Question

The components of an electronic system dissipating \(220 \mathrm{~W}\) are located in a 1 -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 fully developed turbulent flow in the channel. Evaluate the properties of air at a bulk mean temperature of \(35^{\circ} \mathrm{C}\). Is this a good assumption?

Short answer

Answer: The exit temperature of the air is 43.02 °C, and the highest component surface temperature is 49.82 °C.

Step-by-step solution

Calculate the total heat transfer to the air

First, we need to calculate the total heat transfer to the air. The electronic system dissipates 220 W of heat, of which 85% is transferred to the air and 15% is lost through the outer surfaces of the duct. Therefore, the heat transfer to the air (Q) can be calculated as: Q = 0.85 * 220 W = 187 W

Calculate the Mass Flow Rate of the Air

We are given the volumetric flow rate of the air (0.65 m³/min). To find the mass flow rate, we need the density (ρ) of air, which can be found through the given mean temperature (35°C). The properties of air are: density (ρ) = 1.164 kg/m³, specific heat capacity (c_p) = 1007 J/(kg K), and dynamic viscosity (µ) = 1.983e-5 kg/(m s). Now, we can calculate the mass flow rate (m_dot) as: m_dot = ρ * (volumetric flow rate) = 1.164 kg/m³ * (0.65 m³/min * (1/60) min/s) = 0.01253 kg/s

Calculate the Exit Temperature of the Air

Now we will calculate the exit temperature of the air (T_exit) using the energy balance equation: Q = m_dot * c_p * (T_exit - T_inlet) Solving for T_exit, we get: T_exit = (Q / (m_dot * c_p)) + T_inlet = (187 W / (0.01253 kg/s * 1007 J/(kg K))) + 27 °C = 43.02 °C

Calculate the Reynolds Number

To check the assumption of fully developed turbulent flow in the channel, we will first calculate the Reynolds number (Re) for the channel, which is given by the formula: Re = (ρ * D_h * u) / µ Where D_h is the hydraulic diameter of the channel and u is the average velocity of the air. The hydraulic diameter D_h can be calculated as: D_h = (2 * width * height) / (width + height) = (2 * 0.16 m * 0.16 m) / (0.16 m + 0.16 m) = 0.16 m The average velocity u can be calculated as: u = (volumetric flow rate) / (cross-sectional area) = (0.65 m³/min * (1/60) min/s) / (0.16 m * 0.16 m) = 3.3594 m/s Now, calculating Re: Re = (1.164 kg/m³ * 0.16 m * 3.3594 m/s) / 1.983e-5 kg/(m s) = 30237.35 Since Re > 4000, it can be considered a turbulent flow, and the assumption of fully developed turbulent flow is reasonable.

Calculate the Highest Component Surface Temperature

To find the highest component surface temperature, we can use the heat transfer equation: Q = h * A * (T_surface - T_bulk) Here, h is the convection heat transfer coefficient, A is the surface area, T_surface is the highest component surface temperature, and T_bulk is the bulk air temperature. We can estimate h for turbulent flow in a square duct using the Dittus-Boelter equation: h = 0.023 * (Re^0.8) * (Pr^0.3) * (k/D_h) Where Pr is the Prandtl number and k is the thermal conductivity of air. We have Pr = 0.7 and k = 0.0262 W/(m K) for air at 35°C. Calculating h: h = 0.023 * (30237.35^0.8) * (0.7^0.3) * (0.0262 W/(m K) / 0.16 m) ≈ 72.50 W/(m² K) Now we can calculate T_surface: T_surface = (Q / (h * A)) + T_bulk Assuming an average surface area of the components, A = 0.16 m * 1 m: T_surface = (187 W / (72.50 W/(m² K) * 0.16 m²)) + 35 °C = 49.82 °C The exit temperature of the air is 43.02 °C and the highest component surface temperature is 49.82 °C. The assumption of fully developed turbulent flow is reasonable, as the Reynolds number is greater than 4000.