Help-Calculate the conductivity of insulating material.

Question

Example 3.38. An electric cable of 12 mm diameter is insulated to increase the current capacity. Due to insulation the current carrying capacity is increased by 15% without increasing cable surface temperature above 70°C. The environmental temperature is 30°C. Assume that the heat transfer coefficient from the bare or insulated cable is 14 W/m2.K. Calculate the conductivity of insulating material.

Answer 1

To solve this problem, we need to use the formula for the heat transfer rate from a cylindrical surface:

q = h × A × ΔT

where:

q = heat transfer rate (W) h = heat transfer coefficient (W/m2.K) A = surface area of the cylinder (m2) ΔT = temperature difference between the surface and the environment (K)

We can assume that the cable is infinitely long, so the surface area is given by:

A = 2πrL

where:

r = radius of the cable (m) L = length of the cable (m)

We can also assume that the insulation has a uniform thickness, so the radius of the insulated cable is:

r' = r + δ

where:

δ = insulation thickness (m)

The current carrying capacity of the cable is proportional to its cross-sectional area, so we can use the following formula to calculate the new radius of the cable:

r'' = r × √(1 + 0.15)

The new surface area of the cable is:

A'' = 2πr''L

The temperature difference between the surface of the cable and the environment is:

ΔT = 70 - 30 = 40 K

The heat transfer coefficient is given as:

h = 14 W/m2.K

Substituting these values into the heat transfer rate formula, we get:

q = h × A'' × ΔT

Solving for A'', we get:

A'' = q / (h × ΔT)

Substituting the values given in the problem, we get:

A'' = (0.15 × 2πr2L) / (14 × 40)

Simplifying this expression, we get:

r''2 = r2 × (1 + 0.15) r'' = r × √(1 + 0.15) r'' = 1.049r

A'' = 2πr''L A'' = 2π(1.049r)(L)

Substituting this value of A'' into the heat transfer rate formula, we get:

q = h × A'' × ΔT q = 14 × 2π(1.049r)(L) × 40

We know that the heat transfer rate is equal to the electrical power dissipated in the cable:

q = I2R

where:

I = current flowing in the cable (A) R = resistance of the cable (Ω)

The resistance of the cable can be calculated using the formula:

R = ρL / A

where:

ρ = resistivity of the cable material (Ω.m) L = length of the cable (m) A = cross-sectional area of the cable (m2)

Since we know the diameter of the cable, we can calculate its cross-sectional area:

A = πr2

We also know that the current carrying capacity of the cable has increased by 15%, so the new current is:

I'' = I × 1.15

Substituting these values into the resistance formula, we get:

R = ρL / A R = ρL / (πr2)

Substituting the value of I'' into the power formula, we get:

q = I''2R

Substituting the value of q into the heat transfer rate formula, we get:

I''2R = h × A'' × ΔT

Solving for ρ, we get:

ρ = I''2Rπr2 / (h × A'' × ΔT × L)

Substituting the values given in the problem, we get:

ρ = (I × 1.15)2