clausius statement, and vice versa.

REVERSIBLE AND IRREVERSIBLE PROCESSES

Two familiar

reversible processes.

Reversible processes deliver the most and consume

the least work.

Reversible process: A process that can be reversed without leaving any trace

on the surroundings.

Irreversible process: A process that is not reversible.

• All the processes occurring in nature are irreversible.

• Why are we interested in reversible processes?

• (1) they are easy to analyze and (2) they serve as

idealized models (theoretical limits) to which actual

processes can be compared.

• Some processes are more irreversible than others.

• We try to approximate reversible processes. Why?

Irreversibilities

Friction

renders a

process

irreversible.

Irreversible

compression

expansion

processes.

(a) Heat

transfer

through a

temperature

difference is

irreversible,

and (b) the

reverse

process is

impossible.

• The factors that cause a process to be

irreversible are called irreversibilities.

• They include friction, unrestrained expansion,

mixing of two fluids, heat transfer across a finite

temperature difference, electric resistance,

inelastic deformation of solids, and chemical

reactions.

• The presence of any of these effects renders a

process irreversible.

Internally and Externally Reversible Processes

A reversible process

involves no internal and

external irreversibilities.

• Internally reversible process: If no irreversibilities occur within the boundaries of

the system during the process.

• Externally reversible: If no irreversibilities occur outside the system boundaries.

• Totally reversible process: It involves no irreversibilities within the system or its

surroundings.

• A totally reversible process involves no heat transfer through a finite temperature

difference, no nonquasi-equilibrium changes, and no friction or other dissipative

effects.

Totally and internally reversible heat

transfer processes.

PERPETUAL-MOTION MACHINES

A perpetual-motion machine that

violates the first law (PMM1).

A perpetual-motion machine that

violates the second law of

thermodynamics (PMM2).

Perpetual-motion machine: Any device that violates the first or the second

A device that violates the first law (by creating energy) is called a PMM1.

A device that violates the second law is called a PMM2.

Despite numerous attempts, no perpetual-motion machine is known to have

worked. If something sounds too good to be true, it probably is.

Proof of the first Carnot principle.

Week 7

THE CARNOT CYCLE

REVERSIBLE AND IRREVERSIBLE PROCESSES

Two familiar

reversible processes.

Reversible processes deliver the most and consume

the least work.

Reversible process: A process that can be reversed without leaving any trace

on the surroundings.

Irreversible process: A process that is not reversible.

• All the processes occurring in nature are irreversible.

• Why are we interested in reversible processes?

• (1) they are easy to analyze and (2) they serve as

idealized models (theoretical limits) to which actual

processes can be compared.

• Some processes are more irreversible than others.

• We try to approximate reversible processes. Why?

Irreversibilities

Friction

renders a

process

irreversible.

Irreversible

compression

expansion

processes.

(a) Heat

transfer

through a

temperature

difference is

irreversible,

and (b) the

reverse

process is

impossible.

• The factors that cause a process to be

irreversible are called irreversibilities.

• They include friction, unrestrained expansion,

mixing of two fluids, heat transfer across a finite

temperature difference, electric resistance,

inelastic deformation of solids, and chemical

reactions.

• The presence of any of these effects renders a

process irreversible.

Internally and Externally Reversible Processes

A reversible process

involves no internal and

external irreversibilities.

• Internally reversible process: If no irreversibilities occur within the boundaries of

the system during the process.

• Externally reversible: If no irreversibilities occur outside the system boundaries.

• Totally reversible process: It involves no irreversibilities within the system or its

surroundings.

• A totally reversible process involves no heat transfer through a finite temperature

difference, no nonquasi-equilibrium changes, and no friction or other dissipative

effects.

Totally and internally reversible heat

transfer processes.

CARNOT

Reversible Isothermal Expansion (process 1-2, TH = constant)

Reversible Adiabatic Expansion (process 2-3, temperature drops from TH to TL)

Reversible Isothermal Compression (process 3-4, TL = constant)

Reversible Adiabatic Compression (process 4-1, temperature rises from TL to TH)

Execution of

the Carnot

cycle in a

closed

system.

P-V diagram of the Carnot cycle. P-V diagram of the reversed

Carnot cycle.

The Reversed Carnot Cycle

The Carnot heat-engine cycle is a totally reversible cycle.

Therefore, all the processes that comprise it can be reversed,

in which case it becomes the Carnot refrigeration cycle.

THE CARNOT

PRINCIPLES

1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs.

2. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same.

All reversible heat

engines operating

between the same two

reservoirs have the

same efficiency.

THE THERMODYNAMIC TEMPERATURE SCALE

The arrangement of heat

engines used to develop

the thermodynamic

temperature scale.

A temperature scale that is

independent of the

properties of the

substances that are used

to measure temperature is

called a thermodynamic

temperature scale.

Such a temperature scale

offers great conveniences

in thermodynamic

calculations.

For reversible cycles, the

heat transfer ratio QH /QL

can be replaced by the

absolute temperature ratio

TH /TL.

A conceptual experimental setup

to determine thermodynamic

temperatures on the Kelvin

scale by measuring heat

transfers QH and QL.

This temperature scale is

called the Kelvin scale,

and the temperatures on

this scale are called

absolute temperatures.

THE CARNOT HEAT ENGINE

The Carnot

heat engine

is the most

efficient of

all heat

engines

operating

between the

same high-

and low-

temperature

reservoirs.

No heat engine can have a higher

efficiency than a reversible heat engine

operating between the same high- and

low-temperature reservoirs.Any heat

engineCarnot heat

engine

The Quality of Energy

The fraction of heat that

can be converted to work

as a function of source

temperature.

The higher the temperature

of the thermal energy, the

higher its quality.

How do you increase the

thermal efficiency of a Carnot

heat engine? How about for

actual heat engines?

