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 = 0.045 W/m K. ANALYSIS: (a) From the thermal circuit, the heat loss is T,i T,o q= 1 1 L L L 1 + + + + A hi kg ka kg h o $ $ 0 C 10 C q = 1 1 0.007 m 0.007 m 0.007 m 1 + + + + 1.4 W m K 0.045 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 < 0.5 0.015 0.715 0.015 0.0315 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 1.749 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. 30 7 Heat loss, q(w) 4 1 18 15 10 8 46 64 8 100 Double pane Triple pane Outside convection coefficient, ho(w/m^.k) Continued...
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.
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. SCHEMATIC: 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 $ ( 85 5-4 ) C( 10 m ) Pc,max = 4 ( 0.00 / 38) 0.5 10 ( 1/1000) + + m K/W 60 10 4 $ C m Pc,max = 8.4 10-6 + 0.5 10 4 + 10 3 m K/W Pc,max = 5.7 W. < COMMENTS: The dominant resistance is that due to convection 3Rconv > Rt,c >> Rcond8.
PROBLEM 3.48 KNOWN: Saturated steam conditions in a pipe with prescribed surroundings. FIND: (a) Heat loss per unit length from bare pipe and from insulated pipe, (b) Pay back period for insulation. SCHEMATIC: Steam Costs: $4 for 10 9 J Insulation Cost: $100 per meter Operation time: 7500 h/yr ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional heat transfer, (3) Constant properties, (4) Negligible pipe wall resistance, (5) Negligible steam side convection resistance (pipe inner surface temperature is equal to steam temperature), (6) Negligible contact resistance, (7) T sur = T. PROPERTIES: Table A-6, Saturated water (p = 0 bar): T sat = T s = 486K; Table A-3, Magnesia, 85% (T 39K): k = 0.058 W/m K. ANALYSIS: (a) Without the insulation, the heat loss may be expressed in terms of radiation and convection rates, s 4 sur 4 ( s ) W q = επ Dσ T T + h π D T T q =0.8π 0.m 5.67 10 8 486 4 98 4 K 4 m 4 K W +0 ( π 0.m ) ( 486-98) K m K q = ( 1365+36) W/m=377 W/m. < With the insulation, the thermal circuit is of the form Continued..
PROBLEM 3.48 (Cont.) From an energy balance at the outer surface of the insulation, q cond = q conv + q rad Ts,i Ts,o = h π D 4 4 o ( Ts,o T ) + εσπ Do ( Ts,o Tsur ) ln D o / D i / π k ( 486 Ts,o ) K W = 0 π ( 0.3m)( Ts,o 98K) ln( 0.3m/0.m) m K π ( 0.058 W/m K) -8 W +0.8 5.67 10 π 4 4 4 ( 0.3m 4 )( Ts,o 98 ) K. m K By trial and error, we obtain T s,o 305K in which case ( 486-305) K ( ) q = ln 0.3m/0.m = 163 W/m. π 0.055 W/m K (b) The yearly energy savings per unit length of pipe due to use of the insulation is Savings Energy Savings Cost = Yr m Yr. Energy Savings J s h $4 = ( 377 163) 3600 7500 Yr m s m h Yr 10 9 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 = 0.6 Yr = 3.1 mo. < COMMENTS: Such a low pay back period is more than sufficient to justify investing in the insulation. <
PROBLEM 3.87 KNOWN: Energy generation in an aluminum-clad, thorium fuel rod under specified operating conditions. FIND: (a) Whether prescribed operating conditions are acceptable, (b) Effect of q and h on acceptable operating conditions. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in r-direction, () Steady-state conditions, (3) Constant properties, (4) Negligible temperature gradients in aluminum and contact resistance between aluminum and thorium. PROPERTIES: Table A-1, Aluminum, pure: M.P. = 933 K; Table A-1, Thorium: M.P. = 03 K, k 60 W/m K. ANALYSIS: (a) System failure would occur if the melting point of either the thorium or the aluminum were exceeded. From Eq. 3.53, the maximum thorium temperature, which exists at r = 0, is qr T(0) = o + Ts = TTh,max 4k where, from the energy balance equation, Eq. 3.55, the surface temperature, which is also the aluminum temperature, is Hence, qr T o s = T + = TAl h 8 3 $ 7 10 W m 0.015 m $ TAl = Ts = 95 C + = 70 C = 993K 14,000 W m K 8 3 7 10 W m 0.015m TTh,max = + 993K = 1449 K 4 60W m K Although T Th,max < M.P. Th and the thorium would not melt, T al > M.P. Al and the cladding would melt under the proposed operating conditions. The problem could be eliminated by decreasing q, increasing h or using a cladding material with a higher melting point. (b) Using the one-dimensional, steady-state conduction model (solid cylinder) of the IHT software, the following radial temperature distributions were obtained for parametric variations in q and h. < Continued...
