Heat Transfer_K. a. Gavhane

Scilab Textbook Companion for Heat Transfer by K. A. Gavhane 1 Created by Deepak Bachelor of Technology Chemical Engineering DCRUST,Murthal College Teacher Ms. Sunanda Cross-Checked by Lavitha Pereira May 24, 2016

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

Book Description Title: Heat Transfer Author: K. A. Gavhane Publisher: Nirali Prakashan, Pune Edition: 10 Year: 2010 ISBN: 8190639617

1

Scilab numbering policy used in this document and the relation to the above book. Exa Example (Solved example) Eqn Equation (Particular equation of the above book) AP Appendix to Example(Scilab Code that is an Appednix to a particular

Example of the above book) For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means a scilab code whose theory is explained in Section 2.3 of the book.

2

Contents List of Scilab Codes

4

2 Conduction

5

3 Convection

48

4 Radiation

96

5 Heat Exchangers

114

6 Evaporation

149

3

List of Scilab Codes Exa 2.1 Exa 2.2 Exa 2.3 Exa 2.4 Exa 2.5 Exa 2.7 Exa 2.8 Exa 2.9 Exa 2.10 Exa 2.11 Exa 2.12 Exa 2.13

Thickness of insulation . . . . . . . . . . . . . . . . . . Heat loss per metre . . . . . . . . . . . . . . . . . . . Heat Loss . . . . . . . . . . . . . . . . . . . . . . . . . Heat loss . . . . . . . . . . . . . . . . . . . . . . . . . Heat loss . . . . . . . . . . . . . . . . . . . . . . . . . Heat loss . . . . . . . . . . . . . . . . . . . . . . . . . Loss per area . . . . . . . . . . . . . . . . . . . . . . . Heat loss . . . . . . . . . . . . . . . . . . . . . . . . . Heat Passed . . . . . . . . . . . . . . . . . . . . . . . Insulated pipe . . . . . . . . . . . . . . . . . . . . . . Composite brick . . . . . . . . . . . . . . . . . . . . . Heat flow in a pipe . . . . . . . . . . . . . . . . . . . .

Exa 2.15 Exa 2.16 Exa 2.17 Exa 2.18 Exa 2.19 Exa 2.20 Exa 2.21 Exa 2.22 Exa 2.23 Exa 2.24 Exa 2.25 Exa 2.26 Exa 2.27 Exa 2.28 Exa 2.29 Exa 2.30

Thickness of insulation . . . . . . . . . . . . . . . . . . 16 Reduction in heat loss in insulated pipe . . . . . . . . 17 Heat loss in a pipe . . . . . . . . . . . . . . . . . . . . 18 Arrangements for heat loss . . . . . . . . . . . . . . . 19 Insulation thickness . . . . . . . . . . . . . . . . . . . 20 Heat loss in furnace . . . . . . . . . . . . . . . . . . . 20 Rate of heat loss in pipe . . . . . . . . . . . . . . . . . 21 Heat loss from insulated steel pipe . . . . . . . . . . . 22 Heat loss from furnace . . . . . . . . . . . . . . . . . . 23 Rate of heat loss . . . . . . . . . . . . . . . . . . . . . 24 Thickness of insulation . . . . . . . . . . . . . . . . . . 25 Heat loss per metre . . . . . . . . . . . . . . . . . . . 25 Mineral wool insulation . . . . . . . . . . . . . . . . . 26 Furnace wall . . . . . . . . . . . . . . . . . . . . . . . 27 Thickness of insulating brick . . . . . . . . . . . . . . 28 Heat flow through furnace wall . . . . . . . . . . . . . 29 4

5 6 7 8 9 10 10 11 12 13 14 15

Exa 2.31 Exa 2.32 Exa 2.33 Exa 2.34 Exa 2.36 Exa 2.37

Heat loss in pipe . . . . . . . . . . . . . . . . . . . . . 30 Heat flux through layers . . . . . . . . . . . . . . . . . 31 Conductive conductance furnace wall . . . . . . . . . . 31 Critical radius of insulation . . . . . . . . . . . . . . . 33 Critical radius of pipe . . . . . . . . . . . . . . . . . . 34 Time required for steel ball . . . . . . . . . . . . . . . 34

Exa 2.38 Exa 2.39 Exa 2.40 Exa 2.41 Exa 2.42 Exa 2.43 Exa 2.44 Exa 2.45 Exa 2.46 Exa 2.47 Exa 2.49 Exa 2.50 Exa 2.51 Exa 3.1 Exa 3.2

Steel ball quenched . . . . . . . . . . . . . . . . . . . . 35 Ball plunged in a medium . . . . . . . . . . . . . . . . 36 Slab temperature suddenly lowered . . . . . . . . . . . 37 Flow over a flat plate . . . . . . . . . . . . . . . . . . 38 Stainless steel rod immersed in water . . . . . . . . . . 38 Chromel alumel thermocouple . . . . . . . . . . . . . . 39 Thermocouple junction . . . . . . . . . . . . . . . . . 40 Batch reactor . . . . . . . . . . . . . . . . . . . . . . . 41 Heat dissipation by aluminium rod . . . . . . . . . . . 42 Aluminium fin efficiency . . . . . . . . . . . . . . . . . 43 Pin fins . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Metallic wall surrounded by oil and water . . . . . . . 45 Brass wall . . . . . . . . . . . . . . . . . . . . . . . . . 46 Boundary layer thickness . . . . . . . . . . . . . . . . 48 Boundary layer thickness of plate . . . . . . . . . . . . 49

Exa 3.4 3.3 Exa Exa 3.5 Exa 3.6 Exa 3.7 Exa 3.8 Exa 3.9 Exa 3.10 Exa 3.11 Exa 3.12 Exa 3.13 Exa 3.14 Exa 3.15 Exa 3.16 Exa 3.17 Exa 3.18 Exa 3.19

Thickness of hydrodynamic Flat plate boundary layer . .boundary . . . . . . layer . . . . ... . .. .. . . 50 50 Rate of heat removed from plate . . . . . . . . . . . . 51 Heat removed from plate . . . . . . . . . . . . . . . . 52 Local heat transfer coefficient . . . . . . . . . . . . . . 53 Width of plate . . . . . . . . . . . . . . . . . . . . . . 54 Heat transferred in flat plate . . . . . . . . . . . . . . 55 Rate of heat transferred in turbulent flow . . . . . . . 56 Heat transfer from plate in unit direction . . . . . . . 57 Heat lost by sphere . . . . . . . . . . . . . . . . . . . 58 Heat lost by sphere . . . . . . . . . . . . . . . . . . . 58 Percent power lost in bulb . . . . . . . . . . . . . . . . 59 Heat lost by cylinder . . . . . . . . . . . . . . . . . . . 60 Heat transfer in tube . . . . . . . . . . . . . . . . . . . 60 Heat transfer coefficient . . . . . . . . . . . . . . . . . 61 Heat transfer coefficient in heated tube . . . . . . . . . 62 h of water flowing in tube . . . . . . . . . . . . . . . . 63 5

Exa 3.20 Exa 3.21 Exa 3.22 Exa 3.23 Exa 3.24 Exa 3.25

Overall heat transfer coefficient . . . . . . . . . . . . . 64 Number of tubes in exchanger . . . . . . . . . . . . . . 65 Convective film coefficient . . . . . . . . . . . . . . . . 67 Length of tube . . . . . . . . . . . . . . . . . . . . . . 68 Cooling coil . . . . . . . . . . . . . . . . . . . . . . . . 69 Outlet temperature of water . . . . . . . . . . . . . . . 69

Exa 3.26 Exa 3.27 Exa 3.28 Exa 3.29 Exa 3.30 Exa 3.31 Exa 3.32 Exa 3.33 Exa 3.34 Exa 3.35 Exa 3.36 Exa 3.37 Exa 3.38 Exa 3.39 Exa 3.40

Inside heat transfer coefficient . . . . . . . . . . . . . . 70 Film heat transfer coefficient . . . . . . . . . . . . . . 71 Area of exchanger . . . . . . . . . . . . . . . . . . . . 72 Natural and forced convection . . . . . . . . . . . . . 73 Natural convection . . . . . . . . . . . . . . . . . . . . 74 Free convection in vertical pipe . . . . . . . . . . . . . 75 Heat loss per unit length . . . . . . . . . . . . . . . . 76 Free convection in pipe . . . . . . . . . . . . . . . . . 77 Free convection in plate . . . . . . . . . . . . . . . . . 77 Heat transfer from disc . . . . . . . . . . . . . . . . . 78 Rate of heat input to plate . . . . . . . . . . . . . . . 79 Two cases in disc . . . . . . . . . . . . . . . . . . . . . 80 Total heat loss in a pipe . . . . . . . . . . . . . . . . . 82 Heat loss by free convection . . . . . . . . . . . . . . . 83 Heat loss from cube . . . . . . . . . . . . . . . . . . . 83

Exa 3.41 Exa 3.42 Exa 3.43 Exa 3.44 Exa 3.45 Exa 3.46 Exa 3.47 Exa 3.48 Exa 3.49 Exa 3.50 Exa 3.51 Exa 3.52 Exa 3.53 Exa 4.1 Exa 4.2 Exa 4.3 Exa 4.4

Plate exposed to heat. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. 84 Nucleate poolboiling 86 Peak Heat flux . . . . . . . . . . . . . . . . . . . . . . 86 Stable film pool boiling . . . . . . . . . . . . . . . . . 87 Heat transfer in tube . . . . . . . . . . . . . . . . . . . 88 Nucleat boiling and heat flux . . . . . . . . . . . . . . 89 Dry steam condensate . . . . . . . . . . . . . . . . . . 89 Laminar Condensate film . . . . . . . . . . . . . . . . 90 Saturated vapour condensate in array . . . . . . . . . 91 Mass rate of steam condensation . . . . . . . . . . . . 92 Saturated tube condensate in a wall . . . . . . . . . . 93 Condensation rate . . . . . . . . . . . . . . . . . . . . 94 Condensation on vertical plate . . . . . . . . . . . . . 94 Heat loss by radiaiton . . . . . . . . . . . . . . . . . . 96 Radiation from unlagged steam pipe . . . . . . . . . . 96 Interchange of radiation energy . . . . . . . . . . . . . 97 Heat loss in unlagged steam pipe . . . . . . . . . . . . 97 6

Exa 4.5 Exa 4.6 Exa 4.7 Exa 4.8 Exa 4.9 Exa 4.10

Loss from horizontal pipe . . . . . . . . . . . . . . . . 98 Heat loss by radiation in tube . . . . . . . . . . . . . . 99 Net radiant interchange . . . . . . . . . . . . . . . . . 99 Radiant interchange between plates . . . . . . . . . . . 100 Heat loss from thermflask . . . . . . . . . . . . . . . . 100 Diwar flask . . . . . . . . . . . . . . . . . . . . . . . . 101

Exa 4.11 Exa 4.12 Exa 4.13 Exa 4.15 Exa 4.16 Exa 4.17 Exa 4.18 Exa 4.19 Exa 4.20 Exa 4.21 Exa 4.22 Exa 4.23 Exa 4.24 Exa 5.1 Exa 5.2

Heat flow due to radiation . . . . . . . . . . . . . . . . 102 Heat exchange between concentric shell . . . . . . . . 102 Evaporation in concenric vessels . . . . . . . . . . . . 103 infinitely long plates . . . . . . . . . . . . . . . . . . . 104 Heat exchange between parallel plates . . . . . . . . . 105 Thermal radiation in pipe . . . . . . . . . . . . . . . . 106 Heat transfer in concentric tube . . . . . . . . . . . . 107 Heat exchange between black plates . . . . . . . . . . 107 Radiation shield . . . . . . . . . . . . . . . . . . . . . 108 Heat transfer with radiaiton shield . . . . . . . . . . . 109 Radiaition shape factor . . . . . . . . . . . . . . . . . 110 Radiation loss in plates . . . . . . . . . . . . . . . . . 111 Concentric tube . . . . . . . . . . . . . . . . . . . . . 112 Harpin exchanger . . . . . . . . . . . . . . . . . . . . 114 Length of pipe . . . . . . . . . . . . . . . . . . . . . . 116

Exa 5.4 5.3 Exa Exa 5.5 Exa 5.6 Exa 5.7 Exa 5.8 Exa 5.9 Exa 5.10 Exa 5.11 Exa 5.12 Exa 5.13 Exa 5.14 Exa 5.15 Exa 5.16 Exa 5.17 Exa 5.18 Exa 5.19

Double pipe exchanger. . .. . ...... ...... ...... ...... .. 119 117 Parallel flow heat arrangement Counter flow exchanger . . . . . . . . . . . . . . . . . 120 LMTD approach . . . . . . . . . . ...... ..... 121 Shell and tube exchanger . . . . . . . . . . . . . . . . 122 Order of Scale resistance . . . . . . . . . . . . . . . . . 123 Length of tube required . . . . . . . . . . . . . . . . . 124 Suitability of Exchanger . . . . . . . . . . . . . . . . . 126 Number of tubes required . . . . . . . . . . . . . . . . 127 Shell and tube heat exchanger . . . . . . . . . . . . . 129 Length of pipe in Exchanger . . . . . . . . . . . . . . 131 Dirt factor . . . . . . . . . . . . . . . . . . . . . . . . 132 Heat transfer area . . . . . . . . . . . . . . . . . . . . 135 Oil Cooler . . . . . . . . . . . . . . . . . . . . . . . . . 136 Countercurrent flow heat exchanger . . . . . . . . . . 137 Vertical Exchanger . . . . . . . . . . . . . . . . . . . . 138 Countercurrent Heat Exchanger . . . . . . . . . . . . . 140 7

Exa 5.20 Exa 5.21 Exa 5.22 Exa 5.23 Exa 5.24 Exa 5.25

Number of tube side pass . . . . . . . . . . . . . . . . 141 Number of tubes passes . . . . . . . . . . . . . . . . . 142 Outlet temperature for hot and cold fluids . . . . . . . 143 Counterflow concentric heat exchanger . . . . . . . . . 145 Number of tubes required . . . . . . . . . . . . . . . . 146 Parallel and Countercurrent flow . . . . . . . . . . . . 147

Exa 6.1 Exa 6.2 Exa 6.3 Exa 6.4 Exa 6.5 Exa 6.6 Exa 6.7 Exa 6.8 Exa 6.9 Exa 6.10 Exa 6.11 Exa 6.12 Exa 6.13 Exa 6.14 Exa 6.15

Boiling point Elevation . . . . . . . . . . . . . . . . . 149 Capacity of evaporator . . . . . . . . . . . . . . . . . . 149 Economy of Evaporator . . . . . . . . . . . . . . . . . 150 Steam economy . . . . . . . . . . . . . . . . . . . . . . 151 Evaporator economy . . . . . . . . . . . . . . . . . . . 153 Single effect Evaporator . . . . . . . . . . . . . . . . . 154 Single effect evaporator reduced pressure . . . . . . . . 155 Mass flow rate . . . . . . . . . . . . . . . . . . . . . . 156 Heat load in single effect evaporator . . . . . . . . . . 156 Triple effect evaporator . . . . . . . . . . . . . . . . . 157 Double effect evaporator . . . . . . . . . . . . . . . . . 158 lye in Triple effect evaporator . . . . . . . . . . . . . . 161 Triple effect unit . . . . . . . . . . ...... ..... 164 Quadruple effect evaporator . . . . . . . . . . . . . . . 165 Single effect Calendria . . . . . . . . . . . . . . . . . . 167

Exa 6.16

Single effect evaporator . . . . . . . . . . . . . . . . .

8

169

Chapter 2 Conduction

Scilab code Exa 2.1 Thickness of insulation

clc ; clear ; printf ( ”Ex am pl e 2. 1 \n Page no . 2.18 \ n Part −(a)” ) A=1; //sq me tr e printf ( ”Ar ea of heat tra nsf er ,A=%f m ˆ2 \ n ” ,A ) Q=450; // W/ sq mt re printf ( ”Rat e of heat l os s / unit area= %f W/mˆ2 \ n ” ,Q ) dT=400; // K insulati on printf ( ”T e m p er a t u r e diff erenc e across layer \ t , dT=%f K\ n ” ,dT) 10 k=0.11 //W/(m.K) 11 printf ( ”For as bes tos , k=%f \n ” ,k ) 12 //Q=(k ∗ A∗dT)/x 13 x=(k*A*dT)/Q 14 X=x*1000; 15 16 // for f i r e clay insulati on 17 k=0.84; // W/(m.K) 18 printf ( ”For f i r e cl ay in su la ti on , k=%f W/( m.K ) \n ” ,k); 19 x=(k*A*dT)/Q; 1 2 3 4 5 6 7 8 9

20 X=x*1000;

9

21 printf ( ”A ns . (A ) . Thickness of as bes to s i s : %f m=%f mm \ n ” ,x,X) 22 printf ( ”A ns .(B) Thi ck nes s of f i r e clay insu lati on is : %f m =%f mm \ n ” ,x,X)

Scilab code Exa 2.2 Heat loss per metre

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

clc ; printf ( ”Ex am pl e 2. 2 , \ nP ag e no .2 .1 8 \ n ” ) ; L =1 // m printf ( ”Le ng th of pp ip e ,L = %f m \ n ” ,L); r1=(50/2) // in m m r1=r1/1000 / / in m printf ( ”r1=%f m \n ” ,r1); r2=(25+3)/1000 // m printf ( ”r2=%f m \n ” ,r2) rm1=(r2-r1)/ log (r2/r1); printf ( ”rm1= %f m\ n” ,rm1) k1=45 //W/(m.K) R1=(r2-r1)/(k1*(2*%pi*rm1*L)) // K/W printf ( ”Th er ma l re si st an ce of wall pipe= R1=%f K/W \ n ” ,R1); printf ( ”F or inner lagging : \ n” ) ; k2=0.08 //W/(m.K) ri1=0.028 //m ri2=(ri1+r1) // m rmi1=(ri2-ri1)/ log (ri2/ri1) R2=(ri2-ri1)/(k2*2*%pi*rmi1*L) printf ( ”T hermal re si st an ce of inner laggi ng= R2=%f K/ W” ,R2); printf ( ”F or ou te r lagg ing : \ n ” ) ; k3=0.04 //W/(m.K) ro1=0.053 //m

26 ro2=(ro1+0.04)

// m 10

27 rmo1=(ro2-ro1)/ log (ro2/ro1) 28 R3=(ro2-ro1)/(k3*2*%pi*rmo1*L) 29 printf ( ”T hermal re si st an ce of W\n ” ,R3); 30 R=R1+R2+R3 31 Ti=550 // K // in si de 32 33 34 35

inner

laggi ng= R2=%f K/

To=330 //K // out sid e dT=Ti-To; //Te mp er at ur e di ff er en ce Q=dT/R printf ( ”Ra te of heat lo ss per me tr e of pip e ,Q=%f W/m ” ,Q )

Scilab code Exa 2.3 Heat Loss

1 2 3 4 5 6 7 8 9 10

clear ; clc ; printf ( ”Exam ple 2. 3 ” )

//Given r1=44 // [mm] r1=r1/1000 // [m] r2=0.094 // [m] r3=0.124 // [ m] T1=623 //Te mp era tu re at oute r surface T3=313 //Te mp era tu re at oute r surface

ins ula tio n [ K ] //T hermal condu ctivi ty of 1 . . in [W/m.K] 12 k2=0.064 //T hermal condu ctivi ty of 2 [W/m.K] 13 l =1 // L en g th of p i p e [m] 14 rm1=(r2-r1)/ log (r2/r1) // log mean insul ation lay er 1 [m] 15 rm2=(r3-r2)/ log (r3/r2) // log mean insulation laye r 2[ m] 11 k1=0.087

16 // Pu tti ng values

in follow

ing 11

eqn :

of wall in [K ] of oute r insulati

on layer

insulati

on layer

radius

of

radius

of

17 Q= (T 1-T3)/((r2-r1)/(k %pi*rm2*l)); 18 printf ( ”H eat lo ss per

1*2*%pi*rm

me te r pipe

1*l)+(r 3-r2)/(k2*2*

is %f W/m”

,Q )

Scilab code Exa 2.4 Heat loss

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

clc ; clear ;

//Ex am pl e 2. 4 printf ( ”Exam ple 2. 4 ” ) //Given A =1 / /H eat transfer ar ea [ s q m] x1=0.229 // thic knes s of f i r e bri ck in [m] x2=0.115 // thi ck ne ss of insul ating br ic k in [m] x3=0.229 // thi ck ne ss of bui ld ing br ic k in [m] k1=6.05 // th er mal conductivity of f i r brick [W/( m. K) ] k2=0.581 // th er ma l condu ctivi ty of insulati ng bric k [W/m.K] k3=2.33 // th er ma l conduc tivit y of buildin g bric k [W/m.K] T1=1223 // inside te mp er at ur e [ K ] T2=323 // Outsid e temperat ure [K ] dT=T1-T2 // Overal l temp dr op [K ] R1=(x1/k1*A) // th er ma l resist ance 1 R2=(x2/k2*A) / / Thermal resist ance 2 R3=(x3/k3*A) //T hermal res ist anc e 3 Q=dT/(R1+R2+R3) //w/SQ m Ta=-((Q*R1)-T1) //fro m Q1=Q=(T1 −Ta) /( x1/k1 ∗A) // Sim ila rly

13 14 15 16 17 18 19 20 21 22 Tb=(Q*R3)+T2; 23 printf ( ” In ter fac e temp erat ure : \ n i −Be tw ee n K \ n i i −Betw een IB −PB=%fK”,Ta,Tb);

12

FB−IB=%f

Scilab code Exa 2.5 Heat loss

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

clc ; clear ;

//Ex am pl e 2. 5 printf ( ”Ex am pl e 2.5 \ nP ag e 2. 23 ” ) //Given A=1; // le t [ sq m ] x1=0.23; // thickness of f i r brick layer [m] x2=0.115; // [m] x3=0.23; // [m] T1=1213; //T em pe ra tu re of furnace [K ] T2=318; //Te mp er at ur e of fur nace [K ] dT=T1-T2; // [K] k1=6.047; //W/(m.K ) ( f i r e br ic k ) //W/( m. K) ( in su la ti ng bri ck ) k2=0.581; k3=2.33; //W/( m. K) ( bu il di ng bri ck ) Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3)) //H eat lost unit

pe r

Ar ea in Watt

R1=(x1/k1) //Th erma l r es is ta nc e R1=0.04 //Approximate R2=(x2/k2) R2=0.2025 //Approximate R3=(x3/k3) R3=0.1 //Approximate Ta=T1-((dT*R1)/(R1+R2+R3)) Tb=((dT*R3)/(R1+R2+R3))+T2 Tb=565 //Approximation printf ( ” \ nAns wer : He at lo ss per unit area is %f W=%f J / s \n ” ,Q_by_A ,Q_by_A ); 28 printf ( ” \ nAnsw er : \ n Ta=%f K = Tempe ratur e at the

inter face

b et w ee n f i r e bric k and insula

\ n T b=%d K Tem pe ra tu re at the 13

ting

bric k

in te rf ac e be tw ee n

insulating

and bui ldin g bri ck \ n” ,Ta,Tb)

Scilab code Exa 2.7 Heat loss

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

clc printf ( ”Ex am pl e 2. 7 , Pa ge no 2/26 printf ( ”Part −(a ) \ n” ) ; A=1; / / sq me tr e x1=114 // mm x1=114/1000 // me tr e k1=0.138 // W/(m.K) R1= x1/ (k1*A ) x2=229 //mm x2= x2/10 00 // me tr e k2=1.38 // W/m. K R2=x2/(k2*A) dT=1033-349

\n ” ) ;

//H eat los s Q=dT/(R1+R2)

16 printf ( ”ANSWER: He at l o s s from 1 sq metre wal l=%f W” ,Q); 17 printf ( ”Part (b ) \n ” ) ; 18 // cont act resi sta nce= cr 19 cr=0.09 //K/W 20 R=R1+R2+cr 21 Q=dT/R 22 printf ( ”ANSWER: Heat l o s s from 1 sq metre when re si st an ce presen t=%f W” ,Q);

Scilab code Exa 2.8 Loss per area

1 clear ; 2 clc ;

14

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

//Ex am pl e 2. 8 printf ( ”Ex am pl e 2. 8 \n ” ) // Given : x1=0.02 // [m] x2=0.01 // [m] x3=0.02 // [m] k1=0.105 //W/ (m. k ) k3=k1 //W/(m.K) k2=0.041 //W/(m.K) T1=303 T2=263 // [K] dT=T1-T2 Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3)) R=0.625 //K/W Tx=293 //K Rx=0.9524 //K/W x=R*(T1-Tx)/(dT*Rx) x=x*100 //mm printf ( ”T he te mpe rat ure of 293 K wi ll be re ach ed at

poi nt %f mm fro m th e ou te rm os t wal l surface th e ice −box” ,x )

Scilab code Exa 2.9 Heat loss

1 2 3 4 5 6 7 8 9 10

clc printf ( ”Ex am pl e 2. 9 , Pa ge 2.2 8 \ n ” ) ;

//Given ID=50 //mm; dT=(573-303); printf ( ” I n t er n a l diame ter , ID=%f mm” ,ID); r1=ID/2 //mm r1=r1/1000 // met re s OD=150 // mm printf ( ” Out er di am et er ,O D=%f mm” ,OD);

11 r2=OD/2

// mm 15

of

12 13 14 15 16 17

// m //Th er ma l cond ucti vit y k=17.45 // W/(m.K) // Solu tion printf ( ”Q/A=dT / ( r 2 −r1 ) /k \ n ” ) ;

18 19 20 21

A2=4*%pi*(r2^2); A = sqrt (A1*A2) Q=(A*k*dT)/(r2-r1) printf ( ”ANSWER: \ nHeat

r2=75/1000

A1=4*%pi*(r1^2);

l o s s=Q=%f W” ,Q);

Scilab code Exa 2.10 Heat Passed

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

clear ; clc ;

//Exa mp le 2.1 0 printf ( ”Ex am pl e 2. 10 ” ) A = 1 //sq m x1=0.15 x2=0.01 x4=0.15 T1=973 // [K] T2=288 // [K] dT=T1-T2 // [K]

//T hermal cond ucti vit ies k1=1.75 k2=16.86 k3=0.033 k4=5.23

// in ab se nc e of air

gap ,s um of th er ma l resis tance s

sR=(x1/k1*A)+(x2/k2*A)+(x4/k4*A) Q= dT/s R printf ( ”H eat lo st pe r sq me te r is %d W/sq m ”

// When heat

22 Q1=1163

,Q);

loss , Q=1 16 3, the n new re si st an ce =sR1

// [W/ sq m] 16

23 24 25 26 27

sR1=dT/Q1

//w id t h of air

gap be w t h en

w=(sR1-sR)*k3*A // [m] w=w*1000 // in [mm] printf ( ”W idth of air ga p is %f mm”

,w);

Scilab code Exa 2.11 Insulated pipe

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

clear ; clc ;

//Exa mp le 2.1 1 printf ( ”Ex am pl e 2. 11 ” ) ; d1=300 // [mm] r1=d1/2 // [mm] r1=r1/1000 // [m] r2=r1+0.05 // [m] r3= r2+0 .04 // [m] x1=0.05 // [m] x2=0.04 // [m] k1=0.105 //W/(m.K) k2=0.07 //W/(m.K) rm1= (r 2-r 1) / log (r2/r1); rm2=(r3-r2)/ log (r3/r2); L =1 // l et A1=%pi*rm1*L // l e t L=1 R1=x1/(k1*A1); A2=%pi*rm2*L R2=x2/(k2*A2) T1=623 // [K] T2=323 // [K] dT=T1-T2 // [K]

// [m] // [m]

//Pa rt a Q_ by _L= dT/(R 1+R 2) //He at los s printf ( ”He at lo ss is %f W/m” ,Q_by_L);

27 //Part

b: 17

28 P=2*%pi*(r1+x1+x2) // [m] 29 Q_by_L_peri=Q_by_L/P // [W/ sq m ] 30 31 printf ( ”H eat lost p e r s q m e t e r of ou te r insulation i s %f W/sq m” ,Q_by_L_peri); 32 R1=x1/(k1*A1) 33 sR=0.871+0.827 34 dT1=dT*R1/sR 35 printf ( ”Te mp era tu re be twe en two layers =%f K” ,( T1-dT1) );

of ins ulat ion

Scilab code Exa 2.12 Composite brick

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

//Exa mp le 2.1 2 clear ; clc ; printf ( ”Ex am pl e 2.1 2 \ n ” )

//Given x1=0.01 // [m] x2=0.15 // [m] x3=0.15 // [m] T1=973 // [K] T2=423 // [K] dT=T1-T2;

//T hermal cond ucti vit ies k1=16.86 // [W/m. K ] k2=1.75 // [W/m. K ] k3=5.23 // [W/m. K ] k_air=0.0337 // [W/m.K ] A =1 // [ sq m] sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)) Q=dT/sigma_R //H eat f lo w in [W Tm= Q*x 3/k3 //T em pe ra tu re dro p in magn esit e brick

// In te rf ac e tempera ture =iT

22 iT=T2+Tm

// [K] 18

//wi th air ga p for to 1163 per sq m 24 x_by_k= sigma_xbyk -sigm a_R //x / k for air 23 sigm a_xb yk= A*dT/11 63

red uci ng

h e a t loss

25 t=x_by_k*k_air 26 t=t*1000; 27 printf ( ”W idth of the

ai r gap is %f mm”

,t);

Scilab code Exa 2.13 Heat flow in a pipe

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

//Exa mp le 2.1 3 printf ( ”Ex am pl e 2.1 3 \ n” ) ; L =1

//a ss um e

[m]

k1=43.03

// [W/ (m.K)

k2=0.07

/ /(W/m. K)

T1=423

// ins id e temper atur e [K]

T2=305

// [K ]

r1=0.0525

// [mm]

r2=0.0575;

// [m]

r3=0.1075 // [m] // r3 =r3 /1000; //[m ] Q=(2*%pi*L*(T1-T2))/((( log (r2/r1))/k1)+(( /k2)); //H eat los s pe r me tre

21 22 printf ( ”H eat flow per 23 24 printf ( ”Pa rt 2 \ n ” ) ; 25 //T=Te mp er at ur e of

me tr e of pipe

outer

su rfa ce 19

log (r3/r2))

is %f W/m”

,Q);

26 T=T1-(Q* log (r2/r1))/(k1*2*%pi*L); 27 28 printf ( ”T e m p er a t u r e at out er surface %f K” ,T); 29 30 printf ( ” \ nPar t i i i \ n ” ) ; 31 32 33 34 35 36 37

id=0.105

// insi de dia met re in

A=%pi*id*1

// inside

C=Q/(A*(T1-T2));

of ste el pi pe :

[m]

ar ea in [ sq m] // conduc tance

printf ( ”Co nd uc ta nc e pe r m length area i s %f W/K” ,C )

per length ba se d on insi de

Scilab code Exa 2.15 Thickness of insulation

1 //Exa mp le 2.1 5 2 printf ( ”Ex am pl e 2.1 5 \ n” ) 3 A =1 // [ sq m ] 4 x1=0.1 //m 5 x2=0.04 6 k1=0.7 7 k2=0.48 8 sigma=x1/(k1*A)+x2/(k2*A) //K/W 9 //Q=4.42 ∗dT 10 //Q=dT/sigma 11 //with roc kwoo l ins ul ati on added , Q da sh=0 .7 5 ∗Q 12 k3=0.065 // W/(m.K) 13 // Q dash= dT/s igm a+x3/ k3 ∗A 14 //On sol ving Q and Q da sh w e get 15 x3=((1/(0.75*4.42))-sigma)*k3 // [m] 16 x3=x3*1000 // [mm] 17 printf ( ”Thi ck ne ss of ro ck wo ol insulat ion requ ired =

%f mm” ,x3) 20

Scilab code Exa 2.16 Reduction in heat loss in insulated pipe

1 2 3 4 5 6 7 8 9 10

clc ; clear ;

//Ex am pl e 2.1 6 , Pa ge no 2.36 d1=40; // Diam ete r of pipe [mm] r1=(d1/2)/1000 // Ou ts id e radius in [m] t1=20; // Insul ation 1 thickness in [mm] t1=t1/1000 // [m] // Insulation 2 thicknes s i n [m] t2=t1; r2=r1+t1; // ra di us after 1 st insul ation in [ m] r3=r2+t2; //R a d i u s after s e co nd in sulation in [ m ]

11 12 // Sinc e Scilab

do es no t hand les sym bol ic con sta nts , we wi ll as su me some values : 13 // (1 ) 14 printf ( ”L e t t he laye r M −1 be ne ar er t o t he surf ace ” 15 L=1; // [m] 16 T1=10; //Te mp er a t u re of inne r surface of pi pe [K] 17 T2=5; //Te mp era tu re of oute r surface of insulation [K] 18 k=1; //T he rm al cond uct ivit y 19 k1=k; //Fo r M −1 material 20 k2=3*k; //F or material M −2 21 Q1=(T1-T2)/( log (r2/r1)/(2*%pi*L*k1)+ %pi*L*k2)) 22 23 // (2 ) 24 printf ( ”L e t t h e lay er of ma te ri al M th e surface ” ) ; 25 Q2=(T1-T2)/( log (r2/r1)/(2*%pi*L*k2)+

%pi*L*k1))

21

log (r3/r2)/(2*

−2 b e ne ar er t o log (r3/r2)/(2*

)

\n Fo r dummy va ri ab le s un it y . . . \ nFor any value of k ,T 1 and T2,Q 1 is al wa ys le ss th an Q 2” ,Q1,Q2); 27 printf ( ” \n So ,M−1 ne ar th e surface is advisab le ( i . e Arra ng emen t on e will result i , ess he at loss \n ) ”) 26 printf ( ”Q1=%f and Q2= %f

