For the BC at x = 0, Equation (2), use Equation (1) to find (4)

Transcription

1 PROBLEM 3.1 KNOWN: One-dimensional, plane wall separating hot and cold fluids at T,1 and T,, respectively. FIND: Temperature distribution, T(x), and heat flux, q x, in terms of T,1, T,, h1, h, k and L. ASSUMPTIONS: (1) One-dimensional conduction, () 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., T x = C x+ C. (1) 1 The constants of integration, C 1 and C, are determined by using surface energy balance conditions at x = 0 and x = L, Equation.3, and as illustrated above, dt dt k h1 T,1 T( 0 ) k h T( L) T,. dt = = x=0 dx (,3) x=l For the BC at x = 0, Equation (), use Equation (1) to find k( C1+ 0) = h1 T,1 ( C1 0+ C) (4) and for the BC at x = L to find k C1+ 0 = h C1L+ C T,. (5) Multiply Eq. (4) by h and Eq. (5) by h 1, and add the equations to obtain C 1. Then substitute C 1 into Eq. (4) to obtain C. The results are ( T,1 T, ) ( T,1 T, ) C 1 = C = + T,1 1 1 L 1 1 L k + + h1 h1 h k + + h1 h k ( T,1 T, ) x 1 T( x ) = + + T, L k h1 + + h1 h k From Fourier s law, the heat flux is a constant and of the form dt ( T,1 T, ) q x = k = k C 1=+. dx 1 1 L + + h1 h k

2 PROBLEM 3. 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, T s,i and T s,o, and (b) T s,i and T s,o as a function of the outside air temperature T,o and for selected values of outer convection coefficient, h o. ASSUMPTIONS: (1) Steady-state conditions, () 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 and 3.1, $ $ T 40 C 10 C,i T,o q = = 1 L m h o k hi 65 W m K 1.4 W m K 30 W m K 50 C q = = 968W m m K W Hence, with q h ( T T ) $ = i,i,o, the inner surface temperature is q 968W m Ts,i = T,i h = 40 C = 7.7 C i 30 W m K $ $ q = ho Ts,o T,o find Similarly for the outer surface temperature with q 968W m Ts,o = T,o h = 10 C = 4.9 C o 65 W m K $ $ (b) Using the same analysis, T s,i and T s,o have been computed and plotted as a function of the outside air temperature, T,o, for outer convection coefficients of h o =, 65, and 100 W/m K. As expected, T s,i and T s,o are linear with changes in the outside air temperature. The difference between T s,i and T s,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 h o = W/m K, T s,i - T s,o, is too small to show on the plot. Continued..

3 PROBLEM 3. (Cont.) Surface temperatures, Tsi or Tso (C) Outside air temperature, Tinfo (C) Tsi; ho = 100 W/m^.K Tso; ho = 100 W/m^.K Tsi; ho = 65 W/m^.K Tso; ho = 65 W/m^.K Tsi or Tso; ho = W/m^.K COMMENTS: (1) The largest resistance is that associated with convection at the inner surface. The values of T s,i and T s,o could be increased by increasing the value of h i. () The IHT Thermal Resistance Network Model was used to create a model of the window 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 q1 = (T - T1) / R1 q3 = (T3 - T) / R3 q43 = (T4 - T3) / R43 // Nodal energy balances q1 + q1 = 0 q - q1 + q3 = 0 q3 - q3 + 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 = Tinfo // Outside air temperature, C //q1 = // Heat rate, W T = Tso // Outer surface temperature, C q = 0 // Heat rate, W; node, no external heat source T3 = Tsi // Inner surface temperature, C q3 = 0 // Heat rate, W; node, no external heat source T4 = Tinfi // Inside air temperature, C //q4 = // Heat rate, W // Thermal Resistances: R1 = 1 / ( ho * As ) R3 = L / ( k * As ) R43 = 1 / ( hi * As ) // Convection thermal resistance, K/W; outer surface // Conduction thermal resistance, K/W; glass // Convection thermal resistance, K/W; inner surface // Other Assigned Variables: Tinfo = -10 // Outside air temperature, C ho = 65 // Convection coefficient, W/m^.K; outer surface L = // Thickness, m; glass k = 1.4 // Thermal conductivity, W/m.K; glass Tinfi = 40 // Inside air temperature, C hi = 30 // Convection coefficient, W/m^.K; inner surface As = 1 // Cross-sectional area, m^; unit area

4 PROBLEM 3.3 KNOWN: Desired inner surface temperature of rear window with prescribed inside and outside air conditions. FIND: (a) Heater power per unit area required to maintain the desired temperature, and (b) Compute and plot the electrical power requirement as a function of T,o for the range -30 T,o 0 C with h o of, 0, 65 and 100 W/m K. Comment on heater operation needs for low h o. If h ~ V n, where V is the vehicle speed and n is a positive exponent, how does the vehicle speed affect the need for heater operation? ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional heat transfer, (3) Uniform heater flux, q h, (4) Constant properties, (5) Negligible radiation effects, (6) Negligible film resistance. PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m K. ANALYSIS: (a) From an energy balance at the inner surface and the thermal circuit, it follows that for a unit surface area, T,i Ts,i Ts,i T,o + q h = 1h Lk+ 1h i o $ $ Ts,i T,o T,i T 15 C 10 C $ $ s,i 5 C 15 C q h = = Lk+ 1h m 1 1 o 1hi W m K 65 W m K 10 W m K = ( ) = qh W m 170 W m (b) The heater electrical power requirement as a function of the exterior air temperature for different exterior convection coefficients is shown in the plot. When h o = W/m K, the heater is unecessary, since the glass is maintained at 15 C by the interior air. If h ~ V n, we conclude that, with higher vehicle speeds, the exterior convection will increase, requiring increased heat power to maintain the 15 C condition Heater power (W/m^) Exterior air temperature, Tinfo (C) h = 0 W/m^.K h = 65 W/m^.K h = 100 W/m^.K COMMENTS: With q h = 0, the inner surface temperature with T,o = -10 C would be given by

