Question

A long roll of 2-m-wide and \(0.5\)-cm-thick 1-Mn manganese steel plate coming off a furnace at \(820^{\circ} \mathrm{C}\) is to be quenched in an oil bath \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(45^{\circ} \mathrm{C}\). The metal sheet is moving at a steady velocity of \(15 \mathrm{~m} / \mathrm{min}\), and the oil bath is \(9 \mathrm{~m}\) long. Taking the convection heat transfer coefficient on both sides of the plate to be \(860 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the sheet metal when it leaves the oil bath. Also, determine the required rate of heat removal from the oil to keep its temperature constant at \(45^{\circ} \mathrm{C}\).

Short answer

Answer: The required rate of heat removal from the oil bath is 24,029,400 W.

Step-by-step solution

Calculate the surface area of the steel plate in contact with the oil bath

The steel plate has a width of 2 meters, and the oil bath is 9 meters long. They are in contact on both sides of the plate. To calculate the surface area of the steel plate that is in contact with the oil, we multiply the width of the plate by the length of the oil bath and multiply the result by 2 (to account for both sides). So the surface area will be: \(A = 2\times (2\ \text{m}\times 9\ \text{m}) = 36\ \text{m}^2\).

Determine the temperature difference between the steel plate and the oil bath

The initial temperature of the steel plate is given as \(820^{\circ}\mathrm{C}\), and the oil bath is maintained at \(45^{\circ}\mathrm{C}\). So, the temperature difference is: \(\Delta T = 820^{\circ}\mathrm{C} - 45^{\circ}\mathrm{C} = 775\ \mathrm{K}\).

Calculate the rate of heat transfer from the steel plate to the oil bath

Using the convection heat transfer coefficient \(h = 860\ \mathrm{W} / \mathrm{m}^2 \cdot \mathrm{K}\), the surface area \(A = 36\ \text{m}^2\), and the temperature difference \(\Delta T = 775\ \mathrm{K}\), the rate of heat transfer can be calculated as: \(q = h \times A \times \Delta T = 860\ \frac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{K}}\times 36\ \mathrm{m}^2 \times 775\ \mathrm{K} = 24,\!029,\!400\ \mathrm{W}\).

Calculate the final temperature of the sheet metal as it leaves the oil bath

Since the sheet metal is moving at a steady velocity of \(15\ \mathrm{m} / \mathrm{min}\) and the length of the oil bath is \(9\ \text{m}\), it takes the steel plate \(T_{\text{contact}} = \frac{9\ \text{m}}{15\ \text{m/min}} = 0.6\ \text{min}\) to traverse the oil bath. We are given the specific heat capacity of the metal as \(c_{p} = 2.0\ \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\). To find the final temperature of the sheet metal, we can calculate the amount of heat transferred from the steel plate to the oil bath: \(Q_{\text{transferred}} = mc_{p}(T_{\text{final}} - T_{\text{initial}} )\) Since we know the heat transfer rate and the time spent in contact with the oil bath, we can find the heat transferred: \(Q_{\text{transferred}} = q \times T_{\text{contact}} = 24,\!029,\!400\ \mathrm{W} \times 0.6\ \text{min} \times \frac{60\ \text{s}}{1\ \text{min}} = 864,\!105,\!600\ \mathrm{J}\) Due to lack of information about the mass of metal sheet which goes through the bath, we can not determine the sheet metal's final temperature.

Determine the rate of heat removal from the oil bath

To maintain the oil bath temperature constant at \(45^{\circ}\mathrm{C}\), the required rate of heat removal from the oil has to be equal to the heat transfer rate between the sheet metal and the oil bath. So the required rate of heat removal is \(24,\!029,\!400\ \mathrm{W}\). In conclusion, it is not possible to determine the final temperature of the sheet metal as it leaves the oil bath due to insufficient information provided about the mass of the steel plate that comes in contact with the oil bath. However, the required rate of heat removal from the oil to keep its temperature constant is \(24,\!029,\!400\ \mathrm{W}\).

Key concepts

Convection Heat Transfer

Understanding the mechanism of convection heat transfer is fundamental to solving problems related to temperature regulation in various materials and fluids. Convection is one of the three modes of heat transfer, alongside conduction and radiation, and it involves the movement of heat by the motion of fluids, which include liquids and gases.

When a fluid such as oil is heated, it expands and becomes less dense. This less dense fluid then rises and is replaced by cooler, denser fluid, which in turn is heated and rises. This cycle creates a circulating motion, referred to as a convection current. The heat from the steel plate in our exercise is transferred to the surrounding oil predominantly through convection.

In the context of the exercise, the steel plate coming off a furnace has a higher temperature and transfers heat to the cooler oil bath. This transfer is facilitated by the convection heat transfer coefficient, usually denoted as h, which quantifies the rate at which heat is transferred from the surface of a solid to a fluid or from one fluid to another. In our exercise, h is provided as 860 W/m²·K, indicating the effectiveness of the oil in absorbing heat from the steel plate.

Temperature Difference

The temperature difference, often represented by the symbol \(\Delta T\), is a driver for heat transfer; the greater the difference, the higher the potential for heat to transfer. It's like a ball at the top of a hill; the steeper the hill (or the greater the temperature difference), the faster the ball (or heat) will roll down (or transfer).

In the exercise, we have a significant temperature differential: the steel plate is at a scorching 820°C while the oil bath is a relatively cool 45°C, creating a \(\Delta T\) of 775 K. This large temperature difference suggests a rapid rate of heat transfer from the steel to the oil, but it also challenges the oil's capacity to absorb heat without a rise in its own temperature.

Understanding and accurately calculating the temperature difference are crucial for determining the rate of heat transfer in heat exchange systems, as well as designing appropriate cooling mechanisms to control temperatures effectively.

Rate of Heat Transfer

The rate of heat transfer, typically denoted as \(q\), is the amount of heat being transferred per unit time. It is an essential aspect of thermal engineering calculations, reflecting how quickly a material or system gains or loses heat. In our textbook example, the rate of heat transfer is calculated from the convection heat transfer coefficient, the surface area of the sheet metal in contact with the oil, and the temperature difference between the two.

The formula to calculate the rate of heat transfer in the context of convection is given by \(q = h \times A \times \Delta T\), where \(h\) is the convection heat transfer coefficient, \(A\) is the contact area, and \(\Delta T\) is the temperature difference between the two surfaces. Through the exercise, we determine this rate to be 24,029,400 W, which reflects the intense heat taken on by the oil to ensure the steel plate cools.

This value isn't just an academic exercise, as it directly relates to how the system must be managed: An oil bath, or any cooling system, must be capable of removing the same rate of heat to maintain its temperature. This ties directly to the practical application, such as in industrial processes where precise temperature control is critical for product quality and system stability.