Can we use

C unit for

temperature

THE CARNOT REFRIGERATOR

AND HEAT PUMP

No refrigerator can have a higher COP

than a reversible refrigerator operating

between the same temperature limits.

Any refrigerator or heat pump

Carnot refrigerator or heat pump

Summary• The Carnot cycle

The reversed Carnot cycle

• The Carnot principles

• The thermodynamic temperature scale

• The Carnot heat engine

The quality of energy

• The Carnot refrigerator and heat pump

WEEK 9

ENTROPY

Objectives

• Apply the second law of thermodynamics to processes.

• Define a new property called entropy to quantify the second-

law effects.

• Establish the increase of entropy principle.

• Calculate the entropy changes that take place during

processes for pure substances, incompressible substances,

and ideal gases.

• Examine a special class of idealized processes, called

isentropic processes, and develop the property relations for

these processes.

• Introduce and apply the entropy balance to various systems.

ENTROPY AND THE SECOND LAW

The second law of thermodynamics can be stated in 3

equivalent ways:

(1)Heat flows spontaneously from a hotter to a colder object,

but not vice versa.

(2)No heat engine that cycles continuously can change all its

heat-in to useful work-out.

(3) If a system undergoes spontaneous change, it will change in

such a way that its entropy will increase or, at best, remain

constant.

The second law tells us: the manner in which the spontaneous change

will occur.

The first law tells us: the conservation of energy.

The second law deals with the dispersal of energy.

Entropy (S) is a state variable and a

characteristic for a system in

equilibrium . By this is meant that S

is always the same for the system

when it is in equilibrium state.

Entropy is a measure of disorder.

A state that can occur in only one way is a state of

high order. For example : one arrangement of its

molecules.

But a state that can occur in many ways is a more

disordered state. One way to associate a number

with disorder is to take the disorder of a state as

being proportional to W, the number of ways a state

can occur.

Spontaneous processes in systems that

contain many molecules always occur in a

direction from a

State that

can exist

in only a

few ways

State that

can exist

in many

Hence when left to themselves, systems

retain their original state of order or else

increase their disorder.

The most probable state of a system is

the state with the largest entropy.

It is also the state with the most disorder

and the state that can occur in the

largest number of ways.

ENTROPY

Evolution of the Different Perspectives of

the Second Law

· some events are naturally impossible

· Kelvin-Planck statement (no perfect engine)

· Clausius Statement(no perfect refrigerator)

· Carnot Principles (Carnot efficiency is the limit)

· Clausius Inequality (concept of entropy)

The 1st Law deals with the property energy and its

conservation.

The 2nd Law leads to the definition of a new property

called Entropy.

Entropy is somewhat abstract concept and is difficult to

give a physical description of it.

It is best understood by investigating its uses in

commonly encountered processes and problems.

Clausius Inequality

n Another corollary of the 2nd Law.

n Now we will deal with increments of heat

and work, Q and W, rather than Q and W.

n We will employ the symbol

which means to integrate over all the

parts of the cycle .

Q The cyclic integral of the heat transfer

Corollary – hal, akibat atau kesimpulan

yang pasti berlaku mengikut urutan

logik.

S. kJ/K

Any heat transfer to or from a system can be

considered to consist of differential amounts

of heat transfer.

Then the cyclic integral of Q/T can be

viewed as

‘The Differential Amounts of Heat Transfer

DIVIDED by the TEMPERATURE T at the

boundary.

Clausius Inequality is expressed as;

valid for all cycles, reversible or irreversible:

always less than or equal to zero.

German Physicist

R.J.E Clausius –

(1822-1888)

The cyclic integral of Q/T

System Considered in the

Development of Clausius

Inequality

At constant

thermodynamic

temperature TR

(absolute).

Heat received by the cyclic device from

the reservoir and supplies heat

to the system.

Work produced by the

reversible cyclic device.

Dashed lines.

Work produced by the system as a

result of this heat transfer.

Energy balance of combined system:

[or total work of the combined system]

WC = Wrev + Wsys

Hence from first law for combined system

Change in total energy of combined system

Q – W = dU

Q – dU = W

Considering that the cyclic device is a reversible one

But for a cyclic process, dE = 0

WC net work for the combined cycle.

WC is produced by the combined cycle by

exchange heat with a single reservoir

this is against the Kelvin-Planck Statement

therefore WC cannot be work output (must be

negative)

-since TR is always positive

This equation is the Clausius Inequality

Let say the system & the cyclic device are reversible

the cycle of the combined system is internally reversible

(all quantities will have the same magnitude but the

opposite sign)

Wc , which cannot be positive earlier, now cannot be

negative.

Therefore it follows that WC int rev = 0

Let’s Look at a Simple Reversible Cycle

V is a property must be a property

We’ll define a new property, entropy, as:

s = S/m : intensive property

A Special Case: Internally Reversible

Isothermal Heat Transfer Processes

This equation is particularly useful for determining

the entropy changes of thermal energy reservoirs.

Reversible Heat Engine

For a reversible heat engine operating between a high

temperature reservoir at TH and a low-temperature reservoir

at TL,HIGH TEMPERATURE RESERVOIR

LOW TEMPERATURE RESERVOIR

REVERSIBLE

for REVERSIBLE HEAT ENGINE

IRREVERSIBLE HEAT ENGINE

For an irreversible heat engine operating between

the same reservoirs (as in reversible HE):

HIGH TEMPERATURE RESERVOIR

LOW TEMPERATURE RESERVOIR

IRREVERSIBLE

From The Carnot principle

(Wnet)irr < (Wnet)rev

and (QL)irr > (QL)rev

can write : (QL)irr = (QL)rev + Q diff

where Qdiff = (QL)irr - (QL)rev and Qdiff>0

The Increase of

Entropy Principle

The Increase of Entropy Principle

From Clausius inequality

for reversible processfor irreversible process

S = S2 – S : entropy change of the system

=for reversible process

entropy transfer with heat

For irreversible process

or S = S2 – S1 = + Sgen

Sgen is always positive or zero

Depends on process (not a property)

If entropy transfer is zero, S = Sgen

For isolated system (adiabatic closed system), where heat transfer

is zero,

the entropy of an isolated system during a process always

increase or in the limited case of a reversible process, remains

constant

---INCREASE OF ENTROPY PRINCIPLE----

Isolated system – no interaction with surrounding

-may consist of subsystems

The entropy change of an isolated

system is the sum of the entropy

changes of its components, and is

never less than zero.

