218 Fouriees Law and Thermal Conductivity of Materials ha i areas (0.1 m*) with di (6.14 A fat beater is sandwiched between two solids of equal areas (0 (funmat conductivities and thicknesses. The heater operates at a uniform tempera provides a constant power of 290 W. The extemal surface temperature of each soli 500K, and there is perfect thermal contact at each internal interface '2) Caleuate the hea Mux through each slid ') What isthe operating temperature ofthe heater? “Thermal Condoaiviy, Thickness, sid Wm K" am 4 s o a 9 30 7 HEAT TRANSFER AND THE ENERGY EQUATION ‘Wetave designed this chapter to introduce the reader t0 three interwoven topics. First, we develop differential equations in terms of temperature in space (and with time if transient ‘anions apply) for several simple problems, by writing energy balances for unit volumes. solutions, we integrate the differential equations to ascertain the FY consis, and then apply boundary and inital conditions to obtain tbe articular solution, The general procedure is similar to that followed in Chapter 2 for ‘ining the velocity profiles. ‘cond, several ofthe examples are concerned with heat transfer to and from moving ‘hits, We dal only with laminar convestion, but hs enables the reader to become involved inte fundamentals of heat transfer with convection, ‘Third, we bring tothe readers attention more general forms ofthe equation of enersy, leaing to Tables 7-3-7.5 which may be used in a manner similar to the general momentum ceutions given in Chapter 2. ‘71 HEAT TRANSFER WITH FORCED CONVECTION IN A TUBE CCosider laminar flow ina circular tube of radius R, as depicted in Fig. 7.1. Ifthe tube and {the fid exchange heat, then clearly the Nuid's temperature isa function of toth the rand 15> cliatons, A suitable unit volume i a ring-shaped element, Ar thick and a2 high. Energy eters and leaves this rng by thermal conduction; also, aunt mass of fluid, which enters ‘with an enthalpy, mast leave with a different entalpy. "Let us now develop the enerey Inlance for the unit volume Rate of energy in by conduction across surface atr 2er ace, Rate of energy out by conduction across surface at r + ar I(r + ANAL we Rate of energy in by conduction across surface at 2 2ar Arg,220 Heat Transfer and the Energy Equation ig 7. Elemental cteaar ing wed tw develop the ‘frei energy bau for lami ue No Energy out by conduction across surface atz + Az Dar Arie Energy in dve to fluid flow (enthalpy) across surfice at pve art, [Energy out due to fluid flow across surface at 2 + 42 0 :2a0 Arty Here His the enthalpy per unit mass, and u i the velocity in the z-direction. At se state, the energy balance requires equal inputs and outputs, IF we divide all tems 2e Ar Az, we obtain adecan Pale, Milevas Mle, 4, Hla se” Mle carapace xem ema Now Ar and are allowed to approach 20, aa) 8 aH ee eo IF Gs the beat capacity, then Also ~K@Ti@8) co) anda, ‘Substituting Es. (7.3) and (7.4) into Eq. (7.2) yields an energy equation writen i tems of cover ar), a a] °F eat Transfer and the Energy Equation 221 F ecan tes simply te egy bate (Eg. 5) sce, exe for every slow Han eal erm (ecDOTIC hneglgbi even boughs (O70) aa Wid Eg, 9 esac a a fia a” 2G, |r 2] a a elas ow: veloy ton hore pid seit ee in of bl] wb-lle aber cers boundary coniions, a fully develope! temperature profile exists when Gi Tidy T) is unique Function of WR, independent of. Then Ta FOR, oo co) temperature of fluid atthe wall, and T, = mean temperaure of the fuid. A. fully developed temperature profil is analogous to fully developed flow. This is exemplified ty Fig. 7.2. where the lnuid lowing in the zditestion encounters the heated section of the tke Over a finite interval downsteam from this poal, the temperature profile changes fom sniform o fully developed Fora fully developed temperature profile, an important corollary arses; namely, the heat trast coefficient is uniform along the pipe. We realize tis by employing te definition ofthe beat transfer coefiient bated on the mean temperature of the Mud: yaw 7.10) T a he here isthe fax evaluated at the wall (r=). Because the derivative in Eq, (7.10) has ‘uniga value a the wall, independent of :, his therefore