Heat Transfer_K. a. Gavhane

Scilab Textbook Companion for Heat Transfer by K. A. Gavhane1 Created by Deepak Bachelor of Technology Chemical Engineering DCRUST,Murthal College Teacher Ms. Sunanda Cross-Checked by Lavitha Pereira May 24, 2016

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

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

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

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

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Contents List of Scilab Codes

4

2 Conduction

5

3 Convection

48

4 Radiation

96

5 Heat Exchangers

114

6 Evaporation

149

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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 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 . . . . . . . . . . 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 . . . . . . . . . . . . Thickness of insulation . . . . . . . . . . Reduction in heat loss in insulated pipe Heat loss in a pipe   . . . . . . . . . . . . Arrangements for heat loss . . . . . . . Insulation thickness . . . . . . . . . . . Heat loss in furnace . . . . . . . . . . . Rate of heat loss in pipe . . . . . . . . . Heat loss from insulated steel pipe   . . . Heat loss from furnace   . . . . . . . . . . Rate of heat loss   . . . . . . . . . . . . . Thickness of insulation . . . . . . . . . . Heat loss per metre . . . . . . . . . . . Mineral wool insulation . . . . . . . . . Furnace wall . . . . . . . . . . . . . . . Thickness of insulating brick . . . . . . Heat flow through furnace wall   . . . . . 4

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

Exa 2.31 Exa 2.32 Exa 2.33 Exa 2.34 Exa 2.36 Exa 2.37 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 Exa 3.3 Exa 3.4 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

Heat loss in pipe   . . . . . . . . . . . . . . . Heat flux through layers  . . . . . . . . . . . Conductive conductance furnace wall . . . . Critical radius of insulation   . . . . . . . . . Critical radius of pipe   . . . . . . . . . . . . Time required for steel ball . . . . . . . . . Steel ball quenched . . . . . . . . . . . . . . Ball plunged in a medium   . . . . . . . . . . Slab temperature suddenly lowered . . . . . Flow over a flat plate . . . . . . . . . . . . Stainless steel rod immersed in water . . . . Chromel alumel thermocouple . . . . . . . . Thermocouple junction . . . . . . . . . . . Batch reactor  . . . . . . . . . . . . . . . . . Heat dissipation by aluminium rod   . . . . . Aluminium fin efficiency  . . . . . . . . . . . Pin fins   . . . . . . . . . . . . . . . . . . . . Metallic wall surrounded by oil and water   . Brass wall . . . . . . . . . . . . . . . . . . . Boundary layer thickness . . . . . . . . . . Boundary layer thickness of plate . . . . . . Thickness of hydrodynamic boundary layer Flat plate boundary layer   . . . . . . . . . . Rate of heat removed from plate . . . . . . Heat removed from plate . . . . . . . . . . Local heat transfer coefficient   . . . . . . . . Width of plate . . . . . . . . . . . . . . . . Heat transferred in flat plate . . . . . . . . Rate of heat transferred in turbulent flow   . Heat transfer from plate in unit direction . Heat lost by sphere . . . . . . . . . . . . . Heat lost by sphere . . . . . . . . . . . . . Percent power lost in bulb . . . . . . . . . . Heat lost by cylinder . . . . . . . . . . . . . Heat transfer in tube . . . . . . . . . . . . . Heat transfer coefficient   . . . . . . . . . . . Heat transfer coefficient in heated tube . . . h of water flowing in tube   . . . . . . . . . . 5

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30 31 31 33 34 34 35 36 37 38 38 39 40 41 42 43 44 45 46 48 49 50 50 51 52 53 54 55 56 57 58 58 59 60 60 61 62 63

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

Overall heat transfer coefficient   . . . . Number of tubes in exchanger . . . . . Convective film coefficient   . . . . . . . Length of tube   . . . . . . . . . . . . . Cooling coil . . . . . . . . . . . . . . . Outlet temperature of water . . . . . . Inside heat transfer coefficient . . . . . Film heat transfer coefficient . . . . . Area of exchanger . . . . . . . . . . . Natural and forced convection . . . . Natural convection   . . . . . . . . . . . Free convection in vertical pipe   . . . . Heat loss per unit length . . . . . . . Free convection in pipe . . . . . . . . Free convection in plate   . . . . . . . . Heat transfer from disc . . . . . . . . Rate of heat input to plate . . . . . . Two cases in disc . . . . . . . . . . . . Total heat loss in a pipe   . . . . . . . . Heat loss by free convection   . . . . . . Heat loss from cube . . . . . . . . . . Plate exposed to heat . . . . . . . . . Nucleate poolboiling   . . . . . . . . . . Peak Heat flux   . . . . . . . . . . . . . Stable film pool boiling . . . . . . . . Heat transfer in tube . . . . . . . . . . Nucleat boiling and heat flux   . . . . . Dry steam condensate   . . . . . . . . . Laminar Condensate film . . . . . . . Saturated vapour condensate in array Mass rate of steam condensation . . . Saturated tube condensate in a wall . Condensation rate . . . . . . . . . . . Condensation on vertical plate . . . . Heat loss by radiaiton   . . . . . . . . . Radiation from unlagged steam pipe   . Interchange of radiation energy   . . . . Heat loss in unlagged steam pipe   . . . 6

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64 65 67 68 69 69 70 71 72 73 74 75 76 77 77 78 79 80 82 83 83 84 86 86 87 88 89 89 90 91 92 93 94 94 96 96 97 97

Exa 4.5 Exa 4.6 Exa 4.7 Exa 4.8 Exa 4.9 Exa 4.10 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 Exa 5.3 Exa 5.4 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

Loss from horizontal pipe . . . . . . . . Heat loss by radiation in tube . . . . . . Net radiant interchange . . . . . . . . . Radiant interchange between plates . . . Heat loss from thermflask   . . . . . . . . Diwar flask   . . . . . . . . . . . . . . . . Heat flow due to radiation . . . . . . . . Heat exchange between concentric shell Evaporation in concenric vessels . . . . infinitely long plates   . . . . . . . . . . . Heat exchange between parallel plates . Thermal radiation in pipe  . . . . . . . . Heat transfer in concentric tube . . . . Heat exchange between black plates . . Radiation shield . . . . . . . . . . . . . Heat transfer with radiaiton shield   . . . Radiaition shape factor . . . . . . . . . Radiation loss in plates . . . . . . . . . Concentric tube . . . . . . . . . . . . . Harpin exchanger . . . . . . . . . . . . Length of pipe . . . . . . . . . . . . . . Double pipe heat exchanger   . . . . . . . Parallel flow arrangement . . . . . . . . Counter flow exchanger . . . . . . . . . LMTD approach   . . . . . . . . . . . . . Shell and tube exchanger . . . . . . . . Order of Scale resistance . . . . . . . . . Length of tube required   . . . . . . . . . Suitability of Exchanger  . . . . . . . . . Number of tubes required   . . . . . . . . Shell and tube heat exchanger . . . . . Length of pipe in Exchanger . . . . . . Dirt factor . . . . . . . . . . . . . . . . Heat transfer area . . . . . . . . . . . . Oil Cooler . . . . . . . . . . . . . . . . . Countercurrent flow heat exchanger . . Vertical Exchanger  . . . . . . . . . . . . Countercurrent Heat Exchanger . . . . . 7

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98 99 99 100 100 101 102 102 103 104 105 106 107 107 108 109 110 111 112 114 116 117 119 120 121 122 123 124 126 127 129 131 132 135 136 137 138 140

Exa 5.20 Exa 5.21 Exa 5.22 Exa 5.23 Exa 5.24 Exa 5.25 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 Exa 6.16

Number of tube side pass . . . . . . . . . Number of tubes passes . . . . . . . . . . Outlet temperature for hot and cold fluids Counterflow concentric heat exchanger   . . Number of tubes required   . . . . . . . . . Parallel and Countercurrent flow   . . . . . Boiling point Elevation . . . . . . . . . . Capacity of evaporator . . . . . . . . . . . Economy of Evaporator . . . . . . . . . . Steam economy . . . . . . . . . . . . . . . Evaporator economy   . . . . . . . . . . . . Single effect Evaporator   . . . . . . . . . . Single effect evaporator reduced pressure . Mass flow rate . . . . . . . . . . . . . . . Heat load in single effect evaporator . . . Triple effect evaporator . . . . . . . . . . Double effect evaporator . . . . . . . . . . lye in Triple effect evaporator   . . . . . . . Triple effect unit   . . . . . . . . . . . . . . Quadruple effect evaporator  . . . . . . . . Single effect Calendria   . . . . . . . . . . . Single effect evaporator . . . . . . . . . .

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141 142 143 145 146 147 149 149 150 151 153 154 155 156 156 157 158 161 164 165 167 169

Chapter 2 Conduction

Scilab code Exa 2.1   Thickness of insulation

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clc ; clear ; printf ( ” E xa mp le 2 . 1 \n Page no . 2 . 1 8 \ n P a r t −( a ) ” ) A=1;   // s q m et re printf ( ” A re a o f h e a t t r a n s f e r , A=%f mˆ 2\ n ” ,A) Q=450;   / / W/ s q m tr e printf ( ” R a te o f h e a t l o s s / u n i t a r e a =%f W/mˆ 2\ n ” ,Q ) dT=400; // K printf ( ” Te m p er a t u r e d i f f e r e n c e a c r o s s i n s u l a t i o n l a y e r \ t , dT=%f K\n ” ,dT) k=0.11   //W/(m.K)   n ” ,k) printf ( ” F o r a s b e s t o s , k=%f \

  //Q=(k∗ A∗d T ) / x   x=(k*A*dT)/Q   X=x*1000;

// f o r f i r e c l a y i n s u l a t i o n   k=0.84;   // W/(m.K)   printf ( ” F o r f i r e c l a y i n s u l a t i o n , k=%f W/ (m . K) \n ” ,k);   x=(k*A*dT)/Q;   X=x*1000;

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21   printf ( ” Ans . ( A) . T h i c k n e s s o f a s b e s t o s i s : %f m=%f mm \n ” ,x,X) 22   printf ( ” Ans . ( B ) T h ic kn e ss o f f i r e c l a y i n s u l a t i o n i s : %f m =%f mm\n ” ,x,X)

Scilab code Exa 2.2  Heat loss per metre

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16 17 18 19 20 21 22

             

23 24 25 26

       

clc ; printf ( ” E xa mp le 2 . 2 , \ n Pa ge n o . 2 . 1 8 \ n ” ); L =1 // m printf ( ” L en gt h o f p pi pe , L = %f m\n ” ,L); r1=(50/2)   / / i n mm r1=r1/1000   / / i n 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 ( ” T he rm al r e s i s t a n c e o f w a l l p i p e =R1=%f K/W\n ” ,R1); printf ( ” For i n n e r l a g g i n g : \ 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 ( ” Therma l r e s i s t a n c e o f i n n e r l a g g i n g =R2=%f K/ W” ,R2); o u t er l a g g i n g : \ n ” ) ; printf ( ” For k3=0.04   //W/(m.K) ro1=0.053 //m ro2=(ro1+0.04) // m

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27   rmo1=(ro2-ro1)/ log (ro2/ro1) 28   R3=(ro2-ro1)/(k3*2*%pi*rmo1*L) 29   printf ( ” Therma l r e s i s t a n c e o f i n n e r l a g g i n g =R2=%f K/ W\n ” ,R3); 30   R=R1+R2+R3 31   Ti=550 / /K / / i n s i d e 32   To=330 //K // o u t s i d e 33   dT=Ti-To;   // T em pe ra tu re d i f f e r e n c e 34   Q=dT/R 35   printf ( ” R at e o f h e a t l o s s p e r m et re o f p i pe , Q=%f W/m ” ,Q)

Scilab code Exa 2.3  Heat Loss

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

  clear ; clc ;   printf ( ” E x a mp l e 2 . 3 ” )

  //Given   r1=44   // [mm] // [m]   r1=r1/1000 / / [m]   r2=0.094 / / [m]   r3=0.124 // T em per at ure a t o u t er s u r f a c e o f w a l l i n [ K]   T1=623 // T em per at ure a t o u t er s u r f a c e o f o u t er   T3=313 i n su l at i on [K] // Thermal c o n d u c t i v i t y o f i n s u l a t i o n l a y e r   k1=0.087 1 . . i n [W/m . K ] // Thermal c o n d u c t i v i t y o f i n s u l a t i o n l a y e r   k2=0.064 2 [W/m . K ] // L en g th o f p i p e [m] l =1 // l o g mean r a d i u s o f     rm1=(r2-r1)/ log (r2/r1) i n s u l a t i o n l a y e r 1 [m] // l o g mean r a d i u s o f     rm2=(r3-r2)/ log (r3/r2) i n s u l a t i o n l a y e r 2 [m] // P ut t in g v a l u e s i n f o l l o w i n g eqn : 11

17 Q = ( T1 - T 3 ) / ( ( r 2 - r 1 ) / ( k1 * 2 * % p i * r m1 * l ) + ( r3 - r 2 ) / ( k 2 * 2 * %pi*rm2*l)); 18   printf ( ” Hea t l o s s p e r m et er p i p e i s %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 ;

  / / E xa mp le 2 . 4   printf ( ” E x a mp l e 2 . 4 ” )   //Given [ s q m] A =1   // Heat t r a n s f e r a re a // t h i c k n e s s o f f i r e b r ic k i n [m]   x1=0.229 // t h i c k n e s s o f i n s u l a t i n g b r i c k i n [m]   x2=0.115 // t h i c k n e s s o f b u i l d i n g b r i c k i n [m]   x3=0.229 // t he rm a l c o n d u c t i v i t y o f f i r b r i c k [W/ (m.   k1=6.05 K) ] // t he rm al c o n d u c t i v i t y o f i n s u l a t i n g b r i c k   k2=0.581 [W/m.K] // t he rm al c o n d u ct i v i t y o f b u i l d i n g b r i c k   k3=2.33 [W/m.K] // i n s i d e t em pe ra tu re [ K ]   T1=1223   T2=323   / / O u t s i de t e m p e r a tu r e [ K] / / O v e r a l l temp d ro p [ K]   dT=T1-T2   R1=(x1/k1*A)   / / t he rm al r e s i s t a n c e 1   R2=(x2/k2*A)   // Thermal r e s i s t a n c e 2   R3=(x3/k3*A)   // Thermal r e s i s t a n c e 3   Q=dT/(R1+R2+R3)   //w/SQ m   Ta=-((Q*R1)-T1)   // fro m Q1=Q=(T1−Ta ) / ( x 1 / k 1 ∗A) // Sim ila rly

13 14 15 16 17 18 19 20 21 22   Tb=(Q*R3)+T2; 23   printf ( ” I n t e r f a c e t e m pe r a tu r e : \ n i −B et we en K \ n i i −B e t we e n I B−PB=%fK” ,Ta,Tb);

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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 ;

  / / E xa mp le 2 . 5   printf ( ” E xa mp le 2 . 5 \ n Pa ge 2 . 2 3 ” )   //Given   A=1; / / l e t [ s q m] // t h i c k n e s s o f f i r b r i c k l a y e r [m]   x1=0.23; / / [m]   x2=0.115; // [m]   x3=0.23; / / Te mp er at ur e o f f u r n a c e [ K]   T1=1213; // T em pe ra tu re o f f u rn a c e [ K]   T2=318;   dT=T1-T2; // [K] //W/(m.K) ( f i r e br ic k )   k1=6.047; / /W/ (m .K) ( i n s u l a t i n g b r i c k )   k2=0.581; / /W/ (m .K) ( b u i l d i n g b r i c k )   k3=2.33; // Heat l o s t p er   Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3)) u n i t A rea i n Watt

// T herm al r e s i s t a n c e R1=(x1/k1)   //Approximate R1=0.04 R2=(x2/k2)   //Approximate R2=0.2025 R3=(x3/k3)   //Approximate R3=0.1 Ta=T1-((dT*R1)/(R1+R2+R3)) Tb=((dT*R3)/(R1+R2+R3))+T2   //Approximation Tb=565 printf ( ” \ nAn swer : H ea t l o s s p e r u n i t a r e a i s %f W=%f   J / s \n ” ,Q_by_A ,Q_by_A ); 28   printf ( ” \ nAnsw er : \ n Ta=%f K =T e m p er a t u re a t t h e                    

i n t e r f a c e b et w ee n f i r e b r i c k and i n s u l a t i n g b r i c k \n Tb=%d K T e mp er at ur e a t t h e i n t e r f a c e b et we en 13

i n s u l a t i n g and b u i l d i n g b r ic k \n” ,Ta,Tb)

Scilab code Exa 2.7  Heat loss

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clc printf ( ” E xa mp le 2 . 7 , P ag e n o 2 / 2 6 \n ” ) ; printf ( ” P a r t −(a ) \n” ); A=1;   // s q m et re x1=114   / / mm x1=114/1000   / / m et re k1=0.138   // W/(m.K) R 1 = x 1 /( k 1 * A) x2=229   //mm / / m et re x 2 = x 2 / 1 00 0 // W/m. K k2=1.38 R2=x2/(k2*A) dT=1033-349

  / / Heat l o s s

  Q=dT/(R1+R2) f r o m 1 s q m e t r e w a l l =%f W”   printf ( ”ANSWER : H ea t l o s s ,Q); 17   printf ( ” P a r t ( b ) \n ” ) ; 18 // c o n ta c t r e s i s t a n c e =c r 19   cr=0.09   //K/W 20   R=R1+R2+cr 21   Q=dT/R 22   printf ( ”ANSWER : H e a t l o s s f r o m 1 s q m e t r e w hen r e s i s t a n c e p r e s e n t=%f W” ,Q);

Scilab code Exa 2.8  Loss per area

1   clear ; 2 clc ;

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  / / E xa mp le 2 . 8   printf ( ” E xa mp le 2 . 8 \n ” ) // Given : // [m]   x1=0.02 // [m]   x2=0.01 // [m]   x3=0.02 //W/ (m. k )   k1=0.105   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 //K Tx=293 Rx=0.9524   //K/W x=R*(T1-Tx)/(dT*Rx) x=x*100   //mm printf ( ” The t em p er a tu r e o f 29 3 K w i l l be r ea c he d a t

p o in t %f mm fro m t he o ut er mo st w a ll s u r f a c e o f   t he i c e −box” ,x)

Scilab code Exa 2.9  Heat loss

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clc   printf ( ” E xa mp le 2 . 9 , P ag e 2 . 2 8 \ n ” );

  //Given   ID=50   //mm;              

dT=(573-303); printf ( ” I n t e r n a l d i a m e t e r , I D=%f mm” ,ID); r1=ID/2   //mm r1=r1/1000   / / m e tr es OD=150   / / mm printf ( ” Out er di am et er ,OD=%f mm” ,OD); r2=OD/2   / / mm

15

12 13 14 15 16 17 18 19 20 21

  r2=75/1000 // m

  // T he rm al c o n d u c t i v i t y   k=17.45   // W/(m.K) // Solu tion   printf ( ”Q/A=dT/ ( r2−r 1 ) / k \n ” );   A1=4*%pi*(r1^2);   A2=4*%pi*(r2^2); A= sqrt ( A 1 * A 2 )   Q=(A*k*dT)/(r2-r1)   printf ( ”ANSWER: \ nHeat

l o s s=Q=%f W” ,Q);

Scilab code Exa 2.10   Heat Passed

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  clear ; clc ;

  / / E x am pl e 2 . 1 0   printf ( ” E xa mp le 2 . 1 0 ” ) A= 1   / / s q m            

x1=0.15 x2=0.01 x4=0.15 // [K] T1=973 // [K] T2=288 // [K] dT=T1-T2

  / / Thermal c o n d u c t i v i t i e s        

k1=1.75 k2=16.86 k3=0.033 k4=5.23

// i n a bs en ce o f a i r gap , sum o f t he rm al r e s i s t a n c e s   sR=(x1/k1*A)+(x2/k2*A)+(x4/k4*A)   Q = d T / sR   printf ( ” Hea t l o s t p er s q m et er i s %d W/ s q m” ,Q);

/ /When h e a t l o s s ,Q= 11 63 , t h en new r e s i s t a n c e =sR1 // [W/ sq m]   Q1=1163 16

23 24 25 26 27

  sR1=dT/Q1

  / / wi dt h o f a i r gap be w t h e n // [m]   w=(sR1-sR)*k3*A // in [mm]   w=w*1000   printf ( ” Width o f a i r g ap i s %f mm” ,w);

Scilab code Exa 2.11  Insulated pipe

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  clear ; clc ;

  / / E x am pl e 2 . 1 1   printf ( ” E xa mp le 2 . 1 1 ” ) ; // [mm]   d1=300 // [mm]   r1=d1/2 // [m]   r1=r1/1000 // [m]   r2=r1+0.05 r3 = r2 + 0.0 4 // [m] // [m]   x1=0.05 // [m]   x2=0.04   k1=0.105   //W/(m.K)   k2=0.07   //W/(m.K)                

// [m] // [m]

rm1 = ( r2 - r1 )/ log (r2/r1); rm2=(r3-r2)/ log (r3/r2); // l et L =1 / / l e t L=1 A1=%pi*rm1*L R1=x1/(k1*A1); A2=%pi*rm2*L R2=x2/(k2*A2) // [K] T1=623 // [K] T2=323 // [K] dT=T1-T2

  / / P ar t a // H eat l o s s Q _b y_ L = dT / ( R1 + R2 )   printf ( ” H ea t l o s s i s %f W/m” ,Q_by_L); //Part b : 17

28   P=2*%pi*(r1+x1+x2) // [m] 29   Q_by_L_peri=Q_by_L/P / / [W/ s q m] 30 31   printf ( ” Heat l o s t p e r s q m e t e r o f o u t er i n s u l a t i o n i s %f W/ s q m” ,Q_by_L_peri); 32   R1=x1/(k1*A1) 33   sR=0.871+0.827 34   dT1=dT*R1/sR 35   printf ( ” T em per at ur e b etw een two l a y e r s o f i n s u l a t i o n =%f K” , (T 1 - d T 1 ) ) ;

Scilab code Exa 2.12   Composite brick

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  / / E x am pl e 2 . 1 2   clear ; clc ;   printf ( ” E xa mp le 2 . 1 2 \ n ” )

  //Given            

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

  / / Thermal c o n d u c t i v i t i e s // [W/m.K ]   k1=16.86 // [W/m. K ]   k2=1.75 // [W/m. K ]   k3=5.23 // [W/m.K ]   k_air=0.0337 // [ sq m] A =1   sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)) // Heat f l o w i n [W   Q=dT/sigma_R   T m = Q * x3 / k 3 / / Te mp er at ur e d r op i n m a g ne s i te b r i c k

/ / I n t e r f a c e t e m p e r at u r e=i T // [K]   iT=T2+Tm 18

// w it h a i r g ap f o r r e du c in g heat l o s s to 1163 per sq m 24 x_by_k= sigma_xbyk -sigm a_R // x/ k f o r a i r 23

s i g ma _ x by k = A * d T / 1 16 3

25   t=x_by_k*k_air 26   t=t*1000; 27   printf ( ” Width o f t h e a i r gap i s %f mm” ,t);

Scilab code Exa 2.13  Heat flow in a pipe

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  / / E x am pl e 2 . 1 3   printf ( ” E xa mp le 2 . 1 3 \n” ); L =1

/ / as su me

// [W/ (m.K)

  k1=43.03   k2=0.07

[m]

 

  T1=423   T2=305

/ / (W/m. K) / / i n s i d e t e m p e ra t u re [ K ] // [ K]

  r1=0.0525

// [mm]

  r2=0.0575;

// [m]

  r3=0.1075   // [m]

/ / r 3=r 3 / 1 0 0 0 ;

/ / [ m]

  Q=(2*%pi*L*(T1-T2))/((( log (r2/r1))/k1)+(( log (r3/r2)) // Heat l o s s p er m etr e /k2));

21 22   printf ( ” Hea t f l o w p e r m et re o f p i p e i s %f W/m” ,Q); 23 24   printf ( ” P ar t 2\ n ” ); 25   / /T=T em pe ra tu re o f o u t e r s u r f a c e

19

26   T=T1-(Q* log (r2/r1))/(k1*2*%pi*L); 27 28   printf ( ” Te m p er a t u r e a t o u te r s u r f a c e o f s t e e l p i p e : %f K” ,T); 29 30   printf ( ” \ n P a rt i i i \ n ” ) ; 31   id=0.105 // i n s i d e d i am e tr e i n [m] 32 33   A=%pi*id*1 // i n s i d e a re a i n [ s q m] 34 35   C=Q/(A*(T1-T2)); / / c o n d u ct a n c e p e r l e n g t h 36 37   printf ( ” C on du ct an ce p er m l e n g t h b as ed on i n s i d e a r e a i s %f W/K” ,C )

Scilab code Exa 2.15   Thickness of insulation

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  / / E x am pl e 2 . 1 5   printf ( ” E xa mp le 2 . 1 5 \n” ) / / [ s q m] A =1 //m   x1=0.1        

x2=0.04 k1=0.7 k2=0.48 sigma=x1/(k1*A)+x2/(k2*A)

  //K/W

  //Q=4.42∗dT   //Q=dT/sigma // w i t h r o ck w o ol i n s u l a t i o n added , Q d as h = 0. 75∗Q // W/(m.K)   k3=0.065 // Q dash=dT/s igm a+x3/ k3∗A //On s o l v i n g Q and Q d ash we g e t // [m]   x3=((1/(0.75*4.42))-sigma)*k3 // [mm]   x3=x3*1000   printf ( ” T h ic kn es s o f r oc kw oo l i n s u l a t i o n r e q u i r e d= %f mm” ,x3) 20

Scilab code Exa 2.16  Reduction in heat loss in insulated pipe

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clc ;   clear ;              

/ / E xa mp le 2 . 1 6 , P ag e n o 2 . 3 6 / / D i a me t er o f p i p e [mm] d1=40; // O ut si de r a d i u s i n [m] r1=(d1/2)/1000 // I n s u l a t i o n 1 t h i c k n e s s i n [mm] t1=20; // [m] t1=t1/1000 // I n s u l a t i o n 2 t h i c k n e s s in [m] t2=t1; // r a d i u s a f t e r 1 s t i n s u l a t i o n i n [m] r2=r1+t1; // R a d i u s a f t e r s e c on d i n s u l a t i o n i n [m r3=r2+t2; ]

11 12 // S i n ce S c i l a b d oe s n ot h a n dl e s s y mb o li c c o ns t an t s ,

we w i l l a ss um e some v a l u e s : 13 / / ( 1 ) 14   printf ( ” Le t t h e l a y e r M−1 b e n e a r e r t o t h e s u r f a c e ” ) 15   L=1; // [m] 16   T1=10; // T em per a t u re o f i n n er s u r f a c e o f p i p e [K] 17   T2=5; // T em per at ure o f o u t er s u r f a c e o f   i n s u l a t i o n [K] 18   k=1; / / Th er ma l c o n d u c t i v i t y 19   k1=k; / / F or M−1 m a t e r i a l 20   k2=3*k; // For m a t e r i a l M−2 21   Q1=(T1-T2)/( log (r2/r1)/(2*%pi*L*k1)+ log (r3/r2)/(2* %pi*L*k2)) 22 23 / / ( 2 ) 24   printf ( ” Le t t h e l a y e r o f m a t e r i a l M−2 be n e a r e r t o t he s u r f a c e ” ); 25   Q2=(T1-T2)/( log (r2/r1)/(2*%pi*L*k2)+ log (r3/r2)/(2* %pi*L*k1))

21

26   printf ( ”Q1=%f and Q2= %f  \n F or dummy v a r i a b l e s

un it y . . . \ nFo r any v a l u e o f k , T1 and T2 , Q1 i s a lw ay s l e s s t ha n Q2” ,Q1,Q2); 27   printf ( ” \n S o , M−1 n ea r t he s u r f a c e i s a d v i s a b l e ( i . e Arra ng emen t on e w i l l r e s u l t i , e s s h ea t l o s s \ n ) ” ) ; 28   per_red=(Q2-Q1)*100/Q2 29   printf ( ” P er ce nt r e d u ct i o n i n h ea t l o s s i s %f p e r c e n t ” ,per_red) 30   printf ( ” \nNOTE : S l i g h t v a r i a t i o n i n a n sw e rs due t o

l e s s p r e c i s e c a l c u l a t i o n i n book . I f p e rf o rm e d m an ua ll y , t h i s a ns we r s t a n d s t o b e c o r r e c t ” )

Scilab code Exa 2.17  Heat loss in a pipe

1 // Example2 .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 ( r 3 / r 2 ) // [W/m] )/k2+( log (r4/r3))/k3) 13   printf ( ”Heat l o s s=%f W/m” ,Q_by_L1) 14 / / Ca se 2 15   Q_by_L2=2*%pi*(T1-T2)/(( log (r2/r1))/k1+( log ( r 3 / r 2 ) )/k3+( log (r4/r3))/k2) 16 perct=(Q_by_L2 -Q_by_L1) *100/Q_by_L1 17   printf ( ” I f o r d e r i s c ha n ge d t h en h e a t l o s s =%f W/m” ,Q_by_L2)

22

18

 

printf ( ” \ n l o s s

o f h e a t i s i n c r e a s ed e d b y %f % f p er e r ce c e nt nt b y p u tt t t i ng n g m a t er e r i a l w it i t h h i g h e r t he h e rm r m al al c o n d u ct c t i v i t y n ea e a r t he h e p ip i p e s u r f a c e ” ,perct)

Scilab code Exa 2.18   Arrangements for heat loss

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clc ;   clear ;

/ / E xa xa mp mp le le 2 . 1 8 , P ag ag e n o 2 . 3 8   //Given // Assume : // [m] L =1 / / [m] Ou utt s i d e r a d i u s o d p i pe pe   r1=0.10 / / i n n er e r i n s u l a i t o n [m]   ia=0.025 // Outer r a d i u s o f i n n e r i n s u l a t i o n / / O u t e r r a d i u s o f o u te te r i n s u l a t i o n / /C /CASE 1 : ’ a ’ n e a r t h e p i p e s u r f a c e / / l e t k 1= 1=1 / / T he h e rm r m al a l c o n d u c t i v i t y o f A [W [W/m .K .K ]   k1=1;   // and k2= k2=3k1= 3k1=33 / / T he h e rm r m al a l c o n d u c t i v i t y o f B [W [W/m .K .K ]   k2=3;   // Let dT=1

  r2=r1+ia   r3=r2+ia

  dT=1   Q1=dT/( l o g (r2/r1)/(2*%pi*k1*L)+ l o g (r3/r2)/(2*%pi*k2* L)) 21   Q1=22.12   / / A p p r o x i m a t e 22 / /C /CASE 2 : ’ b ’ n e ar ar t h e p i p e s u r f a c e 23   Q2=dT/( l o g (r2/r1)/(2*%pi*k2*L)+ l o g (r3/r2)/(2*%pi*k1* L)) 24   Q2=24.39   //Approximation 25   printf ( ”ANSWER− ”ANSWER−( i ) \ nQ1=%f W \ nQ2 nQ2= %f W \ nQ1 i s l e s s

t ha h a n Q2 . i . e a r ra r a ng n g e me m e nt n t A n e ar ar t h e p i p e s u r f a c e a n d B as a s o u t er e r l a y e r g i v e s l e s s h e a t l o s s \ s \ n ” ,Q1, 23

Q2); 26   percent=(Q2-Q1)*100/Q1;

/ / p e r ce ce n t r e d u c t i o n i n

h ea ea t l o s s 27   printf ( ” \nANSWER− nANSWER−( i i ) \ n Pe P e rc r c en e n t r e d u c ti t i o n i n h eeaa t l o s s ( w it i t h n e ar a r t h e p i p e s u r f a c e ) =% f p e r c e n t ” , percent);

Scilab code Exa 2.19   Insulation thickness

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clc   clear xa mp mp le le 2 . 1 9 . P ag ag e n o . 2 . 3 9 ” )   printf ( ” E xa

  //Given            

x1=0.224 / / m k1=1.3   // W/(m.K) k2=0.346   // W/(m.K) T1=1588 / / K T 2 = 2 99 99 / / K e t re r e / / h e at at l o s s QA=1830   / / W/ s q m et

// sol uti on ”Q/A=(T1−T 2 ) / x 1 / k 1 + x 2 / k 2 ” ) ;   printf ( ”Q/A=(T1−   x2=k2*((T1-T2)*1/(QA)-(x1/k1))   x2=x2*1000; =% f mm” ,x2)   printf ( ” T h i c k n e s s o f i n s u l a t i o n =%

Scilab code Exa 2.20  Heat loss in furnace

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  / / E x am am pl pl e 2 . 2 0   //Given // f o r c l a y // [W/ (m.K) ]   k1=0.533 / / f o r r ed ed b r i c k // [W [W/m. /m.K K]   k2=0.7 24

7   / / C a se se 1 8   //Area A =1 9   x1=0.125 // [m] 10   x2=0.5 // [m] 11 // Resi sta nces 12   r1=x1/(k1*A)   / / R e s o f f i r e c l a y [ K/ K/W] 13   r2=x2/(k2*A) / / R es es o f r e d b r i c k [ K/ K/W] 14   r=r1+r2 15   / / T e m p e r a t u r e s 16   T1=1373 // [K] 17   T2=323 // [K] 18   Q=(T1-T2)/r // [W/ sq m] 19   Tdash=T1-Q*r1 // [K] 20   / / C a s e 2 21 / / H ea e a t l o s s m u s t r em e m ai a i n u nc n c h an a n ge g e d , T h i ck c k n e ss ss o f   22 23 24 25

   

26 27 28 29 30

         

r e d b r i c k a l s o r ed e d uc u c es es t o i t s h a l f   // [m] x3=x2/2 / / [ K/ K/W] W] r3=x3/(k2*A) // [K] T dd d d = T 2 + ( Q * r3 r3 ) / / T h i c k n e s s o f d i a t o m i t e b e x 2 , km km b e me m ea n conductivity // [K] Tm=(Tdash+Tdd)/2 // [W [W// (m.K ] km=0.113+(0.00016*Tm) // [m] x2=km*A*(Tdash-Tdd)/Q / / [m [mm] m] x2=x2*1000 =% f mm” ,x2) printf ( ” T h i c k n e s s o f d i a t o m i t e l a y e r =%

