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of the spheres and convection coefficients. FIND: (a) Show in a qualitative manner, on T-t coordinates, temperatures at the center and the outer surface for each sphere; explain features of the curves; (b) Time required for the outer surface of each sphere to reach 415K, (c) Energy gained by each bath during process of cooling spheres to a surface temperature of 415K. SCHEMATIC: Sphere A Sphere B ro (mm) 150 15 \u3c1 (kg/m3) 1600 400 c (J/kg\u22c5K) 400 1600 k (W/m\u22c5K) 170 1.7 h (W/m2\u22c5K) 5 50 ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Uniform properties, (3) Constant convection coefficient. ANALYSIS: (a) From knowledge of the Biot number and the thermal time constant, it is possible to qualitatively represent the temperature distributions. From Eq. 5.10, with Lc = ro/3, find ( ) ( ) ( ) 2 3 A o 2 B 5 W/m K 0.150m/3 Bi 1.47 10 170 W/m K h r / 3 Bi k 50 W/m K 0.015m/3 Bi 1. (1)\u2212\u22c5= = × \u22c5 = \u22c5 = 0.147 7 W/m K (2)= \u22c5 The thermal time constant for a lumped capacitance system from Eq. 5.7 is ( ) ( ) 3 A 2 s 1600 kg/m 0.150m 400 J/kg K1 Vc 6400s hA 3 5 W/m K \u3c4 \u3c1 \u3c4 × \u22c5= = = × \u22c5 \uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb (3) ( )3o B 2 400 kg/m 0.015m 1600 J/kg K r c 64s 3h 3 50 W/m K \u3c1 \u3c4 \u3c4 × \u22c5 = = = × \u22c5 (4) When Bi << 0.1, the sphere will cool in a spacewise isothermal manner (Sphere A). For sphere B, Bi > 0.1, hence gradients will be important. Note that the thermal time constant of A is much larger than for B; hence, A will cool much slower. See sketch for these features. (b) Recognizing that BiA < 0.1, Sphere A can be treated as spacewise isothermal and analyzed using the lumped capacitance method. From Eq. 5.6 and 5.7, with T = 415 K ( ) i i T T exp t/ T T \u3b8 \u3c4 \u3b8 \u221e \u221e \u2212 = = \u2212 \u2212 (5) Continued \u2026.. PROBLEM 5.64 (Cont.) A A i T T 415 320 t ln 6400s ln 10,367s 2.88h. T T 800 320 \u3c4 \u221e \u221e \uf8ee \uf8f9 \u2212 \u2212\uf8ee \uf8f9 = \u2212 = \u2212 = =\uf8ef \uf8fa \uf8ef \uf8fa \u2212 \u2212\uf8f0 \uf8fb\uf8f0 \uf8fb < Note that since the sphere is nearly isothermal, the surface and inner temperatures are approximately the same. Since BiB > 0.1, Sphere B must be treated by the Heisler chart method of solution beginning with Figure D.8. Using ( )2 -1o B B 50 W/m K 0.015mhrBi 0.44 or Bi 2.27, k 1.7 W/m K \u22c5 × \u2261 = = = \u22c5 find that for r/ro = 1, ( ) ( ) ( )o o o o 1, t T r , t T 415 320 0.8. \u3b8 \u3b8 \u3b8 \u3b8 \u221e\u2212 \u2212 = = = (6) Using Eq. (6) and Figure D.7, find the Fourier number, ( )( ) ( ) ( ) oo 2i i o T r , t T / 0.8 415 320 K/0.8 t0.25 Fo 1.3. T T 800 320 K r \u3b8 \u3b1 \u3b8 \u221e \u221e \u2212 \u2212 = = = = = \u2212 \u2212 ( )22o B 6 2 1.3 0.015mFo r t 110s 1.8 min 2.656 10 m / s\u3b1 \u2212 = = = = × < where \u3b1 = k/\u3c1c = 1.7 W/m\u22c5K/400 kg/m3 × 1600 J/kg\u22c5K = 2.656×10-6m2/s. (c) To determine the energy change by the spheres during the cooling process, apply the conservation of energy requirement on a time interval basis. Sphere A: ( ) ( )in out AE E E Q E E t E 0 .\u2212 = \u2206 \u2212 = \u2206 = \u2212 ( ) ( ) ( ) [ ]33A iQ cV T t T 1600kg/m 400J/kg K 4/3 0.150m 415 800 K\u3c1 \u3c0\uf8ee \uf8f9= \u2212 = × \u22c5 × \u2212\uf8f0 \uf8fb 6 AQ 3.483 10 J.= × < Note that this simple expression is a consequence of the spacewise isothermal behavior. Sphere B: ( ) ( )in out BE E E Q E t E 0 .\u2212 = \u2206 \u2212 = \u2212 For the nonisothermal sphere, the Groeber chart, Figure D.9, can be used to evaluate QB. With Bi = 0.44 and Bi2Fo = (0.44)2×1.3 = 2.52, find Q/Qo = 0.74. The energy transfer from the sphere during the cooling process, using Eq. 5.44, is ( )B o iQ 0.74 Q 0.74 cV T T\u3c1 \u221e\uf8ee \uf8f9= = \u2212\uf8f0 \uf8fb ( ) ( ) ( )33BQ 0.75 400kg/m 1600J/kg K 4/3 0.015m 800 320 K 3257 J.