Polyvinylchloride automotive body panels \((k=\) \(\left.0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=1.05 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, \rho=1714 \mathrm{~kg} / \mathrm{m}^{3}\right), 3-\mathrm{mm}\) thick, emerge from an injection molder at \(120^{\circ} \mathrm{C}\). They need to be cooled to \(40^{\circ} \mathrm{C}\) by exposing both sides of the panels to \(20^{\circ} \mathrm{C}\) air before they can be handled. If the convective heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and radiation is not considered, the time that the panels must be exposed to air before they can be handled is (a) \(1.6 \mathrm{~min}\) (b) \(2.4 \mathrm{~min}\) (c) \(2.8 \mathrm{~min}\) (d) \(3.5 \mathrm{~min}\) (e) \(4.2 \mathrm{~min}\)

Short Answer

Expert verified
a) 1.8 min b) 2.4 min c) 3.2 min d) 4.0 min Answer: b) 2.4 min

Step by step solution

01

Calculate Biot Number

To compute the Biot number, we need to determine the resistances for convection and conduction. We can calculate the conduction resistance (R_cond) using the thickness (L) of the panel and thermal conductivity (k) as follows: \[R_\text{cond} = \frac{L}{k} = \frac{0.003}{0.092}\] Now, we find the convection resistance (R_conv) using the heat transfer coefficient (h) and the area (A) of the panel: \[R_\text{conv} = \frac{1}{hA} = \frac{1}{30A}\] Now, we can compute the Biot number using these resistances: \[Bi = \frac{R_\text{cond}}{R_\text{conv}} = \frac{0.003}{0.092} \cdot \frac{1}{30A} \cdot A = 0.000103\] Since the Biot number is less than 0.1, we can approximate this problem as one-dimensional.
02

Calculate Fourier Number

First, let's determine the required temperature drop, which is from 120°C to 40°C (_ {drop}=80^{\circ}C). Now, let's replace T(x,t) in the formula by T_final = 40°C to get: \[\frac{80}{100} = erf(\frac{0.003}{2\sqrt{\alpha t}})\] Now, from a table of error functions (erf), we find erf(0.8) = 0.7421. So, \[0.7421 = \frac{0.003}{2\sqrt{\alpha t}}\] Now, let's find the thermal diffusivity (α) using the formula: \[\alpha = \frac{k}{\rho c_p} = \frac{0.092}{1714*1.05*10^3} = 5.028 \times 10^{-8}m^{2}/s\] Now, plug this value of α back into our equation: \[0.7421 = \frac{0.003}{2 \sqrt{5.028 \times 10^{-8} t}}\]
03

Find the time to cool the panel

Now, to find the time (t), we need to solve the equation for t: \[t = \frac{(0.003/(2\cdot 0.7421))^2}{(5.028 \times 10^{-8})}\] \[t = 151.28 s\] Now, to find the time in minutes, we divide the time by 60: \[t = \frac{151.28}{60} = 2.52 \mathrm{~min}\] The given options provided do not have an exact match for 2.52 min, but we can choose the closest one, which is (b) \(2.4 \mathrm{~min}\).

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Key Concepts

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

Biot Number
The Biot number (Bi) is a dimensionless quantity used in heat transfer calculations to determine the relationship between the conductive heat resistance within an object and the convective heat transfer resistance on the surface of the object. In simpler terms, it helps us assess whether the temperature gradient within the object is significant when compared to the temperature difference between the object's surface and the surrounding environment.

The formula to calculate the Biot number is given by: \[Bi = \frac{hL}{k}\]where \(h\) is the convective heat transfer coefficient, \(L\) is the characteristic length of the object (typically its thickness for flat plates or diameter for cylindrical and spherical objects), and \(k\) is the thermal conductivity of the material. A low Biot number, typically less than 0.1, suggests that the temperature within the material can be considered uniform, thus simplifying the analysis to a one-dimensional problem.
Fourier Number
The Fourier number (Fo) is another critical dimensionless number in heat transfer, indicating how heat diffuses through a material. It is a ratio that compares the rate of heat conduction with the rate of thermal energy storage. A higher Fourier number implies faster heat diffusion. This number is particularly useful when examining transient heat conduction problems.

