Second Law of Thermodynamics

Dr. Rohit Singh Lather

Limitations of First Law of Thermodynamics• Heat flows from a system of higher temperature to a system of lower temperature and never from

lower temperature system to higher temperature system

• From first law of thermodynamics, “Heat lost and heat gain must be equal in both theprocesses”

• According to first law, it is assumed that the energy transfer can take place in eitherdirection, it does not specify the direction of energy transfer

Heat is transferred from hotter side to the colder side of the rod

But never from colder to hotter side by itself by itself

hotter side colder side

Why does energy travel always

from higher value to lower value?

• All work can be converted into heat but all heat cannot be converted into work

- For example: In internal combustion engine, all heat generated from combustion of fuelis not converted into work, but a portion of input heat has to be rejectedto exhaust gases, oil, cooling water

Image source: http://teamspeed.com

QExhaust

QFuel

QFriction

QCooling Water

QOil

- Example : In power plants, all heat generated from combustion of coal is not converted into work, but a portion of input heat has to be rejected in the condenser

Image source: http://google.com

What First Law of Thermodynamics Tells Us – Interpretations

First law of thermodynamics is a necessary condition but not sufficient condition for a process to take place

First law of thermodynamics states “work can be converted into heat and heat into work”

First law makes no distinction between forms of energy, silent about the possibility of energy conversion

First law is not sufficient to predict weather a system will or will not under go a particular change

What Second Law of Thermodynamics Tells Us – Interpretations

Second law of thermodynamics indicates that, “all heat cannot be converted into work”

According to second law of thermodynamics, “heat will only be transferred from high temperature to lower temperature and not vice versa“

“First law of thermodynamics”, is a quantitative statement and “Second law of thermodynamics”, is a qualitative statement

Second law states that whether it is possible for energy transfer to proceed along a particular direction or not

A cycle can only occur if it satisfies both the first law and second law of thermodynamics

Energy Reservoirs

MER

WT

Wnet

Boiler

Turbine

Condenser

Pump

TERH(SOURCE)

TERL(SINK)

Wp

Q1

Q2

Thermal Energy reservoirs (TER): is defined as a largebody of infinite heat capacity, which is capable ofabsorbing or rejecting an unlimited quantity of heatwithout suffering appreciable change in itsthermodynamic coordinates. All process are quasi-static

Constant Temperature

Constant Temperature

Mechanical Energy reservoirs (MER): is a large bodyenclosed by an adiabatic impermeable wall capable ofstoring work as kinetic energy or potential energy. Allprocess are quasi-static

Heat Engine • Heat Engine is a device which working in a cycle converts energy in form of heat into work

- Heat engines convert heat to work

• There are several types of heat engines, but they are characterized by the following:- They all receive heat from a high-temperature source (oil furnace, nuclear reactor, etc.) - They convert part of this heat to work- They reject the remaining waste heat to a low-temperature sink- They operate in a cycle

Heat Engine

Boiler

Pump

Condenser

QB / Q1

WT

WP

WT - WPNet work output of the

system during cyclic process Turbine

Boiler

Turbine

Condenser

Pump

Water

Steam

Water Steam

Water

Heat Source: Furnace

Heat Sink: Lake/RiverQC / Q2

QB – QC = WT - WP

𝛈 ="#$%#$'()*+,%#$-./$ =WT − WP

Q1

𝛈 = 30% - 40%

=Q1 – Q2Q1

Q2 = 60%- 70% Q1

=1 - Q2Q1

T1 = 500∘C

W

Q – W = Q2

Q1

T2 = 20∘C

Source

Sink(atmosphere)

HeatEngine

Q. Is it possible to save the rejected heat QC in a power cycle? Answer: NO, because without the cooling in 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 in idealized cycle

