# Chapter 2 One-dimensional Steady State Conduction

CHAPTER 2 ONE-DIMENSIONAL STEADY STATE CONDUCTION 2.1 Examples of One-dimensional Conduction: 2.1.1 Plate with Energy Generation and Variable Conductivity 1 Example 2.1: Plate with internal energy generation q and a variable k k ko (1 T ) 0o C Find temperature distribution. 0 (1) Observations Variable k Symmetry Energy generation Rectangular system Specified temperature at boundaries q 0o C

x L Fig. 2.1 2 (2) Origin and Coordinates Use a rectangular coordinate system (3) Formulation (i) Assumptions One-dimensional Steady Isotropic Stationary Uniform energy generation 3 (ii) Governing Equation Eq. (1.7): d dT k q 0 dx dx (2.1) k ko (1 T )

(a) d dT q (1 T ) 0 dx dx ko (b) (a) into eq. (2.1) (iii) Boundary Conditions. Two BC are needed: 4 T (0) 0 (c)

T ( L) 0 (d) 2 q 2 T T x C1 x C 2 2 2ko (e) (4) Solution Integrate (b) twice BC (c) and (d) qL C1 , 2ko C 2 0 (f) 5 (f) into (e)

2 qL x T T ko 2 x 1 L 0 (g) Solving for T 1 1 qL x x T 2 1 ko L

(h) Take the negative sign 1 T 1 qL x x 1 2 ko L (i) 6 (5) Checking Dimensional check Boundary conditions check Limiting check: q 0, T 0 Symmetry Check: 1 2

dT 1 1 qLx x 2 ( 1) dx 2 ko L qL 2 x ( )( 1) (j) ko L Setting x = L/2 in (j) gives dT/dx = 0 7 Quantitative Check Conservation of energy and symmetry: qAL q(0) 2 (k) qAL q( L)

2 (l) Fouriers law at x = 0 and x = L dT (0) qAL q(0) ko 1 T (0) dx 2 (m) 8 dT ( L) qAL q( L) ko 1 T ( L) dx 2 (n) (6) Comments Solution to the special case: k = constant: Set 0

2.1.2 Radial Conduction in a Composite Cylinder with Interface Friction 9 Example 2.2: Rotating shaft in sleeve, frictional heat at interface, convection on outside. Conduction in radial direction. Determine the temperature distribution in shaft and sleeve. h, T Rs Ro T1 r 0 (1) Observations Composite cylindrical wall Cylindrical coordinates Radial conduction only qi T2

sleeve Fig. 2.2 10 Steady state: Energy generated = heat conducted through the sleeve No heat is conducted through the shaft Specified flux at inner radius of sleeve, convection at outer radius (2) Origin and Coordinates Shown in Fig. 2.2 11 (3) Formulation (i) Assumptions One-dimensional radial conduction Steady Isotropic Constant conductivities No energy generation Perfect interface contact Uniform frictional energy flux Stationary 12

(ii) Governing Equation Shaft temperature is uniform. For sleeve: Eq. (1.11) d dT1 r 0 dr dr (2.2) (iii) Boundary Conditions Specified flux at Rs : dT1 ( Rs ) qi k1 dr (a) 13 Convection at Ro : dT1 ( Ro ) k1 h[T1 ( Ro ) T ] dr

(b) (4) Solution Integrate eq. (2.2) twice T1 C1 ln r C 2 (c) BC give C1 and C2 qi Rs C1 k1 (d) 14 and qiRs k1 C 2 T ln Ro k1 hRo

(e) (d) and (e) into (c) qi Rs T1 ( r ) T k1 Ro k1 ln r hR o (f) hRo / k = Biot number Shaft temperature T2: Use interface boundary condition T2 ( r ) T2 ( Rs ) T1 ( Rs ) (g) 15 Evaluate (f) at r = Rs and use (g) qi Rs T2 ( r ) T k1

