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The Second Law of ThermodynamicsPowerPoint Presentation

The Second Law of Thermodynamics

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The Second Law of Thermodynamics

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Consider the following process:

mgh

A rock spontaneously rises by lowering

its temperature such that mCpDT = mgh

so that DU = 0.

The Second Law of Thermodynamics

Since energy is conserved, this type

of process is not forbidden by the

first law, but we know this never

happens! There is a natural flow of

things or direction for spontaneous

processes to occur.

We also know that various forms of

work can be completely converted

into heat, e.g., rubbing of two rocks

together in a heat reservoir such that

they undergo no temperature change

so that W = Q and DU = 0.

In general work of any kind can be done on a system in contact

with a reservoir giving rise to a flow of heat without altering

the state of a system, W = Q. Work can be converted entirely in

to heat by a suitable dissipative process.

Heat Engines - The conversion of heat into work

In order to convert heat into work we require a machine that will

consume heat and produce work. The machine itself must not suffer any

permanent change; it must play a passive role in that following the process

it must return to its initial state. The machine must pass through a cycle.

def.

Thermal efficiency,

Applying the first law to the operation of the machine or engine, W = Q1- Q2

where Q2 corresponds to any heat rejected from the engine,

Q1

W

Heat engine

system

Q2

p

Q1

T1

A

W

Q1

W

B

T1

D

T2

Q2

C

T2

V

Q2

Carnot Cycle

The Carnot cycle is a reversible cycle operating between twotemperatures.

A B: Isothermal expansion adsorbing heat Q1.

B C: Adiabatic expansion decreasing T from T1 to T2.

C D: Isothermal compression rejecting heat Q2.

D A: Adiabatic compression increasing T from T2 to T1.

*Note that if the cycle is operated in reverse refrigerator.

Carnot cycle for a gas

Since all the steps are reversible DU = 0, W = Q1 - Q2and

Kelvin Statement Clausius Statement

The 2nd Law

Kelvin Statement - No process is possible whose sole result is the complete

conversion of heat into work.This addresses the efficiency

of conversion.

Clausius Statement - No process is possible whose sole result is the transfer of heat

from a colder to a hotter body.Spontaneity of processes and the

irreversibility of nature.

Carnot’s Theorem:

No engine operating between two given reservoirs can be more efficient than a

Carnot engine operating between the same two reservoirs.

T1

QC1

QH1

WC = QC1 - QC2

C

H

WH = QH1 - QH2

QH2

QC2

T2

Proof of Carnot’s Theorem

Assume the existence of a Hypothetical engine such that,

Since the Carnot engine is reversible we can drive it backwards using the mechanical energy from H. The Carnot cycle can be adjusted (adiabats) so that in one cycle it uses exactly as much work as H produces.

Now consider C and H as

a Composite Engine.

Composite Engine

T1

QC1

QH1

This composite engine

produces no net work but

simply extracts heat from a

cold reservoir and delivers

an amount of heat,

to a hotter reservoir.

WC = QC1 - QC2

C

H

WH = QH1 - QH2

QH2

QC2

T2

Carnot’s Theorem:

Proof of Carnot’s Theorem

This is a violation of the Clausius statement

of the second law!

Qn+2

Consider a series of Carnot Engines

Tn+2

Cn+2

Wn+2

Qn+1

Tn+1

Wn+1

Cn+1

Qn

Tn

Cn

Wn

Qn-1

Tn-1

Corollary: All reversible engines operating between the same temperature reservoirs are equally efficient.

Thus the efficiency of any reversible engine operating between the same reservoirs

are equally efficient. For any reversible engine,

For the composite engine,

Qn+2

Tn+2

Cn+2

Wn+2

Qn+1

Tn+1

Wn+1

Cn+1

Qn

Tn

Cn

Wn

Qn-1

Tn-1

Then,

This can only be true if the f’s factorize such that

Therefore:The ratio of the temperatures of the reservoirs is equal

to the ratios of heat exchanged by a reversible

engine operating between the same reservoirs.

According to Carnot’s theorem and its corollary we can make the following statements:

Therefore . Taking the heat entering the system as positive, we can say

For any closed cycle, , where the equality necessarily holds

for a reversible cycle.

def:

We can now define a new variable, the entropy S, by the relation

dS = for an infinitesimal reversible change. This definition

holds for reversible changes only. For a finite reversible change of state, the change in entropy is given by,

Qrev

T

Entropy

Entropy in Irreversible Processes

Since entropy is a state function, the change in entropy accompanying a state

change must always be the same regardless of how the state change occurs. Only

when the state change occurs reversibly is the entropy change related to the

heat transfer by the equation

B

X

Consider an irreversible change AB. Construct

any reversible path R thus forming an irreversible

cycle ABRA. For the irreversible cycle the

Clausius theorem says,

R

A

x

Determination of the change in

entropy for an irreversible change

Taking the integral in two parts,

i.e.,

by definition of entropy. Thus

But

Thus we have this general result for a differential

irreversible change.

or

For a thermally isolated system Q = 0 and we have the

general result known as the law of increase of entropy.

Adiabatic Free expansion of an Ideal Gas

(Joule expansion - no Q or W exchanged

with surroundings)

This is an irreversible process, but we can

always use the combined Law and integrate

from the initial to final state by a convenient

reversible path.