PROBLEM 3.87 (Cont.) 1600 1500 100 Temperature, T(K) 1400 1300 100 1100 1000 900 800 0 0.00 0.004 0.006 0.008 0.01 0.01 0.014 Radius, r(m) h = 10000 W/m^.K, qdot = 7E8 W/m^3 h = 10000 W/m^.K, qdot = 8E8 W/m^3 h = 10000 W/m^.K, qdot = 9E9 W/m^3 Temperature, T(K) 1000 800 600 400 0 0.00 0.004 0.006 0.008 0.01 0.01 0.014 Radius, r(m) qdot = E8, h = 000 W/m^.K qdot = E8, h = 3000 W/m^.K qdot = E8, h = 5000 W/m^.K qdot = E8, h = 10000 W/m^.K For h = 10,000 W/m K, which represents a reasonable upper limit with water cooling, the temperature of the aluminum would be well below its melting point for q = 7 10 8 W/m 3, but would be close to the melting point for q = 8 10 8 W/m 3 and would exceed it for q = 9 10 8 W/m 3. Hence, under the best of conditions, q 7 10 8 W/m 3 corresponds to the maximum allowable energy generation. However, if coolant flow conditions are constrained to provide values of h < 10,000 W/m K, volumetric heating would have to be reduced. Even for q as low as 10 8 W/m 3, operation could not be sustained for h = 000 W/m K. The effects of q and h on the centerline and surface temperatures are shown below. 000 Centerline temperature, T(0) (K) 000 1600 100 800 400 0 1E8.8E8 4.6E8 6.4E8 8.E8 1E9 Energy generation, qdot (W/m^3) h = 000 W/m^.K h = 5000 W/m^.K h = 10000 W/m^.K Surface temperature, Ts (K) 1600 100 800 400 0 1E8.8E8 4.6E8 6.4E8 8.E8 1E9 Energy generation, qdot (W/m^3) h = 000 W/m^.K h = 5000 W/m^.K h = 10000 W/m^.K For h = 000 and 5000 W/m K, the melting point of thorium would be approached for q 4.4 10 8 and 8.5 10 8 W/m 3, respectively. For h = 000, 5000 and 10,000 W/m K, the melting point of aluminum would be approached for q 1.6 10 8, 4.3 10 8 and 8.7 10 8 W/m 3. Hence, the envelope of acceptable operating conditions must call for a reduction in q with decreasing h, from a maximum of q 7 10 8 W/m 3 for h = 10,000 W/m K. COMMENTS: Note the problem which would arise in the event of a loss of coolant, for which case h would decrease drastically.
