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Uniform internal heat generation at q = 5 times 107 W/m3 is occurring in a cylindrical nuclear reactor fuel rod of 50-mm diameter, and under steady-state conditions the temperature distribution is of the form T(r) = a + br2, where T is in degrees Celsius and r is in meters, while a = 800 degree C and b = -4.167 times 105 degree C/m2. The fuel rod properties are k = 30 W/m K, p = 1100 kg/m3, and cp = 800 J/kg K. What is the rate of heat transfer per unit length of the rod at r = 0 (the centerline) and at r = 25 mm (the surface)? If the reactor power level is suddenly increased to q2 = 108 W/m3, what is the initial time rate of temperature change at r = 0 and r = 25 mm?
Uniform internal heat generation at q = 5 times 107 W/m3 is occurring in a cylindrical nuclear reactor fuel rod of 50-mm diameter, and under steady-state conditions the temperature distribution is of the form T(r) = a + br2, where T is in degrees Celsius and r is in meters, while a = 800 degree C and b = -4.167 times 105 degree C/m2. The fuel rod properties are k = 30 W/m K, p = 1100 kg/m3, and cp = 800 J/kg K. What is the rate of heat transfer per unit length of the rod at r = 0 (the centerline) and at r = 25 mm (the surface)? If the reactor power level is suddenly increased to q2 = 108 W/m3, what is the initial time rate of temperature change at r = 0 and r = 25 mm?
Anuclear reactorfuel rod is a cylindrical fuel element containing enricheduraniumused in nuclear reactors. Its diameter is usually around 50mm. The initial time rate of temperature change at r = 25 mm is 3.17 x 10-4 K/s when the reactor power level is suddenly increased to q2 = 108 W/m3.When there is a uniform internal heat generation, thetemperaturedistributionis of the form T(r) = a + br2, where T is in degrees Celsius and r is in meters. We have to determine the rate of heat transfer per unit length of the rod at r = 0 (the centerline) and at r = 25 mm (the surface) and theinitialtime rateof temperature change at r = 0 and r = 25 mm when the reactor power level is suddenly increased to q2 = 108 W/m3.We know that theheat generatedper unit volume of the rod, q = 5 x 107 W/m3.The temperature distribution is T(r) = a + br2. Here a = 800 °C and b = -4.167 x 105 °C/m2. Also, we are given fuel rod properties such as k = 30 W/mK, p = 1100 kg/m3, and cp = 800 J/kgK.The rate of heat transfer per unit length of the rod at r = 0 (the centerline) can be calculated as follows:q = - kA * (dT / dr)r=0 ...(i)where dT / dr is thetemperature gradientand A is the cross-sectional area of the rod. At r = 0, the rate of heat transfer per unit length of the rod can be calculated by substituting the given values in equation(i).A = πr2 = π (0.025)2 = 1.963 x 10-3 m2dT/dr = 2br = 2(-4.167 x 105)(0) = 0q = - (30 x 1.963 x 10-3 x 0) = 0 W/mThe rate of heat transfer per unit length of the rod at r = 25 mm (the surface) can be calculated as follows:A = πr2 = π (0.050/2)2 = 1.963 x 10-3 m2dT/dr = 2br = 2(-4.167 x 105)(0.025) = - 0.208 W/m2Kq = - (30 x 1.963 x 10-3 x (-0.208)) = 0.0123 W/mInitial time rate of temperature change at r = 0 can be calculated using the following formula:q = ρVcp(dT/dt)r=0 ...(ii)where V is the volume of the rod, ρ is the density of the rod, cp is the specific heat of therodand dT/dt is the initial time rate of temperature change.At r = 0, substituting the values in equation (ii),dQ/dt = ρVcp(dT/dt)r=0ρ = 1100 kg/m3V = πr2h = π (0.025)2 (1) = 1.963 x 10-3 m3cp = 800 J/kgKq2 = 108 W/m3dQ/dt = ρVcp(dT/dt)r=0 = 1100 x 1.963 x 10-3 x 800 x dT/dt = 1.409 W/mdT/dt = 1.409 / (1100 x 1.963 x 10-3 x 800) = 1.06 x 10-4 K/sThe initial time rate of temperature change at r = 25 mm can be calculated by using the sameformulaas in equation (ii) and substituting the given values.V = πr2h = π (0.025)2 (1) = 1.963 x 10-3 m3dQ/dt = ρVcp(dT/dt)r=25ρ = 1100 kg/m3V = πr2h = π (0.050/2)2 (1) =...