Elements of Mechanical Engineering - R. K. Rajput

Scilab Textbook Companion for Elements of Mechanical Engineering by R. K. Rajput1 Created by Vatsal Shah B.TECH Mechanical Engineering Institute of Technology,Nirma University College Teacher None Cross-Checked by Bhavani Jalkrish September 25, 2014

1 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: Elements of Mechanical Engineering Author: R. K. Rajput Publisher: Laxmi Publications, New Delhi. Edition: 1 Year: 2009 ISBN: 978-81-318-0677-7

1

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

2 Fuels and Combustion

8

3 Properties of Gases

11

4 Properties of Steam

23

5 Heat Engines

51

6 Steam Boilers

80

7 Internal Combustion Engines

96

10 Air Compressors

115

13 Transmission of Motion and Power

123

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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 Exa Exa

2.1 2.2 3.1 3.2 3.3 3.4 3.5 3.6 3.8 3.10 3.11 3.12 3.13 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15

Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example

1 2 1 2 3 4 5 6 8 10 11 12 13 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

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8 9 11 11 12 13 14 14 15 17 18 20 21 23 23 24 26 27 28 29 30 30 32 34 35 36 37 38

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

4.16 4.17 4.18 4.19 4.20 4.21 4.22 4.23 4.24 4.25 4.26 4.27 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21 6.1 6.2 6.3 6.4 6.5

Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example

16 17 18 19 20 21 22 23 24 25 26 27 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 1 2 3 4 5

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38 39 40 41 42 43 44 45 46 47 48 49 51 53 54 55 57 58 60 61 62 63 65 66 67 68 69 70 73 74 75 76 76 80 81 82 83 84

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

6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13 6.14 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 7.15 7.16 7.17 10.1 10.2 10.3 10.4 10.5 10.6 13.1 13.2 13.3 13.4 13.5 13.6

Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example Example

6 7 8 9 10 11 12 13 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 2 3 4 5 6 1 2 3 4 5 6

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85 86 88 89 90 91 92 94 95 96 97 97 98 98 99 101 102 103 104 105 107 108 109 110 111 113 115 116 117 118 119 121 123 124 124 125 126 127

Exa Exa Exa Exa Exa Exa Exa

13.7 13.8 13.9 13.10 13.11 13.12 13.13

Example Example Example Example Example Example Example

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128 130 131 132 133 134 135

Chapter 2 Fuels and Combustion

Scilab code Exa 2.1 Example 1 1 2 3 4 5 6 7 8 9 10 11 12

clc clear //DATA GIVEN c =88; //% o f c a r b o n i n c o a l h =4.2; //% o f h y d r o g e n i n c o a l Wf =0.848; // w e i g h t o f c o a l i n g Wfw =0.027; // w e i g h t o f f u s e w i r e i n calorimeter in g W =1950; // w e i g h t o f w a t e r i n calorimeter in g We =380; // w a t e r e q u i v a l e n t o f calorimeter Dt =3.06; // o b s e r v e d t e m p e r a t u r e r i s e ( t2 −t 1 ) i n deg c e l s i u s tc =0.017; // c o o l i n g c o r r e c t i o n i n deg celsius cfw =6700; // c a l o r i f i c v a l u e o f f u s e wire in J/g

13 14 //CALCULATIONS 15 ctr =( Dt ) + tc ;

// c o r r e c t e d temp . r i s e 8

16 Hw =( W + We ) *4.18*[ ctr ];

// h e a t r e c i e v e d by w a t e r i n

J // h e a t g i v e n o u t by f u s e

17 Hfw = Wfw * cfw ;

wire in J // h e a t p r o d u c e d due t o combustion of f u e l in J 19 HCV = Hcf / Wf ; // h i g h e r c a l o r i f i c v a l u e o f f u e l i n kJ / kg 20 Ms =9* h /100; // steam p r o d u c e d p e r kg o f coal 21 LCV = HCV -2465* Ms ; // l o w e r c a l o r i f i c v a l u e o f f u e l i n kJ / kg 18 Hcf = Hw - Hfw ;

22 23

printf ( ’ The H i g h e r i s : %5 . 1 f kJ / kg . 24 printf ( ’ The Lower i s : %5 . 1 f kJ / kg .

c a l o r i f i c value o f f u e l , H.C.V. \n ’ , HCV ) ; c a l o r i f i c value o f f u e l , L .C.V. \n ’ , LCV ) ;

Scilab code Exa 2.2 Example 2 1 clc 2 clear 3 //DATA GIVEN 4 V1 =0.08; 5 6 7 8 9

i n mˆ3 Pg =5.2; cm o f w a t e r Pb =75.5; o f Hg Ww =28; g a s i n kg Tg =13; celsius Twi =10; i n l e t i n deg c e l s i u s

// g a s b u r n t i n c a l o r i m e t e r // p r e s s u r e o f g a s s u p p l y i n // b a r o m e t e r r e a d i n g i n cm // w e i g h t o f w a t e r h e a t e d by // t e m p e r a t u r e o f g a s i n deg // t e m p e r a t u r e o f w a t e r a t

9

10 Two =23.5;

// t e m p e r a t u r e o f w a t e r a t

o u t l e t i n deg c e l s i u s 11 Ms =0.06; // steam c o n d e n s e d i n kg 12 13 //CALCULATIONS 14 // by u s i n g g e n e r a l g a s e q u a t i o n , r e d u c i n g t h e volume

to S .T.P. // p1 ∗V1/T1=p2 ∗V2/T2 p1 = Pb +( Pg /13.6) ; T1 = Tg +273; p2 =76; T2 =15+273; V2 = p1 * V1 * T2 / T1 / p2 ; Hw = Ww *4.18*( Two - Twi ) ; kJ 22 HCV = Hw / V1 ; f u e l i n kJ /mˆ3 23 LCV = HCV -2465* Ms / V1 ; f u e l i n kJ /mˆ3

15 16 17 18 19 20 21

// i n cm o f Hg // i n K // i n cm o f Hg // i n K // i n mˆ3 // h e a t r e c i e v e d by w a t e r i n // h i g h e r c a l o r i f i c v a l u e o f // l o w e r c a l o r i f i c v a l u e o f

24 25

printf ( ’ The C a l o r i f i c v a l u e s o f f u e l p e r mˆ3 o f g a s a t 15 deg c e l s i u s and 76 cm o f Hg p r e s s u r e a r e : \n ’ ) ; 26 printf ( ’ The H i g h e r c a l o r i f i c v a l u e o f f u e l , H . C . V . i s : %5 . 1 f kJ /mˆ 3 . \n ’ , HCV ) ; 27 printf ( ’ The Lower c a l o r i f i c v a l u e o f f u e l , L . C . V . i s : %5 . 1 f kJ /mˆ 3 . \n ’ , LCV ) ;

10

Chapter 3 Properties of Gases

Scilab code Exa 3.1 Example 1 1 clc 2 clear 3 //DATA GIVEN 4 Q = -50;

// h e a t r e j e c t e d t o

c o o l i n g w a t e r i n kJ / kg // work i n p u t i n kJ /

5 W = -100;

kg 6 7 // u s i n g F i r s t Law o f Thermodynamics , Q=(u2−u1 )+W 8 Du =Q - W ; // ( u2−u1 ) c h a n g e i n

i n t e r n a l e n e r g y i n kJ / kg 9 // s i n c e Du i s +ve , t h e r e i s g a i n i n i n t e r n a l e n e r g y 10 11

printf ( ’ The GAIN i n i n t e r n a l e n e r g y i s : %2 . 0 f kJ / kg . \n ’ , Du ) ;

Scilab code Exa 3.2 Example 2

11

1 clc 2 clear 3 //DATA GIVEN 4 u1 =450;

// i n t e r n a l e n e r g y a t b e g i n n i n g o f t h e e x p a n s i o n i n kJ / kg 5 u2 =220; // i n t e r n a l e n e r g y a f t e r e x p a n s i o n i n kJ / kg 6 W =120; // work done by t h e a i r d u r i n g e x p a n s i o n i n kJ / kg 7 8 // u s i n g F i r s t Law o f Thermodynamics , Q=(u2−u1 )+W 9 Q =( u2 - u1 ) + W ; // h e a t f l o w i n kJ / kg 10 // s i n c e Q i s −ve , t h e r e i s r e j e c t i o n o f h e a t 11 12 printf ( ’ The h e a t REJECTED by a i r i s : %3 . 0 f kJ / kg . \n

’ ,( - Q ) ) ;

Scilab code Exa 3.3 Example 3 1 clc 2 clear 3 //DATA GIVEN 4 m =0.3;

// mass o f n i t r o g e n

i n kg // p r e s s u r e i n MPa // t e m p e r a t u r e b e f o r e

5 p1 =0.1; 6 T1 =40+273; 7 8 9 10 11 12

compression in K p2 =1; T2 =160+273; compression in K W = -30; t h e c o m p r e s s i o n i n kJ / kg Cv =0.75

// p r e s s u r e i n MPa // t e m p e r a t u r e a f t e r // work done d u r i n g // i n kJ /kgK

// u s i n g F i r s t Law o f Thermodynamics , Q=(u2−u1 )+W 12

13 // ( u2−u1 )=m∗Cv ∗ ( T2−T1 ) 14 Du = m * Cv *( T2 - T1 ) ; 15 Q = Du + W ; // h e a t f l o w i n kJ / kg 16 // s i n c e Q i s −ve , t h e r e i s r e j e c t i o n o f h e a t 17 18 printf ( ’ The h e a t REJECTED by a i r i s : %1 . 0 f kJ . \n ’

,( - Q ) ) ;

Scilab code Exa 3.4 Example 4 1 clc 2 clear 3 //DATA GIVEN 4 // i n i t i a l s t a t e 5 p1 =0.105;

// p r e s s u r e o f g a s i n

MPa // volume o f g a s i n m

6 V1 =0.4;

ˆ3 7 // f i n a l s t a t e 8 p2 =0.105; MPa 9 V2 =0.20; ˆ3

// p r e s s u r e o f g a s i n // volume o f g a s i n m

10 11 Q = -42.5;

// h e a t t r a n s f e r r e d

i n kJ 12 p = p1 ; 13 14 // p r o c e s s used − ISOBARIC ( C o n s t a n t p r e s s u r e ) 15 W12 = p *( V2 - V1 ) *1000; // work i n kJ 16 // u s i n g F i r s t Law o f Thermodynamics , Q=(u2−u1 )+W 17 Du =Q - W12 ; // ( u2−u1 ) c h a n g e i n

i n t e r n a l e n e r g y i n kJ 18 // s i n c e Du i s −ve , t h e r e i s d e c r e a s e i n i n t e r n a l energy 13

19 20

printf ( ’ The DECREASE i n i n t e r n a l e n e r g y i s : %2 . 1 f kJ . \n ’ ,( - Du ) ) ;

Scilab code Exa 3.5 Example 5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

clc clear //DATA GIVEN // p a r t −1 // p r e s s u r e=p1 , t e m p e r a t u r e=T1 // p a r t −2 // p r e s s u r e=p2 , t e m p e r a t u r e=T2 // Acc . F i r s t Law o f Thermodynamics , Q=(u2−u1 )+W // when p a r t i t i o n moved DQ =0; DW =0; DU = DQ - DW ; //DU=0

printf ( ’ CONCLUSION : \n ’ ) ; printf ( ’ Acc . t o F i r s t Law o f Thermodynamics , \ n ’ ); 18 printf ( ’ When p a r t i o n moved , t h e r e i s c o n s e r v a t i o n o f i n t e r n a l e n e r g y . \n ’ ) ;

Scilab code Exa 3.6 Example 6 1 clc 2 clear 3 //DATA GIVEN 4 // i n i t i a l s t a t e

14

// i n i t i a l

5 p1 =10^5;

pressure

o f a i r i n Pa // volume o f a i r i n m

6 v1 =1.8;

ˆ3/ kg // i n i t i a l

7 T1 =25+273;

temperature of a i r in K 8 // f i n a l s t a t e 9 p2 =5*10^5; a i r i n Pa 10 T2 =25+273; of air in K

// f i n a l p r e s s u r e o f // f i n a l t e m p e r a t u r e

11 12 // p r o c e s s used − ISOTHERMAL ( C o n s t a n t t e m p e r a t u r e ) 13 W12 =[ p1 * v1 * log ( p1 / p2 ) ]/1000; // work i n kJ / kg 14 // s i n c e W i s −ve , work i s s u p p l i e d t o t h e a i r 15 16 // s i n c e t e m p e r a t u r e i s c o n s t a n t 17 Du =0; // ( u2−u1 ) c h a n g e i n

i n t e r n a l e n e r g y i n kJ / kg 18 19 // u s i n g F i r s t Law o f Thermodynamics , Q=(u2−u1 )+W 20 Q = Du + W12 ; 21 // s i n c e Q i s −ve , t h e r e i s r e j e c t i o n o f h e a t from

system to s u r r o u n d i n g s 22 23

printf ( ’ ( i ) The Work done on t h e a i r i s : %3 . 1 f kJ / kg . \n ’ ,( - W12 ) ) ; 24 printf ( ’ ( i i ) The c h a n g e i n i n t e r n a l e n e r g y i s : %1 . 0 f kJ / kg . \n ’ ,( Du ) ) ; 25 printf ( ’ ( i i i ) The Heat REJECTED i s : %3 . 1 f kJ / kg . \n ’ ,( - Q ) ) ;

Scilab code Exa 3.8 Example 8 1 clc

15

2 clear 3 //DATA GIVEN 4 p1 =4*10^5;

// i n i t i a l

pressure

i n N/mˆ2 // i n i t i a l volume i n

5 V1 =0.2; 6 7 8

mˆ3 T1 =130+273; // i n i t i a l temperature in K p2 =1.02*10^5; // f i n a l p r e s s u r e a f t e r a d i a b a t i c e x p a n s i o n i n N/mˆ2 Q23 =72.5; // i n c r e a s e i n e n t h a l p y d u r i n g c o n s t a n t p r e s s u r e p r o c e s s i n kJ Cp =1; // i n kJ /kgK Cv =0.714; // i n kJ /khK

9 10 11 12 //gamma f o r a i r , g 13 g = Cp / Cv ; 14 R =( Cp - Cv ) *1000; 15 16 // f o r r e v e r s i b l e a d i a b a t i c 17 // p1 ∗ ( V1ˆ g )=p2 ∗ ( V2ˆ g ) 18 V2 = V1 *( p1 / p2 ) ^(1/ g ) ;

p r o c e s s 1−2

ˆ3 19 // ( T2/T1 ) =(p2 / p1 ) ˆ ( ( g −1) / g ) ; 20 T2 = T1 *( p2 / p1 ) ^(( g -1) / g ) ;; K 21 22 23 24 25 26 27 28 29 30 31 32

// f i n a l volume i n m

// f i n a l temp . T2 i n

// mass i n kg

m = p1 * V1 / R / T1 ;

// f o r c o n s t a n t p r e s s u r e p r o c e s s 2−3 // Q23=m∗Cp ∗ ( T3−T2 ) ; T3 = Q23 / m / Cp + T2 ; //V2/T2=V3/T3 V3 = V2 / T2 * T3 ; // Work done by t h e p a t h 1−2−3, W123=W12+W23 W12 =( p1 * V1 - p2 * V2 ) /( g -1) ; W23 = p2 *( V3 - V2 ) ; 16

33 34 35

W123 = W12 + W23 ;

// i f t h e a b o v e p r o c e s s e s a r e r e p l a c e d by a s i n g l e r e v e r s i b l e p o l y t r o p i c p r o c e s s g i v i n g t h e same work b e t w e e n i n i t i a l and f i n a l s t a t e s , 36 //W13=W123=(p1V1−p3V3 ) / ( n −1) 37 p3 = p2 ; 38 n =1+( p1 * V1 - p3 * V3 ) / W123 ; // i n d e x o f e x p a n s i o n , n 39 40

printf ( ’ ( i ) The T o t a l Work done i s : %5 . 0 f Nm o r J . \n ’ , W123 ) ; 41 printf ( ’ ( i i ) The v a l u e o f i n d e x o f e x p a n s i o n , n i s : %1 . 3 f . \n ’ ,n ) ; 42 43 44

//NOTE: // t h e r e i s s l i g h t v a r i a t i o n i n a n s w e r s o f t h e book due t o r o u n d i n g o f f o f t h e v a l u e s

Scilab code Exa 3.10 Example 10 1 clc 2 clear 3 //DATA GIVEN 4 // i n i t i a l s t a t e 5 p1 =10^5; 6 7 8 9 10

// i n i t i a l

o f g a s i n Pa V1 =0.45; g a s i n mˆ3 T1 =80+273; temperature of gas in K // f i n a l s t a t e p2 =5*10^5; g a s i n Pa V2 =0.13; 17

pressure

// i n i t i a l volume o f // i n i t i a l

// f i n a l p r e s s u r e o f // f i n a l volume o f

g a s i n mˆ3 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

//gamma f o r a i r , g g =1.4; R =294.2

// J /kgK

m = p1 * V1 / R / T1 ;

// mass i n kg

// p1 ∗ ( V1ˆ n )=p2 ∗ ( V2ˆ n ) n = log ( p1 / p2 ) / log ( V2 / V1 ) ;

// i n d e x n

// I n a p o l y t r o p i c p r o c e s s // ( T2/T1 ) =(V1/V2 ) ˆ ( n −1) ; T2 = T1 *( V1 / V2 ) ^( n -1) ;

// temp . T2 i n K

Cv = R /( g -1) ; Du = m * Cv *( T2 - T1 ) /1000; i n t e r n a l e n e r g y i n kJ

// i n c r e a s e i n

27 28 // u s i n g F i r s t Law o f Thermodynamics , Q=(u2−u1 )+W 29 //W12=(p1 ∗V1−p2 ∗V2 ) / ( n −1)=mR( T2−T1 ) / ( n −1) 30 W12 = m * R *( T1 - T2 ) /( n -1) /1000; 31 Q = Du + W12 ; 32 // s i n c e Q i s −ve , t h e r e i s r e j e c t i o n o f h e a t from

system to s u r r o u n d i n g s 33 34

printf ( ’ ( i ) The Mass o f t h e g a s i s : %1 . 3 f kg . \n ’ ,( m)); 35 printf ( ’ ( i i ) The i n d e x n i s : %1 . 3 f . \n ’ ,( n ) ) ; 36 printf ( ’ ( i i i ) The c h a n g e i n i n t e r n a l e n e r g y i s : %2 . 1 f kJ . \n ’ ,( Du ) ) ; 37 printf ( ’ ( i v ) The Heat REJECTED i s : %2 . 2 f kJ . \n ’ ,( Q));

Scilab code Exa 3.11 Example 11 18

1 clc 2 clear 3 //DATA GIVEN 4 // i n i t i a l s t a t e 5 p1 =1.02; 6 7 8 9

// i n i t i a l

o f a i r in bar V1 =0.015; a i r i n mˆ3 T1 =22+273; temperature of a i r in K // f i n a l s t a t e p2 =6.8; a i r in bar //Law o f a d i a b a t i c c o m p r e s s i o n ,

10 11 12 //gamma f o r a i r , g 13 g =1.4 14 R =0.287; 15 16 // I n a a d i a b a t i c p r o c e s s 17 // ( T2/T1 ) =(p2 / p1 ) ˆ ( ( g −1) / g ) ; 18 T2 = T1 *( p2 / p1 ) ^(( g -1) / g ) ;;

pressure

// i n i t i a l volume o f // i n i t i a l

// f i n a l p r e s s u r e o f pVˆ g=C


K 19 20 // p1 ∗ ( V1ˆ g )=p2 ∗ ( V2ˆ g ) 21 V2 = V1 *( p1 / p2 ) ^(1/ g ) ;

// f i n a l volume i n m

ˆ3 22 23 m = p1 *10^5* V1 /10^3/ R / T1 ; // mass i n kg 24 25 //W=(p1 ∗V1−p2 ∗V2 ) / ( g −1)=mR( T2−T1 ) / ( g −1) 26 W = m * R *( T1 - T2 ) /( g -1) ; 27 // s i n c e W i s −ve , t h e work i s done on t h e a i r 28 29 printf ( ’ ( i ) The F i n a l t e m p e r a t u r e i s : %3 . 2 f deg .

c e l s i u s . \n ’ ,( T2 -273) ) ; 30 printf ( ’ ( i i ) The F i n a l Volume i s : %1 . 5 f mˆ 3 . \n ’ , V2 ); 19

31

printf ( ’ ( i i i ) The Work done on t h e a i r i s : %1 . 3 f kJ . \n ’ ,( - W ) ) ;

Scilab code Exa 3.12 Example 12 1 clc 2 clear 3 //DATA GIVEN 4 m =0.44; 5 T1 =180+273;

// mass o f a i r i n kg // i n i t i a l

temperature of a i r in K // f i n a l t e m p e r a t u r e

6 T2 =15+273;

of air in K // work done d u r i n g

7 W12 =52.5;

t h e p r o c e s s i n kJ 8 //V2/V1=3 9 Vr =3; V2/V1

// volume r a t i o , Vr=

10 11 //Law o f a d i a b a t i c e x p a n s i o n , pVˆ g=C 12 13 // I n an a d i a b a t i c p r o c e s s 14 // ( T2/T1 ) =(V1/V2 ) ˆ ( g −1) ; 15 g =1+[( log ( T2 / T1 ) / log (1/ Vr ) ) ];

//gamma

f o r a i r , g=Cp/Cv 16 17 //W12=(p1 ∗V1−p2 ∗V2 ) / ( n −1)=mR( T2−T1 ) / ( g −1) 18 R = W12 / m /( T1 - T2 ) *( g -1) ; 19 //R=Cp−Cv 20 21 Cv = R /( g -1) ; 22 Cp = g * Cv ; 23 24 printf ( ’ ( i ) The v a l u e o f Cv i s : %1 . 3 f kJ /kgK . \n ’ ,

Cv ) ; 20

25 26 27 28

printf ( ’ ( i i ) The v a l u e o f Cp i s : %1 . 3 f kJ /kgK . \n ’ , Cp ) ; //NOTE: // t h e r e i s s l i g h t v a r i a t i o n i n a n s w e r s o f t h e book due t o r o u n d i n g o f f o f t h e v a l u e s

Scilab code Exa 3.13 Example 13 1 clc 2 clear 3 //DATA GIVEN 4 m =1; 5 6 7 8

// mass o f e t a h n e g a s

i n kg M =30; of ethane p1 =1.1; in bar T1 =27+273; temperature in K p2 =6.6; bar Cp =1.75;

// m o l e c u l a r w e i g h t // i n i t i a l

pressure

// i n i t i a l // f i n a l p r e s s u r e i n

9 // i n kJ /kgK 10 11 //Law o f c o m p r e s s i o n , pVˆ1.3=C 12 n =1.3; 13 14 // C h a r a c t e r i s t i c g a s c o n s t a n t , R = U n i v e r s a l g a s 15 16 17 18 19 20

c o n s t a n t ( Ro ) / M o l e c u l a r w e i g h t (M) Ro =8314; R = Ro / M /1000; // kJ /kgK //R=Cp−Cv Cv = Cp - R ; g = Cp / Cv ;

//gamma g 21

21 22 // I n a p o l y t r o p i c p r o c e s s 23 // ( T2/T1 ) =(p2 / p1 ) ˆ ( ( n −1) / n ) ; 24 T2 = T1 *( p2 / p1 ) ^(( n -1) / n ) ;;


K 25 26 //W=(p1 ∗V1−p2 ∗V2 ) / ( n −1)=mR( T2−T1 ) / ( g −1) 27 W = m * R *( T1 - T2 ) /( n -1) ; 28 29 Q =[( g - n ) /( g -1) ]* W ; // h e a t f l o w i n kJ / kg 30 31 printf ( ’ The Heat SUPPLIED i s : %2 . 1 f kJ / kg . \n ’ ,( Q ) )

;

22

Chapter 4 Properties of Steam

Scilab code Exa 4.1 Example 1 1 clc 2 clear 3 //DATA GIVEN 4 Ms =50;

// mass o f d r y steam

i n kg // mass o f w a t e r i n

5 Mw =1.5;

s u s p e n s i o n i n kg 6 7

// d r y n e s s f r a c t i o n , x=( mass o f d r y steam ) / ( mass o f d r y steam +mass o f w a t e r i n s u s p e n s i o n ) 8 x = Ms /( Ms + Mw ) ;

9 10

printf ( ’ The D r y n e s s f r a c t i o n ( Q u a l i t y ) o f steam i s : %1 . 3 f . ’ ,x ) ;


23

2 clear 3 //DATA GIVEN 4 V =0.6;

// volume o f t h e

v e s s e l i n mˆ3 // p r e s s u r e i n b a r // mass o f l i q u i d and

5 p =0.5; 6 M =3;

w a t e r v a p o u r i n kg 7 8 v=V/M; 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

// s p e c i f i c volume i n

mˆ3/ kg // At 5 bar , from steam t a b l e s vg =0.375; vf =0.00109; vfg = vg - vf ; // v=vg −(1−x ) v f g x =( v - vg ) / vfg +1; vapour

