Q1. [50 points] Zr alloy cladding undergoes corrosion when exposed to the reactor environment. According to...

 

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Q1. [50 points] Zr alloy cladding undergoes corrosion when exposed to the reactor environment. According to a simple model, the growth of the oxide scale is initially cubic, then changing to a linear regime at a critical oxide thickness. A Zircaloy component undergoing corrosion under this model is immersed in water at 340C for the first year (i.e., in the first fuel cycle) and at 320C in the following time. Cubic to linear transition oxide thickness is 2.08 m. During the cubic growth time, the oxide thickness satisfies 8 = Ket/3 where t is time in unit of days, and 8 is the oxide thickness in unit of m, and 4533 K = 535 exp (T(K)) where T is the temperature in unit of Kelvin. During the linear growth time, the oxide thickness satisfies S=80+ Kx(t - to) 13800 K = 6107 exp(-T(K) where t, 8, and T have the same meaning as the previous equation. to and do is the time and oxide thickness when the cubic oxidation growth changes to linear growth, respectively. (1) [25] Calculate the time at which the critical oxide thickness 2.08 m would be reached. Does this occur in the first fuel cycle? (2) [25] Calculate the final oxide thickness if the cladding is run for one more year after the oxide transition. Q1. [50 points] Zr alloy cladding undergoes corrosion when exposed to the reactor environment. According to a simple model, the growth of the oxide scale is initially cubic, then changing to a linear regime at a critical oxide thickness. A Zircaloy component undergoing corrosion under this model is immersed in water at 340C for the first year (i.e., in the first fuel cycle) and at 320C in the following time. Cubic to linear transition oxide thickness is 2.08 m. During the cubic growth time, the oxide thickness satisfies 8 = Ket/3 where t is time in unit of days, and 8 is the oxide thickness in unit of m, and 4533 K = 535 exp (T(K)) where T is the temperature in unit of Kelvin. During the linear growth time, the oxide thickness satisfies S=80+ Kx(t - to) 13800 K = 6107 exp(-T(K) where t, 8, and T have the same meaning as the previous equation. to and do is the time and oxide thickness when the cubic oxidation growth changes to linear growth, respectively. (1) [25] Calculate the time at which the critical oxide thickness 2.08 m would be reached. Does this occur in the first fuel cycle? (2) [25] Calculate the final oxide thickness if the cladding is run for one more year after the oxide transition.