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As Per Lastest Syllabus of Anna University

FOR B.E IV SEMESTER MECHANICAL ENGINEERING STUDENTSFOR B.E IV SEMESTER MECHANICAL ENGINEERING STUDENTS

R - 2017

ENGINEERING - I

Dr. S. Ramachandran, M.E., Ph.D.,

Dr . A. Anderson, M.E., Ph.D.,

Dr. J. Jayaprabakar, M.E., Ph.D.,

Professors - Mech

Sathyabama Institute of Science and Technology

Chennai - 119

THERMAL

thFirst Edition : 16 , November 2018

300/-

978-93-88084-15-4

www.srbooks.orgwww.airwalkbooks.com

Cell: 9600003081, 9600003082

`

ISBN: 978-93-88084-15-4

R 2017 – ANNA UNIVERSITY SYLLABUS

ME8493 THERMAL ENGINEERING – I LTPC 3003

UNIT I: GAS AND STEAM POWER CYCLES 9

Air Standard Cycles – Otto, Diesel, Dual, Brayton – Cycle Analysis,Performance and Comparison-Rankine, reheat and regenerative cycle.

UNIT II: RECIPROCATING AIR COMPRESSOR 9

Classification and comparison, working principle, work of compression– with and withoutclearance, Volumetric efficiency, Isothermal efficiency andIsentropic efficiency. Multistage aircompressor with Intercooling. Workingprinciple and comparison of Rotary compressors withreciprocating aircompressors.

UNIT III: INTERNAL COMBUSTION ENGINES AND COMBUSTION 9

IC engine – Classification, working, components and their functions.Ideal and actual : Valve andport timing diagrams, p-v diagrams – two stroke& four stroke, and SI & CI engines – comparison.

Geometric, operating, and performance comparison of SI and CIengines. Desirable propertiesand qualities of fuels. Air-fuel ratio calculation– lean and rich mixtures. Combustion in SI & CIEngines – Knocking –phenomena and control.

UNIT IV: INTERNAL COMBUSTION ENGINE PERFORMANCE AND SYSTEMS 9

Performance parameters and calculations. Morse and Heat Balancetests. Multipoint Fuel Injectionsystem and Common Rail Direct lnjectionsystems. Ignition systems – Magneto, Battery andElectronic. Lubrication andCooling systems. Concepts of Supercharging and Turbocharging –EmissionNorms.

UNIT V: GAS TURBINES 9

Gas turbine cycle analysis – open and closed cycle. Performance andits improvement – Regenerative, Intercooled, Reheated cycles and theircombinations. Materials for Turbines.

TOTAL:45 PERIODS

Table of Contents

Unit I: Gas And Steam Power Cycles

1.1 Gas Power Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1

1.2 Different Types of Air Standard Cycles . . . . . . . . . . . . . . . . . . . . . . . 1.1

(i) Internal combustion Engine cycle . . . . . . . . . . . . . . . . . . . . . . . . 1.1

(ii) External combustion engine cycle . . . . . . . . . . . . . . . . . . . . . . . 1.2

1.2.1 Air Standard Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

1.3 Otto Cycle (or) Constant Volume Cycle . . . . . . . . . . . . . . . . . . . . . . 1.2

Work done in otto cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5

Mean effective pressure (M.E.P) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6

Expression For Mean Effective Pressure of Otto cycle . . . . . . . . . 1.6

1.4 Diesel Cycle or Constant Pressure Cycle . . . . . . . . . . . . . . . . . . . . . . 1.7

Air Standard Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10

Work done in Diesel cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10

Expression for Mean Effective Pressure Pm for Diesel cycle . 1.11

1.5 Dual Combustion Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.49

Total Heat Supplied Qs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.50

Total Heat Rejected Qr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.50

Total Work done: W. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.50

Efficiency of Dual Combustion Cycle: . . . . . . . . . . . . . . . . . . . . 1.50

Work done in dual cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.52

Expression for mean effective pressure Pm. . . . . . . . . . . . . . . . . 1.53

1.6 Performance And Comparison of Otto, Diesel and Dual CombustionCycles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.77

(i) Efficiency Versus Compression Ratio . . . . . . . . . . . . . . . . . . . . 1.77

(ii) Comparison of Otto, Dual and Diesel cycle on P V and T Sdiagram for constant compression ratio and heat supplied. . . . . 1.78

(iii) Comparison of Otto and Diesel cycle for same Maximumpressure and Heat supplied . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.78

Contents C.1

(iv) Comparison of Otto, Diesel and Dual cycle for same Maximumpressure and Heat rejection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.79

1.7 Gas Turbine Closed Cycle (or) Brayton Cycle (or) Joule Cycle . 1.79

(i) Air standard efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.81

1.7.1 Optimum pressure ratio for maximum specific work output 1.82

1.8 Standard Rankine Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.99

1.8.1 The Ideal Rankine Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.99

1.8.2 Efficiency of Standard Rankine Cycle . . . . . . . . . . . . . . . . 1.101

1.9 Rankine Cycle - Reheating Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . 1.114

1.9.1 Advantages (or) Effects of Re-heating . . . . . . . . . . . . . . . . 1.116

1.10 Rankine Cycle - Regenerative Cycle (Bleeding Cycle) . . . . . . . 1.129

Energy - balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.131

1.10.1 Advantages of Regenerative cycle . . . . . . . . . . . . . . . . . . . 1.132

Unit II: Reciprocating Air Compressor

2.1 Introduction to Air Compressors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1

2.1.1 Applications of Compressed Air . . . . . . . . . . . . . . . . . . . . . . . 2.1

2.2 Classification of Air Compressors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1

2.3 Single Acting Reciprocating Air Compressor. . . . . . . . . . . . . . . . . . . 2.3

2.4 Double Acting Air Compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3

2.5 Single Stage Compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4

2.6 Multi Stage Compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4

2.7 Working Principle Of Reciprocating Air Compressors . . . . . . . . . . . 2.4

2.7.1 Workdone during isothermal compression PV c withoutclearance volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6

2.7.2 Workdone during polytropic compression [ PVn constant ]without clearance volume. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7

2.7.3 Workdone During Isentropic Compression PV constantWithout Clearance Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8

2.8 Minimum Workdone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9

2.9 Power Required to Run The Compressor . . . . . . . . . . . . . . . . . . . . . . 2.9

C.2 Thermal Engineering - I - www.airwalkbooks.com

2.9.1 Clearance Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.13

2.10 Workdone by Reciprocating Air Compressor with Clearance Volume. 2.14

2.11 Isothermal Efficiency of a Reciprocating Air Compressor . . . . . . 2.15

2.12 Isentropic Efficiency of a Reciprocating Air Compressor . . . . . . 2.15

2.13 Volumetric Efficiency in a Reciprocating Air Compressor . . . . . 2.16

2.13.1 Factors affecting volumetric efficiencies . . . . . . . . . . . . . . 2.17

2.14 Important Technical Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17

1. Volumetric Efficiency vol . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.17

2. Clearance Ratio ‘k’ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.18

3. Volume and Volume Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.18

4. Indicated Power (I.P) and Brake Power (B.P) . . . . . . . . . . . . 2.18

5. Free Air Delivered (FAD) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.19

6. Mean Effective Pressure m.e.p. . . . . . . . . . . . . . . . . . . . . . . . . 2.19

7. Isothermal efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.19

8. Swept Volume Vs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.19

9. Piston speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.20

2.15 Single Stage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.20

2.16 Two Stage Compression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.61

2.16.1. Complete (or) Perfect Intercooling . . . . . . . . . . . . . . . . . . 2.61

2.16.2 Incomplete (or) Imperfect Intercooling. . . . . . . . . . . . . . . . 2.62

2.16.3 Workdone when perfect and imperfect intercooling.. . . . . 2.62

2.16.4 Minimum Work Required for 2 Stages and multi stages. 2.62

2.17 Multistage Reciprocating Compressors . . . . . . . . . . . . . . . . . . . . . . 2.63

2.17.3 Conditions for Minimum Workdone . . . . . . . . . . . . . . . . . . 2.66

2.17.4 Intermediate Pressures for Z Stage compressor Runningunder Ideal Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.67

2.17.5 Heat Rejected per stage per kg of air. . . . . . . . . . . . . . . . 2.67

2.17.6 The Change in Entropy in First Stage Compression . . . . 2.68

2.17.7 For Two Stage Compression . . . . . . . . . . . . . . . . . . . . . . . . 2.68

Contents C.3

2.18 Rotary Compressors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.103

2.18.1 Different Types of Rotary Positive Displacement Compressor 2.103

2.18.2 Different Types of Rotary Non-positive DisplacementCompressors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.103

2.19 Roots Blower . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.103

2.19.1 Back Flow of Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.103

2.20 Vane Type Blower Compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.107

2.21 Centrifugal Compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.114

2.21.1 Principle of Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.115

2.22 Velocity and Pressure Variation. . . . . . . . . . . . . . . . . . . . . . . . . . . 2.116

2.23 Axial Flow Compressors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.117

2.23.1 Working Principles of a Compressor Stage. . . . . . . . . . . 2.118

2.23.2 Stage Velocity Triangles. . . . . . . . . . . . . . . . . . . . . . . . . . . 2.119

2.23.3 Blade Loading, Flow Coefficients and Specific Work . . 2.121

2.24 Losses in Axial Flow Compressor Stage . . . . . . . . . . . . . . . . . . . 2.122

1. Profile losses on the surface of the blades:. . . . . . . . . . . . . . 2.122

2. Skin friction loss on the annulus walls: . . . . . . . . . . . . . . . . . 2.122

3. Secondary flow losses: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.123

2.25 Surging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.123

2.26 Stalling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.124

2.27 Comparison Between Reciprocating and Centrifugal Compressors . 2.125

2.28 Comparison Between Reciprocating and Rotary Air Compressors . 2.126

2.29 Comparison Between Centrifugal and Axial Flow Compressors 2.127

Unit III: Internal Combustion Engines And Combustion

3.1 IC Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1

3.1.1 Comparison between External and Internal CombustionEngines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1

3.2 Basic Terms Connected with I.C. Engines . . . . . . . . . . . . . . . . . . . . . 3.2

3.3 Classification of IC Engines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4

3.4 Four Stroke SI (Petrol) Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6


3.5 Four Stroke CI Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7

3.6 Working of Two Stroke Cycle Engine . . . . . . . . . . . . . . . . . . . . . . . . 3.9

3.7 Working of Two Stroke Petrol Engine. . . . . . . . . . . . . . . . . . . . . . . 3.10

3.8 IC Engine Components – Functions and Materials . . . . . . . . . . . . . 3.11

1. Cylinder. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.12

2. Cylinder head . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.13

3. Piston. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.13

4. Piston rings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.14

5. Gudgeon pin (Or Wrist pin or Piston pin) . . . . . . . . . . . . . . . 3.14

3. Connecting Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.14

7. Crank. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.14

8. Crankshaft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.14

9. Engine Bearings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.15

10. Crank Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.15

11. Fly wheel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.15

12. Governor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.15

13. Valves and Valve Operating Mechanisms . . . . . . . . . . . . . . . . 3.16

3.9 Ideal and Actual Cycles Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.16

3.9.1 Actual Cycle (Actual Indicator Diagram) for 4-stroke PetrolEngine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.17

3.9.1.1 Valve Timing Diagram for a 4-Stroke Cycle Petrol Engine 3.18

3.9.2 Actual Cycle for Four Stroke Cycle Diesel Engine (DieselCycle) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.20

3.9.2.2 Valve Timing Diagram of 4-Stroke Diesel Engine . . . . . 3.20

3.9.3 Port Timing Diagram of Two Stroke Cycle Petrol Engine 3.22

3.9.4 Port timing Diagram for Two Stroke Cycle Diesel Engine 3.23

3.10 Comparison of Four-stroke and Two Stroke Cycle Engines . . . . 3.25

3.10.1 Advantages of two stroke cycle engines over four stroke cycleEngines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.26

3.10.2 Disadvantages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.26

Contents C.5

3.10.3 Comparison between SI Engine and CI Engine . . . . . . . . 3.27

3.11 Geometric, Operating and Performance Comparison of SI and CIEngines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.28

3.12 Desirable Properties and Qualities of Fuels . . . . . . . . . . . . . . . . . . 3.29

3.12.1 Octane Number (ON) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.33

3.12.2 Cetane Number (CN) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.35

3.12.2.1 Fuel Rating for CI Engine . . . . . . . . . . . . . . . . . . . . . . . . 3.35

3.13 Air-Fuel Ratio Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.36

3.13.1 Stoichiometric air-fuel ratio. . . . . . . . . . . . . . . . . . . . . . . . . 3.36

3.13.2 Composition of Dry Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.36

3.13.3 Calculation of Stoichiometric air-fuel ratio . . . . . . . . . . . . 3.37

3.14 Lean and Rich Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.39

3.15 Stages of Combustion in SI Engines . . . . . . . . . . . . . . . . . . . . . . . 3.41

3.15.1 Combustion Phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.43

3.16 Normal Combustion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.43

Factors affecting normal combustion in S.I Engines . . . . . . . . . . 3.44

1. Induction pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.44

2. Engine speed. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.44

3. Ignition timing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.44

4. Fuel choice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.44

5. Combustion chamber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.45

6. Compression ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.45

7. Mixture strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.45

3.17 Combustion Chamber for SI Engines . . . . . . . . . . . . . . . . . . . . . . . 3.45

3.17.1 Types of combustion chambers . . . . . . . . . . . . . . . . . . . . . . 3.46

1. Overhead valve (or) I - Head combustion chamber . . . . . . . . 3.46

2. T - Head combustion chamber. . . . . . . . . . . . . . . . . . . . . . . . . . 3.47

3. L-head combustion chamber . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.47

3.18 Stages of Combustion in CI Engines . . . . . . . . . . . . . . . . . . . . . . . 3.48

1. Ignition Delay period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.49


(i) Physical delay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.50

(ii) Chemical delay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.50

Factors Affecting Delay Period. . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.50

2. Period of Rapid Combustion (or) Uncontrolled Combustion . 3.51

3.19 Factors That Affect Delay Period in Diesel Engine . . . . . . . . . . . 3.51

1. Compression ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.51

2. Intake temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.51

3. Intake pressure (super charging) . . . . . . . . . . . . . . . . . . . . . . . . 3.52

4. Engine size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.52

5. Fuel temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.52

3.19.1 Effect of variables on the Delay period . . . . . . . . . . . . . . 3.52

3.20 Need for Air Motion in Diesel Engine . . . . . . . . . . . . . . . . . . . . . 3.53

3.21 Combustion Chamber Design For Compression Ignition Engine. 3.54

3.21.1 Open combustion chamber. . . . . . . . . . . . . . . . . . . . . . . . . . 3.54

3.21.2 Divided combustion chamber (or) Indirect combustionchamber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.56

(a) Turbulent combustion chamber . . . . . . . . . . . . . . . . . . . . . . . . . 3.56

(b) Pre combustion chamber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.57

(c) Energy-cell chamber. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.59

3.21.3 Open and Divided combustion chambers. . . . . . . . . . . . . . 3.60

3.22 Abnormal Combustion in SI Engines . . . . . . . . . . . . . . . . . . . . . . . 3.61

1. Pre-ignition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.61

2. Knocking (or) Detonation (or) Pinking . . . . . . . . . . . . . . . . . . . 3.62

3.22.1 Theories of Detonation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.63

3.22.2 Effects of Detonation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.63

3.22.3 Control of detonation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.63

3.22.4. Effects of Engine variable on Detonation/Knocks . . . . . . 3.64

3.23 The Phenomenon of Knock in SI Engine . . . . . . . . . . . . . . . . . . . 3.64

3.23.1 Effects of knocking in SI Engine . . . . . . . . . . . . . . . . . . . . 3.66

3.23.2 Factors Affecting Knock in SI Engines . . . . . . . . . . . . . . . 3.67

Contents C.7

3.23.3 Anti-Knock Additives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.69

3.23.4 Factors Affecting Detonation and Remedies . . . . . . . . . . . 3.70

3.24 Knocking in CI Engine (or) Diesel Knock . . . . . . . . . . . . . . . . . . 3.70

3.24.1 Phenomenon of knock in CI engine . . . . . . . . . . . . . . . . . . 3.71

3.24.2 Comparison of knock on SI and CI Engines . . . . . . . . . . 3.73

3.24.3 Characteristics Tending to Reduce Detonation . . . . . . . . . 3.74

Unit IV: Internal Combustion Engine Performance and Systems

4.1 Performance Testing of I.C. Engines . . . . . . . . . . . . . . . . . . . . . . . . . 4.1

1. Indicator Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1

2. Indicated Power (IP) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2

3. Mean Effective Pressure Pm. . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2

4. Measurement of Cylinder Pressure . . . . . . . . . . . . . . . . . . . . . . . 4.2

4.1.1 Transducer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3

5. Brake Power (BP) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5

6. Retardation Test - Different Arrangements used to find BrakePower . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5

7. Friction Power (FP) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7

8. Specific Fuel Consumption (S.F.C) in kg/kW hr . . . . . . . . . . . . 4.7

9. Mechanical Efficiency mech . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8

10. Thermal Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8

11. Indicated Thermal Efficiency indicated . . . . . . . . . . . . . . . . . . 4.8

12. Brake Thermal Efficiency Brake . . . . . . . . . . . . . . . . . . . . . . . 4.8

13. Relative Efficiency (or) Efficiency Ratio . . . . . . . . . . . . . . . . . . 4.9

14. Volumetric Efficiency volumetric . . . . . . . . . . . . . . . . . . . . . . . 4.9

4.2 Retardation Test Using Dynamometer. . . . . . . . . . . . . . . . . . . . . . . . . 4.9

4.2.1 Hydraulic Dynamometer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.10

4.2.2 Eddy Current Dynamometer . . . . . . . . . . . . . . . . . . . . . . . . . 4.12

4.3 Performance Curves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.12

Problems on Performance Calculations of IC Engines . . . . . . . . 4.13


4.4 Morse Test - Measurement of Indicated Power of MulticylinderEngine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.16

4.5 Measurement of Air Consumption. . . . . . . . . . . . . . . . . . . . . . . . . . . 4.25

4.5.1 Problem in measurement of air supplied. . . . . . . . . . . . . . . 4.27

4.5.2 Fuel consumption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.28

4.5.2.1 Volumetric Flow. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.28

Burette method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.28

4.5.2.2 Gravimetric fuel flow measurement . . . . . . . . . . . . . . . . . . 4.31

4.6 Measurement of Heat Carried Away by Cooling Water . . . . . . . . 4.32

4.7 Measurement of Heat Carried Away by Exhaust Gas . . . . . . . . . . 4.33

4.8 Heat Balance Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.33

4.9 Petrol Fuel Injection System (Gasoline Injection System) . . . . . . . 4.67

4.9.1 Monopoint Fuel Injection System . . . . . . . . . . . . . . . . . . . . . 4.69

4.10 (Multiport Fuel - Injection System) Multi Point Fuel InjectionSystem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.70

1. D-MPFI system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.72

2. L-MPFI system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.73

Comparison between petrol injection & carbureted fuel supplySystem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.74

4.11 Common Rail Direct Injection (CRDI) System . . . . . . . . . . . . . . . 4.75

4.12 Ignition Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.78

4.12.1 Spark Plug . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.78

4.13 Battery or Coil Ignition System. . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.80

4.14 Magneto Ignition System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.82

4.15 Electronic Ignition Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.84

4.15.1 Types of Electronic ignition systems: . . . . . . . . . . . . . . . . . 4.85

1. Capacitance Discharge Ignition System . . . . . . . . . . . . . . . . . . 4.85

2. Transistorized Assisted Contact (TAC) Ignition System. . . . . . 4.86

4.16 Lubrication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.88

4.16.1 Functions of lubrication . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.88

Contents C.9

4.16.2 Lubrication systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.89

1. Mist lubrication (Petroil system) . . . . . . . . . . . . . . . . . . . . . . . . 4.89

2. Wet Sump Lubrication System . . . . . . . . . . . . . . . . . . . . . . . . . . 4.90

(i) Splash system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.90

(ii) Pressure-feed system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.92

(iii) Splash and Pressure system. . . . . . . . . . . . . . . . . . . . . . . . . . . 4.92

3. Dry Sump Lubrication System . . . . . . . . . . . . . . . . . . . . . . . . . . 4.93

4.17 Cooling Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.94

4.17.1 Functions of Cooling Systems . . . . . . . . . . . . . . . . . . . . . . . 4.95

4.17.2 Types of Cooling System . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.95

1. Air-cooled System. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.95

(i) Cooling Fins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.95

(ii) Baffles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.96

2. Liquid (or) Water Cooled Systems. . . . . . . . . . . . . . . . . . . . . . . 4.98

(i) Direct or non-return system. . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.98

(ii) Thermosyphon or Natural Circulation System . . . . . . . . . . . . 4.99

(iii) Forced or Pump Circulation System . . . . . . . . . . . . . . . . . . . 4.99

4.17.3 Comparison of Liquid and Air-cooling System . . . . . . . . 4.103

4.18 Scavenging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.104

Cross Flow Scavenging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.105

Loop or Back Flow Scavenging . . . . . . . . . . . . . . . . . . . . . . . . . . 4.105

Uniflow Scavenging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.105

4.19 Supercharging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.105

4.19.1 Supercharger: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.106

4.19.2 Supercharging in SI system . . . . . . . . . . . . . . . . . . . . . . . . 4.108

4.19.3 Supercharging of C.I engines . . . . . . . . . . . . . . . . . . . . . . 4.109

4.19.4 Effects of supercharging on performance of the engine. 4.111

4.20 Turbocharging. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.111

4.20.1 Turbolag. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.113

4.21 Emission Norms (Euro and BS) . . . . . . . . . . . . . . . . . . . . . . . . . . 4.113


4.21.1 Euro norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.113

4.21.2 BS Norms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.114

Parameters determining Emission from vehicles. . . . . . . . . . . . . 4.114

Unit V: Gas Turbine

5.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1

5.2 Simple Gas Turbine Plant. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1

5.3 Components of Gas Turbine Plant . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2

5.4 Parameter of Performance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2

5.5 Classification of Gas Turbine Plants . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3

5.6 Gas Turbine Cycle Analysis (Brayton Cycle) . . . . . . . . . . . . . . . . . . 5.3

5.6.1 Gas turbine Open cycle and closed cycle . . . . . . . . . . . . . . . 5.3

5.7 Performance and its Improvements . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5

5.8 Regeneration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5

5.9 Reheating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6

5.10 Intercooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7

5.11 Combined Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8

1. The intercooled cycle with regeneration . . . . . . . . . . . . . . . . . . . 5.8

2. The intercooled cycle with reheat cycle . . . . . . . . . . . . . . . . . . . 5.9

3. Regeneration and Reheated Cycle: . . . . . . . . . . . . . . . . . . . . . . 5.10

4. The Brayton cycle with intercooling, reheating and regeneration 5.11

5.12 Improvements of Basic Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.28

(a) After-burning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.28

(b) Injecting Refrigerants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.28

(c) Increasing the turbine inlet (or firing) temperatures . . . . . . . 5.28

5.13 Materials for Turbines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.29

5.13.1 Turbine blade materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.30

1. Titanium alloys: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.30

2. Nickel alloys: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.31

5.13.2 Factors to be considered in the selection of materials . . 5.32

Contents C.11

UNIT I

GAS AND STEAM POWER CYCLES

Air Standard Cycles – Otto, Diesel, Dual, Brayton – Cycle Analysis,Performance and Comparison – Rankine, Reheat and regenerative cycle.