𝑭𝒓𝒐𝒎𝒇𝒊𝒓𝒔𝒕𝑳𝒂𝒘𝒐𝒇𝑻𝒉𝒆𝒓𝒎𝒐𝒅𝒚𝒏𝒂𝒎𝒊𝒄𝒔𝒇𝒐𝒓𝒂𝒄𝒚𝒄𝒍𝒊𝒄𝒑𝒓𝒐𝒄𝒆𝒔𝒔∮𝝏𝑸 = ∮𝝏𝑾

Q1 – Q2 = WT - WP

Refrigerator

Throttle

Compressor

QR/Q2

Refrigerator

W

• In nature, heat flows from high-temperature regions to low-temperature ones• The reverse process, however, cannot occur by itself• The transfer of heat from a low- temperature region to a high-temperature one requires special

devices called refrigerators

Refrigerators are cyclic devices, and the working fluids used in the cycles are called refrigerant

Condenser

QC/Q1

Win

Cold Environment

Warm House

Objectives of Refrigerator & Heat Pump

HeatPumpWin

Q1

Q2

Refrigerated Space

Warm Environment

Refrigerator

Desired Output

Condenser

Evaporator

Expansion Valve

Compressor Wc

T < T atm.

T = T atm.

Desired Output

T = T atm.

T > T atm.

Q1

Q2

𝑭𝒓𝒐𝒎𝒇𝒊𝒓𝒔𝒕𝑳𝒂𝒘𝒐𝒇𝑻𝒉𝒆𝒓𝒎𝒐𝒅𝒚𝒏𝒂𝒎𝒊𝒄𝒔𝒇𝒐𝒓𝒂𝒄𝒚𝒄𝒍𝒊𝒄𝒑𝒓𝒐𝒄𝒆𝒔𝒔∮𝝏𝑸 = ∮𝝏𝑾

W = Qc - QR = Q1 - Q2

COPR =𝑯𝒆𝒂𝒕𝑹𝒆𝒎𝒐𝒗𝒆𝒅𝒃𝒚𝒕𝒉𝒆𝑹𝒆𝒇𝒓𝒊𝒈𝒆𝒓𝒂𝒕𝒐𝒓

𝑾𝒐𝒓𝒌𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅 =Q2W

= QRQC − QR

• The performance of refrigerators and heat pumps isexpressed in terms of the coefficient of performance (COP)

COPR =𝑩𝒆𝒏𝒊𝒇𝒕𝑪𝒐𝒔𝒕

W

T2 = - 4∘C

T1 = 35∘C

Refrigerator

HeatEngine

Sink(atmosphere)

Q1

= Q2Q1 − Q2

In a refrigerator, the desired effect is the amount of heat removed Q2 from the space being heated

Q2

W

Q1

Q2

T2 = 4∘C

T1 = 25∘CHeated Space

Atmoshpere

Heat Pump

HeatPump

• Heat pumps transfer heat from a low-temperature medium to a high-temperature• Refrigerators and heat pumps are essentially the same devices; they differ in their objectives

only• Refrigerator is to maintain the refrigerated space at a low temperature• On the other hand, a heat pump absorbs heat from a low-temperature source and supplies the

heat to a warmer medium

In a heat Pump, the desired effect is the amount of heat supplied Q1 to the space being heated

COPHP =𝑯𝒆𝒂𝒕𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅𝒃𝒚𝒕𝒉𝒆𝑯𝒆𝒂𝒕𝑷𝒖𝒎𝒑

𝑾𝒐𝒓𝒌𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅 =Q1W

= Q1Q1 − Q2

=1 + Q2Q1 − Q2

=1 + COPR

The COP of a heat pump operating as a heat pump is higher that the COP of the same machine operating as a refrigerator by unity

• The performance of air conditioners and heat pumps is often expressed in terms of the energyefficiency ratio (EER) or seasonal energy efficiency ratio (SEER) determined by followingcertain testing standards

- SEER : is the ratio the total amount of heat removed by an air conditioner or heat pump during anormal cooling season (in Btu) to the total amount of electricity consumed (in watt-hours, Wh), andit is a measure of seasonal performance of cooling equipment

- EER : is a measure of the instantaneous energy efficiency, and is defined as the ratio of the rateof heat removal from the cooled space by the cooling equipment to the rate of electricityconsumption in steady operation