A system and its surroundings can be viewed as two subsystems of an

isolated system

since isolated system involves no entropy transfer

Sgen = Stotal = Ssys + Ssur 0

Note : equality for the

reversible process

Note : equality is for the reversible process

Since no process is reversible, some entropy is

generated during a process –entropy of the universe

is continuously increasing.

Sgen > 0 : irreversible process

Sgen = 0 : reversible process

Sgen < 0 : impossible process

The entropy of an isolated system will increase until

the entropy of the system reaches a maximum value

– state of equilibrium.

The increase of

entropy principle

Entropy Increase Principle

S = S2 – S1 = where Sgen 0

Isolated system

+ Sgen

Notes:

•A process can take place in the direction that complies with

the increase of entropy princ. , Sgen≥0

•The entropy of the universe is continuously increasing,

more disorganized and is approaching chaotic.

•The Sgen is due to the existence of irreversibilities. For a

device, higher irreversibilities lower the efficiency.

Example.

Source 800 K

Sink 200K

Q = 2000 kJ

Show that the heat can not transfer from the low-temperature

sink to the high-temperature source based on the increase

entropy principle.

∆S(source) = Q/T = 2000 kJ/ 800 K = 2.5 kJ/K

∆S(sink) = Q/T = - 2000 kJ/ 500 K = - 4 kJ/K

Sgen = ∆S(source) + ∆S(sink) = (2.5 – 4) kJ/K = -1.5 kJ/K < 0 ,

…. Impossible

•Based on the entropy increase principle

•heat cannot be transferred from low-temp to high-temp (without

external work)

Source 800 K

Sink 500K

Q = 2000 kJSgen = ∆S(source) + ∆S(sink)

Entropy generation, Sgen (for closed system)

System

inHeat

entropy balance for any system undergoing any process

Sin – Sout + Sgen = Ssystem

For closed system

12systemgen

k SSΔSST

adiabatic closed system: Sgen = Ssystem

system + surroundings: Sgen = S = Ssystem +Ssurr

or Sin – Sout + Sgen= Ssystem

---- entropy balance37

Entropy Balance

System

Esystem

Ssystem

systemtheof

entropytotal

theinChange

generated

entropy

leaving

entropy

entering

entropy

Entropy Balance:

the entropy change of a system during a

process is equal to the net entropy transfer

through the system boundary and the entropy

generated within the system.

Mechanisms of entropy transfer

•heat transfer (for closed and open system)

•mass flow (for open system only)

Tb = 400KQ = 500 kJ

Sheat = Q/T

= 1.25 kJ/kg

system

If T is constant

Entropy transfer by heat transfer (for closed system)

If T is not constant

Entropy transfer by work : Swork = 0

Entropy : a measure of molecular disorder or molecular randomness

Sgas > Sliquid> Ssolid 40

Random motion

Oderly motion

Molecular disorders or disorganized form of energy(heat)

Organized form of energy (work)

Energy : quantity is preserved during process ( 1st Law)

quality is bound to decrease (2nd Law)

decrease in quality is accompanied by an increase in entropy

When energy is transferred as heat (Q) it is accompanied by

entropy transfer (Q/T)

When energy is transferred as work (W) there is no entropy

transfer

1st Law: no distinction between heat transfer and work

2nd Law: gives the distinction between heat transfer and work

an energy interaction that is accompanied by entropy transfer is

heat transfer;

an energy interaction that is not accompanied by entropy

transfer is work.

Q/TsurrSgen

Q/Tsys

System surrounding

Heat transfer

Entropy transfer

Some Remarks about Entropy

The entropy change of a

system can be negative,

but the entropy generation

cannot.

1. Processes can occur in a certain direction

only, not in any direction. A process must

proceed in the direction that complies with

the increase of entropy principle, that is,

Sgen ≥ 0. A process that violates this

principle is impossible.

2. Entropy is a nonconserved property, and

there is no such thing as the conservation of

entropy principle. Entropy is conserved

during the idealized reversible processes

only and increases during all actual

processes.

3. The performance of engineering systems is

degraded by the presence of irreversibilities,

and entropy generation is a measure of the

magnitudes of the irreversibilities during that

process. It is also used to establish criteria

for the performance of engineering devices.

THE INCREASE OF ENTROPY PRINCIPLE

A cycle composed of a

reversible and an

irreversible process.

The equality holds for an internally

reversible process and the inequality

for an irreversible process.

Some entropy is generated or created during an irreversible process,

and this generation is due entirely to the presence of irreversibilities.

The entropy generation Sgen is always a positive quantity or zero.

Can the entropy of a system during a process decrease?

ENTROPY

The system considered in

the development of the

Clausius inequality.

Clausius

inequality

The equality in the Clausius inequality holds

for totally or just internally reversible cycles

and the inequality for the irreversible ones.

Formal

definition

of entropy

Entropy is an extensive

property of a system.

The net change

in volume (a

property) during

a cycle is

always zero.

The entropy change between two

specified states is the same whether

the process is reversible or irreversible.

A quantity whose cyclic

integral is zero (i.e., a

property like volume)

A Special Case: Internally Reversible

Isothermal Heat Transfer Processes

This equation is particularly useful for determining

the entropy changes of thermal energy reservoirs.

The entropy change of an isolated

system is the sum of the entropy

changes of its components, and is

never less than zero.

A system and its surroundings

form an isolated system.