Ro k1 ln R hR s o h, T (5) Checking Dimensional check Boundary conditions check Limiting check: qi 0 (h) Rs Ro T1 r 0 qi T2 sleeve

Fig. 2.2 (6) Comments Conductivity of shaft does not play a role 16 Problem can also be treated formally as a composite cylinder. Need 2 equations and 4 BC. 2.1.1 Composite Wall with Energy Generation Example 2.1: Plate 1 generates heat at q. Plate 1 To x L2 k2 is sandwiched between two plates. Outer surfaces of two plates at To . q L1 L2 0

To k1 k2 Find the temperature distribution in the three plates. Fig. 2.3 17 (1) Observations Composite wall Use rectangular coordinates Symmetry: Insulated center plane To x L2 k2 q

L1 L2 0 k1 k2 To Fig. 2.3 Heat flows normal to plates Symmetry and steady state: Energy generated = Energy conducted out (2) Origin and Coordinates Shown in Fig. 2.3 18 (3) Formulation (i) Assumptions Steady One-dimensional Isotropic Constant conductivities Perfect interface contact Stationary

(ii) Governing Equations Two equations: 19 2 d T1 dx q 0 (a) k 2 2 d T2 dx 2 0 (b) (iii) Boundary Conditions Four BC: Symmetry:

dT1 (0) 0 dx (c) dT1 ( L1 / 2) dT2 ( L1 / 2) k1 k2 dx dx (d) Interface: 20 T1 ( L1 / 2) T2 ( L1 / 2) (e) Outer surface: T2 ( L1 / 2 L2 ) To (f)

(4) Solution Integrate (a) twice q 2 T1 ( x ) x Ax B 2k1 (g) Integrate (b) T2 ( x ) Cx D (h) 21 Four BC give 4 constants: Solutions (g) and (h) become qL21 1 k1 L2 x 2 T1 ( x ) To 2 2k1 4 k2 L1 L1 (i) qL21 1 L2 x T2 ( x ) To

2k2 2 L1 L1 (j) (5) Checking qL2 : Dimensional check: units of k 22 q( W/m 3 ) L2 ( m 2 ) k ( W/m o C) oC Boundary conditions check Quantitative check: 1/2 the energy generated in center plate = Heat conducted at x L1 / 2 L1 dT1 ( L1 / 2) q k1 2 dx

(k) 23 (i) into (k) dT1 ( L1 / 2) L1 k1 q dx 2 Similarly, 1/2 the energy generated in center plate = Heat conducted out L1 dT2 ( L1 / 2 L2 ) q k2 2 dx (l) (j) into (l) shows that this condition is satisfied. Limiting check: (i) If q 0, then T1 ( x ) T2 ( x ) To . (ii) If L1 0 then T1 ( x ) To . 24

(6) Comments Alternate approach: Outer plate with a specified flux at x L1 / 2 and a specified temperature at x L1 / 2 L2 . 2.2 Extended Surfaces - Fins 2.2.1 The Function of Fins Newton's law of cooling: q s hAs (Ts T ) (2.3) 25 Options for increasing q s : Increase h Lower T Increase As Examples of Extended Surfaces (Fins): Thin rods on condenser in back of refrigerator Honeycomb surface of a car radiator Corrugated surface of a motorcycle engine Disks or plates used in baseboard radiators 26

2.2.2 Types of Fins (a) constant area straight fin (b) variable area straight fin (c) pin fin (d) annular fin Fig. 2.5 27 Terminology and types Fin base Fin tip Straight fin Variable cross-sectional area fin Spine or pin fin Annular or cylindrical fin 2.2.3 Heat Transfer and Temperature Distribution in Fins Heat flows axially and laterally (two-dimensional) Temperature distribution is two-dimensional 28 2.2.4 The Fin Approximation

Neglect lateral temperature variation T T ( x ) r Criterion: Biot number = Bi T x T h,T Fig . 2.6 Bi = h /k << 1 (2.4) / k Internal resis tan ce Bi 1 / h external resis tan ce 29 2.2.5 The Fin Heat Equation: Convection at