//mˆ3/ kg //mˆ3/ kg

// q u a l i t y o f t h e

// mass and volume o f l i q u i d Mliq = M *(1 - x ) ; Vliq = Mliq * vf ; // mass and volume o f v a p o u r Mvap = M * x ; Vvap = Mvap * vg ; printf ( ’ ( i ) The Mass and Volume o f l i q u i d i s : \n ’ ) ; printf ( ’ M l i q . i s : %1 . 3 f kg . \n ’ , Mliq ) ; printf ( ’ V l i q . i s : %1 . 4 f mˆ 3 . \n ’ , Vliq ) ; printf ( ’ ( i i ) The Mass and Volume o f v a p o u r i s : \n ’ ) ; printf ( ’ Mvap . i s : %1 . 3 f kg . \n ’ , Mvap ) ; printf ( ’ Vvap . i s : %1 . 4 f mˆ 3 . \n ’ , Vvap ) ;


24

1 clc 2 clear 3 //DATA GIVEN 4 V =0.05;

// volume o f v e s s e l i n

mˆ3 5 Mf =10; kg 6 T =245; celsius 7 8 9 10 11 12 13 14 15 16 17 18 19 20

// mass o f l i q u i d i n // temp . i n deg

// from steam t a b l e s , c o r r e s p o n d i n g t o 245 deg celsius Psat =36.5; // b a r vf =0.001239; //mˆ3/ kg vg =0.0546; //mˆ3/ kg hf =1061.4; // kJ / kg hfg =1740.2; // kJ / kg sf =2.7474; // kJ /kgK sfg =3.3585; // kJ /kgK // volume o f l i q u i d // volume o f v a p o u r // mass o f v a p o u r // t o t a l mass o f

Vf = Mf * vf ; Vg =V - Vf ; Mg = Vg / vg ; m = Mf + Mg ; mixture

21 22 x = Mg /( Mg + Mf ) ;

// q u a l i t y o f t h e

mixture 23 vfg = vg - vf ; 24 v = vf + x * vfg ;

// s p e c i f i c volume

25 26 h = hf + x * hfg ; 27 28 s = sf + x * sfg ; 29 30 u =h - Psat *10^5* v /10^3;

// s p e c i f i c e n t h a l p y // s p e c i f i c e n t r o p y // s p e c i f i c

energy 31

25

internal

32 33 34 35

printf ( ’ ( i ) The P r e s s u r e i s : %2 . 1 f b a r . \n ’ , Psat ) ; printf ( ’ ( i i ) The mass m i s : %2 . 3 f kg . \n ’ ,m ) ; printf ( ’ ( i i i ) The S p e c i f i c volume v i s : %1 . 6 f mˆ3/ kg . \n ’ ,v ) ; 36 printf ( ’ ( i v ) The S p e c i f i c e n t h a l p y h i s : %4 . 2 f kJ / kg . \n ’ ,h ) ; 37 printf ( ’ ( v ) The S p e c i f i c e n t r o p y s i s : %1 . 4 f kJ / kgK . \n ’ ,s ) ; 38 printf ( ’ ( v i ) The S p e c i f i c i n t e r n a l e n e r g y u i s : %4 . 2 f kJ / kg . \n ’ ,u ) ; 39 40 41

//NOTE: // t h e r e i s s l i g h t v a r i a t i o n i n a n s w e r s o f book due t o r o u n d i n g o f f o f t h e v a l u e s i n t h e book

Scilab code Exa 4.4 Example 4 1 clc 2 clear 3 //DATA GIVEN 4 Mw =2;

// mass o f w a t e r t o be

c o n v e r t e d t o steam i n kg 5 Tw =25; deg c e l s i u s 6 p =5; 7 x =0.9; 8 9 // At 5 bar , from steam t a b l e s 10 hf =640.1; 11 hfg =2107.4; 12 13 h = hf + x * hfg ;

// temp . o f w a t e r i n // p r e s s u r e // d r y n e s s f r a c t i o n

// kJ / kg // kJ / kg // s p e c i f i c e n t h a l p y (

a b o v e 0 deg c e l s i u s ) // s e n s i b l e h e a t

14 hs =1*4.18*( Tw -0) ;

26

a s s o c i a t e d w i t h i kg o f w a t e r 15 hnet =h - hs ; // n e t q u a n t i t y o f h e a t t o be s u p p l i e d p e r kg o f w a t e r 16 Htotal = Mw * hnet ; // t o t a l amount o f h e a t t o be s u p p l i e d 17 18

printf ( ’ The T o t a l amount o f h e a t t o be s u p p l i e d i s : %4 . 2 f kJ . ’ , Htotal ) ;

Scilab code Exa 4.5 Example 5 1 clc 2 clear 3 //DATA GIVEN 4 m =4.4;

// mass o f steam t o be

p r o d u c e d i n kg // p r e s s u r e o f steam // temp . o f steam i n

5 p =6; 6 Tsup =250;

deg . c e l s i u s // temp . o f w a t e r i n

7 Tw =30;

deg c e l s i u s // s p e c i f i c h e a t o f

8 Cps =2.2;

steam i n kJ / kg 9 10 // At 6 bar , from steam t a b l e s 11 Ts =158.8; // deg . c e l s i u s 12 hf =670.4; // kJ / kg 13 hfg =2085; // kJ / kg 14 // s i n c e t h e g i v e n temp . 250 deg c e l s i u s i s g r e a t e r

t h a n 1 5 8 . 8 deg c e l s i u s , steam i s s u p e r h e a t e d 15 16

hsup = hf + hfg + Cps *( Tsup - Ts ) ; // e n t h a l p y o f 1 kg s u p e r g e a t e d steam r e c k o n e d from 0 deg . c e l s i u s 17 hs =1*4.18*( Tw -0) ; // s e n s i b l e h e a t a s s o c i a t e d w i t h i kg o f w a t e r 27

hnet = hsup - hs ; // n e t q u a n t i t y o f h e a t t o be s u p p l i e d p e r kg o f w a t e r 19 Htotal = m * hnet ; // t o t a l amount o f h e a t t o be s u p p l i e d 18

20 21

printf ( ’ The T o t a l amount o f h e a t t o be s u p p l i e d i s : %4 . 1 f kJ . ’ , Htotal ) ;

Scilab code Exa 4.6 Example 6 1 clc 2 clear 3 //DATA GIVEN 4 V =0.15;

// volume o f wet steam

i n mˆ3 5 p =4;

// p r e s s u r e o f wet

steam i n b a r 6 x =0.8;

// d r y n e s s f r a c t i o n

7 8 9 10 11 12 13 14 15 16

// At 4 bar , from steam t a b l e s vg =0.462; hf =604.7; hfg =2133;

//mˆ3/ kg // kJ / kg // kJ / kg // d e n s i t y i n kg /mˆ3 // mass o f 0 . 1 5 mˆ3 o f

rho =1/( x * vg ) ; m = rho * V ; steam

Htotal =( rho *1) *( hf + x * hfg ) ; // t o t a l h e a t o f 1 mˆ3 o f steam which h a s a mass o f r h o ( 2 . 7 0 5 6 ) kg

17 18

printf ( ’ ( i ) The Mass o f 0 . 1 5 mˆ3 o f steam i s : %1 . 4 f kg . \n ’ ,m ) ; 19 printf ( ’ ( i i ) The T o t a l h e a t o f 1 mˆ3 o f steam which h a s a mass o f 2 . 7 0 5 6 kg i s : %4 . 2 f kJ . \n ’ , Htotal ) 28

;

Scilab code Exa 4.7 Example 7 1 clc 2 clear 3 //DATA GIVEN 4 m =1000; 5 6 7 8 9 10 11 12 13 14 15 16

// mass o f steam

g e n e r a t e d i n kg / h r p =16; in bar x =0.9; Tsup =380+273; s u p e r h e a t e d steam i n K Tfw =30; i n deg . c e l s i u s Cps =2.2; steam i n kJ / kg

// p r e s s u r e o f steam // d r y n e s s f r a c t i o n // temp . o f // temp . o f f e e d w a t e r // s p e c i f i c h e a t o f

// At 16 bar , from steam t a b l e s Ts =201.4+273; // i n K hf =858.6; // kJ / kg hfg =1933.2; // kJ / kg

Hs = m *[( hf + x * hfg ) -1*4.187*( Tfw -0) ]; // h e a t s u p p l i e d t o f e e d w a t e r p e r h r t o p r o d u c e wet steam 17 Ha = m *[(1 - x ) * hfg + Cps *( Tsup - Ts ) ]; // h e a t a b s o r b e d by s u p e r h e a t e r p e r h o u r

18 19

printf ( ’ ( i ) The Heat s u p p l i e d t o f e e d w a t e r p e r h o u r t o p r o d u c e wet steam i s : %4 . 2 f ∗ 1 0 ˆ 3 kJ . \n ’ ,( Hs /1000) ) ; 20 printf ( ’ ( i i ) The Heat a b s o r b e d by s u p e r h e a t e r p e r h o u r i s : %3 . 2 f ∗ 1 0 ˆ 3 kJ . \n ’ ,( Ha /1000) ) ; 29

Scilab code Exa 4.8 Example 8 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

clc clear // At 0 . 7 5 b a r . From steam t a b l e s , // At 100 deg c e l s i u s T1 =100; // deg c e l s i u s hsup1 =2679.4; // kJ / kg // At 150 deg c e l s i u s T2 =150; // deg c e l s i u s hsup2 =2778.2; // kJ / kg Cps1 =( hsup2 - hsup1 ) /( T2 - T1 ) ; // At 0 . 5 b a r . From steam t a b l e s , // At 300 deg c e l s i u s T3 =300; // deg c e l s i u s hsup3 =3075.5; // kJ / kg // At 400 deg c e l s i u s T4 =400; // deg c e l s i u s hsup4 =3278.9; // kJ / kg Cps2 =( hsup4 - hsup3 ) /( T4 - T3 ) ;

printf ( ’ ( i ) The mean s p e c i f i c h e a t f o r s u p e r h e a t e d steam \n ( At 0 . 7 5 bar , b e t w e e n 100 and 150 deg c e l s i u s ) i s : %1 . 3 f . \n ’ , Cps1 ) ; 23 printf ( ’ ( i i ) The mean s p e c i f i c h e a t f o r s u p e r h e a t e d steam \n ( At 0 . 5 bar , b e t w e e n 300 and 400 deg c e l s i u s ) i s : %1 . 3 f . \n ’ , Cps2 ) ;


1 clc 2 clear 3 //DATA GIVEN 4 m =1.5;

// mass o f steam i n

c o o k e r i n kg 5 p1 =5; in bar 6 x1 =1; f r a c t i o n o f steam 7 x2 =0.6; f r a c t i o n o f steam 8 9 10 11 12 13 14 15 16

17 18 19 20 21

// p r e s s u r e o f steam // i n i t i a l d r y n e s s // f i n a l d r y n e s s

// At 5 bar , from steam t a b l e s Ts1 =151.8+273; hf1 =640.1; hfg1 =2107.4; vg1 =0.375;

// i n K // kJ / kg // kJ / kg //mˆ3/ kg

V1 = m * vg1 ; // volume o f p r e s s u r e c o o k e r i n mˆ3 u1 =( hf1 + hfg1 ) -( p1 *10^5) *( vg1 *10^ -3) ; // i n t e r n a l e n e r g y o f steam p e r kg a t i n i t i a l p o i n t 1 //V1=V2 //V1=m∗[(1 − x2 ) ∗ v f 2+x2 ∗ vg2 ] // v f 2 is negligible vg2 = V1 / x2 /1.5; // from steam t a b l e s c o r e e s p o n d i n g t o vg2 = 0 . 6 2 5 mˆ3/ kg p2 =2.9; Ts2 =132.4+273; // i n K hf2 =556.5; // kJ / kg hfg2 =2166.6; // kJ / kg

22 23 24 25 26 27 u2 =( hf2 + x2 * hfg2 ) -( p2 *10^5) * x2 *( vg2 *10^ -3) ;

// i n t e r n a l e n e r g y o f steam p e r kg a t f i n a l p o i n t 2

28

31

hnet = u2 - u1 ; // h e a t t r a n s f e r r e d a t c o n s t a n t volume p e r kg 30 Htotal = m * hnet ; // t o t a l heat t r a n s f e r r e d 31 //−ve s i g n i n d i c a t e s t h a t h e a t h a s b e e n r e j e c t e d 32 Hrej = -1* Htotal ;

29

33 34

printf ( ’ ( i ) The P r e s s u r e a t new s t a t e i s : %1 . 1 f b a r . \n ’ , p2 ) ; 35 printf ( ’ The T e m p e r a t u r e a t new s t a t e i s : %3 . 1 f deg . c e l s i u s o r %3 . 1 f K . \n ’ ,( Ts2 -273) , Ts2 ) ; 36 printf ( ’ ( i i ) The T o t a l h e a t t o be REJECTED i s : %4 . 2 f kJ . ’ , Hrej ) ;

Scilab code Exa 4.10 Example 10 1 clc 2 clear 3 //DATA GIVEN 4 V =0.9;

// c a p a c i t y o f

s p h e r i c a l v e s s e l i n mˆ3 // p r e s s u r e o f steam

5 p1 =8;

in bar 6 x1 =0.9; o f steam 7 p2 =4; a f t e r blow o f f i n b a r 8 p3 =3; steam i n b a r

// d r y n e s s f r a c t i o n // p r e s s u r e o f steam // f i n a l p r e s s u r e o f

9 10 // At 8 bar , from steam t a b l e s 11 hf1 =720.9; 12 hfg1 =2046.5; 13 vg1 =0.240; 14

32

// kJ / kg // kJ / kg //mˆ3/ kg


15 m1 = V /( x1 * vg1 ) ;

t h e v e s s e l i n kg 16 17 h1 = hf1 + x1 * hfg1 ; 18 19 20 21 22 23 24 25

// e n t h a l p y o f steam b e f o r e b l o w i n g o f f ( p e r kg ) // e n t h a l p y o f steam b e f o r e b l o w i n g o f f ( p e r kg ) = e n t h a l p y o f steam a f t e r b l o w i n g o f f ( p e r kg ) h2 = h1 ; // h2=h f 2+x2 ∗ h f g 2 // At 4 bar , from steam t a b l e s hf2 =604.7; // kJ / kg hfg2 =2133; // kJ / kg vg2 =0.462; //mˆ3/ kg x2 =( h2 - hf2 ) / hfg2 ; // d r y n e s s f r a c t i o n at 2

26 27 m2 = V /( x2 * vg2 ) ;


t h e v e s s e l i n kg 28 m = m1 - m2 ; blown o f f i n kg 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

// mass o f steam

// As i t i s c o n s t a n t volume c o o l i n g , x2 ∗ vg2 ( a t 4 b a r ) =x3 ∗ vg3 ( a t 3 b a r ) // At 3 bar , from steam t a b l e s hf3 =561.4; // kJ / kg hfg3 =2163.2; // kJ / kg vg3 =0.606; //mˆ3/ kg x3 = x2 * vg2 / vg3 ; h3 = hf3 + x3 * hfg3 ; // h e a t l o s t d u r i n g c o o l i n g , Q l o s t=m( u3−u2 ) u2 = h2 - p2 *10^5* x2 * vg2 *10^ -3; u3 = h3 - p3 *10^5* x3 * vg3 *10^ -3; Qlost = m *( u3 - u2 ) ; printf ( ’ ( i ) The Mass o f o f steam blown o f f f kg . \n ’ ,m ) ; 33

i s : %1 . 3

printf ( ’ ( i i ) The D r y n e s s f r a c t i o n o f steam i n t h e v e s s e l a f t e r c o o l i n g i s : %1 . 4 f . \n ’ , x3 ) ; 46 printf ( ’ ( i i i ) The Heat l o s t d u r i n g c o o l i n g i s : %3 . 2 f kJ . \n ’ ,( - Qlost ) ) ; 45

47 48 49 50

//NOTE: // The a n s w e r s o f m1 , x3 a r e INCORRECT i n t h e book , // t h u s , t h e a n s w e r s o f m, x3 and Q l o s t a r e INCORRECT i n t h e book 51 // w h i l e , t h e v a l u e s o b t a i n e d h e r ( i n s c i l a b ) a r e CORRECT.

Scilab code Exa 4.11 Example 11 1 clc 2 clear 3 //DATA GIVEN 4 p =8; 5 6 7 8 9 10 11

// p r e s s u r e o f steam

in bar x =0.8;


// At 8 bar , from steam t a b l e s vg =0.240; hfg =2046.5;

//mˆ3/ kg // kJ / kg

We = p *10^5* x * vg /1000; d u r i n g e v a p o r a t i o n i n kJ 12 LHi = x * hfg - We ; h e a t i n kJ 13 14

// e x t e r n a l work done // I n t e r n a l l a t e n t

printf ( ’ ( i ) The E x t e r n a l work done d u r i n g e v a p o r a t i o n i s : %3 . 1 f kJ . \n ’ , We ) ; 15 printf ( ’ ( i i ) The I n t e r n a l l a t e n t h e a t i s : %4 . 1 f kJ . \n ’ , LHi ) ;

34

Scilab code Exa 4.12 Example 12 1 clc 2 clear 3 //DATA GIVEN 4 p =10;

// p r e s s u r e o f steam ,

p1=p2 i n b a r // d r y n e s s f r a c t i o n // volume o f steam i n

5 x1 =0.85; 6 V1 =0.15;

mˆ3 // temp . o f steam i n K // s p e c i f i c h e a t o f

7 Tsup2 =300+273; 8 Cps =2.2;

steam i n kJ /kgK 9 10 // At 10 bar , from steam t a b l e s 11 vg1 =0.194; 12 hfg1 =2013.6; 13 Ts1 =179.9+273; 14 m = V1 /( x1 * vg1 ) ;

i n kg hnet =(1 - x1 ) * hfg1 + Cps *( Tsup2 - Ts1 ) ; p e r kg o f steam 16 Htotal = m * hnet ; supplied 15

17 18 19 20 21 22 23

//mˆ3/ kg // kJ / kg // i n K // mass o f steam // h e a t s u p p l i e d // t o t a l h e a t

// E x t e r n a l work done d u r i n g t h e p r o c e s s We=p ∗ ( vsup2− x ∗ vg1 ) // s i n c e p1=p2=p , // vg1 / Ts1=v s u p 2 / Tsup2 vsup2 = vg1 * Tsup2 / Ts1 ; We = p *10^5*( vsup2 - x1 * vg1 ) *10^ -3; hp = We / hnet ; //% o f t o t a l h e a t s u p p l i e d ( p e r kg ) which a p p e a r s a s e x t e r n a l work 35

24 25

printf ( ’ ( i ) The T o t a l h e a t s u p p l i e d i s : %3 . 1 f kJ . \ n ’ , Htotal ) ; 26 printf ( ’ ( i i ) The P e r c e n t a g e o f t o t a l h e a t s u p p l i e d ( p e r kg ) which a p p e a r s a s e x t e r n a l work i s : %2 . 1 f p e r c e n t . \n ’ ,( hp *100) ) ;

Scilab code Exa 4.13 Example 13 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc clear //DATA GIVEN p =18; x =0.85;

// p r e s s u r e o f steam // d r y n e s s f r a c t i o n

// At 18 bar , from steam t a b l e s hf =884.6; hfg =1910.3; vg =0.110; uf =883; ug =2598;

// kJ / kg // kJ / kg //mˆ3/ kg // kJ / kg // kJ / kg // s p e c i f i c volume o f

v = x * vg ; wet steam 15 h = hf + x * hfg ; o f wet steam 16 u =(1 - x ) * uf + x * ug ; e n e r g y o f wet steam

// s p e c i f i c e n t h a l p y // s p e c i f i c

17 18

internal

printf ( ’ ( i ) The S p e c i f i c volume v i s : %1 . 4 f mˆ3/ kg . \n ’ ,v ) ; 19 printf ( ’ ( i i ) The S p e c i f i c e n t h a l p y h i s : %4 . 2 f kJ / kg . \n ’ ,h ) ; 20 printf ( ’ ( i i i ) The S p e c i f i c i n t e r n a l e n e r g y u i s : %4 . 2 f kJ / kg . \n ’ ,u ) ; 36

Scilab code Exa 4.14 Example 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

16 17 18 19

clc clear //DATA GIVEN p =7; h =2550;

// p r e s s u r e o f steam // e n t h a l p y o f steam

// At 7 bar , from steam t a b l e s hf =697.1; hfg =2064.9; vg =0.273; uf =696; ug =2573;

// kJ / kg // kJ / kg //mˆ3/ kg // kJ / kg // kJ / kg

hg = hf + hfg ; // At 7 bar , hg =2762 kJ / kg , h e n c e s i n c e a c t u a l e n t h a l p y i s g i v e n a s 2 5 5 0 kJ / kg , t h e steam must be i n wet v a p o u r s t a t e // s p e c i f i c e n t h a l p y o f wet steam , h=h f+x ∗ h f g x =( h - hf ) / hfg ; // d r y n e s s f r a c t i o n v = x * vg ; // s p e c i f i c volume o f wet steam u =(1 - x ) * uf + x * ug ; // s p e c i f i c i n t e r n a l e n e r g y o f wet steam

20 21

printf ( ’ ( i ) The D r y n e s s f r a c t i o n x i s : %1 . 3 f . \n ’ ,x ); 22 printf ( ’ ( i i ) The S p e c i f i c volume v i s : %1 . 4 f mˆ3/ kg . \n ’ ,v ) ; 23 printf ( ’ ( i i i ) The S p e c i f i c i n t e r n a l e n e r g y u i s : %4 . 2 f kJ / kg . \n ’ ,u ) ;

37

Scilab code Exa 4.15 Example 15 1 clc 2 clear 3 //DATA GIVEN 4 p =120; 5 v =0.01721;

// p r e s s u r e o f steam // s p e c i f i c volume o f

steam 6 7 // At 120 bar , from steam t a b l e s 8 vg =0.0143; //mˆ3/ kg 9 // s i n c e vg
mˆ3/ kg 11 T =350; 12 h =2847.7; o f steam 13 u =h - p *10^5* v /10^3; e n e r g y o f steam

// deg . c e l s i u s // s p e c i f i c e n t h a l p y // s p e c i f i c

14 15

internal

printf ( ’ ( i ) The T e m p e r a t u r e i s : %3 . 0 f deg c e l s i u s . \n ’ ,T ) ; 16 printf ( ’ ( i i ) The S p e c i f i c e n t h a l p y h i s : %4 . 1 f kJ / kg . \n ’ ,h ) ; 17 printf ( ’ ( i i i ) The S p e c i f i c i n t e r n a l e n e r g y u i s : %4 . 2 f kJ / kg . \n ’ ,u ) ;

Scilab code Exa 4.16 Example 16 1 clc 2 clear 3 //DATA GIVEN

38

// p r e s s u r e o f steam // s p e c i f i c e n t h a l p y

4 p =140; 5 h =3001.9;

o f steam 6 7 // At 140 bar , from steam t a b l e s 8 hg =2642.4; 9 // s i n c e hg
kJ / kg // deg . c e l s i u s // s p e c i f i c volume o f

11 T =400; 12 v =0.01722;

steam // s p e c i f i c

13 u =h - p *10^5* v /10^3;

internal

e n e r g y o f steam 14 15

printf ( ’ ( i ) The T e m p e r a t u r e i s : %3 . 0 f deg c e l s i u s . \n ’ ,T ) ; 16 printf ( ’ ( i i ) The S p e c i f i c volume v i s : %1 . 5 f mˆ3/ kg . \n ’ ,v ) ; 17 printf ( ’ ( i i i ) The S p e c i f i c i n t e r n a l e n e r g y u i s : %4 . 2 f kJ / kg . \n ’ ,u ) ;

Scilab code Exa 4.17 Example 17 1 clc 2 clear 3 4 p1 =10;

// p r e s s u r e i n

bar // At 10 b a r and 300 deg c e l s i u s , from steam t a b l e s o f s u p e r h e a t e d steam 6 hsup =3051.2 // kJ / kg 7 Tsup =300+273; // temp . o f steam in K 8 // At 10 b a r and 300 deg c e l s i u s , from steam t a b l e s 5

39

o f d r y s a t u r a t e d steam 9 Ts =179.9+273 in K 10 vg =0.194;

// temp . o f steam //mˆ3/ kg

11 12 //By vg / Ts = vsup / Tsup 13 vsup = vg * Tsup / Ts ; 14 u1 = hsup - p1 *10^5* vsup /10^3; 15 16 p2 =1.4; 17 18 19 20 21 22 23 24

in bar x2 =0.8; fraction // At 1 . 4 bar , from steam t a b l e s hf2 =458.4; hfg2 =2231.9; vg2 =1.236; h2 = hf2 + x2 * hfg2 ; wet steam ( a f t e r e x p a n s i o n ) u2 = h2 - p2 *10^5* x2 * vg2 /10^3; e n e r g y o f t h i s steam Du = u2 - u1 ; i n t e r n a l e n e r g y p e r kg

