Hot air at \(80^{\circ} \mathrm{C}\) is blown over a 2-m \(\times 4\) - \(\mathrm{m}\) flat surface at \(30^{\circ} \mathrm{C}\). If the average convection heat transfer coefficient is \(55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer from the air to the plate, in \(\mathrm{kW}\).

Short Answer

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Question: Determine the rate of heat transfer from the hot air to the flat surface, in kW, given its dimensions (2 m by 4m), temperature difference between the hot air and the surface (80°C and 30°C), and the average convection heat transfer coefficient (55 W/m²K). Answer: The rate of heat transfer is 22 kW.

Step by step solution

01

Calculate the surface area

To calculate the surface area A of the flat surface, we can use the formula: \(A = l \times w\) where: - \(l\) is the length of the surface (2 m) - \(w\) is the width of the surface (4 m) So, the surface area A is: \(A = 2 \times 4 = 8 \, \mathrm{m}^2\)
02

Calculate the rate of heat transfer

Now, we can find the rate of heat transfer \(Q\) using the convection heat transfer formula, which is: \(Q = hA(T_{1}-T_{2})\) where: - \(h = 55 \, \mathrm{W/m^2K}\) - \(A = 8 \, \mathrm{m}^2\) - \(T_{1} = 80^{\circ} \mathrm{C}\) - \(T_{2} = 30^{\circ} \mathrm{C}\) Plugging in the given values, we get: \(Q = 55 (8)(80 - 30)\) \(Q = 55 \times 8 \times 50\) \(Q = 22000 \, \mathrm{W}\) Finally, we must convert the rate of heat transfer from watts (W) to kilowatts (kW): \(Q = \frac{22000}{1000} \, \mathrm{kW}\) \(Q = 22 \, \mathrm{kW}\) Therefore, the rate of heat transfer from the air to the plate is 22 kW.

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Key Concepts

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

Heat Transfer Rate
Understanding heat transfer rate is crucial for grasping the principles of thermodynamics and how energy moves in the form of heat. The heat transfer rate, denoted as Q, refers to the amount of heat energy transferred per unit of time. It is typically measured in watts (W) or kilowatts (kW), as was seen in the given exercise.

The general formula to calculate the heat transfer rate due to convection is:
\[Q = h \times A \times (T_1 - T_2)\]
Here, h represents the convection heat transfer coefficient, A is the surface area through which the heat is being transferred, and (T_1 - T_2) is the temperature difference between the hot and cold mediums, in this case, hot air and the flat surface. The larger the temperature difference or the surface area, or the higher the convection coefficient, the greater the rate of heat transfer.

In our example, the temperature of the hot air and the flat surface differ, allowing heat to transfer from the warmer medium to the cooler one, through the process we know as convection. Calculating the rate of this heat transfer is not only essential in solving physics problems but also in practical applications such as designing heating systems, understanding weather patterns, and engineering thermal management in electronic devices.
Convection Coefficient
The convection heat transfer coefficient, symbolized as h, is a measure of the convective heat transfer between a solid surface and a fluid, such as air or water. It is a reflection of how well heat is being convected and is expressed in units of W/m2·K.

A higher convection coefficient indicates that heat will transfer more efficiently between the fluid and the surface. The value of the convection coefficient depends on factors such as the type of fluid, its velocity, its properties (like viscosity and thermal conductivity), and the surface geometry.

In the context of the exercise, the convection coefficient was given as \(55 \text{ W/m}^2\cdot\text{K}\), signifying the heat transfer effectiveness of air flowing over the surface. It's worth noting that finding the coefficient in the real-world situation might require experimentation or complex calculations using fluid dynamics and heat transfer theories.
Surface Area Calculation
Surface area calculation is essential for determining the extent of a surface that is exposed to a particular process, such as convection heat transfer. In the example, the surface area A of a flat surface is calculated simply by multiplying its length l by its width w, following the formula:\[ A = l \times w \]
This straightforward calculation assumes that the surface is perfectly rectangular and flat, with the entire area being uniformly exposed to the convective flow.

The result of this calculation directly affects the heat transfer rate. In practical scenarios, surface area determination can become more complicated if the surface is irregular or comprises different materials with varying thermal properties. Designers and engineers often need to take these into account when planning heat dissipation in systems, ensuring that surfaces are adequate in size to manage the heat generated effectively.

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Most popular questions from this chapter

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\), and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Heat is lost through a brick wall \((k=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is \(4 \mathrm{~m}\) long, \(3 \mathrm{~m}\) wide, and \(25 \mathrm{~cm}\) thick at a rate of \(500 \mathrm{~W}\). If the inner surface of the wall is at \(22^{\circ} \mathrm{C}\), the temperature at the midplane of the wall is (a) \(0^{\circ} \mathrm{C}\) (b) \(7.5^{\circ} \mathrm{C}\) (c) \(11.0^{\circ} \mathrm{C}\) (d) \(14.8^{\circ} \mathrm{C}\) (e) \(22^{\circ} \mathrm{C}\)

A \(3-\mathrm{m}^{2}\) black surface at \(140^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(35^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(16 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat loss from the surface is (a) \(5105 \mathrm{~W}\) (b) \(2940 \mathrm{~W}\) (c) \(3779 \mathrm{~W}\) (d) \(8819 \mathrm{~W}\) (e) \(5040 \mathrm{~W}\)

Consider a person whose exposed surface area is \(1.7 \mathrm{~m}^{2}\), emissivity is \(0.5\), and surface temperature is \(32^{\circ} \mathrm{C}\). Determine the rate of heat loss from that person by radiation in a large room having walls at a temperature of \((a) 300 \mathrm{~K}\) and (b) \(280 \mathrm{~K}\). Answers: (a) \(26.7 \mathrm{~W}\), (b) \(121 \mathrm{~W}\)

A cylindrical resistor element on a circuit board dissipates \(1.2 \mathrm{~W}\) of power. The resistor is \(2 \mathrm{~cm}\) long, and has a diameter of \(0.4 \mathrm{~cm}\). Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a 24-hour period, \((b)\) the heat flux, and \((c)\) the fraction of heat dissipated from the top and bottom surfaces.

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