Scilab code Exa 2.21  Rate of heat loss in pipe

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// Exaample Exaample22 .2 1 //Given / /c / co mmo n b r i c k W/ ( ( m. m . K) K) k1=0.7 / / g yp y p su su m l a y e r [W/ (m ( m . K) K) k2=0.48 R o ck c k wo w o ol ol [W/m . K] K] k3=0.065 / / Ro / / He He a t l o s s w i t h i n s u l a t i o b w i l l b e 2 0% o f w i th t h ou ou t 25

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

 

insulation // s q m A =1 // [m] x1=0.1 x2=0.04   // [m] R1=x1/(k1*A)   //K/W R2=x2/(k2*A)   //K/W   //K/W R=R1+R2 //R3=x3/(k3 ∗A)

         

QbyQd=0.2 sigRbyRd=QbyQd x3=(R/QbyQd-R)/15.4 //m // [mm] x3=x3*1000 printf ( ” T h i c k n es s o f r o ck w oo l i n s u l a t i o n =%f mm” ,x3)

         

Scilab code Exa 2.22  Heat loss from insulated steel pipe

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clc ;   clear ;

  / / E x am pl e 2 . 2 2 / / Steam t e m pe r a tu r e i n [ K]   Ts=451; // A i r t em pe ra t ur e i n [ K]   Ta=294; // I n t e r n a l d i am et e r o f p i pe [mm]   Di=25; // [m]   Di=Di/1000; // O uter d ia me te r o f pi p e [mm]   od=33; // [m]   od=od/1000;   hi=5678; / / I n s i d e h e a t t r a n s f e r c o e f f i c i e n t [W/ (mˆ 2 . K) ] [W/ ( s q   ho=11.36; / / O u t s i d e h e a t t r a n s f e r c o e f f i c i e n t m.K) ] // T hi ck ne ss o f s t e e l p ip e [m]   xw=(od-Di)/2; // k f o r s t e e l i n W/ (m. K)   k2=44.97; / / k f o r r o c k wo o l i n W/ (m. K)   k3=0.175; // t h i c k n e s s o f i n s u l a t i o n i n [   ti=38/1000; m] // [m]   r1=Di/2; 26

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

           

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

/ / Ra te o f h e at l o s s = dT / ( s ig ma R ) // [m]   L=1; // [ K/W]   R1=1/(hi*%pi*Di*L);   R2=xw/(k2*%pi*Dm1*L);   R3=(r3-r2)/(k3*%pi*Dm2*L); // [mm] D o =( o d + 2* t i ) ;   R4=1/(ho*%pi*Do*L);   sigma_R=R1+R2+R3+R4;

// [m]

  / / Heat l o s s        

// [K] dT=Ts-Ta; / / H ea t l o s s [W/m ] Q=dT/sigma_R; printf ( ” \ nAns : R at e o f h e a t l o s s i s %f W/m” ,Q); printf ( ” \n NOTE: S l i g h t v a r i a t i o n i n f i n a l a ns we r due t o l a c k o f p r e c i s i o n i n c a l c u l a t i o n o f R1 , R2 , R3 and R4 . I n book an a pp ro xi ma te v a l u e s o f t h e s e i s t a k e n \n ” )

Scilab code Exa 2.23  Heat loss from furnace

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clc ;

  / / E x am pl e 2 . 2 3 // [K]   T1=913 // [K]   T=513 // [K]   T2=313   //Q=(T1−T ) / ( x / ( k ∗A) )   //Q=(T−T2 ) / ( 1 / ( h∗A) )   //x=2k/h   //Q=(T1−T2) /(x/(kA)+1/(h ∗A) ) // Th er ef or e ,Q=hA/ 3 ∗ ( T1−T2 ) 27

11 12 13 14 15 16 17 18 19 20

  / / With i n c r e a s e i n t h i c k n e s s ( 1 00%)   //x1=4∗k / h   //Q2=(T1−T2 ) / ( x 1 / k ∗A+1/(h∗A) )   //Q2=(h∗A) /5 ) ∗ ( T1−T2 )   //Now   h=1;   //Assume // Assume f o r c a l c u l a t i o n   A=1;   Q1=(h*A/3)*(T1-T2)   Q2=((h*A)/5)*(T1-T2)   percent=(Q1-Q2)*100/Q1

/ / P e rc e nt r e d u c t i o n i n

h ea t l o s s 21   printf ( ” \ n T he r ef o re , P e r ce n t ag e r e d u c t i o n i n h e at l o s s i s %d p e rc e n t ” ,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 19

clc ;   clear ;   printf ( ” E xa mp le 2 . 2 4 \ n Pa ge no . 2 . 4 7 ” );

                         

// giv en L =1 //m thp=2 / / T h i ck n e ss o f p i p e ; i n mm thi=10 // T h ic kn e ss o f i n s u l a t i o n ; i n mm T1=373 //K T2=298 //K id=30 //mm r1=id/2 //mm r2=r1+thp //mm r3=r2+thi //mm // I n S . I u n i t s r1=r1/1000 //m r2=r2/1000 //m r3=r3/1000 //m k1=17.44 //W/(m.K) 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)+( log (r2/r1))/k1+(( log (r3/r2))/k2)+(1/(0.02*ho))) 24   printf ( ”ANSWER: \n R at e o f h e a t l o s s , Q=%f W” ,Q);

Scilab code Exa 2.25   Thickness of insulation

1 2 3 4 5 6 7 8

clc ;   clear ;

  / / E xa mp lr 2 . 2 5 // [W/ sq m.K ]   h = 8. 5 ; // [K]   d T = 17 5 ; // [m]   r2=0.0167; // [W/m]   Q_by_l=h*2*%pi*r2*dT // For i n s u l a t i n g m a t e r i a l i n   k = 0 .0 7 ;

[W/m.

K] 9 // f o r i n s u l a t e d p ip e −−50% r e d uc t i o n i n h ea t l o s s 10   Q_by_l1=0.5*Q_by_l // [ w/m] 11   deff ( ’ [ x]= f ( r 3 ) ’ , ’ x=Q b y l 1 −dT/ (( l o g ( r3 / r2 ) ) /(2 ∗ %p i ∗ k ) + 1 / ( 2∗ %p i ∗ r 3 ∗h ) ) ’ ) 12 13 14 15 16

/ / by t r i a l and e r r o r method we g e t : r3 = fsolve (0.05,f) // t h i c k n e s s o f i n s u l a t i o n i n [m]   t=r3-r2   printf (  ’ \n Hence , r e q u i r e d t h i c k n e s s o f i n s u l a t i o n

%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   / / E x am pl e 2 . 2 6

29

is

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

                         

// C a l cu l a t e h ea t l o s s p er m et r e l e n g t h //Given / / i n t e r n a l d i a m et e r i n [m] id=0.1 / / o u t e r d i a m e te r i n [m] od=0.12 // T em p e r a tu re o f f l u i d [K] T1=358 // T em p e r a tu re o f s u r r o u n d in g [K] T2=298 // t h i c k n e s s o f i n s u l a t i o n [m] t=0.03 // [W/m. K ] k1=58 / /W/ (m. K) i n s u l a t i n g m a t e r i a l k2=0.2 // i n s i d e h ea t t r a n s f e r c o e f f [W/ s q m .K ] h1=720 //W/ sq m.K h2=9 // [m] r1=id/2 // [m] r2=od/2 // [m] r3=r2+t / / Hea t l o s s p e r m et er=Q by L

  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 ( ” Hea t l o s s p e r m et re l e n g t h o f p i p e=%f W” , Q_by_L)

Scilab code Exa 2.27   Mineral wool insulation

1 2 3 4 5 6 7 8 9

clc ;   clear ;

  / / E x am pl e 2 . 2 6 // Given :

// [K] // [K] // [K] // O ut si de h ea t t r a n s f e r 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; // I n t e r n a l r a d i u s [m] 12   r2=0.05; / / O ut er r a d i u s [m ]        

T1=573; T2=323; T3=298; h1=29;

30

13   k 1 = 58 ; // [W/m.K ] 14   k2=0.052; // [W/m.K ] 15   //Q=(T1−T2 ) / ( 1 / ( r 1 ∗ h 1 ) +l o g ( r 2 / r 1 ) / k 1+ l o g ( r 3 / r 2 ) / k 2 )

=(T2−T3 ) / ( 1 / ( r 3 ∗ h2 ) ) 16   deff ( ’ [ x]= f ( r 3 ) ’ , ’x=(T1−T2 ) / ( 1 / ( r 1 ∗ h 1 ) +l o g ( r 2 / r 1 ) / k 1 + l o g ( r 3 / r 2 ) / k 2 ) −(T2−T3 ) / ( 1 / ( r 3 ∗ h2 ) ) ’ ) 17 / / by t r i a l and e r r o r method : 18 19 20 21 22

r3 = fsolve (0.05,f)   t=r3-r2

// T hi ck ne ss o f i n s u l a t i o n i n [m]   //Q=h2 ∗2∗ %p i ∗ r 3 ∗L ∗ ( T2−T3 ) // [W/m]   Q_by_l=h2*2*%pi*r3*(T2-T3)   printf ( ” \n T h ic k n e s s s o f i n s u l a t i o n i s %d mm \n R at e o f h ea t l o s s p er u n i t l e ng t h i s %f W/m” , round ( t *1000),Q_by_l);

Scilab code Exa 2.28   Furnace wall

1 2 3 4 5

clc ;   clear ;

  / / E x am pl e 2 . 2 8   / / C a l c u l a te h ea t l o s s p er s q m and t em p er at u re o f   o u ts i de s u r f ac e   //Given A =1 / / a s su me [ s q m ] // [m]   x1=0.006 // [m]   x2=0.075 // [m]   x3=0.2 // [W/m. K ]   k1=39 // [W/m.K ]   k2=1.1 // [W/m.K ]   k3=0.66 //W/ sq m .K   h0=65 //K   T1=900 //K   T2=300

6 7 8 9 10 11 12 13 14 15 16 17   sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A));

31

18 //To c a l c u l a t e h ea t l o s s / s q m a r ea 19   Q=(T1-T2)/sigma_R // [W/ sq m] 20   printf ( ” Heat l o s s p er s q m et re a r ea i s : %f W/ s q m” ,Q ); 21   //Q/A=T−T2 / ( 1 / h 0 ) , w he re T=Temp o f o u t s i d e s u r f a c e 22 // So , T=T2+Q/ (A∗ h0 ) 23   T=Q/(A*h0)+T2 // [K] 24   printf ( ” Te m p er a t u r e o f u t s i d e s u r f a c e o f f u rn a c e i s : %f K ( %f d e g r e e C ) ” ,T,T-273)

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 ;

  / / E x am pl e 2 . 2 9   // D et e r mi ne n e c e s s a r y t h i c k n e s s o f i n s u l a t i o n b r i c k   //Given / / A ss um e [ s q m ] A =1 // [m]   x1=0.003 // [m]   x3=0.008 // [W/m. K ]   k1=30 // [W/m.K ]   k2=0.7 // [W/m. K ]   k3=40 // [K]   T1=363 // [K]   T=333 // [K]   T2=300 //W/ sq m.K   h0=10   //Q=(T1−T2 ) / ( x 1 / ( k 1∗A)+x2/(k2 ∗A)+x3/(k3 ∗A)+1/(h0 ∗A) ) // Also ,Q=(T−T2) /( 1/ ( h0 ∗A) ) / / So , ( T1−T2 ) / ( ( x 1 / ( k 1 ∗A)+x2/(k2 ∗A)+x3/(k3 ∗A)+1/(h0 ∗ A) )=(T−T2) /( 1/ ( h0 ∗A) ) 20 // or , x2=k2 ∗A ( ( T 1−T2 ) / ( ( T−T2 ) ∗ h0 ∗A) −1/(h0 ∗A)−x 1 / ( k 1 ∗A )−x 3 / ( k 3 ∗A) ) 21   x2=k2*A*((T1-T2)/((T-T2)*h0*A)-1/(h0*A)-x1/(k1*A)-x3

32

/(k3*A));   // [m] 22   printf ( ” T h i c k n e s s o f i n s u l a t i n g mm” ,x2*1000);

b r i c k r e q u i r e d i s %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 26

  clear ; clc ;

  / / E x am pl e 2 . 3 0   //Given // [W/ sq m.K)   hi=75 //m   x1=0.2 // [m]   x2=0.1 // [m]   x3=0.1 // [K]   T1=1943   k1=1.25   //W/m.K   k2=0.074   // /W/m. K   k3=0.555   //W/m.K //K   T2=343 / / a s su me [ s q m ] A =1   sigma_R=1/(hi*A)+x1/(k1*A)+x2/(k2*A)+x3/(k3*A);

  / / Heat l o s s p e r i s q m // [W]   Q=(T1-T2)/sigma_R / / i f T=t e m p e r at u r e b et we en c hr om e b r i c k and k o a l i n b r i c k t he n   //Q=(T1−T ) / ( 1 / ( h i ∗A)+x1/(k1 ∗A) ) // o r T=T1−(Q∗ ( 1 / ( h i ∗A)+x1/(k1 ∗A) ) ) // [K]   T=T1-(Q*(1/(hi*A)+x1/(k1*A)));   printf ( ” Te m p er a t u r e a t i n n er s u r f a c e o f m id dl e l a y e r =%f K( % f d e g r e e C ) ” ,T,T-273); / / i f Tdash=t e m pe r a tu r e a t t h e o u t e r s u r f a c e o f   middel layer , then   //Q=(Tdash−T2 ) / ( x 3 / ( k 1∗A) ) // o r Tdash=T2+(Q∗ x 3 / ( k 3 ∗A) ) 33

27   Tdash=T2+(Q*x3/(k3*A)) // [K] 28   printf ( ” Te m p er a t u r e a t o u te r s u r f a c e o f m id dl e l a y e r =%f K ( %f d e g r e e 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 ;

  / / E x am pl e 2 . 3 1 // C a l c u l a te : ( a ) Heat l o s s p er u n i t l e n g th // ( b ) R ed uc ti on i n h ea t l o s s   //Given //W/ sq m.K   hi=10 //W/ s q .m. K   h0=hi //m   r1=0.09 //m   r2=0.12 // t h i c k n e s s o f i n s u l a t i o n [m]   t=0.05   k1=40   //W/m.K   k2=0.05   //W/m.K //K   T1=473 //K   T2=373

  Q_by_L=2*%pi*(T1-T2)/(1/(r1*hi)+ log (r2/r1)/k1+1/(r2* h0));   //W/m 18   printf ( ” Ans ( a ) H ea t l o s s =%f W/m ” ,Q_by_L) 19 // A f t e r a d d i t i o n o f i n s u l a t i o n : 20   r3=r2+t; // r a di u s o f o u t er s u r f a c e o f i n s u l a i t o n 21   Q_by_L1=2*%pi*(T1-T2)/(1/(r1*hi)+ log (r2/r2)/k1+ log ( // W r3/r2)/k2+1/(r3*h0)); 22 Red= Q_by_L -Q_b y_L1 // R ed uc it on i n he at l o s s i n [W

/m] 23   percent_red=(Red/Q_by_L)*100

/ /% R e d uc t io n i n

h ea t l o s s 24   printf ( ” Ans ( b ) P er ce nt r e d u ct i on i n h ea t l o s s i s %f   p e r c e n t ” ,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 ;

  / / E x am pl e 2 . 3 2 / / D e t e r mi n e : i −Heat f l u x a c r o s s t he l a y e r s and / / i i − I n t e r f a c i a l t em pe ra tu re b et w ee n t he l a y e r s   //Given                    

//K T1=798 //K T2=298 //m x1=0.02 //m x2=x1 k1=60   //W/m.K k2=0.1   //W/m.K //W/ sq m.K hi=100 //W/ sq m.K h0=25 //W/ sq m Q_by_A=(T1-T2)/(1/hi+x1/k1+x2/k2+1/h0); printf ( ”Ans ( i )− Heat f l u x a c r o s s t h e l a y e r s i s %f W / s q m ” ,Q_by_A);

19 / / I f T i s t h e

i n t e r f a c i a l t e m p e r a t u r e b e tw e en s t e e l p l a t e and i n s u l a t i n g m a t e r ia l 20   //Q by A=(T−T2) /( x2/ k2+1/h0 ) 21   T=Q_by_A*(x2/k2+1/h0)+T2 22   printf ( ”Ans−( i i )− I n t e r f a c i a l

t em pe ra t ur e b et we en l a y e r s i s %f K ( %f d eg r e e C) ” ,T,T-273);

Scilab code Exa 2.33  Conductive conductance furnace wall

1

35

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 ;

  / / E x am pl e 2 . 3 3   / / De te rm in e T em pe ra tu re a t t he o u t er s u r f a c e o f w a l l and c o n v e c t i ve c on du ct an ce on t he o u t er w a l l / / T em pe ra tu re o f h o t g a s : //K   T1=2273 / / A mb ie nt a u r t e m p e r a t u r e : //K   T4=318 // Heat f lo w by r a d i a t i o n f rom g a s e s t o i n s i d e s u r fa c e o f w al l : // [W/ sq m]   Qr1_by_A=23260 / / He at t r a n s f e r c o e f f i c i e n t on i n s i d e w a l l : //W/ sq m.K   hi=11.63 // Thermal c o n d u c t i v i t y o f w a l l : //W. s q m/K   K=58 // Heat fl o w by r a d i a t i o n from e x t e r n a l s u r f a c e t o a m b ie n t : //W/ sq m.   Qr4_by_A=9300 / / I n s i d e W al l t e m p er a t u r e : //K   T2=1273   Qr1=Qr1_by_A // s q m A =1

//W

for

//W/ sq m   Qc1_by_A=hi*(T1-T2) / / f o r A=1 s q m   Qc1=Qc1_by_A // T herm al r e s i s t a n c e : / /K/W p e r s q m   R=1/K   //Now Q=(T2−T3) /R, i . e / / E x t e r n a l w a l l t em p T3=T2−Q∗R   / /Q e n t e r i n g w a l l= //W   Q_enter=Qr1+Qc1 //K   T3=T2-Q_enter*R   //Approximate   T3=673   / / Heat l o s s due t o c o n v ec t i o n : //W/ sq m Qc4_by_A =Q_enter -Q r4_by_A 36

37 38 39 40 41

  //Qc4 by A=h0 ∗ ( T3−T4 ) / / o r h 0=Q c 4 b y A / ( T3−T4 ) //W/ sq m.K   h0=Qc4_by_A/(T3-T4) // Res ult   printf ( ” C o n v e c t i v e c o n d u ct a n c e i s : %f W/ s q 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 ;

  / / E x am pl e 2 . 3 4   //Given // [K]   T1=473 // [K]   T2=293   k=0.17   //W/(m.K) //W/ ( sq m.K) h =3 //W/ sq m.K   h0=h //m   rc=k/h // I n s i d e r a d i u s o f i n s u l a i t o n [mm]   r1=0.025   q_by_l1=2*%pi*(T1-T2)/( log (rc/r1)/k+1/(rc*h0))

//

Heat t r a n s f e r w i th i n s u l a t i o n i n W/m 13 / / Wi th ou t i n s u l a t i o n : 14   q_by_l2=h*2*%pi*r1*(T1-T2)   / /W/m 15 inc=(q_by_l1 -q_by_l2) *100/q_by_l2 // I n c r e a s e o f   h ea t t r a n s f e r 16   printf ( ”When c o v e r e d w i th i n s u l a t i o n , \ n h ea t l o s s =%f   W \n When w i th o ut i n s u l a t i o n , h e at l o s s = %f W \n p e r c e n t i n c r e a s e =%f p e r c e n t ” ,q_by_l 1 ,q_b y_l2 ,inc ); 17   k=0.04 / / F i b r e g l a s s i n s u l a i t o n W/ ( s q m. K) 18   rc=k/h // C r i t i c a l r a d i u s o f i n s u l a i t o n 19   printf ( ” I n t h i s c as e t h e a vl ue o f r c=%f m i s l e s s

t h a n t he o u t s i d e r a d i u s o f p i p e ( %f ) , \ n So a dd it on o f any f i b r e g l a s s would c au se a d e cr e a s e i n t h e h e a t t r a n s f e r \n ” ,rc,r1) 37

Scilab code Exa 2.36  Critical radius of pipe

1 2 3 4 5

  clear ; clc ;

  / / E x am pl e 2 . 3 6   / / C a l cu l a t e t he h ea t l o s s p er m e t r e o f p i p e and o u t er s u r f a c e t em p er a tu re   //Given / / Th er ma l c o n d u c t i v i t y i n [W/ s q m .K ] k =1 // Het t r a n s f e r c o e f f i n W/ s q m.K h =8   rc=k/h // C r i t i c a l r a di u s i n m //K   T1=473 //K   T2=293 // O uter r a d i u s =i n n er r a d i u s i n [m]   r1=0.055

6 7 8 9 10 11 12 13   Q_by_L=2*%pi*(T1-T2)/( log (rc/r1)/k+1/(rc*h)) 14   printf ( ” Heat l o s s p er m et er o f p i pe i s %f W/m” , Q_by_L) 15   / / For o u te r s u r f a c e 16   / / Q b y L = 2∗%p i ∗ (T−T2) /(1 / rc ∗h ) 17 / / i m p l i e s t h a t , T=T2+Q b y L / ( r c ∗2∗ %pi) 18   T=T2+Q_by_L/(rc*2*%pi*h) //K 19   printf ( ” O ut er s u r f a c e t em p er a tu r e i s : %f K( %f d e g re e C) ” ,T,T-273)

Scilab code Exa 2.37  Time required for steel ball

1 2 clc ; 3   clear ; 4   / / E x am pl e 2 . 3 7

38

5   / / C a l c u l a t e t h e t i m e r e qu i r e d f o r a b a l l t o a t t a i n a 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

 

t em p er a tu re o f 4 23 K //Given k_steel=35   //W/m.K Cp_steel=0.46   / / k J / ( k g ∗K) Cp_steel=Cp_steel*1000   / / J / ( k g ∗K) //W/ sq m.K h=10 //kg/ cubi c m rho_steel=7800 dia=50   //mm //m dia=dia/1000 // r a d i u s i n m R=dia/2 // Area i n s q m A=4*%pi*R^2 / / Volume i n c u b i c m et er V=A*R/3

                      Nbi=h*(V/A)/k_steel

  / / A s N bi < 0 .1 0 , i n t e r n a l temp g r a d i e n t i s n e g l i g i b l e //K   T=423 //K   T0=723 //K   T_inf=373   / / ( T−T i n f ) / ( T 0−T i n f ) =e ˆ (−h∗A t / r h o ∗Cp∗V)

  t=-rho_steel*Cp_steel*R* log ((T-T_inf)/(T0-T_inf)) // s /(3*h); 24   printf ( ” Time r e q u i r e d f o r a b a l l t o a t t a i n a t em p er a tu r e o f 42 3 K i s %f s= %f h ” ,t,t/(3600))

Scilab code Exa 2.38   Steel ball quenched

1 2 3 4 5 6 7 8 9

clc ;   clear ;

  / / E x am pl e 2 . 3 8   //Given   dia=50   //mm //m   dia=dia/1000 // r a d i u s i n m   r=dia/2 //W/ sq m.K   h=115 39

10 11 12 13 14 15 16 17 18 19

               

rho=8000 / / k g / c u b i c m //kJ/ kg .K Cp=0.42 Cp=Cp*1000   / / J / ( k g ∗K) // Area i n s q m A=4*%pi*r^2 / / Volume i n c u b i c m V=A*r/3 //K T=423 //K T_inf=363 //K T0=723

  / / ( T−T i n f ) / ( T 0−T i n f ) =e ˆ ( −3 ht /( rho ∗Cp∗ r ) )   t=-rho*Cp*r* log ((T-T_inf)/(T0-T_inf))/(3*h);

//

Time i n s e c o nd s 20   printf ( ” Time t a ke n by c e n t r e o f b a l l t o r ea ch a t em p er a tu r e o f 42 3 K i s %f s (=%f m in ut es ” ,t,t /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 18

clc ;   clear ;

  / / E x am pl e 2 . 3 9   //Given //W/ sq m.K   h=11.36   k=43.3   //w/(m.K) / / r a d i u s i n mm   r=25.4 // ra d i u s i n m   r=r/1000 // Area o f s p h er e [ s q m]   A=4*%pi*r^2 / / Volume i n [ c u b i c m]   V=A*r/3 //kg/ cubic m   rho=7849 //J/ kg .K   Cp=0.4606*10^3 t =1   / / h o u r // seco nds   t=t*3600 // [K]   T_inf=394.3 // [K]   T0=700   / / (T−T i n f ) / ( T 0−T i n f )=e ˆ ( −3∗h∗ t / r h o ∗Cp∗V) 40

19   T=T_inf+(T0-T_inf)*(%e^((-h*A*t)/(rho*Cp*V))); 20   printf ( ” T em per at ur e o f b a l l a f t e r 1 h= %f K ( %f   d e g r e e 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 ;

  / / E x am pl e 2 . 4 0   //Given   rho=9000; / / k g / c u b i c m //kJ /( kg .K)   Cp=0.38; //J /( kg .K)   Cp=Cp*1000   k=370;   //W/m.K //W/ sq m.K   h=90;   l=400;   //mm // l e n g th o f c op pe r s l a b   l = l / 1 00 0 ; // t h i c k n e s s i n [m]   t = 5 / 10 00 ; // A rea o f s l a b   A=2*l^2 / / Volume i n [ c u b i c m]   V=t*l^2 // [m]   L_dash=V/A   / / f o r s l ab o f t h i ck n es s 2x   //L dash=x // [m]   L _ d a sh = 0 .0 2 5 ; //<   0 . 1 0   Nbi=h*L_dash/k   var=h*A/(rho*Cp*V)

  / / A s N bi < 0 . 10 , we c an a p pl y lumped c a p a c i t y a n a l y s i s // [K]   T=363 // [K]   T_inf=303 // [K]   T0=523

  t=-( log ((T-T_inf)/(T0-T_inf)))/var   printf ( ” Time a t w hi ch s l a b t e mp e r at u r e b ec om es 3 63 K i s %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 17 18 19 20 21 22

clc ;   clear ;

  / / E x am pl e 2 . 4 1   //Given / / k g / c u b i c m e te r   rho=9000 //kJ /( kg .K)   Cp=0.38 //J/ kg .K   Cp=Cp*1000   k=370   //W/(m.K) //K   T0=483 //K   T_inf=373 //K   delta_T=40 //K   T=T0-delta_T / / t i me i n [ m i n ut e s ] t =5 // [ sec on ds ]   t=t*60 //A=2A . . . . . Two f a c e s //V=A. 2 x // 2x=t h i c k n e s s o f s l a b =30 mm=0 .0 3 // [m]   x=0.015 // t h i c k n e s s o f s l a b   th=2*x

m

  h=-rho*Cp*x* log ((T-T_inf)/(T0-T_inf))/t   printf ( ” H e a t t r a n s f e r c o e f f i c i e n t i s : %f W/ ( s q m . K) ” ,h )

Scilab code Exa 2.42  Stainless steel rod immersed in water

1 2   clear ; 3 clc ; 4   / / E x am pl e 2 . 4 2

42

5   //Given 6   rho=7800 / / [ kg p e r c u b i c m] 7   h=100 //W/ ( s q m. K) C o n ve c ti v e h e a t t r a n s f e r

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

//J /( kg .K)   Cp=460   k=40   //W/(m.K) // [m] l e n g t h o f r o d L =1   D=10   //mm / / d i a m e t e r i n [ m]   D=D/1000 // r a i d u s i n [m]   R=D/2   / / For c y l i n d r i c a l r od : / / Area i n [ s q m]   A=2*%pi*R*L / / Volume i n [ c u b i c m ]   V=%pi*R^2*L // [m]   L_dash=V/A / / B i o t n um be r   Nbi=h*L_dash/k / / N bi < 0 .1 0 , H ence lumped h ea t c a p av i t y i s // [K]   T=473 // [K]   T_inf=393 // [K]   T0=593

  t=-rho*Cp*V* log ((T-T_inf)/(T0-T_inf))/(h*A)   printf ( ” Time r e q u i r e d t o r e ac h t em p er a tu re %f i s %f   s ” ,T,t);

Scilab code Exa 2.43   Chromel alumel thermocouple

1 2 3 4 5 6 7 8 9 10

pos sibl e

  clear ; clc ;

  / / E x am pl e 2 . 4 3   //Given // [ kg/ cu bi c m]   rho=8600 //kJ /( kg .K)   Cp=0.42 //J /( kg .K)   Cp=Cp*1000 // [mm]   dia=0.71 / / [ d i a i n m]   dia=dia/1000 43

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

/ / r a d i u s [m] / / c o n v e c t i v e c o e f f W/ ( s q m .K)   / / L e t l e n g t h =L=1 // [m] L =1

  R=dia/2   h=600

       

A=2*%pi*R*L; V=%pi*(R^2)*L; tao=(rho*Cp*V)/(h*A); printf ( ” Time c o n st a n t o f t he t he rm oc ou pl e i s %f s ” , tao);

// at   t=tao

/ /From

( T−T i n f ) / ( T 0−T i n f )=e ˆ (− t / t a o ) / / R a t i o o f t h e r mo c o u p l e   ratio=%e^(-t/tao) d i f f e r e n c e t o i n i t i a l t em pe ra t ur e d i f f e r e n c e 23   printf ( ” At t he end o f t he t im e p e r i od t=t ao=%f s , T em pe ra tu re d i f f e r e n c e b /n t h e t h er m oc o up l e and t h e g a s s t re a m w ou ld b e %f o f t h e i n i t i a l t em p er a tu r e d i f f e r e n c e ” ,tao,ratio); 24   printf ( ” \n I t s h ou l d be r eo r d e r e d a f t e r %f s ” ,4*tao) ;

Scilab code Exa 2.44   Thermocouple junction

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

clc ;   clear ;

  / / E x am pl e 2 . 4 4 //kg/ cubic m   rho=8000 //J /( kg .K)   Cp=420 / / f o r h o t s t re a m W/ ( s q m .K)   h_hot=60 // [mm]   dia=4   t=10;   r=dia/(2*1000)

// r a d i u s i n [m]

  / / F or s p h e r e   V=(4/3)*%pi*r^3

/ / Volume i n [ c u b i c m] 44

13   A=4*%pi*r^2 / / Volu me i n [ s q m ] 14   tao=rho*Cp*V/(h_hot*A) / / Time c o n s t a n t i n [ s ] 15   ratio=%e^(-t/tao) // %eˆ(− t / t a o ) = (T−T− i n f ) /( T0−

T inf ) 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

// [K]   T_inf=573 // [K]   T0=313   T=T_inf+ratio*(T0-T_inf)   //ANS−[ i ]   printf ( ” \n Answer : Time c o n s t a n t o f t h er m oc o up l e i s % f s ” ,tao); //IN STILL AIR : //W/( sq m .K)   h_air=10          

// [ s ] tao_air=rho*Cp*V/(h_air*A) // [ s ] t_air=20 ratio_air=%e^(-t_air/tao_air) // [K] T_inf_air=303 T0_air=T; T_air=T_inf_air+ ratio_air*(T0_air -T_inf_air)

  //ANS−[ i i ]   printf ( ” Te m p er a t u r e a t t a in e d by j u n c t i o n 20 s a f t e r r em ov in g f ro m t h e h ot a i r s tr ea m i s : %d K” , round ( T_air))

Scilab code Exa 2.45   Batch reactor

1 2 3 4 5 6 7 8 9

clc ;   clear ;

  / / E x am pl e 2 . 4 5            

T_inf=390; U=600; Ac=1; Av=10 m=1000; Cp=3.8*10^3;

// [K] // [W/ sq m.K ] // [ sq m] // V e s s e l a r ea i n [ s q m] // [ kg ] // [ J/ kg .K] 45

10 11 12 13

// [K] // [K] // [W/ sq m.K ] / / Heat g a in e d fr om t he s team=Rate o f i n c r e a s e o f   i n t e r n a l e ne rg y //U∗A∗ ( T i n f  −T)=m∗Cp∗dT deff ( ’ [ x] = f ( t ) ’ , ’ x= l o g ( ( T i n f  −To) /( T in f −T) )−U∗Ac∗ t / (m∗Cp) ’ ); // [ in s ] t= fsolve (1,f); // [ in s ] t= round (t )

  To=290;   T=360;   h=8.5

 

14 15  

16 17 18   Ts=290; 19   printf ( ” \ nTime t ak en t o h ea t t he r e a c t a n t s o v er t he same t em p er a tu r e r an ge i s %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 i n f  −Tmax)−h∗Av ∗ (Tmax− Ts) ’ ) 23   T_max= fsolve (1,fx) 24   printf ( ” \nANS : In CASE 1\ nTime t ak en t o h ea t t he r e a c t a n t s = %f s . i e %f h \n ” ,t,t/3600); 25   printf ( ” \nANS : In CASE 2 \n Time t ak en t o h ea t t he r e a c t a n t s = %f s \n ” ,t1); 26   printf ( ” \nANS . : Maximum t e m p e r a t u r e a t w h i c h t em pe ra tu re ca n be r a i s e d i s %f K\n ” ,T_max);

Scilab code Exa 2.46  Heat dissipation by aluminium rod

1 2 3 4 5 6 7

clc ;   clear ;

  / / E x am pl e 2 . 4 6 // [mm]   dia=3 // [m]   dia=dia/1000 / / r a d i u s i n [ m]   r=dia/2 46