The increase

of entropy

principle

Summary

• Entropy

• Clausius Inequality

• The Increase of entropy principle

WEEK 10

ENTROPY (continuation)

ENTROPY CHANGE OF PURE SUBSTANCES

The entropy of a pure substance

is determined from the tables

(like other properties).

Schematic of the T-s diagram for water.

Entropy is a property, and thus the

value of entropy of a system is fixed

once the state of the system is fixed.

Entropy change

from the definition of entropy

TdsQ or 2

TdsQhence

On a T-S diagram, the area under the curve

represents the heat transfer

The Tds equation

1st law for a closed system : Q - W = dU

and W = PdV

substituting: Q = dU + PdV

for a reversible process Q = TdS

TdS = dU + PdV …….1st TdS eq.

Enthalpy : H = U + PV

dH = d( U + PV )

= dU + PdV + VdP

= TdS + VdP

or TdS = dH – VdP………2nd TdS eq.

Entropy change for an incompressible substance(incompressible: dV 0)

from Q = dU + PdV

divide by T: Q/T = dU/T + PdV/T

dS = dU/T + PdV/T

change in internal energy :

dU = CV dT

dS = Cv dT / T + 0

12 lnT

dTCSSS VV

Example:

Aluminum at 100oC is placed in a large,

insulated tank having 10 kg of water at a

temperature of 30oC. If the mass of the

aluminum is 0.5 kg, find,

•the final temperature of the aluminum and

•the entropy change of the aluminum

•the entropy change of the water

•the total entropy change of the universe

(Given specific heat capacity of water and aluminium are

4.177 kJ/kgK and 0.941 kJ/kgK respectively)

Answer:

Aluminum

System: Constant volume, adiabatic and no work done

From the conservation of energy (1st law)

Q – W = Usys = UW + UAL

But Q = W = 0

at equilibrium (T2)W=(T2)AL= T2

Entropy change for water and aluminum

Sgen>0 : irreversible process

Total Entropy change of the universe

Reversible and irreversible processes: Calculation of entropy

Irreversible process (large temperature differences).

Heating water from 293K to 373K

Heat source

293K 293K373K 373K

break the irreversible process down into a series of reversible steps

293.1K

Heat source

293K 293.1K

293.2K

Heat source

293.2K

Heat source

373K373 K

dTmcdS

dTcds avav ;

When the water is at a temperature T and it’s heated to T + T,

the heat entering (reversibly) is

dQ = mcavT.

the entropy change of the water at each reversible step is:

Therefore the total change in entropy

dTmcdSS

dTcdss avav

Examples:

Calculate the change in the total entropy of the Universe for the

following processes:

dQSBlock

(1) A block of mass 1 kg, temperature 100°C and heat capacity 100 JK-1

is placed in a lake whose temperature is 10°C.

The change in entropy of the block, SBlock, is given by:

SBlok= - 27.61 JK-1

(2) The same block at 10°C is dropped into the lake from a height of 10 m.

Slake = mgh/Tlake = 1 x 9.81 x 10/283 = +0.35 JK-1

Entropy Change of an IDEAL GAS

Ts Diagrams

Integrate for the

area under the

process curve:

Two Special Cases:

For a Carnot Cycle, the Ts diagram

S1=S4 S2=S3 S (kJ/kgK)

T (oC)

S1=S4 S2=S3 S (kJ/kgK)

T (oC)

Remember each process:

1-2: Isothermal expansion

2-3: Adiabatic expansion (Isentropic)

3-4: Isothermal compression4-1: Adiabatic compression (Isentropic)

Also, since Q = TOS for an isothermal, internally

reversible process:

1-2: QH = area under 1-2 line and S = QH/TH

3-4: QL = area under 3-4 line and S = QL/TL

In Process 2-3, 4-1: s = 0 (Isentropic)

TdS = dU + PdV

TdS = dH – VdP

dS = dU/T + P/T dV

dS = 1/TdH – V/T dP

From TdS equation

ideal gas : u = (T)

h = (T)

du = cv(T) dT

dh = cP(T) dT

Pv = RT

by integration,

need to know the function of cv(T) and cp(T) in order

to complete the integration.

Cases with constant specific heats

The integration can be simplified

Ideal Gases in Isentropic Processes (s = 0)Assuming, Constant Specific Heats:

Using ideal gas relationships: cP = cV + R and k = cP/cV

constantTVorV

constantPVorV

constantTPorP

T k)/k(1

1)/k(k

const.S1

note: valid for ideal gas,

isentropic process,

constant specific heats.

Example:Air is compressed from an initial state of 100 kPa

and 300 K to 500 kPa and 360 K. Determine the

entropy change.

(cp = 1.003 kJ/kgK; Rair = 0.287 )

Answer:

Assume cp constant

Tcss p

/279.0100

500ln)287.0(

360ln003.1

)ln()ln(1

P1 = 100 kPa; P2 = 500 kPa

T1 = 300 K; T2 = 360 K

Summary of Entropy Change

Pure Substances:

Any process

s = s2 – s1

[kJ/(kgK)] (from table)

Isentropic process

s2 = s1

Incompressible substances:

Any process:

Isentropic process: T2 = T1

Ideal gasesconstant specific heat

THE ENTROPY CHANGE OF IDEAL GASES

From the first T ds relation From the second T ds relation

THE END…AT LAST!!

Week 11

GAS POWER CYCLES

Thermodynamic cycle :

1. power cycle – for power output (heat engine)

2. refrigeration cycle – for refrigeration effect

(refrigerator, air conditioner, heat pump)

1. gas cycle - working fluid remains in gaseous phase

2. vapour cycle – vapour phase in one part & fluid

phase in another

1. closed cycle – working fluid is returned to initial state

& recirculated

2. open cycle – working fluid is replaced by fresh one

Heat engine:

•internal combustion – burning the fuel inside the system

•external combustion – heat from the external sources

Compressor

Combustor

Turbine

Exhaust

BASIC CONSIDERATIONS IN THE ANALYSIS

OF POWER CYCLES

Modeling is a

powerful

engineering tool

that provides

great insight and

simplicity at the

expense of some

loss in accuracy.

