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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 at a surface are in parallel instead of being in series because both mechanisms occur simultaneously and independently from each other, allowing for separate pathways for heat flow. The total rate of heat transfer at the surface is the summation of the heat transferred through convection and radiation.

Step by step solution

01

Define Convection Resistance

Convection resistance represents the resistance to heat transfer by convection between a solid surface and the surrounding fluid. In other words, it describes the difficulty for heat to transfer from the surface to the fluid (or vice versa) due to the movement of fluid particles.
02

Define Radiation Resistance

Radiation resistance represents the resistance to heat transfer by radiation between a hot surface and its cooler surroundings. It accounts for the heat transfer through thermal radiation, which is the emission of electromagnetic waves (usually infrared) from a heated object.
03

Analyze Heat Transfer Mechanisms

In a real-world scenario, both convection and radiation heat transfer mechanisms occur at the same time at a surface. The heat transfer by convection is usually conducted from a solid surface to the surrounding fluid (or vice versa) and depends on the temperature difference between the surface and the fluid. On the other hand, the heat transfer by radiation is between the surface and the surroundings (typically air) and relies on the temperature difference between the surface and the surroundings.
04

Explain Parallel Resistances

When two resistances are in parallel, they allow for the heat to transfer across them simultaneously. In the case of convection and radiation resistances, both mechanisms occur at the same time and have separate pathways for heat flow. The total rate of heat transfer at the surface (Q) is the summation of the heat transferred through convection (Q_conv) and radiation (Q_rad): \[Q = Q_{conv} + Q_{rad}\]
05

Explain Why Resistances are Not in Series

For resistances to be in series, the heat transfer must pass through both resistances in sequence, one after the other. However, in the case of convection and radiation, the heat transfer does not go through a specific order. Both mechanisms transfer heat simultaneously and independently from each other. In conclusion, the convection and radiation resistances at a surface are in parallel instead of being in series because both mechanisms occur simultaneously and independently from each other, allowing for separate pathways for heat flow.

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Key Concepts

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

Understanding Convection Resistance
Convection resistance refers to the difficulty that heat experiences as it tries to move between a solid surface and the fluid around it. Whether it's air, water, or any other fluid, this process involves heat being transferred as the fluid particles move. The formula for convection resistance is usually expressed as \[ R_{conv} = \frac{1}{h A} \]where \(h\) is the convection heat transfer coefficient, and \(A\) is the area of the surface. This equation reflects how the efficiency of heat transfer increases with a higher convection coefficient and a larger area. Convection can occur naturally, due to buoyancy effects, or be forced, by fans or winds. Understanding this resistance helps in designing systems that efficiently use airflow, like heating or cooling systems, to manage temperature.
  • Higher convection coefficient = less resistance.
  • Larger surface area = less resistance.
Exploring Radiation Resistance
Radiation resistance is quite different from convection resistance as it involves heat transfer via electromagnetic waves. This kind of transfer doesn't require a medium (like air or water) and can even occur in a vacuum. When dealing with radiation resistance, we usually focus on thermal radiation, which is mostly in the infrared spectrum. The resistance to radiation is defined by the equation: \[ R_{rad} = \frac{1}{\epsilon \sigma A (T_s^4 - T_{sur}^4)} \]where \(\epsilon\) is the emissivity of the surface, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the area, \(T_s\) is the surface temperature and \(T_{sur}\) is the surrounding temperature. The emissivity value serves as an indicator of how effectively a surface can emit energy as radiation. Radiation resistance demonstrates how much a surface's ability to radiate heat is impacted by its emissivity and the temperature differences. This is crucial in applications where high temperatures and significant energy emissions are involved.
  • Lower emissivity = higher resistance.
  • Greater temperature difference = less resistance.
The Concept of Parallel Resistances
When convection and radiation resistances are described as parallel, it means they allow for heat to be transferred at the same time through separate avenues. In this arrangement, both heat transfer mechanisms function independently and do not affect each other directly, allowing concurrent heat flow. The total heat transfer rate \(Q\) is therefore the sum of the heat transferred by convection \(Q_{conv}\) and by radiation \(Q_{rad}\):\[ Q = Q_{conv} + Q_{rad} \]This formula underscores the benefit of having these mechanisms in parallel; they can each contribute to the heat transfer without one being a bottleneck to the other. It's similar to having two open gates allowing cars to flow through simultaneously, instead of in a single line.
  • Parallel means simultaneous transfer.
  • Independent paths lead to efficient flow.
Understanding Heat Transfer Mechanisms
Heat transfer mechanisms can be fascinating because they describe how energy moves from one place to another, like from the sun to the Earth's surface or from a radiator to the air in a room. In practical applications, convection and radiation often occur together, making it necessary to grasp how both work and interact. Convection involves heat moving via fluid motion, often visible in heating systems or weather patterns, creating a "conveyor belt" of energy. Radiation allows heat transfer through electromagnetic waves, crucial for instances like warmth from a campfire or sunlight. Both mechanisms have their unique characteristics and operate simultaneously when heat is transferred from a surface to its surroundings. By understanding these basic principles, engineers and scientists can design more efficient systems that use energy wisely, ensure comfort, or protect equipment from overheating. This concept also shows why knowing each mechanism individually is important as it enables better control in practical scenarios.
  • Convection: heat flow through moving fluids.
  • Radiation: heat flow through electromagnetic waves.

