Fundamentals of Heat and Mass Transfer - CH03 (001-050), Exercícios de Mecatrônica

Baixe Fundamentals of Heat and Mass Transfer - CH03 (001-050) e outras Exercícios em PDF para Mecatrônica, somente na Docsity! PROBLEM 3.1 KNOWN: One-dimensional, plane wall separating hot and cold fluids at T and T,1 ,2∞ ∞ , respectively. FIND: Temperature distribution, T(x), and heat flux, ′′qx , in terms of T T h,1 ,2 1∞ ∞, , , h2 , k and L. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties, (4) Negligible radiation, (5) No generation. ANALYSIS: For the foregoing conditions, the general solution to the heat diffusion equation is of the form, Equation 3.2, ( ) 1 2T x C x C .= + (1) The constants of integration, C1 and C2, are determined by using surface energy balance conditions at x = 0 and x = L, Equation 2.23, and as illustrated above, ( ) ( )1 ,1 2 ,2 x=0 x=L dT dT k h T T 0 k h T L T . dt dx∞ ∞     − = − − = −      (2,3) For the BC at x = 0, Equation (2), use Equation (1) to find ( ) ( )1 1 ,1 1 2k C 0 h T C 0 C∞ − + = − ⋅ +  (4) and for the BC at x = L to find ( ) ( )1 2 1 2 ,2k C 0 h C L C T .∞ − + = + −  (5) Multiply Eq. (4) by h2 and Eq. (5) by h1, and add the equations to obtain C1. Then substitute C1 into Eq. (4) to obtain C2. The results are ( ) ( ),1 ,2 ,1 ,2 1 2 ,1 1 1 2 1 2 T T T T C C T 1 1 L 1 1 L k h h h k h h k ∞ ∞ ∞ ∞ ∞ − − = − = − +     + + + +        ( ) ( ),1 ,2 ,1 1 1 2 T T x 1 T x T . k h1 1 L h h k ∞ ∞ ∞ −   = − + +    + +    < From Fourier’s law, the heat flux is a constant and of the form ( ),1 ,2 x 1 1 2 T TdT q k k C . dx 1 1 L h h k ∞ ∞−′′ = − = − = +   + +    < PROBLEM 3.2 KNOWN: Temperatures and convection coefficients associated with air at the inner and outer surfaces of a rear window. FIND: (a) Inner and outer window surface temperatures, Ts,i and Ts,o, and (b) Ts,i and Ts,o as a function of the outside air temperature T∞,o and for selected values of outer convection coefficient, ho. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible radiation effects, (4) Constant properties. PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) The heat flux may be obtained from Eqs. 3.11 and 3.12, ( ),i ,o 2 2o i 40 C 10 CT T q 1 L 1 1 0.004 m 1 h k h 1.4 W m K65 W m K 30 W m K ∞ ∞ − −−′′ = = + + + + ⋅⋅ ⋅ $ $ ( ) 2 2 50 C q 968 W m 0.0154 0.0029 0.0333 m K W ′′ = = + + ⋅ $ . Hence, with ( )i ,i ,oq h T T∞ ∞′′ = − , the inner surface temperature is 2 s,i ,i 2i q 968 W m T T 40 C 7.7 C h 30 W m K ∞ ′′ = − = − = ⋅ $ $ < Similarly for the outer surface temperature with ( )o s,o ,oq h T T∞′′ = − find 2 s,o ,o 2o q 968 W m T T 10 C 4.9 C h 65 W m K ∞ ′′ = − = − − = ⋅ $ $ < (b) Using the same analysis, Ts,i and Ts,o have been computed and plotted as a function of the outside air temperature, T∞,o, for outer convection coefficients of ho = 2, 65, and 100 W/m 2⋅K. As expected, Ts,i and Ts,o are linear with changes in the outside air temperature. The difference between Ts,i and Ts,o increases with increasing convection coefficient, since the heat flux through the window likewise increases. This difference is larger at lower outside air temperatures for the same reason. Note that with ho = 2 W/m 2⋅K, Ts,i - Ts,o, is too small to show on the plot. Continued ….. ,i s,i i ,i ,o i o T T 1 h 0.10 0.846, T T 1 h L k 1 h 0.118 ∞ ∞ ∞ − = = = − + + or ( )s,iT 25 C 0.846 35 C 4.6 C= − = −$ $ $ . PROBLEM 3.4 KNOWN: Curing of a transparent film by radiant heating with substrate and film surface subjected to known thermal conditions. FIND: (a) Thermal circuit for this situation, (b) Radiant heat flux, oq′′ (W/m 2), to maintain bond at curing temperature, To, (c) Compute and plot oq′′ as a function of the film thickness for 0 ≤ Lf ≤ 1 mm, and (d) If the film is not transparent, determine oq′′ required to achieve bonding; plot results as a function of Lf. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat flow, (3) All the radiant heat flux oq′′ is absorbed at the bond, (4) Negligible contact resistance. ANALYSIS: (a) The thermal circuit for this situation is shown at the right. Note that terms are written on a per unit area basis. (b) Using this circuit and performing an energy balance on the film-substrate interface, o 1 2q q q′′ ′′ ′′= + o o 1 o cv f s T T T T q R R R ∞− −′′ = + ′′ ′′ ′′+ where the thermal resistances are 2 2 cvR 1 h 1 50 W m K 0.020 m K W′′ = = ⋅ = ⋅ 2 f f fR L k 0.00025 m 0.025 W m K 0.010 m K W′′ = = ⋅ = ⋅ 2 s s sR L k 0.001m 0.05 W m K 0.020 m K W′′ = = ⋅ = ⋅ ( ) [ ] ( ) ( ) 2 2o 2 2 60 20 C 60 30 C q 133 1500 W m 2833W m 0.020 0.010 m K W 0.020 m K W − −′′ = + = + = + ⋅ ⋅ $ $ < (c) For the transparent film, the radiant flux required to achieve bonding as a function of film thickness Lf is shown in the plot below. (d) If the film is opaque (not transparent), the thermal circuit is shown below. In order to find oq′′ , it is necessary to write two energy balances, one around the Ts node and the second about the To node. . The results of the analyses are plotted below. Continued... PROBLEM 3.4 (Cont.) 0 0.2 0.4 0.6 0.8 1 Film thickness, Lf (mm) 2000 3000 4000 5000 6000 7000 R ad ia nt h ea t f lu x, q ''o ( W /m ^2 ) Opaque film Transparent film COMMENTS: (1) When the film is transparent, the radiant flux is absorbed on the bond. The flux required decreases with increasing film thickness. Physically, how do you explain this? Why is the relationship not linear? (2) When the film is opaque, the radiant flux is absorbed on the surface, and the flux required increases with increasing thickness of the film. Physically, how do you explain this? Why is the relationship linear? (3) The IHT Thermal Resistance Network Model was used to create a model of the film-substrate system and generate the above plot. The Workspace is shown below. // Thermal Resistance Network Model: // The Network: // Heat rates into node j,qij, through thermal resistance Rij q21 = (T2 - T1) / R21 q32 = (T3 - T2) / R32 q43 = (T4 - T3) / R43 // Nodal energy balances q1 + q21 = 0 q2 - q21 + q32 = 0 q3 - q32 + q43 = 0 q4 - q43 = 0 /* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points at which there is no external source of heat. */ T1 = Tinf // Ambient air temperature, C //q1 = // Heat rate, W; film side T2 = Ts // Film surface temperature, C q2 = 0 // Radiant flux, W/m^2; zero for part (a) T3 = To // Bond temperature, C q3 = qo // Radiant flux, W/m^2; part (a) T4 = Tsub // Substrate temperature, C //q4 = // Heat rate, W; substrate side // Thermal Resistances: R21 = 1 / ( h * As ) // Convection resistance, K/W R32 = Lf / (kf * As) // Conduction resistance, K/W; film R43 = Ls / (ks * As) // Conduction resistance, K/W; substrate // Other Assigned Variables: Tinf = 20 // Ambient air temperature, C h = 50 // Convection coefficient, W/m^2.K Lf = 0.00025 // Thickness, m; film kf = 0.025 // Thermal conductivity, W/m.K; film To = 60 // Cure