Chapter 4: Problem 54
A fluc passing hot exhaust gases has a square cross section, \(300 \mathrm{~mm}\) to a side. The walls are constructed of refractory brick \(150 \mathrm{~mm}\) thick with a thermal conductivity of \(0.85 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Calculate the heat loss from the flue per unit length when the interior and exterior surfaces are maintained at 350 and \(25^{\circ} \mathrm{C}\), respectively. Use a grid spacing of \(75 \mathrm{~mm}\).
Short Answer
Step by step solution
Understand the problem
Identify dimensions and material properties
Apply Fourier's Law of Heat Conduction
Calculate the cross-sectional area for heat transfer
Calculate the temperature difference
Compute heat loss for one wall
Compute total heat loss for all walls
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fourier's Law of Heat Conduction
- \(q\) represents the rate of heat transfer.
- \(k\) is the thermal conductivity of the material.
- \(A\) is the cross-sectional area through which heat flows.
- \(\Delta T\) is the temperature difference between the two surfaces.
- \(d\) is the thickness of the material the heat is passing through.
Thermal Conductivity
Temperature Gradient
Heat Loss Calculation
- The thermal conductivity \(k\) is 0.85 W/m·K.
- The area \(A\) is 0.3 m² (per unit length along the flue).
- The temperature difference \(\Delta T\) is 325 K.
- The thickness \(d\) is 0.15 m.