Some Interesting Examples

Isothermal expansion of an Ideal Gas

Combined 1st & 2nd Law

Isothermal dissipation of Work

Electrical work is dissipated isothermally by heat flow into

a reservoir. There is no entropy change of the system because

it’s thermodynamic coordinates do not change. The

reservoir adsorbs Q = W units of heat at temperature T so its

entropy change is

Diathermal Walls

P1, V1

Reservoir (T)

Adiabatic dissipation of Work

Electrical work is dissipated in a thermally isolated

system maintained at fixed pressure. The T of the system increases irreversibly. The coordinates of the system change from P,T1 to P,T2 . The entropy change

can be calculated by

Composite system

Adiabatic Walls

P, T1

The thermal efficiency is

The maximum efficiency of any power cycle is

No good!

Examples

An inventor claims to have developed

a power cycle capable of delivering a

net work output of 410 kJ for an energy

input by heat transfer of 1000 kJ. The

system undergoing the cycle receives

heat transfer from hot gases at

T = 500 K and discharges energy by

heat transfer to the atmosphere at

T = 300 K. Evaluate this claim.

We can use these forms to determine

the entropy change of an ideal gas

subjected to changes in p, v, T.

Using the definition of entropy

Since enthalpy is defined as,

H = U + pV

Rearranging these equations and

writing them on a unit mass basis,

We already know that for an ideal gas,

Different Forms of the Combined 1st and 2nd Law

It is convenient to define a quantity such that

Def:

is necessarily a positive quantity called entropyproduction.

Entropy Production

B

X

R

A

x

Determination of the change in

entropy for an irreversible change

Recall the determination of entropy change for an irreversible process.

The 2nd term is the entropy production term.

Entropy Production

If the end states are fixed the entropy change

on the left hand side of this equation can be

determined.

Rewriting the expression

The 2 terms on the RHS of the

equation are path dependant.

we obtain,

The 1st term on the RHS of the

equation is the entropy transfer associated

with heat transfer. The direction or sign of

the entropy transfer is the same as heat

transfer.

Here Qj/Tj is the amount of entropy

transferred to the portion of the

boundary at temperature Tj .

By contrast the change in the entropy

of the system can be positive, negative

or zero:

On a time rate basis for a closed

system

The entropy balance can be expressed

in various forms. If heat transfer takes

place along several locations on the

boundary of the system where the temperatures

do not vary with position or time,

Entropy Production

can not be less than zero

At constant pressure the work is simply,

Table A-2

Examples

Water initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly.

The water undergoes a process to the corresponding saturated vapor during which the

piston moves freely in the cylinder. If the change of state is brought about by heating

the water as it undergoes an internally reversible process at constant pressure and

temperature determine the work and the heat transfer per unit of mass in kJ/kg

Examples

Since the process is reversible and occurs at constant temperature

This could also have been calculated our old way

For the system,

For the reservoir

Examples

The figure shows a system receiving heat

Q from a reservoir. By definition the reservoir

is free of irreversibilities, but the system is not,

fluid friction, etc. Let’s determine the

entropy change of the system and that of the

reservoir.

Examples

Water initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly.

The water undergoes a process to the corresponding saturated vapor during which the

piston moves freely in the cylinder. There is no heat transfer with the surroundings.

If the change in state is brought about by the action of a paddle wheel, determine the

net work per unit mass in kJ/kg and the amount of entropy produced per unit mass in

kJ/kg-K.

As the volume of the system increases during this process, there is an energy transfer

by work from the system during the expansion as well as an energy transfer by work

to the system done by the paddle wheel. The net work is evaluate from the change

in internal energy.

The entropy produced is evaluated by applying the entropy balance,

On a unit mass basis,

Examples

From the 1st Law, U = -W.

On a unit mass basis we have,

The minus sign indicates that the work input by stirring is greater in magnitude than the

work done by the water as it expands.

isentrop

T

isenthalp

isochor

isobar

isotherm

S

Entropy Diagrams

T

S

Carnot cycle on a T – S

diagram.

Area representation of heat transfer

for an internally reversible process

of a closed system.

Entropy Diagrams

CCW – refrigeration cycle

CW – power cycle

Potential Function in terms of S and p, Enthalpy

Potential Function in terms of T and p,

Gibbs Free Energy

Lengendre Transform subtract a -d(pV) term

from dU

Lengendre Transform add a -d(TS) term

to dH

d ( H-TS ) = Vdp - SdT

where G = (H-TS) is the Gibbs Free Energy

and G = G (p, T)

dU + d(pV) = TdS - pdV + d(pV)

d(U + pV) = TdS + Vdp

where H = (U + pV) is the Enthalpy

and H = H (S, p)

Thermodynamic Potentials

Combined 1st and 2nd Law

dU = TdS - pdV

Enthalpy is a function of S and p

Four Fundamental Thermodynamic Potentials

dU = TdS - pdV

dH = TdS + Vdp

dG= Vdp - SdT

dA = -pdV - SdT

The appropriate thermodynamic potential

to use is determined by the constraints

imposed on the system. For example,

since entropy is hard to control (adiabatic

conditions are difficult to impose) G and A

are more useful. Also in the case of solids

p is a lot easier to control than V so G is

the most useful of all potentials for solids.

Potential Function in terms of T and V,

Helmholtz Free Energy

Lengendre Transform

subtract a -d(TS) term from dU

d(U-TS) = -pdV - SdT = dA

where A = A(V, T) is the

Helmholtz Free Energy

The Maxwell relations are useful in that

the relate quantities that are difficult or

impossible to measure to quantities that

can be measured.

Some important bits of information

For a mechanically isolated system kept at constant temperature and volume

the A = A(V, T) never increases. Equilibrium is determined by the state of

minimum A and defined by the condition, dA = 0.

For a mechanically isolated system kept at constant temperature and pressure

the G = G(p, T) never increases. Equilibrium is determined by the state of

minimum G and defined by the condition, dG = 0.

Consider a system maintained at constant p. Then