/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Convection ovens operate on the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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.

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 91影视!

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.

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

Single fuel cells such as the one of Example \(1.5\) can be scaled up by arranging them into a fuel cell stack. A stack consists of multiple electrolytic membranes that are sandwiched between electrically conducting bipolar plates. Air and hydrogen are fed to each membrane through fiw channels within each bipolar plate, as shown in the sketch. With this stack arrangement, the individual fuel cells are connected in series, electrically, producing a stack voltage of \(E_{\text {stack }}=N \times E_{c}\), where \(E_{c}\) is the voltage produced across each membrane and \(N\) is the number of membranes in the stack. The electrical current is the same for each membrane. The cell voltage, \(E_{c}\), as well as the cell efficiency, increases with temperature (the air and hydrogen fed to the stack are humidified to allow operation at temperatures greater than in Example 1.5), but the membranes will fail at temperatures exceeding \(T \approx 85^{\circ} \mathrm{C}\). Consider \(L \times w\) membranes, where \(L=w=100 \mathrm{~mm}\), of thickness \(t_{m}=0.43 \mathrm{~mm}\), that each produce \(E_{c}=0.6 \mathrm{~V}\) at \(I=60 \mathrm{~A}\), and \(\dot{E}_{c g}=45 \mathrm{~W}\) of thermal energy when operating at \(T=80^{\circ} \mathrm{C}\). The external surfaces of the stack are exposed to air at \(T_{\infty}=25^{\circ} \mathrm{C}\) and surroundings at \(T_{\text {sur }}=30^{\circ} \mathrm{C}\), with \(\varepsilon=0.88\) and \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the electrical power produced by a stack that is \(L_{\text {stack }}=200 \mathrm{~mm}\) long, for bipolar plate thickness in the range \(1 \mathrm{~mm}

A solar flux of \(700 \mathrm{~W} / \mathrm{m}^{2}\) is incident on a flat-plate solar collector used to heat water. The area of the collector is \(3 \mathrm{~m}^{2}\), and \(90 \%\) of the solar radiation passes through the cover glass and is absorbed by the absorber plate. The remaining \(10 \%\) is reflected away from the collector. Water flows through the tube passages on the back side of the absorber plate and is heated from an inlet temperature \(T_{i}\) to an outlet temperature \(T_{o}\). The cover glass, operating at a temperature of \(30^{\circ} \mathrm{C}\), has an emissivity of \(0.94\) and experiences radiation exchange with the sky at \(-10^{\circ} \mathrm{C}\). The convection coefficient between the cover glass and the ambient air at \(25^{\circ} \mathrm{C}\) is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Perform an overall energy balance on the collector to obtain an expression for the rate at which useful heat is collected per unit area of the collector, \(q_{11}^{\prime \prime}\). Determine the value of \(q_{u r^{\prime \prime}}\). (b) Calculate the temperature rise of the water, \(T_{o}-T_{i}\), if the flow rate is \(0.01 \mathrm{~kg} / \mathrm{s}\). Assume the specific heat of the water to be \(4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (c) The collector efficiency \(\eta\) is defined as the ratio of the useful heat collected to the rate at which solar energy is incident on the collector. What is the value of \(\eta\) ?

The concrete slab of a basement is \(11 \mathrm{~m}\) long, \(8 \mathrm{~m}\) wide, and \(0.20 \mathrm{~m}\) thick. During the winter, temperatures are nominally \(17^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the rate of heat loss through the slab? If the basement is heated by a gas furnace operating at an efficiency of \(\eta_{f}=0.90\) and natural gas is priced at \(C_{g}=\$ 0.02 / \mathrm{MJ}\), what is the daily cost of the heat loss?

In the thermal processing of semiconductor materials, annealing is accomplished by heating a silicon wafer according to a temperature-time recipe and then maintaining a fixed elevated temperature for a prescribed period of time. For the process tool arrangement shown as follows, the wafer is in an evacuated chamber whose walls are maintained at \(27^{\circ} \mathrm{C}\) and within which heating lamps maintain a radiant flux \(q_{s}^{\prime \prime}\) at its upper surface. The wafer is \(0.78 \mathrm{~mm}\) thick, has a thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and an emissivity that equals its absorptivity to the radiant flux \(\left(\varepsilon=\alpha_{l}=0.65\right.\) ). For \(q_{s}^{\prime \prime}=3.0 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\), the temperature on its lower surface is measured by a radiation thermometer and found to have a value of \(T_{w, l}=997^{\circ} \mathrm{C}\). To avoid warping the wafer and inducing slip planes in the crystal structure, the temperature difference across the thickness of the wafer must be less than \(2^{\circ} \mathrm{C}\). Is this condition being met?

1.42 One method for growing thin silicon sheets for photovoltaic solar panels is to pass two thin strings of high melting temperature material upward through a bath of molten silicon. The silicon solidifies on the strings near the surface of the molten pool, and the solid silicon sheet is pulled slowly upward out of the pool. The silicon is replenished by supplying the molten pool with solid silicon powder. Consider a silicon sheet that is \(W_{\mathrm{si}}=85 \mathrm{~mm}\) wide and \(t_{\mathrm{si}}=150 \mu \mathrm{m}\) thick that is pulled at a velocity of \(V_{\mathrm{si}}=20 \mathrm{~mm} / \mathrm{min}\). The silicon is melted by supplying electric power to the cylindrical growth chamber of height \(H=350 \mathrm{~mm}\) and diameter \(D=300 \mathrm{~mm}\). The exposed surfaces of the growth chamber are at \(T_{s}=\) \(320 \mathrm{~K}\), the corresponding convection coefficient at the exposed surface is \(h=8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the surface is characterized by an emissivity of \(\varepsilon_{s}=0.9\). The solid silicon powder is at \(T_{\mathrm{s}, i}=298 \mathrm{~K}\), and the solid silicon sheet exits the chamber at \(T_{\text {si, } o}=420 \mathrm{~K}\). Both the surroundings and ambient temperatures are \(T_{\infty}=T_{\text {sur }}=298 \mathrm{~K}\). (a) Determine the electric power, \(P_{\text {elec }}\), needed to operate the system at steady state. (b) If the photovoltaic panel absorbs a time-averaged solar flux of \(q_{\text {sol }}^{\prime \prime}=180 \mathrm{~W} / \mathrm{m}^{2}\) and the panel has a conversion efficiency (the ratio of solar power absorbed to electric power produced) of \(\eta=0.20\), how long must the solar panel be operated to produce enough electric energy to offset the electric energy that was consumed in its manufacture?
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Nuclear Physics

Read Explanation

Translational Dynamics

Read Explanation

Solid State Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Atoms and Radioactivity

Read Explanation
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.