8 9 10 11 12 13 14

             

k=150   //W/(m.K) //W/ ( sq m.K) h=300 // [K] T0=413 // [K] T_inf=288 / / Area i n [ s q m] A=%pi*(r^2) // [W/ sq m.K ] P=%pi*dia // Heat d i s s i p a t e d Q=(T0-T_inf)* sqrt (h*P*k*A)

i n [W

] 15   printf ( ” Heat d i s s i p a t e d by t he r o d i s %f 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 ;

  / / E x am pl e 2 . 4 7   //Given   k=200   //W/(m.K) //W/( sq m.K)   h=15 // [K]   T0=523 // [K]   T_inf=288            

theta_0=T0-T_inf // di am et er [mm] dia=25 // dia met er [m] dia=dia/1000 // r a d i u s i n [m] r=dia/2 // [m] P=%pi*dia // [ sq m] A=%pi*r^2

// For i n s u l a t e d f i n : m= sqrt (h*P/(k*A)) // l e n g t h o f r od i n [mm]   L=100 // l e n g t h o f r o d i n [m]   L=L/1000   Q=theta_0* tanh (m*L)* sqrt (h*P*k*A)

// H eat l o s s

  //ANSWER−1   printf ( ” Heat l o s s by t h e i n s u l a t e d r o d i s %f W \n ” ,Q )

47

23 nf = tanh (m*L)/(m*L) 24   25   26 27 28   29   30  

// Fi n e f f i c i e n c y

f or

i n su l at e d f i n //ANSWER−2 printf ( ” Fi n e f f i c i e n c y i s %f p er ce nt \n ” ,nf*100) / /At t h e end o f t h e f i n : t h e t a / t h e t a 0 =( c o sh [m( L−x ) ] / co sh (mL) ) / / a t x=L , t h e t a / t h e t a 0 = 1/ ( c o s h ( mL ) // [K] T=T_inf+(T0-T_inf)*(1/ cosh (m*L)) //ANSWER−3 printf ( ” Te m p er a t u r e a t t he end o f t he f i n i s %f K \n ” ,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 21

clc ;   clear ;

  / / E x am pl e 2 . 4 9   //Given   k=300   //W/(m.K) //W. ( sq m.K)   h=20 // [m]   P=0.05 / / [ s q cm ] A =2 // [ sq m]   A=A/10000 // [K]   T0=503 // [K]   T_inf=303 // [K]   theta_0=T0-T_inf m= sqrt (h*P/(k*A))

  / /CASE 1 :

6 F i ns o f 1 00 mm l e n g t h // L en g th of f i n i n [m]

  L1=0.1 Q= sqrt (h*P*k*A)*theta_0* tanh (m*L1)

  / / For 6 f i n s // f o r 6 f i n s [W]   Q=Q*6   / /CASE 2 : 10 f i n s o f 60 mm l e n g th // [mm]   L2=60 48

// [W]

22 23 24 25

// [m]   L2=L2/1000 // [W] Q2 = sqrt (h*P*k*A)*theta_0* tanh (m*L2); // For 10 f i n s   Q2=Q2*10   printf ( ”As ,Q f o r 10 f i n s o f 60 mm l e n g t h ( %f W) i s more t ha n Q f o r 6 f i n s o f 1 00 mm l e n g t h ( %f W) . \ n Th e a g r e e m e n t −−>10 f i n s o f 60 mm l en g t h i s more e f f e c t i v 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 25

clc ;   clear ;

  / / E x am pl e 2 . 5 0   //Given //W/( sq m.K)   h_oil=180 //W/ ( sq m.K)   h_air=15 // [K]   T_oil=353 // [K]   T_air=293 // [K]   delta_T=T_oil-T_air; // Conduc tivity in [W/ (m .K) ]   k=80 // [m]   for_section=11*10^-3 // [mm]   L=25 // [m]   L=L/1000 // [m] Width , . . l e t W =1 // [mm] t =1 // [m]   t=t/1000 // [m]   A=W*t   P=2*t // s q m   Af=2*L*W N =1 Ab= for_section -A

// [ sq m]   / /CASE 1 : Fi n on o i l s i d e o n ly m= sqrt (h_oil*P/(k*A))   nf_oil= tanh (m*L)/(m*L)   Ae_oil=Ab+nf_oil*Af*N

// [ sq m] 49

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

// [W] 27   printf ( ” I n o i l s i d e , Q=%f W\n” ,Q); 28   //CASE 2 : Fi n o n a i r s i d e o nl y 29 30 31 32 33

       

m= sqrt (h_air*P/(k*A)) nf_air= tanh (m*L)/(m*L)   //Approximation nf_air=0.928 // [ sq m] Ae_air=Ab+nf_air*Af*N Q=delta_T/(1/(h_oil*for_section)+1/(h_air*Ae_air))

// [W] 34   printf ( ” I n a i r s i d e , Q=%f W” ,Q); 35   printf ( ” \n From a bo ve r e s u l t s we s e e t h at more h ea t t r a n s f e r t ak es p l a c e i f f i n s a re p ro vi de d on t h e a i r s i d 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 18

clc ;   clear ;

  / / E x am pl e 2 . 5 1   //Given / / T he rm al c o n d u c t i v i t y [W/ (m .K) ]   k=75 // [K]   T_water=363 // [K]   T_air=303 // delta T dT= T_water -T_air / / f o r w a t e r [W/ ( s q m . K) ]   h1=150 / / f o r a i r [W/ ( s q m .K) ]   h2=15 / / Width o f w a l l [ m]   W=0.5 // [m]   L=0.025 / / B as e A re a [ s q m]   Area=W^2 // [mm] t =1 // [m]   t=t/1000 // [mm]   pitch=10 // [m]   pitch=pitch/1000 50

19   N=W/pitch // [ No o f f i n s ] 20 / / C a l c u l a t i o n s 21   A=N*W*t / / T o t a l c r o s s − s e c t i o n a l

a r e a of f i n s

[ sq m] 22 23 24 25 26 27 28 29 30 31 32 33 34 35

[ s q m]

//CASE 1 : HEAT TRANSFER WITHOUT FINS // [ sq m]   A1=Area // [ sq m]   A2=A1 // [W]   Q=dT/(1/(h1*A1)+1/(h2*A2));   printf ( ” \ nWithout f i n s ,Q=%f W\n ” ,Q);   / /CASE 2 : F i ns on t h e w at er s i d e   P=2*(t+W);   A=0.5*10^-3; m= sqrt (h1*P/(k*A)) // E f f e c i e n c y on w a te r s i d e   nfw= tanh (m*L)/(m*L) // E f f e c t i v e a re a on t h e w a te r   Aew=Ab+nfw*Af*N

side 36 37 38 39 40 41 42 43 44 45 46

// [ sq m] // S u r f a c e a r e a o f f i n s

  Ab=Area-A   Af=2*W*L

[ s q m]

// [W]   Q=dT/(1/(h1*Aew)+1/(h2*A2));   printf ( ” \n With f i n s on w a te r s i d e , Q=%f W \n ” ,Q); //CASE 3 : FINS ON THE AIR SIDE        

m= sqrt (h2*P/(k*A)) // Effec iency nf_air= tanh (m*L)/(m*L) // E f f e c t i v e a r ea on a i r s i d e Aea=Ab+nf_air*Af*N // [W] Q=dT/(1/(h1*A1)+1/(h2*Aea)); printf ( ” \n With F i n s o n A i r s i d e , Q=%f W \n ” ,Q)

//BOTH SIDE : // [W]   Q=dT/(1/(h1*Aew)+1/(h2*Aea));   printf ( ” \n With F i n s o n b o th s i d e , 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 21

clc ;   clear ;

  / / E xa mp le 3 . 1 //N. s /mˆ2   / /At d i s t a n c e y fro m s u r f a c e   //ux=a+by+cyˆ2+dyˆ3   / / At y =0 , ux=0 t h e r e f o r e a=0 // i . e tao=0 / /At e dg e o f b ou nd ar y l a y e r , i e y=d e l   / / u x= u i n f   // At y=o , c=0 // At y=de l , ux=b ∗ d el+d∗ d e l ˆ 3   mu=10^-3

/ / T h e r e f o r e , b=−3∗d∗ d e l ˆ 3   //d=−u i n f / ( 2∗ d e l ˆ 2 )   //b=3∗ u i n f / ( 2∗ d e l ) // For v e l o c i t y p r o f i l e , we h ave : / / d e l / x = 4 . 6 4 ∗ ( Nr e x ) ˆ( −1/2) / / E v al u a te N r e x 52

22 23 24 25 26 27 28 29 30

         

// [mm] x=75; // [m] x=x/1000; // [m/ s ] u_inf=3; / / [ kg /mˆ 3 ] f o r a i r rho=1000 / / R e y n o l d n um be r Nre_x=u_inf*rho*x/mu // S u b s t i t u t i n g t he v al ue , we g e t

// [m]   del=x*4.64*(Nre_x^(-1/2))   printf ( ” \ nBoundary l a y e r t h i c k n e s s i s d e l=%f m o r %f   mm” ,del,del*1000); 31   printf ( ” \nWrong u n i t s i n a n sw er o f boo k , m and mm a r e w r on g l y i n t e r c h a n g e d ” );

Scilab code Exa 3.2  Boundary layer thickness of plate

1 2 3 4 5 6 7 8 9

clc ;   clear ;

// Example3 . 2   //Given // s q m / s   mu=15*10^-6 v =2   //m/s // [m] l e ng t h o f p l a t e L =2   Nre_x=3*10^5 / / c r i t i c a l l e n g t h a t w h ih c t h e   xc=Nre_x*mu/v t r a n s i t i o n t ak es p l ac e // S i n ce xc i s l e s s t h a n 2 m. T h e r e fo r e t h e f lo w i s laminar // a t any d i s t a n c e x , . i t i s c a l c u l a t e d fro m // d el /x =4. 64/ ( s q r t (NRe , x ) ) // At x=L=2 m

10 11 12 13 14 15 16 17

       

Nre_l=v*L/mu del_l=4.64*L/ sqrt (Nre_l) // [mm] del_l=del_l*1000 printf ( ” Boundary l a y e r t h i c k n e s s i s %f mm” ,del_l);

53

a t t h e t r a i l i n g e d ge

Scilab code Exa 3.3  Thickness of hydrodynamic boundary layer

1 2 3 4 5 6 7 8 9 10

clc ;   clear ;

  / / E xa mp le 3 . 3   //Given // K in em at i c v i s c o s i t y   mu=15*10^-6 // [m]   x=0.4 // [m/ s ]   u_inf=3

i n [ sq m / s ]

/ / At x = 0 . 4 m, Nre_x =u_inf *x/ mu ;   printf ( ” S i n c e Nre , x ( %f ) i s

L e ss t ha n 3 ∗ 1 0 ˆ 5 , . . t h e b ou nda ry l a y e r i s l a mi n ar ” ,Nre_x); 11   del=4.64*x/ sqrt (Nre_x) // [m] 12   del=del*1000 // [mm] 13   printf ( ” \ n T h i c kn e s s o f b ou nd ar y l a y e r a t x=%f m =%f   mm\n ” ,x,del); 14   Cf_x=0.664/ sqrt (Nre_x); 15   printf ( ” L o c a l s k i n 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 8

clc ;   clear ;

  / / E xa mp le 3 . 4          

mu=1.85*10^-5 P=101.325; M_avg=29; R=8.31451; T=300;

// [ kg /(m. s ) ] / / P r e s s u r e i n [ kPa ] // Avg m o l ec u l ar wt o f a i r / / Gas c o n s t a n t // [K] 54

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

29

30

  rho=P*M_avg/(R*T)   u_inf=2

// [ kg/mˆ 3 ] / / V i s c o s i t y i n [m/ s ]

/ / At x =20 cm = 0 .2 m        

// [m] x=0.2; // [ Reyn old s number ] Nre_x=rho*u_inf*x/mu / / [ B o un da ry l a y e r ] del_by_x=4.64/ sqrt ( N r e _ x ) // [m] del=del_by_x*x / / d e l = d e l ∗ 1 0 0 0

/ / [mm]

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

// < 3∗10ˆ5

  / / Boundary l a y e r i s l a mi n ar   del_by_x=4.64/ sqrt ( N r e _ x )   del1=del_by_x*x

// [m] / / [mm] //Del

/ / d e l 1 =d e l 1 ∗ 1 0 0 0

    d=del1-del   function m_dot=f(y),m_dot= u_inf*(1.5*(y/d) -0.5*(y/d) ^3)*rho, endfunction   m_dot= intg (0,d,f)   printf ( ” \ nBoundary l a y e r t h i c k n e s s a t d i s t a n c e 20 cm f ro m l e a d i n g e d ge i s %f m=%f mm\n ” ,del,del*1000) ;   printf ( ” \ nBoundary l a y e r t h i c k n e s s a t d i s t a n c e 40 cm f ro m l e a d i n g e d ge i s %f m=%f mm\n ” ,del1,del1 *1000);   printf ( ” \ nThus , Mass f l o w r a t e e n t e r i n g t h e b ou nd ar y l a y e r i s %f kg / s ” ,m_dot);

Scilab code Exa 3.5  Rate of heat removed from plate

1 2 clc ; 3   clear ; 4   / / E xa mp le 3 . 5

55

5 6 7 8 9 10 11 12

  //Given // K in em at ic v i s c o s i t y i n sq m/ s mu=3.9*10^-4 / / T he rm al c o n d u c t i v i t y i n W/ (m .K) k=36.4*10^-3 Npr=0.69 // [m/ s ] u_inf=8 // L eng h t o f p l a t e i n [m] L =1   Nre_l=u_inf*L/mu

       

  / / S in ce N r e l i s l e s s t h a n 3 ∗ 1 0ˆ 5 , t he f l ow i s l am in ar o ve r t h e e n t i r e l e n g th o f p l a te 13   Nnu=0.664* sqrt (Nre_l)*Npr^(1.0/3.0)   //=hL/k 14 15 16 17 18 19 20 21

             

22   23 24   25  

/ /w/ s q m . K h=k*Nnu/L / / A p pr o xi m at i on [W/ s q m .K ] h=3.06 // [K] T_inf=523 // [K] Tw=351 // Width o f p l a t e [m] W=0.3 // A rea i n [ s q m] A=W*L // Rate o f h ea t r em ov al f rom o ne Q=h*A*(T_inf-Tw) s i d e i n [W] printf ( ” \ nRate o f h ea t r em ov al i s %f W\n ” ,Q) / / fr om two s i d e : // [W] Q=2*Q printf ( ” \n % f W h e at s h o ul d b e rem oved c o n t i n o u s l y f ro m t h e p l a t e ” ,Q);

Scilab code Exa 3.6  Heat removed from plate

1 2 3 4 5 6 7 8

clc ;   clear ;

  / / E xa mp le 3 . 6        

/ / P r e s s u r e i n [ kPa ] P1=101.325 // K i ne ma ti c v i s c o s i t y mu1=30.8*10^-6 // [W/ (m.K) ] k=36.4*10^-3 Npr=0.69 56

in [ sq m /s ]

9 10 11 12 13 14 15 16

         

17 18 19 20

 

   

 

/ / V e l o c i t y i n [m/ s ] //kJ /( kg .K) // L en g t h o f p l a te i n [m] / / Width i n [m ] // A rea i n [ s q m] / / A t c o n s t a n t t e m p e r a t u r e : mu1 /mu2=P2 / P1 // [ kPa ] P2=8 // K i n em at ic v i s c o s i t y a t P2 i n [ s q mu2=mu1*P1/P2 m/s ] //Reynold ’ s no . Nre_l=u_inf*L/mu2 / / S in ce t h i s i s l e s s th a n 3 ∗ 1 0 ˆ 5

u_inf=8 Cp=1.08 L=1.5 W=0.3 A=L*W

  Nnu=0.664* sqrt (Nre_l)*(Npr^(1.0/3.0))   h=Nnu*k/L / / H ea t t r a n s f e r c o e f f f i c i e n t

i n [W/ s q m .

K] 21 22 23 24

       

/ / A p p r o xi m a t io n i n [W/ s q m .K ] h=2.5 // [K] T_inf=523 // [K] Tw=353 / / He at r em ov ed fr om bo th s i d e s Q=2*h*A*(T_inf-Tw)

i n [W] 25   printf ( ” Rate o f h ea t removed from b ot h s i d e s o f   p l a t e i s %f W” ,Q);

Scilab code Exa 3.7  Local heat transfer coefficient

1 2 3 4 5 6 7 8 9 10 11

clc ;   clear ;

  / / E xa mp le 3 . 7                

//kg/ cubi c m rho=0.998 // [ sq m/ s ] v=20.76*10^-6 // [ kJ/ kg .K] Cp=1.009 // [W/m.K ] k=0.03 // [m/ s ] u_inf=3 // [m] x=0.4 // [m] w=1.5 // R e y no ld s no a t x =0 .4 m Nre_x=u_inf*x/v 57

12   / / S in ce

t h i s i s l e s s th a n 3 ∗ 1 0 ˆ5 . The f l ow i s l a mi n ar u pt o x = 0. 4 m 13   mu=rho*v // [ kg /(m. s ) ] 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

           

// [ kJ/ kg .K] Cp=1.009 // [ J/ kg .K] Cp=Cp*1000 k=0.03   //W/(m.K) Npr=Cp*mu/k Nnu_x=0.332*( sqrt (Nre_x))*(Npr^(1.0/3.0)) // [W/ (m.K) ] hx=Nnu_x*k/x

  // A ve r a g e v a l u e i s t w i c e t h i s v a l u e // [W/ (m.K) ]   h=2*hx   //Approximation   h=10.6 // A rea i n [ s q m]   A=x*w // [ k ]   Tw=407 // [K]   T_inf=293 // [W]   Q=h*A*(Tw-T_inf)   / /From b ot h s i d e s o f t he p l a t e : // [W]   Q=2*Q   printf ( ” The h e at t r a n s f e r r e d from b o t h s i d e s o f t he p l a t e i s %d W” , round (Q));

Scilab code Exa 3.8  Width of plate

1 2 3 4 5 6 7 8 9 10 11

clc ;   clear ;

  / / E xa mp le 3 . 8              

// [ kg/ cu bi c m] rho=0.998 // [ sq m/ s ] v=20.76*10^-6 // [W/m.K ] k=0.03 Npr=0.697 // [m] fro m l e a d i n g e d ge o f t he p l a t e x=0.4 // [m/ s ] u_inf=3 / / R e yn o ld nu mebr a t x = 0 . 40 m Nre_x=u_inf*x/v

  / / S in ce t h i s i s l e s s th a n 3 ∗ 1 0 ˆ 5 58

12 13 14 15

  / / t h e r e f o r e f l o w i s l a m in ar and

16 17 18 19 20 21 22 23 24 25

 

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

       

   

/ / Av er ag e h e at t a r n s f e r value // [W/ sq m.K ] h=2*hx // Given : // [W] Q=1450 // [K] Tw=407 // [K] T_inf=293 // [m] L=0.4 //Q=h∗w∗L ∗ (Tw−T i n f ) //L=Q/(h∗w∗ (Tw−T i n f ) )

c o ef f i c i en t is twice thi s

// [m]   w=Q/(h*L*(Tw-T_inf))   printf ( ” \n W idth o f p l a t e i s %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 ;

  / / E xa mp le 3 . 9              

// V i s c o s i t y f o r a i r [ s q m. / s ] v=17.36*10^-6 / / f o r a i r . . [ W/ (m . K) ] k=0.0275 // [ kJ /( kg .K) ] Cp=1.006 // f o r a i r Npr=0.7 // [m/ s ] u_inf=2 // [m] x=0.2 // R ey no ld s number a t x =0.2 m Nre_x=u_inf*x/v

  / / S in ce t h i s i s l e s s th a n 3 ∗ 1 0 ˆ 5   Nnu_x=0.332* sqrt (Nre_x)*(Npr^(1.0/3.0)) // [W/( sq m. K]   hx=Nnu_x*k/x

  / / Av e r a g e v a lu e o f h e a t t r a n s f e r c o e f f i s t wi ce t h i s value 16   h=2*hx // [W/ sq m. K) ] 59

17 18 19 20 21 22 23 24 25 26

  //Approximation   h=12.3 // w i dt h i n [m] w =1 // [ s q m] Ar ea o f p l a t e   A=x*w // [K]   Tw=333 // [K]   T_inf=300 / / Hea t f l o w i n [W]   Q=h*A*(Tw-T_inf)   printf ( ” \nANSWER: \ nHeat f l o w i s : %f W\n ” ,Q ) // From b ot h s i d e s o f p l a t e : // [W]   Q=2*Q   printf ( ” \nANSWER\n H eat f l o w from b o t h s i d e s o f   p l a t e i s %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 ;

  / / E x am pl e 3 . 1 0              

// [ sq m./ s ] v=16.96*10^-6 // [ kg/ cu bi c m] rho=1.128 / / P r a n d t l n um be r Npr=0.699 // [W/m.K ] k=0.0276 // [m/ s ] u_inf=15 // [m] L=0.2 // Reynold ’ s number Nre_l=L*u_inf/v

  / / S in ce t h i s i s l e s s th a n 3 ∗ 1 0 ˆ 5 , t h e b ou nd ar y l a y e r i s l am in ar o ve r e n t i r e l e n g t h          

Nnu=0.664* sqrt (Nre_l)*(Npr^(1.0/3.0)) // [W/ sq m.K ] h=Nnu*k/L // A rea i n [ s q m] A=L^2 // [K] Tw=293 // [K] T_inf=333

// Rate o f h ea t t r a n s f e r from BOTH s i d e s i s : // [W]   Q=2*h*A*(T_inf-Tw)   printf ( ” Rate o f h ea t t r a n s f e r fro m bo th s i d e s o f   p l a t e i s %f W\n ” ,Q); 60

20 / / i i −With t u r b u l e nt b oun da ry l a y e r fro m t he l e a d i n g

edg e : 21   h=k*0.0366*(Nre_l^(0.8))*(Npr^(1.0/3.0))/L

// [

W/( sq m.K) ] 22   // Heat t r a n s f e r fro m b o t h s i d e s i s : 23   Q=2*h*A*(T_inf-Tw) // [W] 24   printf ( ” \ nThese c a l c u l a t i o n s s h o t ha t t he t ha t t r a n s f e r r a t e i s a p p ro xi m a te ly d ou bl ed i f   b ou n da ry l a y e r i s t u r bu l e nt fro m t he l e a d i n g e dg e \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 ;

  / / E x am pl e 3 . 1 1                                    

// [ kg /(m. s ) ] mu=1.906*10^-5 k=0.02723   //W/m.K // [ kJ /( kg .K) ] Cp=1.007 // [ kg/ cu bi c m] rho=1.129 Npr=0.70 Mavg=29 // [m/ s ] u_inf=35 // [m] L=0.75 // [K] Tm=313 // [ kPa ] P=101.325 // Reynold ’ s number > 5∗10ˆ5 Nre_l=rho*u_inf*L/mu Nnu=0.0366*Nre_l^(0.8)*Npr^(1.0/3.0); // [W/ s m.K ] h=Nnu*k/L // [ sq m] A=1*L // [K] Tw=333 // [K] T_inf=293 // [W] Q=h*A*(Tw-T_inf); printf ( ” Heat t r a n s f e r from t he p l a t e i s %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 ;

  / / E x am pl e 3 . 1 2                          

/ / s q m/ s v=18.23*10^-6 // [W/m.K ] k=0.02814 // [m] D=0.012 // [m] r=0.006 // [m/ s ] u_inf=4 // Reynold ’ s number Nre=D*u_inf/v Nnu=0.37*Nre^(0.6); h=Nnu*(k/D) // Area o f s p h er e i n [ s q m] A=4*%pi*r^2 // [K] Tw=350 // [K] T_inf=300 // Heat l o s t b y s p h er e i n [W] Q=h*A*(Tw-T_inf) printf ( ” \n Heat l o s t by s p he r e i s %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 ;

  / / E x ma pl e 3 . 1 3                

// [ sq m./ s ] v=15.69*10^-6 // [W/m.K ] k=0.02624 / / P r a n d t l n um be r Npr=0.708 // kg/m. s mu=2.075*10^-5 // [m/ s ] u_inf=4 / / [m/ s ] mu_inf=1.8462*10^-5 // [K] Tw=350 // [K] T_inf=300 62

vel oci ty

// [m] D=0.012 // R a di us i n [m] r=D/2 // Reynold ’ s numbe Nre=u_inf*D/v 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 / / Area i n [ s q m] 19   Q=h*A*(Tw-T_inf); 20   printf ( ” \n Heat l o s t by t he s p he re i s %f W” ,Q); 12 13 14 15

       

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 ;

  / / E x am pl e 3 . 1 4            

// [ sq m/ s ] v=2.08*10^-5 k=0.03   //W/(m.K) / / P r a n d t l n um be r Npr=0.697 // [m] D=0.06 // [m/ s ] u_inf=0.3 / / R e y n o l d s n um be r Nre=D*u_inf/v

  / / A ve ra ge n u s s e l t number i s g i v e n by :                  

Nnu=0.37*(Nre^0.6); //W/ sq m.K h=Nnu*k/D // [K] Tw=400 // [K] T_inf=300 // [m] D=0.06 // [m] r=0.03 / / Area i n [ s q m] A=4*%pi*r^2 // [W] Q=h*A*(Tw-T_inf) // P er ce nt o f h ea t l o s t by f o r c e d per=Q*100/100

convection 20   printf ( ” Heat t r a n s f e r r a t e i s %f W, And p e r ce n t ag e o f   power l o s t by c o n v e c t i o i s : %f p e r c e n t ” ,Q,per); 63

Scilab code Exa 3.15  Heat lost by cylinder

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

clc ;   clear ;

  / / E x am pl e 3 . 1 5 // v e l o c i t y i n [m/ s ]   u_inf=50 // [ kg /(m. s ) ]   mu=2.14*10^-5 // [ kg/ cu bi c m]   rho=0.966 // [W/ (m.K) ]   k=0.0312 / / P r a n d t l n um be r   Npr=0.695 / / D ia me te r i n [m]   D=0.05 // Reynold ’ s number   N r e = D * u _ in f * r h o / m u ;   printf ( ”%f” ,Nre)   Nnu=0.0266*Nre^0.805*Npr^(1/3); h= Nnu *k /D ;   / /W/ sq m.K   //Approximation   h=171.7   printf ( ” \n%f” ,h) // [K]   Tw=423 // [K]   T_inf=308

  // Heat l o s s p e r u ni t l en g t h i s : // [W]   Q_by_l=h*%pi*D*(Tw-T_inf);   printf ( ” Heat l o s t p e r u n i t l e n g t h o f c y l i n d e r i s %f   W( ap pr ox ) ” , round (Q_by_l));

Scilab code Exa 3.16  Heat transfer in tube

1 2 clc ; 3   clear ; 4   / / E x am pl e 3 . 1 6

64

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

                 

/ / s q m/ s v=20.92*10^-6 k=3*10^-2   //W/(m.K) Npr=0.7 // [m/ s ] u_inf=25 // [mm] d=50 // [m] d=d/1000 // Reynold ’ s number Nre=u_inf*d/v // [K] Tw=397 // [K] T_inf=303

  / / Ca se 1 : C i r c u l a r t ub e          

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

/ / Ca se 2 : S q ua r e t u be / / A re a i n [ s q mm] A=50*50 / / P e r i m e t e r [mm ] P=2*(50+50) // [mm] l=4*A/P // [m] l=l/1000

          Nnu=0.102*(Nre^0.675)*(Npr^(1.0/3.0)) //W/( sq m.K)   h=Nnu*k/d // [ sq m]   A=4*l*l   Q=h*A*(Tw-T_inf) // [W/m]   Q_by_l2=Q/l   printf ( ” \ nRate o f h ea t f l o w fro m t he s q ua r e p i pe=%f  

W/m \n w hich i s more t h a n t ha t from t he c i r c u l a r p i p e w hi ch i s e q u a l t o %f W/m” ,Q_by_l2 ,Q _by_l1 );

Scilab code Exa 3.17   Heat transfer coefficient

65

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

clc ;   clear ;

  / / E x am pl e 3 . 1 7 // V i s c o s i t y o f f l o w i n g f l u i d [ N. s / s q m]   mu=0.8 // D en si ty o f f l o w i n f f l u i d [ g / c u bi c cm ]   rho=1.1 / / D e n s i t y i n [ kg / c u b i c m ]   rho=rho*1000 / / S p e c i f i c h e at [ kJ / kg . K]   Cp=1.26 // in [ J /( kg .K) ]   Cp=Cp*10^3 // [W/ (m.K) ]   k=0.384 // V i s c o s i t y a t w a l l t em p er at u re [ N. s / s q m]   mu_w=1 // [m] L =5 / / V o l um e tr i c f l o w r a t e i n [ c u b i c cm/ s ]   vfr=300 / / [ c u b i c m/ s ]   vfr=vfr*10^-6 // Mass f l o w r a t e o f f l o w i n f f l u i d [ kg   mfr=vfr*rho /s ] / / I n s i d e d i a m e t e r i n [mm]   Di=20 // [m]   Di=Di/1000 // A rea o f c r o ss −s e c t i o n [ s q m ]   Area=(%pi/4)*Di^2 // V e l o c t i y i n [m/ s ]   u=vfr/Area // Reynold ’ s number   Nre=Di*u*rho/mu // As r ey no ld ’ s number i s l e s s t ha n 2 10 0 , he f l o w i s laminar / / P r a n d t l n um be r   Npr=Cp*mu/k

22 23   Nnu=1.86*(Nre*Npr*Di/L)^(1.0/3.0)*(mu/mu_w)^(0.14) 24   hi=Nnu*k/Di / / i n s i d e h e a t t r a n s f e r c o e f f i c i e n t [W

/sq m.K] 25   printf ( ” I n s i d e h e a t t r a n s f e r c o e f f i c i e n t i s %f W/ ( s q m . K ) ” ,hi); 26 // Note : 27   printf ( ” \n The a ns we r g i v en i n book . . i e 1 22 5 i s wrong . p l e a s e r ed o t he c a l c u l a t i o n o f l a s t l i n e m an ua ll y t o c he ck \n ” );

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 18 19 20 21 22 23

clc ;   clear ;

  / / E x am pl e 3 . 1 8 / / Mass f l o w r a t e i n [ kg / h ]   m=5500 // [ kg/ s ]   m=m/3600 // D en si t y o f f l u i d i n [ g /cm ˆ 3 ]   rho=1.07 // [ kg/mˆ 3 ]   rho=rho*1000 / / V o l u m e tr i c f l o w r a t e i n [ mˆ 3/ s ]   vfr=m/rho / / D i a m e te r o f t u b e [mm]   Di=40 // [m]   Di=Di/1000 // Area o f c r o ss −s e c t i o n i n [ s q m]   A=(%pi/4)*Di^2 // V e l o c it y o f f l o w i n g f l u i d [m/ s ]   u=vfr/A / / D e n s i t y i n [ k g /mˆ 3 ]   rho=1070 / / V i s c o s i t y i n [ k g /m . s ]   mu=0.004   Nre=Di*u*rho/mu   //Approx   Nre=12198

           

24  

/ / S i n c e t h i s r e yn o ld ’ s number i s l e s s t ha n 1 0 00 0 , t h e f lo w i s t u rb u le n t / / S p e c i f i c h e at i n [ kJ / kg . K] Cp=2.72 // S p e c i f i c h ea t i n [ J / kg . K] Cp=Cp*10^3 / / t h e rm a l c o n d u c t i v i t y i n [W/m. K] k=0.256 / / P r a n d t l n um be r Npr=Cp*mu/k / / N u s s e l t n um be r Nnu=0.023*(Nre^0.8)*(Npr^0.4) // Ins id e heat tr an sf er c o e f f i c i e n t in hi=k*Nnu/Di [W/mˆ2.K] printf ( ” I n s i d e h e a t t r a n s f e r c o e f f i c i e n t i s %f W/ s q m.K” ,hi);

Scilab code Exa 3.19  h of water flowing in tube

1 2 clc ; 3   clear ; 4   / / E x am pl e 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:              

/ / D e n s i t y o f w a te r [ k g /mˆ 3 ] rho=984.1 // S p e c i f i c h ea t i n [ J /kg . K] Cp=4187 / / V i s c o s i t y a t 3 31 K[ Pa . s ] mu=485*10^-6 // [W/ (m.K) ] k=0.657 // V i s c o s i t y a t 2 97 K [ Pa . s ] mu_w=920*10^-6 // Solu tion / / D i a m e te r i n [mm] D=16 // D i am et er i n [m] D=D/1000 // V e l o c i t y i n [m/ s ] u =3 // [ kg /mˆ 3 ] rho=984.1 / / R e y n o l d s n um ber Nre=D*u*rho/mu

      Nre= round (Nre) / / P r a n d t l n um be r   Npr=Cp*mu/k

/ / D i t t u s −B o e l t e r e q ua t i on ( i ) / / n u s s e l t num ber   Nnu=0.023*(Nre^0.8)*(Npr^0.3) / / H e a t t r a n s f e r c o e f f i c i e n t [W/m ˆ 2 . K ]   h=k*Nnu/D   printf ( ” \nANSWER−( i ) \nBy D i t t u s −B o e l t e r e q ua t i on we ge t h=%f W/ sq m.K\n\n\n ” ,h);

25 26 / / s i e d e r −t a t e e q u a t i o n ( i i ) 27   Nnu=0.023*(Nre^0.8)*(Npr^(1.0/3.0))*((mu/mu_w)^0.14)

/ / N u s s e l t n um be r 28   h=k*Nnu/D / / H e a t t r a n s f e r c o e f f i c i e n t i n [W/ s q m . K] 29   printf ( ” \ nAnswer −( i i ) \n−By S i e d e r −Ta te e q u a t i on we ge t h=%f W/ sq m.K\n ” ,h); 30   printf ( ” \nNOTE : C a l c u l a t i o n m i st a ke i n b oo k i n p a r t 2 i e s i e d e r t a t e 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