25 26

// new p r e s s u r e // d r y n e s s

// kJ / kg // kJ / kg //mˆ3/ kg // e n t h a l p y o f // i n t e r n a l // c h a n g e i n

printf ( ’ ( i ) The I n t e r n a l e n e r g y o f s u p e r h e a t e d steam a t 10 b a r i s : %4 . 1 f kJ / kg . \n ’ , u1 ) ; 27 printf ( ’ ( i i ) The Change i n i n t e r n a l e n e r g y p e r kg i s : %2 . 1 f kJ . \n ’ , Du ) ; 28 printf ( ’ ( N e g a t i v e s i g n i n d i c a t e s DECREASE i n i n t e r n a l energy . ) ’ );

Scilab code Exa 4.18 Example 18 1 clc 2 clear

40

3 //DATA GIVEN 4 m =1; 5 p =20;

// mass o f steam i n kg // p r e s s u r e o f steam

in bar // temp . o f steam i n K // d r y n e s s f r a c t i o n // s p e c i f i c h e a t o f

6 Tsup =400+273; 7 x =0.9; 8 Cps =2.3;

steam i n kJ /kgK 9 10 11 12 13 14 15 16 17

// At 20 bar , from steam t a b l e s Ts =212.4+273; hf =908.6; hfg =1888.6; vg =0.0995; hsup = hf + hfg + Cps *( Tsup - Ts ) ;

// i n K // kJ / kg // kJ / kg //mˆ3/ kg // kJ / kg

// Assume s u p e r h e a t e d steam t o b e h a v e a s a p e r f e c t g a s from t h e commencement o f s u p e r h e a t i n g and t h u s obey C h a r l e ’ s Law 18 //By vg / Ts=vsup / Tsup 19 vsup = vg * Tsup / Ts ; 20 usup = hsup - p *10^5* vsup *10^ -3; // i n t e r n a l e n e r g y o f 1 kg o f s u p e r h e a t e d steam i n kJ / kg

21 22 h = hf + x * hfg ; 23 u =h - p *10^5* x * vg *10^ -3;

// i n t e r n a l e n e r g y o f 1 kg o f wet steam i n kJ / kg

24 25

printf ( ’ ( i ) The I n t e r n a l e n e r g y o f 1 kg o f s u p e r h e a t e d steam a t 400 deg c e l s i u s i s : %4 . 2 f kJ / kg . \n ’ , usup ) ; 26 printf ( ’ ( i i ) The I n t e r n a l e n e r g y o f 1 kg o f wet steam w i t h d r y n e s s f r a c t i o n 0 . 9 i s : %4 . 2 f kJ / kg . \n ’ ,u ) ;

41


// p r e s s u r e i n t h e

b o i l e r s and main i s 20 b a r // t e m p e r a t u r e o f steam i n b o i l e r w i t h s u p e r h e a t e r i n deg . c e l s i u s 6 Tm =250; // t e m p e r a t u r e o f steam i n t h e main i n deg . c e l s i u s 7 Cps =2.25; // s p e c i f i c h e a t o f steam i n kJ / kg

5 Tbs =350;

8 9 10 11 12 13 14 15 16 17 18 19

// At 20 bar , from steam t a b l e s Ts =212.4; hf =908.6; hg =2797.2; hfg =1888.6;

// deg . c e l s i u s // kJ / kg // kJ / kg // kJ / kg

// B o i l e r B1−20 bar , 350 deg . c e l s i u s h1 = hg + Cps *( Tbs - Ts ) ; // Main −20 bar , 250 deg c e l s i u s hm =2*[ hg + Cps *( Tm - Ts ) ]; 2 kg o f steam i n t h e steam main

// t o t a l h e a t o f

20 21 // B o i l e r B2−20 bar , 22 // h2=h f+x2 ∗ h f g 23 // h2=hm−h1 24 x2 =(( hm - h1 ) - hf ) / hfg ; 25 26 printf ( ’ The Q u a l i t y o f steam i n t h e B o i l e r w i t h o u t

s u p e r h e a t e r i s : %1 . 3 f . \n ’ , x2 ) ;

42

Scilab code Exa 4.20 Example 20 1 clc 2 clear 3 //DATA GIVEN 4 m =1;

// mass o f wet

steam i n kg // p r e s s u r e o f

5 p =6;

steam i n b a r // d r y n e s s

6 x =0.8;

fraction 7 8 // At 6 bar , from steam t a b l e s 9 Ts =158.8+273; 10 hfg =2085; 11 swet =4.18* log ( Ts /273) + x * hfg / Ts ;

// i n K // kJ / kg // e n t r o p y o f

wet steam i n kJ /kgK 12 13 14 15 16

printf ( ’ The Entropy o f wet steam i s : %1 . 4 f kJ /kgK . ’ , swet ) ; //NOTE; // t h e e x a c t a n s i s 5 . 7 7 9 4 , w h i l e i n TB i t a s 5 . 7 8 6 5 kJ /kgK

i s given

Scilab code Exa 4.21 Example 21 1 clc 2 clear 3 //DATA GIVEN 4 p1 =10;

// i n i t i a l

steam i n b a r 5 Tsup =250; o f steam i n deg c e l s i u s 6 p2 =0.2; 43

pressure of

// i n i t i a l t e m p e r a t u r e // f i n a l p r e s s u r e o f

steam i n b a r 7 x2 =0.9; f r a c t i o n o f steam 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

// f i n a l d r y n e s s

// At 10 bar , from steam t a b l e s hsup =3263.9; // kJ / kg ssup =7.465; // kJ /kgK h1 = hsup ; s1 = ssup ; // At 0 . 2 bar , from steam t a b l e s hf2 =251.5; hfg2 =2358.4; sf2 =0.8321; sg2 =7.9094; h2 = hf2 + x2 * hfg2 ; sfg2 =( sg2 - sf2 ) ; s2 = sf2 + x2 * sfg2 ;

// kJ / kh // kJ / kg // kJ /kgK // kJ /kgK

// d r o p i n e n t h a l p y // c h a n g e i n e n t r o p y

Dh = h1 - h2 ; Ds = s1 - s2 ;

printf ( ’ ( i ) The Drop i n e n t h a l p y i s : %3 . 1 f kJ / kg . \ n ’ , Dh ) ; 28 printf ( ’ ( i i ) The c h a n g e (DECREASE) i n e n t r o p y i s : %1 . 4 f kJ /kgK . ’ , Ds ) ;

Scilab code Exa 4.22 Example 22 1 clc 2 clear 3 //DATA GIVEN 4 m =1; 5 p =12;

// mass o f steam i n kg // p r e s s u r e o f steam

in bar 44

// temp . o f steam i n K // s p e c i f i c h e a t o f

6 Tsup =250+273; 7 Cps =2.1;

steam i n kJ / kg 8 9 // At 12 bar , from steam t a b l e s 10 Ts =188+273; // i n K 11 hfg =1984.3; // kJ / kg 12 ssup =4.18* log ( Ts /273) + hfg / Ts + Cps * log ( Tsup / Ts ) ;

// e n t r o p y o f wet steam i n kJ /kgK 13 14

printf ( ’ The Entro py o f 1 kg o f s u p e r h e a t e d steam a t 12 b a r and 250 deg c e l s i u s i s : %1 . 3 f kJ / kg . \n ’ , ssup ) ;



in bar // mass o f w a t e r i n

5 Mwt =50; 6 7 8 9

t h e t a n k i n kg t1 =20; deg . c e l s i u s Ms =3; c o n d e n s e d i n kg t2 =40; celsius We =1.5; t a n k i n kg

// i n i t i a l temp . i n // amount o f steam // f i n a l temp . i n deg . // w a t e r e q u i v a l e n t o f

10 11 // At 5 bar , from steam t a b l e s 12 hf =640.1; 13 hfg =2107.4;

45

// i n kJ / kg // i n kJ / kg

14 15 Mw = Mwt + We ;

// t o t a l mass o f w a t e r i n kg 16 // h e a t l o s t by steam = h e a t g a i n e d by w a t e r 17 //Ms [ ( h f+x h f g ) − 1 ∗ 4 . 1 8 ∗ ( t2 −0) ]=Mw[ 1 ∗ 4 . 1 8 ∗ ( t2 −t 1 ) ] 18 x =[ Mw *[1*4.18*( t2 - t1 ) ]/ Ms +1*4.18*( t2 -0) - hf ]/ hfg ; // d r y n e s s f r a c t i o n 19 20

printf ( ’ The D r y n e s s f r a c t i o n o f steam , x i s : %1 . 4 f . ’ ,x ) ;

Scilab code Exa 4.24 Example 24 1 clc 2 clear 3 //DATA GIVEN 4 p =1.1; 5 6 7 8 9 10


in bar x =0.95; Mwt =90; t h e t a n k i n kg t1 =25; deg . c e l s i u s Mt =12.5; c =0.42; m e t a l i n kJ /kgK t2 =40; celsius

// d r y n e s s f r a c t i o n // mass o f w a t e r i n // i n i t i a l temp . i n // mass o f t a n k i n kg // s p e c i f i c h e a t o f // f i n a l temp . i n deg .

11 12 m1 = Mwt ; 13 m2 = Mt * c ;

// w a t e r e q u i v a l e n t o f vessel 14 M = m1 + m2 ; // t o t a l mass o f w a t e r i n kg 15 // At 1 . 1 bar , from steam t a b l e s 46

16 hf =428.8; // i n kJ / kg 17 hfg =2250.8; // i n kJ / kg 18 // h e a t l o s t by steam = h e a t g a i n e d by w a t e r 19 //Ms [ ( h f+x h f g ) − 1 ∗ 4 . 1 8 ∗ ( t2 −0) ]=M[ 1 ∗ 4 . 1 8 ∗ ( t2 −t 1 ) ] 20 Ms = M *[1*4.18*( t2 - t1 ) ]/[( hf + x * hfg ) -1*4.18*( t2 -0) ];

// mass o f steam c o n d e n s e d i n kg 21 22

printf ( ’ The Mass o f steam c o n d e n s e d , Ms i s : %1 . 3 f kg . ’ , Ms ) ;

Scilab code Exa 4.25 Example 25 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

clc clear //DATA GIVEN // c o n d i t i o n o f steam b e f o r e p1 =8; // c o n d i t i o n o f steam a f t e r p2 =1; T2 =115+273; Tsup2 = T2 ; // At 1 bar , Ts2 =99.6+273; Cps =2.1;

throttling // p r e s s u r e i n b a r throttling // p r e s s u r e i n b a r // temp . i n deg . c e l s i u s

// kJ /kgK

// As t h r o t t l i n g i s a c o n s t a n t e n t h a l p y p r o c e s s , // h1=h2 . . . . . h f 1+x1 ∗ h g f 1=h f 2+h f g 2+Cps ( Tsup2−Ts2 ) // At 8 bar , from steam t a b l e s , hf1 =720.9; hfg1 =2046.5; // At 1 bar , from steam t a b l e s , hf2 =417.5; hfg2 =2257.9;

47

24 x1 =[ hf2 + hfg2 + Cps *( Tsup2 - Ts2 ) - hf1 ]/ hfg1 ;

//

dryness fra ction 25 26

printf ( ’ The D r y n e s s f r a c t i o n o f steam i n t h e main , x1 i s : %1 . 2 f . ’ , x1 ) ;

Scilab code Exa 4.26 Example 26 1 clc 2 clear 3 //DATA GIVEN 4 Mw =2;

// mass o f w a t e r

s e p a r a t e d o u t i n kg // amount o f steam ( c o n d e n s a t e ) d i s c h a r g e d from t h r o t t l i n g c a l o r i m e t e r i n kg Tsup3 =110+273; // temp . o f steam a f e t r t h r o t t l i n g in K p1 =12; // i n i t i a l p r e s s u r e o f steam i n b a r p3 =(760+5) /1000*1.3366; // f i n a l p r e s s u r e o f steam i n b a r ( 1 mm o f Hg = 1 . 3 3 6 6 b a r ) Cps =2.1; // kJ /kgK

5 Ms =20.5;

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

p2 = p1 ; // At p1=p2=12 bar , from steam t a b l e s hf2 =798.4; // i n kJ / kg hfg2 =1984.3; // i n kJ / kg // At p3=1 bar , from steam t a b l e s Ts3 =99.6+273; // i n Tsup3 =110+273; // i n hf3 =417.5; // i n hfg3 =2257.9; // i n

48

K K kJ / kg kJ / kg

22 // h2=h3 . . . . . h f 2+x2 ∗ h g f 2=h f 3+h f g 3+Cps ( Tsup3−Ts3 ) 23 x2 =[ hf3 + hfg3 + Cps *( Tsup3 - Ts3 ) - hf2 ]/ hfg2 ; //

d r y n e s s f r a c t i o n x2 24 25 x1 =( x2 * Ms ) /( Mw + Ms ) ;

//

d r y n e s s f r a c t i o n o f steam s u p p l i e d , x1 26 27

printf ( ’ The Q u a l i t y o f steam s u p p l i e d , x1 i s : %1 . 2 f . ’ , x1 ) ;

Scilab code Exa 4.27 Example 27 1 clc 2 clear 3 //DATA GIVEN 4 p1 =15; 5 6 7 8

// p r e s s u r e o f steam sample in bar p3 =1; // p r e s s u r e o f steam at e x i t in bar Tsup3 =150+273; // t e m p e r a t u r e o s steam a t t h e e x i t i n K Mw =0.5; // d i s c h a r g e from s e p a r a t i n g c a l o r i m e t e r i n kg / min Ms =10; // d i s c h a r g e from t h r o t t l i n g c a l o r i m e t e r i n kg / min

9 10 p2 = p1 ; 11 // At p1=p2=15 bar , from steam t a b l e s 12 hf2 =844.7; // i n kJ / kg 13 hfg2 =1945.2; // i n kJ / kg 14 15 // At p3=1 b a r and 150 deg . c e l s i u s , from steam

tables 16 hsup3 =2776.4;

// i n kJ / kg

17

49

18 // h2=h3 . . . . . h f 2+x2 ∗ h g f 2=hsup3 19 x2 =[ hsup3 - hf2 ]/ hfg2 ;


x2 20 21 x1 =( x2 * Ms ) /( Mw + Ms ) ;

// q u a l i t y o f steam

s u p p l i e d , x1 22 23

printf ( ’ The Q u a l i t y o f steam s u p p l i e d , x1 i s : %1 . 3 f . ’ , x1 ) ;

50

Chapter 5 Heat Engines

Scilab code Exa 5.1 Example 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

clc clear //DATA GIVEN Ms =10000/3600; // r a t e o f steam f l o w i n kg / s // i n l e t t o t u r b i n e p1 =60; // p r e s s u e i n b a r T1 =380; // temp . i n deg . c e l s i u s // e x i t from t u r b i n e , i n l e t t o c o n d e n s e r p2 =0.1; // p r e s s u e i n b a r x2 =0.9; // q u a l i t y v2 =200; // v e l o c i t y i n m/ s // e x i t from c o n d e n s e r , i n l e t t o pump p3 =0.09; // p r e s s u e i n b a r // i t i s s a t u r a t e d // e x i t from pump , i n l e t t o b o i l e r p4 =70; // p r e s s u e i n b a r // e x i t from b o i l e r , 51

22 23 24 25 26 27 28 29 30

p5 =65; T5 =400;

// p r e s s u e i n b a r // temp . i n deg . c e l s i u s

// f o r c o n d e n s e r , t1 =20; t2 =30;

// i n l e t temp . i n deg . c e l s i u s // e x i t temp . i n deg . c e l s i u s

// At 60 b a r and 380 deg . c e l s i u s , from steam t a b l e s h1 =3043.0+(3177.2 -3043.0) /(400 -350) *30; //By interpolation

31 32 // At 0 . 1 bar , from steam t a b l e s 33 hf2 =191.8; 34 hfg2 =2392.8; 35 h2 = hf2 + x2 * hfg2 ; 36 Pt = Ms *( h1 - h2 )

// i n kJ / kg // i n kJ / kg // power o u t p u t o f

t h e t u r b i n e i n kW 37 38 // At 70 bar , from steam t a b l e s 39 hf4 =1267.4; // i n kJ / kg 40 // At 60 b a r and 380 deg . c e l s i u s , from steam t a b l e s 41 ha =(3177.2+3158.1) /2; //By i n t e r p o l a t i o n 42 43 44 45

b e t w e e n 60 and 70 deg c e l s i u s Q1 = Ms *3600*( ha - hf4 ) ; per hour i n the b o i l e r // At 0 . 0 9 bar , from steam t a b l e s hf3 =183.3; Q2 = Ms *3600*( h2 - hf3 ) ; per hour i n the c o n d e n s e r

// h e a t t r a n s f e r

// i n kJ / kg // h e a t t r a n s f e r

46 47

// h e a t l o s t by steam=h e a t g a i n e d by t h e c o o l i n g water 48 //Q2=Mw∗ 4 . 1 8 ∗ ( t2 −t 1 ) 49 Mw = Q2 /4.18/10; // mass o f c o o l i n g water c i r c u l e t e d per hour i n c o n d e n s e r 50 51 52

// ( p i ) /4∗ dˆ2=Ms∗ x2 ∗ vg2 // d=d i a m e t e r o f t h e p i p e c o n n e c t i n g t u r b i n e w i t h 52

condenser 53 C =200; steam i n m/ s 54 vg2 =14.67; at 0 . 1 bar 55 d =( Ms * x2 * vg2 /( %pi /4) / C ) ^0.5; 56 57 58 59 60

61

// v e l o c i t y o f // s p e c i f i c volume

printf ( ’ ( i ) The Power o u t p u t o f t u r b i n e i s : %4 . 0 f kW. \n ’ , Pt ) ; printf ( ’ ( i i ) The Heat t r a n s f e r p e r h o u r i n t h e B o i l e r i s : %3 . 2 e kJ / h . \n ’ , Q1 ) ; printf ( ’ The Heat t r a n s f e r p e r h o u r i n t h e C o n d e n s e r i s : %3 . 2 e kJ / h . \n ’ , Q2 ) ; printf ( ’ ( i i i ) The Mass o f c o o l i n g w a t e r c i r c u l a t e d p e r h o u r i n t h e c o n d e n s e r i s : %3 . 2 e kg / h r . \n ’ , Mw ); printf ( ’ ( i v ) The D i a m e t e r o f t h e p i p e c o n n e c t i n g t u r b i n e w i t h c o n d e n s e r i s : %1 . 3 f m o r %3 . 0 f mm. \ n ’ ,d ,( d *1000) ) ;

62 63 64

//NOTE: // a n s o f Mw( 1 . 1 1 6 ∗ 1 0 ˆ 7 ) i s g i v e n i n c o r r e c t i n t h e book . 65 // t h e c o r r e c t a n s o f Mw i s = 5 . 1 7 ∗ 1 0 ˆ 5 kg / h .

Scilab code Exa 5.2 Example 2 1 clc 2 clear 3 //DATA GIVEN 4 p1 =15;

// steam s u p p l y p r e s s u r e i n

bar 5 x1 =1; 6 p2 =0.4;

// q u a l i t y o f steam // c o n d e n s e r p r e s s u r e

7

53

8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

// At 0 . 1 5 bar , from steam t a b l e s T1 =198.3+273; // i n K hg1 =2789.9; // i n kJ / kg sg1 =6.4406; // i n kJ /kgK // At 0 . 4 bar , from steam t a b l e s T2 =75.9+273; // i n K hf2 =317.7; // i n kJ / kg hfg2 =2319.2; // i n kJ / kg sf2 =1.0261; // i n kJ /kgK sfg2 =6.6448; // i n kJ /kgK ETAcarnot =( T1 - T2 ) / T1 ; // C a r n o t e f f i c i e n c y // ETArankine=A d i a b a t i c o r i s e n t r o p i c h e a t d r o p / h e a t supplied // ETArankine =(hg1−h2 ) / ( hg1−h f 2 ) // a s t h e steam e x p a n d s i s e n t r o p i c a l l y , s 1=s 2 // s g 1=s f 2+x2 ∗ s f g 2 x2 =( sg1 - sf2 ) / sfg2 ; h2 = hf2 + x2 * hfg2 ; ETArankine =( hg1 - h2 ) /( hg1 - hf2 ) ; // Rankine efficiency

27 28

printf ( ’ ( i ) The C a r n o t e f f i c i e n c y i s : %1 . 4 f o r %2 . 2 f p e r c e n t . \n ’ , ETAcarnot ,( ETAcarnot *100) ) ; 29 printf ( ’ ( i i ) The Rankine e f f i c i e n c y i s : %1 . 4 f o r %2 . 2 f p e r c e n t . \n ’ , ETArankine ,( ETArankine *100) ) ;

Scilab code Exa 5.3 Example 3 1 2 3 4 5 6

clc clear //DATA GIVEN p1 =20; T1 =360+273; p2 =0.08;

// b o i l e r p r e s s u r e i n b a r // temp . i n K // b o i l e r p r e s s u r e i n b a r 54

7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

// At 20 b a r and 360 deg . c e l s i u s , from steam t a b l e s h1 =3159.3; // i n kJ / kg sg1 =6.9917; // i n kJ /kgK // At 0 . 0 8 bar , from steam t a b l e s hf2 =173.88; // i n kJ / kg hf3 = hf2 ; sf2 =0.5926; // i n kJ /kgK s3 = sf2 ; hfg2 =2403.1; // i n kJ / kg sg2 =8.2287; // i n kJ /kgK vf2 =0.001008; //mˆ3/ kg sfg2 =7.6361; // i n kJ /kgK // a s t h e steam e x p a n d s i s e n t r o p i c a l l y , s 1=s 2 // s g 1=s f 2+x2 ∗ s f g 2 x2 =( sg1 - sf2 ) / sfg2 ; h2 = hf2 + x2 * hfg2 ; // Wnet=Wturbine−Wpump //Wpump=h f 4 −h f 3=v f 3 ( p1−p2 ) Wp = vf2 *( p1 - p2 ) *100; hf4 = Wp + hf3 ; Wt = h1 - h2 ; Wnet = Wt - Wp ; Q1 = h1 - hf4 ; // i n kJ / kg ETAcycle = Wnet / Q1 ; // c y c l e e f f i c i e n c y

printf ( ’ ( i ) The Net work p e r kg o f steam i s : %3 . 2 f kJ / kg . \n ’ , Wnet ) ; 37 printf ( ’ ( i i ) The C y c l e e f f i c i e n c y i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAcycle ,( ETAcycle *100) ) ;


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

clc clear // g i v e n steam t a b l e e x t r a c t p1 =80; // i n b a r t1 =295.1; // i n deg . c e l s i u s vf1 =0.001385; //mˆ3/ kg vg1 =0.0235; //mˆ3/ kg hf1 =1317; // i n kJ / kg hfg1 =1440.5; // i n kJ / kg hg1 =2757.5; // i n kJ / kg sf1 =3.2073; // i n kJ /kgK sfg1 =2.5351; // i n kJ /kgK sg1 =5.7424; // i n kJ /kgK p2 =0.1; t2 =45.84; vf2 =0.0010103; vg2 =14.68 hf2 =191.9; hf3 = hf2 ; hfg2 =2392.3; hg2 =2584.2; sf2 =0.6488; sfg2 =7.5006; sg2 =8.1494;

// i n b a r // i n deg . c e l s i u s //mˆ3/ kg //mˆ3/ kg // i n kJ / kg

ETAt =0.9; ETAp =0.8; efficiency

// steam t u r b i n e e f f i c i e n c y // c o n d e n s a t e pump

// i n // i n // i n // i n // i n

kJ / kg kJ / kg kJ /kgK kJ /kgK kJ /kgK

P1 =80; // i n b a r T1 =600; // i n deg c e l s i u s // At 80 b a r and 600 deg c e l s i u s v1 =0.486; //mˆ3/ kg h1 =3642; // kJ / kg s1 =7.0206; // kJ / kg /K // a s t h e steam e x p a n d s i s e n t r o p i c a l l y , s 1=s 2 56

38 // s g 1=s f 2+x2 ∗ s f g 2 39 x2 =( s1 - sf2 ) / sfg2 ; 40 h2 = hf2 + x2 * hfg2 ; 41 Wta = ETAt *( h1 - h2 ) ; 42 43 44 45 46 47 48 49 50 51

// a c t u a l t u r b i n e work i n kJ / kg Wp = vf2 *( p1 - p2 ) *10^5/10^3; //pump work i n kJ / kg Wpa = Wp / ETAp ; // a c t u a l pump work i n kJ / kg Wnet = Wta - Wpa ; // s p e c i f i c work i n kJ / kg // ETAthermal=Wnet/Q1 //Q1=h1−h f 4 hf4 = hf3 + Wpa ; Q1 = h1 - hf4 ; ETAth = Wnet / Q1 ;

printf ( ’ ( i ) The S p e c i f i c work ( Wnet ) i s : %4 . 2 f kJ / kg . \n ’ , Wnet ) ; 52 printf ( ’ ( i i ) The Thermal e f f i c i e n c y i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAth ,( ETAth *100) ) ;

Scilab code Exa 5.5 Example 5 1 2 3 4 5 6 7 8 9 10 11 12 13 14

clc clear //DATA GIVEN p1 =28; p2 =0.06;