5 T,i Ts,i 1hi 0.10 = = = 0.846, T,i T,o 1hi+ Lk+ 1ho or s,i $ $ $ T = 5C C = 4.6C.

6 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, q o (W/m ), to maintain bond at curing temperature, T o, (c) Compute and plot q o as a function of the film thickness for 0 L f 1 mm, and (d) If the film is not transparent, determine q o required to achieve bonding; plot results as a function of L f. ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional heat flow, (3) All the radiant heat flux q o 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, T q o = q1 + q o T To T1 q o = + R cv + R f Rs where the thermal resistances are R cv = 1 h = 1 50 W m K = 0.00 m K W R f = Lf kf = m 0.05 W m K = m K W R s = Ls ks = 0.001m 0.05 W m K = 0.00 m K W ( ) [ ] ( ) $ $ 60 0 C C q o = + = ( ) W m = 833 W m m K W 0.00 m K W (c) For the transparent film, the radiant flux required to achieve bonding as a function of film thickness L f is shown in the plot below. (d) If the film is opaque (not transparent), the thermal circuit is shown below. In order to find q o, it is necessary to write two energy balances, one around the T s node and the second about the T o node.. The results of the analyses are plotted below. Continued...

7 PROBLEM 3.4 (Cont.) 7000 Radiant heat flux, q''o (W/m^) Film thickness, Lf (mm) 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? () 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 q1 = (T - T1) / R1 q3 = (T3 - T) / R3 q43 = (T4 - T3) / R43 // Nodal energy balances q1 + q1 = 0 q - q1 + q3 = 0 q3 - q3 + 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 T = Ts // Film surface temperature, C q = 0 // Radiant flux, W/m^; zero for part (a) T3 = To // Bond temperature, C q3 = qo // Radiant flux, W/m^; part (a) T4 = Tsub // Substrate temperature, C //q4 = // Heat rate, W; substrate side // Thermal Resistances: R1 = 1 / ( h * As ) R3 = Lf / (kf * As) R43 = Ls / (ks * As) // Convection resistance, K/W // Conduction resistance, K/W; film // Conduction resistance, K/W; substrate // Other Assigned Variables: Tinf = 0 // Ambient air temperature, C h = 50 // Convection coefficient, W/m^.K Lf = // Thickness, m; film kf = 0.05 // Thermal conductivity, W/m.K; film To = 60 // Cure temperature, C Ls = // Thickness, m; substrate ks = 0.05 // Thermal conductivity, W/m.K; substrate Tsub = 30 // Substrate temperature, C As = 1 // Cross-sectional area, m^; unit area

8 PROBLEM 3.5 KNOWN: Thicknesses and thermal conductivities of refrigerator wall materials. Inner and outer air temperatures and convection coefficients. FIND: Heat gain per surface area. ASSUMPTIONS: (1) One-dimensional heat transfer, () Steady-state conditions, (3) Negligible contact resistance, (4) Negligible radiation, (5) Constant properties. ANALYSIS: From the thermal circuit, the heat gain per unit surface area is T,o T,i q = ( 1/hi) + ( L p /kp) + ( L i/ki) + ( L p /kp) + ( 1/ho) 5 4 C q = 1/ 5W / m K 0.003m / 60 W / m K 0.050m / W / m K ( ) + ( ) + ( ) 1 C q = = 14.1 W / m m K / W COMMENTS: Although the contribution of the panels to the total thermal resistance is negligible, that due to convection is not inconsequential and is comparable to the thermal resistance of the insulation.

9 PROBLEM 3.6 KNOWN: Design and operating conditions of a heat flux gage. FIND: (a) Convection coefficient for water flow (T s = 7 C) and error associated with neglecting conduction in the insulation, (b) Convection coefficient for air flow (T s = 15 C) and error associated with neglecting conduction and radiation, (c) Effect of convection coefficient on error associated with neglecting conduction for T s = 7 C. ASSUMPTIONS: (1) Steady-state, () One-dimensional conduction, (3) Constant k. ANALYSIS: (a) The electric power dissipation is balanced by convection to the water and conduction through the insulation. An energy balance applied to a control surface about the foil therefore yields Hence, Pelec = q conv + q cond = h Ts T + k Ts Tb L Pelec k ( Ts Tb ) L 000 W m 0.04 W m K( K) 0.01m h = = Ts T K ( 000 8) W m h = = 996 W m K K If conduction is neglected, a value of h = 1000 W/m K is obtained, with an attendant error of ( )/996 = 0.40% (b) In air, energy may also be transferred from the foil surface by radiation, and the energy balance yields Hence, 4 4 Pelec = q conv + q rad + q cond = h( Ts T ) + εσ ( Ts Tsur ) + k( Ts Tb ) L 4 4 ( ) Pelec εσ Ts Tsur k Ts T L h = Ts T W m W m K K 0.04 W m K (100 K) / 0.01m = 100 K ( ) W m = = 14.5 W m K 100 K Continued...

