Internal combustion engine m l mathur and r p sharmapdf PDF

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Internal combustion engine book The Intercontinental Exchange is an American Fortune 500 company formed in 2000 that operates global exchanges, clearing houses and provides mortgage technology, data and listing services. The company owns exchanges for financial and commodity markets, and oper...

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Scilab Textbook Companion for Internal Combustion Engine by M. l. Mathur and R. P. Sharma1 Created by Manish Yadav B.E. Mechanical Engineering M.I.T.S. Gwalior, M.P. College Teacher None Cross-Checked by Bhavani June 2, 2016

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Funded by a grant from the National Mission on Education through ICT, http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilab codes written in it can be downloaded from the ”Textbook Companion Project” section at the website http://scilab.in

Book Description Title: Internal Combustion Engine Author: M. l. Mathur and R. P. Sharma Publisher: Dhanpat Rai Publications, New Delhi Edition: 8 Year: 2010 ISBN: 9788189928469

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Scilab numbering policy used in this document and the relation to the above book. Exa Example (Solved example) Eqn Equation (Particular equation of the above book) AP Appendix to Example(Scilab Code that is an Appednix to a particular Example of the above book) For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means a scilab code whose theory is explained in Section 2.3 of the book.

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Contents List of Scilab Codes

4

1 Introduction

5

2 Air Standard Cycles

11

3 Fuel Air Cycles

34

5 Combustion in SI Engines

45

7 Comparison of SI and CI Engines

47

8 Fuels

50

10 Air Capacity of Four Stroke Engines

61

11 Carburetion

63

12 Fuel Injection

75

14 Engine Friction and Lubrication

80

15 Engine Cooling

82

16 Two Stroke Engines

84

17 Supercharging

86

18 Testing and Performance

93

3

26 Gas Turbines

117

27 Testing of Internal Combustion Engines According to Indian and International Standards 131

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List of Scilab Codes Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

1.1 1.2 1.3 1.4 1.5 1.6 1.7 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19

Exa 2.20

Calculation of cubic capacity and clearance volume . . Calculation of brake power and friction power . . . . . Calculation of mechanical efficiency . . . . . . . . . . Calculations on four stroke petrol engine . . . . . . . . Calculations on SI engine . . . . . . . . . . . . . . . . Calculations on diesel engine . . . . . . . . . . . . . . Calculations on two stroke CI engine . . . . . . . . . . Calculations on Carnot engine . . . . . . . . . . . . . Calculations on the Carnot cycle . . . . . . . . . . . . Calculation of air standard efficiency of Otto cycle . . Calculations on constant volume cycle . . . . . . . . . Calculations on Otto cycle . . . . . . . . . . . . . . . Calculations on Otto cycle . . . . . . . . . . . . . . . Calculations on Otto cycle . . . . . . . . . . . . . . . Calculations on Otto cycle . . . . . . . . . . . . . . . Calculations on Otto cycle . . . . . . . . . . . . . . . Calculations on Otto cycle . . . . . . . . . . . . . . . Calculations on diesel cycle . . . . . . . . . . . . . . . Calculations on diesel cycle . . . . . . . . . . . . . . . Calculations on diesel cycle . . . . . . . . . . . . . . . Calculations on diesel cycle . . . . . . . . . . . . . . . Calculations on dual combustion cycle . . . . . . . . . Calculations on dual combustion cycle . . . . . . . . . Calculations on dual combustion cycle . . . . . . . . . Calculations on dual combustion cycle . . . . . . . . . Calculations for comparision of Otto cycle and Diesel cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . Calculations for Otto cycle and Limited pressure cycle 5

5 6 6 7 8 9 9 11 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 30

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

2.21 2.22 3.1 3.2 3.3 3.4 3.5 3.6 5.1 7.1 7.2 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 10.1 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 12.1 12.2 12.3 12.4 12.5 14.1 14.2 15.1