To comprehend what the Fourier number means, consider it to be a measure of how 'aged' the temperature profile in a material is at a given time during heating or cooling. It is defined by the formula: \[Fo = \frac{\alpha t}{L^2}\]where \(\alpha\) is the thermal diffusivity, \(t\) is the time, and \(L\) is the characteristic length. In the context of cooling the automotive body panels, the Fourier number enables us to calculate the time required for the panels to cool down to an acceptable temperature for handling.
Thermal Diffusivity
The concept of thermal diffusivity \(\alpha\) is essential in the analysis of transient heat conduction. It is a material property that describes the rate at which heat spreads through a material. Thermal diffusivity depends on the thermal conductivity \(k\), the density \(\rho\), and the specific heat capacity \(c_p\) of the material. The thermal diffusivity is found using the relation: \[\alpha = \frac{k}{\rho \cdot c_p}\]A material with high thermal diffusivity will quickly adapt its temperature to its surroundings because it allows heat to diffuse rapidly. Conversely, a material with low thermal diffusivity will take longer to reach thermal equilibrium. Therefore, the rate at which the automotive body panels cool down is directly associated with the thermal diffusivity of the polyvinylchloride, which determines how quickly the heat can be distributed away from the hot areas towards the cooler areas until a uniform temperature is reached throughout the panels.

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Most popular questions from this chapter

A potato may be approximated as a 5.7-cm-diameter solid sphere with the properties \(\rho=910 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.25 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), \(k=0.68 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.76 \times 10^{-1} \mathrm{~m}^{2} / \mathrm{s}\). Twelve such potatoes initially at \(25^{\circ} \mathrm{C}\) are to be cooked by placing them in an oven maintained at \(250^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(95 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer to the potatoes during a 30-min period is (a) \(77 \mathrm{~kJ}\) (b) \(483 \mathrm{~kJ}\) (c) \(927 \mathrm{~kJ}\) (d) \(970 \mathrm{~kJ}\) (e) \(1012 \mathrm{~kJ}\)

In Betty Crocker's Cookbook, it is stated that it takes \(5 \mathrm{~h}\) to roast a \(14-\mathrm{lb}\) stuffed turkey initially at \(40^{\circ} \mathrm{F}\) in an oven maintained at \(325^{\circ} \mathrm{F}\). It is recommended that a meat thermometer be used to monitor the cooking, and the turkey is considered done when the thermometer inserted deep into the thickest part of the breast or thigh without touching the bone registers \(185^{\circ} \mathrm{F}\). The turkey can be treated as a homogeneous spherical object with the properties \(\rho=75 \mathrm{lbm} / \mathrm{ft}^{3}, c_{p}=0.98 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\), \(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and \(\alpha=0.0035 \mathrm{ft}^{2} / \mathrm{h}\). Assuming the tip of the thermometer is at one- third radial distance from the center of the turkey, determine \((a)\) the average heat transfer coefficient at the surface of the turkey, \((b)\) the temperature of the skin of the turkey when it is done, and \((c)\) the total amount of heat transferred to the turkey in the oven. Will the reading of the thermometer be more or less than \(185^{\circ} \mathrm{F} 5\) min after the turkey is taken out of the oven?

Plasma spraying is a process used for coating a material surface with a protective layer to prevent the material from degradation. In a plasma spraying process, the protective layer in powder form is injected into a plasma jet. The powder is then heated to molten droplets and propelled onto the material surface. Once deposited on the material surface, the molten droplets solidify and form a layer of protective coating. Consider a plasma spraying process using alumina \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=3970 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) powder that is injected into a plasma jet at \(T_{\infty}=15,000^{\circ} \mathrm{C}\) and \(h=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The alumina powder is made of particles that are spherical in shape with an average diameter of \(60 \mu \mathrm{m}\) and a melting point at \(2300^{\circ} \mathrm{C}\). Determine the amount of time it would take for the particles, with an initial temperature of \(20^{\circ} \mathrm{C}\), to reach their melting point from the moment they are injected into the plasma jet.

A long cylindrical wood \(\log (k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.28 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is \(10 \mathrm{~cm}\) in diameter and is initially at a uniform temperature of \(15^{\circ} \mathrm{C}\). It is exposed to hot gases at \(550^{\circ} \mathrm{C}\) in a fireplace with a heat transfer coefficient of \(13.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surface. If the ignition temperature of the wood is \(420^{\circ} \mathrm{C}\), determine how long it will be before the log ignites. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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) \(8.5 \mathrm{~min}\)

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