Advanced Thermodynamics: Lecture 5 Shivasubramanian Gopalakrishnan - - PowerPoint PPT Presentation

▶
advanced thermodynamics lecture 5
SMART_READER_LITE
LIVE PREVIEW

Advanced Thermodynamics: Lecture 5 Shivasubramanian Gopalakrishnan - - PowerPoint PPT Presentation

Apr 27, 2023 19 likes •518 views

Advanced Thermodynamics: Lecture 5 Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661 Heat engines They receive heat from a high-temperature source (solar energy, oil furnace,

slide-1
SLIDE 1

Advanced Thermodynamics: Lecture 5

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-2
SLIDE 2

Heat engines

They receive heat from a high-temperature source (solar energy, oil furnace, nuclear reactor, etc.). They convert part of this heat to work (usually in the form of a rotating shaft). They reject the remaining waste heat to a low-temperature sink (the atmosphere, rivers, etc.). They operate on a cycle.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-3
SLIDE 3

Heat engines

Wnet,out Low-temperature SINK Qout Qin HEAT ENGINE High-temperature SOURCE

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-4
SLIDE 4

Schematic of a Power Plant

  • System boundary

Boiler Pump Turbine Qout Win Wout Qin Energy source (such as a furnace) Energy sink (such as the atmosphere) Condenser

FIGURE 6–10

Total work output Wnet,out = Wout − Win

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-5
SLIDE 5

Thermal Efficiency

Defined as Thermal Efficiency = Net work output Total heat input

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-6
SLIDE 6

Thermal Efficiency

Defined as Thermal Efficiency = Net work output Total heat input ηth = Wnet,out Qh

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-7
SLIDE 7

Thermal Efficiency

Defined as Thermal Efficiency = Net work output Total heat input ηth = Wnet,out Qh Since Wnet,out = QH − QL

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-8
SLIDE 8

Thermal Efficiency

Defined as Thermal Efficiency = Net work output Total heat input ηth = Wnet,out Qh Since Wnet,out = QH − QL ηth = 1 − QL QH

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-9
SLIDE 9

Second Law: Kelvin Planck 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.

HEAT ENGINE Wnet,out = 100 kW QH = 100 kW QL = 0 Thermal energy reservoir · · · Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-10
SLIDE 10

Second Law: Kelvin Planck 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.

HEAT ENGINE Wnet,out = 100 kW QH = 100 kW QL = 0 Thermal energy reservoir · · ·

In other words No heat engine can have a thermal efficiency of 100 percent, or as for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-11
SLIDE 11

Refrigeration and heat pumps

  • CONDENSER

EXPANSION VALVE 120 kPa –25°C 120 kPa –20°C 800 kPa 30°C 800 kPa 60°C COMPRESSOR QL QH Wnet,in Surrounding medium such as the kitchen air Refrigerated space EVAPORATOR

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-12
SLIDE 12

Refrigeration schematic

  • Warm environment

at TH > TL Cold refrigerated space at TL R Wnet,in QH QL Required input Desired

  • utput

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-13
SLIDE 13

Coefficient of performance: Refrigeration

Defined as COPR = Desired output Required input = QL Wnet,in

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-14
SLIDE 14

Coefficient of performance: Refrigeration

Defined as COPR = Desired output Required input = QL Wnet,in Since Wnet,in = QH − QL

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-15
SLIDE 15

Coefficient of performance: Refrigeration

Defined as COPR = Desired output Required input = QL Wnet,in Since Wnet,in = QH − QL COPR = QL QH − QL = 1

QH QL − 1

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-16
SLIDE 16

Heat Pump schematic

Warm heated space at TH > TL Cold environment at TL HP Wnet,in QH QL Required input Desired

  • utput

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-17
SLIDE 17

Coefficient of performance: Heat Pump

Defined as COPHP = Desired output Required input = QH Wnet,in

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-18
SLIDE 18

Coefficient of performance: Heat Pump

Defined as COPHP = Desired output Required input = QH Wnet,in Since Wnet,in = QH − QL

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-19
SLIDE 19

Coefficient of performance: Heat Pump

Defined as COPHP = Desired output Required input = QH Wnet,in Since Wnet,in = QH − QL COPHP = QH QH − QL = 1 1 − QL

QH

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-20
SLIDE 20

Coefficient of performance: Heat Pump

Defined as COPHP = Desired output Required input = QH Wnet,in Since Wnet,in = QH − QL COPHP = QH QH − QL = 1 1 − QL

QH

COPHP = COPR + 1

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-21
SLIDE 21

Second Law: Clausius Statement

It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body.