; 28 per_red=(Q2-Q1)*100/Q2 29 printf ( ”Pe rc en t re duc ti on in he at loss is %f pe rc en t ” ,per_red) 30 printf ( ” \nNOTE: Sli ght var iat ion in ans wer s due to

less precise calculation in book . If p er f or m ed ma nu al ly , thi s an sw er stands to be cor rect ” )

Scilab code Exa 2.17 Heat loss in a pipe

1 // Examp le2 .1 7 2 T1=523 // [K] 3 T2=323 // [K] 4 r1=0.05 // [m] 5 r2=0.055 // [m] 6 r3=0.105 // [m] 7 r4=0.155 // [m] 8 k1=50 // [W/ (m.K) ] 9 k2=0.06 // [W/ (m.K) ] 10 k3=0.12 //W/(m.K) 11 // CASE 1 12 Q_by_L1=2*%pi*(T1-T2)/(( log (r2/r1))/k1+( log (r3/r2) )/k2+( log (r4/r3))/k3) // [W/m] 13 printf ( ”Heat l o s s=%f W/m” ,Q_by_L1) 14 //C as e 2 15 Q_by_L2=2*%pi*(T1-T2)/(( log (r2/r1))/k1+( log (r3/r2) )/k3+( log (r4/r3))/k2) 16 perct=(Q_by_L 2 -Q_by_L1) *100/Q_by_L1 17 printf ( ” If order is ch ang ed the n heat lo ss= %f W/m”

,Q_by_L2)

22

18

printf ( ” \n loss

of h e a t is inc re ase d by % f pe rc en t by put tin g mate rial wi t h hig her th er ma l conduc tivit y ne ar th e pi pe surface ” ,perct)

Scilab code Exa 2.18 Arrangements for heat loss

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

clc ; clear ;

//Ex am pl e 2.1 8 , Pa ge no 2.38 //Given // Assume : L =1 // [m] r1=0.10 // [m] Ou t s i d e ra di us od p ip e ia=0.025 // inne r insulai ton [m] //O u ter r adi us of in ne r insulation //O u ter ra diu s of out er insulation // CASE 1: ’ a ’ near the pipe su rfa ce // l et k1 =1 k1=1; //Th er ma l co nd uct iv ity of A[W/m. K] //and k2= 3k1= 3 k2=3; //Th er ma l con du cti vi ty of B[ W/m. K] // Let dT=1 r2=r1+ia r3=r2+ia

dT=1 Q1=dT/( log (r2/r1)/(2*%pi*k1*L)+ log (r3/r2)/(2*%pi*k2* L)) 21 Q1=22.12 //Approximate 22 // CASE 2: ’b ’ nea r the pipe surf ace 23 Q2=dT/( log (r2/r1)/(2*%pi*k2*L)+ log (r3/r2)/(2*%pi*k1* L)) 24 Q2=24.39 //Approximation 25 printf ( ”ANSWER −( i ) \nQ1=%f W \nQ2= %f W \nQ1 is less

th an Q2. i . e arr an gem en t A nea r the pipe and B a s ou te r lay er giv es less 23

h e a t loss

surf ace \ n ” ,Q1,

Q2); 26 percent=(Q2-Q1)*100/Q1;

// perc ent reduction in he at loss 27 printf ( ” \nANSWER −( i i ) \ nP er ce nt reduct ion in he at lo ss ( wi th nea r the pipe surf ace )=%f percent ” , percent);

Scilab code Exa 2.19 Insulation thickness

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

clc clear printf ( ”Ex am pl e 2. 19 . Pa ge no .2 .3 9 ” )

//Given x1=0.224 // m k1=1.3 // W/(m.K) k2=0.346 // W/(m.K) T1=1588 // K T2= 29 9 // K QA=1830 // W/ sq me tr e // hea t lo ss

// sol uti on printf ( ”Q/A=(T1−T2)/x1/k1+x2/k2”

); x2=k2*((T1-T2)*1/(QA)-(x1/k1)) x2=x2*1000; printf ( ”Thickness of in su la ti on= %f mm” ,x2)

Scilab code Exa 2.20 Heat loss in furnace

1 2 3 4 5 6

//Exa mp le 2.2 0 //Given // for clay k1=0.533 // [W/ (m.K) ] // for re d bric k k2=0.7 // [W/m. K ] 24

7 //Cas e 1 8 A =1 //Area 9 x1=0.125 // [m] 10 x2=0.5 // [m] 11 // Resi sta nces 12 r1=x1/(k1*A) //Res

of f i r e cla y [K /W]

13 r2=x2/(k2*A) //Re s of red bri ck [K /W] 14 r=r1+r2 15 //Temperatures 16 T1=1373 // [K] 17 T2=323 // [K] 18 // [W/ sq m] Q=(T1-T2)/r 19 Tdash=T1-Q*r1 // [K] 20 //Case2 21 // He at lo ss must re ma in un cha ng ed , Thic knes s of

r e d br ic k al so re du ce s t o its 22 23 24 25

26 27 28 29 30

hal f

x3=x2/2 // [m] r3=x3/(k2*A) // [ K/W] Td d= T2+(Q*r 3) // [K]

// Thickness of diatom ite be x2 ,k m be m ean conductivity Tm=(Tdash+Tdd)/2 km=0.113+(0.00016*Tm) // [K] // [W/ (m.K ] x2=km*A*(Tdash-Tdd)/Q // [m] x2=x2*1000 // [mm] printf ( ”Thickness of diatomi te la yer= %f mm” ,x2)

Scilab code Exa 2.21 Rate of heat loss in pipe

1 2 3 4 5

// Exaample 2 .2 1 //Given k1=0.7 // common bri ck W/((m .K ) k2=0.48 //gy ps um l a ye r [W/( m.K ) k3=0.065 //R oc kw oo l [W/m.K ]

6 //H eat loss

w i t h insulatiob

will 25

be 20% of

w it ho ut

insulation 7 8 9 10 11

A =1 //sq m x1=0.1 // [m] x2=0.04 // [m] R1=x1/(k1*A) R2=x2/(k2*A)

//K/W //K/W

12 13 14 15 16 17 18

R=R1+R2

//K/W

//R3=x3/(k3

∗A)

QbyQd=0.2 sigRbyRd=QbyQd x3=(R/QbyQd-R)/15.4 //m // [mm] x3=x3*1000 printf ( ”Thickne ss of roc kwo ol in su lat io n =%f mm” ,x3)

Scilab code Exa 2.22 Heat loss from insulated steel pipe

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

clc ; clear ;

//Exa mp le 2.2 2 Ts=451; //S team temp erat ure in [K ] Ta=294; //Air te mp er atu re in [K ] Di=25; // Interna l dia me ter of pip e [mm] Di=Di/1000; // [m] od=33; //Ou ter di am et er of p ip e [mm] od=od/1000; // [m] hi=5678; // In si de heat tr a n s f er c o e f f i c i e n t [W/( mˆ2 . K) ] ho=11.36; // Outs ideh eat t ra n s fe r c o e f f i c i e n t [W/( sq m.K) ] xw=(od-Di)/2; // Th ic kn es s of steel pi pe [m] k2=44.97; //k for st eel in W/( m.K ) k3=0.175; //k for rockw ool in W/( m.K ) ti=38/1000; // th ick ne ss of insul ation in [ m]

16 r1=Di/2;

// [m] 26

// [m]

17 18 19 20 21 22

r2=od/2; rm1=(r2-r1)/ r3=r2+ti; rm2=(r3-r2)/ Dm1=2*rm1; Dm2=2*rm2;

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

//R at e of hea t lo ss = d T/( si gm a R ) L=1; // [m] R1=1/(hi*%pi*Di*L); // [ K/W]

// [m]

log (r2/r1);

// [m] // [m] // [m] // [m]

log (r3/r2);

R2=xw/(k2*%pi*Dm1*L); R3=(r3-r2)/(k3*%pi*Dm2*L); // [mm] Do= (od+2 *ti) ; R4=1/(ho*%pi*Do*L); sigma_R=R1+R2+R3+R4;

// [m]

//H eat los s dT=Ts-Ta; // [K] Q=dT/sigma_R; //He at lo ss [W/m] printf ( ” \ nAns : Ra te of heat lo ss is %f W/m” ,Q); printf ( ” \n NOTE: Sligh t variat ion in fi na l an sw er due

t o la ck of prec ision in calcu latio n of R 1,R 2,R 3 and R 4. In book an ap pr ox im at e values of these is taken \ n ” )

Scilab code Exa 2.23 Heat loss from furnace

1 2 3 4 5 6 7 8 9

clc ;

//Exa mp le 2.2 3 T1=913 // [K] T=513 // [K] T2=313 // [K] //Q=(T1 −T) /( x/( k ∗A) ) //Q=(T−T2) /(1 /( h ∗A) ) //x=2k/h //Q=(T1 −T2) /(x/(kA)+1/(h

∗A) )

10 // Th er ef or e ,Q=hA/ 3 ∗ ( T1−T2 )

27

11 12 13 14 15 16

//W ith inc rea se in thick ness (10 0%) //x1=4 ∗k / h //Q2=(T1 −T2) /( x1/k ∗A+1/(h ∗A) ) //Q2=(h ∗A) /5 ) ∗ ( T1−T2 ) //Now h=1; //Assume

17 18 19 20

A=1; //A ssume for calcu Q1=(h*A/3)*(T1-T2) Q2=((h*A)/5)*(T1-T2) percent=(Q1-Q2)*100/Q1

lati on

// Per cen t reduction in he at loss 21 printf ( ” \ nTh ere for e , Perc enta ge reduction in hea t loss is %d per ce nt ” ,percent);

Scilab code Exa 2.24 Rate of heat loss

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; clear ; printf ( ”Ex am pl e 2.2 4 \ n Pag e no . 2.47 ” ) ;

// giv en L =1 //m thp=2 // Thic knes s of pipe ; in m m thi=10 // Thi ck nes s of insu lati on ; in m m T1=373 //K T2=298 //K id=30 //mm r1=id/2 //mm r2=r1+thp //mm r3=r2+thi //mm //In S . I units r1=r1/1000 //m r2=r2/1000 //m r3=r3/1000 //m k1=17.44 //W/(m.K)

19 k2=0.58 //W/(m.K)

28

20 21 22 23

hi=11.63 //W/( sq m.K) ho=11.63 //W/( sq m.K)

// Solu tion

Q=(2*%pi*L*(T1-T2))/(1/(r1*hi)+( (r3/r2))/k2)+(1/(0.02*ho))) 24 printf ( ”ANSWER: \ n Ra te of heat

log (r2/r1))/k1+((

log

lo ss ,Q=%f W” ,Q);

Scilab code Exa 2.25 Thickness of insulation

1 2 3 4 5 6 7 8

clc ; clear ;

//Ex am pl r 2.2 5 h=8 .5 ; // [W/ sq m.K ] dT=1 75 ; // [K] r2=0.0167; // [m] Q_by_l=h*2*%pi*r2*dT // [W/m] k=0. 07 ; //F or insulati ng mate rial

K] 9 // for insulated

pi p e −−50% redu ction

in

[W/m.

in he at los s

10 Q_by_l1=0.5*Q_by_l // [ w/m] 11 deff ( ’ [ x]= f ( r 3 ) ’ , ’x =Q by l1 −dT/ (( l o g ( r3 / r2 ) ) /(2 ∗ %p i ∗ k)+1/(2 ∗ %p i ∗ r 3 ∗ h ) ) ’ ) 12 13 //b y t r i a l and err or method we get : 14 r3 = fsolve (0.05,f) 15 t=r3-r2 // thic knes s of insulation in [m] 16 printf ( ’ \n Hence , requi red thickness of insulat ion is

%f m=%f mm or %d m” , t , t ∗ 10 00 , rou nd ( t ∗ 10 00 ) ) ;

Scilab code Exa 2.26 Heat loss per metre

1 2 //Exa mp le 2.2 6

29

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

// Calc ulat e he at loss pe r me t r e len gth //Given id=0.1 // in ter na l diame ter in [m] od=0.12 // outer diamet er in [m] T1=358 //T em p e r a tu re of fluid [K] T2=298 //Te m p e r a tur e of su rr ou ndi ng t=0.03 k1=58 k2=0.2 h1=720 h2=9 r1=id/2 r2=od/2 r3=r2+t

//t hi c k ne s s o f in su la ti on // [W/m. K ] //W/( m.K ) in su la ti ng materia // inside he at transfe r coeff //W/ sq m.K // [m] // [m] // [m] //H eat lo ss per me te r=Q by L

[K] [m] l [W/sq m .K ]

Q_by_L=(T1-T2)/(1/(2*%pi*r1*h1)+ log (r2/r1)/(2*%pi*k1 ) + log (r3/r2)/(2*%pi*k2)+1/(2*%pi*r3*h2)); //W/m 19 printf ( ”H eat lo ss per me tr e length of pipe =%f W” , Q_by_L)

Scilab code Exa 2.27 Mineral wool insulation

1 2 3 4 5 6 7 8 9

clc ; clear ;

//Exa mp le 2.2 6 // Given :

// [K] // [K] // [K] // Ou ts id e he at transfer c o e f f i c i e n t s [W/ s q m. K] 10 h2=12; // [W/ sq m.K] 11 r1=0.047; // Internal radi us [m] T1=573; T2=323; T3=298; h1=29;

12 r2=0.05;

//Ou te r radi us [m] 30

13 k1=5 8 ; // [W/m. K ] 14 k2=0.052; // [W/m. K ] 15 //Q=(T1 −T2) /(1 /( r1 ∗ h1)+ lo g ( r2 / r1 )/ k1+log ( r3 / r2 ) /k2 )

=(T2−T3) /(1 /( r3 ∗ h2 ) ) 16 deff ( ’ [ x]= f ( r 3 ) ’ , ’x=(T1 −T2) /(1 /( r1 ∗ h1)+ lo g ( r2 / r1 ) /k1 +log ( r3 / r2 )/ k2) −(T2−T3) /(1 /( r3 ∗ h2 ) ) ’ ) 17 18 19 20 21 22

//b y t r i a l and err or method : r3 = fsolve (0.05,f) t=r3-r2

// Th ic kn es s of insula tion in [m] //Q=h2 ∗2∗ %p i ∗ r 3 ∗L ∗ ( T2−T3 ) Q_by_l=h2*2*%pi*r3*(T2-T3) // [W/m] \n Ra te printf ( ” \n Thi ck ne sss of insulat ion is %d mm of he at loss pe r uni t len gth is %f W/m” , round ( t *1000),Q_by_l);

Scilab code Exa 2.28 Furnace wall

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

clc ; clear ;

//Exa mp le 2.2 8 // Calculat e he at los s pe r sq m and te mpe ra tur e of out sid e surfa ce //Given A =1 //ass um e [ sq m] x1=0.006 // [m] x2=0.075 // [m] x3=0.2 // [m] k1=39 // [W/m. K ] k2=1.1 // [W/m. K ] k3=0.66 // [W/m. K ] h0=65 //W/ sq m .K T1=900 //K T2=300 //K sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A));

31

18 //To calculat e he at loss /sq m a re a 19 Q=(T1-T2)/sigma_R // [W/ sq m] 20 printf ( ”H eat los s pe r sq me tr e are a is : %f W/sq m ” ); 21 //Q/A=T−T2/(1/h0) , wh er e T=Temp of outs ide sur fac e 22 // So , T=T2+Q/ (A ∗ h0 ) 23 T=Q/(A*h0)+T2 // [K] 24 printf ( ”T e m p er a t u r e of utside surface %f K (% f degree C)” ,T,T-273)

,Q

of fur nac e is :

Scilab code Exa 2.29 Thickness of insulating brick

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clear ; clc ;

//Exa mp le 2.2 9 //De t e rm in e nec essa ry thicknes s of insulat ion bric k //Given A =1 //As su me [ sq m] x1=0.003 // [m] x3=0.008 // [m] k1=30 // [W/m. K ] k2=0.7 // [W/m. K ] k3=40 // [W/m. K ] T1=363 // [K] T=333 // [K] T2=300 // [K] h0=10 //W/ sq m.K //Q=(T1 −T2) /(x1 /( k1 ∗A)+x2/(k2 ∗A)+x3/(k3 ∗A)+1/(h0 ∗A) ) // Also ,Q =(T −T2) /( 1/ ( h0 ∗A) ) //S o , (T 1 −T2) /( ( x1/ ( k1 ∗A)+x2/(k2 ∗A)+x3/(k3 ∗A)+1/(h0 ∗ A) )=(T −T2) /( 1/ ( h0 ∗A) ) 20 // or , x2=k2 ∗A((T1 −T2) /( (T−T2 ) ∗ h0 ∗A) −1/(h0 ∗A)−x1/(k1 ∗A )−x3/(k3 ∗A) ) 21

x2=k2*A*((T1-T2)/((T-T2)*h0*A)-1/(h0*A)-x1/(k1*A)-x3

32

/(k3*A)); // [m] 22 printf ( ” Th ic kn es sof mm” ,x2*1000);

insulating

bri ck req uir ed is %f

Scilab code Exa 2.30 Heat flow through furnace wall

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

clear ; clc ;

//Exa mp le 2.3 0 //Given hi=75 // [W/ sq m.K) x1=0.2 //m x2=0.1 // [m] x3=0.1 // [m] T1=1943 // [K] k1=1.25 //W/m.K k2=0.074 /W/m. // K k3=0.555 //W/m.K T2=343 //K A =1 //ass um e [ sq m] sigma_R=1/(hi*A)+x1/(k1*A)+x2/(k2*A)+x3/(k3*A);

/ /H eat loss

per i sq m // [W] // i f T=tempera ture be tw ee n ch ro me bric k and koa lin brick th en //Q=(T1 −T) /( 1/ ( hi ∗A)+x1/(k1 ∗A) ) // o r T=T1 −(Q∗ (1/ ( hi ∗A)+x1/(k1 ∗A) ) ) T=T1-(Q*(1/(hi*A)+x1/(k1*A))); // [K] printf ( ”T e m p er a t u r e at inne r surface of mi dd le layer =%f K (%f degre e C)” ,T,T-273); // i f Tdash=temp erat ure at the outer surf ace of middel layer , then //Q=(Tdash −T2) /(x3 /( k1 ∗A) ) Q=(T1-T2)/sigma_R

26 // o r Tdash= T2+(Q ∗ x3/(k3 ∗A) )

33

27 Tdash=T2+(Q*x3/(k3*A)) // [K] 28 printf ( ”T e m p er a t u r e at out er surface of mi dd le layer =%f K ( %f degree C)” ,Tdash,Tdash -273);

Scilab code Exa 2.31 Heat loss in pipe

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

clear ; clc ;

//Exa mp le 2.3 1 // Calculat e :( a) Heat los s pe r unit //(b) Re du ct io n in he at los s //Given hi=10 //W/ sq m.K h0=hi //W/ s q .m. K r1=0.09 //m r2=0.12 //m t=0.05 // thic knes s of insulation k1=40 //W/m.K k2=0.05 //W/m.K T1=473 //K T2=373 //K

lengt h

[m]

Q_by_L=2*%pi*(T1-T2)/(1/(r1*hi)+ log (r2/r1)/k1+1/(r2* h0)); /W / /m 18 printf ( ”A ns (a ) He at l os s=%f W/m ” ,Q_by_L) 19 // Af te r addit ion of insulat ion : 20 r3=r2+t; // rad ius of ou te r surfa ce of insulaiton 21 Q_by_L1=2*%pi*(T1-T2)/(1/(r1*hi)+ log (r2/r2)/k1+ log ( r3/r2)/k2+1/(r3*h0)); // W 22 Red= Q_by_L -Q_b y_L1 //Re du ci to n in h ea t lo ss in [ W

/m] //% Redu cti on in he at loss 24 printf ( ” Ans (b ) Pe rc en t re duc tio n in he at loss is %f 23 percent_red=(Red/Q_by_L)*100

perce nt ” ,percent_red) 34

Scilab code Exa 2.32 Heat flux through layers

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear ; clc ;

//Exa mp le 2.3 2 //Determ ine : i −Heat flux acr oss th e layers and // i i − In ter fa cia l te mp er at ur e be t w ee n th e layers //Given T1=798 //K T2=298 //K x1=0.02 //m x2=x1 //m k1=60 //W/m.K k2=0.1 //W/m.K hi=100 //W/ sq m.K h0=25 //W/ sq m.K Q_by_A=(T1-T2)/(1/hi+x1/k1+x2/k2+1/h0); //W/ sq m printf ( ”An s ( i ) − Heat flu x ac ro ss t h e laye rs is %f W /sq m” ,Q_by_A);

19 // I f Tis

the i n t e r f a c i a l temperature plat e and insulating mat eri al 20 //Q by A=(T −T2) /( x2/ k2+ 1/h 0 )

bet wee n s t e el

21 T=Q_by_A*(x2/k2+1/h0)+T2 22 printf ( ”Ans −( i i ) − In te rf ac ia l te mp er atu re be tw ee n layers is %f K (% f de gr ee C )” ,T,T-273);

Scilab code Exa 2.33 Conductive conductance furnace wall

1

35

2 3 4 5 6

clc ; clear ;

//Exa mp le 2.3 3 //D et er mi ne Te mp er at ur e at th e oute r surface of wall and convectiv e co nd uc ta nc e on th e oute r wall

7 //Te mp er at ur e of hot gas : 8 T1=2273 //K 9 //Am bi en t aur temperature : 10 T4=318 //K 11 //H eat fl ow by radia tion f rom gas es to inside

surf ace

of wa ll :

12 Qr1_by_A=23260 // [W/ sq m] 13 //H ea t tr an sf er c o e f f i c i e n t on in si de wall : 14 hi=11.63 //W/ sq m.K 15 //T hermal conductivity of wall : 16 K=58 //W. s q m/K 17 //H eat f low by rad iation f rom exter nal surface

to ambi ent : 18 Qr4_by_A=9300 //W/ sq m. 19 // Ins ide Wa ll tempe rature : 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

T2=1273

//K

Qr1=Qr1_by_A A =1 //sq m

//W for

Qc1_by_A=hi*(T1-T2) //W/ sq m Qc1=Qc1_by_A // fo r A=1 sq m

//Th erma l re si st an ce : // K/W per sq m //Now Q=(T2 −T3) /R, i . e // Exte rna l wal l te mp T3=T2 −Q∗R // Q entering wall = Q_enter=Qr1+Qc1 //W T3=T2-Q_enter*R //K T3=673 //Approximate //H eat los s due to conve ction : Qc4_by_A =Q_enter -Q r4_by_A //W/ sq m R=1/K

36

37 38 39 40 41

//Qc4 by A=h0 ∗ ( T3−T4 ) // or h0=Qc4 by A /(T 3 −T4 ) h0=Qc4_by_A/(T3-T4) //W/ sq m.K // Res ult printf ( ”Convective conduc tance is : %f W/sq m.K ” ,h0)

Scilab code Exa 2.34 Critical radius of insulation

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

clc ; clear ;

//Exa mp le 2.3 4 //Given T1=473 // [K] T2=293 // [K] k=0.17 //W/(m.K) h =3 //W/ ( sq m.K) h0=h //W/ sq m.K rc=k/h //m r1=0.025 // Inside

radi us of insulai

q_by_l1=2*%pi*(T1-T2)/(

ton

[mm]

log (rc/r1)/k+1/(rc*h0))

Heat transfe r w it h insulati 13 //W it ho ut in su lat io n :

//

on in W/m

14 q_by_l2=h*2*%pi*r1*(T1-T2) /W/m / 15 inc=(q_by_ l1 -q_by_l2) *100/q_by_l2

// Increase of he at transfer 16 printf ( ”When covered wit h ins ula tio n , \ n he at los s=%f W \n When wit hou t insu latio n , hea t lo ss= % f W \n percent inc rea se =%f percent ” ,q_by_l 1 ,q_b y_l2 ,inc ); 17 k=0.04 // Fibre gl ass in sul ai ton W /( sq m.K ) 18 rc=k/h // Critical ra di us of insul aiton 19 printf ( ”I n this ca se t h e av lu e of rc =%f m is less

t h a n th e outsi de radi us of pi pe (% f) , \ n So ad di to n of any fibre glass would ca us e a dec re ase in t h e h e a t trans fer

\ n ” ,rc,r1) 37

Scilab code Exa 2.36 Critical radius of pipe

1 2 3 4 5

clear ; clc ;

//Exa mp le 2.3 6 // Calc ulat e th e he at loss pe r m e t r e of pi pe and oute r surface te mpe rat ur e //Given //T he rm al condu ctiv ity in [W/sq m. K] k =1 h =8 //H et transfer coeff in W/sq m.K rc=k/h // Critical rad ius in m T1=473 //K T2=293 //K r1=0.055 //Ou ter radius =inne r radi us in [m]

6 7 8 9 10 11 12 13 Q_by_L=2*%pi*(T1-T2)/( log (rc/r1)/k+1/(rc*h)) 14 printf ( ”H eat los s pe r me te r of pip e is %f W/m” , Q_by_L) 15 //F or out er surface 16 //Q by L=2 ∗%p i ∗ (T−T2) /(1 / rc ∗ h ) 17 // im pl ie s that , T=T2+Q by L /( rc ∗ 2∗ %pi) 18 T=T2+Q_by_L/(rc*2*%pi*h) //K 19 printf ( ”Ou te r surface te mpe rat ure is : %f K (% f degr ee C) ” ,T,T-273)

Scilab code Exa 2.37 Time required for steel ball

1 2 clc ; 3 clear ; 4 //Exa mp le 2.3 7

38

5 // Ca lc ula te t he t i m e req uir ed

te mpe rat ur e of 42 3 K 6 //Given 7 k_steel=35 //W/m.K 8 Cp_steel=0.46 //kJ/(kg 9 Cp_steel=Cp_steel*1000 10 11 12 13 14 15 16 17 18 19 20 21 22 23

for a ball

∗K) //J/(kg

t o att ain a

∗K)

h=10 //W/ sq m.K rho_steel=7800 //kg/ cubi c m dia=50 //mm dia=dia/1000 //m R=dia/2 // radius in m //A rea in s q m A=4*%pi*R^2 V=A*R/3 //V olume in cubic me te r Nbi=h*(V/A)/k_steel

//As Nb i

0. 10 , internal temp gradi ent is neg lig ibl e //K //K //K //(T −T in f ) /(T0 −T in f )=eˆ( −h∗ At/rho ∗Cp∗V) <

T=423 T0=723 T_inf=373

t=-rho_steel*Cp_steel*R* /(3*h); // s

log ((T-T_inf)/(T0-T_inf))

24 printf ( ” Time qu ir423 ed Kfor a te mpe rat urere of i s a%fball s= %t fo h”att ain ,t,t/(3600))

Scilab code Exa 2.38 Steel ball quenched

1 2 3 4 5 6 7 8

clc ; clear ;

//Exa mp le 2.3 8 //Given dia=50 //mm dia=dia/1000 //m r=dia/2 // radius in m

9 h=115

//W/ sq m.K 39

10 11 12 13 14 15

rho=8000 //kg/ cubic m Cp=0.42 //kJ/ kg .K Cp=Cp*1000 //J/(kg ∗K) A=4*%pi*r^2 //A rea in sq m V=A*r/3 //V olume in cubic m T=423 //K

16 17 18 19

T_inf=363 T0=723

//K //K //(T −T in f ) /(T0 −T in f )=eˆ( −3 ht /( rho ∗Cp∗ r ) )

log ((T-T_inf)/(T0-T_inf))/(3*h); // Time in secon ds 20 printf ( ” Time t ak e n by ce ntr e of ball t o re ac h a te mpe rat ure of 423 K i s %f s (=%f mi nu te s” ,t,t t=-rho*Cp*r*

/60);

Scilab code Exa 2.39 Ball plunged in a medium

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

clc ; clear ;

//Exa mp le 2.3 9 //Given h=11.36 //W/ sq m.K k=43.3 //w/(m.K) r=25.4 // radi us in mm r=r/1000 // r ad iu s in m A=4*%pi*r^2 //A rea of sphe re [ sq m] V=A*r/3 //V olume in [ cubic m] rho=7849 //kg/ cubic m Cp=0.4606*10^3 //J/ kg .K t =1 //hour t=t*3600 // seco nds T_inf=394.3 // [K] T0=700 // [K]

18 // (T −T in f ) /(T0 −T in f )=eˆ( −3∗ h∗ t/rho ∗Cp∗V)

40

19 T=T_inf+(T0-T_inf)*(%e^((-h*A*t)/(rho*Cp*V))); 20 printf ( ”Te mp era tu re of bal l afte r 1 h = %f K (% f degree C)” ,T,T-273)

Scilab code Exa 2.40 Slab temperature suddenly lowered

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

clc ; clear ;

//Exa mp le 2.4 0 //Given rho=9000; //kg/ cubi c m Cp=0.38; //kJ /( kg .K) Cp=Cp*1000 //J /( kg .K) k=370; //W/m.K h=90; //W/ sq m.K l=400; //mm l=l/10 00 ; // lengt h of co pp er slab t=5/1 00 0 ; // thickness in [m] A=2*l^2 //Ar ea of slab V=t*l^2 //V olume in [ cubic m] L_dash=V/A // [m] // for sla b of thic kne ss 2 x //L dash=x L_das h =0. 025 ; // [m] Nbi=h*L_dash/k //< 0.10 var=h*A/(rho*Cp*V)

//As Nb i

<

0.1 0 ,w e ca n app ly lumped capacity

ana lys is

T=363 // [K] T_inf=303 // [K] T0=523 // [K] t=-( log ((T-T_inf)/(T0-T_inf)))/var printf ( ”T ime at wh ic h slab tem pera ture be co me s 36 3 K is %f s” ,t ) 27 printf ( ”CALCULATION MISTAKE IN BOOK IN LAST LINE ” )

41

Scilab code Exa 2.41 Flow over a flat plate

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

clc ; clear ;

//Exa mp le 2.4 1 //Given rho=9000 //kg/ cubic met er Cp=0.38 //kJ /( kg .K) Cp=Cp*1000 //J/ kg .K k=370 //W/(m.K) T0=483 //K T_inf=373 //K delta_T=40 //K T=T0-delta_T //K t =5 //tim e in [ minu tes ] t=t*60 // [ sec on ds ] //A=2A . . . . . Two f a c e s

17 //V=A. 2 x 18 //2 x=thi ck nes s of sl ab= 30 mm=0 .0 3 m 19 x=0.015 // [m] 20 th=2*x // thicknes s of slab 21 h=-rho*Cp*x* log ((T-T_inf)/(T0-T_inf))/t 22 printf ( ”Heat tr a n s f er c o e f f i c i e n t i s : %f W/( sq m.K ) ” ,h )

Scilab code Exa 2.42 Stainless steel rod immersed in water

1 2 clear ; 3 clc ; 4 //Exa mp le 2.4 2

42

5 //Given 6 rho=7800 7 h=100

// [ kg per cubic m ] //W/( sq m.K ) Conv ect ive heat

tr an sfe r

coeff 8 Cp=460 9 k=40 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

//J /( kg .K) //W/(m.K)

L =1 // [m] length ofrod D=10 //mm D=D/1000 // diam eter in [m ] R=D/2 // raidus in [m]

//F or cyl ind ric al ro d : //A rea in [ sq m] A=2*%pi*R*L V=%pi*R^2*L //V olume i n [ cubic m] L_dash=V/A // [m] Nbi=h*L_dash/k // Bio t nu mb er //N b i < 0. 10 ,He nce lumped he at capa vity T=473 // [K] T_inf=393 // [K] T0=593 // [K]

is

t=-rho*Cp*V* log ((T-T_inf)/(T0-T_inf))/(h*A) printf ( ”T ime required to rea ch te mpe rat ur e %f is %f

s ” ,T,t);

Scilab code Exa 2.43 Chromel alumel thermocouple

1 2 3 4 5 6 7 8 9

pos sibl e

clear ; clc ;

//Exa mp le 2.4 3 //Given rho=8600 // [ kg/ cu bi c m] Cp=0.42 //kJ /( kg .K) Cp=Cp*1000 //J /( kg .K) dia=0.71 // [mm]

10 dia=dia/1000

// [ dia

in m ] 43

11 12 13 14 15 16

// radi us [m] // conve ctive co ef f W/( sq m. K) // Let leng th =L=1 L =1 // [m] R=dia/2 h=600

A=2*%pi*R*L; V=%pi*(R^2)*L;

17 tao=(rho*Cp*V)/(h*A); 18 printf ( ”T ime cons tant of th e th er mo co up le is %f s” tao); 19 // at 20 t=tao 21 // From (T −T in f ) /(T0 −T in f )=eˆ( − t/ tao ) 22 ratio=%e^(-t/tao) // Ratio of therm ocouple

,

dif fer enc e to i n i t i a l te mp er atu re dif fer enc e 23 printf ( ”A t th e end of th e ti me perio d t=ta o=%f s ,

Te mp er at ur e di ff er en ce b/ n the the rmo cou ple and the gas str eam wo ul d be %f of the i n i t i a l te mpe rat ure dif fer enc e ” ,tao,ratio); 24 printf ( ” \n It s ho u ld be re or de re d after %f s” ,4*tao) ;

Scilab code Exa 2.44 Thermocouple junction

1 2 3 4 5 6 7 8 9 10 11

clc ; clear ;

//Exa mp le 2.4 4 rho=8000 //kg/ cubic m Cp=420 //J /( kg .K) h_hot=60 // for hot str eam W/( sq m. K) dia=4 // [mm] t=10; r=dia/(2*1000)

// radius in

[m]

//Fo r sphere

12 V=(4/3)*%pi*r^3

//V olume in 44

[ cubic

m]

13 A=4*%pi*r^2 //V olum e in [ sq m] 14 tao=rho*Cp*V/(h_hot*A) // T ime constant in [ s ] 15 ratio=%e^(-t/tao) // %eˆ( − t / tao )=( T−T− i n f ) /( T0 −

T inf ) 16 T_inf=573 17 T0=313

// [K] // [K]

