/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 The electronic components of a c... [FREE SOLUTION] | 91Ó°ÊÓ

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The electronic components of a computer consume \(0.1 \mathrm{~kW}\). of electrical power. To prevent overheating, cooling air is supplied by a 25-W fan mounted at the inlet of the electronics enclosure. At steady state, air enters the fan at \(20^{\circ} \mathrm{C}, 1\) bar, and exits the electronics enclosure at \(35^{\circ} \mathrm{C}\). There is no significant energy transfer by heat from the outer surface of the enclosure to the surroundings, and the effects of kinetic and potential energy can be ignored. Determine the volumetric flow rate of the entering air, in \(\mathrm{m}^{3} / \mathrm{s}\).

Short Answer

Expert verified
The volumetric flow rate of the entering air is approximately 0.0071 \text{m}^3/\text{s\.

Step by step solution

01

Identify Known Values

List all the given values in the problem: \(P_{electronics} = 0.1 \text{ kW}\), \(P_{fan} = 25 \text{ W} = 0.025 \text{ kW}\), \(T_{in} = 20^{\text{°C}}\), \(T_{out} = 35^{\text{°C}}\), \(P_{atm} = 1 \text{ bar}\).
02

Calculate Total Power Input

Sum the electrical power consumed and the power of the fan:\[P_{total} = P_{electronics} + P_{fan} = 0.1 \text{ kW} + 0.025 \text{ kW} = 0.125 \text{ kW}\]
03

Apply Energy Balance

Use the first law of thermodynamics for a control volume assuming steady state:\[Q_{in} = \text{Total Power Input} = 0.125 \text{ kW}\]This power must be equal to the rate of energy change of the air.
04

Relate Energy to Mass Flow Rate

Relate the rate of energy transfer to the mass flow rate \(\dot{m}\) by:\[Q_{in} = \dot{m} c_p (T_{out} - T_{in})\]Where \(c_p\) is the specific heat capacity of air, typically \c_p = 1.005 \text{kJ/kg·K}\.
05

Rearrange to Solve for Mass Flow Rate

Rearrange the equation to solve for \(\dot{m}\):\[\dot{m} = \frac{Q_{in}}{c_p (T_{out} - T_{in})} = \frac{0.125 \text {kW}}{1.005 \text{kJ/kg·K} (35 - 20) \text{K}}\]\[\dot{m} = \frac{0.125}{1.005 \times 15} \text {kg/s}\]\[\dot{m} ≈ 0.0083 \text {kg/s}\]
06

Convert Mass Flow Rate to Volumetric Flow Rate

Use the ideal gas law to convert mass flow rate to volumetric flow rate \(\dot{V}\):\[\dot{V} = \frac{\dot{m} R T_{in}}{P_{atm}}\]With \R = 287 \text {J/kg·K}\ for air and converting \P_{atm}\ to \Pa\ (1 bar = 100000 Pa):\[\dot{V} = \frac{0.0083 \text{kg/s} \times 287 \text{J/kg·K} \times (20 + 273) \text{K}}{100000 \text{Pa}}\]\[\dot{V} \approx 0.0071 \text{m}^3/\text{s}\]

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a principle of conservation of energy. It states that the energy in an isolated system is constant. Energy cannot be created or destroyed, only transformed from one form to another. In this problem, the electronic components and fan generate electrical power. This power is converted into thermal energy, which must be balanced within the system to avoid overheating.

The formula embodying the First Law for a control volume, assuming steady state, is: \[ \text{Total Energy In} = \text{Total Energy Out} \] In our case: \[ Q_{in} = P_{total} \] Understanding this concept is crucial because it tells us the total energy input of the system must equal the energy taken away by the airflow.
Energy Balance
Energy balance involves making sure that the total energy entering a system equals the total energy leaving it. This ensures there is no accumulation of unaccounted energy within the system.

In the given problem, the electronic components consume 0.1 kW of power, and the fan adds 0.025 kW. Summing these gives the total energy input: \[P_{total} = 0.1 \text{ kW} + 0.025 \text{ kW} = 0.125 \text{ kW} \]

This total energy input must be equal to the rate of energy change in the air, which is calculated using the specific heat capacity and temperature difference of the air in and out of the enclosure.
Ideal Gas Law
The Ideal Gas Law relates the pressure, volume, and temperature of an ideal gas with the amount of gas present: \[ PV = nRT \] Where:
  • P is pressure
  • V is volume
  • n is the number of moles
  • R is the universal gas constant
  • T is temperature in Kelvin
In this exercise, it's used to convert mass flow rate to volumetric flow rate. We rearrange the ideal gas law to solve for the volumetric flow rate \[ \frac{{\text{Mass Flow Rate} \times R \times T_{\text{in}}}} {P_{\text{atm}}} \] Here, we must ensure temperature is in Kelvin and pressure in Pascals. This allows us to find the volumetric flow rate in m³/s, which is crucial for understanding how much air is flowing through the system.
Specific Heat Capacity
Specific heat capacity (\text{cp}) is a material's ability to absorb heat energy. It is the amount of energy required to raise the temperature of 1 kg of a substance by 1 degree Celsius. For air, the specific heat capacity is typically 1.005 kJ/kg·K.

In our problem, we use \text{cp} to relate the input energy to the temperature change of air flowing into and out of the system: \[ Q_{in} = \text{Mass Flow Rate} \times \text{cp} \times \text{Temperature Change} \] Rearranging this equation allows us to solve for the mass flow rate: \[ \text{Mass Flow Rate} = \frac{Q_{in}}{\text{cp} \times (T_{\text{out}} - T_{\text{in}})} \] Understanding \text{cp} is essential because it tells us how much energy is required to achieve the desired cooling effect.

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Most popular questions from this chapter

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