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Heat is being transferred conductivity from a cylindrical nuclear reactor fuel rod of 50 mm diameter to water at 75°C, under steady state condition, the rate of heat generation within the fuel element is 106 W/m3 and the convective heat transfer coefficient is 1 kW/m2K, the outer surface temperature of the fuel element would be
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- 700 K
- 625 K
- 360 K
- 400 K
Correct Option: C
By applying energy balance, we get
Rate of heat generation = Rate of convective heat transfer
qG × Volume of fuel rod = h0 A (Ts – T0)
| qQ × | d2 × 1 = h0 × πdl(Ts – T0) | 4 |
| = h0(Ts – T0) | 4 |
| (Ts – 75) | 4 |
Ts – 75 = 12.5
Ts = 87.5°C or 360.5K
