2. 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 the
processes”
• According to first law, it is assumed that the energy transfer can take place in either
direction, 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?
3. • 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 fuel
is not converted into work, but a portion of input heat has to be rejected
to exhaust gases, oil, cooling water
Image source: http://teamspeed.com
QExhaust
QFuel
QFriction
QCooling Water
QOil
4. - 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
5. 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
6. 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
7. Energy Reservoirs
MER
WT
Wnet
Boiler
Turbine
Condenser
Pump
TERH
(SOURCE)
TERL
(SINK)
Wp
Q1
Q2
Thermal Energy reservoirs (TER): is defined as a large
body of infinite heat capacity, which is capable of
absorbing or rejecting an unlimited quantity of heat
without suffering appreciable change in its
thermodynamic coordinates. All process are quasi-static
Constant Temperature
Constant Temperature
Mechanical Energy reservoirs (MER): is a large body
enclosed by an adiabatic impermeable wall capable of
storing work as kinetic energy or potential energy. All
process are quasi-static
8. 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
9. Heat Engine
Boiler
Pump
Condenser
QB / Q1
WT
WP
WT - WP
Net work output of the
system during cyclic process
Turbine
Boiler
Turbine
Condenser
Pump
Water
Steam
Water Steam
Water
Heat Source: Furnace
Heat Sink: Lake/River
QC / Q2
10. QB – QC = WT - WP
𝛈 =
"#$%#$ '()*
+,%#$ -./$
=
WT − WP
Q1
𝛈 = 30% - 40%
=
Q1 – Q2
Q1
Q2 = 60%- 70% Q1
= 1 -
Q2
Q1
T1 = 500∘C
W
Q – W = Q2
Q1
T2 = 20∘C
Source
Sink
(atmosphere)
Heat
Engine
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
𝑭𝒓𝒐𝒎 𝒇𝒊𝒓𝒔𝒕 𝑳𝒂𝒘 𝒐𝒇 𝑻𝒉𝒆𝒓𝒎𝒐𝒅𝒚𝒏𝒂𝒎𝒊𝒄𝒔 𝒇𝒐𝒓 𝒂 𝒄𝒚𝒄𝒍𝒊𝒄 𝒑𝒓𝒐𝒄𝒆𝒔𝒔
∮𝝏𝑸 = ∮𝝏𝑾
Q1 – Q2 = WT - WP
11. 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
12. Win
Cold Environment
Warm House
Objectives of Refrigerator & Heat Pump
Heat
Pump
Win
Q1
Q2
Refrigerated Space
Warm Environment
Refrig
erator
Desired Output
Condenser
Evaporator
Expansion
Valve
Compressor Wc
T < T atm.
T = T atm.
Desired Output
T = T atm.
T > T atm.
Q1
Q2
13. 𝑭𝒓𝒐𝒎 𝒇𝒊𝒓𝒔𝒕 𝑳𝒂𝒘 𝒐𝒇 𝑻𝒉𝒆𝒓𝒎𝒐𝒅𝒚𝒏𝒂𝒎𝒊𝒄𝒔 𝒇𝒐𝒓 𝒂 𝒄𝒚𝒄𝒍𝒊𝒄 𝒑𝒓𝒐𝒄𝒆𝒔𝒔
∮𝝏𝑸 = ∮𝝏𝑾
W = Qc - QR
= Q1 - Q2
COPR =
𝑯𝒆𝒂𝒕 𝑹𝒆𝒎𝒐𝒗𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝑹𝒆𝒇𝒓𝒊𝒈𝒆𝒓𝒂𝒕𝒐𝒓
𝑾𝒐𝒓𝒌 𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅
=
Q2
W
=
QR
QC − QR
• The performance of refrigerators and heat pumps is
expressed in terms of the coefficient of performance (COP)
COPR =
𝑩𝒆𝒏𝒊𝒇𝒕
𝑪𝒐𝒔𝒕
W
T2 = - 4∘C
T1 = 35∘C
Refrigerator
Heat
Engine
Sink
(atmosphere)
Q1
=
Q2
Q1 − Q2
In a refrigerator, the desired effect is the amount of heat
removed Q2 from the space being heated
Q2
14. W
Q1
Q2
T2 = 4∘C
T1 = 25∘C
Heated Space
Atmoshpere
Heat Pump
Heat
Pump
• 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 =
𝑯𝒆𝒂𝒕 𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝑯𝒆𝒂𝒕 𝑷𝒖𝒎𝒑
𝑾𝒐𝒓𝒌 𝑺𝒖𝒑𝒑𝒍𝒊𝒆𝒅
=
Q1
W
=
Q1
Q1 − Q2
= 1 +
Q2
Q1 − 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
15. • The performance of air conditioners and heat pumps is often expressed in terms of the energy
efficiency ratio (EER) or seasonal energy efficiency ratio (SEER) determined by following
certain testing standards
- SEER : is the ratio the total amount of heat removed by an air conditioner or heat pump during a
normal cooling season (in Btu) to the total amount of electricity consumed (in watt-hours, Wh), and
it 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 rate
of heat removal from the cooled space by the cooling equipment to the rate of electricity
consumption 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 cooling
One ton = 12,000 Btu = 211 kJ/min
16. • 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 produce
1.2 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
17. W
Heat
Engine
Second Law of Thermodynamics: Kelvin Plank’s Statement
- 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
- In other words, no heat engine can have a thermal efficiency of
