elements of mechanical engineering - r. k. rajput
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7/21/2019 Elements of Mechanical Engineering - R. K. Rajput
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Scilab Textbook Companion for
Elements of Mechanical Engineering
by R. K. Rajput1
Created byVatsal Shah
B.TECHMechanical Engineering
Institute of Technology,Nirma UniversityCollege Teacher
NoneCross-Checked byBhavani Jalkrish
September 25, 2014
1Funded by a grant from the National Mission on Education through ICT,http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilabcodes written in it can be downloaded from the ”Textbook Companion Project”section at the website http://scilab.in
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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
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Scilab numbering policy used in this document and the relation to theabove 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 particularExample of the above book)
For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 meansa 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 2.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . 8Exa 2.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . 9Exa 3.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . 11Exa 3.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . 11Exa 3.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . 12Exa 3.4 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . 13Exa 3.5 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . 14Exa 3.6 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . 14Exa 3.8 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . 15Exa 3.10 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . 17Exa 3.11 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . 18Exa 3.12 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . 20Exa 3.13 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . 21
Exa 4.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . 23Exa 4.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . 23Exa 4.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . 24Exa 4.4 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . 26Exa 4.5 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . 27Exa 4.6 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . 28Exa 4.7 Example 7 . . . . . . . . . . . . . . . . . . . . . . . . 29Exa 4.8 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . 30Exa 4.9 Example 9 . . . . . . . . . . . . . . . . . . . . . . . . 30Exa 4.10 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . 32Exa 4.11 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . 34
Exa 4.12 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . 35Exa 4.13 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . 36Exa 4.14 Example 14 . . . . . . . . . . . . . . . . . . . . . . . . 37Exa 4.15 Example 15 . . . . . . . . . . . . . . . . . . . . . . . . 38
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Exa 4.16 Example 16 . . . . . . . . . . . . . . . . . . . . . . . . 38
Exa 4.17 Example 17 . . . . . . . . . . . . . . . . . . . . . . . . 39Exa 4.18 Example 18 . . . . . . . . . . . . . . . . . . . . . . . . 40Exa 4.19 Example 19 . . . . . . . . . . . . . . . . . . . . . . . . 41Exa 4.20 Example 20 . . . . . . . . . . . . . . . . . . . . . . . . 42Exa 4.21 Example 21 . . . . . . . . . . . . . . . . . . . . . . . . 43Exa 4.22 Example 22 . . . . . . . . . . . . . . . . . . . . . . . . 44Exa 4.23 Example 23 . . . . . . . . . . . . . . . . . . . . . . . . 45Exa 4.24 Example 24 . . . . . . . . . . . . . . . . . . . . . . . . 46Exa 4.25 Example 25 . . . . . . . . . . . . . . . . . . . . . . . . 47Exa 4.26 Example 26 . . . . . . . . . . . . . . . . . . . . . . . . 48Exa 4.27 Example 27 . . . . . . . . . . . . . . . . . . . . . . . . 49
Exa 5.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . 51Exa 5.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . 53Exa 5.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . 54Exa 5.4 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . 55Exa 5.5 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . 57Exa 5.6 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . 58Exa 5.7 Example 7 . . . . . . . . . . . . . . . . . . . . . . . . 60Exa 5.8 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . 61Exa 5.9 Example 9 . . . . . . . . . . . . . . . . . . . . . . . . 62Exa 5.10 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . 63
Exa 5.11 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . 65Exa 5.12 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . 66Exa 5.13 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . 67Exa 5.14 Example 14 . . . . . . . . . . . . . . . . . . . . . . . . 68Exa 5.15 Example 15 . . . . . . . . . . . . . . . . . . . . . . . . 69Exa 5.16 Example 16 . . . . . . . . . . . . . . . . . . . . . . . . 70Exa 5.17 Example 17 . . . . . . . . . . . . . . . . . . . . . . . . 73Exa 5.18 Example 18 . . . . . . . . . . . . . . . . . . . . . . . . 74Exa 5.19 Example 19 . . . . . . . . . . . . . . . . . . . . . . . . 75Exa 5.20 Example 20 . . . . . . . . . . . . . . . . . . . . . . . . 76Exa 5.21 Example 21 . . . . . . . . . . . . . . . . . . . . . . . . 76
Exa 6.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . 80Exa 6.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . 81Exa 6.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . 82Exa 6.4 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . 83Exa 6.5 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . 84
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Exa 6.6 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . 85
Exa 6.7 Example 7 . . . . . . . . . . . . . . . . . . . . . . . . 86Exa 6.8 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . 88Exa 6.9 Example 9 . . . . . . . . . . . . . . . . . . . . . . . . 89Exa 6.10 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . 90Exa 6.11 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . 91Exa 6.12 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . 92Exa 6.13 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . 94Exa 6.14 Example 14 . . . . . . . . . . . . . . . . . . . . . . . . 95Exa 7.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . 96Exa 7.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . 97Exa 7.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . 97
Exa 7.4 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . 98Exa 7.5 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . 98Exa 7.6 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . 99Exa 7.7 Example 7 . . . . . . . . . . . . . . . . . . . . . . . . 101Exa 7.8 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . 102Exa 7.9 Example 9 . . . . . . . . . . . . . . . . . . . . . . . . 103Exa 7.10 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . 104Exa 7.11 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . 105Exa 7.12 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . 107Exa 7.13 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . 108
Exa 7.14 Example 14 . . . . . . . . . . . . . . . . . . . . . . . . 109Exa 7.15 Example 15 . . . . . . . . . . . . . . . . . . . . . . . . 110Exa 7.16 Example 16 . . . . . . . . . . . . . . . . . . . . . . . . 111Exa 7.17 Example 17 . . . . . . . . . . . . . . . . . . . . . . . . 113Exa 10.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . 115Exa 10.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . 116Exa 10.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . 117Exa 10.4 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . 118Exa 10.5 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . 119Exa 10.6 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . 121Exa 13.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . 123
Exa 13.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . 124Exa 13.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . 124Exa 13.4 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . 125Exa 13.5 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . 126Exa 13.6 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . 127
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Exa 13.7 Example 7 . . . . . . . . . . . . . . . . . . . . . . . . 128
Exa 13.8 Example 8 . . . . . . . . . . . . . . . . . . . . . . . . 130Exa 13.9 Example 9 . . . . . . . . . . . . . . . . . . . . . . . . 131Exa 13.10 Example 10 . . . . . . . . . . . . . . . . . . . . . . . . 132Exa 13.11 Example 11 . . . . . . . . . . . . . . . . . . . . . . . . 133Exa 13.12 Example 12 . . . . . . . . . . . . . . . . . . . . . . . . 134Exa 13.13 Example 13 . . . . . . . . . . . . . . . . . . . . . . . . 135
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Chapter 2
Fuels and Combustion
Scilab code Exa 2.1 Example 1
1 clc
2 clear
3 //DATA GIVEN4 c = 8 8 ; //% o f c ar bo n i n c o a l5 h = 4 . 2 ; //% o f h yd ro ge n i n c o a l6 W f = 0 . 8 4 8 ; // w e ig ht o f c o al i n g
7 W f w = 0 . 0 2 7 ; // w ei gh t o f f u s e w ir e i nc a l o r i m e t e r i n g
8 W = 1 9 5 0 ; // w e ig h t o f w at er i nc a l o r i m e t e r i n g
9 W e = 3 8 0 ; // w at er e q u i v al e n t o f c a l o r i m e t e r
10 D t = 3 . 0 6 ; // o b s er v e d t e mp e r at u r e r i s e( t2−t 1 ) i n deg c e l s i u s
11 t c = 0 . 0 1 7 ; // c o o l i n g c o r r e c t i o n i n degc e l s i u s
12 c f w = 6 7 0 0 ; / / c a l o r i f i c v a l u e o f f u s ew ir e i n J / g13
14 //CALCULATIONS15 c t r = ( D t ) + t c ; // c o r r e c t e d temp . r i s e
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16 H w = ( W + W e ) * 4 . 1 8 * [ c t r ] ; // h ea t r e c i e v e d by w a te r i n
J17 H f w = W f w * c f w ; // h ea t g i ve n o ut by f u s ew i re i n J
18 H c f = H w - H f w ; / / h e a t p r od u ce d d ue t oc o m b us t i o n o f f u e l i n J
19 H C V = H c f / W f ; / / 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 M s = 9 * h / 1 0 0 ; // s te am p ro du ce d p e r kg o f c o a l
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
2223 printf ( ’ The H i g he 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 / k g . \n ’ , H C V ) ;
24 printf ( ’ The L ower 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 / k g . \n ’ , L C V ) ;
Scilab code Exa 2.2 Example 2
1 clc2 clear
3 //DATA GIVEN4 V 1 = 0 . 0 8 ; // g a s b ur nt i n c a l o r i m e t e r
i n m ˆ 35 P g = 5 . 2 ; // p r e s s ur e o f g as s up pl y i n
cm o f w at er6 P b = 7 5 . 5 ; / / b a ro m et er r e a d i n g i n cm
o f Hg7 W w = 2 8 ; / / w e ig h t o f w at er h e at e d by
g as i n kg8 T g = 1 3 ; // t em p er at u re o f g as i n degc e l s i u s
9 T w i = 1 0 ; / / t e mp e r at u r e o f w at er a ti n l e t i n deg c e l s i u s
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10 T w o = 2 3 . 5 ; / / t e mp e r at u r e o f w at er a t
o u t l e t i n deg c e l s i u s11 M s = 0 . 0 6 ; / / st ea m c o nd e ns e d i n kg12
13 //CALCULATIONS14 / / by u s i ng g e n e r a l g as e qu at io n , r e d uc i n g t he volume
t o S . T . P .15 //p1∗V1/T1=p2∗V2/T216 p 1 = P b + ( P g / 1 3 . 6 ) ; / / i n cm o f Hg17 T 1 = T g + 2 7 3 ; / / i n K18 p 2 = 7 6 ; / / i n cm o f Hg19 T 2 = 1 5 + 2 7 3 ; / / i n K
20 V 2 = p 1 * V 1 * T 2 / T 1 / p 2 ; // in mˆ321 H w = W w * 4 . 1 8 * ( T w o - T w i ) ; // h ea t r e c i e v e d by w at er i n
kJ22 H C V = H w / V 1 ; / / 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 /mˆ 323 LCV=HCV -2465* Ms/V1; / / 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 /mˆ324
25 printf ( ’ The C a l o r i f i c v al u e s o f f u e l p e r mˆ3 o f g a sa 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 re :
\n ’) ;
26 printf ( ’ The H i gh er 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 k J /mˆ 3 . \n ’ , H C V ) ;
27 printf ( ’ The L ow er 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 ’ , L C V ) ;
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Chapter 3
Properties of Gases
Scilab code Exa 3.1 Example 1
1 clc
2 clear
3 //DATA GIVEN4 Q = - 5 0 ; // h ea t r e j e c t e d t o
c o o l i n g w a te r i n kJ /kg5 W = - 1 0 0 ; // w ork i n p u t i n kJ /
kg6
7 / / u s i n g F i r s t Law o f T he rm od yn am ic s , Q=( u2−u1 )+W8 D u = Q - W ; / / ( u 2−u1 ) c ha ng e i n
i n t e r n a l e ne rg y i n kJ /kg9 // s i n c e Du i s +ve , t h er e i s g ai n i n i n t e r n a l e ne rg y
10
11 printf ( ’ The GAIN i n i n t e r n a l e n er g y i s : %2 . 0 f kJ / kg .\n ’ , D u ) ;
Scilab code Exa 3.2 Example 2
11
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1 clc
2 clear3 //DATA GIVEN4 u 1 = 4 5 0 ; // i n t e r n a l e ne rg y a t
b e gi n ni n g o f t he e xp an si on i n kJ /kg5 u 2 = 2 2 0 ; // i n t e r n a l e ne rg y
a f t e r e xp an si on i n kJ /kg6 W = 1 2 0 ; / / wor k d on e by t h e
a i r d ur in g e xp an si on i n kJ /kg7
8 / / u s i n g F i r s t Law o f T he rm od yn am ic s , Q=( u2−u1 )+W9 Q = ( u 2 - u 1 ) + W ; // h e at f l o w i n kJ / kg
10 // s i n c e Q i s −ve , t he r e i s r e j e c t i o n o f h e a t11
12 printf ( ’ T he h e a t REJECTED b y a i r i s : %3 . 0 f k J / k g . \n’ , ( - Q ) ) ;
Scilab code Exa 3.3 Example 3
1 clc
2 clear3 //DATA GIVEN4 m = 0 . 3 ; // mass o f n i t r o g e n
i n kg5 p 1 = 0 . 1 ; / / p r e s s u r e i n MPa6 T 1 = 4 0 + 2 7 3 ; / / t e m p e ra t u r e b e f o r e
c o mp r es s io n i n K7 p 2 = 1 ; / / p r e s s u r e i n MPa8 T 2 = 1 6 0 + 2 7 3 ; // t e mp e ra t u re a f t e r
c o mp r es s io n i n K
9 W = - 3 0 ; / / wor k d on e d u r i n gt he c o mp r es s io n i n kJ / kg10 C v = 0 . 7 5 // i n k J /kgK11
12 / / u s i n g F i r s t Law o f T he rm od yn am ic s , Q=( u2−u1 )+W
12
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13 / / ( u 2−u1 )=m∗Cv∗ ( T2−T1 )
14 D u = m * C v * ( T 2 - T 1 ) ;15 Q = D u + W ; // h e at f l o w i n kJ / kg16 // s i n c e Q i s −ve , t he r e i s r e j e c t i o n o f h e a t17
18 printf ( ’ T he h e a t REJECTED b y a i r i s : %1 . 0 f k J . \n ’, ( - Q ) ) ;
Scilab code Exa 3.4 Example 4
1 clc
2 clear
3 //DATA GIVEN4 // i n i t i a l s t a t e5 p 1 = 0 . 1 0 5 ; // p r e s s ur e o f g as i n
MPa6 V 1 = 0 . 4 ; // v olume o f g as i n m
ˆ37 // f i n a l s t a t e8 p 2 = 0 . 1 0 5 ; // p r e s s ur e o f g as i n
MPa9 V 2 = 0 . 2 0 ; // v olume o f g as i n m
ˆ310
11 Q = - 4 2 . 5 ; // h ea t t r a n s f e r r e di n kJ
12 p = p 1 ;
13
14 / / p r o c e s s u se d− ISOBARIC ( C onst an t pr e s su r e )15 W 1 2 = p * ( V 2 - V 1 ) * 1 0 0 0 ; // w ork i n kJ
16 / / u s i n g F i r s t Law o f T he rm od yn am ic s , Q=( u2−u1 )+W17 D u = Q - W 1 2 ; / / ( u 2−u1 ) c ha ng e i ni n t e r n a l e ne rg y i n kJ
18 / / s i n c e Du i s −ve , t h e re i s d e c r e a s e i n i n t e r n a le n e r g y
13
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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 k J. \n ’ , ( - D u ) ) ;
Scilab code Exa 3.5 Example 5
1 clc
2 clear
3 //DATA GIVEN
4 / / p a r t −15 // pr e s su r e =p1 , t e m pe r at u r e =T16 / / p a r t −27 // pr e s su r e =p2 , t e m pe r at u r e =T28
9 / / A cc . F i r s t Law o f T he rm o dy na mi cs , Q=( u2−u1 )+W10 / / when p a r t i t i o n moved11 D Q = 0 ;
12 D W = 0 ;
13 D U = D Q - D W ;
14 //DU=0
1516 printf ( ’ CONCLUSION : \n ’ ) ;
17 printf ( ’ Acc . t o F i r s t Law o f T hermodynamics , \n ’ ) ;
18 printf ( ’ When p ar t i o n moved , t h er e i sc o n s er v a t i o n o f i n t e r n a l e ne rg y . \n ’ ) ;
Scilab code Exa 3.6 Example 6
1 clc
2 clear
3 //DATA GIVEN4 // i n i t i a l s t a t e
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5 p 1 = 1 0 ^ 5 ; // i n i t i a l pr e s su r e
o f a i r i n Pa6 v 1 = 1 . 8 ; // v olume o f a i r i n mˆ 3 / k g
7 T 1 = 2 5 + 2 7 3 ; // i n i t i a lt em pe r at ur e o f a i r i n K
8 // f i n a l s t a t e9 p 2 = 5 * 1 0 ^ 5 ; // f i n a l p r e s s u re o f
a i r i n Pa10 T 2 = 2 5 + 2 7 3 ; // f i n a l t e mp e r at u re
o f a i r i n K11
12 / / p r o c e s s u se d− ISOTHERMAL ( Co ns ta nt te mp er at u re )13 W 1 2 = [ p 1 * v 1 * log ( p 1 / p 2 ) ] / 1 0 0 0 ; / / wor k i n kJ / kg14 / / s i n c e W i s −ve , work i s s u pp l ie d t o t h e a i r15
16 / / s i n c e t em p er a tu re i s c o n st a n t17 D u = 0 ; / / ( u 2−u1 ) c ha ng e i n
i n t e r n a l e ne rg y i n kJ /kg18
19 / / u s i n g F i r s t Law o f T he rm od yn am ic s , Q=( u2−u1 )+W20 Q = D u + W 1 2 ;
21 // s i n c e Q i s −ve , t he r e i s r e j e c t i o n o f h e a t froms ys te m t o s u r r ou n d i ng s
22
23 printf ( ’ ( i ) The Work d on e on t h e a i r i s : %3 . 1 f kJ /kg . \n ’ , ( - W 1 2 ) ) ;
24 printf ( ’ ( i i ) The c ha ng e i n i n t e r n a l e ne rg y i s : %1 . 0f k J / kg . \n ’ , ( D u ) ) ;
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
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2 clear
3 //DATA GIVEN4 p 1 = 4 * 1 0 ^ 5 ; // i n i t i a l pr e s su r ei n N/mˆ2
5 V 1 = 0 . 2 ; // i n i t i a l volume inmˆ3
6 T 1 = 1 3 0 + 2 7 3 ; // i n i t i a lt e mp e r at u r e i n K
7 p 2 = 1 . 0 2 * 1 0 ^ 5 ; // f i n a l p r e s su r ea f t e r a d i a b a t i c e xp a ns i on i n N/mˆ2
8 Q 2 3 = 7 2 . 5 ; // i n c r e a s e i ne nt ha l p y d ur in g c o ns t a n t p r e s su r e p r o c es s i n kJ
9 C p = 1 ; // i n k J /kgK10 C v = 0 . 7 1 4 ; // in kJ/khK11
12 / /gamma f o r a i r , g13 g = C p / C v ;
14 R = ( C p - C v ) * 1 0 0 0 ;
15
16 // f o r r e v e r s i b l e a d i a b a t i c p r o c e s s 1−217 //p1 ∗ (V1ˆg)=p2∗ ( V 2ˆg)18 V 2 = V 1 * ( p 1 / p 2 ) ^ ( 1 / g ) ; // f i n a l volume i n m
ˆ319 // ( T2/T1) =( p2/p1) ˆ( ( g−1) /g ) ;20 T 2 = T 1 * ( p 2 / p 1 ) ^ ( ( g - 1 ) / g ) ; ; // f i n a l temp . T2 i n
K21
22 m = p 1 * V 1 / R / T 1 ; // m ass i n kg23
24 / / f o r c o ns t an t p r e s su r e p r o c es s 2−325 //Q23=m∗Cp∗( T3−T2) ;26 T 3 = Q 2 3 / m / C p + T 2 ;
27 //V2/T2=V3/T3
28 V 3 = V 2 / T 2 * T 3 ;29
30 / /Work d on e by t h e p at h 1−2−3, W123=W12+W2331 W 1 2 = ( p 1 * V 1 - p 2 * V 2 ) / ( g - 1 ) ;
32 W 2 3 = p 2 * ( V 3 - V 2 ) ;
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33 W 1 2 3 = W 1 2 + W 2 3 ;
3435 // i f t he abov e p r o c e s s e s a re 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 oc e ss g i v i n g t h e samewo rk b et we en 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 p 3 = p 2 ;
38 n = 1 + ( p 1 * V 1 - p 3 * V 3 ) / W 1 2 3 ; / / i n d e x o f e x p an s i o n, n
39
40 printf ( ’ ( i ) The T o t a l Work d on e i s : %5 . 0 f Nm o r J .\n ’ , W 1 2 3 ) ;
41 printf ( ’ ( i i ) The v al ue o f i nd ex o f e xp an si on , n i s :%1. 3 f . \n ’ , n ) ;
42
43 //NOTE:44 / / 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 ns we r s o f t he book
due t o r ou nd in g o f f o f t he v a lu e s
Scilab code Exa 3.10 Example 10
1 clc
2 clear
3 //DATA GIVEN4 // i n i t i a l s t a t e5 p 1 = 1 0 ^ 5 ; // i n i t i a l pr e s su r e
o f g a s i n Pa6 V 1 = 0 . 4 5 ; // i n i t i a l volume o f
g a s i n mˆ 37 T 1 = 8 0 + 2 7 3 ; // i n i t i a l
t em pe ra tu re o f g as i n K8 // f i n a l s t a t e9 p 2 = 5 * 1 0 ^ 5 ; // f i n a l p r e s s u re o f
g as i n Pa10 V 2 = 0 . 1 3 ; // f i n a l volume o f
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g a s i n mˆ 3
1112 / /gamma f o r a i r , g13 g = 1 . 4 ;
14 R = 2 9 4 . 2 //J/kgK15
16 m = p 1 * V 1 / R / T 1 ; // m ass i n kg17
18 //p1 ∗ (V1ˆn)=p2 ∗ (V2ˆn)19 n = log ( p 1 / p 2 ) / log ( V 2 / V 1 ) ; / / i n d e x n20
21 / / I n a p o l y t r o p i c p r oc e ss
22 // (T2/T1) =(V1/V2) ˆ( n−1) ;23 T 2 = T 1 * ( V 1 / V 2 ) ^ ( n - 1 ) ; / / temp . T2 i n K24
25 C v = R / ( g - 1 ) ;
26 D u = m * C v * ( T 2 - T 1 ) / 1 0 0 0 ; // i n c r e a s e i ni n t e r n a l e ne rg y i n kJ
27
28 / / u s i n g F i r s t Law o f T he rm od yn am ic s , Q=( u2−u1 )+W29 //W12=(p1∗V1−p2 ∗V2) /( n−1)=mR( T2−T1) /( n−1)30 W 1 2 = m * R * ( T 1 - T 2 ) / ( n - 1 ) / 1 0 0 0 ;
31 Q = D u + W 1 2 ;
32 // s i n c e Q i s −ve , t he r e i s r e j e c t i o n o f h e a t froms ys te m t o s u r r ou n d i ng s
33
34 printf ( ’ ( i ) The Mass o f t he g as i s : %1 . 3 f kg . \n ’ ,(
m ) ) ;
35 printf ( ’ ( i i ) The i n d ex n i s : %1 . 3 f . \n ’ ,(n)) ;
36 printf ( ’ ( i i i ) The c ha ng e i n i n t e r n a l e n er g y i s : %2 . 1f kJ . \n ’ , ( D u ) ) ;
37 printf ( ’ ( iv ) The Heat REJECTED i s : %2. 2 f kJ . \n ’ ,( -
Q ) ) ;
Scilab code Exa 3.11 Example 11
18
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1 clc
2 clear3 //DATA GIVEN4 // i n i t i a l s t a t e5 p 1 = 1 . 0 2 ; // i n i t i a l pr e s su r e
o f a i r i n ba r6 V 1 = 0 . 0 1 5 ; // i n i t i a l volume o f
a i r i n mˆ37 T 1 = 2 2 + 2 7 3 ; // i n i t i a l
t em pe r at ur e o f a i r i n K8 // f i n a l s t a t e9 p 2 = 6 . 8 ; // f i n a l p r e s s u re o f
a i r i n ba r10 / / Law o f a d i a b a t i c c o m p r e s s i o n , pVˆ g=C11
12 / /gamma f o r a i r , g13 g = 1 . 4
14 R = 0 . 2 8 7 ;
15
16 / / I n a a d i a b a t i c p r o c e s s17 // ( T2/T1) =( p2/p1) ˆ( ( g−1) /g ) ;18 T 2 = T 1 * ( p 2 / p 1 ) ^ ( ( g - 1 ) / g ) ; ; // f i n a l temp . T2 i n
K19
20 //p1 ∗ (V1ˆg)=p2∗ ( V 2ˆg)21 V 2 = V 1 * ( p 1 / p 2 ) ^ ( 1 / g ) ; // f i n a l volume i n m
ˆ322
23 m = p 1 * 1 0 ^ 5 * V 1 / 1 0 ^ 3 / R / T 1 ; // m ass i n kg24
25 //W=(p1∗V1−p2∗V2) /( g−1)=mR( T2−T1) /( g−1)26 W = m * R * ( T 1 - T 2 ) / ( g - 1 ) ;
27 / / s i n c e W i s −ve , t he work i s done on t he a i r
2829 printf ( ’ ( i ) The F i n a l t e m pe r a tu r e i s : %3 . 2 f d eg .
