One of the thermoelectric modules of Example 3.13 is...

One of the thermoelectric modules of Example $3.13$ is installed between a hot gas at $T_{\infty, 1}=450^{\circ} \mathrm{C}$ and a cold gas at $T_{\infty, 2}=20^{\circ} \mathrm{C}$. The convection coefficient associated with the flowing gases is $h=h_{1}=h_{2}=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ while the electrical resistance of the load is $R_{e, l o s d}=4 \Omega$. (a) Sketch the equivalent thermal circuit and determine the electric power generated by the module for the situation where the hot and cold gases provide convective heating and cooling directly to the module (no heat sinks). (b) Two heat sinks $(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$; see sketch), each with a base thickness of $L_{b}=4 \mathrm{~mm}$ and fin length $L_{f}=20 \mathrm{~mm}$, are soldered to the upper and lower sides of the module. The fin spacing is $3 \mathrm{~mm}$, while the solder joints each have a thermal resistance of $R_{t, c}^{\prime \prime}=2.5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}$. Each heat sink has $N=11$ fins, so that $t=2.182 \mathrm{~mm}$ and $S=5.182 \mathrm{~mm}$, as determined from the requirements that $W=(N-1) S+t$ and $S-t=3 \mathrm{~mm}$. Sketch the equivalent thermal circuit and determine the electric power generated by the module. Compare the electric power generated to your answer for part (a). Assume adiabatic fin tips and convection coefficients that are the same as in part (a).

One of the thermoelectric modules of Example $3.13$ is installed between a hot gas at $T_{\infty, 1}=450^{\circ} \mathrm{C}$ and a cold gas at $T_{\infty, 2}=20^{\circ} \mathrm{C}$. The convection coefficient associated with the flowing gases is $h=h_{1}=h_{2}=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ while the electrical resistance of the load is $R_{e, l o s d}=4 \Omega$.
(a) Sketch the equivalent thermal circuit and determine the electric power generated by the module for the situation where the hot and cold gases provide convective heating and cooling directly to the module (no heat sinks).
(b) Two heat sinks $(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$; see sketch), each with a base thickness of $L_{b}=4 \mathrm{~mm}$ and fin length $L_{f}=20 \mathrm{~mm}$, are soldered to the upper and lower sides of the module. The fin spacing is $3 \mathrm{~mm}$, while the solder joints each have a thermal resistance of $R_{t, c}^{\prime \prime}=2.5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}$. Each heat sink has $N=11$ fins, so that $t=2.182 \mathrm{~mm}$ and $S=5.182 \mathrm{~mm}$, as determined from the requirements that $W=(N-1) S+t$ and $S-t=3 \mathrm{~mm}$. Sketch the equivalent thermal circuit and determine the electric power generated by the module. Compare the electric power generated to your answer for part (a). Assume adiabatic fin tips and convection coefficients that are the same as in part (a).

Key Concepts

Heat Sink Design and Fin Optimization

Heat sink design focuses on enhancing heat dissipation through increased surface area and improved conduction pathways, often using fins to augment the natural convective process. The effectiveness of a heat sink is influenced by its material properties, geometric dimensions, and arrangement of fins. Optimizing fin geometry, such as length, spacing, and thickness, is crucial to minimize thermal resistance and improve the overall thermal management in systems that require efficient cooling or heating.

Convective Heat Transfer

Convective heat transfer is the process of heat exchange between a surface and a moving fluid, governed by Newton's law of cooling. The convection coefficient, or heat transfer coefficient, quantifies the efficiency of this transfer. This concept is critical when assessing the thermal performance of systems exposed to flowing gases or liquids, where it determines how effectively heat is removed or supplied to the surfaces involved.

Thermal Resistance and Interface Resistances

Thermal resistance represents the opposition to heat flow and is analogous to electrical resistance in circuits. It includes contributions from conduction through materials, convection at boundaries, and contact resistances at interfaces, such as solder joints. Understanding and calculating these resistances is vital for predicting temperature gradients and optimizing the coupling between thermal and electrical domains in energy conversion systems.

Thermoelectric Energy Conversion

Thermoelectric energy conversion involves generating electrical power directly from a temperature difference across a thermoelectric module, utilizing the Seebeck effect. In this process, carriers in the semiconductor material are driven by a thermal gradient, creating a voltage that can perform electrical work. This concept is essential for understanding how thermal energy is transformed into electricity in systems where temperature differences are inherent.

Equivalent Thermal Circuit Analysis

Equivalent thermal circuit analysis is a method used to model heat transfer processes by analogizing heat flow to electrical current and thermal resistances to electrical resistors. This technique allows the complex conduction and convection paths to be represented in a simplified circuit framework, where series and parallel combinations of thermal resistances help in analyzing temperature distributions and heat flow rates throughout the system.

Transcript

00:01 So as you see in this question, it is to be shown that the work output of the car not engine.

00:06 Okay, car not engine work will be maximum at tp equal to root of t .o.

00:20 Into t .a .f.

00:23 And also, we have to show that the maximum work output of the car not engine is in the case when w equals to ctaf, 1 minus t, root of t0 on root of t a f square when we see this car not engine we have the adabatic combustion chamber here the heat exchanger here so the combustion gas will leave the combustion chamber and enter the heat exchanger it will enter the heat at the adiabatic flame temperature that is t a since the chamber is adibating and the fuel is burned completely.

01:10 So that is there.

01:11 So since the fuel is burned completely, it will have the temperature of daf.

01:17 Now the combustion gases experiences no change.

01:21 So the gases experience no change here in the chemical composition as the flow through the heat exchanges.

01:30 Therefore, we can trade the combustion gases as a gas steam with constant.

01:37 Specific heat right so the combustion gases can be considered as the gas team noting that the lead heat exchanger involves no work interaction the energy balance equation for single stream flow would be if you see here we have q minus m of hg minus hi is equals to m dot c into tp minus t a f so where q is the negative of the heat supplied to the engine and that is q h equal to minus q that is equal to m of c into t a f minus t okay so that is a second equation that we have now we will be seeing that the work output of the car not engine can be expressed by w equal to qh into 1 minus t o upon tp which is equal to m into c into taf minus tp that into 1 minus t o upon tp...

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