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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.

Short Answer

Expert verified
Question: Determine the convection heat transfer coefficient for a plastic rod that has been heated and is cooling in ambient air. Short Answer: Using the given information and the analytical one-term approximation method, we can calculate the dimensionless Biot Number and dimensionless time. Solving the equation θ(t) = 1 - 0.75 * (τ·exp(-Bi²·τ)) for the Biot Number, we can then use the Biot number and the given values of Lc and k to find the convection heat transfer coefficient, h, for this process.

Step by step solution

01

Obtain the given information

The diameter of the rod, D = 2 cm = 0.02 m, the density of the plastic rod, ρ = 1190 kg/m³, specific heat capacity at constant pressure, cp = 1465 J/(kg·K), thermal conductivity, k = 0.19 W/(m·K), initial temperature Ti = 70°C, final temperature Tf = 30°C, the ambient temperature Ta = 25°C, and the cooling time t = 1388 s.
02

Calculate the dimensionless Biot Number

To calculate the Biot number, we need the radius of the rod, r = D/2 = 0.01 m. The characteristic length for a cylinder is Lc = r/2 = 0.005 m. Now, we can calculate the Biot number, Bi = (h·Lc)/k, where h is the convection heat transfer coefficient. We will solve for h later.
03

Calculate the dimensionless time

First, we need to calculate the thermal diffusivity, α = k/(ρ·cp) = 0.19 W·m/K / (1190 kg/m³ · 1465 J/kg·K). Next, we calculate the dimensionless time, τ = (α·t) / (Lc²) = (α·1388 s) / (0.005 m)².
04

Find the analytical one-term approximation

We use the analytical one-term approximation for cylinders, given by: θ(t) = (T(t) - Ta) / (Ti - Ta) = 1 - 0.75 * (τ·exp(-Bi²·τ)) where θ(t) is the dimensionless temperature. The final temperature T(t) is 30°C, so substitute this into the formula to find θ(t): θ(t) = (30°C - 25 °C) / (70°C - 25°C)
05

Solve for the Biot Number

Now, we will substitute the values of θ(t), τ, and the analytical one-term approximation into the equation and then solve for the Bi: θ(t) = 1 - 0.75 * (τ·exp(-Bi²·τ)) Now we solve this equation for Bi.
06

Calculate the convection heat transfer coefficient

We can now use the Biot number to find the convection heat transfer coefficient, h: Bi = (h·Lc)/k Solve this equation for h using the calculated Bi and given values of Lc and k. After finding the value of h, we have successfully determined the convection heat transfer coefficient for this process.

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Most popular questions from this chapter

A small chicken $(k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad \alpha=0.15 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) can be approximated as an \(11.25-\mathrm{cm}\)-diameter solid sphere. The chicken is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is to be cooked in an oven maintained at \(220^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). With this idealization, the temperature at the center of the chicken after a 90 -min period is (a) \(25^{\circ} \mathrm{C}\) (b) \(61^{\circ} \mathrm{C}\) (c) \(89^{\circ} \mathrm{C}\) (d) \(122^{\circ} \mathrm{C}\) (e) \(168^{\circ} \mathrm{C}\)

Consider a spherical shell satellite with outer diameter of \(4 \mathrm{~m}\) and shell thickness of \(10 \mathrm{~mm}\) that is reentering the atmosphere. The shell satellite is made of stainless steel with properties of $\rho=8238 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=468 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\(, and \)k=13.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. During the reentry, the effective atmosphere temperature surrounding the satellite is \(1250^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $130 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}$. If the initial temperature of the shell is \(10^{\circ} \mathrm{C}\), determine the shell temperature after $5 \mathrm{~min}$ of reentry. Assume heat transfer occurs only on the satellite shell.

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 \(3^{\circ} \mathrm{C}\).

In a meat processing plant, 2-cm-thick steaks $\left(k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\( and \)\left.\alpha=0.91 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\( that are initially at \)25^{\circ} \mathrm{C}$ are to be cooled by passing them through a refrigeration room at \(-11^{\circ} \mathrm{C}\). The heat transfer coefficient on both sides of the steaks is \(9 \mathrm{~W} / \mathrm{m}^{2}\). K. If both surfaces of the steaks are to be cooled to \(2^{\circ} \mathrm{C}\), determine how long the steaks should be kept in the refrigeration room. Solve this problem using the analytical one-term approximation method.

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 \(k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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