PROBEM 5.88 KNOWN: Initial temperature of fire clay bric which is cooled by convection. FIND: Center and corner temperatures after 50 minutes of cooling. ASSUMPTIONS: () Homogeneous medium with constant properties, () Negligible radiation effects. PROPERTIES: Table A-3, Fire clay bric (900K): ρ = 050 g/m 3, =.0 W/m K, c p = 960 J/g K. α = 0.5 0-6 m /s. ANAYSIS: From Fig. 5.(h), the center temperature is given by T 0,0,0,t = P 0,t P 0,t P3 0,t where P, P and P3 must be obtained from Fig. D.. h = 0.03m: Bi = =.50 Fo = =.70 h = 0.045m: Bi = =.5 Fo = = 0.756 h 3 3 = 0.0m: Bi3 = = 5.0 Fo3 = = 0.53 3 Hence from Fig. D., T( 0,0,0,t) T P 0, t 0. P 0, t 0.50 P3 0, t 0.85. Hence, and the center temperature is 0. 0.50 0.85 = 0.094 T( 0,0,0,t) 0.094( 600 33) K + 33K = 434K. < Continued..
PROBEM 5.88 (Cont.) The corner temperature is given by where T,, 3,t (,t) P(,t) = P( 0,t ), etc. o = P,t P,t P 3,t and similar forms can be written for and 3. From Fig. D., (,t ) (,t) (,t 3 ) 0.55 0.43 0.5. o o o Hence, and or ( ) ( ) P, t 0.55 0. = 0. P, t 0.43 0.50 = 0. P 3, t 0.85 0.5 = 0. T,, 3,t 0. 0. 0. = 0.0056 T,, 3, t 0.0056 600 33 K + 33K. The corner temperature is then T(,, 3, t) 30K. < COMMENTS: () The foregoing temperatures are overpredicted by ignoring radiation, which is significant during the early portion of the transient. () Note that, if the time required to reach a certain temperature were to be determined, an iterative approach would have to be used. The foregoing procedure would be used to compute the temperature for an assumed value of the time, and the calculation would be repeated until the specified temperature were obtained.
PROBEM 5.89 KNOWN: Cylindrical copper pin, 00mm long 50mm diameter, initially at 0 C; end faces are subjected to intense heating, suddenly raising them to 500 C; at the same time, the cylindrical surface is subjected to a convective heating process (T,h). FIND: (a) Temperature at center point of cylinder after a time of 8 seconds from sudden application of heat, (b) Consider parameters governing transient diffusion and justify simplifying assumptions that could be applied to this problem. ASSUMPTIONS: () Two-dimensional conduction, () Constant properties and convection heat transfer coefficient. ( + ) PROPERTIES: Table A-, Copper, pure T 500 0 C/ 500K : ρ = 8933 g/m 3, c = 407 J/g K, = 386 W/m K, α = /ρc = 386 W/m K/8933 g/m 3 407 J/g K =.064 0-4 m /s. ANAYSIS: () The pin can be treated as a two-dimensional system comprised of an infinite cylinder whose surface is exposed to a convection process (T,h) and of a plane wall whose surfaces are maintained at a constant temperature (T e ). This configuration corresponds to the short cylinder, Case (i) of Fig. 5., ( r,x,t) = C( r,t) P( x,t ). () i For the infinite cylinder, using Fig. D.4, with -3 4 m o.064 0 8s hr 00 W/m K 5 0 m 3 Bi = = = 6.47 0 and Fo = = s =.36, 385 W/m K -3 ro 5 0 m ( 0,8s) find C( 0,8s) =. i cyl For the infinite plane wall, using Fig. D., with 4 h -.064 0 m / s 8s Bi = or Bi 0 and Fo = = = 0.34, find ( 0,8s) P 0,8s = 0.5. i wall Combining Eqs. () and (3) with Eq. (), find -3 ( 50 0 m) () (3) 0, 0,8s T 0,0,8s = 0.5 = 0.5 i < T 0,0,8s = T + 0.5 = 500 + 0.5 0 500 = 60 C. (b) The parameters controlling transient conduction with convective boundary conditions are the Biot and Fourier numbers. Since Bi << 0. for the cylindrical shape, we can assume radial gradients are negligible. That is, we need only consider conduction in the x-direction.