1.1 GAS POWER CYCLESA cycle is defined as a repeated series of operations occurring in a

certain order as shown in Fig (1.1). It may be repeated by repeating theprocesses in the same order. The cycle may be of imaginary perfect engineor actual engine. The former is called ideal cycle and latter is actual cycle.

The method of producing mechanical power by transferring heat froma reservoir to a working fluid is done through a thermodynamic cycle. A partof heat received by fluid is rejected to the sink. These cycles are termed aspower cycles. Power cycles are classified as (i) Vapour power cycles (ii) Gaspower cycles.

In vapour power cycles, the working fluid is steam. In gas power cyclesthe working fluid is a gas. The gas is normally air or the products ofcombustion of fuel and air. In gas power cycles the supply of heat is bymeans of the combustion of a suitable mixture of air and fuel within theconfines of the engines called I.C engines. (Internal combustion engines).

1.2 DIFFERENT TYPES OF AIR STANDARD CYCLES

Air standard cycles are classified as follows

(i) Internal combustion Engine cycleCombustion (Burning of fuel) takes place inside the cylinder

Example: Petrol engine (otto cycle), Diesel engine (Diesel cycle), Dual cycle

CYCLECYCLE

1 2

3 1

2

3

4

Fig. 1.1 Cycle

(ii) External combustion engine cycleCombustion (Burning of fuel) takes place outside the cylinder.

Example: Gas turbine cycle (or) Joule cycle (or) Brayton cycle.

1.2.1 Air Standard Efficiency

What is an air standard efficiency? (AU. May/June 2014)The efficiency of engine using air as a working medium is known as

“Air standard efficiency”.

To compare the effects of diferent cycles, the calorific value of fuelsshould be neglected. To acheive this, air is considered as a working substancein the engine cylinder and the efficiency obtained by using air as workingmedium is called air standard efficiency or ideal efficiency.

Assumptions made in the Analysis of air standard efficiency

State the Assumptions made in air standard cycle analysis?(AU. May/June 2016)

1. The gas in the engine cylinder is a perfect gas i.e., it obeys the gaslaws and has constant specific heats.

2. The physical constants of the gas in the cylinder are same as thoseof air at moderate temperatures.

3. The compression and expansion processes are adiabatic and they takeplace without internal friction i.e. processes are isentropic

4. No chemical reaction takes place in the cylinder

5. The cycle is considered as closed with the same ‘air’ alwaysremaining in the cylinder to repeat the cycle.

1.3 OTTO CYCLE (OR) CONSTANT VOLUME CYCLEThe present day petrol engine operates on the principle of otto cycle.

This cycle was introduced in practical form by scientist Otto in 1876. AnI.C. Engine does not undergo a cycle change, but it is assumed here that theworking medium is pure air which does not undergo any chemical change.The air is simply heated and cooled to undergo a cycle. The working fluidis also assumed to be a perfect gas (having constant specific heats). Furtherit is assumed that the ideal indicator diagram is strictly followed.

The cycle consists of two reversible adiabatic (isentropic) processesand two constant volume processes.

The P-V and T-S diagrams are shown in Fig (1.2)

The otto cycle consists of

1.2 Thermal Engineering - I - www.airwalkbooks.com

Process 1-2: Isentropic compression Q 0

Air is compressed isentropically.

Process 2-3: Constant volume heat addition [Qs mCv T3 T2]

Heat is added (spark ignition takes place) while volume remainsconstant.

Process 3-4: Isentropic Expansion Q 0

Expansion of air takes place isentropically

Process 4-1: Constant volume heat rejection [Qr mCv T4 T1]

Heat is rejected to the surroundings at constant volume.

Net work done Wnet Qs Qr Heat supplied Heat rejected

air standard thermal Wnet

Qs

Qs Qr

Qs

1 Qr

Qs 1

mCv T4 T1mCv T3 T2

1 T4 T1

T3 T2 ... (1)

Comparison ratio, r Volume at the beginning of compression

Volume at the end of compression

V1

V2

V4

V3

P

v

pv =C

W E

W E

W CW C

Q 1

Q 2

1

4

3

2s=c

s=c

s

2 4

3

Q 1

Q 2

v=c

v=c

(a ) (b )

O TTO CY CLE

Fig 1.2 O tto cyclev = v2 3 v = v1 4

O TTO CY CLE

www.srbooks.org Gas and Steam Power Cycles 1.3

1-2 Isentropic process:

T2

T1 V1

V2

1

r 1

3-4 Isentropic process

T3

T4 V4

V3

1

V1

V2

1

r 1 . . . V1 V4

V2 V3

T2

T1

T3

T4

T3

T2

T4

T1

T3

T2 1

T4

T1 1

T3 T2

T2

T4 T1

T1

T4 T1

T3 T2

T1

T2 V2

V1

1

Substitute T4 T1

T3 T2 V2

V1

1

in equation 1, we get

1 V2

V1

1

1 1

r 1

otto 1 1

r 1 [Note: r

V1

V2

v1

v2]

This expression in which 1.4 is called air standard efficiency ofthe Otto cycle.

How does the change in compression ratio affect air standard efficiency ofan ideal otto cycle. (AU. Apr/May 2008)

So the efficiency is the function of compression ratio only. If thecompression ratio is more, then efficiency will be more. However, we cannotincrease the compression ratio beyond the limit since detonation (noisy anddestructive combustion problem) occurs in petrol engine at higher compressionratio.


Work done in otto cycleWe know that the work done per kg in otto cycle is given by

W P3V3 P4V4

1

P2V2 P1V1

1 ...(2)

Also P3P4

P2

P1 r

...(3)

Also, P3P2

P4

P1 rP Pressure ratio

Also we know V1 rV2 and V4 rV3 . . .

V1

V2

V4

V3 r

substituting these in (2) we have

W 1

1 P4V4

P3V3P4V4

1 P1V1

P2V2

P1V1 1

1

1 P4V4

P3

P4r 1

P1V1

P2

P1r 1

. . .

V1

V2

V4

V3 r

V1

1 P4

r

r 1

P1

r

r 1

. . . from 3

P3

P4

P2

P1 r

V1

1 [P4 r 1 1] P1 r 1 1

V1

1 [r 1 1 P4 P1]

On multiplying throughout by P1, we get

Work done W P1V1

1 [r 1 1 rP 1]

...(4) . . .

P4

P1 rP


Mean effective pressure (M.E.P)Define mean effective pressure. Show that on a P.V diagram

(AU. May/June 2016, FAQ)Mean effective pressure is defined as the constant pressure to which

the cylinder is subjected to, so that the same work is developed.

Expression For Mean Effective Pressure of Otto cycleMean effective pressure is the ratio of net work done to the swept

volume. Therefore

Mean effective pressure (M.E.P) Pm Net work doneSwept volume

W

V1 V2

From (2) we have Pm

P3V3 P4V4

1

P2V2 P1V1

1V1 V2

Now from (4) we have Pm

P1V1

1 [r 1 1 rP 1]

V1 V2 ...(5)

Now V1 V2 V1 V1

r . . .

V1

V2 r

V1 V2 V1 r 1

r ...(6)

P

v

pv =C

1

4

3

2

s=c

s=c

=

= W in H atched area

W in U nhatched a rea

Fig 1.3


Substituting (6) in (5) we get

Pm P1V1

1 [r 1 1 rP 1]

1

V1 r 1

r

Therefore we get,

Mean effective pressure

Pm P1r [r 1 1 rP 1]

1 r 1

P1 rP 1 r rr 1 1

1.4 DIESEL CYCLE OR CONSTANT PRESSURE CYCLE

Derive an expression for air standard efficiency of diesel cycle and thendeduce it for mean effective pressure. (AU. Nov/Dec 2010)

Dr. Rudholph Diesel introduced diesel cycle in 1897. In case of ottocycle, heat is supplied to working fluid at constant volume, while in case ofDiesel cycle heat is supplied at constant pressure. Diesel cycle consists of thefollowing four operations.

Mention the Four thermodynamic processes involved in Diesel cycle?(AU. Apr/May 2008)

Process 1 2: Adiabatic compression

Process 2 3: Constant pressure heat addition

Process 3 4: Adiabatic expansion

Process 4 1: Heat rejection at constant volume

The limitation on compression ratio in the S.I. engine can be overcomeby compressing air alone, instead of the fuel-air mixture, and then injecting fuelinto the cylinder in spray form when combustion is desired. The temperature ofair after compression must be high enough so that the fuel sprayed into the hotair burns spontaneously. The rate of burning can, to some extent, be controlledby the rate of injection of fuel. An engine operating in this way is called acompression ignition (C.I.) engine. The sequence of processes in C.I. engine is,shown in Fig. 1.4.

Process 1-2, Intake: The air valve is open. the piston moves out admittingair into the cylinder at constant pressure.

Process 2-3, Compression: The air is then compressed by the piston to theminimum volume with all valves closed.


Process 3-4, Fuel injection and combustion: The fuel valve is open, fuelis sprayed into the hot air, and combustion take place at constant pressure.

Process 4-5, Expansion: The combustion products expand, by doing workon the piston which moves out the piston to the maximum volume.

Process 5-6, Blow-down: The exhaust valve opens, and the pressure dropsto the initial pressure.

Pressure 6-1, Exhaust: With the exhaust valve open, the piston movestowards the cylinder cover, driving away the combustion products from thecylinder at constant pressure.

The above processes constitute an engine cycle, which is completed infour strokes of the piston or two revolutions of the crank shaft.

Fig. 1.5 shows the air standard cycle, called the Diesel cycle,corresponding to the C.I. engine, as described above. The cycle is composed of:

P

v

12 ,6

5

43

(b)(a)

Fuel valve

A ir(in le t) va lve

Exhaust va lve

CI Eng ine Indicator d iagramFig 1.4

1 1

4

4

33

2

2

W cW c

W E

W E

pv =cg

Q2 Q 2

Q 1

Q 1

v=c

v=c

p=c

s

T

v

P

(a ) (b )Diesel cycleFig 1.5


Two reversible adiabatic, one reversible isobar, and one reversibleisochore.

Air is compressed reversibly and adiabatically in process 1 2. Heatis then added to it from an external source reversibly at constant pressure inprocess 2 3. Air then expands reversibly and adiabatically in process

3 4. Heat is rejected reversibly at constant volume in process 4 1, and thecycle repeats itself.

For m kg of air in the cylinder, the efficiency analysis of the cyclecan be made as given below.

Heat supplied Q1 Q2 3 m Cp T3 T2

Heat rejected Q2 Q4 1 m Cv T4 T1

Efficiency 1 Q2

Q1 1

m Cv T4 T1

m Cp T3 T2

1 T4 T1

T3 T2 ...(1)

Let compression ratio, r V1

V2

v1

v2;

cutoff ratio V3

V2

v3

v2

Now, Process 12 compression

T2

T1 V1

V2

1

r 1 or T2 T1 r 1

Process 2-3 (constant pressure)T3

T2

V3

V2 or T3 T2 T1 r 1

Process 3-4 (Expansion)

T3

T4 V4

V3

1

r

1

. . .

V4

V3

V1

V3

V1V2

V2V3

r


Now T4 T3

r/ 1

T1 r 1

r/ 1 T1

Substituting valves of T2, T3, T4 in (1) we get

diesel 1

T1 T1

T1 r 1 T1 r 1

Air Standard Efficiency

diesel 1

1

r 1 1 1

1

r 1 1 1

...(2)

As 1, 1

1 1

is also greater than unity. Therefore, the

efficiency of the Diesel cycle is less than that of the Otto cycle for the samecompression ratio.

Work done in Diesel cycleThe net work done in diesel cycle in terms of P, V is given as follows:

The work done

W P2 V3 V2 P3V3 P4V4

1

P2V2 P1V1

1

P2 V2 V2 P3V2 P4rV2

1

P2V2 P1rV2

1

. . .

V3

V2 ; V3 V2 ;

V1

V2 r ; V1 rV2 ; V1 V4 V4 rV2

W P2V2 1 P3V2 P4rV2

1

P2V2 P1rV2

1

V2

1 [P2 1 1 P3 P4r P2 P1r]

W P2V2

1 1 1

P4

P3 r 1

P1

P2 r

[. . . P3 P2]


P2V2

1 [ 1 1 r1 1 r1 ]

. . .

P4

P3 V3

V4

r

r r

W P1V1r 1

1 [ 1 1 r1 1 r1 ]

. . .

P2

P1 V1

V2

or P2 P1 r and V1

V2 r or V2 V1r 1

Work done W P1V1r 1

1 [ 1 r1 [ 1]]

...(3)

Expression for Mean Effective Pressure Pm for Diesel cycle

We know that Mean Effective Pressure Pm is given by

Pm Net work doneSwept volume

W

V1 V2 ...(4)

V1 V2 V1 1

1r V1

r 1

r ...(5)

Substituting (3), (5) in (4) we get

Pm P1V1r 1

1 V1 r 1

r

[ 1 r1 [ 1]]


Pm P1r

1 r 1 [ 1 r1 1]


Problem 1.1: In an air standard otto cycle, the condition of air at beginningof compression is 1 bar, 300 K, compression ratio 6, the maximumtemperature is 1000C. Determine Thermal efficiency th

Solution:

Beginning ofCompression

P1 1 bar

T1 300 K

th 1 1

r 1 1

1

60.4 0.51164 or 51.16% [. . . 1.4]

Problem 1.2: Calculate the ideal air standard cycle efficiency based on theOtto cycle for a petrol engine with a cylinder bore of 50 mm, a stroke of

75 mm and a clearance volume of 21.3 cm3 (AU. Apr/May 2008)

SolutionD 50 mm; L 75 mm

otto 1 1

r 1

r compression ratio Vs Vc

Vc

Compressor ratio r 6

Tmax T3 1000C 1273 K

where Vs Swept volume and

Vc Clearance volume 21.3 cm3

P

v

Pv =CW E

W E

W CW C

Q 1

Q 21

4

3

2s=c

s=c

s

2 4

3

Q 1

Q 2

v=c

v=c

(a) (b)

OTTO CYCLE

v2 = v3 v1 = v4


Vs 4

D2 L 4

0.052 0.075

Vs 1.473 10 4 m3

Vc 21.3 10 6 m3

r Vs Vc

Vc

1.473 10 4 21.3 10 6

21.3 10 6

r 7.92

otto 1 1

r 1 1

1

7.920.4 0.5629

otto 56.29%

Problem 1.3: In an air standard otto cycle the condition of air at thebeginning of compression is 1 bar, 300 K. The temperature at the beginning,end of burning are 400C, 1000C. Determine (a) compression ratio (b)

Thermal efficiency th.

Solution:

Beginning ofcompression

P1 1 bar

T1 300 K

P

v

pv =C

W EW E

W CW C

Q 1

Q 21

4

3

2s=c

s=c

s

2 4

3

Q 1

Q 2

v=c

V=c

(a) (b)v2 = v3 v1 = v4


Beginning, endof burning

T2 400C 673 K

T3 1000C 1273 K

1 2 isentepic

T2

T1 V1V2

1

673300

1 1.4 1

V1

V2

r V1

V2

v1v2

7.53762

(a) Compression ratio r 7.53762

(b) Thermal efficiencyth 1 1

r 1 1

1

7.537021.4 1 1 0.44576

th 0.55423 or 55.423%

Problem 1.4: In an air standard otto cycle the pressure at the beginningand end of compression are 1 bar, 15 bar. The maximum pressure is 30 bar.Determine (a) Compression ratio (b) Thermal efficiency (c) MEP

Otto cycle

P

v

pv =CgW E

W E

W CW C

Q 1

Q 21

4

3

2s=c

s=c

s

2 4

3

Q 1

Q 2

V=c

V=c

(a ) (b )v2 = v3 v1 = v4


Solution:

Begining and endcompression

P1 1 bar

P2 15 bar

Maximum pressure P3 30 bar

(1) - (2) isentropic process

P2

P1 V1

V2

151

V1

V2

1.4

151 1.4

V1

V2

(i) V1

V2 r 6.919348 (Compression ratio)

(ii) Thermal efficiency

th 1 1

r 1 1

1

6.9193481.4. 1 0.5387

th 53.87%

(iii) Mean effective pressure MEP

Pm P1 rp 1

r 1 1 r

r

rp P3

P2

3015

2

MEP 1 2 1

6.919348 1 1.4 1 [6.9193481.4 6.919348]

MEP 3.41281 bar

Problem 1.5: The condition of air at the beginning of compression of anotto cycle are 1 bar, 300 K. The max. temperature is 1400 K. The exhausttemperature is 700 K. Determine (a) compression ratio (b) max. pressure(c) th (Thermal efficiency)

Solution:

Beginning of compression P1 1 bar; T1 300K;


Tmax T3 1400 K ; T4 700 K

T2, P2, r are not given. so we cannot go by the usual method

P4 V4 R T4

P1 V1 R T1 [. . . ideal gas]

P1V1 R T1 ideal gas

V1 V4

P4

P1

T4T1

P4

1

700300

P4 2.33333 bar

P1V1 R T1

V1 R T1

P1

287 300

1 105 0.861 m3/kg

V1 0.861 m3/kg

Isentropic Process

T3

T4 V4

V3

1

1400700

1/0.4

V4

V3

V4

V3 5.65685

0.861V3

V3 0.15220 m3/kg.

V2 V3

r V1

V2

0.8610.15220

5.65685

P

v

Pv =CW E

W C

Q 1

Q 21

4

3

2s=c

s=c

OTTO CYCLE

v2 = v3 v1 = v4

W E

W C

s

2 4

3

Q 1

v=c

v=c

OTTO CYCLE


(i) Compression ratio r 5.65685


th 1 1

r 1 1

1

5.656850.4 0.5 50%

Pmax P3 ; P3V3 R T3 P3 R T3

V3

287 14000.15220

P3 26.39947 bar

P2

P1 V1

V2

0.8610.15220

1.4

P2 11.3137 bar

rp P3P2

26.3994711.3137

2.33333

rp 2.33333

(iii) Mean Efffective Pressure (MEP)

MEP P1 r r rp 1 1 r 1

1 5.656851.4 5.65685 2.33333 1

1.4 1 5.65685 1

MEP 4.04912 bar

Problem 1.6: The condition of air at the beginning of compression of andiesel cycle are 1 bar, 300 K. The max. temperature is 1400 K. The exhausttemperature is 700 K. Determine (a) compression ratio (b) maximum pressure(c) th (d) MEP.

Solution

Beginning of compression P1 1bar, T1 300K ; T3 Tmax 1400K

T4 Exhaust. temperature 700K ; T2, P2 r are not given.

so we cannot go by the usual method

P4 V4 R T4 1

P1V1 R T1 2

P4P1

T4

T1


P4P1

700300

P4 2.33333 bar

P1V1 R T1

V1 R T1

P1

287 300

1 105 0.861 m3/kg

V1 V4 0.861 m3/kg

Process 3 4 Isentropic

T3T4

V4

V3

1

T3

T4

1/ 1

V4

V3

1400700

11.4 1

0.861

V3

V3 0.15220 m3/kg

P3 R T3

V3

287 14000.15220

P3 26.39865 bar

P3 P2 Pmax 26.39865 bar

Process 1 2 Isentropic

P2

P1 V1

V2

P2P1

1/

V1

V2 26.39865

1

1/1.4

V1

V2

(i) Compression ratio

r V1

V2 10.36133

1

4

3

2

W c

W E

Q 2

Q 1

v=c

v=c

p=c

s

T

Diesel cycle

Diesel cycle

1

4

32

W c

W E Pv =c

Q 2

Q 1

v

P


P2

P1

1

T2

T1

T2

300 26.39865

1

0.41.4

T2 764.341534 K

(ii) Thermal efficiency th

th 1 T4 T1 T3 T2

1 700 300

1.4 1400 764.341534 0.550522 55.052%

Qs CpT3 T2 1005 1400 764.341534 638836.7583 J/kg

W Qsth 638836.7583 0.550522

W 351693.6899 J/kg

(iii) Mean effective pressure

MEP WVs

W

V1 V2

351693.68990.861 V2

r V1

V2 V2 V1/r 0.083 m3/kg

351693.68990.861 0.083

452048.4446 Pa

MEP 4.5205 bar Mean effective pressureProblem 1.7: For air standard diesel cycle the following data is availableCompression ratio 16

Heat added/kg 2500 kJ/kg

Lowest temperature in the cycle 300 K

Lowest pressure in the cycle 1 barCalculate:1. Pressure and temperature at each point in the cycle2. Thermal efficiency3. Mean effective pressure if air flow rate of 0.25 kg/secAssume Cp 1 kJ/kg K and Cv 0.714 kJ/kg K. (AU. April/May 2011)


Compression ratio r 16

P1 V1 mRT1

V1 mRT1

P1

1 0.287 300 103

1 105

V1 0.861 m3

r V1

V2

0.861V2

V2 0.861

16 0.0538 m3

T2 T1 Rc 1 300 161.4 1 909.43 K

P2 P1 Rc 1 161.4 48.5 bar.

