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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?

Short Answer

Expert verified
Question: 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. Answer: (a) The heat transfer from the outer surfaces of the duct to the ambient air by natural convection is 80.22 W. (b) The average temperature of the duct is 36.93掳C.

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 (m鲁/s) for SI units. Flow_rate = 22 cfm * (0.0283168 m鲁/cuft) * (1/60 min) = 0.01042 m鲁/s
02

Find the velocity of the air inside the duct

We are given the dimensions of the duct: 6 in 脳 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)虏 = 0.0232 m虏 Velocity = Flow_rate / Cross-sectional area = 0.01042 m鲁/s / 0.0232 m虏 = 0.449 m/s
03

Calculate properties of air at film temperature

The film temperature is given as 100掳F; convert it to Celsius. Film_temperature = (100掳F - 32) * (5/9) = 37.78掳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 (蟻) = 1.146 kg/m鲁 Dynamic viscosity (碌) = 2.03 脳 10鈦烩伒 kg/(m路s) Specific heat (cp) = 1006 J/(kg路K) Thermal conductivity (k) = 0.0273 W/(m路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, 谓 = 碌/蟻: 谓 = 2.03 脳 10鈦烩伒 kg/(m路s) / 1.146 kg/m鲁 = 1.77 脳 10鈦烩伒 m虏/s Re = (0.449 m/s * 0.1524 m) / 1.77 脳 10鈦烩伒 m虏/s = 3912 Pr = cp * 碌 / k = (1006 J/(kg路K) * 2.03 脳 10鈦烩伒 kg/(m路s)) / 0.0273 W/(m路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 10鲁 < Re < 5 脳 10鈦. 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路K) / 0.1524 m = 10.345 W/(m虏路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 * 螖T 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 m虏 Q_in = 10.345 W/(m虏路K) * 0.1857 m虏 * (100掳F - 85掳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 m虏
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 * 尾 * 螖T * Duct_width^3) / 谓^2 Gravitational acceleration, g = 9.81 m/s虏 Coefficient of thermal expansion, 尾 鈮 1/T_film = 1/310.93 K = 3.217 脳 10鈦宦 K鈦宦 螖T = 100掳F - 80掳F = 11.11 K (converted to Celsius) Gr = (9.81 m/s虏 * 3.217 脳 10鈦宦 K鈦宦 * 11.11 K * 0.1524 m^3) / (1.77 脳 10鈦烩伒 m虏/s)^2 = 4.807 脳 10鈦 Now, calculate h_out: h_out = 0.0273 W/(m路K) * (4.807 脳 10鈦 * 0.715)^0.25 / 0.1524 m = 9.733 W/(m虏路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 * 螖T = 9.733 W/(m虏路K) * 0.7416 m虏 * 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 * 螖T_in = h_out * A_out * 螖T_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/(m虏路K) * 0.7416 m虏 * 80掳F * (5/9)) / (9.733 W/(m虏路K) * 0.7416 m虏 - 10.345 W/(m虏路K) * 0.1857 m虏) = 310.08 K Answer (b) The average temperature of the duct is 310.08 K or 36.93掳C. Considering that we evaluated air properties at a film temperature of 37.78掳C and our calculated average temperature is close enough, our initial assumption appears to be reasonable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electronic Cooling
Electronic cooling is crucial for maintaining the optimal performance and longevity of electronic components. As electronic devices generate heat during operation, it is essential to dissipate this heat effectively to prevent overheating. In this context, electronic cooling techniques enable thermal management that ensures devices function within their safe operating temperatures.
This process involves removing excess heat and transferring it away from the components. The most common methods include air or liquid-based systems, with air being the most prevalent due to its simplicity. This exercise exemplifies air cooling, using a duct system to channel forced convection to extract heat from electronic components.
Efficient electronic cooling is vital for devices such as computers, servers, and other electronics, where compact designs increase the risk of accumulated heat, potentially leading to malfunctions or diminished performance.
Forced Convection
Forced convection is a method where a fluid, usually air or liquid, is forced over a surface to improve heat transfer. This technique is especially useful in situations where natural convection alone is inadequate to displace the heat.
Key points about forced convection include:
  • Increased Heat Transfer: By using fans or pumps, forced convection enhances the rate of heat removal.
  • Control over Cooling: It offers better control over the cooling process by varying the speed of the fluid.
  • Application in Electronics: In electronic cooling, like in the duct system, forced convection helps maintain reliable operation by quickly moving heat away from critical components.
In the given exercise, air enters a duct to maintain cool temperatures, effectively dispersing heat generated by electronic components.
Natural Convection
In natural convection, the movement of fluid is primarily driven by buoyancy forces arising from temperature differences within the fluid. When a fluid is heated, it expands, becoming less dense, and rises. Cooler fluid then takes its place, continuing the circulation process.
This process is slower than forced convection but is crucial for scenarios where an external aid like a fan is unavailable or impractical. In the exercise, the outer surface of the duct interacts with the ambient air using natural convection, moving heat away from the duct.
Highlights of natural convection include:
  • Energy Efficiency: It doesn't require additional energy usage since it relies on natural processes.
  • Simplicity: The absence of mechanical components like fans makes it more straightforward to implement.
  • Passive Cooling: Useful for systems where passive cooling suffices or where the environment must not be disturbed by fan noise.
This approach can be a valuable supplement to forced convection in electronic cooling solutions, especially for passive heat dissipation.
Dittus-Boelter Equation
The Dittus-Boelter equation is a fundamental formula used in heat transfer calculations. It is applied to determine the convective heat transfer coefficient for fluid flow inside pipes or ducts, and it is particularly beneficial when working with turbulent flow systems.
This equation helps link various parameters, such as:
  • Nusselt Number ( ): Expresses convective heat transfer relative to conductive transfer.
  • Reynolds Number: Measures the ratio of inertial forces to viscous forces, indicating whether the flow is laminar or turbulent.
  • Prandtl Number: Relates the momentum diffusivity to thermal diffusivity of the fluid.
For the problem at hand, with a specific Reynolds number range and the air properties, the Dittus-Boelter equation provided a means to accurately compute the heat transfer coefficient inside the duct. This calculation is vital for assessing how efficiently heat is being evacuated from the electronic components.

