Question

Hot air at atmospheric pressure and \(85^{\circ} \mathrm{C}\) enters a \(10-\mathrm{m}\)-long uninsulated square duct of cross section \(0.15 \mathrm{~m} \times\) \(0.15 \mathrm{~m}\) that passes through the attic of a house at a rate of \(0.1 \mathrm{~m}^{3} / \mathrm{s}\). The duct is observed to be nearly isothermal at \(70^{\circ} \mathrm{C}\). Determine the exit temperature of the air and the rate of heat loss from the duct to the air space in the attic. Evaluate air properties at a bulk mean temperature of \(75^{\circ} \mathrm{C}\). Is this a good assumption?

Short answer

Answer: The exit temperature of the air is 62.11°C, and the rate of heat loss from the duct to the air space in the attic is -0.4927 kW.

Step-by-step solution

Calculate the Reynolds number and the Nusselt number

First, we need to find the Reynolds number and use it to determine the Nusselt number. Let's denote the hydraulic diameter as \(D_{h}\): $$D_{h} = \frac{4 \cdot 0.15^2}{4 \cdot 0.15} = 0.15 \mathrm{m}$$ Next, we need to determine the air properties at a bulk mean temperature of \(75^{\circ} \mathrm{C}\). Using a property table for air, we get the following values: $$k = 0.03014 \mathrm{W / (m \cdot K)}, \ \mu = 2.075 \times 10^{-5} \mathrm{Pa \cdot s}, \ \rho = 0.9482 \mathrm{kg / m^3}, \ C_p = 1.006 \mathrm{kJ / (kg \cdot K)}$$ Now, we can find the Reynolds number (\(Re\)): $$Re = \frac{\rho \cdot u \cdot D_{h}}{\mu} = \frac{0.9482 \cdot 0.1 \cdot 0.15}{2.075 \times 10^{-5}} = 6809$$ The flow is turbulent, so we'll use the Dittus-Boelter equation to find the Nusselt number (\(Nu\)): $$Nu = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3} = 0.023 \cdot 6809^{0.8} \cdot 0.7^{0.3} = 164.9$$

Calculate the convective heat transfer coefficient

Now, we can calculate the convective heat transfer coefficient (\(h\)) using the Nusselt number: $$h = \frac{Nu \cdot k}{D_{h}} = \frac{164.9 \cdot 0.03014}{0.15} = 33.49 \mathrm{W / (m^2 \cdot K)}$$

Apply energy balance to find the exit temperature of the air

We'll set up an energy balance equation to determine the exit temperature (\(T_{out}\)). The energy balance is based on the convective heat transfer rate from the duct surface to the air space in the attic and the heat capacity rate of the air: $$Q = h \cdot A_c \cdot (T_{s} - T_{a}) = \dot{m} \cdot C_p \cdot (T_{out} - T_{in})$$ Where \(T_{s} = 70^{\circ} C\) is the surface temperature of the duct, \(T_{a}\) is the air temperature in the attic, \(A_c\) is the surface area of the duct, \(\dot{m}\) is the mass flow rate, and \(T_{in} = 85^{\circ} C\) is the inlet temperature. We'll make a simplifying assumption that \(T_{a} \approx T_{s}\). The mass flow rate can be found using the following formula: $$\dot{m} = \rho \cdot u \cdot A = 0.9482 \cdot 0.1 \cdot 0.15^2 = 0.02128 \mathrm{kg / s}$$ Now, we can solve for \(T_{out}\): $$33.49 \cdot 10 \cdot 0.6 \cdot (70 - T_{out}) = 0.02128 \cdot 1006 \cdot (T_{out} - 85)$$ Solving for \(T_{out}\) gives us: $$T_{out} = 62.11^{\circ} \mathrm{C}$$

