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Consider an insulated pipe exposed to the atmosphere. Will the critical radius of insulation be greater on calm days or on windy days? Why?

Short Answer

Expert verified
Explain your reasoning. Answer: The critical radius of insulation will be greater on windy days compared to calm days. This is because the increased convection effect on windy days requires a greater thermal resistance due to conduction to balance it, resulting in a larger critical radius of insulation.

Step by step solution

01

Understanding the concept of critical radius of insulation

Critical radius of insulation is the radius at which the thermal resistance due to convection equals the thermal resistance due to conduction. At the critical radius, any addition to the insulation layer would result in an increase in heat transfer, and any decrease would result in a decrease in heat transfer.
02

Understanding how convection affects heat transfer on calm and windy days

On calm days, the air surrounding the pipe is relatively steady. The convection effect is weaker as air particles don't exchange heat very quickly. On windy days, the air surrounding the pipe is constantly moving and being replaced by cooler air. This results in a stronger convection effect and a higher heat transfer rate.
03

Comparing the critical radius of insulation for calm and windy days

On calm days, the weaker convection effect leads to a smaller critical radius of insulation, as the thermal resistance due to conduction increases more slowly. On windy days, the stronger convection effect requires a greater thermal resistance due to conduction to balance it, leading to a larger critical radius of insulation.
04

Conclusion

The critical radius of insulation will be greater on windy days compared to calm days. This is because the increased convection effect on windy days requires a greater thermal resistance due to conduction to balance it, resulting in a larger critical radius of insulation.

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Most popular questions from this chapter

Consider a wall that consists of two layers, \(A\) and \(B\), with the following values: $k_{A}=1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{A}=8 \mathrm{~cm}\(, \)k_{B}=0.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{B}=5 \mathrm{~cm}\(. If the temperature drop across the wall is \)18^{\circ} \mathrm{C}$, the rate of heat transfer through the wall per unit area of the wall is (a) \(56.8 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(72.1 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(114 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(201 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(270 \mathrm{~W} / \mathrm{m}^{2}\)

One wall of a refrigerated warehouse is \(10.0 \mathrm{~m}\) high and $5.0 \mathrm{~m}\( wide. The wall is made of three layers: \)1.0$-cm-thick aluminum \((k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), 8.0\)-cm-thick fiberglass \((k=0.038\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\), and \(3.0\)-cm-thick gypsum board \((k=0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The warehouse inside and outside temperatures are \(-10^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, and the average value of both inside and outside heat transfer coefficients is \(40 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\). (a) Calculate the rate of heat transfer across the warehouse wall in steady operation. (b) Suppose that 400 metal bolts ( $k=43 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\(, each \)2.0 \mathrm{~cm}\( in diameter and \)12.0 \mathrm{~cm}$ long, are used to fasten (i.e., hold together) the three wall layers. Calculate the rate of heat transfer for the "bolted" wall. (c) What is the percent change in the rate of heat transfer across the wall due to metal bolts?

Consider a house with a flat roof whose outer dimensions are $12 \mathrm{~m} \times 12 \mathrm{~m}\(. The outer walls of the house are \)6 \mathrm{~m}$ high. The walls and the roof of the house are made of 20 -cm-thick concrete $(k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$. The temperatures of the inner and outer surfaces of the house are \(15^{\circ} \mathrm{C}\) and $3^{\circ} \mathrm{C}$, respectively. Accounting for the effects of the edges of adjoining surfaces, determine the rate of heat loss from the house through its walls and the roof. What is the error involved in ignoring the effects of the edges and corners and treating the roof as a $12-\mathrm{m} \times 12-\mathrm{m}\( surface and the walls as \)6-\mathrm{m} \times 12-\mathrm{m}$ surfaces for simplicity?

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