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PROBLEM 3.10 KNOWN: Dimensions and surface conditions of a plate thermally joined at its ends to heat sinks at different temperatures. FIND: (a) Differential equation which determines temperature distribution in plate, (b) Temperature distribution and an expression for the heat rate from the plate to the sinks, and (c) Compute and plot temperature distribution and heat rates corresponding to changes in different parameters. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction in x (W,L >> t), (3) Constant properties, (4) Uniform surface heat flux and convection coefficient, (5) Negligible contact resistance. ANALYSIS: (a) Applying conservation of energy to the differential control volume qx + dqo = qx+ dx + dqconv where qx+ dx = qx + ( dqx dx) dx dqconv = h ( T T )( W dx) Hence, dq qx q x + o( W dx) = qx + ( dqx dx) dx + h ( T T )( W dx) + hw( T T ) = q ow. dx Using Fourier s law, qx = k( t W) dt dx, d T d T h q ktw + hw( T T ) q = o o ( T T ) 0 + =. < dx dx kt kt (b) Introducing θ T T, the differential equation becomes d θ h q θ + o = 0. dx kt kt This differential equation is of second order with constant coefficients and a source term. With λ hktand S q o kt, it follows that the general solution is of the form + λx λx θ = Ce 1 + Ce + S λ. (1) Appropriate boundary conditions are: θ(0) = To T θo θ(l) = TL T θl (,3) Substituting the boundary conditions, Eqs. (,3) into the general solution, Eq. (1), 0 0 + λl λl θo = Ce 1 + Ce + S λ θl = Ce 1 + Ce + S λ (4,5) To solve for C, multiply Eq. (4) by -e +λl and add the result to Eq. (5), + λl + λl λl + λl θoe + θl = C e + e + S λ e + 1 ( ) ( ) ( θ θ + λl) λ ( + λl ) ( + λl λl) C = L oe S e + 1 e + e (6) Continued...

PROBLEM 3.10 (Cont.) Substituting for C from Eq. (6) into Eq. (4), find + λ { ( L ) + λ ( L + λ C1 o L oe S e 1) ( e L λ = θ θ θ λ + + e L )} S λ (7) Using C 1 and C from Eqs. (6,7) and Eq. (1), the temperature distribution can be expressed as + λx sinh ( λx) + λl sinh ( λx) + λl sinh ( λx) + λl S θ(x) = e e θo θl ( 1 e ) ( 1 e ) sinh( λl) + + sinh( λl) + sinh ( λl) (8) λ < The heat rate from the plate is q q () 0 q () L with A c = W t, are p = x + x and using Fourier s law, the conduction heat rates, λl dθ 0 e λ qx() 0 = kac = kac λe λ θo + θl dx x= 0 sinh( λl) sinh ( λl) + λl 1 e S + λ λ sinh ( λl) λ ( L) ( λ ) λl dθ λl e λcosh λ qx( L) = kac = kac λe λcosh ( λl) θo + θl dx x= L sinh ( λl) sinh L + λl 1 e + λl S + λcosh( λl) λe sinh ( λl) λ (c) For the prescribed base-case conditions listed below, the temperature distribution (solid line) is shown in the accompanying plot. As expected, the maximum temperature does not occur at the midpoint, but slightly toward the x-origin. The sink heat rates are q x() 0 = 17. W q x() L = 3.6 W < < < 300 Temperature, T(x) (C) 00 100 0 0 0 40 60 80 100 Distance, x (mm) q''o = 0,000 W/m^; h = 50 W/m^.K q''o = 30,000 W/m^; h = 50 W/m^.K q''o = 0,000 W/m^; h = 00 W/m^.K q''o = 497 W/m^ with q''x(0) = 0; h = 00 W/m^.K The additional temperature distributions on the plot correspond to changes in the following parameters, with all the remaining parameters unchanged: (i) q o = 30,000 W/m, (ii) h = 00 W/m K, (iii) the value of q o for which q x (0) = 0 with h = 00 W/m K. The condition for the last curve is q o = 497 W/m for which the temperature gradient at x = 0 is zero. Base case conditions are: o q = 0,000 W/m, T o = 100 C, T L = 35 C, T = 5 C, k = 5 W/m K, h = 50 W/m K, L = 100 mm, t = 5 mm, W = 30 mm.