Warm environment Cold refrigerated space R Wnet,in = 0 QH = 5 kJ QL = 5 kJ

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-22
SLIDE 22

Second Law: Clausius Statement

It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body.

Warm environment Cold refrigerated space R Wnet,in = 0 QH = 5 kJ QL = 5 kJ

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-23
SLIDE 23

Equivalence of Kelvin–Planck and Clausius Statements

High-temperature reservoir at TH Low-temperature reservoir at TL HEAT ENGINE ηth = 100% REFRIG- ERATOR QH QH + QL Wnet = QH QL (a) A refrigerator that is powered by

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-24
SLIDE 24

Equivalence of Kelvin–Planck and Clausius Statements

High-temperature reservoir at TH Low-temperature reservoir at TL REFRIG- ERATOR QL QL (b) The equivalent refrigerator

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-25
SLIDE 25

Heat Rejection by a Refrigerator

The food compartment of a refrigerator is maintained at 4C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2 kW, determine (a) the coefficient of performance of the refrigerator and (b) the rate of heat rejection to the room that houses the refrigerator.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-26
SLIDE 26

Heating a House by a Heat Pump

A heat pump is used to meet the heating requirements of a house and maintain it at 20C. On a day when the outdoor air temperature drops to -2C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP

  • f 2.5, determine (a) the power consumed by the heat pump and

(b) the rate at which heat is absorbed from the cold outdoor air.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-27
SLIDE 27

Internally and Externally Reversible Processes

A process is called internally reversible if no irreversibilities

  • ccur within the boundaries of the system during the process.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-28
SLIDE 28

Internally and Externally Reversible Processes

A process is called internally reversible if no irreversibilities

  • ccur within the boundaries of the system during the process.

A process is called externally reversible if no irreversibilities

  • ccur out- side the system boundaries during the process.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-29
SLIDE 29

Internally and Externally Reversible Processes

A process is called internally reversible if no irreversibilities

  • ccur within the boundaries of the system during the process.

A process is called externally reversible if no irreversibilities

  • ccur out- side the system boundaries during the process.

A process is called totally reversible, or simply reversible, if it involves no irreversibilities within the system or its surroundings.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-30
SLIDE 30

Internally and Externally Reversible Processes

A process is called internally reversible if no irreversibilities

  • ccur within the boundaries of the system during the process.

A process is called externally reversible if no irreversibilities

  • ccur out- side the system boundaries during the process.

A process is called totally reversible, or simply reversible, if it involves no irreversibilities within the system or its surroundings. A totally reversible process involves no heat transfer through a finite temperature difference, no non–quasi-equilibrium changes, and no friction or other dissipative effects.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-31
SLIDE 31

Carnot Cycle 1 QH TH = const. TL = const. QL 2 4 3 Wnet,out P V

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-32
SLIDE 32

Reversed Carnot Cycle

1 QH TH = c

  • n

s t . TL = c

  • n

s t . QL 4 2 3 Wnet,in P V

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-33
SLIDE 33

Carnot Principles

The second law of thermodynamics puts limits on the operation of cyclic devices as expressed by the KelvinPlanck and Clausius

  • statements. A heat engine cannot operate by exchanging heat with

a single reservoir, and a refrigerator cannot operate without a net energy input from an external source The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-34
SLIDE 34

Proof: First Carnot Principle

Irreversible HE Reversible HE (or R) Low-temperature reservoir at TL High-temperature reservoir at TH QH QH Wirrev W QL,irrev <

QL,rev

(assumed) QL,rev

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-35
SLIDE 35

Proof: First Carnot Principle

Combined HE + R Wirrev – Wrev QL,rev

– QL,irrev

(b) The equivalent combined system Low-temperature reservoir at TL

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-36
SLIDE 36

Proof: Second Carnot Principle

Low-temperature reservoir at TL = 300 K High-temperature reservoir at TH = 1000 K A reversible HE ηth,A η

th, A = ηth,B = 70%

Another reversible HE ηth,B

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-37
SLIDE 37

Thermodynamic Temperature scale

A temperature scale that is independent of the properties of the substances that are used to measure temperature is called a thermodynamic temperature scale. Energy reservoirs are characterized by their temperatures, the thermal efficiency of reversible heat engines is a function of the reservoir temperatures only. That is, ηrev,th = g(TH, TL)

  • r

QH QL = f (TH, TL)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-38
SLIDE 38

Thermodynamic Temperature scale

Engines A and C are supplied with the same amount of heat Q1 from the high-temperature reservoir at T1. Engine C rejects Q3 to the low-temperature reservoir at T3. Engine B receives the heat Q2 rejected by engine A at temperature T2 and rejects heat in the amount of Q3 to a reservoir at T3.

