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A desktop computer is to be cooled by a fan whose flow rate is \(0.34 \mathrm{m}^{3} / \mathrm{min}\). Determine the mass flow rate of air through the fan at an elevation of \(3400 \mathrm{m}\) where the air density is \(0.7 \mathrm{kg} / \mathrm{m}^{3}\). Also, if the average velocity of air is not to exceed \(110 \mathrm{m} / \mathrm{min}\), determine the diameter of the casing of the fan.

Short Answer

Expert verified
Answer: The mass flow rate of air through the computer fan is 0.238 kg/min, and the diameter of the fan casing should be approximately 62.6 mm.

Step by step solution

01

Determine the mass flow rate

To find the mass flow rate of air through the fan, we can use the following formula: Mass flow rate = Flow rate × Air density Given flow rate = \(0.34 \mathrm{m^3/min}\) Given air density at \(3400 \mathrm{m}\) elevation = \(0.7 \mathrm{kg/m^3}\) Now we can calculate the mass flow rate of air: Mass flow rate = \((0.34 \mathrm{m^3/min}) \times (0.7 \mathrm{kg/m^3})\) Mass flow rate = \(0.238 \mathrm{kg/min}\)
02

Find the cross-sectional area of the fan casing

To find the diameter of the fan casing, we first need to determine the cross-sectional area of the casing. We can use the formula for flow rate to find the area: Flow rate = Area × Average velocity We are given the maximum allowed average velocity: Average velocity = \(110 \mathrm{m/min}\) Now we can solve the formula for the cross-sectional area: Area = Flow rate / Average velocity Area = \((0.34 \mathrm{m^3/min}) / (110 \mathrm{m/min})\) Area = \(0.00309 \mathrm{m^2}\)
03

Calculate the diameter of the fan casing

Now that we have the cross-sectional area of the fan casing, we can find the diameter of the casing. Assuming the fan casing is cylindrical, we can use the formula for the cross-sectional area of a circle: Area = \(\pi \frac{d^2}{4}\) Where \(d\) is the diameter of the casing. Now we can solve for the diameter: \(d^2 = 4 \times \frac{Area}{\pi}\) \(d^2 = 4 \times \frac{0.00309 \mathrm{m^2}}{\pi}\) \(d^2 = 0.00392 \mathrm{m^2}\) Taking the square root of both sides: \(d = \sqrt{0.00392 \mathrm{m^2}}\) \(d = 0.0626 \mathrm{m}\) or \(62.6 \mathrm{mm}\) (approximately) So, the diameter of the casing of the fan should be about 62.6 mm to maintain the given average velocity and flow rate of the air.

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Key Concepts

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

Thermodynamics in Cooling
Thermodynamics is the scientific study of heat and energy and their interaction with matter. In the context of a desktop computer's cooling system, thermodynamics is key to understanding how the heat generated by the computer's components can be transferred to the surrounding air. The first law of thermodynamics, also known as the law of energy conservation, indicates that the heat energy removed from the computer needs to be released into the environment for the system to cool down.

For efficient cooling, the fan must be capable of moving heat away from the computer's internal components and dispersing it into the ambient air. This is achieved by ensuring a specific mass flow rate of air travels through the fan, effectively transferring heat through convection. The calculation of this flow involves knowing both the volume of air moving per unit of time (flow rate) and the air's density.
Air Density and Altitude
Air density, or the mass per unit volume of air, greatly affects cooling systems since the denser the air, the more particles there are to absorb and carry heat away. However, it's crucial to recognize that air density decreases with increasing altitude due to the reduction in atmospheric pressure. At an elevation of 3400 meters, where the air is thinner, fans need to work differently compared to sea level conditions.

Devices designed for use in high-altitude locations must consider this change in density to maintain effectiveness. As demonstrated in our calculation, the provided air density of 0.7 kg/m3 directly impacts the mass flow rate of the air through the cooling system, which is a product of the flow rate and air density.
Understanding Flow Rate
The flow rate is a measure of the volume of fluid (in this case, air) that passes a point in a certain period of time. In the cooling system of a computer, ensuring the right flow rate of air is crucial for maintaining optimal operating temperatures. Too low a flow rate may not provide sufficient cooling, while too high a flow rate can cause unwanted noise or stress on the fan's components.

To calculate the flow rate requirement, we used the provided fan flow rate of 0.34 m3/min, which also influences the calculation for the fan's casing diameter to ensure the average air velocity does not exceed 110 m/min. The mass flow rate, calculated from the flow rate and air density, indicates the amount of mass passing through the fan per unit of time, a critical measure for the system's cooling capacity.
Fan Cooling System Dynamics
A fan cooling system uses the principles of fluid mechanics and thermodynamics to remove excess heat from electronic devices like computers. The dynamics of a fan involve not only the air flow rate but also the distribution of air and the properties of the fan itself – including its shape and size.

The size of the fan's casing has a direct influence on its performance. For a given flow rate and air velocity, the fan's diameter must be sized correctly to meet the desired specifications. If the diameter is too small, the air velocity may exceed safe or efficient levels as we have calculated where the appropriate diameter should be approximately 62.6 mm for our specific conditions. Moreover, the design of the fan blades and the motor's efficiency also play vital parts in the overall cooling effectiveness, requiring a balanced integration of all components for optimal operation.

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Most popular questions from this chapter

A thin-walled double-pipe counter-flow heat exchanger is used to cool oil \(\left(c_{p}=2.20 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) from 150 to \(40^{\circ} \mathrm{C}\) at a rate of \(2 \mathrm{kg} / \mathrm{s}\) by water \(\left(c_{p}=4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) that enters at \(22^{\circ} \mathrm{C}\) at a rate of \(1.5 \mathrm{kg} / \mathrm{s}\). Determine the rate of heat transfer in the heat exchanger and the exit temperature of water.

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