c e l s i u s . \n ’ , ( T2 - 2 7 3 ) ) ;
30 printf ( ’ ( i i ) The F i n a l Volume i s : %1 . 5 f mˆ 3 . \n ’ , V2
) ;
19
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31 printf ( ’ ( i i i ) The Work d on e 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 GIVEN4 m = 0 . 4 4 ; // mass o f a i r i n kg
5 T 1 = 1 8 0 + 2 7 3 ; // i n i t i a lt em pe r at ur e o f a i r i n K6 T 2 = 1 5 + 2 7 3 ; // f i n a l t e mp e r at u re
o f a i r i n K7 W 1 2 = 5 2 . 5 ; / / wor k d on e d u r i n g
t h e p r oc e ss i n kJ8 //V2/V1=39 V r = 3 ; / / v ol um e r a t i o , Vr=
V2/V110
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
1213 // I n an a d i a b a t i c p r o c e s s14 // (T2/T1) =(V1/V2) ˆ( g−1) ;15 g = 1 + [ ( log ( T 2 / T 1 ) / log ( 1 / V r ) ) ] ; //gamma
f o r a i r , g=Cp /Cv16
17 //W12=(p1∗V1−p2 ∗V2) /( n−1)=mR( T2−T1) /( g−1)18 R = W 1 2 / m / ( T 1 - T 2 ) * ( g - 1 ) ;
19 //R=Cp−Cv20
21 C v = R / ( g - 1 ) ;22 C p = g * C v ;
23
24 printf ( ’ ( i ) The v a l u e o f Cv i s : %1 . 3 f kJ /kgK . \n ’ ,
C v ) ;
20
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25 printf ( ’ ( i i ) The v a l u e o f Cp i s : %1 . 3 f kJ /kgK . \n ’ ,
C p ) ;26
27 //NOTE:28 / / 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 ns we r s o f t he book
due t o r ou nd in g o f f o f t he v a lu e s
Scilab code Exa 3.13 Example 13
1 clc2 clear
3 //DATA GIVEN4 m = 1 ; // mass o f e ta hn e g as
i n kg5 M = 3 0 ; / / m o l e c u l a r w e i g ht
o f e th a ne6 p 1 = 1 . 1 ; // i n i t i a l pr e s su r e
i n b ar7 T 1 = 2 7 + 2 7 3 ; // i n i t i a l
t e mp e r at u r e i n K
8 p 2 = 6 . 6 ; // f i n a l p r e s s u re i nb ar
9 C p = 1 . 7 5 ; // i n k J /kgK10
11 / / Law o f c o m p r e s s i o n , pV ˆ 1. 3 =C12 n = 1 . 3 ;
13
14 // C h a r a c t e r i s t i c g as c on st an t , R = U n i ve r sa l g asc o n s t a n t ( Ro ) / M o l e c u l a r w e i g ht (M)
15 R o = 8 3 1 4 ;
16 R = R o / M / 1 0 0 0 ; //kJ/kgK17
18 //R=Cp−Cv19 C v = C p - R ;
20 g = C p / C v ; //gamma g
21
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21
22 / / I n a p o l y t r o p i c p r oc e ss23 // ( T2/T1) =( p2/p1) ˆ( ( n−1)/n) ;24 T 2 = T 1 * ( p 2 / p 1 ) ^ ( ( n - 1 ) / n ) ; ; // f i n a l temp . T2 i n
K25
26 //W=(p1∗V1−p2∗V2) /( n−1)=mR( T2−T1) /( g−1)27 W = m * R * ( T 1 - T 2 ) / ( n - 1 ) ;
28
29 Q = [ ( g - n ) / ( g - 1 ) ] * W ; // h e at f l o w i n kJ / kg30
31 printf ( ’ The H e at SUPPLIED i s : %2. 1 f k J /k g . \n ’ ,(Q))
;
22
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Chapter 4
Properties of Steam
Scilab code Exa 4.1 Example 1
1 clc
2 clear
3 //DATA GIVEN4 M s = 5 0 ; / / mass o f d ry s te am
i n kg5 M w = 1 . 5 ; // mass o f w at er i n
s u s pe n si o n i n kg6
7 / / d r y n e s s f r a c t i o n , x =( ma ss o f d ry s te am ) / ( m as s o f d ry s te am +mass o f w at er i n s u s p e n s i o n )
8 x = M s / ( M s + M w ) ;
9
10 printf ( ’ The D ry ne ss f r a c t i o n ( Q u al i ty ) o f stea m i s :%1. 3 f . ’ , x ) ;
Scilab code Exa 4.2 Example 2
1 clc
23
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2 clear
3 //DATA GIVEN4 V = 0 . 6 ; / / vo lu me o f t h ev e s s e l i n mˆ3
5 p = 0 . 5 ; // p r e s s u r e i n b ar6 M = 3 ; // mass o f l i q u i d and
w at er v ap ou r i n kg7
8 v = V / M ; // s p e c i f i c volume inmˆ3/kg
9 / / At 5 b ar , f ro m s te am t a b l e s10 v g = 0 . 3 7 5 ; //mˆ3/kg
11 v f = 0 . 0 0 1 0 9 ; //mˆ3/kg12 v f g = v g - v f ;
13 //v=vg−(1−x ) v f g14 x = ( v - v g ) / v f g + 1 ; // q u a l i t y o f t he
v apour15
16 / / mass and volume o f l i q u i d17 M l i q = M * ( 1 - x ) ;
18 V l i q = M l i q * v f ;
19
20 / / m as s a nd v ol um e o f v ap o ur21 M v a p = M * x ;
22 V v a p = M v a p * v g ;
23
24 printf ( ’ ( i ) The Mass and Volume o f l i q u i d i s : \n ’ ) ;
25 printf ( ’ Ml i q . i s : %1 . 3 f kg . \n ’ , M l i q ) ;
26 printf ( ’ V l i q . i s : %1 . 4 f mˆ 3 . \n ’ , V l i q ) ;
27 printf ( ’ ( i i ) The Ma ss a nd Volu me o f v a p ou r i s : \n ’ ) ;
28 printf ( ’ Mvap . i s : %1 . 3 f kg . \n ’ , M v a p ) ;
29 printf ( ’ Vvap . i s : %1 . 4 f mˆ 3. \n ’ , V v a p ) ;
Scilab code Exa 4.3 Example 3
24
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1 clc
2 clear3 //DATA GIVEN4 V = 0 . 0 5 ; // volume o f v e s s e l i n
mˆ35 M f = 1 0 ; // mass o f l i q u i d i n
kg6 T = 2 4 5 ; / / temp . i n d eg
c e l s i u s7
8 / / f rom stea m t a b l e s , c o r r es p o n d i ng t o 2 45 degc e l s i u s
9 P s a t = 3 6 . 5 ; / / b a r10 v f = 0 . 0 0 1 2 3 9 ; //mˆ3/kg11 v g = 0 . 0 5 4 6 ; //mˆ3/kg12 h f = 1 0 6 1 . 4 ; / / k J / k g13 h f g = 1 7 4 0 . 2 ; / / k J / k g14 s f = 2 . 7 4 7 4 ; //kJ/kgK15 s f g = 3 . 3 5 8 5 ; //kJ/kgK16
17 V f = M f * v f ; // v olume o f l i q u i d18 V g = V - V f ; / / v ol um e o f v ap o ur19 M g = V g / v g ;
/ / ma ss o f v a po ur20 m = M f + M g ; // t o t a l mass o f m i x t u r e
21
22 x = M g / ( M g + M f ) ; // q u a l i t y o f t hem i x t u r e
23 v f g = v g - v f ;
24 v = v f + x * v f g ; // s p e c i f i c volume25
26 h = h f + x * h f g ; // s p e c i f i c en th al py27
28 s = s f + x * s f g ; // s p e c i f i c en tr op y29
30 u = h - P s a t * 1 0 ^ 5 * v / 1 0 ^ 3 ; / / s p e c i f i c i n t e r n a le n e r g y
31
25
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32
33 printf ( ’ ( i ) The P r e s s u r e i s : %2 . 1 f b ar . \n ’ , P s a t ) ;34 printf ( ’ ( i i ) The ma ss m i s : %2 . 3 f k g . \n ’ , m ) ;
35 printf ( ’ ( i i i ) The S p e c i f i c v ol um e v i s : %1 . 6 f mˆ 3/ k g. \n ’ , v ) ;
36 printf ( ’ ( i v ) The S p e c i f i c e nt ha 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 tr o py 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 ne rg y u i s : %4. 2 f k J / kg . \n ’ , u ) ;
39
40 //NOTE:41 / / 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 ns we r s o f book due
t o r o un di ng o f f o f t h e v a l u es i n t he book
Scilab code Exa 4.4 Example 4
1 clc
2 clear
3 //DATA GIVEN4 M w = 2 ; // mass o f w at er t o be
c o nv e rt e d t o s team i n kg5 T w = 2 5 ; / / temp . o f w at er i n
deg c e l s i u s6 p = 5 ; / / p r e s s u r e7 x = 0 . 9 ; // d r yn e s s f r a c t i o n8
9 / / At 5 b ar , f ro m s te am t a b l e s10 h f = 6 4 0 . 1 ; / / k J / k g
11 h f g = 2 1 0 7 . 4 ; / / k J / k g12
13 h = h f + x * h f g ; / / s p e c i f i c e n t h a l p y (abov e 0 deg c e l s i u s )
14 h s = 1 * 4 . 1 8 * ( T w - 0 ) ; // s e n s i b l e h ea t
26
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a s s o c i a t e d w i t h i kg o f w a t e r
15 h n e t = h - h s ; // n et q u a nt i t y o f h ea t t o be s u pp l i e d p er kg o f w a t e r16 H t o t a l = M w * h n e t ; / / t o t a l amount o f
h ea t t o be s u p p l i e d17
18 printf ( ’ The T ot al amount o f h ea t t o be s u p p l i e d i s :%4. 2 f k J . ’ , H t o t a l ) ;
Scilab code Exa 4.5 Example 5
1 clc
2 clear
3 //DATA GIVEN4 m = 4 . 4 ; // mass o f s tea m t o be
p ro du ce d i n kg5 p = 6 ; / / p r e s s u r e o f s te am6 T s u p = 2 5 0 ; / / temp . o f s te am i n
deg . c e l s i u s7 T w = 3 0 ; / / temp . o f w at er i n
deg c e l s i u s8 C p s = 2 . 2 ; // s p e c i f i c he at of
s te am i n kJ / kg9
10 / / At 6 b ar , f ro m s te am t a b l e s11 T s = 1 5 8 . 8 ; // d eg . c e l s i u s12 h f = 6 7 0 . 4 ; / / k J / k g13 h f g = 2 0 8 5 ; / / k J / k g14 // s i n c e t he g i v en temp . 250 deg c e l s i u s i s g r e a t e r
t han 1 5 8. 8 deg c e l s i u s , steam i s s up e r he at e d
1516 h s u p = h f + h f g + C p s * ( T s u p - T s ) ; // e n th a lp y o f 1 kgs u p er g e at e d stea m r ec ko n ed from 0 d eg . c e l s i u s
17 h s = 1 * 4 . 1 8 * ( T w - 0 ) ; // s e n s i b l e h ea ta s s o c i a t e d w i t h i kg o f w a t e r
27
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18 h n e t = h s u p - h s ; // n et q u a nt i t y o f
h ea t t o be s u pp l i e d p er kg o f w a t e r19 H t o t a l = m * h n e t ; / / t o t a l amount o f h ea t t o be s u p p l i e d
20
21 printf ( ’ The T ot al amount o f h ea t t o be s u p p l i e d i s :%4. 1 f k J . ’ , H t o t a l ) ;
Scilab code Exa 4.6 Example 6
1 clc
2 clear
3 //DATA GIVEN4 V = 0 . 1 5 ; / / v ol um e o f w et s te am
i n m ˆ 35 p = 4 ; // p r e s s u r e o f wet
stea m i n b ar6 x = 0 . 8 ; // d r yn e s s f r a c t i o n7
8 / / At 4 b ar , f ro m s te am t a b l e s
9 v g = 0 . 4 6 2 ; //mˆ3/kg10 h f = 6 0 4 . 7 ; / / k J / k g11 h f g = 2 1 3 3 ; / / k J / k g12
13 r h o = 1 / ( x * v g ) ; / / d e n s i t y i n kg /mˆ 314 m = r h o * V ; // mass o f 0 . 1 5 mˆ3 o f
steam15
16 H t o t a l = ( r h o * 1 ) * ( h f + x * h f g ) ; // t o t a l h ea t o f 1 mˆ3o f s team whi ch h as a mass o f r ho ( 2 . 7 0 5 6 ) kg
1718 printf ( ’ ( i ) The Mass o f 0 . 1 5 mˆ 3 o f s te am 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 mˆ 3 o f s te am w hi chh as a mass o f 2 . 7 05 6 kg i s : %4 . 2 f kJ . \n ’ , H t o t a l )
28
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;
Scilab code Exa 4.7 Example 7
1 clc
2 clear
3 //DATA GIVEN4 m = 1 0 0 0 ; / / ma ss o f s te am
g e n er a t ed i n kg / h r
5 p = 1 6 ; / / p r e s s u r e o f s te ami n b ar6 x = 0 . 9 ; // d r yn e s s f r a c t i o n7 T s u p = 3 8 0 + 2 7 3 ; / / t em p . o f
s u p e r he a t e d s te am i n K8 T f w = 3 0 ; / / temp . o f f e e d w at er
i n deg . c e l s i u s9 C p s = 2 . 2 ; // s p e c i f i c he at of
s te am i n kJ / kg10
11 / /At 1 6 b ar , f ro m s te am t a b l e s
12 T s = 2 0 1 . 4 + 2 7 3 ; / / i n K13 h f = 8 5 8 . 6 ; / / k J / k g14 h f g = 1 9 3 3 . 2 ; / / k J / k g15
16 H s = m * [ ( h f + x * h f g ) - 1 * 4 .1 8 7 * ( T fw - 0 ) ] ; // hea ts u p p li e d t o f e ed w at er p er hr t o p ro du ce wetsteam
17 H a = m * [ ( 1 - x ) * h f g + C p s * ( T s u p - T s ) ] ; // hea ta bs o rb ed by s u p e r h ea t e r p er h ou r
18
19 printf ( ’ ( i ) The Heat s u p p l i e d t o f e e d w at er p er h ou rt o p r od u ce wet s te am i s : %4 . 2 f ∗1 0 ˆ 3 k J . \n ’ ,(Hs
/ 1 0 0 0 ) ) ;
20 printf ( ’ ( i i ) The H ea t a b s o r be d by s u p e r h e a t e r p e rh ou r i s : %3 . 2 f ∗1 0 ˆ 3 k J . \n ’ , ( H a / 1 0 0 0 ) ) ;
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Scilab code Exa 4.8 Example 8
1 clc
2 clear
3
4 / / At 0 . 7 5 b a r . From s te am t a b l e s ,5 / /At 1 00 deg c e l s i u s6 T 1 = 1 0 0 ; // deg c e l s i u s
7 h s u p 1 = 2 6 7 9 . 4 ; / / k J / k g8 / /At 1 50 deg c e l s i u s9 T 2 = 1 5 0 ; // deg c e l s i u s
10 h s u p 2 = 2 7 7 8 . 2 ; / / k J / k g11 C p s 1 = ( h s u p 2 - h s u p 1 ) / ( T 2 - T 1 ) ;
12
13 / / At 0 . 5 b a r . From s te am t a b l e s ,14 / /At 3 00 deg c e l s i u s15 T 3 = 3 0 0 ; // deg c e l s i u s16 h s u p 3 = 3 0 7 5 . 5 ; / / k J / k g17 / /At 4 00 deg c e l s i u s18 T 4 = 4 0 0 ; // deg c e l s i u s19 h s u p 4 = 3 2 7 8 . 9 ; / / k J / k g20 C p s 2 = ( h s u p 4 - h s u p 3 ) / ( T 4 - T 3 ) ;
21
22 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 dsteam \n ( At 0 . 7 5 bar , b e tw e en 100 and 150deg c e l s i u s ) i s : %1 . 3 f . \n ’ , C p s 1 ) ;
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 dsteam \n ( At 0 .5 bar , b e tw e en 300 and 400
deg c e l s i u s ) i s : %1 . 3 f . \n ’ , C p s 2 ) ;
Scilab code Exa 4.9 Example 9
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1 clc
2 clear3 //DATA GIVEN4 m = 1 . 5 ; / / mass o f s te am i n
c oo k e r i n kg5 p 1 = 5 ; / / p r e s s u r e o f s te am
i n b ar6 x 1 = 1 ; // i n i t i a l dr yn es s
f r a c t i o n o f steam7 x 2 = 0 . 6 ; // f i n a l d r yn e ss
f r a c t i o n o f steam8
9 / / At 5 b ar , f ro m s te am t a b l e s10 T s 1 = 1 5 1 . 8 + 2 7 3 ; / / i n K11 h f 1 = 6 4 0 . 1 ; / / k J / k g12 h f g 1 = 2 1 0 7 . 4 ; / / k J / k g13 v g 1 = 0 . 3 7 5 ; //mˆ3/kg14
15 V 1 = m * v g 1 ; //volume o f p r e s s u r e c o ok e r i n mˆ3
16 u 1 = ( h f 1 + h f g 1 ) - ( p 1 * 1 0 ^ 5 ) * ( v g 1 * 1 0 ^ - 3 ) ; //i n t e r n a l e ne rg y o f s tea m p er kg a t i n i t i a l p o i nt
117 //V1=V218 //V1=m∗ [(1 − x2 ) ∗ vf2+x2∗vg2 ] // v f 2
i s n e g l i g i b l e19 v g 2 = V 1 / x 2 / 1 . 5 ;
20
21 / / f ro m s te am t a b l e s c o r e e s p o n d i n g t o v g2 = 0 .6 25 mˆ 3/kg
22 p 2 = 2 . 9 ;
23 T s 2 = 1 3 2 . 4 + 2 7 3 ; / / i n K24 h f 2 = 5 5 6 . 5 ; / / k J / k g
25 h f g 2 = 2 1 6 6 . 6 ; / / k J / k g26
27 u 2 = ( h f 2 + x 2 * h f g 2 ) - ( p 2 * 1 0 ^ 5 ) * x 2 * ( v g 2 * 1 0 ^ - 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 oi nt 2
28
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29 h n e t = u 2 - u 1 ; // hea t
t r a n s f e r r e d a t c o ns t a n t volume p er kg30 H t o t a l = m * h n e t ; //t o t a l h e at t r a n s f e r r e d
31 //−ve s i gn i n d i c a t e s t ha t h ea t h as be en r e j e c t e d32 H r e j = - 1 * H t o t a l ;
33
34 printf ( ’ ( i ) The P r e s su r e a t new s t a t e i s : %1 . 1 f b ar. \n ’ , p 2 ) ;
35 printf ( ’ The Te mper at ur 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 b e REJECTED i s : %4 . 2
f k J . ’ , H r e j ) ;
Scilab code Exa 4.10 Example 10
1 clc
2 clear
3 //DATA GIVEN4 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 mˆ35 p 1 = 8 ; // p r e s s u r e o f s te am
i n b ar6 x 1 = 0 . 9 ; // d r yn e ss f r a c t i o n
o f s tea m7 p 2 = 4 ; // p r e s s u r e o f s te am
a f t e r blow o f f i n b ar8 p 3 = 3 ; // f i n a l p r e s s u re o f
stea m i n b ar9
10 / / At 8 b ar , f ro m s te am t a b l e s11 h f 1 = 7 2 0 . 9 ; / / k J / k g12 h f g 1 = 2 0 4 6 . 5 ; / / k J / k g13 v g 1 = 0 . 2 4 0 ; //mˆ3/kg14
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15 m 1 = V / ( x 1 * v g 1 ) ; // m ass o f s te am i n
t h e v e s s e l i n kg16
17 h 1 = h f 1 + x 1 * h f g 1 ; / / e n t h a l p y o f s te amb e f o r e b lo wi ng o f f ( p er kg )
18 // e n th a lp y o f stea m b e f o r e b lo wi ng o f f ( p er kg ) =e nt ha l p y o f steam a f t e r b lo wi ng o f f ( p er kg )
19 h 2 = h 1 ;
20 //h2=hf2+x2∗ h f g 221 / / At 4 b ar , f ro m s te am t a b l e s22 h f 2 = 6 0 4 . 7 ; / / k J / k g23 h f g 2 = 2 1 3 3 ; / / k J / k g
24 v g 2 = 0 . 4 6 2 ; //mˆ3/kg25 x 2 = ( h 2 - h f 2 ) / h f g 2 ; // d r yn e ss f r a c t i o n
a t 226
27 m 2 = V / ( x 2 * v g 2 ) ; // m ass o f s te am i nt h e v e s s e l i n kg
28 m = m 1 - m 2 ; / / ma ss o f s te amblown o f f i n kg
29
30 / /As i t i s c o ns t an t volume c o ol i ng , x2∗v g 2 ( a t 4 b a r )
=x3∗v g 3 ( a t 3 b a r )31 / / At 3 b ar , f ro m s te am t a b l e s32 h f 3 = 5 6 1 . 4 ; / / k J / k g33 h f g 3 = 2 1 6 3 . 2 ; / / k J / k g34 v g 3 = 0 . 6 0 6 ; //mˆ3/kg35
36 x 3 = x 2 * v g 2 / v g 3 ;
37 h 3 = h f 3 + x 3 * h f g 3 ;
38
39 / / h e at l o s t d u r in g c o o l i n g , Q l o st=m( u3−u2 )40 u 2 = h 2 - p 2 * 1 0 ^ 5 * x 2 * v g 2 * 1 0 ^ - 3 ;
41 u 3 = h 3 - p 3 * 1 0 ^ 5 * x 3 * v g 3 * 1 0 ^ - 3 ;42 Q l o s t = m * ( u 3 - u 2 ) ;
43
44 printf ( ’ ( i ) The Mass o f o f s team blown o f f i s : %1 . 3f kg . \n ’ , m ) ;
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45 printf ( ’ ( i i ) The D ry n es s f r a c t i o n o f steam i n t he
v e s s e l a f t e r c o o l i n g i s : %1 . 4 f . \n ’ , x 3 ) ;46 printf ( ’ ( i i i ) The Hea t l o s t d u r i ng c o o l i n g i s : %3 . 2 f kJ . \n ’ , ( - Q l o s t ) ) ;
47
48 //NOTE:49 //T he ans w e r s of m1, x 3 ar e INCORRECT i n t he book ,50 / / t h u s , t h e a n s w e r s o f m, x 3 a nd Q l o s t a r e INCORRECT
i n t he book51 // w hi le , t he v a l ue s o bt ai ne d h er ( i n s c i l a b ) a re
CORRECT.
Scilab code Exa 4.11 Example 11
1 clc
2 clear
3 //DATA GIVEN4 p = 8 ; / / p r e s s u r e o f s te am
i n b ar5 x = 0 . 8 ; // d r yn e s s f r a c t i o n
67 / / At 8 b ar , f ro m s te am t a b l e s8 v g = 0 . 2 4 0 ; //mˆ3/kg9 h f g = 2 0 4 6 . 5 ; / / k J / k g
10
11 W e = p * 1 0 ^ 5 * x * v g / 1 0 0 0 ; / / e x t e r n a l wo rk d on ed ur in g e v ap o ra t i o n i n kJ
12 L H i = x * h f g - W e ; // I n t e r n a l l a t e n th ea t i n kJ
13
14 printf ( ’ ( i ) The E x t e rn a l work d on e d u r in ge v a p o r a t i o n i s : %3 . 1 f kJ . \n ’ , W e ) ;
15 printf ( ’ ( i i ) The I n t e r n a l l a t e n t h ea t i s : %4 . 1 f kJ .\n ’ , L H i ) ;
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Scilab code Exa 4.12 Example 12
1 clc
2 clear
3 //DATA GIVEN4 p = 1 0 ; / / p r e s s u r e o f steam ,
p 1=p 2 i n b a r5 x 1 = 0 . 8 5 ; // d r yn e s s f r a c t i o n6 V 1 = 0 . 1 5 ; / / vo lu me o f s te am i n
mˆ37 T s u p 2 = 3 0 0 + 2 7 3 ; / / temp . o f s te am i n K8 C p s = 2 . 2 ; // s p e c i f i c he at of
s te a m i n k J /kgK9
10 / /At 1 0 b ar , f ro m s te am t a b l e s11 v g 1 = 0 . 1 9 4 ; //mˆ3/kg12 h f g 1 = 2 0 1 3 . 6 ; / / k J / k g13 T s 1 = 1 7 9 . 9 + 2 7 3 ; / / i n K14 m = V 1 / ( x 1 * v g 1 ) ; / / ma ss o f s te am
i n kg15 h n e t = ( 1 - x 1 ) * h f g 1 + C p s * ( T s u p 2 - T s 1 ) ; / / h e at s u p p l i e d
p er kg o f stea m16 H t o t a l = m * h n e t ; // t o t a l h ea t
s u p p l i e d17
18 / / E x t e r n a l wo rk d on e d u r i n g t h e p r o c e s s We=p ∗( v s u p 2−x∗vg1 )
19 // s i n c e p1=p2=p ,20 //vg1/Ts1=vsup2/Tsup221 v s u p 2 = v g 1 * T s u p 2 / T s 1 ;
22 W e = p * 1 0 ^ 5 * ( v s u p 2 - x 1 * v g 1 ) * 1 0 ^ - 3 ;
23 h p = W e / h n e t ; //% o f t o t a lh ea t s u p p li e d ( p er kg ) which a pp ea rs a s e x t e r na lwork
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24
25 printf ( ’ ( i ) The T ot al h ea t s u p p l i e d i s : %3 . 1 f kJ . \n ’ , H t o t a l ) ;
26 printf ( ’ ( i i ) The P er ce nt ag e o f t o t a l h ea t s u p p l i e d( p er kg ) wh ich a p pe ar 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 ’ , ( h p * 1 0 0 ) ) ;
Scilab code Exa 4.13 Example 13
1 clc2 clear
3 //DATA GIVEN4 p = 1 8 ; / / p r e s s u r e o f s te am5 x = 0 . 8 5 ; // d r yn e s s f r a c t i o n6
7 / /At 1 8 b ar , f ro m s te am t a b l e s8 h f = 8 8 4 . 6 ; / / k J / k g9 h f g = 1 9 1 0 . 3 ; / / k J / k g
10 v g = 0 . 1 1 0 ; //mˆ3/kg11 u f = 8 8 3 ; / / k J / k g
12 u g = 2 5 9 8 ; / / k J / k g13
14 v = x * v g ; // s p e c i f i c volume of w et s t e am
15 h = h f + x * h f g ; // s p e c i f i c en th al pyo f wet s te am
16 u = ( 1 - x ) * u f + x * u g ; / / s p e c i f i c i n t e r n a le n er g y o f wet s te am
17
18 printf ( ’ ( i ) The S p e c i f i c vo lu me v i s : %1 . 4 f mˆ 3/ k g .
\n ’ , v ) ;19 printf ( ’ ( i i ) The S p e c i f i c e nt ha 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 ne rg y u i s : %4. 2 f k J / kg . \n ’ , u ) ;
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Scilab code Exa 4.14 Example 14
1 clc
2 clear
3 //DATA GIVEN4 p = 7 ; / / p r e s s u r e o f s te am5 h = 2 5 5 0 ; / / e n t h a l p y o f s te am6
7 / / At 7 b ar , f ro m s te am t a b l e s8 h f = 6 9 7 . 1 ; / / k J / k g9 h f g = 2 0 6 4 . 9 ; / / k J / k g
10 v g = 0 . 2 7 3 ; //mˆ3/kg11 u f = 6 9 6 ; / / k J / k g12 u g = 2 5 7 3 ; / / k J / k g13
14 h g = h f + h f g ;
15 / /At 7 b ar , hg =2762 kJ / kg , h en ce s i n c e a c t u a le n th a lp y i s g i ve n a s 2 55 0 kJ /kg , t he stea m mustbe i n wet va po ur s t a t e
16 / / s p e c i f i c e n t h a l p y o f w et s te am , h=h f +x∗ h f g17 x = ( h - h f ) / h f g ; // d r yn e s s f r a c t i o n18 v = x * v g ; // s p e c i f i c volume of
w et s t e am19 u = ( 1 - x ) * u f + x * u g ; / / s p e c i f i c i n t e r n a l
e n er g y o f wet s te am20
21 printf ( ’ ( i ) The D ry ne ss 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 v ol ume v i s : %1 . 4 f mˆ 3/ k g
. \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 ne rg y u i s : %4
. 2 f k J / kg . \n ’ , u ) ;
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Scilab code Exa 4.15 Example 15
1 clc
2 clear
3 //DATA GIVEN4 p = 1 2 0 ; / / p r e s s u r e o f s te am5 v = 0 . 0 1 7 2 1 ; // s p e c i f i c volume of
steam6
7 / /At 1 20 ba r , fr om s te am t a b l e s8 v g = 0 . 0 1 4 3 ; //mˆ3/kg9 / / s i n c e vg<v , t he s team i s s u pe r he a te d
10 / / s o fro m s u p er h ea t t a b l e s a t 1 20 b ar and v = 0. 01 72 1mˆ3/kg
11 T = 3 5 0 ; // d eg . c e l s i u s12 h = 2 8 4 7 . 7 ; // s p e c i f i c en th al py
o f s tea m13 u = h - p * 1 0 ^ 5 * v / 1 0 ^ 3 ; / / s p e c i f i c i n t e r n a l
e n er g y o f s te am
1415 printf ( ’ ( i ) The T em pe ra tu re i s : %3 . 0 f d eg c e l s i u s .