uniform along the pipe under the fil developed temperature conditions. Now consider the cate where gs uniform, This represents @ uniform heat ux at the wal, ad could be physically obitined by using an electric heater, depicted in Fig. 7.2 FFurher, since h and gy are constast, Eq. (7.10) specifies that Ty - Tis constant, and an, aT, Gt aats aay222 Neat Transfer and the Energy Equation Heat Transfer and the Energy Equation 223 ving obtained the temperature profile, we can evaluate h. From Eq, (7.10) (Ty) Havin then be evaited. Fist we find Tq 7, by pexorming the [oe -npere Flow q Sue 1 te 2 oun ee Folie [ varrar = second, we determine 9, by evaluating the gradient a the wall sing Ba. (7-16 a +(Z] om ‘when these operations have been catied out, we can determine the heat transfer coefficient ‘The final result, with D asthe diameter, is aD &e B Jeans cevetopes ee Fig. 7.2 Heating 2 fvid in a tube showing the development ofthe temperature profile 4.36. 19) ‘The dimensionless number resulting from this analysis is the Musele number. This inporant dimensionless umber for heat flow with forced convection reappears. as we cuamine other solutions and correlations. For emphasis, then, the Nusselt number is nD nu, = 2. 720) (ote that Tzand 7, themselves are not constants.) Now expand Eq, (7.9) ina general sene 8 ‘where each quantity varies as follows ‘This Nusselt number is for fully developed flow and uniform heat flux with parabolic velocity a at, Ty aay pple, Tee subscript with oo teeate it represent the limiting case of ful veloped a mul"? . Temperature profile. Many situations have been analyzed, some of which are given in Sle 71 and oters in Holman." staan Bas (0-1) are Table 7.1. Nusselt numbers for fully developed laminar flow’ a, at _ My re a ke ong Very Contin gM Geometry Aistribution’ at wall Equation (7.13) is important because it allows Eq. (7.7) 0 be integrated dir tube ‘Parabolic Uniform gy 4.38 atiae = OT 82 hive PaboleUnfom f 3.88 tbe Shp ow Uniform gy 80D Ae (ert lee 019 Creatas be Sugflow Unio #575 wae] f Parallel plates Parabolic Uniform gy an a Paral Fanbolic Unio 760 Tengu Panbotie Unio gr 3.00 Integrating, we set “ean Faabole Unfom i 235 ar} apey) oe or ay “From. M,Rohsenow and H.Y. Choy, Mot nd Morente a (2) 05. ow “Table 7.2. Typial Prandl numbers Subuaee ange of Pron nmber a) Common igus (wae ako. ee) 28D gai meals Son.003 a Heat Transfer and the Energy Equation 227 Equations (727-(7.29) are valid only for Pr > 0.5, and thus do not apply to liquid ‘metals. For liquid metals with uniform wall temperatures as a boundary condition, the results fie approximated by* — {__ 0564 Nu, = (Re, Pr | 2564 7.3) ord For uniform heat Mux at the wall, we present these resus, Pr> 05, Nu, = 0.458 Pr? fe, oy 0.880 0.006 < Pr < 003, Nu, = fer |—°* _] aay) [ + 13IT Pr Bxample 7.1. Airat 1 atm (1.013 x 10* N m*) and 290 K flows parallel toa plate's surface 15 ms*. The plate, 0.3 m long, is at 360 K. Assume that limit flow is sable along the entre length '2) Calculate the thicknesses of the velocity and thermal boundary layers 0.15 m from the leading edge of the plate ‘Calculate the rate of heat transfer from the ent plate per 0.1 m of plate width, solution 1) We calculate the thickness of the momentum boundary layer using Eq. (2.101), foyeves.0) *° Wan Forair, evaluating vat an average boundary-layeremperance of 4(290 + 360) = 325 K, the Kinematic viscosity is 18.4 10* m? s'. Then, Vex _ Sm | 0.15 m 5 Tea <0" Pho aa, (0.1950) Via x 10 Next, we use Bg. (7.29) 215 x 109 m = 2.15 mm. f= 0975 Pr. °B.M, Sparow and J. L. Gregg, J. Aero. Se. 24, 852 (195) i REA. Nickerson and 1. P. Smith, as reported in Rohsenow and Choi, ibid. pane 149.228 Heat Transfer and the Energy Equation ‘The Prandtl number for air, evaluated at 325 K, is 0.708, Then 5, = (2.159(0.9750.708)""" = 2.36 mm + Equation (7.27) suse forthe average heatransfer coefficient, which applies tothe whe plate Nu, = (0.664)(0.203) (2.44 x TOF = 291 “The thermal conductivity for air at 325 K is 28.1 10° War K', Then baa x 109}a90) k ne Eu, = OBL MID - a2 wW mK, re OF and finally @ = MAT, ~ T) = (27.2103 * 0.19860 - 290) = 57.1 W 7.3 HEAT TRANSFER WITH NATURAL CONVECTION In Sections 