Why are the transient temperature charts prepared using nondimensionalized quantities such as the Biot and Fourier numbers instead of the actual variables such as thermal conductivity and time?

Short Answer

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Answer: Transient temperature charts are prepared using nondimensionalized quantities like Biot and Fourier numbers because they offer several advantages such as easier comparison across various conditions and systems, generalization, and easier interpretation. Nondimensionalization simplifies complex equations, reduces the number of variables involved, and allows for a more universally applicable chart. Using Biot and Fourier numbers enables us to create a single chart that can be easily referenced and scaled to actual variables as needed, making analysis and understanding of temperature-dependent processes more accessible.

Step by step solution

01

Define Biot and Fourier numbers

First, let's define the Biot number (Bi) and Fourier number (Fo). The Biot number is a dimensionless parameter that characterizes the ratio of internal thermal resistance to external (convective) thermal resistance, and mathematically it is defined as: \[Bi = \frac{hL}{k}\] where \(h\) is the heat transfer coefficient (W/m²K), \(L\) is the characteristic length (m), and \(k\) is the thermal conductivity (W/mK). The Fourier number, on the other hand, is a dimensionless parameter relating to the rate of heat conduction and the rate of heat storage, and is given by: \[Fo = \frac{\alpha t}{L^2}\] where \(\alpha\) is the thermal diffusivity (m²/s) and \(t\) is the time (s).
02

Explain the importance of nondimensionalization

Nondimensionalization is the process of scaling quantities to remove dimensions, and it's useful in several ways: 1. Simplifying complex equations: By using dimensionless quantities, we can simplify complex equations that involve multiple parameters and dimensions into simpler forms. 2. Easier comparison of different systems: With dimensionless quantities, it's easier to compare different systems that may have widely varying scales or units. This makes it easy to generalize the results for a wider range of applications. 3. Reduced number of variables in an equation: Using nondimensionalized quantities decreases the number of variables in an equation, making it easier to solve and analyze. 4. Universality: Nondimensionalization allows the equation to be valid across different systems, regardless of the initial units being used.
03

Advantages of using Biot and Fourier numbers in transient temperature charts

Using Biot and Fourier numbers in transient temperature charts offers several advantages: 1. Easier comparison: By using Biot and Fourier numbers, we eliminate units associated with thermal conductivity, heat transfer coefficient, characteristic length, and time, enabling easier comparison of transient temperature charts for various conditions and systems. 2. Generalization: Using nondimensionalized quantities like Biot and Fourier numbers allows us to develop a single chart for various conditions and systems that can be easily referenced and scaled to actual variables as needed. 3. Easier interpretation: Transient temperature charts with nondimensionalized quantities are often easier to interpret, as they provide a simple method to look at trends and patterns across various systems and conditions, without involving the complexities of actual variables. In conclusion, preparing transient temperature charts using Biot and Fourier numbers, instead of actual variables like thermal conductivity and time, provide us with a more generalized and easily interpreted chart. The nondimensionalized quantities make it possible to draw comparisons across various systems and conditions, simplifying the analysis and understanding of temperature-dependent processes.

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Most popular questions from this chapter

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.

Stainless steel ball bearings \(\left(\rho=8085 \mathrm{~kg} / \mathrm{m}^{3}, k=\right.\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C}, c_{p}=0.480 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) having a diameter of \(1.2 \mathrm{~cm}\) are to be quenched in water. The balls leave the oven at a uniform temperature of \(900^{\circ} \mathrm{C}\) and are exposed to air at \(30^{\circ} \mathrm{C}\) for a while before they are dropped into the water. If the temperature of the balls is not to fall below \(850^{\circ} \mathrm{C}\) prior to quenching and the heat transfer coefficient in the air is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\), determine how long they can stand in the air before being dropped into the water.

Hailstones are formed in high altitude clouds at \(253 \mathrm{~K}\). Consider a hailstone with diameter of \(20 \mathrm{~mm}\) and is falling through air at \(15^{\circ} \mathrm{C}\) with convection heat transfer coefficient of \(163 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the hailstone can be modeled as a sphere and has properties of ice at \(253 \mathrm{~K}\), determine the duration it takes to reach melting point at the surface of the falling hailstone. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Layers of 6-in-thick meat slabs \(\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(50^{\circ} \mathrm{F}\) are cooled by refrigerated air at \(23^{\circ} \mathrm{F}\) to a temperature of \(36^{\circ} \mathrm{F}\) at their center in \(12 \mathrm{~h}\). Estimate the average heat transfer coefficient during this cooling process. Solve this problem using the Heisler charts. Answer: \(1.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\)

We often cut a watermelon in half and put it into the freezer to cool it quickly. But usually we forget to check on it and end up having a watermelon with a frozen layer on the top. To avoid this potential problem a person wants to set the timer such that it will go off when the temperature of the exposed surface of the watermelon drops to \(3^{\circ} \mathrm{C}\). Consider a 25 -cm- diameter spherical watermelon that is cut into two equal parts and put into a freezer at \(-12^{\circ} \mathrm{C}\). Initially, the entire watermelon is at a uniform temperature of \(25^{\circ} \mathrm{C}\), and the heat transfer coefficient on the surfaces is \(22 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the watermelon to have the properties of water, determine how long it will take for the center of the exposed cut surfaces of the watermelon to drop to \(3^{\circ} \mathrm{C}\).

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