10 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/m K (7.6%), 16 W/m K (10.3%), and 0 W/m K (37.9%). (c) For a fixed value of T s = 7 C, the conduction loss remains at q cond = 8 W/m, which is also the fixed difference between P elec and q conv. Although this difference is not clearly shown in the plot for 10 h 1000 W/m K, it is revealed in the subplot for W/m K Power dissipation, P''elec(W/m^) Power dissipation, P''elec(W/m^) Convection coefficient, h(w/m^.k) Convection coefficient, h(w/m^.k) No conduction With conduction No conduction With conduction Errors associated with neglecting conduction decrease with increasing h from values which are significant for small h (h 100 W/m 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.

11 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). ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, () 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. W/m K. ANALYSIS: The thermal circuit for this situation is Hence, the heat rate is Ts,1 T Ts,1 T q = =. Rtot L/kA+ 1/hA Therefore, L 1 + q k h calm windy =. q windy L 1 + k h calm Applying a surface energy balance to the outer surface, it also follows that q cond = q conv. Continued..

12 Hence, k T T h T T L k T + Ts,1 T hl s, =. k 1+ hl ( s,1 s, ) = ( s, ) PROBLEM 3.7 (Cont.) 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, Ts,1 T Ts,1 T q = = L 1 L k h windy k h calm From these relations, we can now find the results sought: m 1 + q (a) calm 0. W/m K 65 W/m K = = q windy m W/m K 5 W/m K q calm = q windy $ 0. W/m K $ 15 C + 36 C ( 5 W/m K)( m) (b) Ts, calm 0. W/m K 1+ ( 5 W/m K)( m) = =.1 C $ 0. W/m K $ 15 C + 36 C ( 65 W/m K)( 0.003m) Ts, = = 10.8 C windy 0. W/m K 1+ (c) ( 65 W/m K)( 0.003m) ( 0.003/0. + 1/ 5) ( 0.003/ / 65) T = 36 C C = 56.3 C $ $ $ $ $ COMMENTS: The wind chill effect is equivalent to a decrease of T s, by 11.3 C and increase in the heat loss by a factor of (0.553) -1 = 1.81.

13 PROBLEM 3.8 KNOWN: Dimensions of a thermopane window. Room and ambient air conditions. FIND: (a) Heat loss through window, (b) Effect of variation in outside convection coefficient for double and triple pane construction. SCHEMATIC (Double Pane): ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional heat transfer, (3) Constant properties, (4) Negligible radiation effects, (5) Air between glass is stagnant. PROPERTIES: Table A-3, Glass (300 K): k g = 1.4 W/m K; Table A-4, Air (T = 78 K): k a = W/m K. ANALYSIS: (a) From the thermal circuit, the heat loss is T,i T,o q= 1 1 L L L A hi kg ka kg h o $ $ 0 C 10 C q = m m m W m K W m K 1.4 W m K 0.4 m 10 W m K 80 W m K $ $ 30 C 30 C q = = = 9.4 W K W 1.01K W ( ) (b) For the triple pane window, the additional pane and airspace increase the total resistance from 1.01 K/W to K/W, thereby reducing the heat loss from 9.4 to 17. W. The effect of h o on the heat loss is plotted as follows Heat loss, q(w) Double pane Triple pane Outside convection coefficient, ho(w/m^.k) Continued...

14 PROBLEM 3.8 (Cont.) Changes in h o influence the heat loss at small values of h o, for which the outside convection resistance is not negligible relative to the total resistance. However, the resistance becomes negligible with increasing h o, particularly for the triple pane window, and changes in h o have little effect on the heat loss. COMMENTS: The largest contribution to the thermal resistance is due to conduction across the enclosed air. Note that this air could be in motion due to free convection currents. If the corresponding convection coefficient exceeded 3.5 W/m K, the thermal resistance would be less than that predicted by assuming conduction across stagnant air.

15 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, k B. ASSUMPTIONS: (1) Steady-state conditions, () 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 ( ) Ts,i Ts,o C q = = LA LB LC 0.3 m 0.15 m 0.15 m ka kb kc 0 W/m K kb 50 W/m K 580 q= W/m /kB The heat flux may be obtained from $ q =h T Ts,i = 5 W/m K C q =5000 W/m. Substituting for the heat flux from Eq. () into Eq. (1), find = = = kb q 5000 kb = 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. $ (1) ()

16 PROBLEM 3.10 KNOWN: Properties and dimensions of a composite oven window providing an outer surface safeto-touch temperature T s,o = 43 C with outer convection coefficient h o = 30 W/m K and ε = 0.9 when the oven wall air temperatures are T w = T a = 400 C. See Example 3.1. FIND: Values of the outer convection coefficient h o required to maintain the safe-to-touch condition when the oven wall-air temperature is raised to 500 C or 600 C. ASSUMPTIONS: (1) Steady-state conditions, () 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, the surface energy balances at the inner and outer surfaces are used to determine the required value of h o when T s,o = 43 C and T w,i = T a = 500 or 600 C. εσ ( T 4 4 w,i T s,i) hi( Ta Ts,i) T + = T Ts,i Ts,o ( L /k ) + ( L /k ) A A B B s,i s,o = εσ ( T 4 4 ) s,o T w,o + h o ( T s,o T ) L /k + L /k A A B B Using these relations in IHT, the following results were calculated: T w,i, T s ( C) T s,i ( C) h o (W/m K) 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?

