Problem statement:
A wire of diameter d -2mm is heated as a result of passing electric current through it. The wire dissipates heat by convection with h= 125W/m²K into the air. If the wire is covered with insulation of thickness 0.2mm of k = 0.175W/ insulation mK, on further addition of?

Solution:

To determine the effect of adding insulation on the heat dissipation of the wire, we need to calculate the heat transfer rate before and after adding insulation.

Step 1: Calculate the heat transfer rate without insulation
The heat transfer rate from the wire to the air by convection can be calculated using the following formula:

q = h*A*ΔT

Where,
q = heat transfer rate (W)
h = convective heat transfer coefficient (125 W/m²K)
A = surface area of the wire (π*d*L)
ΔT = temperature difference between the wire and the air (assumed to be 50°C)

Since the diameter of the wire is given as 2mm, its radius (r) can be calculated as:

r = d/2 = 1mm

The surface area of the wire can be calculated as:

A = π*d*L = π*(2*r)*L = 2π*r*L

where L is the length of the wire.

Assuming a wire length of 1m, the surface area of the wire is:

A = 2π*0.001*1 = 0.00628 m²

Substituting the given values in the above formula, we get:

q = 125*0.00628*50 = 39.25 W

Therefore, the heat transfer rate from the wire to the air without insulation is 39.25 W.

Step 2: Calculate the heat transfer rate with insulation
After adding insulation to the wire, the heat transfer rate can be calculated using the following formula:

q = k*A*ΔT / (d + 2t)

Where,
k = thermal conductivity of insulation (0.175 W/mK)
t = thickness of insulation (0.2mm = 0.0002m)

The surface area of the wire with insulation can be calculated as:

A = π*(d+2t)*L = π*(2*r+2t)*L

Substituting the given values, we get:

A = π*(2*0.001+2*0.0002)*1 = 0.00754 m²

The heat transfer rate can now be calculated as:

q = 0.175*0.00754*50 / (0.002+2*0.0002) = 46.48 W

Therefore, the heat transfer rate from the wire to the air with insulation is 46.48 W.

Step 3: Compare the results
Comparing the two heat transfer rates, we can see that adding insulation to the wire increases the heat transfer rate by:

Δq = q_insulated - q_uninsulated = 46.48 - 39.25 = 7.23 W

Therefore, adding insulation to the wire increases the rate of heat dissipation by 7.23 W.

Conclusion:
Thus, we can