Surface (1) Objective: Determine fin heat transfer rate. Need temperature distribution. (2) Procedure: Formulate the fin heat equation. Apply conservation of energy. Select an origin and coordinate axis x. Assume Bi 0.1, T T ( x ) Stationary material, steady state 30 y 0 ys dx h, T x dAs C qx

qx dqx dx dx dy (a) dqc (b) dx ds (c ) Fig. 2.7 Conservation of energy for the element dx: E in E g = E out (a) E in q x

(b) dq x Eout q x dx dqc dx (c) 31 dAs C qx qx dq x dx dx dy dqc (b ) dx ds (c )

(b) and (c) into (a) dq x Eg dx dqc dx (d) Fourier's law and Newtons law dT q x kAc dx dqc h(T T )dAs Energy generation (e) (f) 32 E g qAc ( x )dx (g) (e), (f) and (g) into (d)

dT d kAc ( x) dx dx dx h (T T )dAs qAc ( x ) dx 0 (2.5a) Assume constant k d 2T dAs q 1 dAc dT h (T T ) 0 2 A ( x ) dx dx kA ( x ) dx k dx c c (2.5b)

(2.5b) is the heat equation for fins Assumptions: (1) Steady state (2) Stationary 33 (3) Isotropic (4) Constant k (5) No radiation (6) Bi << 1 Ac , dAc / dx , and dAs / dx are determined from the geometry of fin. 2.2.6 Determination of dAs /dx From Fig. 2.7b dAs C ( x ) ds (a) C ( x ) = circumference ds = slanted length of the element 34 For a right triangle

2 2 1/ 2 ds [dx dy s ] (b) (b) into (a) 2 1/ 2 dy s dAs C ( x ) 1 dx dx For dy s / dx << 1 dAs C ( x ) dx (2.6a) (2.6b) 2.2.7 Boundary Conditions Need two BC

35 2.2.8 Determination of Fin Heat Transfer Rate q f : h, T 0 q( 0 ) qs x h, T qs qs Fig. 2.8 Conservation of energy for q 0 : q f q(0) q s Two methods to determine q f (a) : 36

(1) Conduction at base. Fourier's law at x = 0 dT (0) q f q(0) kAc (0) dx (2.7) (2) Convection at the fin surface. Newton's law applied at the fin surface q f q s h[T ( x ) T ]dAs As (2.8) Fin attached at both ends: Modify eq. (2.7) accordingly Fin with convection at the tip: Integral in eq. (2.8) includes tip 37 Convection and radiation at surface: Apply eq. (2.7). Modify eq. (2.8) to include heat exchange by radiation.

2.2.9 Applications: Constant Area Fins with Surface Convection h, T 0 To C x h, T Fig. 2.9 Ac 38 A. Governing Equation Use eq. (2.5b). Set dAc / dx 0 h, T (a) 0 To

y s = constant dy s / dx 0 C x h, T Ac Fig. 2.9 Eq. (2.6a) dAs / dx C (b) (a) and (b) into eq. (2.5b) d 2T hC ( T T

) 0 dx 2 kAc (2.9) 39 Rewrite eq. (2.9) T T (c) hC m kAc (d) 2 Assume T = constant, (c) and (d) into (2.9) 2 d dx 2

2 m 0 (2.10) Valid for: (1) Steady state (2) constant k, Ac and T 40 (3) No energy generation (4) No radiation (5) Bi 1 (6) Stationary fin B. Solution Assume: h = constant ( x ) A1 exp(mx ) A2 exp( mx ) (2.11a) ( x ) B1 sinhmx B2 coshmx (2.11b) 41

C. Special Case (i): Finite length Specified temperature at base, convection at tip Boundary conditions: 0 To x h, T C ht h, T Fig. 2.10 Ac T (0) To dT ( L) k ht [T ( L) T ] dx (0) o (e)

