* Reading Assignments:

3.1 3.1.1

3.2

3.3

3.4 3.4.1 3.4.2

3.5

3.6 3.6.1

4. The Second Law

4.1 Reversible vs Irreversible processes

Reversible: A process for which a system can be restored to its initial state, without leaving a net influence on the system or its environment.

* idealized, frictionless;

* proceeds slowly enough for the system to remain in thermodynamic equilibrium.

Irreversible: Not reversible

* natural;

* proceeds freely, drives the system out of thermodynamic equilibrium;

* interacts with environment, can not be exactly reversed

Example: Gas-piston system under a constant temperature

* Slow expansion and compression

02112 wwpdvwq

* Rapid expansion and compression

02112 wwpdvwq

)( sp pp

)( sp pp )( sp pp

)0( du

4.2 Entropy / Carnot’s Theorem

* Consider the first law,

pdvdTcq v vdpdTcq p , or

Divide them by T and use pv=RT,

,v

dvR

T

dTc

T

qv

p

dpR

T

dTc

T

qp

or

The specific entropy (s) is defined as,

T

qds

which is a state variable, a property of the system.

* The Carnot cycle and Carnot Theorem

1) From state 1 to state 2: Isothermal expansion

121212 , vvTTT

2) From state 2 to state 3: Adiabatic expansion

23122334 , vvTTTT

012 u

1

212

2

11212 lnv

vRTpdvwq

023 q

)( 12342323 TTcuw v

3) From state 3 to state 4: Isothermal compression

41344112 , vvTTTT

343434 , vvTTT

4) From state 4 to state 1: Adiabatic compression

034 u

3

434

4

33434 lnv

vRTpdvwq

041 q

)( 34124141 TTcuw v

The net heat transfer and the net work over the Carnot cycle are:

3

434

1

212 lnln

v

vRT

v

vRTqw

Using Poisson’s equation,

4

3

1

2

1

1

4

1

2

3

34

12 ,v

v

v

v

v

v

v

v

T

T

0ln)(1

23412

v

vTTRqw

Then,

So the system absorbs heat and performs net work in the Carnot cycle,which behaves as a heat engine.

Why is the Carnot cycle reversible?

For the Carnot cycle, we can also have:

0lnln3

4

1

2

34

34

12

12

v

v

v

vR

T

q

T

q

T

q

This relationship also hold for the reversed Carnot cycle.

This is called the Carnot’s Theorem:

revT

qdsds

,0

which shows that the change of entropy is independent of path under a reversible process.

(4.1)

4.3 The Second Law and its Various Forms

To get the second law, we use the Clausius Inequality, i.e.,

0Tq

for a cyclic process,

which indicates during the cycle,

1) heat must be rejected to the environment somewhere during a cycle;

2) heat exchange is larger at high temperature than at low temperature under reversible conditions;

3) the net heat absorbed is smaller under the irreversible condition than under the reversible condition.

Now, consider two cycles as shown in the plot.

For the cycle which contains onereversible process and one irreversible process,

01

2

2

1

irrevrev T

q

T

q (4.2)

For the cycle which have two reversible processes,

01

2

2

1

revrev T

q

T

q (4.3)

The difference between (4.2) and (4.3) gives

irrevrev T

qds

T

q

1

2

1

2

1

2

Because states 1 and 2 are arbitrary, we have the second law,

T

qds

(4.4)

It indicates that the heat absorbed by the system during a processhas an upper limit, which is the heat absorbed during a reversibleprocess.

Combine (4.1) and (4.4), we have

revT

q

T

q

The first law relates the state of a system to work it performs and heat it absorbs.

The second law controls how the systems move to the thermodynamicequilibriums, i.e., the direction of processes.

Several simplified forms of the second law:

1) For an adiabatic process, (4.4) becomes

0dsIf the adiabatic process is reversible, then

0ds

It is also isentropic (s is constant).

2) For an isochoric process, (4.1) becomes

vv

revvv T

dTc

T

dTcds

Because only state variables are involved, it holds for eitherreversible or irreversible processes.

(4.5) and (4.6) show that :

Irreversible work can only increase entropy; heat transfer can either increase or decrease entropy.

(4.5)

(4.6)

4.4 Fundamental Relations / The Maxwell Relations

Combine forms of the first law and the second law,

T

dvp

T

dTc

T

qds v

or pdvduTds

pdvTdsdu

Similarly,

T

dpv

T

dTc

T

qds p

or vdpdhTds

vdpTdsdh

(4.7)

(4.8)

For reversible processes, the equal signs apply and the equationsare called Fundamental Relations,

pdvTdsdu

vdpTdsdh

(4.9)

(4.10)

Because these equations involve only state variables, they do notdepend on path. So they must hold for both reversible and irreversible processes.

These identities describe the change in one state variable in terms of changes in two other state variables.