7.1 and 7.2. we considered heat transfer with forced convection, In forced convection, the Known velocity distribution can be entered into the energy equation, The Situation is more complex in problems of heat transfer with natural convection because the ‘elocities are not known a prior to solving the energy equation, Hence, the velocity ani temperature distributions cannot be treated as separate problems; the temperature ¢istribution, in effect, produces the velocity distribution by causing density differences within the uid ‘Consider the vertical surface in Fig. 7.5; the surface is at 7, and it heats the neighboring ‘uid whose bulk temperature is T. In ths situation, the velocity component w, is quite Small. the fluid moves almost entirely upward, and therefore we write the momentum ‘equation for the x-component only. For steady forced convection over a fat plate, we fanored the pravity force, and no pressure gradient was involved. We cannot ignore these fees in free convection, and therefore the momentum equation for the present case contains these terms mw, ap By ys & Soe os) sy pu HF 8 We apply the so-called Boussinesg approximation in which variations in the density of the fluid ate neglected except in the buoyancy force term that drives the natural convection, ‘Thus, in Bq, (7-33) the density is a reference density at the reference temperature, T. ‘Therefore, the inertial terms become and the pressure gration is approximated 8 ew me Heat Transfer and the Energy Equation 229 oe Fig. 7.5 Thermal and momentum boundary layers for vertical pate natural convection, with g, the buoyancy term is simply oe, = -psell 00 - 7, where the volume expansion coefficient is defined as { ar ‘The momentum balance inthe x-direction then becomes my yy Fit 1.) a4) 2.2 ich is idemiet with Eq. (28D, except for the addon of te buoyancy tem, Eqution 7.34 stows thatthe monentom eqetion mut be coupled tos appropri egy equation, order to test he bunny tem ‘The energy equation applied tothe control volume A-xAy in this case is for flow over a fat plate y identical to that roar er 1 a 038 Tis enation is coupled 10 £4, (7.34) caplet 10 the presence ofthe velocity tems. The termi mi san cc, oie cups wh he ay BCL aty=0, wy =o THh =o, T=. BC2 aty=o, y=20 Heat Tans and he Energy Equation oa : ations are beyond the sope of his et Analyt! solutions of sch couple ieee ef Hse Sethe he test weer em exanine » Sins ay Seen ae bing forth pertinent dimensionless numbers PO ee een he conditions for which the velocity profilin raat comes stn ia the vec rofl aot ata convein ttn re a mame boundary condos, ie veloty Hero at he sac 2 Bah ed amved om the srface. Now employ te dynamic sinlry Srpumentinoducd in Section 3.11 int Eg. (28) is writen for sytem eo eaee ess a 8, OE my Oo aft T. wa Be on aye TN aye a system 2 i sae to system 1 by geometrical and dynam siniuries expressed Wy the k, k h 2 om Re y eee p Now, we replace Yu Le He Ye Bv 6 Eq, (7.36) by thee equivalets in Eq, (7.37) them ‘we write Eq, (7.34) for system 2: Be KK, a + KKK BB{T = Tay 038 oe eee Equation (7.38) if rewritten without all the KS would of course be valid, because it would transform back to Eq. (7.34), Hence, Ke, 0.39) and therefore 401.09) Ths, 8,B\(T% - Ta), 1 we combine Eqs. (7408) and (7-408) we et wi aay eat Transfer and the Energy Equation 231 which are Reynolds mumbers. The combination of Egs. (7.408) and (7.402), yids Billo ~ Tale &aBalTy - Teed Ee) caper 742) ‘The srup of variables represented in Ba. (7.42) could be considered dimensionless umber, but by reflecting on the physical aspect, we realize thatthe velocity ofthe Tid is fot an independent quaniy, bul hat it rather depends onthe uoyan force. Hence, the v's tre eliminated fom Eq, (7-42), and substnting their equivalents from Eq. (7.41), we obtain BP (To Ta)de _— B,0(To ~ Tahb? 7.43) ‘This dimensionless number is important in natural convection problems and is called the Grashof number, Gr. When buoyancy is the only driving force for convection, the velocity profile is determined entirely by the quantities in the Grashof number, and the Reynolds ‘number is superfluous. Recall that for forced convection, as discussed in Section 