Question

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

The volumetric flow rate of the entering air is approximately 0.0069 \(\text{m}^3/\text{s}\).

Step-by-step solution

- Understand the given data

Identify all the given information:- Electrical power consumed by components: \(P_{elec} = 0.1 \text{ kW}\)- Power of the cooling fan: \(P_{fan} = 25 \text{ W} = 0.025 \text{ kW}\)- Inlet air temperature: \(T_{in} = 20^{\bullet}C\)- Pressure of air: \(P = 1 \text{ bar}\)- Outlet air temperature: \(T_{out} = 35^{\bullet}C\)- No significant heat transfer from outer surface- Ignore kinetic and potential energy changes

- Calculate the total power added to the air

Sum the power consumed by electronics and the fan:\(P_{total} = P_{elec} + P_{fan} = 0.1 \text{ kW} + 0.025 \text{ kW} = 0.125 \text{ kW}\)

- Apply the first law of thermodynamics for steady-state analysis

For steady-state, the energy added to the air is equal to the power consumed:\(Q = P_{total} = 0.125 \text{ kW} = 125 \text{ W}\)

- Determine the specific heat capacity of air

The specific heat capacity of air at constant pressure, \(c_p \text{(approx)} = 1005 \text{ J/kg·K}\)

- Calculate the mass flow rate of air

Use the energy balance equation to find the mass flow rate:\(Q = \text{mass flow rate} \times c_p \times \text{temperature difference}$$125 \text{ W} = \text{mass flow rate} \times 1005 \text{ J/kg·K} \times (35 - 20)^{\bullet}C\)Solve for the mass flow rate:\(\text{mass flow rate} = \frac{125 \text{ W}}{1005 \text{ J/kg·K} \times 15 \text{ K}} = 0.0083 \text{ kg/s}\)

- Use the ideal gas law to find the volumetric flow rate

Use the ideal gas law: \(P V = n R T\)Convert mass flow rate to volumetric flow rate: \(PV = \text{mass flow rate} \times R_s \times T$$R_s = \text{specific gas constant for air} = 287 \text{ J/kg·K}\) Insert known values into the equation:\(P V = 0.0083 \text{ kg/s} \times 287 \text{ J/kg·K} \times 293 \text{ K}\) Compute the volumetric flow rate:\(V = \frac{0.0083 \times 287 \times 293}{10^5} = 0.0069 \text{ m}^3/\text{s}\)

Key concepts

Steady-State Analysis

Steady-state analysis is crucial in engineering thermodynamics as it simplifies the complexity of many problems. In such analysis, the conditions within the system remain constant over time. This means that any energy or mass entering the system must be equal to the energy or mass leaving the system. In our exercise, the air enters and exits the electronics enclosure at a constant rate, ensuring that the system is in steady-state. This assumption allows us to ignore transient effects and focus on a simpler energy balance equation.
By considering the system to be in a steady state, we can simplify our calculations and directly apply the first law of thermodynamics.

First Law of Thermodynamics

The first law of thermodynamics, often called the law of energy conservation, is foundational in engineering. It states that energy can neither be created nor destroyed, only transformed from one form to another. For open systems like our electronics enclosure, the first law can be expressed as:\[ Q - W = \frac{dE}{dt} \ \]where \( Q \) is the heat transfer into the system, \( W \) is the work done by the system, and \( \frac{dE}{dt} \) is the rate of change of internal energy. In our steady-state case, \( \frac{dE}{dt} = 0 \) because there is no accumulation of energy inside the system. Therefore, the energy added to the system by the electrical components and the fan must be equal to the energy carried away by the air exiting the enclosure. This relationship helps us determine the necessary parameters such as mass flow rate and volumetric flow rate.

Mass Flow Rate Calculation

Mass flow rate is the amount of mass passing through a given surface per unit time, often measured in kilograms per second (kg/s). To determine the mass flow rate of air in this problem, we use the energy balance equation. Given that the total power added to the air is 125 W, we equate this to the energy carried by the air as it heats up from 20°C to 35°C. Using the formula:\[ Q = \text{mass flow rate} \times c_p \times \text{temperature difference} \ \ \text{mass flow rate} = \frac{125 \text{ W}}{1005 \text{ J/kg·K} \times 15 \text{ K}} = 0.0083 \text{ kg/s} \]This calculation shows that to maintain steady-state, 0.0083 kg of air must be moved through the system every second.

Volumetric Flow Rate Determination

Volumetric flow rate measures the volume of fluid passing through a section per unit time, typically expressed in cubic meters per second (m³/s). To find the volumetric flow rate from the mass flow rate, we can use the ideal gas law relationship: \[ PV = nRT \ \text{or} \ V = \frac{\text{mass flow rate} \times R_s \times T}{P} \]Given the specific gas constant for air \( R_s = 287 \text{ J/kg·K} \) and the entering air temperature of 293 K, the calculation is:\[ V = \frac{0.0083 \times 287 \times 293}{10^5} = 0.0069 \text{ m}^3/\text{s} \]This shows that the volumetric flow rate of the air entering the system is approximately 0.0069 m³/s. This parameter is crucial for designing and assessing the effectiveness of cooling systems.