17 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 T o = 40 C. ASSUMPTIONS: (1) Steady-state conditions, () 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 T,i Ti To T + i + q rad = 0 (1) R cv,i R cd Ti To T,o To + = 0 R cd R cv,o where the thermal resistances are R cv,i = 1/ hi = m K / W (3) R cd = L / k = L / 0.05 m K / W (4) R cv,o = 1/ ho = m K / W (5) Substituting numerical values, and solving Eqs. (1) and () 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. T,i Ti T,o Ti + + q rad = 0 Ti = 98.3 C R cv,o R cd + R cv,i It follows that T i is close to T,i since the wall represents the dominant resistance of the system. () Verify that q i = 50 W/m and q o = 150 W / m. Is the overall energy balance on the system satisfied? ()

18 PROBLEM 3.1 KNOWN: Configurations of exterior wall. Inner and outer surface conditions. FIND: Heating load for each of the three cases. ASSUMPTIONS: (1) Steady-state, () One-dimensional conduction, (3) Constant properties, (4) Negligible radiation effects. PROPERTIES: (T = 300 K): Table A.3: plaster board, k p = 0.17 W/m K; urethane, k f = 0.06 W/m K; wood, k w = 0.1 W/m K; glass, k g = 1.4 W/m K. Table A.4: air, k a = W/m K. ANALYSIS: (a) The heat loss may be obtained by dividing the overall temperature difference by the total thermal resistance. For the composite wall of unit surface area, A = 1 m, T,i T,o q = ( 1hi) + ( Lp kp) + ( Lf kf ) + ( Lw kw) + ( 1ho) A q = $ $ 0 C 15 C m K W 1m $ 35 C q = = 15.0W.33 K W (b) For the single pane of glass, T,i T,o q = ( 1hi) + ( Lg kg) + ( 1ho) A 35 C 35 C q = = = W m K W 1m 0.69 K W $ $ (c) For the double pane window, T,i T,o q = ( 1hi) + ( Lg kg) + ( La ka) + ( 1ho) A 35 C 35 C q = = = 75.9W m K W 1m 0.461K W $ $ COMMENTS: The composite wall is clearly superior from the standpoint of reducing heat loss, and the dominant contribution to its total thermal resistance (8%) is associated with the foam insulation. Even with double pane construction, heat loss through the window is significantly larger than that for the composite wall.

19 PROBLEM 3.13 KNOWN: Composite wall of a house with prescribed convection processes at inner and outer surfaces. FIND: (a) Expression for thermal resistance of house wall, R tot ; (b) Total heat loss, q(w); (c) Effect on heat loss due to increase in outside heat transfer convection coefficient, h o ; and (d) Controlling resistance for heat loss from house. ASSUMPTIONS: (1) One-dimensional conduction, () Steady-state conditions, (3) Negligible contact resistance. $ $ ( = i + o = ) T T T / 0 15 C/=.5 C 300K : PROPERTIES: Table A-3, Fiberglass blanket, 8 kg/m 3, k b = W/m K; Plywood siding, k s = 0.1 W/m K; Plasterboard, k p = 0.17 W/m K. ANALYSIS: (a) The expression for the total thermal resistance of the house wall follows from Eq Lp Lb Ls 1 R tot = hia kpa kba ksa hoa (b) The total heat loss through the house wall is q = T/Rtot = ( Ti T o )/R tot. Substituting numerical values, find m 0.10m Rtot = W/m K 350m 0.17W/m K 350m 0.038W/m K 350m 0.0m W/m K 350m 60W/m K 350m R 5 5 tot [ ] 10 $ = C/W = $ C/W The heat loss is then, -5 $ $ q= C/ C/W=4.1 kw. (c) If h o changes from 60 to 300 W/m K, R o = 1/h o A changes from C/W to C/W. This reduces R tot to C/W, which is a 0.5% decrease and hence a 0.5% increase in q. (d) From the expression for R tot in part (b), note that the insulation resistance, L b /k b A, is 75/830 90% of the total resistance. Hence, this material layer controls the resistance of the wall. From part (c) note that a 5-fold decrease in the outer convection resistance due to an increase in the wind velocity has a negligible effect on the heat loss.

20 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. ASSUMPTIONS: (1) One-dimensional, steady-state conduction (negligible change in wall thermal energy storage over 4h period), () Negligible contact resistance. PROPERTIES: Table A-3, T 300 K: Fiberglass blanket (8 kg/m 3 ), k b = W/m K; Plywood, k s = 0.1 W/m K; Plasterboard, k p = 0.17 W/m K. ANALYSIS: The heat loss may be approximated as 4h T,i T,o Q= dt where R 0 tot 1 1 Lp Lb Ls 1 Rtot = A hi kp kb ks ho m 0.1m 0.0m 1 Rtot = W/m K W/m K 0.1 W/m K 00m 30 W/m K 60 W/m K Rtot = K/W. Hence the heat rate is 1h 4h 1 π π Q = sin t dt sin t dt R tot W 4 πt 1 4 πt 4 Q = t+5 cos 0t+11 cos K h K π π Q = ( 1 1) ( 1+ 1) W h π π { } Q = W h Q=36.18 kw h= 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 $3.6/day.

21 PROBLEM 3.15 KNOWN: Dimensions and materials associated with a composite wall (.5m 6.5m, 10 studs each.5m high). FIND: Wall thermal resistance. ASSUMPTIONS: (1) Steady-state conditions, () 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, k A = W/m K; Hardwood, k B = 0.16 W/m K; Gypsum, k C = 0.17 W/m K; Insulation (glass fiber paper faced, 8 kg/m 3 ), k D = 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 0.008m L A / kaaa = = K/W W/m K 0.65m.5m 0.13m L B/ kbab = = 8.15 K/W 0.16 W/m K 0.04m.5m 0.13m L D/kDAD = =.43 K/W W/m K 0.61m.5m 0.01m L C/ kcac = = K/W W/m K 0.65m.5m The equivalent resistance of the core is 1 1 Req = ( 1/ RB + 1/ RD ) = ( 1/ /.43) = K/W and the total unit resistance is Rtot,1 = RA + Req + RC = K/W. With 10 such units in parallel, the total wall resistance is 1 tot,1 Rtot = 10 1/ R = K/W. COMMENTS: If surfaces parallel to the heat flow direction are assumed adiabatic, the thermal circuit and the value of R tot will differ.