P2 P3 48.5 bar

Heat supplied Qs CP T3 T2

2500 1 T3 T2


T3 2500 909.45

T3 3409.43 K

3409.43909.43

3.75

3.75

V3 V2 T3T2

0.0538 3409.43909.43

V3 0.202 m3

Diesel 1 1

r 1

1

1 1

1 1

161.4 1

11.4

3.751.4 13.75 1

Diesel 54.1 %

Expansion Ratio

Re r

16

3.675 4.27

T4 T3

Re 1

3409

4.271.4 1 1907.70 K

T4 1907.70 K

P4 48.5

4.271.4 6.36 bar.

P4 6.36 bar


Heat Rejected QR Cv T4 T1

0.714 1907.70 300

QR 1147.89 kJ/kg

Wnet QS QR 2500 1147.89

Wnet 1352.11 kJ/kg

Thermal efficiency th

t Wnet

Heat supplied

1352.112500

54.08

t 54.1%

Mean effective pressure When the air flow rate is 0.25 kg/sec is given

Mep Wnet

V1 V2

1352.110.861 0.0538

Mep 1675 kPa Or 16.75 bar

Problem 1.8: In an air standard diesel cycle the condition of air at thebeginning of compression is 1 bar, 300 K. Compression ratio is 15. The heatsupplied is 800 kJ/kg. Determine (a) P, V, T at all points (b) th (c) MEP

(AU. Nov/Dec 2013)(FAQ)

Solution:

Beginning of compression P1 1bar; T1 300K

r 15, Heat supplied ; Qs 800 kJ/kg 800 103 J/kg

P1V1 R T1

V1 287 300

1 105 0.861 m3/kg

V1 V4 0.861 m3/kg.


V1

V2 r

0.861V2

15

V2 0.0574 m3/kg

Process 1 2 isentropic

T2

T1 V1

V2

1

T2

300 150.4

T2 886.25308 K

P2

P1 V1

V2

P2

1 bar 151.4

P2 44.31265 bar

(or) P2V2 R T2

P2 287 886.2508

0.0574

P3 P2 44.31265 bar

Qs cp T3 T2

800 103 1005 T3 886.25308

T3 1682.669207 K

P3 V3 R T3

V3 R T3

P3

287 1682.6692

44.31265 105

V3 0.10898 m3/kg

V1 V4 0.861 m3/kg

1

4

3

2

W c

WE

Q 2

Q 1

v=c

v=c

P=c

s

T

Diesel cycle

1

4

32

W c

W E PV =C

Q 2

Q 1

v

P

v 3 - v2

Fuel cut o ff volum e

Diesel cycle



T4

T3 V3

V4

1

T4

1682.6692 0.108980.861

0.4

T4 736.09896 K

Heat rejected

Qr Cv T4 T1 718 736.09896 300 312.901 103 J/kg

(ii) Thermal efficiency th

th 1 Qr

Qs 1

313 103

800 103

th 0.60875

W Qs th 800 103 0.60875

Work done W 487 103 J/kg

(iii) Mean efficiency pressure (MEP)

MEP W

V1 V2

487 103

0.861 0.0574 606022.897 Pa

MEP 6.06022 bar

Problem 1.9: For air standard Otto cycle the following data is available:Compression ratio 9

Heat added/kg 1200 kJ

Lowest temperature in the cycle 300 K

Lowest pressure in the cycle 1 barCalculate1. Pressure and temperature at each point in the cycle2. Thermal efficiency3. Mean effective pressure if air flow rate of 0.25 kg/secAssume Cp 1 kJ/kg K and Cv 0.714 kJ/kg K. (AU. Apr/May 2011)

P4 R T4

V4

P4 287 736.09896

0.861P4 2.45366 bar


V1 RT1

P1

0.287 103 300

1 105 0.861 m3/kg

V2 V1

Rc

0.8619

0.096 m3/kg

T2 T1 Rc 1 300 91.4 1 722.47 K

P2 P1 Rc 1 91.4 21.67 bar

Heat supplied Qs CV T3 T2

1200 0.714 T3 722.47

T3 2403.14 kJ/kg

T4 T3

Rc 1

2403.14

91.4 1 997.88 K

P4 P3

Rc ? P3 P2

T3

T2

2403.14722.47

21.67

P4 72

91.4 3.32 bar

P3 72 bar


Thermal efficiency t

t 1 1

Rc 1

1 1

91.4 1 0.5847

t 58.47%

Mean effective pressure Pm

Heat Reheated QR CV T4 T1

0.714 997.88 300QR 498.29 kJ/kg

Wnet QS QR 1200 498.29 701.71 kJ/kg

Pm Wnet

V1 V2

701.72 0.25

0.861 0.096 0.25

Pm 917.28 kN/m2 9.17 bar.

Problem 1.10: In an air standard otto cycle the condition of air at thebeginning of compression is 1 bar, 300 K. Compression ratio is 7. The heatsupplied is 800 kJ/kg. Determine (a) P, V, T at all points (b) th (c) MEP

(AU. April/May 2011)(FAQ)

Solution:

Beginning of compression P1 1 bar; T1 300K

r 7, Qs 800 103 kJ/kg

P1V1 R T1

V1 287 300

1 105 0.861 m3/kg

V1 0.861 m3/kg.

V1 V4 0.861 m3/kg.

V1

V2 r

0.861V2

7 0.123 m3/kg

V2 V3 0.123 m3/kg



T2

T1 V1

V2

1

T2

300 0.8610.123

0.4

T2 653.37192 K

P2V2 R T2

P2 287 653.37192

0.123

P2 15.24534 bar

Heat supplied Qs Cv T3 T2

800 103 718 T3 653.37192

T3 1767.578048 K

P3 V3 R T3

P3 287 T3

0.123 P3

287 1767.5780480.123

41.2435 bar

T4

T3 V3V4

1

T4

1768.3545

0.1230.861

0.4

T4 811.9515 K

P4 V4 R T4 P4 287 811.9515

0.861 2.7065 bar

P4 2.7065 bar

Heat rejected Qr Cv T4 T1 718 811.952 300

Qr 367.582 103 J/kg

(b) Thermal Efficiency th

th 1 Qr

Qs 1

367.582 103

800 103 0.5405 54.05%

P

v

Pv = CW E

W C

Q 1

Q 21

4

3

2s=c

s=c

W E

W C

s

2 4

3

Q 1

Q 2

v=cv=

c

OTTO CY CLE

V = V2 3 v = v1 4

OTTO CY CLE


Work doneW Qs th 800 103 0.5405

W 432418 J/kg

(c) Mean effective pressure (MEP)

MEP W

V1 V2

432.42 103

0.861 0.123 5.86

MEP 5.86 bar

Problem 1.11: An engine working on the Otto cycle has an air standardcycle efficiency of 50% and rejects 544 kJ/kg of air. The pressure andtemperature of air at the beginning of compression are 0.1 MPa and 60Crespectively, Compute:(i) The compression ratio of the engine(ii) The work done per kg of air (iii) The pressure and temperature at the end of compression, and (iv) The maximum pressure in the cycle (AU. May 2015)

Givenotto 50%

Qr Cv T4 T1 544 kJkg

P1 0.1 MPa 100kPa

T1 60C 333 K

Otto cycle

1

4

3

2

SV

P T

1

42

3


Solution

(i) Compression ratio (r)

otto 1 1

r 1

0.50 1 1

r0.4

r 5.667

(ii) Work done/ kg of air

otto QS QR

QS

QS 544 0.50 QS

0.50 QS 544

QS 1088 kJkg

W QS QR 1088 544

544 kJkg

(iii) Pressure and temperature at end of compression

T2

T1 r 1

T2 333 5.6670.4

666 K

P2

P1 r

P2 100 5.6671.4 1134.21kPa

(iv) Maximum pressure in the cycle, P3

Heat supplied QS CV T3 T2

1088 0.718 T3 666


T3 1423.6 K

P3

T3

P2

T2

P3 P2 T3

T2 227.27

1423.6

666

2424.5kPa

IMPORTANT POINTS TO BE NOTED

(i) If fuel cut-off volume is given as X%of stroke volume, then

fuel cut off volume = X% of strokevolume

V3 V2 X

100 V1 V2 1

Divide (1) by V2

V3

V2 1

X100

V1

V2 1

V3

V2

cut 1 X

100 r 1

...

V1

V2 r

(ii) The percentage increase in volumeduring constant pressure burning is givenas y%

V3

V2 1

y100

1 y

100

Note: If the volume increases by 100% ,then the is 2.

1 100100

1 1 2

Fig. 1.7

1

4

32

W c

W E PV =C

Q 2

Q 1

v

P Diesel cycle

Fig. 1.6

v 3 - v2

Fuel cut off volum e

1

4

32

W c

W E P v = c

Q 2

Q 1

v

P

v3 - v2

Fuel cut off volum e

Diesel cycle


(iii) In otto cycle the % increase inpressure during constant volumeburning is given as Z%.

P3

P2 1

Z100

rp 1 Z

100

...

P3

P2 rp

Note: If the pressure increases by40% then rp

rp 1 40100

rp 1 0.4

rp 1.4

Problem 1.12: In an air standard diesel cycle the condition of air at thebeginning of compression is 1 bar, 300 K r 15:1 Fuel cut off occurs upto17

th of stroke volume. Determine (a) th (b) MEP.

Solution:

P1 1 bar; T1 300 K; r V1

V2 15

X100

17

1 17

r 1

1 17

15 1

3

(i) Thermal efficiency nth

Thermal efficiency 1

[ 1]

r 1 1 1

[ 31.4 1 ]

1.4 150.4 3 1

1

3.655538.271695

1 0.441933

Thermal efficiency th 0.55806 or 55.806%

P

v

PV =CW E

W C

Q 1

Q 21

4

3

2

s=c

s=c

OTTO CY CLE

Fig. 1.8v = v2 3 v = v1 4


(ii) Mean effective pressure

Pm P1 r [ [ 1] r1 1]

1 r 1

1 151.4 [1.4 3 1 151 1.4 31.4 11.4 1 15 1

Pm 12.36471 bar

Problem 1.13: A spark ignition engine working on ideal Otto cycle has thecompression ratio 6. The initial pressure and temperature of air are 1 barand 37 C. The maximum pressure in the cycle is 30 bar. For unit mass flow,calculate(i) p, V and T at various salient points of the cycle and(ii) The ratio of heat supplied to the heat rejected. Assume 1.4 and

R

8.314 kJ/kmol K. (AU. Nov/Dec 2012)

29 for air; P1 1 105 Pa

m 1 kg ; R

8314 kJ

kmol K

R R

831429

Solution:

Point 1

n m

129

P

v

pv =C�

W E

W E

W CW C

Q 1

Q 21

4

3

2s=c

s=c

s

2 4

3

Q 1

Q 2

v=c

v=c

(a ) (b )

OTTO CYCLE

Fig v = v2 3 v = v1 4

OTTO CYCLE


V1 mRT1

P1 1

831429

310

105

0.889 m3

Point 2

P2 P1 r 105 61.4

12.3 105 N/m2 12.3 bar

V2 V16

0.889

6 0.148 m3

T2 P2 V2

P1 V1 T1

12.3 105 0.148 310

1 105 0.889 634.8 K

361.8 C

Point 3

V3 V2 0.148 m3

P3 30 105 N/m2 30 bar

P3

T3

P2T2

T3 30 105

12.3 105 634.8 1548 K 1275 C

Point 4

P3 V3 P4 V4

P4 P3 V3

V4

30 105 16

1.4

2.44 105 N/m2 2.44 bar

V4 V1 0.889 m3

T4 T1 P4

P1 310

2.44 105

1 105

756.4 K 483.4 C


Cv R

r 1

8.31429 0.4

0.717 kJ/KgK

For unit mass

Heat supplied Cv T3 T2

0.717 1548 635.5 654.3 kJ/kg

Heat rejected Cv T4 T1

0.717 756.4 310 320.1 kJ/kg

Heat suppliedHeat rejected

654.3320.1

2.04

Problem 1.14: In an air standard otto cycle the condition of air at thebeginning of compression is 1 bar, 300 K compression ratio 8. Percentageincrease in pressure during constant volume burning is 100%. Determine (a)Pressure, Volume, temperature at all points. (b) th (c) MEP.

Solution: P1 1 bar ; T1 300 K; r 8

Hint: % increase in pressure is

rp 1 Z

100

rp 1 100100

rp 2

P1V1 R T1

V1 287 300

1 105 0.861 m3/kg


T2

T1 V1

V2

1

T2

300 81.4 1

T2 689.21901 K

P

v

Pv = CW E

W C

Q 1

Q 21

4

3

2s=c

s=c

W E

W C

s

2 4

3

Q 1

Q 2

v=c

v=c

OTTO CY CLE

V = V2 3 v = v1 4

OTTO CY CLE


P2

P1 V1

V2

P2

1 81.4

P2 18.37917 bar

P2

P1 0.861

V2

1.4

P2

P1

1/1.4

0.861

V2

18.37917

1

1/1.4

0.861

V2

8 0.861

V2

V2 0.107625 m3/kg

(or)

P2V2 R T2

V2 287 689.21901

18.37917 105 0.107625 m3/kg.

rp 2 (given)

rp P3

P2 Pressure ratio 2

P3

18.37917 2

P3 36.75834 bar

V3 V2 0.107625 m3/kg

P3V3 R T3

T3 36.75834 105 0.107625

287

T3 1378.43775 K

V1 V4 0.861 m3/kg

V1

V2 r

0.861V2

8

V2 0.1076 m3/kg



T3

T4 V4

V3

1.4 1

1378.43775T4

0.8610.107625

0.4

T4 599.9998 K ~ 600 K

P4V4 R T4

P4 287 600

0.861

P4 2 bar

(i) Thermal efficiency th

th 1 1

r 1

Heat supplied Qs 1 1

80.4

th 0.56472 56.472%


0.718 1378.43775 689.21901 494.5 kJ kg

Qs 494.5 kJ/kg

Workdone W th Qs 494.5 0.56472 279.26219 kJ/kg

(ii) Mean effective pressure (MEP)

MEP W

V1 V2

Cv T1 rp 1V1 V2

r 1 1

718 300 2 10.861 0.107625

80.4 1 3.70943 bar


Problem 1.15: An engine of 250 mm bore and 375 mm stroke works on

otto cycle. The clearance volume is 0.00263 m3. The initial pressure andtemperature are 1 bar and 50C. If the maximum pressure is limited to 25bar. Find the following (i) The air standard efficiency (ii) Mean effectivepressure. (AU. Nov/Dec 2011)

Given: d 250 mm 0.25 m ; L 375 mm 0.375 m ; Vc 0.00263 m3

P1 1 bar ; T1 50 273 323 K ; P3 25 bar

Swept volume

Vs d2

4 L

4

0.252 0.375 0.0184 m3

Compression ratio r Vs Vc

Vc

0.0184 0.002630.00263

8

(i) Air standard efficiency otto

otto 1 1

r 1 1

1

81.4 1 0.565 or 56.5%

(ii) Mean effective pressure MEP

We know that MEP Pm P1r [r 1 1 rp 1]

1 r 1

Process 1-2 is isentropic

P1V1 P2V2

or P2 P1 V1

V2

P1 r1.4

P2 1 81.4 18.38 bar

Pressure ratio rP P3

P2

2518.38

1.36

MEP Pm 1 8 [81.4 1 1 1.36 1]

1.4 1 8 1 1.334 bar

Mean effective pressure MEP 1.334 bar


Problem 1.16: An Otto cycle has a compression ratio of 7. The initialpressure and temperature at the beginning of compression stroke is 1 barand 40C. The heat supplied is 2510 kJ/kg. Find (i) The maximumtemperature and pressure (ii) Workdone per of air (iii) The cycle efficiency(iv) Mean effective pressure. Take is Cv 0.713 kJ/kg K and

R 287 J/kg K (AU. Nov/Dec 2013)

Beginning of compression P1 1 bar; T1 40 273 313K

r 7, Qs 2510 kJ/kg

P1V1 R T1

V1 0.287 313

100 0.898 m3/kg

V1 0.898 m3/kg.

V1 V4 0.898 m3/kg.

V1

V2 r

0.898V2

7 0.1283 m3/kg

V2 V3 0.1283 m3/kg


T2

T1 V1

V2

1

T2

313

0.8980.1283

0.4

T2 681.6 K

P2V2 R T2

P2 0.287 681.6

0.1283 1524.68 kPa

P2 15.25 bar



2510 0.718 T3 681.6

Maximum Temperature T3 4177.42 K

P3 V3 R T3

P3 0.287 T3

0.1283 P3

0.287 4177.420.1283

P3 9344.6 kPa 93.45 bar Maximum pressure

T4T3

V3

V4

1

T4

4177.42

0.12830.898

0.4

T4 1918.18 K

P4 V4 R T4 P4 0.287 1918.18

0.898 613.05 kPa

P4 6.13 bar

Heat rejected Qr Cv T4 T1

Qr 0.718 1918.18 313

Qr 1152.52 kJ/kg

(b) Thermal Efficiency th

th 1 Qr

Qs 1

1152.522510

0.5408 54.08%

Work done per kg of air W Qs th

2510 0.5408

W 1357.41 kJ/kg

(c) Mean effective pressure (MEP)

MEP W

V1 V2

1357.410.898 0.1283

1763.6 kPa

MEP 17.64 bar


Problem 1.17: An engine works on otto cycle with minimum temperature

and pressure of 27C and 105 N/m2. The heat added to cycle is 1500 kJ/kg.Determine (i) P and T at all points (ii) Specific work (iii) Thermalefficiency of cycle if compression ratio is 8:1. TakeCv 0.72 kJ/kg and 1.4

Solution:

Given: P1 105 N/m2 1 bar ; T1 27 273 300 K ;

Qs 1500 kJ/kg; r 8 ; Cv 0.72 kJ/kg ; 1.4

(i) P and T at all points Process 1-2: (Compression)

T2

T1 V1

V2

1

r 1 81.4 1 2.297

Temperature T2 300 2.297 689 K

Also P2 P1 V1

V2

1 81.4 1 81.4 18.379 bar

Process 2-3 (Constant volume)Heat added Qs Cv T3 T2

1500 0.72 T3 689

T3 2772 K

Also P2

T2

P3

T3 or P3 P2

T3

T2 18.379

2772689

73.94 bar

P3 73.94 bar

Process 3-4 (Expansion)

T3

T4 V4

V3

1

r 1 81.4 1 2.297

T4 27722.297

1207 K

P4 P3 V3

V4

73.94 18

1.4

4.023 bar

P4 4.023 bar


(ii) Specific work

Specific work W Heat supplied Heat rejected

Cv T3 T2 Cv T4 T1

0.72 [2772 689 1207 300]

W 847 kJ/kg

(iii) Thermal efficiency th

th 1 1

r 1 1

1

81.4 1 0.5647 or 56.47%

Problem 1.18: Derive an expression for the compression ratio for maximumwork done per kg of air in otto cycle. Also determine the air standardefficiency of the cycle when the cycle develops maximum work withtemperature limits of 310 K and 1220 K with air as working fluid. Alsodetermine expression for Intermediate temperature T2, T4 for maximum

work done. If engine works between 1450 K and 310 K, find maximum powerdeveloped if m 0.38 kg/min

Solution:

We know that work done per kg of fluid in cycle is

W Qs Qr Cv T3 T2 Cv T4 T1 ...(1)

Also T2 T1 V1V2

1

T1 r 1

T3 T4 V4V3

1

T4 r 1

...(2)

Substituting (2) in (1) we get

W Cv T3 T1 r 1

T3

r 1 T1

...(3)

Differentiate (3) w.r.t r and equate to zero.

dWdr

0 dWdr

T1 1 r 2 T3 1 r 0

T3 r T1 r 2


T3

T1 r2 1

Compression ratio r T3T1

12 1

(ii) Air standard efficiency

1.4 [T3 1220 K ; T1 310 K]

From above expression

r T3T1

12 1

1220310

12 1.4 1

5.543

Air standard efficiency

otto 1 1

r 1 1

1

5.541.4 1 0.495

or otto 49.5%

(iii) Intermediate Temperature T2We have derived expression for compression ratio

r T3T1

12 1

we know that T2 T1 r 1 ; T4 T3

r 1

T2 T1 T3

T1

12 1

1

T1 T3

T1

1/2

T1T3

T4 T3

T3

T1

12 1

1

T3

T3

T1

1/2 T3T1

T2 T4 T1T3


(iv) Maximum power developed

Work done W Cv [T3 T2 T4 T1]

T2 T4 T1T3 310 1450 670 K

Work done W 0.72 [1450 670 670 310]

W 301.56 kJ/kg

Work done per second 301.56 0.3860

1.91 kJ/s

Power developed P 1.91 kW

Problem 1.19: A diesel engine has a compression ratio of 10 and heataddition takes place at constant pressure at 8% of stroke. Find the airstandard efficiency of the engine. Take 1.4

Solution:

Given: r 10 ; V3 V2 8

100 Vs ; 1.4

Compression ratio r V1

V2 10

Air standard efficiency diesel 1 1

r 1 1 1

Cut off ratio V3V2

V3 V2 0.08 V1 V2

V3 V2 0.08 10V2 V2

V3

V2

V2

V2 0.08 9 0.72

V3

V2 1 0.72 1.72

diesel 1 1

1.4 101.4 1 1.721.4 1

1.72 1

diesel 0.551 or 55.1%


Problem 1.20: An indicator diagram taken on a Diesel engine shows that

the compression curve follows the law PV1.4 C. At two points lying on thecompression curve at 1/8 th and 7/8 th of the stroke the pressures arerespectively 1.6 bar and 16 bar. Find the compression ratio of the engine.If the cut off occurs at 6% of the stroke, calculate the air standard efficiencyof the engine.