// p r e s s u r e a t 1 i n b a r // p r e s s u r e a t 2 i n b a r

// At 28 bar , from steam t a b l e s h1 =2802; // i n kJ / kg s1 =6.2104; // i n kJ /kgK // At 0 . 0 6 bar , from steam t a b l e s hf2 =151.5; // i n kJ / kg hf3 = hf2 ; hfg2 =2415.9; // i n kJ / kg 57

15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

sf2 =0.521; sf3 = sf2 ; sfg2 =7.809; vf2 =0.001;

// i n kJ /kgK // i n kJ /kgK //mˆ3/ kg

// a s t h e steam e x p a n d s i s e n t r o p i c a l l y , s 1=s 2 // s g 1=s f 2+x2 ∗ s f g 2 x2 =( s1 - sf2 ) / sfg2 ; h2 = hf2 + x2 * hfg2 ; // Wnet=Wturbine−Wpump //Wpump=h f 4 −h f 3=v f 3 ( p1−p2 ) Wp = vf2 *( p1 - p2 ) *10^5/10^3; hf4 = Wp + hf2 ; Wt = h1 - h2 ; Wnet = Wt - Wp ; Q1 = h1 - hf4 ; // i n kJ / kg ETAcycle = Wnet / Q1 ; // c y c l e e f f i c i e n c y wr = Wnet / Wt ; // work r a t i o ssc =3600/ Wnet ; // s p e c i f i c steam c o n s u m p t i o n i n kg /kWh

36 37

printf ( ’ ( i ) The C y c l e e f f i c i e n c y i s : %1 . 4 f o r %2 . 2 f p e r c e n t . \n ’ , ETAcycle ,( ETAcycle *100) ) ; 38 printf ( ’ ( i i ) The Work r a t i o i s : %1 . 3 f kJ / kg . \n ’ , wr ); 39 printf ( ’ ( i i i ) The S p e c i f i c steam c o n s u m p t i o n i n kg / kWh i s : %1 . 3 f kg /kWh . \n ’ , ssc ) ;

Scilab code Exa 5.6 Example 6 1 clc 2 clear 3 //DATA GIVEN

58

4 p1 =35; 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

t u r b i n e in bar x1 =1; p2 =0.2; m =9.5;

// p r e s s u r e a t i n l e t t o

// p r e s s u r e a t e x h a u s t i n b a r // f l o w r a t e i n kg / s

// At 35 bar , from steam t a b l e s hg1 =2802; // i n kJ / kg h1 = hg1 ; sg1 =6.1228; // i n kJ /kgK // At 0 . 2 bar , from steam t a b l e s hf2 =251.5; // i n kJ / kg hf3 = hf2 ; hfg2 =2358.4; // i n kJ / kg vf2 =0.001017; //mˆ3/ kg sf2 =0.8321; // i n kJ /kgK sfg2 =7.0773; // i n kJ /kgK // Wnet=Wturbine−Wpump //Wpump=h f 4 −h f 3=v f 3 ( p1−p2 ) Wp = vf2 *( p1 - p2 ) *10^5/10^3; Wpnet = m * Wp ; hf4 = Wp + hf3 ; // a s t h e steam e x p a n d s i s e n t r o p i c a l l y , s 1=s 2 // s g 1=s f 2+x2 ∗ s f g 2 x2 =( sg1 - sf2 ) / sfg2 ; // d r y n e s s fraction h2 = hf2 + x2 * hfg2 ; Wt = h1 - h2 ; Wtnet = m * Wt ; ETArankine =( h1 - h2 ) /( h1 - hf2 ) ; // Rankine efficiency chf = m *( h2 - hf3 ) ; // c o n d e n s e r heat flow printf ( ’ ( i ) The Pump Work i s : %2 . 2 f kW. \n ’ , Wpnet ) ; 59

printf ( ’ ( i i ) Wtnet ) ; 39 printf ( ’ ( i i i ) .2 f percent 40 printf ( ’ ( i v ) \n ’ , chf ) ; 41 printf ( ’ (v) x2 i s : %1 . 3

38

42 43 44 45 46

The T u r b i n e Work i s : %2 . 2 f kW. \n ’ , The Rankine e f f i c i e n c y i s : %1 . 4 f o r %2 . \n ’ , ETArankine ,( ETArankine *100) ) ; The C o n d e n s e r h e a t f l o w i s : %1 . 3 f kW. The d r y n e s s a t t h e end o f e x p a n s i o n , f o r %2 . 1 f p e r c e n t . \n ’ ,x2 ,( x2 *100) ) ;

//NOTE: // The v a l u e o f x2 i n t h e book i s g i v e n a s 0 . 7 4 7 0 // w h i l e t h e e x a c t a n s i s 0 . 7 4 7 5 5 // and s o t h e v a l u e s o f o t h e r a n s w e r s a r e v a r y i n g by some u n i t s

Scilab code Exa 5.7 Example 7 1 clc 2 clear 3 //DATA GIVEN 4 h12 =840;

h1−h2 ) i n kJ / kg 5 h1 =2940; i n kJ / kg 6 p2 =0.1;

// A d i a b a t i c e n t h a l p y drop , ( // e n t h a l p y o f steam s u p p l i e d // back p r e s s u r e i n b a r

7 8 // At 0 . 1 bar , from steam t a b l e s 9 hf =191.8; // i n kJ / kg 10 // ETArankine =(hg1−h2 ) / ( hg1−h f 2 ) 11 ETArankine =( h12 ) /( h1 - hf ) ; 12 Wuse = h12 ; // u s e f u l work done p e r kg o f

steam i n kJ / kg 13 ssc =1/ Wuse *3600; 14 15

// s p e c i f i c steam c o n s u m p t i o n

printf ( ’ ( i ) The Rankine e f f i c i e n c y 60

i s : %1 . 4 f o r %2 . 2

f p e r c e n t . \n ’ , ETArankine ,( ETArankine *100) ) ; 16 printf ( ’ ( i i ) The S p e c i f i c steam c o n s u m p t i o n i s : %1 . 3 f kg /kWh . \n ’ , ssc ) ;

Scilab code Exa 5.8 Example 8 1 clc 2 clear 3 //DATA GIVEN 4 IP =35;

// power d e v e l o p e d by t h e

e n g i n e i n kW 5 m =284; 6 p1 =15;

// f l o w r a t e i n kg / h // steam i n l e t p r e s s u r e i n

bar 7 p2 =0.14;

// c o n d e n s e r p r e s s u r e i n b a r

8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

// At 35 b a r and 25 deg c e l s i u s from steam t a b l e s h1 =2923.3; // i n kJ / kg s1 =6.709; // i n kJ /kgK // At 0 . 1 4 bar , from steam t a b l e s hf2 =220; // i n kJ / kg hf3 = hf2 ; hfg2 =2376.6; // i n kJ / kg sf2 =0.737; // i n kJ /kgK sfg2 =7.296; // i n kJ /kgK

// a s t h e steam e x p a n d s i s e n t r o p i c a l l y , s 1=s 2 // s g 1=s f 2+x2 ∗ s f g 2 x2 =( s1 - sf2 ) / sfg2 ; // d r y n e s s fraction 23 h2 = hf2 + x2 * hfg2 ; 24 25

ETArankine =( h1 - h2 ) /( h1 - hf2 ) ; efficiency 61

// Rankine

26 27

ETAthermal = IP /( m /3600*( h1 - hf2 ) ) ; efficiency 28 ETArel = ETAthermal / ETArankine ; efficiency

// Thermal // R e l a t i v e

29 30

printf ( ’ ( i ) The F i n a l c o n d i t i o n o f steam i s : %1 . 2 f . \n ’ , x2 ) ; 31 printf ( ’ ( i i ) The Rankine e f f i c i e n c y i s : %1 . 4 f o r %2 . 2 f p e r c e n t . \n ’ , ETArankine ,( ETArankine *100) ) ; 32 printf ( ’ ( i i i ) The R e l a t i v e e f f i c i e n c y i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETArel ,( ETArel *100) ) ;

Scilab code Exa 5.9 Example 9 1 2 3 4 5 6 7 8 9 10

clc clear //DATA GIVEN T1 =400+273; T2 = T1 ; T3 =40+273; T4 = T3 ; W =130; ETAth =( T1 - T3 ) / T1 ; efficiency

// temp . i n K // temp . i n K // work p r o d u c e d i n kJ // E n g i n e t h e r m a l

11 12 //ETAth=Work done / Heat added 13 Ha = W / ETAth ; // Heat added i n kJ 14 Hr = Ha - W ; // Heat r e j e c t e d i n kJ 15 // Heat r e j e c t e d =T3 ( S3−S4 ) 16 S34 = Hr / T3 ; // Entropy c h a n g e d u r i n g

the heat r e j e c t i o n process 17 18

printf ( ’ ( i ) The E n g i n e t h e r m a l e f f i c i e n c y 62

i s : %1 . 3 f

o r %2 . 1 f 19 printf ( ’ ( i i 20 printf ( ’ ( i i i rejection

p e r c e n t . \n ’ , ETAth ,( ETAth *100) ) ; ) The Heat added i s : %3 . 0 f kJ . \n ’ , Ha ) ; ) The Entropy c h a n g e d u r i n g t h e h e a t p r o c e s s i s : %1 . 3 f kJ /K . \n ’ , S34 ) ;

Scilab code Exa 5.10 Example 10 1 clc 2 clear 3 //DATA GIVEN 4 p1 =18; 5 6 7 8 9 10

//maximum p r e s s u r e in bar T1 =410+273; //maximum temperature in K T2 = T1 ; Rac =6; // r a t i o o f i s e n t r o p i c o r a d i a b a t i c c o m p r e s s i o n , V4/V1=6 Rie =1.5; // r a t i o o f i s o t h e r m a l e x p a n s i o n , V2/V1=1.5 V1 =0.18; // volume o f a i r a t b e g i n n i n g o f i s o t h e r m a l e x p a n s i o n i n mˆ3 wc =210; // no . o f c y c l e s per s

11 12 //gamma f o r a i r =1.4 13 g =1.4; 14 15 // f o r i s e n t r o p i c p r o c e s s 4−1 16 // A l s o ( T1/T4 ) =(V4/V1 ) ˆ ( g −1) 17 // ( V4/V1 )=Rac 18 T4 = T1 / Rac ^( g -1) ; 19 T3 = T4 ; 20 // p1 ( V1ˆgamma )=p4 ( V4ˆgamma ) 21 // p4=p1 ∗ ( V1/V4 ) ˆ g 22 // where , ( V4/V1 )=Rac

63

23 p4 = p1 /( Rac ^ g ) ; 24 25 // f o r i s o t h e r m a l p r o c e s s 1−2 26 // p1V1=p2V2 27 //V1/V2=1/ R i e 28 p2 = p1 *(1/ Rie ) ; 29 30 // f o r i s e n t r o p i c p r o c e s s 2−3 31 // p2 ( V2ˆgamma )=p3 ( V3ˆgamma ) 32 //V2/V3=V1/V4=1/Rac 33 p3 = p2 *(1/ Rac ) ^ g ; 34 35 // c h a n g e i n e n t r o p y , DS=S2−S1=mRlog ( V2/V1 )=p1V1 /T1∗

l o g ( V2/V1 ) 36 DS = p1 *10^5* V1 /10^3/ T1 * log ( Rie ) ; 37 38 // Heat s u p p l i e d , Qs=p1 ∗V1∗ l o g ( V2/V1 ) 39 // Qs=T1 ( S2−S1 ) 40 Qs = T1 * DS ; 41 // Qr=p4 ∗V4∗ l o g ( V3/V4 )

// h e a t r e j e c t e d i n kJ 42 // Qr=T4 ( S3−S4 ) , b c s i n c r e a s e i n e n t r o p y d u r i n g h e a t addition i s equal to decrease in entropy during heat r e j e c t i o n 43 Qr = T4 * DS ; 44 45 ETA =( Qs - Qr ) / Qs ;

// mean

thermal e f f i c i e n c y of the c y c l e 46 47 48 49 50 51 52 53

// mean e f f e c t i v e p r e s s u r e o f t h e c y c l e , Pm = work done p e r c y c l e / s t r o k e volume Rv31 = Rac * Rie ; // r a t i o o f v o l u m e s a t 3 and 1 , V3/V1=V3/V2∗V2/V1 // s t r o k e volume , Vs=V3−V1 Vs = V1 *( Rv31 -1) ; J =1; Pm =( Qs - Qr ) *10^3/10^5* J / Vs ;

64

// power o f

54 P =( Qs - Qr ) * wc /60;

t h e e n g i n e i n kW 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69

70 71 72 73 74

printf ( ’ ( i ) The P r e s s u r e and T e m p e r a t u r e a t p o i n t 1 a r e : \ n ’ ); printf ( ’ p1 : %2 . 0 f b a r . \ n ’ , p1 ) ; printf ( ’ T1 : %3 . 0 f K. \ n ’ , T1 ) ; printf ( ’ The P r e s s u r e and T e m p e r a t u r e a t p o i n t 2 a r e : \ n ’ ); printf ( ’ p2 : %2 . 0 f b a r . \ n ’ , p2 ) ; printf ( ’ T2 : %3 . 0 f K. \ n ’ , T2 ) ; printf ( ’ The P r e s s u r e and T e m p e r a t u r e a t p o i n t 3 a r e : \ n ’ ); printf ( ’ p3 : %1 . 2 f b a r . \ n ’ , p3 ) ; printf ( ’ T3 : %3 . 1 f K. \ n ’ , T3 ) ; printf ( ’ The P r e s s u r e and T e m p e r a t u r e a t p o i n t 4 a r e : \ n ’ ); printf ( ’ p4 : %1 . 2 f b a r . \ n ’ , p4 ) ; printf ( ’ T4 : %3 . 1 f K. \ n ’ , T4 ) ; printf ( ’ ( i i ) The Change i n e n t r o p y d u r i n g i s o t h e r m a l e x p a n s i o n i s : %1 . 3 f kJ /K . \n ’ , DS ) ; printf ( ’ ( i i i ) The Mean t h e r m a l e f f i c i e n c y o f t h e c y c l e i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ ,ETA ,( ETA *100) ) ; printf ( ’ ( i v ) The Mean e f f e c t i v e p r e s s u r e i s : %1 . 3 f b a r . \n ’ , Pm ) ; printf ( ’ ( v ) The Power o f t h e e n g i n e w o r k i n g on t h i s c y c l e i s g i v e n by : %3 . 1 f kW. ’ ,P ) ; //NOTE: // t h e r e i s s l i g h t v a r i a t i o n i n a n s w e r s o f book due to rounding o f f of the values


65

1 clc 2 clear 3 //DATA GIVEN 4 //CASE−1 5 // ( T1−T2 ) /T1=1/6 6 //SO , T1 = 1 . 2 ( T2 ) . . . . . . . . . Eqn ( 1 ) 7 8 //CASE−2 9 //T2 REDUCED BY 70 DEG. CELSIUS 10 // {T1−[T2−(70+273) ] } / T1 = 1 / 3 . . . . . . . . . . . . . . Eqn ( 2 ) 11 // 2T1=3T2−1029 12 13 //By Eqn ( 1 ) and ( 2 ) 14 T2 =(70+273) *3/(3 -2*1.2) ; 15 T1 =1.2* T2 ; 16 17 printf ( ’ ( i ) The T e m p e r a t u r e o f t h e S o u r c e , T1 i s : %4

. 0 f K o r %4 . 0 f deg . c e l s i u s . \n ’ ,T1 ,( T1 -273) ) ; 18 printf ( ’ ( i i ) The T e m p e r a t u r e o f t h e Sink , T2 i s : %4 . 0 f K o r %4 . 0 f deg . c e l s i u s . \n ’ ,T2 ,( T2 -273) ) ;

Scilab code Exa 5.12 Example 12 1 clc 2 clear 3 //DATA GIVEN 4 T1 =1990;

// T e m p e r a t u r e o f t h e

heat Source in K 5 T2 =850; heat Sink in K 6 Qs =32.5; min 7 P =0.4; e n g i n e i n kW

// T e m p e r a t u r e o f t h e // h e a t s u p p l i e d i n kJ / // power d e v e l o p e d by t h e

8

66

9 10 11 12 13

ETAcarnot =( T1 - T2 ) / T1 ; // A l s o ETAth=work done / Heat s u p p l i e d ETAth = P / Qs ;

printf ( ’ The E f f i c i e n c y o f c a r n o t c y c l e i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAcarnot ,( ETAcarnot *100) ) ; 14 printf ( ’ The Thermal e f f i c i e n c y o f e n g i n e c l a i m e d by i n v e n t o r i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n\n ’ , ETAth ,( ETAth *100) ) ; 15 16 if ( ETAth > ETAcarnot ) 17 printf ( ’ Thus , The c l a i m o f t h e i n v e n t o r

is

p o s s i b l e . ’ ); 18 19

else printf ( ’ Thus , The c l a i m o f t h e i n v e n t o r i s NOT f e a s i b l e , \n a s no e n g i n e can be more e f f i c i e n t t h a n t h a t w o r k i n g on c a r n o t c y c l e . ’ );

Scilab code Exa 5.13 Example 13 1 clc 2 clear 3 //DATA GIVEN 4 ETAotto =60;

// E f f i c i e n c y o f o t t o

cycle in % // r a t i o o f s p e c i f i c

5 shr =1.5;

heats 6 7 // ETAotto=1−1/( r ) ˆ ( s h r −1) 8 r =(1/(1 - ETAotto /100) ) ^(1/( shr -1) ) ; // c o m p r e s s i o n

ratio 9 10

printf ( ’ The c o m p r e s s i o n r a t i o i s : %1 . 2 f . ’ ,r ) ;

67

Scilab code Exa 5.14 Example 14 1 clc 2 clear 3 //DATA GIVEN 4 D =0.25;

// b o r e o f t h e

engine in m // s t r o k e o f t h e

5 L =0.375; 6 7 8 9

engine in m Vc =0.00263; volume i n mˆ3 p1 =1; p r e s s u r e in bar T1 =50+273; temperature in K p3 =25; p r e s s u r e in bar

// c l e a r a n c e // i n i t i a l // i n i t i a l //maximum

10 11 Vs =( %pi /4) * D ^2* L ; 12 r =( Vs + Vc ) / Vc ;

// s w e p t volume // c o m p r e s s i o n

ratio 13 14 // f o r a i r , gamma=1.4 15 g =1.4; 16 // A i r s t a n d a r d e f f i c i e n c y

o f o t t o c y c l e ETAotto

=1−1/( r ) ˆ ( g −1) 17 ETAotto =1 -1/( r ) ^( g -1) ; 18 19 // f o r a d i a b a t i c p r o c e s s 1−2 20 // p1 ( V1ˆgamma )=p2 ( V2ˆgamma ) 21 // p2=p1 ∗ ( V1/V2 ) ˆ g 22 // where , ( V1/V2 )=r 23 p2 = p1 *( r ^ g ) ;

p r e s s u r e at 2 in bar 68

//

//

24 rp = p3 / p2 ;

pressure ratio 25 Pm = p1 * r *[( r ^( g -1) -1) *( rp -1) ]/[( g -1) *( r -1) ]; // mean

e f f e c t i v e p r e s s u r e in bar 26 27

printf ( ’ ( i ) The A i r s t a n d a r d e f f i c i e n c y o f o t t o c y c l e i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAotto ,( ETAotto *100) ) ; 28 printf ( ’ ( i i ) The Mean e f f e c t i v e p r e s s u r e i s : %1 . 3 f b a r . \n ’ , Pm ) ;

Scilab code Exa 5.15 Example 15 1 clc 2 clear 3 //DATA GIVEN 4 T1 =38+273; 5

// i n i t i a l

temperature in K T3 =1950+273; temperature K rp =15; // f o r a i r , gamma=1.4 g =1.4;

//maximum

6 // p r e s s u r e 7 8 9 10 // f o r a d i a b a t i c c o m p r e s s i o n 1−2 11 // p1 ( V1ˆgamma )=p2 ( V2ˆgamma ) 12 // ( V1/V2 )=r 13 r =( rp ) ^(1/ g ) ; 14 15 // Thermal e f f i c i e n c y ETAth=1−1/( r ) ˆ ( g −1) 16 ETAth =1 -1/( r ) ^( g -1) ; 17 18 // f o r a d i a b a t i c c o m p r e s s i o n 1−2 19 // ( T2/T1 ) =(V1/V2 ) ˆ ( g −1) 20 // ( V1/V2 )=r

69

ratio

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

T2 = T1 * r ^( g -1) ; // f o r a d i a b a t i c e x p a n s i o n 3−4 // ( T3/T4 ) =(V4/V3 ) ˆ ( g −1) // ( V4/V3 )=r T4 = T3 / r ^( g -1) ; // h e a t s u p p l i e d p e r kg o f a i r , Qs=m∗Cv ∗ ( T3−T2 ) R =0.287; Cv = R /( g -1) ; Qs = Cv *( T3 - T2 ) ; // h e a t r e j e c t e d p e r kg o f a i r , Qr=m∗Cv ∗ ( T4−T1 ) Qr = Cv *( T4 - T1 ) ; // work done p e r kg

W = Qs - Qr ; of air

37 38 39

printf ( ’ ( i ) The c o m p r e s s i o n r a t i o i s : %1 . 1 f . \ n ’ ,r ) ; printf ( ’ ( i i ) The Thermal e f f i c i e n c y i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAth ,( ETAth *100) ) ; 40 printf ( ’ ( i i i ) The Work done i s : %3 . 1 f kJ o r %6 . 0 f Nm . ’ ,W ,( W *1000) ) ; 41 42 43

//NOTE: // t h e r e i s s l i g h t v a r i a t i o n i n a n s w e r s i n t h e book because of rounding o f f of the values

Scilab code Exa 5.16 Example 16 1 clc 2 clear 3 //DATA GIVEN 4 V1 =0.45; 5 p1 =1;

// volume i n mˆ3 // i n i t i a l p r e s s u r e 70

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

in bar T1 =30+273; // i n i t i a l temperature in K p2 =11; // p r e s s u r e a t t h e end o f c o m p r e s s i o n s t r o k e i n b a r Qs =210; // h e a t addaed a t c o n s t a n t volume i n kJ wc =210; // no . o f w o r k i n g c y c l e s / min // f o r a i r , gamma=1.4 g =1.4; // f o r a d i a b a t i c c o m p r e s s i o n 1−2 // p1 ( V1ˆgamma )=p2 ( V2ˆgamma ) // ( V1/V2 )=r r =( p2 / p1 ) ^(1/ g ) ; // A l s o ( T2/T1 ) =(V1/V2 ) ˆ ( g −1) // ( V1/V2 )=r T2 = T1 * r ^( g -1) ; // A p p l y i n g g a s l a w s t o p o i n t s 1 and 2 // p1V1 /T1=p2V2 /T2 V2 = T2 / T1 * p1 / p2 * V1 ; // h e a t s u p p l i e d d u r i n g p r o c e s s 2 −3 , Qs=mCv( T3−T2 ) R =287; m = p1 *10^5* V1 / R / T1 ; Cv = R /1000/( g -1) ; T3 = Qs / m / Cv + T2 ; // f o r c o n s t a n t volume p r o c e s s 2−3 // p3 /T3=p2 /T2 p3 = p2 / T2 * T3 ; V3 = V2 ; // f o r a d i a b a t i c e x p a n s i o n 3−4 // p3 ( V3ˆgamma )=p4 ( V4ˆgamma ) 71

39 40 41 42 43 44 45 46 47 48 49 50 51 52

// ( V4/V3 )=r p4 = p3 *(1/ r ) ^( g ) ; // A l s o T3/T4 ) =(V4/V3 ) ˆ ( g −1) // ( V4/V3 )=r T4 = T3 / r ^( g -1) ; V4 = V1 ; // p e r c e n t a g e c l e a r a n c e , pc=Vc/ Vs=V2 / ( V1−V2 ) pc = V2 /( V1 - V2 ) *100; // h e a t r e j e c t e d p e r c y c l e , Qr=Cv ∗ ( T4−T1 ) Qr = m * Cv *( T4 - T1 ) ; // A i r s t a n d a r d e f f i c i e n c y o f o t t o c y c l e ETAotto=(Qs− Qr ) /Qs ETAotto =( Qs - Qr ) / Qs ; // A l t e r n a t i v e l y // ETAotto=1−1/( r ) ˆ ( g −1) ETAotto =1 -1/( r ) ^( g -1) ;