The analysis of many

complex processes can be

reduced to a manageable

level by utilizing some

idealizations.

Most power-producing devices operate on cycles.

Ideal cycle: A cycle that resembles the actual cycle

closely but is made up totally of internally reversible

processes.

Reversible cycles such as Carnot cycle have the

highest thermal efficiency of all heat engines

operating between the same temperature levels.

Unlike ideal cycles, they are totally reversible, and

unsuitable as a realistic model.

Thermal efficiency of heat engines

The idealizations and simplifications in the

analysis of power cycles:

1. The cycle does not involve any friction.

Therefore, the working fluid does not

experience any pressure drop as it flows in

pipes or devices such as heat exchangers.

2. All expansion and compression processes

take place in a quasi-equilibrium manner.

3. The pipes connecting the various

components of a system are well

insulated, and heat transfer /lost through

them is negligible.

Care should be exercised

in the interpretation of the

results from ideal cycles.

On both P-v and T-s diagrams, the area enclosed

by the process curve represents the net work of the

cycle.

On a T-s diagram, the ratio of the

area enclosed by the cyclic curve to

the area under the heat-addition

process curve represents the thermal

efficiency of the cycle. Any

modification that increases the ratio

of these two areas will also increase

the thermal efficiency of the cycle.

increases when TH

increases

increases when TL

decreases

For both ideal and actual cycles:

Thermal efficiency increases with an increase

in the average temperature at which heat is

supplied to the system

with a decrease in the average temperature at

which heat is rejected from the system.

A steady-flow Carnot engine.

P-v and T-s

diagrams of

a Carnot

cycle.

AIR-STANDARD ASSUMPTIONS

The combustion process is replaced by a heat-addition process in ideal cycles.

Air-standard assumptions:

1. The working fluid is air, which

continuously circulates in a closed loop

and always behaves as an ideal gas.

2. All the processes that make up the

cycle are internally reversible.

3. The combustion process is replaced by

a heat-addition process from an

external source.

4. The exhaust process is replaced by a

heat-rejection process that restores the

working fluid to its initial state.

Cold-air-standard assumptions:

When the working fluid is considered to

be air with constant specific heats at room

temperature (25°C).

Air-standard cycle:

A cycle for which the air-standard

assumptions are applicable.

AN OVERVIEW OF RECIPROCATING ENGINES

Nomenclature for reciprocating engines.

• Spark-ignition (SI) engines

• Compression-ignition (CI) engines

Compression ratio

Mean effective

pressure

OTTO CYCLE

The Otto cycle is the ideal cycle for the

spark-ignition (internal combustion)

reciprocating engines, and it consists of

four internal reversible process;

•isentropic compression

•constant volume heat addition

•isentropic expansion

•constant volume heat rejection

Schematic of a two-stroke

reciprocating engine.

The two-stroke engines are

generally less efficient than

their four-stroke counterparts

but they are relatively simple

and inexpensive, and they

have high power-to-weight

and power-to-volume ratios.

diagram

of the

ideal Otto

cycle.

Four-stroke cycle

1 cycle = 4 stroke = 2 revolution

Two-stroke cycle

1 cycle = 2 stroke = 1 revolution

Mechanical

strokes

INTAKE STROKE

Air-fuel is drawn into the

cylinder

COMPRESSION STROKE:

air-fuel is compressed

COMBUSTION:

by spark-pluq, QinPOWER STROKE:

gas expands, delivering work

EXHAUST STROKE:

residual gases are

exhausted

OTTO CYCLE: THE IDEAL CYCLE FOR

SPARK-IGNITION ENGINES

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

ideal cycle

1 – 2 : isentropic compression

2 – 3 : constant volume heat transfer (Qin)

3 – 4 : isentropic expansion

4 – 1 : constant volume heat rejection (Qout)

Thermal efficiency ( ideal gas assumption)

)T(TmC

)T(TmC1

process 1-2 & 3-4 : isentropic

but V2 = V3 & V1 = V4

Thermal efficiency of Otto Cycle

where : r : compression ratio

k : cp/cv

The thermal efficiency of the Otto

cycle increases with the specific

heat ratio k of the working fluid.

Thermal efficiency of the ideal

Otto cycle as a function of

compression ratio (k = 1.4).

In SI engines, the compression ratio is limited by auto-ignition or engine knock.

THE END

Week 12 (Part A)

THE DIESEL CYCLE

THE IDEAL CYCLE

FOR COMPRESSION-IGNITION ENGINES

Automobiles, Power generation,

diesel-electric locomotives and submarines

The Diesel Cycle

-The Diesel cycle is the ideal cycle for compression - ignition

reciprocating engines.

-very similar to the Otto cycle, except that the constant volume

heat-addition process is replaced by a constant pressure heat-

addition process.

-The fuel and the air are compressed separately, and bringing

them together at the time of combustion.

-the fuel is injected into the cylinder which contains

compressed air at a higher temperature than the self-ignition

temperature of the fuel.

-once injected, the fuel ignites on its own.

In diesel engines, only air is

compressed during the compression

stroke, eliminating the possibility of

autoignition (engine knock).

Therefore, diesel engines can be

designed to operate at much higher

compression ratios than SI engines,

typically between 12 and 24.

In diesel engines, the spark plug is replaced

by a fuel injector, and only air is compressed

during the compression process.

• 1-2 isentropic

compression

• 2-3 constant-

pressure heat

addition

• 3-4 isentropic

expansion

• 4-1 constant-

volume heat

rejection.