• Therefore, both EER and SEER have the unit Btu/Wh- 1 kWh = 3.412 Btu (1 Wh = 3.412 Btu, a device that removes 1 kWh of heat from the cooled

space for each kWh of electricity it consumes (COP = 1) will have an EER of 3.412)- Therefore, the relation between EER and COP, EER = 3.412.COPR

Performance of Refrigerators, Air-Conditioners, and Heat Pumps

The heat transfer rate is often given in terms of tones of heating or coolingOne ton = 12,000 Btu = 211 kJ/min

• Air conditioners or heat pumps SEER: 13 to 21, which correspond to COP values of 3.8 to 6.2. - Most air conditioners have an EER between 8 to 12 (COP of 2.3 to 3.5)

• Best performance is achieved using units equipped with variable-speed drives (also called inverters) - Variable-speed compressors and fans allow the unit to operate at maximum efficiency for varying

heating/cooling needs and weather conditions as determined by a microprocessor- In the air-conditioning mode, for example, they operate at higher speeds on hot days and at lower speeds

on cooler days, enhancing both efficiency and comfort

• The EER or COP of a refrigerator decreases with decreasing refrigeration temperature- Therefore, it is not economical to refrigerate to a lower temperature than needed

• The COPs of refrigerators (range): 2.6 to 3.0 for cutting and preparation rooms 2.3 to 2.6 for meat, deli, dairy, and produce1.2 to 1.5 for frozen foods 1.0 to 1.2 for ice cream units

Note: COP of freezers is about half of the COP of meat refrigerators- It costs twice as much to cool the meat products with refrigerated air that is cold

enough to cool frozen foods- It is good energy conservation practice to use separate refrigeration systems to meet

different refrigeration needs

WHeatEngine

Second Law of Thermodynamics: Kelvin Plank’s Statement

- It is impossible for any device that operates on a cycle to receiveheat from a single reservoir and produce a net amount of work

- In other words, no heat engine can have a thermal efficiency of100%

Source (TH)

A heat engine that violates the Kelvin-Planck statement of the second law cannot be built

Thermal efficiency of 100%

Qin

Wnet = Qin

Qout = 0

“It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of work to its surroundings while receiving an energy transfer by heat from a single thermal reservoir”

WHeatPump

Second Law of Thermodynamics: Clausius Statement

- Heat cannot flow from itself from a system low temperature to asystem at high temperature.

- COP = Q/W = Q/0 = ∞ ( a condition not possible)

- The only alternative is that some external work must besupplied to the machine

System (T1)

A heat engine that violates the Kelvin-Planck statement of the second law cannot be built

Qin

Q2

“Heat cannot, of itself, pass from a colder to a hotter body ”

“It is impossible for a self acting machine working in a cyclic process unaided by any externalagency, to convey heat from a body at a lower temperature to a body at a higher temperature”

System (T2)

Perpetual Motion Machine of the Second Kind (PMMK2) • Without violating the first law, a machine can be imagined which would continuously absorb heat

from a single thermal reservoir and would convert this heat completely into work- The efficiency of such a machine would be 100%- This machine is called the perpetual motion machine of the second kind (PMM2)

WHeatEngine

Source (TH)

Qin

Wnet = Qin

Qout = 0

When the thermal energy is equivalent to the work done, this does not violate the law ofconservation of energy. However it does violate the more subtle second law of thermodynamics

Carnot Cycle • The cycle was first suggested by Sadi Carnot, in 1824, which works on reversible cycle

• Any fluid may be used to operate the Carnot cycle, which is performed in an engine cylinder the

head of which is supposed alternatively to be perfect conductor or a perfect insulator of a heat

• Heat is caused to flow into the cylinder by the application of high temperature energy source to

the cylinder head during expansion, and to flow from the cylinder by the application of a lower

temperature energy source to the head during compression

Source, T1

Sink, T2

Working Substance

Adi

abat

ic C

over

Dia

ther

mic

Cov

er

Cylinder Head

Heat Insulation

Heat Insulation

Piston

Piston motion

The assumptions made for describing the working of the Carnot engine are as follows :