\n ’ , T ) ;
16 printf ( ’ ( i i ) The S p e c i f i c e nt ha 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 ne rg y u i s : %4. 2 f k J / kg . \n ’ , u ) ;
Scilab code Exa 4.16 Example 16
1 clc
2 clear
3 //DATA GIVEN
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4 p = 1 4 0 ; / / p r e s s u r e o f s te am
5 h = 3 0 0 1 . 9 ; // s p e c i f i c en th al pyo f s tea m6
7 / /At 1 40 ba r , fr om s te am t a b l e s8 h g = 2 6 4 2 . 4 ;
9 / / s i n c e hg<h , t he s team i s s u pe r he a te d10 / / s o fro m s u p er h ea t t a b l e s a t 1 40 b ar and h = 30 01 .9
k J /k g11 T = 4 0 0 ; // d eg . c e l s i u s12 v = 0 . 0 1 7 2 2 ; // s p e c i f i c volume of
steam
13 u = h - p * 1 0 ^ 5 * v / 1 0 ^ 3 ; / / s p e c i f i c i n t e r n a le n er g y o f s te am
14
15 printf ( ’ ( i ) The T em pe ra tu re i s : %3 . 0 f d eg c e l s i u s .\n ’ , T ) ;
16 printf ( ’ ( i i ) The S p e c i f i c v ol ume v i s : %1 . 5 f mˆ 3/ k g. \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 ne rg y u i s : %4. 2 f k J / kg . \n ’ , u ) ;
Scilab code Exa 4.17 Example 17
1 clc
2 clear
3
4 p 1 = 1 0 ; // p r e s s u r e i nb ar
5 / /At 1 0 b ar and 3 00 deg c e l s i u s , fro m s team t a b l e s
o f s u p e r he a t e d s te am6 h s u p = 3 0 5 1 . 2 / / k J / k g7 T s u p = 3 0 0 + 2 7 3 ; / / temp . o f s te a m
i n K8 / /At 1 0 b ar and 3 00 deg c e l s i u s , fro m s team t a b l e s
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o f dr y s a t u r a t ed stea m
9 T s = 1 7 9 . 9 + 2 7 3 / / temp . o f s te a mi n K10 v g = 0 . 1 9 4 ; //mˆ3/kg11
12 / / By v g / Ts = v s u p / Ts up13 v s u p = v g * T s u p / T s ;
14 u 1 = h s u p - p 1 * 1 0 ^ 5 * v s u p / 1 0 ^ 3 ;
15
16 p 2 = 1 . 4 ; / /new p r e s s u r ei n b ar
17 x 2 = 0 . 8 ; / / d r y n e s s
f r a c t i o n18 // At 1 . 4 bar , fr om stea m t a b l e s19 h f 2 = 4 5 8 . 4 ; / / k J / k g20 h f g 2 = 2 2 3 1 . 9 ; / / k J / k g21 v g 2 = 1 . 2 3 6 ; //mˆ3/kg22 h 2 = h f 2 + x 2 * h f g 2 ; / / e n t h a l p y o f
wet s te am ( a f t e r e x p an s i o n )23 u 2 = h 2 - p 2 * 1 0 ^ 5 * x 2 * v g 2 / 1 0 ^ 3 ; / / i n t e r n a l
e ne rg y o f t h i s steam24 D u = u 2 - u 1 ; / / c ha ng e i n
i n t e r n a l e ne rg y p e r kg25
26 printf ( ’ ( i ) The I n t e r n a l e ne rg y o f s u pe rh e a t eds te am a t 1 0 b ar i s : %4 . 1 f kJ / kg . \n ’ , u 1 ) ;
27 printf ( ’ ( i i ) The Change i n i n t e r n a l e ne rg y p er kgi s : %2 . 1 f kJ . \n ’ , D u ) ;
28 printf ( ’ ( N eg at iv e s i g n i n d i c a t e s DECREASE ini n t e r n a l e ne rg y . ) ’ ) ;
Scilab code Exa 4.18 Example 18
1 clc
2 clear
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Scilab code Exa 4.19 Example 19
1 clc
2 clear
3 //DATA GIVEN4 p = 2 0 ; // p r e s s u r e i n t he
b o i l e r s and m ain i s 20 b a r5 T b s = 3 5 0 ; / / t e m p e r a tu r e o f s te am
i n b o i l e r w i t h s u pe r h e a t er i n deg . c e l s i u s6 T m = 2 5 0 ; / / t e m p e r a tu r e o f s te am
i n t he main i n deg . c e l s i u s7 C p s = 2 . 2 5 ; // s p e c i f i c he a t of
s te am i n kJ / kg8
9 / /At 2 0 b ar , f ro m s te am t a b l e s10 T s = 2 1 2 . 4 ; // d eg . c e l s i u s11 h f = 9 0 8 . 6 ; / / k J / k g12 h g = 2 7 9 7 . 2 ; / / k J / k g13 h f g = 1 8 8 8 . 6 ; / / k J / k g14
15 / / B o i l e r B1−20 bar , 350 deg . c e l s i u s16 h 1 = h g + C p s * ( T b s - T s ) ;
17
18 //Main−20 bar , 250 d eg c e l s i u s19 h m = 2 * [ h g + C p s * ( T m - T s ) ] ; // t o t a l h ea t o f
2 kg o f s team i n t he stea m m ain20
21 / / B o i l e r B2−20 bar ,22 //h2=hf+x2∗ h f g23 //h2=hm−h124 x 2 = ( ( h m - h 1 ) - h f ) / h f g ;
25
26 printf ( ’ The Q ua l i t y o f steam i n t he B o i l e r w it ho ut
s u p e r h e a t e r i s : %1 . 3 f . \n ’ , x 2 ) ;
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Scilab code Exa 4.20 Example 20
1 clc
2 clear
3 //DATA GIVEN4 m = 1 ; // m ass o f wet
s te am i n kg5 p = 6 ; // p r e s s u r e o f
stea m i n b ar6 x = 0 . 8 ; / / d r y n e s s
f r a c t i o n7
8 / / At 6 b ar , f ro m s te am t a b l e s9 T s = 1 5 8 . 8 + 2 7 3 ; / / i n K
10 h f g = 2 0 8 5 ; / / k J / k g11 s w e t = 4 . 1 8 * log ( T s / 2 7 3 ) + x * h f g / T s ; / / e n t r o p y o f
w et s t ea m i n k J /kgK12
13 printf ( ’ The E nt ro py o f we t s te am i s : %1 . 4 f kJ / kgK . ’ ,
s w e t ) ;
14
15 //NOTE;16
// t h e e xa ct an s i s 5 . 7 7 94 , w hi l e i n TB i t i s g iv ena s 5 . 7 8 6 5 kJ / kgK
Scilab code Exa 4.21 Example 21
1 clc
2 clear
3 //DATA GIVEN
4 p 1 = 1 0 ; / / i n i t i a l p r e s s u r e o f s team i n b ar5 T s u p = 2 5 0 ; // i n i t i a l t e m pe r at u r e
o f steam i n deg c e l s i u s6 p 2 = 0 . 2 ; // f i n a l p r e s s u r e o f
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stea m i n b ar
7 x 2 = 0 . 9 ; // f i n a l d r yn e ssf r a c t i o n o f steam8
9 / /At 1 0 b ar , f ro m s te am t a b l e s10 h s u p = 3 2 6 3 . 9 ; / / k J / k g11 s s u p = 7 . 4 6 5 ; //kJ/kgK12 h 1 = h s u p ;
13 s 1 = s s u p ;
14
15 // At 0 . 2 bar , fr om stea m t a b l e s16 h f 2 = 2 5 1 . 5 ; //k J /k h
17 h f g 2 = 2 3 5 8 . 4 ; / / k J / k g18 s f 2 = 0 . 8 3 2 1 ; //kJ/kgK19 s g 2 = 7 . 9 0 9 4 ; //kJ/kgK20 h 2 = h f 2 + x 2 * h f g 2 ;
21 s f g 2 = ( s g 2 - s f 2 ) ;
22 s 2 = s f 2 + x 2 * s f g 2 ;
23
24 D h = h 1 - h 2 ; // d ro p i n e n th a lp y25 D s = s 1 - s 2 ; / / c ha ng e i n e n tr o p y26
27 printf ( ’ ( i ) The Drop i n e n t h al p y i s : %3 . 1 f kJ / kg . \n ’ , D h ) ;
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 k J /kgK . ’ , D s ) ;
Scilab code Exa 4.22 Example 22
1 clc
2 clear3 //DATA GIVEN4 m = 1 ; // mass o f s tea m i n kg5 p = 1 2 ; / / p r e s s u r e o f s te am
i n b ar
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6 T s u p = 2 5 0 + 2 7 3 ; / / temp . o f s te am i n K
7 C p s = 2 . 1 ; // s p e c i f i c he at of s te am i n kJ / kg8
9 / /At 1 2 b ar , f ro m s te am t a b l e s10 T s = 1 8 8 + 2 7 3 ; / / i n K11 h f g = 1 9 8 4 . 3 ; / / k J / k g12 s s u p = 4 . 1 8 * log ( T s / 2 7 3 ) + h f g / T s + C p s * log ( T s u p / T s ) ;
/ / e n t r o py o f w et s te am i n kJ / kgK13
14 printf ( ’ The E ntro py o f 1 kg o f s u pe r he a te d stea m a t12 ba r and 250 deg c e l s i u s i s : %1 . 3 f kJ /kg . \n ’ ,
s s u p ) ;
Scilab code Exa 4.23 Example 23
1 clc
2 clear
3 //DATA GIVEN4 p = 5 ; / / p r e s s u r e o f s te am
i n b ar5 M w t = 5 0 ; // mass o f w at er i n
t he t a nk i n kg6 t 1 = 2 0 ; // i n i t i a l temp . in
deg . c e l s i u s7 M s = 3 ; / / amo unt o f s t ea m
c on d en s ed i n kg8 t 2 = 4 0 ; / / f i n a l temp . i n d eg .
c e l s i u s9 W e = 1 . 5 ; // w at e r e q u i v a l e n t o f
t a nk i n kg10
11 / / At 5 b ar , f ro m s te am t a b l e s12 h f = 6 4 0 . 1 ; / / i n k J / kg13 h f g = 2 1 0 7 . 4 ; / / i n k J / kg
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14
15 M w = M w t + W e ; // t o t a l mass o f w at eri n kg16 // h ea t l o s t by st ea m = h ea t g a in ed by w at er17 //Ms [ ( hf+xhfg ) −1∗4. 18∗( t2 −0) ]=Mw[ 1 ∗ 4 . 1 8 ∗ ( t 2−t1 ) ]18 x = [ M w * [ 1 * 4 . 1 8 * ( t 2 - t 1 ) ] / M s + 1 * 4 . 1 8 * ( t 2 - 0 ) - h f ] / h f g ;
// d r yn e s s f r a c t i o n19
20 printf ( ’ The D ry ne ss 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 GIVEN4 p = 1 . 1 ; / / p r e s s u r e o f s te am
i n b ar5 x = 0 . 9 5 ; // d r yn e s s f r a c t i o n6 M w t = 9 0 ; // mass o f w at er i n
t he t a nk i n kg7 t 1 = 2 5 ; // i n i t i a l temp . in
deg . c e l s i u s8 M t = 1 2 . 5 ; // mass o f t an k i n kg9 c = 0 . 4 2 ; // s p e c i f i c he at of
m e ta l i n kJ / kgK10 t 2 = 4 0 ; / / f i n a l temp . i n d eg .
c e l s i u s11
12 m 1 = M w t ;
13 m 2 = M t * c ; // w at e r e q u i v a l e n t o f v e s s e l14 M = m 1 + m 2 ; // t o t a l mass o f w at er
i n kg15 // At 1 . 1 bar , fr om stea m t a b l e s
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16 h f = 4 2 8 . 8 ; / / i n k J / kg
17 h f g = 2 2 5 0 . 8 ; / / i n k J / kg18 // h ea t l o s t by st ea m = h ea t g a in ed by w at er19 //Ms [ ( hf+xhfg ) −1∗4. 18∗( t2 −0)]=M[1 ∗ 4 . 1 8 ∗ ( t 2−t1 ) ]20 M s = M * [ 1 * 4 . 1 8 * ( t 2 - t 1 ) ] / [ ( h f + x * h f g ) - 1 * 4 .1 8 * ( t 2 - 0 ) ] ;
// m ass o f s te am c o nd e ns e d i n kg21
22 printf ( ’ The Mass o f s te am c o nd e ns e d , Ms i s : %1 . 3 f kg . ’ , M s ) ;
Scilab code Exa 4.25 Example 25
1 clc
2 clear
3 //DATA GIVEN4 / / c o n d i t i o n o f st eam b e f o r e t h r o t t l i n g5 p 1 = 8 ; // p r e s s u r e i n b ar6 // c o n d i t i o n o f st eam a f t e r t h r o t t l i n g7 p 2 = 1 ; // p r e s s u r e i n b ar8 T 2 = 1 1 5 + 2 7 3 ; // temp . i n deg . c e l s i u s
9 T s u p 2 = T 2 ;10 / / At 1 b ar ,11 T s 2 = 9 9 . 6 + 2 7 3 ;
12 C p s = 2 . 1 ; //kJ/kgK13
14 / /As t h r o t t l i n g i s a c on s ta n t e nt ha l p y p r o c e ss ,15 // h1=h2 . . . . . hf 1+x1∗ hgf1=hf2+hfg2+Cps(Tsup2−T s2)16
17 / /At 8 b ar , f ro m s te am t a b l e s ,18 h f 1 = 7 2 0 . 9 ;
19 h f g 1 = 2 0 4 6 . 5 ;20 / /At 1 b ar , f ro m s te am t a b l e s ,21 h f 2 = 4 1 7 . 5 ;
22 h f g 2 = 2 2 5 7 . 9 ;
23
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24 x 1 = [ h f 2 + h f g 2 + C p s * ( T s u p 2 - T s 2 ) - h f 1 ] / h f g 1 ; //
d r y ne s s f r a c t i o n25
26 printf ( ’ The D ry ne ss f r a c t i o n o f stea m i n t he main ,x1 i s : %1 . 2 f . ’ , x 1 ) ;
Scilab code Exa 4.26 Example 26
1 clc
2 clear3 //DATA GIVEN4 M w = 2 ; // m ass o f w at er
s e pa r at e d o ut i n kg5 M s = 2 0 . 5 ; / / amount o f s te am (
c o nd e ns a te ) d i s c h a rg e d fro m t h r o t t l i n gc a l o r i m e t e r i n kg
6 T s u p 3 = 1 1 0 + 2 7 3 ; / / temp . o f s t ea ma f e t r t h r o t t l i n g i n K
7 p 1 = 1 2 ; // i n i t i a l pr e s su r eo f st eam i n ba r
8 p 3 = ( 7 6 0 + 5 ) / 1 0 0 0 * 1 . 3 3 6 6 ; // f i n a l p r e s s u re o f s te am i n b ar ( 1 mm o f Hg =1 .3 36 6 b ar )
9 C p s = 2 . 1 ; //kJ/kgK10
11 p 2 = p 1 ;
12 / / At p 1=p2 =12 b ar , f ro m s te a m t a b l e s13 h f 2 = 7 9 8 . 4 ; / / i n kJ / k g14 h f g 2 = 1 9 8 4 . 3 ; / / i n kJ / k g15
16 / / At p3=1 b ar , f ro m s te am t a b l e s
17 T s 3 = 9 9 . 6 + 2 7 3 ; / / i n K18 T s u p 3 = 1 1 0 + 2 7 3 ; / / i n K19 h f 3 = 4 1 7 . 5 ; / / i n kJ / k g20 h f g 3 = 2 2 5 7 . 9 ; / / i n kJ / k g21
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22 // h2=h3 . . . . . hf 2+x2∗ hgf2=hf3+hfg3+Cps(Tsup3−T s3)
23 x 2 = [ h f 3 + h f g 3 + C p s * ( T s u p 3 - T s 3 ) - h f 2 ] / h f g 2 ; //d ry ne ss f r a c t i o n x224
25 x 1 = ( x 2 * M s ) / ( M w + M s ) ; //d r y ne s s f r a c t i o n o f steam s u pp l i e d , x1
26
27 printf ( ’ The Q u a l it y o f s te am s u p p l i e d , x1 i s : %1 . 2 f .’ , x 1 ) ;
Scilab code Exa 4.27 Example 27
1 clc
2 clear
3 //DATA GIVEN4 p 1 = 1 5 ; // p r e s s u r e o f s te am
s am pl e i n b ar5 p 3 = 1 ; // p r e s s u r e o f s te am
a t e x i t i n b a r6 T s u p 3 = 1 5 0 + 2 7 3 ; / / 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 K7 M w = 0 . 5 ; / / d i s c h a r g e f ro m
s e p a r at i n g c a l o r i m e t e r i n kg /min8 M s = 1 0 ; / / d i s c h a r g e f ro m
t h r o t t l i n g c a l o r i m e t e r i n kg /min9
10 p 2 = p 1 ;
11 / / At p 1=p2 =15 b ar , f ro m s te a m t a b l e s12 h f 2 = 8 4 4 . 7 ; / / i n kJ / k g13 h f g 2 = 1 9 4 5 . 2 ; / / i n kJ / k g
1415 / /At p3=1 b ar and 1 50 d eg . c e l s i u s , f ro m s te amt a b l e s
16 h s u p 3 = 2 7 7 6 . 4 ; / / i n kJ / k g17
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18 // h2=h3 . . . . . hf 2+x2∗ hgf2=hsup3
19 x 2 = [ h s u p 3 - h f 2 ] / h f g 2 ; // d r yn e ss f r a c t i o nx220
21 x 1 = ( x 2 * M s ) / ( M w + M s ) ; // q u a l i t y o f s teams u p p l i ed , x1
22
23 printf ( ’ The Q u a l it y o f s te am s u p p l i e d , x1 i s : %1 . 3 f .’ , x 1 ) ;
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Chapter 5
Heat Engines
Scilab code Exa 5.1 Example 1
1 clc
2 clear
3 //DATA GIVEN4 M s = 1 0 0 0 0 / 3 6 0 0 ; // r a t e o f s team f l ow i n kg / s5 / / i n l e t t o t u r bi n e6 p 1 = 6 0 ; // p r e s s u e i n b ar
7 T 1 = 3 8 0 ; / / temp . i n d eg . c e l s i u s8
9 // e x i t from t ur bi ne , i n l e t t o c on de ns e r10 p 2 = 0 . 1 ; // p r e s s u e i n b ar11 x 2 = 0 . 9 ; / / q u a l i t y12 v 2 = 2 0 0 ; // v e l o c i t y i n m/ s13
14 / / e x i t f ro m c o nd e ns e r , i n l e t t o pump15 p 3 = 0 . 0 9 ; // p r e s s u e i n b ar16 // i t i s s a tu r a t ed
1718 / / e x i t from pump , i n l e t t o b o i l e r19 p 4 = 7 0 ; // p r e s s u e i n b ar20
21 // e x i t from b o i l e r ,
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22 p 5 = 6 5 ; // p r e s s u e i n b ar
23 T 5 = 4 0 0 ; / / temp . i n d eg . c e l s i u s24
25 / / f o r c o n d en s e r ,26 t 1 = 2 0 ; // i n l e t temp . i n deg . c e l s i u s27 t 2 = 3 0 ; // e x i t temp . i n deg . c e l s i u s28
29 // At 60 b ar and 3 80 d eg . c e l s i u s , fro m s team t a b l e s30 h 1 = 3 0 4 3 . 0 + ( 3 1 7 7 . 2 - 3 0 4 3 . 0 ) / ( 4 0 0 - 3 5 0 ) * 3 0 ; //By
i n t e r p o l a t i o n31
32 // At 0 . 1 bar , fr om stea m t a b l e s
33 h f 2 = 1 9 1 . 8 ; / / i n k J / kg34 h f g 2 = 2 3 9 2 . 8 ; / / i n k J / kg35 h 2 = h f 2 + x 2 * h f g 2 ;
36 P t = M s * ( h 1 - h 2 ) // p owe r o ut p ut o f t he t u r b i n e i n kW
37
38 / /At 7 0 b ar , f ro m s te am t a b l e s39 h f 4 = 1 2 6 7 . 4 ; / / i n k J / kg40 // At 60 b ar and 3 80 d eg . c e l s i u s , fro m s team t a b l e s41 h a = ( 3 1 7 7 . 2 + 3 1 5 8 . 1 ) / 2 ; //By i n t e r p o l a t i o n
be tw e en 60 and 70 deg c e l s i u s42 Q 1 = M s * 3 6 0 0 * ( h a - h f 4 ) ; // h ea t t r a n s f e rp e r hour i n t h e b o i l e r
43 // At 0 . 0 9 bar , fro m s tea m t a b l e s44 h f 3 = 1 8 3 . 3 ; / / i n k J / kg45 Q 2 = M s * 3 6 0 0 * ( h 2 - h f 3 ) ; // h ea t t r a n s f e r
p er hour i n t he c on de ns e r46
47 // h ea t l o s t by st ea m=h ea t g a in e d by t he c o o l i n gw a t e r
48 //Q2=Mw∗ 4 . 1 8 ∗ ( t 2−t 1 )
49 M w = Q 2 / 4 . 1 8 / 1 0 ; // mass o f c o o l i n gw at er c i r c u l e t e d p er hour i n c on de ns er
50
51 // ( pi ) /4∗dˆ2=Ms∗x2∗vg252 / /d=d i am e te r o f t he p i pe c o n n ec t i ng t u r b i n e w it h
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c o n d e n s e r
53 C = 2 0 0 ; // v e l o c i t y o f s te am i n m/ s54 v g 2 = 1 4 . 6 7 ; // s p e c i f i c volume
a t 0 . 1 ba r55 d = ( M s * x 2 * v g 2 / ( % p i / 4 ) / C ) ^ 0 . 5 ;
56
57 printf ( ’ ( i ) The Power o u tp ut o f t u r b i n e i s : %4 . 0 f kW. \n ’ , P t ) ;
58 printf ( ’ ( i i ) The Heat t r a n s f e r p er hour i n t heB o i l e r i s : %3 . 2 e kJ / h . \n ’ , Q 1 ) ;
59 printf ( ’ The Heat t r a n s f e r p e r hour i n th e
C o n de n s er i s : %3 . 2 e k J /h . \n ’ , Q 2 ) ;60 printf ( ’ ( i i i ) The Mass o f c o o l i n g w at er c i r c u l a t e d
p er h ou r i n t he c o nd e ns e r i s : %3 . 2 e kg / hr . \n ’ , Mw
) ;
61 printf ( ’ ( i v ) The D ia me te r o f t he p i pe c o n n ec t i ngt u r b i n e w it h c o n d en s e r i s : %1 . 3 f m o r %3 . 0 f mm. \n ’ , d , ( d * 1 0 0 0 ) ) ;
62
63 //NOTE:64 / / a n s o f Mw( 1 . 1 1 6 ∗1 0 ˆ 7 ) i s g iv en i n c o r r e c t i n t he
book .65 // t he c o r r e c t an 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 p 1 = 1 5 ; // s tea m s u pp ly p r e s s u r e i nb ar5 x 1 = 1 ; // q u a l i t y o f s tea m6 p 2 = 0 . 4 ; / / c o n d en s e r p r e s s u r e7
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8 // At 0 . 1 5 bar , fro m s tea m t a b l e s
9 T 1 = 1 9 8 . 3 + 2 7 3 ; / / i n K10 h g 1 = 2 7 8 9 . 9 ; / / i n k J / kg11 s g 1 = 6 . 4 4 0 6 ; // i n k J /kgK12 // At 0 . 4 bar , fr om stea m t a b l e s13 T 2 = 7 5 . 9 + 2 7 3 ; / / i n K14 h f 2 = 3 1 7 . 7 ; / / i n k J / kg15 h f g 2 = 2 3 1 9 . 2 ; / / i n k J / kg16 s f 2 = 1 . 0 2 6 1 ; // i n k J /kgK17 s f g 2 = 6 . 6 4 4 8 ; // i n k J /kgK18
19 E T A c a r n o t = ( T 1 - T 2 ) / T 1 ; // C arnot e f f i c i e n c y
20 / / ETArankine=A d i a b a ti c o r i s e n t r o p i c h e at d ro p / h e ats u p p l i e d
21 //ETArankine=(hg1−h2) /( hg1−hf 2 )22 / / a s t he stea m e xp an ds i s e n t r o p i c a l l y , s 1=s 223 //sg1= sf 2+ x 2∗ s f g 224 x 2 = ( s g 1 - s f 2 ) / s f g 2 ;
25 h 2 = h f 2 + x 2 * h f g 2 ;
26 E T A r a n k i n e = ( h g 1 - h 2 ) / ( h g 1 - h f 2 ) ; / / R a n k i n ee f f i c i e n c y
27
28 printf ( ’ ( i ) The Ca rnot e f f i c i e n c y i s : %1 . 4 f o r %2 . 2f p e r ce n t . \n ’ , E T A c a rn o t , ( E T A c a r n o t * 1 0 0 ) ) ;
29 printf ( ’ ( i i ) The R ank ine 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 ’ , E T A r a n ki n e , ( E T A r a n k i n e * 1 0 0 ) ) ;
Scilab code Exa 5.3 Example 3
1 clc
2 clear3 //DATA GIVEN4 p 1 = 2 0 ; // b o i l e r p r e s s u re i n b a r5 T 1 = 3 6 0 + 2 7 3 ; / / temp . i n K6 p 2 = 0 . 0 8 ; // b o i l e r p r e s s u re i n b a r
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7
8 // At 20 b ar and 3 60 deg . c e l s i u s , fro m s team t a b l e s9 h 1 = 3 1 5 9 . 3 ; / / i n k J / kg10 s g 1 = 6 . 9 9 1 7 ; // i n k J /kgK11
12 // At 0 . 0 8 bar , fro m s tea m t a b l e s13 h f 2 = 1 7 3 . 8 8 ; / / i n k J / kg14 h f 3 = h f 2 ;
15 s f 2 = 0 . 5 9 2 6 ; // i n k J /kgK16 s 3 = s f 2 ;
17 h f g 2 = 2 4 0 3 . 1 ; / / i n k J / kg18 s g 2 = 8 . 2 2 8 7 ; // i n k J /kgK