•Process 1 to 2 is isentropic compression

of the fluid (blue)

•Process 2 to 3 is reversible constant

pressure heating (red)

•Process 3 to 4 is isentropic expansion

(yellow)

•Process 4 to 1 is reversible constant

volume cooling (green)[

Consider: (i) Heat IN

(ii) Heat OUT

(iii) Efficiency

Cutoff ratio

for the same compression ratio

(ratio between the end and start volume

for the combustion phase)

Thermal

efficiency of the

ideal Diesel

cycle as a

function of

compression

and cutoff ratios

(k=1.4).

Otto and Diesel engines have the following

undesirable features:

•The up-and-down movement of the piston

stresses the bearings and limits the speed.

•A large fraction of the fuel energy is lost to

cooling water circulating through the cylinder

housing.

•Fuel combustion is not complete,

particularly near the cool cylinder walls.

•Their complexity requires a large

number of parts.

•The piston seals cause friction and

wear, thus limiting engine life.

•Upon exhaust, there is still significant

pressure in the cylinder which reduces

efficiency and creates significant

throttling noise.

Compression, combustion, and

expansion all occur in a single cylinder

which requires compromises.

For maximum efficiency, the

compression should be cool and the

combustion should be hot. A single

cylinder cannot be both hot and cold at

the same time.

THE END

Week 12 (Part B)

THE REFRIGERATION CYCLE

REFRIGERATION CYCLE

Refrigeration is the withdrawal of heat from a substance or space so that

temperature lower than that of the natural surroundings is achieved.

Refrigeration may be produced by

-thermoelectric means

-vapour compression systems

-expansion of compressed gases

-throttling or unrestrained expansion of gases.

The objective of a refrigerator is to remove heat (QL) from the cold medium

The objective of a heat pump is to supply heat (QH) to a warm medium.

High Temperature Reservoir

CONDENSER

EVAPORATOR

COMPRESSOR

EXPANSION

Low Temperature Reservoir

(Refrigerated Area)

process 1 - 2 :

compression of the fluid

(vapour) by the

compressor where the

temperature and pressure

are elevated.

Process 2 – 3:

condensing if the

vapour at higher

pressure, and the

resultant heat is

dissipated to the

surrounding.

Process 3 – 4 :

expansion of the fluid

by an expansion

valve through which

the fluid pressure is

lowered.

Process 4 – 1 :The

low-pressure fluid

enters the

evaporator and

evaporates by

absorbing heat from

the refrigerated

space, and reenters

the compressor.

The whole cycle is repeated.

process 1 - 2 : compression of the fluid (vapour) by the

compressors where the temperature and pressure are

elevated.

Process 2 – 3: condensing if the vapour at higher

pressure, and the resultant heat is dissipated to the

surrounding.

Process 3 – 4 :expansion of the fluid by an expansion

valve through which the fluid pressure is lowered.

Process 4 – 1 :The low-pressure fluid enters the

evaporator and evaporates by absorbing heat from the

refrigerated space, and reenters the compressor.

The whole cycle is repeated.

Components of a Refrigeration system

•Compressor

•Condenser

•Expansion valve

•Evaporator

An ordinary household refrigerator.

Refrigeration: the transfer of heat from lower

temperature regions to the higher temperature

Refrigerators: devices that produce

refrigeration.

Refrigeration cycle: the cycle on which the

refrigeration operates.

Refrigerants: the working fluids in refrigerators.

Coefficient of performance

Schematic of a Carnot refrigerator and T-s diagram of the reversed Carnot cycle.

Expansion

Carnot Refrigerator and Heat Pump

Operate on reverse Carnot cycle

reasons for using reversed Carnot Refrigeration Cycle:

-easier to compress vapour only and not liquid-vapour mixture.

-cheaper to have irreversible expansion through an expansion

valve.

P-h diagram for ideal refrigeration cycle

Four processes of the ideal

vapour compression

refrigeration cycle

The actual vapour compression cycle.

Schematic and T-s diagram for

the ideal vapour-compression refrigeration cycle.

THE END

Week 14

THERMODYNAMIC

PROPERTY RELATIONS

Objectives• Develop fundamental relations between commonly

encountered thermodynamic properties and express the

properties that cannot be measured directly in terms of

easily measurable properties.

• Develop the Maxwell relations, which form the basis for

many thermodynamic relations.

• Develop the Clapeyron equation and determine the

enthalpy of vaporization from P, v, and T measurements

alone.

• Develop general relations for cv, cp, du, dh, and ds that

are valid for all pure substances.

• Discuss the Joule-Thomson coefficient.

• Develop a method of evaluating the ∆h, ∆u, and ∆s of

real gases through the use of generalized enthalpy and

entropy departure charts.

A LITTLE MATH—PARTIAL

DERIVATIVES AND

ASSOCIATED RELATIONS

The derivative of a function at a

specified point represents the slope of

the function at that point.

The state postulate: The state of

a simple, compressible substance

is completely specified by any two

independent, intensive properties.

All other properties at that state

can be expressed in terms of

those two properties.

The derivative of a function f(x)

with respect to x represents the

rate of change of f with x.

Partial Differentials

Geometric representation of

partial derivative (z/x)y.

The variation of z(x, y) with x when

y is held constant is called the

partial derivative of z with respect

to x, and it is expressed as

The symbol represents differential changes, just like the symbol d. They differ in that the symbol d represents the total differential change of a function and reflects the influence of all variables, whereas represents the partial differential change due to the variation of a single variable.

The changes indicated by d and are identical for independent variables, but not for dependent variables.

Geometric

representation of

total derivative dz for

a function z(x, y).

This is the fundamental relation for the

total differential of a dependent variable

in terms of its partial derivatives with

respect to the independent variables.

Partial Differential Relations

The order of differentiation is immaterial for

properties since they are continuous point

functions and have exact differentials. Thus,

Reciprocity

relation

Cyclic

relation

Demonstration of the

reciprocity relation for the

function z + 2xy 3y2z = 0.