Solution:

We have V1 Vc 78

Vs (at 7/8 th of stroke)...(1)

V2 Vc Vs8

(at 1/8 th of stroke)...(2)

But law of compression

P1V11.4 P2V2

1.4 or V1

V2 P2

P1

11.4

161.6

11.4

5.18

V1

V2 5.18

Now Dividing (1) (2)

V1

V2

Vc 7/8 Vs

Vc Vs/8 5.18

1 7/8 Vs

Vc

1 Vs

8Vc

5.18

1 7 Vs

8 Vc 5.18

5.188

Vs

Vc

solving Vs

Vc 18.39

Compression ratio r 1 VsVc

1 18.39 19.39

Cut off ratio

1Vc

Vc Vs 0.06 1 0.06 18.39 2.10


Air standard efficiency 1 1r

1

1

1

1

1

119.39

1.4 1

1

1.4 2.101.4 1

2.10 1 1 0.362 0.6378 63.78%

Problem 1.21: The mean effective pressure of a diesel engine is 7.5 barand compression ratio is 12.5 bar. Find the percentage cut off of the cycleif its initial pressure is 1 bar.

Solution:

Given: Pm 7.5 bar ; r 12.5 ; P1 1 bar

We know that Pm P1r [ 1 r1 1]

1 r 1

7.5 1 12.51.4 [1.4 1 12.51 1.4 1.4 1]

1.4 1 12.5 1

7.5 7.46 1.4 1.036 0.364 1.4

solving 0.364 1.4 1.4 2.04 0

By trial and error we get 2.24

Percentage cut off 1

r 1 100

2.24 1

12.5 1 100

% cutoff 10.78%

Problem 1.22: A engine working on diesel engine with 200 mm cylinderdiameter and 300 mm stroke. The initial pressure and temperature of air are1 bar and 27C. The cut off is 8% of the stroke. Determine (i) Pressureand temperature at all points (ii) Air standard efficiency (iii) Mean effectivepressure (iv) Power of the engine if the working cycles per minute are 380.Take r 15 and 1.4.

Solution:

Given:

d 0.2 m ; L 0.3 m ; P1 1 bar ; T1 27 273 300 K

cut off 8

100 Vs 0.08 Vs


(i) Pressure and Temperature

Swept volume, Vs d2

4 L

4

0.22 0.3 0.00942 m3

V1 Vs Vc Vs Vs

r 1 Vs

1

1r 1

rr 1

Vs

V1 15

15 1 0.00942 0.0101 m3 V1 0.0101 m3

Also P1V1 mRT1

m P1V1

RT1

1 105 0.0101

287 300 0.0116 kg/cycle

Process 1-2

P1V1 P2V2

or P2 P1 V1

V2

P1 r 1 151.4 44.31

P2 44.31 bar

T2

T1 V1

V2

1

r 1 151.4 1 2.954

T2 2.954 300 886.25 K

V2 Vc Vs

r 1

0.0094215 1

0.000673 m3

P2 P3 44.31 bar

% cut off 1r 1

8

100

115 1

8

100 or 2.12

V3 V2 2.12 0.000673 0.001428 m3

V3 0.001427 m3


Process 2-3

T3 T2 V3

V2 887

0.0014270.000673

1880.44 K

Process 3-4

P3V3 P4V4

or P4 P3 V3

V4

44.31 0.0014270.0101

1.4

P4 2.86 bar [. . . V4 V1]

T4 T3 V3

V4

1

T3 0.0014270.0101

1.4 1

1878 0.457 859.60 K

T4 859 K

V4 V1 0.0101 m3

(ii) Air standard efficiency

diesel 1 1

r 1 1

1

1 1

1.4 151.4 1 2.121.4 1

2.12 1 0.597

diesel 59.7%

(iii) Mean effective pressure Pm

Pm P1r [ 1 r1 1]

1 r 1

1 151.4 [1.4 2.12 1 151 1.4 2.121.4 1]

1.4 1 15 1

Pm 7.42 bar

(iv) Power of engine

Work done W PmVs 7.42 102 0.00942

W 7 kJ/cycle

Power 7 38060

44.28 kJ/s or kW


Problem 1.23: Show that the efficiency of otto cycle is greater than that ofdiesel engine for same compression ratio. (AU. May/June 2013)

Solution:

The PV diagram of diesel & otto cycle are given below.

We know that

otto 1 1

r 1 ; Diesel 1

1

r 1

1

1

1

For same compression ratio

V1

V2

V1

V2 r

If V4

V3 r1 then

V3

V2

rr1

diesel 1 1

r 1

1

rr1

1

rr1

1

...(1)

We have rr1

1 from above equation.

Let r1 r

1’

4 ’

3 ’2 ’

W c

W E PV =C

Q 2

Q 1

v

PDiesel cycle

P

v

PV =CW E

W C

Q 1

Q 21

4

3

2s=c

s=c

OTTO CYCLE

v = v2 3v = v1 4


rr1

r

r

r

r 1

r

1

r

1

1 r

2

r2

...(2)

Similarly rr1

1

r

1 r

1

2! 2

r2 ...(3)

(2), (3) in (1)

Diesel 1 1

r 1

1

r

1

2! 2

r2

r

2

r2

1 1

r 1

r

1

2! 2

r2

r

2

r2

The term inside bracket is greater than 1 and hence the efficiency ofDiesel is less than otto cycle.

1.5 DUAL COMBUSTION CYCLE

Draw the P-V and T-S diagrams of a dual cycle and write the variousprocesses (AU. April/May 2015) (FAQ)

Dual cycle is a combination of otto cycle and diesel cycle. In thiscycle, heat is supplied to the system partly at constant volume and partly atconstant pressure process. This cycle is also called as semi-diesel cycle,limited pressure cycle or mixed cycle.

Dual cycle consists of the following processes.

Process 1-2: Reversible adiabatic (Isentropic) compression.

Process 2-3: Addition of heat (partly) at constant volume.

Process 3-4: Addition of heat (partly) at constant pressure.

Process 4-5: Reversible adiabatic (Isentropic) expansion (work done).

Process 5-1: Heat rejection at constant volume.


Draw the PV diagram for a dual cycle (AU. Nov 2010)

Total Heat Supplied Qs

Qs Heat supplied during constant volume process 23

Heat supplied during constant pressure process 34

Qs 23 Qs 34 mCv T3 T2 mCp T4 T3

Total Heat Rejected Qr

Qr Heat rejected dur ing constant volume process 45

Qr 51 mCv T5 T1

Total Work done: WW Qsupplied Qrejected [ mCv T3 T2 mCp T4 T3 ] [ mCv T5 T1 ]

Efficiency of Dual Combustion Cycle:

Work done

Heat supplied

[ mCv T3 T2 mCp T4 T3 ] [ mCv T5 T1 ]mCv T3 T2 mCp T4 T3

Cv T3 T2 Cp T4 T3 Cv T5 T1

Cv T3 T2 Cp T4 T3

V

Q s 3-4

4

Adiabatic

Adiabatic

1

5Q r

W 1-2

P

3

2W 4-5

Q s 2-3

T

S1

2

3

4

5v = c

p = c

Q r

Q s3-4

W 1-2

W 4-5

Q s 2-3

Fig. 1.9


1 Cv T5 T1

Cv T3 T2 Cp T4 T3

1 T5 T1

T3 T2 T4 T3 . . .

Cp

Cv

Dual 1 T5 T1

T3 T2 T4 T3 ...(1)

Mean Effective pressure W

V1 V2

Work doneSwept volume

Compression ratio r V1

V2

Pressure ratioor

Explosion ratio

p3

p2

T3

T2

Cut-off ratio rcut V4

V3

v4

v3

Process 1-2 (adiabatic compression)

T2

T1 V1

V2

1

r 1

Process 2-3 (constant volume)P3

T3

P2

T2 or

P3

P2

T3T2

known as explosion ratio

or T2 T3

Process 3-4 (constant pressure)V3

T3

V4

T4 or T4 T3

V4

V3 T3

V4

V3 cut off ratio


Process 4-5 (Adiabatic expansion)

T4

T5 V5V4

1

r

1

. . .

V5

V4

V1

V4

V1

V2

V2

V4

V1

V2

V3V4

r

T5 T4 r

1

T3 r

1

Also T1 T2 1r

1

T3

1r

1

T3

l

r 1

substituting values of T1, T2, T4 and T5 in equation (1)

dual 1

.T3 r

1

T3

1

rr 1

T3

T3

T3 T3

dual 1 1

r 1

1 1 1

Work done in dual cycle

Work done, W P3 V4 V3 P4V4 P5V5

1

P2V2 P1V1

1

P3V3 1 P4V3 P5rV3 P2V3 P1rV3

1

W

P3V3 1 1 P4V3

P5

P4 r P2V3

1

P1

P2 r

1

Also we have P5

P4 V4V5

r

and P2

P1 V1

V2

r

P3 P4, V2 V3 ; V5 V1

Substituting in above equation.


W V3 [P3 1 1 P3 r1 P2 1 r1 ]

1

Multiply and divide throughout by P2

P2V2 [ 1 1 r1 1 r1 ]

1

W P1V1r 1 [ 1 1 r 1 1]

1

Expression for mean effective pressure PmWe know that mean effective pressure Pm

Pm Work done

Swept volume

WV1 V2

W

V1 r 1

r

Pm P1V1 r 1 [ 1 1 r 1 1]

1 V1 r 1

r

Pm P1 r [ 1 1 r 1 1]

1 r 1Problem 1.24: The compression ratio for a single-cylinder engine operatingon dual cycle is 9. The maximum pressure in the cylinder is limited to 60bar. The pressure and temperature of the air at the beginning of the cycleare 1 bar and 30C. Heat is added during constant pressure process upto 4percent of the stroke. Assuming the cylinder diameter and stroke length as250 mm and 300 mm respectively. Determine (i) The air standard efficiencyof the cycle, (ii) The power developed if the number of working cycles is 3per second. (AU. May/June 2015)

Solution

Cylinder diameter, D 250 mm 0.25 m

Compression ratio r 9

Stroke length L 300 mm

Initial pressure P1 1 bar


Initial temperature T1 30 273 303 K

Maximum pressure P3 P4 60 bar

Cut-off = 4% of stroke volume

No. of working cycles/sec = 3.

(i) Air standard efficiency

Swept volume, Vs 4

D2 L

4

0.252 0.3

0.0147 m3

Compression ratio r Vs Vc

Vc

9 0.0147 Vc

Vc

9 Vc Vc 0.0147

P(bar)

603 4

0 .04 V S

2

5

1

V C V S

V(m )3

1


8 Vc 0.0147

Vc 0.0147

8 1.8375 10 3 m3

0.0018 m3

V1 Vs Vc 0.0147 0.0018 0.0165 m3

For process 1 - 2 (adiabatic (or) isentropic)

P1 V1Y P2 V2

Y

P2 P1 V1

V2

Y

1 rY 1 91.4

... V1

V2 r

21.67 bar

T2

T1 V1

V2

Y 1

rY 1 91.4 1 90.4

T2

T1 2.408

T2 T1 2.408

303 2.408

T2 729.6 K

For process 2 - 3 (constant volume)T3

P3

T2

P2

T3 T2 P3

P2 729.6

6021.67

2020.11 K 2020 K

Also We know that

1r 1

4

100

19 1

0.04 1.32


For the constant pressure process 3 - 4V4

T4

V3

T3

(or)

T4

T3

V4

V3

T4 T3

2020 1.32

2666.4 K

Also expansion work,

V5

V4

V5

V2

V2

V4

. . . V5 V1

V2 V3

V1V2

V3

V4

r

For adiabatic process 4 - 5

T5

T4 V4V5

1

r

1

T5 T4 r

1

2666.4 1.32

9

1.4 1

1237 K

Also,

P4 V4Y P5 V5

Y


P5 P4 V4

V5

Y

60 r

Y

60 1.32

9

1.4

4.08 bar

Heat supplied, Qs Cv T3 T2 Cp T4 T3

0.71 2020 729.6 1.0 2666.4 2020

1562.58 kJ/kg

Heat rejected, Qr Cv T5 T1

0.71 1237 303 663.14 kJ/kg

air std Qs Qr

Qs

1562.85 663.14

1562.58

0.5756

(or)

57.56%

(ii) Power developed by the engineMass of air in the cycle is given by

m P1 V1

RT1

1 105 0.0165

287 303 0.0189 kg

Workdone/cycle m Qs Qr

0.0189 1562.58 663.14

16.999 kJ

Power developed Workdone/cycle no. of cycles/sec

16.999 3

50.79 kW


Problem 1.25: In an air standard Dual cycle the compression, ratio is 12: 1. The condition of air at the beginning of compression is 1 bar, 300 K,maximum pressure is 6 MPa. Fuel cut off occurs upto 5% of stroke volume.

Determine (a) Pressure, volume, temp at all points (b) (c) MEP.

Solution: r 121

12

Beginning of compressionP1 1 bar ; T1 300 K

P4 P3 Pmax 6 MPa 60 barFuel cut off at 5% of stroke volumeV4 V3 0.05 V1 V2 [Divide by V3]

V4

V3

V3

V3

0.05 V1 V2V3

rcut 1 0.05 V1V2

1 [... V3 V2]

rcut 1 0.05 r 1 rcut 1 0.05 12 1 1.55

rcut 1.55

P1V1 RT1

V1 287 300

1 105 0.861 m3/kg

Isentropic process (1 - 2)

T2

T1 V1V2

1

T2

300 120.4

T2 810.57602 KV1

V2 12

0.86112

V2

V2 0.07175 m3/kg

P2V2 RT2

P2 287 810.57602

0.07175 32.42304 105

P2 32.42304 bar

T

S

2

34

v = c

v = c

p = c

1

5

V

4

1

5

P3

2


Process (2) - (3) constant volume processP3 V3 RT3 ...(i)

P2 V2 RT2 ...(ii)[Divide (i) by (ii)]

P3

P2

T3

T2

60 105

32.42304 105

T3

810.57602

T3 1500 K

V3 V2 0.07175 m3/kg

P4 P3 60 bar given

rcut V4

V3 1.55

V4

0.07175

V4 0.11121 m3/kg

P4 V4 RT4

T4 60 105 0.11121

287 2325 K

V5 V1 0.861 m3/kg


T5

T4 V4

V5

1

T5

2325 0.111210.861

0.4

T5 1025.36675 K

P5V5 RT5

P5 287 1025.36675

0.861 3.41788 bar

(ii) Efficiency

dual 1 T5 T1

T3 T2 T4 T3

1 1025.36675 300

1500 810.57602 1.4 2325 1500 1 0.3932


0.60672 or 60.672 %(iii) Mean effective pressureHeat supplied Qs cv T3 T2 Cp T4 T3

718 1500 810.57602 1005 2325 1500Qs 1323.3742 103 J/kg

Workdone W Qs dual 1323.3742 103 0.60672

W 802.917599 103 J/kg

MEP W

V1 V2

802.917599 103

0.861 0.071755

MEP 10.17317 barProblem 1.26: The swept volume of a diesel engine working on dual cycle

is 0.0053 m3 and clearance volume is 0.00035 m3. The maximum pressure is65 bar. Fuel injection ends at 5 percent of the stroke. The temperature andpressure at the start of the compression are 80C and 0.9 bar. Determine

the air standard efficiency of the cycle. Take for air as 1.4. (AU. May/June 2016)

Solution

Swept volume, Vs 0.0053 m3

Clearance volume, Vc V3 V2 0.00035 m3

Maximum pressure P3 P4 65 bar

Initial temperature T1 80 273 353 K

Initial pressure P1 0.9 bar

To finddual ?

The efficiency of a dual combustion cycle is given by,

dual 1 1

r 1

1 1

1 1 .... (A)

Compression ratio, r V1

V2

Vs VcVc


0.0053 0.00035

0.00035

16.14

[. . . V2 Vc Clearance volume]

Cut-off ratio, V4

V3

5100

Vs V3

V3

0.05 Vs Vc

Vc

. . . V2 V3 C

0.05 0.0053 0.00035

0.00035

1.757 say 1.76

Process 1 - 2 [Compression operation]

P1 V1Y P2 V2

Y

P(bar)

653 4

2

5

1

V = V =V =0.00035mC 3 23

V =0.0053mS3

V(m )3

Ad iabatic

Ad iabatic

0 .9


P2

P1 V1

V2

Y

16.141.4 49.14

P2 P1 49.14 0.9 49.14 44.22 bar

Pressure ratio P3P2

65

44.22 1.47

Subsitute the value of r, and in eqn (A)

dual 1 1

16.141.4 1

1.47 1.761.4 1

1.47 1 1.47 1.4 1.76 1

1 0.328

3.243 10.47 1.564

0.6383 or

dual 63.83%

Problem 1.27: In an air standard duel cycle the compression ratio is 10.The condition of air at the beginning of compression is 1 bar, 17C.Percentage increase in pressure during constant volume burning is 60%. Theheat supplied is 100 kJ/kg. Determine (a) P, V, T at all points (b) Workdone(c) MEP

Solution:Given: r 10

(a) Beginning of compression, P1 1 bar ; T1 17C 290 K

% increase in pressure during constant volume Z 60%

1 z

100

1 60

100 rp 1.6

P1V1 RT1

V1 287 290

1 105 0.8323 m3/kg


Process (1) - (2) isentropic process

T2

T1 V1

V2

1

T2

290 100.4

T2 728.44706 K

r V1

V2 10

0.8323V2

V2 0.08323 m3/kg

P2V2 RT2

P2 287 728.44706

0.08323 25.11886 bar

V2 V3 0.08323 m3/kg

P3

P2 1.6

P3

25.118886 bar

P3 40.19018 bar ; P3 P4 40.19018 bar

P3V3 RT3

T3 40.19018 105 0.08323

287 1165.5153 K

Heat supplied Qs Cv T3 T2 Cp T4 T3

1000 103 718 1165.5153 728.44706 1005 T4 1165.5153

686403.5348 1005 T4 1165.5153

683.32855 T4 1165.5153

T4 1848.84385 K

P4V4 RT4

V4 287 1848.84385

40.19018 105 0.132026 m3/kg

V5 V1 0.8323 m3/kg

V

4

1

5

P 3

2

T

S

2

34

v = c

v = c

p = c

1

5



T5T4

V4V5

1

T51848.84385

0.1320260.8323

0.4

T5 885.22393 K

P5V5 RT5

P5 287 885.22393

0.8323 3.05249 105 Pa

P5 3.05249 bar

Efficiency of cycle

thermal 1 T5 T1

T3 T2 T4 T3

1 885.22393 290

1165.5153 728.44706 1.4 1848.84385 1165.5153 1 0.427073

0.572926 or 57.29%

(b) Work doneWorkdone

W Qs thermal 1000 103 0.572926 572.926 kJ/kg

W 572.926 kJ/kg

(b) Mean effective pressure

MEP W

V1 V2

572.926 103

0.8323 0.08323 7.6485 105 Pa

MEP 7.6485 bar


Problem 1.28: In an engine working on Dual cycle the compression ratiois 9. The conditions of air at the beginning of compression is 1 bar, 90C.Max. pressure is 6.8 MPa. Heat supplied is 1750 kJ/kg. Determine(a) P, V, T at all points (b) thermal (c) MEP (FAQ)

Solution:Given r 9

Beginning of compression

P1 1 bar T1 90C 363 K

Pmax P3 P4 6.8 106 Pa 68 bar

Heat supplied Qs 1750 103 J/kg

P1V1 RT1

V1 287 363

1 105 1.04181

V1 1.04181 m3/kg


T2

T1 V1

V2

1

T2

363 90.4

T2 874.18556 K

V1

V2 9

1.041819

V2

V2 0.11575 m3/kg

P2V2 RT2

P2 287 874.18556

0.11576 21.674 105 Pa

P2 21.674 bar

Process (2) - (3) constant volumeP3V3

P2V2

RT3

RT2

T

S

2

34

v = c

v = c

p = c

1

5

V

4

1

5

P

3

2


P3

P2

T3

T2

6821.674

T3

874.18556

T3 2742.66946 K

V2 V3 0.11575 m3/kg

Heat supplied Qs cv T3 T2 Cp T4 T3

1750 103 718 2742.66946 874.18556 1005 T4 2742.66946

409.3628 103 1005 T4 2742.66946

T4 3150.19838 K

P4V4 RT4

V4 287 3150.19838

68 105 0.13295

V4 0.13295 m3/kg

Process (4) - (5) isentropic

T5

T4 V4V5

1

[V5 V1 1.0418 m3/kg]

T5

3150.19838 0.132951.0418

0.4

T5 1382.60938 K

P5V5 RT5

P5 287 1382.60938

1.0418 3.80887 105 Pa

P5 3.80887 bar

(b) Thermal efficiency

thermal 1 T5 T1

T3 T2 T4 T3


1 1382.60938 363

2742.66946 874.18556 1.4 3150.19838 2742.66946

1 0.418039 0.58196 or 58.19%

Workdonethermal 0.58196

W Qs thermal 1750 0.58196

W 1018.43026 103 J/kg

(c) Mean effective pressure

MEP W

V1 V2

1018.430.26 103

1.0418 0.11575 10.99757 bar

Problem 1.29: In an air standard dual cycle, the pressure and temperatureat the beginning of compression are 1 bar and 57C, respectively. The heatsupplied in the cycle is 1250 kJ/kg, two third of this being added at constantvolume and rest at constant pressure. If the compression ratio is 16, determinethe maximum pressure, temperature in the cycle, thermal efficiency and meaneffective pressure. (AU. Nov/Dec 2011)