19 v f 2 = 0 . 0 0 1 0 0 8 ; //mˆ3/kg20 s f g 2 = 7 . 6 3 6 1 ; // i n k J /kgK21
22 / / a s t he stea m e xp an ds i s e n t r o p i c a l l y , s 1=s 223 //sg1= sf 2+ x 2∗ s f g 224 x 2 = ( s g 1 - s f 2 ) / s f g 2 ;
25 h 2 = h f 2 + x 2 * h f g 2 ;
26
27 //Wnet=Wturbine−Wpump28 //Wpump=hf4−hf 3=vf 3 ( p1−p2 )29 W p = v f 2 * ( p 1 - p 2 ) * 1 0 0 ;
30 h f 4 = W p + h f 3 ;
31 W t = h 1 - h 2 ;
32 W n e t = W t - W p ;
33 Q 1 = h 1 - h f 4 ; / / i n k J / kg34 E T A c y c l e = W n e t / Q 1 ; // c y c l e e f f i c i e n c y35
36 printf ( ’ ( i ) The Net work p er kg o f stea m i s : %3 . 2 f kJ/ kg . \n ’ , W n e t ) ;
37 printf ( ’ ( i i ) The C yc le e f f i c i e n c y i s : %1 . 3 f o r %2 . 1f p e r ce n t . \n ’ , E T A c yc l e , ( E T A c y c l e * 1 0 0 ) ) ;
Scilab code Exa 5.4 Example 4
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1 clc
2 clear3 // g i v en s team t a b l e e x t r a c t4 p 1 = 8 0 ; / / i n b a r5 t 1 = 2 9 5 . 1 ; // i n deg . c e l s i u s6 v f 1 = 0 . 0 0 1 3 8 5 ; //mˆ3/kg7 v g 1 = 0 . 0 2 3 5 ; //mˆ3/kg8 h f 1 = 1 3 1 7 ; / / i n k J / kg9 h f g 1 = 1 4 4 0 . 5 ; / / i n k J / kg
10 h g 1 = 2 7 5 7 . 5 ; / / i n k J / kg11 s f 1 = 3 . 2 0 7 3 ; // in kJ/kgK12 s f g 1 = 2 . 5 3 5 1 ; // in kJ/kgK
13 s g 1 = 5 . 7 4 2 4 ; // in kJ/kgK14
15 p 2 = 0 . 1 ; / / i n b a r16 t 2 = 4 5 . 8 4 ; // i n deg . c e l s i u s17 v f 2 = 0 . 0 0 1 0 1 0 3 ; //mˆ3/kg18 v g 2 = 1 4 . 6 8 //mˆ3/kg19 h f 2 = 1 9 1 . 9 ; / / i n k J / kg20 h f 3 = h f 2 ;
21 h f g 2 = 2 3 9 2 . 3 ; / / i n k J / kg22 h g 2 = 2 5 8 4 . 2 ; / / i n k J / kg23 s f 2 = 0 . 6 4 8 8 ;
// in kJ/kgK24 s f g 2 = 7 . 5 0 0 6 ; // in kJ/kgK25 s g 2 = 8 . 1 4 9 4 ; // in kJ/kgK26
27 E T A t = 0 . 9 ; // steam t u r bi n e e f f i c i e n c y28 E T A p = 0 . 8 ; // c o nde nsa t e pump
e f f i c i e n c y29
30 P 1 = 8 0 ; / / i n b a r31 T 1 = 6 0 0 ; // i n deg c e l s i u s32 / /At 80 b ar and 600 deg c e l s i u s
33 v 1 = 0 . 4 8 6 ; //mˆ3/kg34 h 1 = 3 6 4 2 ; / / k J / k g35 s 1 = 7 . 0 2 0 6 ; //k J /k g/K36
37 / / a s t he stea m e xp an ds i s e n t r o p i c a l l y , s 1=s 2
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38 //sg1= sf 2+ x 2∗ s f g 2
39 x 2 = ( s 1 - s f 2 ) / s f g 2 ;40 h 2 = h f 2 + x 2 * h f g 2 ;
41 W t a = E T A t * ( h 1 - h 2 ) ; // a c t u a l t u r b i n e work i n kJ/ kg
42 W p = v f 2 * ( p 1 - p 2 ) * 1 0 ^ 5 / 1 0 ^ 3 ; / / pump w or k i n k J / k g43 W p a = W p / E T A p ; / / a c t u a l pump w or k i n k J / kg44 W n e t = W t a - W p a ; // s p e c i f i c w or k i n k J /k g45 //ETAthermal=Wnet/Q146 //Q1=h1−h f 447 h f 4 = h f 3 + W p a ;
48 Q 1 = h 1 - h f 4 ;
49 E T A t h = W n e t / Q 1 ;50
51 printf ( ’ ( i ) The S p e c i f i c work ( Wnet ) i s : %4 . 2 f kJ /kg . \n ’ , W n e t ) ;
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 clc
2 clear
3 //DATA GIVEN4 p 1 = 2 8 ; // p r e s s ur e a t 1 i n b ar5 p 2 = 0 . 0 6 ; // p r e s s ur e a t 2 i n b ar6
7 / /At 2 8 b ar , f ro m s te am t a b l e s8 h 1 = 2 8 0 2 ; / / i n k J / kg9 s 1 = 6 . 2 1 0 4 ; // i n k J /kgK
1011 // At 0 . 0 6 bar , fro m s tea m t a b l e s12 h f 2 = 1 5 1 . 5 ; / / i n k J / kg13 h f 3 = h f 2 ;
14 h f g 2 = 2 4 1 5 . 9 ; / / i n k J / kg
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15 s f 2 = 0 . 5 2 1 ; // i n k J /kgK
16 s f 3 = s f 2 ;17 s f g 2 = 7 . 8 0 9 ; // i n k J /kgK18 v f 2 = 0 . 0 0 1 ; //mˆ3/kg19
20
21 / / a s t he stea m e xp an ds i s e n t r o p i c a l l y , s 1=s 222 //sg1= sf 2+ x 2∗ s f g 223 x 2 = ( s 1 - s f 2 ) / s f g 2 ;
24 h 2 = h f 2 + x 2 * h f g 2 ;
25
26 //Wnet=Wturbine−Wpump
27 //Wpump=hf4−hf 3=vf 3 ( p1−p2 )28 W p = v f 2 * ( p 1 - p 2 ) * 1 0 ^ 5 / 1 0 ^ 3 ;
29 h f 4 = W p + h f 2 ;
30 W t = h 1 - h 2 ;
31 W n e t = W t - W p ;
32 Q 1 = h 1 - h f 4 ; / / i n k J / kg33 E T A c y c l e = W n e t / Q 1 ; // c y c l e e f f i c i e n c y34 w r = W n e t / W t ; // work r a t i o35 s s c = 3 6 0 0 / W n e t ; // s p e c i f i c st e am c onsumpt i on
in kg/kWh36
37 printf ( ’ ( i ) The C yc le 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 ’ , E T A c yc l e , ( E T A c y c l e * 1 0 0 ) ) ;
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 s te am c on su mp ti on i n kg /kWh i s : %1. 3 f kg/kWh. \n ’ , s s c ) ;
Scilab code Exa 5.6 Example 6
1 clc
2 clear
3 //DATA GIVEN
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4 p 1 = 3 5 ; // p r e s s u r e a t i n l e t t o
t u r b i n e i n b a r5 x 1 = 1 ;
6 p 2 = 0 . 2 ; // p r e s s ur e a t e xh au st i n ba r7 m = 9 . 5 ; // f l o w r a t e i n kg / s8
9 / /At 3 5 b ar , f ro m s te am t a b l e s10 h g 1 = 2 8 0 2 ; / / i n k J / kg11 h 1 = h g 1 ;
12 s g 1 = 6 . 1 2 2 8 ; // i n k J /kgK13
14 // At 0 . 2 bar , fr om stea m t a b l e s
15 h f 2 = 2 5 1 . 5 ; / / i n k J / kg16 h f 3 = h f 2 ;
17 h f g 2 = 2 3 5 8 . 4 ; / / i n k J / kg18 v f 2 = 0 . 0 0 1 0 1 7 ; //mˆ3/kg19 s f 2 = 0 . 8 3 2 1 ; // i n k J /kgK20 s f g 2 = 7 . 0 7 7 3 ; // i n k J /kgK21
22 //Wnet=Wturbine−Wpump23 //Wpump=hf4−hf 3=vf 3 ( p1−p2 )24 W p = v f 2 * ( p 1 - p 2 ) * 1 0 ^ 5 / 1 0 ^ 3 ;
25 W p n e t = m * W p ;
26 h f 4 = W p + h f 3 ;
27
28 / / a s t he stea m e xp an ds i s e n t r o p i c a l l y , s 1=s 229 //sg1= sf 2+ x 2∗ s f g 230 x 2 = ( s g 1 - s f 2 ) / s f g 2 ; / / d r y n e s s
f r a c t i o n31 h 2 = h f 2 + x 2 * h f g 2 ;
32 W t = h 1 - h 2 ;
33 W t n e t = m * W t ;
34 E T A r a n k i n e = ( h 1 - h 2 ) / ( h 1 - h f 2 ) ; / / R a n k i n e
e f f i c i e n c y35 c h f = m * ( h 2 - h f 3 ) ; / / c o n d e n s e r
h ea t f l ow36
37 printf ( ’ ( i ) The Pump Work i s : %2. 2 f kW. \n ’ , W p n e t ) ;
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38 printf ( ’ ( i i ) The T u r bi n e Work i s : %2 . 2 f kW . \n ’ ,
W t n e t ) ;39 printf ( ’ ( i i i ) The R an ki ne 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 ’ , E T A r a n ki n e , ( E T A r a n k i n e * 1 0 0 ) ) ;
40 printf ( ’ ( i v ) The C o nd en s er h e a t f l o w i s : %1 . 3 f kW .\n ’ , c h f ) ;
41 printf ( ’ ( v ) The d r y n es s a t t he end o f e xp an si on ,x2 i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , x 2 , ( x 2 * 1 0 0 ) ) ;
42
43 //NOTE:44 / /The v al ue o f x2 i n t h e b ook i s g iv en a s 0 . 7 47 045 / / w hi l e t he e x a c t a ns i s 0 . 7 4 75 5
46 // and s o t he v a l ue s o f o t he r a ns we rs a r e v ar y i ng bysome u n i t s
Scilab code Exa 5.7 Example 7
1 clc
2 clear
3 //DATA GIVEN
4 h 1 2 = 8 4 0 ; // A d i a b a t ic e n t ha l p y dr op , (h1−h2 ) i n kJ / kg
5 h 1 = 2 9 4 0 ; // e n t ha l p y o f s te am s u p p l i e di n kJ / kg
6 p 2 = 0 . 1 ; // b ack p r e s s ur e i n b ar7
8 // At 0 . 1 bar , fr om stea m t a b l e s9 h f = 1 9 1 . 8 ; / / i n k J / kg
10 //ETArankine=(hg1−h2) /( hg1−hf 2 )11 E T A r a n k i n e = ( h 1 2 ) / ( h 1 - h f ) ;
12 W u s e = h 1 2 ; // u s e f u l work done p er kg o f s te am i n kJ / kg13 s s c = 1 / W u s e * 3 6 0 0 ; // s p e c i f i c st e am c onsumpt i on14
15 printf ( ’ ( i ) The Ra nki ne e f f i c i e n c y i s : %1 . 4 f o r %2 . 2
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f p e r ce n t . \n ’ , E T A r a n ki n e , ( E T A r a n k i n e * 1 0 0 ) ) ;
16 printf ( ’ ( i i ) The S p e c i f i c s te am c on su mp ti on i s : %1 . 3f kg/kWh. \n ’ , s s c ) ;
Scilab code Exa 5.8 Example 8
1 clc
2 clear
3 //DATA GIVEN
4 I P = 3 5 ; // p owe r d e v el o p ed by t h ee n g i n e i n kW5 m = 2 8 4 ; // f l o w r a t e i n kg /h6 p 1 = 1 5 ; // steam i n l e t p r e s s u re i n
b ar7 p 2 = 0 . 1 4 ; // c o nd e ns e r p r e s s u r e i n b ar8
9 / /At 3 5 b ar and 25 deg c e l s i u s from steam t a b l e s10 h 1 = 2 9 2 3 . 3 ; / / i n k J / kg11 s 1 = 6 . 7 0 9 ; // in kJ/kgK12
13 // At 0 . 1 4 bar , fro m s tea m t a b l e s14 h f 2 = 2 2 0 ; / / i n k J / kg15 h f 3 = h f 2 ;
16 h f g 2 = 2 3 7 6 . 6 ; / / i n k J / kg17 s f 2 = 0 . 7 3 7 ; // i n k J /kgK18 s f g 2 = 7 . 2 9 6 ; // i n k J /kgK19
20 / / a s t he stea m e xp an ds i s e n t r o p i c a l l y , s 1=s 221 //sg1= sf 2+ x 2∗ s f g 222 x 2 = ( s 1 - s f 2 ) / s f g 2 ; / / d r y n e s s
f r a c t i o n23 h 2 = h f 2 + x 2 * h f g 2 ;
24
25 E T A r a n k i n e = ( h 1 - h 2 ) / ( h 1 - h f 2 ) ; / / R a n k i n ee f f i c i e n c y
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26
27 E T A t h e r m a l = I P / ( m / 3 6 0 0 * ( h 1 - h f 2 ) ) ; //Thermale f f i c i e n c y28 E T A r e l = E T A t h e r m a l / E T A r a n k i n e ; / / R e l a t i v e
e f f i c i e n c y29
30 printf ( ’ ( i ) The F i n al c o n d i t i o n o f s tea m i s : %1 . 2 f .\n ’ , x 2 ) ;
31 printf ( ’ ( i i ) The R ank ine 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 ’ , E T A r a n ki n e , ( E T A r a n k i n e * 1 0 0 ) ) ;
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 clc
2 clear
3 //DATA GIVEN4 T 1 = 4 0 0 + 2 7 3 ; / / temp . i n K5 T 2 = T 1 ;
6 T 3 = 4 0 + 2 7 3 ; / / temp . i n K7 T 4 = T 3 ;
8 W = 1 3 0 ; / / wo rk p r od u ce d i n kJ9
10 E T A t h = ( T 1 - T 3 ) / T 1 ; / / E n gi n e t h e rm a le f f i c i e n c y
11
12 //ETAth=Work d one / Heat adde d13 H a = W / E T A t h ; / / Hea t ad ded i n kJ14 H r = H a - W ; // Heat r e j e c t e d i n kJ
15 / / H ea t r e j e c t e d =T3 ( S 3−S4 )16 S 3 4 = H r / T 3 ; / / E nt ro py c ha n ge d u r i n gt h e h e a t r e j e c t i o n p r oc e ss
17
18 printf ( ’ ( i ) The E ng in e t he rm al e f f i c i e n c y i s : %1 . 3 f
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o r %2 . 1 f p e r c e n t . \n ’ ,ETAth ,(ETAth*100));
19 printf ( ’ ( i i ) The H ea t a dd ed i s : %3 . 0 f kJ . \n ’ , H a ) ;20 printf ( ’ ( i i i ) The E nt ro py c ha n ge d u r i n g t h e h e a tr e j e c t i o n p r o c es s i s : %1 . 3 f kJ /K. \n ’ , S 3 4 ) ;
Scilab code Exa 5.10 Example 10
1 clc
2 clear
3 //DATA GIVEN4 p 1 = 1 8 ; //maximum
p r e s s u r e i n ba r5 T 1 = 4 1 0 + 2 7 3 ; //maximum
t e mp e r at u r e i n K6 T 2 = T 1 ;
7 R a c = 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 mp r es s io n , V4/V1=6
8 R i e = 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
9 V 1 = 0 . 1 8 ; // v olume o f a i r
a t b e g i nn i ng o f i s o t he r m al e xp an si on i n mˆ310 w c = 2 1 0 ; // no . o f c y c l e s
p er s11
12 / / gamma f o r a i r = 1 .413 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 4−116 // Als o (T1/T4) =(V4/V1) ˆ( g−1)17 // (V4/V1)=Rac
18 T 4 = T 1 / R a c ^ ( g - 1 ) ;19 T 3 = T 4 ;
20 // p1 ( V1ˆgamma)=p4 (V4ˆgamma)21 //p4=p1 ∗(V1/V4)ˆg22 // where , (V4/V1)=Rac
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23 p 4 = p 1 / ( R a c ^ g ) ;
2425 // f o r i s o t he r ma l p r o c es s 1−226 //p1V1=p2V227 //V1/V2=1/Rie28 p 2 = p 1 * ( 1 / R i e ) ;
29
30 / / f o r i s e n t r o p i c p r o c e s s 2−331 // p2 ( V2ˆgamma)=p3 (V3ˆgamma)32 //V2/V3=V1/V4=1/Rac33 p 3 = p 2 * ( 1 / R a c ) ^ 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 D S = p 1 * 1 0 ^ 5 * V 1 / 1 0 ^ 3 / T 1 * log ( R i e ) ;
37
38 / / H ea t s u p p l i e d , Qs=p 1∗V1∗ lo g (V2/V1)39 //Qs=T1(S2−S1 )40 Q s = T 1 * D S ;
41 //Qr=p4∗V4∗ l o g ( V3/V4 ) // h e a tr e j e c t e d i n kJ
42 //Qr=T4(S3−S4 ) , b cs i n c r e a s e i n e nt ro py d ur in g h ea t
a d di t i o n i s e qu al t o d e cr e a s e i n e nt ro py d ur in gh ea t r e j e c t i o n43 Q r = T 4 * D S ;
44
45 E T A = ( Q s - Q r ) / Q s ; //meant 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
46
47 / / mean e f f e c t i v e p r e s s u r e o f t he c y cl e , Pm = workd on e p e r c y c l e / s t r o k e v ol um e
48 R v 3 1 = R a c * R i e ; // r a t i o o f v ol ume s at 3 and 1 , V3/V1=V3/V2∗V2/V1
49 // st r o k e v olume , Vs=V3−V150 Vs=V1 *(Rv31 -1);
51 J = 1 ;
52 P m = ( Q s - Q r ) * 1 0 ^ 3 / 1 0 ^ 5 * J / V s ;
53
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54 P = ( Q s - Q r ) * w c / 6 0 ; / / p ow er o f
t he e n g in e i n kW55
56 printf ( ’ ( i ) The P r e s s u r e and T em pe ra tu re a t p o i n t 1ar e : \ n ’ ) ;
57 printf ( ’ p1 : %2 . 0 f b a r . \ n ’ , p 1 ) ;
58 printf ( ’ T1 : %3 . 0 f K.\ n ’ , T 1 ) ;
59 printf ( ’ The P re ss ur e and Te mper at ur e a t p o in t2 a r e :\ n ’ ) ;
60 printf ( ’ p2 : %2 . 0 f b a r . \ n ’ , p 2 ) ;
61 printf ( ’ T2 : %3 . 0 f K.\ n ’ , T 2 ) ;
62 printf ( ’ The P re ss ur e and Te mper at ur e a t p o in t
3 a r e :\ n ’ ) ;63 printf ( ’ p3 : %1 . 2 f b a r . \ n ’ , p 3 ) ;
64 printf ( ’ T3 : %3 . 1 f K.\ n ’ , T 3 ) ;
65 printf ( ’ The P re ss ur e and Te mper at ur e a t p o in t4 a r e :\ n ’ ) ;
66 printf ( ’ p4 : %1 . 2 f b a r . \ n ’ , p 4 ) ;
67 printf ( ’ T4 : %3 . 1 f K.\ n ’ , T 4 ) ;
68 printf ( ’ ( i i ) The Ch ang e i n e n t r o p y d u r i n gi s o t h e r m a l e x p an s i o n i s : %1 . 3 f kJ /K . \n ’ , D S ) ;
69 printf ( ’ ( i i i ) The Mean t h er m al 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 ce n t . \n ’, E T A , ( E T A
* 1 0 0 ) ) ;
70 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 bar . \n ’ , P m ) ;
71 printf ( ’ ( v ) The P ower o f t he e ng i ne w o rk in g ont h i s c y c l e i s g i ve n by : %3 . 1 f kW. ’ , P ) ;
72
73 //NOTE:74 / / 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 ns we r s o f book due
t o r o un di ng o f f o f t h e v a l u es
Scilab code Exa 5.11 Example 11
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1 clc
2 clear3 //DATA GIVEN4 //CASE−15 //( T 1−T2)/T1=1/66 //SO , T1 = 1 . 2 ( T2 ) . . . . . . . . . Eqn ( 1 )7
8 //CASE−29 // T2 REDUCED BY 70 DEG. CELSIUS
10 //{T1−[T2−(70+27 3) ]}/ T1 = 1 / 3 . . . . . . . . . . . . . . Eqn ( 2 )11 //2T1=3T2−102912
13 / /By Eqn ( 1 ) and ( 2 )14 T 2 = ( 7 0 + 2 7 3 ) * 3 / ( 3 - 2 * 1 . 2 ) ;
15 T 1 = 1 . 2 * T 2 ;
16
17 printf ( ’ ( i ) The T em pe ra tu re o f t h e S o ur ce , T1 i s : %4. 0 f K o r %4 . 0 f deg . c e l s i u s . \n ’ , T 1 , ( T 1 - 2 7 3 ) ) ;
18 printf ( ’ ( i i ) The T em pe ra tu re o f t h e S in k , T2 i s : %4. 0 f K o r %4 . 0 f deg . c e l s i u s . \n ’ , T 2 , ( T 2 - 2 7 3 ) ) ;
Scilab code Exa 5.12 Example 12
1 clc
2 clear
3 //DATA GIVEN4 T 1 = 1 9 9 0 ; / / T em pe ra tu re o f t h e
h ea t S ou rc e i n K5 T 2 = 8 5 0 ; / / T em pe ra tu re o f t h e
h ea t S in k i n K
6 Q s = 3 2 . 5 ; // h ea t s u p p l i e d i n kJ /mi n7 P = 0 . 4 ; / / po wer d e v el o p e d by t h e
e n g i n e i n kW8
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9 E T A c a r n o t = ( T 1 - T 2 ) / T 1 ;
10 / / A l s o ETAth=w or k d o ne / H ea t s u p p l i e d11 E T A t h = P / Q s ;
12
13 printf ( ’ The E f f i c i e n c y o f c ar no t c y c l e i s : %1 . 3 f o r%2 . 1 f p e r c e n t . \n ’ , E T A c a rn o t , ( E T A c a r n o t * 1 0 0 ) ) ;
14 printf ( ’ The Thermal e f f i c i e n c y o f e ng i ne c la im ed byi n v e n t o r i s : %1 . 3 f o r %2 . 1 f p e r ce n t . \n\n ’ ,ETAth
, ( E T A t h * 1 0 0 ) ) ;
15
16 if ( E T A t h > E T A c a r n o t )
17 printf ( ’ Thus , The c la i m o f t he i n v en t o r i s
p o s s i b l e . ’ ) ;18 else
19 printf ( ’ Thus , The c l ai m o f t he i n v e n t o r i s NOTf e a s i b l e , \n a s no e ng i ne can be moree f f i c i e n t t h an t h a t w o r k in g o n c a r n o t c y c l e . ’) ;
Scilab code Exa 5.13 Example 13
1 clc
2 clear
3 //DATA GIVEN4 E T A o t t o = 6 0 ; // E f f i c i e n c y o f o t t o
c y c l e i n %5 s h r = 1 . 5 ; / / r a t i o o f s p e c i f i c
h e a t s6
7 //ETAotto=1−1/(r ) ̂ ( sh r −1)
8 r = ( 1 / ( 1 - E T A o t t o / 1 0 0 ) ) ^ ( 1 / ( s h r - 1 ) ) ; / / c o m p r e s s i o nr a t i o9
10 printf ( ’ The c o m p re s s i on r a t i o i s : %1 . 2 f . ’ , r ) ;
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Scilab code Exa 5.14 Example 14
1 clc
2 clear
3 //DATA GIVEN4 D = 0 . 2 5 ; // b or e o f t he
e ng i n e i n m5 L = 0 . 3 7 5 ; // s t r o k e o f t he
e ng i n e i n m6 V c = 0 . 0 0 2 6 3 ; / / c l e a r a n c e
v ol um e i n mˆ 37 p 1 = 1 ; // i n i t i a l
p r e s s u r e i n ba r8 T 1 = 5 0 + 2 7 3 ; // i n i t i a l
t e mp e r at u r e i n K9 p 3 = 2 5 ; //maximum
p r e s s u r e i n ba r10
11 V s = ( % p i / 4 ) * D ^ 2 * L ; / / s w e pt v ol um e
12 r = ( V s + V c ) / V c ; / / c o m p r e s s i o nr a t i o
13
14 / / f o r a i r , gamma = 1. 415 g = 1 . 4 ;
16 // A ir s ta nd ar d e f f i c i e n c y o f o t to c y c l e ETAotto=1−1/( r ) ̂ ( g−1)
17 E T A o t t o = 1 - 1 / ( r ) ^ ( g - 1 ) ;
18
19 // f o r a d i a b a t i c p r oc e ss 1−220 // p1 ( V1ˆgamma)=p2 (V2ˆgamma)21 //p2=p1 ∗(V1/V2)ˆg22 // where , (V1/V2)=r23 p 2 = p 1 * ( r ^ g ) ; //
p r e s s u r e a t 2 i n ba r
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24 r p = p 3 / p 2 ; //
p r e s s u r e r a t i o25 P m = p 1 * r * [ ( r ^ ( g - 1 ) - 1 ) * ( r p - 1 ) ] / [ ( g - 1 ) * ( r - 1 ) ] ; //meane f f e c t i v e p r e s s u r e i n b ar
26
27 printf ( ’ ( i ) The A i r s ta nd ar d e f f i c i e n c y o f o t t oc y c l e i s : %1 . 3 f o r %2 . 1 f p e r ce n t . \n ’ ,ETAot to ,(
E T A o t t o * 1 0 0 ) ) ;
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 bar . \n ’ , P m ) ;
Scilab code Exa 5.15 Example 15
1 clc
2 clear
3 //DATA GIVEN4 T 1 = 3 8 + 2 7 3 ; // i n i t i a l
t e mp e r at u r e i n K5 T 3 = 1 9 5 0 + 2 7 3 ; //maximum
t e m p e r a t u r e K
6 r p = 1 5 ; // p r e s s u r e r a t i o7 / / f o r a i r , gamma = 1. 48 g = 1 . 4 ;
9