temperature, C Ls = 0.001 // Thickness, m; substrate ks = 0.05 // Thermal conductivity, W/m.K; substrate Tsub = 30 // Substrate temperature, C As = 1 // Cross-sectional area, m^2; unit area PROBLEM 3.6 (Cont.) If conduction, radiation, or conduction and radiation are neglected, the corresponding values of h and the percentage errors are 18.5 W/m2⋅K (27.6%), 16 W/m2⋅K (10.3%), and 20 W/m2⋅K (37.9%). (c) For a fixed value of Ts = 27°C, the conduction loss remains at condq′′ = 8 W/m 2, which is also the fixed difference between elecP′′ and convq′′ . Although this difference is not clearly shown in the plot for 10 ≤ h ≤ 1000 W/m2⋅K, it is revealed in the subplot for 10 ≤ 100 W/m2⋅K. 0 200 400 600 800 1000 Convection coefficient, h(W/m^2.K) 0 400 800 1200 1600 2000 P ow er d is si pa tio n, P ''e le c( W /m ^2 ) No conduction With conduction 0 20 40 60 80 100 Convection coefficient, h(W/m^2.K) 0 40 80 120 160 200 P ow er d is si pa tio n, P ''e le c( W /m ^2 ) No conduction With conduction Errors associated with neglecting conduction decrease with increasing h from values which are significant for small h (h < 100 W/m2⋅K) to values which are negligible for large h. COMMENTS: In liquids (large h), it is an excellent approximation to neglect conduction and assume that all of the dissipated power is transferred to the fluid. PROBLEM 3.7 KNOWN: A layer of fatty tissue with fixed inside temperature can experience different outside convection conditions. FIND: (a) Ratio of heat loss for different convection conditions, (b) Outer surface temperature for different convection conditions, and (c) Temperature of still air which achieves same cooling as moving air (wind chill effect). SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, (2) Steady-state conditions, (3) Homogeneous medium with constant properties, (4) No internal heat generation (metabolic effects are negligible), (5) Negligible radiation effects. PROPERTIES: Table A-3, Tissue, fat layer: k = 0.2 W/m⋅K. ANALYSIS: The thermal circuit for this situation is Hence, the heat rate is s,1 s,1 tot T T T T q . R L/kA 1/ hA ∞ ∞− −= = + Therefore, windycalm windy calm L 1 k hq . L 1q k h  +  ′′ = ′′  +   Applying a surface energy balance to the outer surface, it also follows that cond convq q .′′ ′′= Continued ….. PROBLEM 3.7 (Cont.) Hence, ( ) ( )s,1 s,2 s,2 s,1 s,2 k T T h T T L k T T hLT . k 1+ hL ∞ ∞ − = − + = To determine the wind chill effect, we must determine the heat loss for the windy day and use it to evaluate the hypothetical ambient air temperature, ′∞T , which would provide the same heat loss on a calm day, Hence, s,1 s,1 windy calm T T T T q L 1 L 1 k h k h ∞ ∞′− −′′ = =    + +       From these relations, we can now find the results sought: (a) 2calm windy 2 0.003 m 1 q 0.2 W/m K 0.015 0.015465 W/m K 0.003 m 1q 0.015 0.04 0.2 W/m K 25 W/m K + ′′ ⋅ +⋅= = ′′ ++ ⋅ ⋅ calm windy q 0.553 q ′′ = ′′ < (b) ( )( ) ( )( ) 2 s,2 calm 2 0.2 W/m K 15 C 36 C 25 W/m K 0.003 m T 22.1 C 0.2 W/m K 1 25 W/m K 0.003 m ⋅− + ⋅  = = ⋅+ ⋅ $ $ $ < ( )( ) ( )( ) 2 s,2 windy 2 0.2 W/m K 15 C 36 C 65 W/m K 0.003m T 10.8 C 0.2 W/m K 1 65 W/m K 0.003m ⋅− + ⋅  = = ⋅+ ⋅ $ $ $ < (c) ( ) ( )( ) 0.003/0.2 1/ 25 T 36 C 36 15 C 56.3 C 0.003/ 0.2 1/ 65∞ + ′ = − + = − + $$ $ < COMMENTS: The wind chill effect is equivalent to a decrease of Ts,2 by 11.3°C and increase in the heat loss by a factor of (0.553) -1 = 1.81. PROBLEM 3.9 KNOWN: Thicknesses of three materials which form a composite wall and thermal conductivities of two of the materials. Inner and outer surface temperatures of the composite; also, temperature and convection coefficient associated with adjoining gas. FIND: Value of unknown thermal conductivity, kB. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible contact resistance, (5) Negligible radiation effects. ANALYSIS: Referring to the thermal circuit, the heat flux may be expressed as ( )s,i s,o CA B BA B C T T 600 20 C q L 0.3 m 0.15 m 0.15 mL L 20 W/m K k 50 W/m Kk k k − − ′′ = = + ++ + ⋅ ⋅ $ 2 B 580 q = W/m . 0.018+0.15/k ′′ (1) The heat flux may be obtained from ( ) ( )2s,iq =h T T 25 W/m K 800-600 C∞′′ − = ⋅ $ (2) 2q =5000 W/m .′′ Substituting for the heat flux from Eq. (2) into Eq. (1), find B 0.15 580 580 0.018 0.018 0.098 k q 5000 = − = − = ′′ Bk 1.53 W/m K.= ⋅ < COMMENTS: Radiation effects are likely to have a significant influence on the net heat flux at the inner surface of the oven. PROBLEM 3.10 KNOWN: Properties and dimensions of a composite oven window providing an outer surface safe- to-touch temperature Ts,o = 43°C with outer convection coefficient ho = 30 W/m 2⋅K and ε = 0.9 when the oven wall air temperatures are Tw = Ta = 400°C. See Example 3.1. FIND: Values of the outer convection coefficient ho required to maintain the safe-to-touch condition when the oven wall-air temperature is raised to 500°C or 600°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in window with no contact resistance and constant properties, (3) Negligible absorption in window material, (4) Radiation exchange processes are between small surface and large isothermal surroundings. ANALYSIS: From the analysis in the Ex. 3.1 Comment 2, the surface energy balances at the inner and outer surfaces are used to determine the required value of ho when Ts,o = 43°C and Tw,i = Ta = 500 or 600°C. ( ) ( ) ( ) ( )s,i s,o4 4 i a s,iw,i s,i A A B B T T T T h T T L / k L / k εσ − − + − = + ( ) ( ) ( ) ( )s,i s,o 4 4s,o w,o o s,oA A B B T T T T h T T L / k L / k εσ ∞ − = − + − + Using these relations in IHT, the following results were calculated: Tw,i, Ts(°C) Ts,i(°C) ho(W/m 2⋅K) 400 392 30 500 493 40.4 600 594 50.7 COMMENTS: Note that the window inner surface temperature is closer to the oven air-wall temperature as the outer convection coefficient increases. Why is this so? PROBLEM 3.11 KNOWN: Drying oven wall having material with known thermal conductivity sandwiched between thin metal sheets. Radiation and convection conditions prescribed on inner surface; convection conditions on outer surface. FIND: (a) Thermal circuit representing wall and processes and (b) Insulation thickness required to maintain outer wall surface at To = 40°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Thermal resistance of metal sheets negligible. ANALYSIS: (a) The thermal circuit is shown above. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. (b) Perform energy balances on the i- and o- nodes finding ,i i o i rad cv,i cd T T T T q 0 R R ∞ − − ′′+ + = ′′ ′′ (1) ,o oi o cd cv,o T TT T 0 R R ∞ −− + = ′′ ′′ (2) where the thermal resistances are 2 cv,i iR 1/ h 0.0333 m K / W′′ = = ⋅ (3) 2 cdR L / k L / 0.05 m K / W′′ = = ⋅ (4) 2 cv,o oR 1/ h 0.0100 m K / W′′ = = ⋅ (5) Substituting numerical values, and solving Eqs. (1) and (2) simultaneously, find L 86 mm= < COMMENTS: (1) The temperature at the inner surface can be found from an energy balance on the i-node using the value found for L. ,i i ,o i rad i cv,o cd cv,i T T T T q 0 T 298.3 C R R R ∞ ∞− − ′′+ + = = ° ′′ ′′ ′′+ It follows that Ti is close to T∞,i since the wall represents the dominant resistance of the system. (2) Verify that 2iq 50 W / m′′ = and 2 oq 150 W / m .′′ = Is the overall energy balance on the system satisfied? PROBLEM 3.14 KNOWN: Composite wall of a house with prescribed convection processes at inner and outer surfaces. FIND: Daily heat loss for prescribed diurnal variation in ambient air temperature. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction (negligible change in wall thermal energy storage over 24h period), (2) Negligible contact resistance. PROPERTIES: Table A-3, T ≈ 300 K: Fiberglass blanket (28 kg/m3), kb = 0.038 W/m⋅K; Plywood, ks = 0.12 W/m⋅K; Plasterboard, kp = 0.17 W/m⋅K. ANALYSIS: The heat loss may be approximated as 24h ,i ,o tot0 T T Q dt where R ∞ ∞−= ∫ p b s tot i p b s o tot 2 2 2 tot L L L1 1 1 R A h k k k h 1 1 0.01m 0.1m 0.02m 1 R 0.17 W/m K 0.038 W/m K 0.12 W/m K200m 30 W/m K 60 W/m K R 0.01454 K/W. = + + + + = + + + + ⋅ ⋅ ⋅⋅ ⋅ =              Hence the heat rate is 12h 24h tot 0 12 1 2 2 Q 293 273 5 sin t dt 293 273 11 sin t dt R 24 24 π π        = − + + − +                ∫ ∫ 12 24 0 12 W 24 2 t 24 2 t Q 68.8 20t+5 cos 20t+11 cos K h K 2 24 2 24 π π π π         = + ⋅                ( ) ( )60 132Q 68.8 240 1 1 480 240 1 1 W h π π     = + − − + − + + ⋅         { }Q 68.8 480-38.2+84.03 W h= ⋅ 8Q=36.18 kW h=1.302 10 J.⋅ × < COMMENTS: From knowledge of the fuel cost, the total daily heating bill could be determined. For example, at a cost of 0.10$/kW⋅h, the heating bill would be $3.62/day. PROBLEM 3.15 KNOWN: Dimensions and materials associated with a composite wall (2.5m × 6.5m, 10 studs each 2.5m high). FIND: Wall thermal resistance. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Temperature of composite depends only on x (surfaces normal to x are isothermal), (3) Constant properties, (4) Negligible contact resistance. PROPERTIES: Table A-3 (T ≈ 300K): Hardwood siding, kA = 0.094 W/m⋅K; Hardwood, kB = 0.16 W/m⋅K; Gypsum, kC = 0.17 W/m⋅K; Insulation (glass fiber paper faced, 28 kg/m 3 ), kD = 0.038 W/m⋅K. ANALYSIS: Using the isothermal surface assumption, the thermal circuit associated with a single unit (enclosed by dashed lines) of the wall is ( ) ( )A A A 0.008m L / k A 0.0524 K/W 0.094 W/m K 0.65m 2.5m = = ⋅ × ( ) ( )B B B 0.13m L / k A 8.125 K/W 0.16 W/m K 0.04m 2.5m = = ⋅ × ( ) ( )D D D 0.13m L /k A 2.243 K/W 0.038 W/m K 0.61m 2.5m = = ⋅ × ( ) ( )C C C 0.012m L / k A 0.0434 K/W. 0.17 W/m K 0.65m 2.5m = = ⋅ × The equivalent resistance of the core is ( ) ( )1 1eq B DR 1/ R 1/ R 1/ 8.125 1/ 2.243 1.758 K/W− −= + = + = and the total unit resistance is tot,1 A eq CR R R R 1.854 K/W.= + + = With 10 such units in parallel, the total wall resistance is ( ) 1tot tot,1R 10 1/ R 0.1854 K/W.−= × = < COMMENTS: If surfaces parallel to the heat flow direction are assumed adiabatic, the thermal circuit and the value of Rtot will differ. PROBLEM 3.16 KNOWN: Conditions associated with maintaining heated and cooled conditions within a refrigerator compartment. FIND: Coefficient of performance (COP). SCHEMATIC: ASSUMPTIONS: (1) Steady-state operating conditions, (2) Negligible radiation, (3) Compartment completely sealed from ambient air. ANALYSIS: The Case (a) experiment is performed to determine the overall thermal resistance to heat transfer between the interior of the refrigerator and the ambient air. Applying an energy balance to a control surface about the refrigerator, it follows from Eq. 1.11a that, at any instant, g outE E 0− =  Hence, elec outq q 0− = where ( )out ,i ,o tq T T R∞ ∞= − . It follows that ( ),i ,o t elec T T 90 25 C R 3.25 C/W q 20 W ∞ ∞− −= = = $ $ For Case (b), heat transfer from the ambient air to the compartment (the heat load) is balanced by heat transfer to the refrigerant (qin = qout). Hence, the thermal energy transferred from the refrigerator over the 12 hour period is ,i ,o out out in t T T Q q t q t t R ∞ ∞−= ∆ = ∆ = ∆ ( ) ( )out 25 5 C Q 12 h 3600s h 266,000 J 3.25 C W − = × = $ $ The coefficient of performance (COP) is therefore out in Q 266,000 COP 2.13 W 125,000 = = = < COMMENTS: The ideal (Carnot) COP is ) ( ) c ideal h c T 278K COP 13.9 T T 298 278 K = = = − − and the system is operating well below its peak possible performance. PROBLEM 3.18 KNOWN: Concrete wall of 150 mm thickness experiences a flash-over fire with prescribed radiant flux and hot-gas convection on the fire-side of the wall. Exterior surface condition is 300°C, typical ignition temperature for most household and office materials. FIND: (a) Thermal circuit representing wall and processes and (b) Temperature at the fire-side of the wall; comment on whether wall is likely to experience structural collapse for these conditions. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Constant properties. PROPERTIES: Table A-3, Concrete (stone mix, 300 K): k = 1.4 W/m⋅K. ANALYSIS: (a) The thermal cirucit is shown above. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. (b) To determine the fire-side wall surface temperatures, perform an energy balance on the o-node. o L o rad cv cd T T T T q R R ∞ − −′′+ = ′′ ′′ where the thermal resistances are 2 2 cv iR 1/ h 1/ 200 W / m K 0.00500 m K / W′′ = = ⋅ = ⋅ 2 cdR L / k 0.150 m /1.4 W / m K 0.107 m K / W′′ = = ⋅ = ⋅ Substituting numerical values, ( ) ( )o o2 2 2 400 T K 300 T K 25,000 W / m 0 0.005 m K / W 0.107 m K / W − − + = ⋅ ⋅ oT 515 C= ° < COMMENTS: (1) The fire-side wall surface temperature is within the 350 to 600°C range for which explosive spalling could occur. It is likely the wall will experience structural collapse for these conditions. (2) This steady-state condition is an extreme condition, as the wall may fail before near steady-state conditions can be met. PROBLEM 3.19 KNOWN: Representative dimensions and thermal conductivities for the layers of fire-fighter’s protective clothing, a turnout coat. FIND: (a) Thermal circuit representing the turnout coat; tabulate thermal resistances of the layers and processes; and (b) For a prescribed radiant heat flux on the fire-side surface and temperature of Ti =.60°C at the inner surface, calculate the fire-side surface temperature, To. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through the layers, (3) Heat is transferred by conduction and radiation exchange across the stagnant air gaps, (3) Constant properties. PROPERTIES: Table A-4, Air (470 K, 1 atm): kab = kcd = 0.0387 W/m⋅K. ANALYSIS: (a) The thermal circuit is shown with labels for the temperatures and thermal resistances. The conduction thermal resistances have the form cdR L / k′′ = while the radiation thermal resistances across the air gaps have the form rad 3rad avg 1 1 R h 4 Tσ ′′ = = The linearized radiation coefficient follows from Eqs. 1.8 and 1.9 with ε = 1 where Tavg represents the average temperature of the surfaces comprising the gap ( )( )2 2 3rad 1 2 avg1 2h T T T T 4 Tσ σ= + + ≈ For the radiation thermal resistances tabulated below, we used Tavg = 470 K. Continued ….. PROBLEM 3.19 (Cont.) Shell Air gap Barrier Air gap Liner Total (s) (a-b) (mb) (c-d) (tl) (tot) ( )2cdR m K / W′′ ⋅ 0.01702 0.0259 0.04583 0.0259 0.00921 -- ( )2radR m K / W′′ ⋅ -- 0.04264 -- 0.04264 -- -- ( )2gapR m K / W′′ ⋅ -- 0.01611 -- 0.01611 -- -- totalR ′′ -- -- -- -- -- 0.1043 From the thermal circuit, the resistance across the gap for the conduction and radiation processes is gap cd rad 1 1 1 R R R = + ′′ ′′ ′′ and the total thermal resistance of the turn coat is tot cd,s gap,a b cd,mb gap,c d cd,tlR R R R R R− −′′ ′′ ′′ ′′ ′′ ′′= + + + + (b) If the heat flux through the coat is 0.25 W/cm 2 , the fire-side surface temperature To can be calculated from the rate equation written in terms of the overall thermal resistance. ( )o i totq T T / R′′ ′′= − ( )22 2 2oT 66 C 0.25 W / cm 10 cm / m 0.1043 m K / W= ° + × × ⋅ oT 327 C= ° COMMENTS: (1) From the tabulated results, note that the thermal resistance of the moisture barrier (mb) is nearly 3 times larger than that for the shell or air gap layers, and 4.5 times larger than the thermal liner layer. (2) The air gap conduction and radiation resistances were calculated based upon the average temperature of 470 K. This value was determined by setting Tavg = (To + Ti)/2 and solving the equation set using IHT with kair = kair (Tavg). PROBLEM 3.21 KNOWN: Thickness, overall temperature difference, and pressure for two stainless steel plates. FIND: (a) Heat flux and (b) Contact plane temperature drop. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3) Constant properties. PROPERTIES: Table A-1, Stainless Steel (T ≈ 400K): k = 16.6 W/m⋅K. ANALYSIS: (a) With 4 2t,cR 15 10 m K/W −′′ ≈ × ⋅ from Table 3.1 and 4 2L 0.01m 6.02 10 m K/W, k 16.6 W/m K −= = × ⋅ ⋅ it follows that ( ) 4 2tot t,cR 2 L/k R 27 10 m K/W;−′′ ′′= + ≈ × ⋅ hence 4 2 -4 2tot T 100 C q = 3.70 10 W/m . R 27 10 m K/W ∆′′ = = × ′′ × ⋅ $ < (b) From the thermal circuit, 4 2 t,cc -4 2s,1 s,2 tot RT 15 10 m K/W 0.556. T T R 27 10 m K/W −′′∆ × ⋅= = = ′′− × ⋅ Hence, ( ) ( )c s,1 s,2T 0.556 T T 0.556 100 C 55.6 C.∆ = − = =$ $ < COMMENTS: The contact resistance is significant relative to the conduction resistances. The value of t,cR′′ would diminish, however, with increasing pressure. PROBLEM 3.22 KNOWN: Temperatures and convection coefficients associated with fluids at inner and outer surfaces of a composite wall. Contact resistance, dimensions, and thermal conductivities associated with wall materials. FIND: (a) Rate of heat transfer through the wall, (b) Temperature distribution. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Negligible radiation, (4) Constant properties. ANALYSIS: (a) Calculate the total resistance to find the heat rate, [ ] A B tot t,c 1 A B 2 tot tot 1 L L 1 R R h A k A k A h A 1 0.01 0.3 0.02 1 K R 10 5 0.1 5 5 0.04 5 20 5 W K K R 0.02 0.02 0.06 0.10 0.01 0.21 W W = + + + +  = + + + + × × × ×  = + + + + = ( ),1 ,2 tot T T 200 40 C q= 762 W. R 0.21 K/W ∞ ∞− −= = $ < (b) It follows that s,1 ,1 1 q 762 W T T 200 C 184.8 C h A 50 W/K∞ = − = − =$ $ A A s,1 2A qL 762W 0.01m T T 184.8 C 169.6 C Wk A 0.1 5m m K × = − = − = × ⋅ $ $ B A t,c K T T qR 169.6 C 762W 0.06 123.8 C W = − = − × =$ $ B s,2 B 2B qL 762W 0.02m T T 123.8 C 47.6 C Wk A 0.04 5m m K × = − = − = × ⋅ $ $ ,2 s,2 2 q 762W T T 47.6 C 40 C h A 100W/K ∞ = − = − = $ $ PROBLEM 3.23 KNOWN: Outer and inner surface convection conditions associated with zirconia-coated, Inconel turbine blade. Thicknesses, thermal conductivities, and interfacial resistance of the blade materials. Maximum allowable temperature of Inconel. FIND: Whether blade operates below maximum temperature. Temperature distribution in blade, with and without the TBC. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, (2) Constant properties, (3) Negligible radiation. ANALYSIS: For a unit area, the total thermal resistance with the TBC is ( ) ( )1 1tot,w o t,c iZr InR h L k R L k h − −′′ ′′= + + + + ( )3 4 4 4 3 2 3 2tot,wR 10 3.85 10 10 2 10 2 10 m K W 3.69 10 m K W− − − − − −′′ = + × + + × + × ⋅ = × ⋅ With a heat flux of ,o ,i 5 2 w 3 2tot,w T T 1300 K q 3.52 10 W m R 3.69 10 m K W ∞ ∞ − − ′′ = = = × ′′ × ⋅ the inner and outer surface temperatures of the Inconel are ( ) ( )5 2 2s,i(w) ,i w iT T q h 400 K 3.52 10 W m 500 W m K 1104 K∞ ′′= + = + × ⋅ = ( ) ( ) ( ) ( )3 4 2 5 2s,o(w) ,i i wInT T 1 h L k q 400 K 2 10 2 10 m K W 3.52 10 W m 1174 K− −∞ ′′= + + = + × + × ⋅ × =   Without the TBC, ( )1 1 3 2tot,wo o iInR h L k h 3.20 10 m K W − − −′′ = + + = × ⋅ , and ( )wo ,o ,i tot,woq T T R∞ ∞′′ ′′= − = (1300 K)/3.20×10-3 m2⋅K/W = 4.06×105 W/m2. The inner and outer surface temperatures of the Inconel are then ( ) ( )5 2 2s,i(wo) ,i wo iT T q h 400 K 4.06 10 W m 500 W m K 1212 K∞ ′′= + = + × ⋅ = ( ) ( )[ ] ( ) ( )3 4 2 5 2s,o(wo) ,i i woInT T 1 h L k q 400 K 2 10 2 10 m K W 4.06 10 W m 1293 K− −∞ ′′= + + = + × + × ⋅ × = Continued... PROBLEM 3.25 KNOWN: Thicknesses and thermal conductivity of window glass and insulation. Contact resistance. Environmental temperatures and convection coefficients. Furnace efficiency and fuel cost. FIND: (a) Reduction in heat loss associated with the insulation, (b) Heat losses for prescribed conditions, (c) Savings in fuel costs for 12 hour period. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Constant properties. ANALYSIS: (a) The percentage reduction in heat loss is tot,wowo with with q wo wo Rq q q R 100% 1 100% 1 100% q q R tot, with ′′  ′′ ′′ ′′−= × = − × = − ×  ′′ ′′ ′′    where the total thermal resistances without and with the insulation, respectively, are w tot,wo cnv,o cnd,w cnv,i o w i L1 1 R R R R h k h ′′ ′′ ′′ ′′= + + = + + ( ) 2 2tot,woR 0.050 0.004 0.200 m K / W 0.254 m K / W′′ = + + ⋅ = ⋅ w ins tot,with cnv,o cnd,w t,c cnd,ins cnv,i t,c o w ins i L L1 1 R R R R R R R h k k h ′′ ′′ ′′ ′′ ′′ ′′ ′′= + + + + = + + + + ( ) 2 2tot,withR 0.050 0.004 0.002 0.926 0.500 m K / W 1.482 m K / W′′ = + + + + ⋅ = ⋅ ( )qR 1 0.254 /1.482 100% 82.9%= − × = < (b) With As = 12 m 2 , the heat losses without and with the insulation are ( ) 2 2wo s ,i ,o tot,woq A T T / R 12 m 32 C / 0.254m K / W 1512 W∞ ∞ ′′= − = × ° ⋅ = < ( ) 2 2with s ,i ,o tot,withq A T T / R 12 m 32 C /1.482 m K / W 259 W∞ ∞ ′′= − = × ° ⋅ = < (c) With the windows covered for 12 hours per day, the daily savings are ( ) ( )6 6wo with g f q q 1512 259 W S t C 10 MJ / J 12h 3600 s / h $0.01/ MJ 10 MJ / J $0.677 0.8η − −− −= ∆ × = × × × = COMMENTS: (1) The savings may be insufficient to justify the cost of the insulation, as well as the daily tedium of applying and removing the insulation. However, the losses are significant and unacceptable. The owner of the building should install double pane windows. (2) The dominant contributions to the total thermal resistance are made by the insulation and convection at the inner surface. PROBLEM 3.26 KNOWN: Surface area and maximum temperature of a chip. Thickness of aluminum cover and chip/cover contact resistance. Fluid convection conditions. FIND: Maximum chip power. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Negligible heat loss from sides and bottom, (4) Chip is isothermal. PROPERTIES: Table A.1, Aluminum (T ≈ 325 K): k = 238 W/m⋅K. ANALYSIS: For a control surface about the chip, conservation of energy yields g outE E 0− =

or ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) c c t,c -4 2 c,max 4 2 4 2 c,max -6 4 3 2 T T A P 0 L/k R 1/ h 85 25 C 10 m P 0.002 / 238 0.5 10 1/1000 m K/W 60 10 C m P 8.4 10 0.5 10 10 m K/W ∞ − − − − − − =  ′′+ +  − =  + × + ⋅   × ⋅= × + × + ⋅ $ $ c,maxP 5.7 W.= < COMMENTS: The dominant resistance is that due to convection R R Rconv t,c cond> >>
. PROBLEM 3.27 KNOWN: Operating conditions for a board mounted chip. FIND: (a) Equivalent thermal circuit, (b) Chip temperature, (c) Maximum allowable heat dissipation for dielectric liquid (ho = 1000 W/m 2⋅K) and air (ho = 100 W/m2⋅K). Effect of changes in circuit board temperature and contact resistance. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible chip thermal resistance, (4) Negligible radiation, (5) Constant properties. PROPERTIES: Table A-3, Aluminum oxide (polycrystalline, 358 K): kb = 32.4 W/m⋅K. ANALYSIS: (a) (b) Applying conservation of energy to a control surface about the chip ( )in outE E 0− =  , c i oq q q 0′′ ′′ ′′− − = ( ) c ,i c ,o c i t,c ob T T T T q 1 h L k R 1 h ∞ ∞− −′′ = + ′′+ + With „„ –q W mc 3 10 4 2 , ho = 1000 W/m 2⋅K, kb = 1 W/m⋅K and 4 2t,cR 10 m K W −′′ = ⋅ , ( ) ( ) 4 2 c c 24 2 T 20 C T 20 C 3 10 W m 1 1000 m K W1 40 0.005 1 10 m K W− − − × = + ⋅+ + ⋅ $ $ ( )4 2 2c c3 10 W m 33.2T 664 1000T 20,000 W m K× = − + − ⋅ 1003Tc = 50,664 Tc = 49°C. < (c) For Tc = 85°C and ho = 1000 W/m2⋅K, the foregoing energy balance yields 2 cq 67,160 W m′′ = < with oq′′ = 65,000 W/m 2 and iq′′ = 2160 W/m 2. Replacing the dielectric with air (ho = 100 W/m 2⋅K), the following results are obtained for different combinations of kb and t,cR′′ . Continued... PROBLEM 3.28 (Cont.) elecP q 0.268 W= = < The convection and radiation resistances are Rcnv = 625 m⋅K/W and Rrad = 345 m⋅K/W, where hr = 7.25 W/m 2⋅K. (b) With the major contribution to the total resistance made by convection, significant benefit may be derived by increasing the value of h. For h = 200 W/m 2⋅K, Rcnv = 12.5 m⋅K/W and Ts,p = 351.6 K, yielding Rrad = 355 m⋅K/W. The effect of radiation is then negligible. COMMENTS: (1) The plate conduction resistance is negligible, and even for h = 200 W/m 2⋅K, the contact resistance is small relative to the convection resistance. However, Rt,c could be rendered negligible by using indium foil, instead of an air gap, at the interface. From Table 3.1, 4 2 t,cR 0.07 10 m K / W, −′′ = × ⋅ in which case Rt,c = 0.035 m⋅K/W. (2) Because Ac < W 2 , heat transfer by conduction in the plate is actually two-dimensional, rendering the conduction resistance even smaller. 0 20 40 60 80 10 0 12 0 14 0 16 0 18 0 20 0 C o nvection coe ffic ien t, h (W /m ^2 .K) 0 0 .5 1 1 .5 2 2 .5 3 3 .5 4 4 .5 P o w e r d is s ip a ti o n , P e le c ( W ) PROBLEM 3.29 KNOWN: Conduction in a conical section with prescribed diameter, D, as a function of x in the form D = ax 1/2 . FIND: (a) Temperature distribution, T(x), (b) Heat transfer rate, qx. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x- direction, (3) No internal heat generation, (4) Constant properties. PROPERTIES: Table A-2, Pure Aluminum (500K): k= 236 W/m⋅K. ANALYSIS: (a) Based upon the assumptions, and following the same methodology of Example 3.3, qx is a constant independent of x. Accordingly, ( )21/2x dT dTq kA k ax / 4dx dxπ   = − = −     (1) using A = πD2/4 where D = ax1/2. Separating variables and identifying limits, 1 1 x Tx 2 x T 4q dx dT. x a kπ = −∫ ∫ (2) Integrating and solving for T(x) and then for T2, ( ) x x 21 2 12 21 1 4q x 4q x T x T ln T T ln . x x a k a kπ π = − = − (3,4) Solving Eq. (4) for qx and then substituting into Eq. (3) gives the results, ( ) ( )2x 1 2 1 2q a k T T /1n x / x4 π= − − (5) ( ) ( ) ( )( ) 1 1 1 2 1 2 ln x/x T x T T T . ln x / x = + − < From Eq. (1) note that (dT/dx)⋅x = Constant. It follows that T(x) has the distribution shown above. (b) The heat rate follows from Eq. (5), ( )2x W 25 q 0.5 m 236 600 400 K/ln 5.76kW. 4 m K 125 π= × × − = ⋅ < PROBLEM 3.30 KNOWN: Geometry and surface conditions of a truncated solid cone. FIND: (a) Temperature distribution, (b) Rate of heat transfer across the cone. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x, (3) Constant properties. PROPERTIES: Table A-1, Aluminum (333K): k = 238 W/m⋅K. ANALYSIS: (a) From Fourier’s law, Eq. (2.1), with ( )2 2 3A= D / 4 a / 4 x ,π π= it follows that x 2 3 4q dx kdT. a xπ = − Hence, since qx is independent of x, 1 1 x Tx 2 3x T 4q dx k dT a xπ = −∫ ∫ or ( ) x x1 x 12 2 4q 1 k T T . a 2xπ   − = − −    Hence x 1 2 2 2 1 2q 1 1 T T . a k x xπ    = + −     < (b) From the foregoing expression, it also follows that ( ) ( ) ( ) ( ) 2 2 1 x 2 2 2 1 -1 x 2 2 -2 a k T T q 2 1/x 1/ x 1m 238 W/m K 20 100 C q 2 0.225 0.075 m π π − − −=  −   ⋅ − = ×  −   $ xq 189 W.= < COMMENTS: The foregoing results are approximate due to use of a one-dimensional model in treating what is inherently a two-dimensional problem. PROBLEM 3.33 KNOWN: Steady-state temperature distribution of convex shape for material with k = ko(1 + αT) where α is a constant and the mid-point temperature is ∆To higher than expected for a linear temperature distribution. FIND: Relationship to evaluate α in terms of ∆To and T1, T2 (the temperatures at the boundaries). SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat generation, (4) α is positive and constant. ANALYSIS: At any location in the wall, Fourier’s law has the form ( )x o dT q k 1 T . dx α′′ = − + (1) Since xq′′ is a constant, we can separate Eq. (1), identify appropriate integration limits, and integrate to obtain ( )2 1 L T x o0 T q dx k 1 T dTα′′ = − +∫ ∫ (2) 2 2 o 2 1 x 2 1 T Tk q T T . L 2 2 α α        ′′ = − + − +           (3) We could perform the same integration, but with the upper limits at x = L/2, to obtain 2 2 o L/2 1 x L/2 1 T T2k q T T L 2 2 α α        ′′ = − + − +          (4) where ( ) 1 2L/2 o T T T T L/2 T . 2 += = + ∆ (5) Setting Eq. (3) equal to Eq. (4), substituting from Eq. (5) for TL/2, and solving for α, it follows that ( ) ( ) o 22 2 1 2 o2 1 2 T . T T / 2 T T / 2 T α ∆=  + − + + ∆  < PROBLEM 3.34 KNOWN: Hollow cylinder of thermal conductivity k, inner and outer radii, ri and ro, respectively, and length L. FIND: Thermal resistance using the alternative conduction analysis method. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) No internal volumetric generation, (4) Constant properties. ANALYSIS: For the differential control volume, energy conservation requires that qr = qr+dr for steady-state, one-dimensional conditions with no heat generation. With Fourier’s law, ( )r dT dT q kA k 2 rL dr dr π= − = − (1) where A = 2πrL is the area normal to the direction of heat transfer. Since qr is constant, Eq. (1) may be separated and expressed in integral form, ( )o o i i r Tr r T q dr k T dT. 2 L rπ = −∫ ∫ Assuming k is constant, the heat rate is ( ) ( ) i o r o i 2 Lk T T q . ln r / r π − = Remembering that the thermal resistance is defined as tR T/q≡ ∆ it follows that for the hollow cylinder, ( )o i t ln r / r R . 2 LKπ = < COMMENTS: Compare the alternative method used in this analysis with the standard method employed in Section 3.3.1 to obtain the same result. PROBLEM 3.35 KNOWN: Thickness and inner surface temperature of calcium silicate insulation on a steam pipe. Convection and radiation conditions at outer surface. FIND: (a) Heat loss per unit pipe length for prescribed insulation thickness and outer surface temperature. (b) Heat loss and radial temperature distribution as a function of insulation thickness. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties. PROPERTIES: Table A-3, Calcium Silicate (T = 645 K): k = 0.089 W/m⋅K. ANALYSIS: (a) From Eq. 3.27 with Ts,2 = 490 K, the heat rate per unit length is ( ) ( ) s,1 s,2 r 2 1 2 k T T q q L ln r r π − ′ = = ( )( ) ( ) 2 0.089 W m K 800 490 K q ln 0.08m 0.06 m π ⋅ −′ = q 603W m′ = . < (b) Performing an energy for a control surface around the outer surface of the insulation, it follows that cond conv radq q q′ ′ ′= + ( ) ( ) ( ) s,1 s,2 s,2 s,2 sur 2 1 2 2 r T T T T T T ln r r 2 k 1 2 r h 1 2 r hπ π π ∞− − −= + where ( )( )2 2r s,2 sur s,2 surh T T T Tεσ= + + . Solving this equation for Ts,2, the heat rate may be determined from ( ) ( )2 s,2 r s,2 surq 2 r h T T h T Tπ ∞′ = − + −   Continued... PROBLEM 3.36 (Cont.) ( ) ( ) ( ) ( ) ( )2 55 20 C q ln 0.417 0.392 1 2 0.026 W m K 0.784 m 2 W m K 2 0.417 m 0.784 mπ π − = + ⋅ ⋅ $ ( ) ( ) ( ) ( ) ( )2 22 2 55 20 C 0.025 m 1 0.026 W m K 4 0.784 m 2 W m K 4 0.784 mπ π − + + ⋅ ⋅ $ ( ) ( ) ( ) ( ) 2 35 C35 C q 48.2 23.1 W 71.3 W 0.483 0.243 K W 1.992 1.036 K W = + = + = + + $ $ The annual energy loss is therefore ( )( )( )3annualQ 71.3W 365days 24 h day 10 kW W 625 kWh−= = With a unit electric power cost of $0.08/kWh, the annual cost of the heat loss is C = ($0.08/kWh)625 kWh = $50.00 Hence, an insulation thickness of δ = 25 mm < will satisfy the prescribed cost requirement. COMMENTS: Cylindrical containers of aspect ratio L/D = 1 are seldom used because of floor space constraints. Choosing L/D = 2, ∀ = πD3/2 and D = (2∀/π)1/3 = 0.623 m. Hence, L = 1.245 m, r1 = 0.312m and r2 = 0.337 m. It follows that q = 76.1 W and C = $53.37. The 6.7% increase in the annual cost of the heat loss is small, providing little justification for using the optimal heater dimensions. PROBLEM 3.37 KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surface and is exposed to a fluid of prescribed h and T∞. Thermal contact resistance between heater and tube wall and wall inner surface temperature. FIND: Heater power per unit length required to maintain a heater temperature of 25°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across heater. ANALYSIS: The thermal circuit has the form Applying an energy balance to a control surface about the heater, ( ) ( ) ( ) ( ) ( ) ( ) ( ) a b o i o o i o t,c 2 q q q T T T T q ln r / r 1/h D R 2 k 25 10 C25-5 C q = ln 75mm/25mm m K 1/ 100 W/m K 0.15m0.01 2 10 W/m K W q 728 1649 W/m π π π π ∞ ′ ′ ′= + − −′ = + ′+  − − ′ +  ⋅ ⋅ × ×+   × ⋅ ′ = + $ $ q =2377 W/m.′ < COMMENTS: The conduction, contact and convection resistances are 0.0175, 0.01 and 0.021 m ⋅K/W, respectively, PROBLEM 3.38 KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surface. Inner and outer wall temperatures. Temperature of fluid adjoining outer wall. FIND: Effect of wall thermal conductivity, thermal contact resistance, and convection coefficient on total heater power and heat rates to outer fluid and inner surface. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across heater, (5) Negligible radiation. ANALYSIS: Applying an energy balance to a control surface about the heater, i oq q q′ ′ ′= + ( ) ( ) o i o o i o t,c T T T T q ln r r 1 2 r h R 2 k π π ∞− −′ = + ′+ Selecting nominal values of k = 10 W/m⋅K, t,cR′ = 0.01 m⋅K/W and h = 100 W/m 2⋅K, the following parametric variations are obtained 0 50 100 150 200 Thermal conductivity, k(W/m.K) 0 500 1000 1500 2000 2500 3000 3500 H ea t r at e (W /m ) qi q qo 0 0.02 0.04 0.06 0.08 0.1 Contact resistance, Rtc(m.K/W) 0 500 1000 1500 2000 2500 3000 H ea t r at e( W /m ) qi q qo Continued... PROBLEM 3.39 (Cont.) and the heat gain per unit length is ,o ,i tot T T 17 C q 7.7 W / m R 2.20 m K / W ∞ ∞− °′ = = = ′ ⋅ COMMENTS: (1) The validity of assuming negligible radiation may be assessed for the worst case condition corresponding to the bare tube. Assuming a tube outer surface temperature of Ts = T∞,i = 279K, large surroundings at Tsur = T∞,o = 296K, and an emissivity of ε = 0.7, the heat gain due to net radiation exchange with the surroundings is ( )( )4 4rad 2 sur sq 2 r T T 7.7 W / m.εσ π′ = − = Hence, the net rate of heat transfer by radiation to the tube surface is comparable to that by convection, and the assumption of negligible radiation is inappropriate. (2) If heat transfer from the air is by natural convection, the value of ho with the insulation would actually be less than the value for the bare tube, thereby further reducing the heat gain. Use of the insulation would also increase the outer surface temperature, thereby reducing net radiation transfer from the surroundings. (3) The critical radius is rcr = kins/h ≈ 8 mm < r2. Hence, as indicated by the calculations, heat transfer is reduced by the insulation. PROBLEM 3.40 KNOWN: Diameter, wall thickness and thermal conductivity of steel tubes. Temperature of steam flowing through the tubes. Thermal conductivity of insulation and emissivity of aluminum sheath. Temperature of ambient air and surroundings. Convection coefficient at outer surface and maximum allowable surface temperature. FIND: (a) Minimum required insulation thickness (r3 – r2) and corresponding heat loss per unit length, (b) Effect of insulation thickness on outer surface temperature and heat loss. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction, (3) Negligible contact resistances at the material interfaces, (4) Negligible steam side convection resistance (T∞,i = Ts,i), (5) Negligible conduction resistance for aluminum sheath, (6) Constant properties, (7) Large surroundings. ANALYSIS: (a) To determine the insulation thickness, an energy balance must be performed at the outer surface, where conv,o radq q q .′ ′ ′= + With ( )conv,o 3 o s,o ,oq 2 r h T T ,π ∞′ = − rad 3q 2 rπ′ = εσ ( ) ( ) ( ) ( )q4 4s,o sur s,i s,o cond,st cond,ins cond,st 2 1 stT T , T T / R R , R n r / r / 2 k ,π′ = ′ ′ ′− − + = " and cond,insR ′ ( )3 2 insn r / r / 2 k ,π= " it follows that ( ) ( ) ( ) ( ) ( ) s,i s,o 4 4 3 o s,o ,o s,o sur 2 1 3 2 st ins 2 T T 2 r h T T T T n r / r n r / r k k π π εσ∞ −  = − + −   +