7.2, the Nusselt number is correlated in the general form Nu = f(Pr.Re), forced convection Correspondingly then, for natural convection, the Nusselt number is correlated as Nu = f(Pr.Gr), natural convection. Returning tthe complete solution of Eqs. (7.34), (7.35) and the appropriate boundary conditions, we present the velocity and temperature distributions (see Fig. 7.6). The curves show that for Pr <1, 8, = 8 but for Pr > 1, 6; < 8, For liquid metals, therefore, 6 is shout equal to & in free convection as contrasted 1 forced convection in which 5, > 6, Corresponding to the temperature profile, shown in Fig. 7.6, the loeal Nusselt number is Nu, __0.676 Pet? nie .861 + PA 0.44) Equation (7.44) applies for a wide range of Pr numbers (0.00835 <= Pr = 1000) for laminar flow conditions, with 10 < Gr,-Pr < 10" Example 7.2 Caleulate the inital heat transfer rate from a plate at 360 K, 0.3 m long X 0.1m wide hung verically in air at 290K. Contrast the results with those of Example 7, Solution. Equation (7.44) should be integrated to obtain the average heat transfer coefficient ‘which can be applied tothe whole plate In Eg (7.44), because h, varies 25 x", then the average hequals $k. lence Nu, defined as ALIkis Nay 0.902 Pe Vora 861+ PI 4s)‘232 Heat Transfer and the Energy Equation 06 oot o mye amc eh ye GPF (7.433) 902 | aOR + PAD Na, Ae evaluate the properies atthe average boundary temperature of 325 K, For ai a 325 K Pr = 0,703 and Biv? = 9.85 x 107 Km” “The Grashof number is or, = Br - TP = 045 x roo - 2900.3") = 1.86 x 10" ‘eat Transfer and the Energy Equation 233 ‘Next, we calculate the product Gr,'Pr to test for laminar flow conditions Gry Pe = (1.86 % 100.705) = 1.31 x10! Since i is between 10° and 10", Eq, (7.45) is valid, When we substiute values of Gr, and Pr into Ea. (7.45), Nu, = 55.8, from which anf os) — sa wa Finally, we evaluate the rate of heat transfer Q. O = NT, ~ TA = (5.23)G60 ~ 2900.1 x 03) = 11.0 W. For Example 7.1, Q was 57.1 W; that is, the rate of heat transfer for forced convection is considerably higher. ‘This is the usual case tis instructive to look at special forms of Eq, (7.48). Fics if Pr = 0.7, then it reduces © Nu, = 0.477 Ge! 7.456) iso happens that for many gases, including air, O,, CO, He (and other inert gases), H, and (CO,, Pris very close to 0.7 and practically constant for temperatures even as high a 1900 K. ‘Thus, we can apply Eq. (7.456) directly to gases. ‘Second, if Pr O (liquid metals) then Eq. (7-59) reduces 10 “GPE 7.450) 0.936 ed Nu 1.4 HEAT CONDUCTION We consider heat conduction through the wall of a hollow solid eylinder. Figure 7.7 depicts the situation, and also locates a suitable unit volume with a thickness ar. From a practical point of view, we may visualize a long cylindrical shaped furnace, and it is desirable 10 falculate the heat loss to the surroundings. Suppose the cylinder is long enough so that end effects ae negligible; in addition, the system is at steady state, so that both the inside and outside surfaces ofthe wall are at some fixed temperatures, T, and T,, respectively. For such system, we develop the energy eqation, Rate of energy in by conduction across surface at r 2derlg|, Rae of energy ou by conduction across surface at r + Ar Darlalese ‘At steady state these are the only terms tht contribute to the energy balance, Thus 2erla,|, «a - 2erlg|, = 0.264 Heat Transfer and the Energy Equation Fig. 7.7 Heat conduction though a slid elidrical wall. The shaded area depits the unit volume, If we divide al terms by 2xlAr, and take the limit as Ar approaches zero, we obtain ara) 7.46) “a ~° tion (7.46) requires that Equation (7.46) req _ am [Note that q, the heat ux, isnot constant in itself. Since g, = -(dTId), Eq, (747) yields als aay) a7 Integrating once again, we find for constant thermal conductivity that 749) By absorbing k in new constants, Eq. (7.49) simplifies even more (0 =GmrtGy i) “The boundary conditions under consideration are BCL atr Te Bc2 ars T=h ‘eat Transfer and th Energy Equation 235 Daeminatn of he constans using he nary coitions yl he tempera distribution T-1 Infor) T=, ~ ini)" os and the heat flux as [Aste heat flows through the wall, it encounters