Two other state variables can be defined,

The Helmholtz function: Tsuf

The Gibbs function: TspvuTshg

Use these definitions in (4.7)-(4.10),

pdvsdTdf

vdpsdTdg

(4.13)

(4.14)

pdvsdTdf

vdpsdTdg

(4.11)

(4.12)

Recall that the condition for a function NdyMdxdz to be exact is,

x

N

y

M

Since the state variables are exact, so for u, from

vs s

p

v

T

(4.15)

* The Maxwell Relations:

pdvTdsdu

We have

Similarly from the fundamental relations we can show that

vT T

p

v

s

pTT

v

p

s

pss

v

p

T

(4.16)

(4.17)

(4.18)

(4.15)-(4.18) are called the Maxwell relations.

* Noncompensated Heat Transfer

For an irreversible process,

pdvTdsdu

To remove the inequality, a term can be added to the right side of the formulation,

qpdvTdsdu

Next, use the second law for a reversible process on du,

qpdvTdsdvpdsT revrev

Finally,

dvppdsTTq revrev )()(

q is called the noncompensated heat transfer, which measuresadditional heat rejection to the environment due to irreversibility.

4.5 Thermodynamic Equilibrium

* Consider an adiabatic process, the second law becomes

0ds

or0ss

For an irreversible condition,

0ds

is the entropy at the initial state. 0s

When reaches the maximum,the state is in thermodynamicequilibrium because the entropy can not increase anymore.

s

* Consider an isentropic-isochoric process,

From (4.7), we have

0du

or 0uu

For an irreversible condition,

0du

0u

When reaches the minimum, the state is in thermodynamicequilibrium because the internal energy can not decrease any further.

u

is the internal energy at the initial state.

* The enthalpy, Helmholtz function and Gibbs function must all decrease as a system reaches thermodynamic equilibrium under certain processes.

4.6 Relationship of Entropy to Potential Temperature

* Use the first law and the equation of state for an ideal gas in the second law,

p

dpR

T

dTc

T

qds p

pdc

RTd

c

ds

pp

lnln

* use log derivatives on the potential temperature,

pc

R

p

pT 0lnln

pdc

RTdd

p

lnlnln

Under the reversible process, we have

lndc

ds

p

(4.20)

Because (4.20) involves only state variables, it is path independentand is valid for both reversible and irreversible processes.

So, the change in entropy can be measured by the change in potential temperature.

So,lnd

c

ds

p

(4.19)

4.7 Implications for Vertical Motion

1) Under adiabatic processes:

0,0 dsandd

Adiabatic conditions require: a. no heat be transferred between the system and environment;

b. no heat exchange between one part of the system and another.

So, these exclude the irreversible turbulent mixing and the irreversibleexpansion work-induced mixing in the system.

Therefore, the adiabatic process is approximately reversible, i.e.,

0,0 dsandd

So, the potential temperature surfaces coincide with Isentropic surfaces . An air parcel will remain on a certainIsentropic surface and undergoes no systematic vertical motion.

)( const)( consts

2) Under diabatic processes:

An air parcel moves across isentropic surfaces followingthe heat transfer with itsenvironment.

Tc

qd

c

ds

pp

ln

3) The displaced motion is sufficiently slow:

The air parcel’s temperaturediffers from the environmentonly infinitesimally;

Rejection of heat during thepoleward moving is balancedby the absorption of heat duringthe equatorward moving.

No net vertical motion, theparcel’s evolution is reversible.

4) The displaced motion is sufficiently fast:

Produce the net heat transfer and a vertical drift of the parcel acrossisentropic surface in acomplete cycle.

1. A dry air parcel undergoes a complete Carnot cycle consisting of the steps indicated in (a)-(d). For each individual step, calculate the mechanical work w (per unit mass) done by the air parcel and the heat q added to the parcel.(a) Adiabatic compression from p1=600 hPa and T1=0oC to a temperature T2 of 25oC;(b) Isothermal expansion to a pressure p3 of 700 hPa;(c) Adiabatic expansion to temperature T4 of 0oC;(d) Isothermal compression back to the original pressure p1.Also, compute(e) the total work done and heat added for the complete cycle, and (f) the efficiency of the cycle.

2. Two hundred grams of mercury at 100oC is added to 100 g of water at 20oC. If the specific heat capacities of water and mercury are 4.18 and 0.14 JK-1g-1, respectively, determine (a) the limiting temperature of the mixture, (b) the change of entropy for the mercury, (c) the change of entropy for the water, and (d) the change of entropy for the system as a whole.

3. During a cloud-free evening, LW heat transfer with the surface causes an air parcel to descend from 900 to 910 mb and its entropy to decrease by 15 J kg-1 K-1. If its initial temperature is 280 K, determine the parcel’s (a) final temperature and (b) final potential temperature.

Meteorology 341

Homework (3)