  / / E x am pl e 3 . 2 0                      

// Mass f l o w a r t e i n [ kg / h ] m_dot=2250 / / S p e c i f i c h e at i n [ kJ / ( k g . K) ] Cp=3.35 // Te m p er a t u r e dr o p f o r o i l [K] dT=316-288.5 // Rate o f h ea t t r a n s f e r i n [ kJ /h ] Q=Cp*m_dot*dT // [ J/ s ] or [W] Q= round (Q*1000/3600) / / I n s i d e d i a me t e r [m] Di=0.04 // O u ts i de d i a mt e r i n [m] Do=0.048 / / f o r s t ea m [W/ s q m . K ] hi=4070 / / Fo r o i l [W/ s q m. K] ho=18.26 // [ sq m .K/W] Rdo=0.123 // [ sq m .K/W] Rdi=0.215 // [W/mˆ 2 .K Uo=1/(1/ho+Do/(hi*Di)+Rdo+Rdi*(Do/Di)) ]

16 17 18 19 20 21

           

Uo=2.3 // [K] dT1=373-288.5 // [K] dT2=373-316 // [K] dTm=(dT1-dT2)/ log (dT1/dT2) // Heat t r a n s f e r a re a in [ mˆ 2 ] Ao=Q/(Uo*dTm) printf ( ” H ea tr t r a n s f e r a r ea i s : %f mˆ2 ” ,Ao);

Scilab code Exa 3.21  Number of tubes in exchanger

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

clc ;   clear ;

  / / E x am pl e 3 . 2 1                

// [W/m.K ] k_tube=111.65 // [ kg /h ] W=4500 / / [ k g / s q m] rho=995.7 // [ kJ /( kg .K) ] Cp=4.174 // [W/ (m.K) ] k=0.617 // K i ne ma ti c v i s c o s i t y v=0.659*10^-6 m_dot=1720   / / k g / h // I n i t i a l temperature in [K] T1=293 69

[ s q m/ s ]

13 14 15 16 17 18 19 20 21

           

/ / F i n a l t e m p e ra t u re i n [ K] T2=318 // [K] dT=T2-T1 // H eat t r a n s f e r r a t e i n [ kJ /h ] Q=m_dot*Cp*dT / / [ J / s ] o r [W] Q=Q*1000/3600 // [m] Di=0.0225 // [m/ s ] u=1.2

  //Nre=Di ∗u∗ r h o /mu o r / / R e y n o l d s n um be r   Nre=Di*u/v // As Nre i s g r e a t e r th an 1 00 00 , D i t tu s B o e l t e r e qu a t i o n i s a p p l i c a b l e //J /( kg .K)   Cp=Cp*10^3 // [ kg /(m. s ) ]   mu=v*rho / / P r a n d t l n um be r   Npr=Cp*mu/k / / D i t t u s −B o el t er e qu a t i o n f o r h e a t i n g i s

22 23 24 25 26   Nnu=0.023*(Nre^0.8)*(Npr^0.4) 27   hi=k*Nnu/Di //Heat t r a n s fe r

coefficient

[W/ ( s q m

.K) ] 28 29 30 31 32 33

           

// [m] Do=0.025 / / Log mean d i am e t e r i n [m ] Dw=(Do-Di)/ log (Do/Di) // [W/ sq m.K] ho=4650 // [W/m. K ] k=111.65 // [m] xw=(Do-Di)/2 // Ove ral l Uo=1/(1/ho+Do/(hi*Di)+xw*Do/(k*Dw))

h e a t t r a n s f e r c o e f f i c i e n t i n W/ (m ˆ 2 . K) 34   T_steam=373 / / Te mp er at ur e o f c o n d e n s in g s te am i n [K] 35 dT1= T_steam -T1 +10 // [K] 36 dT2= T_steam -T2 +10 // [K] 37   dTm=(dT1-dT2)/ log (dT1/dT2) // [K] 38   Ao=Q/(Uo*dTm) / / Area i n [mˆ 2 ] 39 L =4 // l e n g th o f t u be [m] 40   n=Ao/(%pi*Do*L) / / number o f t u b es 41   printf ( ”No . o f t u b es r e q u i r e d =%d\n ” , round ( n ) ) ; 42   printf ( ” \n NOTE: t h e r e i s an e r r o r i n book i n c a l c u l a t i o n o f dT1 a nd dT2 , \ n 3 7 3 −293 i s w r i t t e n a s 9 0 , i n s t e a d o f 8 0 . . . s i m i l a r l y i n dT2 , \ nSo , i n c om p li a nc e w it h t he book , 1 0 i s added t o b oth o f   them” ) 70

Scilab code Exa 3.22  Convective film coefficient

1 2 3 4 5 6 7 8 9 10 11

clc ;   clear ;

  / / E x am pl e 3 . 2 2                

12   13 14 15 16 17

 

     

 

18   19   20   21  

/ / m a ss f lo w r a t e o f w at er [ kg / h ] m_dot=25000 // [ kg /mˆ 3 ] rho=992.2 // [W/m.K ] k=0.634 // [mˆ3/ h ] vfr=m_dot/rho / / P r a n d t l n u m b er l Npr=4.31 // [mm] Di=50 // [m] Di=0.05 // [ K ] a s t h e w a l l i s a t a t e m pe ra tu re o f 1 0 dT=10 K a bo ve t h e b u lk t e mp e r at u r e // V e l o c i t y o f w at er u=(vfr/3600)/(%pi*(Di/2)^2) i n [ m/ s ]   //Approximation u=3.56 //Nre=Di ∗u∗ r ho /mu=Di∗u / v a s v=mu/ r ho // [mˆ2/ s ] v=0.659*10^-6; / / R e y n o l d s n um be r Nre=Di*u/v / /As i t i s l e s s th a n 1 00 00 , t he f lo w i s i n t he t u r b ul e n t r e g i o n f o r h e at t r a n s f e r and D i t tu s B o el t er eqn i s u s e d / / N u s s e l t n um ber Nnu=0.023*(Nre^0.8)*(Npr^0.4); / / H ea t t r a n s f e r c o e f f i c i e t i n [W/ s q m hi=Nnu*k/Di .K] // Heat t r a n s f e r p e r un i t q_by_l=hi*%pi*Di*dT length [kW/m] printf ( ” A v er ag e v a l u e o f c o n v e c t i v e f i l m c o e f f i c i e n t i s h i = %d W/ s q m . K \ nHeat t r a n s f e r r e d p er u n i t le ng th i s Q/L=%f kW/m” , round (hi),q_by_l/1000);

71

Scilab code Exa 3.23  Length of tube

1 2 3 4 5 6 7 8 9 10 11

clc ;   clear ;

  / / E x am pl e 3 . 2 3 vfr = 1200 ; / / Water f l o w r a t e i n [ l / h ] / / D e ns i t y o f w at er i n g / [ c u b i c cm ]   r h o = 0. 98 ; // Mass f l o w r a t e o f w at er [ kg / h ]   m_dot=vfr*rho // [ kg/ s ]   m_dot2=m_dot/3600 // [ J/ kg .K]   C p = 4 . 1 8 7* 1 0 ^ 3 ; Di =0.025 ; / / D ia me te r i n [m] // [ kg /(m. s ) ]   m u = 0 . 00 0 6 ; // A rea o f c r o ss −s e c t i o n i n [m   Ai=%pi*((Di/2)^2) ˆ2] / / R e y n o l d s n um be r   Nre=(Di/mu)*(m_dot2/Ai)   k = 0 .6 3 ;   / / f o r m e ta l w a l l i n [W/ (m. K) ] / / P r a n d t l n um be r   Npr=Cp*mu/k; / / S i n c e Nre > 10000 // t h e r e f o r e , D i t tu s b o e l t e r eqn f o r h e a t i n g i s

12 13 14 15 16 17   Nnu=0.023*(Nre^(0.8))*(Npr^(0.4)) 18 ho =5800 ;   / / F il m h e a t c o e f f i c i e n t W / (mˆ 2 . K) 19   hi=Nnu*k/Di // Heat t r a n s f e r c o e f f c i e n t i n [W/ ( s q

m.K) ] 20 21 22 23 24 25

// [m] D o = 0 .0 28 ; // [m] D i = 0 .0 25 ; // [m] xw=(Do-Di)/2; // [m] Dw=(Do-Di)/ log (Do/Di); k =50 ; / / f o r m e ta l w a l l i n [W/ (m. K) ]   Uo=1/(1/ho+Do/(hi*Di)+xw*Do/(k*Dw));        

/ / i n [W/ s q m

.K] 26 27 28 29 30 31 32 33

               

d T = 3 43 - 3 03 ;   // [K] // [K] d T 1 = 3 93 - 3 0 3 ; // [K] d T 2 = 3 93 - 3 4 3 ; // [K] dTm=(dT1-dT2)/ log (dT1/dT2); // [ in [ kJ/kg .K] ] Cp=Cp/1000; // Rate o f h ea t t r a n s f e r i n [ kJ /h ] Q=m_dot*Cp*dT; / / [ J / s ] o r [W] Q=Q*1000/3600; // Heat t r a n s f e r a re a i n [ s q m] Ao=Q/(Uo*dTm);

72

34 // Al so , . . Ao=%pi∗Do∗L . . i m p l i e s t ha t 35   L=Ao/(%pi*Do) // [m] 36   printf ( ” Len g th o f t u be r e q u i r e d i s %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 ;

  / / E x am pl e 3 . 2 4 / / 1 . F or i n i t i a l c o n d i t i o n s : // [K]   T=360; // [K]   T1=280; // [K]   T2=320; // [K]   dT1=T-T1; // [K]   dT2=T-T2;   //Q1=m1 dot ∗Cp1 ∗ ( T2−T1 ) / / Hea t c a p a c i t y   Cp1=4.187   dTlm=(dT1-dT2)/ log (dT1/dT2)   // [K]   m1_by_UA=dTlm/(Cp1*(T2-T1))

  // For f i n a l c o n d i t i o n s :   //m2 dot=m1 dot   //U2=U1   //A2=5∗A1   deff ( ’ x=f ( t ) ’ , ’x=m1 by UA ∗Cp1 ∗ ( t−T1 ) − 5∗((dT1−(T−t ) ) / l o g ( dT1 / ( T−t ) ) ) ’ )

19 T= fsolve (350.5,f) 20   printf ( ” \ n Ou tl et t em p er a tu r e o f w at er i s %f K\n ” ,T);

Scilab code Exa 3.25   Outlet temperature of water

1 clc ; 2   clear ; 3   / / E x am pl e 3 . 2 5

73

4 5 6 7 8 9

           

// Mass f l o w r a t e o f o i l i n [ g / s ] mo_dot=60 // [ kg/ s ] mo_dot=6*10^-2 // S p e c i f i c h ea t o f o i l i n [ kJ / ( kg . K) ] Cpo=2.0 // [K] T1=420 // [K] T2=320 // Rate o f h ea t f l o w i n [ kJ / s Q=mo_dot*Cpo*(T1-T2) ]

10 11 12 13 14 15 16 17 18

// Mass f lo w r a t e of wa t e r   mw_dot=mo_dot // [K]   t1=290 // [ kJ /( kg .K) ]   Cpw=4.18

// kg / s

  / / For f i n d i n g o u t l e t t em pe ra tu re o f w at er // [K]   t2=t1+Q/(mw_dot*Cpw) // [K]   dT1=T1-t2 // [K]   dT2=T2-t1 // [K]   dTm=(dT1-dT2)/ log (dT1/dT2) / / O i l s i d e h e a t t r a n s f e r c o e f f i c i e n t i n [ kW   ho=1.6 /( sq m.K) ] 19   hi=3.6 // Water s i d e h ea t t r a n s f e r c o e f f i n [kW/ ( s q m.K) ] 20 / / O v e r a l l h e a t t r a n s f e r 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] // [mm] Do=25 // [m] Do=Do/1000 // L en gth o f tu be i n [ m] L=A/(%pi*Do) printf ( ” \ n Ou tl et t em p er a tu r e o f w at er i s %f K \n” , round (t2)); 28   printf ( ” Area o f h ea t t r a n s f e r r e q u i r e d i s %f s q m\n ” ,A); 29   printf ( ” Len g th o f t u be r e q u i r e d i s %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 ;

  / / E x am pl e 3 . 2 6 / / f o r o i l [W/m .K ]   k=0.14   Cp=2.1   / / f o r o i l [ kJ / kg . K] //J/ kg .K   Cp=Cp*10^3 // [mN. s / sq m]   mu=154   mu_w=87 / / ( mn . s / s q m) // [m]   L=1.5 // Mass f l o w r a t e o f o i l [ kg / s ]   m_dot=0.5   Di=0.019   // D i am et er o f t ub e [m] // Mean t e m pe ra t ur e o f o i l [ K]   mean_T=319   mu=mu*10^-3 / / [ N . s / s q m ] o r [ k g / ( m. s ) ] // [ sq m]   A=%pi*(Di/2)^2 // Mass v e l o c i t y i n [ kg / s q m. s ]   G=m_dot/A / / R e y n o l d s n u mb er   Nre=Di*G/mu   / / A s N re < 2100 , t he f l o w i s l a mi n ar   mu_w=mu_w*10^-3 / / [ N . s / s q m ] o r k g / (m . s )   // The s i e d e r t a t e e q u a ti o n i s   hi=(k*(2.0*((m_dot*Cp)/(k*L))^(1.0/3.0)*(mu/mu_w) // H eat t r a n s f e r c o e f f i n [W/ s q m. K ^(0.14)))/Di

] 22   printf ( ” \n The i n s i d e h e at t r a n s f e r c o e f f i c i e n t i s %f W/(mˆ 2.K) ” ,hi); 23 24   printf (  ’ \nNOTE : C a l c u l a t i o n m i st a ke i n l a s t l i n e . i e

i n t he c a l c u l a t i o n o f h i i n book , p l e a s e p er f o rm t he c a l c u l a t i o n m an ua ll y t o c he c k t he a ns we r \n ” )

Scilab code Exa 3.27   Film heat transfer coefficient

1 clc ; 2   clear ; 3   / / E x am pl e 3 . 2 7 4

75

5 6 7 8 9 10 11 12 13

                 

m_dot=0.217   / / Water f l o w r a t e i n [ kg / s ] / / O u t s i d e d i a m e t e r i n [mm] Do=19 // Den sit y rho=1000 / / Wa ll t h i c k n e s s i n [mm] t=1.6 / / i . d o f t u b e i n [mm] Di=Do-2*t // [m] Di=Di/1000 // [m] Do=Do/1000 Ai=%pi*(Di/2)^2   / / C r o s s −s e c t i o n a l a re a i n s q m // Water v e l o c i t y t hr ou gh t u be u=m_dot/(rho*Ai)

[m/

s] 14 15 16 17 18 19

           

20   21   22   23   24  

/ / a pp ro x i n b oo k u=1.12 // a p pr x i n b oo k Di=0.0157 // I n l e t t em p er a tu re o f w at er i n [ K] T1=301 // O u tl e t t em p er at u re o f w at er i n [ K] T2=315 T=(T1+T2)/2 // [K] hi=(1063*(1+0.00293*T)*(u^0.8))/(Di^0.20) / / I n s i d e h e a t t r a n s f e r c o e f f i c i e n t W/ ( s q m . K)   //Approximation hi=5084 printf ( ”%f” ,hi); // I n s i d e h ea t t r a n s f e r c o e f f b a s e d hio=hi*(Di/Do) on o u t s i d e d i am e te r i n W/ ( s q m. K) printf ( ”%f” ,hio); printf ( ” B as ed on o u t s i d e t em p er a tu re , I n s i d e h e at t r a n s f e r c o e f f i c i e n t i s %d W/ (mˆ 2 . K) o r %f kW/ (m ˆ 2 . K ) ” , round ( h i o ) , round (hio)/1000);

Scilab code Exa 3.28   Area of exchanger

1 2 3 4 5 6 7

clc ;   clear ;

  / / E x am pl e 3 . 2 8        

mair_dot=0.90 // [K] T1=283 // [K] T2=366 dT=(T1+T2)/2

// [ kg/ s ]

// [K] 76

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

           

// [mm] Di=12 // [m] Di=Di/1000 // [ kg /( sq m. s ) ] G=19.9 // [mN. s / ( sq m) ] mu=0.0198 mu=mu*10^-3 / / [ N . s / s q m ] o r [ k g / ( m. s ) ] Nre=Di*G/mu   / / R e y n o l d s n um be r

  / / I t i s g r e a t e r th a n 1 0 ˆ 4   k=0.029   //W/(m.K) // [ kJ/ kg .K]   Cp=1 // [ J/ kg .K]   Cp1=Cp*10^3   Npr=Cp1*mu/k / / P a r n d t l n um be r / / D i t t u s −B o e l t e r e qu a ti o n i s // [W/ sq m.K]   hi=0.023*(Nre^0.8)*(Npr^0.4)*k/Di //W/ sq m.K   ho=232   U=1/(1/hi+1/ho) / / O v e r a l l h e a t t r a n s f e r c o e f f i c i e n t [W/mˆ2.K]

23 24 25 26 27 28 29 30 31

           

Q=mair_dot*Cp*(T2-T1)   / / k J / s / / [ J / s ] o r [W] Q=Q*10^3 // [K] T=700 // [K] dT1=T-T2 // [K] dT2=T2-T1 // [K] dTm=(dT1-dT2)/ log (dT1/dT2)

  //Q=U∗A∗dTm   A=Q/(U*dTm)   // A rea i n s q m   printf ( ” Heat t r a n s f e r a r e a o f e q u i p me n t i s %f s q m” , A) ;

Scilab code Exa 3.29  Natural and forced convection

1 2 3 4 5 6

clc ;   clear ;

  / / E x am pl e 3 . 2 9 // [ sq m./ s ]   v=18.41*10^-6 // [W/m.K ]   k=28.15*10^-3   Npr=0.7 / / P r a n d t l n um be r 77

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

           

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

//Kˆ−1

// 1. Free Convection   Ngr=(g*Beta*dT*L^3)/(v^2) / / G r a s h o f n um be r   Npr=0.7 / / P r a n d t l n um be r

/ / N u s s e l t n u mb er   Nnu=0.59*(Ngr*Npr)^(1.0/4.0) / / A v e r a g e h e a t t r a n s f e r c o e f f i c i e n t [W/   h=Nnu*k/L

s q m K] 19   printf ( ” \n I n f r e e c o n ve ct io n , h ea t t r a n s f e r %f W/( sq m.K) \n ” ,h) 20 / / 2 . F o r ce d C o n v e s t i o n 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))

c o e f f , h=

//

N u s s e l t n umb er 24   h=Nnu*k/L // [W/ sq m.K ] 25   printf ( ” \n I n f o r c ed c on v e ct i o n , h ea t t r a n s f e r c o e f f , h=%f W/ ( sq m.K) \n ” ,h ) 26   printf ( ” \n From a b o v e i t i s c l e a r t ha t h ea t t r a n s f e r c o e f f i c i e n t i n f o r c e d c o n v e c t i o n i s much l a r g e r t h a n t ha t i n f r e e c o n v e c t i o n \n ” ) ;

Scilab code Exa 3.30  Natural convection

1 2 3 4 5 6

clc ;   clear ;

  / / E x am pl e 3 . 3 0   k=0.02685   //W/(m.K) // kg /(m. s )   v=16.5*10^-6 78

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

Npr=0.7 / / P r a n d t l n um be r Beta=3.25*10^-3   //Kˆ−1 //m/( s ˆ2 ) g=9.81 Tw=333; // [ k ] // [K] T_inf=283 // [K] dT=Tw-T_inf o f p l a t e [m] L =4   // L en g th / h e ig h t / / G r a s h o f f n um be r   Ngr=(g*Beta*dT*(L^3))/(v^2)

           

  / / L e t c o n s t =N gr ∗Npr   const=Ngr*Npr

// S i c e i t i s

>

10ˆ9

/ / N u s s e l t n um ber   Nnu=0.10*(const^(1.0/3.0)) //W/ ( sq m. K)   h=Nnu*k/L / / Ap pro x i n b oo k   h=4.3 W =7   // w i dt h i n [m] // Area o f h e at t r a n s f e r i n [ s q m]   A=L*W // [W]   Q=h*A*dT   printf ( ” \ nHeat t r a n s f e r r e d i s %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 14

clc ;   clear ;

  / / E x am pl e 3 . 3 1                      

//mˆ2/ s v=18.97*10^-6 k=28.96*10^-3   //W/(m.K) Npr=0.696 / / O u te r d i a m e t e r [mm] D=100 // [m] D=D/1000 / / Fi lm t e mp e r at u r e i n Tf=333 // [K] Tw=373 // [K] T_inf=293 dT=Tw-T_inf // [K] // [Kˆ −1] Beta=1/Tf // [m/ s ˆ 2 ] g=9.81 79

[ K]

15 16 17 18

L =3   // L en gth o f p i pe [m]   Ngr=(g*Beta*dT*(L^3))/(v^2) / / G r a s h o f n um be r   Nra=Ngr*Npr // n u s s e l t number f o r   Nnu=0.10*(Ngr*Npr)^(1.0/3.0)

v e r t i c a l c y li n de r 19   h=Nnu*k/L //W/ ( sq m. K) 20   Q_by_l=h*%pi*D*dT // Heat l o s s p er m et re l e n g t h [W/ m] 21   printf ( ” \n Hence , Hea t l o s s p e r m et re l e n g t h i s %f W/ m \n ” ,Q_by_l);

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 ;

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

// Gras hoff 

number 16 17 18 19 20 21

  Cp1=Cp*1000 // [ J/ kg .K]   Npr=Cp1*mu/k / / P r a n d t l n um be r

  / / A ve ra ge n u s s e l t number i s   Nnu=0.53*(Ngr*Npr)^(1.0/4.0)   h=Nnu*k/D // [W/ sqm .K ] // Heat l o s s p er u n i t l e n g th   Q_by_l=h*%pi*D*dT

80

[W/m

] 22   printf ( ” \ nHeat l o s s p e r u ni t l e n g t h o f t h e h ea t e r i s %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 14 15 16 17 18 19

clc ;   clear ;

  / / E x am pl e 3 . 3 3 // [W/ (m/K) ]   k=0.03406   Beta=2.47*10^-3   //Kˆ−1 / / P r a n d t l n um be r   Npr=0.687 //mˆ2/ s   v=26.54*10^-6 // [m/ s ˆ 2 ]   g=9.81 // [K]   Tw=523 // [K]   T_inf=288   dT=Tw-T_inf // [K] // [m]   D=0.3048   Ngr=(g*Beta*dT*(D^3))/(v^2)   Nra=Ngr*Npr

/ / G r a s h o f n um be r

  / / For Nra l e s s t ha n 1 0ˆ 9 , we h ave f o r h o r i z o n t a l cylinder / / N u s s e l t n um be r   Nnu=0.53*(Nra^(1.0/4.0)) // [W/ sq m.K ]   h=Nnu*k/D   Q_by_l=h*%pi*D*dT;   / /W/m   printf ( ” Heat l o s s o f h ea t t r a n s f e r p er m e te r l en gh 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

  / / E x am pl e 3 . 3 4 / / D e n s i t y i n [ k g /mˆ 3 ]   rho=960.63 / / S p e c i f i c h e at i n [ J / ( k g . K) ]   Cp=4.216*10^3 / / D ia me te r i n [ cm ]   D=16   D=D/100   // [m] / / T he rm al c o n d u c t i v i t y i n [W/m .K ]   k=0.68                    

A=(%pi*(D/2)^2) / / L e ng t h=A/P i n [ m] L=A/(%pi*D) // [Kˆ −1] Beta=0.75*10^-3 // [mˆ2/ s ] alpha=1.68*10^-7 // [m/ s ˆ 2 ] g=9.81 // [K] Tw=403 // [K] T_inf=343 dT=Tw-T_inf // [K] // [mˆ2/ s ] v=0.294*10^-6 Nra=(g*Beta*(L^3)*dT)/(v*alpha)

/ / 1 . F or Top s u r f a c e / / N u s s e l t n um be r   Nnu=0.15*(Nra)^(1.0/3.0) // Heat t r a n s f e r c o e f f f o r t op s u r f a c e i n   ht=Nnu*k/L W/(mˆ2.K)

24 25 26 27 28 29 30

ht = round (ht)

/ / 2 . Fo r b ot to m s u r f a c e / / N u s s e l t n um ber   Nnu=0.27*Nra^(1.0/4.0) // [W/ sq m.K ]   hb=Nnu*k/L hb = round (hb)   Q=(ht+hb)*A*dT;   // [W]   printf ( ” The r a t e o f h ea t i n pu t i s %f W” ,Q)

Scilab code Exa 3.35  Heat transfer from disc

1 2 3 4

clc ;   clear ;

  / / E x am pl e 3 . 3 5 // [mˆ2/ s ]   v=2*10^-5 82

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

                     

Npr=0.7 / / P r a n d t l n um be r // [W/m.K ] k=0.03 // D i am et er i n [ m] D=0.25 // C h a r a c t e r i s t i c l en gt h , l e t L=0.90*D // [K] T1=298 // [K] T2=403 // [K] dT=T2-T1 // [K] Tf=(T1+T2)/2 // [Kˆ −1] Beta=1/Tf / / A re a i n [ s q m] A=%pi*(D/2)^2 // [m/ s ˆ 2 ] g=9.81

[m]

  / / Ca se 1 : Hot s u r f a c e f a c i n g up / / G r a s h o f f n um ber   Ngr=g*Beta*dT*(L^3)/(v^2) / / N u s s e l t n um ber   Nnu=0.15*((Ngr*Npr)^(1.0/3.0)) // [W/ sq m.K ]   h=Nnu*k/L // [W]   Q=h*A*dT   printf ( ” \n Heat t r a n s f e r r e d when h o t s u r f a c e i s f a c i n g up i s %f W\n ” ,Q);

       

// Ca se 2 : For h ot s u r f a c e f a c i n g down // Gra sho f Number Nnu=0.27*(Ngr*Npr)^(1.0/4.0); // [W/ sqm . K] h=Nnu*k/L // [W] Q=h*A*dT printf ( ” \n Heat t r a n s f e r r e d when h o t s u r f a c e i s f a c i n g down i s %f W\n” ,Q);

Scilab code Exa 3.36  Rate of heat input to plate

1 2 3 clc ; 4   clear ; 5   / / E x am pl e 3 . 3 6

83

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

   

rho=960 // [ kg /mˆ 3 ] Beta=0.75*10^-3   // [Kˆ −1] // [W/m.K ] k=0.68 // [mˆ2/ s ] alpha=1.68*10^-7 // [mˆ2/ s ] v=2.94*10^-7 // [ kJ/ kg .K] Cp=4.216 // [K] Tw=403 // [K] T_inf=343 // [K] dT=Tw-T_inf // [m/ s ˆ 2 ] g=9.81 // [m] l=0.8 // [m] W=0.08 / / Area i n [ mˆ 2 ] A=l*W / / P e r i m e t e r i n [m] P=2*(0.8+0.08) // C h a r a c t e r i s t i c d im en si on / l e ng th , L i n L=A/P Nra=g*Beta*L^3*dT/(v*alpha)

[m]

// ( i ) f o r n a t u r a l c o nv e ct i on , h ea t t r a n s f e r fro m t op / u pp er s u r f a c e h ea te d / / N u s s e l t n um be r Nnu=0.15*(Nra^(1.0/3.0)) // [W/mˆ 2 .K ] ht=Nnu*k/L / / A p pr o xi m at i on i n bo ok , I f d on e m a nu a ll y ht=2115.3 then answer d i f f  // ( i i ) F or t he bottom / l o w er s u r f a c e o f t he h ea te d plate / / N u s s e l t n um be r Nnu=0.27*(Nra^(1.0/4.0)) // [W/(mˆ 2 .K) ] hb=Nnu*k/L hb = round (hb)

  // Rate o f h ea t i np ut i s e qu al t o r a t e o f h ea t d i s s i p a t i o n from t h e u p p e r and l ow er s u r f a c e s o f   t he p l a t e 32   Q=(ht+hb)*A*(Tw-T_inf) // [W] 33   printf ( ” \n R ate o f h ea t i np ut i s e qu a l t o h ea t d i s s i p a t i o n =%f W” ,Q);

84

Scilab code Exa 3.37  Two cases in disc

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

clc ;   clear ;

  / / E x am pl e 3 . 3 7   k=0.03   //W/(m.K) / / P r a n d t l n um be r   Npr=0.697 //mˆ2/ s   v=2.076*10^-6   Beta=0.002915   //Kˆ−1 / / [ D i a m et e r i n cm ]   D = 25 ;   D=D/100   // [m] / / Fi lm t e mp e r at u r e i n [ K]   Tf=343 / / Area i n [ mˆ 2 ]   A=%pi*(D/2)^2   P=%pi*D   / / P e r i m e t e r [ m] // [K]   T1=293 // [K]   T2=393 // [m/ s ˆ 2 ]   g=9.81

                 

// Ca se ( i ) HOT SURFACE FACING UPWARD // C h a r a c t e r i s t i c l e ng t h i n [m] L=A/P // [Kˆ −1] Beta=1/Tf; // [K] dT=T2-T1 Ngr=(g*Beta*dT*(L^3))/(v^2) / / G r a s h o f f n um ber Nra=Ngr*Npr / / N u s s e l t n um be r Nnu=0.15*(Nra^(1.0/3.0)) // [W/mˆ 2 .K ] h=Nnu*k/L // [W] Q=h*A*dT printf ( ” \ nHeat t r a n s f e r r e d when d i s c i s h o r i z o n t a l w i th h o t s u r f a c e f a c i n g upward i s %f W\n ” ,Q);

  //Case −( i i ) HOT FACE FACING DOWNWARD / / N u s s e l t n um ber   Nnu=0.27*(Nra^(1/4))   h=Nnu*k/L   //W/(mˆ2.K) // [W]   Q=h*A*dT   printf ( ” \ nHeat t r a n s f e r r e d when d i s c i s h o r i z o n t a l w it h h ot s u r f a c e f a c i n g downward i s %f W\n ” ,Q);

33 34

85

35 36 37 38 39 40 41 42 43 44 45

  //Case −( i i i )−For d i s c v e r t i c a l // C h a r a c t e r i s t i c l e n g th [m]   L=0.25 D= L // di a [m]   A=%pi*((D/2)^2) // [ sq m] / / G r a s h o f f n um be r   Ngr=(g*Beta*dT*(L^3))/(v^2)            

Npr=0.697 Nra=Ngr*Npr / / N u s s e l t n um ber Nnu=0.10*(Nra^(1/3)) // [W/(mˆ 2 .K) ] h=Nnu*k/D // [W] Q=h*A*dT printf ( ” For v e r t i c a l d i s c , h ea t t r a n s f e r r e d Q) ;

i s %f W” ,

Scilab code Exa 3.38  Total heat loss in a pipe

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

clc ;   clear ;

  / / E x am pl e 3 . 3 8

 

                   

v =23.13*10^ -6 ; // [mˆ2 / s ] k =0.0321 ;   // [W/m.K] Beta=2.68*10^-3;   // [Kˆ −1] Tw =443 ; // [K] T_inf =303 ;   // [K] // [K] dT=Tw-T_inf; g = 9 .8 1 ; // [m/ s ˆ 2 ] / / P r a n d t l n um be r Npr=0.688; D = 10 0 ; / / D i a m e t e r [mm ] / / D i a m e te r [ m] D=D/1000 Nra=(g*Beta*dT*(D^3)*Npr)/(v^2) / / N u s s e l t n um be r Nnu=0.53*(Nra^(1.0/4.0)) // [W/(mˆ 2 .K) ] h=Nnu*k/D   //Approximation h=7.93 e=0.90; / / E m i s s i v i t y sigma =5.67*10^ -8 ;

/ /Q=Q c on v+Q r ad

// Total heat l o ss 86

21   / / f o r t o t a l h e a t l o s s p e r m e t e r l en en gt h 22   Q_by_l=h*%pi*D*dT+sigma*e*%pi*D*(Tw^4-T_inf^4)

// [W

/m] 23   printf ( ” T ot o t al a l h ea ea t l o s s %f W/m” ,Q_by_l)

p er e r m e t r e l e n g t h o f p ip ip e i s

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 ;

  / / E x am am pl pl e 3 . 3 9 [W/ (m.K) ]   k=0.035; // [W um be be r N p r =0 =0.684 ; / / P r a n d t l n um   Beta=2.42*10^-3;   // [Kˆ − 1 ] // [mˆ2/ s ]   v=27.8*10^-6; // [K]   Tw=533; // [K] T _ i n f =3 =363 ;   dT=Tw-T_inf // [K] D =0 = 0 . 0 1 ; // [m] [m// s ˆ 2 ]   g=9.81; // [m   Nra=(g*Beta*dT*(D^3))/(v^2)

  // / / Fo For t h i s

1 0 ˆ 5 , we we h av av e f o r s p h e r e re a o f s p h er e r e i n [ mˆ 2 ]   A=4*%pi*(D/2)^2 / / A re um be be r   Nnu=(2+0.43*Nra^(1.0/4.0)) / / N u s s l e t n um   h=Nnu*k/D   //W/(mˆ2.K) // [W [W]]   Q=h*A*dT e a t l o s s i s % f W” , Q )   printf ( ” \ n R a t e o f h ea <

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

  / / E xa xa mp mp e 3 . 4 0 // [mˆ2/ s ] v=17.95*10^-6 // [K] dT=353-293 // [W [W/m. /m.K K] k=0.0283 // [m [m// s ˆ 2 ] g=9.81 / / P r a n d t l n um um be be r Npr=0.698 Cp=1005 / / J / ( k g . K ) / / Fi F i lm l m t e mp m p e r at a t u r e i n [ K] K] Tf=323 // [Kˆ − 1 ] Beta=1/Tf l =1   // [m]   Nra=(g*Beta*dT*(l^3)*Npr)/(v^2)

               

  / / I n t ex e x tb t b o o k r e s u l t o f a b o v e s ta t a te t e me m e nt n t i s w r on o n gl gl y c a l c u l a t e d , So