100%
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”
18. W
Heat
Pump
Second Law of Thermodynamics: Clausius Statement
- Heat cannot flow from itself from a system low temperature to a
system at high temperature.
- COP = Q/W = Q/0 = ∞ ( a condition not possible)
- The only alternative is that some external work must be
supplied 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 external
agency, to convey heat from a body at a lower temperature to a body at a higher temperature”
System (T2)
19. 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)
W
Heat
Engine
Source (TH)
Qin
Wnet = Qin
Qout = 0
When the thermal energy is equivalent to the work done, this does not violate the law of
conservation of energy. However it does violate the more subtle second law of thermodynamics
20. 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
Adiabatic
Cover
Diathermic
Cover
Cylinder Head
Heat Insulation
Heat Insulation
Piston
Piston motion
21. 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
22. Diathermic
Cover
Working
Substance
Adiabatic
Cover
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 high
temperature T1
Q1
Stage 2 – Adiabatic Expansion
(Process 2-3)
- The cylinder becomes a perfect insulator so
that no heat flow takes place
- The fluid expands adiabatically and reversibly
whilst temperature falls from T1 to T2
Source,
T
1
23. Sink,
T
2
Working
Substance
Diathermic
Cover
Adiabatic
Cover
Stage 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 at
constant lower temperature T2
Q2
Stage 4 - Adiabatic Compression
(Process 4-1)
- Cylinder head becomes a perfect insulator so
that no heat flow occurs
- The compression is continued adiabatically and
reversibly during which temperature is raised
from T2 to T1
27. 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)
28. 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 lower
temperature T2 during the
isothermal expansion
(process 2-1)
Q1 is being rejected at
the upper temperature T1
Q1
Q2
29. 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
30. 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
Heat
Engine
Net Work
(W) = Q1 – Q2
High Temperature Reservoir T1
Heat
Pump
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, then
we will automatically violate the Clausius’s statement of the second law (and vice-versa)
No Work
31. Q1
Q2 = 0
Low Temperature Reservoir T2
Heat
Engine
W = Q1
High Temperature Reservoir T1
Heat
Pump
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
32. 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 is
deposited back into the kitchen by the coils on the back!
Second Law of Thermodynamics says that work is needed to
move the heat from cold to hot, so the actual amount of heat
added to the kitchen is MORE than the amount removed from
the refrigerator
33. • The second corollary to the Kelvin-Planck statement holds that “All reversible engines operating
between 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
=
T1
T2
; 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”
34. The amounts of heat rejected by engines B and C must
be the same since engines A and B can be combined into
one reversible engine operating between the same
reservoirs as engine C and thus the combined engine will
have the same efficiency as engine C.
Since the heat input to engine C is the same as the heat
input to the combined engines A and B, both systems
must reject the same amount of heat
35. • 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
36. • 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
37. 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 HEB
WB
38. HEB is reversed
Q1B
Q2A
Sink, T2
HEA
Source, T1
Q1A
Q2B
WA
HB
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
39. • 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
WA
HB
WB E
WA = WB
40. ‘‘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 the
same, the efficiency of a reversible engine is independent of the nature or amount of the working
substance 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
