Give step-by-step solution with explanation and final answer:
6. A solid copper conductor of 11 mm diameter carries a current density of 7 A/mm². The conductor is electrically insulated with a thickn

Question

Give step-by-step solution with explanation and final answer: 
6. A solid copper conductor of 11 mm diameter carries a current density of 7 A/mm². The conductor is electrically insulated with a thickness of rubber insulation such that the wire temperature is kept to the minimum possible. Assuming that the surrounding air is at 30°C, calculate:
i) The thickness of insulation
ii) The wire temperature at the axis
iii) The temperature of the outside surface of the insulation
iv) The wire temperature at the axis with the insulation removed and the new steady state reached.

VIPSolutions · Accepted Answer

Let's solve the problem step by step:

**Given:**
- Diameter of copper conductor, $ d = 11 $ mm $\Rightarrow$ radius $ r_1 = 5.5 $ mm $= 0.0055$ m
- Current density, $ J = 7 $ A/mm² $= 7000$ A/cm² $= 7 \times 10^6 $ A/m²
- Surrounding air temperature, $ T_\infty = 30^\circ C $
- The insulation is of rubber, thickness $ t $ (to be calculated for minimum wire temperature)
- The conductor is solid copper
- We need to find:
  - (i) **Thickness of insulation for minimum wire temperature**
  - (ii) **Wire temperature at the axis**
  - (iii) **Temperature at outside surface of insulation**
  - (iv) **Wire temperature (at axis) if insulation is removed**

We need some extra data:
- **Thermal conductivity of rubber** $ (k) $: assume $ k = 0.15 $ W/m·K
- **Heat transfer coefficient to air** ($ h $): assume $ h = 10 $ W/m²·K
- **Electrical resistivity of copper** ($ \rho $): $ 1.7 \times 10^{-8} \Omega $-m

---

### **Step 1: Calculate Total Current**

Cross-sectional area of the wire:
$$
A = \pi r_1^2 = \pi (0.0055)^2 = 9.503 \times 10^{-5} \text{ m}^2
$$
Total current:
$$
I = J \times A = 7 \times 10^6 \times 9.503 \times 10^{-5} = 665.21 \text{ A}
$$

---

### **Step 2: Heat Generated per Meter Length**

Resistance per meter:
$$
R = \frac{\rho \cdot l}{A} = \frac{1.7 \times 10^{-8} \times 1}{9.503 \times 10^{-5}} = 1.789 \times 10^{-4} \ \Omega
$$
Heat generated per meter:
$$
Q = I^2 R = (665.21)^2 \times 1.789 \times 10^{-4} = 79.07 \ \text{W/m}
$$

---

### **Step 3: Find Thickness of Insulation for Minimum Wire Temperature**

For a wire with insulation, the **critical radius** for insulation is:
$$
r_c = \frac{k}{h}
$$
If the wire radius $ r_1 < r_c $, adding insulation **decreases** the total heat loss (increases wire temperature).
If $ r_1 > r_c $, adding insulation **increases** heat loss (decreases wire temperature).

Given that "the wire temperature is kept to the minimum possible," the thickness should be such that the **outer radius of insulation equals the critical radius**.

$$
r_2 = r_c = \frac{k}{h} = \frac{0.15}{10} = 0.015 \ \text{m} = 15 \ \text{mm}
$$
Therefore,
$$
\text{Thickness of insulation} = r_2 - r_1 = 15 - 5.5 = 9.5 \ \text{mm}
$$

---

### **Step 4: Wire Temperature at the Axis**

We will find temperature drop across:
- The copper wire (from axis to surface, $ r = r_1 $)
- The insulation (from $ r_1 $ to $ r_2 $)
- The convective layer (from $ r_2 $ to air)

The total heat loss per meter = $ Q = 79.07 $ W

#### **(a) Temperature drop in copper:**

Radial heat conduction in solid cylinder, from axis ($ r = 0 $) to surface ($ r = r_1 $):
$$
\Delta T_{Cu} = \frac{Q}{4 \pi k_{Cu} L} r_1^2
$$
Where $ k_{Cu} = 385 $ W/m·K, $ L = 1 $ m

$$
\Delta T_{Cu} = \frac{79.07}{4 \pi \times 385 \times 1} (0.0055)^2 = \frac{79.07 \times 3.025 \times 10^{-5}}{4 \pi \times 385}
$$
Calculate denominator:
$$
4 \pi \times 385 = 4 \times 3.1416 \times 385 = 4 \times 1,209.6 = 4,838.4
$$
Calculate numerator:
$$
79.07 \times (0.0055)^2 = 79.07 \times 3.025 \times 10^{-5} = 0.002393
$$

So,
$$
\Delta T_{Cu} = \frac{0.002393}{4,838.4} = 4.95 \times 10^{-7} \text{ K}
$$

This is **negligible** (as expected for copper).

#### **(b) Temperature drop across insulation:**

Heat conduction through cylindrical insulation:

$$
\Delta T_{ins} = \frac{Q}{2 \pi k L} \ln\left(\frac{r_2}{r_1}\right)
$$

Plugging in values:
$$
\Delta T_{ins} = \frac{79.07}{2 \pi \times 0.15 \times 1} \ln\left(\frac{0.015}{0.0055}\right)
$$
$$
2 \pi \times 0.15 = 0.9425
$$
$$
\ln(0.015/0.0055) = \ln(2.727) = 1.003
$$
$$
\Delta T_{ins} = \frac{79.07}{0.9425} \times 1.003 = 83.93 \times 1.003 = 84.18 \text{ K}
$$

#### **(c) Temperature drop across convection:**

$$
\Delta T_{conv} = \frac{Q}{2 \pi r_2 h L}
$$
$$
2 \pi r_2 h = 2 \pi \times 0.015 \times 10 = 0.9425 \times 0.15 = 0.09425
$$
Wait: $ 2\pi r_2 h = 2 \times 3.1416 \times 0.015 \times 10 = 0.9425 $ (corrected, as above)
$$
\Delta T_{conv} = \frac{79.07}{0.9425} = 83.93 \ \text{K}
$$

---

### **Step 5: Final Temperatures**

Let:
- $ T_0 $ = wire axis temperature
- $ T_1 $ = wire surface temperature ($ r_1 $)
- $ T_2 $ = insulation surface temperature ($ r_2 $)
- $ T_\infty $ = air temperature = 30°C

From the above:

1. $ T_0 - T_1 = \Delta T_{Cu} \approx 0 $
2. $ T_1 - T_2 = \Delta T_{ins} = 84.18 $ K
3. $ T_2 - T_\infty = \Delta T_{conv} = 83.93 $ K

So,
$$
T_0 = T_1 = T_2 + 84.18 = T_\infty + 83.93 + 84.18 = 30 + 168.11 = \boxed{198.1^\circ C}
$$

$$
T_2 = T_\infty + \Delta T_{conv} = 30 + 83.93 = \boxed{113.9^\circ C}
$$

---

### **Step 6: No Insulation Case**

Now, let the wire be bare (no insulation), so heat is lost directly from wire surface ($ r_1 $) to air.

$$
Q = h \cdot (2 \pi r_1 L) \cdot (T_1 - T_\infty)
$$
$$
T_1 = T_\infty + \frac{Q}{2 \pi r_1 h L}
$$
$$
2 \pi r_1 h = 2 \times 3.1416 \times 0.0055 \times 10 = 0.34558
$$
$$
T_1 = 30 + \frac{79.07}{0.34558} = 30 + 228.88 = \boxed{258.9^\circ C}
$$

Th

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