17   Nra=3.95*10^8 18   / / F or o r N ra ra < 1 0 ˆ 9 , f o r a v e r t i c a l

p l a te t e , t he h e a v er e r a ge ge

n u s s e l t number i s 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

       

/ / N u s s e l t n um um be be r Nnu=0.59*Nra^(1.0/4.0) // [W/mˆ 2 .K ] h=Nnu*k/l / / Ap A p pr pr o x i n b oooo k h=2.35 / / A re r e a [ mˆ mˆ 2 ] A=l^2

  / / He H e a t l o s s f o r m 4 v e r t i c a l f a c e s o f 1m∗ 1 m i s / / [W [W]]   Q1=4*(h*A*dT)   // / / Fo F o r t op op s u r f a c e / / P e r i me me t e r i n [m]   P=4*l // [m]   L=A/P              

Nra=(Npr*g*Beta*dT*(L^3))/(v^2) / / N u s s e l t n um um be be r Nnu=0.15*Nra^(1.0/3.0) // [W/mˆ 2 .K ] h=Nnu*k/L   //Approx h=6.7 // [W [W]] Q2=h*A*dT / / T o t a l h e a t l o s s [W] Q_total=Q1+Q2 h e r e fo f o r e t o t a l h ea e a t l o s s i s %d W” , printf ( ” \ n T he 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 20 21 22 23 24 25

clc ;   clear ;

  / / E x am am pl pl e 3 . 4 1 / / D e n s i t y i n [ k g /m /m ˆ 3 ] // [ J/ kg .K] / / [W [W/m. /m.K K] // [N. s /mˆ2 ]

       

rho=0.910; Cp=1.009*1000; k=0.0331; mu=22.65*10^-6;

           

/ / L et e t a= s m a l l e r s i d e / /b / b= b i g g e r s i d e //Qa=ha ∗A∗ dT //Qb=hb ∗A∗ dT //Qb=hb∗ //Qa=1.14 ∗ Qb //Qa=1.14∗ / / G i ve v e n a ∗ b=15 b=15∗∗ 10 ˆ − 4 / /O /On s o l v i n g we g e t : // [m] a=0.03; // [m] b=0.05; / / Ar Ar e a i n [ s q m] A=a*b // [K] Tf=388; // [Kˆ − 1 ] Beta=1/Tf // [K] T1=303; // [K] T2=473; // [K] dT=T2-T1

                  v=mu/rho / /m/ s ˆ 2 [ a c c e l e r a t i o n d u e to to g r a v i t y ]   g=9.81   hb=0.59*(((g*Beta*dT*(b^3))/(v^2))*Cp*mu/k)^(1/4)*(k // [W/ sq m.K ] /b) 26   Qb=hb*A*(dT) // [W [W]] 27 28   Qa=1.14*Qb // [W [W]] 29   printf ( ” \ n Di D i me m e ns n s io i o ns n s o f t he h e p l a t e a r e % f x % f m\ n ” ,a,b ); 30   printf ( ” \ n H e a t t r a n s f e r w h e n t he h e b i g g e r s i d e h el el d v e r t i c a l i s % f W\ n ” ,Qb); 31   printf ( ” \ n H e a t t r a n s f e r w h e n t he h e s m al a l l s i d e h el el d v e r t i c a l i s % 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 ;

  / / E x am pl e 3 . 4 2 // [K]   Ts=373   rho_l=957.9   / / r h o ∗ l [ kg/mˆ3 ] // [ J/ kg .K]   Cpl=4217   mu_l=27.9*10^-5 // [ kg /(m. s ) ] // [ kg/mˆ 3 ]   rho_v=0.5955                        

Csf=0.013 // [ N/m] sigma=5.89*10^-2 Nprl=1.76 lambda=2257 // [ kJ/ kg ] // in [ J/kg ] lambda=lambda*1000 n =1 / / f o r w a te r / / Mass f l o w r a t e [ kg / h ] m_dot=30 // [ kg/ s ] m_dot=m_dot/3600 / / D ia me te r o f pan [ cm ] D=30 D=D/100   // [m] // [m/ s ˆ 2 ] g=9.81 / / Area i n [ s q m] A=%pi*(D/2)^2 // [W/ sq m] Q_by_A=m_dot*lambda/A

  / / For n u c l e a t e b o i l i n g p o i n t we ha ve :   dT=(lambda/Cpl)*Csf*(((Q_by_A)/(mu_l*lambda))* sqrt ( sigma/(g*(rho_l-rho_v))))^(1.0/3.0)*(Nprl^n)   // [K

] 24   Tw=Ts+dT // [K] 25   printf ( ” \n Te mp er at ur e o f t he bottom s u r f a c e p an i s %f W/ ( s q m) ” ,Tw);

Scilab code Exa 3.43  Peak Heat flux

90

o f t he

1 2 3 4 5 6 7 8 9 10 11

clc ;   clear ;

  / / E xa mp le 3 . 4   lambda=2257 // [ kJ/ kg ] // in [ J/kg ]   lambda=lambda*1000   rho_l=957.9   / / r h o ∗ l [ kg/mˆ3 ] // [ kg/mˆ 3 ]   rho_v=0.5955 // [ N/m]   sigma=5.89*10^-2 // [m/ s ˆ 2 ]   g=9.81   // Peak h ea t f l u x i s g i v en by

  Q_by_A_max=(%pi/24)*(lambda*rho_v^0.5*(sigma*g*( rho_l-rho_v))^(1/4))   //W/mˆ2 12   Q_by_A_max=Q_by_A_max/(10^6) //MW/ ( s q m) 13   printf ( ” \n Pea k h e a t f l u x i s %f MW/ s q 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 17

clc ;   clear ;

  / / E x am pl e 3 . 4 4   rho_l=957.9 // [ kg/mˆ 3 ]   lambda=2257 // [ kJ/ kg ] // [ J/ kg ]   lambda=lambda*10^3   rho_v=31.54 // [ kg/mˆ 3 ] // [ kJ/ kg .K]   Cpv=4.64 // [ J/ kg .K]   Cpv=Cpv*10^3   kv=58.3*10^-3 // [W/ (m.K) ] // [m/ s ˆ 2 ]   g=9.81   mu_v=18.6*10^-6 // [ kg /(m. s ) ] // Emis sivit y   e=1.0        

sigma=5.67*10^-8; // [K] Ts=373 // [K] Tw=628 // [K] dT=Tw-Ts

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) / / C o n v e c t i v e h e a t

transfe r coeff

[W/ s q m. K ] //Radiation heat

21   hr=e*sigma*(Tw^4-Ts^4)/(Tw-Ts) 22   23   24   25   26  

t r a n s f e r c o e f f i n [W/ s q m.K ] // Total heat t ra n s fe r c o e f f i c i e n t W h=hc+(3/4)*hr / ( s q m . K) // Heat d i s s i p a t i o n r a t e p e r u n i t Q_by_l=h*%pi*D*dT l e n g t h i n [ kW/m ] printf ( ” \n S t a b l e f i l m b o i l i n g p o i nt h e a t t r a n s f e r c o e f f i c i e n t i s %f W/( sq m.K) ” ,h); // [kW/m] Q_by_l=Q_by_l/1000 printf ( ” \n Heat d i s s i p a t e d p er u n i t l e n g th o f t h e h e a t e r i s %f kW/m” ,Q_by_l);

Scilab code Exa 3.45  Heat transfer in tube

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

clc ;   clear ;

  / / E x ma pl e 3 . 4 5 // [K]   dT=10 // [ kPa ]   P=506.625 // [ Mpa ]   P=P/10^3 / / D i a m e t e r [ mm ]   D=25.4 // [m]   D=D/1000   h=2.54*(dT^3)*(%e^(P/1.551))

// [W/ sq m.K]

  //Q=h∗ %p i ∗D∗L∗dT   / / Heat t r a n s f e r r a te p er m e t e r l e n g th o f t u b e i s // [W/m]   Q_by_l=h*%pi*D*dT   printf ( ” \n R ate o f h ea t t r a n s f e r p e r 1m l e n g t h o f   t u b e i s %f W/m” , round (Q_by_l));

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 ;

  / / E x am pl e 3 . 4 6 // [K]   dT=8 // [ Mpa ]   P=0.17 // [ kPa ]   P=P*1000   h1=2847 // [W/ ( sq m. K) ] // [ kPa ]   P1=101.325 // [W/ sq m.K]   h=5.56*(dT^3)   Q_by_A=h*dT // [W/ sq m] // [W/ sq m.K ]   hp=h*(P/P1)^(0.4)   / / C or re po nd in g h ea t f l u x i s :   Q_by_A1=hp*dT // [W/ sq m] per=(Q_by_A1 -Q_by_A) *100/Q_by_A // P e rc e nt i n c r e a s e

i n heat f l ux 15   printf ( ” \ nHeat f l u x when p r e s s u r e i s 1 0 1 .3 2 5 kPa i s %f W/ sq m\n ” ,Q_by_A); 16   printf ( ” \n P e r c e n t i n c r e a s e i n h e a t f l u x i s %f   p e r c e n t ” , round ( p e r ) ) ;

Scilab code Exa 3.47   Dry steam condensate

1 2 3 4 5 6 7 8

clc ;   clear ;

  / / E x am pl e 3 . 4 7          

// [N. s /mˆ2 ] mu=306*10^-6 k=0.668   // [W/m.K] rho=974 // [ kg /mˆ 3 ] lambda=2225 // [ kJ/ kg ] // [ J/ kg .K] lambda=lambda*10^3 93

9 10 11 12 13 14 15

             

// [m/ s ˆ 2 ] g=9.81 // [K] Ts=373 // [K] Tw=357 // [K] dT=Ts-Tw // [mm] Do=25 // [m] Do=Do/1000 h=0.725*((rho^2*g*lambda*k^3)/(mu*Do*dT))^(1.0/4.0) // [W/ sq m.K ]

16 17 18 19 20

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

  printf ( ” \nMean h e a t t r a n s f e r c o e f f i c i e n t i s %f W/ ( s q m.K) \n ” ,h); 21   printf ( ” \ nHeat t r a n s f e r p er u n i t l e n g th i s %f W/m\n ” ,Q_by_l); 22   printf ( ” \ n C o n de n s at e r a t e p er u n i t l e ng t h i s %f k g/ h ” ,m_dot_byl);

Scilab code Exa 3.48   Laminar Condensate film

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

clc ;   clear ;

  / / E x am pl e 3 . 4 8   rho=960 // [ kh/mˆ 3 ] // [ kg /(m. s ) ]   mu=2.82*10^-4 // [W/ (m.K) ]   k=0.68   lambda=2255 // [ kJ/ kg ] // [ J/ kg ]   lambda=lambda*10^3 / / S a t u r a t i o n t e m pe r a tu r e o f s te am [ K]   Ts=373 // [K]   Tw=371 // [K]   dT=Ts-Tw / / D i m e ns i o n [ m]   L=0.3 // [m/ s ˆ 2 ]   g=9.81   h=0.943*(rho^2*g*lambda*k^3/(L*mu*dT))^(1/4)

94

 

/ /W/

sq m.K 15   A=L^2 // [ sq m] // [W]= [ J/ s ] Q=h*A*(Ts-Tw) / / C o n d e n sa t e r a t e [ k g / s ] m_dot=Q/lambda // [ kg /h ] m_dot=m_dot*3600 printf ( ” \n A ve ra g e h e a t t r a n s f e r c o e f f i c i e n t i s %f W / ( s q m . K) \n” ,h); 20   printf ( ” \ nHeat t r a n s f e r r a t e i s %f J /kg \n ” ,Q); 21   printf ( ” \n S team c o nd e ns a te r a t e p er h ou r i s %f k g/ h \n ” ,m_dot); 16 17 18 19

       

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 ;

  / / E Xa mp le 3 . 4 9 // [ kg/mˆ 3 ]   rho=1174   k=0.069 // [W/ (m.K) ] // [N. s /mˆ2 ]   mu=2.5*10^-4   lambda=132*10^3 // [ J/ kg ] // [m/ s ˆ 2 ]   g=9.81 // [K]   Ts=323 // [K]   Tw=313 // [K]   dT=Ts-Tw / / F or s q u a r e a r r a y , n=4 n =4   // n umber o f t u b e s // [mm]   Do=12 // [m]   Do=Do/1000   h=0.725*(rho^2*lambda*g*k^3/(n*Do*mu*dT))^(1/4)

/ ( s q m . K) 18 // For h ea t t r a n s f e r a r ea c a l c u a l t i o n , n=16 19   A=n*%pi*Do // [ sq m] 20   A=0.603 21   Q=h*A*dT // [W/m]

95

//W

22 23 24 25

       

// [ kg/ s ] m_dot=Q/lambda / / A p p r i xi m a t i o n i n bo ok m_dot=0.049 // [ kg /h ] m_dot=m_dot*3600 printf ( ” \n R ate o f c on d e n s a t i o n p er u n i t l e n g th %f k g / h” ,m_dot);

is

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 26

clc ;   clear ;

  / / E x am pl e 3 . 5 0   rho=960 // [ kg /mˆ 3 ] // [W/m.K ]   k=0.68 // [ kg /(m. s ) ]   mu=282*10^-6 / / Tube w a l l t e m p e r at u r e [ K]   Tw=371 / / S a t u r a t i o n t e m pe r a tu r e i n [ K]   Ts=373 // [K]   dT=Ts-Tw // [ kJ/ kg ]   lambda=2256.9 // [ J/ kg ]   lambda=lambda*10^3 / / F o ra s q u a r e a r r a y w i t h 1 0 0 t u b e s , n =10   Do=0.0125   // [m] // [m/ s ˆ 2 ]   g=9.81   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; / / [ mˆ 2 /m l e n g t h ] A=n*%pi*Do*L // H ea t t r a n s f e r r a t e i n [W/m] Q=h*A*dT ms_dot=Q/lambda // [ kg/ s ] // [ kg /h ] ms_dot=ms_dot*3600 printf ( ” \n Mass r a t e o f stea m c o n de n s at i o n i s %d kg /

96

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

S o l u t i o n i n b o o k . Do i s w r o n g l y t a ke n a s 0 . 0 1 2 i n l i n e s 17 a nd 2 2 o f t he book , A ls o A i s w ro ng ly c a l c u l a t e d \n ” )

Scilab code Exa 3.51  Saturated tube condensate in a wall

1 2 3 4 5 6 7 8 9 10

clc ;   clear ;

  / / E x am pl e 3 . 5 1   rho=975 // [ kg /mˆ 3 ]   k=0.871   // [W/m.K] // [K]   dT=10 // [N. s /mˆ2 ]   mu=380.5*10^-6   lambda=2300 // [ kJ/ kg ] // L at en t h e at o f c o nd e ns a ti o n [   lambda=lambda*1000 J/ kg ] / / O u te r d i a m e t e r [mm]   Do=100 // [m]   Do=Do/1000 // [m/ s ˆ 2 ]   g=9.81 // f o r h o r i z o n t a l t ub e

11 12 13 14 15   h1=0.725*((rho^2*lambda*g*k^3)/(mu*Do*dT))^(1/4) 16 17  

//Average heat t r a n s fe r c o e f f i c i e n t // f o r v e r t i c a l t u b e / / h 2 = 0 . 9 4 3 ∗ ( ( r ho ˆ 2∗ lambda ∗ g ∗ k ˆ3 ) /(mu∗L∗dT) ) ˆ( 1/ 4) //Average heat tr a n s f er c o e f f i c i e n t // For v e r t i c a l t ub e h2=h1 // i m p l i e s t h at

18   19 20   L=(0.943*((rho^2*lambda*g*k^3)^(1/4))/(h1*((mu*dT) // [m] ^(1/4))))^4 21   L=0.29 / / A p p ro x im a te i n b oo k 22   h=0.943*((rho^2*lambda*g*k^3)/(mu*L*dT))^(1/4) // [W

/( sq m.K) ] 97

23 24 25 26 27 28

           

// A rea i n [mˆ 2 ] A=%pi*Do*L // H eat t r a n s f e r r a t e [W] Q=h*A*dT / / [ R at e o f c o n d e n s a t i o n ] i n [ k g / s ] mc_dot=Q/lambda // [ kg /h ] mc_dot=mc_dot*3600 printf ( ” \n T ube l e ng t h i s %f m\n ” ,L); printf ( ” \n Rate o f c on de ms at io n p er h ou r i s %f k g /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 ;

  / / E x am pl e 3 . 5 2 // For h o r i z o n t a l p o s i t i o n [ kg /h ]   m1_dot=50 // [mm]   Do=10 // [m]   Do=Do/1000 // [m] L =1   / / F or 1 00 t u b e s n=10   n=10;

//We know th at // m dot=Q/la mbd a=h∗A∗dT/lambda   / / m dot i s p r o p o r t i o n a l t o h   / / m 1 d ot p ro p t o h1   / / m2 dot p rop n t o h 2   //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 v e r t i c a l p o s i ti o n , Rate o f   c o n d e n s a t i o n i s %f kg /h ” ,m2_dot);

Scilab code Exa 3.53   Condensation on vertical plate

98

// [

1 2 3 4 5 6 7 8 9 10

               

clc ; clear ; rho=975 // [ kg /mˆ 3 ] k=0.671 // [W/ (m.K) ] // [N. s /mˆ2 ] mu=3.8*10^-4 // [K] dT=10 // [ J/ kg ] lambda=2300*10^3 L =1   // [m] // [m/ s ˆ 2 ] g=9.81 h=0.943*((rho^2*lambda*g*k^3)/(mu*L*dT))^(1/4)

/ ( s q m. K)

//W

/ / [W/ s q m. K]

11 12   printf ( ” \n ( i )− A v er a g e h e a t t r a n s f e r c o e f f i c i e n t i s %d W/(mˆ2.K) \n ” , round (h)); 13 14 / / L o c a l h e a t t r a n s f e r c o e f f i c i e n t 15   / / a t 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 )−L o c a l h e a t t r a n s f e r

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

0 . 5 m h e i g h t i s %d W/ ( s q m . K) \n” , round ( h ) ) ; 19   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 )−Fi lm t h i c k n e s s

99

i s %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 ;

  / / E xa mp le 4 . 1 // [ Emi ssi vit y ]   e=0.9 // [W/mˆ 2 .Kˆ 4 ]   sigma=5.67*10^-8 // [K]   T1=377 // [K]   T2=283   Qr_by_a=e*sigma*(T1^4-T2^4) // [W/ sq m]   printf ( ” Heat l o s s by r a d i a t i o n i s %d W/ s q m” , round ( Qr_by_a));

Scilab code Exa 4.2  Radiation from unlagged steam pipe

1 2 3 4 5

clc ;   clear ;

  / / E xa mp le 4 . 2 // Emis sivit y   e=0.9 // [K]   T1=393 100

6 7 8 9

       

// [K] T2=293 // [W/ sq m.K ] sigma=5.67*10^-8 Qr_by_a=e*sigma*(T1^4-T2^4)   / /W/ sq m printf ( ” \n R ate o f h e a t t r a n s f e r by r a d i a t i o n i s %f   W/ sq m” ,Qr_by_a);

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 ;

  / / E xa mp le 4 . 3   L=1;   // [m] e =0.8 ; / / E m i s s i v i t y                  

/ / [ mˆ 2 . K ˆ 4 ] s i g m a = 5 . 67 * 10 ^ - 8 ; // [K] T1=423; // [K] T2=300; // [mm] Do=60; // [m] Do=Do/1000; // [ sq m] A=%pi*Do*L / / App ro x i n b oo k [ mˆ 2 ] A=0.189 // [W/m] Qr=e*sigma*A*(T1^4-T2^4) printf ( ” \n Net r a d i a i t o n r a t e p er 1 m et r e l e n g t h o f   p i p e 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 ;

  / / E xa mp le 4 . 4 // Emis sivit y   e=0.9 L =1   // [m] // [mm]   Do=50 101

7 8 9 10 11 12 13 14 15

                 

// [m] Do=Do/1000 // [W/(mˆ 2 .Kˆ4 ) ] sigma=5.67*10^-8 // [K] T1=415 // [K] T2=290 // [K] dT=T1-T2 // [W/ sq m.K ] hc=1.18*(dT/Do)^(0.25) // A rea i n [ s q m] A=%pi*Do*L Qc=hc*A*dT   / / Hea t l o s s by c o n v e c t i o n W/m // Heat l o s s by r a d i a t i o n Qr=e*sigma*A*(T1^4-T2^4)

p e r l e n g t h W/m 16   Qt=Qc+Qr / / T o t a l h e a t l o s s i n [W/m ] 17   printf ( ” \n T ot al h ea t l o s s by c o n v ec t i o n i s %f W/m” , Qt);

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 ;

  / / E xa mp le 4 . 5                

e=0.85 // [W/ sq m.K ] sigma=5.67*10^-8 // [K] T1=443 // [K] T2=290 // [K] dT=T1-T2 //W/ sq m.K hc=1.64*dT^0.25 // [mm] Do=60 // [m] Do=Do/1000 L =6   // L en g th [ m] // S u r f a ce a r e a o f p i p e i n [ s q m]   A=%pi*Do*L // Rate o f h e a t l o s s by   Qr=e*sigma*A*(T1^4-T2^4)

radiaiton W 15   Qc=hc*A*(T1-T2)   // Rate o f h ea t l o s s by c o n v e c t i o n [

W] 16   Qt=Qr+Qc // T ot al h e at l o s s [W] 17   printf ( ” \n T ot al h ea t l o s s i s %d W” , round ( Q t ) )

102

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 ;

  / / E X am pl e 4 . 6 // [W/mˆ 2 .Kˆ 4 ] sigma=5.67*10^-8 e1=0.79; e2=0.93; T 1 = 50 0 ; // [K] T 2 = 30 0 ; // [K] // [mm] D=70 // [m] D=D/1000 L =3   // [m] // S i de o f c o nd u i t [m]   W=0.3 // [ sq m]   A1=%pi*D*L // A pp ro xim ate c a l c u l a t i o n   A1=0.659              

i n book i n

[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 l o s t by r a d i a t i o n i s %f W” ,Q);

Scilab code Exa 4.7  Net radiant interchange

1 2 3 4 5 6

clc ;   clear ;

  / / E xa mp le 4 . 7   sigma=5.67*10^-8 // [K]   T1=703 // [K]   T2=513

// [W/ sq m.Kˆ 4 ]

103

7 8 9 10

       

e1=0.85 e2=0.75 Q_by_Ar=sigma*(T1^4-T2^4)/(1/e1+1/e2-1) // [W/ sq m] printf ( ” \n Net r a d ia n t i n t er c h an g e p er s q u a r e m et 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 12 13

clc ;   clear ;

  / / E xa mp le 4 . 8   L =3 ; // [m] / / Area i n [ s q m]   A=L^2 // [W/ sq m.Kˆ4 ]   sigma=5.67*10^-8; // [K]   T1=373; // [K]   T2=313;          

e1=0.736; e2=e1; F12=1/((1/e1)+(1/e2)-1) // [W] Q=sigma*A*F12*(T1^4-T2^4) printf ( ” \n Net r a d i a n t i n t e r c h an g e );

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

i s %d W” , round (Q)

9 10 11 12

       

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

// Heat l o s s p er u n it

A re a [W/ s q m] 13   printf ( ” \ nRate o f h e at l o s s when o f s i l v e r e d s u r f a c e i s %f W/ s q m” ,Q_by_A); 14   / /When b ot h t he s u r f a c e s a r e b l a ck 15 16 17 18 19

         

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 When b ot h s u r f a c e s a r e b la c k , R ate o f h e at los s i s %d W/ s q 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 18

clc ;   clear ;

  / / E x am pl e 4 . 1 0                

e1=0.05 e2=e1 A1=0.6944; A2=1; // [K] T1=293 // [K] T2=90 // [W/mˆ 2 .Kˆ 4 ] sigma=5.67*10^-8 / / D ia me te r i n [m] D=0.3

           

F12=1/(1/e1+(A1/A2)*(1/e2-1)) // [W/ sq m] Q_by_A=sigma*F12*(T1^4-T2^4) // [ kJ /h ] Q=Q_by_A*%pi*(D^2) // [ kJ/ h ] Q=Q*3600/1000 / / L a t en t h e a t i n [ kJ / kg ] lambda=21.44 m_dot=Q/lambda   / / k g / h

105

19   printf ( ” \n The l i q u i d / h ” ,m_dot);

o x yg e n w i l l e v a p or a t e a t %f k g

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 ;

// Example4 .1 1                    

sigma=5.67*10^-8   //W/(mˆ2.Kˆ4) e1=0.3; e2=e1; // [m] D1=0.3 // [m] D2=0.5 // [K] T1=90 // [K] T2=313 A1=%pi*D1^2   / / A rea i n [ s q m] A2=%pi*D2^2 / / Area i n [ s q m] Q1=sigma*A1*(T1^4-T2^4)/(1/e1+(A1/A2)*(1/e2-1))

// [

W] 14 Q1 = abs ( Q 1 ) ;   / / A b s o l u t e v a l u e i n [W] 15   printf ( ” \n R ate o f h ea t f lo w due t o r a d i a t i o n i s %f   W” ,Q1); 16 / /When A lu mi ni um i s u s e d 17   e1=0.05 18   e2=0.5 19   Q2=sigma*A1*(T1^4-T2^4)/(1/e1+(A1/A2)*(1/0.3-1)) //

[W] 20 Q2 = abs ( Q 2 )   / / A b s o l u te v a l u e i n [W] 21   Red=(Q1-Q2)*100/Q1 / / P e r c e nt r e d u c t i o n 22   printf ( ” \n R e du ct io n i n h ea t f l ow w i l l be %f p e rc e n t ” ,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 ;

  / / E x am pl e 4 . 1 2                        

// [W/ sq m.Kˆ 4 ] sigma=5.67*10^-8 // [K] T1=77 // [K] T2=303 D1=32   //cm // [m] D1=D1/100 // [ cm ] D2=36 // [m] D2=D2/100 A1=%pi*D1^2 // [ sq m] 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 21

// [ kJ/ h ]   Q=Q*3600/1000 // [ kJ /h ] Q= abs (Q);   lambda=201   / / k J / k g / / E v a p o ra t i o n r a t e i n [ kg / h ]   m_dot=Q/lambda   printf ( ” \n N i tr o ge n e v a p or a t es a t %f k g /h ” ,m_dot);

Scilab code Exa 4.13  Evaporation in concenric vessels

1 2 3 4 5 6 7 8 9 10

clc ;   clear ;

  / / E x am pl e 4 . 1 3 / / I n n e r s p h e r e i d a m e t e r [mm]   D1=250 / / O ut er d i a m e t e r [ m]   D1=D1/1000 // [mm]   D2=350 // [m]   D2=D2/1000 //W/( sq m.Kˆ 4)   sigma=5.67*10^-8   A1=%pi*D1^2 // [ sq m] 107

11 12 13 14 15 16

           

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

// [W] 17   Q=-2.45   18 19 20 21 22

       

//Approximate // [W] Q= abs (Q ) // [ kJ/ h ] Q=Q*3600/1000 lambda=200   / / k J / k g // [ kg /h ] Rate=Q/lambda printf ( ” \n Ra te o f e v a p o r a t i o n i s %f kg / h ( a pp ro x ) ” , Rate);

Scilab code Exa 4.15  infinitely long plates

1 2 3 4 5 6 7 8 9

clc ;   clear ;

  / / E x am pl e 4 . 1 5            

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

// [W/ s q

m] 10   / / Q 1 b y a = s i g m a ∗ (T1ˆ4−T2ˆ 4) /( (1 / e1 ) +(1/ e2 ) −1)=sigma ∗ A∗ (T2ˆ4−T3ˆ 4) /( (1 / e2 ) +(1/ e3 ) −1 ) / / [W/ s q m ] 11 12 13 14

  e2=0.5

/ / S o l v i n g we g e t // [K]   T2=((6/9.5)*((3.5/6)*T3^4+T1^4))^(1/4)   Q1_by_a=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1) // [W/ s q m]

15 red=(Q_by_a -Q1_by_a) *100/Q_by_a

108

16   printf ( ” \ n He at t r a n s f e r

rat e per SHIELD ) due t o r a d i a t i o n i s %f 17   printf ( ” \ n Heat t r a n s f e r r a t e p e r SHIELD ) due t o r a d i a t i o n i s %f

unit W/ s q unit W/ s q

a r e a (WITHOUT m\n ” ,Q_by_a); a r e a (WITH m\n ” ,Q1_by_a)

; 18   printf ( ” \ n Re d uc ti on i n h ea t l o s s ;

i s %f p e rc e nt ” ,red)

Scilab code Exa 4.16  Heat exchange between parallel plates

1 2 3 4 5 6

clc ;   clear ;

  / / E x am pl e 4 . 1 6 // I n s t ea d y s t a te , we ca n w r i t e :   //Qcd=Qdb // sigma (Tcˆ4−Tdˆ4) ∗ /( 1/ ec +1/ed −1)=si gm a (Tdˆ4−Tbˆ4) /( 1/ ed+1/eb −1) 7 / / i . e Td ˆ 4 = 0 . 5∗ (Tcˆ4−Tbˆ4) 8 // Given : 9   Ta=600 // [K] 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

       

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

// For a i r   //(600ˆ4 − Tcˆ 4) /2.25 =( Tcˆ4−Td ˆ 4 ) / 3 . 5   / / 1 . 5 6∗ ( 6 0 0 ˆ 4 − Tcˆ4)=Tcˆ4−Tdˆ4 // P ut t in g v a lu e o f Td i n t er ms o f Tc   / / 1 . 5 6∗ ( 6 0 0 ˆ 4 − 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 // [K] Tc = fsolve (500,f);

// or / / [ K] A pp ro xi ma te a f t e r   Tc=560.94 Td = sqrt ( sqrt (0.5*(Tc^4-300^4))) 109

sol ving // [K]

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

// [W/ s q

m] 26   printf ( ” \ n Ra te o f h e a t e x ch a n ge p e r u n i t a r e a=%f W/m ˆ2 ” ,Q_by_a); 27   printf ( ” \ n S t e a d y s t a t e t e m p e r a t u r e s , T c=%f K , a nd Td= %f K” ,Tc,Td);

Scilab code Exa 4.17  Thermal radiation in pipe

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

clc ;   clear ;

  / / E x am pl e 4 . 1 7 // [W/( sq m.Kˆ 4) ] sigma=5.67*10^-8 e=0.8 // [K] T1=673; // [K] T2=303; // [mm] Do=200 // [m] Do=Do/1000 / / L et [m ] L =1   A1=%pi*Do*L   // [mˆ2/m]

           

  / / CAse 1 : P ip e t o s u r r un d i n g s   Q1=e*A1*sigma*(T1^4-T2^4)   //Approximated   Q1=5600

// [W/m]

//Q1=5600 / / [W/m] a p p r o x i m a t e d i n book f o r c a l c u l a t i o n p ur po se 17 // C o nc e nt r i c c y l i n d e r s e1=0.8; e2=0.91; // [m] D1=0.2 // [m] D2=0.4 Q2=sigma*0.628*(T1^4-T2^4)/((1/e1)+(D1/D2)*((1/e2) // [W/m] le ng th -1)) 23   Red=Q1-Q2 // R ed uc ti on i n h ea t l o s s 24 18 19 20 21 22

         

110

25   printf ( ” \nDue t o t he rm al r a d i a i t o n , L os s o f h ea t t o s u r r o u n d i n g i s %d W/m\n ” , round (Q1)); 26   printf ( ” \nWhen p i p e i s e n c l o s e d i n 1 4 00 mm di a m et e r b r i c k c o nd u it , L o ss o f h e at i s %d W/m\n ” , round ( Q2 )) ; 27   printf ( ” \n R ed uc ti on i n h ea t l o s s i s %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 ;

  / / E x am pl e 4 . 1 8

               

// [W/( sq m.Kˆ 4) ] s i g m a = 5 . 67 * 10 ^ - 8 ; // [K] T1=813; // [K] T2=473; e1=0.87; e2=0.26; D 1 = 0 .2 5 ; // [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 t r a n s f e r by r a d i a i t o n i s %d W/ s q m” , Q_by_a1);

Scilab code Exa 4.19  Heat exchange between black plates

1 2 clc ;

111

3 4 5 6 7 8 9 10 11

  clear ;

  / / E x am pl e 4 . 1 9              

// [W/ sq m.Kˆ 4 ] sigma=5.67*10^-8 // [ sq m] A1=0.5*1 F12=0.285 T1=1273 // / [K] // [K] T2=773 // [W] Q=sigma*A1*F12*(T1^4-T2^4) printf ( ” \n N et r a d i a n t h ea t e xc ha ng e b et wee n p l a t e s 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 ;

  / / E x am pl e 4 . 2 0          

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

// [W/ sq m.Kˆ 4 ]

  / / Heat t r a n s f e r w i th ou t s h i e l d :   Q_by_a=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1)

m] 12 13 14 15 16 17 18 19 20 21

// Heat t r a n s f e r w i th s h i e l d : //Resistance 1   R1=(1-e1)/e1   F13=1;   R2=1/F13

//Resistance 2

  e3=0.05   R3=(1-e3)/e3

//Resistance 3

112

// [W/ s q

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

  R4=(1-e3)/e3

//Resistance 4

  F32=1;   R5=1/F32

//Resistance 5

  R6=(1-e2)/e2

// R e s i s t a n c e 6 // T ot al r e s i s t a n c e

  Total_R=R1+R2+R3+R4+R5+R6

  Q_by_as=sigma*(T1^4-T2^4)/Total_R Red=(Q_by_a -Q_by_as) *100/Q_by_a

// [W/ sq m] / / R ed u ci to n i n

h ea t t r a n f e r due t o s h i e l d 34 35   printf ( ” \n R ed uc t i o n i n h ea t t r a n s f e r

r a te a s a r e s u l t o f r a d i a i o t n s h i e l d i s %f p er ce nt ” ,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 16

clc ;   clear ;