P1 1 bar, T1 57 273 330 K

Q2 3 23

1250 833.3 kJ/kg

P1 V1 m RT1

V1 1 0.287 330

100

0.9471 m3/kg


Q3 4 13

1250 466.7 kJ/kg

r 16

T2

T1 V1

V2

1

Q2 3 833.33 kJ/kg

T2 T1 161.4 1 330 160.4

T2 1000.37 K

Heat supplied at const volume Q23 CV T3 T2

833.33 0.718 T3 1000.37

T3 2161 K

P1 V1 P2 V2

P2 V1

V2

P1 161.4 1 48.5 bar

Take const volume process (2-3)P3T3

P2

T2

P3 P2

T2 T3

48.51000.37

2161

P3 104.76 bar

Thermal efficiency th

th 1 1

r 1

1 1 1

V4

V3,

P3P2

V1

V2 r 16

V2 0.9471

16 0.0592 m3/kg


Q3 4 13

1250 416.67kJ/kg

Q3 4 CP T4 T3

1875 1 T4 2161

T4 2575.6 K

V4

V3

T4

T3

2575.62161

1.19

1.19

P3

P2

104.7648.5

2.16

2.16

th 1 1

161.4 1

2.16 1.191.4 12.16 1 2.16 1.4 1.19 1

th 66.66 %

Take Process 5 1P5T5

P1

T1

P4 V4 P5 V5

P5P4

V4

V5

1r

1.4

116

1.4

P5 104.76

116

1.4

2.15 bar

2.15T5

1

330

T5 709.5 K


Heat rejectedQR Q5 1 CV T5 T1

0.718 709.5 330

QR Q5 1 272.5 kJ/kg

Mean effective pressure Pm

pm Workdone

Swept Volume

Qs QR

V1 V2 833.44 466.7 272.5

0.95 0.059

pm 1153.4 kPa 11.534 bar

Problem 1.30: The condition of air at the beginning of compression of aDual cycle are 1.03 bar, 35C & 150 litres. The volume of the compressionis 10 litres and 42 kJ of heat is added at constant volume, 63 kJ of heat isadded at constant pressure. Determine (a) Air standard efficiency (b) MEP(c) Clearance percentage (d) Cut off percentage. (FAQ)

Solution:

Beginning of compression P1 1.03 bar 1.03 105 Pa ; T1 35C 308 K;

V1 150 lit 0.150 m3 ; V2 10 lit 0.01 m3

Compression ratio r V1V2

15

P1v1 RT1

v1 287 308

1.03 105 0.85821

v1 0.85821 m3/kg

m V1

v1 0.17478 kg

mv2 V2 ; v2

10 10 3

0.17478 0.057214

v2 0.057214 m3/kg

V

4

1

5

P3

2


m cv T3 T2 42 103 J constant volume heat addition

m Cp T4 T3 63 103 J constant pressure heat addition


T2

T1 V1

V2

1

T2

308 150.4

T2 909.88649 K

m cv T3 T2 42 103 J

0.17478 718 T3 909.88649 42 103

T3 909.88649 334.91581

T3 1244.8023 K

m Cp T4 T3 63 103

0.17478 1005 T4 1244.8023 63 103

T4 1244.8023 358.83836

T4 1603.64066 K

For constant pressure process,

P4v4

P3v3

RT4

RT3

v3 v2 0.057214 m3/kg

v4

v3

T4

T3

v4

0.057214

1603.640661244.8023

v4 0.073708 m3/kg

T

S

2

34

v = c

v = c

p = c

1

5


Cut off ratio V4

V3 1.28828

Expansion ratio rexp r

15

1.28828 11.64336

V5

V4


T5

T4 V4

V5

1

T5

1603.64066

1

11.643360.4

T5 600.72503 K

Mean Effective Pressure

thermal 1 T5 T1

T3 T2 T4 T3

1 600.72503 308

1244.8023 909.88649 1.4 1603.64066 1244.8023

1 0.3496085

Air standard efficiency

ther 0.6503939 or 65.039%

Heat supplied per kg qs Qs

m

42000 630000.17478

qs 600749.5 J/kg

Workdone per kg W qs thermal 600749.5 0.6503939

W 390723.8 J/kg

MEP W

v1 v2

390723.80.858213 0.057214

4.87795 105 Pa

MEP 4.87795 bar

To find clearance and cut off percentage

Clearance % Clearance volume

Stroke volume

V2

V1 V2


Divide by V2; 1

V1V2

1

1

r 1

Clearance % 1

15 1 0.071428 7.1%

Cut off % cut off volumestroke volume

V4 V3

V1 V2

Divide by V2;

V4

V2

V3

V2

V1

V2 1

1

r 1

1.28828 1

15 1

0.02059 2.059%

Problem 1.31: An air standard dual cycle has a compression ratio of 18,and compression begins at 1 bar, 40 C. The maximum pressure is 85 bar.The heat transferred to air at constant pressure is equal to that at constantvolume. Estimate:(i) The pressures and temperatures at the cardinal points of the cycle.(ii) The cycle efficiency and(iii) The mean effective pressure of the cycle. (AU. Nov/Dec 2012)

Solution: T1 273 40 313 K

T2

T1 V1

V2

1

180.4

T2 994.6 K

P2 P1 V1

V2

1.0 181.4 57.20 bar

T3 T2 P3P2

994.6 85

57.20 1478 K

Q2 3 Cv T3 T2 0.718 1478 994.6

347.08 kJ/kg

V3 V2V4

V2

V4

V3


Q2 3 Q3 4 Cp T4 T3 347.08 kJ/kg

Total Qs 347.08 347.08 694.2

T4 347.081.005

1478 1823.35K

V4

V3

T4T3

1823.35

1478 1.23

V5

V4

V1

V2

V3

V4

181.23

14.63

T5 T4 V4V5

1

1823.35 1

14.630.4 623.41 K

P5 P1 T5

T1 1.0

611313

1.99 bar

cycle 1 Q2

Q1 1

CV T5 T1CV T3 T2 Cp T4 T3

1 0.718 623.41 313346.215 346.215

1 0.3218

cycle 67.8%

V1 RT1

P1

0.287 kJ/kg K 313 K

102 kN/m2 0.898 m3

Fig.(a) V

4

1

5

P

3

2

T

S

2

34

v = c

v = c

p = c

1

5

Fig.(b)


V1 V2 V1 V1

18

1718

V1

Wnet Q1 cycle 694.2 0.678 471 kJ/kg

M.E.P Wnet

V1 V2

471178

0.898

246.7 kN/m2

2.467 bar

Problem 1.32: The compression ratio of an mixed cycle is 10. The conditionof air at the beginning of the cycle is 1 bar, 27C. The max pressure of thecycle is limited to 7 MPa and the heat supplied is 1675 kJ/kg. Determinethermal and MEP.

Solution:Beginning of cycle

P1 1 bar T1 27C 300 K

rc 10

Pmax P4 P3 7 MPa 70 bar

Heat supplied per kg

Qs 1675 103 J/kg


T2

T1 V1

V2

1

T2

300 100.4

T2 753.56592 K

P2

P1 V1

V2

P21

101.4

P2 25.11886 bar

V

4

1

5

P3

2

T

S

2

34

v = c

v = c

p = c

1

5


Process (2) - (3) constant volume processP3V3

P2V2

RT3

RT2

P3

P2

T3

T2

7025.1188

T3

753.5659

T3 2100 K

qs Cv T3 T2 Cp T4 T3

1675 103 718 2100 753.5659 1005 T4 2100

T4 2805.73785 K

P1v1 RT1

v1 287 300

1 105 0.861 m3/kg

v1 0.861 m3/kg

v1

v2 10

0.86110

v2

v2 0.0861 m3/kg v3

P4v4 RT4

v4 287 2805.757385

70 105 0.115036 m3/kg

v1 v5 0.861 m3/kg ; r 10

rcut v4

v3

0.1150360.0861

1.336


T5

T4 v4

v5

1


T5

2805.757385 0.115036

0.861

0.4

T5 1254.24 K

Thermal efficiency

1 T5 T1

T3 T2 T4 T3

1 1254.24 300

2100 753.56592 1.4 2805.757385 2100 1 0.408758

0.59124 or 59.124%

W thermal qs 0.59124 1675 103

W workdone per kg;

qs Heat supplied per kg

W 990.32897 kJ/kg


MEP W

v1 v2

990.32897 103

0.861 0.0861 12.78 bar

1.6 PERFORMANCE AND COMPARISON OF OTTO, DIESELAND DUAL COMBUSTION CYCLES

Otto, Diesel and Dual cycles can be compared on the basis of thefollowing variables.

(i) Efficiency Versus Compression RatioThe following graph Fig 1.10 shows the variation of air standard

efficiency of the Otto, Diesel and Dualcycles at various compression ratios witha given cut off for dual and diesel cycle.The air standard efficiency is clearlyincreases with the increase in compressionratio. The order of efficiencies of the threecycles are otto Dual Diesel

80

70

60

50

40

300 4 8 12 16 20

Effi

cien

cy

(%)

Otto

Dual combustion

Diesel cycle

S .I. Eng ine C.I. EngineCompress ion ratio(r)

Fig. 1.10


(ii) Comparison of Otto, Dual and Diesel cycle on P V and T SdiagramFor constant compression ratio and heatsupplied.

The P V and T S diagram of thethree cycles are shown in the Fig 1.11 (a)and (b) for same compression ratio r andheat supplied Qs

Otto cycle 1 2 3 4 1

Dual cycle 1 2 2 3 4 1

Diesel cycle 1 2 3 4 1

We know that the efficiency ofcycle is given as

1 Heat rejectedHeat supplied

The heat rejected by cycle isrepresented by the area under the line 4 to1 on T S diagram. The heat is rejected atsame specific volume process 4 to 1. Theleast heat rejected will have maximumefficiency. Thus Otto cycle has greaterefficiency and Diesel cycle has leastefficiency. The Efficiencies of the threecycles are otto dual diesel for constant compression

ratio and heat supplied.

(iii) Comparison of Otto and Diesel cycle for sameMaximum pressure and Heat supplied

The P V and T S diagram for otto and dieselcycles for constant maximum pressure and heat suppliedare shown in the Fig 1.12 (a) and (b).

Since same constant maximum pressure the point3, 3 lie on a constant pressure line.



P

3 ’’O tto

3 ’

3

4

4 ’

4 ’’

2’

2

1

Dual

D iese l

O VFig. 1.11(a)

3 ’’3 ’

32 ’

4’4 ’’

42

1

T

Dua l

O tto

Diese l

O

S

Fig. 1.11(b)

2

2 ’

1

44 ’

3 ’3T

OS

Fig. 1.12 (b)

VFig. 1.12 (a)

4

2 ’

P2 3 3 ’

4 ’

D iesel

O tto


On T S diagram the area under the line 4 1 is less for Diesel cyclethan Otto cycle and hence diesel cycle is more efficient than otto cycle forthe condition of maximum pressure and heat supplied.

Diesel Otto (For maximum pressure and heat supplied)

(iv) Comparison of Otto, Diesel and Dual cycle for same Maximumpressure and Heat rejection

The P V and T S diagram areshown in Fig 1.13 (a) and (b) for constantmaximum pressure and maximumtemperature.

For the same QR, the higher the Qs,

the higher the cycle efficiency. Therefore,


Dual cycle 1 2 3 3 4 1


We know that Air Standard Efficiency

1 Heat rejectedHeat supplied

1 constant

Qs

Qs (diesel) Area under 2 3

Qs (dual) Area under 2 3 3

Qs (otto) Area under 2 3

Qs (otto) < Qs (dual) < Qs (diesel) so we have

diesel dual otto

(for constant maximum pressure and heat rejection)

1.7 GAS TURBINE CLOSED CYCLE (OR) BRAYTON CYCLE (OR) JOULE CYCLE

Brayton cycle is the air standard cycle for gas turbine power plant.Refer the Fig.1.14. In this plant, air is compressed isentropically and heat isadded at constant pressure process. Then air expands in the turbineisentropically and then heat is rejected at constant pressure and the cyclerepeats.

P

2 ’

2

4

2 ’’ 3 ’ 3

1

VFig. 1.13(a)

3

Fig. 1.13(b)

2 ’’

2 ’

2

1

3 ’

4

T

V


Brayton cycle is called as External Combustion Engines Cycle, becausethe burning takes place outside the cylinder (or turbine).

Brayton cycle consists of following processes.

Process (1) - (2) Isentropic compression in compressor Q 0

Process (2) - (3) Constant pressure heating in combustion chamber[Qs Cp T3 T2]

Process (3) - (4) Isentropic expansion in turbine Q 0

Process (4) - (1) Constant pressure heat rejection exhaust.[Qr Cp T4 T1]

Heat supplied Qs Cp T3 T2 J/kg

Heat rejected Qr Cp T4 T1 J/kg

Network done Wnet Qs Qr

V

P

P = P2 3

P = P1 4

14

32

Fig. 1.14 (a)

P

S =CS =C

P =C

P =CC oo ling

c ha m be r

2 4

s

3

1

S =C

S =C

P=CP=C

T

Fig. 1.14 (b)

C T

C om b us tion C ha m be r

C oo lingC ha m be r

1

2 3

4

A tm o sp h e rica ir

F ue l

C - C om p re ss or T - Tu rb ine

Schematic diagram of Brayton Cycle.Fig.1.14 (c)


(i) Air standard efficiency

thermal Network doneHeat supplied

Wnet

Qs

Qs Qr

Qs

1 Qr

Qs 1

Cp T4 T1Cp T3 T2

air standard or thermal 1 T4 T1T3 T2 ...(1)

rp Pressure ratio during compression P2

P1

P3

P4

Process (1) - (2) isentropic proces (3) - (4) isentropic process

T2

T1 P2

P1

1

T3

T4 P3

P4

1

T2

T1 rp

1

T3

T4 rp

1

T2 T1 rp 1

T3 T4 rp

1

Substitute T2 and T3 in eqn (1), we get

Air standard or thermal

1 T4 T1

T4 rp

1

T1 rp

1

1 T4 T1

rp 1

[T4 T1]

air standard or thermal 1 1

rp 1

WT Turbine work Cp T3 T4 J/kg ;

WC Compressor work Cp T2 T1 J/kg ;

Fig:11.5

2 4

s

3

1

C om pressor w orkS =C

S =CTurbine w ork

P=CP=C

T


Wnet Net work WT WC

Wnet Qs Qr Cp T3 T2 Cp T4 T1

W Cp [T3 T4 T2 T1]

[Steady flow turbine work W R 1

T1 T2 Cp T1 T2]

m

a mass rate flow of air, kg/sec;

m

f mass rate of flow of fuel, kg/sec.

m

a

m

f Air fuel ratio

WT Turbine power m

a Cp T3 T4 in Watts.

WC Compression power m

a Cp T2 T1 in Watts.

WNet

Net power WT WC in Watts

Qs Heat supplied m

a Cp T3 T2 in Watts

m

f CV where CV Calorific value

1.7.1 Optimum pressure ratio for maximum specific work outputWork output during the cycle

W Heat received/cycle Heat rejected/cycle

W WT WC mCp [T3 T4 T2 T1]

W mCp

T3

1

T4

T3 T1

T2

T1 1

W mCp

T3

1

1

rpz T1 rp

z 1

where T3

T4

T2

T1 rp

1 rp

z and 1

z

W R T3

1

1

rpz T1 rp

z 1

where R mCp constant


Differentiating above equation and equating to zero.

dWdrp

R T3

zrp z 1

T1 zrpz 1

0

zT3

rpz 1

T1z rpz 1

(or) rp2z

T3T1

or rp T3T1

1/2z

T3T1

2 1

is the expression of pressure ratio for maximum workdone

Work Ratio: Work ratio is defined as ratio of net work output to the workdone by the turbine.

Work ratio WT WC

WT

Work ratio m Cp [T3 T4 T2 T1]

mCp T3 T4

Work ratio 1 T2 T1

T3 T4 1

T1T3

T2

T1 1

1 T4

T3

Work ratio 1 T1

T3

rp

1r 1

1 1

rp 1

r

1 T1

T3 rp

1r

Problem 1.33: In an air standard Brayton cycle air at 1 bar, 20C issupplied to a compressor where pressure ratio is 4.5. The maximumtemperature is 1000 K. Determine. (a) thermal (b) net work (c) work ratio

Solution:P1 1 bar

T1 20C 293 K


Tmax T3 1000 K

P2

P1

P3

P4 4.5

1-2 isentropic process

T2

T1 P2

P1

1

T2 293 4.50.4/1.4

T2 450.297 K

T3

T4 P3

P4

1

1000T4

4.50.4/1.4

T4 650.68 K

thermal 1 1

rp 1

1 1

4.50.4/1.4 0.35

thermal 0.35

(b) Turbine work WT Cp T3 T4 1005 1000 650.58 1005 349.32

WT 350891.94 J/kg

Compressor work Wc Cp T2 T1 1005 450.297 293

Wc 158005.5695 J/kg

Wnet WT Wc 350891.94 158005.5695 192886.3705 J/kg

Work ratio Wnet

WT

192886.3705350891.84

0.54956

Work ratio 0.54956

V

P

1 4

32

S=CS=C

P=C

P=C

2 4

s

3

1

S=C

S=C

P=CP=C

T


Problem 1.34: In an air standard Brayton cycle air at 300 K is suppliedto a compressor whose pressure ratio is 5. Mass rate of flow of air is 3kg/sec, air fuel ratio is 80 : 1. CV of fuel is 42 MJ/kg. Determine (a) network,net power; (b) work ratio; (c) max. temperature; (d) thermal

Solution:

T1 300 K ; rp 5 P2P1

P3P4

m

a 3 kg/sec ; Air fuel ratio m

a

m

f 80:1

CV 42 106 J/kg

m

a

m

f

801

m

f 0.0375 kg/sec

T2T1

P2

P1

1

T2

300 50.4/1.4

T2 475.14 K

Heat supplied

m

a Cp T3 T2 m

f CV Qs

3 1005 T3 475.14 0.0375 42 106

T3 997.79 K

T3

T4 rp

11

997.79T4

50.4/1.4

T4 629.99 K

(a) thermal 1 1

rp

1

1 1

50.4/1.4 0.368614

thermal 0.368614 or 36.8614%

V

P

14

32

S=CS=C

P=C

P=C

2 4

3

1

S=C

S=CP=C

P=C

T

S


(b) Work ratioTurbine work WT Cp T3 T4 1.005 997.79 629.99 369.45 kJ/kg

Compressor work

Wc Cp T2 T1 1.005 475.14 300 Wc 175.93 kJ/kg

Wnet WT Wc 193.52 kJ/kg

Work ratio WWT

193.52 103

369.45 103 0.5238

Work ratio 0.5238

Turbine power

WT m

a Cp T3 T4 3 1.005 997.79 629.99

WT 1108.37 kW

Compressor power

Wc m

a Cp T2 T1 3 1.005 475.14 300 527.8

Wc 527.8 kW

Net power WT Wc 580.57 kW

Problem 1.35: In an air standard joule cycle air at 300 K is supplied toa compressor where pressure ratio is 6. Max. temp is 1000 K. m

a 2.5 kg/s,

CV 42 MJ/kg. Determine (a) Air fuel ratio; (b) net power; (c) Work

ratio; (d) thermal

Solution:T1 300 K

rp P2P1

P3P4

6

T3 Tmax 1000 K

m

a 2.5 kg/s

CV 42 MJ/kg

T2T1

P2

P1

1


T2300

60.4/1.4 ; T2 500.55 K

T3T4

P3P4

1 or 1000

T4 6

0.41.4

T4 599.340 K

WT Cp T3 T4

1.005 1000 599.34

WT 402.46 kJ/kg

Wc Cp T2 T1

1.005 500.55 300

Wc 201.45 kJ/kg

Wnet WT Wc 201.01 kJ/kg

Work ratio Wnet

WT

201.01402.46

0.4994

Work ratio 0.4994

Air fuel ratioQs m

a Cp T3 T2 m

f CV

m

f 2.5 1005 1000 500.55

42 106 0.0299 kg/s

ma

mf 2.5

0.0299 83.72

WT m

a Cp T3 T4 1006.157 kW

Wc m

a Cp T2 T1 503.631 kW

Wnet WT Wc 502.526 kW

th 1 1

rp 1

1 1

60.4/1.4 0.4006 40.06 %

2 4

3

1

S=C

S=CP=C

P=C

T

S

V

P

14

32

S=CS=C

P=C

P=C


Problem 1.36: In an air standard Brayton cycle, the pressure ratio is 6.The condition of air at the beginning of compression is 1 bar, 300 K. Airfuel ratio is 60 : 1. The calorific value of fuel is 42 kJ/kg. Determine (a)Maximum temperature; (b) Work ratio; (c) Net work; (d) th

Solution:

Pressure ratio 6 P2

P1

P3

P4 ; P1 1 bar ; T1 300 K

m

a

m

f

601

CV 42 106 J/kg

(1)-(2) isentropic process

T2T1

P2

P1

1

T2300

60.4/1.4

T2 500.55 K

Qs m

a Cp T3 T2 m

f CV

Divide by m

f

m

a

m

f Cp T3 T2 CV

60 1005 T3 500.55 42 106

T3 1197.41 K

(3) - (4) Isentropic process

T3

T4 P3

P4

1

1197.4172T4

60.4/1.4

T4 717.65 K

WT Cp T3 T4 1005 1197.41 717.65

V

P

14

32

S=CS=C

P=C

P=C

2 4

3

1

S=C

S=C

P=C

P=C

T

S


481919 J/kg

Wc Cp T2 T1 1005 500.55 300

201455 J/kg

(b) Work ratio Wnet

WT

WT WcWT

0.58197

the 1 1

rp 1

1 1

60.4/1.4 0.4006

th 0.4006 or 40.06%

Problem 1.37: In an air standard Brayton cycle air at 1 bar, 20C issupplied to compressor whose pressure ratio is 4.5, max temp 1000 K,calorific value 42 MJ/kg. Determine (a) Net work; (b) Work ratio;(c) th; (d) Air fuel ratio

Solution:P1 1 bar

T1 20 C 293 K

rp P2

P1

P3

P4 4.5

T3 1000 K,

CV 42 106 J/kg


T2

T1 P2

P1

1

T2

293 4.50.4/1.4

T2 450.29 K

V

P

14

32

S=CS=C

P=C

P=C

2 4

3

1

S=C

S=C

P=CP=C

T

S



T3

T4 rp

0.4/1.4 1000

T4 4.50.4/1.4

T4 650.68 K

WT Cp T3 T4 1005 1000 650.68

350891.34 J/kg

Wc Cp T2 T1 1005 450.29 293

158005.57 J/kg

Wnet WT Wc 192885.76 J/kg

Work ratio Wnet

WT 0.5497

th 1 1

rp 1

1 1

4.50.4/1.4 0.349319

Heat supplied Qs m

a Cp T3 T2 m

f CV

Divide by m

f, we get m

a

m

f Cp T3 T2 CV

m

a

m

f

CVCp T3 T2

42 106

1005 1000 450.29

Air fuel ratio ma

mf 76.0627

1.7.2 Turbine and compressor efficiencyMost of the compressible flow turbomachines, such as turbines,

compressors and blowers are assumed to be adiabatic machines (ie) Qs 0.

