91Ó°ÊÓ

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2}+\mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the valie of the maximum temperature when the thickness is used?

Step by step solution

01

Determine Heat Generated in the Cable

The heat generated in the cable due to electrical resistance can be determined using the formula for power loss in a resistor: \[ P = I^2 imes R \]where:- \( I = 700 \) A is the current,- \( R = 6 \times 10^{-4} \) Ω/m is the resistance per meter.Substituting the values, the power loss per meter of cable is:\[ P = (700)^2 \times (6 \times 10^{-4}) = 294 \text{ W/m} \]

02

Calculate Surface Temperature without Insulation

The surface temperature of the cable can be found using the equation for heat transfer by convection and radiation: \[ P = h A (T_s - T_{ ext{env}}) \]where:- \( P = 294 \text{ W/m} \) is the power loss,- \( h = 25 \, \text{W/m}^2 \cdot K \) is the heat transfer coefficient,- \( A = \pi d \) is the surface area of the cable per unit length with \( d = 5 \times 10^{-3} \) m (diameter),- \( T_s \) is the surface temperature,- \( T_{\text{env}} = 30^{\circ} C \) is the environment temperature.First, calculate \( A \): \[ A = \pi \times 5 \times 10^{-3} = 0.0157 \text{ m}^2 \text{/m} \]Substitute values into the heat transfer equation and solve for \( T_s \): \[ 294 = 25 \times 0.0157 \times (T_s - 30) \]\[ T_s = \frac{294}{0.0157 \times 25} + 30 \approx 66.9^{\circ} C \]

03

Calculate Surface Temperature with Insulation

With a thin coating of insulation, the heat transfer equation is modified to account for the contact resistance:\[ P = \frac{T_{ ext{cable}} - T_{ ext{env}}}{R_c + \frac{1}{h A}} \]where:- \( R_c = 0.02 \text{ m}^2 K/W \) is the contact resistance.Rearranging for \( T_{\text{cable}} \):\[ T_{\text{cable}} = P \bigg(R_c + \frac{1}{h A}\bigg) + T_{\text{env}} \]Substituting known values:\[ T_{\text{cable}} = 294 \times (0.02 + \frac{1}{25 \times 0.0157}) + 30 \approx 73.4^{\circ} C \]

04

Calculate Insulation Surface Temperature

The surface temperature of the insulation can be found similar to the surface temperature of cable:\[ P = h A (T_{\text{insulation}} - T_{env}) \]And we have:\[ T_{\text{insulation}} = \frac{P}{h A} + T_{\text{env}} \]Using the value for surface area and power loss calculated earlier:\[ T_{\text{insulation}} = \frac{294}{25 \times 0.0157} + 30 \approx 60.1^{\circ} C \]

05

Find Insulation Thickness for Minimum Temperature

The critical thickness of insulation can be determined using:\[ r_{\text{crit}} = \frac{k}{h} \]where:- \( k = 0.5 \text{ W/m} \cdot K \) is the thermal conductivity of the insulation.Substitute the given values:\[ r_{\text{crit}} = \frac{0.5}{25} \approx 0.02 \text{ m} \]Thus, the critical thickness for minimum surface temperature is roughly 20 mm or 0.02 m.

06

Calculate Maximum Temperature with Critical Thickness

Now calculate the maximum insulation temperature using the critical thickness. The formula for the temperature at the inner surface of insulation is the same as previously used:\[ T_{\text{max}} = \frac{P}{h ( \pi (d+2r_{\text{crit}}))} + T_{\text{env}} \]Replacing \( d + 2r_{\text{crit}} \) with \( 0.05 + 2 \times 0.02 = 0.09 \text{ m} \):\[ T_{\text{max}} = \frac{294}{25 \times \pi \times 0.09} + 30 \approx 37.3^{\circ} C \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrical Resistance

Electrical resistance is a measure of how much an object opposes the flow of electric current. It is expressed in ohms (Ω). In this exercise, the stainless steel cable has a resistance of \(6 \times 10^{-4}\ \text{Ω/m}\). This resistance causes electrical energy to be converted into heat, which is also known as Joule heating or resistive heating.

• The power loss due to this resistance can be calculated using the formula \( P = I^2 \times R \).
• In the case of the cable, with a current \( I = 700 \text{ A} \), the power loss per meter is \( 294 \text{ W/m} \).

This power loss is significant because it determines how much heat the cable generates, affecting the cable's temperature.

Thermal Insulation

Thermal insulation is used to reduce heat transfer between objects or environments with different temperatures. In this exercise, thin electrical insulation is applied to the cable. This insulation adds a layer of resistance to heat flow, expressed here as a contact resistance \( R_c = 0.02 \ \text{m}^2 \cdot \text{K/W} \).

• When insulation is applied, the heat transfer equation must be modified to account for this additional thermal resistance.
• The equation used is \( P = \frac{T_{\text{cable}} - T_{\text{env}}}{R_c + \frac{1}{h A}} \), where \( h \) is the heat transfer coefficient.

Thus, the insulation significantly affects both the cable's exterior temperature and the overall thermal resistance.

Critical Thickness

The concept of critical thickness of insulation is important when discussing thermal insulation. It refers to the thickness where additional insulation starts to result in an increase in heat loss due to increased surface area outweighing the resistance provided by the insulation.
The formula used for critical thickness \( r_{\text{crit}} \) is \( r_{\text{crit}} = \frac{k}{h} \), where \( k \) is the thermal conductivity of the material.

• For this exercise, \( k = 0.5 \ \text{W/m} \cdot \text{K} \) and the heat transfer coefficient \( h = 25 \ \text{W/m}^2 \cdot \text{K} \).
• Substituting these gives a critical thickness of approximately 0.02 m or 20 mm.

Understanding this aspect is crucial in insulation design, ensuring that insulation reduces heat loss efficiently without inadvertently causing a temperature rise.

Surface Temperature Calculation

Surface temperature calculation involves determining the temperature of an object's surface when it is subjected to heat transfer. It considers both convective and radiative heat losses.
To calculate the surface temperature of the cable, the equation \( P = h A (T_s - T_{\text{env}}) \) is used. Here's a breakdown:

• \( P \) is the power loss per meter due to resistance, which is 294 W/m.
• \( h \) is the heat transfer coefficient.
• \( A \) represents the surface area per meter of the cable.
• \( T_s \) is the surface temperature we aim to find, while \( T_{\text{env}} \) is the surrounding environment's temperature.

For the bare cable, applying rearrangement of the formula yields a surface temperature \( T_s \) of about 66.9°C. When insulated, similar calculations help find the temperatures accounting for different thermal resistances. Understanding this allows for precise control of temperatures in systems subject to heat transfer.