\u3c0= × × \u22c5 \u2212 = < COMMENTS: (1) In summary: Sphere Bi = hro/k \u3c4 s\ufffd \ufffd t(s) Q(J) A 4.41×10-3 6400 10,370 3.48×106 B 0.44 64 110 3257 PROBLEM 5.65 KNOWN: Spheres of 40-mm diameter heated to a uniform temperature of 400°C are suddenly removed from an oven and placed in a forced-air bath operating at 25°C with a convection coefficient of 300 W/m2\u22c5K. FIND: (a) Time the spheres must remain in the bath for 80% of the thermal energy to be removed, and (b) Uniform temperature the spheres will reach when removed from the bath at this condition and placed in a carton that prevents further heat loss. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction in the spheres, (2) Constant properties, and (3) No heat loss from sphere after removed from the bath and placed into the packing carton. PROPERTIES: Sphere (given): \u3c1 = 3000 kg/m3, c = 850 J/kg\u22c5K, k = 15 W/m\u22c5K. ANALYSIS: (a) From Eq. 5.52, the fraction of thermal energy removed during the time interval \u2206t = to is ( ) ( )3o 1 1 11 o Q 1 3 / sin cosQ \u3b8 \u3b6 \u3b6 \u3b6 \u3b6 \u2217 \uf8ee \uf8f9= \u2212 \u2212\uf8f0 \uf8fb (1) where Q/Qo = 0.8. The Biot number is 2 oBi hr / k 300 W / m K 0.020m /15 W / m K 0.40= = \u22c5 × \u22c5 = and for the one-term series approximation, from Table 5.1, 1 11.0528 rad C 1.1164\u3b6 = = (2) The dimensionless temperature o ,\u3b8 \u2217 Eq. 5.31, follows from Eq. 5.50. ( )2o 1 1C exp Fo\u3b8 \u3b6\u2217 = \u2212 (3) where 2o oFo t / r .\u3b1= Substituting Eq. (3) into Eq. (1), solve for Fo and to. ( ) ( ) ( )2 31 1 1 11 1 o Q 1 3 C exp Fo / sin cosQ \u3b6 \u3b6 \u3b6 \u3b6 \u3b6\uf8ee \uf8f9= \u2212 \u2212 \u2212\uf8f0 \uf8fb (4) oFo 1.45 t 98.6 s= = < (b) Performing an overall energy balance on the sphere during the interval of time to \u2264 t \u2264 \u221e, in out f iE E E E E 0\u2212 = \u2206 = \u2212 = (5) where Ei represents the thermal energy in the sphere at to, ( ) ( ) ( )i o iE 1 0.8 Q 1 0.8 cV T T\u3c1 \u221e= \u2212 = \u2212 \u2212 (6) and Ef represents the thermal energy in the sphere at t = \u221e, ( )f avgE cV T T\u3c1 \u221e= \u2212 (7) Combining the relations, find the average temperature ( ) ( )( )avg icV T T 1 0.8 T T 0\u3c1 \u221e \u221e\uf8ee \uf8f9\u2212 \u2212 \u2212 \u2212 =\uf8f0 \uf8fb avgT 100 C= ° < PROBLEM 5.66 KNOWN: Diameter, density, specific heat and thermal conductivity of Pyrex spheres in packed bed thermal energy storage system. Convection coefficient and inlet gas temperature. FIND: Time required for sphere to acquire 90% of maximum possible thermal energy and the corresponding center temperature. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional radial conduction in sphere, (2) Negligible heat transfer to or from a sphere by radiation or conduction due to contact with adjoining spheres, (3) Constant properties. ANALYSIS: With Bi \u2261 h(ro/3)/k = 75 W/m2\u22c5K (0.0125m)/1.4 W/m\u22c5K = 0.67, the approximate solution for one-dimensional transient conduction in a sphere is used to obtain the desired results. We first use Eq. (5.52) to obtain o .\u3b8\u2217 ( ) ( ) 3 1 o o1 1 1 Q1 Q3 sin cos \u3b6 \u3b8 \u3b6 \u3b6 \u3b6 \u2217 \uf8eb \uf8f6 = \u2212\uf8ec \uf8f7\uf8ee \uf8f9\u2212 \uf8ed \uf8f8\uf8f0 \uf8fb With Bi \u2261 hro/k = 2.01, 1 2.03\u3b6 \u2248 and C1 \u2248 1.48 from Table 5.1. Hence, ( ) ( ) 3 o 0.1 2.03 0.837 0.155 5.3863 0.896 2.03 0.443 \u3b8\u2217 = = =\uf8ee \uf8f9\u2212 \u2212\uf8f0 \uf8fb The center temperature is therefore ( )o g,i i g,iT T 0.155 T T 300 C 42.7 C 257.3 C= + \u2212 = ° \u2212 ° = ° < From Eq. (5.50c), the corresponding time is 2 o o 2 11 r t ln C \u3b8 \u3b1\u3b6 \u2217\uf8eb \uf8f6\uf8ec \uf8f7= \u2212 \uf8ec \uf8f7\uf8ed \uf8f8 where ( )3 7 2k / c 1.4 W / m K / 2225 kg / m 835 J / kg K 7.54 10 m / s.\u3b1 \u3c1 \u2212= = \u22c5 × \u22c5 = × ( ) ( ) ( ) 2 27 2 0.0375m ln 0.155 /1.48 t 1,020s 7.54 10 m / s 2.03\u2212 = \u2212 = × < COMMENTS: The surface temperature at the time of interest may be obtained from Eq. (5.50b). With r 1,\u2217 = ( ) ( )o 1s g,i i g,i 1 sin 0.155 0.896T T T T 300 C 275 C 280.9 C 2.03 \u3b8 \u3b6 \u3b6 \u2217 ×\uf8eb \uf8f6 = + \u2212 = ° \u2212 ° = °\uf8ec \uf8f7\uf8ed \uf8f8 < PROBLEM 5.67 KNOWN: Initial temperature and properties of a solid sphere. Surface temperature after immersion in a fluid of prescribed temperature and convection coefficient. FIND: (a) Time to reach surface temperature, (b) Effect of thermal diffusivity