THE MAXWELL RELATIONSThe equations that relate the partial derivatives of properties P, v, T, and s

of a simple compressible system to each other are called the Maxwell

relations. They are obtained from the four Gibbs equations by exploiting the

exactness of the differentials of thermodynamic properties.

Helmholtz function

Gibbs function

Maxwell relations

Maxwell relations are extremely valuable in thermodynamics because they provide a means of determining the change in entropy, which cannot be measured directly, by simply measuring the changes in properties P, v, and T.

These Maxwell relations are limited to simple compressible systems.

CLAPEYRON

EQUATION

The slope of the saturation curve

on a P-T diagram is constant at a

constant T or P.

The Clapeyron equation enables

us to determine the enthalpy of

vaporization hfg at a given

temperature by simply

measuring the slope of the

saturation curve on a P-T

diagram and the specific volume

of saturated liquid and saturated

vapor at the given temperature.

General form of the Clapeyron

equation when the subscripts 1

and 2 indicate the two phases.

Clapeyron

equation

At low pressures

Treating vapor

as an ideal gas

The Clapeyron equation can be simplified for liquid–vapor and solid–

vapor phase changes by utilizing some approximations.

Substituting these equations into the

Clapeyron equation

Integrating between two saturation states

The Clapeyron–Clausius

equation can be used to

determine the variation of

saturation pressure with

temperature.

It can also be used in the

solid–vapor region by

replacing hfg by hig (the

enthalpy of sublimation) of

the substance.

Clapeyron–

Clausius equation

GENERAL RELATIONS FOR du, dh, ds, cv, AND cp

• The state postulate established that the state of a simple

compressible system is completely specified by two independent,

intensive properties.

• Therefore, we should be able to calculate all the properties of a

system such as internal energy, enthalpy, and entropy at any state

once two independent, intensive properties are available.

• The calculation of these properties from measurable ones depends

on the availability of simple and accurate relations between the two

groups.

• In this section we develop general relations for changes in internal

energy, enthalpy, and entropy in terms of pressure, specific volume,

temperature, and specific heats alone.

• We also develop some general relations involving specific heats.

• The relations developed will enable us to determine the changes in

these properties.

• The property values at specified states can be determined only after

the selection of a reference state, the choice of which is quite

arbitrary.

Internal Energy Changes

Enthalpy Changes

Entropy Changes

relation

The volume expansivity (also called the coefficient of

volumetric expansion) is a measure of the change in

volume with temperature at constant pressure.

Conclusions from Mayer relation:

1. The right hand side of the equation is

always greater than or equal to zero.

Therefore, we conclude that

2. The difference between cp and cv

approaches zero as the absolute

temperature approaches zero.

3. The two specific heats are identical for

truly incompressible substances since v

constant. The difference between the two

specific heats is very small and is usually

disregarded for substances that are nearly

incompressible, such as liquids and solids.

The internal energies and specific heats of ideal

gases and incompressible substances depend

on temperature only.

THE JOULE-THOMSON COEFFICIENT

The temperature of a fluid may increase,

decrease, or remain constant during a

throttling process.The development of an h = constant

line on a P-T diagram.

The temperature behavior of a fluid during a throttling (h = constant) process is

described by the Joule-Thomson coefficient

The Joule-Thomson coefficient

represents the slope of h = constant

lines on a T-P diagram.

Constant-enthalpy lines of a substance

on a T-P diagram.

A throttling process proceeds along

a constant-enthalpy line in the

direction of decreasing pressure,

that is, from right to left.

Therefore, the temperature of a

fluid increases during a throttling

process that takes place on the

right-hand side of the inversion line.

However, the fluid temperature

decreases during a throttling

process that takes place on the left-

hand side of the inversion line.

It is clear from this diagram that a

cooling effect cannot be achieved

by throttling unless the fluid is

below its maximum inversion

temperature.

This presents a problem for

substances whose maximum

inversion temperature is well below

room temperature.

The temperature of

an ideal gas remains

constant during a

throttling process

since h = constant

and T = constant lines

on a T-P diagram

coincide.

THE ∆h, ∆u, AND ∆s OF REAL GASES

• Gases at low pressures behave as ideal gases and obey

the relation Pv = RT. The properties of ideal gases are

relatively easy to evaluate since the properties u, h, cv,

and cp depend on temperature only.

• At high pressures, however, gases deviate considerably

from ideal-gas behavior, and it becomes necessary to

account for this deviation.

• In Chap. 3 we accounted for the deviation in properties P,

v, and T by either using more complex equations of state

or evaluating the compressibility factor Z from the

compressibility charts.

• Now we extend the analysis to evaluate the changes in

the enthalpy, internal energy, and entropy of nonideal

(real) gases, using the general relations for du, dh, and

ds developed earlier.

Enthalpy Changes of Real Gases

An alternative process path to evaluate

the enthalpy changes of real gases.

The enthalpy of a real gas, in

general, depends on the pressure

as well as on the temperature.

Thus the enthalpy change of a real

gas during a process can be

evaluated from the general relation

for dh

For an isothermal process dT = 0,

and the first term vanishes. For a

constant-pressure process, dP = 0,

and the second term vanishes.

Using a superscript asterisk (*) to denote an ideal-gas state, we can express

the enthalpy change of a real gas during process 1-2 as

The difference between h and h* is called the

enthalpy departure, and it represents the

variation of the enthalpy of a gas with pressure at

a fixed temperature. The calculation of enthalpy

departure requires a knowledge of the P-v-T

behavior of the gas. In the absence of such data,

we can use the relation Pv = ZRT, where Z is the

compressibility factor. Substituting,

Enthalpy

departure

factor

The values of Zh are presented in graphical form as a function of PR

(reduced pressure) and TR (reduced temperature) in the generalized

enthalpy departure chart.