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Most popular questions from this chapter

Chilled water enters a thin-shelled 5-cm-diameter, 150-mlong pipe at \(7^{\circ} \mathrm{C}\) at a rate of \(0.98 \mathrm{~kg} / \mathrm{s}\) and leaves at \(8^{\circ} \mathrm{C}\). The pipe is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the pipe is to be insulated with glass wool insulation \((k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in order to decrease the temperature rise of water to \(0.25^{\circ} \mathrm{C}\), determine the required thickness of the insulation.

A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-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 \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), 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 1.4-m-diameter spherical steel tank filled with iced water at \(0^{\circ} \mathrm{C}\) is buried underground at a location where the thermal conductivity of the soil is \(k=0.55 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the tank center and the ground surface is \(2.4 \mathrm{~m}\). For ground surface temperature of \(18^{\circ} \mathrm{C}\), determine the rate of heat transfer to the iced water in the tank. What would your answer be if the soil temperature were \(18^{\circ} \mathrm{C}\) and the ground surface were insulated?

The boiling temperature of nitrogen at atmospheric pressure at sea level ( 1 atm pressure) is \(-196^{\circ} \mathrm{C}\). Therefore, nitrogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at \(-196^{\circ} \mathrm{C}\) until it is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of \(198 \mathrm{~kJ} / \mathrm{kg}\) and a density of \(810 \mathrm{~kg} / \mathrm{m}^{3}\) at 1 atm. Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and \(-196^{\circ} \mathrm{C}\). The tank is exposed to ambient air at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air if the tank is \((a)\) not insulated, \((b)\) insulated with 5 -cm-thick fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and (c) insulated with 2 -cm-thick superinsulation which has an effective thermal conductivity of \(0.00005 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Consider a house that has a 10-m \(\times 20-\mathrm{m}\) base and a 4 -m-high wall. All four walls of the house have an \(R\)-value of \(2.31 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\). The two \(10-\mathrm{m} \times 4-\mathrm{m}\) walls have no windows. The third wall has five windows made of \(0.5-\mathrm{cm}\)-thick glass \((k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), 1.2 \mathrm{~m} \times 1.8 \mathrm{~m}\) in size. The fourth wall has the same size and number of windows, but they are doublepaned with a \(1.5-\mathrm{cm}\)-thick stagnant air space \((k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) enclosed between two \(0.5\)-cm-thick glass layers. The thermostat in the house is set at \(24^{\circ} \mathrm{C}\) and the average temperature outside at that location is \(8^{\circ} \mathrm{C}\) during the seven-month-long heating season. Disregarding any direct radiation gain or loss through the windows and taking the heat transfer coefficients at the inner and outer surfaces of the house to be 7 and \(18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine the average rate of heat transfer through each wall. If the house is electrically heated and the price of electricity is \(\$ 0.08 / \mathrm{kWh}\), determine the amount of money this household will save per heating season by converting the single-pane windows to double-pane windows.

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