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Thermodynamics: Work, Heat, and the Second Law, Lecture notes of Thermodynamics

Indian Institute of Science Thermodynamics

Work and Heat, Second Law of Thermodynamics, Carnot Engine

Typology: Lecture notes

2018/2019

Uploaded on 08/07/2019

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Work and Heat

Consider an isolated system composed of two identical bodies Aand B,

say equal masses of water, which can exchange heat. Initially, TA= 310

Kand TB= 290 K. Being isolated, no heat exchange is possible between

environment and the system.

What will happen after a while?

First law of thermodynamics i.e. conservation of energy (and mass)

admits the following two possibilities.

1. TA→320 K, TB→280 K : Thermal energy lost by Bis gained by A

2. TA→300 K, TB→300 K : Thermal energy lost by Ais gained by B

How to explain that case 1 is never observed in nature?

A few other phenomena, which may satisfy energy conservation but not

observed in nature :

Iwater running uphill without an external assist

ICO2and H2Oreacting spontaneously to form CH4and O2

Iair separating into its constituents spontaneously

Discover Lecture notes of Thermodynamics Indian Institute of Science

Partial preview of the text

Download Thermodynamics: Work, Heat, and the Second Law and more Lecture notes Thermodynamics in PDF only on Docsity!

Work and Heat

Consider an isolated system composed of two identical bodies A and B,

say equal masses of water, which can exchange heat. Initially, TA = 310

K and TB = 290 K. Being isolated, no heat exchange is possible between

environment and the system.

What will happen after a while?

First law of thermodynamics i.e. conservation of energy (and mass)

admits the following two possibilities.

1. TA → 320 K, TB → 280 K : Thermal energy lost by B is gained by A

2. TA → 300 K, TB → 300 K : Thermal energy lost by A is gained by B

How to explain that case 1 is never observed in nature?

A few other phenomena, which may satisfy energy conservation but not

observed in nature :

I water running uphill without an external assist

I CO 2 and H 2 O reacting spontaneously to form CH 4 and O 2

I air separating into its constituents spontaneously

An equivalent situation arises in Work Heat.

Without work hot Q

−→ cold is possible but cold

Q

−→ hot is impossible.

All work can be converted into heat but not all heat into work.

E.g. Two stones when rubbed against each other, the work done against

friction is transformed into internal energy leading to rise of temperature

of the stones. The stones quickly attain the same temperature with the

surroundings, which is acting as reservoir whose temperature increase is

insignificant. The stones thus return to their initial state and the next

cycle of rubbing begins.

The net result of the process is conversion of mechanical work into heat

i.e. Q = W. This transformation can continue indefinitely.

Opposite process, namely, conversion of heat into work, should continue

indefinitely without involving any changes in the state of the system.

Seems like isothermal expansion of ideal gas fits the bill. Since dU = 0,

we get W = Q i.e. complete conversion of heat into work! But just for

once i.e. volume increases and pressure decreases until it reaches the

atmospheric pressure and the process stops. It just cannot go back again

spontaneously from where it started, namely, high pressure and smaller

volume!!

E extracts heat Q 1 from a hot body and converts fully into work

W = Q 1 , just sufficient to operate one cycle of R. The R extracts Q 2

from a cold body and deliver Q 1 + Q 2 to hot body in each cycle.

Hot Body Body Body Body

Cold (^) Hot Cold

R Q 2

Q 2 + Q 1 Q 2 ≡ Q 2

Q 1 E

E+R W =

Q

1

The above process is equivalent to the composite E+R transferring Q 2

from cold to hot body without any assist. This renders Clausius

statement false. The reverse can also be proved.

An engineer would want to harvest as much as possible of thermal energy

and convert it into mechanical energy i.e. work. Kelvin-Planck statement

of second law simply says we cannot turn all heat into work. But it lets

us have some! Trick is to employ the cold body differently – harvest a

part of thermal energy for work and reject the rest into the cold body.

This way we can satisfy the second law. The schematic of such a

realizable heat engine is shown below. Henceforth, we will call the hot

and cold body as reservoirs in the sense that their thermal states remain

unchanged upon any amount of heat extraction and dumping.

Hot Reservoir

Cold QH QL Reservoir Cyclic Engine ∆U = 0

W^ =^ Q

H^ −

QL

The first law for this system is,

∆U = QH − QL − W

For cyclic process, ∆U = 0 and hence W = QH − QL.

