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1. Calculate the maximum linear heat rate in a rod if the bulk coolant temperature is 280C and the centerline temperature is 2,800C. Assume the following properties (15 points): = 0.40[cm] W 1;=0.39[cm] ho =0.75 cm-K 1 = 0.47[cm] w ky = 0.032 cm-K ke=0.12 h=4.2 cm-K cm-K W W Ts=400C U=lomis Too =200c KD=25mm properties of Nak(25-75.1.) at flim temperature of Tg = 300c -6m2 P= 775 kg Im?, V = 0,308 X 100m2, Pr = 0.0108 k = 22.1 W/m kg cp=1000. 6 3 1kg k Reynold number for flow - 0.413 Re= PUD Re=775 x 168 0.025 0.308 X166 X 775 Re=129 8701.3 using correlation suggested by Ishiguro et al. (1979) N4 = 1.125 [Rexp8]0.413 hD = 1254 Rex pr]0,413 hxo.025 = 1.125[1298701. 3 X 0.0108] 22.1 h=51320.18 W/mK Heat generation rate & jx FoL=h(10L) ( Ts-Tow j * 1X0.025 = ( X ) " 51320.18 ( 1 X0,025) (400-200) = 1642245602 W/m3