18 T=T_inf+ratio*(T0-T_inf) 19 //ANS −[ i ] 20 printf ( ” \n Answer : Time constant of the rmo cou ple is %f s” ,tao); 21 22 //IN STILL AIR : 23 h_air=10 //W/( sq m .K) 24 tao_air=rho*Cp*V/(h_air*A) // [ s ] 25 t_air=20 // [ s ] 26 ratio_air=%e^(-t_air/tao_air) 27 T_inf_air=303 // [K] 28 T0_air=T; 29 T_air=T_inf_air+ ratio_air*(T0_ai r -T_inf_air) 30 //ANS −[ i i ] 31 printf ( ”T e m p er a t u r e attai ned by junct ion 20 s after

re mo vi ng fr om the T_air))

ho t air

st re am is :%d K”

, round (

Scilab code Exa 2.45 Batch reactor

1 2 3 4 5 6 7 8

clc ; clear ;

//Exa mp le 2.4 5 T_inf=390; U=600; Ac=1; Av=10 m=1000;

9 Cp=3.8*10^3;

// [K] // [W/ sq m.K ] // [ sq m] // Vessel are a in [ sq m] // [ kg ] // [ J/ kg .K] 45

// [K] // [K] // [W/ sq m.K ] //H eat gai ned fro m th e st eam=Rate of increa internal en er gy 14 //U∗A∗ ( T inf −T)=m∗Cp∗dT 10 11 12 13

To=290; T=360; h=8.5

se of

15 deff ( ’ [ x] = f ( t ) ’ , ’x =log (( T inf −To) /( T in f −T) )−U∗Ac∗ t / (m∗Cp) ’ ) ; 16 t = fsolve (1,f); // [ in s ] 17 t = round ( t ) // [ in s ] 18 Ts=290; 19 printf ( ” \ nTime ta ke n to he at th e reactants ove r th e same te mpe rat ure ra ng e is %f h” ,t); 20 function t1=g(T),t1=m*Cp/(U*Ac*(T_inf-T)-h*Av*(T-Ts) ) , endfunction 21 t1 = intg (To,T,g); 22 deff ( ’ [m]= fx ( Tmax) ’ , ’m=U∗Ac ∗ ( T inf −Tmax)−h ∗Av ∗ (Tmax− Ts) ’ ) 23 T_max= fsolve (1,fx) 24 printf ( ” \nANS : In CASE 1 \ nTime ta ke n to he at th e rea cta nts = % f s . ie %f h \n ” ,t,t/3600); 25 printf ( ” \nANS 2 \n T ime ta ke n to he at th e \n ” ,t1); reacta nts =: %In f sCASE 26 printf ( ” \nANS . : Maximum tem per atu re at which te mp er at ur e ca n be raised is %f K \n ” ,T_max);

Scilab code Exa 2.46 Heat dissipation by aluminium rod

1 2 3 4 5 6

clc ; clear ;

//Exa mp le 2.4 6 dia=3 // [mm] dia=dia/1000

7 r=dia/2

// [m] // radi us in [m ] 46

8 9 10 11 12 13

k=150 //W/(m.K) h=300 //W/ ( sq m.K) T0=413 // [K] T_inf=288 // [K] A=%pi*(r^2) //A rea in [ sq m] P=%pi*dia // [W/ sq m.K ]

14 Q=(T0-T_inf)*

]

sqrt (h*P*k*A)

15 printf ( ”H eat dissipated

//H eat dissipated

by th e r od is %f W”

in [ W ,Q )

Scilab code Exa 2.47 Aluminium fin efficiency

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

clc ; clear ;

//Exa mp le 2.4 7 //Given k=200 //W/(m.K) h=15 //W/( sq m.K) T0=523 // [K] T_inf=288 // [K] theta_0=T0-T_inf dia=25 // di am et er [mm] dia=dia/1000 // dia met er [m] r=dia/2 // radius in [m] P=%pi*dia // [m] A=%pi*r^2 // [ sq m]

//F or insulated

fin :

m = sqrt (h*P/(k*A)) L=100 // length of ro d in [mm] L=L/1000 // len gth of ro d in [m] Q=theta_0* tanh (m*L)* sqrt (h*P*k*A)

/ ANSWER −1 printf ( ”H eat loss

by t he insula

)

47

ted

//He at los s r o d is %f W

\n ” ,Q

//F i n efficiency fo r ins ula ted fin 24 / ANSWER −2 25 printf ( ”F i n efficiency is %f pe rc en t \ n ” ,nf*100) 26 // At the end of the fi n : theta / theta 0= (cos h [m(L −x ) ] / co sh (m L) ) 23 nf = tanh (m*L)/(m*L)

27 // at x=L , th et a / th et a 0 =1 /(c osh (mL) 28 T=T_inf+(T0-T_inf)*(1/ cosh (m*L)) // [K] 29 / ANSWER −3 30 printf ( ”T e m p er a t u r e at th e end of th e fin is %f K ” ,T )

Scilab code Exa 2.49 Pin fins

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

clc ; clear ;

//Exa mp le 2.4 9 //Given k=300 //W/(m.K) h=20 //W. ( sq m.K) P=0.05 // [m] A =2 // [ sq c m ] A=A/10000 // [ sq m] T0=503 // [K] T_inf=303 // [K] theta_0=T0-T_inf // [K] m = sqrt (h*P/(k*A))

// CASE 1 :

6 Fin s of 10 0 mm length in [ m]

L1=0.1 //Le n g th o f fin Q = sqrt (h*P*k*A)*theta_0*

tanh (m*L1)

//F or 6 fins Q=Q*6 // for 6 fin s [W] // CASE 2: 10 fi ns of 60 mm lengt h

21 L2=60

// [mm] 48

// [W]

\n

22 23 24 25

L2=L2/1000 // [m] Q2 = sqrt (h*P*k*A)*theta_0* tanh (m*L2); Q2=Q2*10 //F or 10 fins printf ( ”As ,Q for 10 fin s of 60 mm length

// [W]

( %f W) is more th an Q for 6 fin s of 10 0 mm length (% f W) . \n The agreement −−>10 fin s of 60 mm le ngt h is more effe ctiv e ” ,Q2,Q);

Scilab code Exa 2.50 Metallic wall surrounded by oil and water

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

clc ; clear ;

//Exa mp le 2.5 0 //Given h_oil=180 //W/( sq m.K) h_air=15 //W/ ( sq m.K) T_oil=353 // [K] T_air=293 // [K] delta_T=T_oil-T_air; // [K] k=80 // Conduc tivity in [W/( m. K) ] for_section=11*10^-3 // [m] L=25 // [mm] L=L/1000 // [m] W =1 // [m] Width , . . l e t t =1 // [mm] t=t/1000 // [m] A=W*t // [m] P=2*t Af=2*L*W //sq m N =1 Ab= for_section -A

// CASE 1:

// [ sq m] Fin o n oi l side onl y

m = sqrt (h_oil*P/(k*A)) nf_oil= tanh (m*L)/(m*L)

25 Ae_oil=Ab+nf_oil*Af*N

// [ sq m] 49

26

Q=delta_T/(1/(h_oil*Ae_oil)+1/(h_air*for_section))

27 28 29 30

printf ( ”In

// [W] //CASE 2:

o i l side ,Q=%f W \ n” ,Q); Fi n o n air side on ly

m = sqrt (h_air*P/(k*A)) nf_air= tanh (m*L)/(m*L)

31 nf_air=0.928 //Approximation 32 Ae_air=Ab+nf_air*Af*N // [ sq m] 33 Q=delta_T/(1/(h_oil*for_section)+1/(h_air*Ae_air))

// [W] 34 printf ( ” In ai r sid e ,Q=%f W” ,Q); 35 printf ( ” \n F rom ab ov e res ul ts we see

transf er ta ke s pl ac e i f fins air side ” ) ;

tha t more he at ar e pr ov id e d on t h e

Scilab code Exa 2.51 Brass wall

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

clc ; clear ;

//Exa mp le 2.5 1 //Given k=75 //Th er ma l con du ct ivi ty [W/( m. K) ] T_water=363 // [K] T_air=303 // [K] dT= T_water -T_air // delta T h1=150 // fo r water [W/( sq m.K ) ] h2=15 // fo r ai r [W/( sq m. K) ] W=0.5 //W idth of wall [m ] L=0.025 // [m] Area=W^2 //Ba se Ar ea [ sq m] t =1 // [mm] t=t/1000 // [m] pitch=10 // [mm]

18 pitch=pitch/1000

// [m] 50

19 N=W/pitch // [N o of fi ns ] 20 // Calcul ation s 21 A=N*W*t // Total cross − se ct io na l a r e a o f f in s

[ sq m] 22 Ab=Area-A 23 Af=2*W*L 24 25 26 27 28 29 30 31 32 33 34 35

// [ sq m] //S u r f a c e a r e a of fins

[ s q m]

//CASE 1 : HEAT TRANSFER WITHOUT FINS A1=Area // [ sq m] A2=A1 // [ sq m] Q=dT/(1/(h1*A1)+1/(h2*A2)); // [W] printf ( ” \ nWithout f i n s ,Q=%f W \ n ” ,Q); // CASE 2: Fin s on the wa te r side P=2*(t+W); A=0.5*10^-3; m = sqrt (h1*P/(k*A)) nfw= tanh (m*L)/(m*L) // Effeciency on wat er side Aew=Ab+nfw*Af*N // Effective ar ea o n th e w at e r

side

[s q m]

36 Q=dT/(1/(h1*Aew)+1/(h2*A2)); // [W] 37 printf ( ” \n With fi ns on wat er side ,Q=%f W 38 39 40 41 42 43 44 45 46

\n ” ,Q);

//CASE 3 : FINS O N THE AIR SIDE m = sqrt (h2*P/(k*A)) nf_air= tanh (m*L)/(m*L) // Effec iency Aea=Ab+nf_air*Af*N // Effective are a o n air side Q=dT/(1/(h1*A1)+1/(h2*Aea)); // [W] printf ( ” \n With Fins on Air side ,Q=%f W \n ” ,Q ) //BOTH SIDE : Q=dT/(1/(h1*Aew)+1/(h2*Aea)); // [W] printf ( ” \n With Fins on bot h side ,Q=%f W \ n” ,Q);

51

in

Chapter 3 Convection

Scilab code Exa 3.1 Boundary layer thickness

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

clc ; clear ;

//Ex am pl e 3. 1 //N. s /mˆ2 // At distan ce y fro m surface //ux=a+by+cyˆ2+dyˆ3 //A t y= 0,u x=0 th er ef or e a=0 // i . e tao=0 // At ed ge of bo un da ry layer , ie y=del //ux =u in f // At y=o , c=0 // At y=de l , ux=b ∗ d el+ d∗ de l ˆ3 mu=10^-3

// Therefore , b= −3∗d∗ de l ˆ3 //d=−u inf /(2 ∗ de l ˆ2) //b=3 ∗ u in f /(2 ∗ de l ) //F or velo city pro file ,w e ha ve : // de l /x =4.64 ∗ ( Nr e x ) ˆ( −1/2)

21 // Eva luat e N re x

52

22 23 24 25 26 27

x=75; // [mm] x=x/1000; // [m] u_inf=3; // [m/ s ] rho=1000 // [ kg/ mˆ3] for ai r Nre_x=u_inf*rho*x/mu // Reynold nu mb er

28 // Substituting th e va lu e ,w e get 29 del=x*4.64*(Nre_x^(-1/2)) // [m] 30 printf ( ” \ nBoundary layer thickness is del =%f m or %f mm” ,del,del*1000); 31 printf ( ” \nWrong uni ts in ans we r of book ,m a nd mm are wro ngly interchan ged ” ) ;

Scilab code Exa 3.2 Boundary layer thickness of plate

1 2 3 4 5 6 7 8 9

clc ; clear ;

// Examp le3 . 2 //Given mu=15*10^-6 //sq m /s v =2 //m/s L =2 // [m] len gth of plate Nre_x=3*10^5 xc=Nre_x*mu/v // c r i t i c a l leng th at whi hc the

transition 10 // Si nc e xc is 11 12 13 14 15 16 17

ta ke s pla ce less t h a n 2 m. Th er ef or e t he fl ow is

laminar // at any distanc e x , . it is calculated // d el /x =4. 64/ ( s q r t (NRe , x ) ) // At x=L=2 m Nre_l=v*L/mu del_l=4.64*L/ sqrt (Nre_l) del_l=del_l*1000 // [mm] printf ( ”B oundary layerthickness

i s %f mm” ,del_l); 53

fro m

at the

tra il in g e dg e

Scilab code Exa 3.3 Thickness of hydrodynamic boundary layer

1 clc ; 2 clear ; 3 //Ex am pl e 3. 3 4 //Given 5 mu=15*10^-6 //Ki ne ma t ic visc osit y in [ s q m /s ] 6 x=0.4 // [m] 7 u_inf=3 // [m/ s ] 8 //A t x=0.4 m , 9 Nre_x= u_inf* x /m u ; 10 printf ( ” Since Nre , x (% f) is Les s th an 3 ∗ 10 ˆ5 ,.. the bo un dar y layer is lam ina r” ,Nre_x); 11 del=4.64*x/ sqrt (Nre_x) // [m] 12 del=del*1000 // [mm] 13 printf ( ” \ nThick ness of bo un da ry lay er at x=%f m =%f mm\ n ” ,x,del); 14 Cf_x=0.664/ sqrt (Nre_x); 15 printf ( ” Loc al sk in f r i c t i o n c o e f f i c i e n t i s : %f” ,Cf_x );

Scilab code Exa 3.4 Flat plate boundary layer

1 2 3 4 5 6 7

clc ; clear ;

//Ex am pl e 3. 4 mu=1.85*10^-5 P=101.325; M_avg=29; R=8.31451;

8 T=300;

// [ kg /(m. s ) ] // Pres sure in [ kPa ] //A vg mole cula r wt of air //G as constant // [K] 54

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

// [ kg/m ˆ 3 ] // Visc osit y in [m/s ]

rho=P*M_avg/(R*T) u_inf=2

//A t x= 20 cm =0. 2 m x=0.2; // [m] Nre_x=rho*u_inf*x/mu // [ Reyn old s num ber ] del_by_x=4.64/ sqrt (Nre_x) // [ Bou nd ar y la ye r ] del=del_by_x*x

// [m]

// de l=del ∗ 1 0 0 0

/ / [mm]

//At x=0 .4 ; // [m] Nre_x=(rho*u_inf*x)/mu

//B oundary layer

// < 3∗10ˆ5

is lam ina r

del_by_x=4.64/ sqrt (Nre_x) del1=del_by_x*x

// [m]

// del 1= del 1 ∗ 1 0 0 0

/ / [mm]

d=del1-del //Del function m_dot=f(y),m_dot= u_inf*(1.5*(y/d) -0. 5*(y/d) ^3)*rho, endfunction 27 m_dot= intg (0,d,f) 28 printf ( ” \ nBoundary layer thickness at distance 20 c m ; fr om lead ing edg e is %f m=%f mm \ n ” ,del,del*1000) 29 printf ( ” \ nBoundary layer thickness at distance 40 c m fr om lead ing edg e is %f m=%f mm \ n ” ,del1,del1 *1000); 30 printf ( ” \ nThus , Mass flow rate ente rin g the bo un da ry laye r is %f k g/s” ,m_dot);

Scilab code Exa 3.5 Rate of heat removed from plate

1 2 clc ; 3 clear ; 4 //Ex am pl e 3. 5

55

5 6 7 8 9 10

//Given mu=3.9*10^-4 //Ki ne ma ti c v isco sity in s q m/s k=36.4*10^-3 //Th er ma l con du cti vit y in W/( m. K) Npr=0.69 u_inf=8 // [m/ s ] L =1 //Le ng h t of plat e in [ m]

11 Nre_l=u_inf*L/mu 12 // Si nc e Nr e l is

less t h a n 3 ∗ 10 ˆ5 , th e flo w is la mi na r ov er th e entire le ngt h of plat e 13 Nnu=0.664* sqrt (Nre_l)*Npr^(1.0/3.0) //=hL/k 14 15 16 17 18 19 20 21

// w/ sq m.K h=k*Nnu/L h=3.06 //App rox ima tio n [W/sq m. K] T_inf=523 // [K] Tw=351 // [K] W=0.3 //W idth of plate [m] A=W*L //Ar ea in [ sq m] Q=h*A*(T_inf-Tw) // R ate of he at re mo va l fr om on e side

in [W]

22 printf ( ” \ nRate of he at re mo va l is %f W \n ” ,Q ) 23 //f ro m two side : 24 Q=2*Q // [W] 25 printf ( ” \n %f W hea t shou ld be remo ved contin fr om the plate ” ,Q);

ously

Scilab code Exa 3.6 Heat removed from plate

1 2 3 4 5 6 7

clc ; clear ;

//Ex am pl e 3. 6 P1=101.325 mu1=30.8*10^-6 k=36.4*10^-3

// Pres sur e in [ kPa ] //Kin em at ic vis cos ity // [W/ (m.K) ]

8 Npr=0.69

56

in [ sq m /s ]

9 10 11 12 13 14

// Velocity in [m/s ] //kJ /( kg .K) //Le n g t h of plat e in [ m] //W idth in [ m] //Ar ea in [ sq m] //At co ns ta nt tem per atu re : mu1/ mu2=P2/P 1 u_inf=8 Cp=1.08 L=1.5 W=0.3 A=L*W

15 P2=8 // [ kPa ] 16 mu2=mu1*P1/P2

//Ki ne ma ti c visc osit y at P2 in

[ sq

m/s ] 17 18 19 20

Nre_l=u_inf*L/mu2

/ / Si nc e this Nnu=0.664* h=Nnu*k/L

is less

//Reynold ’ s no . th a n 3 ∗ 10ˆ5

sqrt (Nre_l)*(Npr^(1.0/3.0))

// He at t ra n s fe r c o e f f f i c i e n t in [W/sq m.

K] 21 22 23 24

h=2.5 //Approx T_inf=523 // [K] Tw=353 // [K] Q=2*h*A*(T_inf-Tw)

imati on in [W/sq m. K]

//H ea t re mo ve d f ro m b ot h s id es in [W] 25 printf ( ”R ate of he at removed from bo th side s of plate is %f W” ,Q);

Scilab code Exa 3.7 Local heat transfer coefficient

1 2 3 4 5 6 7 8 9 10

clc ; clear ;

//Ex am pl e 3. 7 rho=0.998 //kg/ cubi c m v=20.76*10^-6 // [ sq m/ s ] Cp=1.009 // [ kJ/ kg .K] k=0.03 // [W/m. K ] u_inf=3 // [m/ s ] x=0.4 // [m] w=1.5 // [m]

11 Nre_x=u_inf*x/v

//Reyn ol ds n o at x= 0. 4 m 57

∗ 10ˆ 5.T he flo w is lam ina r

12 / / Si nc e this

is less th a n 3 up to x=0 .4 m 13 mu=rho*v // [ kg /(m. s ) ] 14 15 Cp=1.009 16 Cp=Cp*1000 17 18 19 20 21 22 23 24 25 26 27 28 29 30

// [ kJ/ kg .K] // [ J/ kg .K]

k=0.03 //W/(m.K) Npr=Cp*mu/k Nnu_x=0.332*( sqrt (Nre_x))*(Npr^(1.0/3.0)) hx=Nnu_x*k/x // [W/ (m.K) ]

//Av e r a g e va lu e is tw ic e this va lu e // [W/ (m.K) ] h=2*hx h=10.6 //Approximation A=x*w //Ar ea in [ sq m] Tw=407 // [ k ] T_inf=293 // [K] Q=h*A*(Tw-T_inf) // [W] // From bo th side s of th e plate : Q=2*Q // [W] printf ( ” The hea t transferred from b o t h sides pla te is %d W” , round (Q));

of th e

Scilab code Exa 3.8 Width of plate

1 2 3 4 5 6 7 8 9 10

clc ; clear ;

//Ex am pl e 3. 8 rho=0.998 // [ kg/ cu bi c m] v=20.76*10^-6 // [ sq m/ s ] k=0.03 // [W/m. K ] Npr=0.697 x=0.4 // [m] fro m leadi ng e dg e of th e plate u_inf=3 // [m/ s ] Nre_x=u_inf*x/v //Rey nol d n um ebr at x=0.4

11 / / Si nc e this

is less

∗ 10ˆ5

th a n 3 58

0 m

12 13 14 15

// therefore

17 18 19 20 21 22 23 24 25

// Given :

fl ow is la mi na r and

Nnu_x=0.332* hx=Nnu_x*k/x

sqrt (Nre_x)*(Npr^(1.0/3.0));

// [W/ sq m.K] //A ve ra ge hea t tar ns fer c o e f f i c i e n t is twice value 16 h=2*hx // [W/ sq m.K ]

thi s

Q=1450 // [W] Tw=407 // [K] T_inf=293 // [K] L=0.4 // [m]

//Q=h∗w∗L ∗ (Tw−T inf ) //L=Q/(h ∗w∗ (Tw−T in f ) ) w=Q/(h*L*(Tw-T_inf)) // [m] printf ( ” \n W idth of plate is %f m”

,w);

Scilab code Exa 3.9 Heat transferred in flat plate

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

clc ; clear ;

//Ex am pl e 3. 9 v=17.36*10^-6 // Vi sc os it y fo r air [ s q m./ s ] k=0.0275 // fo r a i r . . [W/( m.K ) ] Cp=1.006 // [ kJ /( kg .K) ] Npr=0.7 // for air u_inf=2 // [m/ s ] x=0.2 // [m] Nre_x=u_inf*x/v //Re yn ol ds number at x= 0.2 m

∗ 10ˆ5

/ / Si nc e this

is less

Nnu_x=0.332* hx=Nnu_x*k/x

sqrt (Nre_x)*(Npr^(1.0/3.0))

th a n 3

// [W/( sq m. K] / /A v e r a g e v al ue of h e a t transf er coeff value

16 h=2*hx

// [W/ sq m. K) ] 59

is tw ic e this

17 18 19 20 21 22

h=12.3 //Approximation w =1 //wid th in [m] A=x*w // [ sq m] Are a of plate Tw=333 // [K] T_inf=300 // [K] Q=h*A*(Tw-T_inf) //H eat flow in [ W]

23 24 25 26

printf ( ” \nANSWER: \ nHeat flow is : %f W \ n ” ,Q ) //F rom bo th side s of plate : Q=2*Q // [W] printf ( ” \nANSWER \ n He at fl ow from b o t h sides of plate is %f W” ,Q);

Scilab code Exa 3.10 Rate of heat transferred in turbulent flow

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; clear ;

//Exa mp le 3.1 0 v=16.96*10^-6 // [ sq m./ s ] rho=1.128 // [ kg/ cu bi c m] Npr=0.699 // Pra ndt l nu mb er k=0.0276 // [W/m. K ] u_inf=15 // [m/ s ] L=0.2 // [m] Nre_l=L*u_inf/v // Reynold ’ s num be r

/ / Si nc e this is less th a n 3 ∗ 10ˆ5 , the bo un da ry lay er is la mi na r ov e r entire le ng th Nnu=0.664* sqrt (Nre_l)*(Npr^(1.0/3.0)) h=Nnu*k/L // [W/ sq m.K ] A=L^2 //Ar ea in [ sq m] Tw=293 // [K] T_inf=333 // [K]

//R ate of he at tran sfer

from BOTH s ides

Q=2*h*A*(T_inf-Tw) // [W] printf ( ”R ate of he at transfer

pla te is %f W \n ” ,Q); 60

is :

fro m bo th sides

of

20 // i i −With turbulen

t bo und ar y layer

from th e leading

edg e : 21

h=k*0.0366*(Nre_l^(0.8))*(Npr^(1.0/3.0))/L

W/( sq m.K) ] 22 //H eat transfer

// [

fro m b o t h sides is : // [W]

23 Q=2*h*A*(T_inf-Tw)

24 printf ( ” \ nThese calcul

ation s s ho th at th e th at transfe r rate is ap pr ox im at el y do ub le d i f b ou n da ry layer is turb ulen t fro m th e leadin g ed ge \n ” ) ;

Scilab code Exa 3.11 Heat transfer from plate in unit direction

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

clc ; clear ;

//Exa mp le 3.1 1 mu=1.906*10^-5 // [ kg /(m. s ) ] k=0.02723 //W/m.K Cp=1.007 // [ kJ /( kg .K) ] rho=1.129 // [ kg/ cu bi c m] Npr=0.70 Mavg=29 u_inf=35 // [m/ s ] L=0.75 // [m] Tm=313 // [K] P=101.325 // [ kPa ] Nre_l=rho*u_inf*L/mu // Reynold ’ s num be r > 5∗10ˆ5 Nnu=0.0366*Nre_l^(0.8)*Npr^(1.0/3.0); h=Nnu*k/L // [W/ s m.K ] A=1*L // [ sq m] Tw=333 // [K] T_inf=293 // [K] Q=h*A*(Tw-T_inf); // [W] printf ( ”H eat transfer from th e plate is %f W” ,Q);

61

Scilab code Exa 3.12 Heat lost by sphere

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

clc ; clear ;

//Exa mp le 3.1 2 v=18.23*10^-6 //sq m/s k=0.02814 // [W/m. K ] D=0.012 // [m] r=0.006 // [m] u_inf=4 // [m/ s ] Nre=D*u_inf/v // Reynold ’ s Nnu=0.37*Nre^(0.6); h=Nnu*(k/D) A=4*%pi*r^2 //A rea of sphe Tw=350 // [K] T_inf=300 // [K] Q=h*A*(Tw-T_inf) //H eat lo printf ( ” \n Heat lost by sph er

num be r

re in [ sq m]

st by sphe re in [W] e is %f W” ,Q);

Scilab code Exa 3.13 Heat lost by sphere

1 2 3 4 5 6 7 8 9 10 11

clc ; clear ;

//Exm ap le 3.1 3 v=15.69*10^-6 // [ sq m./ s ] k=0.02624 // [W/m. K ] Npr=0.708 // Pra ndt l nu mb er mu=2.075*10^-5 // kg/m . s u_inf=4 // [m/ s ] mu_inf=1.8462*10^-5 // [m/s ] vel oci ty Tw=350 // [K] T_inf=300 // [K]

62

12 D=0.012 // [m] 13 r=D/2 //Rad iu s in [m] 14 Nre=u_inf*D/v // Reynold ’ s numbe 15 Nnu=2+(0.4*Nre^(1.0/2.0)+0.06*Nre^(2.0/3.0))*Npr ^(0.4)*(mu_inf/mu)^(1.0/4.0) 16 h=Nnu*k/D // [W/ sq m.K ] 17 18 A=4*%pi*r^2 //A rea in [ sq m] 19 Q=h*A*(Tw-T_inf); 20 printf ( ” \n Heat lost by th e sph er e is %f W”

,Q);

Scilab code Exa 3.14 Percent power lost in bulb

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; clear ;

//Exa mp le 3.1 4 v=2.08*10^-5 // [ sq m/ s ] k=0.03 //W/(m.K) Npr=0.697 // Pra ndt l nu mb er D=0.06 // [m] u_inf=0.3 // [m/ s ] Nre=D*u_inf/v //Reynolds nu mb er

//Av er ag e nus sel t number is given

by :

Nnu=0.37*(Nre^0.6); h=Nnu*k/D //W/ sq m.K Tw=400 // [K] T_inf=300 // [K] D=0.06 // [m] r=0.03 // [m] A=4*%pi*r^2 //A rea in [ sq m] Q=h*A*(Tw-T_inf) // [W] per=Q*100/100 // Pe rc en t of he at lost

convection 20 printf ( ”H eat tran sfer power lost

rate

by forced

is %f W,A nd perc enta ge of

by con ve ct io is : %f pe rc en t ” 63

,Q,per);

Scilab code Exa 3.15 Heat lost by cylinder

1 2 clc ; 3 clear ; 4 //Exa mp le 3.1 5 5 u_inf=50 // velo city in [m/s ] 6 mu=2.14*10^-5 // [ kg /(m. s ) ] 7 rho=0.966 // [ kg/ cu bi c m] 8 k=0.0312 // [W/ (m.K) ] 9 Npr=0.695 // Pra ndt l nu mb er 10 D=0.05 //Di am et er in [ m] 11 Nre=D*u_i nf*rho/mu ; // Reynold ’ s num be r 12 printf ( ”%f” ,Nre) 13 Nnu=0.0266*Nre^0.805*Npr^(1/3); 14 h =N nu* k / D ; //W/ sq m.K 15 h=171.7 //Approximation 16 printf ( ” \ n%f” ,h ) 17 Tw=423 // [K] 18 T_inf=308 // [K] 19 / /H eat loss p e r un it le ng th is : 20 Q_by_l=h*%pi*D*(Tw-T_inf); // [W] 21 printf ( ”H eat lost p e r un it le ng th of cyl inde r is %f W( ap pr ox ) ” , round (Q_by_l));

Scilab code Exa 3.16 Heat transfer in tube

1 2 clc ; 3 clear ; 4 //Exa mp le 3.1 6

64

5 6 7 8 9 10

v=20.92*10^-6 //sq m/s k=3*10^-2 //W/(m.K) Npr=0.7 u_inf=25 // [m/ s ] d=50 // [mm] d=d/1000 // [m]

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Nre=u_inf*d/v // Reynold ’ s num be r Tw=397 // [K] T_inf=303 // [K]

26 27 28 29 30 31 32 33 34

//C a se 1: Circu lar

tu be

Nnu=0.0266*Nre^(0.805)*Npr^(1.0/3.0); h=Nnu*k/d // [W/ sq m.K ] A=%pi*d //A rea in [ sq m] Q=h*A*(Tw-T_inf) // [W] Q_by_l1=h*%pi*d*(Tw-T_inf) // [W/m]

//C as e 2: Squ are tub e A=50*50 //Ar ea i n [ sq mm] P=2*(50+50) // Per ime ter [mm] l=4*A/P //// [mm] l=l/1000 [m] Nnu=0.102*(Nre^0.675)*(Npr^(1.0/3.0)) h=Nnu*k/d //W/( sq m.K) A=4*l*l // [ sq m] Q=h*A*(Tw-T_inf) Q_by_l2=Q/l // [W/m] printf ( ” \ nRate of he at flow

from th e squ are pip e=%f W/m \ n wh ich is more t h a n th at from th e circ ula r pipe wh ic h is equal to %f W/m” ,Q_by_l2 ,Q _by_l1 );

Scilab code Exa 3.17 Heat transfer coefficient

65

1 2 3 4 5 6

clc ; clear ;

//Exa mp le 3.1 7 mu=0.8 // Viscosity of flow ing flui d [N . s/sq m] rho=1.1 // De ns it y of flowinf flui d [ g/ cub ic c m]

7 8 9 10 11 12 13 14 15

rho=rho*1000 // Density in [ kg/ cubic m] Cp=1.26 // Sp ec if ic hea t [ kJ/k g .K ] Cp=Cp*10^3 // in [ J /( kg .K) ] k=0.384 // [W/ (m.K) ] mu_w=1 // Viscosit y at wall te mpe ra tur e [N . s/sq m] // [m] L =5 vfr=300 // Volu met ric flow rate in [ cubic c m/s ] vfr=vfr*10^-6 // [ cu bi c m / s ] mfr=vfr*rho //M ass fl ow rate of flowinf fluid [ kg

16 17 18 19 20

Di=20 // In si de diameter in [mm] Di=Di/1000 // [m] Area=(%pi/4)*Di^2 //Ar ea of cros s −section u=vfr/Area // Veloctiy in [m/s ] Nre=Di*u*rho/mu // Reynold ’ s num be r

/s ]

21 //A laminar s re yn ol d ’ s number is

[ s q m]

le ss th an 21 00 , he flow

is

22 Npr=Cp*mu/k // Pra ndt l nu mb er 23 Nnu=1.86*(Nre*Npr*Di/L)^(1.0/3.0)*(mu/mu_w)^(0.14) 24 hi=Nnu*k/Di // in si de heat tr a ns fe r c o e f f i c i e n t [W

/sq m.K] 25 printf ( ” In si de heat tr an sf er c o e f f i c i e n t i s %f W/( sq m.K)” ,hi); 26 // Note : 27 printf ( ” \n The an sw er give n in book . . ie 12 25 is

wrong . ple ase re do th e calculation ma nu al ly to ch ec k \ n ” ) ;

of last

line

Scilab code Exa 3.18 Heat transfer coefficient in heated tube

66

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

clc ; clear ;

//Exa mp le 3.1 8 m=5500 //M ass flow rate in [ kg/h ] m=m/3600 // [ kg/ s ] rho=1.07 // De ns ity of fl ui d in [ g/ cmˆ3] rho=rho*1000 // [ kg/m ˆ 3 ] vfr=m/rho // Volumet ric flow rate in [mˆ3 / s ] Di=40 //Diamet er of tube [ mm] Di=Di/1000 // [m] A=(%pi/4)*Di^2 //A rea of cros s −secti on in [ s q m ] // Ve lo ci ty o f f lo wi ng fluid [m/s ] u=vfr/A rho=1070 // Density in [ kg/ mˆ3] mu=0.004 // Vis cos ity in [ kg/ m. s ] Nre=Di*u*rho/mu Nre=12198 //Approx

// Since thi s rey nol d ’ s number is le ss th an 100 00 , the fl ow is tur bul en t 18 Cp=2.72 // Sp ec if ic hea t in [ kJ/k g .K ] 19 Cp=Cp*10^3 // Spe cif ic he at in [ J/k g .K ] 20 k=0.256 // ther mal con duct ivit y in [W/m.K ] 21 Npr=Cp*mu/k // Pra ndt l nu mb er 22 Nnu=0.023*(Nre^0.8)*(Npr^0.4) // Nus se lt nu mb er 23 hi=k*Nnu/Di // Ins id e heat tr an sf er c o e f f i c i e n t in

[W/mˆ2.K] 24 printf ( ” In si de m.K” ,hi);

heat

tr an sf er

c o e f f i c i e n t i s %f W/sq

Scilab code Exa 3.19 h of water flowing in tube

1 2 clc ; 3 clear ; 4 //Exa mp le 3.1 9 5

67

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

//DATA: rho=984.1 // Density of wat er [ kg/ mˆ3] Cp=4187 // Sp eci fic he at in [ J/ kg .K ] mu=485*10^-6 // Visc osi ty at 33 1 K [P a. s ] k=0.657 // [W/ (m.K) ] mu_w=920*10^-6 // Viscosit y at 29 7 K [ Pa. s ]

// Solu tion D=16 //Diamet er in [ mm] D=D/1000 //Dia me te r in [ m] u =3 // Velocity in [m/s ] rho=984.1 // [ kg /mˆ 3 ] // Reynolds nu mb er Nre=D*u*rho/mu Nre= round (Nre) Npr=Cp*mu/k // Pra ndt l nu mb er

// Ditt us −Boelter

equ atio n ( i )

Nnu=0.023*(Nre^0.8)*(Npr^0.3) // nu ss el t n umb er h=k*Nnu/D //Heat t r a n s f e r c o e f f i c i e n t [W/mˆ2.K] printf ( ” \nANSWER −( i ) \nBy Dittus −Boelter equ atio n we ge t h=%f W/ sq m.K \n\ n\ n ” ,h);

25 26 // si ede r −tate equation ( ii ) 27 Nnu=0.023*(Nre^0.8)*(Npr^(1.0/3.0))*((mu/mu_w)^0.14)

// Nus se lt nu mb er //Heat tr a n s f er

28 h=k*Nnu/D

c o e f f i c i e n t in [W/sq m.