larger areas, so that the ux itself decreases, ‘The heat low Q, however, is constant (as it must befor steady state), and is given by Q = gi2erty re @, 1) 0733) ‘This problem, elementary as iti, demonstrates an interesting enginering characteristic. Suppose we use the cylindrical wall as the insulation of a furnace wall. As increasing thicknesses of insulation are added, the outside layer, because of its greater are, offers less resistance to heat flow than an inner layer of the same thickness, Thus, from a ost point of view, te expense of additional insulation ean become greater than the savings associated with reduction in heat losses [Example 7.3 As part of a proposed contimous annealing process, a rod passes through & cylindrical furnace chamber 101 mm inside diameter and 15.2m long. The inside surface temperatre ofthe furnace wall under operating conditions is predicted tobe about 920 K and the outside surface about 310 K. If itis decided that a heat oss of 73 KW is an accepiable figure, then which ofthe following insultions would you use? k.Wint KY Cost Sperm? Insulation A 070 350 Insulation 2 035 880 Solution, Equation (7.53) can be weiten ts [| - Fe For A, then : :] = 220.70905 21620 — 310) - 9 459, 7 (3 x 10) Fea wah r= 50303, wot n= 11.4 mn. Sinbey hr 2 kg cn ot ee »[s] ra [$38] cosy = aan@® 26 ‘Heat Transfer and the Energy Equation to that r, = 66.8 mm. We calculate the volume of insulation and the corresponding cos. (883° 2) mm | 1 m [3 s05))mm' | 1mt_|is2 9S _ ssre9 '$80.34, ‘The obvious choice is B Cost A = Inthe same manner, Cost B 1.5. THE GENERAL ENERGY EQUATION . tn Sections 7,-7.4, we determined temperate ditibtons nd eat fxes fr some spe ras gvclopnepetne ery tne in diffeenl form In this gecton we ESE tra eneepy equation, which canbe reduced to sole specific mei oo ee eee volume Axayaz in Figs. 2.4 and 2.5; we apie law of cone a Of eoery othe Ti conti within ths vole at any given ime (eae |- See (SEs. [sais] kinetic energy in by convection by fluid “This statement of the law of energy conservation is not completely general, because other forms of energy tansport, e., radiation, and sources such as electrical Soule heating, are ‘ot included, ‘The rate of accumulation of internal and kinetic energy within the unit volume is simply a axaya $ (pu + tov). ass) where U isthe internal energy per unit mass of ful and wis he magnitude ofthe ol Tuid ay pe velocity "The ner rate of internal and Kinetic energies in by convection is vu + sol od + axae(pu + fo], - bu + der), a aytz{uoU + For], «sam sie opus, as Ina similar manner, the net rate of enery in by conduction is aydelle = aulecad * 2824 = ly a) 2x87UGle = leva” 5? “The work done by the fluid consists of work against gravity, work against pressure, and work maint viscous forces. The ate of doing work against the three componens of fraViy “pA xAyAZ(H,B, + YB * %8i) ca ‘eat Transfer and the Ene * Energy Equation 237 ‘The rte of doing work against the pressure atthe six faes of the unit volume ayae{Py ~ (Puy|) + xa2(Pu)], 4, ~ ea} saxo fPril Culp. 039 Poet cae tet ener tae eee sybthratle sas = tend) © 8282( 6% om 1h) + AxdyfrMelesar ~ Failed (7.60) Similar expressions may be written for the work agaist the y- and z-directed viscous forces BYAZ{raMi ls was ~ FoYylat * S282LtH5Ly< ay ~ ty) + Axtrlrotlecas~ tani ast) AyAefrte can teed * B28Ah lw ~ Hal) + A xdylr adele cae > tatld (7.6) Subsintng these express ito Ea, (7.50, ving lines Ar, nd spon er, ohn oe form teeny cation Hou tee) = [eal t U + do) =F ufou + tor) + Zoo + ioe] %, Ha] tbat Ua + 18) oes nda : ce ee a % rade * Toy + wi] 783)x te rr an he nergy Eatin sn ma a oe rr ofthe gr prot or Newnan Pal of son. wand cof e implies that C, vettone py and note that the constany Rectangular coordinates OT eo Hy, ov ® er, aT eeaT ym aT wey te Fsin0 26 aja) tal [ ont ee nd 2{e), 1 o + et alete + rina a aes |) we ts wih vicosdspation andy wn ‘Note: ‘The terms contained in Were ge velocity gradients. Hea Transfer and the Energy Equation 243, Example 7.4 Refer back to Fig. 7.3 and the system described in Section 7.2. Using [Table 7-4 oF 7.5 derive the energy equation, Solution, Table 7. is selected because the Mud has