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Warm air is blown over the inner surface of an automobile windshield to defrost ice accumulated on the outer surface of the windshield. Consider an automobile windshield \(\left(k_{w}=\right.\) \(0.8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R})\) with an overall height of 20 in and thickness of \(0.2\) in. The outside air ( \(1 \mathrm{~atm})\) ambient temperature is \(8^{\circ} \mathrm{F}\) and the average airflow velocity over the outer windshield surface is \(50 \mathrm{mph}\), while the ambient temperature inside the automobile is \(77^{\circ} \mathrm{F}\). Determine the value of the convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield, necessary to cause the accumulated ice to begin melting. Assume the windshield surface can be treated as a flat plate surface.

Short Answer

Expert verified
Based on the given information, determine the convection heat transfer coefficient of the warm air blowing over the inner surface of the windshield to initiate melting of the ice accumulated on the outer surface. Overall temperature difference across the windshield: ΔT = Tin - Tout = 77°F - 8°F = 69°F

Step by step solution

01

Determine the overall temperature difference across the windshield

Given the ambient temperature inside the automobile as \(T_{in} = 77^{\circ}F\) and the outside ambient temperature as \(T_{out} = 8^{\circ}F\). The overall temperature difference across the windshield is given by: \(\Delta T = T_{in} - T_{out}\)

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Key Concepts

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

Heat Transfer
Heat transfer is a phenomenon wherein thermal energy moves from one object or material to another due to a temperature difference between them. It occurs in three primary forms: conduction, convection, and radiation. In the context of defrosting an automobile windshield, convection plays an essential role.

Convection is the movement of heat through fluids (which include liquids and gases) accompanied by the motion of the fluid itself. When warm air is blown over the inner surface of a windshield, it causes the transfer of heat from the air to the windshield. This process is designed to raise the temperature of the windshield above the freezing point of water, resulting in the ice melting.

In creating content for students, it's important to acknowledge that understanding these heat transfer mechanisms can deeply influence the effectiveness of real-world applications, such as the defrosting process in vehicles.
Thermal Conductivity
Thermal conductivity, denoted as 'k', is a material's ability to conduct heat. It is often measured in terms of energy transferred per unit time over a unit area with a unit temperature gradient. In the exercise, the thermal conductivity of the windshield is given as \(0.8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R}\).

The higher the thermal conductivity, the more efficiently a material can transfer heat. For a windshield, this means that thermal conductivity will affect how quickly the heat from the warm air inside the car can defrost the ice on the outer surface. Materials like metals generally have high thermal conductivity, while insulating materials like glass have lower thermal conductivity, which must be taken into account when designing defrosting systems.
Defrosting Process
The defrosting process refers to the removal of frost or ice from a surface, which can be crucial for an automobile windshield to ensure visibility and safety. The process can be initiated by increasing the temperature of the surface above the melting point of ice, usually achieved through convection heat transfer. Warm air is blown across the surface, causing the ice to absorb heat and undergo a phase change from solid to liquid.

To ensure a smooth defrosting process, understanding the properties of the windshield material, such as its thermal conductivity, and the effectiveness of the heating mechanism, is important. Effective defrosting systems also depend on an adequately high convection heat transfer coefficient, which determines the rate at which heat is transferred from the warm air to the windshield surface.
Temperature Difference
Temperature difference is a driving force for heat transfer. In the given exercise, it's crucial to determine the temperature difference across the windshield to understand how the defrosting will occur. The difference between the inside ambient temperature (\(77^{\text{\small \degree}}F\)) and the outside ambient temperature (\(8^{\text{\small \degree}}F\)) gives us the overall temperature gradient that affects heat transfer. This gradient, \(\Delta T = T_{in} - T_{out}\), represents the 'push' needed to move thermal energy from the warmer interior to the colder exterior.

A larger temperature difference typically results in a higher rate of heat transfer, assuming other conditions remain constant. Hence, for efficient defrosting, the temperature of the warm air inside the automobile should not only be above the melting point of ice but should also create a sufficient temperature difference to facilitate quick melting.

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Most popular questions from this chapter

Reconsider Prob. 7-67E. Using EES (or other) software, investigate the effects of air temperature and wind velocity on the rate of heat loss from the arm. Let the air temperature vary from \(20^{\circ} \mathrm{F}\) to \(80^{\circ} \mathrm{F}\) and the wind velocity from \(10 \mathrm{mph}\) to \(40 \mathrm{mph}\). Plot the rate of heat loss as a function of air temperature and of wind velocity, and discuss the results.

Consider a house that is maintained at a constant temperature of \(22^{\circ} \mathrm{C}\). One of the walls of the house has three singlepane glass windows that are \(1.5 \mathrm{~m}\) high and \(1.8 \mathrm{~m}\) long. The glass \((k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\) thick, and the heat transfer coefficient on the inner surface of the glass is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Now winds at \(35 \mathrm{~km} / \mathrm{h}\) start to blow parallel to the surface of this wall. If the air temperature outside is \(-2^{\circ} \mathrm{C}\), determine the rate of heat loss through the windows of this wall. Assume radiation heat transfer to be negligible. Evaluate the air properties at a film temperature of \(5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Oil at \(60^{\circ} \mathrm{C}\) flows at a velocity of \(20 \mathrm{~cm} / \mathrm{s}\) over a \(5.0\)-m-long and \(1.0-\mathrm{m}\)-wide flat plate maintained at a constant temperature of \(20^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the oil to the plate if the average oil properties are: \(\rho=880 \mathrm{~kg} / \mathrm{m}^{3}, \mu=0.005 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\).

Consider laminar flow over a flat plate. Will the friction coefficient change with distance from the leading edge? How about the heat transfer coefficient?

A 12 -ft-long, \(1.5-\mathrm{kW}\) electrical resistance wire is made of \(0.1\)-in-diameter stainless steel \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\). The resistance wire operates in an environment at \(85^{\circ} \mathrm{F}\). Determine the surface temperature of the wire if it is cooled by a fan blowing air at a velocity of \(20 \mathrm{ft} / \mathrm{s}\). For evaluations of the air properties, the film temperature has to be found iteratively. As an initial guess, assume the film temperature to be \(200^{\circ} \mathrm{F}\).

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