Calculations for comparision of Atkinson and Otto cycle Calculations on Joule cycle . . . . . . . . . . . . . . . Effect of variable specific heat on efficiency . . . . . . Effect of variable specific heat on maximum pressure . Calculations on diesel engine . . . . . . . . . . . . . . Calculations on dual combustion cycle . . . . . . . . . Effect of molecular contraction . . . . . . . . . . . . . Calculations on Otto cycle . . . . . . . . . . . . . . . Calculation of optimum spark timing . . . . . . . . . . Calculations for comparison of SI and CI engine . . . . Calculations for comparison of SI and CI engine . . . . Calculation of lowest calorific value . . . . . . . . . . . Calculation of relative fuel air ratio by volume . . . . Calculations on Petrol engine . . . . . . . . . . . . . . Calculation of mass of air . . . . . . . . . . . . . . . . C7H16 in Petrol engine . . . . . . . . . . . . . . . . . Incomplete combustion of Petrol . . . . . . . . . . . . Analysis of fuel from exhaust gas analysis . . . . . . . Orsat analysis . . . . . . . . . . . . . . . . . . . . . . Calculations on gas engine . . . . . . . . . . . . . . . . Calculations on SI engine . . . . . . . . . . . . . . . . Calculation of the throat diameter . . . . . . . . . . . Calculation of throat diameter and orifice diameter . . Calculation of suction at throat . . . . . . . . . . . . . Calculation of the diameter of fuel jet . . . . . . . . . Calculations on carburettor . . . . . . . . . . . . . . . Calculations on carburettor . . . . . . . . . . . . . . . Change in air fuel ratio at altitude . . . . . . . . . . . Calculation of air fuel ratio . . . . . . . . . . . . . . . Effect of air cleaner . . . . . . . . . . . . . . . . . . . Calculation of quantity of fuel injected . . . . . . . . . Calculation of orifice area . . . . . . . . . . . . . . . . Calculation of orifice diameter . . . . . . . . . . . . . Calculations on spray penetration . . . . . . . . . . . Calculations on diesel engine fuel pump . . . . . . . . Calculation of saving in fuel . . . . . . . . . . . . . . . Variation of bsfc with speed . . . . . . . . . . . . . . . Comparison of cooling water required . . . . . . . . . 6

32 33 34 35 36 37 39 41 45 47 48 50 51 52 53 54 55 57 58 59 61 63 64 65 66 67 69 70 71 73 75 76 76 77 78 80 81 82

Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa Exa

16.1 17.1 17.2 17.3 17.4 18.1 18.2 18.3 18.4 18.5 18.6 18.7 18.8 18.9 18.10 18.11 18.12 18.13 18.14 18.15 18.16 26.1 26.2 26.3 26.4 26.5 26.6 26.7 26.8 26.9 26.10 27.1 27.2 27.3 27.4 27.5 27.6 27.7

Calculations on 2 stroke IC engine . . . . . . . . . . . Estimation of increase in brake power . . . . . . . . . Supercharged diesel engine . . . . . . . . . . . . . . . Normally aspirated and supercharged engine . . . . . . Supercharged four stroke oil engine . . . . . . . . . . . Calculations on petrol engine . . . . . . . . . . . . . . Calculations on Gas engine . . . . . . . . . . . . . . . Calculations on oil engine . . . . . . . . . . . . . . . . Calculations on oil engine . . . . . . . . . . . . . . . . Calculations on six cylinder petrol engine . . . . . . . Calculations on two stroke engine . . . . . . . . . . . . Calculations by Morse test . . . . . . . . . . . . . . . Calculations on six cylinder diesel engine . . . . . . . Calculations on six cylinder petrol engine . . . . . . . Calculations on gas engine . . . . . . . . . . . . . . . . Calculations from indicator diagram . . . . . . . . . . Calculations on diesel engine . . . . . . . . . . . . . . Calculations on four stroke engine . . . . . . . . . . . Calculations on petrol engine . . . . . . . . . . . . . . Hit and miss governing . . . . . . . . . . . . . . . . . Calculations on two stroke engine . . . . . . . . . . . . Calculations on Brayton cycle . . . . . . . . . . . . . . Calculations on Joule cycle . . . . . . . . . . . . . . . Calculations for zero efficiency . . . . . . . . . . . . . Calculations on gas turbine . . . . . . . . . . . . . . . Calculations on gas turbine . . . . . . . . . . . . . . . Calculations on gas turbine with heat exchanger . . . Calculations on compound gas turbine . . . . . . . . . Calculations on automotive gas turbine . . . . . . . . Calculations on Helium gas turbine . . . . . . . . . . . Calculations on closed cycle gas turbine . . . . . . . . Calculations on non supercharged CI engine . . . . . . Calculations on turbocharged CI engine . . . . . . . . Calculations on turbocharged CI engine . . . . . . . . Simulating site ambient conditions . . . . . . . . . . . Calculations on unsupercharged SI engine . . . . . . . Calculations on turbocharged CI engine . . . . . . . . Calculations on turbocharged CI engine . . . . . . . . 7

