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perform the calculations of Part (b). Because the model is based on the exact solution, Eq. 5.47a, it is accurate for values of Fo < 0.2, as well as Fo > 0.2. Although in principle, the model may be used to calculate the thermal conductivity for the conditions of Part (a), convergence is elusive and may only be achieved if the initial guesses are close to the correct results. 0 5 1 0 1 5 2 0 Air ve lo c ity, V(m /s ) 3 0 3 5 4 0 4 5 5 0 Ce nt er lin e te m pe ra tu re , T o(C ) 0 5 0 0 1 0 0 0 1 5 0 0 Tim e , t(s ) 2 5 5 0 7 5 1 0 0 Ce nte rlin e tem pe rat ur e, To (C) V=3 m /s V=1 0 m /s V=2 0 m /s PROBLEM 5.57 KNOWN: Diameter, initial temperature and properties of stainless steel rod. Temperature and convection coefficient of coolant. FIND: Temperature distributions for prescribed convection coefficients and times. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties. ANALYSIS: The IHT model is based on the exact solution to the heat equation, Eq. 5.47. The results are plotted as follows For h = 100 W/m2\u22c5K, Bi = hro/k = 0.1, and as expected, the temperature distribution is nearly uniform throughout the rod. For h = 1000 W/m2\u22c5K (Bi = 1), temperature variations within the rod are not negligible. In this case the centerline-to-surface temperature difference is comparable to the surface-to-fluid temperature difference. For h = 5000 W/m2\u22c5K (Bi = 5), temperature variations within the rod are large and [T (0,t) \u2013 T (ro,t)] is substantially larger than [T (ro,t) - T\u221e]. COMMENTS: With increasing Bi, conduction within the rod, and not convection from the surface, becomes the limiting process for heat loss. h =1 0 0 W /m ^2 -K 0 0 .2 0 .4 0 .6 0 .8 1 D im e n s io n le s s ra d iu s , r* 2 5 7 5 1 2 5 1 7 5 2 2 5 2 7 5 3 2 5 Te m pe ra tu re , C t=0 t=1 0 0 s t=5 0 0 s h=1000 W /m ^2 -K 0 0 .2 0 .4 0 .6 0 .8 1 D im ens ion les s rad iu s , r* 25 75 125 175 225 275 325 Te m pe re at u re , C t=0 s t=1 0 s t=5 0 s h=5000 W /m ^2 -K 0 0 .2 0 .4 0 .6 0 .8 1 D im ens ion les s rad iu s , r* 25 75 125 175 225 275 325 Te m pe ra tu re , C t=0 s t=1 s t=5 s t=2 5 s PROBLEM 5.58 KNOWN: A ball bearing is suddenly immersed in a molten salt bath; heat treatment to harden occurs at locations with T > 1000K. FIND: Time required to harden outer layer of 1mm. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Constant properties, (3) Fo \u2265 0.2. ANALYSIS: Since any location within the ball whose temperature exceeds 1000K will be hardened, the problem is to find the time when the location r = 9mm reaches 1000K. Then a 1mm outer layer will be hardened. Begin by finding the Biot number. ( )2o 5000 W/m K 0.020m/2h rBi 1.00. k 50 W/m K \u22c5 = = = \u22c5 Using the one-term approximate solution for a sphere, find ( )1 12 11 1 1Fo ln / C sin r . r \u3b8 \u3b6 \u3b6 \u3b6 \u2217 \u2217 \u2217 \uf8ee \uf8f9 = \u2212 \uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb From Table 5.1 with Bi = 1.00, for the sphere find \u3b61 15708= . rad and C1 = 1.2732. With r* = r/ro = (9mm/10mm) = 0.9, substitute numerical values. ( ) ( ) ( ) ( )2 1000 1300 K1 1Fo ln /1.2732 sin 1.5708 0.9 rad 0.441. 300 1300 K 1.5708 0.91.5708 \uf8ee \uf8f9 \u2212 \u2212 = × =\uf8ef \uf8fa \u2212 ×\uf8ef \uf8fa\uf8f0 \uf8fb From the definition of the Fourier number with \u3b1 = k/\u3c1c, 22 2o o 3 r c 0.020m kg J t Fo Fo r 0.441 7800 500 / 50 W/m K 3.4s. k 2 kg Km \u3c1 \u3b1 \uf8ee \uf8f9 = = \u22c5 = × × \u22c5 =\uf8ef \uf8fa \u22c5\uf8f0 \uf8fb < COMMENTS: (1) Note the very short time required to harden the ball. At this time it can be easily shown the center temperature is T(0,3.4s) = 871 K. (2) The Heisler charts can also be used. From Fig. D.8, with Bi-1 = 1.0 and r/ro = 0.9, read \u3b8/\u3b8o = 0.69(±0.03). Since i iT T 1000 1300 300K T T 1000K\u3b8 \u3b8\u221e \u221e= \u2212 = \u2212 = \u2212 = \u2212 = \u2212 it follows that o o i i o i i i 0.30. Since , then 0.69 \u3b8 \u3b8\u3b8 \u3b8 \u3b8 \u3b8 \u3b8 \u3b8 \u3b8 \u3b8 \u3b8 \u3b8 = = \u22c5 = and ( )o i/ 0.30 / 0.69 0.43 0.02 .