/*! 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

Measurements show that steady-state conduction through a plane wall without heat generation produced a convex temperature distribution such that the midpoint temperature was \(\Delta T_{o}\) higher than expected for a linear temperature distribution. Assuming that the thermal conductivity has a linear dependence on temperature, \(k=k_{o}(1+\alpha T)\), where \(\alpha\) is a constant, develop a relationship to evaluate \(\alpha\) in terms of \(\Delta T_{o}, T_{1}\), and \(T_{2}\).

Determine the parallel plate separation distance \(L\), above which the thermal resistance associated with molecule-surface collisions \(R_{t, m-s}\) is less than \(1 \%\) of the resistance associated with molecule-molecule collisions, \(R_{t, m-m}\) for (i) air between steel plates with \(\alpha_{t}=0.92\) and (ii) helium between clean aluminum plates with \(\alpha_{t}=0.02\). The gases are at atmospheric pressure, and the temperature is \(T=300 \mathrm{~K}\).

A very long rod of \(5-\mathrm{mm}\) diameter and uniform thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is subjected to a heat treatment process. The center, 30 -mm-long portion of the rod within the induction heating coil experiences uniform volumetric heat generation of \(7.5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The unheated portions of the rod, which protrude from the heating coil on either side, experience convection with the ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that there is no convection from the surface of the rod within the coil. (a) Calculate the steady-state temperature \(T_{o}\) of the rod at the midpoint of the heated portion in the coil. (b) Calculate the temperature of the rod \(T_{b}\) at the edge of the heated portion.

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.

Consider the flat plate of Problem \(3.112\), but with the heat sinks at different temperatures, \(T(0)=T_{o}\) and \(T(L)=T_{L}\), and with the bottom surface no longer insulated. Convection heat transfer is now allowed to occur between this surface and a fluid at \(T_{\infty}\), with a convection coefficient \(h\). (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. (c) For \(q_{o}^{\prime \prime}=20,000 \mathrm{~W} / \mathrm{m}^{2}, T_{o}=100^{\circ} \mathrm{C}, T_{L}=35^{\circ} \mathrm{C}\), \(T_{\infty}=25^{\circ} \mathrm{C}, k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(L=100 \mathrm{~mm}, t=5 \mathrm{~mm}\), and a plate width of \(W=\) \(30 \mathrm{~mm}\), plot the temperature distribution and determine the sink heat rates, \(q_{x}(0)\) and \(q_{x}(L)\). On the same graph, plot three additional temperature distributions corresponding to changes in the following parameters, with the remaining parameters unchanged: (i) \(q_{o}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), (ii) \(h=200\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (iii) the value of \(q_{o}^{\prime \prime}\) for which \(q_{x}(0)=0\) when \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
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