Question

Stainless steel ball bearings \(\left(\rho=8085 \mathrm{~kg} / \mathrm{m}^{3}, k=\right.\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C}, c_{p}=0.480 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) having a diameter of \(1.2 \mathrm{~cm}\) are to be quenched in water. The balls leave the oven at a uniform temperature of \(900^{\circ} \mathrm{C}\) and are exposed to air at \(30^{\circ} \mathrm{C}\) for a while before they are dropped into the water. If the temperature of the balls is not to fall below \(850^{\circ} \mathrm{C}\) prior to quenching and the heat transfer coefficient in the air is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\), determine how long they can stand in the air before being dropped into the water.

Short answer

Answer: The stainless steel ball bearings can remain in the air for approximately 39.2 seconds before their temperature drops to 850°C.

Step-by-step solution

1. Write down the Newton's Law of Cooling formula

Newton's Law of Cooling can be represented as: \(\frac{dT}{dt}=-hA\frac{(T-T_{\infty})}{\rho Vc_p}\), where \(\frac{dT}{dt}\) is the rate of change in temperature, \(h\) is the heat transfer coefficient, \(A\) is the area through which heat transfer occurs, \(T\) is the temperature of the object, \(T_{\infty}\) is the ambient temperature, \(\rho\) is the density of the material, \(V\) is the volume of the object, and \(c_p\) is the specific heat.

2. Rewrite the cooling formula in terms of diameter and temperature

We know the diameter of ball bearings, so we can find the surface area and volume expressions using the diameter \(d\). Surface area \(A = 4 \pi (\frac{d}{2})^2 = \pi d^2\) Volume \(V = \frac{4}{3} \pi (\frac{d}{2})^3 = \frac{1}{6} \pi d^3\) Now substitute these expressions of \(A\) and \(V\) into the cooling formula: \(\frac{dT}{dt} = -h\pi d^2 \frac{(T-T_{\infty})}{\rho (\frac{1}{6} \pi d^3)c_p}\)

3. Simplify the formula for the rate of change in temperature

Now we can simplify the formula for \(\frac{dT}{dt}\): \(\frac{dT}{dt} = -\frac{6h}{d} \frac{(T-T_{\infty})}{\rho c_p}\)

4. Solve the cooling formula for time

Now we need to find the time taken for the balls to cool from \(900^{\circ} \mathrm{C}\) to \(850^{\circ} \mathrm{C}\). Integrate both sides of the equation between the initial (900°C) and final (850°C) temperatures and their respective times (0 and t) as limits: \(\int_{900}^{850} \frac{dT}{(T-T_{\infty})} = -\int_{0}^{t} \frac{6h}{d\cdot \rho c_p} dt\) Solve the integral and we get: \(-\ln \frac{(850-T_{\infty})}{(900-T_{\infty})}= \frac{6ht}{d\rho c_p}\)

5. Solve the equation for t

Rearrange the equation and solve for time \(t\): \(t = \frac{d\rho c_p}{6h} \ln \frac{900-T_{\infty}}{850-T_{\infty}}\)

6. Plug in the given values

Now substitute all given values into the equation: \(t = \frac{0.012 \times 8085 \times 480}{6\times 125} \ln \frac{900-30}{850-30}\)

7. Calculate the time to cool down

Finally, calculate the time it takes for the balls to cool down from 900°C to 850°C: \(t = 0.0109 \mathrm{~hour}\) or \(t \approx 39.2\mathrm{~seconds}\) The stainless steel balls can remain in the air for approximately 39.2 seconds before being dropped into the water without their temperature falling below 850°C.

Key concepts

Newton's Law of Cooling

Newton's Law of Cooling is a principle that describes the rate of heat exchange between a body and its surrounding environment. It helps predict how quickly an object's temperature will change when exposed to air or liquid at a different temperature. The law is expressed with the formula:\[ \frac{dT}{dt} = -hA \frac{(T-T_{\infty})}{\rho Vc_p} \]where:

• \( \frac{dT}{dt} \) is the rate of temperature change.
• \( h \) is the heat transfer coefficient.
• \( A \) is the surface area for heat transfer.
• \( T \) is the object's temperature.
• \( T_{\infty} \) is the ambient temperature.
• \( \rho \) is the material's density.
• \( V \) is the object's volume.
• \( c_p \) is the specific heat capacity.

By simplifying this formula, as demonstrated in the exercise solution, you can determine how long it takes for an object to cool to a specific temperature. This law is crucial in processes like cooling, heating, and heat treatment, where precise temperature control is essential.

Quenching Process

The quenching process is a rapid cooling method used to alter the properties of metal, such as hardness or strength. It involves heating the metal to a high temperature and then cooling it quickly, usually in water or oil. Quenching is a common technique in metallurgy and manufacturing to achieve desired structural characteristics. The primary goal of quenching is to trap certain phases in the metal structure by preventing the slow formation of other phases during cooling. This is often used in tools and machinery components to improve wear resistance and toughness. Quenching requires understanding the cooling dynamics involved, as the rate of cooling and the medium used significantly influence the final properties of the metal. In our exercise, the stainless steel balls are quenched in water to maintain specific temperature levels before they undergo this cooling process to control their physical properties adequately.

Thermal Conductivity

Thermal conductivity, represented by the symbol \( k \), is a material's ability to conduct heat. It defines how easily heat passes through a material and is crucial in processes involving heat transfer such as heating, cooling, or thermal insulation. Materials with high thermal conductivity quickly transfer heat, while materials with low thermal conductivity maintain heat longer.In the context of the exercise, thermal conductivity helps understand how heat moves through the stainless steel balls. The value \( k = 15.1 \text{ W/m} \cdot {}^{\circ} \text{C} \) tells us how efficiently the balls disseminate heat during the quenching process. This property determines how fast the internal temperature of the balls changes when exposed to different environments.

Heat Transfer Coefficient

The heat transfer coefficient, denoted as \( h \), reflects how effectively heat can move between a solid surface and a fluid in contact with it. It is a measure of the convective heat transfer capability of a process and is a key parameter in designing cooling and heating systems.In our exercise, the heat transfer coefficient is given as \( 125 \text{ W/m}^2 \cdot {}^{\circ} \text{C} \). This value indicates how rapidly heat is exchanged between the stainless steel balls and the surrounding air. A higher coefficient means that the balls lose heat to the air more effectively, leading to quicker cooling. Understanding and calculating the heat transfer coefficient is essential for controlling the cooling or heating rate in various thermal processes, ensuring that industrial processes maintain efficiency and safety.