  / / E x am pl e 4 . 2 1   e1=0.3   e2=0.8

  / / L e t s i gm a ∗ (T1ˆ4−T2 ˆ 4 ) =z = 1 ( c o n s t )   z=1;   / / L e t //W/ sq m   Q_by_A=z/(1/e1+1/e2-1)   / / Heat t r a n s f e r w i t h r a d i a t i o n s h i e l d   e3=0.04   F13=1;   F32=1;

// The r e s i s t a n c e s a re :   R1=(1-e1)/e1   R2=1/F13

113

17 18 19 20 21 22 23

           

R3=(1-e3)/e3 R4=R3 R5=1/F32 R6=(1-e2)/e2 // T ot al r e s i s t a n c e R=R1+R2+R3+R4+R5+R6 Q_by_As=z/R   / / w h er e z=s i g m a ∗ (T1ˆ4−T2ˆ 4) //W/ sq m // Per cen t red=(Q_by_A -Q_by_As) *100/Q_by_A

r e d u c t i o n i n h ea t t r a n s f e r 24   printf ( ” \n The h ea t t r a n s f e r i s r ed uc ed by %f   p e rc e n t due t o s h i e l d ” ,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 22

clc ;   clear ;

  / / E x am pl e 4 . 2 2                        

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

// R e s i s t a n c e i n t he n et wo rk a re c a l c u l a t e d a s :          

R1=1-e1/(e1*A1) R2=1-e2/(e2*A2) R3=1/(A1*F12) R4=1/(A1*F13) R5=1/(A2*F23)

114

23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

  R6=0

/ / G i v en (1− e 3 ) / e 3 ∗A3=0

  //Also //W/ sq m   Eb1=sigma*T1^4 // [W/ sq m]   Eb2=sigma*T2^4   Eb3=sigma*T3^4 // [W/ sq m] // Equations are :   / / ( E b 1−J 1 ) / 2 + ( J 2−J 1 ) / 7 . 0 1 8 + ( Eb3−J 1 ) / 2 . 7 9 7 = 0   / / ( J 1−J 2 ) / 7 . 0 1 8 + ( Eb3−J 2 ) / 2 . 7 9 7 + ( Eb2−J 2 ) / 2 = 0 / /On s o l v i n g we g e t : // [W/ sq m]   J1=33515 // [W/ sqm ]   J2=15048 // [W/ sq m]   J3=Eb3 // [W/ sq m]   Q1=(Eb1-J1)/((1-e1)/(e1*A1)) // [W/ sq m]   Q2=(Eb2-J2)/((1-e2)/(e2*A2))   Q3=(J1-J3)/(1/(A1*F13))+(J2-J3)/(1/(A2*F23))

// [W

/sq m] 40   printf ( ” \n T o ta l h ea t l o s t by p l a t e 1 i s %f W/ s q m\n ” ,Q1); 41   printf ( ” \n T o ta l h ea t l o s t by p l a t e 2 i s %f W/ s q m\n ” ,Q2); 42   printf ( ” \nThe n et e ne rg y l o s t by b o th p l a t e s must be a b s o r b e d by t h e room , \ n %f=%f” ,Q3,Q1+Q2)

Scilab code Exa 4.23  Radiation loss in plates

1 2 3 4 5 6 7 8

clc ;   clear ;

  / / E x am pl e 4 . 2 3          

sigma=5.67*10^-8 e1=0.7; e2=0.7; // [K] T1=866.5 // [K] T2=588.8

// [W/ sq m.Kˆ 4 ]

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 w it h n s h e l l s =1/( n+1)        

21  

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/ // [W/ sq m] es2)-(n+1)) printf ( ” \n New R a d i a it o n l o s s i s %f W/ s q 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 ;

  / / E x am pl e 4 . 2 4 // 1 .WITHOUT SHIELD              

sigma=5.67*10^-8 e1=0.12; e2=0.15; // [K] T1=100 // [K] T2=300 // [m] r1=0.015 // [m] r2=0.045 // [m] L =1 // [ sq m]   A1=2*%pi*r1*L   Q_by_L=2*%pi*r1*sigma*(T1^4-T2^4)/(1/e1+(r1/r2)*(1/ // [W/m] e2-1))

15 //−ve s a i g n

i n d i c a t e s t ha t t h e n e t h e a t f lo w i s i n th e r a d i a l inward d i r e c t i o n 116

16 17 18 19 20

// 2 .WITH CYLINDRICAL RADIATION SHIELD        

e3=0.10; e4=0.05; // [m] r3=0.0225 Qs_by_L=2*%pi*r1*sigma*(T1^4-T2^4)/(1/e1+r1/r2*(1/e2 // [W/ sq m] -1)+(r1/r3)*(1/e3+1/e4-1)) 21   red=( abs ( Q _ b y _ L ) - abs (Qs_by_L))*100/ abs (Q_by_L) //

p e rc e n t r e d u c t io n i n h ea t g a i n 22 23 24 25 26 27 28 29 30 31 32 33 34 35

/ / R a d i a t i o n n e tw o r k a p p r oa c h // [ sq m]   A3=2*%pi*r3 // [ 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)

  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 With c y l i n d r i c a l r a d i a i t o n s h i e l d Heat

g a i ne d by f l u i d p er 1 m l en gh o f t ub e i s %f W/m\n ” ,Qs_by_L); 37   printf ( ” \ n Pe rc en t r e d uc t i o n i n h ea t g a i n i s %f   p e r c e n t \n ” ,red); 38   printf ( ” \ nWith r a d i a i t o n n e t wo r k a p p r oa c h %f W/ sqm ” ,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 21

clc ;   clear ;

  / / E xa mp le 5 . 1 // [mm]   Di=35 // [m]   Di=Di/1000 // [mm]   Do=42 // [m]   Do=Do/1000 / / f o r b e n ze n e // [ kg /h ]   mb_dot=4450 // [ kJ /( kg .K) ]   Cpb=1.779 // [K]   t2=322 // [K]   t1=300 / / f o r b e nz e ne i n [ kJ / h ]   Q=mb_dot*Cpb*(t2-t1)   / / For t o u l e n e // [K]   T1=344 // [K]   T2=311 // [ kJ/ kg .K]   Cpt=1.842 // [ kg /h ]

  mt_dot=Q/(Cpt*(T1-T2))

118

22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

  Q=Q*1000/3600

// [W]

/ / Hot f l u i d ( t o l u e n e ) / / C o ld f l u i d ( b e n z e n e ) // [K]   dT1=22 // [K]   dT2=11   dTlm=(dT1-dT2)/( log (dT1/dT2))

// [K]

           

/ / C lo d f l u i d : I n n e r p i pe , b e nz e ne // [m] Di=0.035 / / F lo w a r e a [ s q m ] Ai=(%pi/4)*Di^2 / / Mass v e l o c i t y [ kg /mˆ 2 . h ] Gi=mb_dot/Ai // [ kg /mˆ 2 . s ] Gi=Gi/3600 // [ kg /(m. s ) ] mu=4.09*10^-4 / / R e y n o l d s n um be r Nre=Di*Gi/mu

           

// [ J /( kg .K) ] Cp=Cpb*10^3 // [W/m.K ] k=0.147 / / P r a n d t l n um be r Npr=Cp*mu/k // [W/ sq m.K ] hi=(k/Di)*0.023*(Nre^0.8)*(Npr^0.4) // [W/ sq m.K ] hio=hi*Di/Do // O ut si de d i a o f i n s i d e p ip e D1=0.042 [mm]

43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

                               

// I n s i d e d i a o f o u t s i de p i p e [m] D2=0.0525 // [m] De=(D2^2-D1^2)/D1   //Approximated De=0.0236 / / F lo w a r e a [ s q m ] aa=%pi*(D2^2-D1^2)/4 / / Mass v e l o c i t y i n [ kg /mˆ 2 . h ] Ga=mt_dot/aa // [ kg /mˆ 2 . s ] Ga=Ga/3600 // [ kg /(m. s ) ] mu=5.01*10^-4 / / R e y n o l d s n um be r Nre=De*Ga/mu / / P r a n d t l n um be r Npr=Cp*mu/k // [W/ sq m.K ] ho=(k/De)*0.023*(Nre^0.8)*(Npr^0.3) // [W/ sq m.K ] Uc=1/(1/ho+1/hio) / / F o u l i n g f a c t o r [ mˆ 2 . K/W] Rdi=1.6*10^-4 / / F o u l i n g f a c t o r [ mˆ 2 . K/W] Rdo=1.6*10^-4   / / (mˆ2 .K/W) Rd=Rdi+Rdo // [W/ sq m.K ] Ud=1/(1/Uc+Rd) // s q m A=Q/(Ud*dTlm) 119

59 60 61 62 63

         

// [ sq m] ex=0.136 //m l=A/ex // T ot al l e n g t h o f o ne h ar p in o f 6m [m] tl=12 printf ( ”b%f” ,l); printf ( ” \n\ R e qu i r ed s u r f a c e i s f u l f i l l e d by c o n n ec t i ng %d( t h r e e ) 6m h a r pi n s i n s e r i e s  \ n ” , round (l/tl))

Scilab code Exa 5.2  Length of pipe

1 2 3 4

clc ;   clear ;

  / / E xa mp le 5 . 2 // Mass f l o w r a t e o f a c id

  ma_dot=300*1000/24

in [ kg/h] // Mass f l ow r a t e o f  

5   mw_dot=500*1000/24

w a te r i n [ kg / h ] 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

                       

// [ kJ/ kg .K] Cp1=1.465 // [K] T1=333 // [K] T2=313 // [ kJ /h ] Q=ma_dot*Cp1*(T1-T2) // [W] Q=Q*1000/3600 // [ kJ/ kg .K] Cp2=4.187 // [K] t1=288 // [K] t2=(Q/(mw_dot*Cp2))+t1 // [K] dT1=T1-t2 // [K] dT2=T2-t1 dTlm=(dT1-dT2)/ log (dT1/dT2)   // [K] / / A p pr o xi m at i on i n [ K] dTlm=32.26 //Inner pipe

         

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

m_dot=12500 Di=0.075 Ai=(%pi/4)*Di^2 G=ma_dot/Ai G=G/3600

120

24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

                                         

// [ kg /m. s ] mu=0.0112   //W/(m.K) k=0.302 / / R e y n o l d n um um be be r Nre=Di*G/mu / / P r a n d t l n um um be be r Npr=Cp1*10^3*mu/k //W // W/ sq m.K hi=(k/Di)*0.023*(Nre^0.8)*(Npr^0.3) // [m] Do=0.1 //W // W/ sq m.K hio=hi*Di/Do // [m] D1=0.1 // [m] D2=0.125 // [m] De=(D2^2-D1^2)/D1 // [ sq m] Aa=(%pi/4)*(D2^2-D1^2) // [ kg/ kg/mˆ2 mˆ2 . h ] Ga=mw_dot/Aa // [ kg/ sq m. s ] Ga=Ga/3600 // [ kg /m. s ] mu=0.0011 / / R e y n o l d s n um um be be r Nre=De*Ga/mu / / f o r w a te te r k=0.669 / / P r a n d t l n um um be be r Npr=Cp2*10^3*mu/k // [W/ sq m.K ] ho=(k/De)*0.023*(Nre^0.8)*Npr^0.4 // [m] xw=(Do-Di)/2 // [m] Dw=(Do-Di)/ l o g (Do/Di) / / t h er e r m al al c o n d u c t i v i t y kw=46.52 o f w a l l i n [W/m . K] K]

45   Uc=1/(1/ho+1/hio+xw*Do/(kw*Dw)) // [W/ sq m.K ] 46   Ud=Uc / / As As d i r t f a c t o r v a l u es es

a r e n ot o t g i v en en 47   Ud=195.32  

//Approximation // [ sq m]

48   A=Q/(Ud*dTlm) 49 50   L=A/(%pi*Do) // [ sq m] 51   printf ( ” \ nArea =%f mˆ2 , \ n Le L e ng n g th t h f o p i p e r e q u i r e d =% f   m( app rox ) ” ,A,L)

Scilab code Exa 5.3  Double pipe heat exchanger

1 clc ;

121

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

  clear ;

  / / E xa x a mp m p le le 5 . 3 // [ kg /h ]   m e _ d ot ot = 5 5 0 0 ; // [ kg/ s ]   me_dot1=me_dot/3600 / / I . D o f i n n er e r p i p e i n [m] Di =0.037 ; // [ sq m]   Ai=(%pi/4)*Di^2 // [ kg/ sq m. s ]   G=me_dot1/Ai / / [ Pa . s ] o r [ k g / ( m. m. s ) ] m u =3 = 3 . 4 * 1 0 ^ -3 -3 ; / / R e y n o l d s n um um be be r   Nre=Di*G/mu Cp =2.68 ; // [ kJ/ kg .K] // [ J/ kg kgK K]   Cp1=Cp*10^3 k =0.248 ;   // [W/m.K] / / P r a n d t l n um um be be r   Npr=Cp1*mu/k / / Nr Nr e i s

                       

28   29   30   31 32 33 34 35 36 37

             

g r e a t e r t ha h a n 1 0 , 0 0 0 , U s e D i tt tt u s −B o e l t e r eqn : / / N u s s e l t n um um be be r Nnu=0.023*(Nre^0.8)*(Npr^0.3) // [W/ sq m.K ] hi=k*Nnu/Di // [K] T2=358 // [K] T1=341 // [ kJ/ kg .K] Cp2=1.80 // [K] t2=335 // [K] t1=303 // [ kg /h ] mt_dot=me_dot*Cp*(T2-T1)/(Cp2*(t2-t1)) // [ kg/ s ] mt_dot=mt_dot/3600 // [m] D1=0.043 / / I n s i d e d ia i a o f o u t er e r p ip ip e D2=0.064 / / E q u i v al a l e n t d i a me m e t er er [m De=(D2^2-D1^2)/D1 ] // [ sq m] Aa=%pi/4*(D2^2-D1^2) //kg /( sq m. s ) Ga=mt_dot/Aa // V i sc o si t y o f   mu2=4.4*10^-4 t o l u e n e Pa . s / / F o r t o l u e n e [W/m . K] K] k2=0.146 //J/ kg .K Cp2=1.8*10^3 / / R e y n o l d s n um um be be r Nre=De*Ga/mu2 / / P r a n d t l n um um be be r Npr=Cp2*mu2/k2 / / N u s s e l t n um um be be r Nnu=0.023*Nre^0.8*Npr^0.4 //W // W/( sq m.K) ho=k2*Nnu/De // [m] Dw=(D1-Di)/ l o g (D1/Di) 122

38   x=0.003 / / Wa Wa l l t h i c k n e s s i n [m] 39   Uo=1/(1/ho+(1/hi)*(D1/Di)+(x*D1/(46.52*Dw))) // [

W/ sq m.K] 40 41 42 43 44 45 46

             

// [K] dT1=38 // [K] dT2=23 // [K] dTlm=(dT1-dT2)/ l o g (dT1/dT2) // [ kJ/ s ] Q=me_dot1*Cp*(T2-T1) // [ J/ s ] Q=Q*1000 // [m] L=Q/(Uo*%pi*D1*dTlm) T o ta t a l l e ng n g g th t h o f d ou o u bl b l e p i p e h ea ea t printf ( ” \ n To e x ch c h a n ggee r i s % 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 ;

  / / E xa x a mp m p le le 5 . 4                      

// [ kg /h ] mc_dot=1000 // [ kg/ s ] mc_dot=mc_dot/3600 // [ kg /h ] mh_dot=250 // [ kg/ s ] mh_dot=mh_dot/3600 // [ J /( kg .K) ] Cpc=4187 / / [W [W/K /K]] Cph=3350 / / [W [W/K /K]] w=mc_dot*Cpc / / [W [W/K /K]] l=mh_dot*Cph C=mh_dot*Cph/(mc_dot*Cpc) // [W/ sq m.K ] U=1160 / / He Heat t r a n s f e r s u r f a c e A=0.25

f o r ex e x c ha h a ng n g er er i n

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

// E f f e c t i v e n e s s o f  

h e a t e x c h a ng ng e r 17   T1=393 / / I n l e t t em e m p er e r at a t u re r e i n [K [K ] 18   t1=283 / / C o o l i n g w a te t e r [ K] K] 19   T2=T1-E*(T1-t1) / / O ut u t le l e t T o f h ot ot l i q u i d 123

20 21   t2=C*(T1-T2)+t1 // [K] 22   printf ( ” \n\ n E f f e c t i v e n e s s o f h ea t e xc ha ng er i s %f \   n” ,E); 23   printf ( ” \ n Ou tl et t em pe ra tu re o f h o t l i q u i d i s %f \   n”, T2); 24   printf ( ” \ n Ou tl et t em p er a tu r e o f w at er i s %f \   n ” ,t2)

Scilab code Exa 5.5  Counter flow exchanger

1 2 3 4 5

clc ;   clear ;

  / / E xa mp le 5 . 5   Cpc=4187

// S p e c i f i c h ea t o f w a te r i n

[ J /( kg .K) ] 6   Cph=2000

// Sp h ea t o f o i l i n [ J / ( kg . K

)] // [ kg/ s ] // [ kg/ s ] // [W/K] // [W/K]   / / Heat c a p a c i t y o f r a t e o f h o t f l u i d i s s m a l l e r th a n 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

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] / / A p pr o xi m at ed i n b oo k

20   T2=291.83

124

w it h ou t p r e c i s e c a l c u l a t i o n // [K] t2=C*(T1-T2)+t1 // [W] Q=mh_dot*Cph*(T1-T2)   n ” ,E); printf ( ” \n\ n E f f e c t i v e n e s s o f e xc ha ng er i s %f \ printf ( ” \ n Ou tl et t em pe r at ur e o f o i l i s %f K\n ” ,T2); printf ( ” \ n Ou tl et t em p er a tu r e o f w at er i s %f K\n ” ,t2) ; 26   printf ( ” \ nRate o f h ea t t r a n s f e r i s %f W” ,Q); 21 22 23 24 25

         

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 24

clc ;   clear ;

  / / E xa mp le 5 . 6   printf ( ” \nLMTD Ap pr oa ch \n ” )                                

// [ J /( kg .K) ] Cph=4187 // Hot s i d e f l o w r a t e [ kg / s ] mh_dot=600/3600 // [ kg/ s ] mc_dot=1500/3600 // [ J/ kg .K] Cpc=Cph // [K] T1=343 // [K] T2=323 // [W] Q=mh_dot*Cph*(T1-T2) // [K] t1=298 t2=(mh_dot*Cph*(T1-T2))/(mc_dot*Cpc)+t1   // [K] // [K] dT1=45 // [K] dT2=17 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // H eat t r a n s f e r c o e f f i n [W/ s q m. K] hi=1600 // [W/ sq m.K] ho=hi // [W/ sq m.K ] U=1/(1/hi+1/ho) // [ sq m] A=Q/(U*dTlm)

  printf ( ” \ n E f f e c t i v e n e s s −NTU ap pro ac h \n” );

125

25 26 27 28 29 30 31 32 33 34 35

//hot water : // [W/K] // [W/K]   / / Heat c a p a c i t y r a t e o f h o t f l u i d i s s ma ll //   C=mh_dot*Cph/(mc_dot*Cpc) // E f f e c t i v e n e s s   E=(T1-T2)/(T1-t1) // f o r p a r a l e l l f lo w :   h=mh_dot*Cph   c=mc_dot*Cpc

       

ntu=- log (1-E*(1+C))/(1+C) // [ sq m] A2=(ntu*mh_dot*Cph)/U // [K] t2=C*(T1-T2)+t1 printf ( ” \n By LMTD a p p r oa c h a r e a o f h e a t e x c h a n ge r i s %f s q m\n ” ,A); 36   printf ( ” \nBy Ntu a pp ro ac h A rea o f h e at e x ch a n ge r i s %f s q m\n ” ,A); 37   printf ( ” \n O u t le t t e m pe r a tu r e o f c o l d w at er=%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 16

clc ;   clear ;

  / / E xa mp le 5 . 7                          

// [ kg/ s ] mw_dot=10 // [ kJ /( kg .K) ] Cpw=4.187 // [K] t2=318 // [K] t1=295 // [ kJ/ s ] Q=mw_dot*Cpw*(t2-t1) //W Q=Q*1000 // [K] dT1=98 // [K] dT2=75 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [W/ sq m.K] hi=850 / / I n s i d e d i a [m] id=0.027 / / O u t s i de d i a [ m] od=0.031 / / [W/ s q m . K ] hio=hi*id/od 126

17 18 19 20 21 22

// 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]   ho=6000 // [W/ sq m.K ]   Uo=1/(1/ho+1/hio) // [ sq m]   Ao=Q/(Uo*dTlm) / / L e ng t h [ m] L =4 // [ No . o f t u b es ]   n=Ao/(%pi*od*L)   printf ( ” \n Number o f t u b es r e q u i r e d = %d\n ” , round (n) );

Scilab code Exa 5.8  Order of Scale resistance

1 2 3 4

clc ;   clear ;

  / / E xa mp le 5 . 8   mdot=7250;

// N it r o b e nz en e i n s h e l l i n [ kg /

h] 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

                                     

// [ kJ /( kg .K) ] Cp=2.387; //Pa . s m u = 7 *1 0^ - 4 ; // [W/m.K ] k=0.151; // [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) // [ kJ /h ] Q=mdot*Cp*(T1-T2) // [W] Q=Q*1000/3600 // no o f t u be s n=166; // [m] L=5; // [m] Do=0.019; // [m] Di=0.015 // [ sq m] Ao=n*%pi*Do*L // [W/ sq m.K ] Uo=Q/(Ao*dTlm) Ud=Uo // 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

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

       

// S h e l l s i d e c r o s s f l o w a re a // [m] B=0.15 // [m] id=0.45 // [ sq m] as=id*C_dash*B/Pt // As t h e r e a re two s h e l l p as se s , a re a p er p as s i s : // [ sq m] as_dash=as/2

  / / E q ui v al e nt d i am e te r o f s h e l l   De=4*(Pt^2-(%pi/4)*Do^2)/(%pi*Do)

  / / Mass v e l o c i t y on s h e l l s i d e // [ kg/mˆ2 . h ]   Gs=mdot/as_dash // [ kg /mˆ 2 . s ]   Gs=Gs/3600 //Pa. s   mu=7*10^-4 //J/ kg .K   Cp=Cp*1000 / / R e y n o l d n um be r   Nre=De*Gs/mu / / P r a n d t l s n um be r   Npr=Cp*mu/k   Nnu=0.36*Nre^0.55*Npr^(1.0/3.0) / / N u s s e l t s n um ber // [W/ sq m .K]   hi=1050 // [W/ sq m.K ]   ho=Nnu*k/De // [W/ sq m K]   Uo=1/(1/ho+(1/hi*(Do/Di)))   Uc=Uo     Rd=(Uc-Ud)/(Uc*Ud)   printf ( ” \n F o u li n g f a c t o r =S c l a e W\n ” ,Rd);

//mˆ2.K/W r e s i s t a n c e =%f mˆ 2 .K/

Scilab code Exa 5.9  Length of tube required

1 2 3 4

// [m]

clc ;   clear ;

  / / E xa mp le 5 . 9     k=0.628

//W/(m.K) 128

5 6 7 8 9 10 11

             

// [ kg/mˆ 3 ] rho=980 // kg /(m. s ) mu=6*10^-4 //kJ /( kg .K) Cpw=4.187 //J /( kg .K) Cp=Cpw*10^3 // [mm] Di=25 // [m] Di=Di/1000 // Mass f l o w r a t e o f   mw_dot=1200*10^-3*rho water [ kg/h ] // [ kg/ s ] // I n s i d e a re a o f t u be

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

i n sq m 14 15 16 17 18 19 20 21 22 23 24

             

25   26 27 28 29 30 31 32 33 34 35 36 37

// kg/mˆ2 . s / / R e y n o l d s n um ber / / P r a n d d t l n um be r // In si de heat tr a ns fe r c o e f f i c i e n t / / N u s s e l t n um be r Nnu=0.023*Nre^0.8*Npr^0.4 // [W/ sq m.K ] hi=Nnu*k/Di // [W// sq m.K] ho=6000 // [m] Do=0.028 // [m] Dw=(Do-Di)/ log (Do/Di) // [m] x=(Do-Di)/2 // t he rm al c o n d u c t i v i t y o f m et al k2=348.9 i n [W/m.K ] // [W/ s q Uo=1/((1/ho)+(1/hi)*(Do/Di)+(x/k2)*(Do/Dw)) m.K] // [K] t1=303 // [K] t2=343 // [ kJ/ h ] Q=mw_dot*Cpw*(t2-t1) // [W] Q=Q*1000 // [K] Ts=393

  G=mw_dot/Ai   Nre=Di*G/mu   Npr=Cp*mu/k

                     

// [K] dT1=Ts-t1 // [K] dT2=Ts-t2 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [ sq m] Ao=Q/(Uo*dTlm)   //Length L=Ao/(%pi*Do) printf ( ” \n t h e r e f o r e l en g t h o f t u b e r e q u i r e d i s %f m \n ” ,L); 129

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 32

clc ;   clear ;

  / / E x am pl e 5 . 1 0                          

/ / [ kg / h ] o f n i t r o b e n z e n e m_dot=7250 // [ kJ/ kg .K] Cp=2.387; // [ kg /m. s ] mu=7*10^-4; // [W/m.K ] k=0.151; vis=1; / /LMTD c o r r e c t i o n f a c t o r Ft=0.9; // [K] T1=400 // [K] T2=317 // [K] t1=333 // [K] t2=300 // [K] dT1=T1-t1 // [K] dT2=T2-t2 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2)

  / / For n i t r o b e n z e n e // [ kJ/ h ]   Q=m_dot*Cp*(T1-T2) // [W]   Q=Q*1000/3600 //No . o f t u be s   n=170 // [m] L =5 // [m]   Do=0.019 // [m]   Di=0.015 // [ sq m]   Ao=n*%pi*Do*L // [W/ sq m.K ]   Uo=Q/(Ao*Ft*dTlm) // [W/ sq m.K]   Ud=Uo // B a f f l e s p a c in g [m]   B=0.15 / /Tube p i t c h i n [m]   Pt=0.025 / / C l e a ra n c e i n [m]   C_dash=Pt-Do // [m]   id=0.45

130

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

  / / S h e l l s i d e c r o s s f l o w a re a // [ sq m]

  as=id*C_dash*B/Pt

  / / E q ui v al e nt d i am e te r o f s h e l l   De=4*(Pt^2-(%pi/4)*(Do^2))/(%pi*Do)

// [m]

  / / Mass v e l o c i t y on s h e l l            

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

side // [ kg /(m. h ) ] // [ kg /mˆ 2 . s ] // [ kg /m. s ] // [ J/ kg .K] / / R e y n o l d s n um be r / / P r a n d t l n um be r

/ /From e m p i r i c a l e qn : //   mu_w=mu          

     

59  

Nnu=0.36*Nre^0.55*Npr^(1/3) // [W/ sq m.K ] ho=Nnu*k/De / / G i v e n [W/ s q m . K ] hi=1050 // [W/ sq m.K ] Uo=1/(1/ho+(1/hi)*(Do/Di)) //W/ sq m.K Uc=Uo

// S u i t a b i l i t y o f h ea t e xc ha ng er // [W/ sq m.K] Rd_given=9*10^-4 // [W/ sq m.K ] Rd=(Uc-Ud)/(Uc*Ud) printf ( ” \n Rd c a l c u l a t e d ( % f W/mˆ 2 . K) i s mazimum a l l o w a b l e s c a l e r e s i s t a n c e \ n ” ,Rd); printf ( ” \n\nAs Rd c a l c u l a t e d ( % f W/ s q m . K) (OR 1 . 1 ∗ 1 0 ˆ − 3 ) i s m or e t h an Rd g i v e n ( %f W/ s q m, K) , t h e g iv en h ea t e xc ha ng er i s s u i t a b l e \ n ” ,Rd,Rd_given) ;

Scilab code Exa 5.11  Number of tubes required

1 clc ;

131

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

  clear ;

  / / E x am am pl pl e 5 . 1 1                    

/ / w a t er e r i n [ k g /h /h ] mw_dot=1720; // [K] t 1 = 29 29 3 ; // [K] t 2 = 31 31 8 ; // [ kJ/ kg .K] Cpw=4.28; // [ kJ /h ] Q=mw_dot*Cpw*(t2-t1) / /W Q=Q*1000/3600 // [ kJ / kg ] lambda=2230; // [K] d T 1 = 90 90 ; // [K] d T 2 = 65 65 ; // [K] dTlm=(dT1-dT2)/ l o g (dT1/dT2)

/ / C a l c u l a t i o n o f i n s i d e h e at at t r a n s f e r c o e f f i c i e n t // [m]   Di=0.0225; // [m [m// s ] \   u = 1. 1. 2 ; // [ kg/m kg/mˆˆ 3 ]   r h o = 9 95 9 5 .7 .7 ; // [m [m// s ]   v=0.659*10^-6 // [ kg /m. s ]   mu=v*rho / / r e y n o l d s n um um be be r   Nre=Di*u*rho/mu // [ J/ kg .K]   Cp=Cpw*1000 // [ kJ/ h .m.K ]   k = 2 .5 .5 4 ; // [W [W/m. /m. K ]   k=k*1000/3600 / / P r a n d t l n um um be be r   Npr=Cp*mu/k / / N u s s e l t n um um be be r   Nnu=0.023*Nre^0.8*Npr^0.4 // [W/ sq m.K ]   hi=k*Nnu/Di // [ kJ/h .mˆ2 .K]   ho=19200 // [W [W//mˆ 2 .K ]   ho=ho*1000/3600 // [m]   Do=0.025 // [m]   Dw=(Do-Di)/ l o g (Do/Di) // [m]   x=(Do-Di)/2 / / F or o r t ub u b e w a l l m a t e r i a l [ k J /h / h . m . K] K]   kt=460 // [W [W/m. /m. K ]   kt=kt*1000/3600 // [   Uo=1/(1/ho+(1/hi)*(Do/Di)+(x/kt)*(Do/Dw)) W/ sq m.K] 36   //Q=Uo∗ //Q=Uo ∗ Ao Ao∗∗ dTlm 37   Ao=Q/(Uo*dTlm) / / [ s q m] m] 38 L =4 / /T / Tu b e l e n g t h i n [ m ] 132

39   n=Ao/(%pi*Do*L) / / [ Number o f t u b e s ] 40 n = round ( n )   //Approximate 41   printf ( ” \ n N u m b e r o f t u b e s r e u i r e d = %d” ,n);

Scilab code Exa 5.12  Shell and tube heat exchanger

1 2 3 4

clc ;   clear ;

  / / E x am am pl pl e 5 . 1 2   t1=290

/ / I n l e t t em e m pe p e ra r a tu t u re r e o f c o o l i n g w a te te r

[K] //Heat tr a n s f er c o e f f i c i e n t based on i n s i d e a r eeaa i n [W/ s q m . K] K] / / [ k J / kg k g ] L At A t en en t h e a t o f b e n ze ze n e lambda=400 / / [ t / h ] C o nd n d e ns n s a ti t i o n r a t e o f b en e n ze ze n e mb_dot=14.4 vapour / / S p e c i f i c h ea ea t Cpw=4.187 / / Wi With n o S c a l e

5   ho=2250 6  7 

8  9  10 11   Q=mb_dot*1000*lambda

/ / He H e a t d u ty t y o f c on o n d en e n s er er i n

[ kJ/ h ] 12   Q=(Q/3600)*1000 / / [W [W]] 13   / / S h e l l a n d t ub u b e t yp y p e o f h ea e a t e xc x c ha h a ng n g er er 14 15 16 17

       

18   19   20   21  

i s u s ed ed a s a

s i n g l e p as a s s s u r f a c e c on o n de d e ns n s er er / / I . D o f t u b e [ m] m] Di=0.022 / / L en e n gt g t h o f e ac a c h t ub ub e i n [ m] L=2.5 / / Nu Nu m b e r o f t u b e s n=120 / / Ar A r e a o f h ea e a t t r a n s f e r p er er metre A=%pi*Di*L l e n g t h i n [ mˆ 2/ 2 /m ] / / T o ta t a l a re r e a o f h ea ea t t r a n s f e r i n [ m A=n*A ˆ2] re a o f e a ch t u b e Ai=(%pi/4)*Di^2   / / C r o s s − s e c t i o n a l a re i n [ mˆ 2 ] / / T ot o t al a l a re r e a o f f l o w i n [mˆ 2 ] Ai=n*Ai / / V e l oc o c t y o f w at a t er e r [ ms ms ˆ − 1 ] u=0.75 133

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

  V=u*Ai   rho=1000   mw_dot=V*rho

/ / V o l um u m e tr t r i c f l o w o f w at a t er er / / [ D e ns n s i t y o f w a t er e r i n [ k g /m /m ˆ 3 ] ] / / Ma M a s s f l o w r a t e o f w at a t er er i n [ k g / s ]

  / / H ea ea t b a l a n c e   //Q=mw dot ∗Cp Cpw w ∗ ( t2 t 2 −t 1 )            

// [K] t2=Q/(mw_dot*Cpw*1000)+t1 / / C o nd n d en e n si s i ng n g b en e n ze z e ne n e t e m pe p e r a tu t u r e i n [ K] K] T=350 // [K] dT1=T-t1 // [K] dT2=T-t2   //LMTD dTlm=(dT1-dT2)/ l o g (dT1/dT2) // [W/mˆ 2 .K ] U=Q/(A*dTlm) U = round ( U )

/ / N e g l e c t i n g r e s i s t a n c e , we w e h av av e : // [W/mˆ 2 .K ]   hi=1/(1/U-1/ho) // h i i s p r o po r t io n a l to u ˆ 0. 8   //Constant   C=hi/(u^0.8)   / / Wi Wi t h S c a l e   Rd=2.5*10^-4

// [m [mˆ2 ˆ2 K. /W]

