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Chapter 1: Problem 54

Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to be baked in an oven when the convection feature is disabled. For this situation, the free convection coefficient associated with the cake and its pan is \(h_{\mathrm{fr}}=3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The oven air and wall are at temperatures \(T_{\infty}=T_{\text {sur }}=180^{\circ} \mathrm{C}\). Determine the heat flux delivered to the cake pan and cake batter when they are initially inserted into the oven and are at a temperature of \(T_{i}=24^{\circ} \mathrm{C}\). If the convection feature is activated, the forced convection heat transfer coefficient is \(h_{\mathrm{fo}}=27 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the heat flux at the batter or pan surface when the oven is operated in the convection mode? Assume a value of \(0.97\) for the emissivity of the cake batter and pan.

Short Answer

Expert verified
In summary, the heat flux delivered to the cake pan and batter when operating in free convection mode is \(468\,\text{W}/\text{m}^{2}\) and in forced convection mode, it is \(4212\,\text{W}/\text{m}^{2}\).

Step by step solution

01

Calculate the heat flux in free convection mode

To find the heat flux in free convection mode, we'll need to use the free convection heat transfer coefficient, which is given as \(h_{fr} = 3\,\text{W}/\text{m}^{2}\cdot\text{K}\). The formula for heat flux, \(q\), in terms of the heat transfer coefficient, \(h\), is: \[q = h \cdot A \cdot (T_{sur} - T_{i})\] where \(A\) is the surface area of the cake pan and batter. We are not explicitly given the surface area in this problem, however, since we are asked to calculate the heat flux density, we can simplify the equation to: \[q' = h \cdot (T_{sur} - T_{i})\] We are given \(T_{sur} = 180^{\circ}\text{C}\) and \(T_i = 24^{\circ}\text{C}\). Now, we can plug in the known values and solve for \(q'\): \[q_{fr}' = h_{fr} \cdot (T_{sur} - T_i)\]
02

Calculate the heat flux in free convection mode

\[q_{fr}' = 3\,\text{W}/\text{m}^{2}\cdot\text{K} \cdot (180^{\circ}\text{C} - 24^{\circ}\text{C})\] \[q_{fr}' = 3\,\text{W}/\text{m}^{2}\cdot\text{K} \cdot (156\text{K})\] \[q_{fr}' = 468\,\text{W}/\text{m}^{2}\] So, the heat flux delivered to the cake pan and batter when operating in free convection mode is 468 W/m虏.
03

Calculate the heat flux in forced convection mode

To find the heat flux in forced convection mode, we'll need to use the forced convection heat transfer coefficient, which is given as \(h_{fo} = 27\,\text{W}/\text{m}^{2}\cdot\text{K}\). We can use the same formula for heat flux density as in Step 1. We have already calculated the temperature difference in the first step: \[q_{fo}' = h_{fo} \cdot (T_{sur} - T_i)\]
04

Calculate the heat flux in forced convection mode

\[q_{fo}' = 27\,\text{W}/\text{m}^{2}\cdot\text{K} \cdot (180^{\circ}\text{C} - 24^{\circ}\text{C})\] \[q_{fo}' = 27\,\text{W}/\text{m}^{2}\cdot\text{K} \cdot (156\text{K})\] \[q_{fo}' = 4212\,\text{W}/\text{m}^{2}\] So, the heat flux delivered to the cake pan and batter when operating in forced convection mode is 4212 W/m虏. In summary, the heat flux delivered to the cake pan and batter when operating in free convection mode is 468 W/m虏 and in forced convection mode, it is 4212 W/m虏.

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Key Concepts

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

Convection Heat Transfer
Convection heat transfer is a fundamental mechanism by which thermal energy moves from one place to another through the movement of fluids, which can be liquids or gases. It is distinct from other forms of heat transfer like conduction, which involves direct contact between materials, and radiation, which involves heat transfer through electromagnetic waves.

Understanding convection is essential because it applies to various real-life scenarios, from industrial processes to everyday activities like heating a room or baking a cake. The basic idea is that warmer areas of a fluid will rise because they are less dense, while cooler, denser fluid will sink. This creates a circular motion known as a convection current.

In our context, the cake being baked is surrounded by air, which acts as the fluid transferring heat. The oven's air absorbs heat from the oven walls, rises, cools down after losing heat to the cake, and then sinks. In this way, thermal energy is transferred from the oven walls to the cake, heating it up.
Forced Convection
Forced convection occurs when an external force, such as a fan or a pump, propels the fluid, enhancing the heat transfer process. Unlike free convection, where the movement of the fluid relies on natural buoyancy forces, forced convection can significantly increase the efficiency of heat exchange by continually moving the fluid and introducing a fresh, warmer (or cooler) layer to the surface of the object being heated (or cooled).