(f) (h) 42 d ( L) k ht ( L) dx (i) Two BC give B1 and B2 ( x ) T ( x ) T o To T (2.12) cosh m L x ht mk sinh m L x cosh mL ht mk sinh mL Eq. (2.7) gives q f (To T )[sinh mL ( ht /mk )cosh mL] q f [k Ac C h] cosh mL ( ht /mk ) sinh mL 1/ 2

43 (2.13) C. Special Case (ii): Finite length Specified temperature at base, insulated tip BC at tip: d ( L ) 0 dx (j) Set ht 0 eq. (2.12) ( x ) T ( x ) T cosh m( L - x ) o To T cosh mL Set ht 0 eq. (2.13) q f kAc Ch 12 To T tanh mL

(2.14) (2.15) 44 2.2.10 Corrected Length Lc Insulated tip: simpler solution Simplified model: Assume insulated tip, compensate by increasing length by Lc The corrected length is Lc Lc L Lc (2.16) The correction increment Lc depends on the geometry of the fin: Increase in surface area due to Lc = tip area Circular fin: 2 ro 2 ro Lc 45 Lc ro / 2 Square bar of side t

Lc t / 4 2.2.11 Fin Efficiency f Definition f qf qmax (2.17) qmax hAs To T As = total surface area 46 Eq. (2.17) becomes f qf (2.18) h As (To T ) 2.1.12 Moving Fins Examples:

Extrusion of plastics Drawing of wires and sheets Flow of liquids Tsur h, T x To U dx (a) dqr (b) m h qx dqc dh m( h dx ) dx

dq x qx dx dx dx Fig. 2.11 47 Heat equation: Assume: dqr dqc dh m ( h dx ) dx m h q Steady state Constant area Constant velocity U Surface convection and radiation

qx x dx dq x dx dx (b) Fig. 2.11 Conservation of energy for element dx dq d h q x m h q x x dx m h m dqc dqr dx dx (a) m UAc

(b) 48 dh c p dT (c) Fouriers and Newtons laws dT q x kAc dx (d) dqc h (T T )dAs (e) dAs C dx (f) 4 dqr (T 4 Tsur ) dAs (g)

(b)-(g) into (a) assume constant k d 2T dx 2 c p U dT k hC 4 4 (T T ) (T Tsur ) 0 dx kAc k 49 (2.19) Assumptions leading to eq. (2.19): (1) Steady state (2) Constant U, k, P and , (3) Isotropic (4) Gray body

(5) Small surface enclosed by a much larger surface and (6) Bi << 1 50 2.2.13 Application of Moving Fins h, T x U W Example 2.4 Plastic sheet leaves furnace at To . furnace insulated bottom To Fig. 2.12 Sheet is cooled at top by convection. Assume: (1) Steady state (2) Bi < 0.1 (3) No radiation (4) No heat transfer from bottom Determine the temperature distribution in the sheet.

51 t Solution (1) Observations furnace Constant area fin Temperature is one-dimensional Convection at surface Specified furnace temperature Fin is semi-infinite Constant velocity h, T x U W t insulated bottom To Fig. 2.12 (2) Origin and Coordinates 52 (3) Formulation

(i) Assumptions (1) One-dimensional (2) Steady state (3) Isotropic (4) Constant pressure (5) Constant U, k, P and , (6) Negligible radiation (ii) Governing Equation Eq. (2.19) d 2T dx 2 c p U dT k hC (T T ) 0 dx kAc (2.20) 53

Ac Wt C W 2t (a) (b) (a) and (b) into eq. (2.20) 2 d T dT 2 2b m T c 2 dx dx (c) where b c pU 2k h (W 2t ) h(W 2t )

T , m , c kW t kWt 2 (d) (iii) Boundary Conditions T (0) To T () finite (e) (f) 54 Eq. (c) is: Linear Second order Constant coefficients (4) Solution: Eq. (A-6b), Appendix A T C1 exp( bx b 2 m 2 x ) C 2 exp( bx 2

