Question

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 following parameters are associated with each of the spheres and their cooling processes Round numbers off to the nearest 100 to look up tables Fluid h&T 320K h&T=320K Sphere, A T-800 K Sphere, T =800 K Parameter Sphere B Sphere A 03 0.03 1600 400 Diameter, D [m] Density.p [kg/m] Mass, M [kg] Thermal diffusivity, a [m/s] 225 0.006 2.65 x 10+ 2.65 x 10+ Specific heat, cU/kg-K] 400 1600 Thermal conductivity, [W/m-K] 1.7 170 5 Convection coefficient, h[W/m2-K] 100 (1) Calculate the temperature at the center of her after 2 minutes from the start of the quenching process (2) Determine the amount of heat transfer from sphere A during this time period in foule. (3) Calculate the temperature at the center of sphereafter 2 minutes from the start of the quenching process. () Determine the amount of heat transfer from sphere during this time period in Loulan

Answer 1

SPHERE A FLUID Le T320K . SPHERE, To= 8ook [:L=YA JA'S aar Biz hle - hx W k - K A SX 6 ) 70 l - Yo =%) so = 1.47x183 col Bi<0.1 it is A LUMPED BODY. 7CGpvecoxte page (%) = 1600x 400 x(03) h T- Too = Et=2 min . Too Too = 2x60 120 - 120 Sa ) T-320 e 6400 800 - 320 T: 791.08 K AS is KYMO IT IS A LUMPED BODY. TEMPERATU, SAME EVERY WHERE · TEMPERATURE AT THE CENTRE OF SPEERE A is 791.08k..

2) AT AMOUNT OF HEAT TRANSFER - Qfital = Q dt hast I has [Tor Tool é evc. dt. = hag (to-too) [e-bost a 120 E-mc [To-Tor) Tehoten] = -22.5 x 400 [800 - 300] [e L-1] =-22.5x 400 [480][0.9814--]. = 80245.35 J So AMOUNT OF HEAT TRANSFER FROM SPHERE A 15 80245.35 J. Blog SPHERE-B I Too = 320k. SPHERE - 800k 100 x box V 1.7. Biz hemen h*V) = log (ood) = 0.294 >0.1 So LUMPED BODY CONDITION is NOT VALID : So WE NEED TO SOLVE BY HEISLER CHART.

FOR SOLID SPHERE le=R 1-7 Biz hex) si te ñe 100x10.03) = 1.133 2.65xió x 120 2.65x120x16bó (0.032) 2.25x154 E = Lt = 1.413 0.06 Tc-Too 1.133 I 1.41 Fo 2 -> So FROM HEISLER CHART Tez Tot (To_Tos) 0.06. 3 320 +(800-320) 0.06 = 348.8k Te = 348.8k| → CENTER TEMPERATURE FOR SPEHERE B IS 348.8k

Al Al Bit to = 1 h Lefx at = (1.133 * (1.413) 1.1 Bi.fo - Il Bi - Hag=0.882 V133 solo Tego Bio FROM teed 11 Bilfo THIS CHART, 20.94 = PcUC To-Too) = MC(To-Tool Q = 0.006x 16001 800 - 320) 4608 J Q = 0.94x Qo -0.94x 4608 Q - 4331.5231 is So HEAT TRANSFER FROM SPHERE “B" 4331-52J