22 PROBLEM 3.16 KNOWN: Conditions associated with maintaining heated and cooled conditions within a refrigerator compartment. FIND: Coefficient of performance (COP). ASSUMPTIONS: (1) Steady-state operating conditions, () 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, E g E out = 0 Hence, q elec q out = 0 where qout = ( T,i T,o ) Rt. It follows that ( ) $ T,i T,o 90 5 C $ Rt = = = 3.5 C/W qelec 0W For Case (b), heat transfer from the ambient air to the compartment (the heat load) is balanced by heat transfer to the refrigerant (q in = q out ). Hence, the thermal energy transferred from the refrigerator over the 1 hour period is T,i T,o Qout = qout t = qin t = t Rt ( ) $ 5 5 C Qout = ( 1 h 3600s h) = 66,000 J $ 3.5 C W The coefficient of performance (COP) is therefore Qout 66, 000 COP = = =.13 Win 15, 000 COMMENTS: The ideal (Carnot) COP is Tc 78K COP) = = = 13.9 ideal Th Tc ( 98 78) K and the system is operating well below its peak possible performance.

23 PROBLEM 3.17 KNOWN: Total floor space and vertical distance between floors for a square, flat roof building. FIND: (a) Expression for width of building which minimizes heat loss, (b) Width and number of floors which minimize heat loss for a prescribed floor space and distance between floors. Corresponding heat loss, percent heat loss reduction from floors. ASSUMPTIONS: Negligible heat loss to ground. ANALYSIS: (a) To minimize the heat loss q, the exterior surface area, A s, must be minimized. From Fig. (a) where Hence, A s = W + 4WH = W + 4WNfHf N f = Af W A s = W + 4WAf Hf W = W + 4Af Hf W The optimum value of W corresponds to or das 4AfH = W f = 0 dw W Wop = ( Af Hf ) 1/3 The competing effects of W on the areas of the roof and sidewalls, and hence the basis for an optimum, is shown schematically in Fig. (b). (b) For A f = 3,768 m and H f = 4 m, 1/3 Wop = 3,768m 4m = 64m Continued..

24 PROBLEM 3.17 (Cont.) Hence, and A f 3,768m Nf = = = 8 W ( 64 m) 4 3,768m 4 m q = UAs T = 1W m K ( 64 m) + 5 C = 307, 00 W 64m $ For N f =, W = (A f /N f ) 1/ = (3,768 m /) 1/ = 18 m 4 3,768m 4 m $ q = 1W m K ( 18 m) + 5 C = 51,000 W 18m % reduction in q = (51, ,00)/51,000 = 40% COMMENTS: Even the minimum heat loss is excessive and could be reduced by reducing U.

25 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. ASSUMPTIONS: (1) Steady-state conditions, () 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. T To TL T + q o rad = R cv R cd where the thermal resistances are R cv = 1/ hi = 1/ 00 W / m K = m K / W R cd = L / k = m /1.4 W / m K = m K / W Substituting numerical values, ( ) ( ) 400 T o K o + 5,000 W / m 300 T K = m K / W m K / W To = 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. () This steady-state condition is an extreme condition, as the wall may fail before near steady-state conditions can be met.

26 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 T i =.60 C at the inner surface, calculate the fire-side surface temperature, T o. ASSUMPTIONS: (1) Steady-state conditions, () 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): k ab = k cd = 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 resistances across the air gaps have the form 1 1 R rad = = h 3 rad 4σ Tavg R cd = L/k while the radiation thermal The linearized radiation coefficient follows from Eqs. 1.8 and 1.9 with ε = 1 where T avg represents the average temperature of the surfaces comprising the gap h 3 rad = σ T1 + T T + T 4σTavg ( 1 ) For the radiation thermal resistances tabulated below, we used T avg = 470 K. Continued..

27 ( ) ( ) ( ) Rcd m K/W PROBLEM 3.19 (Cont.) Shell Air gap Barrier Air gap Liner Total (s) (a-b) (mb) (c-d) (tl) (tot) Rrad m K/W Rgap m K/W R total From the thermal circuit, the resistance across the gap for the conduction and radiation processes is = + R gap R cd R rad and the total thermal resistance of the turn coat is R tot = R cd,s + R gap,a b + R cd,mb + R gap,c d + R cd,tl (b) If the heat flux through the coat is 0.5 W/cm, the fire-side surface temperature T o can be calculated from the rate equation written in terms of the overall thermal resistance. q = To T i /R tot T o = 66 C W / cm 10 cm / m m K / W To = 37 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. () The air gap conduction and radiation resistances were calculated based upon the average temperature of 470 K. This value was determined by setting T avg = (T o + T i )/ and solving the equation set using IHT with k air = k air (T avg ).

28 PROBLEM 3.0 KNOWN: Materials and dimensions of a composite wall separating a combustion gas from a liquid coolant. FIND: (a) Heat loss per unit area, and (b) Temperature distribution. ASSUMPTIONS: (1) One-dimensional heat transfer, () Steady-state conditions, (3) Constant properties, (4) Negligible radiation effects. PROPERTIES: Table A-1, St. St. (304) ( T 1000K ): k = 5.4 W/m K; Table A-, Beryllium Oxide (T 1500K): k = 1.5 W/m K. ANALYSIS: (a) The desired heat flux may be expressed as ( ) T,1 T, C q= = 1 L A L B m + + R.K t,c + + h ka kb h W q =34,600 W/m. (b) The composite surface temperatures may be obtained by applying appropriate rate q=h 1 T,1 T s,1, it follows that equations. From the fact that q $ 34,600 W/m $ Ts,1 = T,1 = 600 C 1908 C. h1 50 W/m K q= k /L T T, it also follows that With ( A A)( s,1 c,1) L Aq $ 0.01m 34,600 W/m $ Tc,1 = Ts,1 = 1908 C = 189 C. ka 1.5 W/m K Similarly, with ( ) q= Tc,1 T c, /Rt,c $ m K W $ Tc, = Tc,1 Rt,cq =189 C , 600 = 16 C W m $ Continued..

29 and with ( B B)( c, s,) q= k /L T T, PROBLEM 3.0 (Cont.) L Bq $ 0.0m 34,600 W/m $ Ts, = Tc, = 16 C = C. kb 5.4 W/m K The temperature distribution is therefore of the following form: COMMENTS: (1) The calculations may be checked by recomputing q from ( ) $ q =h Ts, T, = 1000W/m K C=34,600W/m () The initial estimates of the mean material temperatures are in error, particularly for the stainless steel. For improved accuracy the calculations should be repeated using k values corresponding to T 1900 C for the oxide and T 115 C for the steel. (3) The major contributions to the total resistance are made by the combustion gas boundary layer and the contact, where the temperature drops are largest.

30 PROBLEM 3.1 KNOWN: Thickness, overall temperature difference, and pressure for two stainless steel plates. FIND: (a) Heat flux and (b) Contact plane temperature drop. ASSUMPTIONS: (1) One-dimensional heat transfer, () Steady-state conditions, (3) Constant properties. PROPERTIES: Table A-1, Stainless Steel (T 400K): k = 16.6 W/m K. ANALYSIS: (a) With R 4 t,c m K/W from Table 3.1 and L 0.01m = = m K/W, k 16.6 W/m K it follows that hence 4 R tot = L/k + R t,c 7 10 m K/W; $ T 100 C q = = = W/m. R tot m K/W (b) From the thermal circuit, Hence, T 4 c R t,c m K/W = = = T -4 s,1 Ts, R tot 7 10 m K/W Tc = Ts,1 Ts, = C = 55.6 C. $ $ COMMENTS: The contact resistance is significant relative to the conduction resistances. The value of R t,c would diminish, however, with increasing pressure.

31 PROBLEM 3. 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. ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional heat transfer, (3) Negligible radiation, (4) Constant properties. ANALYSIS: (a) Calculate the total resistance to find the heat rate, 1 LA LB 1 Rtot = + + Rt,c + + ha 1 kaa kba ha K Rtot = W K K Rtot = [ ] = 0.1 W W ( ) T,1 T, C q= = = 76 W. Rtot 0.1 K/W (b) It follows that q $ 76 W $ Ts,1 = T,1 = 00 C = C h1a 50 W/K $ qla $ 76W 0.01m $ TA = Ts,1 = C = C k W AA 0.1 5m m K $ K $ TB = TA qrt,c = C 76W 0.06 = 13.8 C W qlb $ 76W 0.0m $ Ts, = TB = 13.8 C = 47.6 C k W BA m m K q 76W T, = Ts, 47.6 C 40 C ha = $ 100W/K = $

32 PROBLEM 3.3 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. ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, () Constant properties, (3) Negligible radiation. ANALYSIS: For a unit area, the total thermal resistance with the TBC is R 1 1 tot,w h o ( L k ) Zr R t,c ( L k ) In h = i R tot,w = ( ) m K W = m K W With a heat flux of T,o T,i 1300 K 5 q w = = = W m R 3 tot,w m K W the inner and outer surface temperatures of the Inconel are 5 Ts,i(w) = T,i + q w hi = 400 K W m 500 W m K = 1104 K In T s,o(w) = T,i + 1 h i + L k q w = 400 K m K W W m = 1174 K Without the TBC, R tot,wo h o ( L k ) In h i m K W q wo T,o T,i R tot,wo (1300 K)/ m K/W = W/m. The inner and outer surface temperatures of the Inconel are then = + + =, and 5 = = Ts,i(wo) = T,i + q wo hi = 400 K W m 500 W m K = 11 K [ ] In T s,o(wo) = T,i + 1 h i + L k q wo = 400 K m K W W m = 193 K Continued...

33 PROBLEM 3.3 (Cont.) 1300 Temperature, T(K) Inconel location, x(m) With TBC Without TBC Use of the TBC facilitates operation of the Inconel below T max = 150 K. COMMENTS: Since the durability of the TBC decreases with increasing temperature, which increases with increasing thickness, limits to the thickness are associated with reliability considerations.

34 PROBLEM 3.4 KNOWN: Size and surface temperatures of a cubical freezer. Materials, thicknesses and interface resistances of freezer wall. FIND: Cooling load. ASSUMPTIONS: (1) Steady-state, () One-dimensional conduction, (3) Constant properties. PROPERTIES: Table A-1, Aluminum 04 (~67K): k al = 173 W/m K. Table A-1, Carbon steel AISI 1010 (~95K): k st = 64 W/m K. Table A-3 (~300K): k ins = W/m K. ANALYSIS: For a unit wall surface area, the total thermal resistance of the composite wall is Lal Lins L R st tot = + R t,c + + R t,c + kal kins kst m 4 m K 0.100m 4 m K m Rtot = W / m K W W / m K W 64 W / m K R tot = m K / W Hence, the heat flux is Ts,o Ts,i 6 C W q = = = 10.9 R tot.56 m K / W m and the cooling load is q = A s q = 6 W q = 54m 10.9 W / m = 590 W COMMENTS: Thermal resistances associated with the cladding and the adhesive joints are negligible compared to that of the insulation.

