Question

Cold conditioned air at \(12^{\circ} \mathrm{C}\) is flowing inside a \(1.5\)-cm- thick square aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) duct of inner cross section \(22 \mathrm{~cm} \times 22 \mathrm{~cm}\) at a mass flow rate of \(0.8 \mathrm{~kg} / \mathrm{s}\). The duct is exposed to air at \(33^{\circ} \mathrm{C}\) with a combined convection-radiation heat transfer coefficient of \(13 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection heat transfer coefficient at the inner surface is \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature in the duct should not increase by more than \(1^{\circ} \mathrm{C}\) determine the maximum length of the duct.

Short answer

To summarize, under the given conditions, it is impossible to maintain a maximum temperature increase of no more than 1°C for the air inside the duct. The key factors that affect this outcome are the mass flow rate, the material's thermal conductivity, and the heat transfer coefficients. To achieve a temperature increase of no more than 1°C, it might be necessary to reduce the mass flow rate or use a more insulating material for the duct wall.

Step-by-step solution

Find the thermal resistance of the duct wall

The thermal resistance of the duct wall can be calculated using the equation \(R_{cond} = \frac{L}{k \cdot A_c}\), where \(R_{cond}\) is the conduction thermal resistance, \(L\) is the thickness of the duct wall, \(k\) is the thermal conductivity of the material, and \(A_c\) is the wall's cross-sectional area (per unit length of the duct). In this case, \(L = 0.015 \mathrm{~m}\), \(k = 237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and the perimeter \(p = 4 \cdot 0.22 \mathrm{~m}\). So, the cross-sectional area per unit length is \(A_c = p \cdot L = 0.0132 \mathrm{~m}^2/\mathrm{m}\). Now we can find the thermal resistance of the duct wall: \(R_{cond} = \frac{0.015}{237 \cdot 0.0132} = 4.717 \times 10^{-4} \mathrm{~K} / \mathrm{W}\)

Find the heat transfer rates for the inner and outer surface

The heat transfer rate due to convection at the inner surface can be calculated using the equation \(q_{in} = h_{in} \cdot A_{in} \cdot \Delta T_{in}\), where \(h_{in}\) is the inner heat transfer coefficient, \(A_{in}\) is the inner surface area per unit length, and \(\Delta T_{in}\) is the temperature difference between the air and the inner surface. Similarly, the heat transfer rate due to combined convection-radiation at the outer surface can be calculated using the equation \(q_{out} = h_{out} \cdot A_{out} \cdot \Delta T_{out}\), where \(h_{out}\) is the outer heat transfer coefficient, \(A_{out}\) is the outer surface area per unit length, and \(\Delta T_{out}\) is the temperature difference between the outer surface and the surrounding air.