Calculate the rate of heat loss from the duct

Now, we can find the rate of heat loss (\(Q\)) from the duct to the air space in the attic: $$Q = \dot{m} \cdot C_p \cdot (T_{out} - T_{in}) = 0.02128 \cdot 1.006 \cdot (62.11 - 85) = -0.4927 \mathrm{kW}$$

Check if the bulk mean temperature assumption is reasonable

Finally, let's see if our bulk mean temperature assumption of \(75^{\circ} C\) is reasonable. We can find the average of the inlet and exit temperatures of the air: $$T_{avg} = \frac{T_{in} + T_{out}}{2} = \frac{85 + 62.11}{2} = 73.56^{\circ} \mathrm{C}$$ Since \(T_{avg}\) is close to \(75^{\circ} \mathrm{C}\), our assumption about the bulk mean temperature is reasonable. In conclusion, the exit temperature of the air is \(62.11^{\circ} \mathrm{C}\) and the rate of heat loss from the duct to the air space in the attic is \(-0.4927 \mathrm{kW}\).

Key concepts

Reynolds Number

Understanding the flow of fluids, whether in industrial processes or in everyday scenarios like air moving through a duct, is critical for numerous applications. One concept that helps describe the fluid flow is the Reynolds number (Re). The Reynolds number is a dimensionless quantity that engineers use to predict the flow pattern in different fluid flow situations. It's calculated based on the ratio of inertial forces to viscous forces and it indicates whether the flow will be laminar or turbulent.

In layman's terms, when the Reynolds number is low (typically less than 2000), the fluid flows in smooth layers or 'laminar flow.' On the other hand, a high Reynolds number (usually greater than 4000) suggests a chaotic, swirly kind of flow called 'turbulent flow'. Turbulent flow is what was observed in the textbook example where the air passing through a duct had a Reynolds number of 6809. This turbulent flow increases the efficiency of heat transfer, which is often desirable in heat exchanger applications.

Why is Turbulent Flow Useful?

Turbulent flow enhances the mixing of fluid particles and thus improves the heat transfer between the fluid and the surface it flows over. So, in cases where we want to either heat up or cool down the fluid quickly, achieving turbulent flow can be beneficial.

Nusselt Number

When it comes to heat transfer, we are often concerned with how effectively a fluid transfers heat to a surface or vice versa. This is where the Nusselt number (Nu) comes into play. The Nusselt number is another dimensionless parameter that offers insight into the convective heat transfer occurring within a fluid. It essentially characterizes the ratio of convective to conductive heat transfer across the fluid layer.

A higher Nusselt number corresponds to more efficient convective heat transfer, as was desired in our duct example, where we computed a Nusselt number of 164.9. This higher number is indicative of effective heat transfer from the duct to its surroundings, which would correspond to a quick temperature adjustment of the air flowing within the duct.

Applications of the Nusselt Number

Engineers and scientists use the Nusselt number for designing various thermodynamic systems, such as heaters, radiators, and air-conditioners. Knowing the Nusselt number helps them in selecting the right materials and shapes to optimize energy use.

Convective Heat Transfer Coefficient

The convective heat transfer coefficient (\( h \)) is a measure of the heat transfer between a solid surface and a fluid per unit surface area per unit temperature difference. It’s a crucial variable in the field of thermodynamics, allowing us to quantify the heat transfer rate for convective heat transfer situations. In our example involving air flowing through a duct, the heat transfer coefficient was calculated to be 33.49 W/(m²·K). This value indicates how much heat will be transferred by convection per square meter of duct surface area for each degree of temperature difference between the duct surface and the flowing air.

The determination of this coefficient is vital for thermal engineering tasks since it directly influences the design and performance of heating and cooling systems.

Improving Heat Transfer

By manipulating factors like fluid velocity, surface texture, and duct shape, engineers can enhance the convective heat transfer coefficient, thereby improving the system's overall heat transfer efficiency. For the attic duct example, understanding the convective heat transfer coefficient helps insulate the system adequately or modify airflow to control how much heat the duct loses to its surroundings.