You are asked to solve the temperature distribution from center to outer surface of the cladding by using finite difference method. Note that, conductivity of all regions are temperature dependent. Hint: You may start with initially guested conductivity properties. Table 1 (Doto) Fuel diameter Cladding thickness Water Bulk Temperature Conductivity of Fuel Conductivity of Clad Volumetric heat generation rate Heat Transfer Coefficient of Water The Coefficients Table ) (The Constant Numbers for d te Th kf ke q(r)" Figure 1 hw 90*** 0.8190 cm 0.0654 cm 316 C [1] ke 91" + rdr = 55.65 W/cmK Casel 388 W/cm Case2 388x1.5 W/cm Case3 388x2 W/cm rk + q*** dr dr 1 E W A + BT + 72 exp(-7) mk 0.113 +2.25x10-5T +0.725x10-T (-)* = 0.75 qo" 0.50 qo hw[T(R) - T]: @r = R eq. 1 [2] BC's eq.3 W cmk eg.201

You are asked to solve the temperature distribution from center to outer surface of the cladding by using finite difference method. Fipare 1 r1drdrkdrdT+q=0 Note that, conductivity of all regions are temperature dependent. BC 's Hint:You may start with initially guested conductivity properties. kcdrdTc=hw[Tc(R)Tb:Gr=R ITuble r yoarat