PROBLEM 3.10 KNOWN: Length, diameter, base temperature and environmental conditions associated with a brass rod. FIND: Temperature at specified distances along the rod. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction, (3) Constant properties, (4) Negligible radiation, (5) Uniform convection coefficient h. $ PROPERTIES: Table A-1, Brass ANALYSIS: Evaluate first the fin parameter T = 110 C : k = 133 W/m K. 1/ 1/ 1/ 1/ hp h π D 4h 4 30 W/m K m = = ka c k π D = = /4 kd 133 W/m K 0.005m -1 m 13.43 m. = Hence, m L = (13.43) 0.1 = 1.34 and from the results of Example 3.8, it is advisable not to make the infinite rod approximation. Thus from Table 3.4, the temperature distribution has the form cosh m( L x) + ( h/mk) sinh m( L x) θ = θb cosh ml + ( h/mk) sinh ml Evaluating the hyperbolic functions, cosh ml =.04 and sinh ml = 1.78, and the parameter h 30 W/m K = = 0.0168, mk 13.43m -1 133 W/m K with θ b = 180 C the temperature distribution has the form ( ) + ( ) $ cosh m L x 0.0168 sinh m L x θ = 180 C..07 The temperatures at the prescribed location are tabulated below. x(m) cosh m(l-x) sinh m(l-x) θ T( C) x 1 = 0.05 1.55 1.19 136.5 156.5 < x = 0.05 1.4 0.75 108.9 18.9 < L = 0.10 1.00 0.00 87.0 107.0 < COMMENTS: If the rod were approximated as infinitely long: T(x 1 ) = 148.7 C, T(x ) = 11.0 C, and T(L) = 67.0 C. The assumption would therefore result in significant underestimates of the rod temperature.
PROBLEM 3.136 KNOWN: Copper heat sink dimensions and convection conditions. FIND: (a) Maximum allowable heat dissipation for a prescribed chip temperature and interfacial chip/heat-sink contact resistance, (b) Effect of fin length and width on heat dissipation. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional heat transfer in chip-heat sink assembly, (3) Constant k, (4) Negligible chip thermal resistance, (5) Negligible heat transfer from back of chip, (6) Uniform chip temperature. ANALYSIS: (a) For the prescribed system, the chip power dissipation may be expressed as Tc T qc = Rt,c + Rcond,b + Rt,o R 6 t,c 5 10 m K W where Rt,c = = = 0.0195 K W Wc ( 0.016m) Lb 0.003m Rcond,b = = = 0.093K W kwc 400 W m K( 0.016m) The thermal resistance of the fin array is Rt,o = ( ηohat ) 1 NA where η f o = 1 ( 1 ηf ) At and At = NAf + Ab = N( 4wLc) + ( Wc Nw ) Continued...
PROBLEM 3.136 (Cont.) With w = 0.5 mm, S = 0.50 mm, L f = 6 mm, N = 104, and L 3 c Lf + w 4 = 6.063 10 m, it follows that A 6 f = 6.06 10 m and A 3 t = 6.40 10 m. The fin efficiency is tanh ml η c f = mlc where m= ( hp kac ) 1/ = ( 4h kw) 1/ = 45 m -1 and ml c = 1.49. It follows that η f = 0.608 and η o = 0.619, in which case 3 Rt,o = 0.619 1500 W m K 6.40 10 m = 0.168 K W and the maximum allowable heat dissipation is $ ( 85 5) C ( + + ) qc = = 76W 0.0195 0.093 0.168 K W (b) The IHT Performance Calculation, Extended Surface Model for the Pin Fin Array has been used to determine q c as a function of L f for four different cases, each of which is characterized by the closest allowable fin spacing of (S - w) = 0.5 mm. Case w (mm) S (mm) N A 0.5 0.50 104 B 0.35 0.60 711 C 0.45 0.70 5 D 0.55 0.80 400 < Maximum heat rate, qc(w) 340 330 30 310 300 90 80 70 6 7 8 9 10 Fin length, Lf(mm) w = 0.5 mm, S = 0.50 mm, N =104 w = 0.35 mm, S = 0.60 mm, N = 711 w = 0.45 mm, S = 0.70 mm, N = 5 w = 0.55 mm, S = 0.80 mm, N = 400 With increasing w and hence decreasing N, there is a reduction in the total area A t associated with heat transfer from the fin array. However, for Cases A through C, the reduction in A t is more than balanced by an increase in η f (and η o ), causing a reduction in R t,o and hence an increase in q c. As the fin efficiency approaches its limiting value of η f = 1, reductions in A t due to increasing w are no longer balanced by increases in η f, and q c begins to decrease. Hence there is an optimum value of w, which depends on L f. For the conditions of this problem, L f = 10 mm and w = 0.55 mm provide the largest heat dissipation.