PROBLEM 3.104 KNOWN: Thermal conductivity, diameter and length of a wire which is annealed by passing an electrical current through the wire. FIND: Steady-state temperature distribution along wire. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction along the wire, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient h. ANALYSIS: Applying conservation of energy to a differential control volume, Hence, qx + E g dqconv qx+dx = 0 dqx ( π ) dx ( π ) ( ) g ( π ) qx+dx = qx + dx qx = k D / 4 dt/dx dq conv = h D dx T T E = q D / 4 dx. ( d ) T π ( π ) ( π ) ( ) k D / 4 dx+q D / 4 dx h Ddx T T = 0 dx or, with θ T T, d θ 4h q θ + = 0 dx kd k The solution (general and particular) to this nonhomogeneous equation is of the form mx -mx q θ = C 1 e + C e + km where m = (4h/kD). The boundary conditions are: dθ 0 m C 0 0 1 e mc e C1 C dx = = = x=0 q q/km θ L = 0 = C1 e + e + C 1= = C ml -ml km e + e ml -ml ( ) ( ) The temperature distribution has the form q e mx -mx + e q cosh mx T= T 1 = T 1. km e ml +e -ml km cosh ml COMMENTS: This process is commonly used to anneal wire and spring products. To check the result, note that T(L) = T(-L) = T. <

PROBLEM 3.119 KNOWN: Long, aluminum cylinder acts as an extended surface. FIND: (a) Increase in heat transfer if diameter is tripled and (b) Increase in heat transfer if copper is used in place of aluminum. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction, (3) Constant properties, (4) Uniform convection coefficient, (5) Rod is infinitely long. PROPERTIES: Table A-1, Aluminum (pure): k = 40 W/m K; Table A-1, Copper (pure): k = 400 W/m K. ANALYSIS: (a) For an infinitely long fin, the fin heat rate from Table 3.4 is ( ) 1/ qf = M = hpkac θb π ( ) ( ) 1/ 1/ q 3/ f = h π D k π D / 4 θb = hk D θb. where P = πd and A c = πd /4 for the circular cross-section. Note that q f α D 3/. Hence, if the diameter is tripled, qf( 3D) = 3 3/ = 5. qf D ( ) and there is a 40% increase in heat transfer. < (b) In changing from aluminum to copper, since q f α k 1/, it follows that 1/ q 1/ f ( Cu) kcu 400 = = = 1.9 qf ( A1) k A1 40 and there is a 9% increase in the heat transfer rate. < COMMENTS: (1) Because fin effectiveness is enhanced by maximizing P/A c = 4/D, the use of a larger number of small diameter fins is preferred to a single large diameter fin. () From the standpoint of cost and weight, aluminum is preferred over copper.

PROBLEM 3.10 KNOWN: Length, diameter, base temperature and environmental conditions associated with a brass rod. FIND: Temperature at specified distances along the rod. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient h. $ PROPERTIES: Table A-1, Brass ( ) ANALYSIS: Evaluate first the fin parameter T = 110 C : k = 133 W/m K. 1/ 1/ 1/ 1/ hp h π D 4h 4 30 W/m K m = = ka c k π D = = /4 kd 133 W/m K 0.005m -1 m 13.43 m. = Hence, m L = (13.43) 0.1 = 1.34 and from the results of Example 3.8, it is advisable not to make the infinite rod approximation. Thus from Table 3.4, the temperature distribution has the form cosh m( L x) + ( h/mk) sinh m( L x) θ = θb cosh ml + ( h/mk) sinh ml Evaluating the hyperbolic functions, cosh ml =.04 and sinh ml = 1.78, and the parameter h 30 W/m K = = 0.0168, mk 13.43m -1 133 W/m K ( ) with θ b = 180 C the temperature distribution has the form ( ) + ( ) $ ( ) cosh m L x 0.0168 sinh m L x θ = 180 C..07 The temperatures at the prescribed location are tabulated below. x(m) cosh m(l-x) sinh m(l-x) θ T( C) x 1 = 0.05 1.55 1.19 136.5 156.5 < x = 0.05 1.4 0.75 108.9 18.9 < L = 0.10 1.00 0.00 87.0 107.0 < COMMENTS: If the rod were approximated as infinitely long: T(x 1 ) = 148.7 C, T(x ) = 11.0 C, and T(L) = 67.0 C. The assumption would therefore result in significant underestimates of the rod temperature.

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