Question

265 A person is found dead at 5 p.m. in a room where temperature is 20°C. The (1 temperature of the body is measured to be 25°C when found, and the heat transfer coefficient is estimated to be 8 W/m2.K. Modelling the human body a 30 cm diameter, 1.70 m long cylinder.calculate actual time in hours) of death of the person. Take thermophysical properties of the body :k = 6.08 * W/m.K, p = 900 kg/m3,C = 4000 J/kg.K إجابت A plane wall (k = 45 W/m.K), 10 cm thick,has heat generation at a uniform (2 rate of 8000 KW/m3.The two sides of the wall are maintained at 180°C and 120°C. Neglect end effects; calculate the magnitude and position of * .maximum temperature إجابتك 255 One end of a long rod 3 cm in diameter is inserted into a furnace with the (3 outer end projecting into the outside air. Once the steady state is reached the temperature of the rod is measured at two points, 15 cm apart and found to be 140°C and 100°C, when the atmospheric air is at 30°C with convection coefficient of 20 W/m2.K. Calculate the thermal conductivity of *.the rod material جانتے

Answer 1

Soinu givens Characteristic length (4) = V A 1 Le = 18²2 27142+ +277182 28+ 1 Li- 0:0689 m 2/5 +2.7 water properties = 37+20= 28.5% Tavgi where 37° is normaal temperature of the human body 2 K= 6.08 W/mik, p=900 10g/m3 ,(= 4000 PR h = q W/m2K Time constant て二 P (p V h As 900 X4000 x .0689 8 2= 31005 see. Ti-Too T-Tco e ha baxt Too = 20°C Ti=37°C T= 2roc X 37-20 3loos 25-20 A = 37943.16 see: lit - 10.54hr A actual time of death of the pearson 2.86 AM

120°C Som - 18000 y = 8000 W k = 45 W/mik x loom. d²T - 6 02 -1-24 +C m T-- qn2 tnt (2 2K apply B.ca N=0, T-180 T=/180 = 2 120- at n=0.1m T= 120°c -(.)2+(1-1) + 180 - 60 = 1 1 1 1 192 +16 x 12 =-60 45X2 16, = 28:89] How V18 oIG.2 8000x103 Ta -qnz 2K + 8288,894 + 180 dI for maxm Temperature - - Pan +8288.89 = 0 ZK 8288.89 KK u= q

e-15m = 0.6363 = e-0.15m n= 8288.89 x 45 -0.0466 m 8000 X 103 Tro gooux 103 710466248288.8970466) +180 2xur max Tmax = 373.23°C at location of n=0.0466m U n = 4.66 cm. (Bom left side] Too = 30%, d=0.03m, h= 20 W/m²K T-too mre To-To at x=0.15m T= 140°C 0.15m 140-30 To-To T= 100°C at r = .15+15=0.3 m 100-30 To-To ogn ♡ ! -e-0.3m 0 70 e-0.3m 190 3.013

m-hl 1/KA = 3.013 ph - 9.078 KASTO Trxh k x na 29.078 4 4h kd 9.078 4X20 KX.03 - 9.07% K = 293.74 W/mk dod