35 PROBLEM 3.5 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 1 hour period. ASSUMPTIONS: (1) Steady-state, () One-dimensional heat transfer, (3) Constant properties. ANALYSIS: (a) The percentage reduction in heat loss is q wo q with q with R tot,wo Rq = 100% = 1 100% = 1 100% q wo q wo R tot, with where the total thermal resistances without and with the insulation, respectively, are 1 Lw 1 R tot,wo = R cnv,o + R cnd,w + R cnv,i = + + ho kw hi R tot,wo = m K / W = 0.54 m K / W 1 Lw Lins 1 R tot,with = R cnv,o + R cnd,w + R t,c + R cnd,ins + R cnv,i = + + R t,c + + ho kw kins hi R tot,with = m K / W = 1.48 m K / W Rq = ( /1.48) 100% = 8.9% (b) With A s = 1 m, the heat losses without and with the insulation are q wo = As ( T,i T,o )/ R tot,wo = 1 m 3 C / 0.54m K / W = 151 W q with = As ( T,i T,o )/ R tot,with = 1 m 3 C /1.48 m K / W = 59 W (c) With the windows covered for 1 hours per day, the daily savings are ( qwo qwith ) 6 ( ) W 6 S = t Cg 10 MJ / J = 1h 3600 s / h $0.01/ MJ 10 MJ / J = $0.677 η 0.8 f 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. () The dominant contributions to the total thermal resistance are made by the insulation and convection at the inner surface.

36 PROBLEM 3.6 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. ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional heat transfer, (3) Negligible heat loss from sides and bottom, (4) Chip is isothermal. PROPERTIES: Table A.1, Aluminum (T 35 K): k = 38 W/m K. ANALYSIS: For a control surface about the chip, conservation of energy yields or E g E out = 0 ( Tc T ) A ( L/k) + R + ( 1/ h) Pc = 0 t,c $ ( ) C( 10 m ) Pc,max = 4 ( 0.00 / 38) ( 1/1000) + + m K/W $ C m Pc,max = m K/W Pc,max = 5.7 W. COMMENTS: The dominant resistance is that due to convection 3Rconv > Rt,c >> Rcond8.

37 PROBLEM 3.7 KNOWN: Operating conditions for a board mounted chip. FIND: (a) Equivalent thermal circuit, (b) Chip temperature, (c) Maximum allowable heat dissipation for dielectric liquid (h o = 1000 W/m K) and air (h o = 100 W/m K). Effect of changes in circuit board temperature and contact resistance. ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction, (3) Negligible chip thermal resistance, (4) Negligible radiation, (5) Constant properties. PROPERTIES: Table A-3, Aluminum oxide (polycrystalline, 358 K): k b = 3.4 W/m K. ANALYSIS: (a) (b) Applying conservation of energy to a control surface about the chip ( E E = 0) q c qi q o = 0 Tc T,i Tc T,o q c = + 1hi + ( Lk) + R b t,c 1ho With q c W m, h o = 1000 W/m K, k b = 1 W/m K and 4 ( ) m K W, in out 4 R t,c = 10 m K W, $ $ 4 Tc 0 C Tc 0 C 3 10 W m = m K W W m = ( 33.T c T c 0, 000 ) W m K 1003T c = 50,664 T c = 49 C. (c) For T c = 85 C and h o = 1000 W/m K, the foregoing energy balance yields q c = 67,160 W m with q o = 65,000 W/m and q i = 160 W/m. Replacing the dielectric with air (h o = 100 W/m K), the following results are obtained for different combinations of k b and R t,c. Continued...

38 PROBLEM 3.7 (Cont.) k b (W/m K) R t,c (m K/W) q i (W/m ) q o (W/m ) q c (W/m ) COMMENTS: 1. For the conditions of part (b), the total internal resistance is m K/W, while the outer resistance is m K/W. Hence ( ) ( ) q T o c T,o R o = = = 30. q i Tc T,i R i and only approximately 3% of the heat is dissipated through the board.. With h o = 100 W/m K, the outer resistance increases to 0.01 m K/W, in which case q o qi = Ri R o = /0.01 = 3.1 and now almost 5% of the heat is dissipated through the board. Hence, although measures to reduce R i would have a negligible effect on q c for the liquid coolant, some improvement may be gained for air-cooled conditions. As shown in the table of part (b), use of an aluminum oxide board increase q i by 19% (from 159 to 574 W/m ) by reducing R i from to m K/W. 4 Because the initial contact resistance ( R t,c = 10 m K W) is already much less than R i, any reduction in its value would have a negligible effect on q i. The largest gain would be realized by increasing h i, since the inside convection resistance makes the dominant contribution to the total internal resistance.

39 PROBLEM 3.8 KNOWN: Dimensions, thermal conductivity and emissivity of base plate. Temperature and convection coefficient of adjoining air. Temperature of surroundings. Maximum allowable temperature of transistor case. Case-plate interface conditions. FIND: (a) Maximum allowable power dissipation for an air-filled interface, (b) Effect of convection coefficient on maximum allowable power dissipation. ASSUMPTIONS: (1) Steady-state, () Negligible heat transfer from the enclosure, to the surroundings. (3) One-dimensional conduction in the base plate, (4) Radiation exchange at surface of base plate is with large surroundings, (5) Constant thermal conductivity. PROPERTIES: Aluminum-aluminum interface, air-filled, 10 µm roughness, 10 5 N/m contact pressure (Table 3.1): R 4 t,c = m K / W. ANALYSIS: (a) With all of the heat dissipation transferred through the base plate, Ts,c T Pelec = q = Rtot (1) 1 tot t,c cnd cnv rad where R = R + R + ( 1/R ) + ( 1/R ) R t,c L 1 1 Rtot = + + A () c kw W h+ hr and hr εσ ( Ts,p Tsur )( Ts,p Tsur ) = + + (3) To obtain T s,p, the following energy balance must be performed on the plate surface, Ts,c Ts,p q = = q cnv + qrad = hw Ts,p T + hr W Ts,p Tsur Rt,c + Rcnd ( ) (4) With R t,c = m K/W/ 10-4 m = K/W, R cnd = m/(40 W/m K m ) = K/W, and the prescribed values of h, W, T = T sur and ε, Eq. (4) yields a surface temperature of T s,p = K = 84.6 C and a power dissipation of Continued..