WA Thermal energy reservoir at T1

  • Rev. HE

A Thermal energy reservoir at T3

  • Rev. HE

B Q1 Q2 Q2 Q3 T2 WB WC

  • Rev. HE

C Q1 Q3

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-39
SLIDE 39

Thermodynamic Temperature scale

Applying the above equation to all three engines Q1 Q2 = f (T1, T2) Q2 Q3 = f (T2, T3) Q1 Q3 = f (T1, T3)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-40
SLIDE 40

Thermodynamic Temperature scale

Applying the above equation to all three engines Q1 Q2 = f (T1, T2) Q2 Q3 = f (T2, T3) Q1 Q3 = f (T1, T3) Considering the identity Q1 Q3 = Q1 Q2 · Q2 Q3

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-41
SLIDE 41

Thermodynamic Temperature scale

Applying the above equation to all three engines Q1 Q2 = f (T1, T2) Q2 Q3 = f (T2, T3) Q1 Q3 = f (T1, T3) Considering the identity Q1 Q3 = Q1 Q2 · Q2 Q3 f (T1, T3) = f (T1, T2) · f (T2, T3) A careful examination of this equation reveals that the left-hand side is a function of T1 and T3, and therefore the right-hand side must also be a function of T1 and T3 only, and not T2.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-42
SLIDE 42

Thermodynamic Temperature scale

This implies that the value of the product on the right-hand side of this equation is independent of the value of T2. This condition will be satisfied only if the function f has the following form: f (T1, T2) = φ(T1) φ(T2) and f (T2, T3) = φ(T2) φ(T3)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-43
SLIDE 43

Thermodynamic Temperature scale

This implies that the value of the product on the right-hand side of this equation is independent of the value of T2. This condition will be satisfied only if the function f has the following form: f (T1, T2) = φ(T1) φ(T2) and f (T2, T3) = φ(T2) φ(T3) Q1 Q3 = f (T1, T3) = φ(T1) φ(T3)

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-44
SLIDE 44

Thermodynamic Temperature scale

This implies that the value of the product on the right-hand side of this equation is independent of the value of T2. This condition will be satisfied only if the function f has the following form: f (T1, T2) = φ(T1) φ(T2) and f (T2, T3) = φ(T2) φ(T3) Q1 Q3 = f (T1, T3) = φ(T1) φ(T3) Lord Kelvin first proposed taking φ(T) = T to define a thermodynamic temperature scale as QH QL

  • rev

= TH TL

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-45
SLIDE 45

Carnot heat engine

The hypothetical heat engine that operates on the reversible Carnot cycle is called the Carnot heat engine. The thermal efficiency of any heat engine, reversible or irreversible, is given as ηth = 1 − QL QH

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-46
SLIDE 46

Carnot heat engine

The hypothetical heat engine that operates on the reversible Carnot cycle is called the Carnot heat engine. The thermal efficiency of any heat engine, reversible or irreversible, is given as ηth = 1 − QL QH For reversible heat engines, the heat transfer ratio in the above relation can be replaced by the ratio of the absolute temperatures

  • f the two reservoirs,Then the efficiency of a Carnot engine

becomes ηth = 1 − TL TH

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-47
SLIDE 47

Analysis of a Carnot Heat Engine

A Carnot heat engine receives 500 kJ of heat per cycle from a high-temperature source at 652C and rejects heat to a low-temperature sink at 30C. Determine (a) the thermal efficiency

  • f this Carnot engine and (b) the amount of heat rejected to the

sink per cycle.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-48
SLIDE 48

A Questionable Claim for a Refrigerator

An inventor claims to have developed a refrigerator that maintains the refrigerated space at 35F while operating in a room where the temperature is 75F and that has a COP of 13.5. Is this claim reasonable?

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

slide-49
SLIDE 49

Heating a House by a Carnot Heat Pump

A heat pump is to be used to heat a house during the winter. The house is to be maintained at 21C at all times. The house is estimated to be losing heat at a rate of 135,000 kJ/h when the

  • utside temperature drops to -5C. Determine the minimum power

required to drive this heat pump.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661