sm5-012.pdfsm5-024.pdfsm5-086.pdfsm5-092.pdfHW6a2.pdfHW6b.pdfPROBLEM 5.12KNOWN: Diameter, density, specific heat and thermal conductivity of aluminum spheres used inpacked 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 thecorresponding center temperature. Potential advantage of using copper in lieu of aluminum.SCHEMATIC:Aluminum sphereD = 75 mm, T = 25 CioGasT Cg,io= 300h = 75 W/m -K2= 2700 kg/m3k = 240 W/m-Kc = 950 J/kg-KASSUMPTIONS: (1) Negligible heat transfer to or from a sphere by radiation or conduction due tocontact with other spheres, (2) Constant properties.ANALYSIS: To determine whether a lumped capacitance analysis can be used, first compute Bi =h(ro/3)/k = 75 W/m2K (0.025m)/150 W/mK = 0.013 < 0.1. Hence, the lumped capacitanceapproximation may be made, and a uniform temperature may be assumed to exist in the sphere at anytime. From Eq. 5.8a, achievement of 90% of the maximum possible thermal energy storagecorresponds to tiQ0.90 1 exp t /cV    where3 2t sVc / hA Dc / 6h 2700kg / m 0.075m 950 J / kg K / 6 75 W / m K427s.           Hence,tt ln 0.1 427s 2.30 984s    <From Eq. (5.6), the corresponding temperature at any location in the sphere isg,i i g,iT 984s T T T exp 6ht / Dc    2 3T 984s 300 C 275 C exp 6 75 W / m K 984s / 2700kg / m 0.075m 950J / kg K          T 984 s 272.5 C <Obtaining the density and specific heat of copper from Table A-1, we see that (c)Cu 8900 kg/m3400 J/kgK = 3.56  106J/m3K > (c)Al= 2.57  106J/m3K. Hence, for an equivalent spherediameter, the copper can store approximately 38% more thermal energy than the aluminum.COMMENTS: Before the packed bed becomes fully charged, the temperature of the gas decreasesas it passes through the bed. Hence, the time required for a sphere to reach a prescribed state ofthermal energy storage increases with increasing distance from the bed inlet.PROBLEM 5.24KNOWN: Initial and final temperatures of a niobium sphere. Diameter and properties of the sphere.Temperature of surroundings and/or gas flow, and convection coefficient associated with the flow.FIND: (a) Time required to cool the sphere exclusively by radiation, (b) Time required to cool thesphere exclusively by convection, (c) Combined effects of radiation and convection.SCHEMATIC:T Csuro= 25= 8600 kg/m3= 0.1 or 0.6Niobiumk = 63 W/m-Kc = 290 J/kg-KT = 25 Coooh = 200 W/m -K2Inert gasD = 10 mmT Cio= 900T Cfo= 300ASSUMPTIONS: (1) Uniform temperature at any time, (2) Negligible effect of holding mechanismon heat transfer, (3) Constant properties, (4) Radiation exchange is between a small surface and largesurroundings.ANALYSIS: (a) If cooling is exclusively by radiation, the required time is determined from Eq.(5.18). With V = D3/6, As,r= D2, and  = 0.1,    338 2 48600 kg / m 290J/ kg K 0.01m298 573 298 1173t ln ln298 573 298 117324 0.1 5.67 10 W / m K 298K    1 1573 11732 tan tan298 298              t 6926s 1.153 0.519 2 1.091 1.322 1190s 0.1      <If  = 0.6, cooling is six times faster, in which case,t 199s 0.6  <(b) If cooling is exclusively by convection, Eq. (5.5) yields 3i2f8600 kg / m 290J / kg K 0.010mcD T T 875t ln ln6h T T 2751200W / m K       t 24.1s<(c) With both radiation and convection, the temperature history may be obtained from Eq. (5.15). 3 2 4 4surdTD /6 c D h T T T Tdt          Integrating numerically from Ti= 1173 K at t = 0 to T = 573K, we obtaint 21.0s<Continued …PROBLEM 5.24 (Cont.)Cooling times corresponding to representative changes in  and h are tabulated as followsh(W/m2K) | 200 200 20 500 | 0.6 1.0 0.6 0.6t(s) | 21.0 19.4 102.8 9.1For values of h representative of forced convection, the influence of radiation is secondary, even for amaximum possible emissivity of 1.0. Hence, to accelerate cooling, it is necessary to increase h.However, if cooling is by natural convection, radiation is significant. For a representative naturalconvection coefficient of h = 20 W/m2K, the radiation flux exceeds the convection flux at the surfaceof the sphere during early to intermediate stages of the transient.COMMENTS: (1) Even for h as large as 500 W/m2K, Bi = h (D/6)/k = 500 W/m2K (0.01m/6)/63W/mK = 0.013 < 0.1 and the lumped capacitance model is appropriate. (2) The largest value of hrcorresponds to Ti=1173 K, and for  = 0.6 Eq. (1.9) yields hr= 0.6  5.67  10-8W/m2K4(1173 +298)K (11732+ 2982)K2= 73.3 W/m2K.0 20 40 60 80 100Cooling time (s)010000200003000040000500006000070000Heat fluxes (W/m^2.K)Convection flux(h=20 W/m^2.K)Radiation flux (eps=0.6)PROBLEM 5.86KNOWN: Thermophysical properties and initial temperature of thick steel plate. Temperature ofwater jets used for convection cooling at one surface.FIND: Time required to cool prescribed interior location to a prescribed temperature.SCHEMATIC:T = 25 CoooTooTs=Steel, T = 300ioC= 7800 kg/m3c = 480 J/kg-Kk = 50 W/m-KT(0.025 m, t) Co= 50x0.025 mWater jets,ASSUMPTIONS: (1) One-dimensional conduction in slab, (2) Validity of semi-infinite mediumapproximation, (3) Negligible thermal resistance between water jets and slab surface (Ts= T), (4)Constant properties.ANALYSIS: The desired cooling time may be obtained from Eq. (5.60). With T(0.025m, t) = 50C, si sT x,t T 50 25 Cx0.0909 erfT T 300 25 C2 t         x0.08072 t  2225 20.025mxt 1793s0.0261 1.34 10 m /s0.0807 4  <where  = k/c = 50 W/mK/(7800 kg/m3 480 J/kgK) = 1.34  10-5m2/s.COMMENTS: (1) Large values of the convection coefficient (h ~ 104W/m2K) are associated withwater jet impingement, and it is reasonable to assume that the surface is immediately quenched to thetemperature of the water. (2) The surface heat flux may be determined from Eq. (5.61). In principle,the flux is infinite at t = 0 and decays