84 86 87 89 90 93 94 96 98 99 101 103 104 106 108 110 110 112 114 115 116 117 118 119 120 121 122 123 125 126 128 131 132 133 134 135 136 137

Chapter 1 Introduction

Scilab code Exa 1.1 Calculation of cubic capacity and clearance volume 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

// C a l c u l a t i o n o f c u b i c c a p a c i t y and c l e a r a n c e vo lume clc , clear // G iv en : n =4 // Number o f c y l i n d e r s d =68/10 // B ore i n cm l =75/10 // S t r o k e i n cm r =8 // C o mp re s s i on r a t i o // S o l u t i o n : V_s =( %pi /4) * d ^2* l // Swept vol ume o f on e c y l i n d e r i n cmˆ3 c u b i c _ c a p a c i t y = n * V _s // C u b i c c a p a c i t y i n cmˆ3 // S i n c e , r = ( V c + V s ) / V c V_c = V_s /( r -1) // C l e a r a n c e vol ume i n cmˆ3 // R e s u l t s : printf ( ” \n The c u b i c c a p a c i t y o f t h e e n g i n e = %. 1 f cmˆ3 ” , c u b i c _ c a p a c i t y ) printf ( ” \n The c l e a r a n c e v ol u me o f a c y l i n d e r , V c = %. 1 f cmˆ3\ n\n ” ,V_c )

8

Scilab code Exa 1.2 Calculation of brake power and friction power // C a l c u l a t i o n o f b r a k e power and f r i c t i o n p ow er clc , clear // G iv en : ip =10 // I n d i c a t e d p ow er i n kW eta _m =80 // M e c h a n i c a l e f f i c i e n c y i n p e r c e n t // S o l u t i o n : // S i n c e , et a m = bp / i p bp =( et a_ m /100 ) * ip // B ra k e power i n kW fp =ip - bp // F r i c t i o n p ower i n kW // R e s u l t s : printf ( ” \n The b r a k e p ow er d e l i v e r e d , bp = %d kW\n ” , bp ) 12 printf ( ” The f r i c t i o n power , f p = %d kW\n\n” , fp )

1 2 3 4 5 6 7 8 9 10 11

Scilab code Exa 1.3 Calculation of mechanical efficiency 1 // C a l c u l a t i o n o f m e c h a n i c a l e f f i c i e n c y 2 clc , clear 3 // G iv en : 4 bp =100 // Bra k e power a t f u l l l o a d i n kW 5 fp =25 // F r i c t i o n a l power i n kW ( p r i n t i n g e r r o r ) 6 // S o l u t i o n : 7 eta _m = bp /( bp + fp ) // M e c h a n i c a l e f f i c i e n c y a t f u l l 8 9 10 11 12 13 14

load // ( a ) At h a l f l o a d bp = bp /2 // B ra ke pow er a t h a l f l o a d i n kW eta _m1 = bp /( bp + fp ) // M e c h a n i c a l e f f i c i e n c y a t h a l f load // ( b ) At q u a r t e r l o a d bp = bp /2 // B ra ke pow er a t q u a r t e r l o a d i n kW eta _m2 = bp /( bp + fp ) // M e c h a n i c a l e f f i c i e n c y a t q u a r t e r load // R e s u l t s : 9

printf ( ” \n The m e c h a n i c a l e f f i c i e n c y a t f u l l l o a d , et a m = %d p e r c e n t ” ,eta_m *100) 16 printf ( ” \n The m e c h a n i c a l e f f i c i e n c y , \ n ( a ) At h a l f l o a d , et a m = %. 1 f p e r c e n t \n ( b ) At q u a r t e r l oa d , et a m = %d p e r c e n t \n\n ” , eta_ m1 *100 , eta _m2 *10 0) 17 // Data i n t h e book i s p r i n t e d wrong 15

Scilab code Exa 1.4 Calculations on four stroke petrol engine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

// C a l c u l a t i o n s on f o u r s t r o k e p e t r o l e n g i n e clc , clear // G iv en : bp =35 // B rak e p ow er i n kW eta _m =80 // M e c h a n i c a l e f f i c i e n c y i n p e r c e n t bsfc =0 .4 // Bra k e s p e c i f i c f u e l c on s u mp t i on i n kg /kWh A_F =14 /1 // Ai r − f u e l r a t i o CV =43 00 0 // C a l o r i f i c v a l u e i n kJ / kg // S o l u t i o n : // ( a ) ip = bp *1 00/ eta_ m // I n d i c a t e d pow er i n kW // ( b ) fp =ip - bp // F r i c t i o n a l pow er i n kW // ( c ) // S i n c e , 1 kWh = 360 0 kJ eta _bt =1/( bsfc * CV / 3600 ) // Bra ke t h e r m a l e f f i c i e n c y // ( d ) eta _i t = et a_ bt / e ta_ m * 100 // I n d i c a t e d t h er ma l efficiency // ( e ) m_f = bsfc * bp // F u el c on s u m p t i on i n kg / h r // ( f ) m_a = A_F * m_f // A i r c on s u m p t i on i n kg / h r // R e s u l t s : printf ( ” \n ( a ) The i n d i c a t e d power , i p = %. 2 f kW\n ( b ) The f r i c t i o n power , f p = %. 2 f kW”, ip , fp ) 10

printf ( ” \n ( c ) The b ra k e t h er m a l e f f i c i e n c y , e t a b t = %. 1 f p e r c e n t \n ( d ) The i n d i c a t e d t h e r m a l e f f i c i e n c y , e t a i t = %. 1 f p e r c e n t ” , eta _bt *100 , eta _i t *1 00) 26 printf ( ” \n ( e ) The f u e l c on s u mp t i on p er hour , m f = % . 1 f kg / h r \n ( f ) The a i r c on s u mp t i on p er hour , m a = %d kg / hr \n\n ” , m_f , m_a )

25

Scilab code Exa 1.5 Calculations on SI engine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

// C a l c u l a t i o n s on S I e n g i n e clc , clear // G iv en : F_A =0.07/1 // Fu el −a i r r a t i o bp =75 // B rak e p ow er i n kW eta _b t =20 // Bra k e t h e r m a l e f f i c i e n c y i n p e r c e n t rho _a =1. 2 // D e n s i t y o f a i r i n kg /mˆ3 rho _f =4* rh o_a // D e n s i t y o f f u e l v ap ou r i n kg /mˆ3 CV =43 70 0 // C a l o r i f i c v a l u e o f f u e l i n kJ / kg // S o l u t i o n : m_f = bp *3 600 /( e ta_b t * CV /10 0) // F u e l c o n s u mp t i on i n kg / hr m_a = m_f / F_A // A i r c on s u m p t i on i n kg / hr V_a = m_a / rho _a // Volume o f a i r i n mˆ3/ h r V_f = m_f / rho _f // Volume o f f u e l i n mˆ3/ h r V_ m ix tur e = V_f + V_a // M i x t u re v ol u me i n mˆ3/ h r // R e s u l t s : printf ( ” \n The a i r co ns um pt ion , m a = %. 1 f kg / h r ” , m_a ) printf ( ” \n The v ol u me o f a i r r e q u i r e d , V a = %. 1 f m ˆ3 / h r ” , V_a ) printf ( ” \n The v ol u me o f m i x t u r e r e q u i r e d = %. 1 f m ˆ3/ h r \n\n ” , V_m ixt ur e ) // ( p r i n t i n g e r r o r ) // Answer i n t h e book i s p r i n t e d wrong