53 54 55 56 57 58 // mean e f f e c t i v e 59 W = Qs - Qr ;

p r e s s u r e , Pm=W/ Vs // work done p e r kg

of air 60 Vs = V1 - V2 ; 61 Pm = W *10^3/10^5/ Vs ; 62 63

// power d e v e l o p e d , P=work done p e r c y c l e ∗ no . o f c y c l e s per s 64 P = W *( wc /60) ; 65 66 67 68 69 70 71 72

printf ( ’ ( i ) The P r e s s u r e , T e m p e r a t u r e and Volumes at s a l i e n t p o i n t s i n the c y c l e a r e : \ n ’ ); printf ( ’ At p o i n t 1 a r e : \ n ’ ) ; printf ( ’ p1 : %1 . 1 f b a r . \ n ’ , p1 ) ; printf ( ’ V1 : %1 . 2 f mˆ 3 . \ n ’ , V1 ) ; printf ( ’ T1 : %3 . 0 f K. \ n ’ , T1 ) ; printf ( ’ At p o i n t 2 a r e : \ n ’ ) ; printf ( ’ p2 : %2 . 2 f b a r . \ n ’ , p2 ) ; 72

printf ( ’ V2 : %1 . 3 f mˆ 3 . \ n ’ , V2 ) ; printf ( ’ T2 : %3 . 0 f K. \ n ’ , T2 ) ; printf ( ’ At p o i n t 3 a r e : \ n ’ ) ; printf ( ’ p3 : %2 . 2 f b a r . \ n ’ , p3 ) ; printf ( ’ V3 : %1 . 3 f mˆ 3 . \ n ’ , V3 ) ; printf ( ’ T3 : %4 . 0 f K. \ n ’ , T3 ) ; printf ( ’ At p o i n t 4 a r e : \ n ’ ) ; printf ( ’ p4 : %1 . 2 f b a r . \ n ’ , p4 ) ; printf ( ’ V4 : %1 . 2 f mˆ 3 . \ n ’ , V4 ) ; printf ( ’ T4 : %3 . 1 f K. \ n ’ , T4 ) ; printf ( ’ ( i i ) The P e r c e n t a g e c l e a r a n c e i s : %2 . 2 f p e r c e n t . \n ’ , pc ) ; 84 printf ( ’ ( i i i ) The A i r s t a n d a r d e f f i c i e n c y o f t h e c y c l e i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAotto ,( ETAotto *100) ) ; 85 printf ( ’ ( i v ) The Mean e f f e c t i v e p r e s s u r e i s : %1 . 3 f b a r . \n ’ , Pm ) ; 86 printf ( ’ ( v ) The Power d e v e l o p e d i s : %3 . 1 f kW. ’ ,P ) ;

73 74 75 76 77 78 79 80 81 82 83

87 88 89

//NOTE: // t h e r e i s s l i g h t v a r i a t i o n i n a n s w e r s i n t h e book because of rounding o f f of the values

Scilab code Exa 5.17 Example 17 1 clc 2 clear 3 //DATA GIVEN 4 r =15; // c o m p r e s s i o n r a t i o 5 //V3−V2=a / 1 0 0 ∗ Vs . . . . . . . . . . . . Vs=s t r o k e volume=V1−V2 6 //V3 =1.84 V2 7 c =6; // h e a t a d d i t i o n t a k e s

place at ’a ’ percent of s t r o k e 8 // f o r a i r , gamma=1.4 9 g =1.4; 73

10 11

// A i r s t a n d a r d e f f i c i e n c y o f d i e s e l c y c l e E T A d i e s e l =1 −[1/( r ) ˆ ( g −1) ] [ ( r h o ˆ g −1) / ( rho −1) ] 12 // r h o=c u t o f f r a t i o =V3/V2 13 rho = c /100*( r -1) +1; 14 ETAdiesel =1 -[1/ g /( r ) ^( g -1) ]*[( rho ^g -1) /( rho -1) ]; 15 16

printf ( ’ The A i r s t a n d a r d e f f i c i e n c y o f d i e s e l c y c l e i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAdiesel ,( ETAdiesel *100) ) ;

Scilab code Exa 5.18 Example 18 1 clc 2 clear 3 //DATA GIVEN 4 L =0.25; 5 6 7 8 9 10 11 12 13 14

// s t r o k e o f t h e

engine in m D =0.15; // d i a m e t e r o f cylinder in m V2 =0.0004; // c l e a r a n c e volume i n mˆ3 Vs =( %pi /4) * D ^2* L ; // s w e p t volume i n mˆ3 Vt = Vs + V2 ; // t o t a l c y l i n d e r volume i n mˆ3 c =5; // f u e l i n j e c t i o n takes place at ’ c ’ percent of s t r o k e V3 = V2 + c /100* Vs ; // volume a t p o i n t o f cut − o f f i n mˆ3 rho = V3 / V2 ; // cut − o f f r a t i o r =( Vs + V2 ) / V2 ; // c o m p r e s s i o n ratio // f o r a i r , gamma=1.4 74

15 g =1.4; 16 17 // A i r s t a n d a r d

e f f i c i e n c y o f d i e s e l c y c l e ETAdiesel =1 −[1/( r ) ˆ ( g −1) ] [ ( r h o ˆ g −1) / ( rho −1) ] 18 ETAdiesel =1 -[1/ g /( r ) ^( g -1) ]*[( rho ^g -1) /( rho -1) ]; 19 20

printf ( ’ The E f f i c i e n c y o f d i e s e l e n g i n e i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAdiesel ,( ETAdiesel *100) ) ;


clc clear //DATA GIVEN r =14; // c o m p r e s s i o n r a t i o // f u e l cut − o f f i s d e l a y e d from 5−8% // f o r a i r , gamma=1.4 g =1.4;

// when f u e l i s cut − o f f a t 5% c1 =5; rho1 = c1 /100*( r -1) +1; // E f f i c i e n c y o f d i e s e l e n g i n e E T A d i e s e l =1 −[1/( r ) ˆ ( g −1) ] [ ( r h o ˆ g −1) / ( rho −1) ] 13 ETAdiesel1 =1 -[1/ g /( r ) ^( g -1) ]*[( rho1 ^g -1) /( rho1 -1) ]; 14 15 // when f u e l i s cut − o f f 16 c2 =8; 17 rho2 = c2 /100*( r -1) +1; 18 // E f f i c i e n c y o f d i e s e l

a t 8%

e n g i n e E T A d i e s e l =1 −[1/( r ) ˆ ( g −1) ] [ ( r h o ˆ g −1) / ( rho −1) ] 19 ETAdiesel2 =1 -[1/ g /( r ) ^( g -1) ]*[( rho2 ^g -1) /( rho2 -1) ]; 20 21 22

ETAloss =( ETAdiesel1 - ETAdiesel2 ) *100;

75

23

printf ( ’ The P e r c e n t a g e l o s s i n e f f i c i e n c y due t o d e l a y i n f u e l cut − o f f i s : %1 . 1 f p e r c e n t . \n ’ , ETAloss ) ;

Scilab code Exa 5.20 Example 20 1 clc 2 clear 3 //DATA GIVEN 4 Pm =7.5;

// mean e f f e c t i v e

p r e s s u r e in bar 5 r =12.5; 6 p1 =1; bar

// c o m p r e s s i o n r a t i o // i n i t i a l p r e s s u r e i n

7 8 // f o r a i r , gamma=1.4 9 g =1.4; 10 11 // mean e f f e c t i v e p r e s s u r e , Pm=p1 ∗ r ˆ g ∗ [ g ∗ ( rho −1)−r

ˆ(1 − g ) ∗ ( r h o ˆ g −1) ] / [ ( g −1) ∗ ( r −1) ] 12 // we g e t , 0 . 3 4 6 ( r h o ) ˆ 1 . 4 − 1 . 4 ( r h o ) +2.04 13 //By t r i a l and e r r o r method , we g e t 14 rho =2.24; 15 16 co =( rho -1) /( r -1) *100; //% cut − o f f 17 18 printf ( ’ The P e r c e n t a g e cut − o f f o f t h e c y c l e

. 2 f p e r c e n t . \n ’ , co ) ;


76

i s : %2

3 //DATA GIVEN 4 D =0.2; 5 6 7 8 9 10

// b o r e o f t h e

engine in m L =0.3; engine in m p1 =1; p r e s s u r e in bar T1 =27+273; temperature in K c =8; s t r o k e volume r =15; ratio wc =380; per s

// s t r o k e o f t h e // i n i t i a l // i n i t i a l // cut − o f f % o f // c o m p r e s s i o n // no . o f c y c l e s

11 12 Vs =( %pi /4) * D ^2* L ;

// s w e p t volume i n

mˆ3 13 V1 = Vs *(1+1/( r -1) ) ; 14 // f o r a i r , gamma=1.4 15 g =1.4;

// i n mˆ3

16 17 R =287; 18 m = p1 *10^5* V1 / R / T1 ;

// mass o f a i r

in

t h e c y l i n d e r i n kg / c y c l e 19 20 // f o r a d i a b a t i c p r o c e s s 1−2 21 // p1 ( V1ˆgamma )=p2 ( V2ˆgamma ) 22 // p2=p1 ∗ ( V1/V2 ) ˆ g 23 // where , ( V1/V2 )=r 24 p2 = p1 *( r ^ g ) ; 25 26 27 28 29 30

p r e s s u r e at 2 in bar // A l s o ( T2/T1 ) =(V1/V2 ) ˆ ( g −1) // ( V1/V2 )=r T2 = T1 * r ^( g -1) ; V2 = Vs /( r -1) ; Vc = V2 ; p3 = p2 ; 77

//

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51

// cut − o f f r a t i o , c =( rho −1) / ( r −1) rho = c /100*( r -1) +1; V3 = rho * V2 ; // a l t e r n a t i v e l y V3 = c /100* Vs + Vc ; // f o r c o n s t a n t p r e s s u r e p r o c e s s 2−3 //V3/T3=V2/T2 T3 = T2 / V2 * V3 ; // f o r i s e n t r o p i c p r o c e s s 3−4 // p3 ( V3ˆgamma )=p4 ( V4ˆgamma ) // ( V4/V)=V4/V2∗V2/V3=V1/V2∗V2/V3=r / r h o p4 = p3 *(( rho / r ) ^ g ) ; // A l s o ( T4/T3 ) =(V3/V4 ) ˆ ( g −1) // ( V4/V)=V4/V2∗V2/V3=V1/V2∗V2/V3=r / r h o T4 = T3 *(( rho / r ) ^( g -1) ) ; V4 = V1 ;

// A i r s t a n d a r d e f f i c i e n c y o f d i e s e l c y c l e E T A d i e s e l =1 −[1/( r ) ˆ ( g −1) ] [ ( r h o ˆ g −1) / ( rho −1) ] 52 ETAdiesel =1 -[1/ g /( r ) ^( g -1) ]*[( rho ^g -1) /( rho -1) ]; 53 54

// mean e f f e c t i v e p r e s s u r e , Pm=p1 ∗ r ˆ g ∗ [ g ∗ ( rho −1)−r ˆ(1 − g ) ∗ ( r h o ˆ g −1) ] / [ ( g −1) ∗ ( r −1) ] ; 55 Pm = p1 * r ^ g *[ g *( rho -1) -r ^(1 - g ) *( rho ^g -1) ]/[( g -1) *( r -1) ]; 56 57 P = Pm *10^5* Vs /10^3*( wc /60) ;

// Power o f

t h e e n g i n e i n kW 58 59 60 61 62 63

printf ( ’ ( i ) The P r e s s u r e , T e m p e r a t u r e and Volumes at s a l i e n t p o i n t s i n the c y c l e a r e : \ n ’ ); printf ( ’ At p o i n t 1 a r e : \ n ’ ) ; printf ( ’ p1 : %1 . 1 f b a r . \ n ’ , p1 ) ; printf ( ’ V1 : %1 . 4 f mˆ 3 . \ n ’ , V1 ) ; printf ( ’ T1 : %3 . 0 f K. \ n ’ , T1 ) ; 78

printf ( ’ At p o i n t 2 a r e : \ n ’ ) ; printf ( ’ p2 : %2 . 2 f b a r . \ n ’ , p2 ) ; printf ( ’ V2 : %1 . 7 f mˆ 3 . \ n ’ , V2 ) ; printf ( ’ T2 : %3 . 1 f K. \ n ’ , T2 ) ; printf ( ’ At p o i n t 3 a r e : \ n ’ ) ; printf ( ’ p3 : %2 . 2 f b a r . \ n ’ , p3 ) ; printf ( ’ V3 : %1 . 6 f mˆ 3 . \ n ’ , V3 ) ; printf ( ’ T3 : %4 . 1 f K. \ n ’ , T3 ) ; printf ( ’ At p o i n t 4 a r e : \ n ’ ) ; printf ( ’ p4 : %1 . 3 f b a r . \ n ’ , p4 ) ; printf ( ’ V4 : %1 . 4 f mˆ 3 . \ n ’ , V4 ) ; printf ( ’ T4 : %3 . 2 f K. \ n ’ , T4 ) ; printf ( ’ ( i i ) The T h e o r i t i c a l a i r s t a n d a r d e f f i c i e n c y o f d i e s e l c y c l e i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAdiesel ,( ETAdiesel *100) ) ; 77 printf ( ’ ( i i i ) The Mean e f f e c t i v e p r e s s u r e i s : %1 . 3 f b a r . \n ’ , Pm ) ; 78 printf ( ’ ( i v ) The Power d e v e l o p e d i s : %2 . 2 f kW. ’ ,P ) ;

64 65 66 67 68 69 70 71 72 73 74 75 76

79

Chapter 6 Steam Boilers

Scilab code Exa 6.1 Example 1 1 clc 2 clear 3 //DATA GIVEN 4 LCV =44700; 5 afrn =20;

mixture 6 afrd =1; mixture 7 Cpg =1.08; kgK 8 T1 =38+273;

//LCV o f f u e l i n kJ // a i r p a r t s =20 i n a i r f u e l // f u e l p a r t s =1 i n a i r f u e l // avg s p e c i f i c h e a t i n kJ / // b o i l e r room temp . i n K

9 10 // h e a t o f c o m b u s t i o n=h e a t o f g a s e s 11 // 1∗44700=Mg∗Cpg ∗ ( T2−T1 ) 12 T2 = afrd * LCV /( afrn + afrd ) / Cpg + T1 ; 13 14 printf ( ’ The Maximum temp . T2 a t t a i n e d

in the f u r n a c e o f t h e b o i l e r i s : \ n %5 . 0 f K e l v i n ’ , T2 ) ; 15 printf ( ’ o r %5 . 0 f d e g r e e c e l s i u s . \ n ’ ,( T2 -273) ) ;

80

Scilab code Exa 6.2 Example 2 1 clc 2 clear 3 //DATA GIVEN 4 Ms =5.4;

// mass o f steam u s e d i n kg /

kWh 5 p =50; 6 Tsup =350;

// p r e s s u r e o f steam i n b a r // temp . o f steam i n deg

celsius 7 eta =82; 8 Tfw =150;

; sius 9 C =28100; kJ 10 rate =500;

// b o i l e r e f f i c i e n c y i n % // f e e d w a t e r temp . i n deg c e l // c a l o r i f i c v a l u e o f c o a l i n // c o s t o f c o a l / t o n n e i n Rs

11 12

// b o i l e r e f f i c i e n c y i s g i v e n by , e t a=Ms ∗ ( hsup−h f 1 ) / ( Mf∗C) 13 // from steam t a b l e , a t 45 b a r and 350 deg c e l s i u s , hsup = 3 0 6 8 . 4 kJ / kg 14 h =3068.4; // e n t h a l p y a t 45 b a r and 350 deg c e l s i u s 15 hf1 =4.18*( Tfw -0) ; // h f 1 a t 150 deg c e l s i u s i n kJ / kg 16 17 // s u b s . t h e s e i n eq . o f b o i l e r 18 Mf = Ms *( h - hf1 ) /(( eta /100) * C ) ;

r e q u i r e d i n kg /kWh 19 cost =( Mf /1000) * rate *100; i n p a i s a /kWh 20 21

efficiency // mass o f c o a l // c o s t o f c o a l

printf ( ’ ( i ) The mass o f c o a l r e q u i r e d i s : %5 . 3 f kg /kWh . \n ’ , Mf ) ; 81

22 23 24 25 26

printf ( ’ ( i i ) The T o t a l c o s t o f f u e l ( c o a l ) i s : %2 . 1 f p a i s a /kWh . \n ’ , cost ) ; //NOTE: i n t e x t book // i n q u e s t i o n p r e s s u r e i s g i v e n a s =50 b a r // but from steam t a b l e e n t h a l p y i s f o u n d a t 45 b a r

Scilab code Exa 6.3 Example 3 1 clc 2 clear 3 //DATA GIVEN 4 Mc =1250; 5 6 7 8 9

// q u a n t i t y o f c o a l i n kg

consumed i n 24 h o u r s Mw =13000; e v a p o r a t e d i n kg MEPs =7; o f steam i n b a r Tfw =40; celsius h =2570.7; b a r i n kJ / kg C =30000; i n kJ / kg

// mass o f w a t e r // mean e f f e c t i v e p r e s s u r e // f e e d w a t e r temp . i n deg // e n t h a l p y o f steam a t 7 // c a l o r i f i c v a l u e o f c o a l

10 11 Ma = Mw / Mc ; 12 13 14 15 16 17

// mass o f w a t e r a c t u a l l y e v a p o r a t e d p e r kg o f f u e l hf1 =4.18*( Tfw -0) ; hfg =2257; // i n kJ / kg Me = Ma *( h - hf1 ) / hfg ; // i n kg eta = Ma *( h - hf1 ) / C ; // b o i l e r e f f i c i e n c y

printf ( ’ ( i ) The e q u i v a l e n t e v a p o r a t i o n p e r kg o f c o a l , Me i s : %5 . 3 f kg . \n ’ , Me ) ; 18 printf ( ’ ( i i ) The e f f i c i e n c y o f b o i l e r , e t a i s : %1 82

. 3 f o r %2 . 1 f p e r c e n t .

’ ,eta , eta *100) ;

Scilab code Exa 6.4 Example 4 1 clc 2 clear 3 //DATA GIVEN 4 p =12; 5 6 7 8 9 10 11 12 13 14 15

bar Ms =40000; i n kg x =0.85; Tfw =30; deg c e l s i u s Mc =4000; C =33400; i n kJ / kg

// mean steam p r e s s u r e i n // mass o f steam g e n e r a t e d // mean d r y n e s s f r a c t i o n // mean f e e d w a t e r temp . i n // mass o f c o a l u s e d i n kg // c a l o r i f i c v a l u e o f c o a l

// from steam t a b l e , c o r r e s p o n d i n g t o 12 bar , hf =798.4; // i n kJ / kg hfg =1984.3; // i n kJ / kh h = hf + x * hfg ; // i n k j / kg hf1 =4.18*( Tfw -0) ; // h e a t o f f e e d w a t e r i n kJ / kg

16 17 Fe =( h - hf1 ) /2257;

e v a p o r a t i o n , Fe 18 Ma = Ms / Mc ; 19 Me = Ma *( h - hf1 ) /2257; ) 20 eta = Ma *( h - hf1 ) / C ;

// f a c t o r o f e q u i v a l e n t // p e r kg o f f u e l // ( kg o f steam ) / ( kg o f f u e l // e f f i c i e n c y o f b o i l e r

21 22

printf ( ’ ( i ) The F a c t o r o f e q u i v a l e n t t e m e r a t u r e , Fe i s : %5 . 3 f \n ’ , Fe ) ; 23 printf ( ’ ( i i ) The E q u i v a l e n t e v a p o r a t i o n from and 83

a t 100 deg c e l s i u s , Me i s : %5 . 2 f ( kg o f steam ) / ( kg o f c o a l ) . \ n ’ , Me ) ; 24 printf ( ’ ( i i i ) The E f f i c i e n c y o f b o i l e r i s : %5 . 4 f ’ , eta ) ; 25 printf ( ’ o r %5 . 2 f p e r c e n t . \n ’ , eta *100) ;

Scilab code Exa 6.5 Example 5 1 clc 2 clear 3 //DATA GIVEN 4 M =18000;

// mass o f steam g e n e r a t e d i n

kg / h r 5 p =12.5; 6 x =0.97; 7 Tfw =105;

// steam p r e s s u r e i n b a r // q u a l i t y o f steam // f e e d w a t e r temp . i n deg

celsius 8 Mf =2040; // r a t e o f c o a l f i r i n g i n kg / hr 9 C =27400; // h i g h r e r c a l o r i f i c v a l u e ( HCV) o f c o a l i n kJ / kg 10 11 12 13 14 15

// from steam t a b l e , c o r r e s p o n d i n g t o 1 2 . 5 bar , hf =806.7; // i n kJ / kg hfg =1977.4; // i n kJ / kg h = hf + x * hfg ; // i n kJ / kg hf1 =4.18*( Tfw -0) ; // h e a t o f f e e d w a t e r i n kJ / kg

16 17 // h e a t r a t e o f t h e b o i l e r = h e a t s u p p l i e d p e r h o u r 18 heatrate = M *( h - hf1 ) // h e a t r a t e o f b o i l e r 19 Ma = M / Mf ; // i n kg p e r kg o f f u e l 20 Me = Ma *( h - hf1 ) /2257; // ( kg o f steam ) / ( kg o f f u e l ) 21 eta = Ma *( h - hf1 ) / C ; // t h e r m a l e f f i c i e n c y 22

84

printf ( ’ ( i ) The Heat r a t e o f b o i l e r i s : %1 . 4 e kJ / h . \n ’ , heatrate ) ; 24 printf ( ’ ( i i ) The E q u i v a l e n t e v a p o r a t i o n , Me i s : %5 . 3 f ( kg o f steam ) / ( kg o f f u e l ) . \n ’ , Me ) ; 25 printf ( ’ ( i i i ) The Thermal e f f i c i e n c y i s : %5 . 4 f ’ , eta ); 26 printf ( ’ o r %5 . 2 f p e r c e n t . \n ’ , eta *100) ;

23


clc clear //DATA GIVEN Mw =5940; // mass o f w a t e r e v a p o r a t e d kg / h r Mc =675; // mass o f c o a l b u r n t i n kg / h r C =31600; // l o w e r c a l o r i f i c v a l u e (LCV) o f c o a l i n kJ / kg p1 =14; // p r e s s u r e o f steam a t b o i l e r stop v a l v e in bar Te1 =32; // temp . o f f e e d w a t e r e n t e r i n g e c o n o m i s e r i n deg c e l s i u s Te2 =115; // temp . o f f e e d w a t e r l e a v i n g e c o n o m i s e r i n deg c e l s i u s x =0.96; // d r y n e s s f r a c t i o n o f steam entering superheater Tsup =260; // temp . o f steam l e a v i n g s u p e r h e a t e r i n deg c e l s i u s Cp =2.3 // s p e c i f i c h e a t o f s u p e r h e a t e d steam

13 14 hf1 =4.18*( Te2 - Te1 ) ;

// h e a t u t i l i s e d by 1 kg o f f e e d w a t e r i n e c o n o m i s e r 15 // from steam t a b l e , c o r r e s p o n d i n g t o 14 bar , 16 Ts =195; 17 hf =830.1; 85

18 hfg =1957.7; 19 hboiler =( hf + x * hfg ) - hf1 ; 20 21 22 23 24 25 26 27 28 29 30 31 32

// h e a t u t i l i s e d by 1 kg o f f e e d w a t e r i n b o i l e r hsuperheater =(1 - x ) * hfg + Cp *( Tsup - Ts ) ; // h e a t u t i l i s e d by 1 kg o f f e e d w a t e r i n s u p e r h e a t e r Ma = Mw / Mc ; // i n kg p e r kg o f f u e l Pe = hf1 / C * Ma *100; //% o f h e a t u t i l i s e d in economiser Pb = hboiler / C * Ma *100; //% o f h e a t u t i l i s e d in b o i l e r Ps = hsuperheater / C * Ma *100; //% o f h e a t u t i l i s e d in superheater htotal = hf1 + hboiler + hsuperheater ; // t o t a l h e a t a b s o r b e d i n kg o f w a t e r eta = Ma * htotal / C ; // o v e r a l l e f f i c i e n c y of b o i l e r plant printf ( ’ ( i ) The P e r c e n t a g e o f h e a t u t i l i s e d i n E c o n o m i s e r i s : %5 . 2 f p e r c e n t . \ n ’ , Pe ) ; printf ( ’ The P e r c e n t a g e o f h e a t u t i l i s e d i n B o i l e r i s : %5 . 2 f p e r c e n t . \ n ’ , Pb ) ; printf ( ’ The P e r c e n t a g e o f h e a t u t i l i s e d i n S u p e r h e a t e r i s : %5 . 2 f p e r c e n t . \ n ’ , Ps ) ; printf ( ’ ( i i ) The O v e r a l l E f f i c i e n c y o f b o i l e r p l a n t i s : %5 . 4 f ’ , eta ) ; printf ( ’ o r %5 . 2 f p e r c e n t . \n ’ , eta *100) ;

Scilab code Exa 6.7 Example 7 1 clc 2 clear 3 //DATA GIVEN 4 C =29915;

// c a l o r i f i c v a l u e o f c o a l i n kJ /

kg 86

5 Mw =9.1;

// mass o f f e e d w a t e r p e r kg o f

d r y c o a l i n kg // e q u i v a l e n t e v a p o r a t i o n f r a o m and a t 100 deg c e l s i u s p e r kg o f d r y c o a l i n kg Te =12; // temp . o f f e e d w a t e r t o e c o n o m i s e r i n deg c e l s i u s Tb =105; // temp . o f f e e d w a t e r t o b o i l e r i n deg c e l s i u s Ta =13; // temp . o f a i r Tfg =370; // temp . o f f l u e g a s e s e n t e r i n g economiser Mfg =18.2; // mass o f f l u e g a s e s e n t e r i n g e c o n o m i s e r p e r kg o f c o a l Cp =1.046; // mean s p e c i f i c h e a t o f f l u e gases

