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Chapter 4: Problem 178

Conduct the following experiment at home to determine the combined convection and radiation heat transfer coefficient at the surface of an apple exposed to the room air. You will need two thermometers and a clock. First, weigh the apple and measure its diameter. You can measure its volume by placing it in a large measuring cup halfway filled with water, and measuring the change in volume when it is completely immersed in the water. Refrigerate the apple overnight so that it is at a uniform temperature in the morning and measure the air temperature in the kitchen. Then take the apple out and stick one of the thermometers to its middle and the other just under the skin. Record both temperatures every \(5 \mathrm{~min}\) for an hour. Using these two temperatures, calculate the heat transfer coefficient for each interval and take their average. The result is the combined convection and radiation heat transfer coefficient for this heat transfer process. Using your experimental data, also calculate the thermal conductivity and thermal diffusivity of the apple and compare them to the values given above.

Short Answer

Expert verified
Question: Calculate the average combined convection and radiation heat transfer coefficient for an apple, its thermal conductivity, and thermal diffusivity using the experimental data. Briefly compare these values to the given values.

Step by step solution

01

Carry out the experiment

To perform the experiment, first gather the necessary equipment: - Two thermometers - A clock - A weighing scale - A large measuring cup - A ruler or measuring tape Then follow these steps: 1. Weigh the apple and measure its diameter. 2. Measure the apple's volume by placing it in a measuring cup halfway filled with water and noting the change in water volume. 3. Refrigerate the apple overnight so that it is at a uniform temperature. 4. Measure the air temperature in the kitchen. 5. Take the apple out and stick one thermometer in its middle and the other just under the skin. 6. Record both temperatures every \(5 \mathrm{~min}\) for an hour.
02

Calculate heat transfer coefficient for each interval

Now that we've gathered the data, we can use it to calculate the heat transfer coefficient for each interval. We will need the formula for combined convection and radiation heat transfer, which is: \(h_{cr} = \frac{Q_{total}}{A(T_{s} - T_{ambient})}\) Where: - \(h_{cr}\) is the combined convection and radiation heat transfer coefficient, - \(Q_{total}\) is the total heat transfer, - \(A\) is the surface area of the apple, - \(T_{s}\) is the temperature of the skin thermometer during the experiment, - \(T_{ambient}\) is the air temperature in the kitchen. For each time interval (\(every \ 5 \mathrm{~min}\)), plug in the values and calculate the heat transfer coefficient, \(h_{cr}\).
03

Calculate average heat transfer coefficient

Once you have the heat transfer coefficients for each interval, take their average to obtain the final combined convection and radiation heat transfer coefficient: \(h_{cr,avg} = \frac{1}{n}\sum_{i=1}^{n} h_{cr,i}\) Where: - \(h_{cr,avg}\) is the average combined convection and radiation heat transfer coefficient, - \(n\) is the number of intervals, - \(h_{cr,i}\) is the heat transfer coefficient for each interval.
04

Calculate thermal conductivity and thermal diffusivity of the apple

Using the experimental data, we can now calculate the thermal conductivity and thermal diffusivity of the apple. Thermal conductivity, \(k\) can be calculated using: \(k = \frac{Q}{A\frac{\Delta T}{L}}\) Where: - \(Q\) is the heat transfer, - \(A\) is the surface area of the apple, - \(\Delta T\) is the temperature difference between the center and skin of the apple, - \(L\) is the distance between the center and skin of the apple. Thermal diffusivity, \(\alpha\) can be calculated using: \(\alpha = \frac{k}{\rho c_p}\) Where: - \(\rho\) is the density of the apple, - \(c_p\) is the specific heat capacity of the apple. Finally, compare these calculated values with the given values to check how close or far they are. Note: To obtain the values of \(\rho\) and \(c_p\), you need to refer to a material property table for apples or search for them in literature.

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Key Concepts

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

Convection and Radiation Heat Transfer Coefficient
Understanding heat transfer is crucial in various fields, from cooking to aerospace. Specifically, the convection and radiation heat transfer coefficient, denoted as \(h_{cr}\), represents the rate at which heat is transferred from a surface to its environment due to both convection and radiation.

Consider an apple coming to room temperature, as in the exercise. Heat loss occurs due to air flowing over its surface (convection) and by emitting heat as infrared radiation (radiation). By conducting the described experiment and calculating the change in heat over time, we obtain \(h_{cr}\) which is a combination of these two modes of heat transfer.

The formula \(h_{cr} = \frac{Q_{total}}{A(T_{s} - T_{ambient})}\), used in the solution, incorporates the total heat transfer (\(Q_{total}\)), the surface area of the apple (\(A\)), the apple's surface temperature (\(T_{s}\)), and the ambient air temperature (\(T_{ambient}\)). After taking multiple readings at set intervals, we average these to achieve a more accurate coefficient reflecting the average scenario.
Thermal Conductivity
Thermal conductivity, represented as \(k\), is a fundamental property of materials that describes how well a substance can conduct heat. It is determined by the material's internal structure, which can vary significantly among different substances.