( ) ( ) ( ) ( ) ( )2 8 2 4 4 4 43 3 2 848 323 K 2 r 6 W / m K 323 300 K 0.20 5.67 10 W / m K 323 300 K n r / 0.18n 0.18 / 0.15 35 W / m K 0.10 W / m K π π −− = ⋅ − + × × ⋅ − + ⋅ ⋅   "" A trial-and-error solution yields r3 = 0.394 m = 394 mm, in which case the insulation thickness is ins 3 2t r r 214mm= − = < The heat rate is then ( ) ( ) ( ) 2 848 323 K q 420 W / m n 0.18 / 0.15 n 0.394 / 0.18 35 W / m K 0.10 W / m K π − ′ = = + ⋅ ⋅

< (b) The effects of r3 on Ts,o and q′ have been computed and are shown below. Conditioned ….. PROBLEM 3.40 (Cont.) Beyond r3 ≈ 0.40m, there are rapidly diminishing benefits associated with increasing the insulation thickness. COMMENTS: Note that the thermal resistance of the insulation is much larger than that for the tube wall. For the conditions of Part (a), the radiation coefficient is hr = 1.37 W/m, and the heat loss by radiation is less than 25% of that due to natural convection ( radq 78 W / m,′ = )conv,oq 342 W / m .′ = 0 .2 0 .2 6 0 .3 2 0 .3 8 0 .4 4 0 .5 Ou te r rad ius o f ins u la tion , m 40 80 120 160 200 240 O ut er s ur fa ce te m pe ra tu re , C Ts ,o 0 .2 0 .26 0 .32 0 .38 0 .44 0 .5 Ou te r ra d ius o f in s u la tio n , m 0 5 00 1 00 0 1 50 0 2 00 0 2 50 0 H e a t r a te s , W /m To ta l he a t ra te C o nve ctio n h ea t ra te R ad ia tio n he a t ra te PROBLEM 3.42 (Cont.) ( ) s,i 2 s,i s,2 cond 2 1 i T T T T q R n r / r / 2 kπ − − ′ = = ′  ( )( )s,i2 0.25 W / m K T 307.8K 4 W n 3 π ⋅ − =  s,iT 310.6K 37.6 C= = ° < As shown below, the effect of increasing the insulation thickness is to reduce, not increase, the surface temperatures. This behavior is due to a reduction in the total resistance to heat transfer with increasing r2. Although the convection, h, and radiation, ( )( )2 2r s,2 sur s,2 surh T T T T ,εσ= + + coefficients decrease with increasing r2, the corresponding increase in the surface area is more than sufficient to provide for a reduction in the total resistance. Even for an insulation thickness of t = 4 mm, h = h + hr = (7.1 + 5.4) W/m 2⋅K = 12.5 W/m2⋅K, and rcr = k/h = 0.25 W/m⋅K/12.5 W/m 2⋅K = 0.020m = 20 mm > r2 = 5 mm. The outer radius of the insulation is therefore well below the critical radius. 0 1 2 3 4 In s u la tio n th ickn e s s , m m 3 0 3 5 4 0 4 5 5 0 S u rf a ce te m p e ra tu re s , C In n e r s u rfa ce te m p e ra tu re , C Ou te r s u rfa ce te m p e ra tu re , C PROBLEM 3.43 KNOWN: Diameter of electrical wire. Thickness and thermal conductivity of rubberized sheath. Contact resistance between sheath and wire. Convection coefficient and ambient air temperature. Maximum allowable sheath temperature. FIND: Maximum allowable power dissipation per unit length of wire. Critical radius of insulation. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional radial conduction through insulation, (3) Constant properties, (4) Negligible radiation exchange with surroundings. ANALYSIS: The maximum insulation temperature corresponds to its inner surface and is independent of the contact resistance. From the thermal circuit, we may write ( ) ( ) in,i in,i g cond conv in,o in,i in,o T T T T E q R R n r / r / 2 k 1/ 2 r hπ π ∞ ∞− −′ ′= = = ′ ′+   + 
 where in,i in,o in,ir D / 2 0.001m, r r t 0.003m,= = = + = and in,i maxT T 50 C= = ° yields the maximum allowable power dissipation. Hence, ( ) ( ) ( )g,max 2 50 20 C 30 C E 4.51W / m n 3 1 1.35 5.31 m K / W 2 0.13 W / m K 2 0.003m 10 W / m Kπ π − ° °′ = = = + ⋅+ × ⋅ ⋅  " < The critical insulation radius is also unaffected by the contact resistance and is given by cr 2 k 0.13W / m K r 0.013m 13mm h 10 W / m K ⋅= = = = ⋅ < Hence, rin,o < rcr and g,maxE′ could be increased by increasing rin,o up to a value of 13 mm (t = 12 mm). COMMENTS: The contact resistance affects the temperature of the wire, and for g,maxq E′ ′=  4.51W / m,= the outer surface temperature of the wire is Tw,o = Tin,i + t,cq R 50 C′ ′ = ° ( )4.51W / m+ ( ) ( )4 23 10 m K / W / 0.002m 50.2 C.π−× ⋅ = ° Hence, the temperature change across the contact resistance is negligible. PROBLEM 3.44 KNOWN: Long rod experiencing uniform volumetric generation of thermal energy, q,
concentric with a hollow ceramic cylinder creating an enclosure filled with air. Thermal resistance per unit length due to radiation exchange between enclosure surfaces is radR .′ The free convection coefficient for the enclosure surfaces is h = 20 W/m 2⋅K. FIND: (a) Thermal circuit of the system that can be used to calculate the surface temperature of the rod, Tr; label all temperatures, heat rates and thermal resistances; evaluate the thermal resistances; and (b) Calculate the surface temperature of the rod. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial conduction through the hollow cylinder, (3) The enclosure surfaces experience free convection and radiation exchange. ANALYSIS: (a) The thermal circuit is shown below. Note labels for the temperatures, thermal resistances and the relevant heat fluxes. Enclosure, radiation exchange (given): radR 0.30 m K / W′ = ⋅ Enclosure, free convection: cv,rod 2r 1 1 R 0.80 m K / W h D 20 W / m K 0.020mπ π ′ = = = ⋅ ⋅ × × cv,cer 2i 1 1 R 0.40 m K / W h D 20 W / m K 0.040mπ π ′ = = = ⋅ ⋅ × × Ceramic cylinder, conduction: ( ) ( )o i cd n D / D n 0.120 / 0.040 R 0.10 m K / W 2 k 2 1.75 W / m Kπ π ′ = = = ⋅ × ⋅   The thermal resistance between the enclosure surfaces (r-i) due to convection and radiation exchange is enc rad cv,rod cv,cer 1 1 1 R R R R = + ′ ′ ′ ′+ 1 enc 1 1 R m K / W 0.24 m K / W 0.30 0.80 0.40 − ′ = + ⋅ = ⋅ +  The total resistance between the rod surface (r) and the outer surface of the cylinder (o) is ( )tot enc cdR R R 0.24 0.1 m K / W 0.34 m K / W′ ′ ′= + = + ⋅ = ⋅ Continued ….. PROBLEM 3.45 (Cont.) The heat extraction, and hence the performance of the evaporator coil, decreases with increasing frost layer thickness due to an increase in the total resistance to heat transfer. Although the convection resistance decreases with increasing δ, the reduction is exceeded by the increase in the conduction resistance. (c) The time tm required to melt a 2 mm thick frost layer may be determined by applying an energy balance, Eq. 1.11b, over the differential time interval dt and to a differential control volume extending inward from the surface of the layer. in st latE dt dE dU= = ( )( ) ( ),o f sf sfh 2 rL T T dt h d h 2 rL drπ ρ ρ π∞ − = − ∀ = − ( ) m 1 2 t r ,o f sf0 r h T T dt h drρ∞ − = −∫ ∫ ( ) ( ) ( )( ) ( ) 3 5 sf 2 1 m 2,o f 700 kg m 3.34 10 J kg 0.002 mh r r t h T T 2 W m K 20 0 C ρ ∞ ×− = = − ⋅ − $ mt 11,690s 3.25 h= = < COMMENTS: The tube radius r1 exceeds the critical radius rcr = k/h = 0.4 W/m⋅K/100 W/m2⋅K = 0.004 m, in which case any frost formation will reduce the performance of the coil. PROBLEM 3.46 KNOWN: Conditions associated with a composite wall and a thin electric heater. FIND: (a) Equivalent thermal circuit, (b) Expression for heater temperature, (c) Ratio of outer and inner heat flows and conditions for which ratio is minimized. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant properties, (3) Isothermal heater, (4) Negligible contact resistance(s). ANALYSIS: (a) On the basis of a unit axial length, the circuit, thermal resistances, and heat rates are as shown in the schematic. (b) Performing an energy balance for the heater, in outE E=