6 Me =9.6; 7 8 9 10 11 12

13 14 hb = Me *2257; 15 16 17 18 19 20 21 22 23 24 25 26 27

// h e a t s u p p l i e d f o r steam g e n e r a t i o n i n kJ ETAb = hb / C ; // b o i l e r e f f i c i e n c y hflue = Mfg * Cp *( Tfg - Ta ) ; // h e a t i n t h e f l u e g a s e p e r kg o f d r y c o a l e n t e r i n g e c o n o m i s e r he = Mw *4.184*( Tb - Te ) ; // h e a t u t i l i s e d i n economiser ETAe = he / hflue ; // e c o n o m i s e r e f f i c i e n c y htotal = hb + he ; // t o t a l h e a t a b s o r b e d i n kg o f w a t e r ETA = htotal / C ; // b o i l e r p l a n t e f f i c i e n c y printf ( ’ ( i ) The B o i l e r e f f i c i e n c y i s : %5 . 3 f ’ , ETAb ) ; printf ( ’ o r %2 . 1 f p e r c e n t . \n ’ , ETAb *100) ; printf ( ’ ( i i ) The E c o n o m i s e r e f f i c i e n c y i s : %5 . 3 f ’ , ETAe ) ; printf ( ’ o r %2 . 2 f p e r c e n t . \n ’ , ETAe *100) ; printf ( ’ ( i i i ) The O v e r a l l E f f i c i e n c y o f b o i l e r p l a n t i s : %5 . 3 f ’ , ETA ) ; printf ( ’ o r %2 . 1 f p e r c e n t . \n ’ , ETA *100) ;

87


// r a t e o f steam p r o d u c t i o n i n kg

/ hr 5 x =1; 6 p =10; 7 Tfw =110;

celsius 8 Mf =225; 9 C =30100; kg 10 Puc =10;

// q u a l i t y o f steam // steam p r e s s u r e i n b a r // f e e d w a t e r temp . i n deg // r a t e o f c o a l f i r i n g i n kg / h r // c a l o r i f i c v a l u e o f c o a l i n kJ / //% o f u n b u r n t c o a l

11 12 // from steam t a b l e , 13 h =2776.2; 14 hf1 =4.18*( Tfw -0) ; 15 16 17 18 19

c o r r e s p o n d i n g t o 10 bar , // i n kJ / kg // h e a t c o n t a i n e d i n 1 kg o f f e e d w a t e r b e f o r e e n t e r i n g b o i l e r i n kJ / kg htotal =h - hf1 // t o t a l h e a t g i v e n t o p r o d u c e 1 kg o f steam i n b o i l e r i n kJ / kg Mc = Mf *(100 - Puc ) /100; // mass o f c o a l a c t u a l l y b u r n t i n kg Ma = Ms / Mc ; // ( kg o f steam ) / ( kg o f fuel ) ETAb = Ma *( h - hf1 ) / C ; // t h e r m a l e f f i c i e n c y o f boiler ETAc =( Ms / Mf ) *( h - hf1 ) / C ; // t h e r m a l e f f i c i e n c y o f b o i l e r and g r a t e combined

20 21

printf ( ’ ( i ) The Thermal e f f i c i e n c y o f t h e b o i l e r i s : %5 . 3 f ’ , ETAb ) ; 22 printf ( ’ o r %5 . 2 f p e r c e n t . \n ’ , ETAb *100) ; 88

printf ( ’ ( i i ) The Thermal e f f i c i e n c y o f t h e b o i l e r and g r a t e combined i s : %5 . 3 f ’ , ETAc ) ; 24 printf ( ’ o r %5 . 2 f p e r c e n t . \n ’ , ETAc *100) ;

23

Scilab code Exa 6.9 Example 9 1 clc 2 clear 3 //DATA GIVEN 4 Ma =7.5;

// mass o f steam g e n e r a t e d p e r kg

of coal 5 p =11; 6 Tfw =70;

// steam p r e s s u r e i n b a r // temp . o f f e e d w a t e r temp . i n

deg c e l s i u s 7 eta =75; 8 Fe =1.15; 9 Cps =2.3; kgK

// e f f i c i e n c y o f b o i l e r i n % // f a c t o r o f e v a p o r a t i o n // s p e c i f i c h e a t o f steam i n kJ /

10 11 12 13 14 15 16 17

// from steam t a b l e , c o r r e s p o n d i n g t o 11 bar , hf =781.4; // i n kJ / kg hfg =1998.5; // i n kJ / kg Ts =184.1+273; // i n K hf1 =4.18*( Tfw -0) ;

// F a c t o r o f e v a p o r a t i o n , Fe =[{ h f+h f g+Cps ∗ ( Tsup−Ts ) }− hf1 ]/2257 18 Tsup =[ Fe *2257+ hf1 - hf - hfg ]/ Cps + Ts ; // Tsup i n K 19 x =( Tsup - Ts ) ; // d e g r e e o f s u p e r h e a t i n deg . c e l s i u s 20 21 // B o i l e r e f f i c i e n c y e t a=Ma∗ ( h−h f 1 ) /C ; 22 h =[ hf + hfg + Cps *( Tsup - Ts ) ]; 23 C = Ma *( h - hf1 ) /( eta /100) ;

v a l u e o f c o a l i n kJ / kg 89

// c a l o r i f i c

//

24 Me = Ma *( h - hf1 ) /2257;

E q u i v a l e n t e v a p o r a t i o n im kg 25 26

printf ( ’ ( i ) The T e m p e r a t u r e o f steam g e n e r a t i o n , Tsup i s : %5 . 1 f K\n ’ , Tsup ) ; 27 printf ( ’ The D e g r e e o f s u p e r h e a t i s : %5 . 1 f deg c e l s i u s . \ n ’ ,x ) ; 28 printf ( ’ ( i i ) The c a l o r i f i c v a l u e o f c o a l , C i s : %5 . 0 f kJ / kg . \n ’ ,C ) ; 29 printf ( ’ ( i i i ) The E q u i v a l e n t e v a p o r a t i o n , Me i s : %5 . 3 f kg . \n ’ , Me ) ;

Scilab code Exa 6.10 Example 10 1 clc 2 clear 3 //DATA GIVEN 4 p =13; 5 ds =77; 6 7 8 9 10 11 12 13

// steam p r e s s u r e i n b a r // d e g r e e o f s u p e r h e a t i n

deg . c e l s i u s Tfw =85; // temp . o f f e e d w a t e r i n deg . c e l s i u s Mw =3000; // mass o f w a t e r e v a p o r a t e d i n kg / h r Mc =410; // c o a l f i r e d Mash =40; // mass o f a s h i n kg / h r Pca =9.6; //% o f c o m b u s t i b l e i n ash Pm =4.5; //% o f m o i s t u r e i n c o a l C =30500; // c a l o r i f i c v a a l u e o f d r y c o a l p e r kg Cps =2.1; // s p e c i f i c h e a t o f s u p e r h e a t e d steam i n kJ /kgK

14 15

90

16 17 18 19 20 21 22

// from steam t a b l e , c o r r e s p o n d i n g t o 13 bar , hf =814.7; // i n kJ / kg hfg =1970.7; // i n kJ / kg Ts =191.6; // i n deg . s e l s i u s h = hf + hfg + Cps *( ds ) ; hf1 =4.18*( Tfw -0) ; htotal =h - hf1 ; // t o t a l h e a t s u p p l i e d t o p r o d u c e 1 kg o f steam

23 24 Mc1 = Mc *(1 - Pm /100) ; 25 Ma = Mw / Mc1 ; 26 ETAb = Ma *( h - hf1 ) / C ;

// mass o f d r y c o a l i n kg

// e f f i c i e n c y o f b o i l e r plant including superheater

27 28 29 30 31 32 33 34

Mcom = Mash * Pca /100; // Mass o f c o m b u s t i b l e i n ash per hr // t h e c o m b u s t i b l e p r e s e n t i n a s h i s p r a c t i c a l l y c a r b o n and i t s v a l u e may be t a k e n a s 3 3 8 / 6 0 kJ / kg // h e a t a c t u a l l y s u p p l i e d p r h r=h e a t o f d r y c o a l −h e a t o f combustible in ash Hsupp = Mc1 *C - Mcom *33860; // h e a t a c t u a l l y s u p p l i e d pr hr Huse = Mw *( h - hf1 ) ; // h e a t u s e f u l l y u t i l i s e d i n b o i l e r pr hr // e f f i c i e n c y o f b o i l e r

ETAc = Huse / Hsupp ; and f u r n a c e combined

35 36

printf ( ’ ( i ) The E f f i c i e n c y o f b o i l e r p l a n t i n c l u d i n g s u p e r h e a t e r i s : %5 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAb ,( ETAb *100) ) ; 37 printf ( ’ ( i i ) The E f f i c i e n c y o f t h e b o i l e r and f u r n a c e combined i s : %5 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAc ,( ETAc *100) ) ;

91


// mass o f steam g e n e r a t e d i n

kg / h r 5 Mf =700;

// r a t e o f c o a l f i r i n g i n kg /

hr 6 C =31402;

// h i g h e r c a l o r i f i c v a l u e (HCV

) o f c o a l i n kJ / kg 7 x =0.92; 8 p =12; 9 Tfw =45; celsius

// q u a l i t y o f steam // steam p r e s s u r e i n b a r // f e e d w a t e r temp . i n deg

10 11 12 13 14 15

// from steam t a b l e , c o r r e s p o n d i n g t o 12 bar , hf =798.4; // i n kJ / kg hfg =1984.3; // i n kJ / kg h = hf + x * hfg ; // i n kJ / kg hf1 =4.18*( Tfw -0) ; // h e a t o f f e e d w a t e r i n kJ / kg 16 Ma = Ms / Mf ; // i n kg p e r kg o f f u e l 17 Me = Ma *( h - hf1 ) /2257; // ( kg o f steam ) / ( kg o f f u e l ) 18 eta = Ma *( h - hf1 ) / C ; // t h e r m a l e f f i c i e n c y 19 20

printf ( ’ ( i ) The E q u i v a l e n t e v a p o r a t i o n , Me i s : %5 . 3 f ( kg o f steam ) / ( kg o f c o a l ) . \n ’ , Me ) ; 21 printf ( ’ ( i i ) The B o i l e r e f f i c i e n c y i s : %5 . 3 f o r %2 . 1 f p e r c e n t . \n ’ ,eta , eta *100) ;


92

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

//DATA GIVEN hsup =3373.7; // e n t h a l p y o f steam ( a t 100 bar , 5 0 0 deg . c e l s i u s ) i n kJ / kg hf1 =677; // e n t h a l p y o f f e e d w a t e r ( a t i n l e t temp . 160 deg . c e l s i u s ) i n kJ / kg hf =1407.65; // e n n t h a l p y o f s a t u r a t e d l i q u i d a t 100 b a r i n kJ / kg hg =2724.7; // e n n t h a l p y o f s a t u r a t e d v a p o u t a t 100 b a r i n kJ / kg Ms =100000; // r a t e o f steam g e n e r a t i o n i n kg / h r eta =88; // e f f i c i e n c y o f steam generation C =21000; // c a l o r i f i c v a l u e o f f u e l i n kJ / kg // e t a =( h e a t a b s o r b e d by steam p e r h r ) / ( h e a t added by f u e l per hour ) m = Ms *( hsup - hf1 ) /( C *( eta /100) ) ; // f u e l b u r n i n g r a t e i n kg / h r htotal = hsup - hf1 ; // t o t a l h e a t s u p p l i e d t o steam f o r m a t i o n Pec =( hf - hf1 ) / htotal ; //% o f h e a t absorbed in economiser Pev =( hg - hf ) / htotal ; //% o f h e a t absorbed in evaporator Ps =( hsup - hg ) / htotal ; //% o f h e a t absorbed in superheater

18 19

printf ( ’ ( i ) The F u e l b u r n i n g r a t e , m i s : %5 . 1 f kJ / h . \n ’ ,m ) ; 20 printf ( ’ ( i i ) The P e r c e n t a g e o f h e a t a b s o r b e d i n e c o n o m i s e r i s : %5 . 4 f o r %5 . 2 f p e r c e n t . \ n ’ ,Pec ,( Pec *100) ) ; 21 printf ( ’ The P e r c e n t a g e o f h e a t a b s o r b e d i n e v a p o r a t o r i s : %5 . 4 f o r %5 . 2 f p e r c e n t . \ n ’ ,Pev ,( Pev *100) ) ; 22 printf ( ’ The P e r c e n t a g e o f h e a t a b s o r b e d i n 93

s u p e r h e a t e r i s : %5 . 4 f o r %5 . 2 f p e r c e n t . \ n ’ ,Ps ,( Ps *100) ) ;

Scilab code Exa 6.13 Example 13 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

clc clear //DATA GIVEN //BOILER Mw =2060; // mass o f f e e d w a t e r Mc =227; // mass o f c o a l s u p p l i e d i n kg / h r C =30000; // c a l o r i f i c v a l u e o f c o a l i n kJ / kg hs =2750; // e n t h a l p y o f steam p r o d u c e d i n kJ / kg hfw =398; // e n t h a l p y o f f e e d w a t e r //ECONOMISER Twin =15; // temp . o f f e e d w a t e r e n t e r i n g e c o n o m i s e r i n deg c e l s i u s Twout =95; // temp . o f f e e d w a t e r l e a v i n g e c o n o m i s e r i n deg c e l s i u s Tgout =18; // a t m o s p h e r i c temp . Tgin =370; // temp . o f e n t e r i n g f l u e g a s e s Mfg =4075; // mass o f f l u e g a s e s // a s s u m i n g Cpw and Cpg , Cpw =4.187; Cpg =1.01; ETAb = Mw *( hs - hfw ) /( Mc * C ) ; // e f f i c i e n c y o f

21 22 23

boiler ETAe = Mw * Cpw *( Twout - Twin ) /( Mfg * Cpg *( Tgin - Tgout ) ) ; // e f f i c i e n c y o f e c o n o m i s e r printf ( ’ ( i ) The B o i l e r e f f i c i e n c y 94

i s : %5 . 4 f o r %2 . 2

f p e r c e n t . \n ’ , ETAb ,( ETAb *100) ) ; 24 printf ( ’ ( i i ) The E c o n o m i s e r e f f i c i e n c y i s : %5 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAe ,( ETAe *100) ) ;

Scilab code Exa 6.14 Example 14 1 clc 2 clear 3 //DATA GIVEN 4 Tfw =50;

celsius p =5; x =0.95; Mc =600; C =30400; kg 9 Ms =4800; i n kg / h r

5 6 7 8

10 11 12 13 14 15 16 17 18 19 20

// mean f e e d

w a t e r temp . i n deg

// mean steam p r e s s u r e i n b a r // d r y n e s s f r a c t i o n o f steam // c o a l c o n s u m p t i o n kg / h r // c a l o r i f i c v a l u e o f c o a l i n kJ / // f e e d w a t e r s u p p l i e d t o b o i l e r

// from steam t a b l e , c o r r e s p o n d i n g t o 12 bar , hf =640.1; // i n kJ / kg hfg =2107.4; // i n kJ / kh h = hf + x * hfg ; // i n k j / kg hf1 =4.18*( Tfw -0) ; Ma = Ms / Mc ; // i n kg p e r kg o f f u e l Me = Ma *( h - hf1 ) /2257; // ( kg o f steam ) / ( kg o f f u e l ) printf ( ’ The E q u i v a l e n t e v a p o r a t i o n from and a t 100 deg c e l s i u s , Me i s : %5 . 3 f ( kg o f steam ) / ( kg o f c o a l ) . \ n ’ , Me ) ;

95

Chapter 7 Internal Combustion Engines

Scilab code Exa 7.1 Example 1 1 clc 2 clear 3 //DATA GIVEN 4 Pmi =6;

// mean e f f e c t i v e

p r e s s u r e in bar // e n g i n e s p e e d i n R . P .

5 N =1000;

M. 6 D =0.11;

// d i a m e t e r o f p i s t o n

in m 7 L =0.14; 8 n =1; 9 k =1; cylinder

// s t r o k e l e n g t h i n m // no . o f c y l i n d e r s // f o r 2− s t r o k e

10 11 //INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗ k ∗ 1 0 ) /6 kW 12 A =( %pi /4) *( D ^2) ; 13 IP =( n * Pmi * L * A * N * k *10) /6; 14 15 printf ( ’ The I n d i c t e d Power d e v e l o p e d i s : %2 . 1 f kW. ’ ,

IP ) ;

96

Scilab code Exa 7.2 Example 2 1 clc 2 clear 3 //DATA GIVEN 4 //L=1.5D 5 n =4; 6 P =14.7;

// no . o f c y l i n d e r s // power d e v e l o p e d i n

kW 7 N =1000; M. 8 Pmi =5.5; p r e s s u r e in bar 9 k =0.5; cylinder

// e n g i n e s p e e d i n R . P . // mean e f f e c t i v e // f o r 4− s t r o k e

10 11 //INDICTED POWER, I . P. = ( n∗PMI∗ l ∗A∗N∗ k ∗ 1 0 ) /6 kW 12 //A=( p i / 4 ) ∗Dˆ 2 , 13 //L=1.5D, 14 D =((6* P ) /(10* k * N * n * Pmi *1.5*( %pi /4) ) ) ^(1/3) ;

bore diameter in m 15 L =1.5* D ; length of stroke in m

// //

16 17

printf ( ’ The Bore d i a m e t e r i s : %5 . 2 f mm. \ n ’ ,( D *1000) ) ; 18 printf ( ’ The S t o k e l e n g t h i s : %5 . 2 f mm. \ n ’ ,( L *1000) ) ;


97

2 clear 3 //DATA GIVEN 4 Db =0.6;

// d i a m e t e r o f b r a k e

wheel in m // d i a m e t e r o f r o p e i n

5 d =0.026;

m 6 W =200; brake in N 7 S =30; reading in N 8 N =450; M.

// dead l o a d on t h e // s p r i n g b a l a n c e // e n g i n e s p e e d i n R . P .

9 10 // Brake Power , B . P. = (W−S ) ( p i ) (Db+d )N/ ( 6 0 ∗ 1 0 0 0 ) kW 11 BP =( W - S ) *( %pi ) *( Db + d ) * N /(60*1000) ; 12 13 printf ( ’ The Brake Power , B . P . i s : %2 . 1 f kW. \ n ’ , BP ) ;

Scilab code Exa 7.4 Example 4 1 clc 2 clear 3 //DATA GIVEN 4 T =175;

// t o r q u e due t o b r a k e l o a d

i n Nm 5 N =500; // e n g i n e s p e e d i n R . P .M. 6 7 // Brake Power , BP = ( 2 ∗ p i )NT/ ( 6 0 ∗ 1 0 0 0 ) kW 8 BP = (2* %pi ) * N * T /(60*1000) ; 9 10 printf ( ’ The Brake Power , B . P . i s : %4 . 2 f kW. \ n ’ , BP ) ;


1 clc 2 clear 3 //DATA GIVEN 4 D =0.3; 5 6 7 8 9 10 11 12 13 14 15 16 17 18

// b o r e o f e n g i n e

cylinder in m L =0.45; N =300; M. Pmi =6; p r e s s u r e in bar NBL =1.5; i n kN Db =1.8; drum d =0.02; n =1; k =0.5; cylinder

// s t r o k e l e n g t h i n m // e n g i n e s p e e d i n R . P . // mean e f f e c t i v e // Net b r a k e l o a d (W−S ) // d i a m e t e r o f b r a k e // b r a k e r o p e d i a m e t e r // no . o f c y l i n d e r s // f o r 4− s t r o k e

//INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗ k ∗ 1 0 ) /6 kW A =( %pi /4) *( D ^2) ; IP =( n * Pmi * L * A * N * k *10) /6; BP = NBL *( %pi ) *( Db + d ) * N /(60) ; eta = BP / IP ; // m e c h a n i c a l efficiency

19 20

printf ( ’ ( i ) The I n d i c t e d Power , I . P . i s : %5 . 2 f kW. \n ’ , IP ) ; 21 printf ( ’ ( i i ) The Brake Power , B . P . i s : %5 . 2 f kW. \n ’ , BP ) ; 22 printf ( ’ ( i i i ) M e c h a n i c a l e f f i c i e n c y i s : %5 . 4 f o r %5 . 2 f p e r c e n t . \ n ’ ,eta ,( eta *100) ) ;


99

1 clc 2 clear 3 //DATA GIVEN 4 D =0.2; 5 6 7 8 9 10 11 12 13 14 15

// d i a m e t e r o f e n g i n e cylinder in m L =0.350; // l e n g t h o f s t r o k e i n m Pmico =6.5; // mean e f f e c t i v e p r e s s u r e on c o v e r s i d e i n b a r Pmicr =7; // mean e f f e c t i v e p r e s s u r e on c r a n k s i d e i n b a r N =420; // e n g i n e s p e e d i n R . P . M. Drod =0.02; // d i a m e t e r o f p i s t o n rod in m W =1370; // dead l o a d on t h e brake in N S =145; // s p r i n g b a l a n c e reading in N Db =1.2; // d i a m e t e r o f b r a k e wheel in m d =0.02; // d i a m e t e r o f r o p e i n m n =1; // no . o f c y l i n d e r s k =0.5; // f o r 4− s t r o k e cylinder

16 17 //INDICTED POWER , I . P. = ( n∗Pmi∗ l ∗A∗N∗ k ∗ 1 0 ) /6 kW 18 Aco =( %pi /4) *( D ^2) ; // a r e a o f

c y l i n d e r om c o v e r end i n mˆ2 // a r e a o f

19 Acr =( %pi /4) *( D ^2 - Drod ^2) ;

c y l i n d e r om c r a n k end i n mˆ2 20 IPco =( n * Pmico * L * Aco * N * k *10) /6; s i d e i n kW 21 IPcr =( n * Pmicr * L * Acr * N * k *10) /6; s i d e i n kW 22 IPtotal = IPco + IPcr ; 23

100

// IP on c o v e r end // IP on c r a n k end // IP t o t a l i n kW

24 // Brake Power , B . P. = (W−S ) ( p i ) (Db+d )N/ ( 6 0 ∗ 1 0 0 0 ) kW 25 BP =( W - S ) *( %pi ) *( Db + d ) * N /(60*1000) ; 26 27 eta = BP / IPtotal ; // m e c h a n i c a l

efficiency 28 29

printf ( ’ M e c h a n i c a l e f f i c i e n c y i s : %5 . 4 f o r %5 . 2 f p e r c e n t . \ n ’ ,eta ,( eta *100) ) ;

Scilab code Exa 7.7 Example 7 1 2 3 4 5 6

clc clear //DATA GIVEN IP =30; BP =26; N =1000; . 7 F =0.35; h o u r i n kg /BP/ h 8 C =43900; f u e l u s e d i n kJ / kg

// i n d i c t e d power i n kW // Brake Power i n kW // e n g i n e s p e e d i n R . P .M // f u e l p e r b r a k e power // c a l o r i f i c v a l u e o f

9 10 Fc = F * BP ; 11 12 13 14 15 16

// f u e l c o n s u m p t i o n p e r

hour Mf = Fc /3600; ETAti = IP /( Mf * C ) ; eficiency ETAtb = BP /( Mf * C ) ; efficiency ETAm = BP / IP ;

// I n d i c t e d t h e r m a l // Brake t h e r m a l // M e c h a n i c a l e f f i c i e n c y

printf ( ’ ( i ) The I n d i c t e d t h e r m a l e f i c i e n c y i s : %5 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAti ,( ETAti *100) ) ; 17 printf ( ’ ( i i ) The Brake t h e r m a l e f f i c i e n c y i s : %5 . 3 f 101

o r %2 . 1 f p e r c e n t . \n ’ , ETAtb ,( ETAtb *100) ) ; 18 printf ( ’ ( i i i ) M e c h a n i c a l e f f i c i e n c y i s : %5 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAm ,( ETAm *100) ) ;

Scilab code Exa 7.8 Example 8 1 clc 2 clear 3 //DATA GIVEN 4 Db =0.75;

// d i a m e t e r o f b r a k e

pulley in m // d i a m e t e r o f r o p e i n

5 d =0.05;

m // dead l o a d on t h e

6 W =400; 7 8 9 10 11 12

brake in N S =50; reading in N Fc =4.2; kg / h r N =1000; i n R . P .M. C =43900; f u e l u s e d i n kJ / kg n =1; k =0.5; cylinder

// s p r i n g b a l a n c e // f u e l c o n s u m p t i o n i n // r a t e d e n g i n e s p e e d // c a l o r i f i c v a l u e o f // no . o f c y l i n d e r s // f o r 4− s t r o k e

