Will a hot horizontal plate whose back side is insulated cool faster or slower when its hot surface is facing down instead of up?

Short Answer

Expert verified
Answer: The hot plate cools down faster when the hot surface is facing up. This is due to the more efficient natural convection in this configuration, which causes a higher rate of heat loss. In both orientations, the rate of heat loss due to thermal radiation remains the same.

Step by step solution

01

Understand the different cooling mechanisms on the hot plate

In this problem, the hot plate experiences two main cooling mechanisms: convection (heat loss to the surrounding fluid like air) and radiation (heat loss in the form of electromagnetic waves). To compare the cooling rates when the hot surface is facing up and down, we need to analyze these two mechanisms.
02

Analyze convection when the hot surface is facing up

When the hot surface is facing up, natural convection takes place as the warm air adjacent to the hot surface becomes less dense and rises. This creates a continuous flow of cooler air around the hot surface causing the plate to cool. In this case, the buoyancy force promotes the flow of cooler air, resulting in a higher rate of heat loss from the plate.
03

Analyze convection when the hot surface is facing down

When the hot surface is facing down, the warm air molecules adjacent to the hot surface rise, but they are trapped close to the hot surface because of the opposing buoyancy force. As a result, the air in direct contact with the hot surface stays relatively warm, and the rate of heat removal from the plate is slower.
04

Analyze radiation when the hot surface is facing up and down

Thermal radiation is emitted in all directions from the hot surface, regardless of its orientation. Thus, the rate of heat loss due to radiation will be the same in both cases, whether the hot surface is facing up or down. The small differences in radiation intensity due to emissivity and the angle at which the radiation is emitted is negligible for this exercise.
05

Compare the cooling rates

Considering the heat loss due to both convection and radiation, we can conclude that when the hot surface of the horizontal plate is facing up, the cooling rate will be faster. This is because the plate will lose heat more efficiently through natural convection when the hot surface is facing up compared to when it is facing down, even though the radiation heat loss is the same for both scenarios.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Natural Convection
Natural convection is a process where a fluid (such as air or water) moves and transfers heat as a result of temperature differences within the fluid. When heat is added to a fluid, it expands and becomes less dense than the surrounding cooler fluid. This difference in density creates a buoyancy force that causes the warmer, less dense fluid to rise, while cooler, denser fluid descends to replace it. Imagine placing a heated plate in a room; the air in contact with the plate warms up, becomes lighter, and flows upward, followed by the cooler air which descends to warm up next. This cycle forms a convection current.

Understanding natural convection is essential in numerous applications such as heating and cooling systems, weather patterns, and industrial processes. In our textbook exercise, we specifically look at how the orientation of a hot plate affects the natural convection cooling mechanism, with the plate cooling more effectively when the hot surface is facing up due to the promotion of buoyancy-driven air flow.
Thermal Radiation
Thermal radiation is the emission of electromagnetic waves from all objects that have a temperature above absolute zero. All bodies emit this energy according to their temperature, and it occurs without the need for a medium, meaning it can even happen in a vacuum. A warm object, like our hot plate, emits this radiation in all directions. The intensity of this radiation depends on the object's temperature and its emissivity, which is the measure of an object's ability to emit thermal radiation compared to a perfect black body. Thermal radiation is particularly significant in space applications and many manufacturing processes.

In the context of our exercise, regardless of whether the hot plate is facing up or down, thermal radiation contributes equally to its cooling because radiation is not influenced by the plate’s orientation. This was affirmed in the steps of the solution, where the conclusion is that radiant heat loss is the same in both scenarios.
Heat Transfer Analysis
Heat transfer analysis is the assessment of how heat moves from one body or substance to another, or within a body. There are three core mechanisms of heat transfer: conduction (through direct contact), convection (through fluid flows), and radiation (emission of electromagnetic waves). When analyzing a system, an engineer might consider factors such as material properties, surface temperatures, geometry, and surrounding environmental conditions to predict how quickly heat will be lost or gained.