, it follows that ( ) ( ) ( ) ( ) ( ) h ,i h ,o h 2 i o 1 12 1 3 2 i 1 o 3 B A T T T T q 2 r q q ln r r ln r r h 2 r h 2 r 2 k 2 k π π π π π ∞ ∞ − − − − ′′ ′ ′= + = + + + < (c) From the circuit, ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 i 1h ,oo B 1 3 2i h ,i o 3 A ln r r h 2 rT Tq 2 k ln r rq T T h 2 r 2 k π π π π − ∞ −∞ +−′ = × ′ − + < To reduce o iq q′ ′ , one could increase kB, hi, and r3/r2, while reducing kA, ho and r2/r1. COMMENTS: Contact resistances between the heater and materials A and B could be important. PROBLEM 3.47 KNOWN: Electric current flow, resistance, diameter and environmental conditions associated with a cable. FIND: (a) Surface temperature of bare cable, (b) Cable surface and insulation temperatures for a thin coating of insulation, (c) Insulation thickness which provides the lowest value of the maximum insulation temperature. Corresponding value of this temperature. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3) Constant properties. ANALYSIS: (a) The rate at which heat is transferred to the surroundings is fixed by the rate of heat generation in the cable. Performing an energy balance for a control surface about the cable, it follows that gE q=
or, for the bare cable, ( )( )2 e i sI R L=h D L T T .π ∞′ − With ( ) ( )22 4eq =I R 700A 6 10 / m 294 W/m,−′ ′ = × Ω = it follows that ( ) ( )s 2i q 294 W/m T T 30 C+ h D 25 W/m K 0.005mπ π ∞ ′ = + = ⋅ $ sT 778.7 C.= $ < (b) With a thin coating of insulation, there exist contact and convection resistances to heat transfer from the cable. The heat transfer rate is determined by heating within the cable, however, and therefore remains the same. ( ) s s t,c t,c i i i i s t,c T T T T q= 1 R 1R h D L D L h D L D T T q = R 1/ h π π π π ∞ ∞ ∞ − −= ′′ + + − ′ ′′ + and solving for the surface temperature, find ( ) 2 2 s t,c i q 1 294 W/m m K m K T R T 0.02 0.04 30 C D h 0.005m W Wπ π∞  ′ ⋅ ⋅ ′′= + + = + +       $ sT 1153 C.= $ < Continued ….. PROBLEM 3.48 (Cont.) From an energy balance at the outer surface of the insulation, ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) cond conv rad s,i s,o 4 4 o s,o o s,o sur o i s,o s,o2 -8 4 4 4 s,o2 4 q q q T T h D T T D T T ln D / D / 2 k 486 T K W 20 0.3m T 298K ln 0.3m/0.2m m K 2 0.058 W/m K W +0.8 5.67 10 0.3m T 298 K . m K π εσπ π π π π ∞ ′ ′ ′= + − = − + − − = − ⋅ ⋅ × × − ⋅ By trial and error, we obtain Ts,o ≈ 305K in which case ( ) ( ) ( ) 486-305 K q = 163 W/m. ln 0.3m/0.2m 2 0.055 W/m Kπ ′ = ⋅ < (b) The yearly energy savings per unit length of pipe due to use of the insulation is ( ) 9 Savings Energy Savings Cost Yr m Yr. Energy Savings J s h $4 3727 163 3600 7500 Yr m s m h Yr 10 J Savings $385 / Yr m. Yr m = × ⋅ = − × × × ⋅ ⋅ = ⋅ ⋅ The pay back period is then Insulation Costs $100 / m Pay Back Period = Savings/Yr. m $385/Yr m = ⋅ ⋅ Pay Back Period = 0.26 Yr = 3.1 mo. < COMMENTS: Such a low pay back period is more than sufficient to justify investing in the insulation. PROBLEM 3.49 KNOWN: Temperature and convection coefficient associated with steam flow through a pipe of prescribed inner and outer diameters. Outer surface emissivity and convection coefficient. Temperature of ambient air and surroundings. FIND: Heat loss per unit length. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constant properties, (4) Surroundings form a large enclosure about pipe. PROPERTIES: Table A-1, Steel, AISI 1010 (T ≈ 450 K): k = 56.5 W/m⋅K. ANALYSIS: Referring to the thermal circuit, it follows from an energy balance on the outer surface that ,i s,o s,o ,o s,o sur conv,i cond conv,o rad T T T T T T R R R R ∞ ∞− − −= + + or from Eqs. 3.9, 3.28 and 1.7, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ,i s,o s,o ,o 4 4 o s,o sur i i o i o o s,o s,o 1 12 2 8 2 T T T T D T T 1/ D h ln D / D / 2 k 1/ D h 523K T T 293K ln 75/60 0.6m 500 W/m K 0.075m 25 W/m K 2 56.5 W/m K +0.8 0.075m 5.67 10 W/m K επ σ π π π π π π π ∞ ∞ − − − − − = + − + − − = × × ⋅ + × × ⋅ × ⋅ × × × ⋅ 4 4 4 4s,o s,o s,o 8 4 4 s,o T 293 K 523 T T 293 1.07 10 T 293 . 0.0106+0.0006 0.170 −  −   − −  = + × −   From a trial-and-error solution, Ts,o ≈ 502K. Hence the heat loss is ( ) ( )4 4o o s,o ,o o s,o surq = D h T T D T Tπ επ σ∞′ − + − ( ) ( ) ( )2 8 4 4 4 2 4 W q = 0.075m 25 W/m K 502-293 0.8 0.075m 5.67 10 502 243 K m K π π −′ ⋅ + × − ⋅     q =1231 W/m+600 W/m=1831 W/m.′ < COMMENTS: The thermal resistance between the outer surface and the surroundings is much larger than that between the outer surface and the steam. PROBLEM 3.50 KNOWN: Temperature and convection coefficient associated with steam flow through a pipe of prescribed inner and outer radii. Emissivity of outer surface magnesia insulation, and convection coefficient. Temperature of ambient air and surroundings. FIND: Heat loss per unit length ′q and outer surface temperature Ts,o as a function of insulation thickness. Recommended insulation thickness. Corresponding annual savings and temperature distribution. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3) Constant properties, (4) Surroundings form a large enclosure about pipe. PROPERTIES: Table A-1, Steel, AISI 1010 (T ≈ 450 K): ks = 56.5 W/m⋅K. Table A-3, Magnesia, 85% (T ≈ 365 K): km = 0.055 W/m⋅K. ANALYSIS: Referring to the thermal circuit, it follows from an energy balance on the outer surface that ,i s,o s,o ,o s,o sur conv,i cond,s cond,m conv,o rad T T T T T T R R R R R ∞ ∞− − −= + ′ ′ ′ ′ ′+ + or from Eqs. 3.9, 3.28 and 1.7, ( ) ( ) ( ) ( ) ( ) ( )( ) ,i s,o s,o ,o s,o sur 12 21 i 2 1 s 3 2 m 3 o 3 s,o sur s,o sur T T T T T T 1 2 r h ln r r 2 k ln r r 2 k 1 2 r h 2 r T T T T π π π π π εσ ∞ ∞ − − − − = + + + + +   This expression may be solved for Ts,o as a function of r3, and the heat loss may then be determined by evaluating either the left-or right-hand side of the energy balance equation. The results are plotted as follows. Continued...