The components of an electronic system dissipating \(180 \mathrm{~W}\) are located in a 4-ft-long horizontal duct whose cross section is 6 in \(\times 6\) in. The components in the duct are cooled by forced air, which enters at \(85^{\circ} \mathrm{F}\) at a rate of \(22 \mathrm{cfm}\) and leaves at \(100^{\circ} \mathrm{F}\). The surfaces of the sheet metal duct are not painted, and thus radiation heat transfer from the outer surfaces is negligible. If the ambient air temperature is \(80^{\circ} \mathrm{F}\), determine (a) the heat transfer from the outer surfaces of the duct to the ambient air by natural convection and \((b)\) the average temperature of the duct. Evaluate air properties at a film temperature of \(100^{\circ} \mathrm{F}\) and 1 atm pressure. Is this a good assumption?

Step by step solution

01

Calculate the volumetric flow rate of air

Given the flow rate is 22 cubic feet per minute (cfm), we can convert it to cubic meters per second (m³/s) for SI units. Flow_rate = 22 cfm * (0.0283168 m³/cuft) * (1/60 min) = 0.01042 m³/s

02

Find the velocity of the air inside the duct

We are given the dimensions of the duct: 6 in × 6 in. We'll convert these to meters and then calculate the cross-sectional area of the duct. Finally, we'll use the volumetric flow rate to find the average velocity of the air inside the duct. Duct width = 6 in * (0.0254 m/in) = 0.1524 m Duct height = 0.1524 m Cross-sectional area = Duct width * Duct height = (0.1524 m)² = 0.0232 m² Velocity = Flow_rate / Cross-sectional area = 0.01042 m³/s / 0.0232 m² = 0.449 m/s

03

Calculate properties of air at film temperature

The film temperature is given as 100°F; convert it to Celsius. Film_temperature = (100°F - 32) * (5/9) = 37.78°C By referring to the air property tables, evaluate the properties of air at this film temperature and 1 atm pressure. Assume these properties: density, dynamic viscosity, specific heat, and thermal conductivity. Density (ρ) = 1.146 kg/m³ Dynamic viscosity (µ) = 2.03 × 10⁻⁵ kg/(m·s) Specific heat (cp) = 1006 J/(kg·K) Thermal conductivity (k) = 0.0273 W/(m·K)

04

Determine the Reynolds number and Prandtl number inside the duct

Calculate the Reynolds number (Re) and Prandtl number (Pr) of the flow inside the duct using the properties calculated in Step 3. Re = (Velocity * Duct width) / Kinematic viscosity First, find the kinematic viscosity, ν = µ/ρ: ν = 2.03 × 10⁻⁵ kg/(m·s) / 1.146 kg/m³ = 1.77 × 10⁻⁵ m²/s Re = (0.449 m/s * 0.1524 m) / 1.77 × 10⁻⁵ m²/s = 3912 Pr = cp * µ / k = (1006 J/(kg·K) * 2.03 × 10⁻⁵ kg/(m·s)) / 0.0273 W/(m·K) = 0.715

05

Calculate the heat transfer coefficient inside the duct

We can use the Dittus-Boelter equation for the heat transfer coefficient (h) inside the duct since our Reynolds number is within the range of 0.4 < Pr < 4,000 and 10³ < Re < 5 × 10⁴. Nu = 0.0275 * Re^0.8 * Pr^n, where Nu = h * Duct_width / k, and the exponent n is 0.4 for heating and 0.3 for cooling. Nu = 0.0275 * (3912)^0.8 * 0.715^0.4 = 57.68 h_in = Nu * k / Duct_width = 57.68 * 0.0273 W/(m·K) / 0.1524 m = 10.345 W/(m²·K)

06

Calculate the heat transfer from the components to the air

Now we can find the heat transfer from the components to the air (Q_in) using the heat transfer coefficient and the temperature difference between incoming and outgoing air. Q_in = h_in * A_in * ΔT A_in = Duct_width * Duct_length Duct_length = 4 ft * 0.3048 m/ft = 1.2192 m A_in = 0.1524 m * 1.2192 m = 0.1857 m² Q_in = 10.345 W/(m²·K) * 0.1857 m² * (100°F - 85°F) * (5/9) K = 180 W

07

Calculate the outer surface area of the duct

To find the heat transfer to the ambient air by natural convection, we need to find the outer surface area of the duct. A_out = 2 * (Duct_width + Duct_height) * Duct_length A_out = 2 * (0.1524 m + 0.1524 m) * 1.2192 m = 0.7416 m²

08

Calculate the heat transfer coefficient outside the duct

The heat transfer coefficient (h_out) for the outer surfaces of the duct can be determined using the following formula: h_out = k * (Gr * Pr)^0.25 / l_ci l_ci is the characteristic length for a vertical plate, which in our case is equal to the width of the duct. First, we need to calculate the Grashof number (Gr): Gr = (g * β * ΔT * Duct_width^3) / ν^2 Gravitational acceleration, g = 9.81 m/s² Coefficient of thermal expansion, β ≈ 1/T_film = 1/310.93 K = 3.217 × 10⁻³ K⁻¹ ΔT = 100°F - 80°F = 11.11 K (converted to Celsius) Gr = (9.81 m/s² * 3.217 × 10⁻³ K⁻¹ * 11.11 K * 0.1524 m^3) / (1.77 × 10⁻⁵ m²/s)^2 = 4.807 × 10⁷ Now, calculate h_out: h_out = 0.0273 W/(m·K) * (4.807 × 10⁷ * 0.715)^0.25 / 0.1524 m = 9.733 W/(m²·K)

09

Calculate the heat transfer from the outer surfaces of the duct

Now, we'll determine the heat transfer (Q_out) from the outer surfaces of the duct to the ambient air by natural convection. Q_out = h_out * A_out * ΔT = 9.733 W/(m²·K) * 0.7416 m² * 11.11 K = 80.22 W Answer (a): The heat transfer from the outer surfaces of the duct to the ambient air by natural convection is 80.22 W.

10

Calculate the average temperature of the duct

Finally, we can find the average temperature (T_avg) of the duct by using the heat transfer coefficients and their corresponding areas. Q_total = h_in * A_in * ΔT_in = h_out * A_out * ΔT_out Solving for T_avg: T_avg = (Q_total - (h_out * A_out * T_ambient)) / (h_out * A_out - h_in * A_in) T_avg = (180 W - 9.733 W/(m²·K) * 0.7416 m² * 80°F * (5/9)) / (9.733 W/(m²·K) * 0.7416 m² - 10.345 W/(m²·K) * 0.1857 m²) = 310.08 K Answer (b) The average temperature of the duct is 310.08 K or 36.93°C. Considering that we evaluated air properties at a film temperature of 37.78°C and our calculated average temperature is close enough, our initial assumption appears to be reasonable.

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