Within the context of our experiment, the thermal conductivity of the apple characterizes how readily heat flows from the warmer inner part to the cooler outer surface. The formula \(k = \frac{Q}{A\frac{\Delta T}{L}}\) ties together the heat transfer (\(Q\)), the surface area (\(A\)), the temperature difference between the apple's center and surface (\(\Delta T\)), and the distance between these two points (\(L\)).

By determining thermal conductivity, we can compare the apple's ability to conduct heat with other materials. This could, for instance, inform decisions about food storage or processing.
Thermal Diffusivity
Thermal diffusivity (\(\alpha\)) measures how quickly a material can adjust its temperature to its surroundings. It is related to thermal conductivity but also accounts for the material's density (\(\rho\)) and specific heat capacity (\(c_p\)). The formula \(\alpha = \frac{k}{\rho c_p}\) reveals how thermal diffusivity is influenced by these properties.

High thermal diffusivity indicates that the material can quickly even out temperature discrepancies, a useful trait in industries needing fast temperature changes like metallurgy or culinary arts. In our exercise, by calculating the thermal diffusivity of the apple, we learn about its thermal responsiveness. This information sheds light on the rate at which the apple can reach thermal equilibrium with the surrounding air, ultimately providing insights into the quality and freshness of fruits based on their thermal properties.
Heat Transfer Formula
The heat transfer formula is a mathematical representation that quantifies the energy exchange due to temperature differences. A key aspect of our experiment is calculating the amount of heat transfer over time using the heat transfer formula embedded in the solution.

The formula \(Q = A\frac{\Delta T}{L}\) is foundational when calculating both the thermal conductivity and diffusivity of a material. Here \(Q\) represents the heat transfer which we can derive from the experiment, \(\Delta T\) stands for the temperature difference, \(A\) is the exposed surface area, and \(L\) is the characteristic length of heat travel.

Understanding how to apply this formula can help students delve into more complex heat transfer problems, developing a holistic comprehension of thermal processes in various contexts.

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Most popular questions from this chapter

A long iron \(\operatorname{rod}\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=80.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=23.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) with diameter of \(25 \mathrm{~mm}\) is initially heated to a uniform temperature of \(700^{\circ} \mathrm{C}\). The iron rod is then quenched in a large water bath that is maintained at constant temperature of \(50^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(128 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the time required for the iron rod surface temperature to cool to \(200^{\circ} \mathrm{C}\). Solve this problem using analytical one- term approximation method (not the Heisler charts).

Consider a 1000-W iron whose base plate is made of \(0.5-\mathrm{cm}\)-thick aluminum alloy \(2024-\mathrm{T} 6\left(\rho=2770 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\right.\) \(\left.875 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \alpha=7.3 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\). The base plate has a surface area of \(0.03 \mathrm{~m}^{2}\). Initially, the iron is in thermal equilibrium with the ambient air at \(22^{\circ} \mathrm{C}\). Taking the heat transfer coefficient at the surface of the base plate to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and assuming 85 percent of the heat generated in the resistance wires is transferred to the plate, determine how long it will take for the plate temperature to reach \(140^{\circ} \mathrm{C}\). Is it realistic to assume the plate temperature to be uniform at all times?

Consider a 7.6-cm-diameter cylindrical lamb meat chunk \(\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\), \(\left.\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\). Such a meat chunk intially at \(2^{\circ} \mathrm{C}\) is dropped into boiling water at \(95^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The time it takes for the center temperature of the meat chunk to rise to \(75{ }^{\circ} \mathrm{C}\) is (a) \(136 \mathrm{~min}\) (b) \(21.2 \mathrm{~min}\) (c) \(13.6 \mathrm{~min}\) (d) \(11.0 \mathrm{~min}\) (e) \(8.5 \mathrm{~min}\)

Layers of 23 -cm-thick meat slabs \((k=0.47 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(7^{\circ} \mathrm{C}\) are to be frozen by refrigerated air at \(-30^{\circ} \mathrm{C}\) flowing at a velocity of \(1.4 \mathrm{~m} / \mathrm{s}\). The average heat transfer coefficient between the meat and the air is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the size of the meat slabs to be large relative to their thickness, determine how long it will take for the center temperature of the slabs to drop to \(-18^{\circ} \mathrm{C}\). Also, determine the surface temperature of the meat slab at that time.

4-115 A semi-infinite aluminum cylinder \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\left.\alpha=9.71 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~cm}\) is initially at a uniform temperature of \(T_{i}=115^{\circ} \mathrm{C}\). The cylinder is now placed in water at \(10^{\circ} \mathrm{C}\), where heat transfer takes place by convection with a heat transfer coefficient of \(h=140 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the cylinder \(5 \mathrm{~cm}\) from the end surface 8 min after the start of cooling. 4-116 A 20-cm-long cylindrical aluminum block \((\rho=\) \(2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(\left.9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right), 15 \mathrm{~cm}\) in diameter, is initially at a uniform temperature of \(20^{\circ} \mathrm{C}\). The block is to be heated in a furnace at \(1200^{\circ} \mathrm{C}\) until its center temperature rises to \(300^{\circ} \mathrm{C}\). If the heat transfer coefficient on all surfaces of the block is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long the block should be kept in the furnace. Also, determine the amount of heat transfer from the aluminum block if it is allowed to cool in the room until its temperature drops to \(20^{\circ} \mathrm{C}\) throughout.
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