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Spherical glass beads coming out of a kiln are allowed to cool in a room temperature of \(30^{\circ} \mathrm{C}\). A glass bead with a diameter of $10 \mathrm{~mm}\( and an initial temperature of \)400^{\circ} \mathrm{C}$ is allowed to cool for \(3 \mathrm{~min}\). If the convection heat transfer coefficient is \(28 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the center of the glass bead using the analytical one-term approximation method. The glass bead has properties of $\rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\(, \)c_{p}=750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$, and \(k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
Answer: The temperature at the center of the spherical glass bead after cooling for 3 minutes is 318.42°C.

Step by step solution

01

Convert dimensions and time to SI units

First, we need to convert the dimensions and time given in SI units. Diameter (d) = 10 mm = 0.01 m Cooling Time (t) = 3 min = 180 s
02

Calculate the Biot number (Bi)

The Biot number represents the relative importance of internal temperature gradients to external convection. It's given by the formula: Bi = hLc/k where h is the convection heat transfer coefficient, Lc is the characteristic length, and k is the thermal conductivity. For a sphere, the characteristic length (Lc) is equal to the radius (r) divided by 3. Lc = (d/2)/3 = 0.005/3 m Bi = (28 W/m²·K)(0.005/3 m)/(0.7 W/m·K) = 0.0667
03

Calculate the Fourier number (Fo)

The Fourier number represents the ratio of heat conduction rate to the heat storage rate. It's given by the formula: Fo = αt/Lc^2 where α is the thermal diffusivity and t is the cooling time. The thermal diffusivity (α) can be calculated as: α = k/(ρ·cp) Now, we can calculate the α using the given values: α = (0.7 W/m·K)/(2800 kg/m³ * 750 J/kg·K) = 3.3333×10⁻⁷ m²/s Next, we compute the Fourier number (Fo): Fo = (3.3333×10⁻⁷ m²/s)(180 s)/(0.005/3 m)² = 0.71998
04

Obtain the one-term approximation solution

For the analytical one-term approximation method to solve for the temperature at the center of the sphere (Tc), we need the equation: Tc = T∞ + (Ti - T∞) × Φ where T∞ is the room temperature, Ti is the initial temperature, and Φ is the dimensionless center temperature, which is given by: Φ = (3/2) * e^(-Bi×ξ₁²×Fo)×ξ₁*sin(ξ₁)/(ξ₁+cos(ξ₁)) The first eigenvalue, ξ₁, for a sphere can be found in heat transfer tables or through numerical methods and it is approximately 3.52. Now, we can calculate the dimensionless center temperature: Φ = (3/2) * e^(-0.0667×3.52²×0.71998)×3.52*sin(3.52)/(3.52+cos(3.52)) = 0.75284 Finally, we can find the temperature at the center of the glass bead (Tc): Tc = 30°C + (400°C - 30°C) × 0.75284 = 318.42°C
05

The result

The temperature at the center of the spherical glass bead after cooling for 3 minutes using the analytical one-term approximation method is 318.42°C.

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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 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 at \(25^{\circ} \mathrm{C}\). 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 the analytical one-term approximation method.

The Biot number can be thought of as the ratio of (a) the conduction thermal resistance to the convective thermal resistance (b) the convective thermal resistance to the conduction thermal resistance (c) the thermal energy storage capacity to the conduction thermal resistance (d) the thermal energy storage capacity to the convection thermal resistance (e) none of the above

How does refrigeration prevent or delay the spoilage of foods? Why does freezing extend the storage life of foods for months?

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 a timer so that it will go off when the temperature of the exposed surface of the watermelon drops to \(3^{\circ} \mathrm{C}\). Consider a \(25-\mathrm{cm}\)-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 \(3^{\circ} \mathrm{C}\).

Carbon steel balls $\left(\rho=7833 \mathrm{~kg} / \mathrm{m}^{3}, k=54 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\(, \)c_{p}=0.465 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\(, and \)\left.\alpha=1.474 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right) 8 \mathrm{~mm}$ in diameter are annealed by heating them first to \(900^{\circ} \mathrm{C}\) in a furnace and then allowing them to cool slowly to \(100^{\circ} \mathrm{C}\) in ambient air at \(35^{\circ} \mathrm{C}\). If the average heat transfer coefficient is $75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine how long the annealing process will take. If 2500 balls are to be annealed per hour, determine the total rate of heat transfer from the balls to the ambient air.

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