constant properties. Flow is two mensional in rectangular coordmates so Eq. (A) i the best choice. Before proceeding, tell hat there ae two velocity components, and Also, itis a good idea wo qualitatively ‘Mech the temperature field. Having done so, you should recognize that T = Thy). Now, fee ean proceed 1 simpliy Eq, (A) in Table 7.5, Noe that C, = C,, ar . FF = 0 because there is steady state ar a Z = 0 because v, = 0 and T= Tex) or Fa = 0 becuse 7 = Thx) er in {Jae nero because we select icous esting. Wear ft with ar, at) | [er , ar at * ay? F-45 Except for Auids with very low Pr numbers, we can ignore conduction in the direction of flow; hence ar wn? nd we fly write wa er “eH Tt ay PROBLEMS 1 For laminar flow, calculate the results given in Table 7.1 for Nu. for slug flow (4, uniform) and uniform beat fux in a cirevlar mube 12. A liquid film at Ty lows down a vertical wall at higher temperature, Consider beat teansfer from the wall tothe liquid for such contact times thatthe liquid temperanre changes ‘appreciably only in the immediate vicinity ofthe wall. (See figure on next page.) 2) Show thatthe energy equation can be writen (sate assumptions): ar_ ar oa EE 1) The energy equation contains v, What would you use for? ©) Write appropriate boundary conditions.(2) ‘244, eat Transfer and the Energy Equation 7.2 om.) re a 73. A gap of thickness L exists beeween two 77! Tora pies of pros sold Fis foced Fo ph be tone pine an mt rer la. AsuTe t tt ee voy Vin mina fw with sight seamless he pp. I a cay sate wi he wpet an Tower tes a Tao Ty eT ae opens cregy equation and band) canon tbe Wid in he 2 Mike an prep re he gap. 6) Dee an egaion forthe Bat Mux ass ie e yoo Ft 7.4 A liquid of constant density and viscosity flows upward in the annulus (R, Sr = Ry) between two very Tong and concentric cylinders. Assume that both he flow and the temperature are fully developed. ‘The inner eylinder is electrically heated and supplies constant and uniform flux, qy, 10 the liquid. The ‘utes eylinder is maintained ata constant temperature, Te 4) Solve for v, By Write the energy equation and state your Fay enters assumptions aT 9) Write appropriate boundary conditions. 5 sic at 03 m and 368K flows pra 104 fa pata 310 K, 2) Cale Fe a Sang cue to here he mont Bouday fy KDE 6m Co ya cating eg, ashe a Doda ye ES? Oe a em he eng ce at ane es eri) Dif paw 10) nie? Heat Transfer and the Energy Equation 245 116 Consider natural convection between parallel vertical =~ plates maintained at 7) and Ty, respectively. Assume that the plates are very long and the convection is fully Goveloped. For constant properties: a) Write the enegy — , 2— 2s] equation and boundary” condiions for temperture, ts b) Write the momentum equation with the Bousinesg approximation and boundary condition for velocity 1.7 The surface temperature ofa vertical pate is maintained at 390 K. At 0.24 m from the bottom of the plate, calculate the heat tansfer coefficient to: 8) ait a 290 K; b) helium at wok. 7.8 Liquid meta Nows through a channel witha rectangular cross-section. Two walls are perfectly insulated and two are ata constant temperature of T,. The metal has temperature Tyas itemtrs the channel, and 7, > Tj. Assume steady state, fully developed flow and no salidifcation. Enters uT cane INSULATED 2) Write the energy equation in rms of temperature for constant thermal properties >) Write the boundary conditions, aa 79 Consider the creeping flow ofa fuid about a rigid sphere as illustrated by Fig. 2.9, The sphere is maintained at, and the uid approaches from below witha temperature T, and velocity V_._ a) Write the energy equation which applies to the Mud inthe vicinity Of the sphere. Assume steady-state conttions. b) Write appropriate boundary corditions for part 2). c) What other equations or results would you use in order to Solve the system Aescribed by parts a) and b)? 