Question

Carbon-steel balls \(\left(\rho=7833 \quad \mathrm{kg} / \mathrm{m}^{3} \text { and } c_{p}=\right.\) \(\left.0.465 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right) 8 \mathrm{mm}\) in diameter are annealed by heating them first to \(900^{\circ} \mathrm{C}\) in a furnace and then allowing them to cool slowly to \(100^{\circ} \mathrm{C}\) in ambient air at \(35^{\circ} \mathrm{C}\). If 2500 balls are to be annealed per hour, determine \((a)\) the rate of heat transfer from the balls to the air and ( \(b\) ) the rate of entropy generation due to heat loss from the balls to the air.

Short answer

Answer: (a) The rate of heat transfer from the balls to the air is 1950 kJ/hour, and (b) the rate of entropy generation due to heat loss from the balls to the air is 4.54 kJ/(hour⋅K).

Step-by-step solution

List the given information

We are given: - Density of carbon-steel balls: \(\rho = 7833 \, \mathrm{kg / m^3}\) - Specific heat capacity: \(c_p = 0.465 \, \mathrm{kJ/(kg \cdot ^\circ C)}\) - Diameter of the balls: \(d = 8 \, \mathrm{mm}\) - Initial temperature: \(T_{initial} = 900^{\circ} \mathrm{C}\) - Final temperature: \(T_{final} = 100^{\circ} \mathrm{C}\) - Ambient air temperature: \(T_{ambient} = 35^{\circ} \mathrm{C}\) - 2500 balls to be annealed per hour

Calculate the mass of one ball

Using the density and diameter of the balls, we can find the mass of one ball. First, we need to find the volume of one ball using the formula for the volume of a sphere (\(V = \frac{4}{3}\pi r^3\)), and then multiply by the density: \(r = \frac{d}{2} = 4 \, \mathrm{mm}\) \(V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.004 \, \mathrm{m})^3 \approx 2.68 \times 10^{-7} \, \mathrm{m^3}\) \(m = \rho \cdot V = 7833 \, \mathrm{kg/m^3} \cdot 2.68 \times 10^{-7} \, \mathrm{m^3} \approx 2.1 \times 10^{-3} \, \mathrm{kg}\)

Calculate the heat transfer rate for one ball

Since we have the specific heat, mass, and the temperature difference, we can find the heat transfer per ball using the formula: \(\dot{Q} = m c_p \Delta T\) \(\Delta T = T_{initial} - T_{final} = 900^{\circ}\mathrm{C} - 100^{\circ}\mathrm{C} = 800^{\circ}\mathrm{C}\) \(\dot{Q}_{ball} = (2.1 \times 10^{-3} \, \mathrm{kg}) (0.465 \, \mathrm{kJ/(kg \cdot ^\circ C)})(800^{\circ}\mathrm{C}) \approx 0.78 \, \mathrm{kJ}\)

Calculate the heat transfer rate for 2500 balls

Since we have the heat transfer rate for one ball, we can multiply that value by the total number of balls to find the heat transfer rate for 2500 balls: \(\dot{Q}_{total} = (0.78 \, \mathrm{kJ})\cdot (2500 \, \text{balls}) \approx 1950 \, \mathrm{kJ/hour}\) (a)

Calculate the entropy generation rate

To find the entropy generation rate, we can use the heat transfer rate and the temperatures of the two reservoirs (balls and ambient air). The formula for the entropy generation rate is: \(\dot{S} = \frac{\dot{Q}}{T_{cold}} - \frac{\dot{Q}}{T_{hot}}\) Convert the temperatures to Kelvin: \(T_{cold} = T_{ambient} + 273.15 = 35^{\circ}\mathrm{C} + 273.15 \, \mathrm{K} = 308.15 \, \mathrm{K}\) \(T_{hot} = T_{initial} + 273.15 = 900^{\circ}\mathrm{C} + 273.15 \, \mathrm{K} = 1173.15 \, \mathrm{K}\) Now, we can find the entropy generation rate: \(\dot{S} = \frac{1950\, \mathrm{kJ/hour}}{308.15\, \mathrm{K}} - \frac{1950\, \mathrm{kJ/hour}}{1173.15\, \mathrm{K}}\) \(\dot{S} \approx 4.54\, \mathrm{kJ /(hour \cdot K)}\) (b) In this problem, we have found (a) the rate of heat transfer from the balls to the air to be 1950 kJ/hour, and (b) the rate of entropy generation due to heat loss from the balls to the air to be 4.54 kJ/(hour⋅K).

Key concepts

Specific Heat Capacity

Specific heat capacity, denoted as \(c_p\), is a property that determines the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). In our exercise, the specific heat capacity of carbon-steel balls is given as 0.465 kJ/kg⋅°C. This value is essential for calculating the heat transfer rate, as it directly correlates with the energy needed to raise or lower the temperature of the steel balls during the annealing process.

When we apply the formula \(\dot{Q} = m c_p \Delta T\), where \(m\) is the mass and \(\Delta T\) is the change in temperature, we can quantify the heat exchange for a single temperature shift. In practical applications, knowing the specific heat capacity allows engineers to predict the thermal response of materials under various conditions, which is crucial for processes like annealing.

Entropy Generation

Entropy generation is a concept from the second law of thermodynamics indicating the amount of energy that becomes unavailable for doing useful work. It's a measure of the increase in disorder or randomness within a system. In the problem, we calculate entropy generation due to heat transfer from the hot steel balls to the colder ambient air.

The formula used in our solution, \(\dot{S} = \frac{\dot{Q}}{T_{cold}} - \frac{\dot{Q}}{T_{hot}}\), involves the heat transfer rate \(\dot{Q}\) and the absolute temperatures of the cold (ambient air) and hot (steel balls) reservoirs in Kelvin. Entropy generation is important as it signifies the inefficiency of a process. A higher rate of entropy generation implies more wasted energy, thus making a process less efficient. This is a crucial consideration in the design of thermal systems where energy conservation is key.

Temperature Difference

The temperature difference, \( \Delta T \), is the driving force for heat transfer in thermal processes. It's simply the difference between the temperatures of two systems or regions. In our exercise, the temperature difference is between the initial temperature of the steel balls \( (900^\circ C) \) and the ambient air temperature \( (35^\circ C) \) after cooling.

A larger \( \Delta T \) usually means a higher rate of heat transfer, which can be calculated using \( \Delta T = T_{initial} - T_{final} \). For our problem, this difference is crucial for determining the rate at which heat is lost to the environment (air) and is used in calculating both the heat transfer rate and the entropy generation rate. Understanding how temperature difference affects heat transfer is vital for designing efficient cooling or heating processes.

Annealing Process

The annealing process is a heat treatment method that alters the physical and sometimes chemical properties of a material to increase its ductility and reduce its hardness, making it more workable. During annealing, materials such as the carbon-steel balls mentioned in our exercise are heated to a specific temperature and then allowed to cool slowly. This controlled process of heating and cooling changes the microstructure of the material, relieving internal stresses and improving its overall properties.

Furthermore, annealing is often meticulously planned and executed to achieve the desired material characteristics, and understanding the principles of heat transfer and thermodynamics, like specific heat capacity, entropy generation, and temperature difference, are fundamental to mastering the annealing process. This is why calculating the heat transfer rate and entropy generation during cooling, as we did in the exercise, is not just academic but also has real-world utility in materials engineering and manufacturing processes.