Question

\(\begin{array}{lllllll} & \text { A } & \text { long } & \text { roll } & \text { of } & \text { 2-m-wide } & \text { and } & \text { 0.5-cm-thick }\end{array}\) 1-Mn manganese steel plate \(\left(\rho=7854 \mathrm{kg} / \mathrm{m}^{3} \text { and } c_{p}=\right.\) \(\left.0.434 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) coming off a furnace at \(820^{\circ} \mathrm{C}\) is to be quenched in an oil bath at \(45^{\circ} \mathrm{C}\) to a temperature of \(51.1^{\circ} \mathrm{C}\) If the metal sheet is moving at a steady velocity of \(10 \mathrm{m} / \mathrm{min}\) determine the required rate of heat removal from the oil to keep its temperature constant at \(45^{\circ} \mathrm{C}\).

Short answer

To keep the temperature of the oil constant at \(45^{\circ}\mathrm{C}\), the rate of heat removal needs to be \(4356.1 \mathrm{kJ/s}\).

Step-by-step solution

Calculate the volume and mass of the steel plate

The volume of the steel plate can be calculated using the formula: $$V = W \times T \times L$$ Where \(W\) is the width of the plate (2 m), \(T\) is the thickness of the plate (converted to meters), and \(L\) is the length of the plate, which can be determined from the steady velocity: $$L = v \times t$$ Since the velocity, \(v = 10\) m/min, in order to calculate the mass per unit time, we can consider a length of the steel corresponding to 1 minute of travel: $$L = 10 \mathrm{m/min} \times 1 \mathrm{min} = 10 \mathrm{m}$$ The thickness should be converted to meters: $$T = 0.5 \mathrm{cm} \times \frac{1 \mathrm{m}}{100 \mathrm{cm}} = 0.005 \mathrm{m}$$ So, the volume of the steel plate can be calculated as: $$V = 2 \mathrm{m} \times 0.005 \mathrm{m} \times 10 \mathrm{m} = 0.1 \mathrm{m}^3$$ Now, we can calculate the mass of the steel plate using its density, \(\rho = 7854 \mathrm{kg/m}^3\): $$m = \rho \times V = 7854 \mathrm{kg/m}^3 \times 0.1 \mathrm{m}^3 = 785.4 \mathrm{kg}$$

Calculate the temperature difference

The temperature difference between the initial and final temperatures of the steel plate is: $$\Delta{T} = T_\mathrm{final} - T_\mathrm{initial} = 51.1^{\circ}\mathrm{C} - 820^{\circ}\mathrm{C} = -768.9^{\circ}\mathrm{C}$$

Calculate the heat transfer rate

Now we can calculate the heat transfer rate using the formula: $$q = mc_p\Delta{T}$$ Plug in the values of \(m\), \(c_p\), and \(\Delta{T}\): $$q = 785.4 \mathrm{kg} \times 0.434 \mathrm{kJ/kg}^{\circ}\mathrm{C} \times (-768.9^{\circ}\mathrm{C}) = -261366.03 \mathrm{kJ}$$

Convert the heat transfer rate into the rate of heat removal

Since we are calculating the rate of heat removal per minute, we should divide the calculated heat transfer rate by 1 minute (60 seconds): $$q_\mathrm{removal} = \frac{-261366.03 \mathrm{kJ}}{60 \mathrm{s}} = -4356.1 \mathrm{kJ/s}$$ Therefore, the rate of heat removal from the oil to keep its temperature constant at \(45^{\circ}\mathrm{C}\) is \(4356.1 \mathrm{kJ/s}\).

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In the context of our exercise, thermodynamics helps explain how heat transfers from a hot steel plate to the cooler oil in the quenching process.

The exercise involves calculating the rate of heat transfer, which is a fundamental concept in thermodynamics. It encompasses understanding how energy is converted from one form to another and balancing energy flows to achieve a desired temperature change. In this case, the aim is to maintain the oil at a constant temperature while it absorbs heat from the steel plate.

Specific Heat

Specific heat, symbolized by \(c_p\), represents the amount of heat per unit mass required to raise the temperature of a substance by one degree Celsius. It's a property that determines how a material responds to heating or cooling and hence is critical in calculations involving temperature change.

In the problem, using the specific heat of manganese steel, we calculate the heat that needs to be extracted to lower its temperature from the furnace level until it closely matches the oil's temperature. This vital thermodynamic property aids in determining the thermal energy exchange rate as the steel goes through the quenching process.

Material Thermal Properties

Material thermal properties, including density (\(\rho\)) and specific heat (\(c_p\)), greatly influence how heat is transferred in a material. Density is a measure of how much mass is contained in a given volume, which directly affects the mass of the steel plate calculated in the exercise.

These thermal properties, combined with the steel's dimensions and temperature change, determine the overall energy balance when the steel interacts with the quenching medium. Understanding these properties helps engineers and scientists predict how different materials will behave under thermal processes.

Quenching Process

The quenching process entails rapidly cooling a hot material to alter its physical and sometimes chemical properties. This process is essential for materials like steel to enhance qualities such as hardness and strength.

In the exercise, quenching is achieved by submerging the hot steel plate into an oil bath. Quenching typically requires controlling the rate at which the material cools to prevent warping or introducing residual stresses. Calculating the rate of heat removal is crucial to maintain the oil bath’s temperature, which affects the quality of the quenching and the final properties of the steel.