Question

5.61 Spheres A and B are initially at 800 K, and they are simultaneously quenched in large constant temperature baths, each having a temperature of 320 K. The follow- ing parameters are associated with each of the spheres and their cooling processes. Sphere A Sphere B Diameter (mm) Density (kg/m) Specific heat (kJ/kg .K) Thermal conductivity (W/ m K ) Convection coefficient (W/m²K) 300 1600 0.400 170 5 30 400 1.60 1.70 50 (a) Show in a qualitative manner. on T versus i coordi- nates, the temperatures at the center and at the sur- face for each sphere as a function of time. Briefly explain the reasoning by which you determine the relative positions of the curves. (b) Calculate the time required for the surface of each sphere to reach 415 K (c) Determine the energy that has been gained by each of the baths during the process of the spheres cool- ing to 415K

Answer 1

Answeg ! Assumptions :- 0 10 godicel conduction. 2) unifoom pospesties. ☺ constant convection coefficient 800k =Ti fluid . To=320k LT(20,4)= 415K @ forms the icnowledge of Biot number & Thermal time constant, it is possible to qualitatively represent dhe tempestese distojbution, Hese Les zo : Big = 5*(0.150/3) = 1:47 x 103 170 Bi = hp 2013) : 50 x(0.01573) . Die 1.7 = 0.247 + The thermal time constant for a dumped capacitance system is ; 1. 8. 2.c Tarhana (PVC) + Coghi i Big = 9 CA = 1800 T = 1600 XO:150 X 400 3x5 = 6400 sec 400 X 0.015 X1600 = 3 x 50 = 64 sec.

when Biccob, the Sphere will cool in a Spacewise isothermal mannes ( sphere- A). for sphere B, Bi >0.1, hence gradients will be important. Note that the theomal time constent of A is much larger than for B; Hence, Ą will cool much slower, as shown below. Tlet) (2,4) Sphere- B - Tlo,t)) sphere A Spacewise isothermal. time O Recognizing that Bin <0.1, Sphere A can be foealed as Space wise isothermae of analysed using the Jumped copacitonce method with T = 415K © - I-To = exp(-+10) Ti-Too 415-320 to = -ta [on The]--.0 [m - =-6400 On -100 800 - 320 ta = 10367 see = 2.88h - Answer. - Since Big >0.1, Sphere-B must be treated by exact method of solution. We assume that the one term approximation is valid & check dater, Hence;

Big = hizo - sox0.015 = 0.441 from Table 5.1 ¿, = 0 1.0992 , 4 = 1.1278 0*=1. F) = T (2. t) - Too 415- 320 Ti - Too 800-320 = 0.1979 Ond equation 5.536, can be solved for 3 ;. O* = * E-2 / sin (8..2*) = 0.1979*1.0992x1 sin (1.0992) = 0.2442 + Then equation s.53 c, can be solved for Fo ; F = - (07/G.) - en (0-2442/1-1278) 1. fo = 1.266 to = 6 - Episode 1.266 X 400 X 1600 X - 1.7 - = 107 see ::Ito = 107 See Answer. + és Note theet one team approximation occuesale, since fo >0.2

© Applying conservation of eneegy requirement on a time interval basis, Sphere A : Ein - Eout = DE ;-QA = AE = E(+) - E60) QA = 8.cv [TC+)-T;] = 1600x400 [415-800]****(0-150) QA = 3.483 x 1065 → for Sphere B: Ein - Eout = AE - 98 = E(+) – E(0) for noniso thermal sphere, eq. 5:55,used to evaluate qe = 1-3. [Sin C 2.) – [4SCE)] = 1-300 (sin C109927–1092 Casc1991) = 0.784 - Q8 = 0.78496 = 0.784[8.C.VCT; -To)] Q3 = 0.784 X400 X1600 x 64-TX(0-015) (800–320) 11Q8 = 34055