K] 29 printf ( ” \ nAnswer −( i i ) \n−By Sieder −Tat e equatio n we ge t h=%f W/ sq m.K \ n ” ,h); 30 printf ( ” \nNOTE: Calcula tion mis tak e in bo ok in part 2 ie sied er ta te e qn \ n” )

Scilab code Exa 3.20 Overall heat transfer coefficient

1 clc ; 2 clear ;

68

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

//Exa mp le 3.2 0 m_dot=2250 //M ass flow arte in [ kg/h ] Cp=3.35 // Sp ec if ic hea t in [ kJ/(kg .K ) ] dT=316-288.5 //T e m p er a t u r e dr o p for oil [K] Q=Cp*m_dot*dT //R ate of he at transfe r in [ kJ/ h ] Q = round (Q*1000/3600) // [ J/ s ] or [W] Di=0.04 // Ins ide diam eter [ m] Do=0.048 // Out sid e diam ter in [m] hi=4070 // fo r ste am [W/ sq m.K] ho=18.26 //F or oi l [W/sq m.K ] Rdo=0.123 // [ sq m .K/W] // [ sq m .K/W] Rdi=0.215 Uo=1/(1/ho+Do/(hi*Di)+Rdo+Rdi*(Do/Di))

// [W/mˆ 2 .K

] 16 17 18 19 20 21

Uo=2.3 dT1=373-288.5 // [K] dT2=373-316 // [K] dTm=(dT1-dT2)/ log (dT1/dT2) // [K] Ao=Q/(Uo*dTm) //H eat transfer ar ea i n [m ˆ2] printf ( ”He at r tra nsfe r are a is : %f m ˆ2” ,Ao);

Scilab code Exa 3.21 Number of tubes in exchanger

1 2 3 4 5 6 7 8 9 10 11

clc ; clear ;

//Exa mp le 3.2 1 k_tube=111.65 // [W/m. K ] W=4500 // [ kg /h ] rho=995.7 // [ kg/ sq m ] Cp=4.174 // [ kJ /( kg .K) ] k=0.617 // [W/ (m.K) ] v=0.659*10^-6 //Kin em at ic vis cos ity m_dot=1720 //kg/h

12 T1=293

// I n i t i a l temperature 69

in [K]

[ sq m/s ]

13 14 15 16 17 18

T2=318 // Final temper atur e in [K ] dT=T2-T1 // [K] Q=m_dot*Cp*dT //He at tra nsfe r rate Q=Q*1000/3600 // [ J/ s ] or [W] Di=0.0225 // [m] u=1.2 // [m/ s ]

in [ kJ/ h ]

19 //Nre=Di ∗u ∗ rho /mu or 20 Nre=Di*u/v //Reynolds nu mb er 21 //A s Nre is greater tha n 10 00 0 , Ditt us Boelter 22 23 24 25 26 27

eq ua ti on is applic able //J /( kg .K) // [ kg /(m. s ) ] // Pra ndt l nu mb er // Ditt us −Boe lte r eq ua ti on for he at in g is Cp=Cp*10^3 mu=v*rho Npr=Cp*mu/k

Nnu=0.023*(Nre^0.8)*(Npr^0.4) hi=k*Nnu/Di //Heat t r a n s fe r c o e f f i c i e n t

[W/( sq m

.K) ] 28 29 30 31

Do=0.025 // [m] Dw=(Do-Di)/ log (Do/Di) ho=4650 // [W/ sq m.K] k=111.65 // [W/m. K ]

//L og mean dia meter in

32 xw=(Do-Di)/2 // [m] 33 Uo=1/(1/ho+Do/(hi*Di)+xw*Do/(k*Dw))

// Ove ral l t r a n s fe r c o e f f i c i e n t in W/( mˆ2.K ) T_steam=373 //T em pe ra tu re of condensi ng st ea m in [K] dT1= T_steam -T1 +10 // [K] dT2= T_steam -T2 +10 // [K] dTm=(dT1-dT2)/ log (dT1/dT2) // [K] Ao=Q/(Uo*dTm) //A rea in [mˆ2] L =4 // lengt h of tub e [m] n=Ao/(%pi*Do*L) //n umber of tube s printf ( ”No. of tube s required= %d \ n ” , round (n)); printf ( ” \n NOTE: the re is an error in book in calc ulat ion of dT1 an d d T2, \ n 373 −293 is wr it te n as 90 , instead of 8 0 . . . sim ila rly in dT2, \ nSo , i n co mpl ian ce wi th th e book ,10 is added to bo th of them” ) heat

34 35 36 37 38 39 40 41 42

[m]

70

Scilab code Exa 3.22 Convective film coefficient

1 2 3 4 5 6 7 8 9 10 11

clc ; clear ;

//Exa mp le 3.2 2 m_dot=25000 // mas sfl ow rate of wa te r [ kg/h ] rho=992.2 // [ kg /mˆ 3 ] k=0.634 // [W/m. K ] vfr=m_dot/rho // [mˆ3/ h ] Npr=4.31 // Prand tl numbe rl Di=50 // [mm] Di=0.05 // [m] dT=10 // [ K ] as th e wa ll is at a te mp er at ur e of 10

K ab ov e the bul k tem pera ture // Velocity of

12 u=(vfr/3600)/(%pi*(Di/2)^2)

wa te r

in [m/s ] 13 14 15 16 17

18 19 20 21

//Approximation //Nre=Di ∗u ∗ r ho /mu=Di ∗u /v as v=mu/rh o v=0.659*10^-6; // [mˆ2/ s ] Nre=Di*u/v //Reynolds nu mb er // As it is less th a n 10 00 0 , th e fl ow is in th e tu rbu le nt re gio n for h ea t transfer and Di tt us Boe lte r eqn is u s e d Nnu=0.023*(Nre^0.8)*(Npr^0.4); // Nu ss elt nu mb er hi=Nnu*k/Di //He at tr an sf er c o e f f i c i e t in [W/sq m .K] q_by_l=hi*%pi*Di*dT //H eat transfe r per u nit length [kW/m] printf ( ”Ave ra ge value of conv ectiv e fil m c o e f f i c i e n t i s hi= % d W/sq m.K \ nHeat transferred pe r uni t le ng th i s Q/L=%f kW/m” , round (hi),q_by_l/1000); u=3.56

71

Scilab code Exa 3.23 Length of tube

1 2 3 4 5 6 7 8 9 10 11

clc ; clear ;

//Exa mp le 3.2 3 ; //W ater flow

vfr=1 200

rate

in [ l /h ]

rho=0 .9 8 ; // Den sity of wa te r in g /[ cubic cm] m_dot=vfr*rho //M ass flow rate of wa te r [ kg/h ] m_dot2=m_dot/3600 // [ kg/ s ] Cp=4.187 *10^3 ; // [ J/ kg .K] Di= 0.025 ; //Di am et er in [m] // [ kg /(m. s ) ] mu=0.0 006 ; Ai=%pi*((Di/2)^2) //Ar ea of cros s −sec tio n in [m

ˆ2] 12 Nre=(Di/mu)*(m_dot2/Ai) //Reynolds nu mb er 13 k=0. 63 ; // for met al wall in [W/( m.K ) ] 14 Npr=Cp*mu/k; // Pra ndt l nu mb er 15 // Since Nre > 10000 16 // therefor e , Di tt us boelte r eqn for he at in g is 17 Nnu=0.023*(Nre^(0.8))*(Npr^(0.4)) 18 ho= 5800 ; //Fi lm heat coe ffic ien tW /( mˆ2.K ) 19 hi=Nnu*k/Di m.K) ] 20 21 22 23 24 25

//H eat transfe

r coe ffci ent

Do=0. 02 8 ; // [m] Di=0. 02 5 ; // [m] xw=(Do-Di)/2; // [m] Dw=(Do-Di)/ log (Do/Di); // [m] k =50 ; // for met al wall in [W/( m.K ) ] Uo=1/(1/ho+Do/(hi*Di)+xw*Do/(k*Dw));

in [W/( sq

// in [W/ sq m

.K] 26 27 28 29 30 31 32 33

dT=34 3-30 3 ; // [K] dT1=39 3-303 ; // [K] dT2=39 3-343 ; // [K] dTm=(dT1-dT2)/ log (dT1/dT2); // [K] Cp=Cp/1000; // [ in [ kJ/kg .K] ] Q=m_dot*Cp*dT; //R ate of he at transfer in [ kJ/ h ] Q=Q*1000/3600; // [ J/ s ] or [W] Ao=Q/(Uo*dTm); //H eat transfer ar ea in [ sq m]

72

34 // Al so , . . Ao=%pi ∗Do∗L .. i m p l i e s th a t 35 L=Ao/(%pi*Do) // [m] 36 printf ( ”L en g th of t ub e requ ired is %f m”

, round (L));

Scilab code Exa 3.24 Cooling coil

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; clear ;

//Exa mp le 3.2 4 // 1. Fo r i n i t i a l co nd it io ns : T=360; // [K] T1=280; // [K] T2=320; // [K] dT1=T-T1; // [K] dT2=T-T2; // [K] //Q1=m1 dot ∗Cp1 ∗ ( T2−T1 ) Cp1=4.187 //H eat capacity dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] m1_by_UA=dTlm/(Cp1*(T2-T1))

/ /F or final co ndi ti ons : //m2 dot=m1 dot //U2=U1 //A2=5 ∗A1 deff ( ’ x=f ( t ) ’ , ’x=m1 by UA ∗Cp1 ∗ ( t−T1 ) − 5∗((dT1 −(T−t ) ) / lo g (d T1/ (T−t ) ) ) ’ )

19 T = fsolve (350.5,f) 20 printf ( ” \ nO ut le t te mpe rat ure

of wa te r is %f K \n ” ,T);

Scilab code Exa 3.25 Outlet temperature of water

1 clc ; 2 clear ; 3 //Exa mp le 3.2 5

73

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

mo_dot=60 //M ass flow rate of o i l i n [ g/s ] mo_dot=6*10^-2 // [ kg/ s ] Cpo=2.0 // Speci fic he at of oi l in [ kJ/(k g . K) ] T1=420 // [K] T2=320 // [K] Q=mo_dot*Cpo*(T1-T2) //R ate of he at flow in [ kJ/s

]

mw_dot=mo_dot //M ass fl ow r ate o f w a t e r t1=290 // [K] Cpw=4.18 // [ kJ /( kg .K) ]

//F or finding

outlet

//k g/s

te mp er at ur e of wa te r

// [K] t2=t1+Q/(mw_dot*Cpw) dT1=T1-t2 // [K] dT2=T2-t1 // [K] dTm=(dT1-dT2)/ log (dT1/dT2) // [K] ho=1.6 // Oil sid e heat tr an sf er c o e f f i c i e n t in

[kW /( sq m.K) ] 19 hi=3.6 //W ater side he at tra nsfe r coe ff in [kW/( sq m.K) ] 20 // Ove ral l heat tr an sf er c o e f f i c i e n t i s : 21 U=1/(1/ho+1/hi) // [kW/(mˆ2.K) ] 22 23 24 25 26 27

A=Q/(U*dTm) // [ sq m] Do=25 // [mm] Do=Do/1000 // [m] L=A/(%pi*Do) //Le ng th of t ub e in [m ] printf ( ” \ nO ut le t te mpe rat ure of wa te r is %f K \ n” , round (t2)); 28 printf ( ”A rea of he at transfer req uir ed is %f sq m \ n ” ,A); 29 printf ( ”L en g th of t ub e requ ired is %f m” ,L )

Scilab code Exa 3.26 Inside heat transfer coefficient

1

74

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

clc ; clear ;

//Exa mp le 3.2 6 k=0.14 // fo r o i l [W/m. K] Cp=2.1 // for oi l [ kJ/k g .K ] Cp=Cp*10^3 //J/ kg .K mu=154 // [mN. s / sq m] mu_w=87 // (mn. s / sq m) L=1.5 // [m] m_dot=0.5 //M ass flow rate of oi l [ kg/s ] Di=0.019 //Dia me te r of tu be [m] //M ean temp er atu re of oi l [K ] mean_T=319 mu=mu*10^-3 // [N. s / sq m] or [ kg/(m . s ) ] A=%pi*(Di/2)^2 // [ sq m] G=m_dot/A //M ass vel oc ity in [ kg/sq m . s ] Nre=Di*G/mu // Reynolds num be r

//As Nr e

2100, th e flow is lam ina r // [N. s/ sq m] or kg/( m. s ) //T he sieder tate eq uat ion is <

mu_w=mu_w*10^-3

hi=(k*(2.0*((m_dot*Cp)/(k*L))^(1.0/3.0)*(mu/mu_w) ^(0.14)))/Di //He at tra nsfe r coe ff in [W/sq m.K

] 22 printf ( ” \n The in sid e hea t tr an sfe r c o e f f i c i e n t is %f W/(m ˆ 2.K) ” ,hi); 23 24 printf ( ’ \nNOTE: Calcula tion mis tak e in la st li ne . ie in th e calculation of hi in book , ple ase pe r f or m th e calculati on ma nu al ly to ch e ck th e an sw e r \ n ” )

Scilab code Exa 3.27 Film heat transfer coefficient

1 clc ; 2 clear ; 3 //Exa mp le 3.2 7 4

75

5 6 7 8 9 10

m_dot=0.217 //W ater flow rate in [ kg/s ] Do=19 // Outside diameter in [mm] rho=1000 // Den sit y t=1.6 //W al l thi cknes s in [mm] Di=Do-2*t // i . d of tube in [mm] Di=Di/1000 // [m]

11 Do=Do/1000 // [m] 12 Ai=%pi*(Di/2)^2 //Cross −section al 13 u=m_dot/(rho*Ai) //W ater v eloc ity

ar ea in s q m th ro ug h tub e

[m/

s] 14 15 16 17 18 19

u=1.12 //ap pr ox in bo ok //app rx in bo ok Di=0.0157 T1=301 // Inl et te mpe rat ur e of wa te r in [K ] T2=315 // Out let te mpe ra tur e of wa te r in [K ] T=(T1+T2)/2 // [K] hi=(1063*(1+0.00293*T)*(u^0.8))/(Di^0.20) // Ins id e

heat

t r a n s f e r c o e f f i c i e n t W/( sq m.K )

20 hi=5084 //Approximation 21 printf ( ”%f” ,hi); 22 hio=hi*(Di/Do) // Insid e he at transfer

on outside

dia met er

coeff

based

in W/( sq m .K )

23 printf printf ( ( ”Ba ”%f”se,hio); 24 d on outsid

e te mpe rat ur e , Insi de hea t t r a n s fe r c o e f f i c i e n t i s %d W/( mˆ2 .K ) or %f kW/( m ˆ2.K)” , round (hio), round (hio)/1000);

Scilab code Exa 3.28 Area of exchanger

1 2 3 4 5 6

clc ; clear ;

//Exa mp le 3.2 8 mair_dot=0.90 T1=283 // [K] T2=366 // [K]

7 dT=(T1+T2)/2

// [ kg/ s ]

// [K] 76

8 9 10 11 12 13

Di=12 // [mm] Di=Di/1000 // [m] G=19.9 // [ kg /( sq m. s ) ] mu=0.0198 // [mN. s / ( sq m) ] mu=mu*10^-3 // [N. s / sq m] or [ kg/(m . s ) ] Nre=Di*G/mu // Reynolds nu mb er

14 15 16 17 18 19 20 21 22

// It is greater th an 1 0 ˆ 4 k=0.029 //W/(m.K) Cp=1 // [ kJ/ kg .K] Cp1=Cp*10^3 // [ J/ kg .K] Npr=Cp1*mu/k // Par ndt l nu mb er // Ditt us −Boel ter eq uat ion is hi=0.023*(Nre^0.8)*(Npr^0.4)*k/Di ho=232 //W/ sq m.K U=1/(1/hi+1/ho) // Over all heat

// [W/ sq m.K] tr an s fe r c o e f f i c i e n t

[W/mˆ2.K] 23 24 25 26 27

Q=mair_dot*Cp*(T2-T1) //kJ/s Q=Q*10^3 // [ J/ s ] or [W] T=700 // [K] dT1=T-T2 // [K] dT2=T2-T1 // [K]

28 dTm=(dT1-dT2)/ log (dT1/dT2) // [K] 29 //Q=U ∗A∗dTm 30 A=Q/(U*dTm) / /Ar ea in sq m 31 printf ( ”H eat transfer ar ea of e q u i pm e n t is %f s q m” A);

Scilab code Exa 3.29 Natural and forced convection

1 2 3 4 5

clc ; clear ;

//Exa mp le 3.2 9 v=18.41*10^-6 k=28.15*10^-3

6 Npr=0.7

// [ sq m./ s ] // [W/m. K ]

// Pra ndt l nu mb er 77

,

7 8 9 10 11 12

Beta=3.077*10^-3 g=9.81 //m/sˆ2 Tw=350 // [K] T_inf=300 // [K] dT=Tw-T_inf // [K] L=0.3 // [m]

13 14 15 16 17 18

// 1. Free

//Kˆ−1

Convection

Ngr=(g*Beta*dT*L^3)/(v^2) Npr=0.7 // Pra ndt l nu mb er Nnu=0.59*(Ngr*Npr)^(1.0/4.0) //Average heat h=Nnu*k/L

// Gras hof nu mb er

// Nu ss elt num be r t r a n s fe r c o e f f i c i e n t [W/

sq m K ] 19 printf ( ” \n In free

co nv ec ti on , he at transfer %f W /( sq m.K) \n ” ,h ) 20 // 2. Forc ed Convestion 21 u_inf=4 // [m/ s ] 22 Nre_l=u_inf*L/v 23 Nnu=0.664*(Nre_l^(1/2))*(Npr^(1.0/3.0))

Nus selt 24 h=Nnu*k/L

coeff , h=

//

nu mbe r // [W/ sq m.K ]

25 printf ” \n( sq In m.K) force \dn ”co h=%f( W/ ,hnv ) ec ti on , he at transfe r coeff 26 printf ( ” \n F rom a b o v e it is clear th at he at transfer

c o e f f i c i e n t in forc ed convectio t h a n th at in free co nv ec ti on

Scilab code Exa 3.30 Natural convection

1 2 3 4 5

clc ; clear ;

//Exa mp le 3.3 0 k=0.02685 //W/(m.K)

6 v=16.5*10^-6

// kg /(m. s ) 78

n is much la rg er \n ”);

,

7 8 9 10 11 12

Npr=0.7 // Pra ndt l nu mb er Beta=3.25*10^-3 //Kˆ −1 g=9.81 //m /( s ˆ2 ) Tw=333; // [ k ] T_inf=283 // [K] dT=Tw-T_inf // [K]

13 14 15 16 17 18 19 20 21 22 23 24

L =4 //Le n g th/ hei ght of plate Ngr=(g*Beta*dT*(L^3))/(v^2)

//Let

[m] // Gra sho ff nu mb er

con st= Ng r ∗ Npr

const=Ngr*Npr

// Si ce it is

>

10ˆ9

// Nu ss elt nu mb er Nnu=0.10*(const^(1.0/3.0)) h=Nnu*k/L //W/ ( sq m. K) h=4.3 //A pp rox in bo ok W =7 //wid th in [m] A=L*W //A rea of h ea t transfer in [ s q m] Q=h*A*dT // [W] printf ( ” \ nHeat transferred is %d W \ n ” ,Q )

Scilab code Exa 3.31 Free convection in vertical pipe

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

clc ; clear ;

//Exa mp le 3.3 1 v=18.97*10^-6 //mˆ2/ s k=28.96*10^-3 //W/(m.K) Npr=0.696 D=100 //Out er diameter [mm] D=D/1000 // [m] Tf=333 //F il m tem pera ture in [K ] Tw=373 // [K] T_inf=293 // [K] dT=Tw-T_inf // [K] Beta=1/Tf // [Kˆ −1]

14 g=9.81

// [m/ s ˆ 2 ] 79

15 16 17 18

L =3 //Le ng th of pip e [m] Ngr=(g*Beta*dT*(L^3))/(v^2) Nra=Ngr*Npr Nnu=0.10*(Ngr*Npr)^(1.0/3.0)

vertical 19 h=Nnu*k/L

// Gras hof nu mb er // nus selt

number f or

cyl ind er //W/ ( sq m. K)

20 Q_by_l=h*%pi*D*dT

m]

//H eat los s pe r me tr e length

21 printf ( ” \n Hence , Heat lo ss m \ n ” ,Q_by_l);

per me tr e length

[W/

is %f W/

Scilab code Exa 3.32 Heat loss per unit length

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

clc ; clear ;

//Exa mp le 3.3 2 k=0.630 //W/(m.K Beta=3.04*10^-4 //Kˆ −1 rho=1000 //kg/mˆ3 mu=8.0*10^-4 // [ kg /(m. s ) ] Cp=4.187 //kJ /( kg .K) g=9.81 // [m/( s ˆ2 ) ] Tw=313 // [K] T_inf=298 // [K] dT=Tw-T_inf // [K] D=20 // [mm] D=D/1000 // [m] Ngr=9.81*(rho^2)*Beta*dT*(D^3)/(mu^2)

// Gras hoff

number 16 17 18 19 20

// [ J/ kg .K] // Pra ndt l nu mb er //Av er ag e nus sel t number is Cp1=Cp*1000 Npr=Cp1*mu/k

Nnu=0.53*(Ngr*Npr)^(1.0/4.0) h=Nnu*k/D // [W/ sqm .K ]

21 Q_by_l=h*%pi*D*dT

//H eat los s pe r unit 80

lengt h [W/m

] 22 printf ( ” \ nHeat loss p e r un it le ng th of t h e he at er is %f W/m” ,Q_by_l);

Scilab code Exa 3.33 Free convection in pipe

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

clc ; clear ;

//Exa mp le 3.3 3 k=0.03406 // [W/ (m/K) ] Beta=2.47*10^-3 //Kˆ −1 Npr=0.687 // Pra ndt l nu mb er v=26.54*10^-6 //mˆ2/ s g=9.81 // [m/ s ˆ 2 ] Tw=523 // [K] T_inf=288 // [K] dT=Tw-T_inf // [K] D=0.3048 // [m] // Gras hof nu mb er

Ngr=(g*Beta*dT*(D^3))/(v^2)

14 Nra=Ngr*Npr 15 //F or Nra le ss

th an 10 ˆ9 ,w e ha ve for

horizo

ntal

cylinder 16 17 18 19

Nnu=0.53*(Nra^(1.0/4.0)) // Nus se lt nu mb er h=Nnu*k/D // [W/ sq m.K ] Q_by_l=h*%pi*D*dT; /W/m printf ( ”H eat loss of he at transfer pe r m e te r le ng h i s %f W/m” ,Q_by_l);

Scilab code Exa 3.34 Free convection in plate

1 2 clc ; 3 clear ;

81

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

//Exa mp le 3.3 4 rho=960.63 // Density in [ kg/ mˆ3] Cp=4.216*10^3 // Sp ec if ic hea t in [ J/(kg .K ) ] D=16 //Di am et er in [c m] D=D/100 // [m] k=0.68 //Th er ma l con du ct ivi ty in [W/m. K] A=(%pi*(D/2)^2) L=A/(%pi*D) //Len gth =A/P in [m ] Beta=0.75*10^-3 // [Kˆ −1] alpha=1.68*10^-7 // [mˆ2/ s ] g=9.81 // [m/ s ˆ 2 ] // [K] Tw=403 T_inf=343 // [K] dT=Tw-T_inf // [K] v=0.294*10^-6 // [mˆ2/ s ] Nra=(g*Beta*(L^3)*dT)/(v*alpha)

// 1. Fo r Top su rf ac e Nnu=0.15*(Nra)^(1.0/3.0) ht=Nnu*k/L //H eat transf

// Nu ss elt nu mb er er coeff for to p sur fac e in

W/(mˆ2.K) 24 25 26 27 28 29 30

ht2. = round // For bo(ht) tt om su rfa ce Nnu=0.27*Nra^(1.0/4.0) // Nu ss elt nu mb er hb=Nnu*k/L // [W/ sq m.K ] hb = round (hb) Q=(ht+hb)*A*dT; // [W] printf ( ”T he rate of he at inp ut is %f W” ,Q )

Scilab code Exa 3.35 Heat transfer from disc

1 clc ; 2 clear ; 3 //Exa mp le 3.3 5 4 v=2*10^-5

// [mˆ2/ s ] 82

// Pra ndt l nu mb er // [W/m. K ] //Dia me te r in [m ] // Characteristic // [K] // [K]

5 6 7 8 9 10

Npr=0.7 k=0.03 D=0.25 L=0.90*D T1=298 T2=403

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

dT=T2-T1 // [K] Tf=(T1+T2)/2 // [K] Beta=1/Tf // [Kˆ −1] A=%pi*(D/2)^2 //Ar ea in [ sq m ] g=9.81 // [m/ s ˆ 2 ]

/ /C a se 1: Hot surface

le ng th , let

facing

[m]

up

Ngr=g*Beta*dT*(L^3)/(v^2) // Gra sho ff nu mb er Nnu=0.15*((Ngr*Npr)^(1.0/3.0)) // Nu ss elt nu mb er h=Nnu*k/L // [W/ sq m.K ] Q=h*A*dT // [W] printf ( ” \n Heat transfer red when h o t surfa ce is faci ng up is %f W \ n ” ,Q);

23 24 25 26 27 28 29

//C as e 2: For ho t surface facing down Nnu=0.27*(Ngr*Npr)^(1.0/4.0); // Gra sho f Number h=Nnu*k/L // [W/ sqm . K] Q=h*A*dT // [W] printf ( ” \n Heat transfer red when h o t surfa ce is facing down is %f W \ n” ,Q);

Scilab code Exa 3.36 Rate of heat input to plate

1 2 3 clc ; 4 clear ; 5 //Exa mp le 3.3 6

83

6 7 8 9 10 11

rho=960 // [ kg /mˆ 3 ] Beta=0.75*10^-3 // [Kˆ −1] k=0.68 // [W/m. K ] alpha=1.68*10^-7 // [mˆ2/ s ] v=2.94*10^-7 // [mˆ2/ s ] Cp=4.216 // [ kJ/ kg .K]

12 13 14 15 16 17 18 19 20 21 22 23

Tw=403 // [K] T_inf=343 // [K] dT=Tw-T_inf // [K] g=9.81 // [m/ s ˆ 2 ] l=0.8 // [m] // [m] W=0.08 A=l*W //A rea in [m ˆ2] P=2*(0.8+0.08) // Perimeter in [m] L=A/P // Characteristic di me ns io n/len gt h ,L in Nra=g*Beta*L^3*dT/(v*alpha)

[m]

//( i ) for natural up pe r surface

con vec tio n , he at tra nsfe r from to p/ he at ed 24 Nnu=0.15*(Nra^(1.0/3.0)) // Nus se lt nu mb er 25 ht=Nnu*k/L // [W/mˆ 2 .K ] 26 ht=2115.3 //Appd irox then answer f f ima tio n in boo k , If do ne man ual ly 27 //( i i )Fo r th e bottom/lowe r surface of th e he at ed

plate 28 29 30 31

// Nus se lt nu mb er

Nnu=0.27*(Nra^(1.0/4.0)) hb=Nnu*k/L // [W/(mˆ 2 .K) ] hb = round (hb)

/ /R ate of he at in pu t is eq ua l to rat e of he at dissipation from t he u p p e r and lo we r surface th e plate 32 Q=(ht+hb)*A*(Tw-T_inf) // [W] 33 printf ( ” \n R ate of he at in pu t is eq ua l t o he at di ss ip at io n =%f W” ,Q);

84

s of

Scilab code Exa 3.37 Two cases in disc

1 2 3 4

clc ; clear ;

//Exa mp le 3.3 7 k=0.03 //W/(m.K)

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Npr=0.697 // Pra ndt l nu mb er v=2.076*10^-6 //mˆ2/ s Beta=0.002915 //Kˆ −1 D=2 5 ; // [ Diame ter in cm] D=D/100 // [m] //F il m tem pera ture in [K ] Tf=343 A=%pi*(D/2)^2 //A rea in [m ˆ2] P=%pi*D // Perimeter [m ] T1=293 // [K] T2=393 // [K] g=9.81 // [m/ s ˆ 2 ]

20 21 22 23 24 25 26

dT=T2-T1 // [K] Ngr=(g*Beta*dT*(L^3))/(v^2) Nra=Ngr*Npr Nnu=0.15*(Nra^(1.0/3.0)) h=Nnu*k/L // [W/mˆ 2 .K ] Q=h*A*dT // [W] printf ( ” \ nHeat transferred

// Ca se ( i ) HOT SURFACE FACING UPWARD len gth in [m]

L=A/P // Characteristic Beta=1/Tf; // [Kˆ −1]

w it h h ot surface 27 28 29 30 31 32

facing

// Gra sho ff nu mb er // Nus se lt nu mb er

when disc is horizontal upward is %f W \n ” ,Q);

//Case −( i i ) HOT FACE FACING DOWNWARD Nnu=0.27*(Nra^(1/4)) // Nu ss elt nu mb er h=Nnu*k/L //W/(mˆ2.K) Q=h*A*dT // [W] printf ( ” \ nHeat transferred when disc is horizontal wi th ho t surface facing downward is %f W \n ” ,Q);

33 34

85

35 36 37 38 39 40 41 42 43 44 45

//Case −( i i i ) −For dis c vertical L=0.25 // Chara cteri stic lengt h [m] D = L // di a [m] A=%pi*((D/2)^2) // [ sq m] Ngr=(g*Beta*dT*(L^3))/(v^2) // Gra sho ff nu mb er Npr=0.697 Nra=Ngr*Npr Nnu=0.10*(Nra^(1/3)) // Nu ss elt nu mb er h=Nnu*k/D // [W/(mˆ 2 .K) ] Q=h*A*dT // [W] printf ( ”F or ver tica l di sc , he at transferred Q);

is %f W”

Scilab code Exa 3.38 Total heat loss in a pipe

1 clc ; 2 clear ; 3 //Exa mp le 3.3 8 4 v= 23.13*10 ^- 6 ; // [m ˆ2 / s ] 5 k= 0. 0321 ; // [W/m.K] 6 Beta=2.68*10^-3; // [Kˆ −1] 7 Tw= 443 ; // [K] 8 T_inf= 303 ; // [K] 9 dT=Tw-T_inf; // [K] 10 g=9. 81 ; // [m/ s ˆ 2 ] 11 Npr=0.688; // Pra ndt l nu mb er 12 D=1 00 ; // Diameter [mm] 13 D=D/1000 //Diamet er [m ] 14 Nra=(g*Beta*dT*(D^3)*Npr)/(v^2) 15 Nnu=0.53*(Nra^(1.0/4.0)) // Nus se lt nu mb er 16 h=Nnu*k/D // [W/(mˆ 2 .K) ] 17 h=7.93 //Approximation 18 e=0.90; // Emis sivit y 19 sigma= 5.67*10^- 8 ; 20 // Q=Q co nv+Q ra d

// Total

heat 86

l o ss

,

21 / / for tota l h e a t loss p e r m e t er le ng th 22 Q_by_l=h*%pi*D*dT+sigma*e*%pi*D*(Tw^4-T_inf^4)

// [W

/m] 23 printf ( ”To ta l he at loss %f W/m” ,Q_by_l)

pe r m e t r e le ng th of pi pe is

Scilab code Exa 3.39 Heat loss by free convection

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; clear ;

//Exa mp le 3.3 9 k=0.035; // [W/ (m.K) ] Npr= 0.684 ; // Pra ndt l nu mb er Beta=2.42*10^-3; // [Kˆ −1] v=27.8*10^-6; // [mˆ2/ s ] Tw=533; // [K] T_inf= 363 ; // [K] dT=Tw-T_inf // [K] D= 0.01 ; // [m] g=9.81; // [m/ s ˆ 2 ] Nra=(g*Beta*dT*(D^3))/(v^2)

//F or this

<

10ˆ5,w e ha ve for

sphere

A=4*%pi*(D/2)^2 //Ar ea of sphe re in [mˆ2] Nnu=(2+0.43*Nra^(1.0/4.0)) // Nus sl et nu mb er h=Nnu*k/D //W/(mˆ2.K) Q=h*A*dT // [W] printf ( ” \ nRate of he at loss is %f W” ,Q )

Scilab code Exa 3.40 Heat loss from cube

1 2 clc ; 3 clear ;