Efficiency of turbine Actual turbine work output

Isentropic turbine work output

T T3 T4T3 T4

The compressor efficiency is given by


W ide al W a ct

P 3

= C (T - T )p 3 4

T - T3 4

T - T3 ’4

= C (T - T )p 3 ’4

P =P4

4’

4

4 ’

3

T

s

T T2 1’ T T2 1

T

2 ’

P =P2

2’

P 1

2

1

W ide al

W a ctua l

= C (T - T )p 2 1

= C (T - T )p 2 ’ 1

S


c Isentropic compression work

Actual compression work

C T2 T1

T2 T1

1.7.3 Combustion Chamber - Combustion efficiencyThe compressed air from the compressor enter the combustion chamber

at a pressure P2, temperature T2 and velocity C2. During combustion, the

enthalpy of the air-fuel mixture increases. The mass of air-fuel mixture comingout from the combustion chamber is

m m

a m

f

where m

a Mass of air per second

m

f Mass of fuel per second

The efficiency of the combustion chamber,

CB increase in enthalpy of gasesenergy supplied in the fuel

m

a m

f h03 m

a h02

m

f Qf

The combustion efficiency can also be defined as

CB theoretical air fuel ratio

actual air fuel ratio

ff

where f m

f

m

a

Problem 1.38: A gas turbine unit receives air at 100 kPa and 300 K andcompresses it adiabatically to 620 kPa with the efficiency of the compressor88%. The fuel has a heating value of 44180 kJ/kg and the air fuel ratio is0.017 kg fuel/kg air. The turbine internal efficiency is 90%. Calculate thecompressor work, turbine work and thermal efficiency. TakeCp 1.005 kJ/kgK, 1.4 for air and Cp 1.147 kJ/kg, 1.33 for products

of combustion. (JNTUH/June-July 2014/Set 1)

Solution: P1 100 kPa 1 bar; comp 88%; T1 300 K; P2 P3 620 kPa 6.2 bar;


Turbine 90%, C.V 44180 kJ/kg ; Cpa 1.005 kJ/kgK; 1.4 ;

Cpg 1.147 kJ/kgK; n 1.333

From isentropic process 1 2,

T2

T1 P2

P1

1

T2 T1 6.21

1.4 11.4

300 6.20.4/1.4

T2 505.26 K

Since the compressor efficiency is 88%,

comp T2 T1

T2 T1 ; 0.88

505.26 300

T2 300

T2 533.25 K

In the heat addition process 2 3, the heat balance is given by

Heat added m

f C.V

m

a m

f Cpg T3

m

a Cpa T2

m

f C.V

1

m

f

m

a Cpg

T3 Cpa

T2

m

f

m

a

C.V

0.017 44180 1 0.017 1.147 T3

1.005 533.25

751.06 1.1664 T3 535.91

T3 1103.36 K

In expansion process 3 4 , T4

T3 P4

P3

n 1n

T4 T3 P4

P3

n 1n

T4 1103.36

16.2

0.331.33

T4 701.64 K

T

S

1

2

3

4

4 ’2 ’

620 kPa

100 kPa


Actual temperature at turbine inlet based on internal turbine efficiency,

turbine T3 T4T3 T4

; 0.9

1103.36 T41103.36 701.64

T4 741.8 K

The compressor work is given by,

Wc CPa T2 T1

1.005 533.25 300

234.42 kJ/kg of air

and

Turbine work is given by,

WT Cpg T3 T4

1.147 1103.36 741.8

414.7 kJ/kg of air

Net work done, Wnet WT WC 414.7 234.42

180.28 kJ/kg of air

But Heat supplied, Qs m

f

m

a

C.V 0.017 44180

751.06 kJ/kg of air

Since, Thermal efficiency WnetQs

180.28751.06

th

24%

Problem 1.39: A simple gas turbine cycle works with a pressure ratio of8. The compressor and turbine inlet temperatures are 300 K and 800 K

respectively. If the volume flow rate of air is 250 m3/s, compute the poweroutput and thermal efficiency. (GATE - 2014)

Solution:

P2P1

8; T1 300 K; T3 800 K; { V

a 250 m3/s }


Process 1 2 isentropic compression

T2

T1 P2

P1

1

T2 300 8

1.4 11.4

T2 543.75 K

Process 3 4 isentropic expansion

T3

T4 P3

P4

1

T4

T3

P3

P4

1

800

80.41.4

T4 441.37 K

Since

Work done on the compressor,Wc CpT2 T1 1.005 543.75 300

244.96 kJ/kg

Work done by the turbine, WT Cp T3 T4 1.005 800 441.37

360.42 kJ/kg

Wnet WT WC 360.42 244.96 115.46 kJ/kg

Heat supplied Cp T3 T2 1.005 800 543.75 257.53 kJ/kg

Since P RT

1 P1

R T1

1 105

287 300 1.161 kg/m3

Power output Wnet V

a 1 115.46 250 1.161

33512.265 kW

Thermal efficiency Wnet

heat supplied

115.46257.53

44.83%


Problem 1.40: In an oil-gas turbine installation, it is taken at pressure of1 bar and 27C and compressed to a pressure of 4 bar. The oil with acalorific value of 42000 kJ/kg is burnt in the combustion chamber to raisethe temperature of air at 550C. If the air flows at the rate of 1.2 kg/s, findthe net power of the installation. Also find the air fuel ratio. TakeCp 1.05 kJ/kgK. (GATE - 2012)

Given:P1 1 bar; T1 27 C 300 K; P2 4 bar ; CV 42000 kJ/kg,

T3 550C 823 K ; m

a 1.2 kg/s; Cp 1.05 kJ/kgK

Solution:

Since, T2

T1 P2

P1

1

T2 T1

P2

P1

1

300 41

0.41.4

T2 446 K

Since m

f C.V m

a Cp T3 T2

m

f m

a Cp T3 T2CV

1.2 1.05 823 446

42000

mf 0.01131 kg/s

Since airfuel ratio, m

a

m

f

1.20.01131

106.1008 kg of air/ kg of fuel

Since T3

T4 P3

P4

1

P2

P1

1

T4 T3 P1

P2

1

T4 823 14

0.41.4

T4 554 K

Since,

Work done on the compressor, Wc


m

a Cpa T2 T1 1.2 1.05 446 300

Wc 183.96 kJ/kg

Since,

Work done by the turbine WT

m

f m

a Cp T3 T4 1.21131 1.05 823 554

WT 342.13 kJ/kg

Net work Wnet

WT WC 342.13 183.96

Wnet 158.17 kW

Problem 1.41: In an open cycle gas turbine plant, air enters at 1.5 barand 25C and leaves the compressor at 5.5 bar. Maximum cycle temperature

is 675 C, pressure loss in the combustion chamber is 0.2 bar. Efficiency of

compressor is 0.85, turbine is 0.80, combustion is 0.90, 1.4 and

Cp 1.022 kJ/kgK for air and flue gases. Neglecting mass of the fuel,

determine (a) Quantity of air circulation, if power developed is 1050 kW(b) Heat supplied per kg of air (c) Thermal efficiency of the cycle (GATE - 2016)

Solution:

P1 1.5 bar; P2 5.5 bar; P4 1.5 bar; P3 5.5 0.2

P3 5.3 bar; T1 25 C 273 298 K; T3 675 C 273

T3 948 K ; c 0.85; T 0.80 ; com 0.90; Cp 1.022 kJ/kgK 1.4

The Quantity of air circulation, ma

Process 1 2 isentropic compression

T2

T1 P2P1

1

5.51.5

1.4 11.4

T2 298 5.51.5

0.41.4

T2 431.95 K


Since

c T2 T1

T2 T1

0.85 431.95 298

T2 298

T2 455.58 K

Process 3 4 isentropic expansion

T4

T3 P4

P3

1

1.55.3

0.41.4

T4 948 1.55.3

0.41.4

T4 660.97 K

Since T T3 T4T3 T4

0.80 948 T4

948 660.97 T4 718.37 K

Work done on compressor, Wc Cp T2 T1 1.022 455.58 298

Wc 161.046 kJ/kg

Work done by Turbine, WT Cp T3 T4 1.022 948 718.37

WT 234.68 kJ/kg

Wnet WT WC 234.68 161.046 73.635 kJ/kg

The power developed, P m

a Wnet 1050 m

a 73.635

ma 14.259 kg

(ii) Heat supplied per kg of air

Cp T3 T2

com

1.022 948 455.580.9

559.17 kJ/kg

(iii) Thermal efficiency of cycle

thermal Wnet

Heat supplied

73.635559.17

0.1316 13.16%

T

S1

2

3

4

4 ’

T =948K3

T =298K1

5 .5bar 5 .3bar

1 .5bar


1.8 STANDARD RANKINE CYCLEDraw the p-V, T-S, h-S diagrams and theoretical lay out for Rankine cycleand hence deduce the expression for its efficiency. (AU. Nov/Dec 2015)

1.8.1 The Ideal Rankine CycleMany of the impracticalities associated with the Carnot cycle can be

eliminated by superheating the steam in the boiler and condensing itcompletely in the condenser, in the cycle called Rankine cycle, which is theideal cycle for vapour power plants. The ideal Rankine cycle does not involveany internal irreversiblities and consists of the following four processes:

1-2 Isentropic expansion in a turbine

2-3 Constant pressure heat rejection in a condenser

3-4 Isentropic compression in a pump

4-1 Constant pressure heat addition in a boiler

Refer Fig 1.18 and Fig.1.19

Water enters the pump at state 3 as saturated liquid and is compressedisentropically to the operating pressure of the boiler. The water temperatureincreases somewhat during this isentropic compression process due to a slightdecrease in the specific volume of the water. The vertical distance betweenstates 3 and 4 on the Ts diagram is greatly exaggerated for clarity. Waterenters the boiler as a compressed liquid at state 4 and leaves as a superheated

w pump.in

Pum p

3 C onden ser

Tu rbine

Bo iler

w turb,out

41

2q out

q in

Fig. 1 .18 The sim ple ideal Rankine cycle


vapour at state 1. The boiler is basically a large heat exchanger where theheat originating from combustion gases, nuclear reactors, or other sourceswith the section where the steam is superheated, is often called the steamgenerator.

The superheated vapour at state 1 enters the turbine, where it expandsisentropically and produces work by rotating the shaft connected to an electricgenerator. The pressure and the temperature of the steam drop during thisprocess to the values at state 2, where steam enters the condenser. At thisstate, steam is usually a saturated liquid-vapour mixture with a high quality.Steam is condensed at constant pressure in the condenser, which is basicallya large heat exchanger, by rejecting heat to a cooling medium such as a lake,a river, or the atmosphere. Steam leaves the condenser as saturated liquid andenters the pump, completing the cycle.

Fig. 1.19(a) Schem aticd iagram

(b) T-S d iagram


1.8.2 Efficiency of Standard Rankine Cycle

BoilerRefer Fig. 1.19. Take point (4) to (1): Feed water is passing to the

boiler. Heat is added to the water in the boiler. The water gets heated andbecomes dry saturated steam (or) super heated steam.

TurbineTake point (1) to (2): The high pressure steam is expanding in the turbine,

thus work is produced. i.e. The turbine rotates. The steam leaves the turbine aslow pressure steam.

Condenser: Condenser is used to convert the low pressure steam into water.Take point (2) to (3). The low pressure steam is passing through condenserwhere heat is liberated from the steam. So the steam becomes water. To coolthe steam, separate cooling water is circulated through condenser from thecooling tower. This cooling water and the steam will not mix together inmost of the condensers.

PumpTake point (3) to (4) The water leaving condenser is pumped to the

boiler by pump. Usually, pump work is neglected since it is very small workwhen compared to turbine work output.

1 - 2 Turbine workTurbine work output (Isentropic expansion in Turbine)

WT h1 h2 kJ / kg

Turbine power mh1 h2 kW

where m Mass flow rate of steam in kg/sec.

h1 and h2 can be taken from steam table for P1 and P2 respectively.

P1 high pressure (or) boiler pressure (or) inlet to turbine pressure

P2 low pressure (or) condenser pressure

Also, we can use Mollier diagram to find h1 and h2.

2 - 3 Constant pressure condensationQ2 = Heat rejected h2 h3 kJ / kg

Q2 in kW mh2 h3 kW

h3 hf at low pressure P2.


3 - 4 Pump WorkWpump Wp h4 h3 kJ / kg

vf P1 P2 kJ / kg

where vf for P2 from steam table and

P1 and P2 are in kPa

Pump power mWp

Define specific steam consumption, specific heat rate and work ratio.(Nov/Dec 2012 - AU, May/June 2012 - AU)

Net WorkWnet WT Wp

WT If Wp is negligible

Thermal Efficiency: It is ratio of network to the heat supplied.

cycle or rankine or thermal Wnet

Qsupply

4 - 1 Heat Supplied in Boiler: Qsupply

(Constant pressure heat supply)

Qsupply h1 h4 kJ / kg

Qsupplyin kW m h1 h4 kW Specific heat rate

Specific steam consumption. ‘SSC’ (or) steam flow rate per kW

3600Wnet

kg

kWhr

Work ratio Wnet

WT

Problem 1.42: A steam turbine receives steam at 15 bar and 350C andexhausts to the condenser at 0.06 bar. Determine the thermal efficiency ofthe ideal rankine cycle operating between these two limits. Neglect the pumpwork.

Solution

P1 15 barBoiler pressure

; t1 350C; P2 0.06 barCondenser pressure

Work ratio Wnet

WT


From Mollier diagram,

h1 3147.5 kJ ; h2 2188.1 kJ / kg ; h3 hf for 0.06 bar 151.5 kJ / kg

[From steam (pressure) table, For 0.06 bar]

h4 h3 Wp h3 151.5 kJ / kg [ . . . Wp is negligible ]

Wnet WT h1 h2 3147.5 2188.1 959.4 kJ / kg

Qs h1 h4 h1 h3 3147.5 151.5 2996 kJ / kg[ . . . h4 h3 ]

rankine Wnet

Qs

959.42996

32.023%

h

h =3147.5 kJ/kg1

M ollier diagram

1

350 Co

15bar

0.06bar

2

h =2188.1 kJ/kg2

s


Problem 1.43: In a rankine cycle, the steam flows to turbine as saturatedsteam at a pressure of 35 bar and the exhaust pressure is 0.2 bar. Determine(using steam table only) (i) pump work (ii) the turbine work (iii) the rankineefficiency (iv) the condenser heat flow (v) the dryness fraction at the end ofexpansion. The mass flow rate of steam is 9.5 kg/sec. (AU. Nov/Dec 2011)(AU. Nov/Dec 2014)

Solution:

P1 35 bar dry saturated;

P2 0.2 bar;

m 9.5 kg / sec.

To Find h1

h1 hg for 35 bar 2802 kJ / kg

To Find x2

s1 sg for 35 bar = 6.123 kJ/kg K

Isentropic expansion. So, s1 s2

So, s2 6.123 kJ / kg K

But at 0.2 bar , sg 7.909 kJ / kg K

Since s2 6.123

sg 7.909

,

it is wet steam at exit of turbine.

So, s2 sf x2sfg for 0.2 bar

6.123 0.832 x27.077

x2 0.748

Dryness fraction at the exit of turbine x2 0.748

To Find h2

hf 251.5 kJ/kg; hfg 2358.4 kJ/kg for 0.2 bar

h2 hf x2hfg for 0.2 bar

251.5 0.7482358.4 2014.72 kJ / kg

p =p =35bar1 4

1

4

p =p =0.2bar2 33 2

S

T

35 bar

0 .2 bar1

2

2802

2014.7

h

x = 0.7482

S


To Find h3

h3 hf3 251.5 kJ / kg for 0.2 bar.

To Find h4

vf3 0.001017 m3 / kg for 0.2 bar

h4 h3 Wp

Wp vf3P1 p2 0.00101735 0.2 102

3.54 kJ / kg

h4 251.5 3.54 255.04 kJ / kg

To Find Pump WorkWp 3.54 kJ / kg

Pump power mWp 3.54 9.5 33.63 kW

To Find Turbine WorkWT h1 h2 2802 2014.72 787.28 kJ / kg

Turbine power mWT 9.5 787.28 7479.2 kW

To Find Rankine Efficiency

rankine Wnet

Qs

Qs h1 h4 2802 255.04 2546.96 kJ / kg

Wnet WT Wp 787.28 3.54 783.74 kJ / kg

rankine 783.742546.96

0.30772 30.772%

To Find Condenser Heat Flow Qrej

Qrej h2 h3 2014.72 251.5 1763.22 kJ / kg

Qrej in kW mh2 h3 9.5 1763.22 16751 kW

Extra:

To find Ratio of pump work and Turbine work:WP

WT

3.54787.28

5 10 3


Problem 1.44: Consider a steam power plant operating on the ideal Rankinecycle. Steam enters the turbine at 3 MPa and 623 K and is condensed inthe condenser at a pressure of 10 kPa. Determine (i) the thermal efficiencyof this power plant, (ii) the thermal efficiency if steam is superheated to873 K instead of 623 K, and (iii) the thermal efficiency if the boiler pressureis raised to 15 MPa while the turbine inlet temperature is maintained at873 K (AU. Nov/Dec 2009)

Solution:(i) P1 3 MPa; T1 623 K; P2 10 kPa;

From steam table at P2 10 kPa, hf2

191.83 kJ / kg

From Mollier chart,

h1 3120 kJ / kg ; h2 2130 kJ / kg

Thermal efficiency of the plant

h1 h2

h1 hf2

3120 2130

3120 191.83 33.8%

(ii) P1 3 MPa ; T1 873 K ;

P2 10 kPa

From mollier chart,

h1 3690 kJ / kg ; h2 2390 kJ / kg


h1 h2

h1 hf2

3690 2390

3690 191.83

0.3716 or 37.16 %

h

s

350 Co

10kPa

3MP a

3120 kJ 1

kg

22130 kJ

kg

3690kJkg

600 Co

1

h

s

3mP

a

10kPa

2390kJkg 2

x=1


(iii) P1 15 MPa; T1 873 K;

P2 10 kPa

From Mollier chart,

h1 3600 kJ / kg;

h2 2120 kJ / kg


h1 h2

h1 hf2

3600 2120

3600 191.83 43.42%

Problem 1.45: A simple Rankine cycle works between pressure of 30 barand 0.04 bar, the initial condition of steam being dry saturated. Calculatethe cycle efficiency, work ratio and specific steam consumption. (JNTUK/May - June 015/Set 2)

Solution

P1 P4 30 bar; P2 P3 0.04 bar

From steam tables, corresponding to P1 30 bar

sg1 6.184 kJkgK; hg1 2802.3 kJkg

From steam tables, corresponding to P2 0.04 bar

hf2 121.4 kJkg; hfg2

2433.1 kJkg; sf2 0.423 kJkgK;

sfg2 8.053 kJkgK; vf 0.001004 m3/kg

To find x2

Since s1 s2

s1 sg1 sf2 x2 sfg2

6.184 0.423 x2 8.053

x2 0.715

h2 hf2 x2 hfg2

121.4 0.715 2433.1

h2 1861.06 kJ/kg

h

s

2

1

600 Co

15M Pa

10kPa

x=1

3600kJ

kg

2120kJkg

1

4

p =p2 3

3 2


Since the compression work,

h4 h3 vf P1 P2

h4 h3 vf P1 P2

Since, h3 hf2

h4 121.4 0.001004 30 0.04 102

[. . . 1 bar = 100 kpa] h4 124.407 kJ/kgK

The turbine work,WT h1 h2

2802.3 1861.06

WT 941.24 kJ/kg

The pump work,WP v P1 P2

h4 h3

124.407 121.4

WP 3.007 kJ/kg

rankine Wnet

Qs

WT WP

h1 h4

941.24 3.007

2802.3 124.407

0.35

rankine cycle 35%

Since work ratio,

WnetWT

WT WP

WT

941.24 3.007

941.24

Work ratio 0.996


Specific steam consumption: (SSC)

3600Wnet

3600

WT WP

3600

941.24 3.007

3.836 kg/kW hr

Problem 1.46: Dry saturated steam at 15 bar is supplied to a rankine cyclewhere exhaust pressure is 1 bar. Find (a) thermal (rankine), steam

consumption per kW, carnot. (b) If the exhaust pressure is reduced to 0.2

bar by introducing a jet condenser, then determine % increase in rankineefficiency and % decrease in SSC.

Solution

Given: P1 P4 15 bar; P2 P3 1 bar; Initially dry saturated.

rankine Wnet

Qsupply

WT Wp

Qs

WT h1 h2

Wp h4 h3 vfP1 P2

Qs h1 h4

Find h1, h2, h3 and h4

h1 hg for P1 15 bar 2790 kJ / kg from steam table (or) from

Mollier chart, hs diagram15 bar line will cut saturated curve at (1). Draw vertical line from (1).

