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## CHAPTER 2 ONE-DIMENSIONAL STEADY STATE CONDUCTION

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**CHAPTER 2ONE-DIMENSIONAL STEADY STATE CONDUCTION**2.1 Examples of One-dimensional Conduction: 2.1.1 Plate with Energy Generation and Variable Conductivity**Example 2.1:Plate with internal energy generation**and a variable k ¢ ¢ ¢ q o 0 C o 0 C x 0 L Fig. 2.1 Find temperature distribution. (1) Observations • Variable k • Symmetry • Energy generation • Rectangular system • Specified temperature at boundaries**(2) Origin and Coordinates**Use a rectangular coordinate system (3) Formulation (i) Assumptions • One-dimensional • Steady • Isotropic • Stationary • Uniform energy generation**(2.1)**(a) (b) (ii) Governing Equation Eq. (1.7): (a) into eq. (2.1) (iii) Boundary Conditions. Two BC are needed:**(c)**(d) (e) (f) , (4) Solution Integrate (b) twice BC (c) and (d)**(g)**(h) (i) (f) into (e) Solving for T Take the negative sign**• Limiting check:**(j) (5) Checking • Dimensional check • Boundary conditions check • Symmetry Check: Setting x = L/2 in (j) gives dT/dx = 0**(k)**(l) (m) • Quantitative Check Conservation of energy and symmetry: Fourier’s law at x = 0 and x = L**(n)**k = constant:Set (6) Comments Solution to the special case: 2.1.2 Radial Conduction in a Composite Cylinder with Interface Friction**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. (1) Observations • Composite cylindrical wall • Cylindrical coordinates • Radial conduction only**• 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**(3) Formulation**(i) Assumptions • One-dimensional radial conduction • Steady • Isotropic • Constant conductivities • No energy generation • Perfect interface contact • Uniform frictional energy flux • Stationary**(2.2)**Specified flux at (a) (ii) Governing Equation Shaft temperature is uniform. For sleeve: Eq. (1.11) (iii) Boundary Conditions**(c)**(d) Convection at (b) (4) Solution Integrate eq. (2.2) twice BC give C1 and C2**= Biot number**(g) (e) (f) and (d) and (e) into (c) Shaft temperature T2: Use interface boundary condition**(h)**• Limiting check: Evaluate (f) at r = Rsand use (g) (5) Checking • Dimensional check • Boundary conditions check (6) Comments • Conductivity of shaft does not play a role**Example 2.1:Plate 1 generates heat at**Plate 1 is sandwiched between two plates. Outer surfaces of two plates at • Problem can also be treated formally as a composite cylinder. Need 2 equations and 4 BC. 2.1.1 Composite Wall with Energy Generation Find the temperature distribution in the three plates.**(1) Observations**• Composite wall • Use rectangular coordinates • Symmetry: Insulated center plane • Heat flows normal to plates • Symmetry and steady state: Energy generated = Energy conducted out (2) Origin and Coordinates Shown in Fig. 2.3**(3) Formulation**(i) Assumptions • Steady • One-dimensional • Isotropic • Constant conductivities • Perfect interface contact • Stationary (ii) Governing Equations Two equations:**(a)**(b) (c) (d) (iii) Boundary Conditions Four BC: Symmetry: Interface:**(g)**(h) (e) (f) Outer surface: (4) Solution Integrate (a) twice Integrate (b)**• Dimensional check: units of**(i) (j) Four BC give 4 constants: Solutions (g) and (h) become (5) Checking**1/2 the energy generated in center plate = Heat**conducted at (k) • Boundary conditions check • Quantitative check:**(l)**(i) If then (ii) If then (i) into (k) Similarly, 1/2 the energy generated in center plate = Heat conducted out (j) into (l) shows that this condition is satisfied. • Limiting check:**Alternate approach: Outer plate with a specified**flux at and a specified temperature at (2.3) (6) Comments 2.2 Extended Surfaces - Fins 2.2.1 The Function of Fins Newton's law of cooling:**Options for increasing**• Lower • Increase • Increase h 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**(a)**constant area variable area (b) straight fin straight fin Fig. 2.5 (c) pin fin (d) annular fin 2.2.2 Types of Fins**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**r**T ¥ d x T h , T ¥ Fig . 2 . 6 Bi = h /k << 1 (2.4) 2.2.4 The Fin Approximation Neglect lateral temperature variation Criterion: Biot number = Bi**•Assume**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. •Stationary material, steady state**(a)**(b) (c) Conservation of energy for the element dx:**(d)**(e) (f) (b) and (c) into (a) Fourier's law and Newton’s law Energy generation**(2.5a)**(2.5b) (g) (e), (f) and (g) into (d) Assume constant k • (2.5b) is the heat equation for fins • Assumptions: (1) Steady state (2) Stationary**•**and are determined from thegeometry of fin. (a) = circumference (3) Isotropic (4) Constant k (5) No radiation (6) Bi << 1 2.2.6 Determination of dAs /dx From Fig. 2.7b ds= slanted length of the element**(b)**(2.6a) << 1 For (2.6b) For a right triangle (b) into (a) 2.2.7 Boundary Conditions Need two BC**2.2.8 Determination of Fin Heat Transfer Rate**Conservation of energy for Two methods to determine (a)**(2.8)**(2.7) (1) Conduction at base. Fourier's law at x = 0 (2) Convection at the fin surface. Newton's law applied at the fin surface • Fin attached at both ends: Modify eq. (2.7) accordingly • Fin with convection at the tip: Integral in eq. (2.8) includes tip**• 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**(a)**= constant (b) (2.9) A. Governing Equation Use eq. (2.5b). Set Eq. (2.6a) (a) and (b) into eq. (2.5b)**(c)**(d) • Assume = constant, (c) and (d) into (2.9) (2.10) (2) constant k, and Rewrite eq. (2.9) Valid for: (1) Steady state**(2.11a)**(2.11b) (3) No energy generation (4) No radiation (5) Bi1 (6) Stationary fin B. Solution Assume: h = constant**(e)**(f) (h) C. Special Case (i): • Finite length • Specified temperature at base, convection at tip Boundary conditions:**(i)**(2.12) Eq. (2.7) gives (2.13) Two BC give B1 and B2**(j)**Set eq. (2.12) (2.14) Set eq. (2.13) (2.15) C. Special Case (ii): • Finite length • Specified temperature at base, insulated tip BC at tip:**• Simplified model: Assume insulated tip, compensate**by increasing length by • The corrected length is (2.16) Increase in surface area due to = tip area 2.2.10 Corrected Length Lc • Insulated tip: simpler solution • The correction incrementLcdepends on the geometry of the fin: Circular fin:**2.2.11 Fin Efficiency**(2.17) = total surface area Square bar of side t Definition**(2.18)**Eq. (2.17) becomes 2.1.12 Moving Fins Examples: • Extrusion of plastics • Drawing of wires and sheets • Flow of liquids**(a)**(b) Heat equation: Assume: • Steady state • Constant area • Constant velocity U • Surface convection and radiation Conservation of energy for element dx**(c)**(d) (e) (f) (g) (2.19) Fourier’s and Newton’s laws (b)-(g) into (a) assume constant k**(2) Constant U, k, Pand**Assumptions leading to eq. (2.19): (1) Steady state (3) Isotropic (4) Gray body (5) Small surface enclosed by a much larger surface and (6) Bi << 1