Question

Hot air at \(60^{\circ} \mathrm{C}\) leaving the furnace of a house enters a 12-m-long section of a sheet metal duct of rectangular cross section \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\) at an average velocity of \(4 \mathrm{~m} / \mathrm{s}\). The thermal resistance of the duct is negligible, and the outer surface of the duct, whose emissivity is \(0.3\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in the basement, with a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine \((a)\) the temperature at which the hot air will leave the basement and \((b)\) the rate of heat loss from the hot air in the duct to the basement. Evaluate air properties at a bulk mean temperature of \(50^{\circ} \mathrm{C}\). Is this a good assumption?

Short answer

#Answer# To summarize the solution, we have used the following steps: 1. Calculate the surface area of the duct: \(A_{duct} = 9.6 m^2\) 2. Determine the convection and radiation heat transfer rates: \(q_{conv}\) and \(q_{rad}\) 3. Find the total heat transfer rate: \(q_{total} = q_{conv} + q_{rad}\) 4. Use the energy balance equation to find the temperature of the air leaving the basement: \(T_{out} = T_{in} - \frac{q_{total}}{m \times c_p}\) 5. Calculate the rate of heat loss from the hot air to the basement: \(Q_{loss} = q_{conv} + q_{rad}\) With the given parameters, we can now calculate the values of \(q_{total}\), \(q_{conv}\), \(q_{rad}\), \(Q_{loss}\), \(T_{out}\), and \(m\).

Step-by-step solution

Given parameters and Preliminary Calculations

The given parameters are: - Hot air temperature entering the furnace: \(T_{in} = 60^{\circ}C\) - Duct length: \(L = 12 m\) - Duct cross-sectional dimensions: \(a \times b = 20 cm \times 20 cm\) - Hot air average velocity: \(V = 4 m/s\) - Emissivity of the outer duct surface: \(\varepsilon = 0.3\) - Cold air temperature: \(T_{sur} = 10^{\circ}C\) - Convection heat transfer coefficient: \(h = 10 \frac{W}{m^2 \cdot K}\) We will evaluate air properties at a bulk mean temperature of \(50^{\circ}C\). Now, let's find the duct surface area: \(A_{duct} = 2 \times (a \times L + b \times L)\)

Determine Surface Area of Duct

Since the cross-section of the duct is rectangular, the area can be calculated as the sum of the areas of the four sides of the duct: \(A_{duct} = 4 \times a \times L = 4 \times (0.2 m \times 12 m) = 9.6 m^2\)

Calculate the Total Rate of Heat Transfer from Hot Air

The total rate of heat transfer from the hot air inside the duct can be found by considering both convection and radiation: \(q_{total} = q_{conv} + q_{rad}\) The heat transfer due to convection is given by: \(q_{conv} = h \times A_{duct} \times \Delta T\) And the heat transfer due to radiation is given by: \(q_{rad} = \varepsilon \times \sigma \times A_{duct} \times (T_{in}^4 - T_{sur}^4)\) where \(\sigma\) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \frac{W}{m^2 \cdot K^4}\), and \(\Delta T\) is the difference in temperature between the hot air and the surroundings.

Find the Temperature at which Hot Air Leaves the Basement

Now, to find the temperature at which the hot air leaves the basement (the temperature of the air exiting the duct), we will use the energy balance equation: \(q_{total} = m \times c_p \times (T_{in} - T_{out})\) where m is the mass flow rate of the hot air, \(c_p\) is the specific heat of air at constant pressure, and \(T_{out}\) is the temperature of the air leaving the basement. Rearrange the equation to solve for \(T_{out}\): \(T_{out} = T_{in} - \frac{q_{total}}{m \times c_p}\)

Calculate the Rate of Heat Loss from Hot Air to Basement

The rate of heat loss from the hot air to the basement can be found by summing up the heat transfer from the hot air due to convection and radiation: \(Q_{loss} = q_{conv} + q_{rad}\) Now all given parameters can be used to calculate the various quantities mentioned above. Calculate \(q_{total}\), \(q_{conv}\), \(q_{rad}\), \(Q_{loss}\), \(T_{out}\) and \(m\).

Key concepts

Convection Heat Transfer

Convection heat transfer is a fundamental principle in thermal physics, describing how heat moves through fluids, such as air and water. When hot air enters a duct, it carries energy that can be transferred to the cooler surroundings. This occurs through the process of convection, where the heat is transferred by the motion of the fluid particles. In the problem at hand, as the hot air at a temperature of 60 degrees Celsius passes through a duct, it loses heat to the cold air in the basement, which is at 10 degrees Celsius.

The heat loss due to convection is quantified by Newton's law of cooling, which in mathematical terms is expressed as: \( q_{conv} = h \times A_{duct} \times (T_{in} - T_{sur}) \), where \( h \) is the convection heat transfer coefficient, \( A_{duct} \) is the surface area through which the heat transfer takes place, \( T_{in} \) is the temperature of the hot air entering the duct, and \( T_{sur} \) is the temperature of the cold basement air. The coefficient \( h \) depends on various factors, including the properties of the fluid and the flow conditions.

In our scenario with the duct, the air's physical movement—blown by the house's heating system—amplifies heat transfer. As the hot air comes into contact with the cooler duct walls, it transfers its energy, cooling down as it flows along the duct's length.

Thermal Radiation

In contrast to convection, thermal radiation does not require a medium to transfer heat. It is the process by which energy is emitted from a surface as electromagnetic waves. In the context of the given problem, the surface of the duct emits thermal radiation due to its temperature being higher than that of the surrounding basement air.

The amount of energy emitted by radiation is determined by the Stefan-Boltzmann law: \( q_{rad} = \varepsilon \times \sigma \times A_{duct} \times (T_{in}^4 - T_{sur}^4) \), where \( \varepsilon \) is the emissivity of the duct surface, \( \sigma \) is the Stefan-Boltzmann constant, and \( T_{in} \) and \( T_{sur} \) are the absolute temperatures of the hot air and the cold air in Kelvin, respectively.

Emissivity is a measure of how effectively a surface emits thermal radiation and ranges from 0 (perfect reflector) to 1 (perfect emitter). The duct's emissivity is given as 0.3, which means the duct does not emit radiation as efficiently as a perfect black body. The energy lost through radiation adds to the total energy being transferred from the air inside the duct to the basement.

Energy Balance Equation

The energy balance equation serves as a foundational concept in thermodynamics and heat transfer. It demonstrates the conservation of energy within a system by balancing the energy entering, leaving, and being stored within the system. In our heat transfer problem, the energy balance equation is used to predict the final temperature of the air as it leaves the duct and the rate of heat loss from this air to the surrounding basement environment.

Mathematically, the energy balance for the air flowing through the duct is given by: \( q_{total} = m \times c_p \times (T_{in} - T_{out}) \), where \( m \) is the mass flow rate of the air, \( c_p \) is the specific heat capacity at constant pressure, \( T_{out} \) is the exit temperature of the air, and \( q_{total} \) is the total rate of heat transfer which includes both convection and radiation effects.To determine the exit temperature and heat loss, we rearrange the equation to solve for \( T_{out} \) and then sum the individual heat transfer rates from convection and radiation to find \( q_{total} \). This requires accounting for the properties of the flowing air, such as its density and specific heat, which depend on the mean temperature of the air, assumed to be 50 degrees Celsius in the exercise. By inserting the values from the given problem, students will be able to calculate the exact temperature and heat loss, understanding how energy balance is vital to such calculations.