The thermal efficiency η of the engine is defined as,

η = work output heat input

W

QH

⇒ η = 1 −

QL

QH

Clearly, if QL = 0 we get η = 1 and our engine does a perfect job in

converting all heat into work. But this violates second law, so we must

reject some heat i.e. QL 6 = 0 ⇒ η < 1.

Let us look into the workings of a couple of engines that utilize the above

process, as shown in the schematic diagram.

The processes of Otto cycle is plotted in the PV -diagram below.

Work

1

2

3

4

Adiabatic

P 05

QL

QH

V 2 V 1

Assuming CV to be constant along 2 → 3 and 4 → 1, we have

QH =

∫ T 3

T 2

CV dT = CV (T 3 − T 2 ) and QL = −

∫ T 1

T 4

CV dT = CV (T 4 − T 1 )

Two adiabatic processes during compression and power stroke give us

T 1 V 1 γ −^1 = T 2 V 2 γ−^1 T 4 V 1 γ −^1 = T 3 V 2 γ−^1

T 1

T 4

T 2

T 3

T 4 − T 1

T 3 − T 2

T 4

T 3

T 1

T 2

Hence, the thermal efficiency of an idealizes gasoline engine is

η = 1 −

QL

QH

T 4 − T 1

T 3 − T 2

T 1

T 2

Real life operating efficiency of gasoline engine is typically 20 – 30%.

Diesel engine

The cyclic process of Diesel engine i.e. diesel cycle is plotted in the

PV -diagram below. First successful engine using liquid fuel.

1

2 3

4 P 0 5 V 2 V 3 V 1

QH

QL

Adiabatic Adiabatic

Work

Assuming CP to be constant along 2 → 3 and CV along 4 → 1

QH =

∫ T 3

T 2

CP dT = CP (T 3 − T 2 ) and QL = −

∫ T 1

T 4

CV dT = CV (T 4 − T 1 )

The two adiabatic processes give

T 1 V 1 γ −^1 = T 2 V 2 γ−^1 T 4 V 1 γ −^1 = T 3 V 3 γ−^1

T 1

T 4 =^

T 2

T 3

V 2

V 3

)γ− 1

Therefore, the efficiency of Diesel engine is

η = 1 −

QL

QH

CV

CP

T 4 − T 1

T 3 − T 2

γ

r (^) Cγ − 1 rC − 1

T 1

T 2

, where rC =

V 3

V 2

Carnot’s theorem

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

Carnot engine operating between the same two reservoirs.

Let us examine how the efficiency of a hypothetical engine E ′^ is

restricted by Clausius statement of second law.

Suppose the engine E ′^ (efficiency η′) and a Carnot engine C , (efficiency

ηC ) are operating between the same heat reservoirs at temperatures

TH , TL. The schematics of the processes are shown below,

Hot reservoir

Cold reservoir

E′^ C

TH

TL

Q′ H

Q′ L

QH

QL

W ′ W

Hot reservoir

Cold reservoir

E′^ C

TH

TL

Q′ H

Q′ L

QH

QL

W ′ W

The engines are so adjusted that W ′^ = |W |, where

W ′^ = Q H′ − Q L′ and W = QH − QL

Now, suppose η′^ > ηC ,

η′^ > ηC →

W ′

Q H′

W

QH

implying

W

Q′ H

W

QH

⇒ QH > Q H′

At this point, we take the formal advantage of reversibility of Carnot

engine to reverse all processes and create a refrigerator i.e. moving QL

from cold reservoir and dumping QH in the hot one. With W as input

work, the second law is hold valid. Because of the assumption W ′^ = |W |,

W ′^ = |W | ⇒ QH − Q H′ = QL − Q L′

the composite engine E ′^ + C is extracting heat QL − Q′ L = QH − Q H′ from

cold reservoir in each cycle and delivering exactly the same amount of

heat QH − Q H′ to hot reservoir without work input from any other device.

This is in contradiction with Clausius statement of second law. Hence the

assumption η′^ > ηC cannot be valid. This proves the Carnot theorem.

The only other possibility is η′^ = ηC. For that, QL − Q′ L = QH − Q H′ = 0 i.e.

the composite device achieves nothing, no heat transfer and no work

performed. The conclusion is,

η′^ ≤ ηC

Corollary : All Carnot engines operating between the same two reservoirs

have the same efficiency (independent of working substance).

