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Chapter 3: Problem 86

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.

Short Answer

Expert verified
The maximum allowable electric current for the air heater is approximately \(13.81\,\text{A}\). The corresponding length of the Nichrome wire that may be used in the heater is about \(50\,\text{m}\), and the power rating of the heater is approximately \(1516.49\,\text{W}\).

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 Ac = 蟺DL, where D is the diameter and L is the length of the wire. The thermal energy transfer rate due to convection, Pc, is then determined as Pc = hAc(Tmax - T鈭�), 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 (蔚) and the temperature of both the surface and its surroundings.

The rate of heat loss due to radiation, Pr, can be expressed using the Stefan-Boltzmann law, described by the equation Pr = 蔚蟽Ac(Tmax4 - Tsur4), where 蟽 is the Stefan-Boltzmann constant and Ac 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 Pe = 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.

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Most popular questions from this chapter

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heat \(\sin k\) to the chip surface in order to increase the surface area available for convection heat transfer. Because of the ease with which it may be manufactured (by taking orthogonal sawcuts in a block of material), an attractive option is to use a heat sink consisting of an array of square fins of width \(w\) on a side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fin width designated as the fin pitch \(S\). The method by which the heat sink is joined to the chip would determine the interfacial contact resistance, \(R_{t, c^{*}}^{n}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a dielectric liquid with \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat \(\operatorname{sink}\) is fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and its characteristic dimensions are \(w=0.25 \mathrm{~mm}\), \(S=0.50 \mathrm{~mm}, L_{f}=6 \mathrm{~mm}\), and \(L_{b}=3 \mathrm{~mm}\). The prescribed values of \(w\) and \(S\) represent minima imposed by manufacturing constraints and the need to maintain adequate flow in the passages between fins. (a) If a metallurgical joint provides a contact resistance of \(R_{t, c}^{\prime \prime}=5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q_{c} ?\) Assume all of the heat to be transferred through the heat sink. (b) It may be possible to increase the heat dissipation by increasing \(w\), subject to the constraint that \((S-w) \geq 0.25 \mathrm{~mm}\), and/or increasing \(L_{f}\) (subject to manufacturing constraints that \(L_{f} \leq 10 \mathrm{~mm}\) ). Assess the effect of such changes.

A thin flat plate of length \(L\), thickness \(t\), and width \(W \geqslant L\) is thermally joined to two large heat sinks that are maintained at a temperature \(T_{o}\). The bottom of the plate is well insulated, while the net heat flux to the top surface of the plate is known to have a uniform value of \(q_{o}^{\prime \prime}\) (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks.

In a test to determine the friction coefficient \(\mu\) associated with a disk brake, one disk and its shaft are rotated at a constant angular velocity \(\omega\), while an equivalent disk/shaft assembly is stationary. Each disk has an outer radius of \(r_{2}=180 \mathrm{~mm}\), a shaft radius of \(r_{1}=20 \mathrm{~mm}\), a thickness of \(t=12 \mathrm{~mm}\), and a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). A known force \(F\) is applied to the system, and the corresponding torque \(\tau\) required to maintain rotation is measured. The disk contact pressure may be assumed to be uniform (i.e., independent of location on the interface), and the disks may be assumed to be well insulated from the surroundings. (a) Obtain an expression that may be used to evaluate \(\mu\) from known quantities. (b) For the region \(r_{1} \leq r \leq r_{2}\), determine the radial temperature distribution \(T(r)\) in the disk, where \(T\left(r_{1}\right)=T_{1}\) is presumed to be known. (c) Consider test conditions for which \(F=200 \mathrm{~N}\), \(\omega=40 \mathrm{rad} / \mathrm{s}, \tau=8 \mathrm{~N} \cdot \mathrm{m}\), and \(T_{1}=80^{\circ} \mathrm{C}\). Evaluate the friction coefficient and the maximum disk temperature.

A particular thermal system involves three objects of fixed shape with conduction resistances of \(R_{1}=1 \mathrm{~K} / \mathrm{W}\), \(R_{2}=2 \mathrm{~K} / \mathrm{W}\) and \(R_{3}=4 \mathrm{~K} / \mathrm{W}\), respectively. An objective is to minimize the total thermal resistance \(R_{\text {tot }}\) associated with a combination of \(R_{1}, R_{2}\), and \(R_{3}\). The chief engineer is willing to invest limited funds to specify an alternative material for just one of the three objects; the alternative material will have a thermal conductivity that is twice its nominal value. Which object (1, 2, or 3 ) should be fabricated of the higher thermal conductivity material to most significantly decrease \(R_{\text {tot }}\) ? Hint: Consider two cases, one for which the three thermal resistances are arranged in series, and the second for which the three resistances are arranged in parallel.

A rod of diameter \(D=25 \mathrm{~mm}\) and thermal conductivity \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) protrudes normally from a furnace wall that is at \(T_{w}=200^{\circ} \mathrm{C}\) and is covered by insulation of thickness \(L_{\text {ins }}=200 \mathrm{~mm}\). The rod is welded to the furnace wall and is used as a hanger for supporting instrumentation cables. To avoid damaging the cables, the temperature of the rod at its exposed surface, \(T_{o}\), must be maintained below a specified operating limit of \(T_{\max }=100^{\circ} \mathrm{C}\). The ambient air temperature is \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the convection coefficient is \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Derive an expression for the exposed surface temperature \(T_{o}\) as a function of the prescribed thermal and geometrical parameters. The rod has an exposed length \(L_{o}\), and its tip is well insulated. (b) Will a rod with \(L_{o}=200 \mathrm{~mm}\) meet the specified operating limit? If not, what design parameters would you change? Consider another material, increasing the thickness of the insulation, and increasing the rod length. Also, consider how you might attach the base of the rod to the furnace wall as a means to reduce \(T_{o}\).
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