2 b m x) B.C. (f) C1 0 (2.21) c m2 (g) 55 B.C. (e) c C 2 T0 2 m (h) (d), (g) and (h) into (2.21) c pU T ( x ) T exp

To T 2k c pU 2 h(W 2t ) ( 2k ) k W t x (2.22) (5) Checking Dimensional check: Each term in the exponential in eq. (2.22) is dimensionless 56 Boundary conditions check: Eq.(2.22) satisfies (e) and (f). Limiting checks: (i) If h 0 : (ii) If U : T ( x ) To T ( x ) To (6) Comments (i) Temperature decays exponentially (ii) Motion slows decay 57 2.2.14 Variable Area Fins

Ac Ac ( x ) Example: Cylindrical or annular fin Governing equation: Usually has variable coefficients Case (i) : The Annular Fin h, T h, T dr Fig. 2.13 d 2T dAs 1 dAc dT h ( T T ) 0

dr dr 2 Ac ( r ) dr dr kAc ( r ) 58 (2.23) r Ac ( r ) 2 r t (a) dAc / dr 2 t (b) Eq.(2.6a) gives dAs h, T h, T / dr dr Fig. 2.13 2 1/ 2

dy s dAs C ( r ) 1 dr dr For y s = constant: dy s / dr 0 C ( r ) 2( 2 r ) (c) 59 r Eq. (2.6a): dAs 2( 2 r ) dr (d) (a), (b) and (d) into eq. (2.23) d 2T 1 dT

( 2 h / kt )( T T ) 0 dr 2 r dr (2.24) Case (ii): Triangular straight fin Fin equation: Constant k, eq. (2.5b) Ac 2W y s ( x ) (e) y L h, T 0

h, T ys x dx Fig. 2.14 60 t y L h, T 0 h, T ys x t dx

Fig. 2.14 Eq. (2.6a): y s ( x / L)( t / 2) (f) Ac (W t / L) x (g) dAc W t / L dx (h) dAs 2W [1 (dy s / dx )2 ]1 / 2 2W [1 ( t / 2 L)2 ]1 / 2 dx (i) 61 (g), (h) and (i) into eq. (2.5b) d 2T 1 dT 2 1/ 2

1 / x (T T ) 0 2hL / kt 1 t / 2 L 2 dx x dx (2.25) Equations (2.24) and (2.25) are: Linear Second order Variable coefficients 62 2.3 Bessel Differential Equations and Bessel Functions 2.3.1 General form of Bessel Equations 2 d y 2 dy x (1 2 A) x 2 B x

2 dx dx 2 (2.26) C 2 D 2 x 2C B 2 x 2 B(1 2 A) x A2 C 2 n 2 y 0 Note the following: (1) Eq. (2.26) is linear, second order with variable coefficients (2) A, B, C, D, and n are constants 63 (3) n is called the order of the differential equation (4) D can be real or imaginary 2.3.2 Solution: Bessel Functions Form: Infinite power series solutions General solution: depends on the constants n and D

(1) n is zero or integer, D is real y( x ) x A exp(B x ) C1 J n ( D x C ) C 2 Yn ( D x C ) (2.27) where C1 ,C 2 = constant of integration 64 J n ( Dx C ) = Bessel function of order n of the first kind Yn ( Dx C ) = Bessel function of order n of the second kind Note the following: (i) The term ( D x C ) is the argument of the Bessel function (ii) Values of Bessel functions are tabulated (2) n is neither zero nor a positive integer, D is real y( x ) x A exp(B x ) C1 J n ( D x C ) C 2 J n ( D x C )

(2.28) 65 (3) n is zero or integer, D is imaginary A C C y( x ) x exp(B x ) C1 I n ( p x ) C 2 K n ( p x ) (2.29) where D p , i is imaginary = 1 i I n = modified Bessel function of order n of the first kind K n = modified Bessel function of order n of the second kind 66

