The surface temperature of the uranium fuel rod can be calculated using the rate of heat generation and convective cooling. Substituting the given values into the formula, we find that the surface temperature of the rod is 308 K. Option 1 is correct.
The surface temperature of the uranium fuel rod can be calculated using the rate of heat generation and convective cooling. The rate of heat generation is given as 4x10⁷ W/m³ and the convective heat transfer coefficient is 1000 W/m².K.
The rate of heat transfer is given by Q = hxAxdT, where Q is the rate of heat transfer, h is the convective heat transfer coefficient, A is the surface area of the rod, and dT is the temperature difference.
Given that the radius of the rod is 5 mm and the temperature of the cooling liquid is 25°C, we can calculate the surface temperature using the formula T_surface = T_liquid + (Q / (hxA)). Substituting the values, we get T_surface = 25 + (4x10⁷ / (1000 x pi x (0.005)^2)).
Calculating this value gives T_surface = 308 K. Therefore, the surface temperature of the rod is 308 K.
Hence, 1. is the correct option.
--The given question is incomplete, the complete question is
"A cylindrical uranium fuel rod of radius 5 mm in a nuclear reactor is generating heat at the rate of 4× 107 W/m3. The rod is cooled by a liquid (convective heat transfer coefficient 1000 W/m2.K) at 25°C. At steady state, the surface temperature (in K) of the rod is 1 308 2. 398 3. 418 4. 448."--