An everyday example of forced convection is a convection oven, which uses a fan to circulate hot air and cook food more evenly and quickly. In the exercise we're examining, activating the convection feature in the oven increases the heat transfer coefficient to a higher value because the forced movement of air over the batter's surface leads to more efficient heat transfer.

Advantages of Forced Convection

  • Enhances heat transfer rate.
  • Provides uniform heating or cooling.
  • Can be controlled and maintained with the help of mechanical equipment.
The exercise demonstrates that with forced convection, the cake pan and batter experience a much higher heat flux, thus allowing the cake to bake faster.
Free Convection
Free convection, or natural convection, occurs when fluid motion is caused by buoyancy forces that result from density variations due to temperature differences within the fluid. There's no mechanical aid, and the movement of heat depends on the temperature gradients and the physical properties of the fluid.

In the baking scenario, when the convection feature of the oven is disabled, the air inside the oven moves solely due to the heat-induced density changes. This movement is inherently less efficient than forced convection because there is no external mechanism to enhance the heat transfer. The coefficient of heat transfer (\(h_{fr}\) in the exercise) is significantly lower in free convection, which results in a lower heat flux. This means it will take a longer time for the cake to receive the same amount of heat as in forced convection.

Characteristics of Free Convection

  • Occurs without external assistance.
  • Depends heavily on the geometry and orientation of the heated surface.
  • Typically a slower process compared to forced convection.
This distinction clearly shows why the cake would bake more slowly without the fan, with the heat flux calculated to be much lower in free convection mode.

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Most popular questions from this chapter

A rectangular forced air heating duct is suspended from the ceiling of a basement whose air and walls are at a temperature of \(T_{\infty}=T_{\text {sur }}=5^{\circ} \mathrm{C}\). The duct is \(15 \mathrm{~m}\) long, and its cross section is \(350 \mathrm{~mm} \times 200 \mathrm{~mm}\). (a) For an uninsulated duct whose average surface temperature is \(50^{\circ} \mathrm{C}\), estimate the rate of heat loss from the duct. The surface emissivity and convection coefficient are approximately \(0.5\) and \(4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (b) If heated air enters the duct at \(58^{\circ} \mathrm{C}\) and a velocity of \(4 \mathrm{~m} / \mathrm{s}\) and the heat loss corresponds to the result of part (a), what is the outlet temperature? The density and specific heat of the air may be assumed to be \(\rho=1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=1008 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

The roof of a car in a parking lot absorbs a solar radiant flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\), and the underside is perfectly insulated. The convection coefficient between the roof and the ambient air is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Neglecting radiation exchange with the surroundings, calculate the temperature of the roof under steadystate conditions if the ambient air temperature is \(20^{\circ} \mathrm{C}\). (b) For the same ambient air temperature, calculate the temperature of the roof if its surface emissivity is \(0.8\). (c) The convection coefficient depends on airflow conditions over the roof, increasing with increasing air speed. Compute and plot the roof temperature as a function of \(h\) for \(2 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

In one stage of an annealing process, 304 stainless steel sheet is taken from \(300 \mathrm{~K}\) to \(1250 \mathrm{~K}\) as it passes through an electrically heated oven at a speed of \(V_{s}=10 \mathrm{~mm} / \mathrm{s}\). The sheet thickness and width are \(t_{s}=8 \mathrm{~mm}\) and \(W_{s}=2 \mathrm{~m}\), respectively, while the height, width, and length of the oven are \(H_{o}=2 \mathrm{~m}\), \(W_{o}=2.4 \mathrm{~m}\), and \(L_{o}=25 \mathrm{~m}\), respectively. The top and four sides of the oven are exposed to ambient air and large surroundings, each at \(300 \mathrm{~K}\), and the corresponding surface temperature, convection coefficient, and emissivity are \(T_{s}=350 \mathrm{~K}, h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(\varepsilon_{s}=0.8\). The bottom surface of the oven is also at \(350 \mathrm{~K}\) and rests on a \(0.5\)-m-thick concrete pad whose base is at \(300 \mathrm{~K}\). Estimate the required electric power input, \(P_{\text {elec }}\), to the oven.

A freezer compartment is covered with a 2 -mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient air at \(20^{\circ} \mathrm{C}\) and a coefficient of \(h=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).
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