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PotatoMerchant
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Homework Statement
Homework Equations
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Rate of Heat Transfer for a Two-Layer System
- Thread starter PotatoMerchant
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SUMMARY
The discussion focuses on calculating the rate of heat transfer in a two-layer system using the correct equation for heat flow. The appropriate formula for a single layer is given as 2πrLq = Q = 2πkL(ΔT/Δlnr), where Q represents total heat flow in watts, and q is heat flow per unit area. The analogy to resistors in series is emphasized, with ΔT representing the voltage drop across each layer and Q as the current flow. The overall temperature difference across both layers is specified as 155°C.
PREREQUISITES- Understanding of heat transfer principles
- Familiarity with thermal conductivity (k) and its role in heat flow
- Knowledge of logarithmic functions as they apply to heat transfer equations
- Basic concepts of electrical circuits, specifically series resistors
- Study the derivation of the heat transfer equation for concentric cylinders
- Learn about thermal resistance and its application in multi-layer systems
- Explore numerical methods for solving heat transfer problems in complex geometries
- Investigate the impact of material properties on heat transfer rates
Students in thermal engineering, mechanical engineers, and professionals involved in heat transfer analysis will benefit from this discussion.
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- #2
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It doesn't look like this has been done correctly. To begin with, the starting equation is incorrect.
For a single layer, it should read:
2πrLq =Q=2πkL\frac{ΔT}{Δlnr}
where Q is the total heat flow (watts) and q is the heat flow per unit area. This equation applies to each of the concentric layers. The problem is analogous to having two resistors in series, where ΔT is like the voltage drop across each layer and Q is the current flow through the resistors. You have to solve for the current flow, given the overall voltage drop across both layers (155 C).
For a single layer, it should read:
2πrLq =Q=2πkL\frac{ΔT}{Δlnr}
where Q is the total heat flow (watts) and q is the heat flow per unit area. This equation applies to each of the concentric layers. The problem is analogous to having two resistors in series, where ΔT is like the voltage drop across each layer and Q is the current flow through the resistors. You have to solve for the current flow, given the overall voltage drop across both layers (155 C).
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