Question

The components of an electronic system dissipating \(180 \mathrm{~W}\) are located in a \(1-\mathrm{m}\)-long horizontal duct whose cross section is \(16 \mathrm{~cm} \times 16 \mathrm{~cm}\). The components in the duct are cooled by forced air, which enters at \(27^{\circ} \mathrm{C}\) at a rate of \(0.65 \mathrm{~m}^{3} / \mathrm{min}\). Assuming 85 percent of the heat generated inside is transferred to air flowing through the duct and the remaining 15 percent is lost through the outer surfaces of the duct, determine \((a)\) the exit temperature of air and \((b)\) the highest component surface temperature in the duct. As a first approximation assume flow is fully developed in the channel. Evaluate properties of air at a bulk mean temperature of \(35^{\circ} \mathrm{C}\). Is this a good assumption?

Short answer

The exit temperature of the air is found to be 310.70 K, and the highest component surface temperature in the duct is determined to be 312.67 K. The flow is verified to be laminar and fully developed, as the Reynolds number is less than 4000.

Step-by-step solution

Conversion of the physical properties

First, let's convert the given physical properties into SI units: Length of duct: \(1 \mathrm{~m}\) Cross-sectional area of duct: \(A = 16 \times 10^{-4} \mathrm{~m}^2 \times 16 \times 10^{-4} \mathrm{~m}^2 = 2.56 \times 10^{-3} \mathrm{~m}^2\) Heat generation rate: \(Q = 180 \mathrm{~W}\) Inlet temperature of air: \(T_{in} = 27 + 273.15 = 300.15 \mathrm{~K}\) Rate of heat transferred to the air: \(85\% \times Q = 0.85Q\) Loss through outer surface: \(15\% \times Q = 0.15Q\) Air flow rate: \(\dot{m} = 0.65 \mathrm{~m}^{3} / \mathrm{min} = 0.65/60 = 0.01083 \mathrm{~m}^{3} / \mathrm{s}\)

Calculation of mass flow rate

To find the mass flow rate of air, we need to find its density at the given average temperature, \(T_{avg}=35^{\circ} \mathrm{C}\). Using the ideal gas law and air properties at \(35^{\circ} \mathrm{C}\), we can find the mass flow rate as \(\dot{m} = \rho V\). The density of air at \(35^{\circ} \mathrm{C}\) is approximately \(\rho = 1.146 \mathrm{~kg/m^3}\). So, mass flow rate is: \(\dot{m} = \rho V = 1.146 \times 0.01083 = 0.0124 \mathrm{~kg/s}\)

Calculation of specific heat capacity

Now, we need to find the specific heat capacity of air at \(35^{\circ} \mathrm{C}\). The specific heat capacity of air at this temperature is approximately \(c_p = 1007 \mathrm{~J/kg \cdot K}\).

Calculation of the exit temperature

We can find the exit temperature by considering the conservation of energy equation and dividing the total heat absorbed by the mass flow rate and specific heat capacity of the air: \(T_{out} = T_{in} + \frac{0.85Q}{\dot{m} c_p} = 300.15 + \frac{0.85 \times 180}{0.0124 \times 1007} = 300.15 + 10.55 = 310.70 \mathrm{~K}\)

Calculation of the highest component surface temperature

Knowing that 15% of the heat is lost through the outer surfaces of the duct, we can deduce that this heat is the energy required to raise the component surface temperature above the exit air temperature. Thus, we can use the conservation of energy principle to calculate the temperature rise: \(T_{component} = T_{out} + \frac{0.15Q}{\dot{m} c_p} = 310.70 + \frac{0.15 \times 180}{0.0124 \times 1007} = 310.70 + 1.97 = 312.67 \mathrm{~K}\) So, the highest component surface temperature is 312.67 K.