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Most popular questions from this chapter

An average person generates heat at a rate of \(240 \mathrm{Btu} / \mathrm{h}\) while resting in a room at \(70^{\circ} \mathrm{F}\). Assuming onequarter of this heat is lost from the head and taking the emissivity of the skin to be \(0.9\), determine the average surface temperature of the head when it is not covered. The head can be approximated as a 12 -in-diameter sphere, and the interior surfaces of the room can be assumed to be at the room temperature.

In a plant that manufactures canned aerosol paints, the cans are temperature- tested in water baths at \(55^{\circ} \mathrm{C}\) before they are shipped to ensure that they withstand temperatures up to \(55^{\circ} \mathrm{C}\) during transportation and shelving (as shown in Fig. P9-44 on the next page). The cans, moving on a conveyor, enter the open hot water bath, which is \(0.5 \mathrm{~m}\) deep, \(1 \mathrm{~m}\) wide, and \(3.5 \mathrm{~m}\) long, and move slowly in the hot water toward the other end. Some of the cans fail the test and explode in the water bath. The water container is made of sheet metal, and the entire container is at about the same temperature as the hot water. The emissivity of the outer surface of the container is 0.7. If the temperature of the surrounding air and surfaces is \(20^{\circ} \mathrm{C}\), determine the rate of heat loss from the four side surfaces of the container (disregard the top surface, which is open). The water is heated electrically by resistance heaters, and the cost of electricity is \(\$ 0.085 / \mathrm{kWh}\). If the plant operates \(24 \mathrm{~h}\) a day 365 days a year and thus \(8760 \mathrm{~h}\) a year, determine the annual cost of the heat losses from the container for this facility.

Consider a 3-m-high rectangular enclosure consisting of two surfaces separated by a \(0.1-\mathrm{m}\) air gap at \(1 \mathrm{~atm}\). If the surface temperatures across the air gap are \(30^{\circ} \mathrm{C}\) and \(-10^{\circ} \mathrm{C}\), determine the ratio of the heat transfer rate for the horizontal orientation (with hotter surface at the bottom) to that for vertical orientation.

When neither natural nor forced convection is negligible, is it correct to calculate each independently and add them to determine the total convection heat transfer?

Two concentric cylinders of diameters \(D_{i}=30 \mathrm{~cm}\) and \(D_{o}=40 \mathrm{~cm}\) and length \(L=5 \mathrm{~m}\) are separated by air at \(1 \mathrm{~atm}\) pressure. Heat is generated within the inner cylinder uniformly at a rate of \(1100 \mathrm{~W} / \mathrm{m}^{3}\), and the inner surface temperature of the outer cylinder is \(300 \mathrm{~K}\). The steady-state outer surface temperature of the inner cylinder is (a) \(402 \mathrm{~K}\) (b) \(415 \mathrm{~K}\) (c) \(429 \mathrm{~K}\) (d) \(442 \mathrm{~K}\) (e) \(456 \mathrm{~K}\) (For air, use \(k=0.03095 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7111, v=\) \(\left.2.306 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\)

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