PROBEM 5.90 KNOWN: Cylindrical-shaped meat roast weighing.5 g, initially at 6 C, is placed in an oven and subjected to convection heating with prescribed (T,h). FIND: me required for the center to reach a done temperature of 80 C. ASSUMPTIONS: () Two-dimensional conduction in x and r directions, () Uniform and constant properties, (3) Properties approximated as those of water. PROPERTIES: Table A-6, Water, liquid ( = + ) T 80 6 C/ 35K : ρ = /v f = /.009 0-3 m 3 /g = 99. g/m 3, c p,f = 479 J/g K, = 0.634 W/m K, α = /ρc =.53 0-7 m /s. ANAYSIS: The dimensions of the roast are determined from the requirement r o = and nowledge of its weight and density, /3 /3 M.5 g M = ρv = ρ π r o or ro = = 0.07m. πρ = = () π 99. g/m 3 The roast corresponds to Case (i), Figure 5., and the temperature distribution may be T( x,r,t) expressed as the product of one-dimensional solutions, = P( x,t) C( r,t ), where P(x,t) and C(r,t) are defined by Eqs. 5.65 and 5.66, respectively. For the center of the cylinder, T( 0,0,t) T ( 80 75) C = = 0.56. () T ( 6 75) C In terms of the product solutions, T( 0,0,t) T T( 0,t) T T( 0,t) T 0.56 = = (3) T T T wall i T cylinder For each of these shapes, we need to find values of o / i such that their product satisfies Eq. (3). For both shapes, h r o h 5 W/m K 0.07m Bi = = = =.68 or Bi - 0.6 0.634 W/m K Fo = /r o = / =.53 0 7 m / s t/ ( 0.07m) = 3.00 0 5 t. Continued..
PROBEM 5.90 (Cont.) A trial-and-error solution is necessary. Begin by assuming a value of Fo; obtain the respective o / i values from Figs. D. and D.4; test whether their product satisfies Eq. (3). Two trials are shown as follows: Trial Fo t(hrs) o / i) wall o / i) o o cyl i w i cyl 0.4 3.68 0.7 0.50 0.36 0.3.75 0.78 0.68 0.53 For Trial, the product of 0.53 agrees closely with the value of 0.56 from Eq. (). Hence, it will tae approximately ¾ hours to roast the meat.
PROBEM 4.47 KNOWN: Steady-state temperatures (K) at three nodes of a long rectangular bar. FIND: (a) Temperatures at remaining nodes and (b) heat transfer per unit length from the bar using nodal temperatures; compare with result calculated using nowledge of q. & ASSUMPTIONS: () Steady-state, -D conduction, () Constant properties. ANAYSIS: (a) The finite-difference equations for the nodes (,,3,A,B,C) can be written by inspection using Eq. 4.39 and recognizing that the adiabatic boundary can be represented by a symmetry plane. q& x 5 0 7 W/m 3 ( 0.005m) Tneighbors 4+ q& x / = 0 and = = 6.5K. 0 W/m K Node A (to find T ): T + TB 4TA+ q& x / = 0 T = ( 374.6+ 4 398.0 6.5) K= 390.K < Node 3 (to find T 3 ): T c + T + TB+ 300K 4T3+ q& x / = 0 T3 = ( 348.5+ 390.+ 374.6+ 300+ 6.5) K = 369.0K < 4 Node (to find T ): 300+ T C+ T 4T+ q& x / = 0 T = ( 300+ 348.5+ 390.+ 6.5) = 36.4K < 4 (b) The heat rate out of the bar is determined by calculating the heat rate out of each control volume around the 300K nodes. Consider the node in the upper left-hand corner; from an energy balance E& in E& out+ E& g = 0 or q a = q a,in+ E & g where E& g = qv. & Hence, for the entire bar q bar = qa + qb + q c+ q d+ q e+ q f, or yt 300 x y TC 300 x x y q = q y q y q bar + & + + + + x & a x & b c TC 300 y T3 300 y xtb 300 x y x + q& x + x + q x + + q y & d y & e y. f Substituting numerical values, find q bar = 7,50.5 W/m. From an overall energy balance on the bar, q = E& 7 3 bar g = qv/ & l = q& ( 3 x y) = 5 0 W/m 60.005m = 7,500 W/m. < As expected, the results of the two methods agree. Why must that be?