91Ó°ÊÓ

An air heater may be fabricated by coiling Nichrome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter \(D=\) \(1 \mathrm{~mm}\), electrical resistivity \(\rho_{e}=10^{-6} \Omega \cdot \mathrm{m}\), thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and emissivity \(\varepsilon=0.20\). The heater is designed to deliver air at a temperature of \(T_{\infty}=50^{\circ} \mathrm{C}\) under flow conditions that provide a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for the wire. The temperature of the housing that encloses the wire and through which the air flows is \(T_{\text {sur }}=50^{\circ} \mathrm{C}\). If the maximum allowable temperature of the wire is \(T_{\max }=1200^{\circ} \mathrm{C}\), what is the maximum allowable electric current \(I\) ? If the maximum available voltage is \(\Delta E=110 \mathrm{~V}\), what is the corresponding length \(L\) of wire that may be used in the heater and the power rating of the heater? Hint: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this assumption.

Step by step solution

01

Find the maximum heat generated by the wire.

We need to find the electric power generated by the wire. By Ohm's law, we can write the power as: \[P_\text{e} = I^2 R\] Where \(I\) is the current and \(R\) is the resistance of the wire. Since our goal is to find the maximum allowable current, let's express the resistance \(R\) in terms of wire length \(L\) and resistivity \(\rho_e\), \[R = \frac{\rho_e L}{A} = \frac{\rho_e L}{\pi (D/2)^2}\] Now, substituting the resistance back to the power equation, we get: \[P_\text{e} = I^2 \frac{\rho_e L}{\pi(D/2)^2}\]

02

Find the heat dissipated by convection and radiation.

The heat dissipated by the wire can be broken into two components - convection and radiation. Find the heat transfer by convection: \[P_\text{c} = h A_c (T_\text{max} - T_\infty)\] Where \(A_c = \pi D L\) is the surface area for convection. Find the heat transfer by radiation: \[P_\text{r} = \varepsilon \sigma A_c (T_\text{max}^4 - T_\text{sur}^4)\] Where \(\sigma\) is the Stefan-Boltzmann constant, which equals \(5.67 \times 10^{-8}\, \text{W} / (\text{m}^2 \cdot \text{K}^4)\). Now, sum up these two components, since they both dissipate heat: \[P_\text{total} = P_\text{c} + P_\text{r} = h \pi D L (T_\text{max} - T_\infty) + \varepsilon \sigma \pi D L (T_\text{max}^4 - T_\text{sur}^4)\]

03

Equating the electrical power and total power heat dissipation to find the maximum current.

Since the electrical power generated by the wire should be equal to the total heat dissipated by convection and radiation, we can equate them: \[I^2 \frac{\rho_e L}{\pi (D/2)^2} = h \pi D L (T_\text{max} - T_\infty) + \varepsilon \sigma \pi D L (T_\text{max}^4 - T_\text{sur}^4)\] Simplify the equation and solve for the current \(I\): \[I = \sqrt{\frac{h \pi^2 D^3 (T_\text{max} - T_\infty) + \varepsilon \sigma \pi^2 D^3 (T_\text{max}^4 - T_\text{sur}^4)}{\rho_e}}\] Substitute the given values to find the maximum allowable current \(I\): \[I = \sqrt{\frac{250 \pi^2 (0.001)^3 (1200 - 50) + 0.20 (5.67 \times 10^{-8}) \pi^2 (0.001)^3 ((1200 + 273)^4 - (50 + 273)^4)}{10^{-6}}}\] \[I \approx 13.81\,\text{A}\] So, the maximum allowable electric current is about \(13.81\,\text{A}\).

04

Calculate the length of the Nichrome wire and the power rating of the heater.

Using Ohm's law, we can find the resistance of the wire: \[R = \frac{\Delta E}{I} = \frac{110\,\text{V}}{13.81\,\text{A}} \approx 7.96\,\Omega\] Calculate the length of the wire and substituting the given values: \[L = \frac{\pi (D/2)^2}{\rho_e} R\] \[L = \frac{\pi (0.001/2)^2}{10^{-6}} (7.96) \approx 50\,\text{m}\] Now, find the power rating of the heater: \[P_\text{total} = I^2 R \approx (13.81)^2 (7.96) \approx 1516.49\,\text{W}\] So, the corresponding length of the wire that may be used in the heater is around \(50\,\text{m}\) and the power rating of the heater is approximately \(1516.49\,\text{W}\).

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Key Concepts

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

Convection Heat Transfer

Convection is a mechanism of heat transfer where thermal energy is carried away by a fluid, such as air or water, which is in motion. Accompanying an exercise like the one provided, where air flows over a Nichrome wire, the heat from the wire is transferred to the air via convection. The rate at which heat is dissipated can be calculated using a convection coefficient, represented by the symbol 'h'. This coefficient depends on factors such as fluid velocity, its properties, the surface geometry, and the temperature difference between the surface and the fluid.

For a cylindrical wire, the convective surface area is given by the formula A_c = πDL, where D is the diameter and L is the length of the wire. The thermal energy transfer rate due to convection, P_c, is then determined as P_c = hA_c(T_max - T_∞), signifying that higher convection coefficients and larger temperature differences will lead to greater heat transfer rates.

Radiation Heat Transfer

Radiation heat transfer refers to the emission of electromagnetic waves that carry energy away from the emitting surface. Unlike convection, radiation does not require a medium to transfer heat and can occur in a vacuum. In our exercise, the Nichrome wire also loses heat through radiation, which depends on the surface's emissivity (ε) and the temperature of both the surface and its surroundings.

The rate of heat loss due to radiation, P_r, can be expressed using the Stefan-Boltzmann law, described by the equation P_r = εσA_c(T_max⁴ - T_sur⁴), where σ is the Stefan-Boltzmann constant and A_c is the convective surface area, the same as for convection heat transfer. Materials with a higher emissivity will radiate more heat, and the temperature difference to the fourth power greatly influences the radiation heat loss.

Electrical Power Generation

The generation of electrical power in the context of the Nichrome wire heater is a conversion of electrical energy into thermal energy. This is described by Joule heating, where an electric current passing through a resistor (in this case, the Nichrome wire) converts the electrical energy into heat. The formula P_e = I²R succinctly demonstrates this relationship, where I is the electric current and R is the resistance of the wire.

This principle is widely utilized in electrical power appliances, where the control of current and resistance allows for the precise generation of desired amounts of heat. In industrial applications, such as electrical power generation plants, electromechanical conversions (like in turbines and generators) are used instead to convert mechanical energy into electrical power.

Thermal Conductivity

Thermal conductivity is an intrinsic physical property of a material that quantifies its ability to conduct heat. Within the given exercise, the thermal conductivity (k) of Nichrome governs how efficiently thermal energy is distributed through the wire. Materials with high thermal conductivity, such as metals, transfer heat quickly, whereas insulators, with low thermal conductivity, do so much slower.

In our scenario, the assumption is made that the temperature variations within the Nichrome wire are negligible, implying that heat conduction along the wire is efficient enough that it maintains a uniform temperature. However, for materials with low thermal conductivity, temperature gradients can occur, leading to differential heating along the source which can greatly affect the efficiency and performance of the heating element.