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Why are the convection and the radiation resistances at a surface in parallel instead of being in series?

Short Answer

Expert verified
Answer: Convection and radiation resistances are considered to be in parallel at a surface because heat can transfer simultaneously through both modes of heat transfer, allowing for more efficient and simultaneous heat flow. In contrast, series resistances would require heat to transfer sequentially through each resistance, creating a more limited and slower heat flow process.

Step by step solution

01

Understanding Convection and Radiation Resistances

Convection and radiation are two different methods of heat transfer. Convection resistance is the resistance to heat transfer between a heat source and a fluid or gas in contact with the heat source (such as air). Radiation resistance is the resistance to heat transfer between a heat source and other surfaces or objects within its surroundings without direct contact (for example, through thermal radiation). These resistances represent the difficulty and efficiency of heat transfer through these different modes.
02

Interactions of Convection and Radiation Resistances

At a surface, both convection and radiation resistances exist simultaneously since heat is being transferred from the surface to its surroundings through both convection (to the surrounding fluid or gas) and radiation (to the surrounding objects).
03

Parallel versus Series Resistances

Resistances can be present in either parallel or series, depending on how they affect the flow of heat. In series resistances, heat must pass through one resistance and then the other, meaning the heat transfer is sequential and the sum of the resistances is the total resistance. In parallel resistances, heat transfer can occur simultaneously through both resistances, and the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances.
04

Convection and Radiation Resistances in Parallel

Convection and radiation resistances at a surface are considered to be in parallel because heat can transfer simultaneously through both modes of heat transfer. The surface does not have to release heat exclusively through convection before releasing it through radiation or vice versa. Instead, heat can be transferred through both convection and radiation at the same time.
05

Calculating Total Resistance

To calculate the total resistance in a system with convection and radiation resistances in parallel, you can use the following formula: \[\frac{1}{R_{total}} = \frac{1}{R_{convection}} + \frac{1}{R_{radiation}}\] R_total is the total resistance, R_convection is the convection resistance, and R_radiation is the radiation resistance. In summary, convection and radiation resistances are considered to be in parallel at a surface because heat can transfer simultaneously through both modes of heat transfer. This is contrary to series resistances, where heat transfer would need to occur sequentially through each resistance.

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Most popular questions from this chapter

Steam at \(280^{\circ} \mathrm{C}\) flows in a stainless steel pipe $(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ whose inner and outer diameters are \(5 \mathrm{~cm}\) and \(5.5 \mathrm{~cm}\), respectively. The pipe is covered with \(3-\mathrm{cm}\)-thick glass wool insulation $(k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\(. Heat is lost to the surroundings at \)5^{\circ} \mathrm{C}$ by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of $22 \mathrm{~W} / \mathrm{m}^{2}$. . Taking the heat transfer coefficient inside the pipe to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation.

A 2.2-mm-diameter and 14-m-long electric wire is tightly wrapped with a \(1-\mathrm{mm}\)-thick plastic cover whose thermal conductivity is $k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire, and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2}\), \(\mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

A 5 -m-diameter spherical tank is filled with liquid oxygen $\left(\rho=1141 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1.71 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\( at \)-184^{\circ} \mathrm{C}$. It is observed that the temperature of oxygen increases to \(-183^{\circ} \mathrm{C}\) in a 144-hour period. The average rate of heat transfer to the tank is (a) \(124 \mathrm{~W}\) (b) \(185 \mathrm{~W}\) (c) \(246 \mathrm{~W}\) (d) \(348 \mathrm{~W}\) (e) \(421 \mathrm{~W}\)

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially that of the surrounding air, i.e., \(20^{\circ} \mathrm{C}\). Its width is \(5.0 \mathrm{~cm}\); thickness is \(1.0 \mathrm{~mm}\); thermal conductivity is $200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(; and base temperature is \)40^{\circ} \mathrm{C}$. The heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2}\). . Estimate the fin temperature at a distance of \(5.0 \mathrm{~cm}\) from the base and the rate of heat loss from the entire fin.

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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