This vertical line will cut the 1 bar line at (2).

h2 2340 kJ / kg

h3 hf (for 1 bar from steam table) = 417.5 kJ/kg

vf 0.001043 m3 / kg (for 1 bar)

h4 h3 vfP1 P2

h4 h3 vfP1 P2 417.5 0.00104315 1 102

418.9602 kJ / kg


[15 and 1 bar is multiplied by 102 to make kPa . . . 1 bar 100 kPa ]

To Find WT, Wp and Qs

WT h1 h2 2790 2340 450 kJ / kg

Wp vfP1 P2 0.00104315 1 102 1.4602 kJ / kg

Qs h1 h4 2790 418.9602 2371.04 kJ / kg

Wnet WT Wp 450 1.4602 448.54 kJ / kg

To Find rankine

rankine Wnet

Qs

WT Wp

Qs

450 1.46022371.04

18.92%

To Find Specific Steam Consumption (SSC)

SSC 3600Wnet

3600

448.54 8.03 kg / kWhr

To Find carnot

carnot Tmax Tmin

Tmax

For 15 bar, tsat Tmax 198.3C 273 471.3 K

For 1 bar , tsat Tmin 99.63C 273 372.63 K

h =27901

h =23402

1

2

s

h

15 bar

1 bar

sa tu ra ted curve

s =s1 2

h - h1 4

p =p1 41

23

4

s =s1 2

W =T h -h1 2

p =p2 3

p=

p 1

4

p

4

3 w

=h

-h

s

T


carnot 471.3 372.63

471.3 20.936%

Case (b)When P2 0.2 bar ; h2 2120 kJ / kg

h3 hffor 0.2 bar 251.5 kJ

vf3 0.001017 m3 / kg (for 0.2 bar)

h4 h3 vf3P1 P2

h4 251.5 0.00101715 0.2 102 253.005 kJ / kg

WT h1 h2 2790 2120 670 kJ / kg

Wp vf3P1 P2 0.00101715 0.2 102 1.51 kJ / kg

Wnet WT Wp 670 1.51 668.5 kJ / kg

Qs h1 h4 2790 253.005 2536.995 kJ / kg

New rankine

rankine Wnet

Qs

668.52536.995

26.35%

Increase in rankine efficiency 26.35 18.92

18.92 39.3%

T

S

p =p =15bar1 4

4

32p =p =0.2bar2 3

1

S

h =27901

2

1

h

0.2bar

15ba r

h =21202


New SSC

SSC 3600Wnet

3600668.5

5.39 kg / kWhr

Decrease in SSC 8.03 5.39

8.03 32.94%

Problem 1.47: Steam at 20 bar, 360 C is expanded in a steam turbine to0.08 bar. It then enters a condenser, where it is condensed to saturated liquidwater. The pump feeds back the water into the boiler.1. Assuming ideal processes find the net-work and cycle efficiency per kgof steam2. If the pump and the turbine have 80% efficiency, find the percentagereduction in the net-work and cycle efficiency. (AU. May/June 2012)

Solution

Boiler pressure P1 20 bar 360C; Condenser pressure P2 0.08 bar

At 20 bar and 360C

h1 3159.3 kJ / kg

s1 6.9917 kJ / kgK

h3 hf P2 173.88 kJ / kg

s3 sf P2 0.5926 kJ / kg K

hfgP2 2403.1 kJ / kg

sg p2 8.2287 kJ / kgK

vf P2 0.001008 m3 / kg

sfg P2 7.6361 kJ / kg K

Now, s1 s2

6.9917 sf P2 x2 sfgP2

0.5926 x2 7.636

T1

P = 2 0bar1

360 Co

P =0 .08 bar2

4

3 2

S

5


x2 6.9917 0.5926

7.6361

x2 0.838

h2 hf P2 x2 hfgP2

173.88 0.838 2403.1 2187.68 kJ / kg

Network, Wnet

WNet Wturbine Wpump

Wpump h4 hf3 vf P2 P1 P2 0.001008 20 0.08 100

2.008 kJ / kg

h4 2.008 hf P2 2.008 173.88

175.89 kJ / kg

Wturbine h1 h2 3159.3 2187.68 971.62 kJ / kg

Wnet WT WP 971.62 2.008 969.61 kJ / kg

Q1 h1 h4 3159.3 175.89 2983.41 kJ / kg

cycle Wnet

Q1

969.612983.41

0.325 or 32.5 %

Case (2)

Wpump 2.008

0.8 2.51 kJ / kg

WT 0.8 971.62 777.296 kJ / kg

Wnet WT WP 777.296 2.51 774.8 kg / kg

% Reduction in network 969.61 774.8

969.61 100 20.1%

h4 2.51 173.88 176.4 kJ / kg

Q1 h1 h4 3159.3 176.4 2983 kJ / kg

Wnet

Q1

774.82983

25.9 %


1.9 RANKINE CYCLE - REHEATING CYCLEWhat are the ways to improve Rankine cycle efficiency?

(AU. Apr/May 2011)If the dryness fraction of steam leaving the turbine is less than 0.85,

then corrosion and erosion of turbine blades occur. To avoid this situation,reheat is used.

Refer Fig. 1.20 and Fig.1.21 In the reheat cycle, theexpansion of steam takes placein one (or) more turbines.Steam is expanded in the HP(High pressure) turbine first,then it is reheated. Thereheated steam is againexpanded in the LP (LowPressure) turbine. Reheat cyclegives small increase in cycleefficiency. It increases the network output.

Refer Fig. 1.21.

Condenser

HPT LPT

Pum p

Reheater

6

4

1

2

3

5

Boiler6

5

3

4

Fig. 1.20

Fig. 1.21


P1 - Boiler pressure; P2 P3 = Reheat pressure; P4 - Condenser

pressure; T1 - boiler temp (or) superheat temperature; T3 - Reheat temperature.

Note:

If T3 is not given, then we can assume T3 T1

Refer Fig. 1.22

h1, h2, h3, h4 - Take from mollier chart h s diagram) (or) from steam table.

h5 hf for condenser pressure.

Wp h6 h5 vf P1 P2 100

[ vf sp. volume of fluid at condenser pressure]

WT h1 h2 h3 h4

Wp h6 h5

Wnet net work WT WP

Qs heat supplied h1 h6 h3 h2

thermal Wnet

Qs

1

2

3

4

5

6

P =C1

P =C4

T 1T3

Fig. 1.22


1.9.1 Advantages (or) Effects of Re-heatingWhat are the effects Reheating? (Nov/Dec 2016 - AU)

(i) Due to reheating, networkdone increases

(ii) Due to reheating, heat supply increases

(iii) Due to reheating, thermal efficiency increases

(iv) Due to reheating, the turbine exit steam dryness fraction increases -so moisture decreases - so blade errosion becomes minimum - solife of the turbine will be increased.

Problem 1.48: Steam at 90 bar, 480C is supplied to a steam turbine. Thesteam is reheated to its original temperature by passing it through a reheaterat 12 bar. The condenser pressure is 0.07 bar. Steam flow rate is 1 kg/sec.Determine (a) network output; (b) thermal [neglect the pressure loss in

reheating and boiler. The expansion is isentropic] (AU. Nov/Dec 2013)

Given:

P1 boiler pressure 90 bar T1 boiler temperature 480C

T3 Reheat temperature T1 given 480C

P4 condenser pressure 0.07 bar

From mollier chart,

h1 3330 kJ / kg ; h2 2805 kJ / kg ; h3 3440 kJ / kg ; h4 2360 kJ / kg

h5 hf for 0.07 163.4 kJ/kg ; PR Reheat pressure;

Pb Cooler Pressure; Pc Condenser Pressure

h6 h5 vf Pb pc 100 vf for condenser pressure

h6 163.4 0.001007 90 0.07 100

h6 163.4 9.0625 172.4625

WT h1 h2 h3 h4 3330 2805 3440 2360 1605 kJ / kg

Wp h6 h5 172.4625 163.4 9.0625 kJ / kg

Wnet WT Wp 1595.9375 kJ / kg

Qs h1 h6 h3 h2 3330 172.4625 3440 2805

3792.5375 kJ / kg


thermal Wnet

Qs

1595.93753792.5375

0.4208 42.08%

EXTRA

Compare network done and efficiency of the above cycle with that of thesimple cycle.

The simple cycle is shown in Fig B.

T =4801oC

P=12

R

P =0.07c

P=90

b

1

2

3

4

5

6

T =480 C1o

P =0 .07 ba r

c

P=90 ba r

b

1

2

3

4


h1 3330 kJ / kg ; h2 2030 kJ / kg ; h3 hf for condenser pressure 163.4 kJ / kg

h4 h3 vf Pb Pc 100 ; [vf for condenser pressure Pc]

h4 172.4559

WT h1 h2 1300 kJ / kg

WP h4 h3 9.0559 kJ / kg


Qs h1 h4 3157.5441 kJ / kg

thermal Wnet

Qs

1290.9443157.5441

thermal 40.884 %

Note: reheat cycle

Wnet 1595.9375 kJ / kg

thermal 42.08%

Because of reheating, work output increases and also thermalefficiency.

Problem 1.49: In an ideal reheat cycle, the steam enters the turbine at 30bar and 500C. After expansion to 5 bar, the steam is reheated to 500Cand then expanded to the condenser pressure of 0.1 bar. Determine the cyclethermal efficiency, mass flow rate of steam. Take power output as 100 MW. (AU. May/June 2007) (FAQ)

Solution:

From Mollier chart


h6 hf4 from steam tables, at 0.1bar 191.8 kJ / kg

cycle h1 h2 h3 h4h1 h6 h3 h2

3470 2940 3490 25603470 191.8 3490 2940

1460

3828.2 0.3813 38.13%

Mass flow rate of steam m


Wnet m Power out put

h1 h2 h3 h4

m 100 103 kW

m

100 103

[3470 2940 3490 2560] 68.49 kg / s

Steam flow rate 3600

h1 h2 h3 h4

3600

3470 2940 3490 2560

2.45 kg / kWh

Problem 1.50: In the reheat cycle steam at 150 bar, 550C enters into theHP turbine. The condenser pressure is 0.1 bar. The moisture content atcondenser inlet is 5%. Determine (a) reheat pressure; (b) cycle efficiency;(c) steam flow rate per kW.

Solution:

Hint: condenser inlet is condition (4) x4 0.95

Assumption: T3 T1, From Mollier Chart

h1 3455 kJ / kg ; h2 2785 kJ / kg h3 3590 kJ / kg ; h4 2460 kJ / kg

30bar

5bar

h

1

2

3

0.1bar

500 C o

4

S

30bar

5bar1

2

3

5

6

4

S

T


h5 hf for condenser pressure 191.8 kJ / kg

h6 h5 vf P1 P4 100 ; [vf is for condenser pressure 0.1 bar]

h6 191.8 0.001010 150 0.1 100

h6 206.9399 kJ / kg

(a) P3 reheat pressure 12.75 bar

WT h1 h2 h3 h4 3455 2785 3590 2460

WT 1800 kJ / kg

Wp h6 h5 206.9399 191.8 15.1399 kJ / kg


Qs h1 h6 h3 h2

3455 206.9399 3590 2785 4053.06

Qs 4053.06 kJ / kg

(b) thermal Wnet

Qs

1784.864053.06

0.44037 44.037 %

(c) SSC 1

Wnet 3600

11784.8601

3600 2.01696 kg

kW hr

T 1

0.1 bar

P =150bar

b

T 3 550 Co

x =0.954

1

2

3

4

5

6

P =C


Problem 1.51: Consider a steam power plant operating on the ideal reheatRankine cycle. Steam enters the high-pressure turbine at 15 MPa and 878 Kand is condensed in the condenser at a pressure of 10 kPa. If the moisturecontent of the steam at the exit of the low-pressure turbine is not to exceed10.4 percent, determine (i) the pressure at which the steam should bereheated and (ii) the thermal efficiency of the cycle. Assume the steam isreheated to the inlet temperature of the high-pressure turbine. (AU. Nov/Dec 2009)

Solution:P1 15 MPa ; T1 873 K ;

P4 10 kPa

x4 100 10.4 89.6% 0.896

From Mollier chart,

h1 3600 kJ / kg ;

h2 3120 kJ / kg,

P2 P3 38 bar

h3 3680 kJ / kg

h4 2350 kJ / kg

h5 hf at 10 kPa from steam table 191.83 kJ / kg

Pump work Wp

Wp vf P1 P4 100 0.001 150 0.1 100

14.99 kJ / kg Vf1 0.001, at 150 bar Wp h6 h5 14.99

h6 h5 14.99 191.8 14.99 206.79 kJ / kg

Heat supplied Qs

Qs h1 h6 h3 h2 3600 206.79 3680 3120

3953.21 kJ / kg

Turbine work WT

h

15MPa

1

2

T =873K1

6

5

4

x=0 .896

10kP a

s

3


WT h1 h2 h3 h4 3600 3120 3680 2350

1810 kJ / kg

Net work done Wnet

Wnet WT WP 1810 14.99 1795.01 kJ / kg

(i) The pressure at which steam should be reheated, P2 P3 38 bar


WnetQs

1795.013953.21

45.4%

Problem 1.52: In a reheat rankine cycle the condenser pressure is 7.5 kPa.The boiler temperature and reheat temperature are 500C. The moisturecontent at any stage should not exceed 15 %. Determine (a) Boiler pressure;(b) reheat pressure; (c) work done; (d) thermal

Solution

x2 x4 0.85 [moisture is 15%]

Condenser Pressure Pc P4 7.5 KPa 100

0.075 bar

P =0.075 bar

c

P b

x =x =0.852 4

T1 T 3 500 Co= =1

2

3

4

5

6


From mollier chart


h5 hf for Pc 168.65 kJ / kg

h6 h5 vf P1 P4 100

h6 168.65 0.0010075 340 0.075 100

h6 203.9049 kJ / kg

(a) Boiler pressure P1 340 bar

(b) Reheat pressure P2 38 bar

WT h1 h2 h3 h4

2990 2540 3460 2215 1695 kJ / kg

Wp h6 h5 203.9049 168.65 35.2549 kJ / kg

(c) Wnet WT Wp 1695 35.2549 1659.7451 kJ / kg

Qs h1 h6 h3 h2 2990 203.9049 3460 2540

Qs 3706.0951 kJ / kg

(d) the Wnet

Qs

1659.743706.095

44.78%

Problem 1.53: Steam at a pressure of 10 MPa, 500C is supplied to a reheatrankine cycle. After expansion in the HPT, the steam is reheated at anoptimum pressure to an optimum temprature. The moisture content at LPTexit should not exceed 15%. Network done is 1600 kJ / kg. Determine (a) heatsupply per kg; (b) thermal Condenser pressure is 7 kPa.

Soltuion:

Hint: Assume T1 T3 because steam is reheated to optimum temperature.

P2 is not given, reheat temperature is not given.

Pc P4 Condenser pressure = 0.07 bar.

The moisture content in LP turbine is 15%.

(ie) x4 0.85


h1 3374.6 kJ / kg ; h4 hf x4 hfg 163.4 0.85 2409.2 2211.22 kJ / kg ;

h5 hf for 0.07 bar 163.4 kJ / kg

h6 h5 vf [vf for Pc]

P1 P4 100 163.4 0.001007 100 0.07 100

h6 173.4629 kJ / kg

WT h1 h2 h3 h4

Wp h6 h5 173.4629 163.4 10.0629 kJ / kg

Wnet WT WP ; WT Wnet Wp 1600 10.063

WT 1610.063 kJ / kg

WT h1 h2 h3 h4 h1 h4 h3 h2

1610.063 3374.6 2211.2 h3 h2 ; h3 h2 446.7 kJ / kg

Qs h1 h6 h3 h2 3374.6 173.46295 446.7

Qs 3647.8 kJ / kg

thermal WnetQs

1600

3647.8 0.4386

th 43.86%

P =0.075 bar

c

P b

x =x =0 .852 4

T1 T 3500 Co= =1

2

3

4

5

6


Problem 1.54: A steam power plant running on Rankine cycle has steamentering HP turbine at 20 MPa, 500C and leaving LP turbine at 90%dryness. Considering condenser pressure of 0.005 MPa and reheatingoccurring up to the temperature of 500C determine,(i) The pressure at which steam leaves HP turbine (ii) The thermalefficiency (iii) Work done (AU. Apr/May 2011)

Solution:

Steam expands from state point 1 (200 bar, 500C) to state point 2 inthe high pressure turbine.

From Molliter chart, the following enthalpy values may be read as

h1 3220 kJ / kg. mark point 1 for 200 bar and 500C

Since the condenser is 0.0005 MPa = 0.05 bar and the dryness fractionat the exit of Low pressure turbine is x4 0.9, Point 4 is located on the

molliter chart. From the point 4, constant entropy line is drawn to reach point3 (ie) 500C line.

P3 14 bar , h4 2320 kJ / kg , h3 3480 kJ / kg

Since the reheat pressure is constant P2 P3 following along the P3

line and the constant entropy line from “1” it meets at point “2”

h2 2640 kJ / kg ; x2 0.92


Pressure of steam leaving from H.P turbine

P2 14 bar

Thermal efficiency th Wnet

Qs

Work doneHeat supplied

But work done H.P turbine work L.P turbine work

h1 h2 h3 h4

3220 2640 3480 2320

Work done 1740 kJ / kg

Heat supplied Qs h1 h6 h3 h2

But h6 h5 hf4 137.8 kJ / kg

Qs 3220 137.8 3480 2640

Qs 3922.2 kJ / kg

th 1740

3922.2 0.4436

th 44.36%


Problem 1.55: In a reheat rankine cycle, steam at 3 MPa, 450C is supplied

to a HPT. The reheat temperature is 450C. Condenser pressure is 4 kPa.The HPT expansion is limited with dry amd saturation. Determine (a) reheatpressure; (b) net work; (c) th

Solution

From Mollier Chart h4 2510 kJ / kg ; h1 3345 kJ / kg ; h2 2720 kJ / kg ; h3 3385 kJ / kg

h5 hf for Pc P4 121.4 kJ / kg

h6 h5 vf P1 P4 100

h6 121.4 0.001004 30 0.04 100

h6 124.40798 kJ / kg

Reheat pressure 2.5 bar from Mollier chart

WT h1 h2 h3 h4

3345 2720 3385 2510 1500 kJ / kg

Wp h6 h5 124.40798 121.4 3.00798 kJ / kg

Wnet WT Wp 1500 3.00798 1496.99 kJ / kg

P=30 b

a r

b

P =0 .04 bar

c

d ry & sa tu ra tion

450 Co

1

2

3

4

5

6


Qs h1 h6 h3 h2

3345 124.40798 3385 2720 3885.59 kJ / kg

th Wnet

Qs

1496.993885.59

0.3853

th 38.53%

Problem 1.56: A steam power plant uses the Reheat cycle. Steam inlet toturbine 150 bar, 550C. Reheat at 40 bar to 550C Condenser pressure at0.1 bar. Using mollier diagram, find (i) the dryness fraction of steam at exitof turbine (ii) cycle efficiency (iii) Specific Steam Consumption SSC.

Solution:

From mollier chart,

h1 3465 kJ / kg; h2 3065 kJ / kg; h3 3565 kJ / kg

h4 2300 kJ / kg; x4 0.88

h5 hf at 0.1 bar from steam table 191.8 kJ / kg

Pump work Wp

Wp vf P1 P4 100 0.001 [ 150 0.1 100]

14.99 kJ / kg

Wp h6 h5 14.99 kJ / kg

h6 h5 14.99 191.8 14.99 206.79 kJ / kg

p =1 5 0 b a r1

p = 0 .1 b a r3

p = 4 .0 b a r2

55 0 Co

6

5 4

2

31

x4

T

S

(a)

p= 15 0 ba r

1

p =0 .1 b a r3

p = 40 ba r

2

55 0 Co

6

54

2

31

x4

S(b)

h


Heat supplied Qs

Qs h1 h6 h3 h2 3465 206.79 3565 3065

3758.21 kJ / kg

Turbine Work WT

WT h1 h2 h3 h4 3465 3065 3565 2300

1665 kJ / kg

Net work done Wnet

Wnet WT WP 1665 14.99 1650.01 kJ / kg

cycle Wnet

Qs

1650.013758.21

0.439 43.9%

Specific steam consumption

SSC 3600Wnet

3600

1650.01 2.182 kg / kWhr

1.10 RANKINE CYCLE - REGENERATIVE CYCLE (BLEEDING CYCLE)

Fig. 1.23 Regeneration


P =cb

P=c

R

P =cc

1kg

(1-m)

(1-m)

(1-m)kg

1kg

m kg

S

T

1

2

56

7

34

(1-m)

(1-m)kg

1kg

kg

Fig. 1.24 Regeneration cycle in h-s and T-s diagram

P b

P C

m

(1-m)kg

P R

1kg


Assume 1 kg of steam is expanded in the turbine. Before completeamount of steam is expanded, some amount of steam (m kg) is extracted(this process is called bleeding) and utilized for heating the feedwater. Soremaining amount of steam 1 m kg is completely expanded in the turbineand condensing in condenser as shown in Fig. 1.23.

In the regenerator, m kg of high temperature steam and 1 m kg ofcondensate are passing. Heat transfer from steam to condensate (feed water)takes place. So this process increases the enthalpy of feed water. Thus theheat supplied to boiler will be reduced.

Re-generation means heating the feed water by steam taken from theturbine. The steam is exhausted (or) bled from the turbine at several locations(before exhaust) and is supplied to regenerator (feed water heater) to heat thefeed water. Extracting the steam in the turbine before exhaust is called bleeding.

Energy - balanceAssume 1 kg of working fluid is circulated Refer Fig 1.24.

Energy in = energy out

m mass of steam bled

mass of steam circulated

Energy entering regenerator = Energy leaving regenerator.

m h2 1 m h5 1 h6

m h2 h5 mh5 h6

m h2 h5 h6 h5

mass of steam bledmass of steam circulated

m h6 h5

h2 h5

kg of steam bledkg of steam circulated

Take h1, h2, h3 from mollier diagram h s diagram) (or) from steam table.

h4 hf for condenser pressure P3 P4

h6 hf for regenerator pressure P6 P2 P5

h5 h4 vf P6 P4 102 [ . . . P is in bar; P 102 kPa]

[vf for condenser pressure Pc P3 P4]

h7 h6 vf Pb PR 102 [PR for regenerator pressure and vf for PR]


WT 1 h1 h2 1 m h2 h3

WP h5 h4 1 m 1 h7 h6

Wnet WT WP

Qs h1 h7

thermal Wnet

Qs

Note:Mass rate of steam bled m m

[ . . . m kg of steam circulated / s ]

m

mass of steam bledmass of steam circulated

kg of steam bled

kg of steam circulated

kg of steam circulatedsec

So mass rate of steam bled kg of steam bled

sec m m

1.10.1 Advantages of Regenerative cycleWhat are the advantages of Regenerative cycle? (Apr/May 2011 - AU)

1. Heat supplied to boiler become reduced.

2. The heating process in the boiler approaches the reversible process.

3. Since feed water temperature is high, the ranges of temperature in theboiler is minimum. It reduces the thermal stresses produced in the boiler.