87

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

//Ex ampe 3. 40 v=17.95*10^-6 // [mˆ2/ s ] dT=353-293 // [K] k=0.0283 // [W/m. K ] g=9.81 // [m/ s ˆ 2 ] Npr=0.698 // Pra ndt l nu mb er Cp=1005 //J /( kg .K) Tf=323 //F il m tem pera ture in Beta=1/Tf // [Kˆ −1] l =1 // [m] Nra=(g*Beta*dT*(l^3)*Npr)/(v^2)

[K ]

/ /In te xt bo ok resul t of a b o v e st at em en t is w ro ng ly calc ulat ed , So

17 Nra=3.95*10^8 18 // Fo r Nr a < 10ˆ9, for

nussel

a ve rti ca l plat e , th e ave rag e

t number is

19 20 21 22

Nnu=0.59*Nra^(1.0/4.0) // Nu ss elt h=Nnu*k/l // [W/mˆ 2 .K ] h=2.35 //A pp rox in bo ok A=l^2 //Ar ea [m ˆ2]

23 24 25 26 27 28 29 30 31 32 33 34

/Q1=4*(h*A*dT) /H eat loss form // 4 [W] vertical //F or to p surface P=4*l // Perim eter in [m] L=A/P // [m]

nu mb er

fac es of 1m

∗1m is

Nra=(Npr*g*Beta*dT*(L^3))/(v^2) Nnu=0.15*Nra^(1.0/3.0) // Nu ss elt nu mb er h=Nnu*k/L // [W/mˆ 2 .K ] h=6.7 //Approx Q2=h*A*dT // [W] Q_total=Q1+Q2 // Total heat lo ss [W] printf ( ” \n Th er ef or e total he at loss is %d W” Q_total);

88

,

Scilab code Exa 3.41 Plate exposed to heat

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; clear ;

//Exa mp le 3.4 1 rho=0.910; Cp=1.009*1000; k=0.0331; mu=22.65*10^-6;

// Density

in [ kg/ mˆ3]

// [ J/ kg .K] // [W/m. K ] // [N. s /mˆ2 ] side

//Le t a=smaller // b=bigger side //Qa=ha ∗A∗dT //Qb=hb ∗A∗dT //Qa=1.14 ∗Qb //Giv en a ∗ b=15 ∗10 ˆ −4 // On sol ving we get : a=0.03; // [m] b=0.05; // [m] A=a*b //A rea in [ sq m] Tf=388; // [K] Beta=1/Tf // [Kˆ −1]

20 21 22 23 24 25

T1=303; [K] T2=473; ////[K] dT=T2-T1 // [K] v=mu/rho g=9.81 //m/s ˆ2[ acce lera tion due t o gravity ] hb=0.59*(((g*Beta*dT*(b^3))/(v^2))*Cp*mu/k)^(1/4)*(k /b) // [W/ sq m.K ] 26 Qb=hb*A*(dT) // [W] 27 28 Qa=1.14*Qb // [W] 29 printf ( ” \ nD im en si on s of th e plate are %fx%f m \n ” ,a,b ); 30 printf ( ” \ nHeat transfer when th e bigge r side he ld verti cal is %f W \n ” ,Qb); 31 printf ( ” \ nHeat transfer when th e sma ll side he ld verti cal is %f W \n ” ,Qa);

89

Scilab code Exa 3.42 Nucleate poolboiling

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

clc ; clear ;

//Exa mp le 3.4 2 Ts=373 // [K] rho_l=957.9 //rho ∗ l [ kg/ mˆ3 ] Cpl=4217 // [ J/ kg .K] mu_l=27.9*10^-5 // [ kg /(m. s ) ] rho_v=0.5955 // [ kg/m ˆ 3 ] Csf=0.013 sigma=5.89*10^-2 // [ N/m] Nprl=1.76 lambda=2257 // [ kJ/ kg ] lambda=lambda*1000 // in [ J/kg ] n =1 // for wat er m_dot=30 //M ass flow rate [ kg/h ] m_dot=m_dot/3600 // [ kg/ s ] D=30 //Di am et er of p an [c m] D=D/100 // [m] g=9.81 // [m/ s ˆ 2 ] A=%pi*(D/2)^2 //A rea in [ sq m] Q_by_A=m_dot*lambda/A // [W/ sq m]

//F or nucleate

boi ling

point w e hav e :

dT=(lambda/Cpl)*Csf*(((Q_by_A)/(mu_l*lambda))* sigma/(g*(rho_l-rho_v))))^(1.0/3.0)*(Nprl^n)

sqrt (

// [K

] 24 Tw=Ts+dT // [K] 25 printf ( ” \n Tem pe ra tu re of th e bottom surface pa n i s %f W/( sq m)” ,Tw);

Scilab code Exa 3.43 Peak Heat flux

90

of th e

1 2 3 4 5 6 7 8 9 10 11

clc ; clear ;

//Ex am pl e 3. 4 lambda=2257 // [ kJ/ kg ] lambda=lambda*1000 // in [ J/kg ] rho_l=957.9 //rho ∗ l [ kg/ mˆ3 ] rho_v=0.5955 // [ kg/m ˆ 3 ] sigma=5.89*10^-2 // [ N/m] g=9.81 // [m/ s ˆ 2 ]

//P eak he at flux

is gi ve n by

Q_by_A_max=(%pi/24)*(lambda*rho_v^0.5*(sigma*g*( //W/mˆ2 rho_l-rho_v))^(1/4)) 12 Q_by_A_max=Q_by_A_max/(10^6) //MW/ ( s q m) 13 printf ( ” \n Peak heat flu x is %f MW/sq m ” ,Q_by_A_max) ;

Scilab code Exa 3.44 Stable film pool boiling

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

clc ; clear ;

//Exa mp le 3.4 4 rho_l=957.9 // [ kg/m ˆ 3 ] lambda=2257 // [ kJ/ kg ] lambda=lambda*10^3 // [ J/ kg ] rho_v=31.54 // [ kg/m ˆ 3 ] Cpv=4.64 // [ kJ/ kg .K] Cpv=Cpv*10^3 // [ J/ kg .K] kv=58.3*10^-3 // [W/ (m.K) ] g=9.81 // [m/ s ˆ 2 ] mu_v=18.6*10^-6 // [ kg /(m. s ) ] e=1.0 // Emis sivit y sigma=5.67*10^-8; Ts=373 // [K] Tw=628 // [K]

17 dT=Tw-Ts

// [K] 91

18 D=1.6*10^-3 // [m] 19 T=(Tw+Ts)/2 // [K] 20 hc=0.62*((kv^3)*rho_v*(rho_l-rho_v)*g*(lambda+0.40* Cpv*dT)/(D*mu_v*dT))^(1.0/4.0) // Convective

transfe

r coeff

// Radiation

21 hr=e*sigma*(Tw^4-Ts^4)/(Tw-Ts)

transfe

r coeff

22 h=hc+(3/4)*hr

heat

[W/sq m. K ] heat

in [W/sq m.K ] // Total heat t ra n s fe r c o e f f i c i e n t W

/( sq m.K ) //H eat dissipatio n rat e pe r uni t leng th in [kW/m] 24 printf ( ” \n Sta bl e fi lm boilin g po in t h e a t transf er c o e f f i c i e n t i s %f W/( sq m.K) ” ,h); 25 Q_by_l=Q_by_l/1000 // [kW/m] 26 printf ( ” \n H eat dissipate d pe r un it le ngt h of t he he at er i s %f kW/m” ,Q_by_l); 23 Q_by_l=h*%pi*D*dT

Scilab code Exa 3.45 Heat transfer in tube

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

clc ; clear ;

//Exm ap le 3.4 5 dT=10 // [K] P=506.625 // [ kPa ] P=P/10^3 // [ Mpa ] D=25.4 // Diameter [mm] D=D/1000 // [m] // [W/ sq m.K] //Q=h∗ %p i ∗D∗L∗dT / /H eat transfer rat e pe r m e t e r le ngt h of t u b e is Q_by_l=h*%pi*D*dT // [W/m] printf ( ” \n Ra te of he at transfer p er 1 m l en gt h of tube i s %f W/m” , round (Q_by_l)); h=2.54*(dT^3)*(%e^(P/1.551))

92

Scilab code Exa 3.46 Nucleat boiling and heat flux

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

clc ; clear ;

//Exa mp le 3.4 6 dT=8 // [K] P=0.17 // [ Mpa ] P=P*1000 // [ kPa ] h1=2847 // [W/ ( sq m. K) ] P1=101.325 // [ kPa ] h=5.56*(dT^3) // [W/ sq m.K] Q_by_A=h*dT // [W/ sq m] hp=h*(P/P1)^(0.4) // [W/ sq m.K ] // Co rr ep on di ng he at flux is : Q_by_A1=hp*dT // [W/ sq m] per=(Q_by_ A1 -Q_by_A) *100/Q_by_A

// Per cen t inc rea se

in h e a t flu x 15 printf ( ” \ nHeat flux w hen pre ssu re

is 10 1. 32 5 kPa is

%f W/ sq m \ n ” ,Q_by_A); 16 printf ( ” \n Pe r c e nt inc rea se in h e a t fl ux perc ent ” , round (per));

Scilab code Exa 3.47 Dry steam condensate

1 2 3 4 5 6 7 8

clc ; clear ;

//Exa mp le 3.4 7 mu=306*10^-6 // [N. s /mˆ2 ] k=0.668 // [W/m.K] rho=974 // [ kg /mˆ 3 ] lambda=2225 // [ kJ/ kg ] lambda=lambda*10^3 // [ J/ kg .K]

93

is %f

9 10 11 12 13 14

g=9.81 // [m/ s ˆ 2 ] Ts=373 // [K] Tw=357 // [K] dT=Ts-Tw // [K] Do=25 // [mm] Do=Do/1000 // [m]

15

h=0.725*((rho^2*g*lambda*k^3)/(mu*Do*dT))^(1.0/4.0)

16 17 18 19 20

Q_by_l=h*%pi*Do*dT // [W/m] m_dot_byl=(Q_by_l/lambda) // [ kg/ s ] m_dot_byl=m_dot_byl*3600 // [ kg /h ]

// [W/ sq m.K ]

printf ( ” \nMean heat tr a n s f er c o e f f i c i e n t i s %f W/( sq m.K) \ n ” ,h); 21 printf ( ” \ nHeat tran sfer pe r unit lengt h is %f W/m \ n ” ,Q_by_l); 22 printf ( ” \ n C o nd e n sa t e rate pe r uni t len gth is %f kg / h ” ,m_dot_byl);

Scilab code Exa 3.48 Laminar Condensate film

clc ; clear ;

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

//Exa mp le 3.4 8 rho=960 // [ kh/ mˆ 3 ] mu=2.82*10^-4 // [ kg /(m. s ) ] k=0.68 // [W/ (m.K) ] lambda=2255 // [ kJ/ kg ] lambda=lambda*10^3 // [ J/ kg ] Ts=373 // Saturation temp erat ure of st ea m [K ] Tw=371 // [K] dT=Ts-Tw // [K] L=0.3 //Dimen sion [m ] g=9.81 // [m/ s ˆ 2 ]

14

h=0.943*(rho^2*g*lambda*k^3/(L*mu*dT))^(1/4)

94

/W/ /

sq m.K 15 16 17 18 19

A=L^2 // [ sq m] Q=h*A*(Ts-Tw) // [W]= [ J/ s ] m_dot=Q/lambda //Condens ate rat e [ kg/s ] m_dot=m_dot*3600 // [ kg /h ] printf ( ” \n Av er age heat tr an sf er c o e f f i c i e n t is %f W

/( sq m.K ) \ n” ,h); 20 printf ( ” \ nHeat transfer rat e is %f J/ kg \ n ” ,Q); 21 printf ( ” \n St eam con den sat e rate pe r ho ur is %f kg /h \ n ” ,m_dot);

Scilab code Exa 3.49 Saturated vapour condensate in array

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

clc ; clear ;

//EX ample 3. 49 rho=1174 // [ kg/m ˆ 3 ] k=0.069 // [W/ (m.K) ] mu=2.5*10^-4 // [N. s /mˆ2 ] lambda=132*10^3 // [ J/ kg ] g=9.81 // [m/ s ˆ 2 ] Ts=323 // [K] Tw=313 // [K] dT=Ts-Tw // [K] //Fo r square array , n=4 n =4 //nu mber of tubes Do=12 // [mm] Do=Do/1000 // [m] h=0.725*(rho^2*lambda*g*k^3/(n*Do*mu*dT))^(1/4)

/( sq m.K ) 18 //F or he at tran sfer are a calcualtio 19 A=n*%pi*Do // [ sq m] 20 A=0.603 21 Q=h*A*dT // [W/m]

95

n , n=16

//W

22 23 24 25

m_dot=Q/lambda // [ kg/ s ] m_dot=0.049 // Apprix imation in b oo k m_dot=m_dot*3600 // [ kg /h ] printf ( ” \n Ra te of co nd en sa ti on pe r un it le ngt h is %f kg/h ” ,m_dot);

Scilab code Exa 3.50 Mass rate of steam condensation

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

clc ; clear ;

//Exa mp le 3.5 0 rho=960 // [ kg /mˆ 3 ] k=0.68 // [W/m. K ] mu=282*10^-6 // [ kg /(m. s ) ] Tw=371 //T ube wall tempera ture [K ] Ts=373 // Saturation temp erat ure in [K ] dT=Ts-Tw // [K] lambda=2256.9 // [ kJ/ kg ] lambda=lambda*10^3 // [ J/ kg ] //For a square array with 100 tubes , n= 10 Do=0.0125 // [m] g=9.81 // [m/ s ˆ 2 ] n=10 h=0.725*(((rho^2)*g*lambda*(k^3)/(mu*n*Do*dT)) ^(1.0/4.0)) //W/ ( sq m.K) L =1 // [m] //n=100 n=100; A=n*%pi*Do*L // [mˆ2/ m le ng th ] Q=h*A*dT //He at tra ns fer rate in [ W/m] ms_dot=Q/lambda // [ kg/ s ] ms_dot=ms_dot*3600 // [ kg /h ]

26 printf ( ” \n M ass rate

of steam cond ensa tion 96

is %d k g/

h \n ” , round (ms_dot)); 27 28 printf ( ” \n NOTE:ERROR in

So lu ti on in book . Do i s w r o n g l y t ak e n as 0. 01 2 in lines 17 an d 22 of th e book , Al so A is wr on gl y calc ula ted \ n ” )

Scilab code Exa 3.51 Saturated tube condensate in a wall

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

clc ; clear ;

//Exa mp le 3.5 1 rho=975 // [ kg /mˆ 3 ] k=0.871 // [W/m.K] dT=10 // [K] mu=380.5*10^-6 // [N. s /mˆ2 ] lambda=2300 // [ kJ/ kg ] lambda=lambda*1000 // La te nt hea t of con den sat ion J/ kg ] Do=100 //Out er diameter [mm] Do=Do/1000 // [m] g=9.81 // [m/ s ˆ 2 ] // for horizontal tu be

[

h1=0.725*((rho^2*lambda*g*k^3)/(mu*Do*dT))^(1/4)

//Average

heat

t r a n s fe r c o e f f i c i e n t

16 // for vertica l t u b e 17 //h2=0.943 ∗ (( rh o ˆ2 ∗ lambda ∗ g ∗ k ˆ3 ) /(m u∗L∗dT) ) ˆ( 1/ 4)

//Average

heat

tr a n s f er

coefficient

18 h2=h1 //F or vertic al tu be 19 // implies tha t 20 L=(0.943*((rho^2*lambda*g*k^3)^(1/4))/(h1*((mu*dT) ^(1/4))))^4 // [m] 21 L=0.29 //Appr oxi mat e in bo ok 22 h=0.943*((rho^2*lambda*g*k^3)/(mu*L*dT))^(1/4)

/( sq m.K) ] 97

// [W

23 24 25 26 27 28

A=%pi*Do*L //Ar ea in [mˆ2] Q=h*A*dT //He at tra nsfe r rate [ W] mc_dot=Q/lambda // [ Ra te of condensation ] in [ kg/s ] mc_dot=mc_dot*3600 // [ kg /h ] printf ( ” \n T ube len gth is %f m \n ” ,L); printf ( ” \n Rate of co nd em sa ti on pe r ho ur is %f kg/ h” ,mc_dot);

Scilab code Exa 3.52 Condensation rate

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

clc ; clear ;

//Exa mp le 3.5 2 m1_dot=50 // For horizo Do=10 // [mm] Do=Do/1000 // [m] L =1 // [m] //Fo r 10 0 tubes n=10

ntal

posit ion [ kg/ h ]

n=10;

// We kn ow th at // m dot=Q/la mbd a=h ∗A∗dT/lambda / /m dot is pr op or ti on al t o h //m1 do t pr op to h1 //m 2 dot p rop n to h2 //m1 dot/m2 dot=h1/h2 // or : m2_dot=m1_dot/((0.725/0.943)*(L/(n*Do))^(1/4))

kg/ h ] 18 printf ( ” \n For vert ica l posit ion , Rate of condensationis %f kg/ h” ,m2_dot);

Scilab code Exa 3.53

Condensation on vertical plate 98

// [

1 2 3 4 5 6

clc ; clear ; rho=975 // [ kg /mˆ 3 ] k=0.671 // [W/ (m.K) ] mu=3.8*10^-4 // [N. s /mˆ2 ] dT=10 // [K]

7 lambda=2300*10^3 // [ J/ kg ] 8 L =1 // [m] 9 g=9.81 // [m/ s ˆ 2 ] 10 h=0.943*((rho^2*lambda*g*k^3)/(mu*L*dT))^(1/4)

/( sq m.K )

//W

//[ W/sq m.K ]

11 12 printf ( ” \n ( i ) − Ave rage heat tr a ns fe r c o e f f i c i e n t i s %d W/(mˆ2.K) \n ” , round (h)); 13 14 // Loca l heat tr a n s f er c o e f f i c i e n t 15 // at x= 0. 5 //[ m] 16 x=0.5 // [m] 17 h=((rho^2*lambda*g*k^3)/(4*mu*dT*x))^(1/4) // [W/ s q

m.K] 18 printf ( ” \n ( i i ) −Local 19

heat

tr an sf er

c o e f f i c i e n t at

0.5 m height is %d W/( sq m.K ) \ n” , round (h)); delta=((4*mu*dT*k*x)/(lambda*rho^2*g))^(1/4)

// [m

] 20 delta=delta*10^3 // [mm] 21 printf ( ” \n ( i i i ) −Fil m thick ness

99

is %f mm” ,delta);

Chapter 4 Radiation

Scilab code Exa 4.1 Heat loss by radiaiton

1 2 3 4 5 6 7 8 9

clc ; clear ;

//Ex am pl e 4. 1 e=0.9 // [ Emi ssi vit y ] sigma=5.67*10^-8 // [W/mˆ 2 .Kˆ 4 ] T1=377 // [K] T2=283 // [K] Qr_by_a=e*sigma*(T1^4-T2^4) // [W/ sq m] printf ( ”H eat los s by radiati on is %d W/sq m ” Qr_by_a));

Scilab code Exa 4.2 Radiation from unlagged steam pipe

1 2 3 4

clc ; clear ;

//Ex am pl e 4. 2 // Emis sivit y

e=0.9

5 T1=393

// [K] 100

, round (

6 7 8 9

T2=293 // [K] sigma=5.67*10^-8 // [W/ sq m.K ] Qr_by_a=e*sigma*(T1^4-T2^4) //W/ sq m printf ( ” \n R ate of h e a t transf er by rad iat ion W/ sq m” ,Qr_by_a);

is %f

Scilab code Exa 4.3 Interchange of radiation energy

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

clc ; clear ;

//Ex am pl e 4. 3 L=1; // [m] e= 0.8 ; // Emis sivi ty sigma=5.6 7*1 0^-8 ; // [m ˆ2 .Kˆ4] T1=423; // [K] T2=300; // [K] Do=60; // [mm] Do=Do/1000; // [m] A=%pi*Do*L // [ sq m] A=0.189 //A ppr ox in bo ok [m ˆ2] Qr=e*sigma*A*(T1^4-T2^4) // [W/m] printf ( ” \n N et radiai ton rat e pe r 1 m et r e le ng th of pi pe i s %d W/m” , round (Qr));

Scilab code Exa 4.4 Heat loss in unlagged steam pipe

1 2 3 4 5 6

clc ; clear ;

//Ex am pl e 4. 4 e=0.9 // Emis sivit y L =1 // [m] Do=50 // [mm]

101

7 8 9 10 11 12

Do=Do/1000 // [m] sigma=5.67*10^-8 // [W/(mˆ 2 .Kˆ4 ) ] T1=415 // [K] T2=290 // [K] dT=T1-T2 // [K] hc=1.18*(dT/Do)^(0.25) // [W/ sq m.K ]

13 A=%pi*Do*L //Ar ea in [ sq m ] 14 Qc=hc*A*dT //H eat lo ss by convection W/m 15 Qr=e*sigma*A*(T1^4-T2^4) //H eat loss b y radiation

per len gth W /m 16 Qt=Qc+Qr // Total heat lo ss in [W/m] 17 printf ( ” \n To ta l he at los s by conve ction Qt);

is %f W/m”

Scilab code Exa 4.5 Loss from horizontal pipe

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

clc ; clear ;

//Ex am pl e 4. 5 e=0.85 sigma=5.67*10^-8 // [W/ sq m.K ] T1=443 // [K] T2=290 // [K] dT=T1-T2 // [K] hc=1.64*dT^0.25 //W/ sq m.K Do=60 // [mm] Do=Do/1000 // [m] L =6 // Le ngt h [m ] A=%pi*Do*L // Sur fac e ar ea of pi pe in [ sq m] Qr=e*sigma*A*(T1^4-T2^4) // Rate of h e at loss b

y

radiaiton W 15 Qc=hc*A*(T1-T2)

// Rate of he at loss

by co nv ec ti on [

W] 16 Qt=Qr+Qc

// To ta l h ea t lo ss

17 printf ( ” \n To ta l he at loss

[W]

is %d W” 102

, round (Qt))

,

Scilab code Exa 4.6 Heat loss by radiation in tube

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

clc ; clear ;

//EXa mple 4. 6 sigma=5.67*10^-8 // [W/mˆ 2 .Kˆ 4 ] e1=0.79; e2=0.93; T1=5 00 ; // [K] T2=3 00 ; // [K] D=70 // [mm] D=D/1000 // [m] L =3 // [m] W=0.3 // Sid e of con duit [m] A1=%pi*D*L // [ sq m] A1=0.659 //Ap pr ox ima te cal cula tion

in book in

[mˆ2] 16 A2=4*(L*W) // [ sq m] 17 Q=sigma*A1*(T1^4-T2^4)/(1/e1+((A1/A2)*(1/e2-1)))

// [W] 18 printf ( ” \n Heat lost

by radia tion

Scilab code Exa 4.7 Net radiant interchange

1 2 3 4 5

clc ; clear ;

//Ex am pl e 4. 7 sigma=5.67*10^-8 T1=703 // [K]

6 T2=513

// [W/ sq m.Kˆ 4 ]

// [K] 103

is %f W”

,Q);

7 e1=0.85 8 e2=0.75 9 Q_by_Ar=sigma*(T1^4-T2^4)/(1/e1+1/e2-1) // [W/ sq m] 10 printf ( ” \n Net radi ant inte rcha nge pe r sq ua re me t r e i s %d W/ sq m” , round (Q_by_Ar));

Scilab code Exa 4.8 Radiant interchange between plates

1 2 3 4 5 6 7 8 9 10 11

clc ; clear ;

//Ex am pl e 4. 8 L= 3 ; // [m] A=L^2 //A rea in [ sq m] sigma=5.67*10^-8; // [W/ sq m.Kˆ4 ] T1=373; // [K] T2=313; // [K] e1=0.736; e2=e1; F12=1/((1/e1)+(1/e2)-1)

12 Q=sigma*A*F12*(T1^4-T2^4) // [W] 13 printf ( ” \n Net radiant intercha nge );

Scilab code Exa 4.9 Heat loss from thermflask

1 2 3 4 5 6 7 8

clc ; clear ; sigma=5.67*10^-8 e1=0.05 e2=0.05

// [W/ sq m.Kˆ 4 ]

// A1=A2=1 ( l e t ) A1=1; A2=A1;

104

is %d W”

, round ( Q )

9 10 11 12

F12=1/(1/e1+(A1/A2)*(1/e2-1)) T1=368 // [K] T2=293 // [K] Q_by_A=sigma*F12*(T1^4-T2^4)

Ar ea [W/sq m] 13 printf ( ” \ nRate of h ea t loss 14 15 16 17 18 19

//H eat loss pe r uni t when of silvered

// When bo th th e surfa ces

are blac k

e1=1; e2=1; F12=1/(1/e1+(A1/A2)*(1/e2-1)) // [W/ sq m] Q_by_A=sigma*F12*(T1^4-T2^4) printf ( ” \n W hen bo th sur fac es are bl ack , Ra te of hea t los s is %d W/sq m” , round (Q_by_A));

Scilab code Exa 4.10 Diwar flask

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

surfa ce

i s %f W/ sq m” ,Q_by_A);

clc ; clear ;

//Exa mp le 4.1 0 e1=0.05 e2=e1 A1=0.6944; A2=1; T1=293 // [K] T2=90 // [K] sigma=5.67*10^-8 // [W/mˆ 2 .Kˆ 4 ] D=0.3 //Di am et er in [ m] F12=1/(1/e1+(A1/A2)*(1/e2-1)) Q_by_A=sigma*F12*(T1^4-T2^4) Q=Q_by_A*%pi*(D^2) // [ kJ /h ] Q=Q*3600/1000 // [ kJ/ h ] lambda=21.44 // Late nt heat

18 m_dot=Q/lambda

//kg/h 105

// [W/ sq m]

in [ kJ/k g ]

19 printf ( ” \n T he liquid / h ” ,m_dot);

o x yg e n will

ev apo ra te at % f kg

Scilab code Exa 4.11 Heat flow due to radiation

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

clc ; clear ;

// Examp le4 .1 1 sigma=5.67*10^-8 //W/(mˆ2.Kˆ4) e1=0.3; e2=e1; D1=0.3 // [m] D2=0.5 // [m] T1=90 // [K] T2=313 // [K] A1=%pi*D1^2 //Ar ea in [ sq m ] A2=%pi*D2^2 //A rea in [ sq m ] Q1=sigma*A1*(T1^4-T2^4)/(1/e1+(A1/A2)*(1/e2-1))

// [

W] 14 Q1 = abs (Q1); // Absolute value in [W] 15 printf ( ” \n R ate of he at fl ow due to radia tion W” ,Q1); 16 // When Al um in iu m i s used 17 e1=0.05 18 e2=0.5 19 Q2=sigma*A1*(T1^4-T2^4)/(1/e1+(A1/A2)*(1/0.3-1))

is %f

[W] 20 Q2 = abs (Q2) // Absolut e value in [W] 21 Red=(Q1-Q2)*100/Q1 // Percen t reduc tion 22 printf ( ” \n Red uc ti on in he at flo w will be % f per ce nt ” ,Red);

Scilab code Exa 4.12 Heat exchange between concentric shell

106

//

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

clc ; clear ;

//Exa mp le 4.1 2 sigma=5.67*10^-8 T1=77 // [K]

// [W/ sq m.Kˆ 4 ]

T2=303 // [K] D1=32 //cm D1=D1/100 // [m] D2=36 // [ cm ] D2=D2/100 // [m] // [ sq m] A1=%pi*D1^2 A2=%pi*D2^2 // [ sq m] e1=0.03; e2=e1; Q=sigma*A1*(T1^4-T2^4)/(1/e1+(A1/A2)*(1/e2-1))

// [W

] 17 18 19 20

Q=Q*3600/1000 // [ kJ/ h ] Q = abs (Q); // [ kJ /h ] lambda=201 //kJ/kg m_dot=Q/lambda // Evapor ation

rate

in [ kg/h ]

21 printf ( ” \n Nit rog en evapo rate s at %f kg/ h”

Scilab code Exa 4.13 Evaporation in concenric vessels

1 2 3 4 5 6 7 8 9

clc ; clear ;

//Exa mp le 4.1 3 D1=250 // Inner spher e idameter [mm] D1=D1/1000 //Ou te r diameter [m ] D2=350 // [mm] D2=D2/1000 // [m] sigma=5.67*10^-8 //W/( sq m.Kˆ 4)

10 A1=%pi*D1^2

// [ sq m] 107

,m_dot);

11 12 13 14 15 16

A2=%pi*D2^2 // [ sq m] T1=76 // [K] T2=300 // [K] e1=0.04; e2=e1; Q=sigma*A1*(T1^4-T2^4)/((1/e1)+(A1/A2)*((1/e2)-1))

17 18 19 20 21 22

Q=-2.45 //Approximate Q = abs ( Q ) // [W] Q=Q*3600/1000 // [ kJ/ h ] lambda=200 //kJ/kg // [ kg /h ] Rate=Q/lambda printf ( ” \n Rat e of evaporation Rate);

// [W]

is %f kg/h( ap pr ox )”

,

Scilab code Exa 4.15 infinitely long plates

1 2 3 4 5 6 7 8 9

clc ; clear ;

//Exa mp le 4.1 5 sigma=5.67*10^-8 // [W/(mˆ 2 .Kˆ4 ) ] e1=0.4 e3=0.2 T1=473 // [K] T3=303 // [K] Q_by_a=sigma*(T1^4-T3^4)/((1/e1)+(1/e3)-1)

// [W/ s q m] 10 //Q1 by a=sigma ∗ (T1ˆ4 −T2ˆ 4) /( (1 / e1 ) +(1 / e2 ) −1)=sigma ∗ A∗ (T2ˆ4 −T3ˆ 4) /( (1 / e2 ) +(1 / e3 ) −1) // [W/ sq m] 11 e2=0.5 12 // Solving we get 13 T2=((6/9.5)*((3.5/6)*T3^4+T1^4))^(1/4) 14 Q1_by_a=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1)

m] 15 red=(Q_by_ a -Q1_by_a) *100/Q_by_a

108

// [K] // [W/ s q

16 printf ( ” \ nH ea t tr an sf er

rat e per SHIELD) due to rad iati on is %f 17 printf ( ” \ nH eat tra ns fer rate per SHIELD) due to rad iati on is %f

unit W/sq unit W/sq

area (WITHOUT m \ n ” ,Q_by_a); area (WITH m \ n ” ,Q1_by_a)

; 18 printf ( ” \ nR e du ct io n in he at loss

is %f per cen t ”

,red)

;

Scilab code Exa 4.16 Heat exchange between parallel plates

1 2 3 4 5 6

clc ; clear ;

//Exa mp le 4.1 6 //In ste ady stat e ,w e can write : //Qcd=Qdb // sigma (Tc ˆ4 −Tdˆ4) ∗ /( 1/ ec +1/e d −1)=si gm a (Tdˆ 4−Tbˆ4) /( 1/ ed+1 /eb −1) 7 // i . e Tdˆ4=0.5 ∗ (Tcˆ4 −Tbˆ4) 8 // Given : 9 Ta=600 // [K] 10 11 12 13 14 15 16 17 18 19 20 21 22 23

eA=0.8; eC=0.5; eD=0.4; sigma=5.67*10^-8

//F or air //(600ˆ4 − Tcˆ 4) /2.25 =( Tcˆ4 −Tdˆ4 ) /3 .5 //1.56 ∗ (600ˆ4 − Tcˆ4)=Tcˆ4 −Tdˆ4 // Pu tti ng val ue of Td in te rm s of Tc //1.56 ∗ (600ˆ4 − Tcˆ4)=Tcˆ4 − 0 . 5 ∗ (Tcˆ4 −300ˆ4) function y=f(Tc) y=1.56*(600^4-Tc^4)-Tc^4+0.5*(Tc^4-300^4) endfunction Tc = fsolve (500,f); // [K]

// or Tc=560.94

// [K ] Ap pr ox im at e af ter sol ving

24 Td = sqrt ( sqrt (0.5*(Tc^4-300^4)))

109

// [K]

25

Q_by_a=sigma*(Ta^4-Tc^4)/(1/eA+1/eC-1)

// [W/ s q

m] 26 printf ( ” \ nR at e of heat exc hang e per unit area =%f W/m ˆ2 ” ,Q_by_a); 27 printf ( ” \ nSteady st at e temper atures ,Tc =%f K, an d Td= %f K” ,Tc,Td);

Scilab code Exa 4.17 Thermal radiation in pipe

1 clc ; 2 clear ; 3 //Exa mp le 4.1 7 4 sigma=5.67*10^-8 // [W/( sq m.Kˆ 4) ] 5 e=0.8 6 T1=673; // [K] 7 T2=303; // [K] 8 Do=200 // [mm] 9 Do=Do/1000 // [m] 10 L =1 //Le t [m] 11 A1=%pi*Do*L // [mˆ2/m] 12 //C Ase 1: Pi pe to surru ndings 13 14 Q1=e*A1*sigma*(T1^4-T2^4) // [W/m] 15 Q1=5600 //Approximated 16 //Q1=5600 // [W/m] a p p r o x i m a t e d in book for 17 18 19 20 21 22

calculat ion pu rp os e // Con cen tric cyli nders

e1=0.8; e2=0.91; D1=0.2 // [m] D2=0.4 // [m] Q2=sigma*0.628*(T1^4-T2^4)/((1/e1)+(D1/D2)*((1/e2) -1)) // [W/m] le ng th 23 Red=Q1-Q2 //Re du ct io n in he at l oss 24

110

25 printf ( ” \ nDue to th er ma l radiaiton , Lo ss of he at to surrou nding i s %d W/m \n ” , round (Q1)); 26 printf ( ” \nWhen pipe is enclosed in 1 40 0 mm d iame ter brick con dui t , Los s of hea t is %d W/m \ n ” , round ( Q2 )); 27 printf ( ” \n Re du ct io n in he at los s is %d W/m \n ” , round