  //1/U=1/hi+1/ho+Rd //U=hi //U =hi /( 1+ 3. 38 ∗ u ˆ 0 . 8 )   //mw dot=rho ∗ u ∗ A i / / [ kg / s ]   / / Le L e t t 2 b e t he h e o u t l e t t em e m pe p e ra r a tu t u re r e o f w a te te r   //Q=mw dot ∗Cp Cpw w ∗ ( t2 t 2 −t 1 ) // t2= t2=Q/( Q/( mw dot ∗Cpw)+t1   dT1=60

  //dT2=T //dT2=T− 73 / u ) −( t 1 + 8 . 3 73   / / d T l m = 8 . 3 7 3 / ( u∗ u ∗ l o g ( 6 0 ∗ u / ( 6 0∗ 0 ∗ u − 8.3 73) ) )   //Q= / /Q=U U∗A∗ dTlm / / 1 . 8 9 = ( ( u ˆ − 0 . 2 ) / ( 1 + 3 . 3 8 ∗ u ˆ 0 . 8 ) ) ∗ ( 1 / l o g ( ( 6 0 ∗ u ) /60 ∗ u −8.373) / / I f we a s s u m e v a l u e s o f u g r e a t e r t h a n 0 . 7 5 m/ m/ s / / Fo Fo r u = 3 . 8 / / [ ms ˆ − 1 ] // ] msˆ − 1 ]   u=3.8   printf ( ” \ n W a t e r v e l o c i t y m u s t b e 3 . 8 8 msˆ −1 ” ) ; 134

Scilab code Exa 5.13  Length of pipe in Exchanger

1 2 3 4 5

clc ;   clear ;

  / / E x am pl e 5 . 1 3 // [ kg/ s ] // Heat c a p a c i t y o f w at er i n [ J /

  mh_dot=1.25   Cpw=4.187*10^3

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 ] // Rate o f h ea t t r a n s f e r from

[ kJ / s ] // [W] / / T em pe ra tu re o f c o n d e n s i n g

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

         

// I n l e t t em p er a tu re o f w at er [ K] t1=290 / / O u t l e t t e m p e ra t u r e o f w a te r [ K] t2=310 // [K] dT1=Ts-t1 // [K] dT2=Ts-t2 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2)

/ / He at r em ov ed f ro m v ap o ur = H ea t g a i n e d // [ kg/ s ]   mw_dot=Q/(Cpw*(t2-t1)) // [kW/ sq m.K]   hi=2.5 // [W/ sq m.K ]   hi=hi*1000 // [m]   Do=0.025 // [m]   Di=0.020 // I n s i d e h ea t t r a n s f e r   hio=hi*(Di/Do) c o s f f i c i e n t r e f e r r e d t o o u t s i d e d i a i n [W/ s q m. K] // O u ts id e h ea t t r a n b s f e r   ho=0.8 c o e f f i c i e n t i n [kW/ sq m.K ] // [W/ sq m.K ]   ho=ho*1000 // [W/ sq m.K ]   Uo=1/(1/ho+1/hio)   / /Ud i s 8 0% o f Uc // [W/ sq m.K ]   Ud=(80/100)*Uo // [ sq m]   Ao=Q/(Ud*dTlm) 135

// [m] // O u ts id e a r ea o f p i pe

28 L =1 29   A=%pi*Do*L

p er m l en g t h o f p ip e // T ot al l e n g th o f p i p i n g

30   len=Ao/A

required . 31 32 33 34

       

rho=1000 V=mw_dot/rho v=0.6 a=V/v

// [ kg/mˆ3 ] // [mˆ3/ s ] // [m/ s ] / / C r o s s − s e c t i o n a l a re a

 

f o r f l o w p as s [ s q m] 35   a1=(%pi*Di^2)/4 // [ sq m] 36 // f o r s i n g l e p a s s on t u be s i d e f l u i d ( w at er ) 37 n= round (a/a1) //No . o f t u be s

p er p a ss // L en gth o f e ac h t ub e i n

38   l=len/n

[m] 39 // For two p a s s e s on w at er s i d e : 40   tn=2*n / / T o ta l no o f t u b es 41   l2=len/tn // L en gth o f e ac h t ub e i n [m] 42 // For f o u r p a s s es on w at er s i d e / t ub e s i d e 43   tn2=4*n / / T o ta l no . o f t u b es 44   l3=len/tn2 // L en gth o f e ac h t ub e i n [m] 45 46   printf ( ” \nNo . tn2,l3);

o f t u b e s=%d , \ n L e ng t h o f t u b e=%f m” ,

Scilab code Exa 5.14   Dirt factor

1 2 3 4 5

clc   clear

  / / E x am pl e 5 . 1 4 // P r o p e r t i e s o f c ru de o i l : 136

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

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

Cpc =1.986 ;   mu1=2.9*10^-3; k1 =0.136 ; rho1 =824

// [ kg/mˆ 3 ]

;

/ / P r o p e r t i e s o f b otto m p r od u ct : // [ kJ/ kg .K] Cp2 =2.202 ; // [ kg/mˆ 3 ] rho2 =867 ; // [N. s /sq m] mu2 = 5.2*10^ - 3 ; // [W/ sq m.K] k2 =0.119 ; mc_dot =135000

// B a si s : c r u i d o i l f l o w

;

r a t e i n [ kg / h ] 19 m_dot =106000

/ / Bottom p r o d u ct f l o w

;

r a t e i n n [ kg / h ] 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

         

// [K] t1 =295 ; // [K] t2 =330 ; // [K] T1 =420 ; // [K] T2 =380 ; // [K] dT1=T1-t2 // [K] dT2=T2-t1 dTlm=(dT1-dT2)/ log (dT1/dT2)   Q=mc_dot*Cpc*(t2-t1) Q=Q*1000/3600

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

// S h e l l s i d e c a l c u l a t i o n s : Pt =25 Pt = Pt /1000 B =0.23 Do =0.019

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

; ; ; ;

d ia me te r f o r s qu a r e p i t c h // C l ea r an c e i n [

35   c_dash=Pt-Do

m] // [m] // C ro ss f l ow

36 id =0.6 ; 37   as=id*c_dash*B/Pt

a re a o f s h e l l [ s q m] 38 // s i n c e t h e r e i s a C a l c u l a i t o n m is ta ke , we t ak e : 137

39   as=0.0353; 40   Gs=m_dot/as

// S h e l l s i d e

mass v e l o c i t y i n [ kg / s q 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)

/ / N u s s e l t n um be r 47 48 49 50 51 52 53 54 55

// [W/ sq m.K]

  ho=Nnu*k2/De

/ / Tube s i d e h e a t t r a n s f e r n =324 ; n_p =324/2 ; t =2.1 ; t= t /1000 ;   Di=Do-2*t   A=(%pi/4)*(Di^2)

  on e t ub e i n [ s q m]

56   A_p=n_p*A

coefficient : //No . o f t u be s / /No . o f t u b es p e r p a s s / / T h i c k n e s s i n [mm] // [m] / / I . d o f t ub e i n [m] / / C r o s s − s e c t i o n a l a re a o f   // T ot al a r e a f o r f l o w p er

p as s i n [ s q m ] 57 58 59 60 61 62 63 64 65 66

                   

// [ kg/ sq m h ] G=mc_dot/A_p // [ kg/ sq m. s ] G=G/3600 / / R e y n o l o d s n um be r Nre=Di*G/mu1 / / P r a n d t l n um be r N p r = 4 2. 35 ; / / N u s s e l t n um be r Nnu=0.023*(Nre^0.8)*(Npr^0.4) // [W/ sq m.K ] hi=Nnu*k1/Di // [W/ sq m.K] hio=hi*Di/Do // [W/ sq m.K] Uo=1/(1/ho+1/hio) Uc=Uo / / L en gt h o f   L = 4 .8 8 ; t ub e i n [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 The c a l c u l a t i o n

o f l i n e no . 3 6 t o c a l c u l a t e d a s i s w ro ng ly do ne i n Book by p r i n t i n g 0 . 0 3 5 3 , , . . w hi ch i s wrong \n ” ); 71   printf ( ” \nRd=%f K/w, or 7. 34 ∗1 0 ˆ − 4 whi ch i s l e s s th a n t h e p ro vi de d , s o t h i s i f i n s t a l l e d w i l l n o t g i ve r e q u i r e d t em p er a ru es w it ho ut f r e q u e n t c l e a n i n g \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 25

clc ;   clear ;

  / / E x am pl e 5 . 1 5 //CASE I :                          

// [ J/ kg .K] Cp=4*10^3; // [K] t1=295; // [K] t2=375; // S p e c i f i c g r a v i t y o f l i q u i d sp=1.1; // Flow o f l i q u i d i n [ mˆ3 / s ] v1=1.75*10^-4; // [ kg/mˆ 3 ] rho=sp*1000 // [ kg/ s ] m_dot=v1*rho // [W] Q=m_dot*Cp*(t2-t1) // [K] T=395; // [K] dT1=T-t1 // [K] dT2=T-t2 dTlm=(dT1-dT2)/ log (dT1/dT2)   // [K] // [W/K] U1A=Q/dTlm

  //CASE−I I          

v2=3.25*10^-4 T2=370 m_dot=v2*rho Q=m_dot*Cp*(T2-t1) dT1=T-t1

/ / F lo w i n [ mˆ 3 / s ] // [K] // [ kg/ s ] // [W] // [K] 139

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

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

  dT2=T-T2   dTlm=(dT1-dT2)/ log (dT1/dT2)   U2A=Q/dTlm

// s i n c e u i s p ro pn t o v // h i =C∗ v ˆ 0 . 8   U2_by_U1=U2A/U1A

// H eat t r a n s f e r c o e f f f o r c o nd e ns i ng s team i n [W/ s q m. K] C= poly (0 , ”C” ) / / L e t C=1 a n d v=v 1   //C=1;   / / = 1 . 7 5∗10ˆ −4 mˆ3/ s   v=v1;   ho=3400

  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)

//

/ / S i n c e U2 = 1 . 6 U1 / /On s o l v i n g we g e t :          

C=142497 v=v1 hi=C*v^0.8 U1=1/(1/ho+1/hi) A=U1A/U1

// // Heat t r a n s f e r a r e a i n [ s q

m] 54   printf ( ” \n O v e r a l l h e at t r a n s f e r

c o e f f i c i e n t i s %f W / s q m . K a n d \n\ nHeat t r a n s f e r a re a i s %f s q m” ,U1,

A) ;

Scilab code Exa 5.16  Oil Cooler

140

1 2 3 4 5

clc ;   clear ;

  / / E x am pl e 5 . 1 6 // [ kg/ s ] // S p e c i f i c h ea t o f  

  mo_dot=6*10^-2   Cpo=2*10^3

o i l i n [ J /kg . K] // S p e c i f i c h ea t o f w a te r

6   Cpw=4.18*10^3

in [ J/kg .K] // [K] // [K] / / [ K] Water e n t e r i n g

7   T1=420 8   T2=320 9   T=290

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

// [ J/ s ] = [W] / / H ea t g i v e n o u t =H ea t g a i n e d // [K] t2=Q/(mo_dot*Cpw)+T // [K] dT1=T1-t2 // [K] dT2=T2-T dTlm=(dT1-dT2)/ log (dT1/dT2)   // [K] // [W/ sq m.K ] hi=1.6*1000 // [W/ sq m.K ] ho=3.6*1000 U=1/(1/ho+1/hi) // [W/ sq m.K] // [ sq m] A=Q/(U*dTlm) // [m] D=0.025 // [m] L=A/(%pi*D) printf ( ” \n Len gt h o f t ub e r e q u i r e d = %f m” ,L);

  Q=mo_dot*Cpo*(T1-T2)

                       

Scilab code Exa 5.17  Countercurrent flow heat exchanger

1 2 3 4 5 6 7

clc ;   clear ;

  / / E x am pl e 5 . 1 7        

mb_dot=1.25 Cpb=1.9*10^3 Cpw=4.187*10^3 T1=350

/ / B en ze ne i n [ kg / s ] / / Fo r b e n ze n e i n [ J / kg . K] // in [ J/kg .K] // [K] 141

8 9 10 11 12 13 14 15

               

// [K] T2=300 // [W] Q=mb_dot*Cpb*(T1-T2) // [K] t1=290 // [K] t2=320 // [K] dT1=T1-t2 // [K] dT2=T2-t1 dTlm=(dT1-dT2)/ log (dT1/dT2)   // [K] / / Minimum f l o w mw_dot=Q/(Cpw*(t2-t1))

ra te of 

w at er i n [ kg / s ] 16 17 18 19 20 21 22 23 24 25 26 27

// [W/ sq m.K ] hi=850 // [W/ sq m.K ] ho=1700 // [m] Do=0.025 // [m] Di=0.022 / / T h i c k n e s s i n [m] x = ( Do - D i ) / 2 // [W/ sq m.K ] hio=hi*(Di/Do) // [m] Dw=(Do-Di)/ log (Do/Di) // [W/m.K ] k=45 Uo=1/((1/ho)+(1/hio)+(x/k)*(Do/Dw)) // [W/ sq m.K] // [ sq m] Ao=Q/(Uo*dTlm) // L e ng th i n [m] L =1 // O ut si de s u r f a c e   area=%pi*Do*L

                   

a re a o f t u b e p er i m l en g t h // T ot al l e n g th o f  

28   Tl=Ao/area

t ub in g r e q u i r e d i n [m] 29   printf ( ” \ n To t al l e n g t h o f t u b in g r e q u i r e d =%d m” , round (Tl));

Scilab code Exa 5.18   Vertical Exchanger

1 2 3 4

clc ;   clear ;

  / / E x am pl e 5 . 1 8   m_dot=4500

// B en zen e c o n de n s at i o n r a t e i n [

kg/ h ] 5   lambda=394

// L at en t h ea t o f c o n de n s at i o n o f   142

6 7 8 9 10 11 12 13 14

         

 

b e n ze n e i n [ kJ / kg ] // [ kJ /h ] Q=m_dot*lambda // [W] Q=Q*1000/3600 // [ kJ/ kg .K] Cpw=4.18 // [K] t1=295 // [K] t2=300 / / For w at er :

// [ kg/ s ]   mw_dot=Q/(Cpw*1000*(t2-t1)) // [ kg /mˆ3   rho=1000 // V o lu m et ri c f l ow r a t e i n [m   V=mw_dot/rho ˆ3 / s ]

15   u=1.05 16   A = V / u

  r e q u i r e d i n [ s q m]

17 18 19 20 21 22 23 24 25 26 27

// [m/ s ] / / C r o s s − s e c t i o n a l a re a

/ / F or t u b e :            

x=1.6 x=x/1000 Do=0.025 Di=Do-2*x A1=(%pi*Di^2)/4 n=A/A1 n= round (n )   L=2.5   Ao=n*%pi*Do*L

in [ sq 28   Ts=353

/ / t h i c k n e s s i n [mm] // [m] // [m] // [m] / / Of o ne t ub e [ s q m] //No . o f t ub es r e u i r e d // L en gth o f t ub e i n [m] // S u rf a c e a re a f o r h ea t t r a n s f e r

m] / / C on de ns in g temp o f b e nz en e i n

[K] 29 30 31 32 33 34 35 36 37 38

             

// I n l e t t em p er a tu re i n [ K] T1=295 / / O u t le t t e mp e ra t u re i n [ K] T2=300 // [K] dT1=Ts-T1 // [K] dT2=Ts-T2 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [W/ s q mK] Uo=Q/(Ao*dTlm) // [W/ sq m.K ] Ud=Uo //OVERALL HEAT TRANSFER COEFFCIENT : // I n s i d e s i d e : 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] // [m

43   Dw=(Do-Di)/ log (Do/Di)

] 44 45 46 47 48 49 50 51 52

  k=45

/ / F or t u b e i n [W/ (m . ) ]

/ / O u t s i de o f t u be :            

// [ kg/ s ] // [ kg /(m. s ) ] // [W/ (m.K) ] // [ kg/mˆ 3 ] // [N. s/ sq m] // [m/ s ˆ 2 ]

mdot_dash=1.25/n M=mdot_dash/(%pi*Do) k=0.15 rho=880 mu=0.35*10^-3 g=9.81

A c c e l e r a t i o n due t o g r a v i t y 53   hm=(1.47*((4*mdot_dash)/mu)^(-1/3))/(mu^2/(k^3*rho // [W/ sq m.K] ^2*g))^(1/3) 54   ho=hm // [W/ sq m.K ] 55   k=45 // [W/m] 56   Uo=1/(1/ho+1/hio+(x*Do)/(k*Dw)) 57   //Uo=1/(1/ho+1/hio+(x∗D o / ( k ∗Dw) ) ) // O v e ra l l h e a t

t r a n s f e r c o e f f i c i e n t i n [W/ s q m . K ] 58   Uc=Uo // [W/ sq m.K] 59 60   Rd=(Uc-Ud)/(Uc*Ud)

  //Maximum a l l o w a b l e s c l a e r e s i s t a n c e i n [K/W] 61   printf ( ” \n Uc ( %f ) i s i n e x c e s s o f Ud ( %f ) , t h e r e f o r e we a l l o w f o r r e a s o n a b l e s c a l e r e s i s t a n c e , \ nRd=%f  K/W\n ” ,Uc,Ud,Rd); 62   printf ( ” \n No . o f t u b es = %d ” ,n)

Scilab code Exa 5.19  Countercurrent Heat Exchanger

144

1 2 3 4 5

clc ;   clear ;

  / / E x am pl e 5 . 1 9 / / Water f l o w r a t e i n [ kg / s ] / / Hea t c a p a c i t y o f w at er [ kJ / kg

  mw_dot=5;   Cpw=4.18;

.K] 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

                             

// [K] t1=303; // [K] t2=343; // [ kJ/ s ] Q=mw_dot*Cpw*(t2-t1) // [W] Q=Q*1000; // [K] T1=413; // [K] T2=373; // [K] dT1=T1-t2 // [K] dT2=T2-t1 // / [K] dTlm=dT1 // [W/ sq m.K ] hi=1000; // [W/ sq m.K ] ho=2500; / / F o u l i n g f a c t o r [ mˆ 2 . K/KW] Rd=1/(0.714*1000) // [W/ sq m.K ] U=1/(1/hi+1/ho+Rd) // [ sq m] A=Q/(U*dTlm) printf ( ” \ nHeat t r a n s f e r a re a i s %f s q m” ,A);

Scilab code Exa 5.20  Number of tube side pass

1 2 3 4 5 6 7 8 9 10 11

clc ;   clear ;

  / / E x am pl e 5 . 2 0          

Cpo=1.9 Cps=1.86 ms_dot=5.2 T1=403 T2=383

// H ea t c a p a c i t y f o r o i l [ kJ / kg . K] / / Hea t c a p a c i t y f o r s te am [ kJ / kg . K] // Mass f l o w r a t e i n [ kg / s ] // [K] // [K] // [ kJ/ s ] // [W]

  Q=ms_dot*Cps*(T1-T2)   Q=Q*1000

145

12 13 14 15 16 17

           

// [K] t1=288; // [K] t2=358; // [K] dT2=T1-t2 // [K] dT1=T2-t1 //LMTD i n [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // Overall heat U = 27 5 ;

t r a n s f e r c o e f f c i e n t i n [W// s q m. K ] 18   Ft=0.97 / /LMTD c o r r e c t i o n factor 19   A=Q/(U*Ft*dTlm) // [ sq m] 20   printf ( ” \ nHeat e xc ha n ge r s u r f a c e a r ea i s %f s q m” ,A ) ;

Scilab code Exa 5.21  Number of tubes passes

1 2 3 4 5 6

clc ;   clear ;

  / / E x am pl e 5 . 2 1   mc_dot=3.783;   mh_dot=1.892;   Cpc=4.18;

/ / C ol d w a te r f l o w r a t e [ kg / s ] / / Hot w at er f l o w r a t e [ kg / s ] // Sp h ea t o f c o l d w at er [ kJ / ( kg .

K) ] 7 8 9 10

       

T1=367; t2=328; t1=311; Cph=4.18;

// [K] // [K] // [K] // S p e c i f i c he at o f h o t w a te r [ kJ

/( kg .K) ] / / D e n s i t y [ k g /mˆ 3 ] // D ia me te r o f t ub e i n [m] // O ve ra l h ea t t r a n s f e r c o e f f i c i e n t i n [W/ sq m.K ] // [K] T2=T1-mc_dot*Cpc*(t2-t1)/(mh_dot*Cph) // [ kJ/ s ] Q=mc_dot*Cpc*(t2-t1) // [W] Q=Q*1000 // F or c o u n t e r f l o w h e at e x ch a n ge r

11   rho=1000; 12   D=0.019; 13   U = 1 45 0 ; 14 15 16 17

     

 

146

18 19 20 21 22 23 24

             

dT1=T1-t2 dT2=17; dTlm=(dT1-dT2)/ log (dT1/dT2)   lmtd=dTlm Ft=0.88 A=Q/(U*dTlm) u=0.366;

// [K] // [K]   // [K] / /LMTD / /LMTD c o r r e c t i o n f a c t o r // [ sq m] / / V e l o c i t y t hr o ug h t u b es

[msˆ −1] / / T o ta l f l o w Area i n [ s q

25   Ai=mc_dot/(rho*u)

m] //No . o f t u be s / / P e r m l e n g t h [ m] / /S . S p e r t ub e p e r 1 m

26   n=Ai/((%pi/4)*(D^2)) 27 L =1 28   sa=%pi*D*L

length 29   L=A/(n*%pi*D) // L en gth o f t u be s i n [m] 30   printf ( ” \nThe l e n g t h i s more t h a n a l l o w a b l e 2 . 4 4 m l e n g t h , s o we must u s e mo re t ha n o ne t u be \n ” ); 31 32 // For 2 p a s s e s on t he t ub e s i d e 33   A=Q/(U*Ft*lmtd) // [ sq m] 34   L=A/(2*n*%pi*D) / / Le ng th i n [m] 35   printf ( ” \n T h is l e ng t h i s w i t h in 2 . 44 m r eq ui re me nt , s o t h e d es i g n c h o i c e i s \n\n ” ) ; 36   printf ( ” \ nType o f h ea t e xc ha n ge r : 1−2 S h e l l and t ub e h e at e x ch a n ge r \n” ) 37   printf ( ” \nNo o f t u b es p e r p a s s= %d\n ” , round ( n ) ) ; 38   printf ( ” \ n Le ng th o f t u be p e r p a s s=%f m\n ” ,L);

Scilab code Exa 5.22  Outlet temperature for hot and cold fluids

1 2 3 4

clc ;   clear ;

  / / E x am pl e 5 . 2 2   mh_dot=16.67;

// Mass f lo w r a t e o f h o t f l u i d i n

[ kg/ s ] 147

5   mc_dot=20;

// Mass f lo w r a t e o f c ol d f l u i d i n

[ kg/ s ] 6   Cph=3.6;

// Sp h ea t o f h o t f l u i d i n [ kJ /kg

.K] 7   Cph=Cph*1000;

// Sp h ea t o f h o t f l u i d i n [ J / kg

.K] 8   Cpc=4.2;

// Sp h ea t o f c o l d f l u i d i n [ kJ / (

kg .K) ] 9   Cpc=Cpc*1000;

// Sp h e a t o f c ol d f l u i d in [ J / (

kg .K) ] // O v e r a l l h ea t t r a n s f e r c o e f f i c i e n t i n [W/ sq m.K ] 11   A=100; // S u r f a ce a r ea i n [ s q m] 12   mCp_h=mh_dot*Cph // [ J/ s ] or [W/K] 13   mCp_c=mc_dot*Cpc // [ J/ s ] or [W/K] 14   mCp_small=mCp_h // [W/K] 15   C=mCp_small/mCp_c / / C a pa c it y r a t i o 16   ntu=U*A/mCp_small   //NTU 17   T1=973; // Hot f l u i d i n l e t t em pe ra tu re i n [K] 18   t1=373; // Cold f l u i d i n l e t t em pe ra tu re i n [K ] 19   / / C a se 1 : C o u n t e r c u r r e n t f l o w a r r an g e me n t 20   E=(1-%e^(-(1-C)*ntu))/(1-C*%e^(-(1-C)*ntu)) // Effectiveness 21   / /W=T1−T2/(T1−t 1 ) t h er e f or e : 22   T2=T1-E*(T1-t1) // [K] 23   printf ( ” \ n Ex it t em pe ra tu re o f h ot f l u i d i s %d K” , 10   U=400;

round (T2)); 24   t2=mCp_h*(T1-T2)/(mCp_c)+t1

/ / [ From e n e r g y

b a la n c e eqn i n ] [ K ] 25   printf ( ” \ n Ex it t em p er at u re o f c o l d f l u i d i s %d K(%d C) \n ” , round ( t 2 ) , round (t2-273)); 26 27 // Ca se 2 : P a r a l l e l f l o w a rr an ge me nt 28   E1=(1-%e^(-(1+C)*ntu))/(1+C) 29 // I n t he t ex tb o k h er e i s a c a l c u l a t i o n

t he v a l u e o f E i s t ak ne a s E= 0 .9 7 148

m is ta ke , and

30 31   T2=T1-E1*(T1-t1) // [K] 32   t2=mCp_h*(T1-T2)/(mCp_c)+t1

/ / [ From e n e r g y

b a la n c e eqn i n ] [ K ] 33   printf ( ” \ n E x it t e m p e r at u r e o f Hot w a te r=%f K\n ” ,T2); 34   printf ( ” \ n E xi t t e mp e r at u r e o f c o l d w at er=%f K\n ” ,t2) ;

Scilab code Exa 5.23  Counterflow concentric heat exchanger

1 2 3 4 5 6 7 8

clc ;   clear ;

  / / E x am pl e 5 . 2 3          

Cpo=2131; Cpw=4187; mo_dot=0.10; mw_dot=0.20; U=380;

// Sp he at o f o i l i n [ J /kg . K ] / /Sp he a t o f w a te r i n [ J / kg . K] // O i l f l ow r a t e i n [ kg / s ] // Water f l o w r a t e i n [ kg / s ] // O v er a l l h ea t t r a n s f e r c o e f f i n [W/

sq m.K] 9 10 11 12 13 14 15 16 17 18 19 20

       

/ / I n i t i a l temp o f o i l [ K ] T1=373; // F i na l t em pe ra tu re o f o i l [ K ] T2=333; / / Water e n t e r t e mp e r at u r e i n [ K] t1=303; // [K] t2=t1+mo_dot*Cpo*(T1-T2)/(mw_dot*Cpw) // 1 .LMTD met hod

             

// [K] dT1=T1-t2 // [K] dT2=T2-t1 // [K] dTlm=(dT1-dT2)/ log (dT1/dT2) // [K] lmtd=dTlm; // [ J/ s ] Q=mo_dot*Cpo*(T1-T2) // [ sq m] A=Q/(U*dTlm) / / I n n e r t ub ve d i a me t er [ Do=0.025; m] / / Le ng th i n [m]

21   L=A/(%pi*Do) 22 23 // 2 .NTU method

149

24   mCp_c=mw_dot*Cpw // [W/K] 25   mCp_h=mo_dot*Cpo // [W/K] 26   printf ( ” \n I n t ex tb oo k t h i s v al ue o f mCp h i s

w r o n g l y c a l c u l a t e d a s 2 31 .1 s o we w i l l t a k e t h i s o nl y f o r c a l c u l a t i o n \ n ” ) ; 27   mCp_h=231.1; // [W/K] 28 // mCp h i s s m a l l e r 29 30 31 32

  C=mCp_h/mCp_c   E=(T1-T2)/(T1-t1)

33 34 35 36 37 38

           

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

  / / For c o u n t e r cu r r e n t f l o w   deff ( ’ [ x] = f ( nt u ) ’ , ’ x=E−(1−%eˆ(−(1 −C) ∗ ntu ) ) /(1 −C∗%e ˆ( −(1 −C) ∗ nt u ) ) ’ ) ntu= fsolve (1,f) // [ sq m] A=ntu*mCp_h/U   //Approximately A=0.56 / / Le ng th i n [m] L1=A/(%pi*Do) printf ( ” \nFrom LMTD a pp ro ac h : \ n l e n g t h =%f m\n ” ,L); printf ( ” \nFrom NTU met hod : \ n l e n g t h =%f m\n ” ,L1);

Scilab code Exa 5.24  Number of tubes required

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

clc ;   clear ;

  / / E x am pl e 5 . 2 4 // [W/ sq m.K ]   ho=200; // [W/ sq m.K ]   hi=1500; / / Sp h e a t o f Water i n [ kJ / ( k g . K) ]   Cpw=4.2; / /Sp h e at o f O i l i n [ kJ / ( k g . K) ]   Cpo=2.1; // E f f e c t i v e n e s s   E=0.8; // [W/m.K ]   k=46; // [ kg/ s ]   m_dot=0.167; / / F or o i l [W/K ]   / / m Cp o il i s w r on gl y c a l c u l a t e d a s 7 1 0. 4   mCp_oil=2*m_dot*Cpo*1000

150

15   mCp_water=m_dot*Cpw*1000 / / F o r w a t e r [W/K ] 16   / / m Cp o il i s w r on gl y c a l c u l a t e d a s 7 1 0. 4 17 / /NOTE : The a bo ve two v a l u e s a r e w r on g ly c a l c u l a t e d

i n book a s 7 1 0. 4 18 // s o we t a ke h e re : 19   mCp_small=710.4 // [W/K] 20 / / S i n c e b ot h mCp water and m Cp oi l a r e e q u a l

,

there fore : 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

  C=1;                              

deff ( ’ [ x] = f ( nt u ) ’ , ’ x=E−(ntu /(1+ntu ) ) ’ ); ntu= fsolve (1,f) // I n t e r n a l d i a m et e r i n [mm] id=20; / / E x t e r n a l d i a m e t er i n [mm] od=25; // [W/ sq m.K ] hio=hi*id/od // [mm] Dw=(od-id)/ log (od/id) // [m] Dw=Dw/1000 // [mm] x=(od-id)/2 // [m] x=x/1000 // E x te r na l d i a i n [m] Do=0.025 // L en gth o f t ub e i n [m] L=2.5; // [W/ sq m.K] Uo=1/(1/ho+1/hio+(x/k)*(Dw/Do)) // Heat t r a n s f e r a r e a i n [ s q m] A=ntu*mCp_small/Uo / /No o f t u b es n=A/(%pi*Do*L) printf ( ” \nNo . o f t u be s r e q u i r e d = %d” , round ( n + 1 ) ) ;

Scilab code Exa 5.25  Parallel and Countercurrent flow

1 2 3 4 5 6 7

clc ;   clear ;

  / / E x am pl e 5 . 2 5 // ( i ) P a r a l l e l f l o w // [K]   T1=633; // [K]   t2=303; 151

8 9 10 11 12 13

           

T2=573; t1=400; dT1=T1-t2; dT2=T2-t1; mh_dot=1.2; U=500;

// [K] // [K] // [K] // [K] // [ kg/ s ] // Overal l heat tr an sf er

c o e f f i c i e n t in [

W/ sqm .K ] 14 15 16 17 18 19 20 21 22 23 24

       

/ /Sp . h e at o f o i l J / kg . K Cp=2083; dTlm=(dT1-dT2)/ log (dT1/dT2)   // [K] // [W] Q=mh_dot*Cp*(T1-T2) // [ sq m] A=Q/(U*dTlm)

/ / ( i i ) C ou nt er c u r r e n t f l o w // [K]   dT1=T1-t1; // [K]   dT2=T2-t2;

// [K]   dTlm=(dT2-dT1)/ log (dT2/dT1) // [ sq m]   A1=Q/(U*dTlm)   printf ( ” \ nFor p a r a l l e l f lo w , Area = %f s q m \n F o r c o u n t e r c u r r e n t f l o w , A re a=%f s q m\n ” ,A,A1); 25   printf ( ” \n\ nFor t he same t e r m in a l t e mp e ra t ur e s o f   t h e f l u i d , t h e s u r f a c e a re a f o r t h e c o u n te r f l o w a r r a n g e m e n t \n i s l e s s t h a n t h e r e q u i r e d f o r t h e p a r a l l e l f lo w \n ” )

152

Chapter 6 Evaporation

Scilab code Exa 6.1   Boiling point Elevation

1 2 3 4 5 6 7 8 9

clc ;   clear ;

// Example6 . 1

//B . P o f s o l u t i o n [ K] T=380 //B . P o f w at er [ K] T_dash=373 // B o i l i n g p o i n t e l e v a t i o n i n [ K ] BPE=T-T_dash // S a t u r a t i n g t em p e ra t u re i n [K ] Ts=399 // D r i v i n g f o r c e i n [ K] DF=Ts-T printf ( ” \ n B o i l i n g p oi n t o f e l e v a t i o n o f t h e s o l u t i o n i s %d K \n” ,BPE); 10   printf ( ” \ n Dr iv in g f o r v e f o r h e at t r a n s f e r i s %d K \n ” ,DF)            

Scilab code Exa 6.2  Capacity of evaporator

1 clc ; 2   clear ; 3   / / E xa mp le 6 . 2

153

4   m_dot=10000 5   fr_in=0.04

/ /Weak l i q u o r e n t e r i n g i n [ kg / h ] // F r a c it o n o f c a u s t i c s od a IN i . e 4

% 6   fr_out=0.25

/ / F r a c i t o n o f c a u s t i c s od a OUT i . e 2 5

% 7   / / Le t m da s h d o t be t he kg /h o f t h i c k l i q u o r l e a v i n g 8   mdash_dot=fr_in*m_dot/fr_out // [ kg /h ] 9 10 // O v e r a ll m a t e r i a l b a l an c e 11   / / kg / h o f f e e d=kg / h o f w at er e v a po r a te d +kg / h o f  

t h ic k l i q u o r 12   / / we=w a te r e v a p o r a t e d i n kg / h 13 / / T h e r e f o r e 14   we=m_dot-mdash_dot // [ kg /h ] 15   printf ( ” \n C ap ac it y o f e v a po r a to r i s %d kg /h ” ,we);