11

Scilab code Exa 1.6 Calculations on diesel engine 1 // C a l c u l a t i o n s on d i e s e l e n g i n e 2 clc , clear 3 // G iv en : 4 bp =5 // Bra k e power i n kW 5 eta _i t =30 // I n d i c a t e d t h e r m a l e f f i c i e n c y i n p e r c e n t 6 eta _m =75 // M e c h a n i c a l e f f i c i e n c y i n p e r c e n t ( 7 8 9 10 11 12 13 14 15 16 17

18 19 20

printing error ) // S o l u t i o n : ip = bp *1 00/ eta_ m // I n d i c a t e d pow er i n kW CV =42 00 0 // C a l o r i f i c v a l u e o f d i e s e l ( f u e l ) i n kJ / kg m_f = ip *3 600 /( e ta_i t * CV /10 0) // F u e l c o n s u mp t i on i n kg / hr // D e n s i t y o f d i e s e l ( f u e l ) = 0 . 8 7 kg / l rho_f =0.87 // D e n s i t y o f f u e l i n kg / l V_f = m_f / rho _f // F u el c on s u m p t i on i n l / h r isfc = m_f / ip // I n d i c a t e d s p e c i f i c f u e l c o n s u m p t i o n i n kg /kWh bsfc = m_f / bp // Bra k e s p e c i f i c f u e l c o n s u m p t i o n i n kg / kWh // R e s u l t s : printf ( ” \n The f u e l c on s um pt io n o f en gi n e , m f i n , \ n ( a ) kg / hr = %. 3 f kg / h r \n ( b ) l i t r e s / h r = %. 2 f l / h r ” ,m_f , V_f ) printf ( ” \n\n ( c ) I n d i c a t e d s p e c i f i c f u e l c on s u mp t io n , i s f c = %. 3 f kg /kWh” , i sf c ) printf ( ” \n ( d ) B ra k e s p e c i f i c f u e l c on s u mp t i on , b s f c = %. 3 f kg /kWh\n\n ” , bsfc ) // Data i n t h e book i s p r i n t e d wrong

Scilab code Exa 1.7 Calculations on two stroke CI engine 12

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

// C a l c u l a t i o n s on two s t r o k e C I e n g i n e clc , clear // G iv en : bp =500 0 // Bra k e power i n kW fp =100 0 // F r i c t i o n pow er i n kW m_f =23 00 // F ue l c on s u mp t i o n i n kg / h r A_F =20 /1 // Ai r − f u e l r a t i o CV =42 00 0 // C a l o r i f i c v a l u e o f f u e l i n kJ / kg // S o l u t i o n : // ( a ) ip = bp + fp // I n d i c a t e d pow er i n kW // ( b ) eta _m = bp / ip // M e c h a n i c a l e f f i c i e n c y // ( c ) m_a = A_F * m_f // A i r c on s u m p t i on i n kg / h r // ( d ) eta _it = ip *36 00/ ( m_f * CV ) // I n d i c a t e d t h e r m a l efficiency // ( e ) eta _b t = et a_ it * e ta_ m // Bra k e t h e r m a l e f f i c i e n c y // R e s u l t s : printf ( ” \n ( a ) The i n d i c a t e d power , i p = %d kW” , ip ) printf ( ” \n ( b ) The m e c h a n i c a l e f f i c i e n c y , et a m = %d p e r c e n t ” , eta_m *100) printf ( ” \n ( c ) The a i r c on s um pt i on , m a = %d kg / h r ” , m_a ) printf ( ” \n ( d ) The i n d i c a t e d t h e r m a l e f f i c i e n c y , e t a i t = %. 1 f p e r c e n t \n ( e ) The b r a k e t h e r m a l e f f i c i e n c y , e t a b t = %. 1 f p e r c e n t \n\n ” , eta_i t *100 , eta _b t *10 0)