(4) n is neither zero nor a positive integer, D is imaginary y( x ) x A exp(B x ) C1 I n ( p x C ) C 2 I n ( p x C ) (2.30) 2.3.3 Form of Bessel Functions J n , Yn , J n , I n , I n and K n : Symbols for infinite power series The form of each series depends on n Example: n 2, Bessel function J 2 ( x ) k 2 k 2 ( 1) ( x / 2) J 2 ( x) k! k 2 ! k 0 (2.31) 67

2.3.4 Special Closed-form Bessel Functions: odd integer n 2 For n 1 / 2 2 J1 / 2 ( x ) sin x x (2.32) 2 J 1/ 2 ( x) cos x x (2.33) and 68 For n = 3/2, 5/2, 7/2, use Eq. (2.32) or eq. (2.33) and the recurrence equation: 2k 1

J k 1 / 2 ( x ) J k 1 / 2 ( x) J k 2 / 3 ( x) x k = 1, 2, 3, (2.34) For the modified Bessel functions, n = 1/2 2 I1 / 2 ( x ) sinh x x (2.35) 2 I 1 / 2 ( x) cosh x x (2.36) and 69 For n = 3/2, 5/2, 7/2, . use eq. (2.35) or eq. (2.36) and the recurrence equation 2k 1 I k 1 / 2 ( x )

Ik 1 / 2 ( x) Ik 2 / 3 ( x) x k = 1, 2, 3, (2.37) 2.3.5 Special Relations for n = 0, 1, 2, n (2.38a) n Y n ( x ) ( 1) Yn ( x ) (2.38b) I n ( x ) I n ( x ) (2.38c) K n ( x ) K n ( x ) (2.38d) 70 J n ( x ) ( 1) J n ( x ) 2.3.6 Derivatives and Integrals of Bessel Functions

n mx Z n 1 mx d n x Z n mx dx mx n Z n1 mx Z J , Y , I (2.39) Z K (2.40) n mx Z n1 mx Z J , Y , K (2.41) d n x Z n mx dx

(2.42) mx n Z n1 mx Z I 71 n mZ n 1 mx Z n mx Z J , Y , I d x (2.43) Z n mx dx mZ mx n Z mx Z K n 1 n x (2.44) n mZ n1 mx Z n mx Z J , Y , K

d x (2.45) Z n mx dx mZ mx n Z mx Z I n1 n x (2.46) x n n Z n 1 ( mx )dx (1 / m ) x Z n ( mx ) Z J ,Y , I (2.47) 72 n n

x Z mx dx 1 m x Z n mx n1 Z J , Y , K (2.48) 2.3.7 Tabulation and Graphical Representation of Selected Bessel Functions Table 2.1 x J0(x) Jn(x) I0(x) In(x) 0

1 0 0 0 1 0 Yn(x) Kn(x) - 0 0 73 Fig. 2.15 Graphs of selected Bessel functions 74 2.4 Equidimensional (Euler) Equation 2 d y

dy x a1 x a0 y 0 2 dx dx 2 (2.49) Solution:Depends on roots r1 and r2 r1, 2 (a1 1) (a1 1) 2 4a0 2 (2.50) Three possibilities: (1) Roots are distinct r1 y( x ) C1 x C 2 x r2

(2.51) 75 (2) Roots are imaginary a y( x ) x C1 cos(b log x ) C 2 sin(b log x ) (2.52) (3) One root only y( x ) x r (C1 C 2 log x ) (2.53) 2.5 Graphically Presented Fin Solutions to Fin Heat Transfer Rate q f Fin efficiency f : f qf h As (To T ) (2.18) 76

Fig. 2.16 Fin efficiency of three types of straight fins [5] 77 Fig. 2.17 Fin efficiency of annular fins of constant thickness [5] 78

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