10 / / f o r a d i a b a t i c c o mp r es s io n 1−211 // p1 ( V1ˆgamma)=p2 (V2ˆgamma)12 //(V1/V2)=r13 r = ( r p ) ^ ( 1 / g ) ;
14
15 / / T he rm al e f f i c i e n c y ETAth=1−1/( r ) ˆ( g−1)
16 E T A t h = 1 - 1 / ( r ) ^ ( g - 1 ) ;17
18 / / f o r a d i a b a t i c c o mp r es s io n 1−219 // (T2/T1) =(V1/V2) ˆ( g−1)20 //(V1/V2)=r
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21 T 2 = T 1 * r ^ ( g - 1 ) ;
2223 // f o r a d i a b a t i c e xp an si on 3−424 // (T3/T4) =(V4/V3) ˆ( g−1)25 //(V4/V3)=r26 T 4 = T 3 / r ^ ( g - 1 ) ;
27
28 / / h e at s u p p l i e d p e r kg o f a i r , Qs=m∗Cv∗ ( T3−T2 )29 R = 0 . 2 8 7 ;
30 C v = R / ( g - 1 ) ;
31 Q s = C v * ( T 3 - T 2 ) ;
32
33 // h ea t r e j e c t e d p er kg o f a i r , Qr=m∗Cv∗ ( T4−T1 )34 Q r = C v * ( T 4 - T 1 ) ;
35
36 W = Q s - Q r ; / / wor k d on e p e r kgo f a i r
37
38 printf ( ’ ( i ) The c o mp r es s io n r a t i o i s : %1 . 1 f .\ n ’ , r ) ;
39 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 d o ne i s : %3 . 1 f k J o r %6 . 0 f Nm
. ’, W , ( W * 1 0 0 0 ) ) ;
41
42 //NOTE:43 / / 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 ns we r s i n t he book
b ec au se o f r o un di ng o f f o f t h e v a l u es
Scilab code Exa 5.16 Example 16
1 clc2 clear
3 //DATA GIVEN4 V 1 = 0 . 4 5 ; / / v o lu me i n mˆ 35 p 1 = 1 ; // i n i t i a l pr e s su r e
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i n b ar
6 T 1 = 3 0 + 2 7 3 ; // i n i t i a lt e mp e r at u r e i n K7 p 2 = 1 1 ; // p r e s s u r e a t t he
end o f c om pr es si on s t r o k e i n ba r8 Q s = 2 1 0 ; / / h e a t a d da ed a t
c o n s t a n t vo lu me i n kJ9 w c = 2 1 0 ; // no . o f w or ki ng
c y c l e s / min10
11 / / f o r a i r , gamma = 1. 412 g = 1 . 4 ;
1314 / / f o r a d i a b a t i c c o mp r es s io n 1−215 // p1 ( V1ˆgamma)=p2 (V2ˆgamma)16 //(V1/V2)=r17 r = ( p 2 / p 1 ) ^ ( 1 / g ) ;
18 // Als o (T2/T1) =(V1/V2) ˆ( g−1)19 //(V1/V2)=r20 T 2 = T 1 * r ^ ( g - 1 ) ;
21
22 / / A p pl yi ng g as l aw s t o p o i n ts 1 and 223
//p1V1/T1=p2V2/T224 V 2 = T 2 / T 1 * p 1 / p 2 * V 1 ;
25
26 // h ea t s u p pl i e d d ur in g p r o c e s s 2−3 , Qs=mCv( T3−T2 )27 R = 2 8 7 ;
28 m = p 1 * 1 0 ^ 5 * V 1 / R / T 1 ;
29 C v = R / 1 0 0 0 / ( g - 1 ) ;
30 T 3 = Q s / m / C v + T 2 ;
31
32 // f o r c o n st a n t volume p r o c e s s 2−333 //p3/T3=p2/T2
34 p 3 = p 2 / T 2 * T 3 ;35 V 3 = V 2 ;
36
37 // f o r a d i a b a t i c e xp an si on 3−438 // p3 ( V3ˆgamma)=p4 (V4ˆgamma)
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39 //(V4/V3)=r
40 p 4 = p 3 * ( 1 / r ) ^ ( g ) ;41 // Als o T3/T4) =(V4/V3) ˆ( g−1)42 //(V4/V3)=r43 T 4 = T 3 / r ^ ( g - 1 ) ;
44 V 4 = V 1 ;
45
46 // pe r c e n t a ge c l e a r a nc e , pc=Vc /Vs=V2/( V1−V2 )47 p c = V 2 / ( V 1 - V 2 ) * 1 0 0 ;
48
49 / / h e a t r e j e c t e d p e r c y c l e , Qr=Cv∗( T4−T1 )50 Q r = m * C v * ( T 4 - T 1 ) ;
5152 // A i r s t an d ar 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)/Qs53 E T A o t t o = ( Q s - Q r ) / Q s ;
54 / / A l t e r n a t i v e l y55 //ETAotto=1−1/( r ) ˆ( g−1)56 E T A o t t o = 1 - 1 / ( r ) ^ ( g - 1 ) ;
57
58 / / m ean e f f e c t i v e p r e s s u r e , Pm=W/ Vs59 W = Q s - Q r ; / / wor k d on e p e r kg
o f a i r60 V s = V 1 - V 2 ;
61 P m = W * 1 0 ^ 3 / 1 0 ^ 5 / V s ;
62
63 / / p ow er d e v el o p ed , P=wo rk d on e p e r c y c l e ∗no . o f c y c l e s p e r s
64 P = W * ( w c / 6 0 ) ;
65
66 printf ( ’ ( i ) The P r e s s u r e , T e mp e ra t ur e a nd V ol um esa t s a l i e n t p o i n t s i n t h e c y c l e a re :\ n ’ ) ;
67 printf ( ’ At p o i n t 1 a r e : \ n ’ ) ;
68 printf ( ’ p1 : %1 . 1 f b a r . \ n ’ , p 1 ) ;69 printf ( ’ V1 : %1 . 2 f mˆ 3 .\ n ’ , V 1 ) ;
70 printf ( ’ T1 : %3 . 0 f K.\ n ’ , T 1 ) ;
71 printf ( ’ At p o i n t 2 a r e : \ n ’ ) ;
72 printf ( ’ p2 : %2 . 2 f b a r . \ n ’ , p 2 ) ;
72
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73 printf ( ’ V2 : %1 . 3 f mˆ 3 .\ n ’ , V 2 ) ;
74 printf ( ’ T2 : %3 . 0 f K.\ n ’ , T 2 ) ;75 printf ( ’ At p o i n t 3 a r e : \ n ’ ) ;
76 printf ( ’ p3 : %2 . 2 f b a r . \ n ’ , p 3 ) ;
77 printf ( ’ V3 : %1 . 3 f mˆ 3 .\ n ’ , V 3 ) ;
78 printf ( ’ T3 : %4 . 0 f K.\ n ’ , T 3 ) ;
79 printf ( ’ At p o i n t 4 a r e : \ n ’ ) ;
80 printf ( ’ p4 : %1 . 2 f b a r . \ n ’ , p 4 ) ;
81 printf ( ’ V4 : %1 . 2 f mˆ 3 .\ n ’ , V 4 ) ;
82 printf ( ’ T4 : %3 . 1 f K.\ n ’ , T 4 ) ;
83 printf ( ’ ( i i ) The P e r ce n t ag e c l e a r a n c e i s : %2 . 2 f p e r c e n t . \n ’ , p c ) ;
84 printf ( ’ ( i i i ) The A ir s ta nd ar d e f f i c i e n c y o f t hec y c l e i s : %1 . 3 f o r %2 . 1 f p e r ce n t . \n ’ ,ETAot to ,(
E T A o t t o * 1 0 0 ) ) ;
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 bar . \n ’ , P m ) ;
86 printf ( ’ ( v ) The Power d e v el o p e d i s : %3 . 1 f kW. ’ , P ) ;
87
88 //NOTE:89 / / 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 ns we r s i n t he book
b ec au se o f r o un di ng o f f o f t h e v a l u es
Scilab code Exa 5.17 Example 17
1 clc
2 clear
3 //DATA GIVEN4 r = 1 5 ; / / c o m p re s s i on r a t i o5 //V3−V2=a/100∗Vs . . . . . . . . . . . . Vs=s t r o k e volume=V1−V2
6 //V3=1.84V27 c = 6 ; // h ea t a d d i t i o n t a k esp l a c e a t ’ a ’ p e r c e nt o f s t r o k e
8 / / f o r a i r , gamma = 1. 49 g = 1 . 4 ;
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10
11 // A ir s ta nd ar 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 ) ̂ ( g−1) ] [ ( rh o ˆg −1)/ ( rho −1) ]12 // r h o=c u t o f f r a t i o =V3 /V213 r h o = c / 1 0 0 * ( r - 1 ) + 1 ;
14 E T A d i e s e l = 1 - [ 1 / g / ( r ) ^ ( g - 1 ) ] * [ ( r h o ^ g - 1 ) / ( r h o - 1 ) ] ;
15
16 printf ( ’ The A i r s t an da rd e f f i c i e n c y o f d i e s e l c y c l ei s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ ,ETAdiesel ,(
E T A d i e s e l * 1 0 0 ) ) ;
Scilab code Exa 5.18 Example 18
1 clc
2 clear
3 //DATA GIVEN4 L = 0 . 2 5 ; // s t r o k e o f t he
e ng i n e i n m5 D = 0 . 1 5 ; / / d i a m e t e r o f
c y l i n d e r i n m
6 V 2 = 0 . 0 0 0 4 ; / / c l e a r a n c ev ol um e i n mˆ 3
7 V s = ( % p i / 4 ) * D ^ 2 * L ; / / s w ep t v ol um e i nmˆ3
8 V t = V s + V 2 ; // t o t a l c y l i n d e rv ol um e i n mˆ 3
9 c = 5 ; // f u e l i n j e c t i o nt ak es p l a c e a t ’ c ’ p er ce nt o f s t r o k e
10 V 3 = V 2 + c / 1 0 0 * V s ; / / vo lu me a t p o i n to f c ut−o f f i n mˆ3
11 r h o = V 3 / V 2 ; / / c u t−o f f r a t i o12 r = ( V s + V 2 ) / V 2 ; / / c o m p r e s s i o nr a t i o
13
14 / / f o r a i r , gamma = 1. 4
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15 g = 1 . 4 ;
1617 // A ir s ta nd ar 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 ) ̂ ( g−1) ] [ ( rh o ˆg −1)/ ( rho −1) ]18 E T A d i e s e l = 1 - [ 1 / g / ( r ) ^ ( g - 1 ) ] * [ ( r h o ^ g - 1 ) / ( r h o - 1 ) ] ;
19
20 printf ( ’ The E f f i c i e n c y o f d i e s e l e ng in e i s : %1 . 3 f o r %2 . 1 f p e r c e n t . \n ’ , E T A d i es e l , ( E T A d i e s e l * 1 0 0 ) ) ;
Scilab code Exa 5.19 Example 19
1 clc
2 clear
3 //DATA GIVEN4 r = 1 4 ; / / c o m p re s s i on r a t i o5 / / f u e l c ut−o f f i s d el ay ed from 5−8%6 / / f o r a i r , gamma = 1. 47 g = 1 . 4 ;
8
9 // when f u e l i s cut−o f f a t 5%
10 c 1 = 5 ;11 r h o 1 = c 1 / 1 0 0 * ( r - 1 ) + 1 ;
12 // E f f i c i e n c y o f d i e s e l e ng i ne E TA d ie se l =1−[1/(r ) ̂ ( g−1) ] [ ( rh o ˆg−1)/ ( rho −1) ]
13 E T A d i e s e l 1 = 1 - [ 1 / g / ( r ) ^ ( g - 1 ) ] * [ ( r h o 1 ^ g - 1 ) / ( r h o1 - 1 ) ] ;
14
15 // when f u e l i s cut−o f f a t 8%16 c 2 = 8 ;
17 r h o 2 = c 2 / 1 0 0 * ( r - 1 ) + 1 ;
18 // E f f i c i e n c y o f d i e s e l e ng i ne E TA d ie se l =1−[1/(r ) ̂ ( g
−1) ] [ ( rh o ˆg−1)/ ( rho −1) ]19 E T A d i e s e l 2 = 1 - [ 1 / g / ( r ) ^ ( g - 1 ) ] * [ ( r h o 2 ^ g - 1 ) / ( r h o2 - 1 ) ] ;
20
21 E T A l o s s = ( E T A d i e s e l1 - E T A d i e s e l 2 ) * 1 0 0 ;
22
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23 printf ( ’ The P er ce nt ag e l o s s i n e f f i c i e n c y due t o
d el ay i n f u e l cut−o f f i s : %1 . 1 f p e r ce n t . \n ’ ,E T A l o s s ) ;
Scilab code Exa 5.20 Example 20
1 clc
2 clear
3 //DATA GIVEN
4 P m = 7 . 5 ; // mean e f f e c t i v ep r e s s u r e i n ba r5 r = 1 2 . 5 ; / / c o m p re s s i on r a t i o6 p 1 = 1 ; / / i n i t i a l p r e s s u r e i n
b ar7
8 / / f o r a i r , gamma = 1. 49 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 ∗( r h o −1)−rˆ( 1−g ) ∗ ( r ho ˆg−1) ] / [ ( g−1) ∗ ( r −1) ]
12 / / we g e t , 0 . 3 4 6 ( r h o ) ̂ 1 . 4 −1. 4( rho ) +2.0 413 / /By t r i a l a nd e r r o r m ethod , we g e t14 r h o = 2 . 2 4 ;
15
16 c o = ( r h o - 1 ) / ( r - 1 ) * 1 0 0 ; //% cut−o f f 17
18 printf ( ’ The P e r ce n t ag e c ut−o f f o f t h e c y c l e i s : %2. 2 f p e r c e n t . \n ’ , c o ) ;
Scilab code Exa 5.21 Example 21
1 clc
2 clear
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31
32 / / c u t−o f f r a t i o , c =( rho −1 ) / ( r −1)33 r h o = c / 1 0 0 * ( r - 1 ) + 1 ;
34 V 3 = r h o * V 2 ;
35 // a l t e r n a t i v e l y36 V 3 = c / 1 0 0 * V s + V c ;
37
38 / / f o r c o ns t an t p r e s su r e p r o c es s 2−339 //V3/T3=V2/T240 T 3 = T 2 / V 2 * V 3 ;
41
42 / / f o r i s e n t r o p i c p r o c e s s 3−4
43 // p3 ( V3ˆgamma)=p4 (V4ˆgamma)44 // ( V4/V)=V4/V2∗V2/V3=V1/V2∗V2/V3=r/rho45 p 4 = p 3 * ( ( r h o / r ) ^ g ) ;
46 // Als o (T4/T3) =(V3/V4) ˆ( g−1)47 // ( V4/V)=V4/V2∗V2/V3=V1/V2∗V2/V3=r/rho48 T 4 = T 3 * ( ( r h o / r ) ^ ( g - 1 ) ) ;
49 V 4 = V 1 ;
50
51 // A ir s ta nd ar 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 ) ̂ ( g−1) ] [ ( rh o ˆg −1)/ ( rho −1) ]
52 E T A d i e s e l = 1 - [ 1 / g / ( r ) ^ ( g - 1 ) ] * [ ( r h o ^ g - 1 ) / ( r h o - 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 ∗( r h o −1)−rˆ( 1−g ) ∗ ( r ho ˆg−1) ] / [ ( g−1) ∗ ( r −1) ] ;
55 P m = p 1 * r ^ g * [ g * ( r h o - 1 ) - r ^ ( 1 - g ) * ( r h o ^ g - 1 ) ] / [ ( g - 1 ) * ( r - 1 )
];
56
57 P = P m * 1 0 ^ 5 * V s / 1 0 ^ 3 * ( w c / 6 0 ) ; / / Po we r o f t he e n g in e i n kW
58
59 printf ( ’ ( i ) The P r e s s u r e , T e mp e ra t ur e a nd V ol um es
a t s a l i e n t p o i n t s i n t h e c y c l e a re :\ n ’ ) ;60 printf ( ’ At p o i n t 1 a r e : \ n ’ ) ;
61 printf ( ’ p1 : %1 . 1 f b a r . \ n ’ , p 1 ) ;
62 printf ( ’ V1 : %1 . 4 f mˆ 3 .\ n ’ , V 1 ) ;
63 printf ( ’ T1 : %3 . 0 f K.\ n ’ , T 1 ) ;
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64 printf ( ’ At p o i n t 2 a r e : \ n ’ ) ;
65 printf ( ’ p2 : %2 . 2 f b a r . \ n ’ , p 2 ) ;66 printf ( ’ V2 : %1 . 7 f mˆ 3 .\ n ’ , V 2 ) ;
67 printf ( ’ T2 : %3 . 1 f K.\ n ’ , T 2 ) ;
68 printf ( ’ At p o i n t 3 a r e : \ n ’ ) ;
69 printf ( ’ p3 : %2 . 2 f b a r . \ n ’ , p 3 ) ;
70 printf ( ’ V3 : %1 . 6 f mˆ 3 .\ n ’ , V 3 ) ;
71 printf ( ’ T3 : %4 . 1 f K.\ n ’ , T 3 ) ;
72 printf ( ’ At p o i n t 4 a r e : \ n ’ ) ;
73 printf ( ’ p4 : %1 . 3 f b a r . \ n ’ , p 4 ) ;
74 printf ( ’ V4 : %1 . 4 f mˆ 3 .\ n ’ , V 4 ) ;
75 printf ( ’ T4 : %3 . 2 f K.\ n ’ , T 4 ) ;
76 printf ( ’ ( i i ) The T h e o r i t i c a l a i r s ta nd ar de 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 ’ , E T A d i es e l , ( E T A d i e s e l * 1 0 0 ) ) ;
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 bar . \n ’ , P m ) ;
78 printf ( ’ ( i v ) The Po wer d e v e l o p e d i s : %2 . 2 f kW . ’ , P ) ;
79
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Chapter 6
Steam Boilers
Scilab code Exa 6.1 Example 1
1 clc
2 clear
3 //DATA GIVEN4 L C V = 4 4 7 0 0 ; //LCV o f f u e l i n kJ5 a f r n = 2 0 ; // a i r p a rt s =20 i n a i r f u e l
m i x t u r e
6 a f r d = 1 ; // f u e l p a r t s =1 i n a i r f u e lm i x t u r e
7 C p g = 1 . 0 8 ; / / a vg s p e c i f i c h e a t i n kJ /kgK
8 T 1 = 3 8 + 2 7 3 ; // b o i l e r room temp . i n K9
10 / / h e at o f c om bu st io n=h e at o f g a s e s11 // 1∗44700=Mg∗Cpg ∗ ( T2−T1 )12 T 2 = a f r d * L C V / ( a f r n + a f r d ) / C p g + T 1 ;
13
14 printf ( ’ The Maximum temp . T2 a t t a i n e d i n t h ef ur na c e o f t h e b o i l e r i s : \ n %5 . 0 f K el v in ’ , T 2 ) ;
15 printf ( ’ o r %5 . 0 f d e gr e e c e l s i u s . \ n ’ , ( T2 - 2 7 3 ) ) ;
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Scilab code Exa 6.2 Example 2
1 clc
2 clear
3 //DATA GIVEN4 M s = 5 . 4 ; // mass o f s team u se d i n kg /
kWh5 p = 5 0 ; // p r e s s u r e o f stea m i n b ar6 T s u p = 3 5 0 ; / / temp . o f s te am i n d eg
c e l s i u s7 e t a = 8 2 ; // b o i l e r e f f i c i e n c y i n %8 T f w = 1 5 0 ; / / f e e d w at er temp . i n d eg c e l
; s i u s9 C = 2 8 1 0 0 ; / / c a l o r i f i c v a l u e o f c o a l i n
kJ10 r a t e = 5 0 0 ; // c o s t o f c o a l / t on ne i n Rs11
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 ta=Ms∗( hsup−hf 1 ) /(Mf ∗C)
13 / / f rom stea m t a bl e , a t 45 b ar and 35 0 deg c e l s i u s ,h su p = 3 0 6 8. 4 k J / kg
14 h = 3 0 6 8 . 4 ; / / e n t h a l p y a t45 b a r and 350 deg c e l s i u s
15 h f 1 = 4 . 1 8 * ( T f w - 0 ) ; // h f 1 a t 1 50deg c e l s i u s i n k J /kg
16
17 // s ub s . t he se i n eq . o f b o i l e r e f f i c i e n c y18 M f = M s * ( h - h f 1 ) / ( ( e t a / 1 0 0 ) * C ) ; // mass o f c o a l
r e q u i r e d i n kg /kWh19 c o s t = ( M f / 1 0 0 0 ) * r a t e * 1 0 0 ; // c o s t o f c o a l
i n pa i s a /kWh20
21 printf ( ’ ( i ) The mass o f c o a l r e q ui r e d i s : %5 . 3 f kg/kWh. \n ’ , M f ) ;
81
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22 printf ( ’ ( i i ) The T ot al c o s t o f f u e l ( c o a l ) i s : %2 . 1
f pa i s a /kWh. \n ’ , c o s t ) ;23
24 //NOTE: in te xt book25 / / i n q u e s t i on p r e s s ur e i s g i ve n a s =50 ba r26 / / but from steam t a b l e e nt ha l p y i s f o un d a t 45 b ar
Scilab code Exa 6.3 Example 3
1 clc2 clear
3 //DATA GIVEN4 M c = 1 2 5 0 ; // q u an t it y o f c o a l i n kg
co ns um ed i n 2 4 h o u r s5 M w = 1 3 0 0 0 ; // m ass o f w at er
e va p or a te d i n kg6 M E P s = 7 ; // mean e f f e c t i v e p r e s s ur e
o f steam i n b ar7 T f w = 4 0 ; / / f e e d w a te r temp . i n d eg
c e l s i u s
8 h = 2 5 7 0 . 7 ; // e n th a lp y o f s tea m a t 7b ar i n kJ / kg
9 C = 3 0 0 0 0 ; / / c a l o r i f i c v a l u e o f c o a li n kJ / kg
10
11 M a = M w / M c ; // mass o f w at er a c t u a l l ye va po ra t e d p er kg o f f u e l
12 h f 1 = 4 . 1 8 * ( T f w - 0 ) ;
13 h f g = 2 2 5 7 ; / / i n kJ / k g14 M e = M a * ( h - h f 1 ) / h f g ; / / i n k g
15 e t a = M a * ( h - h f 1 ) / C ; // b o i l e r e f f i c i e n c y16
17 printf ( ’ ( i ) The e q ui v a l e n t e v a p o r a t i o n p er kg o f c o a l , Me i s : %5 . 3 f kg . \n ’ , M e ) ;
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
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. 3 f o r %2 . 1 f p e r ce n t . ’ , e t a , e t a * 1 0 0 ) ;
Scilab code Exa 6.4 Example 4
1 clc
2 clear
3 //DATA GIVEN4 p = 1 2 ; / /mean s te am p r e s s u r e i n
b ar
5 M s = 4 0 0 0 0 ; / / mass o f s tea m g e n e r a t edi n kg6 x = 0 . 8 5 ; // mean d r yn e s s f r a c t i o n7 T f w = 3 0 ; / /mean f e e d wa te r temp . i n
deg c e l s i u s8 M c = 4 0 0 0 ; // mass o f c o a l u se d i n kg9 C = 3 3 4 0 0 ; / / c a l o r i f i c v a l u e o f c o a l
i n kJ / kg10
11 // f rom stea m t a bl e , c o r r es p o n d i ng t o 12 bar ,12 h f = 7 9 8 . 4 ; / / i n k J / kg
13 h f g = 1 9 8 4 . 3 ; / / i n k J / kh14 h = h f + x * h f g ; / / i n k j / k g15 h f 1 = 4 . 1 8 * ( T f w - 0 ) ; // h ea t o f f e e d w a te r i n kJ /
kg16
17 F e = ( h - h f 1 ) / 2 2 5 7 ; // f a c t o r o f e q u i v al e n te v a p o r a t i o n , Fe
18 M a = M s / M c ; // p er kg o f f u e l19 M e = M a * ( h - h f 1 ) / 2 2 5 7 ; // ( kg o f s tea m ) / ( kg o f f u e l
)
20 e t a = M a * ( h - h f 1 ) / C ; // e f f i c i e n c y o f b o i l e r21
22 printf ( ’ ( i ) The F ac to r o f e q u i v a l e n t t em er at ur e , Fei s : %5 . 3 f \n ’ , F e ) ;
23 printf ( ’ ( i i ) The E q ui v a le n t e v a p or a t i on from and
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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 ’ , M e ) ;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 ’ ,
e t a ) ;
25 printf ( ’ o r %5 . 2 f p e r c e n t . \n ’ , e t a * 1 0 0 ) ;
Scilab code Exa 6.5 Example 5
1 clc
2 clear3 //DATA GIVEN4 M = 1 8 0 0 0 ; // mass o f stea m g e n er a t ed i n
k g / h r5 p = 1 2 . 5 ; // s tea m p r e s s u r e i n b ar6 x = 0 . 9 7 ; // q u a l i t y o f stea m7 T f w = 1 0 5 ; // f e e d w at er temp . i n deg
c e l s i u s8 M f = 2 0 4 0 ; // r a t e o f c o a l f i r i n g i n kg /
hr9 C = 2 7 4 0 0 ; / / 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 / kg10
11 // f rom stea m t a bl e , c o r r es p o n d i ng t o 1 2 . 5 bar ,12 h f = 8 0 6 . 7 ; / / i n k J / kg13 h f g = 1 9 7 7 . 4 ; / / i n k J / kg14 h = h f + x * h f g ; / / i n k J / kg15 h f 1 = 4 . 1 8 * ( T f w - 0 ) ; // h ea t of f e e d w a te r i n kJ /
kg16
17 // h ea t r a t e o f t he b o i l e r = h ea t s u pp l i e d p er hour
18 h e a t r a t e = M * ( h - h f 1 ) // h ea t r a te o f b o i l e r19 M a = M / M f ; // i n kg p er kg o f f u e l20 M e = M a * ( h - h f 1 ) / 2 2 5 7 ; // ( kg o f s team ) / ( kg o f f u e l )21 e t a = M a * ( h - h f 1 ) / C ; // t he rm al e f f i c i e n c y22
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23 printf ( ’ ( i ) The Heat r a t e o f b o i l e r i s : %1 . 4 e kJ /h .
