Question

A long 18-cm-diameter bar made of hardwood \(\left(k=0.159 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=1.75 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is exposed to air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(8.83 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the center temperature of the bar is measured to be \(15^{\circ} \mathrm{C}\) after a period of 3-hours, the initial temperature of the bar is (a) \(11.9^{\circ} \mathrm{C}\) (b) \(4.9^{\circ} \mathrm{C}\) (c) \(1.7^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-9.2^{\circ} \mathrm{C}\)

Short answer

Question: Using the lumped system analysis approach and the given information, find the initial temperature of the 18-cm-diameter hardwood bar. Options: a) 23°C b) 27°C c) 32°C d) 36°C Answer: __________

Step-by-step solution

Identify the relevant data

From the problem, we can gather the following information: - Diameter of the bar (D) = 0.18 m - Thermal conductivity (k) = 0.159 W/(m·K) - Thermal diffusivity (α) = 1.75 × 10⁻⁷ m²/s - Ambient temperature (T_∞) = 30°C - Heat transfer coefficient (h) = 8.83 W/(m²·K) - Center temperature after 3 hours (T_c) = 15°C - Time (t) = 3 hours = 10,800 s

Calculate the Biot number

The Biot number (Bi) is the ratio of the heat conduction resistance inside the bar to the heat convection resistance at the surface of the bar, which is given by the formula: Bi = hL/k where L is the characteristic length of the bar. For a cylinder, the characteristic length is given by the formula: L = V/A = (πD² / 4h) / πDh = D/4 Now, we can plug in the values of h, D and k to calculate the Biot number: Bi = (8.83 × (0.18/4)) / 0.159

Check the validity of the lumped system analysis

To be able to use the lumped system analysis, the Biot number must be less than 0.1. If the calculated Biot number is less than 0.1, we can proceed with the lumped system analysis. Otherwise, we need to use other methods.

Apply Newton's Law of Cooling to find the initial temperature

Assuming that the Biot number is less than 0.1, we can use Newton's Law of Cooling to find the initial temperature (T_i) of the bar. The equation is: (T_c - T_∞) / (T_i - T_∞) = e^(-ht / (ρVc_p)) where ρ is the density of the bar, V is the volume of the bar, and c_p is the specific heat of the bar. As we have the thermal diffusivity (α) given, the equation can also be written as: (T_c - T_∞) / (T_i - T_∞) = e^(-(h * t * α * A) / V) Plugging in the values, we have: (15 - 30) / (T_i - 30) = e^(-((8.83 * 10,800 * (1.75 * 10^(-7)) * (π * 0.18)) / (π * (0.18^2) / 4))) Now, solve for T_i to find the initial temperature of the bar and match it with one of the given options. The step-by-step solution provided above should enable the student to understand the problem and the approach required to find the initial temperature of the hardwood bar.

Key concepts

Biot number

In heat transfer analysis, the Biot number (Bi) is a dimensionless quantity that helps in determining the choice of an appropriate heat transfer model. It contrasts the heat transfer resistances within a body and at its surface. When dealing with problems involving heating or cooling of a solid object, such as a wooden bar, the Biot number is crucial to identify whether the temperature within the object can be considered uniform. This determination influences the complexity of the calculation significantly.

To calculate the Biot number, one would typically use the formula:
\[ Bi = \frac{hL}{k} \]
where \( h \) is the heat transfer coefficient, \( L \) is the characteristic length of the body, and \( k \) is the thermal conductivity of the material. For cylindrical objects like the bar in the exercise, the characteristic length is calculated as the diameter divided by four (\( L = \frac{D}{4} \)).

If the Biot number is less than 0.1, it indicates that the temperature within the object is nearly uniform and a simpler model called 'lumped system analysis' can be applied. Conversely, a Biot number greater than 0.1 suggests significant temperature gradients within the object, requiring a more complex heat transfer model. This concept is vital for students to grasp as it informs the subsequent steps in their calculations.

Thermal conductivity

Thermal conductivity (\( k \)) is a property that measures a material's ability to conduct heat. It plays a prominent role in the field of thermodynamics and heat transfer, informing us about how well a substance can transfer heat through it. In the context of the exercise, the thermal conductivity value for hardwood was given as 0.159 W/(m·K), which is relatively low compared to metals. This means that hardwood is not a very good conductor of heat.

The value of thermal conductivity becomes critical when you need to determine the rate at which heat flows through materials. The higher the thermal conductivity, the more efficient the material is at conducting heat. This property is utilized in the formula for the Biot number and in Fourier's law of heat conduction, which describesthe rate of heat flow across a material. Learning to apply the concept of thermal conductivity will help students solve a range of problems in heat transfer.

Newton's Law of Cooling

Newton's Law of Cooling is a principle used to describe the rate of heat loss of a body in relation to the surrounding environment. It's specifically applicable when the heat transfer between the body and its environment is governed by convection. Students should understand that this law holds true for situations where the temperature difference between the object and its environment is relatively small and the rate of cooling is proportional to this temperature difference.

The mathematical expression of Newton's Law of Cooling is:
\[ \frac{T_c - T_{\infty}}{T_i - T_{\infty}} = e^{-\frac{ht}{\rho V c_p}} \]
Where \( T_c \) is the current temperature, \( T_{\infty} \) is the ambient temperature, \( T_i \) is the initial temperature, \( h \) is the heat transfer coefficient, \( t \) is time, \( \rho \) is the density, \( V \) is the volume, and \( c_p \) is the specific heat. In our exercise, this law helps with concluding the initial temperature of the hardwood bar after undergoing cooling for a known time period.

Thermal diffusivity

Thermal diffusivity (\( \alpha \)) is a material-specific parameter that describes how quickly heat is spread through a material. It is calculated as the ratio of thermal conductivity (\( k \)) to the product of density (\( \rho \)) and specific heat capacity (\( c_p \)). Thermal diffusivity is measured in square meters per second (\( m^2/s \)) and is represented in formulas as \( \alpha \).

In the formulas used to solve heat transfer problems, such as in the hardwood bar exercise, thermal diffusivity incorporates both the ability of the material to conduct heat and its capacity to store heat. The equation provided in Step 4 of the solution utilizes thermal diffusivity:
\[ (T_c - T_{\infty}) / (T_i - T_{\infty}) = e^{-\left(\frac{h t \alpha A}{V}\right)} \]
By understanding thermal diffusivity, students can appreciate how quickly different materials respond to changes in temperature. Materials with high thermal diffusivity can adjust their temperature more quickly to match their surroundings, which is key to predicting the temperature changes in thermal systems.