Checking the assumption of fully developed flow

The assumption of fully developed flow can be verified by calculating the corresponding Reynolds number. For this problem, the relevant characteristic length is the hydraulic diameter, \(D_h = \frac{4A}{P}\), where \(P\) is the wetted perimeter. In this case, \(P = 4 \times 0.16 = 0.64 \mathrm{~m}\), so \(D_h = \frac{4 \times 2.56 \times 10^{-3}}{0.64} = 0.016\mathrm{~m}\). We can calculate the air velocity as \(V = \frac{0.01083}{2.56 \times 10^{-3}} = 4.228 \mathrm{~m/s}\). At \(35^{\circ} \mathrm{C}\), the dynamic viscosity of air is about \(\mu = 1.983 \times 10^{-5} \mathrm{~Pa \cdot s}\), and thus the kinematic viscosity is \(\nu = \frac{\mu}{\rho} = 1.730 \times 10^{-5} \mathrm{~m^2/s}\). Finally, the Reynolds number is: \(Re = \frac{VD_h}{\nu} = \frac{(4.228)(0.016)}{1.730 \times 10^{-5}} = 3911\) Since \(Re < 4000\), this indicates that the flow is laminar and fully developed flow assumption can be considered valid.

Key concepts

Forced Convection

Forced convection is a mechanism where fluid motion is generated by an external source, such as a fan or pump, to enhance heat transfer. In the context of the electronic system in the duct, forced air is used to cool components by moving air through the duct.

This process significantly boosts the heat transfer rate compared to natural convection, as the air velocity controlled by the external source can be much higher. As a result, more heat is carried away by the air.

The primary factors affecting forced convection include:

• Air velocity: Faster moving air improves heat transfer.
• Air properties: Specific heat, density, and viscosity of air at given temperatures influence efficiency.
• Surface area: Larger surface areas in contact with the air improve heat dissipation.

In our exercise, the components lose 85% of their generated heat to the forced air, illustrating efficient forced convection. Understanding this mechanism helps us predict how well systems can be cooled under various conditions.

Energy Conservation

In thermodynamics, energy conservation is a fundamental principle stating that energy cannot be created or destroyed, only converted from one form to another. This principle plays a crucial role in analyzing heat transfer systems, like the electronic duct.

For the given problem, the energy generated by electronic components translates into heat. The conservation of energy enables us to calculate how much of this heat is absorbed by the air and how much is lost through the duct’s surface.

Key steps in applying energy conservation:

• Determine energy input: Here, 180 W is produced by components.
• Calculate distribution: 85% absorbed by air, 15% lost.
• Compute air temperature changes: Use heat absorbed to find air exit temperature.

Applying these steps allows for the calculation of key parameters like exit temperature, giving insights into the system's efficiency and effectiveness.

Laminar Flow

Laminar flow is a flow regime characterized by smooth, parallel layers of fluid, with little to no mixing between them. It's the opposite of turbulent flow, which features irregular and chaotic movements.
In this exercise, verifying that the airflow is laminar helps ensure predictable heat transfer rates. Knowing whether the flow is laminar or turbulent informs us about how effectively the air can remove heat from the electronic components.
The benefits of laminar flow in cooling applications include:

• Consistent heat transfer rates.
• Less energy required to maintain flow.
• Reduced risk of hot spots due to uniform air distribution.

With a Reynolds number calculated to be below 4000, the airflow in the problem is confirmed to be laminar. This means the engineer can rely on the system's efficiency and the effectiveness of heat removal under these conditions.

Reynolds Number

The Reynolds number (\[ Re \]) is a dimensionless quantity used to predict flow patterns in different fluid flow situations. It tells us whether flow is laminar or turbulent. In our exercise, calculating the Reynolds number helps verify the assumption of fully developed laminar airflow inside the duct.

The formula for Reynolds number is given by: \[Re = \frac{VD_h}{u}\]where:

• \( V \) is the fluid velocity.
• \( D_h \) is the hydraulic diameter of the duct.
• \( u \) is the kinematic viscosity of the fluid.

With the air velocity, hydraulic diameter, and air’s viscosity known, the Reynolds number was calculated as 3911. Since this value is less than 4000, the flow is deemed laminar.

Understanding the Reynolds number’s role in identifying flow regimes is pivotal for designing effective heat exchange systems, ensuring cooling processes are optimized.