/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 172 Rows of the thermoelectric modul... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 172

Rows of the thermoelectric modules of Example \(3.13\) are attached to the flat absorber plate of Problem 3.108. The rows of modules are separated by \(L_{\text {sep }}=0.5 \mathrm{~m}\) and the backs of the modules are cooled by water at a temperature of \(T_{w}=40^{\circ} \mathrm{C}\), with \(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the electric power produced by one row of thermoelectric modules connected in series electrically with a load resistance of \(60 \Omega\). Calculate the heat transfer rate to the flowing water. Assume rows of 20 immediately adjacent modules, with the lengths of both the module rows and water tubing to be \(L_{\text {row }}=20 W\) where \(W=54 \mathrm{~mm}\) is the module dimension taken from Example 3.13. Neglect thermal contact resistances and the temperature drop across the tube wall, and assume that the high thermal conductivity tube wall creates a uniform temperature around the tube perimeter. Because of the thermal resistance provided by the thermoelectric modules, it is no longer appropriate to assume that the temperature of the absorber plate directly above a tube is equal to that of the water.

Short Answer

Expert verified
In conclusion, the electric power produced by one row of thermoelectric modules is approximately \(47444.98 \, \mathrm{W}\), and the heat transfer rate to the flowing water is approximately \(20595.61 \, \mathrm{W}\).

Step by step solution

01

Find the thermal resistance of a single module

To find the thermal resistance of a single module, we use the equation given in Example 3.13: \[ R_{th} = \frac{1}{h A} \] where \(h = 45 \, \mathrm{W/m^2\cdot K}\) and \(A = 20 \times W\), as there are 20 modules in a row. The vertical sides of each module are \(W = 54 \times 10^{-3} \mathrm{m}\), so each module has an area of \(W^2\): \[ A = 20 W^2 = 20 (54 \times 10^{-3})^2 = 0.5832 \, \mathrm{m^2} \] Now, we can calculate the thermal resistance of a single module: \[ R_{th} = \frac{1}{45 \times 0.5832} = 0.03832 \, \mathrm{K/W} \]
02

Find the temperature of the absorber plate

Now we need to find the temperature of the absorber plate. We are given the temperature of the cooling water \(T_w = 40^{\circ} \mathrm{C}\) and the temperature difference across the module from Example 3.13, \(\Delta T = 30^{\circ} \mathrm{C}\). The temperature of the absorber plate can be found using the thermal resistance and the temperature difference: \[ T_{abs} = T_w + \Delta T / R_{th} = 40 + \frac{30}{0.03832} = 823.44^{\circ} \mathrm{C} \]
03

Find the overall temperature difference for the row of modules

Now we can find the overall temperature difference for the row of modules by multiplying the temperature difference for a single module by the total number of modules in a row: \[ \Delta T_{row} = 20 \Delta T = 20 \times 30 = 600^{\circ} \mathrm{C} \]
04

Estimate the electric power generated

We can estimate the electric power generated by one row of thermoelectric modules as follows: \[ P_{elec} = \frac{\Delta T_{row}^2}{R_{th} R_L} \] where \(R_L = 60 \, \Omega\) is the load resistance. Plugging in the values, we get: \[ P_{elec} = \frac{(600)^2}{0.03832 \times 60} = 47444.98 \, \mathrm{W} \]
05

Calculate the heat transfer rate to the flowing water

Finally, we can calculate the heat transfer rate to the flowing water as follows: \[ q = h A (T_{abs} - T_w) \] Plugging in the values, we get: \[ q = 45 \times 0.5832 \times (823.44 - 40) = 20595.61 \, \mathrm{W} \] In conclusion, the electric power produced by one row of thermoelectric modules is approximately \(47444.98 \, \mathrm{W}\), and the heat transfer rate to the flowing water is approximately \(20595.61 \, \mathrm{W}\).

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Key Concepts

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

Thermal Resistance
Thermal resistance is a concept used to quantify how well a material resists the flow of heat. When we think about thermoelectric modules, like the ones in our exercise, thermal resistance is crucial in understanding how efficiently they can convert temperature differences into electric power. It's similar to electrical resistance but for heat flow.

The calculation in the exercise highlights that the thermal resistance, denoted as \( R_{th} \), can be found using the equation \( R_{th} = \frac{1}{h A} \) where \( h \) is the heat transfer coefficient and \( A \) is the surface area through which heat is being transferred. The lower the thermal resistance, the more effectively heat transfers from the hot to the cold side of the module 鈥� this is essential for the thermoelectric module to generate electricity efficiently.
Heat Transfer Rate
The heat transfer rate tells us how quickly heat energy is being moved from one place to another. In the context of the thermoelectric modules, it's the rate at which heat is absorbed from the hot absorber plate and transferred to the cool water. The step-by-step solution uses the formula \( q = h A (T_{abs} - T_w) \) to calculate this.

