Topic: Heat Transfer
For a current wire of 20 mm diameter exposed to air (h=20 W/m^2K), maximum heat dissipation occurs when thickness of insulation (k=0.5 W/mK) is
- A. 20 mm
- B. 25 mm
- C. 10 mm
- D. none of the above
Answer: D. none of the above
Explanation: The critical radius of insulation for a cylinder is calculated as r_c = k / h = 0.5 / 20 = 0.025 m = 25 mm. Since the wire diameter is 20 mm, its radius is 10 mm. Therefore, the required thickness of insulation to reach the critical radius is 25 mm - 10 mm = 15 mm. This value is not listed in the options.