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A plane wall has a thermal conductivity of 1.15 W/mK. If the inner surface is at 1100°C and the outer surface is at 350°C, then the design thickness (in meter) of the wall to maintain a steady heat flux of 2500 W/m2 should be
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- 0.345 m.
- 0.245 m.
- 1.145 m.
- 1.345 m.
Correct Option: A
| Q = KA | dx |
| q = | = 2500 W/m² | A |
| 2500 = K | dx |
| 2500 = 1.15 × = | x |
x = 0.345 m.
