91Ó°ÊÓ

A long plastic rod of \(30-\mathrm{mm}\) diameter \((k=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\rho c_{p}=1040 \mathrm{~kJ} / \mathrm{m}^{3} \cdot K\right)\) is uniformly heated in an oven as preparation for a pressing operation. For best results, the temperature in the rod should not be less than \(200^{\circ} \mathrm{C}\). To what uniform temperature should the rod be heated in the oven if, for the worst case, the rod sits on a conveyor for 3 min while exposed to convection cooling with ambient air at \(25^{\circ} \mathrm{C}\) and with a convection coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? A further condition for good results is a maximum-minimum icmperature difference of less than \(10^{\circ} \mathrm{C}\). Is this condition satisfied and, if not, what could you do to satisfy it?

Step by step solution

01

Understand the Problem

We need to determine the initial uniform temperature to heat a plastic rod in the oven such that its temperature does not fall below 200°C after 3 minutes of cooling. We also need to ensure that the maximum temperature difference across the rod is less than 10°C.

02

Define Initial Parameters

Let the initial temperature be \(T_0\) and the ambient temperature be \(T_\infty = 25^{\circ} \mathrm{C}\). The rod's diameter is 30 mm, and the convection coefficient is \(h = 8 \mathrm{~W/m^2 \, K}\). The thermal properties are given as \(k = 0.3 \mathrm{~W/m \, K}\) and \(\rho c_p = 1040 \mathrm{~kJ/m^3 \, K}\).

03

Calculate Heat Transfer Coefficient

The Biot number \(Bi\) needs to be calculated using:\[ Bi = \frac{hL_c}{k} \]where \(L_c = \frac{D}{2} = 0.015 \mathrm{~m}\) is the characteristic length of the rod. This will help determine whether temperature gradients inside the rod are significant.

04

Determine Biot Number

Now, let's calculate:\[ Bi = \frac{8 \times 0.015}{0.3} = 0.4 \]This Biot number suggests that the internal temperature gradient cannot be ignored, so the lumped capacitance method isn't applicable.

05

Use the Transient Conduction Equation

We use the one-term approximation for transient heat conduction:\[ \frac{T(t) - T_\infty}{T_0 - T_\infty} = e^{-\mathrm{Fo}(C_1^2 + Bi^2)} \]\[ \mathrm{Fo} = \frac{\alpha t}{L_c^2} \] where \(\alpha = \frac{k}{\rho c_p}\) is the thermal diffusivity.

06

Calculate Thermal Diffusivity and Fourier Number

Calculate \(\alpha\):\[ \alpha = \frac{0.3}{1040} = 2.88 \times 10^{-4} \mathrm{~m^2/s} \]Find \(\mathrm{Fo}\) for \(t = 180 \, \text{s}\):\[ \mathrm{Fo} = \frac{2.88 \times 10^{-4} \times 180}{(0.015)^2} = 2.304 \]

07

Solve for Initial Temperature

Use the transient conduction equation: \[ \frac{T(t) - 25}{T_0 - 25} = e^{-(2.304)(\pi^2)} \]Solve for \(T_0\): Requiring \(T(t) \geq 200\), solve to find \(T_0\) greater than a critical value for when max gradient across is 10C.

08

Analyze Maximum Temperature Difference

Check if the rod's surface reaches 200°C keeping a max/min difference across rod maintained within 10°C using exact conditions.

09

Conclusion and Recommendations

Analyze based on \(T_0 \geq 200 + 10 + 25 = 235\,\text{°C}\). Adjust conditions like time or better insulation if condition is not met.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. It indicates how easily heat flows through a material. In the context of the plastic rod problem, thermal conductivity (\( k \)) is given as 0.3 W/m·K. This means that 0.3 watts of heat energy pass through a one-meter thick piece of the material per degree Celsius temperature difference.

Materials with high thermal conductivity, like metals, transfer heat quickly, while materials with low thermal conductivity, such as plastic, insulate heat more effectively. Thus, the plastic rod has a relatively low thermal conductivity, making it slower to heat up and cool down compared to a metal rod.

• Units of measurement: Watts per meter per degree Kelvin (W/m·K)
• Importance in engineering: Affects design considerations for heating and cooling processes.

Biot Number

The Biot number (Bi) is a dimensionless quantity used in heat transfer calculations to determine whether or not a body can be assumed to conduct heat uniformly. It relates the rate of heat conduction within a body to the rate of heat transfer across the surface to the surroundings.

In the rod problem, the Biot number is calculated using the formula:\[ Bi = \frac{hL_c}{k} \]where \( h \) is the convection heat transfer coefficient, \( L_c \) is the characteristic length, and \( k \) is the thermal conductivity. With a Bi number of 0.4, it indicates that internal temperature gradients are significant, so we cannot ignore conduction effects within the rod.

• If Bi < 0.1, surface temperature differences are negligible (lumped system approach is applicable).
• If Bi > 0.1, temperature gradients within the object need to be considered.

Fourier Number

The Fourier number (\( \,Fo \,\)) is another dimensionless number used in heat transfer analysis. It provides insight into the ability of thermal energy to diffuse through an object in relation to elapsed time. It is especially useful when analyzing transient heat conduction.

For our rod, the Fourier number is calculated by:\[ \mathrm{Fo} = \frac{\alpha t}{L_c^2} \]where \( \alpha \) is the thermal diffusivity, \( t \) is time, and \( L_c \) is the characteristic length. A Fourier number of 2.304 for 180 seconds implies a significant change in temperature profile within the rod over this period.

• Higher Fourier numbers indicate greater heat diffusion and faster temperature equalization.
• Used to predict temperature distribution in transient heat conduction scenarios.

Thermal Diffusivity

Thermal diffusivity is a property that measures how quickly heat diffuses through a material. It combines the effects of thermal conductivity, density, and specific heat capacity. Mathematically, it's expressed as:\[ \alpha = \frac{k}{\rho c_p} \]where \( \alpha \) is the thermal diffusivity, \( k \) is thermal conductivity, \( \rho \) is density, and \( c_p \) is specific heat capacity.

For the plastic rod, thermal diffusivity is 2.88 x 10^-4 m²/s, which is a relatively low value. This suggests that heat spreads slowly through the rod, making it retain temperature differentials longer than materials with higher thermal diffusivity.

• Affects how quickly a material adjusts to changes in temperature.
• Less diffusion means longer time for temperature equilibrium.

Heat Conduction in Solids

Heat conduction in solids involves the transfer of thermal energy through a solid material without any movement of the material itself. This process is driven by temperature differences and can be described by Fourier's law of heat conduction.

In the case of the plastic rod, heat conduction is a crucial factor from initial heating to cooling when removed from the oven. The rate of heat loss and the uniformity of the temperature profile are determined by the rod's material properties, specifically its thermal conductivity and diffusivity.
Key points regarding heat conduction:

• Conduction is the dominant form of heat transfer inside solids.
• Temperature gradients within a body dictate the direction and rate of heat flow.