Question

Carbon steel balls ( \(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}, k=64 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\left.c_{p}=434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(150^{\circ} \mathrm{C}\) are quenched in an oil bath at \(20^{\circ} \mathrm{C}\) for a period of 3 minutes. If the balls have a diameter of \(5 \mathrm{~cm}\) and the convection heat transfer coefficient is \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The center temperature of the balls after quenching will be (Hint: Check the Biot number). (a) \(27.4^{\circ} \mathrm{C}\) (b) \(143^{\circ} \mathrm{C}\) (c) \(12.7^{\circ} \mathrm{C}\) (d) \(48.2^{\circ} \mathrm{C}\) (e) \(76.9^{\circ} \mathrm{C}\)

Short answer

Answer: _____

Step-by-step solution

Calculate the Biot number

First, let's calculate the Biot number (Bi). The Biot number is the ratio of internal thermal resistance to external thermal resistance. It can be calculated using the following formula: \(Bi = \frac{h \cdot L_c}{k}\) Where: \(h\) = convection heat transfer coefficient (\(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)) \(L_c\) = characteristic length (diameter / 6) (diameter is \(5 \mathrm{~cm}\), so we convert it to meters and divide by 6) \(k\) = thermal conductivity of the steel (\(64 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\))

Find the time constant

Next, we need to find the time constant of the heat transfer process (τ), which can be found using the following formula: \(\tau = \frac{\rho \cdot c_p \cdot {!L_c}^2}{3 \cdot k}\) Where: \(ρ\) = density of the steel (\(7830 \mathrm{~kg} / \mathrm{m}^{3}\)) \(c_p\) = specific heat of the steel (\(434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\))

Calculate the temperature ratio

If the Biot number is << 0.1, we can assume that the temperature is uniform throughout the steel ball, as the internal temperature gradient is negligible. It means we can use the following equation: \(\Theta = \frac{T_{c}- T_{\infty}}{T_i - T_{\infty}} = e^{-t/(\rho c_p L_c / h)}\) Where: \(\Theta\)= temperature ratio \(T_i\) = initial temperature (\(150^{\circ} \mathrm{C}\)) \(T_{\infty}\) = temperature of oil bath (\(20^{\circ} \mathrm{C}\)) \(t\) = quenching time (\(3 \mathrm{~min}\), convert it to seconds)

Calculate the center temperature

Finally, having the temperature ratio \((\Theta)\), we can find the center temperature of the steel ball after 3 minutes of quenching: \(T_c = \Theta \cdot (T_i - T_{\infty}) + T_{\infty}\) Use the calculated values from the previous steps to calculate the center temperature \((T_c)\). Compare the result with the given options to find the correct answer.

Key concepts

Biot Number

When studying how heat is transferred in objects during processes like quenching, the Biot number is an essential dimensionless parameter. Its value denotes the ratio of internal to external thermal resistances and is crucial in predicting how uniformly the temperature will change within an object.

The Biot number is calculated using the formula:
\[Bi = \frac{h \cdot L_c}{k}\]
where h represents the convection heat transfer coefficient, L_c is the characteristic length (often the volume over surface area), and k is the thermal conductivity of the material. A small Biot number (Bi < 0.1) indicates that the temperature within the object is reasonably uniform, meaning that we can simplify our calculations by assuming that it changes uniformly throughout the material. Conversely, a large Biot number means that there's a significant temperature gradient inside the object, requiring more complex modeling to accurately predict internal temperatures. Understanding the Biot number helps students and engineers in fields such as materials science and mechanical engineering predict how quickly an object will respond to changes in temperature at its surface.

Convection Heat Transfer Coefficient

Another pivotal concept in heat transfer is the convection heat transfer coefficient, symbolized as h. This coefficient measures the heat transfer rate between a solid surface and a fluid (liquid or gas) per unit area and per unit temperature difference.

The value of the convection heat transfer coefficient depends on various factors, including the properties of the fluid, the flow characteristics, and the surface geometry. For example, in the context of the quenching process mentioned, the convection heat transfer coefficient is a measure of how effectively the oil bath can remove heat from the surface of the steel balls.

The coefficient is in the units of \(W/m^2 \cdot K\) and plays a critical role in calculating the heat loss or gain by an object. Higher values of h imply more efficient convection cooling or heating, leading to quicker changes in an object's surface temperature. Engineers must accurately estimate this coefficient to ensure proper control of the thermal processes they are designing or analyzing.

Thermal Conductivity

Thermal conductivity, denoted by k, is a material-specific property that quantifies the ability of a material to conduct heat. It features prominently in the study of heat transfer since it defines how well heat will flow through a substance.

Materials with high thermal conductivity, such as metals, can transfer heat rapidly, thereby equalizing temperature differences quickly. In contrast, insulating materials with low thermal conductivity inhibit heat flow, maintaining temperature gradients for longer periods.

The unit of thermal conductivity is \(W/m \cdot K\), which indicates the amount of heat (in watts) that will flow through a material with a one-meter thickness for each degree of temperature difference across the material. This concept is integral to solving heat transfer problems like the quenching of steel balls in an oil bath because it helps determine the internal distribution and rate of heat transfer within the object. By combining our knowledge of thermal conductivity with the convection heat transfer coefficient and the Biot number, we can gain a comprehensive understanding of the thermal behaviors exhibited during convection cooling processes.