1. The piston moving in a cylinder does not develop any friction during motion

2. The walls of piston and cylinder are considered as perfect insulators of heat

3. The cylinder head is so arranged that it can be a perfect heat conductor or perfect heat insulator

4. The transfer of heat does not affect the temperature of source or sink

5. Working medium is a perfect gas and has constant specific heat

6. Compression and expansion are reversible

Dia

ther

mic

Cov

er

Working Substance

Adi

abat

ic C

over

Piston motion

Stage 1 – Isothermal Expansion (Process 1-2)

- Hot energy source at temperature T1 is applied- Heat Q1 is taken in whilst the fluid expands

isothermally and reversiblyat constant hightemperature T1

Q1

Stage 2 – Adiabatic Expansion (Process 2-3)

- The cylinder becomes a perfect insulator sothat no heat flow takes place

- The fluid expands adiabatically and reversibly whilst temperature falls from T1 to T2

Sour

ce,

T1

Sink

, T

2

Working Substance

Dia

ther

mic

Cov

erA

diab

atic

Cov

erStage 3 – Isothermal Compression

(Process 3-4) - Cold energy source at temperature T2 is applied- Heat Q2 flows from the fluid whilst it is

compressed isothermally and reversibly atconstant lower temperature T2Q2

Stage 4 - Adiabatic Compression(Process 4-1)

- Cylinder head becomes a perfect insulator sothat no heat flow occurs

- The compression is continued adiabatically andreversibly during which temperature is raisedfrom T2 to T1

Q1 = W1-2 = P1 V1 In 𝑽𝟐𝑽𝟏

Q1 = W1-2 = mRT1 In 𝑽𝟐𝑽𝟏

Q2 = W3-4 = - P3 V3 In 𝑽𝟒𝑽𝟑

Q2 = W3-4 = mRT2 In 𝑽𝟑𝑽𝟒Heat Rejected

Heat Added𝜂 =1 - Q2

Q1

𝜂 = 𝑵𝒆𝒕𝑾𝒐𝒓𝒌𝑶𝒖𝒕𝒑𝒖𝒕

𝑯𝒆𝒂𝒕𝑰𝒏𝒑𝒖𝒕 = 𝑾𝒏𝒆𝒕

𝑸𝟏

𝑭𝒓𝒐𝒎𝒇𝒊𝒓𝒔𝒕𝑳𝒂𝒘𝒐𝒇𝑻𝒉𝒆𝒓𝒎𝒐𝒅𝒚𝒏𝒂𝒎𝒊𝒄𝒔𝒇𝒐𝒓𝒂𝒄𝒚𝒄𝒍𝒊𝒄𝒑𝒓𝒐𝒄𝒆𝒔𝒔∮𝝏𝑸 = ∮𝝏𝑾𝒐𝒓𝑾𝒏𝒆𝒕 = 𝑸𝟏 − 𝑸𝟐

𝜂 =1 - T2T1

(V2/V1) = (V3/V4)

P2 V2𝛾 = P3 V3

𝛾

T1 V2(𝛾- 1)= T2 V3

(𝛾- 1)

P4 V4𝛾 = P1 V1

𝛾

PV= nRT ; PV/T = nR =C

P2V2/T1 = P3V3/T2

P4V4/T2 = P1V1/T1

1

2

3

4

T1 V1(𝛾- 1)= T2 V4

(𝛾- 1)

5

6

Dividing 1/3 and 2/4

Dividing 5/6

=1 - Q2Q1

Tem

pera

ture

Entropy

Q1

Q2T2

T1

4 3

21

Area of the rectangle a-b-c-d represents work output per cycle and it equalsQ1 – Q2 = (T1 – T2).dS

Isotherms

Frictionless Adiabats

S1 =S4 S2 = S3

Efficiency of a Reversible Heat Engine

• From the above expression, it may be noted that as T2 decreases and T1 increases, efficiency of the reversible cycle increases

• Since 𝜂 is always less than unity, T2 is always greater than zero and positive (+ ve)