Scilab code Exa 6.3   Economy of Evaporator

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

clc ;   clear ;

  / / E xm ap le 6 . 3 / / I n i t i a l c o n c e n t r a t i o n ( 5%)   ic=0.05 // F i na l c o n c en t r a ti o n ( 20%)   fc=0.2 //B . P o f w a te r i n [ K ]   T_dash=373 // B o i l i n g p o i nt e l e v a t i o n [K ]   bpe=5 // [ B a s i s ] f e ed t o e v a p o ra t or i n [   mf_dot=5000 kg/ h ] // M a te r ia l b a la n ce o f s o l u t e // [ kg /h ]   mdash_dot=ic*mf_dot/fc // O v e r a ll m a t e r i a l b a l an c e / / Wa te r e v a p o r a t e d [ k g / mv_dot= mf_dot -m dash_dot h] / / L a te n t h e at o f c o n d e n s a t io n   lambda_s=2185 of steam [ kJ/kg ] // L at en t h ea t o f v a p o r i s a t i o n of     lambda_v=2257 154

water [ kJ/kg ] 15 16 17 18 19 20 21 22 23 24 25

       

// [ kJ/ kg ] lambda=lambda_v / / Te mp er at ur e o f t h i c k l i q u o r [ K] T=T_dash+bpe // T e mp er at ur e o f f e e d [ K] Tf=298 / /Sp . h e at o f f e e d i n [ kJ / kg . K] Cpf=4.187 / / Hea t b a l a n ce o v er e v a p o r a t o r=m s d ot

  ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s          

26   27   28  

/ / S te am c o n s u m p t i o n [ k g / h ] / / Economy o f e v a p o r a t o r Eco=mv_dot/ms_dot // S a t u r a ti o n t e mp e ra t ur e o f s team i n [ K ] Ts=399 // T em pe ra tu re d r i v i n g f o r c e [ K] dT=Ts-T // [W/ sq m.K ] U=2350 // Rate o f h e a t t r a n s f e r i n [ Q=ms_dot*lambda_s kJ/kg ] // [ J/ s ] = [W] Q=Q*1000/3600 // Heat t r a n s f e r a r e a i n [ s q A=Q/(U*dT) m] printf ( ” \nANSWER E co no mo y p f e v a p o r a t o r i s % f   \n” ,

Eco); 29   printf ( ” \ nHeat t a r n s f e r m\n ” ,A);

a re a t o be p r o vi d e d = %f s q

Scilab code Exa 6.4   Steam economy

1 2 3 4 5 6

clc ;   clear ;

  / / E xa mp le 6 . 4

// S p e c i f i c h ea t o f f e e d i n kJ / ( kg . K) // L at en t h e at o f c on ds o f h ea t a t 0 . 2 MPa i n [ k J / k g ] 7   lambda=2383 // L at en t he at o f v a p o r i s a t i o n o f   w a te r a t y 3 23 [ kJ / kg 8   ic=0.1 // I n i t i a l c o n c e n t r a t i o n o f s o i l d s i n [%]   Cpf=3.98   lambda_s=2202

155

9   fc=0.5 //Final concentrat ion 10   m_dot=30000 / / Feed t o e v a p o r a t o r i n [ kg / h ] 11 m d a sh _ d ot = i c * m _ do t / f c // Mass f l o w r a t e o f t h i c k

l i q u o r i n [ kg /h ] / / Water e v a po r a te d i n [

12   mv_dot=m_dot-mdash_dot

kg/ h ] 13 14 15 16 17 18 19 20

  / / Ca se 1 : Feed a t 2 93K // [ kg /h ]   mf_dot=30000 // [ kg /h ]   mv_dot=24000 // [ kJ /( kg .K) ]   Cpf=3.98 // S a t u r a ti o n t e mp e ra tu r e o f steam i n [ K ]   Ts=393 // B o i l i n g p o i n t o f s o l u t i o n [ K ]   T=323 // L at en t h ea t o f co n d en s a ti o n [   lambda_s=2202 kJ/kg ] 21   lambda=2383 / / L a te n t h e at o f v a p o r i s a t i o n [ kJ / kg ] 22   Tf=293 / / F ee d t e m p e r a t u r e 23   / / E n th al py b a l a n c e o v er t h e e v a p o r a t o r : 24   ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s 25   26   27   28   29   30   31   32  

/ / S t ea m c o n s u m p t i o n [ k g / h ] // Steam economy eco=(mv_dot/ms_dot) printf ( ” \nWhen Fe ed i n t r o d u c e d a t 2 93 K , S team econ omy i s %f \  n ” ,eco); // [K] dT=Ts-T // [W/ sq m.K] U=2900 / / He at l o a d =R at e o f   Q=ms_dot*lambda_s h ea t t r a n s f e r i n [ kJ /h ] // [ J/ s ] Q=Q*1000/3600 // Heat t r a n s f e r a r e a A=Q/(U*dT) r e q u i r e d [ s q m] printf ( ” \n ANSWER−( i ) \n\n At 29 3 K , Heat t r a n s f e r a re a r e q u i r e d i s %f s q m\n” ,A);

33 34 / / C a se 2 : F ee d a t 3 0 8K 35   Tf=308 / / [ F ee d t e m p e r a t u r e ] [ K ] 36   ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s

/ / St ea m c o n s um p t i on i n [ k g /h ] 37   eco=mv_dot/ms_dot //Economy o f  156

evaporator 38   printf ( ” \n ANSWER−( i i ) \n\n When T=308 K \nEconomy of  e v a p o r a t o r i s %f \  n ” ,eco); 39   Q=ms_dot*lambda_s // [ kJ /h ] 40   Q=Q*1000/3600 // [ J/ s ] 41   A=Q/(U*dT) // Heat t r a n s f e r a re a required [ sq m] 42   printf (  ’ \nANSWER−( i i i ) \n When T=308 K, \ nHeat t r a n s f e r Area r e qu i r e d i s %f s q m\n ” ,A) ;

Scilab code Exa 6.5  Evaporator economy

1 2 3 4 5 6 7 8 9 10

clc ;   clear ;

  / / E xa mp le 6 . 5 // F eed t o t h e e v a p o r a t o r [ kg / h ]   m_dot=5000 / /Cp o f f e e d i n [ kJ / kg . K]   Cpf=4.187 // I n i t i a l co nc en tr at io n   ic=0.10 //Final concentrati on   fc=0.4 / / [ kg / h ] o f t h i c k   mdash_dot=m_dot*ic/fc liquor / / Wa ter e v a p o r a t e d   mv_dot=m_dot-mdash_dot in [ kg/h ] // L a te n t h e at o f c o n d en s i n g   lambda_s=2162 steam [ kJ/kg ] / / P r e s s u r e i n t h e e v a p o r a t o r [ kPa ]   P=101.325 // [K]   bp=373 / / E n th a lp y o f w a te r v a p o r [ kJ / kg ]   Hv=2676 // [ kJ/ kg ]   H_dash=419 // [ kJ/ kg ]   Hf=170

11 12 13 14 15 16 ms_dot=(mv_dot*Hv+ mdash_dot*H_dash -m_dot*Hf)/ / / S te am c o n s um p t i on i n [ k g / h ] lambda_s 17   eco=mv_dot/ms_dot / / S te am e co no my o f  

evaporator // [ kJ/ h ]

18   Q=ms_dot*lambda_s

157

19 20 21 22 23

         

// [W/ sq m.K] U=1750 // [K] dT=34 // [ J/ s ] Q=Q*1000/3600 // [ sq m] A=Q/(U*dT) printf ( ” \n H eat t r a n s f e r a re a t o b e p ro vi de d i s %f   sq m” ,A);

Scilab code Exa 6.6   Single effect Evaporator

1 2 3 4 5 6 7

clc ;   clear ;

  / / E xa mp le 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 ] // F i n a l c o n c e n t r a t i o n [ kg /h ] // B o i l i n g p t o f s a t u r a t i o n i n [ K

] 8   Ts=383

/ / S a t u r a t i o n t e m pe r a tu r e o f  

stea m i n [ K] 9   mdash_dot=ic*mf_dot/fc 10 mv_dot= mf_dot -m dash_dot

// [ kg /h ] // Water e v a po r a t ed i n [

kg/ h ] // [ kJ/ kg ] Hf=125.79 // [ kJ/ kg ] Hdash=419.04 // [ kJ/ kg ] Hv=2676.1 // [ kJ/ kg ] lambda_s=2230.2 ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/ / / Steam f l o w r a t e i n [ kg / h ] lambda_s 16   eco=mv_dot/ms_dot // Steam economy 17   Q=ms_dot*lambda_s // Rate o f h ea t t r a n s f e r 11 12 13 14 15

         

in [ kJ/h ] 18   Q=Q*1000/3600 19   dT=Ts-T 20 21   A=69

// [ J/ s ] // [K] // H ea ti ng a r ea o f e v a po r a to r i n [ s q 158

m] 22   U=Q/(A*dT)

// O v er a l l h ea t t r a n s f e r

c o e f f i n [W/

sq m.K] 23   printf ( ” \ n Ste am e co no my i s %f \   n ” ,eco); 24   printf ( ” \n\ n O v e r a l l h e a t t r a n s f e r c o e f f i c i e n t i s %d W/ sq m.K” , round ( U ) ) ;

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 ;

  / / E xa mp le 6 . 7 / / From p r e v i o u s e x am p le : // [ kg /h ]   mf_dot=5000 // [ kJ/ kg ]   Hf=125.79 // [ kJ/ kg ]   lambda_s=2230.2 // [ kg /h ]   mdash_dot=2500 // [ kJ / kg ]   Hdash=313.93 // [ kg /h ]   mv_dot=2500 // [ kJ/ kg ]   Hv=2635.3   ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/ / / Steam f l o w r a t e i n [ kg / h ] lambda_s // [ kJ/ h ]   Q=ms_dot*lambda_s // [W]   Q=Q*1000/3600 // [W/ sq m.K ]   U=2862 // [K]   dT=35 // [ sq m]   A=Q/(U*dT)   printf ( ” \n The h ea t t r a n s f e r a re a i n t h i s c as e i s %f   s q m\n ” ,A);   printf ( ” \n\nNOTE : T he re i s a c a l c u l a t i o n m i s ta k e i n

t he book a t t he l i n e 1 2 o f t h i s code , m s d o t v a l u e i s w r i t t en a s 2 3 2 0 .1 8 , whi ch i s 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 ;

  / / E xa mp le 6 . 8 / / Feed r a t e i n [ kg / h ] // T akin g t he g i ve n v a l u e s fro m p r e v i o u s exa mp le ( 6 . 6 ) // [ kJ/ kg ] Hf=125.79 // [ kg /h ] ms_dot=3187.56 // [ kJ/ kg ] lambda_s=2230.2 // [ kJ/ kg ] Hdash=419.04 // [ kJ/ kg ] Hv=2676.1

  mf_dot=6000          

mv_dot=(mf_dot*Hf+ ms_dot*lambda_s -6000*Hdash)/(HvHdash) / / Water e v a p o r a t e d i n [ k g /h ] 12   mdash_dot=6000-mv_dot // Mass f l ow r a t e o f  

p r o d u c t [ k g /h ] 13   x=(0.01*mf_dot)*100/mdash_dot

/ /Wt % o f s o l u t e

i n p r od u ct s 14   printf ( ” \ nMass f l o w r a t e o f p ro du ct i s %f k g /h \n\n ” ,  mdash_dot ) ; 15   printf ( ” \n\nThe p ro du ct c o n c e n t r a t i o n i s %f p e r ce n t by w e i gh t \n\n ” ,x);

Scilab code Exa 6.9  Heat load in single effect evaporator

1 2 3 4 5 6 7 8 9 10

clc ;   clear ;

  / / E xa mp le 6 . 9 // F eed t e mp e ra t u re i n [ K]   Tf=298 // [K]   T_dash=373 // [ kJ/ kg .K]   Cpf=4 // F i na l c o n c en t r a ti o n o f s a l t   fc=0.2 // I n i t i a l co nc en tr at io n   ic=0.05 / / [ kg / h ] Fe ed t o e v a p o r a t o r   mf_dot=20000 / / Th ic k l i q u o r [ kg / h ]   mdash_dot=ic*mf_dot/fc 160

/ / Water e v a po r a te d i n [

11 mv_dot= mf_dot -m dash_dot

kg/ h ] 12 13 14 15 16 17 18 19 20 21 22 23 24

           

// [ kJ/ kg ] lambda_s=2185 // [ kJ/ kg ] lambda=2257 // B o i l i n g p o in t r i s e [K ] bpr=7 // B o i l i n g p o i n t o f s o l u t i o n i n [K ] T=T_dash+bpr // T e mp er at ur e o f c o n d en s i n g s te am i n [ K] Ts=39 ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda)/lambda_s

         

/ / St ea m c o n s um p t i on i n [ k g /h ] / / Economy o f e v a p o r a t o r eco=mv_dot/ms_dot // [ kJ /h ] Q=ms_dot*lambda_s // [ J/ s ] Q=Q*1000/3600 printf ( ” \ nHeat l o ad i s %d W o r J / s ” , round ( Q ) ) ; printf ( ” \n\nEconomy o f e v a p o r a t o r i s %f ” ,eco);

  printf ( ” \n\nNOTE : A ga in t h e r e

is a cal cua lt io n m is ta ke i n book a t l i n e 19 o f co de , i t i s w r i t t e n a s 4 0 4 1 5 0 7. 1 i n s t e a d o f 4 04 15 07 1 \n\n ” ) ;

Scilab code Exa 6.10  Triple effect evaporator

1 2 3 4 5 6

clc ;   clear ;

  / / E x am pl e 6 . 1 0 // [K]   Ts=381.3 // [K]   dT=56.6;   U1=2800; // O v er a l l h ea t t r a n s f e r

coeff in f i r s t

effect 7   U2=2200; // O v er a l l h ea t t r a n s f e r

coeff in f i r s t

effect 8   U3=1100; // O v er a l l h ea t t r a n s f e r

coeff in f i r s t

effect 9   dT1=dT/(1+(U1/U2)+(U1/U3)) 10   dT2=dT/(1+(U2/U1)+(U2/U3)) 11   dT3=dT-(dT1+dT2)

161

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

12 13 14 15 16

  //dT1=Ts−T1 d a s h

/ / [K ]

  T1dash=Ts-dT1

  //dT2=T1 dash−T2 d a s h

/ / [K ] // [K] T2_dash =T1dash -dT2   printf ( ” \n\ n B o i l i n g p o in t o f s o l u t i o n i n f i r s t e f f e c t =%f K\n\n” ,T1dash); 17   printf ( ” \n\ n B oi l in g p o i n t o f s o l u t i o n i n s ec o nd e f f e c t =%f K\n\n” ,T2_dash);

Scilab code Exa 6.11  Double effect evaporator

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

clc ;   clear ;

  / / E x am pl e 6 . 1 1                  

13   14   15 16 17 18 19

         

20  

/ / [ k g /h ] o f f e e d mf_dot=10000 // I n i t i a l co nc en tr at io n ic=0.09 // F i n a l c o n c e n t r a t i o n fc=0.47 // [ kg /h ] m1dot_dash=ic*mf_dot/fc / / S te am p r e s s u r e [ kPa . g ] Ps=686.616 // [ kPa ] Ps=Ps+101.325 / / S a t u r a t i o n t e mp e ra t u re i n [ K] Ts=442.7 / /Vacuum i n s e co n d e f f e c t i n [ kPa ] P2=86.660 // O v e r a l l h ea t t r a n s f e r i n f i r s t e f f e c t U1=2326 [W/ sq m.K] // O v e r a l l h ea t t r a n s f e r i n 2 nd e f f e c t [W U2=1744.5 /sqm .K ] // A bs o lu te p r e s s u r e i n s ec on d P2_abs=101.325-P2 e f f e c t [ kPa ] // T em per at ur e i n 2 nd e f f e c t i n [K ] T2=326.3 // [K] dT=Ts-T2 / / Fe ed t e m p e ra t u r e i n [ K] Tf=309 // [K] T=273 // kJ / kg . K S p e c i f i c h e a t f o r a l l Cpf=3.77 c a u s t i c s tr ea ms //Q1=Q2 162

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

  //U1∗A1∗dT1=U2∗A2∗dT2 // [K]   dT2=dT/1.75 // [K]   dT1=(U2/U1)*dT2 // S i nc e t h e r e i s no B . P . R / / Te mp er at ur e i n v ap or s p a ce o f     Tv1=Ts-dT1 f i r s t e f f e c t i n [K] // S eco n d e f f e c t [K ]   Tv2=Tv1-dT2 / / F e ed e n t h a l p y [ k J / k g ]   Hf=Cpf*(Tf-T) // E nt ha lp y o f f i n a l p ro du ct [   H1dash=Cpf*(Tv1-T) kJ/kg ]   //kJ/kg   H2dash=Cpf*(Tv2-T)   / / For s team a t 4 4 2 . 7 K // [ kJ/ kg ]   lambda_s=2048.7   / / For v ap ou r a t 3 9 2 . 8 K // [ kJ/ kg ]   Hv1=2705.22 // [ kJ/ kg ]   lambda_v1=2202.8 // f o r v ap ou r a t 3 2 6 . 3 K : // [ kJ/ kg ]   Hv2=2597.61 // [ kJ/ kg ]   lambda_v2=2377.8 //Overall material balance : // [ kg /h ]

mv_dot= mf_dot -m 1dot_dash

/ / E q u a t i o n 4 b e co m es : / / m v 1 do t ∗ l a m b d a v 1 +m f d o t ∗ Hf=(mv dot−mv1 dot ) ∗Hv2+( m f d o t −mv2 dot ) ∗ H 2 d a s h

44 mv1_dot=(H2dash*( mf_dot -mv_dot)- mf_dot*Hf+mv_dot* Hv2 )/(Hv2+lambda_v1 -H2dash) 45 mv2_dot =mv_dot -mv1_d ot // [ kg /h ] 46 47   / / From e q u a t i o n 2 48 49   m2dot_dash=m1dot_dash+mv1_dot // Fi rs t

e f f e c t m a t e r i a l b a l an c e [ kg /h ] 50 ms_dot=(mv1_dot* Hv1+m1dot_dash* H1dash -m2dot_dash* // [ kg /h ] H2dash)/lambda_s 51 52

163

53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78

  / / Heat t r a n s f e r Area // F i r s t e f f e c t   A1=ms_dot*lambda_s*(10^3)/(3600*U1*dT1)

// [ sq m]

  // S ec on d e f f e c t   lambda_v1=lambda_v1*(10^3/3600)   A2=mv1_dot*lambda_v1/(U2*dT2)

// [ sq m]

/ / S i n c e A1 n o t= A2 //SECOND TRIAL   Aavg=(A1+A2)/2   dT1_dash=dT1*A1/Aavg   dT2_dash=dT-dT1

// [ sq m] // [K] // / [K]

  // T em pe ra tu re d i s t r i b u t i o n // [K]   Tv1=Ts-dT1_dash // [K]   Tv2=Tv1-dT2_dash // [ kJ/ kg ]   Hf=135.66 // [ kJ/ kg ]   H1dash=Cpf*(Tv1-T) // [ kJ/ kg ]   H2dash=200.83   / / Vapour a t 3 8 8 . 5 K // [ kJ/ kg ]   Hv1=2699.8 // [ kJ/ kg ]   lambda_v1=2214.92

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   / / F i r s t e f f e c t Energ y b a la n ce 82 ms_dot=((mv1_dot* Hv1+m1dot_dash* H1dash)-(mf_dot // [ kg /h ]  mv2_dot )* H2dash )/ lambda_s 83 84   // Area o f h ea t t r a n s f e r 85   lambda_s=lambda_s*1000/3600 86   A1=ms_dot*lambda_s/(U1*dT1_dash) // [ sq m] 87 88 // S ec on d e f f e c t :

164

89   A2=mv1_dot*lambda_v1*1000/(3600*U2*dT2_dash)

// [ sq m] 90 91   printf ( ” \nA1 ( %f )=A2 ( %f ) , S o t h e a r e a i n e ac h e f f e c t can be %f s q m\n” ,A1,A2,A2); 92   printf ( ” \ nHeat t r a n s f e r s u r f a c e i n e ac h e f f e c t i s %f   s q m\n ” ,A2); 93   printf ( ” \ nSteam cons umpt ion=%d kg/h \n ” , round (ms_dot) ); 94   printf ( ” \ n Ev ap or at io n i n t he f i r s t e f f e c t i s %d kg /h \n ” , round (mv1_dot)); 95   printf ( ” \ n Ev ap o ra ti on i n 2 nd e f f e c t i s %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 19

clc ;   clear ;

  / / E x am pl e 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 c o nc o f g l y c e r i n e // Fi na L CONC OF GLYCERINE // O v e r a l l g l y c e r i n e b a l a nc e // [ kg /h ] m3dot_dash=(ic/fc)*mf_dot // / [ kg/ h ] mv_dot= mf_dot -m 3dot_dash / / S te am p r e s s u r e [ kPa ] P=313; / / [ f ro m s t ea m t a b l e ] [ K ] Ts=408; // [ P r e s s u r e i n l a s t e f f e c t ] [ kPa ] P1=15.74; / / [ V a po u r t e m p e r a t u r e ] Tv3=328; / / O v e r a l l a p pa r en t [ K] dT=Ts-Tv3 // [K] bpr1 =10 ;

  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; // [K] sum_bpr=bpr1+bpr2+bpr3 // True Overall dT=dT-sum_bpr // [K] dT1=14.5; // [K] dT2=16; // [K] dT3=19.5; // [ kJ /( kg .K) ] Cpf=3.768

// E n t h a lp i e s o f v a r i o u s s tr ea m s // [ kJ/ kg ] Hf=Cpf*(Tf-T) // [ kJ/ kg ] H1=Cpf*(393.5-T) // [ kJ/ kg ] H2=Cpf*(367.5-T) // [ kJ/ kg ] H3=Cpf*(338-T) / / F or s te am a t 4 0K // [ kJ/ kg ] 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 //MATERIAL AND EBERGY BALANCES // F i r s t e f f e c t //Material balance

  / / m 1 d o t d a s h = m f d o t −m v 1 d o t   //m1dot dash=1750+mv2 dot+mv3 dot

  / / E n er gy b a l a n c e / / m s do t ∗ l a m b d a s + m f D o t ∗ hf=mv1 dot ∗Hv1+m1dot dash ∗ H1 50   / / 2 1 6 0∗ m s d o t + 2 23 8∗( mv2 dot+mv3 dot ) =19800500 51 52 53 54 55 56

  // S ec on d e f f e c t / / E n er g y b a l a n c e : / / m v 3 d o t = 8 7 0 9 . 5 4 − 2 . 0 7 6 ∗ m v 2 d o t // T hi rd e f f e c t : 166

57 58 59 60

  //m2dot dash=mv3 dot+m3dot dash   //m2dot dash=mv3 dot+1750   / / From e qn 8 we g e t

61 62 63 64 65 66

  / / From e qn 8 :

  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)   mv3_dot=8709.54-2.076*mv2_dot mv1_dot=mv_dot -(mv2_dot+ mv3_dot)

// [ kg /h ] // [ kg /h ]

  / / From e q u a t i o n 4 :   / / m 1 d o t d a s h = m f d o t −m v 1 d o t   / / m s d o t = ( m v 1 d o t ∗Hv1+m1dot dash ∗H1−m f d o t ∗ H f ) / l am bd a s / / [ kg /h ]

67   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

  / / Heat t r a n s f e r Area i s // [W/ sq m.K]   U1=710 // [W/ sq m.K]   U2=490 // [W/ sq m.K]   U3=454   A1=ms_dot*lambda_s*1000/(3600*U1*dT1)   A2=mv1_dot*lambda_v1*1000/(3600*U2*dT2)   A3=mv2_dot*lambda_v2*1000/(3600*U3*dT3)

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

  / / The d e v i a i t o n i s w i th i n +−10% / / H e n c e maximum A1 a r e a c a n b e r ec o mm en d ed   eco=(mv_dot/ms_dot)

// [ Steam economy ]

// [ kJ/ h ] Qc=mv3_dot*lambda_v3 / / R i s e i n w at er t e mp e r at u r e dT=25 Cp=4.187 mw_dot=Qc/(Cp*dT) printf ( ” \nANSWER\n A rea i n e a c h e f f e c t % f s q m\n ” ,A1) ; 86   printf ( ” \nANSWER \n S tea m e co no my i s % f  \n ” ,eco); 87   printf ( ” \nANSWER C o o l i n g w a t e r r a t e i s %f t / h ” ,  mw_dot /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 ;

  / / E x am pl e 6 . 1 3                        

// [ kJ/ kg .K] Cpf=4.18 // [K] dT1=18 // [K] dT2=17 // [K] dT3=34 // [ kg/ s ] mf_dot=4 // [K] Ts=394 //Bp o f wa te r a t 13 . 1 72 kPa [K ] bp=325 // [K] dT=Ts-bp // [ kJ/ kg ] lambda_s=2200 // [K] T1=Ts-dT1 // [ kJ/ kg ] lambda1=2249 // [ kJ/ kg ] lambda_v1=lambda1

  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 c o nc o f s o l i d s   ic=0.1 // F i n a l c o n c o f s o l i d s   fc=0.5 // [ kg/ s ]   m3dot_dash=(ic/fc)*mf_dot //Total evapora tion mv_dot= mf_dot -m 3dot_dash

i n [ kg / s ] 29 // M a t er i a l b a l an c e o v er f i r s t 30   / / m f d o t = m v 1 d o t m 1 d o t d a s h 31 / / E n er g y b a l a n c e : 168

ef fe ct

32 / / m s do t ∗ l a m b d a s = m f d o t ∗ ( C p f  ∗ ( T1−Tf)+mv1 dot ∗

lambda v1 ) 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 59 60

// M a te r ia l b a la n ce o ve r s ec on d e f f e c t   //m1dot dash=mv2 dot+m2dot dash / / E n th a lp y b a l a n c e : / / m v 1 do t ∗ l a m b d a v 1+m 1 d o t d a s h ( c p ∗ ( T1−T2)=mv2 dot ∗ lambda v2 )   / / M a te r ia l b a la n ce o ve r t h i r d e f f e c t   //m2dot dash=mv3 dot+m3dot+dash / / E n th a lp y b a l a n c e :   / / m v 2 l a m b d a v 2 + m 2 d o t d a s h ∗ cp ∗ ( T2−T3)=mv3 dot ∗ lambda v3                  

294 // [ kg/ s ] 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 ms_dot=(mf_dot*Cpf*(T1-294)+mv1_dot*lambda_v1)/ // [ kg/ s ] lambda_s // Steam economy eco=mv_dot/ms_dot eco= round (eco) printf ( ” \ n St ea m e co no my i s %d\n ” ,eco); // [kW/ sq m.K] U1=3.10 // [kW/ sq m.K] U2=2 // [kW/ sq m.K] U3=1.10

// F i r s t e f f e c t : // [ sq m] // [ sq m] // [ sq m]   // A r e a s a re c a l c u l a t e d w i th a d e v i a t i o n o f +−10%   printf ( ” \ nArea p f h e a t t r a n s f e r i n e a c h e f f e c t i s %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 ;

  / / E x am pl e 6 . 1 4        

// [ kg /h ] mf_dot=1060 // I n i t i a l co nc en tr at io n ic=0.04 //Final concentrat ion fc=0.25 // [ kg /h ] m4dot_dash=(ic/fc)*mf_dot //Total evaporati on= // [ kg /h ]

mv_dot= mf_dot -m 4dot_dash

/ / F ro ms te am t a b l e : // [ kPa . g ]   P1=370 // [K]   T1=422.6 // [ kJ/ kg ]   lambda1=2114.4 // [ kPa . g ]   P2=235 // [K]   T2=410.5 // [ kJ/ kg ]   lambda2=2151.5 // [ kPa . g ]   P3=80 // [K]   T3=390.2 // [ kJ/ kg ]   lambda3=2210.2 // [ kPa . g ]   P4=50.66 // [K]   T4=354.7 // [ kJ/ kg ]   lambda4=2304.6 / / L a t e n t h e a t o f s te a m [ kPa . g ]   P=700 // [ kJ/ kg ]   lambda_s=2046.3   / / FIRST EFFECT / / E n th a lp y b a l a n c e :   / / m s d o t = m f d o t ∗ Cp f ∗ ( T1−Tf)+mv1 dot ∗ lambda1 / / m s d o t = 1 3 4 5 . 3 − 1 . 0 3 3∗ m 1 d o t d a s h //SECOND EFFECT 170

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

  //m1dot dash=m2dot dash+mdot v2 / / E n th a lp y b a l a n c e : / / m 1 d o t d a s h = 5 3 1 . 3 8 + 0 . 5 1 0∗ m 2 d o t d a s h //THIRD EFFECT // Material balance : / / m 2 d o t da s h−m 3 d o t d a s h + m v 3 d o t //FOURTH EFFECT   //m3dot dash=m4dot dash+mv4 dot // [ kg /h ]   mv4dot_dash=169.6 // [ kg /h ]   m3dot_dash=416.7   / / From e q n 4 :   m2dot_dash=-176.84+1.98*m3dot_dash

// [ kg /h ]

  / / From e qn 2 :   m1dot_dash=531.38+0.510*m2dot_dash

// [ kg /h ]

  / / From e qn 1 :   ms_dot=1345.3-1.033*m1dot_dash / / [ kg e v a p o r a t i o n / kg   eco=mv_dot/ms_dot

steam ] 59   printf ( ” \ nS team eco no my i s %f e v a p o r a t i o n / kg s te am ” , eco);

Scilab code Exa 6.15   Single effect Calendria

1 2 3 4 5 6 7

clc ;   clear ;

  / / E x am pl e 6 . 1 5        

// [ kg /h ] m1_dot=5000 // I n i t i a l co nc en tr at io n ic=0.1 // F i n a l c o n c e n t r a t i o n fc=0.5 // [ kg /h ] mf_dot=(fc/ic)*m1_dot 171

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

                     

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

                             

35 36 37 38

       

/ / W a te r e v a p o r a t e d [ k g / h ] mv_dot= mf_dot -m1_dot / / St ea m p r e s s u r e [ kN/ s q m ] P=357 // [K] Ts=412 // [ kJ/ kg ] H=2732 // [ kJ/ kg ] lambda=2143 // [K] bpr=18.5 // [K] T_dash=352+bpr // [ kJ/ kg ] Hf=138 // [ kJ/ kg ] lambda_s=2143 // [ kJ/ kg ] Hv=2659 // [ kJ/ kg ] H1=568 ms_dot=(mv_dot*Hv+m1_dot*H1-mf_dot*Hf)/lambda_s / / Steam c o ns u mp t io n i n k g /h printf ( ” \ nS team c o ns u mp t io n i s %f kg / h\n ” ,ms_dot); printf ( ” \ n Ca p ac it y i s %f k g /h\n ” ,mv_dot);   //Economy eco=mv_dot/ms_dot   n ” ,eco); printf ( ” \ n Ste am e co no my i s %f \ // [K] dT=Ts-T_dash // [W/ sq m.K ] hi=4500 // [W/ sq m.K ] ho=9000 // [m] Do=0.032 // [m] Di=0.028 // [m] x1=(Do-Di)/2 // [m] Dw=(Do-Di)/ log (32/28) // [m] x2=0.25*10^-3 / / L e ng t h [ m] L=2.5 // [W/ sq m.K ] hio=hi*(Di/Do) printf ( ” \n NOTE: I n t ex tb o ok t h i s v a lu e o f h i o i s w r o n g l y c a l c u l a t e d a s 3 9 7 5 . 5 . . So we w i l l t ak e t h i s \ n\n ” ) ; hio=3975.5 / / Tube m a t e r i a l i n [W/ s q m. K] k1=45 / / F o r s c a l e [W/m . K ] k2=2.25 Uo=1/(1/ho+1/hio+(x1*Dw)/(k1*Do)+(x2/k2))

O v e r a l l h ea t t r a n s f e r 39   Q=ms_dot*lambda_s 40   Q=Q*1000/3600 41

c o e f f i n W/ s q m. K // [ kJ /h ] // [W]

172

//

42   A=Q/(Uo*dT) // [ sq m] 43   n=A/(%pi*Do*L) // from A=n∗ %p i ∗Do∗L 44   printf ( ” \n No . o f t ub es r e q u i r e d i s %d” , round ( n ) ) ;

Scilab code Exa 6.16  Single effect evaporator

1 2 3 4 5 6 7 8 9 10 11

clc ;   clear ;

  / / E x am pl e 6 . 1 6                

// [K] bpr=40.6; // [ kJ/ kg .K] Cpf=1.88; // [ kJ/ kg ] Hf=214; // [ kJ/ kg ] H1=505; // [ kg /h ] o f f e e d s o l u t i o n mf_dot=4536; // I n i t i a l co nc ic=0.2; //Final concentrati on fc=0.5; // T hi sc k l i q u o r m1dot_dash=(ic/fc)*mf_dot

flow

a r te [ kg/h ] 12 mv_dot= mf_dot -m 1dot_dash 13   Ts=388.5; // S a t u r a t i o n

// [ kg /H] t e mp e r at u r e o f s te am

i n [ K] 14 15 16 17 18 19 20

       

// b . P o f s o l u t i o n i n [K ] bp=362.5 // [ kJ/ kg ] lambda_s=2214; / / V ap or s p a c e i n [ kPa ] P=21.7; // [ kJ/ kg ] Hv=2590.3;

  // E n th al py b a l a n c e o v er e v a p o r a t o r   ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s

// [ kg/ h 21   printf ( ” \ nS team c o ns u mp t io n i s %f kg / h\n ” ,ms_dot); 22   dT=Ts-bp // [K] 23   U=1560 // [W/ sq m.K ] 24   Q=ms_dot*lambda_s // [ kJ/ h ] 25   Q=Q*1000/3600 // [W] 26   A=Q/(U*dT) // [ sq m] 173