4. Thermal efficiency is increased since the average temperature of heataddition to the cycle is increased.

5. Due to bleeding in the turbine, erosion of turbine due to moisture isreduced.

6. Condenser can be a smaller size.


Problem 1.57: A steam turbine plant equipped with a single regenerativefeed water heating operates with the following data. Initial pressure 16.5 bar; Initial super heat 93C; Extraction pressure 2 bar; exhaust

pressure 0.05 bar. Compare regenerative and non-regenerative cycle for (a)

the; (b) network; (c) SSC.

SolutionGiven data: Pb P1 16.5 bar tsat1 202.9CDegree of superheat 93C t1 202.9 93 295.9C

PR regenerative pressureExtraction Pressere 2 bar

P2 Pc condenser pressure 0.05 bar

Case (a) Non-regenerative (Simple cycle)From mollier chart h1 3035 kJ / kg;

h2 2100 kJ / kg; h3 hf for P2, h3 137.8 kJ / kg

h4 h3 vf P1 P2 100 vf for P2

h4 137.8 0.001005 16.5 0.05 100

h4 139.45 kJ / kg

WT 1 h1 h2 3035 2100 935 kJ / kg

Wp h4 h3 139.4532 137.8 1.6532

Wp 1.6532 kJ / kg

P=16.5 bar

1

P =0.05 2

bar

T 1=295.9 Co1

2

3

4

=296 Co



Qs h1 h4 3035 139.45 2895.568 kJ / kg

thermal Wnet

Qs

933.352895.568

32.23%

SSC 1

Wnet 3600

1933.35

3600 3.85708 kg / kWhr

Case (b) Regenerative cycle

h1 3035 kJ / kg ; h2 2610 kJ / kg ; h3 2100 kJ / kg

h4 hf for Pc 0.05 bar 137.8 kJ / kg

h5 h4 vf4 PR Pc 100

h5 137.8 0.001005 2 0.05 100 137.996 kJ / kg

h6 hf for 2 bar PR

h6 504.7 kJ / kg

h7 h6 vf6 Pb PR 100 [vf6

for PR 2 bar]

h7 504.7 0.001061 16.5 2 100

h7 506.24 kJ / kg

T 1

P =16.5bar= P

b

1

P = 2ba rR

P =0 .05c

m kg

(1-m

)

(1 -m )

1kg

1kg

=2 96 Co

(1 -m )kg

Mollier diagram


WT 1 h1 h2 1 m h2 h3

3035 2610 1 m 2610 2100

m h6 h5

h2 h5

504.7 137.996

2610 137.996 0.148

m 0.148 kg of steam bled


WT 3035 2610 1 0.148 2610 2100 859.345 kJ / kg

Wp h5 h4 1 m 1 h7 h6

137.996 137.8 1 0.148 1 506.24 504.7 1.71 kJ / kg

Wp 1.71 kJ / kg


Qs h1 h7 3035 506.24 2528.76 kJ / kg

th WnetQs

0.3392 33.92%

SSC 1

Wnet 3600 4.198 kg / kWhr

Simple cycle Regenerative cycle

Wnet 933.35 kJ / kg Wnet 857.64 kJ / kg

th 32.116% th 33.92%

Note: Work output slightly decreases and efficiency increases.

Problem 1.58: A steam turbine operates on a simple regenerative cycle.Steam is supplied dry saturated at 40 bar and exhausted to condenserpressure of 0.07 bar. The condensate is pumped to a pressure of 3.5 bar atwhich it is mixed with bled steam from the turbine at 3.5 bar. The resultingwater at saturation is then pumped to the boiler. For the ideal cycle,calculate. (a) the amount of steam bled per kg of supply steam and (b)the of the plant neglecting pump work. (AU. April 2001) (FAQ)


Solution

h1 2800 kJ / kg ; h2 2380 kJ / kg ; h3 1880 kJ / kg

h4 hf for Pc 163.4 kJ / kg

h4 h5 Neglecting pump work

h6 hf for PR 3.5 bar 584.3 kJ / kg

h6 h7 Neglecting pump work

m h6 h5

h2 h5

584.3 163.4

2380 163.4 0.1899

kg of steam bledkg of steam circulated

Pb Boiler Pressure; PR Regenerator Pressure;

PC Condensor Pressure

WT 1 h1 h2 1 m h2 h3

2800 2380 1 0.1899 2380 1880

WT Wnet 825.058 kJ / kg [. . . Wp is neglected]

Qs h1 h7 2800 584.3 2215.7 kJ / kg

th WnetQs

37.237%

P =40barb

P =3.5bar

R

P =0.07c

Mollier diagram


Problem 1.59: An ideal regenerative cycle operates with steam supplied at30 bar and 400C and condenser pressure of 0.10 bar. For this cycle, find(a) WT in kJ / kg; (b) cycle efficiency; (c) steam rate in kg/kW hr. The feed

water heater can be assumed to be direct contact type which operates at 5bar. (Nov/Dec 2005 - AU, Apr/May 2008 - AU) (FAQ)

Solution:

h1 3230 kJ / kg

h2 2800 kJ / kg

h3 2195 kJ / kg

h4 hf for Pc 0.1 bar

191.8 kJ / kg

h5 h4 vf4 PR Pc 100

h5 191.8 0.001010 5 0.1 100

h5 192.295 kJ / kg

h6 hf for PR 5 bar 640.1 kJ / kg

h7 h6 vf6 Pb PR 100

h7 640.1 0.001093 30 5 100

h7 642.83 kJ / kg

m h6 h5

h2 h5

640.1 192.29

2800 192.29 0.1717

kg of steamkg of steam circulated

WT h1 h2 1 m h2 h3

3230 2800 1 0.1717 2800 2195WT 931.107 kJ / kg


WP 1 m h5 h4 h7 h6

1 0.1717192.29 191.8 642.83 640.1

3.14 kJ / kg

30ba r

P=5bar

R

(m kg)

P =0.1c

1kg

(1 -m )

1kg

400 Co

(1 -m )kg

(1 -m )kg

Fig. (b)

1

2

34

5

6

7

T

S

(1-m ) kg

(1-m ) kg

0.1 bar

m kg

5 bar

1kg

1kg

400 Co

Fig. (a)



Qs h1 h7 3230 642.83 2587.168 kJ / kg

th Wnet

Qs 35.87%

SSC 1

Wnet 3600

1927.97

3600 3.88 kJ / kWhr

Problem 1.60: A steam power plant operates on an ideal regenerativeRankine cycle. Steam enters the turbine at 6 MPa and 450 C and iscondensed in the condenser at 20 kPa. Steam is extracted from the turbineat 0.4 MPa to heat the feedwater in an open feedwater heater. Water leavesthe feedwater heater as a saturated liquid. Show the cycle on a T-s diagram,and determine (i) the network output per kilogram of steam flowing throughthe boiler and (ii) the thermal efficiency of the cycle.

(AU. Nov/Dec 2016)

Solution:h1 3301.15 kJ / kg

h2 2660 kJ / kg

h3 2210 kJ / kg

h4 hf for Pc 0.2 bar 251.5 kJ / kg

h5 h4 vf4 PR Pc 100

h5 251.5 0.001017 4 0.2 100

0.38646

h5 251.89 kJ / kg

h6 hf for PR 4 bar 604.7 kJ / kg

h7 h6 vf6 Pb PR 100

h7 604.7 0.001084 60 4 100

h7 610.77 kJ / kg


m h6 h5

h2 h5

604.7 251.89

2660 251.89 0.1465

m 0.1465 kg of steam


60bar

P=4bar

R

(m kg)

P =0 .2c

1kg

(1 -m )

1kg

450 Co

(1 -m )kg

(1 -m )kg

Fig. (b)

1

2

34

5

6

7

T

S

(1 -m ) kg

(1 -m ) kg

20 kPa

m kg

4 ba r

1kg

1kg

45 0 Co

Fig. (a)


WT h1 h2 1 m h2 h3

3301.15 2660 1 0.1465 2660 2210WT 1025.23 kJ / kg

WP 1 m h5 h4 h7 h6

1 0.1465 251.89 251.5 610.77 604.7

0.333 6.07 6.4 kJ / kg

Wnet WT Wp 1025.23 6.4 1018.83 kJ / kg

Qs h1 h7 3301.15 610.77 2690.4 kJ / kg

th Wnet

Qs

1018.832690.4

0.3787 37.87%

SSC 1

Wnet 3600

11018.83

3600 3.53 kg / kWhr

Problem 1.61: A steam turbine plant, working on a single stage ofregenerative feed heating receives steam at 3 MPa and 300C. The turbineexhausts to a condenser at 15 kPa while the bled steam is at 300 kPa.Assuming that the cycle uses actual regenerative cycle. Calculate the thermalefficiency of cycle. Compare this value with a rankine cycle operating betweensame boiler and condenser pressures. (AU. Apr/May 2010)

Solution:Regenerative Cycle

h1 2990 kJ/kg; h2 2540 kJ/kg; h3 2115 kJ / kg ; h4 hf for Pc 226 kJ / kg

h5 h4 vf4 PR Pc 100

h5 226 0.001014 3 0.15 100

h5 226.29 kJ / kg

h6 hf for PR 561.5 kJ / kg

h7 561.5 0.001074 30 3 100

h7 564.4 kJ / kg


m h6 h5

h2 h5

561.5 226.29

2540 226.29

m 0.145 kg of steam bled


WT h1 h2 1 m h2 h3

2990 2540 1 0.145 2540 2115 813.426 kJ / kg

Wp 1 m h5 h4 h7 h6

1 0.145 226.29 226 564.4 561.5 3.15 kJ / kg


Qs h1 h7 2990 564.4 2425.6 kJ / kg

th Wnet

Qs 33.41%

Simple Rankine cycle h1 2990 kJ kg ; h2 2115 kJ / kg ; h3 226 kJ / kg ; h4 229.03 kJ / kg

WT h1 h2 2990 2115 875 kJ / kg

P =30bar

b

P =3barR

(m kg)

P =0.15bar

c

1kg

T =300 C1

O

(1 -m )kg

1kg


Wp h4 h3 229.03 226 3.03 kJ / kg

Wnet WT WP 871.97 kJ / kg

Qs h1 h4 2990 229.03 2760.97 kJ / kg

th Wnet

Qs

871.972760.97

0.3158

th 31.58%

Problem 1.62: Steam enters the turbine at 3 MPa and 400C and iscondensed at 10 kPa. Some quantity of steam leaves the turbine at 0.6 MPaand enters feed water heater. Compute the fraction of the steam extractedper kg of steam and cycle thermal efficiency. (AU. Nov/Dec 2012)

Solution:

From Mollier chart,

h1 3230 kJ / kg

h2 2830 kJ / kg

h3 2195 kJ / kg

h4 hf for 0.1 bar 191.8 kJ / kg

h5 h4 vf4 6 0.1 100

P =30bar

b

P =0.15bar

R

1kg

T =100 C1O1

2

3

4


h5 191.8 0.001010 6 0.1 100

h5 192.4 kJ / kg

h6 hf for 6 bar 670.4 kJ / kg

h7 h6 vf6 30 6 100

h7 670.4 0.001101 24 100

h7 673.04 kJ / kg

Fraction of steam extracted/kg of steam

m h6 h5

h2 h5

670.4 192.4

2830 192.4 0.181

m 0.181 kg

WT h1 h2 1 m h2 h3

3230 2830 1 0.181 2830 2195WT 920.065 kJ / kg

1

2

34

5

6

7

T

S

(1-m ) kg

(1-m ) kg

0.1 bar

m kg

5 bar

1kg

1kg

400 Co

Fig (a)


WP 1 m h5 h4 h7 h6

1 0.181 192.4 191.8 673.04 670.4

WP 3.13 kJkg

Wnet WT WP 920.065 3.13 916.935 kJ / kg

QS h1 h7 3230 673.04 2556.06 kJ / kg

Cycle thermal efficiency

th Wnet

Qs

916.9352556.06

0.3587 or 35.87 %

Problem 1.63: In a steam power plant the condition of steam at inlet tothe steam turbine is 20 bar and 300C and the condenser pressure is 0.1bar. Two feed water heaters operate at optimum temperatures. Determine:(1) The quality of steam at turbine exhaust, (2) network per kg of steam,(3) cycle efficiency, and (4) the steam rate. Neglect pump work. (AU. Nov\Dec 2008) (FAQ)

Solution:

From superheated steam table,

corresponding to pressure 20 bar and 300C

hsup h1 3025 kJ / kgK ; ssup s1 6.770 kJ / kgK s2 s3 s4

tsat at 20 bar 212.4C ; tsat at 0.1 bar 45.8C

tsat 212.4 45.8 166.6C

Temperature rise per heater 166.6

3 55.53C

Temperature at which the first heater

Operates 212.4 55.53 156.87C ~ 150C (assumed) P2 4.76 bar

Temperature at which the second heater operates

156.87 55.53 101.04 C ~ 100 C (assumed)

corresponding to 0.1 bar P4 , h4 2145 kJ / kg

From steam tables, corresponding to 100C t3P3 1.013 bar


hf3 419.1 kJ / kg ; h2 2715 kJ / kg ; h3 2460 kJ / kg ; h4 2145 kJ / kg

From steam tables corresponding to 150C t2 P2 4.76 bar

hf2 632.2 kJ / kg ; h2 2717.47 kJ / kg

Similarlly, For P3 1.013 bar , h3 2460 kJ / kg

Since pump work is neglected, h10 h9, h8 h7, h6 h5. By making an

energy balance for the HP heater

(I-m )kg1

(I- - ) kgm m1 2

m kg1

m kg2

(I- - )kgm m1 2

5

6

7

8

9

10

1 kgP 1

P 2

P 3

P 4

S

T 1 kg

2

3

4

1

h =3025kJ/kg1 1

2

3

41.013bar

0.1bar

4.76bar

20bar

x =0 .8154 s

h


1 m1 h9 h8 m1 h2 h9

h7 hf3 419.1 ; h9 hf2 632.2

Rearranging

m1 h9 h7

h2 h7

hf2 hf3

h2 hf3

632.2 419.1

2715 419.1

213.12296.37

0.093 kg

By making an energy balance for the LP heater

1 m1 m2 h7 h6 m2 h3 h7

1 0.093 m2 419.1 191.8 m2 2457.6 419.1

m2 0.091 kg

WT 1 h1 h2 1 m1 h2 h3 1 m1 m2 h3 h4

3025 2717.47 1 0.093 2717.47 2457.6

1 0.093 0.091 2457.6 2144.2

797.48 kJ / kg

and Q1 h1 h9 3025 632.2 2392.8 kJ / kg

cycle WT WP

Q1

797.482392.8

0.3334 or 33.34%

Steam rate 3600Wnet

3600

797.48 4.51 kJ / kWh


IndexA

Abnormal Combustion in SI Engines3.61

Absorption Dynamometers 4.10

After burning 3.43

After-burning 5.28

Air Standard Efficiency 1.2

Air-cooled System 4.95

Air-Fuel Ratio Calculation 3.36

Anti-Knock Additives 3.69

Axial Flow Compressors 2.117

B

Back Flow of Air 2.103

Baffles 4.96

Bath-tub form 3.46

Battery or Coil Ignition System4.80

Brake Power (BP) 4.5

Brake Thermal Efficiency Brake4.8

BS Norms 4.114

Burette method 4.28

C

Capacitance Discharge Ignition

System 4.85

Centrifugal Compressor 2.114

Cetane Number (CN)3.35

Chemical delay 3.50

Clearance Ratio ‘k’ 2.18

Clearance Volume 2.13

Combined Cycles 5.8

Combustion Phenomenon 3.43

Common Rail Direct Injection (CRDI)

System 4.75

Complete (or) Perfect Intercooling2.61

Composition factor 3.69

Composition of Dry Air 3.36

Conditions for Minimum Workdone2.66

Connecting Rod 3.14

Control of detonation3.63

Cooling Fins 4.95

Cooling Systems4.94

Crank 3.14

Crank Case 3.15

Crankshaft 3.14

Cross Flow Scavenging 4.105

Cycle 1.79

Cylinder 3.12

Cylinder head 3.13

Cylindrical chamber 3.55

D

D-MPFI system 4.72

Density factors 3.67

Desirable Properties and Qualities of

Fuels 3.29

Diesel Cycle or Constant Pressure

Cycle 1.7

Direct or non-return system 4.98

Divided combustion chamber (or)

Indirect combustion chamber3.56

Double Acting Air Compressor2.3

Index I.1

Dry Sump Lubrication System4.93

Dual Combustion Cycle 1.49

E

Eddy Current Dynamometer 4.12

Effects of Detonation3.63

Electronic Ignition Systems 4.84

Emission Norms (Euro and BS)4.113

Energy-cell chamber 3.59

Engine Bearings 3.15

Engine 4.16

Euro norms 4.113

Expression for mean effective pressure

Pm 1.53

F

F - head combustion chamber3.48

Flame propagation stage 3.42

Fly wheel 3.15

Forced or Pump Circulation System4.99

Four Stroke SI (Petrol) Engine3.6

Four Stroke CI Engine 3.7

Free Air Delivered (FAD) 2.19

Friction Power (FP) 4.7

Fuel consumption 4.28

Fuel Rating for CI Engine 3.35

G

Gas Turbine Cycle Analysis (Brayton

Cycle) 5.3

Gas turbine Open cycle and closed

cycle 5.3

Gas Power Cycles 1.1

Gas Turbine Closed Cycle (or) Brayton

Cycle (or) Joule Cycle3.15

Gravimetric fuel flow measurement4.31

Gudgeon pin (Or Wrist pin or Piston

pin) 3.14

H

Heat Balance Sheet 4.33

Hemispherical chamber 3.55

Hydraulic Dynamometer 4.10

I

IC Engine 3.1

Ignition lag stage 3.41

Ignition Systems4.78

Ignition Delay period3.49

Incomplete (or) Imperfect Intercooling2.62

Indicated Thermal Efficiency indicated4.8

Indicated Power (IP) 4.2

Indicated Power (I.P) and Brake Power

(B.P) 2.18

Indicator Diagram 4.1

Injecting Refrigerants5.28

Intercooling 5.7

K

Knocking in CI Engine (or) Diesel

Knock 3.70

Knocking (or) Detonation (or)

Pinking 3.62

L

L-head combustion chamber 3.47

L-MPFI system 4.73

I.2 Thermal Engineering - I - www.airwalkbooks.com

Lean and Rich Mixtures 3.39

Liquid (or) Water Cooled Systems4.98

Loop or Back Flow Scavenging4.105

Lubrication 4.88

Lubrication systems 4.89

M

Magneto Ignition System 4.82

Materials for Turbines 5.29

Mean Effective Pressure m.e.p2.19

Mean Effective Pressure Pm4.2

Measurement of Air Consumption4.25

Mechanical Efficiency mech4.8

Mist lubrication (Petroil system)4.89

Monopoint Fuel Injection System4.69

Morse Test 4.16

Motor Octane Number 3.34

Motor Octane Number 3.26

(Multiport Fuel - Injection System)

Multi Point Fuel Injection

System 4.70

Multi Stage Compressor 2.4

Multistage Reciprocating Compressors2.63

N

Nickel alloys: 5.31

Normal Combustion 3.43

O

Octane value 3.69

Octane Number (ON) 3.33

Open and Divided combustion

chambers 3.60

Open combustion chamber 3.54

Otto Cycle (or) Constant Volume

Cycle 1.2

Overhead valve (or) I - Head

combustion chamber 3.46

P

Performance Curves 4.12

Period of Rapid Combustion (or)

Uncontrolled Combustion3.51

Petrol Fuel Injection System (Gasoline

Injection System) 4.67

Physical delay 3.50

Piston 3.13

Piston rings 3.14

Power 4.5

Pre-ignition 3.61

Pre combustion chamber 3.57

Pressure-feed system 4.92

R

Rankine Cycle - Regenerative Cycle

(Bleeding Cycle) 1.129

Rankine Cycle - Reheating Cycle1.114

Regeneration and Reheated Cycle:5.10

Regeneration 5.5

Reheating 5.6

Relative Efficiency (or) Efficiency

Ratio 4.9

Research Octane Number (RON)3.34

Research Octane Number (RON)3.26

Retardation Test 4.5

Roots Blower 2.103

Index I.3

Rotary Compressors 2.103

S

Scavenging 4.104

Shallow depth chamber 3.54

Simple Gas Turbine Plant5.1

Single Stage 2.20

Single Stage Compressor 2.4

Spark Plug 4.78

Specific Fuel Consumption (S.F.C) in

kg/kW hr 4.7

Splash system 4.90

Splash and Pressure system 4.92

Stage Velocity Triangles 2.119

Stages of Combustion in CI Engines3.48

Stages of Combustion in SI Engines3.41

Stalling 2.124

Standard Rankine Cycle 1.99

Stoichiometric air-fuel ratio 3.36

Supercharger: 4.106

Supercharging 4.105

Surging 2.123

Swept Volume Vs 2.19

T

T - Head combustion chamber3.47

Theories of Detonation 3.63

Thermal Efficiency 4.8

Thermosyphon or Natural Circulation

System 4.99

Time factors 3.68

Titanium alloys: 5.30

Toroidal chamber 3.55

Transducer 4.3

Transistorized Assisted Contact (TAC)

Ignition System 4.86

Turbine blade materials 5.30

Turbocharging 4.111

Turbolag 4.113

Turbulent combustion chamber3.56

Two Stage Compression 2.61

U

Uniflow Scavenging 4.105

V

Vane Type Blower Compressor2.107

Volumetric flow 4.28

Volumetric Efficiency volumetric4.9

Volumetric Efficiency vol 2.17

Volumetric Efficiency in a Reciprocating

Air Compressor2.16

W

Wedge form 3.46

Wet Sump Lubrication System4.90

I.4 Thermal Engineering - I - www.airwalkbooks.com

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