To complete definition of thermodynamic scale, we make an arbitrary

choice that one of the reservoirs is at triple point of water 273.15 K.

T = (273. 15 K )

Q

Qtriple point

Carnot

If a system undergoes a reversible isothermal process without transfer of

heat, temperature at which this process takes place is called an absolute

zero. At absolute zero, an isotherm and an adiabatic are identical.

Since Carnot cycle and its efficiency is independent of working substance,

so is absolute zero. Furthermore, this definition is in terms of purely

macroscopic concepts i.e. no reference is made to atoms, molecules or

kinetic theory. Importantly, thermodynamic temperature is the same as

those in ideal gas equation, kinetic theory and statistical mechanics.

Let us establish the equality of ideal gas and thermodynamic

temperature. Consider a Carnot cycle of an ideal gas operating between

heat reservoirs at temperatures θH and θL. The cycle is depicted on a

pV -diagram in the following slide.

P

V

1

3

4

2

θH

θL

pV (^) = (^) nRθ H pV = nRθ L

pV γ = k

pV (^) γ = k

The equations for the two isothermal processes 2 → 3 and 4 → 1 are,

2 → 3 : pV = nRθH , 4 → 1 : pV = nRθL

From the first law we have δQ = CV dθ + p dV with dθ = 0 (isothermal).

Hence, the heat exchange is calculated from Q =

∫ (^) b

a p dV^ , thus

Q 23 = nRθH ln V V^32 Q 41 = nRθL ln V V^41

⇒ Q^23

Q 41

= θH^ ln(V^3 /V^2 ) θL ln(V 4 /V 1 )

The equations for adiabatic process δQ = 0 is

−CV dθ = p dV = nRθ dV V

θ

CV

dθ θ = nR ln Vf Vi

Entropy

Clausius inequality

Considers heat transfers to a substance as it is taken round an arbitrary

cyclic process exchanging heat with any number of surrounding bodies. It

can be derived by breaking down the process into equivalent interactions

with a large number of Carnot engines E and refrigerators R.

For simplicity, choose a system exchanging heat with 3 different bodies.

Cold reservoir T 0

Hot reservoir Hot reservoir

RA RB

E

Q 1 Q 2

Q 0 A Q 0 Q 0 B

Q 1 A Q 2 B

W WA WB

T 1 T 2

Two Carnot refrigerators RA and RB are adjusted to deliver exactly the

same amount of heat to their hot reservoirs as is extracted from those by

the Carnot engine E , i.e. Q 1 A = Q 1 , Q 2 B = Q 2. It means no heat is

flowing in or out of the hot reservoirs.

No such restrictions on the heat transfers to and from cold reservoir.

Therefore, heat flowing from it is Q 0 A + Q 0 B − Q 0.

The composite system E + RA + RB , with the Engine driving the

Refrigerators, delivers a net work W − (WA + WB ) to the surroundings.

Since the composite system is working in a reversible cycle, net change in

internal energy is zero and, therefore, from first law it follows that

Q 0 A + Q 0 B − Q 0 = W − (WA + WB )

Because the heat flow is only from single cold reservoir, Kelvin-Planck

statement of 2nd law ensures that

W ≤ WA + WB and Q 0 A + Q 0 B − Q 0 ≤ 0

By definition, in a Carnot cycle,

Q 1 A

Q 0 A

= T^1

T 0

⇒ Q 0 A = Q 1 A^ T^0

T 1

= Q 1 T^0

T 1

and similarly Q 0 B = Q 2 T^0 T 2

Entropy – a new state variable

Consider a system performing a reversible cycle from initial state 1 to an

intermediate state 2 and back. The cycle can take various paths to

achieve this. A couple of them are sketched in a pV -diagram below.

p

V

A B

C

2

1

Evaluate

δQ/T = 0 along the paths A – B and A – C,

[∫ 2

1

δQ T

]

A

[∫ 1

2

δQ T

]

B

= 0 and

[∫ 2

1

δQ T

]

A

[∫ 1

2

δQ T

]

C

Upon subtracting one expression from the other gives,

[∫ 2

1

δQ T

]

B

[∫ 1

2

δQ T

]

C

[∫ 2

1

δQ T

]

B

[∫ 1

2

δQ T

]

C

Paths B and C being arbitrary, then

1 δQ/T^ must be path independent.

It, therefore, defines a thermodynamic property of the system i.e. δQ/T is

an exact differential of some state variable.