(Red));

Scilab code Exa 4.18 Heat transfer in concentric tube

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

clc ; clear ;

//Exa mp le 4.1 8

sigma=5.6 7*1 0^-8 T1=813; // [K]

;

// [W/( sq m.Kˆ 4) ]

T2=473; // [K] e1=0.87; e2=0.26; D1=0. 25 ; // [m] D2=0.3; // [m] Q_by_a1=sigma*(T1^4-T2^4)/(1/e1+(D1/D2)*(1/e2-1))

// [W/ sqm ] 16 printf ( ” \n Heat tran sfer Q_by_a1);

by radiaito

n is %d W/sq m”

Scilab code Exa 4.19 Heat exchange between black plates

1 2 clc ;

111

,

3 4 5 6 7 8

clear ;

//Exa mp le 4.1 9 sigma=5.67*10^-8 // [W/ sq m.Kˆ 4 ] A1=0.5*1 // [ sq m] F12=0.285 T1=1273 // / [K]

9 T2=773 // [K] 10 Q=sigma*A1*F12*(T1^4-T2^4) // [W] 11 printf ( ” \n Ne t radiant he at ex ch an ge be tw een plates i s %d W” ,Q);

Scilab code Exa 4.20 Radiation shield

1 2 3 4 5 6 7 8 9 10 11

clc ; clear ;

//Exa mp le 4.2 0 sigma=5.67*10^-8 T1=750 // [K] T2=500 // [K] e1=0.75; e2=0.5;

/ /H eat transfer

// [W/ sq m.Kˆ 4 ]

w it ho ut shield

:

Q_by_a=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1)

m] 12 13 14 15 16 17 18 19 20

//H eat transfe R1=(1-e1)/e1 F13=1; R2=1/F13 e3=0.05 R3=(1-e3)/e3

r w it h shield : // Resistance 1

// Resistance

2

// Resistance 3

21

112

// [W/ s q

22 23 24 25 26 27

R4=(1-e3)/e3 F32=1; R5=1/F32

// Resistance 4

// Resistance

5

// Resistance

R6=(1-e2)/e2

6

28 29 Total_R=R1+R2+R3+R4+R5+R6 // To ta l res ist anc e 30 31 Q_by_as=sigma*(T1^4-T2^4)/Total_R // [W/ sq m] 32 33 Red=(Q_by_ a -Q_by_as) *100/Q_by_a //Re duc it on in

he at tranfer

due t o shield

34 35 printf ( ” \n Re du c t i on in he at transfer

result

of rad iai otn

rat e as a shie ld is %f pe rc en t” ,Red);

Scilab code Exa 4.21 Heat transfer with radiaiton shield

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

clc ; clear ;

//Exa mp le 4.2 1 e1=0.3 e2=0.8

//Let

sig ma ∗ (T1ˆ4 −T2ˆ4 )=z=1(c on st )

z=1; //Let Q_by_A=z/(1/e1+1/e2-1)

/ /H eat transfer

//W/ sq m

w i t h radia tion

e3=0.04 F13=1; F32=1;

//T he resi stan ces

ar e :

R1=(1-e1)/e1

16 R2=1/F13

113

shield

17 18 19 20 21 22

R3=(1-e3)/e3 R4=R3 R5=1/F32 R6=(1-e2)/e2 R=R1+R2+R3+R4+R5+R6 // To ta l resi sta nce Q_by_As=z/R // whe re z=sigma ∗ (T1ˆ4 −T2ˆ 4) //W/ sq m

23 red=(Q_by_ A -Q_by_As) *100/Q_by_A

// Per cen t re du ct ion in he at transfer 24 printf ( ” \n T he he at transfer is re du ce d by % f per ce nt due to shield ” ,red)

Scilab code Exa 4.22 Radiaition shape factor

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

clc ; clear ;

//Exa mp le 4.2 2 sigma=5.67*10^-8; T1=1273 // [K] T2=773 // [K] T3=300 // [K] A1=0.5 // [ sq m] A2=A1; // [ sq m] F12=0.285; F21=F12; F13=1-F12; F23=1-F21; e1=0.2; e2=0.5;

// Res ista nce

in th e ne tw or k ar e calculated

R1=1-e1/(e1*A1) R2=1-e2/(e2*A2) R3=1/(A1*F12) R4=1/(A1*F13)

22 R5=1/(A2*F23)

114

as :

//Give n (1 − e3 ) /e3 ∗A3=0

23 24 25 26 27 28

R6=0

29 30 31 32 33 34 35 36 37 38 39

// Equations are : //(Eb1 −J1)/2+(J2 −J1 ) /7.0 18+( Eb3−J1 ) /2.797=0 //(J1 −J2 ) /7.0 18+( Eb3−J2 ) /2.7 97+( Eb2−J2)/2=0

//Also //W/ sq m // [W/ sq m] // [W/ sq m]

Eb1=sigma*T1^4 Eb2=sigma*T2^4 Eb3=sigma*T3^4

// On sol ving

we get :

// [W/ sq m] J1=33515 J2=15048 // [W/ sqm ] J3=Eb3 // [W/ sq m] Q1=(Eb1-J1)/((1-e1)/(e1*A1)) // [W/ sq m] Q2=(Eb2-J2)/((1-e2)/(e2*A2)) // [W/ sq m] Q3=(J1-J3)/(1/(A1*F13))+(J2-J3)/(1/(A2*F23))

/sq m] 40 printf ( ” \n T ot al he at lost ” ,Q1); 41 printf ( ” \n T ot al he at lost

by pla te 1 is %f W/s q m

\n

by pla te 2 is %f W/s q m

\n

” ,Q2); 42 printf ( ” \ nThe ne t en er gy lost absorbed

by b o th plates must be by the room , \ n %f=%f” ,Q3,Q1+Q2)

Scilab code Exa 4.23 Radiation loss in plates

1 2 3 4 5 6 7

clc ; clear ;

//Exa mp le 4.2 3 sigma=5.67*10^-8 e1=0.7; e2=0.7; T1=866.5 // [K]

8 T2=588.8

// [W

// [W/ sq m.Kˆ 4 ]

// [K] 115

9

Q_by_A=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1)

// [W/ s q

m] 10 11 12 13 14 15 16 17 18 19 20

e1=0.7; e2=e1; e3=e1; e4=e1; e=e1;

// Q wi th n sh el ls =1/(n +1)

n =2 Q_shield=1/(n+1); es1=e1; es2=e1; Q_by_A=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)+2*(1/es1+1/ es2)-(n+1)) // [W/ sq m] 21 printf ( ” \n N ew Radiai ton lo ss is %f W/sq m ” ,Q_by_A);

Scilab code Exa 4.24 Concentric tube

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

clc ; clear ;

//Exa mp le 4.2 4 // 1 .WITHOUT SHIELD sigma=5.67*10^-8 e1=0.12; e2=0.15; T1=100 // [K] T2=300 // [K] r1=0.015 // [m] r2=0.045 // [m] L =1 // [m] A1=2*%pi*r1*L // [ sq m] Q_by_L=2*%pi*r1*sigma*(T1^4-T2^4)/(1/e1+(r1/r2)*(1/ e2-1)) // [W/m]

15 //−ve sa ig n indic ates

t he radial

th at t h e n e t h e a t fl ow is in

i n w a r d direction 116

16 17 18 19 20

// 2 .WITH CYLINDRICAL RADIATION SHIELD e3=0.10; e4=0.05; r3=0.0225 // [m] Qs_by_L=2*%pi*r1*sigma*(T1^4-T2^4)/(1/e1+r1/r2*(1/e2 -1)+(r1/r3)*(1/e3+1/e4-1)) // [W/ sq m]

21 red=( abs (Q_by_L)- abs (Qs_by_L))*100/

per ce nt red ucti on in he at ga in

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

abs (Q_by_L)

//

// Radiati on net work appro ach A3=2*%pi*r3 // [ sq m] // [ sq m] A2=2*%pi*r2 F13=1; F32=1; R1=(1-e1)/(e1*A1) R2=1/(A1*F13) R3=(1-e3)/(e3*A3) R4=(1-e4)/(e4*A3) R5=1/(A3*F32) R6=(1-e2)/(e2*A2)

35

Qs=sigma*(T1^4-T2^4)/((1-e1)/(e1*A1)+1/(A1*F13)+(1e3)/(e3*A3)+(1-e4)/(e4*A3)+1/(A3*F32)+(1-e2)/(e2* A2)) 36 printf ( ” \n W ith cylind rical radia iton shield Heat

ga in ed by flui d pe r 1 m le ng h of tu be is %f W/m \n ” ,Qs_by_L); 37 printf ( ” \ nP er ce nt redu cti on in he at ga in is %f percent \ n ” ,red); 38 printf ( ” \ nWith ra di ai to n netw ork appro ach %f W/s qm ” ,Qs);

117

Chapter 5 Heat Exchangers

Scilab code Exa 5.1 Harpin exchanger

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

clc ; clear ;

//Ex am pl e 5. 1 // [mm] // [m] // [mm] // [m] // fo r benz ene mb_dot=4450 // [ kg /h ] Cpb=1.779 // [ kJ /( kg .K) ] t2=322 // [K] t1=300 // [K] Q=mb_dot*Cpb*(t2-t1) // for Di=35 Di=Di/1000 Do=42 Do=Do/1000

ben zen e in

//F or toulene T1=344 // [K] T2=311 // [K] Cpt=1.842

// [ kJ/ kg .K]

21 mt_dot=Q/(Cpt*(T1-T2))

// [ kg /h ] 118

[ kJ/h ]

22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

// [W]

Q=Q*1000/3600

//H ot fl ui d ( tolue ne ) //Col d f lu i d ( benzene ) dT1=22 // [K] dT2=11 // [K] dTlm=(dT1-dT2)/(

log (dT1/dT2))

// [K]

//Cl od fl ui d : Inner pip e , ben zen e Di=0.035 // [m] Ai=(%pi/4)*Di^2 //Fl ow ar ea [ sq m] Gi=mb_dot/Ai //M ass ve lo ci ty [ kg/ mˆ2. h ] // [ kg /mˆ 2 . s ] Gi=Gi/3600 mu=4.09*10^-4 // [ kg /(m. s ) ] Nre=Di*Gi/mu //Reynolds nu mb er Cp=Cpb*10^3 // [ J /( kg .K) ] k=0.147 // [W/m. K ] Npr=Cp*mu/k // Pra ndt l nu mb er hi=(k/Di)*0.023*(Nre^0.8)*(Npr^0.4) // [W/ sq m.K ] hio=hi*Di/Do // [W/ sq m.K ] D1=0.042 // Ou ts id e di a of inside pi pe

[mm] D2=0.0525 // Inside di a of outsid e pi pe [m] De=(D2^2-D1^2)/D1 // [m] De=0.0236 //Approximated aa=%pi*(D2^2-D1^2)/4 //Fl ow area [ sq m] Ga=mt_dot/aa //M ass velo cit y in [ kg/ mˆ2.h ] Ga=Ga/3600 // [ kg /mˆ 2 . s ] mu=5.01*10^-4 // [ kg /(m. s ) ] Nre=De*Ga/mu //Reynolds nu mb er Npr=Cp*mu/k // Pra ndt l nu mb er ho=(k/De)*0.023*(Nre^0.8)*(Npr^0.3) // [W/ sq m.K ] Uc=1/(1/ho+1/hio) // [W/ sq m.K ] Rdi=1.6*10^-4 // Fouling fa ct or [m ˆ2.K /W] Rdo=1.6*10^-4 // Foul ing fa ct o r [m ˆ2 .K/W] Rd=Rdi+Rdo / (mˆ2 .K/W) / Ud=1/(1/Uc+Rd) // [W/ sq m.K ] A=Q/(Ud*dTlm) //sq m 119

59 60 61 62 63

ex=0.136 // [ sq m] l=A/ex //m tl=12 // To ta l length of on e ha rpi n of 6 m [m] printf ( ”b%f” ,l); printf ( ” \n \ Req uire d su rfa ce is f u l f i l l e d by conne ctin g %d( three ) 6m harp ins in se ri es \n” , round (l/tl))

Scilab code Exa 5.2 Length of pipe

1 2 3 4

clc ; clear ;

//Ex am pl e 5. 2 ma_dot=300*1000/24

//M ass fl ow rate

of aci d

//M ass flo w rate

of

in [ kg/h ] 5 mw_dot=500*1000/24

wat er in [ kg/h ] 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Cp1=1.465 // [ kJ/ kg .K] T1=333 // [K] T2=313 // [K] Q=ma_dot*Cp1*(T1-T2) // [ kJ /h ] Q=Q*1000/3600 // [W] Cp2=4.187 // [ kJ/ kg .K] t1=288 // [K] t2=(Q/(mw_dot*Cp2))+t1 // [K] dT1=T1-t2 // [K] dT2=T2-t1 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] dTlm=32.26 //App rox ima tio n in

// Inner

pipe

m_dot=12500 Di=0.075 Ai=(%pi/4)*Di^2 G=ma_dot/Ai

23 G=G/3600

// [ kg /h ] // [m] // [ sq m] // [ kg/ mˆ2 . h ] // [ kg /mˆ 2 . s ] 120

[K ]

24 25 26 27 28 29

mu=0.0112 // [ kg /m. s ] k=0.302 //W/(m.K) Nre=Di*G/mu // Reynold nu mb er Npr=Cp1*10^3*mu/k // Pra ndt l nu mb er hi=(k/Di)*0.023*(Nre^0.8)*(Npr^0.3) //W/ sq m.K Do=0.1 // [m]

30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

hio=hi*Di/Do //W/ sq m.K D1=0.1 // [m] D2=0.125 // [m] De=(D2^2-D1^2)/D1 // [m] Aa=(%pi/4)*(D2^2-D1^2) // [ sq m] // [ kg/ mˆ2 . h ] Ga=mw_dot/Aa Ga=Ga/3600 // [ kg/ sq m. s ] mu=0.0011 // [ kg /m. s ] Nre=De*Ga/mu // Reynolds nu mb er k=0.669 // for wat er Npr=Cp2*10^3*mu/k // Pra ndt l nu mb er ho=(k/De)*0.023*(Nre^0.8)*Npr^0.4 // [W/ sq m.K ] xw=(Do-Di)/2 // [m] Dw=(Do-Di)/ log (Do/Di) // [m] kw=46.52 // the rma l conduct ivity

of wall in [W/m.K ] 45 Uc=1/(1/ho+1/hio+xw*Do/(kw*Dw)) // [W/ sq m.K ] 46 Ud=Uc //A s dirt factor va lue s ar e no t gi ve n 47 48 49 50 51

Ud=195.32 A=Q/(Ud*dTlm)

//Approximation // [ sq m]

L=A/(%pi*Do) // [ sq m] printf ( ” \ nArea =%f mˆ2 , \nL en gt h fo pipe m( app rox ) ” ,A,L)

Scilab code Exa 5.3 Double pipe heat exchanger

1 clc ;

121

required

=%f

2 clear ; 3 //Ex am pl e 5. 3 4 me_do t=5500 ; // [ kg /h ] 5 me_dot1=me_dot/3600 // [ kg/ s ] 6Di =0.037 ; // I . D of inne r pi pe in [m] 7 Ai=(%pi/4)*Di^2 // [ sq m] 8 G=me_dot1/Ai // [ kg/ sq m. s ] 9 mu= 3.4*10^- 3 ; // [ Pa . s ] or [ kg/(m . s ) ] 10 Nre=Di*G/mu //Reynolds nu mb er 11Cp =2.68 ; // [ kJ/ kg .K] 12 Cp1=Cp*10^3 // [ J/ kgK ] 13 // [W/m.K] k =0.248 ; 14 Npr=Cp1*mu/k // Pra ndt l nu mb er 15 //N re is great er th an 10 ,000 , Use Dit tus −Boelter eqn : 16 Nnu=0.023*(Nre^0.8)*(Npr^0.3) // Nus sel t nu mb er 17 hi=k*Nnu/Di // [W/ sq m.K ] 18 T2=358 // [K] 19 T1=341 // [K] 20 Cp2=1.80 // [ kJ/ kg .K] 21 t2=335 // [K] 22 t1=303 // [K] 23 24 25 26 27

mt_dot=me_dot*Cp*(T2-T1)/(Cp2*(t2-t1)) mt_dot=mt_dot/3600 // [ // kg/[ kg s ] /h ] D1=0.043 // [m] D2=0.064 // Insid e di a of ou te r pi pe De=(D2^2-D1^2)/D1 // Equiva lent diam ete r [m

] // [ sq m] //kg /( sq m. s ) // Vis cos ity of

28 Aa=%pi/4*(D2^2-D1^2) 29 Ga=mt_dot/Aa 30 mu2=4.4*10^-4

toluene 31 32 33 34 35 36 37

Pa. s

k2=0.146 //For to lu en e [W/m.K ] Cp2=1.8*10^3 //J/ kg .K Nre=De*Ga/mu2 //Reynolds nu mb er Npr=Cp2*mu2/k2 // Pra ndt l nu mb er Nnu=0.023*Nre^0.8*Npr^0.4 // Nu ss elt nu mb er ho=k2*Nnu/De //W/( sq m.K) Dw=(D1-Di)/ log (D1/Di) // [m]

122

38 x=0.003 //W all thickness 39 Uo=1/(1/ho+(1/hi)*(D1/Di)+(x*D1/(46.52*Dw)))

in [m] // [

W/ sq m.K] 40 dT1=38 // [K] 41 dT2=23 // [K] 42 dTlm=(dT1-dT2)/ 43 44 45 46

log (dT1/dT2)

// [K]

Q=me_dot1*Cp*(T2-T1) // [ kJ/ s ] Q=Q*1000 // [ J/ s ] L=Q/(Uo*%pi*D1*dTlm) // [m] printf ( ” \ nT ot al len ggt h of do ub le pi pe he at exc hang er is %f m” ,L )

Scilab code Exa 5.4 Parallel flow arrangement

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

clc ; clear ;

//Ex am pl e 5. 4 mc_dot=1000 // [ kg /h ] mc_dot=mc_dot/3600 // [ kg/ s ] mh_dot=250 // [ kg /h ] mh_dot=mh_dot/3600 // [ kg/ s ] Cpc=4187 // [ J /( kg .K) ] Cph=3350 // [W/K] w=mc_dot*Cpc // [W/K] l=mh_dot*Cph // [W/K] C=mh_dot*Cph/(mc_dot*Cpc) U=1160 // [W/ sq m.K ] A=0.25 //H eat transfer surf ace fo r e x ch an ge r in

[ sq m] 15 ntu=U*A/(mh_dot*Cph) // 16 E=(1-%e^(-ntu*(1+C)))/(1+C)

heat 17 T1=393 18 t1=283

// Effectivenes exchan ger // Inl et te mpe ra tur e in [ K] // Cooling wat er [K ]

19 T2=T1-E*(T1-t1)

// Ou tl et T of ho t liquid 123

s of

20 21 t2=C*(T1-T2)+t1 // [K] 22 printf ( ” \n \nEffec tivene ss of he at ex ch an ge r is %f \ n ” ,E); 23 printf ( ” \ nO ut le t te mp er at ur e of h ot liquid is %f \n ” , T2); 24 printf ( ” \ nO ut le t te mpe rat ure

of wa te r is %f \n ” ,t2)

Scilab code Exa 5.5 Counter flow exchanger

1 2 3 4 5

clc ; clear ;

//Ex am pl e 5. 5 Cpc=4187

// Specific

he at of w at e r in

[ J /( kg .K) ] 6 Cph=2000

//S p he at of oi l in [ J/(k g . K

)] // [ kg/ s ] // [ kg/ s ] // [W/K] // [W/K] / /H eat ca pa ci ty of ra te of h o t fluid water 12 U=1075 // [W/ sq m.K] 13 A =1 // [ sq m] 7 8 9 10 11

mc_dot=1300/3600 mh_dot=550/3600 w=mc_dot*Cpc o=mh_dot*Cph

is sm al le r th a n

14 ntu=(U*A)/(mh_dot*Cph) 15 C=mh_dot*Cph/(mc_dot*Cpc) 16 E=(1-%e^(-ntu*(1-C)))/(1-C*%e^(-ntu*(1-C)))

//

Effeciency 17 T1=367 // [K] 18 t1=288 // [K] 19 T2=T1-E*(T1-t1)

// Outlet

temperature

[K] 20 T2=291.83

//App rox ima te d in bo ok 124

wi t ho u t precise 21 22 23 24 25

calculation

t2=C*(T1-T2)+t1 Q=mh_dot*Cph*(T1-T2) printf ( ” \n \nEffec tivene ss of printf ( ” \ nO ut le t te mp e ra tu re printf ( ” \ nO ut le t te mpe rat ure

// [K] // [W] ex ch an ge r is %f \ n ” ,E); of oil is %f K \n ” ,T2); of wa te r is %f K \n ” ,t2)

; 26 printf ( ” \ nRate of he at transfer

is %f W”

,Q);

Scilab code Exa 5.6 LMTD approach

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

clc ; clear ;

//Ex am pl e 5. 6 printf ( ” \nLMTD Ap pr oa ch \n ” ) Cph=4187 // [ J /( kg .K) ] mh_dot=600/3600 //H ot side flow rate [ kg/s ] mc_dot=1500/3600 // [ kg/ s ] Cpc=Cph // [ J/ kg .K] T1=343 // [K] T2=323 // [K] Q=mh_dot*Cph*(T1-T2) // [W] t1=298 // [K] t2=(mh_dot*Cph*(T1-T2))/(mc_dot*Cpc)+t1 // [K] dT1=45 // [K] dT2=17 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] hi=1600 //He at tran sfer coe ff in [W/sq m.K ] ho=hi // [W/ sq m.K] U=1/(1/hi+1/ho) // [W/ sq m.K ] A=Q/(U*dTlm) // [ sq m] printf ( ” \ nEffectiveness

−NTU ap pro ac h \ n” ) ;

24

125

25 26 27 28 29 30

//hot

31 32 33 34 35

// for

water :

// [W/K ] // [W/K ] / /H eat ca pa ci ty ra te of h o t fluid h=mh_dot*Cph c=mc_dot*Cpc

is sm al l

C=mh_dot*Cph/(mc_dot*Cpc) // E=(T1-T2)/(T1-t1) // Effec tive ness

para lell

fl ow :

ntu=- log (1-E*(1+C))/(1+C) A2=(ntu*mh_dot*Cph)/U // [ sq m] t2=C*(T1-T2)+t1 // [K] printf ( ” \n By LMTD appro ach area of heat exchang er is %f s q m \ n ” ,A); 36 printf ( ” \nBy N tu ap pr oa ch Ar ea of hea t exc hang er is %f s q m \n ” ,A); 37 printf ( ” \n Outl et temp erat ure of cold wa te r=%f K \n ” , t2 )

Scilab code Exa 5.7 Shell and tube exchanger

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

clc ; clear ;

//Ex am pl e 5. 7 mw_dot=10 // [ kg/ s ] Cpw=4.187 // [ kJ /( kg .K) ] t2=318 // [K] t1=295 // [K] Q=mw_dot*Cpw*(t2-t1) // [ kJ/ s ] Q=Q*1000 //W dT1=98 // [K] dT2=75 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] hi=850 // [W/ sq m.K] id=0.027 // In sid e dia [m] od=0.031 // Outsid e dia [m ]

16 hio=hi*id/od

// [W/ sq 126

m.K]

17 18 19 20 21 22

ho=6000 // Heat t r a n s f e r c o e f f i c i e n t s [W/ sq m.K] Uo=1/(1/ho+1/hio) // [W/ sq m.K ] Ao=Q/(Uo*dTlm) // [ sq m] L =4 //Len gth [m ] n=Ao/(%pi*od*L) // [N o. of tube s ] printf ( ” \n Number of tube s required = %d \n ” , round ( n ) );

Scilab code Exa 5.8 Order of Scale resistance

1 2 3 4

clc ; clear ;

//Ex am pl e 5. 8 mdot=7250;

// Ni tr ob en ze ne in she ll in [ kg/

h] 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Cp=2.387; // [ kJ /( kg .K) ] mu=7* 10 ^-4 ; //Pa . s k=0.151; // [W/m. K ] T1=400; // [K] T2=317; // [K] t1=305; // [K] t2=345; // [K] dT1=T1-t2 // [K] dT2=T2-t1 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] Q=mdot*Cp*(T1-T2) // [ kJ /h ] Q=Q*1000/3600 // [W] n=166; //n o of tub es L=5; // [m] Do=0.019; // [m] Di=0.015 // [m] Ao=n*%pi*Do*L // [ sq m] Uo=Q/(Ao*dTlm) // [W/ sq m.K ] Ud=Uo

24 // Sh ell

si de heat

tr an sf er

coefficient 127

25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51

Pt=0.025 // [m] C_dash=Pt-(0.5*Do+0.5*Do)

// Shell

side

B=0.15 id=0.45

cross flow ar ea // [m] // [m]

as=id*C_dash*B/Pt

//A s th er e ar e two shell as_dash=as/2

// [ sq m] pa ss es , ar ea pe r pa ss is : // [ sq m]

// Equ iva len t dia met er of sh el l De=4*(Pt^2-(%pi/4)*Do^2)/(%pi*Do)

/ /M ass velocity

on shell

// [m]

side

Gs=mdot/as_dash // [ kg/ mˆ2 . h ] Gs=Gs/3600 // [ kg /mˆ 2 . s ] mu=7*10^-4 //Pa. s Cp=Cp*1000 //J/ kg .K Nre=De*Gs/mu // Reynold nu mb er Npr=Cp*mu/k // Pra ndt ls nu mb er Nnu=0.36*Nre^0.55*Npr^(1.0/3.0) // Nus sel ts nu mb er hi=1050 // .K]] ho=Nnu*k/De // [W/ [W/ sq sq m m.K Uo=1/(1/ho+(1/hi*(Do/Di))) // [W/ sq m K] Uc=Uo Rd=(Uc-Ud)/(Uc*Ud) //mˆ2.K/W printf ( ” \n Foul ing fac tor= Sclae res is ta nc e=%f mˆ2. K/ W\n ” ,Rd);

Scilab code Exa 5.9 Length of tube required

1 clc ; 2 clear ; 3 //Ex am pl e 5. 9 4 k=0.628

//W/(m.K) 128

5 6 7 8 9 10

rho=980 mu=6*10^-4 Cpw=4.187 Cp=Cpw*10^3 Di=25 Di=Di/1000

// [ kg/m ˆ 3 ] // kg /(m. s ) //kJ /( kg .K) //J /( kg .K) // [mm] // [m]

11 mw_dot=1200*10^-3*rho

//M ass flow

12 mw_dot=mw_dot/3600 13 Ai=(%pi*Di^2)/4

// [ kg/ s ] // Inside ar ea of t ub e

water

[ kg/h ]

rate

of

in s q m 14 15 16 17 18 19 20 21 22 23

// kg/ mˆ2 . s // Reynolds nu mb er // Pra nddt l nu mb er // In si de heat tr a ns fe r c o e f f i c i e n t Nnu=0.023*Nre^0.8*Npr^0.4 // Nu ss elt nu mb er hi=Nnu*k/Di // [W/ sq m.K ] ho=6000 // [W// sq m.K] Do=0.028 // [m] Dw=(Do-Di)/ log (Do/Di) // [m] x=(Do-Di)/2 // [m] G=mw_dot/Ai Nre=Di*G/mu Npr=Cp*mu/k

24 k2=348.9 i n [W/m.K ]

// th er ma l conductivity

25

Uo=1/((1/ho)+(1/hi)*(Do/Di)+(x/k2)*(Do/Dw))

26 27 28 29 30 31 32 33 34 35 36 37

t1=303 // [K] t2=343 // [K] Q=mw_dot*Cpw*(t2-t1) // [ kJ/ h ] Q=Q*1000 // [W] Ts=393 // [K]

of me ta l // [W/ s q

m.K]

dT1=Ts-t1 // [K] dT2=Ts-t2 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] Ao=Q/(Uo*dTlm) // [ sq m] L=Ao/(%pi*Do) //Length printf ( ” \n the ref ore le ng th of t u b e re qu ir ed \ n ” ,L);

129

is %f m

Scilab code Exa 5.10 Suitability of Exchanger

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

clc ; clear ;

//Exa mp le 5.1 0 m_dot=7250 Cp=2.387; mu=7*10^-4; k=0.151; vis=1; Ft=0.9; T1=400 T2=317 t1=333 t2=300 dT1=T1-t1 dT2=T2-t2 dTlm=(dT1-dT2)/

// [ kg/h ] of nitro benzen e // [ kJ/ kg .K] // [ kg /m. s ] // [W/m. K ] //LMTD co rr ec ti on fa ct or // [K] // [K] // [K] // [K] // [K] // [K] log (dT1/dT2) // [K]

//F or nitrobenzene // [ kJ/ h ] Q=m_dot*Cp*(T1-T2) Q=Q*1000/3600 // [W] n=170 //No. of tub es L =5 // [m] Do=0.019 // [m] Di=0.015 // [m] Ao=n*%pi*Do*L // [ sq m] Uo=Q/(Ao*Ft*dTlm) // [W/ sq m.K ] Ud=Uo // [W/ sq m.K] B=0.15 // Baffle spaci ng [m] Pt=0.025 // Tube pitch in [m] C_dash=Pt-Do // Clear ance in [m] id=0.45 // [m]

32

130

33 // Shell side cros s fl ow ar ea 34 as=id*C_dash*B/Pt // [ sq m] 35 36 // Equ iva len t dia met er of sh el l 37 De=4*(Pt^2-(%pi/4)*(Do^2))/(%pi*Do) 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

/ /M ass velocity

// [m]

on shell

side // [ kg /(m. h ) ] // [ kg /mˆ 2 . s ] // [ kg /m. s ] // [ J/ kg .K] //Reynolds nu mb er // Pra ndt l nu mb er

Gs=m_dot/as Gs=Gs/3600 mu=7*10^-4 Cp=Cp*1000 Nre=De*Gs/mu Npr=Cp*mu/k

// From emp iri cal

eq n :

mu_w=mu // Nnu=0.36*Nre^0.55*Npr^(1/3) ho=Nnu*k/De // [W/ sq m.K ] hi=1050 //Given [W/ sq m.K] Uo=1/(1/ho+(1/hi)*(Do/Di)) // [W/ sq m.K ] Uc=Uo //W/ sq m.K

// Sui tabi lity

of he at ex ch an ge r

Rd_given=9*10^-4 // [W/ sq m.K] Rd=(Uc-Ud)/(Uc*Ud) // [W/ sq m.K ] printf ( ” \n Rd ca lc ul at ed (%f W/mˆ2 .K ) i s mazimum al lo wa bl e scal e resist ance \ n ” ,Rd); 59 printf ( ” \n \ nAs Rd ca lc ul at ed (%f W/sq m.K ) (OR

1 . 1 ∗ 1 0 ˆ − 3) i s mo re tha n Rd given (% f W/sq m,K ) , the gi ve n he at ex ch an ge r is suitable \ n ” ,Rd,Rd_given) ;

Scilab code Exa 5.11 Number of tubes required

1 clc ;

131

2 3 4 5 6 7

clear ;

//Exa mp le 5.1 1 //wate r in [ kg/ h ] // [K] // [K] // [ kJ/ kg .K]

mw_dot=1720; t1=2 93 ; t2=3 18 ; Cpw=4.28;

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

Di=0.0225; u=1 .2 ; rho=99 5. 7 ; v=0.659*10^-6 mu=v*rho Nre=Di*u*rho/mu Cp=Cpw*1000

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

k=2. 54 ; // [ kJ/ h .m.K k=k*1000/3600 // ][W/m. K ] Npr=Cp*mu/k // Pra ndt l nu mb er Nnu=0.023*Nre^0.8*Npr^0.4 // Nu ss elt hi=k*Nnu/Di // [W/ sq m.K ] ho=19200 // [ kJ/h .mˆ2 .K] ho=ho*1000/3600 // [W/mˆ 2 .K ] Do=0.025 // [m] Dw=(Do-Di)/ log (Do/Di) // [m] x=(Do-Di)/2 // [m] kt=460 //Fo r tu be wall material [ kt=kt*1000/3600 // [W/m. K ] Uo=1/(1/ho+(1/hi)*(Do/Di)+(x/kt)*(Do/Dw))

Q=mw_dot*Cpw*(t2-t1) // [ kJ /h ] Q=Q*1000/3600 //W lambda=2230; // [ kJ / kg ] dT1=9 0 ; // [K] dT2=6 5 ; // [K] // [K] dTlm=(dT1-dT2)/ log (dT1/dT2)

// Calculat

ion

of ins id e hea t tra ns fer c o e f f i c i e n t // [m] // [m/ s ] \ // [ kg/m ˆ 3 ] // [m/ s ] // [ kg /m. s ] // rey no ld s nu mb er // [ J/ kg .K]

W/ sq m.K] 36 //Q=Uo∗Ao∗dTlm 37 Ao=Q/(Uo*dTlm) 38 L =4

// [ sq m ] // Tube length in [m] 132

nu mb er

kJ/ h .m.K ] // [

39 n=Ao/(%pi*Do*L) // [ Number of t ub es ] 40 n = round ( n ) //Approximate 41 printf ( ” \n Number of tubes reu ire d= %d” ,n);

Scilab code Exa 5.12 Shell and tube heat exchanger

1 2 3 4

clc ; clear ;

//Exa mp le 5.1 2 t1=290

// Inlet

te mp er at ur e of cooling

w at er

[K] //Heat tr a n s f er c o e f f i c i e n t based on are a in [W/sq m.K ] lambda=400 // [ kJ/k g ] LA te nt heat of benz ene mb_dot=14.4 // [ t/h ] Con den sat ion rate of be nz ene vapour Cpw=4.187 // Spec ific he at //W ith no Scale