The exercise provided gives us a practical application of heat transfer analysis: understanding the variations in cooling rates of a horizontal plate with different orientations. By dissecting the mechanisms at play, namely convection and radiation, the analysis helps to predict that a plate facing upward will cool faster, primarily due to more effective natural convection.
Buoyancy Force
The buoyancy force is the upward force exerted on an object when it is placed in a fluid, and it is the driving mechanism behind natural convection. This force is a result of pressure differences in the fluid and is directed against the gravitational pull. The Archimedes' principle states that the buoyancy force applied to an object in a fluid is equivalent to the weight of the fluid displaced by the object.

Relating this back to our hot plate scenario, when the hot surface is facing upwards, the warm air expands, displaces cooler air, and experiences a buoyancy force that helps it rise away from the plate, allowing cooler air to take its place and facilitate cooling. However, when the hot surface faces downwards, the warm air rises but cannot easily move away due to the plate obstructing the movement, hence slowing down the cooling process due to reduced buoyancy assistance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a 3-m-high rectangular enclosure consisting of two surfaces separated by a \(0.1-\mathrm{m}\) air gap at \(1 \mathrm{~atm}\). If the surface temperatures across the air gap are \(30^{\circ} \mathrm{C}\) and \(-10^{\circ} \mathrm{C}\), determine the ratio of the heat transfer rate for the horizontal orientation (with hotter surface at the bottom) to that for vertical orientation.

Why are finned surfaces frequently used in practice? Why are the finned surfaces referred to as heat sinks in the electronics industry?

A spherical tank \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with an inner diameter of \(3 \mathrm{~m}\) and a wall thickness of \(10 \mathrm{~mm}\) is used for storing hot liquid. The hot liquid inside the tank causes the inner surface temperature to be as high as \(100^{\circ} \mathrm{C}\). To prevent thermal burns on the skin of the people working near the vicinity of the tank, the tank is covered with a \(7-\mathrm{cm}\) thick layer of insulation \((k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the outer surface is painted to give an emissivity of \(0.35\). The tank is located in a surrounding with air at \(16^{\circ} \mathrm{C}\). Determine whether or not the insulation layer is sufficient to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\) to prevent thermal burn hazards. Discuss ways to further decrease the outer surface temperature. Evaluate the air properties at \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

A \(1.5\)-m-diameter, 4-m-long cylindrical propane tank is initially filled with liquid propane, whose density is \(581 \mathrm{~kg} / \mathrm{m}^{3}\). The tank is exposed to the ambient air at \(25^{\circ} \mathrm{C}\) in calm weather. The outer surface of the tank is polished so that the radiation heat transfer is negligible. Now a crack develops at the top of the tank, and the pressure inside drops to \(1 \mathrm{~atm}\) while the temperature drops to \(-42^{\circ} \mathrm{C}\), which is the boiling temperature of propane at \(1 \mathrm{~atm}\). The heat of vaporization of propane at \(1 \mathrm{~atm}\) is \(425 \mathrm{~kJ} / \mathrm{kg}\). The propane is slowly vaporized as a result of the heat transfer from the ambient air into the tank, and the propane vapor escapes the tank at \(-42^{\circ} \mathrm{C}\) through the crack. Assuming the propane tank to be at about the same temperature as the propane inside at all times, determine how long it will take for the tank to empty if it is not insulated.

Consider an \(L \times L\) horizontal plate that is placed in quiescent air with the hot surface facing up. If the film temperature is \(20^{\circ} \mathrm{C}\) and the average Nusselt number in natural convection is of the form \(\mathrm{Nu}=C \mathrm{Ra}_{L}^{n}\), show that the average heat transfer coefficient can be expressed as $$ \begin{aligned} &h=1.95(\Delta T / L)^{1 / 4} 10^{4}<\mathrm{Ra}_{L}<10^{7} \\ &h=1.79 \Delta T^{1 / 3} \quad 10^{7}<\mathrm{Ra}_{L}<10^{11} \end{aligned} $$

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free