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Two flow passages with different cross-sectional shapes, one circular another square, are each centered in a square solid bar of the same dimension and thermal conductivity. Both configurations have the same length, \(T_{1}\), and \(T_{2}\). Determine which configuration has the higher rate of heat transfer through the square solid bar for \((a) a=1.2 b\) and \((b) a=2 b\).

Short Answer

Expert verified
Answer: In both cases, (a) \(a=1.2b\) and (b) \(a=2b\), the square flow passage configuration has a higher rate of heat transfer than the circular flow passage configuration.

Step by step solution

01

Calculate the cross-sectional areas

Calculate the cross-sectional area of the circular flow passage and square flow passage by using the formulas: \(A_{circle} = \pi (\frac{b}{2})^2\) and \(A_{square} = a^2\) respectively.
02

Comparing the areas for Case (a) \(a=1.2b\)

Substitute the values of \(a=1.2b\) into the equations for the cross-sectional areas and compare them. $$ A_{circle} = \pi (\frac{b}{2})^2 = \frac{\pi b^2}{4} $$ $$ A_{square} = (1.2b)^2 = 1.44b^2 $$ Now, compare \(A_{circle}\) and \(A_{square}\) to determine which is larger: $$ \frac{A_{square}}{A_{circle}} = \frac{1.44b^2}{\frac{\pi b^2}{4}} = \frac{1.44}{\frac{\pi}{4}} \approx 1.83 $$ Since \(A_{square} \approx 1.83 A_{circle}\), the square flow passage has a larger cross-sectional area, and hence a higher rate of heat transfer in case (a).
03

Comparing the areas for Case (b) \(a=2b\)

Substitute the values of \(a=2b\) into the equations for the cross-sectional areas and compare them. $$ A_{circle} = \pi (\frac{b}{2})^2 = \frac{\pi b^2}{4} $$ $$ A_{square} = (2b)^2 = 4b^2 $$ Now, compare \(A_{circle}\) and \(A_{square}\) to determine which is larger: $$ \frac{A_{square}}{A_{circle}} = \frac{4b^2}{\frac{\pi b^2}{4}} = \frac{4}{\frac{\pi}{4}} \approx 5.09 $$ Since \(A_{square} \approx 5.09 A_{circle}\), the square flow passage has a larger cross-sectional area, and hence a higher rate of heat transfer in case (b). To conclude, the square flow passage configuration has a higher rate of heat transfer in both cases, \((a) a=1.2b\) and \((b) a=2b\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross-sectional Area Comparison
Understanding the difference in cross-sectional areas is crucial when analyzing heat transfer in different configurations. Cross-sectional area is simply the surface area of a shape when viewed from a specific direction. In terms of heat transfer, larger areas allow more space for heat flow, which typically leads to a higher rate of heat transfer.
To compare areas:
  • Circular Flow Passage: The area is calculated using the formula \(A_{circle} = \pi \left(\frac{b}{2}\right)^2\). This formula derives from the area of a circle, \(\pi r^2\), where the radius \(r\) is \(\frac{b}{2}\).
  • Square Flow Passage: The area uses the formula \(A_{square} = a^2\). Here, the side length \(a\) is adjusted according to the problem's description.
For both cases presented, substituting the values of \(a\) related to \(b\), indicates that the square passage always results in a larger cross-sectional area compared to the circular passage.
Thermal Conductivity
Thermal conductivity is a measure of how well a material can conduct heat. Different materials have different thermal conductivities, which means that the efficiency of heat transfer can greatly depend on the material of the flow passage.
The quantity of heat that can be conducted through a material is influenced by:
  • The material's inherent thermal conductivity value, often denoted by \(k\).
  • The temperature gradient across the material, as heat flow is driven by temperature differences.
  • The cross-sectional area through which the heat flows.
In this exercise, since both configurations have the same thermal conductivity value, we focus on how the shape and area impact the heat transfer rate. Nonetheless, in real applications, choosing materials with higher thermal conductivity can significantly enhance heat transfer.
Flow Passage Shape
The shape of the flow passage plays a vital role in heat transfer processes. Different shapes can significantly affect the rate at which heat transfers through the material.
Here are some considerations:
  • Circular Shapes: Commonly used due to their structural efficiency, but may not always provide the maximum area for heat transfer compared to other shapes.
  • Square Shapes: As shown in the examples, they tend to offer more surface area for a given dimension, which enhances the rate of heat transfer.
Thus, selecting the optimal flow passage shape needs consideration of both the intended thermal application and the practical design constraints. The choice between square and circular depends not only on the maximal potential area but also the ease of manufacturing and maintenance.
Heat Transfer Rate
The heat transfer rate is an essential metric in many engineering and scientific applications. It refers to the amount of heat energy transferred per unit time through a material. This rate is influenced by multiple factors, including cross-sectional area, thermal conductivity, and shape.
Calculating and comparing the heat transfer rate in different configurations helps determine the optimal design for specific thermal applications. Larger cross-sectional areas will often support higher rates of heat transfer, assuming the material and conditions remain constant.
In the comparison of circular and square passages:
  • The square passage has a significantly larger area, resulting in a proportional increase in the heat transfer rate.
  • As the size of \(a\) increases relative to \(b\), the disparity in cross-sectional areas increases, further amplifying the difference in heat transfer rates.
Ultimately, optimizing heat transfer rates requires balancing the shape and material properties with functional and structural requirements.

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Most popular questions from this chapter

A cylindrical pin fin of diameter \(0.6 \mathrm{~cm}\) and length of \(3 \mathrm{~cm}\) with negligible heat loss from the tip has an efficiency of 0.7. The effectiveness of this fin is (a) \(0.3\) (b) \(0.7\) (c) 2 (d) 8 (e) 14

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum ( \(k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a 10 -cm-thick layer of insulation \((k=\) \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the exterior surface of the insulation is (a) \(13^{\circ} \mathrm{C}\) (b) \(9^{\circ} \mathrm{C}\) (c) \(2^{\circ} \mathrm{C}\) (d) \(-3^{\circ} \mathrm{C}\) (e) \(-12^{\circ} \mathrm{C}\)

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 \(20-\mathrm{cm}-\) 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?

The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?

Superheated steam at an average temperature \(200^{\circ} \mathrm{C}\) is transported through a steel pipe \(\left(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{o}=8.0 \mathrm{~cm}\right.\), \(D_{i}=6.0 \mathrm{~cm}\), and \(L=20.0 \mathrm{~m}\) ). The pipe is insulated with a 4-cm thick layer of gypsum plaster \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulated pipe is placed horizontally inside a warehouse where the average air temperature is \(10^{\circ} \mathrm{C}\). The steam and the air heat transfer coefficients are estimated to be 800 and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate \((a)\) the daily rate of heat transfer from the superheated steam, and \((b)\) the temperature on the outside surface of the gypsum plaster insulation.

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