\u3b8 \u3b8 = = ± From Fig. D.7 at \u3b8o/\u3b8i=0.43, Bi -1 =1.0, read Fo = 0.45 (±0.03) and t = 3.5 (±0.2)s. Note the use of tolerances associated with reading the charts to ±5%. PROBLEM 5.59 KNOWN: An 80mm sphere, initially at a uniform elevated temperature, is quenched in an oil bath with prescribed T¥, h. FIND: The center temperature of the sphere, T(0,t) at a certain time when the surface temperature is T(ro,t) = 150°C. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Initial uniform temperature within sphere, (3) Constant properties, (4) Fo ³ 0.2. ANALYSIS: Check first to see if the sphere is spacewise isothermal. ( ) 2oc c h r / 3hL 1000 W/m K 0.040m/3 Bi 0.26. k k 50 W/m K × ´ = = = = × Since Bic > 0.1, lumped capacitance method is not appropriate. Recognize that when Fo ³ 0.2, the time dependence of the temperature at any point within the sphere will be the same as the center. Using the Heisler chart method, Fig. D.8 provides the relation between T(ro,t) and T(0,t). Find first the Biot number, 2 ohr 1000 W/m K 0.040mBi 0.80. k 50 W/m K × ´ = = = × With Bi -1 = 1/0.80 = 1.25 and r/ro =1, read from Fig. D.8, ( ) ( ) o o T r , t T 0.67. T 0,t T q q ¥ ¥ - = = - It follows that ( ) ( ) [ ]o 1 1 T 0,t T T r , t T 50 C 150 50 C 199 C. 0.67 0.67¥ ¥ é ù= + - = + - =ë û oo o < COMMENTS: (1) There is sufficient information to evaluate Fo; hence, we require that the time be sufficiently long after the start of quenching for this solution to be appropriate. (2)The approximate series solution could also be used to obtain T(0,t). For Bi = 0.80 from Table 5.1, 1 1.5044z = rad. Substituting numerical values, r* = 1, ( ) ( ) ( ) ( ) o 1 o 1 T r , t T 1 1 sin r sin 1.5044 rad 0.663. T 0,t T 1.5044r q z q z * ¥ * * *¥ - = = = = - It follows that T(0,t) = 201°C. PROBLEM 5.60 KNOWN: Steel ball bearings at an initial, uniform temperature are to be cooled by convection while passing through a refrigerated chamber; bearings are to be cooled to a temperature such that 70% of the thermal energy is removed. FIND: Residence time of the balls in the 5m-long chamber and recommended drive velocity for the conveyor. SCHEMATIC: ASSUMPTIONS: (1) Negligible conduction between ball and conveyor surface, (2) Negligible radiation exchange with surroundings, (3) Constant properties, (4) Uniform convection coefficient over ball\u2019s surface. ANALYSIS: The Biot number for the lumped capacitance analysis is ( ) ( )2oc h r / 3 1000 W/m K 0.1m/3hLBi 0.67. k k 50 W/m K \u22c5 \u2261 = = = \u22c5 Since Bi > 0.1, lumped capacitance analysis is not appropriate. In Figure D.9, the internal energy change is shown as a function of Bi and Fo. For 2 o o hrQ 1000 W/m K 0.1m0.70 and Bi 2.0,Q k 50 W/m K \u22c5 × = = = = \u22c5 find Bi2 Fo \u2248 1.2. The Fourier number is ( ) 5 2 3 2 2 o t 2 10 m / s tFo 2.0 10 t r 0.1 m \u3b1 \u2212 \u2212 × × = = = × giving ( )22 -3 -3 3 1.2 / 2.0Fo 1.2 / Bi t 150s. 2.0 10 2.0 10 2.0 10\u2212 = = = = × × × The velocity of the conveyor is expressed in terms of the length L and residence time t. Hence L 5mV 0.033m/s 33mm/s. t 150s = = = = < COMMENTS: Referring to Eq. 5.10, note that for a sphere, the characteristic length is 3 2 o c s o o r4L V/A r / 4 r . 3 3 \u3c0 \u3c0= = = However, when using the Heisler charts, note that Bi \u2261 h ro/k. PROBLEM 5.61 KNOWN: Diameter and initial temperature of ball bearings to be quenched in an oil bath. FIND: (a) Time required for surface to cool to 100°C and the corresponding center temperature, (b) Oil bath cooling requirements. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional