Q1: Maximum heat dissipation occurs from a steel wire (k = 0.5 W/m. k) of 15 mm diameter exposed to air (h = 20W/m² .k), when the insulation thickness is __________ mm.

ANS:B - 25

To determine the insulation thickness that results in maximum heat dissipation from the steel wire, we need to find the critical radius of insulation. The critical radius of insulation (rc​) is the insulation thickness that maximizes heat dissipation. The formula for the critical radius of insulation for a cylindrical shape exposed to convection is: rc​=k/h​ where:

• rc​ is the critical radius of insulation,
• k is the thermal conductivity of the steel wire, and
• ℎh is the convective heat transfer coefficient of the surrounding air.

Given:

• Diameter of the steel wire (D) = 15 mm
• Radius of the steel wire (r) = D/2​ = 7.5 mm
• Thermal conductivity of the steel wire (k) = 0.5 W/m·K
• Convective heat transfer coefficient of air (ℎ) = 20 W/m²·K

First, let's calculate the critical radius of insulation (rc​): rc​=200.5​=0.025m=25mm Therefore, the critical radius of insulation is 25 mm. This means that to achieve maximum heat dissipation from the steel wire, the insulation thickness should be 25 mm.