Question

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor is flowing at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) inside a pipe with \(5 \mathrm{~cm}\) in diameter and \(10 \mathrm{~m}\) in length. The pipe is located in a power plant at \(20^{\circ} \mathrm{C}\), and has a uniform pipe surface temperature of \(100^{\circ} \mathrm{C}\). If the temperature drop between the inlet and exit of the pipe is \(30^{\circ} \mathrm{C}\), and the specific heat of the vapor is \(2190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine the heat transfer coefficient as a result of convection between the pipe surface and the surrounding.

Short answer

Answer: The heat transfer coefficient is 157.3 W/m²·K.

Step-by-step solution

Calculate the heat loss from the pipe due to the temperature drop

Using the specific heat of the vapor, the mass flow rate, and the temperature drop, we can calculate the heat loss from the pipe as follows: \(q = mc\Delta T\) where \(q\) is the heat loss, \(m\) is the mass flow rate (\(0.3\,\mathrm{kg/s}\)), \(c\) is the specific heat of the vapor (\(2190\,\mathrm{J/kg\cdot K}\)), and \(\Delta T\) is the temperature drop (\(30^\circ \mathrm{C}\)). Plugging the given values into the equation, we get: \(q = (0.3\,\mathrm{kg/s}) (2190\,\mathrm{J/kg\cdot K}) (30\,\mathrm{K})\) Calculating the result, we obtain: \(q = 19710\,\mathrm{W}\) (Watts)

Calculate the area of the pipe's surface

Next, we need to find the area of the pipe's surface to determine the heat transfer due to convection. The surface area of a cylinder can be calculated as: \(A = 2\pi rl\) where \(A\) is the area, \(r\) is the radius of the pipe, and \(l\) is the length of the pipe. Given the diameter of the pipe is \(5\,\mathrm{cm}\), we can calculate the radius as: \(r = \frac{d}{2} = \frac{5\,\mathrm{cm}}{2} = 2.5\,\mathrm{cm} = 0.025\,\mathrm{m}\) (converted to meters) Now, we can calculate the surface area of the pipe: \(A = 2\pi (0.025\,\mathrm{m})(10\,\mathrm{m})\) Calculating the result, we obtain: \(A = 1.57\,\mathrm{m^2}\)

Determine the heat transfer coefficient

Finally, we can use the formula for the heat transfer due to convection to find the heat transfer coefficient. The formula is: \(q = hA\Delta T_s\) where \(h\) is the heat transfer coefficient, \(A\) is the area of the pipe's surface, and \(\Delta T_s\) is the temperature difference between the pipe surface and the surroundings. Since we have already calculated the heat loss (\(q\)) and the surface area (\(A\)), we can rearrange the formula to solve for the heat transfer coefficient (\(h\)): \(h = \frac{q}{A\Delta T_s}\) The temperature difference between the pipe surface and the surroundings is: \(\Delta T_s = (100^\circ \mathrm{C}) - (20^\circ \mathrm{C}) = 80\,\mathrm{K}\) Now, we can plug the known values into the formula and solve for \(h\): \(h = \frac{19710\,\mathrm{W}}{(1.57\,\mathrm{m^2})(80\,\mathrm{K})}\) Calculating the result, we obtain: \(h = 157.3\,\mathrm{W/m^2\cdot K}\) Thus, the heat transfer coefficient as a result of convection between the pipe surface and the surrounding is \(157.3\,\mathrm{W/m^2\cdot K}\).

Key concepts

Convection

Convection is one of the primary modes of heat transfer and it occurs when heat is carried away by the movement of fluids or gases. In our exercise involving superheated vapor inside a pipe at a power plant, convection is the process by which heat is transferred from the outer surface of the pipe to the surrounding air.

In technical terms, convection is classified into two types: natural or free convection and forced convection. The former occurs due to buoyancy forces that result from density differences caused by temperature variations in the fluid. The latter, which is more applicable to this exercise, involves the movement of fluid by external means such as a pump or fan. Here, the mass flow rate of the vapor is driven through the pipe, which enhances the heat transfer coefficient, impacting the efficiency of heat exchange.

Role of Convection in Heat Transfer Coefficient

The heat transfer coefficient, a measure of how well heat is transferred by convection, can vary greatly depending on the velocity of the fluid, its properties, and the temperature difference between the fluid and the surroundings. It serves as an indicator of the convection heat transfer capability of the system, and is a crucial value in calculating the energy balance in power plant operations.

Thermal Conductivity

Thermal conductivity represents a material's ability to conduct heat and it is a vital property in the study of thermodynamics and heat transfer. In the context of our exercise, the pipe's material will have its own thermal conductivity value, although it isn't directly provided or required to solve for the heat transfer coefficient due to convection. However, it indirectly affects the overall heat transfer process.

Materials with high thermal conductivity, such as metals, are efficient at transferring heat, whereas materials with low thermal conductivity, such as plastics or ceramics, are less efficient and often serve as insulators. The effectiveness of the pipe in transferring heat from the superheated vapor to its walls before losing it to the surroundings by convection depends on its thermal conductivity.

Understanding Heat Flow

Inside the pipe, the heat initially conducts through the pipe material itself before being convected away. The greater the thermal conductivity of the pipe material, the quicker and more effectively this process occurs. While this property is not the focus of the exercise, knowledge of thermal conductivity is essential when designing and analyzing real-world heat transfer systems in power plants.

Mass Flow Rate

Mass flow rate is a critical concept in thermodynamics and fluid mechanics, indicating the amount of mass passing through a cross-section of a pipe or conduit per unit time. It is typically expressed in units such as kilograms per second (kg/s) and plays a central role in the calculation of heat loss in our exercise.

Understanding the mass flow rate is crucial to determining how much heat is being transported by the vapor. This has direct implications on the heating or cooling processes within any system, including the power plant scenario outlined in the exercise.

Impact on Heat Transfer

A higher mass flow rate generally implies that more heat energy is being carried by the fluid, thus potentially increasing the heat transfer to the surroundings provided the pipe's surface is large enough for effective heat dissipation. Consequently, accurately calculating the mass flow rate is essential for evaluating the cooling needs of the pipe and ensuring optimal operational conditions for the power plant.