Equate the heat transfer rates and solve for the maximum duct length

Now we will equate the heat transfer rates for the inner and outer surfaces, and solve for the maximum duct length: \(q_{in} = q_{out}\) \(h_{in} \cdot A_{in} \cdot \Delta T_{in} = h_{out} \cdot A_{out} \cdot \Delta T_{out}\) Provided that \(A_{in} = A_{out} = A_p\), it can be simplified to: \(\Delta T_{in} = \frac{h_{out}}{h_{in}} \cdot \Delta T_{out}\) Since the temperature of the air should not increase by more than 1°C, \(\Delta T_{in} = 1 \mathrm{~K}\): \(1 = \frac{0.0132 \cdot h_{out}}{0.0132 \cdot h_{in}} \cdot (33 - T_2)\) Now we can find the outer surface's temperature, \(T_2\): \(T_2 = 33 - \frac{h_{in}}{h_{out}} \) \(T_2 = 33 - \frac{75}{13} \approx 27.4^{\circ} \mathrm{C}\) Knowing that the temperature difference between the air and the inner surface is 1°C, we can calculate the inner surface's temperature, \(T_1\): \(T_1 = T_{air} + 1 = 12 + 1 = 13^{\circ} \mathrm{C}\) Now we can use the thermal resistance and the temperature difference between the inner and outer surfaces to find the heat transfer rate: \(q = \frac{T_1 - T_2}{R_{cond}}\) \(q = \frac{13 - 27.4}{4.717 \times 10^{-4}} \approx -3.07 \times 10^4 \mathrm{~W} / \mathrm{m}\) Finally, we can calculate the maximum duct length using the heat transfer rate and the mass flow rate: \(q = \dot{m} \cdot c_p \cdot \Delta T_{air}\) \(L = \frac{\dot{m} \cdot c_p \cdot \Delta T_{air}}{q}\) Assuming the specific heat capacity of air \(c_p = 1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and using the given mass flow rate, \(\dot{m} = 0.8 \mathrm{~kg} / \mathrm{s}\), and the allowed temperature increase, \(\Delta T_{air} = 1 \mathrm{~K}\): \(L = \frac{0.8 \cdot 1005 \cdot 1}{-3.07 \times 10^4} \approx -2.62 \times 10^{-2} \mathrm{~m}\) The negative result indicates that it is impossible to maintain a temperature increase of no more than 1°C with the given conditions. However, either reducing the mass flow rate or using a more insulating material for the duct wall might make it possible to achieve the desired outcome.

Key concepts

Understanding Heat Transfer

When we talk about heat transfer, we are referring to the movement of thermal energy from a region of higher temperature to one of lower temperature. It is an essential concept in thermodynamics and occurs in many real-world applications, from industrial processes to everyday appliances.

There are three main modes of heat transfer: conduction, convection, and radiation. In our exercise, we primarily deal with conduction through the aluminum duct wall and convection at both the inner and outer surfaces of the duct. Heat transfers by conduction within the solid material of the duct wall, and by convection, a process involving fluid motion, between the air inside the duct and the duct surface, as well as between the outer surface of the duct and the surrounding air.

Thermal resistance is a key concept in this context. It quantifies how much resistance a material provides to the flow of heat. The lower the thermal resistance, the more efficiently heat can pass through the material. In contrast, a higher thermal resistance means that the material is a better insulator, impeding the heat flow.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient, represented by the symbol 'h', provides a measure of the convection heat transfer occurring between a solid surface and a fluid moving over it. This coefficient is crucial in calculating how much heat is exchanged by convection.

In the context of our exercise, we have two different convection heat transfer coefficients: one for the inside of the duct (where the cooler air flows) and one for the outside (where the duct is exposed to warmer ambient air). The inner convection coefficient of 75 W/m²·K is a reflection of how the moving air inside the duct takes heat from the duct's inner surface. The outer coefficient of 13 W/m²·K combines the effects of both convection and radiation to reflect the overall heat loss from the duct to the surrounding air.

Understanding the nuances of these coefficients can impact the thermal management decisions significantly. For instance, a high convection coefficient means the surface is effectively transferring heat to the fluid, while a low coefficient indicates that the heat transfer process is less efficient, which may require design modifications to achieve desired heating or cooling effects.

Thermal Conductivity

Thermal conductivity, denoted by the symbol 'k', is an inherent property of materials that describes how well a material can conduct heat. In the exercise, the aluminum duct has a thermal conductivity of 237 W/m·K, indicating that aluminum is an excellent conductor of heat.

The larger the thermal conductivity value, the more quickly and effectively the material can transfer heat. This is often desirable for materials used in heat sinks or cooking utensils, as it allows for rapid heat distribution. In the case of insulation, however, we look for materials with low thermal conductivity because we want to minimize heat transfer to preserve energy.

In thermal resistance calculations, like in our exercise, the thermal conductivity is inversely proportional to the thermal resistance—high conductivity means low thermal resistance and vice versa. This deepens our understanding that while the aluminum duct is good at moving heat, it is not the best insulator for maintaining the cold temperature of the air flowing inside since it allows the heat to be conducted from the warmer exterior to the cooler interior readily.