\n ’ , h e a t r a t e ) ;24 printf ( ’ ( i i ) The E q u i v a le n t e v a p or a t i on , Me i s : %5. 3 f ( k g o f s team ) / ( kg o f f u e l ) . \n ’ , M e ) ;
25 printf ( ’ ( i i i ) The Therma l 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 ’ , e t a * 1 0 0 ) ;
Scilab code Exa 6.6 Example 6
1 clc
2 clear
3 //DATA GIVEN4 M w = 5 9 4 0 ; / / mass o f w at er e v a po r a te d kg / h r5 M c = 6 7 5 ; // mass o f c o a l b ur nt i n kg / hr6 C = 3 1 6 0 0 ; // lo we r c a l o r i f i c va lu e (LCV) of
c o a l i n kJ /kg7 p 1 = 1 4 ; // p r e s s ur e o f steam a t b o i l e r
s to p v al ve i n ba r8 T e 1 = 3 2 ; // temp . o f f e e d w at er e n t e r i n g
e co no mi se r i n deg c e l s i u s9 T e 2 = 1 1 5 ; // temp . o f f e e d w at er l e a v i n g
e co no mi se r i n deg c e l s i u s10 x = 0 . 9 6 ; // d r yn e s s f r a c t i o n o f stea m
e n t e r i n g s u p e r h e a t e r11 T s u p = 2 6 0 ; / /temp . o f s te am l e a v i n g
s u pe r h e a t er i n deg c e l s i u s12 C p = 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
steam13
14 h f 1 = 4 . 1 8 * ( T e 2 - T e 1 ) ; // hea tu t i l i s e d by 1 kg o f f e e d w a t e r i n e co no mi se r15 // f rom stea m t a bl e , c o r r es p o n d i ng t o 14 bar ,16 T s = 1 9 5 ;
17 h f = 8 3 0 . 1 ;
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18 h f g = 1 9 5 7 . 7 ;
19 h b o i l e r = ( h f + x * h f g ) - h f 1 ; // hea tu t i l i s e d by 1 kg o f f e ed w a t e r i n b o i l e r20 h s u p e r h e a t e r = ( 1 - x ) * h f g + C p * ( T s u p - T s ) ; // hea t
u t i l i s e d by 1 kg o f f ee d w a t e r i n s u p e r he at e r21 M a = M w / M c ; / / i n kg p e r
kg o f f u e l22 P e = h f 1 / C * M a * 1 0 0 ; / /% o f h e at
u t i l i s e d i n e co no mi se r23 P b = h b o i l e r / C * M a * 1 0 0 ; / /% o f h e at
u t i l i s e d i n b o i l e r24 P s = h s u p e r h e a t e r / C * M a * 1 0 0 ; / /% o f h e at
u t i l i s e d i n s up e r he a t e r25 h t o t a l = h f 1 + h b o i l e r + h s u p e r h e a t e r ; // t o t a l h ea t
a bs or be d i n kg o f w at er26 e t a = M a * h t o t a l / C ; / / 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 la nt27
28 printf ( ’ ( i ) The P er ce nt ag e o f h ea t u t i l i s e d i nE c on o mi s er i s : %5 . 2 f p e r c e n t . \ n ’ , P e ) ;
29 printf ( ’ The P e r c e n t a g e of he a t u t i l i s e d i nB o i l e r i s : %5 . 2 f p e r ce n t .\ n ’ , P b ) ;
30 printf ( ’ The P e r c e n t a g e of he a t u t i l i s e d i nS u p e r he a t e r i s : %5 . 2 f p e r c e n t .\ n ’ , P s ) ;
31 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 nti s : %5 . 4 f ’ , e t a ) ;
32 printf ( ’ o r %5 . 2 f p e r c e n t . \n ’ , e t a * 1 0 0 ) ;
Scilab code Exa 6.7 Example 7
1 clc2 clear
3 //DATA GIVEN4 C = 2 9 9 1 5 ; / / c a l o r i f i c v a l u e o f c o a l i n kJ /
kg
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5 M w = 9 . 1 ; // mass o f f e ed w a te r p er kg o f
dr y c o a l i n kg6 M e = 9 . 6 ; / / e q u i v a l e n t e v a p o r a t i o n f ra omand a t 100 deg c e l s i u s p e r kg o f dry c o a l i n k g
7 T e = 1 2 ; // temp . o f f e e d w at er t oe co no mi se r i n deg c e l s i u s
8 T b = 1 0 5 ; // temp . o f f e ed w at er t o b o i l e ri n d eg c e l s i u s
9 T a = 1 3 ; / /temp . o f a i r10 T f g = 3 7 0 ; // temp . o f f l u e g a s es e n t e r i n g
e c o n o m i s e r11 M f g = 1 8 . 2 ; // mass o f f l u e g a s e s e n t e r i n g
e co no mi se r p er kg o f c o a l12 C p = 1 . 0 4 6 ; / / mean s p e c i f i c h e a t o f f l u e
g a s e s13
14 h b = M e * 2 2 5 7 ; // h ea t s u p p l i e d f o r s tea mg e n e r a t i o n i n kJ
15 E T A b = h b / C ; // b o i l e r e f f i c i e n c y16 h f l u e = M f g * C p * ( T f g - T a ) ; // h ea t i n t h e f l u e g as e
p er kg o f dr y c o a l e n t e r i n g e co no mi se r17 h e = M w * 4 . 1 8 4 * ( T b - T e ) ; // h ea t u t i l i s e d i n
e c o n o m i s e r18 E T A e = h e / h f l u e ; // e co n om i se r e f f i c i e n c y19 h t o t a l = h b + h e ; // t o t a l h ea t a bs o rb ed i n
kg o f w a te r20 E T A = h t o t a l / C ; // b o i l e r p la nt e f f i c i e n c y21
22 printf ( ’ ( i ) The B o i l e r e f f i c i e n c y i s : %5 . 3 f ’ , E T A b )
;
23 printf ( ’ o r %2 . 1 f p e r c e n t . \n ’ , E T A b * 1 0 0 ) ;
24 printf ( ’ ( i i ) The E co no mi se r e f f i c i e n c y i s : %5 . 3 f ’ ,
E T A e ) ;
25 printf ( ’ o r %2 . 2 f p e r c e n t . \n ’ , E T A e * 1 0 0 ) ;26 printf ( ’ ( i i i ) The O v er a l l E f f i c i e n c y o f b o i l e r p l an t
i s : %5 . 3 f ’ , E T A ) ;
27 printf ( ’ o r %2 . 1 f p e r c e n t . \n ’ , E T A * 1 0 0 ) ;
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Scilab code Exa 6.8 Example 8
1 clc
2 clear
3 //DATA GIVEN4 M s = 2 0 0 0 ; // r a t e o f s tea m p r o du c ti o n i n kg
/ hr5 x = 1 ; // q u a l i t y o f s team6 p = 1 0 ; // steam p r e s s ur e i n b ar7 T f w = 1 1 0 ; // f e e d w at er temp . i n d eg
c e l s i u s8 M f = 2 2 5 ; // r a t e o f c o a l f i r i n g i n kg / h r9 C = 3 0 1 0 0 ; / / c a l o r i f i c v a l u e o f c o a l i n kJ /
kg10 P u c = 1 0 ; / /% o f u nb ur nt c o a l11
12 // f rom stea m t a bl e , c o r r es p o n d i ng t o 10 bar ,13 h = 2 7 7 6 . 2 ; / / i n k J / kg14 h f 1 = 4 . 1 8 * ( T f w - 0 ) ; // h ea t c o n ta i n ed 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 k J /kg15 h t o t a l = h - h f 1 // t o t a l h ea t g i ve n t o
p r o du ce 1 kg o f steam i n b o i l e r i n kJ /kg16 M c = M f * ( 1 0 0 - P u c ) / 1 0 0 ; // mass o f c o a l a c t u a l l y
b ur n t i n kg17 M a = M s / M c ; / / ( k g o f s te am ) / ( k g o f
f u e l )18 E T A b = M a * ( h - h f 1 ) / C ; // t he rm al e f f i c i e n c y o f
b o i l e r19 E T A c = ( M s / M f ) * ( h - h f 1 ) / C ; // t he rm al e f f i c i e n c y o f
b o i l e r and g r a t e co mb ined20
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 ’ , E T A b ) ;
22 printf ( ’ o r %5 . 2 f p e r c e n t . \n ’ , E T A b * 1 0 0 ) ;
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23 printf ( ’ ( i i ) The Thermal e f f i c i e n c y o f t he b o i l e r
and g r a t e co mb in ed i s : %5 . 3 f ’ , E T A c ) ;24 printf ( ’ o r %5 . 2 f p e r c e n t . \n ’ , E T A c * 1 0 0 ) ;
Scilab code Exa 6.9 Example 9
1 clc
2 clear
3 //DATA GIVEN
4 M a = 7 . 5 ; // mass o f s tea m g e n er a t ed p er kgo f c o a l5 p = 1 1 ; // steam p r e s s ur e i n b ar6 T f w = 7 0 ; / /temp . o f f e e d w at er temp . i n
deg c e l s i u s7 e t a = 7 5 ; // e f f i c i e n c y o f b o i l e r i n %8 F e = 1 . 1 5 ; // f a c t o r o f e v ap o ra t i o n9 C p s = 2 . 3 ; / / s p e c i f i c h e a t o f s te a m i n k J /
kgK10
11 // f rom stea m t a bl e , c o r r es p o n d i ng t o 11 bar ,
12 h f = 7 8 1 . 4 ; / / i n k J / kg13 h f g = 1 9 9 8 . 5 ; / / i n k J / kg14 T s = 1 8 4 . 1 + 2 7 3 ; / / i n K15 h f 1 = 4 . 1 8 * ( T f w - 0 ) ;
16
17 / / F a c t o r o f e v a p o r a t i o n , Fe =[{ h f+hf g+Cps∗(Tsup−Ts )}−h f 1 ] / 2 2 5 7
18 T s u p = [ F e * 2 2 5 7 + h f 1 - h f - h f g ] / C p s + T s ; / / Tsup i n K19 x = ( T s u p - T s ) ; / / d e g r e e o f
s up er he at i n deg . c e l s i u s
2021 // B o i l e r e f f i c i e n c y e t a=Ma∗ ( h−hf 1 ) /C ;22 h = [ h f + h f g + C p s * ( T s u p - T s ) ] ;
23 C = M a * ( h - h f 1 ) / ( e t a / 1 0 0 ) ; // c a l o r i f i cv al ue o f c o al i n k J/ kg
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24 M e = M a * ( h - h f 1 ) / 2 2 5 7 ; //
E q ui v al e nt e v a po r a t io n im kg25
26 printf ( ’ ( i ) The T em pe ra tu re o f s te am g e n e r a t i o n ,Tsup i s : %5 . 1 f K\n ’ , T s u p ) ;
27 printf ( ’ The De g r e e o f s u p er h ea t i s : %5 . 1 f d egc 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 k J / 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 k g . \n ’ , M e ) ;
Scilab code Exa 6.10 Example 10
1 clc
2 clear
3 //DATA GIVEN4 p = 1 3 ; // s team p r e s s ur e i n b ar5 d s = 7 7 ; // d e gr e e o f s u p er h ea t i n
deg . c e l s i u s
6 T f w = 8 5 ; // temp . o f f e e d w at er i ndeg . c e l s i u s
7 M w = 3 0 0 0 ; / / mass o f w at ere va p or a te d i n kg / hr
8 M c = 4 1 0 ; // c o a l f i r e d9 M a s h = 4 0 ; // mass o f a sh i n kg / hr
10 P c a = 9 . 6 ; //% o f c o mb u s ti b le i nas h
11 P m = 4 . 5 ; //% o f m oi st ur e i n c o a l12 C = 3 0 5 0 0 ; // c a l o r i f i c v aa l u e o f
dr y c o al p e r k g13 C p s = 2 . 1 ; // s p e c i f i c he at of s u p e r h e a t e d s te am i n kJ / kgK
14
15
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16 // f rom stea m t a bl e , c o r r es p o n d i ng t o 13 bar ,
17 h f = 8 1 4 . 7 ; / / i n k J / kg18 h f g = 1 9 7 0 . 7 ; / / i n k J / kg19 T s = 1 9 1 . 6 ; // i n deg . s e l s i u s20 h = h f + h f g + C p s * ( d s ) ;
21 h f 1 = 4 . 1 8 * ( T f w - 0 ) ;
22 h t o t a l = h - h f 1 ; // t o t a l h ea t s u p pl i e d t op ro du ce 1 kg o f stea m
23
24 M c 1 = M c * ( 1 - P m / 1 0 0 ) ; // mass o f dry c o al i n kg25 M a = M w / M c 1 ;
26 E T A b = M a * ( h - h f 1 ) / C ; // e f f i c i e n c y o f b o i l e r
p l an t i n c l u d i n g s u p er h e at e r27
28 M c o m = M a s h * P c a / 1 0 0 ; // Mass o f c o mb u s ti b l e i na sh p er hr
29 / / t h e c om bu st ib l e p r e s en t i n as h i s p r a c t i c a l l yc ar bo n and i t s v a lu e may b e t ak en a s 3 38 /6 0 kJ / kg
30 // h ea t a c t u a l l y s u p pl i e d pr hr=h ea t o f dr y c oa l−h e a to f c om b us t i bl e i n a sh
31 H s u p p = M c 1 * C - M c o m * 3 3 8 6 0 ; // h ea t a c t u a l l y s u p p l i e dpr hr
32 H u s e = M w * ( h - h f 1 ) ; // h ea t u s e f u l l y u t i l i s e di n b o i l e r pr h r
33
34 E T A c = H u s e / H s u p p ; // e f f i c i e n c y o f b o i l e ra nd f u r n a c e c om bi ne d
35
36 printf ( ’ ( i ) The E f f i c i e n c y o f b o i l e r p la nti n c l u d i n g s u p e r h ea t e r i s : %5 . 3 f o r %2 . 1 f p e r ce n t .\n ’ , E T A b , ( E T A b * 1 0 0 ) ) ;
37 printf ( ’ ( i i ) The E f f i c i e n c y o f t he b o i l e r andf u r n a c e c om bi ned i s : %5 . 3 f o r %2 . 1 f p e r c e n t . \n ’ ,
E T A c , ( E T A c * 1 0 0 ) ) ;
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Scilab code Exa 6.11 Example 11
1 clc
2 clear
3 //DATA GIVEN4 M s = 5 0 0 0 ; // mass o f stea m g e n er a t ed i n
k g / h r5 M f = 7 0 0 ; // r a t e o f c o a l f i r i n g i n kg /
hr6 C = 3 1 4 0 2 ; // h i g h e r c a l o r i f i c va lu e (HCV
) o f c o a l i n k J / kg7 x = 0 . 9 2 ; // q u a l i t y o f stea m
8 p = 1 2 ; // s tea m p r e s s u r e i n b ar9 T f w = 4 5 ; // f e e d w at er temp . i n deg
c e l s i u s10
11 // f rom stea m t a bl e , c o r r es p o n d i ng t o 12 bar ,12 h f = 7 9 8 . 4 ; / / i n k J / kg13 h f g = 1 9 8 4 . 3 ; / / i n k J / kg14 h = h f + x * h f g ; / / i n k J / kg15 h f 1 = 4 . 1 8 * ( T f w - 0 ) ; // h ea t of f e e d w a te r i n kJ /
kg16 M a = M s / M f ;
// i n kg p er kg o f f u e l17 M e = M a * ( h - h f 1 ) / 2 2 5 7 ; // ( kg o f s team ) / ( kg o f f u e l )18 e t a = M a * ( h - h f 1 ) / C ; // t he rm al e f f i c i e n c y19
20 printf ( ’ ( i ) The E q u i v al e n t e v a po r a t io n , Me i s : %5 . 3f ( k g o f s team ) / ( kg o f c o a l ) . \n ’ , M e ) ;
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 ’ , e t a , e t a * 1 0 0 ) ;
Scilab code Exa 6.12 Example 12
1 clc
2 clear
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3 //DATA GIVEN
4 h s u p = 3 3 7 3 . 7 ; / / e n t h a l p y o f s te am ( a t10 0 bar , 5 0 0 deg . c e l s i u s ) i n kJ / kg5 h f 1 = 6 7 7 ; // e n th a lp y o f f e e d w at er
( a t i n l e t temp . 160 deg . c e l s i u s ) i n kJ /kg6 h f = 1 4 0 7 . 6 5 ; // e n n th a l py o f s a t u r a t e d
l i q u i d a t 100 ba r i n kJ /kg7 h g = 2 7 2 4 . 7 ; // e n n th a l py o f s a t u r a t e d
va po ut a t 10 0 b ar i n kJ / kg8 M s = 1 0 0 0 0 0 ; // r a t e o f s te am
g e n e r a t i o n i n kg / hr9 e t a = 8 8 ; // e f f i c i e n c y o f steam
g e n e r a t i o n10 C = 2 1 0 0 0 ; / / c a l o r i f i c v a l u e o f f u e l
i n kJ / kg11
12 / / e t a =( h e a t a b s o r be d by s te am p e r h r ) / ( h e a t a dd ed byf u e l p er h ou r )
13 m = M s * ( h s u p - h f 1 ) / ( C * ( e t a / 1 0 0 ) ) ; / / f u e l b u rn i ngr a t e i n kg / hr
14 h t o t a l = h s u p - h f 1 ; // t o t a l h ea ts u p p l i e d t o stea m f o rm a ti o n
15 P e c = ( h f - h f 1 ) / h t o t a l ; / /% o f h e ata b so r be d i n e c o n om i s e r
16 P e v = ( h g - h f ) / h t o t a l ; / /% o f h e ata bs o rb ed i n e v a po r a to r
17 P s = ( h s u p - h g ) / h t o t a l ; / /% o f h e ata bs o rb ed i n s u p e r h ea t e r
18
19 printf ( ’ ( i ) The F ue l b u rn i ng r a t e , m i s : %5 . 1 f kJ /h. \n ’ , m ) ;
20 printf ( ’ ( i i ) The P er c en t ag e o f h ea t a bs o rb ed i ne c o n om i s e r i s : %5 . 4 f o r %5 . 2 f p e r c e n t .\ n ’ , P e c , (
P e c * 1 0 0 ) ) ;21 printf ( ’ The P e r c e n t a g e o f he a t ab 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 ’ , P e v , (
P e v * 1 0 0 ) ) ;
22 printf ( ’ The P e r c e n t a g e o f he a t ab s o r b e d i n
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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 ’ , P s , ( P s
* 1 0 0 ) ) ;
Scilab code Exa 6.13 Example 13
1 clc
2 clear
3 //DATA GIVEN4 //BOILER
5 M w = 2 0 6 0 ; // mass o f f e e d w at er6 M c = 2 2 7 ; // mass o f c o a l s u p p li e d i n kg / hr7 C = 3 0 0 0 0 ; / / c a l o r i f i c v a l u e o f c o a l i n kJ /
kg8 h s = 2 7 5 0 ; / / e n t ha l p y o f s te am p ro du ce d i n
k J /k g9 h f w = 3 9 8 ; // e n th a lp y o f f e e d w at er
10 //ECONOMISER11 T w i n = 1 5 ; // temp . o f f e e d w at er e n t e r i n g
e co no mi se r i n deg c e l s i u s12 T w o u t = 9 5 ; // temp . o f f e e d w at er l e a v i n g
e co no mi se r i n deg c e l s i u s13 T g o u t = 1 8 ; / / a t m o s p h e r i c t em p .14 T g i n = 3 7 0 ; // temp . o f e n t e r i n g f l u e g a s es15 M f g = 4 0 7 5 ; // mass o f f l u e g a s e s16 // assumi ng Cpw and Cpg ,17 C p w = 4 . 1 8 7 ;
18 C p g = 1 . 0 1 ;
19
20 E T A b = M w * ( h s - h f w ) / ( M c * C ) ;
// e f f i c i e n c y o f
b o i l e r21 E T A e = M w * C p w * ( T w o u t - T w i n ) / ( M f g * C p g * ( T g i n - T g o u t ) ) ;
// e f f i c i e n c y o f e co no mi se r22
23 printf ( ’ ( i ) The B o i l e r e f f i c i e n c y i s : %5 . 4 f o r %2 . 2
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f p e r ce n t . \n ’ , E T A b , ( E T A b * 1 0 0 ) ) ;
24 printf ( ’ ( i i ) The E co no mi se 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 ’ , E T A e , ( E T A e * 1 0 0 ) ) ;
Scilab code Exa 6.14 Example 14
1 clc
2 clear
3 //DATA GIVEN
4 T f w = 5 0 ; / /mean f e e d w at er temp . i n d egc e l s i u s5 p = 5 ; / /mean s te am p r e s s u r e i n b ar6 x = 0 . 9 5 ; // d r yn e s s f r a c t i o n o f stea m7 M c = 6 0 0 ; / / c o a l c o ns u mp t io n kg / h r8 C = 3 0 4 0 0 ; / / c a l o r i f i c v a l u e o f c o a l i n kJ /
kg9 M s = 4 8 0 0 ; // f e ed w at er s u p p l i e d t o b o i l e r
i n kg / h r10
11 // f rom stea m t a bl e , c o r r es p o n d i ng t o 12 bar ,
12 h f = 6 4 0 . 1 ; / / i n k J / kg13 h f g = 2 1 0 7 . 4 ; / / i n k J / kh14 h = h f + x * h f g ; / / i n k j / k g15 h f 1 = 4 . 1 8 * ( T f w - 0 ) ;
16
17 M a = M s / M c ; // i n kg p e r kg o f f u e l18 M e = M a * ( h - h f 1 ) / 2 2 5 7 ; // ( kg o f stea m ) / ( kg o f f u e l )19
20 printf ( ’ The E q ui v a le n t e v a p or a t i on fro m and a t 1 00deg c e l s i u s , Me i s : %5 . 3 f ( k g o f s team ) / ( kg o f
c o al ) .\ n ’ , M e ) ;
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Chapter 7
Internal Combustion Engines
Scilab code Exa 7.1 Example 1
1 clc
2 clear
3 //DATA GIVEN4 P m i = 6 ; // mean e f f e c t i v e
p r e s s u r e i n ba r5 N = 1 0 0 0 ; / / e n g i n e s p ee d i n R . P .
M.6 D = 0 . 1 1 ; // d i am et er o f p i s t o n
i n m7 L = 0 . 1 4 ; // s t r o k e l e ng t h i n m8 n = 1 ; // no . o f c y l i n d e r s9 k = 1 ; / / f o r 2− s t r o k e
c y l i n d e r10
11 // INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗k ∗10) /6 kW12 A = ( % p i / 4 ) * ( D ^ 2 ) ;
13 I P = ( n * P m i * L * A * N * k * 1 0 ) / 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. ’ ,
I P ) ;
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Scilab code Exa 7.2 Example 2
1 clc
2 clear
3 //DATA GIVEN4 //L=1.5D5 n = 4 ; // no . o f c y l i n d e r s6 P = 1 4 . 7 ; / / po wer d e v el o p e d i n
kW7 N = 1 0 0 0 ; / / e n g i n e s p ee d i n R . P .
M.8 P m i = 5 . 5 ; // mean e f f e c t i v e
p r e s s u r e i n ba r9 k = 0 . 5 ; / / f o r 4− s t r o k e
c y l i n d e r10
11 // INDICTED POWER, I . P. = ( n∗PMI∗ l ∗A∗N∗k ∗10) /6 kW12 //A=(p i /4 ) ∗Dˆ2 ,13 //L=1.5D,
14 D = ( ( 6 * P ) / ( 1 0 * k * N * n * P m i * 1 . 5 * ( % p i / 4 ) ) ) ^ ( 1 / 3 ) ; //b or e d ia me te r i n m
15 L = 1 . 5 * D ; //l e n gt h o f s t r o k e i n m
16
17 printf ( ’ The B or e d i a m e t e r i s : %5 . 2 f mm. \ n ’ , ( D * 1 0 0 0 ) )
;
18 printf ( ’ The S t o ke l e n g t h i s : %5 . 2 f mm. \ n ’ , ( L * 1 0 0 0 ) )
;
Scilab code Exa 7.3 Example 3
1 clc
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2 clear
3 //DATA GIVEN4 D b = 0 . 6 ; / / d i a me t er o f b ra k ew he el i n m
5 d = 0 . 0 2 6 ; // d i am et er o f r op e i nm
6 W = 2 0 0 ; // d ead l o ad on t heb ra ke i n N
7 S = 3 0 ; / / s p r i n g b a l a n c er e a d i ng i n N
8 N = 4 5 0 ; / / e n g i n e s p ee d i n R . P .M.
910 // Br ak e P ow er , B. P . = (W−S ) ( p i ) (Db+d )N/ ( 6 0∗1 0 0 0 ) kW11 B P = ( W - S ) * ( % p i ) * ( D b + d ) * N / ( 6 0 * 1 0 0 0 ) ;
12
13 printf ( ’ The B ra k e Power , B . P . i s : %2 . 1 f kW.\ n ’ , B P ) ;
Scilab code Exa 7.4 Example 4
1 clc2 clear
3 //DATA GIVEN4 T = 1 7 5 ; / / t o r q ue d ue t o b ra k e l o a d
in Nm5 N = 5 0 0 ; / / e n g i n e s p e ed i n R . P .M.6
7 / / B r a k e P owe r , BP = ( 2∗ p i )NT/( 60 ∗1 0 0 0 ) kW8 B P = ( 2* % p i )* N * T / (6 0 *1 0 00 ) ;
9
10 printf ( ’ The B ra k e Power , B . P . i s : %4 . 2 f kW.\ n ’ , B P ) ;
Scilab code Exa 7.5 Example 5
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1 clc
2 clear3 //DATA GIVEN4 D = 0 . 3 ; / / b o re o f e n g i n e
c y l i n d e r i n m5 L = 0 . 4 5 ; // s t r o k e l e ng t h i n m6 N = 3 0 0 ; / / e n g i n e s p ee d i n R . P .
M.7 P m i = 6 ; // mean e f f e c t i v e
p r e s s u r e i n ba r8 N B L = 1 . 5 ; / / Net b ra k e l o a d (W−S )
i n kN
9 D b = 1 . 8 ; / / d i a me t er o f b ra k edrum
10 d = 0 . 0 2 ; / / b r ak e r o pe d i a m et e r11 n = 1 ; // no . o f c y l i n d e r s12 k = 0 . 5 ; / / f o r 4− s t r o k e
c y l i n d e r13
14 // INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗k ∗10) /6 kW15 A = ( % p i / 4 ) * ( D ^ 2 ) ;
16 I P = ( n * P m i * L * A * N * k * 1 0 ) / 6 ;
17 B P = N B L * ( % p i ) * ( D b + d ) * N / ( 6 0 ) ;
18 e t a = B P / I P ; / / m e c h a n i c a le f f i c i e n c y
19
20 printf ( ’ ( i ) The I n d i c t e d Power , I . P . i s : %5 . 2 f kW .\n ’ , I P ) ;
21 printf ( ’ ( i i ) The B ra k e Power , B . P . i s : %5 . 2 f kW . \n’ , B P ) ;
22 printf ( ’ ( i i i ) M ec ha ni ca 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 ’ , e t a , ( e t a * 1 0 0 ) ) ;
Scilab code Exa 7.6 Example 6
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1 clc
2 clear3 //DATA GIVEN4 D = 0 . 2 ; / / d i a me t er o f e n g i n e
c y l i n d e r i n m5 L = 0 . 3 5 0 ; // l e ng t h o f s t r o k e i n
m6 P m i c o = 6 . 5 ; // mean e f f e c t i v e
p r e s s u r e on c ov er s i d e i n b a r7 P m i c r = 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 r8 N = 4 2 0 ; / / e n g i n e s p ee d i n R . P .