5 ho=2250

ins ide

6 7

8 9 10 11 Q=mb_dot*1000*lambda

//H eat dut y of co nde nse r in

[ kJ/ h ] 12 Q=(Q/3600)*1000 // [W] 13 // Shell and tu be ty pe of he at ex ch an ge r is

u se d as a pa ss surfa ce co nd en se r 14 Di=0.022 // I .D of tube [m ] 15 L=2.5 //Le ng th of ea ch tu be in [m] 16 n=120 //N umber of tubes 17 A=%pi*Di*L //A rea of he at transfer pe r m e tr e length in [mˆ2 /m] 18 A=n*A // T ot al ar ea of he at transfer in [m ˆ2] 19 Ai=(%pi/4)*Di^2 //Cross −sectional ar ea of e a c h t u b e in [mˆ2] 20 Ai=n*Ai // To ta l ar ea of fl ow in [mˆ2] single

21 u=0.75

// Velo cty of wa te r [m sˆ 133

−1]

22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

V=u*Ai rho=1000 mw_dot=V*rho

// Volu met ric flow of wa te r // [ Den sity of wate r in [ kg/ mˆ3] ] //M ass flow rate of wa te r in [ kg/s ]

//He at balance //Q=mw dot ∗Cpw∗ ( t2 −t 1 )

t2=Q/(mw_dot*Cpw*1000)+t1 // [K] T=350 //Con de ns in g be nz en e temp erat ure dT1=T-t1 // [K] dT2=T-t2 // [K] //LMTD dTlm=(dT1-dT2)/ log (dT1/dT2) U=Q/(A*dTlm) // [W/mˆ 2 .K ] U = round ( U )

in [K ]

// Neglecting

resi stan ce ,w e ha ve : // [W/mˆ 2 .K ] // hi is prop orti onal t o uˆ0 .8 C=hi/(u^0.8) //Constant hi=1/(1/U-1/ho)

//W ith Sca le Rd=2.5*10^-4 //1/U=1/hi+1/ho+Rd // [mˆ2 K. /W] //U=hi /( 1+ 3. 38 ∗ uˆ0.8) //mw dot=rho ∗ u∗ Ai / / [kg / s] //L e t t2 be th e outlet te mp er at ur e of w at er //Q=mw dot ∗Cpw∗ ( t2 −t 1 ) // t2=Q/( mw dot ∗Cpw)+t1 dT1=60

//dT2=T −(t1 +8.37 3/u ) //dTlm=8.373/(u ∗ log (60 ∗ u/(60 ∗ u − 8.3 73) ) ) //Q=U ∗A∗dTlm // 1.89 =(( uˆ − 0.2) /(1+ 3.38 ∗ uˆ0 .8) ) ∗ (1/ lo g ((6 0 ∗ u ) /60 ∗ u − 8.373) 55 // If we assu me val ues of u greater th a n 0.7 5 m /s 56 //F or u =3 .8 / /[ msˆ −1] 57 u=3.8 // ] msˆ −1] 58 printf ( ” \ nWater vel oci ty must be 3.88 msˆ −1” ) ; 134

Scilab code Exa 5.13 Length of pipe in Exchanger

1 2 3 4 5

clc ; clear ;

//Exa mp le 5.1 3 mh_dot=1.25 Cpw=4.187*10^3

// [ kg/ s ] //H eat cap aci ty of wa te r in [ J/

kg .K] 6 lambda=315 7 Q=mh_dot*lambda

va p o u r 8 Q=Q*10^3 9 Ts=345

// [ kJ/ kg ] //R ate of he at transfer f

rom

[kJ /s ] // [W] //Te mp er at ur e of condensing

vapou r [K] 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

// Inl et te mpe rat ur e of wa te r [K ] // Outlet temper ature of wat er [K ] // [K] // [K] log (dT1/dT2) // [K] //H ea t re mo ve d fr om va pou r = He at gained // [ kg/ s ] mw_dot=Q/(Cpw*(t2-t1)) hi=2.5 // [kW/ sq m.K] hi=hi*1000 // [W/ sq m.K ] Do=0.025 // [m] Di=0.020 // [m] hio=hi*(Di/Do) // Inside he at transf er c o s f f i c i e n t re fe rr ed to outs ide dia in [W/sq m.K ] ho=0.8 // Out si de he at tranbs fer c o e f f i c i e n t i n [kW/ sq m.K ] ho=ho*1000 // [W/ sq m.K ] Uo=1/(1/ho+1/hio) // [W/ sq m.K ] // Ud is 80% of Uc Ud=(80/100)*Uo // [W/ sq m.K ] Ao=Q/(Ud*dTlm) // [ sq m] t1=290 t2=310 dT1=Ts-t1 dT2=Ts-t2 dTlm=(dT1-dT2)/

135

// [m] // Out si de are a of pip e

28 L =1 29 A=%pi*Do*L

pe r m le ng th of pi pe // To ta l lengt h of piping

30 len=Ao/A

required . // [ kg/ mˆ3 ]

31 rho=1000 32 V=mw_dot/rho 33 v=0.6 34 a=V/v

for

// [mˆ3/ s ] // [m/ s ] //Cross − sectional

ar ea

fl ow pa ss [ sq m]

35 a1=(%pi*Di^2)/4 // [ sq m] 36 // for sing le pa ss on t ub e side flui d ( wa te r ) 37 n = round (a/a1) //No. of tub es

pe r pas s //Le ng th of ea ch tu be in

38 l=len/n

[m] 39 //F or two passes 40 tn=2*n 41 l2=len/tn

on wa te r side : // Tot al no of tube s //Le ng th of ea ch tu be in

[m] 42 //F or four

passe s on wa te r side /tu be side

43 l3=len/tn2 tn2=4*n 44

// Totngalthnoof. of tube //Le ea ch tu sbe in

[m] 45 46 printf ( ” \nNo. tn2,l3);

of tubes =%d , \nLen gth of tube =%f m ” ,

Scilab code Exa 5.14 Dirt factor

1 2 clc 3 clear 4 //Exa mp le 5.1 4 5 // Propert

ies

of cr ud e oi l : 136

// [ kJ /( kg .K) ] // [N. s/ sq m] // [W/m. K ]

6 Cpc= 1.986 ; 7 mu1=2.9*10^-3; 8k1 =0.136 ; 9 10rho1 =824 ; 11

// [ kg/m ˆ 3 ]

12 // Prope rties of bo ttom pro duc t : 13Cp2 = 2.202 ; // [ kJ/ kg .K] 14 rho2 =867 ; // [ kg/m ˆ 3 ] 15 mu2=5 .2*10^-3 ; // [N. s /sq m] 16 k2 =0.119 ; // [W/ sq m.K] 17 18 mc_ dot= 13 5000 ; // Bas is : cru id

rate

19m_dot= 1060 00

rate

oi l fl ow

in [ kg/h ] inn

t1 20 =295 t2 21 =330 T1 22 =420 T2 23 =380 24 dT1=T1-t2

//B ottom produc t flow

;

[ kg/h ] ; ; ; ;

// [K] // [K] // [K] // [K] // [K]

25 dT2=T2-t1 // [K] 26 dTlm=(dT1-dT2)/ log (dT1/dT2) 27 Q=mc_dot*Cpc*(t2-t1) 28 Q=Q*1000/3600 29 30 // Shell side calculations : Pt 31 =25 ; 32Pt = Pt /1000 ; B33=0.23 ; 34 Do =0.019 ;

di am et er for

// [K] //kJ/h // [W]

// [mm] // [m] // [m] // [m] Ou ts id e

sq ua re pit ch // Cle ara nce in [

35 c_dash=Pt-Do

m] // [m] // Cr os s flo w

id 36 =0.6 ; 37 as=id*c_dash*B/Pt

ar e a of shell [sq m ] the re is a Calc ulaito n mi st ak e ,w e ta ke :

38 // since

137

39 as=0.0353; 40 Gs=m_dot/as

// Shell

side

mass vel oci ty in [ kg/sq m. h ] 41 Gs=Gs/3600; 42 De=4*(Pt^2-(%pi/4)*Do^2)/(%pi*Do) 43 Nre=De*Gs/mu2

// [ kg/ sq m. s ] // [m] //Reynolds

number

// Prand tl

44 Npr=Cp2*1000*mu2/k2

number 45 muw=mu2 // S i n c e mu/muw=1 46 Nnu=0.36*(Nre^0.55)*Npr^(1.0/3.0)*(mu2/muw)^(0.14)

// Nu ss elt nu mb er 47 ho=Nnu*k2/De 48 49 //T ube si de heat 50n =324 ; 51 n_p= 324/2 ; 52 t =2.1 ; 53t = t /1000 ; 54 Di=Do-2*t 55 A=(%pi/4)*(Di^2)

// [W/ sq m.K] tr an sf er c o e f f i c i e n t : //No. of tub es // No. of tube s per pass // Thickness in [mm] // [m] // I . d of tu be in [m] //Cross − secti onal ar e a of

one tu be in [ sq m ] 56 A_p=n_p*A

// To ta l ar ea for

fl ow pe r

pa ss in [ s q m ] 57 58 59 60 61 62 63 64 65 66

G=mc_dot/A_p // [ kg/ sq m h ] G=G/3600 // [ kg/ sq m. s ] Nre=Di*G/mu1 //Reynolods nu mb er Npr=42 .3 5 ; // Pra ndt l nu mb er Nnu=0.023*(Nre^0.8)*(Npr^0.4) // Nus sel t nu mb er hi=Nnu*k1/Di // [W/ sq m.K ] hio=hi*Di/Do // [W/ sq m.K] Uo=1/(1/ho+1/hio) // [W/ sq m.K] Uc=Uo L=4. 88 ; //Le ng th of

tu be in [m] // [ sq m] // [W/ sq m.K] // [mˆ 2 .K/W]

67 Ao=n*%pi*Do*L 68 Ud=Q/(Ao*dTlm) 69 Rd=(Uc-Ud)/(Uc*Ud)

138

70 printf ( ” \n T he calcu latio n of line

no .3 6 t o calculated as is wr on gl y do ne in Book b y printing 0. 03 53 , ,. . wh ic h is wrong \ n ” ) ; 71 printf ( ” \nRd=%f K/w, or 7. 34 ∗ 1 0 ˆ − 4 whi ch is less th a n t he pr ov id e d , so this i f insta lled will n o t giv e required te mpe rar ue s wi th ou t frequent cleaning \n\ n ” ,Rd);

Scilab code Exa 5.15 Heat transfer area

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

clc ; clear ;

//Exa mp le 5.1 5 //CASE I : Cp=4*10^3; // [ J/ kg .K] t1=295; // [K] t2=375; // [K] sp=1.1; // Specific gr av it y of liquid v1=1.75*10^-4; //F low of l iqui d in [m ˆ3/ s ] rho=sp*1000 // [ kg/m ˆ 3 ] m_dot=v1*rho // [ kg/ s ] Q=m_dot*Cp*(t2-t1) // [W] T=395; // [K] dT1=T-t1 // [K] dT2=T-t2 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] U1A=Q/dTlm // [W/K]

//CASE−I I v2=3.25*10^-4 T2=370 m_dot=v2*rho Q=m_dot*Cp*(T2-t1)

//Fl ow in [mˆ3/ s ] // [K] // [ kg/ s ] // [W]

25 dT1=T-t1

// [K] 139

26 27 28 29 30 31

dT2=T-T2 dTlm=(dT1-dT2)/ U2A=Q/dTlm

// [K] log (dT1/dT2)

// [K] // [W/K ]

// since u is p ro pn t o v // h i =C ∗ vˆ0.8

32 U2_by_U1=U2A/U1A 33 34 ho=3400

//He at tra nsfe r coe ff for con den sin g st eam in [W/sq m.K ] 35 C = poly (0 , ”C” ) 36 // Let C=1 and v=v1 37 //C=1; 38 v=v1; //=1.75 ∗10ˆ −4 mˆ3/ s 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53

hi=C*v^0.8 U1=1/(1/ho+1/hi)

//

// When v=v2 v=v2; hi=C*v^0.8 U2=1/(1/ho+1/hi)

//

// Si nc e U2=1.6U 1 // On sol ving we get : C=142497 v=v1 hi=C*v^0.8 U1=1/(1/ho+1/hi) A=U1A/U1

// //H eat transfer

ar ea in [ sq

m] 54 printf ( ” \n Overall

/sq m.K and

hea t tra ns fer c o e f f i c i e n t is %f W \n \ nHeat transfe r ar ea is %f sq m” ,U1,

A);

Scilab code Exa 5.16 Oil Cooler

140

1 2 3 4 5

clc ; clear ;

//Exa mp le 5.1 6 // [ kg/ s ] // Specific

mo_dot=6*10^-2 Cpo=2*10^3

he at of

oi l in [ J/ kg .K ] 6 Cpw=4.18*10^3

// Speci fic

in [ J/kg .K]

he at of w at er

// [K] // [K] // [K ] Water ente rin g

7 T1=420 8 T2=320 9 T=290

temperature 10 11 12 13 14 15 16 17 18

// [ J/ s ] = [W] out = He at gained t2=Q/(mo_dot*Cpw)+T // [K] dT1=T1-t2 // [K] dT2=T2-T // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] hi=1.6*1000 // [W/ sq m.K ] ho=3.6*1000 // [W/ sq m.K ] U=1/(1/ho+1/hi) // [W/ sq m.K] Q=mo_dot*Cpo*(T1-T2)

//He at given

19 D=0.025 A=Q/(U*dTlm) // [ sq m] 20 // [m] 21 L=A/(%pi*D) // [m] 22 printf ( ” \n Leng th of tu be required = %

f m”

,L);

Scilab code Exa 5.17 Countercurrent flow heat exchanger

1 2 3 4 5 6

clc ; clear ;

//Exa mp le 5.1 7 mb_dot=1.25 Cpb=1.9*10^3 Cpw=4.187*10^3

7 T1=350

//Be nz en e in [ kg/s ] //F or benz ene in [ J/k g .K ] // in [ J/kg .K] // [K] 141

8 9 10 11 12 13

T2=300 // [K] Q=mb_dot*Cpb*(T1-T2) // [W] t1=290 // [K] t2=320 // [K] dT1=T1-t2 // [K] dT2=T2-t1 // [K]

14 dTlm=(dT1-dT2)/ log (dT1/dT2) 15 mw_dot=Q/(Cpw*(t2-t1))

// [K] //M inimum flow

ra te of

wa te r in [ kg/s ] 16 17 18 19 20 21 22 23 24 25 26 27

hi=850 // [W/ sq m.K ] ho=1700 // [W/ sq m.K ] // [m] Do=0.025 Di=0.022 // [m] x=(D o-Di) /2 // Thickness in [m] hio=hi*(Di/Do) // [W/ sq m.K ] Dw=(Do-Di)/ log (Do/Di) // [m] k=45 // [W/m. K ] Uo=1/((1/ho)+(1/hio)+(x/k)*(Do/Dw)) // [W/ sq m.K] Ao=Q/(Uo*dTlm) // [ sq m] L =1 //Len gt h in [m] area=%pi*Do*L // Ou ts id e surface

ar ea of t u b e pe r i m le ng th 28 Tl=Ao/area

// To ta l lengt h of tu bi ng requ ired in [m] 29 printf ( ” \ nT ota l length of tubi ng required= %d m” , round (Tl));

Scilab code Exa 5.18

1 2 3 4

Vertical Exchanger

clc ; clear ;

//Exa mp le 5.1 8 m_dot=4500

//Be nz ene cond ensa tion

rate

in [

kg/ h ] 5 lambda=394

// La te nt he at of cond ensa tion 142

of

benz ene in [ kJ/k g ] // [ kJ /h ] // [W] // [ kJ/ kg .K] // [K] // [K]

6 7 8 9 10

Q=m_dot*lambda Q=Q*1000/3600 Cpw=4.18 t1=295 t2=300

11 12 13 14

//F or wa te r :

mw_dot=Q/(Cpw*1000*(t2-t1)) // [ kg/ s ] rho=1000 // [ kg /mˆ3 V=mw_dot/rho // Vol ume tr ic flo w rate

in [m

ˆ3 / s ] // [m/ s ] //Cross − sectional

15 u=1.05 16 A=V/u

requi red 17 18 19 20 21 22 23

ar ea

in [ sq m ]

//Fo r tube : x=1.6 x=x/1000 Do=0.025 Di=Do-2*x A1=(%pi*Di^2)/4

24 n=A/A1 25 n = round ( n ) 26 L=2.5 27 Ao=n*%pi*Do*L

i n [s q

// thi ckn ess in [mm] // [m] // [m] // [m] //O f on e tu be [ sq m] //No. of tu be s reuired //Le ng th of tu be in [m] // Sur fa ce ar ea for he at transfer

m]

28 Ts=353

//Co nd en si ng temp of ben ze ne in

[K] 29 30 31 32 33 34 35 36 37 38

T1=295 T2=300 dT1=Ts-T1 dT2=Ts-T2 dTlm=(dT1-dT2)/ Uo=Q/(Ao*dTlm) Ud=Uo

// Inl et te mpe rat ur e in [K ] // Outl et tem per atur e in [K ] // [K] // [K] log (dT1/dT2) // [K] // [W/ s q mK] // [W/ sq m.K ]

//OVERALL HEAT TRANSFER COEFFCIENT: // Inside side : 143

39 T=(T2+T1)/2 // [K] 40 41 hi=1063*((1+0.00293*T)*u^0.8)/(Di^0.2)

// [W/

sq m.K] // [W

42 hio=hi*(Di/Do)

/sq m.K] 43 Dw=(Do-Di)/

]

44 45 46 47 48 49 50 51 52

log (Do/Di)

k=45

// [m

//Fo r tube

in [W/( m.) ]

// Outsid e of tub e : // [ kg/ s ] // [ kg /(m. s ) ] // [W/ (m.K) ] // [ kg/m ˆ 3 ] // [N. s/ sq m] // [m/ s ˆ 2 ] Acce lerati on due to gravit y

mdot_dash=1.25/n M=mdot_dash/(%pi*Do) k=0.15 rho=880 mu=0.35*10^-3 g=9.81

hm=(1.47*((4*mdot_dash)/mu)^(-1/3))/(mu^2/(k^3*rho ^2*g))^(1/3) // [W/ sq m.K] 54 ho=hm // [W/ sq m.K ] 53

55 k=45 // [W/m] 56 Uo=1/(1/ho+1/hio+(x*Do)/(k*Dw)) 57 //Uo=1/(1/ho+1/hio+(x ∗Do/(k ∗Dw) ) )

// Ov er al l h e a t t r a n s f e r c o e f f i c i e n t in [W/ sq m.K] 58 Uc=Uo // [W/ sq m.K] 59 60 Rd=(Uc-Ud)/(Uc*Ud)

//Maximum allowa ble sclae resist ance in [K/W] 61 printf ( ” \n Uc(% f) is in exce ss of Ud (% f) , therefore we al lo w for re aso nab le scale resis tanc e , \ nRd=%f K/W\n ” ,Uc,Ud,Rd); 62 printf ( ” \n No. of tube s = %d ” ,n )

Scilab code Exa 5.19 Countercurrent Heat Exchanger

144

1 2 3 4 5

clc ; clear ;

//Exa mp le 5.1 9 //W ater flow rate in [ kg/s ] //H eat capacity of wa te r [ kJ/k g

mw_dot=5; Cpw=4.18;

.K] 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

t1=303; // [K] t2=343; // [K] Q=mw_dot*Cpw*(t2-t1) // [ kJ/ s ] Q=Q*1000; // [W] T1=413; // [K] // [K] T2=373; dT1=T1-t2 // [K] dT2=T2-t1 // [K] dTlm=dT1 // / [K] hi=1000; // [W/ sq m.K ] ho=2500; // [W/ sq m.K ] Rd=1/(0.714*1000) // Fou li ng f a ct o r [m ˆ2 .K/KW] U=1/(1/hi+1/ho+Rd) // [W/ sq m.K ] A=Q/(U*dTlm) // [ sq m] printf ( ” \ nHeat transfer ar ea is %f sq m” ,A);

Scilab code Exa 5.20 Number of tube side pass

1 2 3 4 5 6 7 8 9 10

clc ; clear ;

//Exa mp le 5.2 0 Cpo=1.9 Cps=1.86 ms_dot=5.2 T1=403 T2=383

//He at capacity for o il [ kJ/k g .K ] //H eat capacity for st ea m [ kJ/k g .K ] //M ass flow rate in [ kg/s ] // [K] // [K]

Q=ms_dot*Cps*(T1-T2)

11 Q=Q*1000

// [ kJ/ s ] // [W] 145

12 13 14 15 16 17

t1=288; t2=358; dT2=T1-t2 dT1=T2-t1 dTlm=(dT1-dT2)/ U=2 75 ;

transfe

18 Ft=0.97

// [K] // [K] // [K] // [K] log (dT1/dT2)

r coe ffci ent

//LMTD i n [K] // Overall heat

in [W//sq m. K ] //LMTD co r r e c t i o n

factor 19 A=Q/(U*Ft*dTlm) // [ sq m] 20 printf ( ” \ nHeat ex ch ang er surface are a is %f sq m” ;

Scilab code Exa 5.21 Number of tubes passes

1 2 3 4 5 6

clc ; clear ;

//Exa mp le 5.2 1 mc_dot=3.783; mh_dot=1.892; Cpc=4.18;

//Co ld wat er flow ra te [ kg/s ] //H ot wa te r flow rate [ kg/s ] //S p he at of cold wa te r [ kJ/(k g .

K) ] 7 8 9 10

T1=367; t2=328; t1=311; Cph=4.18;

// [K] // [K] // [K] // Speci fic h ea t of h ot w at er [ kJ

/( kg .K) ] // Densi ty [ kg/ mˆ3 ] //Di am et er of tu be in [m] // Ov er al he at tran sfer c o e f f i c i e n t i n [W/ sq m.K ] 14 T2=T1-mc_dot*Cpc*(t2-t1)/(mh_dot*Cph) // [K] 15 Q=mc_dot*Cpc*(t2-t1) // [ kJ/ s ] 16 Q=Q*1000 // [W] 11 rho=1000; 12 D=0.019; 13 U=14 50 ;

17 //Fo r counterflow

hea t exc hang er 146

,A )

18 19 20 21 22 23

dT1=T1-t2 dT2=17; dTlm=(dT1-dT2)/ lmtd=dTlm Ft=0.88 A=Q/(U*dTlm)

// [K] // [K] // [K] /LMTD / //LMTD co rr ec ti on fa ct or // [ sq m]

log (dT1/dT2)

24 u=0.366;

[msˆ −1]

25 Ai=mc_dot/(rho*u)

// Velocity

th rou gh tube s

// Tot al flow

Area in [ sq

m] 26 n=Ai/((%pi/4)*(D^2)) 27 L =1 28 sa=%pi*D*L

//No. of tub es //Per m leng th [m ] // S . S per tu be per 1 m

length //Le ng th of tub es in [m] is more t h a n allow able 2.4 4 m length , so we must use mor e th an on e tub e \n ” ) ;

29 L=A/(n*%pi*D) 30 printf ( ” \ nThe len gth

31 32 //F or 2 passes on th e tu be side 33 A=Q/(U*Ft*lmtd) // [ sq m] 34 L=A/(2*n*%pi*D) //L en gt h in

[m]

35 printf gth n ” )2.4 s o t(h”e\ndeTsihigns len ch oi ce isis wi \thi n\n ; 4 m re qu ir em en t , 36 printf ( ” \ nType of he at ex ch ang er : 1 −2 Shell and tu be hea t exc hang er \ n” ) 37 printf ( ” \nNo of tube s per pass = %d \ n ” , round (n)); 38 printf ( ” \ nL en gt h of tub e per pass =%f m \n ” ,L);

Scilab code Exa 5.22 Outlet temperature for hot and cold fluids

1 2 3 4

clc ; clear ;

//Exa mp le 5.2 2 mh_dot=16.67;

//M ass fl ow rat e of h o t fluid in

[ kg/ s ] 147

//M ass fl ow ra te of co ld fluid

5 mc_dot=20;

in

[ kg/ s ] //S p he at of h o t fluid

6 Cph=3.6;

in [ kJ/ kg

.K] //S p he at of h o t fluid in [

7 Cph=Cph*1000;

J /k g

.K] 8 Cpc=4.2;

//S p he at of co ld fluid

kg .K) ]

//S p h e a t of co ld fluid i

9 Cpc=Cpc*1000;

in [ kJ/( n [ J/ (

kg .K) ] // Ove ral l he at transfer c o e f f i c i e n t i n [W/ sq m.K ] A=100; // Surfac e are a in [ sq m] mCp_h=mh_dot*Cph // [ J/ s ] or [W/K] mCp_c=mc_dot*Cpc // [ J/ s ] or [W/K] mCp_small=mCp_h // [W/K] C=mCp_small/mCp_c // Cap aci ty rat io ntu=U*A/mCp_small //NTU T1=973; //H ot flui d inl et te mp er at ur e in [K] t1=373; //C old flui d inl et te mp er at ur e

10 U=400; 11 12 13 14 15 16 17 18

in e [K1:] Countercurrent 19 //Cas flow arra ngem ent 20 E=(1-%e^(-(1-C)*ntu))/(1-C*%e^(-(1-C)*ntu)) Effectiveness 21 //W=T1 −T2/(T1 −t 1 )

//

the refo re :

22 T2=T1-E*(T1-t1) // [K] 23 printf ( ” \ nE xi t te mp er at ur e of ho t flui d is %d K” round (T2)); 24 t2=mCp_h*(T1-T2)/(mCp_c)+t1 // [ From energy

bal ance

,

eqn in ] [K]

25 printf ( ” \ nE xi t te mpe ra tur e of cold fl ui d is %d K(%d C) \ n ” , round (t2), round (t2-273)); 26 27 //C as e 2: Par all el flow ar ra ng em en t 28 E1=(1-%e^(-(1+C)*ntu))/(1+C) 29 //In th e te xt bo k he re is a calcula tion mi st ak e , and

th e va lu e of E is ta kn e as E = 0 .9 7 148

30 31 T2=T1-E1*(T1-t1) // [K] 32 t2=mCp_h*(T1-T2)/(mCp_c)+t1

bal ance eqn in ] [K] 33 printf ( ” \ nExi t tempera ture 34 printf ( ” \ nEx it tem pera ture

// [ From energy of Hot wat er =%f K \ n ” ,T2); of cold wa te r=%f K \n ” ,t2)

;

Scilab code Exa 5.23 Counterflow concentric heat exchanger

1 2 3 4 5 6 7 8

clc ; clear ;

//Exa mp le 5.2 3 Cpo=2131; Cpw=4187; mo_dot=0.10; mw_dot=0.20; U=380;

//S p h ea t of oi l in [ J/ kg . K ] // Sp h eat of wat er in [ J/k g .K ] // Oil flo w rate in [ kg/s ] //W ater flow rate in [ kg/s ] // Ove ral l he at transfer coeff in [W/

sq m.K] 9 10 11 12 13 14 15 16 17 18 19 20

T1=373; // I n i t i a l temp of o i l [K] T2=333; // Fin al te mp er at ur e of oi l [ K ] t1=303; //W ater enter tem pera ture in [K ] t2=t1+mo_dot*Cpo*(T1-T2)/(mw_dot*Cpw) // [K]

// 1 .LMTD met hod dT1=T1-t2 // [K] dT2=T2-t1 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] lmtd=dTlm; // [K] Q=mo_dot*Cpo*(T1-T2) // [ J/ s ] A=Q/(U*dTlm) // [ sq m] Do=0.025; // Inner tu bv e diam ete r [

m] 21 L=A/(%pi*Do) 22

//L en gt h in [m]

23 // 2 .NTU method

149

// [W/K ] // [W/K ] va lu e of mCp h is w r o n g l y cal cul ate d a s 23 1. 1 s o w e will t a k e this on l y for calcu lation \ n ” ) ; 27 mCp_h=231.1; // [W/K] 24 mCp_c=mw_dot*Cpw 25 mCp_h=mo_dot*Cpo 26 printf ( ” \n I n te xt bo ok this

28 29 30 31 32

//m Cp h is smaller

33 34 35 36 37 38

ntu= fsolve (1,f) A=ntu*mCp_h/U // [ sq m] A=0.56 //Approximately L1=A/(%pi*Do) //L en gt h in [m] printf ( ” \nFr om LMTD a pp ro ac h : \ n len gth= %f m \n ” ,L); printf ( ” \nFrom NTU met hod : \ n len gth= %f m \n ” ,L1);

C=mCp_h/mCp_c E=(T1-T2)/(T1-t1)

// Effeci ency //F or counterc urrent flow deff ( ’ [ x] = f ( nt u ) ’ , ’ x=E−(1−%eˆ( −(1 −C) ∗ ntu ) ) /(1 −C∗%e ˆ( −(1 −C) ∗ nt u ) ) ’ )

Scilab code Exa 5.24 Number of tubes required

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

clc ; clear ;

//Exa mp le 5.2 4 ho=200; // [W/ sq m.K ] hi=1500; // [W/ sq m.K ] Cpw=4.2; //S p heat of Water in [ kJ/( kg .K ) ] Cpo=2.1; // Sp hea t of Oil in [ kJ/(kg .K ) ] E=0.8; // Effe ctiv enes s k=46; // [W/m. K ] m_dot=0.167; // [ kg/ s ] //Fo r o i l [W/K]

mCp_oil=2*m_dot*Cpo*1000

14 / /mC p oi l is

w ro ng ly calculated 150

as 710 .4

15 mCp_water=m_dot*Cpw*1000 //For water [W/K] 16 / /mC p oi l is w ro ng ly calculated as 710 .4 17 //NOTE: The ab ov e two value s are wro ngl y cal cul at ed

in book as 710 .4 18 //so we tak e her e : 19 mCp_small=710.4

// [W/K]

20 // Since

bo th mCp water and mC p o il are equal there fore :

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

,

C=1; deff ( ’ [ x] = f ( nt u ) ’ , ’ x=E−(ntu /(1+ntu ) ) ’ ) ; ntu= fsolve (1,f) id=20; // Int erna l diame ter in [ mm] od=25; // External diamete r in [ mm] hio=hi*id/od // [W/ sq m.K ] Dw=(od-id)/ log (od/id) // [mm] Dw=Dw/1000 // [m] x=(od-id)/2 // [mm] x=x/1000 // [m] Do=0.025 // Ext ern al dia in [m] L=2.5; //Le ng th of tu be in [m]

34 A=ntu*mCp_small/Uo Uo=1/(1/ho+1/hio+(x/k)*(Dw/Do)) 35 //H eat transfer //ar[W/ ea sqin m.K] [ s q m] 36 n=A/(%pi*Do*L) // No of tube s 37 printf ( ” \nNo. of tub es required = %d” , round (n+1));

Scilab code Exa 5.25 Parallel and Countercurrent flow

1 2 3 4 5 6

clc ; clear ;

//Exa mp le 5.2 5 //( i ) Par all el flow T1=633; // [K]

7 t2=303;

// [K] 151

// [K] // [K] // [K] // [K] // [ kg/ s ] // Overal l heat

8 9 10 11 12 13

T2=573; t1=400; dT1=T1-t2; dT2=T2-t1; mh_dot=1.2; U=500;

14 15 16 17 18 19 20 21 22 23 24

Cp=2083; // Sp . hea t of oi l J/k g .K dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] Q=mh_dot*Cp*(T1-T2) // [W] A=Q/(U*dTlm) // [ sq m]

tr an sf er

c o e f f i c i e n t in [

W/ sqm .K ]

//( i i ) Co un te r curren t flow dT1=T1-t1; // [K] dT2=T2-t2; // [K]

dTlm=(dT2-dT1)/ log (dT2/dT1) // [K] A1=Q/(U*dTlm) // [ sq m] printf ( ” \ nFor pa ra ll el fl ow , Area = %f sq m \n For cou nte rcu rre nt flow , Ar ea =%f sq m \ n ” ,A,A1); 25 printf ( ” \n \ nFor th e same termi nal tem per atu res of

t he fluid

, t he surfa ce ar ea for

arrangement \n \n is ” )less parallel fl ow

t he co unt er fl ow

t h a n t h e re qu ir ed fo r t h e

152

Chapter 6 Evaporation

Scilab code Exa 6.1 Boiling point Elevation

1 2 3 4 5 6 7 8 9

clc ; clear ;

// Examp le6 . 1 T=380 //B.P of solut ion [K ] T_dash=373 //B.P of wa te r [K ] BPE=T-T_dash // Boi ling po in t eleva tion in [ K ] Ts=399 // Saturating te mper atur e in [ K] DF=Ts-T // Dr iv in g force in [K ] printf ( ” \nB oi li ng po in t of ele vat ion of t h e solu tio n is %d K \ n” ,BPE); 10 printf ( ” \ nD ri vi ng for ve for h ea t transfer is %d K \n ” ,DF)

Scilab code Exa 6.2 Capacity of evaporator

1 clc ; 2 clear ; 3 //Ex am pl e 6. 2

153

4 m_dot=10000 5 fr_in=0.04

// Weak li quo r enterin g in [ kg/h ] // Fr aci ton of caustic so da IN i . e 4

% // Fraciton of cau sti c so da O UT i . e 25 % 7 //L e t m da s h d o t be th e kg/ h of thic k liquor leavin g 6 fr_out=0.25

8 mdash_dot=fr_in*m_dot/fr_out // [ kg /h ] 9 10 // Overal l material bala nce 11 //k g/h of feed =kg/h of wa te r evap orat ed + kg/h of 12 13 14 15

thi ck liquor //w e=wat er evaporated // There fore

in kg/h

we=m_dot-mdash_dot // [ kg /h ] printf ( ” \n Ca pa ci ty of evap orat or

is %d k g/ h”

,we);

Scilab code Exa 6.3 Economy of Evaporator

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

clc ; clear ;