13 14 15 // Brake Power , B . P. = (W−S ) ( p i ) (Db+d )N/ ( 6 0 ∗ 1 0 0 0 ) kW 16 BP =( W - S ) *( %pi ) *( Db + d ) * N /(60*1000) ; 17 sfc = Fc / BP ; // b r a 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 /kWhr 18 Mf = Fc /3600; 19 ETAtb = BP /( Mf * C ) ; thermal e f f i c i e n c y 102

// Brake

20 21

printf ( ’ ( i ) The Brake s p e c i f i c f u e l c o n s u m p t i o n , s . f . c ( b r a k e ) i s : %5 . 3 f kg /kWh . \n ’ , sfc ) ; 22 printf ( ’ ( i i ) The Brake t h e r m a l e f f i c i e n c y i s : %5 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAtb ,( ETAtb *100) ) ;

Scilab code Exa 7.9 Example 9 1 clc 2 clear 3 //DATA GIVEN 4 n =6; 5 D =0.09; 6 7 8 9 10 11 12

// no . o f c y l i n d e r s // b o r e o f e a c h

cylinder in m L =0.1; m r =7; ETArel =0.55; Fsc =0.3; f u e l c o n s u m p t i o n i n kg /kWh Pmi =8.6; e f f e c t i v e p r e s s u r e in bar N =2500; M. k =0.5; cylinder

// l e n g t h o f s t r o k e i n // c o m p r e s s i o n r a t i o // r e l a t i v e e f f i c i e n c y // i n d i c a t e d s p e c i f i c // i n d i c a t e d mean // e n g i n e s p e e d i n R . P . // f o r 4− s t r o k e

13 14 // A i r s t a n d a r d e f f i c i e n c y , ETAair =1−1/( r ˆ ( gamma−1) ) 15 g =1.4; //gamma o f a i r =1.4 16 ETAair =1 -1/( r ^( g -1) ) ; 17 // 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 , ETArel=ETAthi / ETAair

; 18 ETAthi = ETArel * ETAair ; 19 // I n d i c t e d t h e r m a l e f i c i e n c y , ETAthi=IP / ( Mf∗C) 20 Mf = Fsc /3600; 103

21 // t a k i n g IP =1 , 22 C =1/( ETAthi * Mf ) ; 23 24 25 26

// c a l o r i f i c v a l u e i n kJ / kg //INDICTED POWER , I . P. = ( n∗Pmi∗ l ∗A∗N∗ k ∗ 1 0 ) /6 kW A =( %pi /4) *( D ^2) ; IP =( n * Pmi * L * A * N * k *10) /6; Fc = Fsc * IP ; // t o t a l f u e l c o n s u m p t i o n i n kg / h r

27 28

printf ( ’ ( i ) The C a l o r i f i c v a l u e o f c o a l , C i s : %5 . 0 f kJ / kg . \n ’ ,C ) ; 29 printf ( ’ ( i i ) The F u e l c o n s u m p t i o n i s : %5 . 2 f kg / h . \ n ’ , Fc ) ;

30 31 32

//NOTE: // a n s o f c a l o r i f i c v a l u e h e r e i s e x a c t , w h i l e i n TB i t s rounded o f f v a l u e


clc clear //DATA GIVEN n =4; BP =30; N =2500; M. Pmi =8; p r e s s u r e in bar ETAm =0.8; efficiency ETAthb =0.28; efficiency C =43900; f u e l u s e d i n kJ / kg

// no . o f c y l i n d e r s // Brake Power i n kW // e n g i n e s p e e d i n R . P . // mean e f f e c t i v e // m e c h a n i c a l // b r a k e t h e r m a l // c a l o r i f i c v a l u e o f

104

// f o r 2− s t r o k e

11 k =1;

cylinder 12 13 // m e c h a n i c a l e f f i c i e n c y , ETAm=BP/ IP 14 IP = BP / ETAm ; 15 //INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗ k ∗ 1 0 ) /6 kW 16 //L=1.5D, 17 D =((6* IP ) /(10* k * N * n * Pmi *1.5*( %pi /4) ) ) ^(1/3) ;

//

bore diameter in m //

18 L =1.5* D ;

length of stroke in m 19 // Brake t h e r m a l e f f i c i e n c y , ETAtb=BP/ ( Mf∗C) 20 Mf = BP /( ETAthb * C ) ; f u e l c o n s u m p t i o n i n kg / h r

//

21 22

printf ( ’ ( i ) The Bore d i a m e t e r i s : %5 . 3 f m o r %2 . 0 f mm. \ n ’ ,D ,( D *1000) ) ; 23 printf ( ’ The S t o k e l e n g t h i s : %2 . 0 f mm. \ n ’ ,( L *1000) ) ; 24 printf ( ’ ( i i ) The F u e l c o n s u m p t i o n i s : %5 . 5 f kg / s o r %3 . 2 f kg / h r . \n ’ ,Mf ,( Mf *3600) ) ;

Scilab code Exa 7.11 Example 11 1 clc 2 clear 3 //DATA GIVEN 4 n =6; 5 Pdisp =700;

// no . o f c y l i n d e r s // p i s t o n d i s p p e r

c y l i n d e r i n cmˆ3 6 P =78; kW 7 N =3200; M. 8 Mf =27;

// power d e v e l o p e d i n // e n g i n e s p e e d i n R . P . // mass o f f u e l u s e d i n 105

9 10 11 12 13 14

kg / h r C =44000; f u e l u s e d i n kJ / kg afr =12; Pa =0.9; in bar Ta =32+273; tempertaure in K R =0.287; i n kJ /kgK k =0.5; cylinder

// c a l o r i f i c v a l u e o f // a i r f u e l r a t i o // i n t a k e a i r p r e s s u r e // i n t a k e a i r // g a s c o n s t a n t f o r a i r // f o r 4− s t r o k e

15 16 Ma = afr * Mf ; 17 18 19 20

// mass o f

air // by eq . pa ∗Va=Ma∗R∗Ta Va = Ma * R * Ta / Pa /100; i n t a k e a i r i n mˆ3/ h r Vswept =( Pdisp /10^6) * n *( N /2) *60; s w e p t i n mˆ3/ h r ETAvol = Va / Vswept ; efficiency

// volume o f // volume // v o l u m e t r i c

21 22

// Brake t h e r m a l e f f i c i e n c y , ETAbt=b r a k e work / h e a t s u p p l i e d by t h e f u e l 23 ETAbt = P /( Mf * C /3600) ; 24 // Brake Power , BP = ( 2 ∗ p i )N∗Tb / ( 6 0 ∗ 1 0 0 0 ) kW 25 Tb = P *60/(2* %pi * N ) ; // b r a k e t o r q u e i n kNm 26 27

printf ( ’ ( i ) The V o l u m e t r i c e f f i c i e n c y i s : %5 . 3 f o r %5 . 1 f p e r c e n t . \n ’ , ETAvol ,( ETAvol *100) ) ; 28 printf ( ’ ( i i ) The Brake t h e r m a l e f f i c i e n c y i s : %5 . 4 f o r %5 . 2 f p e r c e n t . \n ’ , ETAbt ,( ETAbt *100) ) ; 29 printf ( ’ ( i i i ) The Brake Torque i s : %5 . 4 f kNm . \n ’ , Tb ) ;

106

Scilab code Exa 7.12 Example 12 1 clc 2 clear 3 //DATA GIVEN 4 //L=1.5D 5 n =6; 6 Vs =1.75; 7 8 9 10

// no . o f c y l i n d e r s // s t r o k e volume i n

litres IP =26.3; kW Ne =504; M. Pmi =6; p r e s s u r e in bar k =0.5; cylinder

// power d e v e l o p e d i n // e n g i n e s p e e d i n R . P . // mean e f f e c t i v e // f o r 4− s t r o k e

11 12 //INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗ k ∗ 1 0 ) /6 kW 13 //L∗A=Vs 14 Na = IP *6/( n * Pmi *( Vs /10^3) * k *10) ; // a c t u a l s p e e d 15 16 17 18 19 20

i n R . P .M Fa = Na * n * k ; // a c t u a l no . o f f i r e s i n one m i n u t e Fe = Ne * n /2; // e x p e c t e d no . o f f i r e s i n one m i n u t e Fm = Fe - Fa ; // m i s f i r e s p e r minute Fmavg = Fm / n ; // avg . no . o f t i m e s e a c h c y l i n d e r m i s f i r e s i n one m i n u t e printf ( ’ The A v e r a g e no . o f t i m e s e a c h c y l i n d e r m i s f i r e s i n one m i n u t e i s : %1 . 0 f . \ n ’ , Fmavg ) ;

107


clc clear //DATA GIVEN D =0.075; // b o r e i n m L =0.09; // s t r o k e l e n g t h i n m n =4; // no . o f c y l i n d e r s erar =39/8; // e n g i n e t o r e a r a x l e r a t i o =39:8 Dw =0.65; // w h e e l d i a m e t e r w i t h tyre f u l l y i n f l a t e d in m Fc =0.227; // p e t r o l c o n s u m p t i o n f o r a d i s t a n c e o f 3 . 2 km a t a s p e e d o f 48 km/ h r Pmi =5.625; // mean e f f e c t i v e p r e s s u r e in bar C =43470; // c a l o r i f i c v a l u e o f f u e l u s e d i n kJ / kg k =0.5; // f o r 4− s t r o k e cylinder

13 14 s =48*1000/60; 15 16 17 18

// s p e e d o f c a r i n m/

min // i f Nt r e v a r e made by t y r e p e r minute , s p e e d=p i ∗Dw ∗Nt Nt = s /( %pi * Dw ) ; //R . P .M. // a s e n g i n e t o r e a r a x l e r a t i o i s 3 9 : 8 Ne = erar * Nt ; // s p e e d o f e n f i n e s h a f t i n R . P .M. //INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗ k ∗ 1 0 ) /6 kW A =( %pi /4) *( D ^2) ; IP =( n * Pmi * L * A * Ne * k *10) /6;

19 20 21 22 23 s = s /1000;

// s p e e d o f c a r i n km/ 108

24 25 26 27

min t =3.2/ s ; // t i m e i n min f o r c o v e r i n g 3 . 2 km // p e t r o l c o n s u m p t i o n f o r a d i s t a n c e o f 3 . 2 km a a t a s p e e d o f 48 km/ h r i s 0 . 2 2 7 kg Mf = Fc /( t *60) ; // f u e l consumed p e r sec ETAthi = IP /( Mf * C ) ; // I n d i c a t e d f u e l efficiency

28 29

printf ( ’ ( i ) The I n d i c a t e d Power d e v e l o p e d i s : %5 . 2 f kW. \n ’ , IP ) ; 30 printf ( ’ ( i i ) 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 i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETAthi ,( ETAthi *100) ) ;

Scilab code Exa 7.14 Example 14 1 clc 2 clear 3 //DATA GIVEN 4 D =0.25; 5 6 7 8 9 10 11

// c y l i n d e r d i a m e t e r i n

m L =0.4; Pmg =7; p r e s s u r e in bar Pmp =0.5; e f f e c t i v e p r e s s u r e in bar N =250; M. NBL =1080; b r a k e (W−S ) i n N Db =1.5; of the brake in m Fc =10; kg 109

// s t r o k e l e n g t h i n m // G r o s s mean e f f e c t i v e // Pumping mean // e n g i n e s p e e d i n R . P . // n e t l o a d on t h e // e f f e c t i v e d i a m e t e r // f u e l u s e d p e r h r i n

// c a l o r i f i c v a l u e o f

12 C =44300;

f u e l u s e d i n kJ / kg // no . o f c y l i n d e r s // f o r 4− s t r o k e

13 n =1; 14 k =0.5;

cylinder 15 16 17 18 19 20 21

//INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗ k ∗ 1 0 ) /6 kW Pm = Pmg - Pmp ; A =( %pi /4) *( D ^2) ; IP =( n * Pm * L * A * N * k *10) /6; BP = NBL *( %pi ) *( Db ) * N /(60*1000) ; ETAm = BP / IP ; // m e c h a n i c a l efficiency 22 Mf = Fc /3600; 23 ETAthi = IP /( Mf * C ) ; // I n d i c a t e d t h e r m a l efficiency 24 25

printf ( ’ ( i ) The I n d i c a t e d Power , I . P . i s : %5 . 2 f kW. \n ’ , IP ) ; 26 printf ( ’ ( i i ) The Brake Power , B . P . i s : %2 . 1 f kW. \n ’ , BP ) ; 27 printf ( ’ ( i i i ) M e c h a n i c a l e f f i c i e n c y i s : %5 . 3 f o r %2 . 1 f p e r c e n t . \ n ’ , ETAm ,( ETAm *100) ) ; 28 printf ( ’ ( i v ) 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 s : %5 . 3 f o r %2 . 1 f p e r c e n t . \ n ’ , ETAthi ,( ETAthi *100) ) ;

Scilab code Exa 7.15 Example 15 1 clc 2 clear 3 //DATA GIVEN 4 ETAthb =30;

// Brake t h e r m a l

e f f i c i e n c y in % 5 afr =20; weight

// a i r f u e l r a t i o by

110

// c a l o r i f i c v a l u e o f

6 C =41800;

f u e l u s e d i n kJ / kg 7 8 9 10 11 12 13 14 15 16 17

// Brake t h e r m a l e f f i c i e n c y , ETAthb=work p r o d u c e d / heat supplied work =( ETAthb /100) * C ; // work p r o d u c e d p e r kg of fuel //STP c o n d i t i o n s r e f e r t o 1 . 0 1 3 2 b a r and 15 deg celsius m = afr ; // mass o f a i r p e r kg of fuel R =287; V = m * R *(15+273) /(1.0132*10^5) ; // volume o f a i r u s e d // Brake mean e f f e c t i v e p r e s s u r e , Pmb=work done / c y l i n d e r volume Pmb =( work *1000) /( V *10^5) ; printf ( ’ The Brake mean e f f e c t i v e p r e s s u r e , Pmb i s : %2 . 2 f b a r . \ n ’ , Pmb ) ;

Scilab code Exa 7.16 Example 16 1 clc 2 clear 3 //DATA GIVEN 4 V1 =0.216; 5 6 7 8 9

// g a s c o n s u m p t i o n i n m

ˆ3/ min P1 =75; mm o f w a t e r T1 =17+273; m =2.84; kg / min Ta =17+273; br =745; mm o f Hg

// g a s t e m p e r a t u r e i n // g a s t e m p e r t a u r e i n K // a i r c o n s u m p t i o n i n // a i r t e m p e r t a u r e i n K // b a r o m e t e r r e a d i n g i n

111

// b o r e o f e n g i n e

10 D =0.25;

cylinder in m // s t r o k e l e n g t h i n m // e n g i n e s p e e d i n R . P .

11 L =0.475; 12 N =240;

M. 13 R =287; i n J /kgK 14 n =1; 15 k =1; cylinder

// g a s c o n s t a n t f o r a i r // no . o f c y l i n d e r s // f o r 2− s t r o k e

16 17 18 19 20 21 22

// p r e s s u r e o f t h e g a s

P1 = br + P1 /13.6; // a t NTP P2 =760; T2 =0+273; // P1∗V1/T1=P2∗V2/T2 V2 = P1 * V1 * T2 /( P2 * T1 ) ; a t NTP i n mˆ3 23 Vg = V2 /( N /2) ; i n mˆ3

//mm o f Hg // i n K // volume o f g a s u s e d // g a s u s e d p e r s t r o k e

24 25 //PV=mRT 26 P2 =1.0132*10^5; 27 V = m * R * T2 / P2 ;

// volume o c c u p i e d by

a i r i n mˆ3/ min // a i r u s e d p e r s t r o k e

28 Va = V /( N /2) ;

in m 29 30

// m i x t u r e o f g a s and

Vmix = Vg + Va ; a i r i n mˆ3

31 32

// ETAvol=( a c t u a l volume o f m i x t u r e drawn p e r s t r o k e a t NTP) / ( s w e p t volume o f s y s t e m ) 33 ETAvol = Vmix /(( %pi /4) * D ^2* L ) ; 34 35

printf ( ’ The V o l u m e t r i c e f f i c i e n c y i s : %3 . 3 f o r %3 . 1 f p e r c e n t . \n ’ , ETAvol ,( ETAvol *100) ) ;

112

Scilab code Exa 7.17 Example 17 1 clc 2 clear 3 //DATA GIVEN 4 t =1;

// d u r a t i o n o f t r i a l i n

hr // r e v o l u t i o n s // no . o f m i s s e d c y c l e s // Net b r a k e l o a d (W−S )

5 N =14000; 6 mc =500; 7 NBL =1470; 8 9 10 11 12 13 14 15 16

in N Pmi =7.5; p r e s s u r e in bar Vg =20000/3600; litres /s C =21; c o n d i t i o n s i n kJ / l i t r e D =0.25; m L =0.4; Cb =4; circumference in m r =6.5; n =1; k =0.5; cylinder

// mean e f f e c t i v e // g a s c o n s u m p t i o n i n //LCV o f g a s a t s i p p l y // c y l i n d e r d i a m e t e r i n // s t r o k e l e n g t h i n m // e f f e c t i v e b r a k e // c o m p r e s s i o n r a t i o // no . o f c y l i n d e r s // f o r 4− s t r o k e

17 18 //gamma f o r a i r , g =1.4 19 g =1.4; 20 21 //INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗ k ∗ 1 0 ) /6 kW 22 Nk =( N *k - mc ) /60; // (N∗ k )−w o r k i n g c y c l e s

/ min 23 A =( %pi /4) *( D ^2) ;

113

24 25 26 27

IP =( n * Pmi * L * A * Nk *10) /6; N = N /60; BP = NBL *( Cb ) * N /(60*1000) ; eta = BP / IP ; efficiency 28 ETAthi = IP /( Vg * C ) ; efficiency

// m e c h a n i c a l // I n d i c a t e d t h e r m a l

29 30 31 32

// r e l a t i v e e f f i c i e n c y , ETArel=ETAthi /ETAas // ETAas=1−1/( r ˆ ( g −1) ) ETAas =1 -1/( r ^( g -1) ) ; // a i r −s t a n d a r d efficiency 33 ETArel = ETAthi / ETAas ; // r e l a t i v e e f f i c i e n c y 34 35 36 37 38 39

printf ( ’ ( i ) The I n d i c a t e d Power , I . P . i s : %5 . 2 f kW . \n ’ , IP ) ; printf ( ’ ( i i ) The Brake Power , B . P . i s : %5 . 2 f kW. \ n ’ , BP ) ; printf ( ’ ( i i i ) M e c h a n i c a l e f f i c i e n c y i s : %5 . 3 f o r %2 . 1 f p e r c e n t . \ n ’ ,eta ,( eta *100) ) ; printf ( ’ ( i v ) 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 i s : %2 . 2 f o r %2 . 0 f p e r c e n t . \n ’ , ETAthi ,( ETAthi *100) ) ; printf ( ’ ( v ) The R e l a t i v e e f f i c i e n c y i s : %2 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , ETArel ,( ETArel *100) ) ;

114

Chapter 10 Air Compressors

Scilab code Exa 10.1 Example 1 1 clc 2 clear 3 //DATA GIVEN 4 V1 =1;

// volume o f a i r

t a k e n i n mˆ3/mim // i n t a k e p r e s s u r e i n

5 p1 =1.013;

bar // i n t a k e t e m p e r a t u r e

6 T1 =15+273; 7

in K p2 =7; // d e l i v e r y p r e s s u r e in bar t =1*60; // t i m e i n s e c o n d s // law o f c o m p r e s s i o n , pVˆ 1 . 3 5 =C n =1.35; R =287;

8 9 10 11 12 13 m = p1 *10^5* V1 / R / T1 ;

// mass o f a i r

d e l i v e r e d i n kg / min 14 15 // ( T2/T1 ) =(p2 / p1 ) ˆ ( ( n −1) / n ) ; 16 T2 = T1 *( p2 / p1 ) ^(( n -1) / n ) ;

115

// d e l i v e r y temp . T2

in K 17 18 W =( n ) /( n -1) * m * R *( T2 - T1 ) /1000;

// i n d i c a t e d work i n

kJ / min 19 20 IP = W / t ;

// i n d i c a t e d power i n

kW 21 22

printf ( ’ The I n d i c a t e d power , IP i s : %1 . 2 f kW. \n ’ , IP ) ;

Scilab code Exa 10.2 Example 2 1 clc 2 clear 3 // c o n t i n u e d from Example 1 4 //DATA GIVEN 5 V =1;

// volume d e a l t w i t h

p e r min a t i n l e t i n mˆ3/mim // volume drawn i n

6 Vc =1/300;

p e r c y c l e , i n mˆ3/ c y c l e // s t r o k e t o b o r e

7 r =1.5;

ratio 8 ETAc =0.85; // m e c h a n i c a l e f f i c i e n c y of the compressor 9 ETAmt =0.90; // m e c h a n i c a l e f f i c i e n c y o f t h e motor t r a n s m i s s i o n 10 11 // c y l i n d e r volume , Vc=( p i / 4 )Dˆ2∗L 12 D =[( Vc *4/ %pi ) / r ]^(1/3) ; // b o r e i n m 13 14 // from e x a m p l e 1 15 Pi =4.23/ ETAc ; // power i n p u t t o t h e

c o m p r e s s o r i n kW // motor power i n kW

16 MP = Pi / ETAmt ;

116

17 18

printf ( ’ ( i ) The C y l i n d e r b o r e , D i s : %3 . 1 f mm. \n ’ ,( D *1000) ) ; 19 printf ( ’ ( i i ) The Motor power i s : %1 . 2 f kW. \n ’ , MP ) ;

Scilab code Exa 10.3 Example 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

clc clear //DATA GIVEN T1 =20+273; p1 =1; p2 =10; Cv =0.718;

// t e m p e r a t u r e i n K // p r e s s u r e i n b a r // p r e s s u r e i n b a r // i n kJ /kgK

// law o f c o m p r e s s i o n , pVˆ1.2=C n =1.2; R =0.287;

// i n kJ /kgK

// ( T2/T1 ) =(p2 / p1 ) ˆ ( ( n −1) / n ) ; T2 = T1 *( p2 / p1 ) ^(( n -1) / n ) ; // temp . T2 i n K m =1; W =( n ) /( n -1) * m * R * T1 *[( p2 / p1 ) ^(( n -1) / n ) -1]; // work done p e r kg o f a i r ( kJ / kg o f a i r )

17 18 //By t h e F i r s t Law o f Thermodynamics 19 // h e a t t r a n s f e r r e d d u r i n g c o m p r e s s i o n , Q=W+DU 20 //Q=(p1V1−p2V2 ) / ( n −1)+Cv ( T2−T1 ) 21 //Q=(T2−T1 ) ∗ [ Cv−R/ ( n−1) ] 22 Q =( T2 - T1 ) *[ Cv - R /( n -1) ]; 23 24 printf ( ’ ( i ) The T e m p e r a t u r e a t t h e end o f

c o m p r e s s i o n i s : %3 . 0 f K o r %3 . 0 f deg . c e l s i u s . \n ’ ,T2 ,( T2 -273) ) ; 25 printf ( ’ ( i i ) The Work done d u r i n g c o m p r e s s i o n p e r 117

kg o f a i r i s : %3 . 2 f kJ / kg o f a i r . \n ’ ,W ) ; 26 printf ( ’ The Heat t r a n s f e r r e d d u r i n g c o m p r e s s i o n p e r kg o f a i r i s : %2 . 2 f kJ / kg o f a i r . \n ’ ,Q ) ; 27 printf ( ’ ( Negative sign i n d i c a t e s heat REJECTION . ) \n ’ ) ;

Scilab code Exa 10.4 Example 4 1 clc 2 clear 3 //DATA GIVEN 4 p1 =1; 5 6 7 8 9 10

// s u c t i o n p r e s s u r e

in bar T1 =20+273; temperature in K p2 =6; in bar T2 =180+273; temperature in K N =1200; c o m p r e s s o r i n R . P .M. Pshaft =6.25; Ma =1.7; d e l i v e r e d i n kg / min D =0.14; L =0.1; R =287;

// s u c t i o n // d i s c h a r g e p r e s s u r e // d i s c h a r g e // s p e e d o f // s h a f t power i n kW // mass o f a i r // d i a m e t e r i n m // s t r o k e i n m // i n kJ /kgK

11 12 13 14 15 Vd =( %pi /4) * D ^2* L * N ;

// d i s p l a c e m e n t volume f o r s i n g l e a c t i n g c o m p r e s s o r i n mˆ3/ min 16 FAD = Ma * R * T1 / p1 /10^5; //mˆ3/ min 17 ETAvol = FAD / Vd *100; // a c t u a l v o l u m e t r i c efficiency 18

118

19 // ( T2/T1 ) =(p2 / p1 ) ˆ ( ( n −1) / n ) ; 20 n =1/[1 -( log ( T2 / T1 ) / log ( p2 / p1 ) ) ];

// i n d e x o f

compression , n 21 22 IP =( n ) /( n -1) * Ma /60* R /1000* T1 *[( p2 / p1 ) ^(( n -1) / n ) -1];

// i n d i c a t e d power i n kW 23 24

Piso = Ma /60* R /1000* T1 * log ( p2 / p1 ) ; // i s o t h e r m a l power 25 ETAiso = Piso / IP *100; // i s o t h e r m a l efficiency 26 27