40 PROBLEM 3.8 (Cont.) Pelec = q = 0.68 W The convection and radiation resistances are R cnv = 65 m K/W and R rad = 345 m K/W, where h r = 7.5 W/m K. (b) With the major contribution to the total resistance made by convection, significant benefit may be derived by increasing the value of h. Power dissipation, Pelec (W) Convection coefficient, h (W/m ^.K) For h = 00 W/m K, R cnv = 1.5 m K/W and T s,p = K, yielding R rad = 355 m K/W. The effect of radiation is then negligible. COMMENTS: (1) The plate conduction resistance is negligible, and even for h = 00 W/m K, the contact resistance is small relative to the convection resistance. However, R t,c could be rendered negligible by using indium foil, instead of an air gap, at the interface. From Table 3.1, R 4 t,c = m K / W, in which case R t,c = m K/W. () Because A c W, heat transfer by conduction in the plate is actually two-dimensional, rendering the conduction resistance even smaller.

41 PROBLEM 3.9 KNOWN: Conduction in a conical section with prescribed diameter, D, as a function of x in the form D = ax 1/. FIND: (a) Temperature distribution, T(x), (b) Heat transfer rate, q x. ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction in x- direction, (3) No internal heat generation, (4) Constant properties. PROPERTIES: Table A-, Pure Aluminum (500K): k= 36 W/m K. ANALYSIS: (a) Based upon the assumptions, and following the same methodology of Example 3.3, q x is a constant independent of x. Accordingly, 1/ dt dt qx = ka = k π ax /4 dx dx (1) using A = πd /4 where D = ax 1/. Separating variables and identifying limits, 4q π a k dx dt. () x x T = x1 x T1 Integrating and solving for T(x) and then for T, 4q x x 4qx x T x = T 1 ln T = T1 ln. (3,4) π a k x1 π a k x1 Solving Eq. (4) for q x and then substituting into Eq. (3) gives the results, π q x = a k( T1 T ) /1n ( x 1/x) (5) 4 ( 1) ln x/x T( x) = T1+ ( T1 T ). ln x 1/ 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), π W 5 qx = 0.5 m 36 ( ) K/ln = 5.76kW. 4 m K 15

42 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. ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction in x, (3) Constant properties. PROPERTIES: Table A-1, Aluminum (333K): k = 38 W/m K. 3 A= D /4 a /4 x, ANALYSIS: (a) From Fourier s law, Eq. (.1), with π ( π ) 4qxdx = kdt. π a x 3 Hence, since q x is independent of x, 4qx x dx T = k dt π a x1 x 3 T 1 or x 4qx 1 ( 1) k T T. π a x = x 1 Hence q x 1 1 T= T 1 +. π a k x x 1 (b) From the foregoing expression, it also follows that π a k T T q 1 x = 1/x 1/ x 1 π -1 ( 1m ) 38 W/m K $ ( 0 100) C qx = - ( 0.5) ( 0.075) m = it follows that qx = 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.

43 PROBLEM 3.31 KNOWN: Temperature dependence of the thermal conductivity, k. FIND: Heat flux and form of temperature distribution for a plane wall. ASSUMPTIONS: (1) One-dimensional conduction through a plane wall, () Steady-state conditions, (3) No internal heat generation. ANALYSIS: For the assumed conditions, q x and A(x) are constant and Eq. 3.1 gives L T1 q x dx = ( o ) 0 k + at dt To 1 a q x = ko( To T1) + ( To T 1 ). L From Fourier s law, q x = ko + at dt/dx. Hence, since the product of (k o +at) and dt/dx) is constant, decreasing T with increasing x implies, a > 0: decreasing (k o +at) and increasing dt/dx with increasing x a = 0: k = k o => constant (dt/dx) a 0: increasing (k o +at) and decreasing dt/dx with increasing x. The temperature distributions appear as shown in the above sketch.

44 PROBLEM 3.3 KNOWN: Temperature dependence of tube wall thermal conductivity. FIND: Expressions for heat transfer per unit length and tube wall thermal (conduction) resistance. ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional radial conduction, (3) No internal heat generation. ANALYSIS: From Eq. 3.4, the appropriate form of Fourier s law is dt dt qr = kar = k( π rl) dr dr dt q r = π kr dr dt q r = π rko( 1+ at ). dr Separating variables, q r dr = ko ( 1+ at) dt π r and integrating across the wall, find q r ro dr To ko ( 1+aT) dt π = ri r T i q T r r o at o ln = ko T + π ri T i q r ro a ln k = o ( To Ti) ( To T π r + i ) i ( To Ti ) a q r = π ko 1+ To + T i. ln r o/ri It follows that the overall thermal resistance per unit length is T ln ( r o/ ri) R t = =. q r a π ko 1+ ( To + Ti) COMMENTS: Note the necessity of the stated assumptions to treating q r as independent of r.