We call the new state variable as Entropy S

1

δQ T

= S 2 − S 1 = ∆S

Unit of entropy is kJ/K, the same as CP , CV.

The differential form of second law (equivalent to dU = δQ − δW ) is,

δQ = T dS

Hence, Q 12 =

1 T dS^ is the heat transfer equivalent of^ W^12 =^

1 p dV^.

Isentropic – a line on which entropy S is constant i.e. Q 12 = 0 implying

the process is adiabatic + reversible.

Now suppose, one of the paths in the process in above diagram is

irreversible (say, A). Then from Clausius inequality,

1

δQir T

2

δQr T

1

δQir T

1

δQr T

Thermodynamics: Work, Heat, and the Second Law, Lecture notes of Thermodynamics

Related documents

Partial preview of the text

Download Thermodynamics: Work, Heat, and the Second Law and more Lecture notes Thermodynamics in PDF only on Docsity!

Work and Heat

Consider an isolated system composed of two identical bodies A and B,

say equal masses of water, which can exchange heat. Initially, TA = 310

K and TB = 290 K. Being isolated, no heat exchange is possible between

environment and the system.

What will happen after a while?

First law of thermodynamics i.e. conservation of energy (and mass)

admits the following two possibilities.

1. TA → 320 K, TB → 280 K : Thermal energy lost by B is gained by A

2. TA → 300 K, TB → 300 K : Thermal energy lost by A is gained by B

How to explain that case 1 is never observed in nature?

A few other phenomena, which may satisfy energy conservation but not

observed in nature :

I water running uphill without an external assist

I CO 2 and H 2 O reacting spontaneously to form CH 4 and O 2

I air separating into its constituents spontaneously

An equivalent situation arises in Work Heat.

−→ cold is possible but cold

−→ hot is impossible.

E.g. Two stones when rubbed against each other, the work done against

friction is transformed into internal energy leading to rise of temperature

of the stones. The stones quickly attain the same temperature with the

surroundings, which is acting as reservoir whose temperature increase is

insignificant. The stones thus return to their initial state and the next

cycle of rubbing begins.

The net result of the process is conversion of mechanical work into heat

i.e. Q = W. This transformation can continue indefinitely.

Opposite process, namely, conversion of heat into work, should continue

indefinitely without involving any changes in the state of the system.

Seems like isothermal expansion of ideal gas fits the bill. Since dU = 0,

we get W = Q i.e. complete conversion of heat into work! But just for

once i.e. volume increases and pressure decreases until it reaches the

atmospheric pressure and the process stops. It just cannot go back again

spontaneously from where it started, namely, high pressure and smaller

volume!!

E extracts heat Q 1 from a hot body and converts fully into work

W = Q 1 , just sufficient to operate one cycle of R. The R extracts Q 2

from a cold body and deliver Q 1 + Q 2 to hot body in each cycle.

The above process is equivalent to the composite E+R transferring Q 2

from cold to hot body without any assist. This renders Clausius

statement false. The reverse can also be proved.

An engineer would want to harvest as much as possible of thermal energy

and convert it into mechanical energy i.e. work. Kelvin-Planck statement

of second law simply says we cannot turn all heat into work. But it lets

us have some! Trick is to employ the cold body differently – harvest a

part of thermal energy for work and reject the rest into the cold body.

This way we can satisfy the second law. The schematic of such a

realizable heat engine is shown below. Henceforth, we will call the hot

and cold body as reservoirs in the sense that their thermal states remain

unchanged upon any amount of heat extraction and dumping.

The first law for this system is,

∆U = QH − QL − W

For cyclic process, ∆U = 0 and hence W = QH − QL.

The thermal efficiency η of the engine is defined as,

W

QH

QL

QH

Clearly, if QL = 0 we get η = 1 and our engine does a perfect job in

converting all heat into work. But this violates second law, so we must

reject some heat i.e. QL 6 = 0 ⇒ η < 1.

Let us look into the workings of a couple of engines that utilize the above

process, as shown in the schematic diagram.

The processes of Otto cycle is plotted in the PV -diagram below.

Assuming CV to be constant along 2 → 3 and 4 → 1, we have

QH =

∫ T 3

∫ T 1

Two adiabatic processes during compression and power stroke give us

T 1

T 4

T 2

T 3

T 4 − T 1

T 3 − T 2

T 4