Question

2. A nuclear fuel rod with diameter of D=40 mm and length L=1m, has properties of k=1 W/mK, c=1600J/kg-K, and p=400 kg/m² (a)Heat is generated uniformly in the rod with q'"' = 2 x 106 W/m. The rod is first cooled in oil with constant temperature To= 400 K and average heat transfer coefficient h=50 W/m2K. Under steady state, determine the surface temperature of the rod Ts. (10 pts) (b)Now the heat generation in the rod is stopped, where q"' = 0 W/m2. Assume the rod has an initial uniform temperature To= Ts as solved in (a) and is still cooled in oil with constant temperature T = 400 K and h =50 W/m2K. Consider two different cases as following: (25 pts) In case 1, with initial temperature To= Ts, calculate the time required for the surface of the rod to reach 500K. In case 2, if the rod is bonded with a very thin tube for mechanical protection, and the bonding between the rod and the tube has a thermal contact resistance of Rt.c= 0.1 m K/W. Now with initial temperature To= Ts, calculate again the time required for the surface of the rod to reach 500K. Case 1 Case 2 Tah Toh Rod Rod Tube

Answer 1

Solution Given data; The rod with diameter D=yomm :R- De comm length L=im. Giron Properties k = 1 whom k C = 1600*tig.k P = 4ookg/m3 =) Heat is generated. uniformly volumetric heat genovate Wur 1009 axido 3 gate, balance Applying Energy Heat generated in rod = Heat transported by fluid loi) q XII OL = h (MPV) Os - Toc) Ts =) Surface temporaine of sod; Too of fluid Temperature look 9x106 X[46] ?C 50(11x40x 1) (T3-400) X To: sook

(1) (a) when 9" 0. transisent heat Condition then Bi - Biot No prom hRo = 1 Bi - к kt e=at 2 Ro DO R2. For suojace temperature to be T. Ge ६८ - T To Ts-Too Jo (&) T: 500k T: 8ook. G=1.2001 @1-2558 To=400k. Jolg) - 0-6456 (from interpolation) -125583€ (0.6458) 1-2071 500-400 Do 400 0.7184 &= kt P(Ro? 0.784 = ext 400X1600%20 R. ti 183.915

clig (b): when Reic = 0.1 mit/w. tot + R TD heff 1-23 hell Bi = hep Ro k 0.0246 --G=1.00625 : T&170-22115 Jola) : 0.987303 (foon interpolation) By applying Same procedere. G= 28.06 kt 28.06. PCR2 t= 7183.18