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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 2-cm-diameter plastic rod has a thermocouple inserted to measure temperature at the center of the rod. The plastic \(\operatorname{rod}\left(\rho=1190 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1465 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(k=0.19 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) was initially heated to a uniform temperature of \(70^{\circ} \mathrm{C}\), and allowed to be cooled in ambient air temperature of \(25^{\circ} \mathrm{C}\). After \(1388 \mathrm{~s}\) of cooling, the thermocouple measured the temperature at the center of the rod to be \(30^{\circ} \mathrm{C}\). Determine the convection heat transfer coefficient for this process. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Consider the freezing of packaged meat in boxes with refrigerated air. How do \((a)\) the temperature of air, (b) the velocity of air, \((c)\) the capacity of the refrigeration system, and \((d)\) the size of the meat boxes affect the freezing time?

Lumped system analysis of transient heat conduction situations is valid when the Biot number is (a) very small (b) approximately one (c) very large (d) any real number (e) cannot say unless the Fourier number is also known.

A stainless steel slab \((k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=\) \(\left.3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) and a copper slab \((k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=117 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) are placed under an array of laser diodes, which supply an energy pulse of \(5 \times 10^{7} \mathrm{~J} / \mathrm{m}^{2}\) instantaneously at \(t=0\) to both materials. The two slabs have a uniform initial temperature of \(20^{\circ} \mathrm{C}\). Using \(\mathrm{EES}\) (or other) software, investigate the effect of time on the temperatures of both materials at the depth of \(5 \mathrm{~cm}\) from the surface. By varying the time from 1 to \(80 \mathrm{~s}\) after the slabs have received the energy pulse, plot the temperatures at \(5 \mathrm{~cm}\) from the surface as a function of time.

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.

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