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?

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: \(\dot{m} = 0.65 \mathrm{~m}^{3} / \mathrm{min} = 0.65/60 = 0.01083 \mathrm{~m}^{3} / \mathrm{s}\)

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: \(\dot{m} = \rho V = 1.146 \times 0.01083 = 0.0124 \mathrm{~kg/s}\)

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 \(c_p = 1007 \mathrm{~J/kg \cdot K}\).

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: \(T_{out} = T_{in} + \frac{0.85Q}{\dot{m} c_p} = 300.15 + \frac{0.85 \times 180}{0.0124 \times 1007} = 300.15 + 10.55 = 310.70 \mathrm{~K}\)

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.

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.