Carnot Heat Pump

• An engine, which consists entirely of reversible processes, can operate in the reverse direction, so that it follows the cycle as shown and operates as a heat pump

Work (W) will be needed to drive the pump

The enclosed area represents this work which is exactly equal to that flowing from it when used as engine

(process 4-3) Q2 is being taken in at the lowertemperature T2 during theisothermal expansion

(process 2-1)Q1 is being rejected atthe upper temperature T1

Q1

Q2

1. It is impossible to perform a frictionless process

2. It is impossible to transfer the heat without temperature potential

3. Isothermal process can be achieved only if the piston moves very slowly to allow heat transfer so that the temperature remains constant- Adiabatic process can be achieved only if the piston moves as fast as possible so that the heat

transfer is negligible due to very short time available- The isothermal and adiabatic processes take place during the same stroke therefore the piston

has to move very slowly for part of the stroke and it has to move very fast during remaining stroke

- This variation of motion of the piston during the same stroke is not possible

Carnot cycle cannot be performed in practice because of the following reasons

Equivalence of Clausius Statement to the Kelvin-Planck Statement

W = Q1 – Q2; Since, there is no heat interaction with the low temperature, it can be eliminated

Q1

Q2

Low Temperature Reservoir T2

HeatEngine

Net Work (W) = Q1 – Q2

High Temperature Reservoir T1

HeatPump

Q1

Q2

A heat pump which requires no work and transfers an amount of Q2 from a low temperature to a higher temperature reservoir

(violation of the Clausius statement)

The combined system of the heat engine and heat pump acts then like a heat engine exchanging heat with a single reservoir, which is the violation of the Kelvin-Planck statement

Heat rejected

Q1 > Q2

The Kelvin’s and Clausius’s statements of the second law are equivalent. i.e. if we violate Kelvin’s statement, thenwe will automatically violate the Clausius’s statement of the second law (and vice-versa)

No Work

Q1

Q2 = 0

Low Temperature Reservoir T2

HeatEngine

W = Q1

High Temperature Reservoir T1

HeatPump

Q1

Q2

A heat engine which converts all heat to work, without rejecting

heat to low temperature(Violation of the Kelvin Plank statement)

The combined system constitutes a device which transfers heat from low temperature reservoir to high temperature without any work from external agency, which is the violation of the Clausius

statement

Q1 > Q2

Violation of Kelvin – Plank Statement leads to violation of Clausius’s statements

W = Q1

Can you beat Second Law

Can you cool your room by leaving the refrigerator door

open ?

The heat removed from the interior of the refrigerator isdeposited back into the kitchen by the coils on the back!

Second Law of Thermodynamics says that work is needed tomove the heat from cold to hot, so the actual amount of heatadded to the kitchen is MORE than the amount removed fromthe refrigerator

• The second corollary to the Kelvin-Planck statement holds that “All reversible engines operatingbetween the same thermal reservoirs have the same 𝜂”- This is independent of any details of the cycle or the materials involved- The thermal efficiency, 𝜂, should depend only on the character of the reservoirs involved

Thermodynamic Temperature

Consider a case of reversible heat engine operating between two reservoirs- Its thermal efficiency is given by 𝜂 = Q1 − Q2

Q1= 1 - Q2

Q1

• The temperature of a reservoir remains uniform and fixed irrespective of heat transfer - This means that reservoir has only one property defining its state and the heat transfer from a reservoir

is some function of that property, temperature. Thus Q = φ (K), where K is the temperature of reservoir

Q1 Q2

= φ(K1)φ(K2)

=> Q1 Q2

= T1T2

; T1 and T2 are the thermodynamic temperatures of the reservoirs

Zero thermodynamic temperature (that temperature to which T2 tends, as the heat transfer Q2 tends to zero) has never been attained and one form of third law of thermodynamics is the statement :

‘‘The temperature of a system cannot be reduced to zero in a finite number of processes”

The amounts of heat rejected by engines B and C mustbe the same since engines A and B can be combined intoone reversible engine operating between the samereservoirs as engine C and thus the combined engine willhave the same efficiency as engine C.