M.9 D r o d = 0 . 0 2 ; // d i am et er o f p i s t o n
r o d i n m10 W = 1 3 7 0 ; // d ead l o ad on t he
b ra ke i n N11 S = 1 4 5 ; / / s p r i n g b a l a n c e
r e a d i ng i n N12 D b = 1 . 2 ; / / d i a me t er o f b ra k e
w he el i n m13 d = 0 . 0 2 ; // d i am et er o f r op e i n
m14 n = 1 ; // no . o f c y l i n d e r s15 k = 0 . 5 ; / / f o r 4− s t r o k e
c y l i n d e r16
17 // INDICTED POWER , I . P. = ( n∗Pmi∗ l ∗A∗N∗k ∗10) /6 kW18 A c o = ( % p i / 4 ) * ( D ^ 2 ) ; / / a r e a o f
c y l i n d e r om c o ve r end i n mˆ219 A c r = ( % p i / 4 ) * ( D ^ 2 - D r o d ^ 2 ) ; / / a r e a o f
c y l i n d e r om c ra nk end i n mˆ220 I P c o = ( n * P m i c o * L * A c o * N * k * 1 0 ) / 6 ; / / IP on c o v e r end
s i d e i n kW21 I P c r = ( n * P m i c r * L * A c r * N * k * 1 0 ) / 6 ; / / IP on c ra n k end
s i d e i n kW22 I P t o t a l = I P c o + I P c r ; // IP t o t a l i n kW23
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24 // Br ak e P ow er , B. P . = (W−S ) ( pi ) (Db+d )N/ ( 6 0∗1 0 0 0 ) kW
25 B P = ( W - S ) * ( % p i ) * ( D b + d ) * N / ( 6 0 * 1 0 0 0 ) ;26
27 e t a = B P / I P t o t a l ; / / m e c h a n i c a le f f i c i e n c y
28
29 printf ( ’ M ec ha ni ca 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 ’ , e t a , ( e t a * 1 0 0 ) ) ;
Scilab code Exa 7.7 Example 7
1 clc
2 clear
3 //DATA GIVEN4 I P = 3 0 ; / / i n d i c t e d p ow er i n kW5 B P = 2 6 ; / / B r a k e P ow er i n kW6 N = 1 0 0 0 ; // e n g i n e s p ee d i n R . P .M
.7 F = 0 . 3 5 ; // f u e l p e r b r ak e po we r
h o u r i n k g /BP/ h
8 C = 4 3 9 0 0 ; // c a l o r i f i c v a l u e of f u e l u se d i n kJ /kg
9
10 F c = F * B P ; / / f u e l c o ns u mp t io n p e rhour
11 M f = F c / 3 6 0 0 ;
12 E T A t i = I P / ( M f * C ) ; / / I n d i c t e d t h er m ale f i c i e n c y
13 E T A t b = B P / ( M f * C ) ; / / B r ak e t h e r m a le f f i c i e n c y
14 E T A m = B P / I P ; // M ec ha ni ca l e f f i c i e n c y15
16 printf ( ’ ( i ) The I n d i c te d t he rm al e f i c i e n c y i s : %5 . 3f o r %2 . 1 f p e r c e n t . \n ’ ,ETAti ,(ETAti*100));
17 printf ( ’ ( i i ) The B ra ke t he rm al e f f i c i e n c y i s : %5 . 3 f
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o r %2 . 1 f p e r c e n t . \n ’ ,ETAtb ,(ETAtb*100));
18 printf ( ’ ( i i i ) M ec ha ni ca 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 ’ , E T A m , ( E T A m * 1 0 0 ) ) ;
Scilab code Exa 7.8 Example 8
1 clc
2 clear
3 //DATA GIVEN
4 D b = 0 . 7 5 ; / / d i a m et e r o f b r ak ep u l l ey i n m5 d = 0 . 0 5 ; // d i am e te r o f r op e i n
m6 W = 4 0 0 ; // d ead l o ad on t he
b ra ke i n N7 S = 5 0 ; / / s p r i n g b a l a n c e
r e a d i ng i n N8 F c = 4 . 2 ; / / f u e l c on su mp ti on i n
k g / h r9 N = 1 0 0 0 ; / / r a t e d e n g i n e s p ee d
i n R . P . M .10 C = 4 3 9 0 0 ; // c a l o r i f i c v a l u e of
f u e l u se d i n kJ /kg11 n = 1 ; // no . o f c y l i n d e r s12 k = 0 . 5 ; / / f o r 4− s t r o k e
c y l i n d e r13
14
15 // Br ak e P ow er , B. P . = (W−S ) ( pi ) (Db+d )N/ ( 6 0∗1 0 0 0 ) kW16 B P = ( W - S ) * ( % p i ) * ( D b + d ) * N / ( 6 0 * 1 0 0 0 ) ;
17 s f c = F c / B P ; / / b r a k es p e c i f i c f u e l c o n s um p t io n i n k g /kWhr18 M f = F c / 3 6 0 0 ;
19 E T A t b = B P / ( M f * C ) ; / / B r a k et he r ma l e f f i c i e n c y
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20
21 printf ( ’ ( i ) The B ra ke s p e c i f i c f u e l c on su mp ti o n , s .f . c ( b r a k e ) i s : %5 . 3 f k g /kWh . \n ’ , s f c ) ;
22 printf ( ’ ( i i ) The B ra ke t he rm al 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 clear3 //DATA GIVEN4 n = 6 ; // no . o f c y l i n d e r s5 D = 0 . 0 9 ; / / b o re o f e ac h
c y l i n d e r i n m6 L = 0 . 1 ; // l e ng t h o f s t r o k e i n
m7 r = 7 ; / / c o m p re s s i on r a t i o8 E T A r e l = 0 . 5 5 ; // r e l a t i v e e f f i c i e n c y9 F s c = 0 . 3 ; / / i n d i c a t e d s p e c i f i c
f u e l c o ns u mp t io n i n kg /kWh
10 P m i = 8 . 6 ; / / i n d i c a t e d meane f f e c t i v e p r e s s u r e i n b ar
11 N = 2 5 0 0 ; / / e n g i n e s p ee d i n R . P .M.
12 k = 0 . 5 ; / / f o r 4− s t r o k ec y l i n d e r
13
14 / / A i r s t a nd 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 of a i r = 1. 416 E T A a i r = 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 , E TA rel=ETAthi / ETA ai r;18 E T A t h i = E T A r e l * E T A a i r ;
19 / / I n d i c t e d t h e rm a l e f i c i e n c y , ETAthi=IP / ( Mf ∗C)20 M f = F s c / 3 6 0 0 ;
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21 / / t a k i n g I P =1 ,
22 C = 1 / ( E T A t h i * M f ) ; // c a l o r i f i c va lu e ink J /k g23 // INDICTED POWER , I . P. = ( n∗Pmi∗ l ∗A∗N∗k ∗10) /6 kW24 A = ( % p i / 4 ) * ( D ^ 2 ) ;
25 I P = ( n * P m i * L * A * N * k * 1 0 ) / 6 ;
26 F c = F s c * I P ; // t o t a l f u e lc o ns u mp t io n i n kg / h r
27
28 printf ( ’ ( i ) The C a l o r i f i c v al ue o f c oa l , C i s : %5 . 0f k J / kg . \n ’ , C ) ;
29 printf ( ’ ( i i ) The F u el c o ns u mp t io n i s : %5 . 2 f k g /h . \
n ’ , F c ) ;30
31 //NOTE:32 / / a ns o f c a l o r i f i c v a l u e h e r e i s e xa ct , w h i l e i n TB
i t s r ou nded o f f v al ue
Scilab code Exa 7.10 Example 10
1 clc2 clear
3 //DATA GIVEN4 n = 4 ; // no . o f c y l i n d e r s5 B P = 3 0 ; / / B r a k e P ow er i n kW6 N = 2 5 0 0 ; / / e n g i n e s p ee d i n R . P .
M.7 P m i = 8 ; // mean e f f e c t i v e
p r e s s u r e i n ba r8 E T A m = 0 . 8 ; / / m e c h a n i c a l
e f f i c i e n c y9 E T A t h b = 0 . 2 8 ; / / b r a k e t h e rm a le f f i c i e n c y
10 C = 4 3 9 0 0 ; // c a l o r i f i c v al ue o f f u e l u se d i n kJ /kg
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11 k = 1 ; / / f o r 2− s t r o k e
c y l i n d e r12
13 // me c han i c a l e f f i c i e n c y , ETAm=BP/IP14 I P = B P / E T A m ;
15 // INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗k ∗10) /6 kW16 //L=1.5D,17 D = ( ( 6 * I P ) / ( 1 0 * k * N * n * P m i * 1 . 5 * ( % p i / 4 ) ) ) ^ ( 1 / 3 ) ; //
b or e d ia me te r i n m18 L = 1 . 5 * D ; //
l e n gt h o f s t r o k e i n m19 / / B r a ke t h e r m a l e f f i c i e n c y , ETAtb=BP / ( Mf ∗C)
20 M f = B P / ( E T A t h b * C ) ; //f u e l c on su mp ti on i n kg / h r
21
22 printf ( ’ ( i ) The B or e d i a me t er i s : %5 . 3 f m o r %2 . 0 f mm. \ n ’ , D , ( D * 1 0 0 0 ) ) ;
23 printf ( ’ The St o k e l en gt h i s : %2 . 0 f mm. \ n ’ ,( L
* 1 0 0 0 ) ) ;
24 printf ( ’ ( i i ) The F ue l c on su mp ti on i s : %5 . 5 f kg / s o r%3 . 2 f k g / h r . \n ’ , M f , ( M f * 3 6 0 0 ) ) ;
Scilab code Exa 7.11 Example 11
1 clc
2 clear
3 //DATA GIVEN4 n = 6 ; // no . o f c y l i n d e r s5 P d i s p = 7 0 0 ; // p i s t o n d i s p p er
c y l i n d e r i n cmˆ3
6 P = 7 8 ; / / po wer d e v el o p e d i nkW7 N = 3 2 0 0 ; / / e n g i n e s p ee d i n R . P .
M.8 M f = 2 7 ; // mass o f f u e l u s e d i n
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k g / h r
9 C = 4 4 0 0 0 ; // c a l o r i f i c v al ue o f f u e l u se d i n kJ /kg10 a f r = 1 2 ; // a i r f u e l r a t i o11 P a = 0 . 9 ; // i n t a k e a i r p r e s s ur e
i n b ar12 T a = 3 2 + 2 7 3 ; // i n t a k e a i r
t e mp e r ta u r e i n K13 R = 0 . 2 8 7 ; // g as c on s t an t f o r a i r
i n k J / kgK14 k = 0 . 5 ; / / f o r 4− s t r o k e
c y l i n d e r
1516 M a = a f r * M f ; / / ma ss o f
a i r17 / / by e q . pa∗Va=Ma∗R∗Ta18 V a = M a * R * T a / P a / 1 0 0 ; / / v ol um e o f
i n t a k e a i r i n mˆ3/ h r19 V s w e p t = ( P d i s p / 1 0 ^ 6 ) * n * ( N / 2 ) * 6 0 ; //v ol ume
s we pt i n mˆ 3/ h r20 E T A v o l = V a / V s w e p t ; / / v o l u m e t r i c
e f f i c i e n c y21
22 / / B ra ke t h e rm a l e f f i c i e n c y , ETAbt=b r a ke wo rk / h e a ts u pp l i e d by t h e f u e l
23 E T A b t = P / ( M f * C / 3 6 0 0 ) ;
24 / / B r a k e P owe r , BP = ( 2∗ p i )N∗T b/( 60∗1 0 0 0 ) k W25 T b = P * 6 0 / ( 2 * % p i * N ) ; / / b r a k e
t o r q u e i n kNm26
27 printf ( ’ ( i ) The V ol um et ri 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 he rm al 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 B r ak e T or qu e i s : %5 . 4 f kNm . \n ’ ,
T b ) ;
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Scilab code Exa 7.12 Example 12
1 clc
2 clear
3 //DATA GIVEN4 //L=1.5D5 n = 6 ; // no . o f c y l i n d e r s6 V s = 1 . 7 5 ; / / s t r o k e v ol um e i n
l i t r e s7 I P = 2 6 . 3 ; / / po wer d e v el o p e d i n
kW8 N e = 5 0 4 ; / / e n g i n e s p ee d i n R . P .
M.9 P m i = 6 ; // mean e f f e c t i v e
p r e s s u r e i n ba r10 k = 0 . 5 ; / / f o r 4− s t r o k e
c y l i n d e r11
12 // INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗k ∗10) /6 kW
13 //L∗A=Vs14 N a = I P * 6 / ( n * P m i * ( V s / 1 0 ^ 3 ) * k * 1 0 ) ; / / a c t u a l s p e ed
i n R . P . M15 F a = N a * n * k ; // a c t u a l no . o f
f i r e s i n o n e m i nu t e16 F e = N e * n / 2 ; / / e x p e c t e d n o .
o f f i r e s i n o ne m in ut e17 F m = F e - F a ; // m i s f i r e s p er
mi nut e18 F m a v g = F m / n ; / / a vg . no . o f
t im es e a c h c y l i n d e r m i s f i r e s i n one mi nut e19
20 printf ( ’ The Av er ag e no . o f t i me s e ac h c y l i n d e rm i s f i r e s i n one m in ut e i s : %1 . 0 f .\ n ’ , F m a v g ) ;
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Scilab code Exa 7.13 Example 13
1 clc
2 clear
3 //DATA GIVEN4 D = 0 . 0 7 5 ; / / b o re i n m5 L = 0 . 0 9 ; // s t r o k e l e ng t h i n m6 n = 4 ; // no . o f c y l i n d e r s7 e r a r = 3 9 / 8 ; // e ng i ne t o r e a r a x l e
r a t i o =3 9: 88 D w = 0 . 6 5 ; / / w he el d i a m et e r w it h
t y r e f u l l y i n f l a t e d i n m9 F c = 0 . 2 2 7 ; / / p e t r o l c o ns u mp t io n
f o r a d i s t a n ce o f 3 . 2 km a t a s pe e d o f 48 km/ hr10 P m i = 5 . 6 2 5 ; // mean e f f e c t i v e
p r e s s u r e i n ba r11 C = 4 3 4 7 0 ; // c a l o r i f i c v al ue o f
f u e l u se d i n kJ /kg12 k = 0 . 5 ; / / f o r 4− s t r o k e
c y l i n d e r13
14 s = 4 8 * 1 0 0 0 / 6 0 ; // s pe ed o f c a r i n m/mi n
15 / / i f Nt r e v a r e made by t y r e p e r mi nu te , s p ee d=p i ∗Dw∗Nt
16 N t = s / ( % p i * D w ) ; //R.P.M.17 // a s e n g i ne t o r e a r a xl e r a t i o i s 3 9: 818 N e = e r a r * N t ; // s pe ed o f e n f i n e
s h a f t i n R . P .M.19 // INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗k ∗10) /6 kW20 A = ( % p i / 4 ) * ( D ^ 2 ) ;
21 I P = ( n * P m i * L * A * N e * k * 1 0 ) / 6 ;
22
23 s = s / 1 0 0 0 ; / / s p ee d o f c a r i n km/
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mi n
24 t = 3 . 2 / s ; // t im e i n min f o rc o v e r in g 3 . 2 km25 / / p e t r o l c o ns ump ti on f o r a d i s t a nc e o f 3 . 2 km a a t a
s pe ed o f 48 km/ h r i s 0 . 2 2 7 kg26 M f = F c / ( t * 6 0 ) ; / / f u e l co ns um ed p e r
s e c27 E T A t h i = I P / ( M f * C ) ; // I n d i c a t e d f u e l
e f f i c i e n c y28
29 printf ( ’ ( i ) The I n d i c a t e d Power d e v el o p e d i s : %5 . 2 f kW. \n ’ , I P ) ;
30 printf ( ’ ( i i ) The I n d i ca t e d t he rm al 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 . 2 5 ; // c y l i n d e r d i am e te r i nm
5 L = 0 . 4 ; // s t r o k e l e ng t h i n m6 P m g = 7 ; // G ro ss mean e f f e c t i v e
p r e s s u re i n b ar7 P m p = 0 . 5 ; // Pumping mean
e f f e c t i v e p r e s s u r e i n b ar8 N = 2 5 0 ; / / e n g i n e s p ee d i n R . P .
M.9 N B L = 1 0 8 0 ; / / n e t l o a d on t h e
b r a ke (W−S ) i n N10 D b = 1 . 5 ; // e f f e c t i v e d ia me te ro f t he b r a ke i n m
11 F c = 1 0 ; // f u e l u s e d p er hr i nkg
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12 C = 4 4 3 0 0 ; // c a l o r i f i c v al ue o f
f u e l u se d i n kJ /kg13 n = 1 ; // no . o f c y l i n d e r s14 k = 0 . 5 ; / / f o r 4− s t r o k e
c y l i n d e r15
16 // INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗k ∗10) /6 kW17 P m = P m g - P m p ;
18 A = ( % p i / 4 ) * ( D ^ 2 ) ;
19 I P = ( n * P m * L * A * N * k * 1 0 ) / 6 ;
20 B P = N B L * ( % p i ) * ( D b ) * N / ( 6 0 * 1 0 0 0 ) ;
21 E T A m = B P / I P ; / / m e c h a n i c a l
e f f i c i e n c y22 M f = F c / 3 6 0 0 ;
23 E T A t h i = I P / ( M f * C ) ; / / I n d i c a t e d t h er m ale f f i c i e n c y
24
25 printf ( ’ ( i ) The I n d i c a t e d Power , I . P . i s : %5 . 2 f kW.\n ’ , I P ) ;
26 printf ( ’ ( i i ) The B ra k e Power , B . P . i s : %2 . 1 f kW . \n’ , B P ) ;
27 printf ( ’ ( i i i ) M ec ha ni ca 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 ’, E T A m , ( E T A m * 1 0 0 ) ) ;
28 printf ( ’ ( i v ) I n d i ca t e d t he rm al 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 GIVEN4 E T A t h b = 3 0 ; / / B r ak e t h e r m a le f f i c i e n c y i n %
5 a f r = 2 0 ; // a i r f u e l r a t i o byw e i g h t
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6 C = 4 1 8 0 0 ; // c a l o r i f i c v al ue o f
f u e l u se d i n kJ /kg7
8 / / B r a ke t h e r m a l e f f i c i e n c y , ETAthb=w or k p r o d uc e d /h ea t s u p p l i e d
9 w o r k = ( E T A t h b / 1 0 0 ) * C ; / / wo rk p r od u ce d p e r k go f f u e l
10 / /STP c o n d i t i o n s r e f e r t o 1 . 01 3 2 b ar and 15 degc e l s i u s
11 m = a f r ; // mass o f a i r p er kgo f f u e l
12 R = 2 8 7 ;
13 V = m * R * ( 1 5 + 2 7 3 ) / ( 1 . 0 1 3 2 * 1 0 ^ 5 ) ; // v olume o f a i r u se d14 / / B ra ke mean e f f e c t i v e p r e s s u r e , Pmb=wo rk d on e /
c y l i n d e r vo lu me15 P m b = ( w o r k * 1 0 0 0 ) / ( V * 1 0 ^ 5 ) ;
16
17 printf ( ’ The B ra ke 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 ’ , P m b ) ;
Scilab code Exa 7.16 Example 16
1 clc
2 clear
3 //DATA GIVEN4 V 1 = 0 . 2 1 6 ; / / g a s c on su mp ti on i n m
ˆ3/min5 P 1 = 7 5 ; // g a s t em p er a tu re i n
mm o f w a t e r6 T 1 = 1 7 + 2 7 3 ; // g a s t em p er ta u re i n K
7 m = 2 . 8 4 ; / / a i r c o ns u mp t io n i nkg/min8 T a = 1 7 + 2 7 3 ; // a i r t em p er ta u re i n K9 b r = 7 4 5 ; // b a ro me te r r e a d in g i n
mm of Hg
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10 D = 0 . 2 5 ; / / b o re o f e n g i n e
c y l i n d e r i n m11 L = 0 . 4 7 5 ; // s t r o k e l e ng t h i n m12 N = 2 4 0 ; / / e n g i n e s p ee d i n R . P .
M.13 R = 2 8 7 ; // g as c on s t an t f o r a i r
i n J / kgK14 n = 1 ; // no . o f c y l i n d e r s15 k = 1 ; / / f o r 2− s t r o k e
c y l i n d e r16
17 P 1 = b r + P 1 / 1 3 . 6 ; // p r e s s ur e o f t he g as
18 // at NTP19 P 2 = 7 6 0 ; //mm o f Hg20 T 2 = 0 + 2 7 3 ; / / i n K21 //P1∗V1/T1=P2∗V2/T222 V 2 = P 1 * V 1 * T 2 / ( P 2 * T 1 ) ; / / vo lu me o f g a s u se d
at NTP i n mˆ323 V g = V 2 / ( N / 2 ) ; // g a s u se d p er s t r o k e
i n m ˆ 324
25 //PV=mRT26 P 2 = 1 . 0 1 3 2 * 1 0 ^ 5 ;
27 V = m * R * T 2 / P 2 ; / / v ol um e o c c u p i e d bya i r i n mˆ 3/ min
28 V a = V / ( N / 2 ) ; // a i r u se d p er s t r o k ei n m
29
30 V m i x = V g + V a ; / / m i xt u re o f g a s anda i r i n mˆ3
31
32 / / ETAvol=( a c t u a l v ol um e o f m i x tu r e drawn p e r s t r o k ea t NTP) / ( s w e p t v o lu m e o f s y s t e m )
33 E T A v o l = V m i x / ( ( % p i / 4 ) * D ^ 2 * L ) ;34
35 printf ( ’ The V ol um et ri c e f f i c i e n c y i s : %3 . 3 f o r %3 . 1f p e r ce n t . \n ’ ,ETAvol ,(ETAvol*100));
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Scilab code Exa 7.17 Example 17
1 clc
2 clear
3 //DATA GIVEN4 t = 1 ; / / d u r a t i o n o f t r i a l i n
hr5 N = 1 4 0 0 0 ; / / r e v o l u t i o n s6 m c = 5 0 0 ; // no . o f m is se d c y c l e s7 N B L = 1 4 7 0 ; / / Net b ra k e l o a d (W−S )
i n N8 P m i = 7 . 5 ; // mean e f f e c t i v e
p r e s s u r e i n ba r9 V g = 2 0 0 0 0 / 3 6 0 0 ; / / g a s c on su mp ti on i n
l i t r e s / s10 C = 2 1 ; //LCV o f g as a t s i p p l y
c o n d i t i o n s i n kJ / l i t r e11 D = 0 . 2 5 ; // c y l i n d e r d i am e te r i n
m
12 L = 0 . 4 ; // s t r o k e l e ng t h i n m13 C b = 4 ; // e f f e c t i v e b ra ke
c i r cu m f er e n ce i n m14 r = 6 . 5 ; / / c o m p re s s i on r a t i o15 n = 1 ; // no . o f c y l i n d e r s16 k = 0 . 5 ; / / f o r 4− s t r o k e
c y l i n d e r17
18 / / gamma f o r a i r , g = 1 .419 g = 1 . 4 ;
20
21 // INDICTED POWER , I . P. = ( n∗PMI∗ l ∗A∗N∗k ∗10) /6 kW22 N k = ( N * k - m c ) / 6 0 ; / / ( N∗k )−w or ki ng c y c l e s
/min23 A = ( % p i / 4 ) * ( D ^ 2 ) ;
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24 I P = ( n * P m i * L * A * N k * 1 0 ) / 6 ;
25 N = N / 6 0 ;26 B P = N B L * ( C b ) * N / ( 6 0 * 1 0 0 0 ) ;
27 e t a = B P / I P ; / / m e c h a n i c a le f f i c i e n c y
28 E T A t h i = I P / ( V g * C ) ; / / I n d i c a t e d t h er m ale f f i c i e n c y
29
30 / / r e l a t i v e e f f i c i e n c y , E TA re l=ETAthi / ETAas31 //ETAas=1−1/(r ˆ( g−1) )32 E T A a s = 1 - 1 / ( r ^ ( g - 1 ) ) ; // ai r −s t a n d a r d
e f f i c i e n c y
33 E T A r e l = E T A t h i / E T A a s ; // r e l a t i v e e f f i c i e n c y34
35 printf ( ’ ( i ) The I n d i c a t e d Power , I . P . i s : %5 . 2 f kW. \n ’ , I P ) ;
36 printf ( ’ ( i i ) The B ra ke Power , B . P . i s : %5 . 2 f kW. \n ’ , B P ) ;
37 printf ( ’ ( i i i ) M ec ha ni ca 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 ’ , e t a , ( e t a * 1 0 0 ) ) ;
38 printf ( ’ ( i v ) The I n di c at e d t he r ma 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));
39 printf ( ’ ( v ) The R e l at 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));
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Chapter 10
Air Compressors
Scilab code Exa 10.1 Example 1
1 clc
2 clear
3 //DATA GIVEN4 V 1 = 1 ; // v olume o f a i r
t a k e n i n mˆ 3 /mim5 p 1 = 1 . 0 1 3 ; // i n t a k e p r e s s ur e i n
b a r6 T 1 = 1 5 + 2 7 3 ; / / i n t a k e t e m p e ra t u r e
i n K7 p 2 = 7 ; // d e l i v e r y p r e s s u r e
i n b ar8 t = 1 * 6 0 ; // t i me i n s e co n ds9 / / l a w o f c o m p r e s s i o n , pV ˆ1 . 35 =C
10 n = 1 . 3 5 ;
11 R = 2 8 7 ;
12
13 m = p 1 * 1 0 ^ 5 * V 1 / R / T 1 ; // mass o f a i rd e l i v e r e d i n kg / min14
15 // ( T2/T1) =( p2/p1) ˆ( ( n−1)/n) ;16 T 2 = T 1 * ( p 2 / p 1 ) ^ ( ( n - 1 ) / n ) ; / / d e l i v e r y temp . T2
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i n K
1718 W = ( n ) / ( n - 1 ) * m * R * ( T 2 - T 1 ) / 1 0 0 0 ; // i n d i c a t e d work i n
kJ/min19
20 I P = W / t ; // i n d i c a t e d p ower i nkW
21
22 printf ( ’ The I n d i c a t e d p ower , IP i s : %1 . 2 f kW . \n ’ ,
I P ) ;
Scilab code Exa 10.2 Example 2
1 clc
2 clear
3 / / c o n t i n u e d f ro m E xa mp le 14 //DATA GIVEN5 V = 1 ; // v olu me d e a l t w it h
p e r min a t i n l e t i n mˆ 3/ mim6 V c = 1 / 3 0 0 ; / / v o lu me d ra wn i n
p e r c y c l e , i n mˆ 3/ c y c l e7 r = 1 . 5 ; // s t r o k e t o b or e
r a t i o8 E T A c = 0 . 8 5 ; / / 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 c om pr es so r9 E T A m t = 0 . 9 0 ; / / 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 m otor t r an s m i s s i o n10
11 / / c y l i n d e r v ol um e , Vc=( p i / 4 ) Dˆ 2∗L12 D = [ ( V c * 4 / % p i ) / r ] ^ ( 1 / 3 ) ; // b o re i n m
1314 / / f r om e xa mp le 115 P i = 4 . 2 3 / E T A c ; // p ower i n pu t t o t he
c o m p r e s so r i n kW16 M P = P i / E T A m t ; / / m o to r p ow er i n kW
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17
18 printf ( ’ ( i ) The C y l i n d e r b or e , D i s : %3 . 1 f mm. \n ’ ,(D * 1 0 0 0 ) ) ;
19 printf ( ’ ( i i ) The M ot or p ow er i s : %1 . 2 f kW . \n ’ , M P ) ;
Scilab code Exa 10.3 Example 3
1 clc
2 clear
3 //DATA GIVEN4 T 1 = 2 0 + 2 7 3 ; / / t e m p e ra t u r e i n K5 p 1 = 1 ; // p r e s s u r e i n b ar6 p 2 = 1 0 ; // p r e s s u r e i n b ar7 C v = 0 . 7 1 8 ; // i n k J /kgK8
9 / / l a w o f c o m p r e s s i o n , pV ˆ1 . 2=C10 n = 1 . 2 ;
11 R = 0 . 2 8 7 ; // i n k J /kgK12
13 // ( T2/T1) =( p2/p1) ˆ( ( n−1)/n) ;
14 T 2 = T 1 * ( p 2 / p 1 ) ^ ( ( n - 1 ) / n ) ; / / temp . T2 i n K15 m = 1 ;
16 W = ( n ) / ( n - 1 ) * m * R * T 1 * [ ( p 2 / p 1 ) ^ ( ( n - 1 ) / n ) - 1 ] ; //work done p er 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 T he rm od yn am ic s19 / / 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+DU20 //Q=(p1V1−p2V2) /( n−1)+Cv(T2−T1 )21 //Q=(T2−T1 ) ∗ [ Cv−R / ( n−1) ]22 Q = ( T 2 - T 1 ) * [ C v - R / ( n - 1 ) ] ;
2324 printf ( ’ ( i ) The T em pe ra tu re a t t h e end o f c o mp r es s io n i s : %3 . 0 f K o r %3 . 0 f deg . c e l s i u s . \n’ , T 2 , ( T 2 - 2 7 3 ) ) ;
25 printf ( ’ ( i i ) The Work d on e d u r i n g c o m p r e s s i o n p e r
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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 gc om pr es si on p er kg o f a i r i s : %2 . 2 f kJ /kg o f a i r .\n ’ , Q ) ;
27 printf ( ’ ( N e g a t i v e s i g n i n d i c a t e s h e a tREJECTION . ) \n ’ ) ;