13

Chapter 2 Air Standard Cycles

Scilab code Exa 2.1 Calculations on Carnot engine 1 2 3 4 5 6 7 8 9 10 11 12 13 14

// C a l c u l a t i o n s on C a rn ot e n g i n e clc , clear // G iv en : T2 =27+273 // T e mp era t u re o f c o o l i n g pond i n K eta =30 // E f f i c i e n c y i n p e r c e n t Q2 =200 // Hea t r e c e i v e d by c o o l i n g pond i n kJ / s // S o l u t i o n : // S i n c e , e t a = ( Q1−Q2 ) /Q1 = ( T1−T2 ) /T1 T1 = T2 /(1 -( eta /100 ) ) // T emp era t u re o f h e a t s o u r c e i n K Q1 = Q2 /(1 -( eta /100 ) ) // Heat s u p p l i e d by s o u r c e i n kJ / s Pow er = round ( Q1 - Q2 ) // Power o f e n g i n e i n kJ / s // R e s u l t s : printf ( ” \n T em pe ra tu re o f h e a t s o u r c e , T1 = %. 1 f d eg re eC ” ,T1 -273 ) printf ( ” \n Power o f e n g i n e = %d kW\n\n ” , Po we r )

Scilab code Exa 2.2 Calculations on the Carnot cycle 14

1 // C a l c u l a t i o n s on t h e C a rn ot c y c l e 2 clc , clear 3 // G iv en : 4 T3 =800+ 273 , T1 =1 5+ 27 3 // T emp e ra t u re o f a h ot and c o l d 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

reservoir in K P3 =210 , P1 =1 //Maximum and minimum p r e s s u r e i n b a r // S o l u t i o n : // R e f e r f i g 2 . 2 1 et a_c ar not =1 -( T1 / T3 ) // E f f i c i e n c y o f C a rn o t c y c l e T4 = T3 // I s o t h e r m a l p r o c e s s 3−4 g =1.4 // S p e c i f i c h e a t r a t i o ( gamma ) P4 = P1 *( T4 / T1 ) ^( g /(g -1) ) // I n i t i a l p r e s s u r e o f i s e n t r o p i c p r o c e s s 4−1 i n b a r R =0.287 // S p e c i f i c g a s c o n s t a n t i n kJ /kgK Q3_4 =R * T3 * log ( P3 / P4 ) // Hea t s u p p l i e d i n kJ / kg W3_4 = Q3_4 //Work s u p p l i e d i n kJ / kg Ne t_ w or k = e ta _ c ar n o t * Q3_4 // Net work o u t p u t i n kJ / kg cv =0. 71 8 // S p e c i f i c h e a t a t c o n s t a n t vol ume i n kJ / kgK W4_1 = cv *( T4 - T1 ) //Work f o r i s e n t r o p i c p r o c e s s i n kJ / kg Gr o ss _wo r k = W3_ 4 + W4_1 // G r os s work s u p p l i e d i n kJ / kg wo r k_ r a ti o = N e t_ w or k / G ro s s _w o r k //Work r a t i o // R e s u l t s : printf ( ” \n The e f f i c i e n c y o f t h e C a rn ot c y c l e , e t a c a r n o t = %. 1 f p e r c e n t ” , et a_c arn o t *10 0) printf ( ” \n The work r a t i o o f t h e C a r no t c y c l e = %. 3 f \n\n ” , wo rk_ ra...