//Ex ma pl e 6. 3 // I n i t i a l co nc en tr at io n (5 %) // Fin al conce ntrat ion (2 0%) //B.P of w at er in [ K ] // Boiling poin t elevation [K] // [ Bas is ] fee d to eva por ato r in [ kg/ h ] // Mat eri al bal anc e of solute mdash_dot=ic*mf_dot/fc // [ kg /h ] // Overal l material bala nce mv_dot= mf_dot -m dash_dot //W at er evaporated [ kg/ h] lambda_s=2185 // Lat ent hea t of condensati on of steam [ kJ/kg ] ic=0.05 fc=0.2 T_dash=373 bpe=5 mf_dot=5000

14 lambda_v=2257

// La te nt he at of vap orisation o 154

f

water 15 16 17 18 19

[ kJ/kg ]

lambda=lambda_v // [ kJ/ kg ] T=T_dash+bpe //T em pe ra tu re of thick liq uor [K ] Tf=298 //Tem pe ra tu re of feed [K ] Cpf=4.187 // Sp . hea t of feed in [ kJ/k g .K ]

//H eat balanc e ove r evaporator

=ms do t

20

ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s

21 22 23 24 25

Eco=mv_dot/ms_dot //E conomy of evapo rato r Ts=399 // Saturat ion tem per atu re of st eam in [K] dT=Ts-T //Te mp er at ur e driving force [K ] // [W/ sq m.K ] U=2350 Q=ms_dot*lambda_s //R ate of h e a t transf er in [

//St ea m consumption

[ kg/h ]

kJ/kg ] 26 Q=Q*1000/3600 27 A=Q/(U*dT)

// [ J/ s ] = [W] //H eat transfer

ar ea in [ sq

m] 28 printf ( ” \nANSWER Ec on om oy pf ev ap or at or i s %f \ n” , Eco); 29 printf ( ” \ nHeat tarnsfer ar ea to be pr ov id ed = %f sq m\n ” ,A);

Scilab code Exa 6.4 Steam economy

1 2 3 4 5 6

clc ; clear ;

//Ex am pl e 6. 4

// Spec ific he at of f eed in kJ/(k g . K) // La te nt hea t of co nd s of he at at 0. 2MPa in [ kJ/kg ] 7 lambda=2383 // La te nt h ea t of vaporisation of wat er aty 32 3 [ kJ/k g 8 ic=0.1 // I n i t i a l concentration of so il ds in Cpf=3.98 lambda_s=2202

[%] 155

// Final concentrat ion //F eed to evaporator in [ kg/h ] m_d ot/fc //M ass flow rate of thic k liquo r in [ kg/ h ] 12 mv_dot=m_dot-mdash_dot //W ater evap orat ed in [ kg/ h ] 9 fc=0.5 10 m_dot=30000 11 mdas h_do t=ic*

13 14 15 16 17 18 19 20

//C as e 1: Feed at 29 3K mf_dot=30000 // [ kg /h ] mv_dot=24000 // [ kg /h ] Cpf=3.98 // [ kJ /( kg .K) ] // Saturat ion tem per at ure of s team in [K] Ts=393 T=323 // Boil ing po in t of solution [K] lambda_s=2202 // La te nt he at of c onde nsat ion [ kJ/kg ] 21 lambda=2383 // Lat ent hea t of vapo risa tion [ kJ/k g ] 22 Tf=293 //Fe ed temperature 23 //Ent ha lp y balance ove r the evaporator : 24

ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s

//Ste am consumpt ion [ kg/h ] // Ste am econ omy

25 eco=(mv_dot/ms_dot)

26 printf ( ”my \nWhen introduc \n d econo is %fFee ” ,eco);

ed at 29 3 K ,St eam

27 dT=Ts-T // [K] 28 U=2900 // [W/ sq m.K] 29 Q=ms_dot*lambda_s //H ea t load

he at transfe

=Ra te of

r in [ kJ/ h ] // [ J/ s ] //H eat transfer

30 Q=Q*1000/3600 31 A=Q/(U*dT)

ar ea required [ sq m ] 32 printf ( ” \n ANSWER−( i ) \ n\n At 293 K, Heat tran sfer ar ea req uir ed is %f sq m \ n” ,A); 33 34 //Cas e2 : Fe ed at 308 K 35 Tf=308 // [ Fe ed tempe ratur e ] [K] 36 ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s

//S te am consu mptio n in [ kg/ h ] //Ec onomy o f

37 eco=mv_dot/ms_dot

156

evaporator 38 printf ( ” \n ANSWER−( i i ) \ n\ n When T=308 K \ nEco nomy of ev apo ra tor is %f \ n ” ,eco); 39 Q=ms_dot*lambda_s // [ kJ /h ] 40 Q=Q*1000/3600 // [ J/ s ] 41 A=Q/(U*dT) //H eat transfe r ar ea

required

[ sq m]

42 printf ( ’ \nANSWER −( i i i )

transfer

\n When T=308 K, \ nHeat Area req uir ed is %f sq m \n ” ,A) ;

Scilab code Exa 6.5 Evaporator economy

1 2 3 4 5 6 7 8 9 10

clc ; clear ;

//Ex am pl e 6. 5 m_dot=5000 //Fe ed to the evaporator [ kg/h ] Cpf=4.187 // Cp of feed in [ kJ/k g .K ] ic=0.10 // I n i t i a l co nc en tr at io n fc=0.4 // Final concentrati on mdash_dot=m_dot*ic/fc // [ kg/h ] of thic k liquor mv_dot=m_dot-mdash_dot //W at er evaporated in [ kg/h ] lambda_s=2162 // Lat ent hea t of conde nsing steam [ kJ/kg ] P=101.325 // Pressure in the evaporator [ kPa ] bp=373 // [K] Hv=2676 //Ent hal py of wat er vapor [ kJ/k g ] H_dash=419 // [ kJ/ kg ] Hf=170 // [ kJ/ kg ]

11 12 13 14 15 16 ms_dot=(mv_dot*Hv+ lambda_s 17 eco=mv_dot/ms_dot

mdash_dot*H_das

h -m_dot*Hf)/

//St ea m consu mptio n in [ kg/h ] //St ea m ec on om y of

evaporator 18 Q=ms_dot*lambda_s

// [ kJ/ h ] 157

19 20 21 22 23

U=1750 // [W/ sq m.K] dT=34 // [K] Q=Q*1000/3600 // [ J/ s ] A=Q/(U*dT) // [ sq m] printf ( ” \n H eat transfer ar ea to be sq m” ,A);

pr ov id ed is %f

Scilab code Exa 6.6 Single effect Evaporator

1 2 3 4 5 6 7

clc ; clear ;

//Ex am pl e 6. 6 mf_dot=5000 ic=0.01 fc=0.02 T=373

// [ kg /h ] // I n i t i a l co nc en tr at io n [ kg/h ] // Final concentration [ kg/ h ] // Boi ling p t of satu rati on in [ K

] 8 Ts=383

// Saturation

temp erat ure of

steam in [K ] 9 mdash_dot=ic*mf_dot/fc 10 mv_dot= mf_dot -m dash_dot

// [ kg /h ] //W ater evap orate d in [

kg/ h ] Hf=125.79 // [ kJ/ kg ] Hdash=419.04 // [ kJ/ kg ] Hv=2676.1 // [ kJ/ kg ] lambda_s=2230.2 // [ kJ/ kg ] ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/ lambda_s //S team flow rate in [ kg/h ] 16 eco=mv_dot/ms_dot // Ste am eco nomy 17 Q=ms_dot*lambda_s //R ate of he at transfer 11 12 13 14 15

in [ kJ/h ] 18 Q=Q*1000/3600 19 dT=Ts-T 20 21 A=69

// [ J/ s ] // [K] // He at in g are a of evap orat or in [ sq 158

m] // Ove ral l he at transfer coeff in [ W/ sq m.K] 23 printf ( ” \ nS tea m ec on om y i s %f \ n ” ,eco); 24 printf ( ” \n \ nOverall heat tr an sf er c o e f f i c i e n t is %d W/ sq m.K” , round (U)); 22 U=Q/(A*dT)

Scilab code Exa 6.7 Single effect evaporator reduced pressure

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

clc ; clear ;

//Ex am pl e 6. 7 //F rom pre vio us exa mpl e : mf_dot=5000 // [ kg /h ] Hf=125.79 // [ kJ/ kg ] lambda_s=2230.2 // [ kJ/ kg ] mdash_dot=2500 // [ kg /h ] Hdash=313.93 // [ kJ / kg ] mv_dot=2500 // [ kg /h ] Hv=2635.3 // [ kJ/ kg ] ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/ lambda_s //S team flow rate in [ kg/h ] Q=ms_dot*lambda_s // [ kJ/ h ] Q=Q*1000/3600 // [W] U=2862 // [W/ sq m.K ] dT=35 // [K] A=Q/(U*dT) // [ sq m] printf ( ” \n T he he a t trans fer ar ea in this ca se is %f s q m\ n ” ,A); printf ( ” \n \nNOTE : Th er e is a cal cul at io n mist ake in

th e book at th e line12 of this code , m s d o t va lu e is writte n as 2320. 18 , whic h is wrong \n \n ” ) ;

159

Scilab code Exa 6.8 Mass flow rate

1 2 3 4 5 6 7 8 9 10 11

clc ; clear ;

//Ex am pl e 6. 8 //F eed rate in

mf_dot=6000

[ kg/h ]

//Ta king th e giv en values from previous Hf=125.79 // [ kJ/ kg ] ms_dot=3187.56 // [ kg /h ] lambda_s=2230.2 // [ kJ/ kg ] Hdash=419.04 // [ kJ/ kg ] // [ kJ/ kg ] Hv=2676.1

exam pl e (6.6 )

mv_dot=(mf_dot*Hf+ ms_dot*lambda _s -6000*Hdash)/(HvHdash) //W ater evaporated in [ kg/ h ] 12 mdash_dot=6000-mv_dot //M ass flo w rate of

product

[ kg/ h ] // Wt % of sol ute

13 x=(0.01*mf_dot)*100/mdash_dot

in pro duc ts 14 printf ( ” \ nMass fl ow rate

of pr od uc t is %f k g/ h

m da sh _d ot ) ; 15 printf ( ” \n \ nThe pr od uc t concentration

by weig ht

\ n\n ” ,

is %f perc ent

\ n\ n ” ,x);

Scilab code Exa 6.9 Heat load in single effect evaporator

1 2 3 4 5 6 7 8 9

clc ; clear ;

//Ex am pl e 6. 9 Tf=298 //Fe ed tem per atur e in [K ] T_dash=373 // [K] Cpf=4 // [ kJ/ kg .K] fc=0.2 // Fin al conce ntrat ion of salt ic=0.05 // I n i t i a l co nc en tr at io n mf_dot=20000 // [ kg/h ] Fee d to evaporator

10 mdash_dot=ic*mf_dot/fc

//T hi ck liqu or [ kg/h ] 160

//W ater evap orat ed in [

11 mv_dot= mf_dot -m dash_dot

kg/ h ] 12 13 14 15

lambda_s=2185 // [ kJ/ kg ] lambda=2257 // [ kJ/ kg ] bpr=7 // Boiling poi nt ris e [K] T=T_dash+bpr // Boiling point of

solution

in [K ]

16 Ts=39 //Tem pe ra tu re of conde nsing st ea m in 17 ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s

[K ]

//S te am consu mptio n in [ kg/ h ] 18 19 20 21 22 23 24

eco=mv_dot/ms_dot //E conomy of evapo rato r Q=ms_dot*lambda_s // [ kJ /h ] // [ J/ s ] Q=Q*1000/3600 printf ( ” \ nHeat loa d is %d W or J/s” , round (Q)); printf ( ” \n \ nEconomy of evaporator is %f ” ,eco); printf ( ” \n \nNOTE: Ag ai n there

is a cal cua lt io n mi st ak e in book at line 19 of co de , it is wr it te n as 40 41 507 .1 inste ad of 40 41 50 71 \n \n ” ) ;

Scilab code Exa 6.10 Triple effect evaporator

1 2 3 4 5 6

clc ; clear ;

//Exa mp le 6.1 0 Ts=381.3 dT=56.6; U1=2800;

// [K] // [K] // Ove rall he at transfe

r coeff

in f i r s t

// Ove rall

he at transfe

r coeff

in f i r s t

// Ove rall

he at transfe

r coeff

in f i r s t

effect 7 U2=2200;

effect 8 U3=1100;

effect // / [K] // / [K]

9 dT1=dT/(1+(U1/U2)+(U1/U3)) 10 dT2=dT/(1+(U2/U1)+(U2/U3)) 11 dT3=dT-(dT1+dT2)

// [K] 161

12 13 14 15 16

//dT1=Ts −T1 d a s h

/ / [K ]

T1dash=Ts-dT1

//dT2=T1 dash −T2 d a s h T2_dash =T1dash -dT2 printf ( ” \n \nBoi ling

/ / [K ] // [K] poi nt of solution

in f i r s t

ef fe ct =%f K \ n\ n” ,T1dash); 17 printf ( ” \n \nBo ili ng po in t of solution ef fe ct =%f K \ n\ n” ,T2_dash);

in se c on d

Scilab code Exa 6.11 Double effect evaporator

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

clc ; clear ;

//Exa mp le 6.1 1 mf_dot=10000 // [ kg/ h ] o f fee d ic=0.09 // I n i t i a l co nc en tr at io n fc=0.47 // Final concentra tion m1dot_dash=ic*mf_dot/fc // [ kg /h ] Ps=686.616 //St ea m pr es su re [ kPa . g ] Ps=Ps+101.325 // [ kPa ] Ts=442.7 // Saturation tem per atur e in [K ] P2=86.660 // Vacuum in sec ond ef fe ct in [ kPa ] U1=2326 // Ove ral l he at transfe r in f i r s t eff ect

[W/ sq m.K] // Ov er all /sqm .K ]

13 U2=1744.5

14 P2_abs=101.325-P2

he at transfer

in 2n d effec t [W

// Ab sol ut e pressure

in se co nd

e f f e c t [ kPa ] 15 16 17 18 19

T2=326.3 dT=Ts-T2 Tf=309 T=273 Cpf=3.77

caustic

//Te mp era tu re in 2n d ef fe ct in [K] // [K] //F ee d temper ature in [K ] // [K] //k J /k g . K Specific h e a t for all st re am s

20 //Q1=Q2

162

21 22 23 24 25

//U1 ∗A1∗dT1=U2∗A2∗dT2 // [K] // [K] // Sin ce the re is no B.P .R Tv1=Ts-dT1 //T em pe ra tu re in va po r spac e of f i r s t eff ect in [ K ] dT2=dT/1.75 dT1=(U2/U1)*dT2

26 Tv2=Tv1-dT2 //Se co n d eff ect [K ] 27 Hf=Cpf*(Tf-T) //Fee d ent halp y [ kJ/kg ] 28 H1dash=Cpf*(Tv1-T) //En th al py of fi na l pr od uc t [

kJ/kg ] 29 30 31 32 33 34 35 36 37 38 39

//kJ/kg //F or st eam at 442.7 K lambda_s=2048.7 // [ kJ/ kg ] //F or va po ur at 392.8 K Hv1=2705.22 // [ kJ/ kg ] lambda_v1=2202.8 // [ kJ/ kg ] // for va po ur at 326.3 K: Hv2=2597.61 // [ kJ/ kg ] lambda_v2=2377.8 // [ kJ/ kg ] H2dash=Cpf*(Tv2-T)

// Overall

material

balance :

40 mv_dot= mf_dot -m 1dot_dash // [ kg /h ] 41 42 //Equation 4 bec ome s : 43 //mv1 d ot ∗ lambda v1+ mf dot ∗ Hf=(mv dot −mv1 dot ) ∗Hv2+(

mf dot −mv2 dot ) ∗ H2 dash 44 mv1_dot=(H2dash*( mf_ dot -mv_dot)- mf_dot*Hf+mv_dot* )/(Hv2+lambda_v1-H2dash) 45 mv2_dot =mv_dot -mv1_d ot // [ kg /h ] 46 47 //F rom equation 2 48 49 m2dot_dash=m1dot_dash+mv1_dot // Fi rs t

ef fe ct

material

bala nce [ kg/ h ]

50 ms_dot=(mv1_dot* Hv1+m1dot_dash* H1d ash -m2dot_dash* H2dash)/lambda_s // [ kg /h ] 51 52

163

Hv2

53 54 55 56 57 58

//H eat tra nsfe r Area // First effec t

59 60 61 62 63 64 65 66 67 68 69 70 71 72 73

A2=mv1_dot*lambda_v1/(U2*dT2)

74 75 76 77 78

// [ sq m]

A1=ms_dot*lambda_s*(10^3)/(3600*U1*dT1)

//Se co nd ef fe ct lambda_v1=lambda_v1*(10^3/3600)

// [ sq m]

// Sin ce A1 not = A 2 //SECOND TRIAL Aavg=(A1+A2)/2 dT1_dash=dT1*A1/Aavg dT2_dash=dT-dT1

// [ sq m] // [K] // / [K]

//Te mp er at ur e di st ri bu ti on Tv1=Ts-dT1_dash // [K] Tv2=Tv1-dT2_dash // [K] Hf=135.66 // [ kJ/ kg ] H1dash=Cpf*(Tv1-T) // [ kJ/ kg ] H2dash=200.83 // [ kJ/ kg ] //V apour at 388.5 K Hv1=2699.8 // [ kJ/ kg ] lambda_v1=2214.92 // [ kJ/ kg ]

mv1_dot=(H2dash*( mf_ dot -mv_dot)- mf_dot*Hf+mv_dot* Hv2 )/(Hv2+lambda_v1-H2dash) 79 mv2_dot =mv_dot -mv1_d ot // [ kg /h ] 80 81 // First ef fe ct Energy bal anc e 82 ms_dot=((mv1_dot* Hv1+m1dot_dash* H1dash)-(mf_do t mv 2_d ot ) * H2dash ) / l amb da _s // [ kg /h ] 83 84 //A rea of he at transfer 85 lambda_s=lambda_s*1000/3600 86 A1=ms_dot*lambda_s/(U1*dT1_dash) // [ sq m] 87 88 //Se co nd ef fe ct :

164

89

A2=mv1_dot*lambda_v1*1000/(3600*U2*dT2_dash)

// [ sq m] 90 91 printf ( ” \nA1(% f) =A2(% f) ,So the area in ea ch ef fe ct can be %f sq m \ n” ,A1,A2,A2); 92 printf ( ” \ nHeat trans fer sur fac e in ea c h effect is %f

s q m\ n ” ,A2); 93 printf ( ” \ nSt eam cons umpt ion= %d kg/h \n ” , round (ms_dot) ); 94 printf ( ” \ nE va po ra ti on in th e f i r s t ef fe ct is %d kg/ h \ n ” , round (mv1_dot)); 95 printf ( ” \ nE va p or at io n in 2 nd effec t is %d kg/ h \n ” , round (mv2_dot));

Scilab code Exa 6.12 lye in Triple effect evaporator

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

clc ; clear ;

//Exa mp le 6.1 2 Tf=353; T=273 mf_dot=10000; ic=0.07; fc=0.4;

// [K] // [K]

//Feed [ kg/h ] // I n i t i a l con c of gl yc er in e // Fi na L CONC OF GLYCERINE // Ove ral l glycerine ba lan ce m3dot_dash=(ic/fc)*mf_dot // [ kg /h ] mv_dot= mf_dot -m 3dot_dash // / [ kg/ h ] P=313; //St ea m pr es su re [ kPa ] Ts=408; // [ fr om ste am ta bl e ] [K] P1=15.74; // [ Pressure in la st ef fe ct ] [ kPa ] Tv3=328; // [ Vap our tem per atu re ] dT=Ts-Tv3 // Overall app are nt [K ] bpr1= 10 ; // [K]

19 bpr2=bpr1;

165

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

bpr3=bpr2; sum_bpr=bpr1+bpr2+bpr3 dT=dT-sum_bpr dT1=14.5; // [K] dT2=16; // [K] dT3=19.5; // [K]

// [K] // True Overall

Cpf=3.768

// [ kJ /( kg .K) ] of various st re ams Hf=Cpf*(Tf-T) // [ kJ/ kg ] H1=Cpf*(393.5-T) // [ kJ/ kg ] H2=Cpf*(367.5-T) // [ kJ/ kg ] // [ kJ/ kg ] H3=Cpf*(338-T) //Fo r st ea m at 40 K lambda_s=2160 // [ kJ/ kg ] Hv1=2692 // [ kJ/ kg ] lambda_v1=2228.3 // [ kJ/ kg ] Hv2=2650.8 // [ kJ/ kg ] lambda_v2=2297.4 // [ kJ/ kg ] Hv3=2600.5 // [ kJ/ kg ] lambda_v3=2370 // [ kJ/ kg ]

// Enthal pies

//MATERIAL // First effec AND t EBERGY BALANCES // Material balance //m1dot dash=mf dot −mv1 dot //m1dot dash=1750+mv2 dot+mv3 dot

//Ene rg y balance //ms d ot ∗ lambda s+mf Dot ∗ hf=mv1 dot ∗Hv1+m1dot dash H1 50 //2160 ∗ ms dot +22 38 ∗( mv2 dot +mv3 dot ) =19800500 51 52 53 54 55 56

//Se co nd ef fe ct //Ene rgy balan ce : //mv3 dot=8709.54 − 2.076 ∗ mv2 dot //Th ir d ef fe ct : 166

∗

57 58 59 60

//m2dot dash=mv3 dot+m3dot dash //m2dot dash=mv3 dot+1750 //F rom eq n 8 we get

61 62 63 64 65 66

//F rom eq n 8:

67

mv2_dot =(8709.54*2600.5+1750*244.92-8790.54*356.1-356.1*1750) /(-2.076*356.1+2297.4+2600.5*2.076)

// [ kg /h ] // [ kg /h ]

mv3_dot=8709.54-2.076*mv2_dot mv1_dot=mv_do t -(mv2_dot+ mv3_dot)

//F rom equation 4: //m1dot dash=mf dot −mv1 dot //ms dot=(mv1 dot ∗Hv1+m1dot dash la mb da s //[ kg/ h ]

∗H1−mf dot ∗ Hf)/

ms_dot=(19800500-2238*(mv2_dot+mv3_dot))/2160

// [ kg/ h ] 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85

/ /H eat transfe

r Area is

U1=710 // [W/ sq m.K] U2=490 // [W/ sq m.K] U3=454 // [W/ sq m.K] A1=ms_dot*lambda_s*1000/(3600*U1*dT1)

// [ sq m]

A2=mv1_dot*lambda_v1*1000/(3600*U2*dT2) A3=mv2_dot*lambda_v2*1000/(3600*U3*dT3)

// [[ sq sq m] m] //

/ /T he deviai ton is wit hin + −10% //Hence maximum A 1 are a can be re com me nde d eco=(mv_dot/ms_dot)

// [ Steam economy ]

Qc=mv3_dot*lambda_v3 // [ kJ/ h ] dT=25 // Rise in wa te r tem pera ture Cp=4.187 mw_dot=Qc/(Cp*dT) printf ( ” \nANSWER \ n Ar ea in e a c h effe ct%f s q m \n ” ,A1) ; 86 printf ( ” \nANSWER \ n St eam ec on om y is%f \n ” ,eco); 87 printf ( ” \nANSWER Cooli ng water rat e i s %f t/h” , mw_do t / 1000)

167

Scilab code Exa 6.13 Triple effect unit

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

clc ; clear ;

//Exa mp le 6.1 3 Cpf=4.18 // [ kJ/ kg .K] dT1=18 // [K] dT2=17 // [K] dT3=34 // [K] mf_dot=4 // [ kg/ s ] Ts=394 // [K] bp=325 //Bp of w at er at 1 3.17 2 kPa [ K] dT=Ts-bp // [K] lambda_s=2200 // [ kJ/ kg ] T1=Ts-dT1 // [K] // [ kJ/ kg ] lambda1=2249 lambda_v1=lambda1 // [ kJ/ kg ] T2=T1-dT2 lambda2=2293 lambda_v2=lambda2

// [K] // [ kJ/ kg ] // [ kJ/ kg ]

T3=T2-dT3 lambda3=2377 lambda_v3=lambda3

// [K] // [ kJ/ kg ] // [ kJ/ kg ]

// I n i t i a l con c of so li ds ic=0.1 fc=0.5 // Fi na l c o n c of solids m3dot_dash=(ic/fc)*mf_dot // [ kg/ s ] mv_dot= mf_dot -m 3dot_dash // Total evapora

in [ kg/s ] 29 // Mate rial bala nce ove r f i r s t ef fe ct 30 //mf dot=mv1 dot m1dot dash 31 //Ene rgy balan ce :

168

tion

32 //ms d ot ∗ lambda s=mf dot

∗ (Cpf ∗ ( T1−Tf)+mv1 dot

∗

lambda v1 ) 33 34 // Mat eri al bal anc e ov er se co nd eff ect 35 //m1dot dash=mv2 dot+m2dot dash 36 //Ent hal py balance : 37 //mv1 d ot ∗ lambda

lambda v2 )

38 39 40 41 42 43

v1 +m1dot dash ( cp ∗ ( T1−T2)=mv2 dot ∗

// Mat eri al bal anc e ov er third //m2dot dash=mv3 dot+m3dot+dash //Ent hal py balance : //mv2 lambda v2+m2dot dash lambda v3

eff ect

∗ cp ∗ ( T2−T3)=mv3 dot ∗

44 45 46 47 48

294 mv2_dot=3.2795/3.079 // [ kg/ s ] mv1_dot =1.053* mv2_dot -0.1305 // [ kg/ s ] mv3_dot=1.026*mv2_dot+0.051 // [ kg/ s ] ms_dot=(mf_dot*Cpf*(T1-294)+mv1_dot*lambda_v1)/ lambda_s // [ kg/ s ]

49 50 51 52 53 54 55 56 57 58 59 60

eco=mv_dot/ms_dot // Ste am econ omy eco= round (eco) printf ( ” \ nS te am ec on om y i s %d \ n ” ,eco); U1=3.10 // [kW/ sq m.K] U2=2 // [kW/ sq m.K] U3=1.10 // [kW/ sq m.K]

// First

eff ect :

// [ sq m] // [ sq m] // [ sq m] //A r e a s ar e calcu lated w it h a dev iati on of + −10% printf ( ” \ nArea p f h e a t trans fer in e a c h effect is %f s q m\ n ” ,A3) A1=ms_dot*lambda_s/(U1*dT1) A2=mv1_dot*lambda_v1/(U2*dT2) A3=mv2_dot*lambda_v2/(U3*dT3)

169

Scilab code Exa 6.14 Quadruple effect evaporator

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

clc ; clear ;

//Exa mp le 6.1 4 // [ kg /h ]

mf_dot=1060

ic=0.04 // I n i t i a l co nc en tr at io n fc=0.25 // Final concentrat ion m4dot_dash=(ic/fc)*mf_dot // [ kg /h ]

// Total

evaporati

on= // [ kg /h ]

mv_dot= mf_dot -m 4dot_dash

//Fr om st ea m ta bl e : P1=370 // [ kPa . g ] T1=422.6 // [K] lambda1=2114.4 // [ kJ/ kg ] P2=235 // [ kPa . g ] T2=410.5 // [K] lambda2=2151.5 // [ kJ/ kg ]

P3=80 // [ kPa .g] T3=390.2 // [K] lambda3=2210.2 // [ kJ/ kg ] P4=50.66 T4=354.7 lambda4=2304.6

// [ kPa . g ] // [K] // [ kJ/ kg ]

P=700 // Latent lambda_s=2046.3

heat of st eam [ kPa . g ] // [ kJ/ kg ]

// FIRST EFFECT //Ent hal py balance : //ms dot=mf dot ∗ Cp f ∗ ( T1−Tf)+mv1 dot //ms dot =1345.3 − 1.033 ∗ m1dot dash //SECOND EFFECT 170

∗ lambda1

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57

//m1dot dash=m2dot dash+mdot v2 //Ent hal py balance : //m1dot dash =531.38+0.510 ∗ m2dot dash //THIRD EFFECT // Material balance : //m2dot d ash −m3dot dash+mv3 dot //FOURTH EFFECT //m3dot dash=m4dot dash+mv4 dot mv4dot_dash=169.6 m3dot_dash=416.7

// [ kg /h ] // [ kg /h ]

//F rom eq n 4: m2dot_dash=-176.84+1.98*m3dot_dash

//F rom eq n 2: m1dot_dash=531.38+0.510*m2dot_dash

// [ kg /h ]

//F rom eq n 1: ms_dot=1345.3-1.033*m1dot_dash

58 eco=mv_dot/ms_dot steam ]

// [ kg evaporat ion /k g tion /k g st ea m”

59 printf ( ” \ nSt eam econ om y is %f evapora eco);

Scilab code Exa 6.15 Single effect Calendria

1 2 3 4 5 6

// [ kg /h ]

clc ; clear ;

//Exa mp le 6.1 5 m1_dot=5000 ic=0.1 fc=0.5

// [ kg /h ] // I n i t i a l co nc en tr at io n // Final concentra tion

7 mf_dot=(fc/ic)*m1_dot

// [ kg /h ] 171

,

8 9 10 11 12 13

mv_dot= mf_dot -m1_dot //Wat er eva po rat ed [ kg/h ] P=357 //S te am pr ess ur e [k N/sq m] Ts=412 // [K] H=2732 // [ kJ/ kg ] lambda=2143 // [ kJ/ kg ] bpr=18.5 // [K]

14 15 16 17 18 19

T_dash=352+bpr // [K] Hf=138 // [ kJ/ kg ] lambda_s=2143 // [ kJ/ kg ] Hv=2659 // [ kJ/ kg ] H1=568 // [ kJ/ kg ] ms_dot=(mv_dot*Hv+m1_dot*H1-mf_dot*Hf)/lambda_s

20 21 22 23 24 25 26 27

printf ( ” \ nSt eam con sum pti on is %f kg/h \n ” ,ms_dot); printf ( ” \ nC apa ci ty is %f kg/ h \ n ” ,mv_dot); eco=mv_dot/ms_dot //Economy printf ( ” \ nS tea m ec on om y i s %f \ n ” ,eco); dT=Ts-T_dash // [K] hi=4500 // [W/ sq m.K ] ho=9000 // [W/ sq m.K ] Do=0.032 // [m]

28 29 30 31 32 33 34

Di=0.028 // [m]// [m] x1=(Do-Di)/2 Dw=(Do-Di)/ log (32/28) // [m] x2=0.25*10^-3 // [m] L=2.5 //Len gth [m ] hio=hi*(Di/Do) // [W/ sq m.K ] printf ( ” \n NOTE: In te xt boo k this val ue of hio

//S team con sum pti on in kg/ h

w r o n g l y calcu lated this \ n\n ” ) ;

as 3975.5..

So we will

35 hio=3975.5 36 k1=45 //T ube mate rial in [W/sq m.K ] 37 k2=2.25 //For sc a l e [W/m.K] 38 Uo=1/(1/ho+1/hio+(x1*Dw)/(k1*Do)+(x2/k2))

Ove ral l he at transfer 39 Q=ms_dot*lambda_s 40 Q=Q*1000/3600 41

coeff in W/sq m. K // [ kJ /h ] // [W]

172

is ta ke

//

42 A=Q/(Uo*dT) 43 n=A/(%pi*Do*L) 44 printf ( ” \n N o.

// [ sq m] // from A=n ∗ %p i ∗Do∗L of tu be s req uir ed is %d” , round (n));

Scilab code Exa 6.16 Single effect evaporator

1 2 3 4 5 6 7 8 9 10 11

clc ; clear ;

//Exa mp le 6.1 6 bpr=40.6; // [K] Cpf=1.88; // [ kJ/ kg .K] Hf=214; // [ kJ/ kg ] H1=505; // [ kJ/ kg ] mf_dot=4536; // [ kg/ h ] of feed solut ion ic=0.2; // I n i t i a l co nc fc=0.5; // Final concentrati on m1dot_dash=(ic/fc)*mf_dot // Th is ck liqu or

flow

a r te [ kg/h ] 12 mv_dot= mf_dot -m 1dot_dash 13 Ts=388.5; // Saturation

// [ kg /H] tem pera ture

of st ea m

in [ K] 14 15 16 17 18 19 20

bp=362.5 //b .P of solutio n in [K ] lambda_s=2214; // [ kJ/ kg ] P=21.7; //Va po r space in [ kPa ] Hv=2590.3; // [ kJ/ kg ]

21 22 23 24 25

printf ( ” \ nSt eam con sum pti on is %f kg/h \n ” ,ms_dot); dT=Ts-bp // [K] U=1560 // [W/ sq m.K ] Q=ms_dot*lambda_s // [ kJ/ h ] Q=Q*1000/3600 // [W]

//Ent ha lp y balance

ove r evaporator

ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s

// [ kg/ h

26 A=Q/(U*dT)

// [ sq m] 173

27 printf ( ” \ nHeat transfer 28 29 // Calcula tions considering

ar ea is %f sq m

\ n ” ,A);

en th al py of sup er he ate d

vapour 30 31 Hv=Hv+Cpf*bpr 32

// [ kJ/ kg ]

ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s

// [ kg/ h ]

33 printf ( ” \n Now, St ea m consu mptio n i s %f kg/h \n ” , ms_do t ) ; 34 eco=mv_dot/ms_dot //Steam eco nomy 35 printf ( ” \ nEconomy of evapo rato r %f \ n ” ,eco); 36 Q=ms_dot*lambda_s // [ kJ /h ] 37 Q=Q*1000/3600 // [w ] 38 A2=Q/(U*dT) //Area 39 printf ( ” \nNow, Ar ea i s %f \ n ” ,A); 40 perc=(A2-A)*100/A //%er ro r in th e he at

transfer

ar ea ent hal py of wa te r vap ou r H v we re ba se d on th e satu rate d v a p ou r at th e pressu re \ nthe error in tr od uc ed is on ly % f pe rc en t \n ” ,perc);

41 printf ( ” \n If

174