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An electrical current is passed through a horizontal copper rod \(2 \mathrm{~mm}\) in diameter and \(30 \mathrm{~cm}\) long, located in an air stream at \(20^{\circ} \mathrm{C}\). If the ends of the rod are also maintained at \(20^{\circ} \mathrm{C}\) and the convective heat transfer coefficient is estimated to be \(30 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), determine the maximum current that can be passed if the midpoint temperature is not to exceed \(50^{\circ} \mathrm{C}\). (i) Ignore thermal radiation. (ii) Include the effect of thermal radiation. For the copper rod, take \(k=386 \mathrm{~W} / \mathrm{m} \mathrm{K}, \varepsilon=0.8\), and electrical resistivity of \(1.72 \times 10^{-8} \Omega \mathrm{m}\).

Step by step solution

01

Understand the System

The rod has a diameter of 2 mm (or 0.002 m) and is 30 cm (or 0.3 m) long. The convective heat transfer coefficient is 30 W/m虏K. The rod's midpoint should not exceed 50掳C when the ambient air is at 20掳C. We must find the maximum current given the thermal and electrical properties of the rod.

02

Calculate Heat Generated

The heat generated by the rod due to electrical resistance can be expressed as \[ q = I^2R \], where \( I \) is the current and \( R \) is the electrical resistance.To calculate \( R \), use the formula \[ R = \frac{\rho L}{A} \], where \( \rho \) is resistivity (1.72 脳 10^{-8} 惟m), \( L \) is the length of the rod, and \( A \) is the cross-sectional area \( (\pi r^2) \) of the rod.

03

Calculate Electrical Resistance

Given the diameter \( d = 0.002 \) m, radius \( r = \frac{0.002}{2} = 0.001 \) m.The cross-sectional area \( A = \pi r^2 = \pi (0.001)^2 \).The resistance \( R = \frac{\rho L}{A} = \frac{1.72 \times 10^{-8} \times 0.3}{\pi (0.001)^2} \).

04

Calculate Heat Loss by Convection

The heat lost by convection from the surface of the rod is \[ q_{conv} = hA_s(T - T_{\infty}) \], where \( h = 30 \) W/m虏K, \( T \) is the rod's midpoint temperature (50掳C), \( T_{\infty} = 20^{\circ} \text{C} \), and \( A_s \) is the surface area of the rod, given by \( A_s = \pi dL \).

05

Equating Heat Generated and Lost (Ignoring Radiation)

Since the system is in steady state, the heat generated by electrical current must equal the heat lost by convection:\[ I^2R = q_{conv} \]. Substitute \( q_{conv} \) and solve for \( I \) to find the maximum current.

06

Consider Thermal Radiation Effect

Include thermal radiation using the Stefan-Boltzmann law: \[ q_{rad} = \varepsilon \sigma A_s (T^4 - T_{\infty}^4) \], where \( \varepsilon = 0.8 \) and \( \sigma = 5.67 \times 10^{-8} \) W/m虏K鈦�.Recalculate the total heat loss \( q_{total} = q_{conv} + q_{rad} \) and equate it to \( I^2R \).

07

Solve for Maximum Current with Radiation

With the inclusion of radiation, solve the equation \( I^2R = q_{conv} + q_{rad} \) for \( I \) to determine the maximum allowable current.

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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 key factor in understanding how heat is generated in electrical systems. It determines how much of the electric energy flowing through a conductor is converted into heat. This occurs as the electrons collide with atoms in the conductor, losing energy as they do so.
Resistance depends on several factors:

• The material of the conductor, which dictates the resistivity (\(\rho\)). Materials like copper have a low resistivity, making them excellent conductors.
• The length of the conductor (\(L\)), with longer conductors exhibiting higher resistance.
• The cross-sectional area (\(A\)). Thinner wires have higher resistance because electrons have less space to move.

In our example, the electrical resistance \(R\) is calculated using the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho = 1.72 \times 10^{-8} \; \Omega\text{m}\), \(L = 0.3\; \text{m}\), and \(A = \pi \left(0.001\, \text{m}\right)^2\). This determines how much heat is generated when the current passes through the copper rod.

Convection

Convection plays a crucial role in how heat is transferred from the copper rod to the surrounding environment. It's a form of heat transfer that occurs in fluids (liquids and gases) due to the movement of the fluid itself.
This process depends on:

• Temperature difference. The greater the difference between the rod's surface temperature and the ambient temperature, the higher the convective heat transfer.
• The convective heat transfer coefficient (\(h\)), which characterizes the efficiency of heat transfer. In this exercise, \(h = 30 \; \text{W/m}^2\text{K}\).
• The surface area of the rod (\(A_s\)), calculated as \(A_s = \pi d L\) where \(d\) is the diameter and \(L\) is the length.

The heat lost through convection is calculated with the formula:
\[ q_{\text{conv}} = h A_s (T - T_\infty) \]
where \(T\) is the rod's midpoint temperature, and \(T_\infty\) is the surrounding air temperature. Convection helps limit the rod's temperature rise, enhancing the safety and stability of electrical systems.

Thermal Radiation

Thermal radiation is another mode of heat transfer that occurs through electromagnetic waves. It doesn't require a medium, unlike conduction or convection, which is why it's significant even in environments like space.
Several factors affect thermal radiation:

• The emissivity (\(\varepsilon\)) of a material, indicating how effectively it emits radiation. Here, \(\varepsilon = 0.8\) for the copper rod means it radiates 80% of the energy a perfect blackbody would.
• The surface area (\(A_s\)) of the rod.
• The difference in the fourth power of the absolute temperatures of the body and its surroundings, as dictated by the Stefan-Boltzmann law.

Using the Stefan-Boltzmann law, the heat loss due to radiation is given by:
\[ q_{\text{rad}} = \varepsilon \sigma A_s (T^4 - T_\infty^4) \]
where \(\sigma = 5.67 \times 10^{-8} \; \text{W/m}^2\text{K}^4\). In our context, including radiation provides a more accurate estimation of the rod's heat loss, crucial for ensuring the current does not raise the midpoint temperature beyond 50掳C.