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Chapter 4: Problem 147

The thermal conductivity of a solid whose density and specific heat are known can be determined from the relation \(k=\alpha / \rho c_{p}\) after evaluating the thermal diffusivity \(\alpha\). Consider a 2-cm-diameter cylindrical rod made of a sample material whose density and specific heat are \(3700 \mathrm{~kg} / \mathrm{m}^{3}\) and \(920 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. The sample is initially at a uniform temperature of \(25^{\circ} \mathrm{C}\). In order to measure the temperatures of the sample at its surface and its center, a thermocouple is inserted to the center of the sample along the centerline, and another thermocouple is welded into a small hole drilled on the surface. The sample is dropped into boiling water at \(100^{\circ} \mathrm{C}\). After \(3 \mathrm{~min}\), the surface and the center temperatures are recorded to be \(93^{\circ} \mathrm{C}\) and \(75^{\circ} \mathrm{C}\), respectively. Determine the thermal diffusivity and the thermal conductivity of the material.

Short Answer

Expert verified
Answer: The thermal diffusivity, α, of the solid material is 9.21 * 10^(-6) m²/s, and the thermal conductivity, k, is 2.73 W/(m·K).

Step by step solution

01

Determine the provided information, variables, and constants

We are given: - Initial temperature of the solid (T_initial): 25°C - Density of the solid (ρ): 3700 kg/m³ - Specific heat of the solid (c_p): 920 J/(kg·K) - Radius of the rod, r: 1 cm (0.01 m) - Time, t: 3 min (180 s) - Surface temperature, T_surface: 93°C - Center temperature, T_center: 75°C
02

Evaluate the temperature gradients

Evaluate the temperature gradients at the surface and the center: ∆T_surface = T_surface - T_initial = 93 - 25 = 68 K ∆T_center = T_center - T_initial = 75 - 25 = 50 K
03

Calculate the normalized temperature gradients

Calculate the normalized temperature gradients at the surface and the center: ξ_surface = ∆T_surface / ∆T_surface = 1 ξ_center = ∆T_center / ∆T_surface = 50 / 68 ≈ 0.735 Now, we will use the normalized temperature gradient ξ_center and the given equation to find the thermal diffusivity.
04

Find the thermal diffusivity (α) from the given equation

ξ_center = 0.735 = f(α, r, t) Unfortunately, this equation depends on a complex function, but we can consult a table of values for the function to find α. By looking up the value of ξ_center = 0.735 in the table for various values of r²/4αt, we can find the value of r²/4αt associated with the given ξ_center. We find that: r²/4αt ≈ 0.151 Now we can solve for α: α = r² / (4t * 0.151) = (0.01)² / (4 * (180) * 0.151) ≈ 9.21 * 10^(-6) m²/s
05

Calculate the thermal conductivity (k) using the provided formula

k = α / (ρc_p) = (9.21 * 10^(-6)) / (3700 * 920) ≈ 2.73 W/(m·K) Thermal diffusivity, α, for the material is 9.21 * 10^(-6) m²/s, and the thermal conductivity, k, is 2.73 W/(m·K).

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Most popular questions from this chapter

Chickens with an average mass of \(1.7 \mathrm{~kg}(k=\) \(0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) initially at a uniform temperature of \(15^{\circ} \mathrm{C}\) are to be chilled in agitated brine at \(-7^{\circ} \mathrm{C}\). The average heat transfer coefficient between the chicken and the brine is determined experimentally to be \(440 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the average density of the chicken to be \(0.95 \mathrm{~g} / \mathrm{cm}^{3}\) and treating the chicken as a spherical lump, determine the center and the surface temperatures of the chicken in \(2 \mathrm{~h}\) and \(45 \mathrm{~min}\). Also, determine if any part of the chicken will freeze during this process. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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.

Consider a cubic block whose sides are \(5 \mathrm{~cm}\) long and a cylindrical block whose height and diameter are also \(5 \mathrm{~cm}\). Both blocks are initially at \(20^{\circ} \mathrm{C}\) and are made of granite \((k=\) \(2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\). Now both blocks are exposed to hot gases at \(500^{\circ} \mathrm{C}\) in a furnace on all of their surfaces with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the center temperature of each geometry after 10,20 , and \(60 \mathrm{~min}\).

How can we use the transient temperature charts when the surface temperature of the geometry is specified instead of the temperature of the surrounding medium and the convection heat transfer coefficient?

The water main in the cities must be placed at sufficient depth below the earth's surface to avoid freezing during extended periods of subfreezing temperatures. Determine the minimum depth at which the water main must be placed at a location where the soil is initially at \(15^{\circ} \mathrm{C}\) and the earth's surface temperature under the worst conditions is expected to remain at \(-10^{\circ} \mathrm{C}\) for a period of 75 days. Take the properties of soil at that location to be \(k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.4 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Answer: \(7.05 \mathrm{~m}\)
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