Scilab code Exa 10.4 Example 4
1 clc2 clear
3 //DATA GIVEN4 p 1 = 1 ; // s u c t i o n p r e s s u r e
i n b ar5 T 1 = 2 0 + 2 7 3 ; / / s u c t i o n
t e mp e r at u r e i n K6 p 2 = 6 ; // d i s c h a r g e p r e s s u r e
i n b ar7 T 2 = 1 8 0 + 2 7 3 ; / / d i s c h a r g e
t e mp e r at u r e i n K
8 N = 1 2 0 0 ; / / s p e ed o f c o m p r e ss o r i n R . P . M.
9 P s h a f t = 6 . 2 5 ; / / s h a f t p ow er i n kW10 M a = 1 . 7 ; // mass o f a i r
d e l i v e r e d i n kg / min11 D = 0 . 1 4 ; / / d i a m e t e r i n m12 L = 0 . 1 ; // s t r o k e i n m13 R = 2 8 7 ; // i n k J /kgK14
15 V d = ( % p i / 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 om p re ss o r i n mˆ3 /min16 F A D = M a * R * T 1 / p 1 / 1 0 ^ 5 ; //mˆ3/min17 E T A v o l = F A D / V d * 1 0 0 ; // 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 y18
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19 // ( T2/T1) =( p2/p1) ˆ( ( n−1)/n) ;
20 n = 1 / [ 1 - ( log ( T 2 / T 1 ) / log ( p 2 / p 1 ) ) ] ; / / i n d e x o f c o m p r e s s i o n , n21
22 I P = ( n ) / ( n - 1 ) * M a / 6 0 * R / 1 0 0 0 * T 1 * [ ( p 2 / p 1 ) ^ ( ( n - 1 ) / n ) - 1 ] ;
/ / i n d i c a t e d p ow er i n kW23
24 P i s o = M a / 6 0 * R / 1 0 0 0 * T 1 * log ( p 2 / p 1 ) ;
/ / i s o t h e r m a l p ow er25 E T A i s o = P i s o / I P * 1 0 0 ;
/ / i s o t h e r m a le f f i c i e n c y
2627 E T A m e c h = I P / P s h a f t * 1 0 0 ;
/ / m e c h a n i c a le f f i c i e n c y
28
29 E T A o v r _ i s o = P i s o / P s h a f t * 1 0 0 ;
// o v e r a l l i s o t h e r ma le d d i c i e n c y
30
31 printf ( ’ ( i ) The a c t ua l V ol um et ri c e f f i c i e n c y i s : %2
. 2 f p e r c e n t . \n ’, E T A v o l ) ;
32 printf ( ’ ( i i ) The I n d i c a t e d Power , I P i s : %1 . 3 f KW.\n ’ , I P ) ;
33 printf ( ’ ( i i i ) The I s o t h er 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 ’ , E T A i s o ) ;
34 printf ( ’ ( i v ) The M ec ha ni ca l e f f i c i e n c y i s : %2 . 1 f p e r c e n t .\ n ’ , E T A m e c h ) ;
35 printf ( ’ ( v ) The O ve ra l l i s o t h e r ma l e f f i c i e n c y i s :%2 . 1 f p e r c e n t . \ n ’ , E T A o v r _ i s o ) ;
Scilab code Exa 10.5 Example 5
1 // 5 (b ) i s a s f o l l o w s :
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2 clc
3 clear4 //DATA GIVEN5 m = 6 . 7 5 ; // mass o f a i r i n kg / min6 p 1 = 1 ; // p r e s s u r e i n b ar7 T 1 = 2 1 + 2 7 3 ; / / temp . i n K8 p 2 = 1 . 3 5 ; // p r e s s u r e i n b ar9 T 2 = 4 3 + 2 7 3 ; / / temp . i n K
10 D T c w = 3 . 3 ; // temp . r i s e o f c o o l i n gw a t e r i n deg . c e l s i u s
11 C p = 1 . 0 0 3 ; //Cp f o r a i r i n kJ /kgK12 / / gamma f o r a i r = 1 .4
13 g = 1 . 4 ;14
15 W = m * C p * ( T 2 - T 1 ) ; / / wo rk i n k J / min16 // I f t he c o mp r es s io n would ha ve b een i s o t r o p i c ,17 //T 2=T1∗( rp ) ˆ[ ( g−1)/g ]18 r p = p 2 / p 1 ;
19 T _ 2 = T 1 * ( r p ) ^ [ ( g - 1 ) / g ] ;
20 Q r = m * C p * ( T _ 2 - T 2 ) ; // h ea t r e j e c t e d t o c o o l i n gw a t e r
21
22 M w = Q r / [ 4 . 1 8 * ( D T c w ) ] ; // mass of c o o l i n g w at er i nkg/min
23
24 printf ( ’ ( i ) The Work i s : %3 . 2 f k J / mi n . \n ’ , W ) ;
25 printf ( ’ ( i i ) The Mass o f c o o l i n g w at er i s : %1 . 2 f kg/ min . \n ’ , M w ) ;
26
27 //NOTE:28 / / i n t he q u e s t i o n c o mp r es s io n p r o c e s s i s m en ti on ed
and p2 i s g i v en a s 0 . 3 5 ba r ( p2<p1 )29 / / whi ch i s wrong and f u r t h e r p2 i s g iv en a s 1 . 3 5 ba r
which i s a l l o w ab l e30 / / so h er e v al ue o f p2 i s t a ke n a s 1 . 3 5 ba r .
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Scilab code Exa 10.6 Example 6
1 clc
2 clear
3 //DATA GIVEN4 V 1 = 1 4 ; // q u an t it y o f a i r t o
b e d e l i v e r e d , i n mˆ 3/ mim5 p 1 = 1 . 0 1 3 ; // i n t a k e p r e s s ur e i n
b a r6 T 1 = 1 5 + 2 7 3 ; / / i n t a k e t e m p e ra t u r e
i n K7 p 2 = 7 ; // d e l i v e r y p r e s s u r e
i n b ar8 N = 3 0 0 ; / / s p e ed o f
c o m p r e ss o r i n R . P . M.9 n = 1 . 3 ; / / c o m p r e s s i o n a nd
e x p an s i o n i n d ex10 R = 0 . 2 8 7 ;
11
12 / / c l e a r a n c e volume , Vc = 0 . 0 5 Vs , Vs=s we pt vo lu me13 // swept volume Vs=V1−V3=V1−Vc=V1−0.05Vs14 //V1=1.05Vs15 V p c = V 1 / N / 2 ;
// (V1−V4 ) vo lum e i n du c ed p e r c y c l e i n mˆ 3
16 //V4/V3=(p2/p1 ) ̂ ( 1/ n )17 c = ( p 2 / p 1 ) ^ ( 1 / n ) ;
18 //V4=c ∗V3=c ∗ 0 . 0 5 V s19 //V1−V4=1.05Vs−c ∗ 0 . 0 5 V s20 V s = V p c / ( 1 . 0 5 ) / ( 1 . 0 5 - c * 0 . 0 5 ) ;
/ / v o lu me s w ep t i n mˆ 321
22 / / u s i n g r e l a t i o n ( T2 /T1 ) =( p2 / p1 ) ˆ ( ( n−1)/n) ;23 T 2 = T 1 * ( p 2 / p 1 ) ^ ( ( n - 1 ) / n ) ;
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/ / d e l i v e r y temp .
T 2 i n K24
25 I P = ( n ) / ( n - 1 ) * p 1 * 1 0 ^ 5 * V p c / 1 0 0 * [ ( p 2 / p 1 ) ^ ( ( n - 1 ) / n ) - 1 ] ;
/ / i n d i c a t e d p ow er i n kW26
27 printf ( ’ ( i ) The Swept volume o f t he c y l i n d e r , Vs i s: %1 . 4 f mˆ 3 . \n ’ , V s ) ;
28 printf ( ’ ( i i ) The d e l i v e r y t em p er a tu r e , Ts i s : %3 . 0 f deg . c e l s i u s . \n ’ , ( T2 - 2 7 3 ) ) ;
29 printf ( ’ ( i i i ) The I n d i c a t e d p ow er , I P i s : %2 . 2 f kW .\n ’ , I P ) ;
3031 //NOTE:32 / / 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 ns we r s i n t ex tb oo k
due t o r ou nd in g o f f o f v a l u es i n book
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Chapter 13
Transmission of Motion and
Power
Scilab code Exa 13.1 Example 1
1 clc
2 clear
3 //DATA GIVEN4 N 1 = 2 4 0 ; // s pe ed o f t he e n g in e
s h a f t i n R . P .M.5 d 1 = 1 . 5 ; // d i am et e r o f p u l l e y on
e ng i n e s h a f t i n m6 d 2 = 0 . 7 5 ; // d i am et e r o f p u l l e y on
ma ch ine s h a f t i n m7 t = 0 . 0 0 5 ; // t h i c k n e s s o f t h e b e l t
i n m8
9 // w it h no s l i p10 // (N2/N1) =(d1+t ) /( d2+t )
11 N 2 = ( d 1 + t ) / ( d 2 + t ) * N 1 ; / / s p ee d o f t h e m ac hi nes h a f t i n R . P .M.12
13 // w i t h s l i p o f 2%14 S = 2 ; // s l i p i n %
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15 // (N2/N1) =(d1+t ) /( d2+t ) ∗((100−S ) / 1 0 0 )
16 N 2 s = ( d 1 + t ) / ( d 2 + t ) * N 1 * ( ( 1 0 0 - S ) / 1 0 0 ) ;17
18 printf ( ’ ( i ) The S pe ed o f m ac hi ne s h a f t , N2 w it h nos l i p i s : %4 . 1 f R . P .M. \n ’ , N 2 ) ;
19 printf ( ’ ( i i ) The S pe ed o f m ac hi ne s h a f t , N2 w it hs l i p o f 2 p e rc e n t i s : %4 . 1 f R .P .M. \n ’ , N 2 s ) ;
Scilab code Exa 13.2 Example 2
1 clc
2 clear
3 //DATA GIVEN4 r 1 = 9 0 0 / 2 0 0 0 ; // r a di u s o f l a r g e r
p u l l ey i n m5 r 2 = 3 0 0 / 2 0 0 0 ; // r a d i us o f s m a l l e r
p u l l ey i n m6 d = 6 ; / / d i s t a n c e b et we en t h e
c e n t r e s o f p u l l e y i n m7
8 / / L e n gt h 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 L c r o s s = ( % p i ) * ( r 1 + r 2 ) + ( r 1 + r 2 ) ^ 2 / d + 2 d ;
10 / / L e n g t h o f o p en b e l t , L op en =( p i ) ( r 1 +r 2 ) +( r 2−r1 ) ̂ 2/ d+2d ;
11 L o p e n = ( % p i ) * ( r 1 + r 2 ) + ( r 2 - r 1 ) ^ 2 / d + 2 d ;
12
13 L r e d = L c r o s s - L o p e n ; / / l e n g t h t o be r e du c ed14 printf ( ’ The L en gth o f t he b e l t t o be r ed uc ed , \n (
t o c h an g e t h e d i r e c t i o n o f r o t a t i o n o f t h ef o l l o w e r p u l l e y s ) i s : %2 . 0 f mm. \n ’ , ( L r ed * 1 0 0 0 ) ) ;
Scilab code Exa 13.3 Example 3
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1 clc
2 clear3 //DATA GIVEN4 T 1 = 1 5 0 0 ; // t e n s i o n on t he t i g h t
s i d e i n N5 T 2 = 1 2 0 0 ; // t e n s i o n on t he s l a c k
s i d e i n N6 v = 8 0 ; // s pe e d o f t h e b e l t i n m
/ s7
8 P = ( T 1 - T 2 ) * v ; / / po wer t r a n s m i t t e d byt h e b e l t i n w a tt s
910 printf ( ’ The Power t r a n s mi t t ed by t he b e l t i s : %2 . 0 f
kW. \n ’ , ( P / 1 0 0 0 ) ) ;
Scilab code Exa 13.4 Example 4
1 clc
2 clear
3 //DATA GIVEN4 v = 5 0 0 ; // s pe e d o f t h e b e l t i n m
/min5 m u = 0 . 3 ; // c o e f f i c i e n t o f
f r i c t i o n6 t h e t a = 1 6 0 ; // a n gl e o f c o nt a ct i n
d e g r e e s7 T 1 = 7 0 0 ; / /maximum t e n s i o n i n t h e
b e l t i n N8
9 // (T1/T2)=e ̂ (mu∗ t h e t a )10 t h e t a = t h e t a * ( % p i ) / 1 8 0 ; // t h e ta c o nv e rt e d i n t or a d i a n s
11 T 2 = T 1 / ( % e ^ ( m u * t h e t a ) ) ; // t e n s i o n on t he s l a c ks i d e i n N
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12 v = v / 6 0 ; // s pe ed o f t he b e l t
c o n v er t e d i n t o m/ s13 P = ( T 1 - T 2 ) * v ; / / po wer t r a n s m i t t e d byt h e b e l t i n w a tt s
14
15 printf ( ’ The Power t r a n s mi t t ed by t he b e l t i s : %2 . 3 f kW. \n ’ , ( P / 1 0 0 0 ) ) ;
Scilab code Exa 13.5 Example 5
1 clc
2 clear
3 //DATA GIVEN4 r 1 = 7 5 0 / 2 0 0 0 ; // r a di u s o f l a r g e r
p u l l ey i n m5 r 2 = 3 0 0 / 2 0 0 0 ; // r a d i us o f s m a l l e r
p u l l ey i n m6 d = 1 . 5 ; / / d i s t a n c e b et we en t h e
c e n t r e s o f p u l l e y i n m7 T m s = 1 4 ; / /maximum s a f e t e n s i o n
i n N/mm8 b = 1 5 0 ; // w i dt h o f t h e b e l t i n
mm9 v = 5 4 0 ; // s pe e d o f t h e b e l t i n m
/min10 m u = 0 . 2 5 ; // c o e f f i c i e n t o f
f r i c t i o n11
12 T 1 = T m s * b ; / /maximum t e n s i o n i n t h eb e l t i n N
13 v = v / 6 0 ; // s pe ed o f t he b e l tc o n v er t e d i n t o m/ s14 // ( i ) f o r open b e l t15 A L P H A o = asin ( ( r 1 - r 2 ) / d ) * 1 8 0 / ( % p i ) ; / / a l p h a i n
d e g r e e s
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16 T H E T A o = 1 8 0 - 2 * A L P H A o ; // a n g l e o f
l ap o r c on t a c t i n deg17 T 2 o = T 1 / ( % e ^ ( m u * ( T H E T A o * % p i / 1 8 0 ) ) ) ; // t e n s i o n ont h e s l a c k s i d e i n N
18 P o = ( T 1 - T 2 o ) * v ; //pow e rt r an s mi t te d by t he b e l t i n w at ts
19
20 // ( i i ) f o r c r o s s b e l t21 A L P H A c = asin ( ( r 1 + r 2 ) / d ) * 1 8 0 / ( % p i ) ; / / a l p h a i n
d e g r e e s22 T H E T A c = 1 8 0 + 2 * A L P H A c ; // a n g l e o f
l ap o r c on t a c t i n deg
23 T 2 c = T 1 / ( % e ^ ( m u * ( T H E T A c * % p i / 1 8 0 ) ) ) ; // t e n s i o n ont h e s l a c k s i d e i n N
24 P c = ( T 1 - T 2 c ) * v ; //pow e rt r an s mi t te d by t he b e l t i n w at ts
25
26 printf ( ’ ( i ) The Maximum P ow er t r a n s m i t t e d by t h eo pe n b e l t i s : %2 . 3 f kW. \n ’ , ( P o / 1 0 0 0 ) ) ;
27 printf ( ’ ( i i ) The Maximum P ower t r a n s m i t t e d by t h ec r o s s b e l t i s : %2 . 3 f kW. \n ’ , ( P c / 1 0 0 0 ) ) ;
Scilab code Exa 13.6 Example 6
1 clc
2 clear
3 //DATA GIVEN4 b = 0 . 2 5 ; // w i dt h o f t h e b e l t i n m5 t = 0 . 0 0 6 ; // t h i c k n e s s o f t h e b e l t
i n m
6 r = 9 0 0 / 2 0 0 0 ; // r a d i us o f t he p u l l e yi n m7 r h o = 1 1 0 0 ; // d e n s i t y o f t he
m a t e r i a l i n kg /mˆ 38 T p = 2 ; // p e r m i s s i b l e t e n s i o n o f
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t h e b e l t i n MN/mˆ 2
9 r a t i o = 2 ; / / r a t i o o f T1 /T2=210 N = 2 0 0 ; // s pe ed o f t he p u l l e y i nR.P.M.
11
12 T m a x = T p * 1 0 ^ 6 * b * t ; / /maximum s a f e t e n s i o no f t h e b e l t
13 / / c e n t r i f u g a l t e n s i o n , Tc=m∗v ˆ214 m = ( b * t ) * 1 * r h o ; // mass o f t he b e l t p er
u n i t m et re l e n g th15 v = 2 * ( % p i ) * ( r + t / 2 ) * N / 6 0 ;
16 T c = m * v ^ 2 ;
1718 T 1 = T m a x - T c ; // t e n s i o n i n t he t i g h t
s i d e i n N19 T 2 = T 1 / r a t i o ; // t e n s i o n i n t he s l a c k
s i d e i n N20 P = ( T 1 - T 2 ) * v ; / / po wer t r a n s m i t t e d by
t h e b e l t i n w a tt s21
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 ’, T c ) ;
24 printf ( ’ ( i i ) The Power t r a n s mi t t e d by t he b e l t i s :%2. 1 f kW. \n ’ , ( P / 1 0 0 0 ) ) ;
Scilab code Exa 13.7 Example 7
1 clc
2 clear
3 //DATA GIVEN4 P = 3 5 ; // p ower r e q u i r e d t o bet r an s mi t te d by t he b e l t i n kW
5 d = 1 . 5 ; // e f f e c t i v e d ia me te r o f p u l l ey i n m
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6 N = 3 0 0 ; // s pe ed o f t he p u l l e y i n
R.P.M.7 t h e t a = 1 1 / 2 4 * 2 * % p i ; // a n gl e o f c o nt a ct i nr a d i a n s
8 m u = 0 . 3 ; // c o e f f i c i e n t o f f r i c t i o n
9 t = 0 . 0 0 9 5 ; // t h i c k n e s s o f t h e b e l ti n m
10 r h o = 1 1 0 0 ; // d e n s i t y o f t hem a t e r i a l i n kg /mˆ 3
11 s i g m a = 2 . 5 ; // p e r m i s s i b l e s t r e s s i nMN/mˆ2
1213 v = % p i * d * N / 6 0 ; // s pe e d o f t h e b e l t i n m
/ s14 //P=(T2−T1 ) ∗v , s o ( T2−T1)=P/v . . . . . . . . . . . . . . . . . . . . ( 1 )15 c = % e ^ ( m u * t h e t a ) ; // so , T2/T1=c . . . . . . . . ( 2 )16 //By e q ua t i on ( 1 ) and ( 2 ) ,17 T 2 = ( P / v * 1 0 0 0 ) / ( c - 1 ) ; // t e n s i o n i n t he s l a c k
s i d e i n N18 T 1 = c * T 2 ; // t e n s i o n i n t he t i g h t
s i d e i n N19
20 //maximum te ns i o n , Tmax=sigma ∗b∗ t =0.23 75∗b∗1 0 ˆ 6 N( 3 )
21 / / 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 )
22 //T1=Tmax−c( 5 )
23 / /By eq n . ( 3 ) , ( 4 ) and ( 5 )24 b = T 1 / ( ( s i g m a * 1 0 ^ 6 * t ) - ( t * 1 * r h o * v ^ 2 ) ) ; // w i dt h o f
t h e b e l t i n m25
26 printf ( ’ The Width o f t h e b e l t i s : %3 . 0 f mm ( s ay 1 50mm) . \n ’ , ( b * 1 0 0 0 ) ) ;
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Scilab code Exa 13.9 Example 9
1 clc
2 clear
3 //DATA GIVEN4 T o = 1 0 0 0 ; / / i n i t i a l t e n s i o n i n t h e
b e l t i n N5 t h e t a = 1 5 0 ; / / a n g l e o f e mb ra ce i n
d e g r e e s6 m u = 0 . 2 5 ; // c o e f f i c i e n t o f
f r i c t i o n7 v = 5 0 0 ; // s pe e d o f t h e b e l t i n m
/min8
9 // I n i t i a l te ns i on , To=(T1+T2) /210 // so , (T1+T2 ) = 2 0 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 )11 t h e t a = t h e t a * ( % p i ) / 1 8 0 ; // t h e ta c o nv e rt e d i n t o
r a d i a n s12 c = % e ^ ( m u * t h e t a ) ; // so , T2/T1=c . . . . . . . . ( 2 )
13 //By e q ua t i on ( 1 ) and ( 2 ) ,14 T 2 = ( T o * 2 ) / ( c + 1 ) ; // t e n s i o n i n t he s l a c k
s i d e i n N15 T 1 = c * T 2 ; // t e n s i o n i n t he t i g h t
s i d e i n N16
17 v = v / 6 0 ; // s pe ed o f t he b e l tc o n v er t e d i n t o m/ s
18 P = ( T 1 - T 2 ) * v ; / / po wer t r a n s m i t t e d byt h e b e l t i n w a tt s
19
20 printf ( ’ ( i ) The T e ns io n i n t he t i g h t s i d e T1 i s : %4. 0 f N . \n ’ , T 1 ) ;
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 ’ , T 2 ) ;
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22 printf ( ’ ( i i ) The Power t r a n s mi t t e d by t he b e l t i s :
%2. 2 f kW. \n ’ , ( P / 1 0 0 0 ) ) ;
Scilab code Exa 13.10 Example 10
1 clc
2 clear
3 //DATA GIVEN4 P = 4 0 0 ; / /maximum v a l u e o f f o r c e
t ha t can be d ev el op ed i n N5 m u = 0 . 2 5 ; // c o e f f i c i e n t o f
f r i c t i o n6 d = 0 . 6 ; / / d i a m et e r o f drum i n m7 // R e fe r t he f i g u r e8 t h e t a = 1 8 0 + 4 5 ; // a n gl e o f c o nt a ct i n
d e g r e e s9 t h e t a = t h e t a * ( % p i ) / 1 8 0 ; // t h e ta c o nv e rt e d i n t o
r a d i a n s10
11 //moments about A , Ma= 0,
12 T 1 = 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 e15 //T1>T2 ( T1/T2 )=e ̂ (mu∗ t h e t a )16 T 2 = T 1 / ( % e ^ ( m u * t h e t a ) ) ;
17 M c a c = ( T 1 - T 2 ) * ( d / 2 ) ; //maximum br ak in gt o r q u et h at can be d ev el op ed 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 e20 //T2>T1 ( T2/T1 )=e ̂ (mu∗ t h e t a )
21 T 2 = T 1 * ( % e ^ ( m u * t h e t a ) ) ;22 M c c = ( T 2 - T 1 ) * ( d / 2 ) ; //maximum br ak in gt o r q u et h at can be d ev el op ed 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 c an b e
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d ev el op ed 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 ’ , M c a c ) ;25 printf ( ’ ( i i ) The Maximum b r a k i n g t o r q u e t h a t c an b ed e ve l op 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 ’ , M c c ) ;
Scilab code Exa 13.11 Example 11
1 clc
2 clear3 //DATA GIVEN4 P t = 8 0 ; / / po wer t o be
t r a n s mi t t e d by t he r op e i n kW5 d = 1 . 5 ; // d ia me te r o f p u l l e y i n
m6 N = 2 0 0 ; // s pe ed o f t he p u l l e y i n
R.P.M.7 a l p h a = 4 5 / 2 ; // s emi a n gl e o f g ro ov e
i n d e gr e e s8 t h e t a = 1 6 0 ; // a n gl e o f c o nt a ct i n
d e g r e e s9 m u = 0 . 3 ; // c o e f f i c i e n t o f
f r i c t i o n10 m = 0 . 6 ; // mass o f e ac h r op e p er
u n i t m et re l e n g th11 T s = 8 0 0 ; // s a f e p u l l i n N12
13 / / c e n t r i f u g a l t e n s i o n , Tc=m∗v ˆ214 v = ( % p i ) * d * N / 6 0 ; / / v e l o c i t y
o f t he r op e i n m/ s
15 T c = m * v ^ 2 ;16
17 T 1 = T s - T c ; // t e n s i o n i nt h e t i g h t s i d e i n N
18 // (T1/T2)=e ̂ (mu∗ t h e t a )
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19 t h e t a = t h e t a * ( % p i ) / 1 8 0 ; / / t h e t a
c on v e rt e d i n t o r a di a n s20 a l p h a = a l p h a * ( % p i ) / 1 8 0 ; // al p hac on v e rt e d i n t o r a di a n s
21 T 2 = T 1 / ( % e ^ ( m u * t h e t a / sin ( a l p h a ) ) ) ; / / t e n s i o n ont h e s l a c k s i d e i n N
22 p = ( T 1 - T 2 ) * v ; //pow e rt r an s mi t te d by t he b e l t i n w at ts
23
24 / / no . o f r o p e s r e q u i r e d , n=T o ta l po wer 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 ac h r o pe
25 n = P t / ( p / 1 0 0 0 ) ;
2627 // I n i t i a l t en s i o n in rope , To=(T1+T2+2Tc) /228 T o = ( T 1 + T 2 + 2 * T c ) / 2 ;
29
30 printf ( ’ ( i ) The Number o f r o p es r e q u i r e d f o r t hed 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 op e , To i s: %3 . 2 f N . \n ’ , T o ) ;
Scilab code Exa 13.12 Example 12
1 clc
2 clear
3 //DATA GIVEN4 T = 7 2 ; / / number o f t e e t h5 P c = 2 6 ; // c i r c u l a r p i t ch i n mm6
7 / / c i r c u l a r p i t ch , Pc=( p i ∗D)/T
8 D = P c * T / ( % p i ) ; // p i t c h d i am et e r i n m9 //Pc∗Pd=(p i )10 P d = ( % p i ) / P c ; // d i a m et r a l p i t c h i n
t e e t h /mm11 // Module , m=D/T
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12 m = D / T ; / / m o d ul e i n mm/ t o o t h
1314 printf ( ’ ( i ) The P i t c h d ia me te rm , D i s : %3 . 2 f mm. \n
’ , D ) ;
15 printf ( ’ ( i i ) The D i a me t ra l p i t ch , Pd i s : %1 . 2 f t e e t h /mm. \n ’ , P d ) ;
16 printf ( ’ ( i i i ) The M odul e , m i s : %1. 2 f mm/ t o ot h . \n ’ ,m
) ;
Scilab code Exa 13.13 Example 13
1 clc
2 clear
3 //DATA GIVEN4 T a = 4 0 ; // number o f t e e t h o f
g e a r A5 T b = 1 0 0 ; // number o f t e e t h o f
g e a r B6 T c = 5 0 ; // number o f t e e t h o f
g e a r C
7 T d = 1 5 0 ; // number o f t e e t h o f g e a r D
8 T e = 5 2 ; // number o f t e e t h o f g e a r E
9 T f = 1 3 0 ; // number o f t e e t h o f g e a r F
10 N a = 1 0 0 0 ; / / s p ee d o f t h e mo to rs h a f t i n R . P .M.
11
12 //(Nf/Na)=(Ta/Tb) ∗ (Tc/Td) ∗ ( T e / T f )
13 N f = ( T a / T b ) * ( T c / T d ) * ( T e / T f ) * N a ; // S peed o f t heo ut pu t s h a f t i n R . P .M.14
15 printf ( ’ The S peed o f t he o ut pu t s h a ft , Nf i s : %3 . 2 f R.P.M. \n ’ , N f ) ;
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