7.10 A very long fiber of glass (radius = R) is extracted from 2 hole in the bottom of a crucible. It is extracted ene with a constant velocity V into a gas at Ta; assume slug tow. as 1) For uniform properties write the energy equation for = temperature inthe fiber. Do not ignore conduction in | the direction of flow, Sher +) b» Wie bounty conitions, Hints ALF = R the \ fun 10 the sirface must equal the ax 1 the : surrounding gas "via h.") U 7.41 Starting. with Eq. (7.44), derive Eq. (7.45) and define the dimensionless numbers in Ey. (7.45),246 Heat Transfer and the Energy Equation 72 +} Determine an expression that gives the heat flow Q (W) through a solid spherical she with inside and outside radi ofr, and r, respectively by Examine the results regarding what happens as the shell thickness becomes larger compared withthe inside radius. 17.14 For the system in Fig. 2.1 develop an expression for the temperature dstibution inthe falling film. Assume fully developed flow, constant properties, and fully developed temperature profile. The fre liquid surface is maintained at T = Ty ad the solid surface at TT, where T, and T, are constants. a) Ignore viscous heating effects. b) Include viscous heating effects aie T-% «x xy where Br w Brinkman number. 17.18 Consider heat conduction though a plane wall of thickness x, and 7, and Tar the ‘lrface temperatures. Derive the steady-state heat flux in terms of T,, T; and Ox ifthe thermal conductivity varies according to k= Alt + a7) where fy and a are constants 7.16 A liquid at a temperature Ty continuously centers the bottom of a small tank, overflows into Artube, and then lows downward as a film on the inside, At some position down the tube (¢ = 0) ‘when the flow is fully developed, the pipe heats the fuid with a uniform flux gp. The heat loss, ==} >| from the liguid’s surface is sufficiently small so that it may be neglected =| ‘) For steady-state laminar flow with constant properties, develop by shell balance or show by reducing an equation in Table 7.5 the pertinent differential energy equation that Applies tothe falling film, 1) Write the boundary conditions for the heat flow. ‘©) What other information must complement parts 2) and b) in order to solve the energy euation? 27 8 CORRELATIONS AND DATA FOR HEAT TRANSFER COEFFICIENTS The problems of beat flow with convection, discussed in the preceding chapter, pertain to sinple systems with laminar flow. Despite the simplicity of laminar low problems, they should not be underestimated. Many simple solutions have been applied to real systems with aproximating assumptions and, besides, the simpler systems provide models for ierpretation of complex systems.‘The more complex nature of turbulent flow and its limited aecessiblty to mathematical treatment requires, however, an empirical approach 10 heat tansfer. On the other hand, the study of turbulent flow is not entirely empirical; it is posible to establish certain theoretical bases for the analyses of turbulent transfer processes 2nd an introduction to this complex area is given in Chapter 16. Figure 8.1 illustrates heat transfer in a bounded fuid, ‘The fui is atictaly subdivided ino three regions: the turbulent core, the transition zone, and the laminae sublayer neat the surface. In the rurbulent core, thermal energy is transferred rapidly due to the eddy (mining) ston of urbulent flow. Conversely, within the laminar sublaer, energy is transferred by canduction alone—a much slower process than the eddy process. I the transition zone, ‘zergy transport by both conduction and by eddies is appreciable. Hence, most of the tual temperature drop between the fluid and the surface is across the laminar sublayer and t ‘umsition zone. Within the rurbulent core, the temperature gradients ae quite shallow, In Chapter 3, it was convenient 1o define a friction factor to deal with momentum transport in fluids in contact with surfaces. Similarly, for energy transport between Muids al surfaces, itis convenient to define a heat transfer coefficient by (aT) en Where the subscript “O" refers to the respective quantities evaluated atthe wall, and is ‘ome temperaure of the Muid. Ifthe fluid is infinie in exten, we ake T; a4 the Abid femperature far removed ftom the surface, and designate it T.. If the fluid flows in a Sonfined space, suchas inside a tube, Ts usually the mixed mean temperature, denoted by Ta itis a temperature that would exist ifthe Nid at a particular cross section were removed ad allowed to mix at