ETAmech = IP / Pshaft *100; // m e c h a n i c a l efficiency

28 29

ETAovr_iso = Piso / Pshaft *100; // o v e r a l l i s o t h e r m a l eddiciency

30 31 32 33 34 35

printf ( ’ ( i ) The a c t u a l V o l u m e t r i c e f f i c i e n c y i s : %2 . 2 f p e r c e n t . \n ’ , ETAvol ) ; printf ( ’ ( i i ) The I n d i c a t e d Power , IP i s : %1 . 3 f KW. \n ’ , IP ) ; printf ( ’ ( i i i ) The I s o t h e r m a l e f f i c i e n c y i s : %2 . 2 f p e r c e n t . \ n ’ , ETAiso ) ; printf ( ’ ( i v ) The M e c h a n i c a l e f f i c i e n c y i s : %2 . 1 f p e r c e n t . \ n ’ , ETAmech ) ; printf ( ’ ( v ) The O v e r a l l i s o t h e r m a l e f f i c i e n c y i s : %2 . 1 f p e r c e n t . \ n ’ , ETAovr_iso ) ;


// 5 ( b ) i s a s f o l l o w s : 119

2 3 4 5 6 7 8 9 10

clc clear //DATA GIVEN m =6.75; p1 =1; T1 =21+273; p2 =1.35; T2 =43+273; DTcw =3.3; w a t e r i n deg . c e l s i u s 11 Cp =1.003; 12 //gamma f o r a i r =1.4 13 g =1.4; 14 15 16 17 18 19 20

// mass o f a i r i n kg / min // p r e s s u r e i n b a r // temp . i n K // p r e s s u r e i n b a r // temp . i n K // temp . r i s e o f c o o l i n g //Cp f o r a i r i n kJ /kgK

W = m * Cp *( T2 - T1 ) ; // work i n kJ / min // I f t h e c o m p r e s s i o n would have b e e n i s o t r o p i c , // T 2=T1 ∗ ( r p ) ˆ [ ( g −1) / g ] rp = p2 / p1 ; T_2 = T1 *( rp ) ^[( g -1) / g ]; Qr = m * Cp *( T_2 - T2 ) ; // h e a t r e j e c t e d t o c o o l i n g water

21 22 Mw = Qr /[4.18*( DTcw ) ];

// mass o f c o o l i n g w a t e r i n

kg / min 23 24 25

printf ( ’ ( i ) The Work i s : %3 . 2 f kJ / min . \n ’ ,W ) ; printf ( ’ ( i i ) The Mass o f c o o l i n g w a t e r i s : %1 . 2 f kg / min . \n ’ , Mw ) ;

26 27 28

//NOTE: // i n t h e q u e s t i o n c o m p r e s s i o n p r o c e s s i s m e n t i o n e d and p2 i s g i v e n a s 0 . 3 5 b a r ( p2
120

Scilab code Exa 10.6 Example 6 1 clc 2 clear 3 //DATA GIVEN 4 V1 =14;

// q u a n t i t y o f a i r t o

be d e l i v e r e d , i n mˆ3/mim // i n t a k e p r e s s u r e i n

5 p1 =1.013; 6 7 8 9

bar T1 =15+273; in K p2 =7; in bar N =300; c o m p r e s s o r i n R . P .M. n =1.3; expansion index R =0.287;

// i n t a k e t e m p e r a t u r e // d e l i v e r y p r e s s u r e // s p e e d o f // c o m p r e s s i o n and

10 11 12 // c l e a r a n c e volume , Vc = 0 . 0 5 Vs , Vs=s w e p t volume 13 // s w e p t volume Vs=V1−V3=V1−Vc=V1 −0.05 Vs 14 //V1 =1.05 Vs 15 Vpc = V1 / N /2;

// ( V1−V4 ) volume i n d u c e d p e r c y c l e i n mˆ3 16 //V4/V3=(p2 / p1 ) ˆ ( 1 / n ) 17 c =( p2 / p1 ) ^(1/ n ) ; 18 //V4=c ∗V3=c ∗ 0 . 0 5 Vs 19 //V1−V4 =1.05 Vs−c ∗ 0 . 0 5 Vs 20 Vs = Vpc /(1.05) /(1.05 - c *0.05) ;

// volume s w e p t i n mˆ3 21 22 // u s i n g r e l a t i o n ( T2/T1 ) =(p2 / p1 ) ˆ ( ( n−1) / n ) ; 23 T2 = T1 *( p2 / p1 ) ^(( n -1) / n ) ;

121

// d e l i v e r y temp . T2 i n K 24 25 IP =( n ) /( n -1) * p1 *10^5* Vpc /100*[( p2 / p1 ) ^(( n -1) / n ) -1];

// i n d i c a t e d power i n kW 26 27

printf ( ’ ( i ) The Swept volume o f t h e c y l i n d e r , Vs i s : %1 . 4 f mˆ 3 . \n ’ , Vs ) ; 28 printf ( ’ ( i i ) The d e l i v e r y t e m p e r a t u r e , Ts i s : %3 . 0 f deg . c e l s i u s . \n ’ ,( T2 -273) ) ; 29 printf ( ’ ( i i i ) The I n d i c a t e d power , IP i s : %2 . 2 f kW. \n ’ , IP ) ; 30 31 32

//NOTE: // t h e r e i s s l i g h t v a r i a t i o n i n a n s w e r s i n t e x t b o o k due t o r o u n d i n g o f f o f v a l u e s i n book

122

Chapter 13 Transmission of Motion and Power

Scilab code Exa 13.1 Example 1 1 clc 2 clear 3 //DATA GIVEN 4 N1 =240;

// s p e e d o f t h e e n g i n e

s h a f t i n R . P .M. 5 d1 =1.5; engine shaft in m 6 d2 =0.75; machine s h a f t i n m 7 t =0.005; in m

// d i a m e t e r o f p u l l e y on // d i a m e t e r o f p u l l e y on // t h i c k n e s s o f t h e b e l t

8 9 // w i t h no s l i p 10 // ( N2/N1 ) =(d1+t ) / ( d2+t ) 11 N2 =( d1 + t ) /( d2 + t ) * N1 ;

// s p e e d o f t h e machine

s h a f t i n R . P .M. 12 13 // w i t h 14 S =2;

s l i p o f 2% // s l i p i n % 123

15 // ( N2/N1 ) =(d1+t ) / ( d2+t ) ∗((100 − S ) / 1 0 0 ) 16 N2s =( d1 + t ) /( d2 + t ) * N1 *((100 - S ) /100) ; 17 18 printf ( ’ ( i ) The Speed o f machine s h a f t , N2 w i t h no

s l i p i s : %4 . 1 f R . P .M. \n ’ , N2 ) ; 19 printf ( ’ ( i i ) The Speed o f machine s h a f t , N2 w i t h s l i p o f 2 p e r c e n t i s : %4 . 1 f R . P .M. \n ’ , N2s ) ;

Scilab code Exa 13.2 Example 2 1 clc 2 clear 3 //DATA GIVEN 4 r1 =900/2000;

// r a d i u s o f l a r g e r

pulley in m // r a d i u s o f s m a l l e r

5 r2 =300/2000;

pulley in m // d i s t a n c e b e t w e e n t h e

6 d =6;

centres of pulley in m 7 8

// Length o f c r o s s b e l t , L c r o s s =( p i ) ( r 1+r 2 ) +( r 1+r 2 ) ˆ2/ d+2d ; 9 Lcross =( %pi ) *( r1 + r2 ) +( r1 + r2 ) ^2/ d +2 d ; 10 // Length o f open b e l t , Lopen =( p i ) ( r 1+r 2 ) +( r2 −r 1 ) ˆ2/ d +2d ; 11 Lopen =( %pi ) *( r1 + r2 ) +( r2 - r1 ) ^2/ d +2 d ; 12 13 14

Lred = Lcross - Lopen ; // l e n g t h t o be r e d u c e d printf ( ’ The Length o f t h e b e l t t o be r e d u c e d , \n ( to change the d i r e c t i o n o f r o t a t i o n o f the f o l l o w e r p u l l e y s ) i s : %2 . 0 f mm. \n ’ ,( Lred *1000) ) ;


1 clc 2 clear 3 //DATA GIVEN 4 T1 =1500;

// t e n s i o n on t h e t i g h t

side in N 5 T2 =1200; side in N 6 v =80; /s

// t e n s i o n on t h e s l a c k // s p e e d o f t h e b e l t i n m

7 8 P =( T1 - T2 ) * v ;

// power t r a n s m i t t e d by

the b e l t in watts 9 10

printf ( ’ The Power t r a n s m i t t e d by t h e b e l t i s : %2 . 0 f kW. \n ’ ,( P /1000) ) ;

Scilab code Exa 13.4 Example 4 1 clc 2 clear 3 //DATA GIVEN 4 v =500;

// s p e e d o f t h e b e l t i n m

/ min 5 mu =0.3; friction 6 theta =160; degrees 7 T1 =700; belt in N

// c o e f f i c i e n t o f // a n g l e o f c o n t a c t i n //maximum t e n s i o n i n t h e

8 9 10

// ( T1/T2 )=e ˆ (mu∗ t h e t a ) theta = theta *( %pi ) /180; radians 11 T2 = T1 /( %e ^( mu * theta ) ) ; side in N

// t h e t a c o n v e r t e d i n t o // t e n s i o n on t h e s l a c k

125

// s p e e d o f t h e b e l t

12 v = v /60;

c o n v e r t e d i n t o m/ s // power t r a n s m i t t e d by

13 P =( T1 - T2 ) * v ;


printf ( ’ The Power t r a n s m i t t e d by t h e b e l t i s : %2 . 3 f kW. \n ’ ,( P /1000) ) ;

Scilab code Exa 13.5 Example 5 1 clc 2 clear 3 //DATA GIVEN 4 r1 =750/2000; 5 6 7 8 9 10

// r a d i u s o f l a r g e r

pulley in m r2 =300/2000; pulley in m d =1.5; centres of pulley in m Tms =14; i n N/mm b =150; mm v =540; / min mu =0.25; friction

// r a d i u s o f s m a l l e r // d i s t a n c e b e t w e e n t h e //maximum s a f e t e n s i o n // w i d t h o f t h e b e l t i n // s p e e d o f t h e b e l t i n m // c o e f f i c i e n t o f

11 12 T1 = Tms * b ;

//maximum t e n s i o n i n t h e belt in N 13 v = v /60; // s p e e d o f t h e b e l t c o n v e r t e d i n t o m/ s 14 // ( i ) f o r open b e l t 15 ALPHAo = asin (( r1 - r2 ) / d ) *180/( %pi ) ; // a l p h a i n degrees 126

16

THETAo =180 -2* ALPHAo ; l a p o r c o n t a c t i n deg 17 T2o = T1 /( %e ^( mu *( THETAo * %pi /180) ) ) ; the s l a c k s i d e in N 18 Po =( T1 - T2o ) * v ; t r a n s m i t t e d by t h e b e l t i n w a t t s 19 20 21

// ( i i ) f o r c r o s s b e l t ALPHAc = asin (( r1 + r2 ) / d ) *180/( %pi ) ; degrees 22 THETAc =180+2* ALPHAc ; l a p o r c o n t a c t i n deg 23 T2c = T1 /( %e ^( mu *( THETAc * %pi /180) ) ) ; the s l a c k s i d e in N 24 Pc =( T1 - T2c ) * v ; t r a n s m i t t e d by t h e b e l t i n w a t t s

// a n g l e o f // t e n s i o n on // power

// a l p h a i n // a n g l e o f // t e n s i o n on // power

25 26

printf ( ’ ( i ) The Maximum Power t r a n s m i t t e d by t h e open b e l t i s : %2 . 3 f kW. \n ’ ,( Po /1000) ) ; 27 printf ( ’ ( i i ) The Maximum Power t r a n s m i t t e d by t h e c r o s s b e l t i s : %2 . 3 f kW. \n ’ ,( Pc /1000) ) ;

Scilab code Exa 13.6 Example 6 1 clc 2 clear 3 //DATA GIVEN 4 b =0.25; 5 t =0.006;

// w i d t h o f t h e b e l t i n m // t h i c k n e s s o f t h e b e l t

in m 6 r =900/2000; in m 7 rho =1100; m a t e r i a l i n kg /mˆ3 8 Tp =2;

// r a d i u s o f t h e p u l l e y // d e n s i t y o f t h e // p e r m i s s i b l e t e n s i o n o f 127

t h e b e l t i n MN/mˆ2 9 ratio =2; 10 N =200; R . P .M. 11 12 13 14

// r a t i o o f T1/T2=2 // s p e e d o f t h e p u l l e y i n

Tmax = Tp *10^6* b * t ; //maximum s a f e t e n s i o n of the b e l t // c e n t r i f u g a l t e n s i o n , Tc=m∗ v ˆ2 m =( b * t ) *1* rho ; // mass o f t h e b e l t p e r u n i t metre l e n g t h v =2*( %pi ) *( r + t /2) * N /60; Tc = m * v ^2;

15 16 17 18 T1 = Tmax - Tc ;

// t e n s i o n i n t h e t i g h t

side in N 19 T2 = T1 / ratio ; side in N 20 P =( T1 - T2 ) * v ; the b e l t in watts

// t e n s i o n i n t h e s l a c k // power t r a n s m i t t e d by

21 22 23

printf ( ’ ( i ) The C e n t r i f u g a l t e n s i o n Tc i s : %3 . 1 f N . \n ’ , Tc ) ; 24 printf ( ’ ( i i ) The Power t r a n s m i t t e d by t h e b e l t i s : %2 . 1 f kW. \n ’ ,( P /1000) ) ;

Scilab code Exa 13.7 Example 7 1 clc 2 clear 3 //DATA GIVEN 4 P =35;

// power r e q u i r e d t o be t r a n s m i t t e d by t h e b e l t i n kW 5 d =1.5; // e f f e c t i v e d i a m e t e r o f pulley in m 128

// s p e e d o f t h e p u l l e y i n

6 N =300; 7 8 9 10 11

R . P .M. theta =11/24*2* %pi ; radians mu =0.3; friction t =0.0095; in m rho =1100; m a t e r i a l i n kg /mˆ3 sigma =2.5; MN/mˆ2

// a n g l e o f c o n t a c t i n // c o e f f i c i e n t o f // t h i c k n e s s o f t h e b e l t // d e n s i t y o f t h e // p e r m i s s i b l e s t r e s s i n

12 13 v = %pi * d * N /60; 14 15 16 17 18 19 20 21 22 23 24 25 26

// s p e e d o f t h e b e l t i n m /s //P=(T2−T1 ) ∗v , s o ( T2−T1 )=P/ v . . . . . . . . . . . . . . . . . . . . ( 1 ) c = %e ^( mu * theta ) ; // so , T2/T1=c . . . . . . . . ( 2 ) //By e q u a t i o n ( 1 ) and ( 2 ) , T2 =( P / v *1000) /( c -1) ; // t e n s i o n i n t h e s l a c k side in N T1 = c * T2 ; // t e n s i o n i n t h e t i g h t side in N //maximum t e n s i o n , Tmax=s i g m a ∗b∗ t = 0 . 2 3 7 5 ∗ b ∗ 1 0 ˆ 6 N (3) // c e n t r i f u g a l t e n s i o n , Tc=m∗ v ˆ 2 = 5 8 0 0 . 5 ∗ b N (4) //T1=Tmax−c (5) //By eqn . ( 3 ) , ( 4 ) and ( 5 ) b = T1 /(( sigma *10^6* t ) -( t *1* rho * v ^2) ) ; // w i d t h o f the b e l t in m printf ( ’ The Width o f t h e b e l t i s : %3 . 0 f mm ( s a y 150 mm) . \n ’ ,( b *1000) ) ;

129

Scilab code Exa 13.8 Example 8 1 clc 2 clear 3 //DATA GIVEN 4 b =0.2; 5 t =0.01;

// w i d t h o f t h e b e l t i n m // t h i c k n e s s o f t h e b e l t

in m 6 Tp =2; t h e b e l t i n MN/mˆ2 7 ratio =1.8; 8 rho =1100; m a t e r i a l i n kg /mˆ3 9 10 11 12 13 14

// p e r m i s s i b l e t e n s i o n o f // r a t i o o f T1/T2 =1.8 // d e n s i t y o f t h e

Tmax = Tp *10^6* b * t ; //maximum s a f e t e n s i o n of the b e l t // we know c e n t r i f u g a l t e n s i o n , Tc=Tmax/3 Tc = Tmax /3; // c e n t r i f u g a l t e n s i o n , Tc=m∗ v ˆ2 m =( b * t ) *1* rho ; // mass o f t h e b e l t p e r u n i t metre l e n g t h v =( Tc / m ) ^0.5;

15 16 17 T1 = Tmax - Tc ;

// t e n s i o n i n t h e t i g h t

side in N // t e n s i o n i n t h e s l a c k

18 T2 = T1 / ratio ;

side in N // power t r a n s m i t t e d by

19 P =( T1 - T2 ) * v ;


printf ( ’ ( i ) The V e l o c i t y o f t h e b e l t i s : %3 . 1 f m/ s . \n ’ ,v ) ; 22 printf ( ’ ( i i ) The Maximum power t r a n s m i t t e d by t h e b e l t i s : %2 . 2 f kW. \n ’ ,( P /1000) ) ; 130

Scilab code Exa 13.9 Example 9 1 clc 2 clear 3 //DATA GIVEN 4 To =1000;

// i n i t i a l

belt in N theta =150; degrees 6 mu =0.25; friction 7 v =500; / min

// a n g l e o f embrace i n

5

8 9 10 11 12 13 14 15

tension in the

// c o e f f i c i e n t o f // s p e e d o f t h e b e l t i n m

// I n i t i a l t e n s i o n , To=(T1+T2 ) /2 // so , ( T1+T2 ) = 2 0 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 ) theta = theta *( %pi ) /180; // t h e t a c o n v e r t e d i n t o radians c = %e ^( mu * theta ) ; // so , T2/T1=c . . . . . . . . ( 2 ) //By e q u a t i o n ( 1 ) and ( 2 ) , T2 =( To *2) /( c +1) ; // t e n s i o n i n t h e s l a c k side in N T1 = c * T2 ; // t e n s i o n i n t h e t i g h t side in N

16 17 v = v /60;

// s p e e d o f t h e b e l t

c o n v e r t e d i n t o m/ s 18 P =( T1 - T2 ) * v ; the b e l t in watts

// power t r a n s m i t t e d by

19 20

printf ( ’ ( i ) The T e n s i o n i n t h e t i g h t s i d e T1 i s : %4 . 0 f N . \n ’ , T1 ) ; 21 printf ( ’ The T e n s i o n i n t h e s l a c k s i d e T2 i s : %3 . 1 f N . \n ’ , T2 ) ; 131

22

printf ( ’ ( i i ) The Power t r a n s m i t t e d by t h e b e l t i s : %2 . 2 f kW. \n ’ ,( P /1000) ) ;

Scilab code Exa 13.10 Example 10 1 clc 2 clear 3 //DATA GIVEN 4 P =400; 5 6 7 8 9

//maximum v a l u e o f f o r c e t h a t can be d e v e l o p e d i n N mu =0.25; // c o e f f i c i e n t o f friction d =0.6; // d i a m e t e r o f drum i n m // R e f e r t h e f i g u r e theta =180+45; // a n g l e o f c o n t a c t i n degrees theta = theta *( %pi ) /180; // t h e t a c o n v e r t e d i n t o radians

10 11 // moments a b o u t A, Ma=0 , 12 T1 = P *1/0.5; 13 14 // ( i ) Drum i s r o t a t i n g a n t i c l o c k w i s e 15 //T1>T2 ( T1/T2 )=e ˆ (mu∗ t h e t a ) 16 T2 = T1 /( %e ^( mu * theta ) ) ; 17 Mcac =( T1 - T2 ) *( d /2) ; //maximum b r a k i n g

t o r q u e t h a t can be d e v e l o p e d i n N 18 19 // ( i ) Drum i s r o t a t i n g c l o c k w i s e 20 //T2>T1 ( T2/T1 )=e ˆ (mu∗ t h e t a ) 21 T2 = T1 *( %e ^( mu * theta ) ) ; 22 Mcc =( T2 - T1 ) *( d /2) ; //maximum b r a k i n g

t o r q u e t h a t can be d e v e l o p e d i n N 23 24

printf ( ’ ( i ) The Maximum b r a k i n g t o r q u e t h a t can be 132

d e v e l o p e d i n a n t i c l o c k w i s e d i r e c t i o n i s : %3 . 0 f Nm . \n ’ , Mcac ) ; 25 printf ( ’ ( i i ) The Maximum b r a k i n g t o r q u e t h a t can be d e v e l o p e d i n c l o c k w i s e d i r e c t i o n i s : %3 . 1 f Nm. \ n ’ , Mcc ) ;

Scilab code Exa 13.11 Example 11 1 clc 2 clear 3 //DATA GIVEN 4 Pt =80; 5 6 7 8 9 10

// power t o be t r a n s m i t t e d by t h e r o p e i n kW d =1.5; // d i a m e t e r o f p u l l e y i n m N =200; // s p e e d o f t h e p u l l e y i n R . P .M. alpha =45/2; // s e m i a n g l e o f g r o o v e in degrees theta =160; // a n g l e o f c o n t a c t i n degrees mu =0.3; // c o e f f i c i e n t o f friction m =0.6; // mass o f e a c h r o p e p e r u n i t metre l e n g t h Ts =800; // s a f e p u l l i n N

11 12 13 // c e n t r i f u g a l t e n s i o n , Tc=m∗ v ˆ2 14 v =( %pi ) * d * N /60;

// v e l o c i t y

o f t h e r o p e i n m/ s 15 Tc = m * v ^2; 16 17 T1 = Ts - Tc ; 18

// t e n s i o n i n

the t i g h t s i d e in N // ( T1/T2 )=e ˆ (mu∗ t h e t a ) 133

19

theta = theta *( %pi ) /180; converted into radians 20 alpha = alpha *( %pi ) /180; converted into radians 21 T2 = T1 /( %e ^( mu * theta / sin ( alpha ) ) ) ; the s l a c k s i d e in N 22 p =( T1 - T2 ) * v ; t r a n s m i t t e d by t h e b e l t i n w a t t s

// t h e t a // a l p h a // t e n s i o n on // power

23 24

// no . o f r o p e s r e q u i r e d , n=T o t a l power t r a n s m i t t e d / Power t r a n s m i t t e d by e a c h r o p e 25 n = Pt /( p /1000) ; 26 27 // I n i t i a l t e n s i o n i n r o p e , To=(T1+T2+2Tc ) /2 28 To =( T1 + T2 +2* Tc ) /2; 29 30 printf ( ’ ( i ) The Number o f r o p e s r e q u i r e d f o r t h e

d r i v e s i s : %1 . 1 f s a y %1 . 0 f . \n ’ ,n , n ) ; 31 printf ( ’ ( i i ) The I n i t i a l t e n s i o n i n t h e r o p e , To i s : %3 . 2 f N . \n ’ , To ) ;


clc clear //DATA GIVEN T =72; Pc =26;

// number o f t e e t h // c i r c u l a r p i t c h i n mm

// c i r c u l a r p i t c h , Pc=( p i ∗D) /T D = Pc * T /( %pi ) ; // p i t c h d i a m e t e r i n m // Pc ∗Pd=( p i ) Pd =( %pi ) / Pc ; // d i a m e t r a l p i t c h i n t e e t h /mm 11 // Module , m=D/T 134

12 m = D / T ; // module i n mm/ t o o t h 13 14 printf ( ’ ( i ) The P i t c h diameterm , D i s : %3 . 2 f mm. \n

’ ,D ) ; printf ( ’ ( i i ) The D i a m e t r a l p i t c h , Pd i s : %1 . 2 f t e e t h /mm. \n ’ , Pd ) ; 16 printf ( ’ ( i i i ) The Module ,m i s : %1 . 2 f mm/ t o o t h . \n ’ ,m ); 15

Scilab code Exa 13.13 Example 13 1 clc 2 clear 3 //DATA GIVEN 4 Ta =40;

// number o f t e e t h o f

gear A // number o f t e e t h o f

5 Tb =100; 6 7 8 9 10

gear B Tc =50; gear C Td =150; gear D Te =52; gear E Tf =130; gear F Na =1000; s h a f t i n R . P .M.

// number o f t e e t h o f // number o f t e e t h o f // number o f t e e t h o f // number o f t e e t h o f // s p e e d o f t h e motor

11 12 // ( Nf /Na ) =(Ta/Tb ) ∗ ( Tc/Td ) ∗ ( Te/ Tf ) 13 Nf =( Ta / Tb ) *( Tc / Td ) *( Te / Tf ) * Na ;

// Speed o f t h e

o u t p u t s h a f t i n R . P .M. 14 15

printf ( ’ The Speed o f t h e o u t p u t s h a f t , Nf i s : %3 . 2 f R . P .M. \n ’ , Nf ) ; 135

136

Elements of Mechanical Engineering - R. K. Rajput

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