Since the heat input to engine C is the same as the heatinput to the combined engines A and B, both systemsmust reject the same amount of heat

• After establishing the concept of a zero thermodynamic temperature, a reference reservoir is chosen and assigned a numerical value of temperature

• Any other thermodynamic temperature may now be defined in terms of reference value and the heat transfers that would occur with reversible engine,

T = Tref.Q

Qref.

• Let us make an arbitrary choice to avoid ratios. We take, for convenience, the temperature of the triple point of water to be 273.15 K. Thus for any system, the local T is

T = 273.3.Q

Qref.

This implies we can connect our heat engine to a reservoir maintained at the triple point temperature of water, and measure the associated Qs for the heat engine

• We would like to drive our efficiency to be as close to unity as possible, nature limits us • Generally, we have little to no control over the environmental temperature TL, so it is a lower bound, usually

around TL ∼ 300 K. And material properties for engines limit TH . For many metals, TH ∼ 1500 K is approaching values where material strength is lost

• So a practical upper bound based on these numbers tell us • η∼ 1 − (300K)/(1500K) = 0.8 is may be the most we can expect. We plot η as a function of TH for fixed TL =

300 K For real systems, with irreversible features, the values are much lower

• The determination of thermodynamic temperature cannot be made in this way as it is not possible to build a reversible engine

• Temperatures are determined by the application of thermodynamic relations to other measurements

• The SI unit of thermodynamic temperature is the kelvin (K)• The relation between thermodynamic temperature and Celsius scale

- Thermodynamic temperature = Celsius temperature + 273.15°- The kelvin unit of thermodynamic temperature is the fraction 1 temperature of ‘Triple point’ of

water

Carnot Theorem “It states that of all engines operating between a given constant temperature source and a

given constant temperature sink, none has a higher efficiency than a reversible engine”

Let HEA be any heat engine and HEB be any reversible heat engine

We have to prove that efficiency of HEB is more than that of HEA

Let us assume that 𝜂A > 𝜂B

Q1A = Q1B = Q1

𝜂A = 𝜂B

𝑾𝑨

𝑸𝟏𝑨= 𝑾𝑩

𝑸𝟏𝑩

WA > WB

Q1B

Q2A

Sink, T2

HEA

Source, T1

Q1A

Q2B

WA HEBWB

HEB is reversed

Q1B

Q2A

Sink, T2

HEA

Source, T1

Q1A

Q2B

WAHB

WB E• Since HEB is a reversible heat engine, the magnitudes of heat and work transfer quantities will

remain the same, but their directions will be reversed

• Since WA > WB, some part of WA (equal to WB) may be fed to drive the reversed heat engine ∃HB.

Since Q1A = Q1B = Q1, the heat discharged by ∃HB may be supplied to HEA• The source may, therefore, be eliminated

• The net result is that HEA and ∃HB together constitute a heat engine which, operating in a cycle

produces net work WA – WB while exchanging heat with a single reservoir at T2

• This violates the Kelvin-Planck statement of the second law Hence the assumption that 𝜂A>𝜂B is

wrong

The combined system of heat pump HEB and engine HEA, becomes a PMM2

Q1B = Q1

Q2A

Sink, T2

HEA

Q1A = Q1

Q2B

WAHB

WB E

WA = WB

‘‘The efficiency of all reversible heat engines operating between the same temperature levels is the same”

Since, the efficiencies of all reversible engines operating between the same heat reservoirs are thesame, the efficiency of a reversible engine is independent of the nature or amount of the workingsubstance undergoing the cycle

• 𝜂A cannot be greater than 𝜂B

• HEA and ∃HB together violate the Kelvin-Planck statement, ∴ 𝜂B > 𝜂A

• Similarly, if we assume 𝜂B > 𝜂A and reverse the engine HEA, we observe that 𝜂B cannot be greater

than 𝜂A, ∴ 𝜂B = 𝜂A

Hopefully, you understand today’s lesson. Otherwise, you’ll end up like this cow.