Zh is used to determine the deviation of the enthalpy of a gas at a

given P and T from the enthalpy of an ideal gas at the same T.

from ideal gas tables

Internal Energy Changes of Real Gases

For a real gas

during a

process 1-2

Using the definition

Entropy Changes of Real Gases

An alternative process path to

evaluate the entropy changes of real

gases during process 1-2.

General relation for ds

Using the approach in the figure

During isothermal process

Entropy departure

Entropy

departure

factor

The values of Zs are presented in graphical form as a function of PR

entropy departure chart.

Zs is used to determine the deviation of the entropy of a gas at a

given P and T from the entropy of an ideal gas at the same P and T.

For a real gas

during a

process 1-2

from the ideal gas relations

Summary• A little math—Partial derivatives and associated relations

Partial differentials

Partial differential relations

• The Maxwell relations

• The Clapeyron equation

• General relations for du, dh, ds, cv,and cp

Internal energy changes

Enthalpy changes

Entropy changes

Specific heats cv and cp

• The Joule-Thomson coefficient

• The ∆h, ∆ u, and ∆ s of real gases

Enthalpy changes of real gases

Internal energy changes of real gases

Entropy changes of real gases

WEEK 15

THERMODYNAMIC

PROPERTY RELATIONS

(cont.)

Internal Energy Changes

Enthalpy Changes

Entropy Changes

Specific Heats cv and cp

relation

The volume expansivity (also called the coefficient of

volumetric expansion) is a measure of the change in

volume with temperature at constant pressure.

Conclusions from Mayer relation:

1. The right hand side of the equation is

always greater than or equal to zero.

Therefore, we conclude that

2. The difference between cp and cv

approaches zero as the absolute

temperature approaches zero.

3. The two specific heats are identical for

truly incompressible substances since v

constant. The difference between the two

specific heats is very small and is usually

disregarded for substances that are nearly

incompressible, such as liquids and solids.

The internal energies and specific heats of ideal

gases and incompressible substances depend

on temperature only.

THE JOULE-THOMSON COEFFICIENT

The temperature of a fluid may increase,

decrease, or remain constant during a

throttling process.The development of an h = constant

line on a P-T diagram.

The temperature behavior of a fluid during a throttling (h = constant) process is

described by the Joule-Thomson coefficient

The Joule-Thomson coefficient

represents the slope of h = constant

lines on a T-P diagram.

Constant-enthalpy lines of a substance

on a T-P diagram.

A throttling process proceeds along

a constant-enthalpy line in the

direction of decreasing pressure,

that is, from right to left.

Therefore, the temperature of a

fluid increases during a throttling

process that takes place on the

right-hand side of the inversion line.

However, the fluid temperature

decreases during a throttling

process that takes place on the left-

hand side of the inversion line.

It is clear from this diagram that a

cooling effect cannot be achieved

by throttling unless the fluid is

below its maximum inversion

temperature.

This presents a problem for

substances whose maximum

inversion temperature is well below

room temperature.

The temperature of

an ideal gas remains

constant during a

throttling process

since h = constant

and T = constant lines

on a T-P diagram

coincide.

THE ∆h, ∆u, AND ∆s OF REAL GASES

• Gases at low pressures behave as ideal gases and obey

the relation Pv = RT. The properties of ideal gases are

relatively easy to evaluate since the properties u, h, cv,

and cp depend on temperature only.

• At high pressures, however, gases deviate considerably

from ideal-gas behavior, and it becomes necessary to

account for this deviation.

• In Chap. 3 we accounted for the deviation in properties P,

v, and T by either using more complex equations of state

or evaluating the compressibility factor Z from the

compressibility charts.

• Now we extend the analysis to evaluate the changes in

the enthalpy, internal energy, and entropy of nonideal

(real) gases, using the general relations for du, dh, and

ds developed earlier.

Enthalpy Changes of Real Gases

An alternative process path to evaluate

the enthalpy changes of real gases.

The enthalpy of a real gas, in

general, depends on the pressure

as well as on the temperature.

Thus the enthalpy change of a real

gas during a process can be

evaluated from the general relation

for dh

For an isothermal process dT = 0,

and the first term vanishes. For a

constant-pressure process, dP = 0,

and the second term vanishes.

Using a superscript asterisk (*) to denote an ideal-gas state, we can express

the enthalpy change of a real gas during process 1-2 as

The difference between h and h* is called the

enthalpy departure, and it represents the

variation of the enthalpy of a gas with pressure at

a fixed temperature. The calculation of enthalpy

departure requires a knowledge of the P-v-T

behavior of the gas. In the absence of such data,

we can use the relation Pv = ZRT, where Z is the

compressibility factor. Substituting,

Enthalpy

departure

factor

The values of Zh are presented in graphical form as a function of PR

enthalpy departure chart.

Zh is used to determine the deviation of the enthalpy of a gas at a

given P and T from the enthalpy of an ideal gas at the same T.

from ideal gas tables

Internal Energy Changes of Real Gases

For a real gas

during a

process 1-2

Using the definition

Entropy Changes of Real Gases

An alternative process path to

evaluate the entropy changes of real

gases during process 1-2.

General relation for ds

Using the approach in the figure

During isothermal process

Entropy departure

Entropy

departure

factor

The values of Zs are presented in graphical form as a function of PR

entropy departure chart.

Zs is used to determine the deviation of the entropy of a gas at a

given P and T from the entropy of an ideal gas at the same P and T.

For a real gas

during a

process 1-2

from the ideal gas relations

Summary• A little math—Partial derivatives and associated relations

Partial differentials

Partial differential relations

• The Maxwell relations

• The Clapeyron equation

• General relations for du, dh, ds, cv,and cp

Internal energy changes

Enthalpy changes

Entropy changes

Specific heats cv and cp

• The Joule-Thomson coefficient

• The ∆h, ∆ u, and ∆ s of real gases

Enthalpy changes of real gases

Internal energy changes of real gases

Entropy changes of real gases

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