Thus, the key factors affecting the heat transfer rate are the temperature difference between the absorber plate and water \( (T_{abs} - T_w) \), the area of the modules \( A \), and the heat transfer coefficient \( h \). This rate has to be optimized for the system to generate the expected amount of electric power, as too much heat transfer can cool down the absorber plate too quickly and reduce the efficiency of power generation.
Electric Power Generation
Electric power generation in thermoelectric modules is the process of converting heat energy into electrical energy. According to the Seebeck effect, a temperature difference across the thermoelectric material generates a voltage, which drives an electric current if the circuit is closed. In our exercise, the electric power produced by a row of thermoelectric modules is estimated through the equation \( P_{elec} = \frac{\Delta T_{row}^2}{R_{th} R_L} \).

Here, \( \Delta T_{row} \) represents the overall temperature difference across the module row, and \( R_L \) is the load resistance connected to it. Generating a significant amount of power requires a substantial temperature difference and minimal resistances, both thermal and electrical, in the system.
Temperature Difference
Temperature difference is a driving force in thermoelectric modules. It's essentially the fuel that powers the thermoelectric effect. The bigger the temperature difference across the module, the more electric voltage is generated, leading to more power. In this context, the module generates power proportional to the square of the temperature difference, as shown in the solution with the formula for electric power generation.

For a thermoelectric generator to be efficient, maintaining a high temperature difference is key. Any decrease in this difference can significantly impact the power output. That's why in practical applications, managing the heat input into the system and the cooling rate is critical to maintain the optimal temperature difference for electricity generation.

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Most popular questions from this chapter

The cross section of a long cylindrical fuel element in a nuclear reactor is shown. Energy generation occurs uniformly in the thorium fuel rod, which is of diameter \(D=25 \mathrm{~mm}\) and is wrapped in a thin aluminum cladding. (a) It is proposed that, under steady-state conditions, the system operates with a generation rate of \(\dot{q}=\) \(7 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3}\) and cooling system characteristics of \(T_{\infty}=95^{\circ} \mathrm{C}\) and \(h=7000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Is this proposal satisfactory? (b) Explore the effect of variations in \(\dot{q}\) and \(h\) by plotting temperature distributions \(T(r)\) for a range of parameter values. Suggest an envelope of acceptable operating conditions.

The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is \(3 \mathrm{~mm}\) thick and whose interior surface is maintained at a temperature of \(36^{\circ} \mathrm{C}\). On a calm day the convection heat transfer coefficient at the outer surface is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), but with \(30 \mathrm{~km} / \mathrm{h}\) winds it reaches \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). In both cases the ambient air temperature is \(-15^{\circ} \mathrm{C}\). (a) What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day? (b) What will be the skin outer surface temperature for the calm day? For the windy day? (c) What temperature would the air have to assume on the calm day to produce the same heat loss occurring with the air temperature at \(-15^{\circ} \mathrm{C}\) on the windy day?

Radioactive wastes \(\left(k_{\mathrm{rw}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{ss}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{o}=0.6 \mathrm{~m}\). Heat is generated volumetrically within the wastes at a uniform rate of \(\dot{q}=10^{5} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\). (a) Evaluate the steady-state outer surface temperature, \(T_{s, o}\) (b) Evaluate the steady-state inner surface temperature, \(T_{s, i^{*}}\) (c) Obtain an expression for the temperature distribution, \(T(r)\), in the radioactive wastes. Express your result in terms of \(r_{i}, T_{s, i}, k_{\mathrm{rw}}\), and \(\dot{q}\). Evaluate the temperature at \(r=0\). (d) A proposed extension of the foregoing design involves storing waste materials having the same thermal conductivity but twice the heat generation \(\left(\dot{q}=2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\right)\) in a stainless steel container of equivalent inner radius \(\left(r_{i}=0.5 \mathrm{~m}\right)\). Safety considerations dictate that the maximum system temperature not exceed \(475^{\circ} \mathrm{C}\) and that the container wall thickness be no less than \(t=0.04 \mathrm{~m}\) and preferably at or close to the original design \((t=0.1 \mathrm{~m})\). Assess the effect of varying the outside convection coefficient to a maximum achievable value of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (by increasing the water velocity) and the container wall thickness. Is the proposed extension feasible? If so, recommend suitable operating and design conditions for \(h\) and \(t\), respectively.

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.

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.
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