Question

Air enters a 0.6-m-diameter fan at \(16^{\circ} \mathrm{C}, 101 \mathrm{kPa}\), and is discharged at \(18^{\circ} \mathrm{C}, 105 \mathrm{kPa}\), with a volumetric flow rate of \(0.35 \mathrm{~m}^{3} / \mathrm{s}\). Assuming ideal gas behavior, determine for steadystate operation (a) the mass flow rate of air, in \(\mathrm{kg} / \mathrm{s}\). (b) the volumetric flow rate of air at the inlet, in \(\mathrm{m}^{3} / \mathrm{s}\). (c) the inlet and exit velocities, in \(\mathrm{m} / \mathrm{s}\).

Short answer

Mass flow rate: 0.444 kg/sVolumetric flow rate at inlet: 0.366 m^3/sInlet velocity: 1.295 m/sExit velocity: 1.239 m/s.

Step-by-step solution

Identify Given Data

First, identify all the given data. The inlet data is: \( T_1 = 16^{\circ} \mathrm{C} = 289 \mathrm{K} \), \( P_1 = 101 \mathrm{kPa} \), and the exit data is: \( T_2 = 18^{\circ} \mathrm{C} = 291 \mathrm{K} \), \( P_2 = 105 \mathrm{kPa} \). The volumetric flow rate at the exit is \( \dot{V} = 0.35 \mathrm{~m}^3 / \mathrm{s} \), and the fan diameter is \( D = 0.6 \mathrm{~m} \).

Determine the Mass Flow Rate

Use the ideal gas law to determine the density of air at both the inlet and exit. The ideal gas law is: \( \rho = \frac{P}{RT} \). Assuming \( R = 287 \mathrm{~J} / (\mathrm{kg \cdot K}) \) for air: Inlet: \( \rho_1 = \frac{P_1}{R T_1} = \frac{101 \mathrm{kPa}}{287 \mathrm{~J} / (\mathrm{kg \cdot K}) \cdot 289 \mathrm{K}} = 1.214 \mathrm{~kg / m}^3 \) Exit: \( \rho_2 = \frac{P_2}{R T_2} = \frac{105 \mathrm{kPa}}{287 \mathrm{~J} / (\mathrm{kg \cdot K}) \cdot 291 \mathrm{K}} = 1.267 \mathrm{~kg / m}^3 \).

Calculate the Mass Flow Rate

The mass flow rate can be calculated using the formula \( \dot{m} = \rho \dot{V} \). Hence, we get: \( \dot{m} = (1.267 \mathrm{~kg/m}^3) \cdot 0.35 \mathrm{~m}^3 / \mathrm{s} \approx 0.444 \mathrm{~kg / s} \).

Calculate Volumetric Flow Rate at Inlet

Using the continuity equation for constant mass flow rate: \( \dot{m} = \rho_1 \dot{V}_1 = \rho_2 \dot{V}_2 \), solve for inlet volumetric flow rate: \( \dot{V}_1 = \frac{\dot{m}}{\rho_1} \approx \frac{0.444 \mathrm{~kg / s}}{1.214 \mathrm{~kg/m}^3} \approx 0.366 \mathrm{~m}^3 / \mathrm{s} \).

Calculate Inlet and Exit Velocities

Finally, use the formula for volumetric flow rate to determine the velocities: \( \dot{V} = A \cdot v \), where \( A = \frac{\pi D^2}{4} \). Hence, we get: \( A = \frac{\pi (0.6 \mathrm{~m})^2}{4} \approx 0.2827 \mathrm{~m}^2 \). Velocity at Inlet: \( v_1 = \frac{\dot{V}_1}{A} \approx \frac{0.366 \mathrm{~m}^3 / \mathrm{s}}{0.2827 \mathrm{~m}^2} \approx 1.295 \mathrm{~m/s} \). Velocity at Exit: \( v_2 = \frac{\dot{V}_2}{A} \approx \frac{0.35 \mathrm{~m}^3 / \mathrm{s}}{0.2827 \mathrm{~m}^2} \approx 1.239 \mathrm{~m/s} \).

Key concepts

ideal gas law

The ideal gas law is a pivotal concept in understanding the behavior of gases under various conditions. It is generally stated as \( PV = nRT \) where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. In our exercise, we use a rearranged form to find the density (\( \rho \)) of air. The formula \( \rho = \frac{P}{RT} \) allows us to relate pressure and temperature directly to the density, which is crucial for calculating mass flow rate. For air, \( R \) is typically taken as 287 J/(kg·K). Knowing this helps us to calculate the densities at both the inlet and exit conditions, which are essential for subsequent steps.

mass flow rate

The mass flow rate (\( \dot{m} \)) represents the amount of mass passing through a given point per unit time. Key to many fluid dynamics problems, it is calculated as \( \dot{m} = \rho \dot{V} \). Here, \( \rho \) is the density, and \( \dot{V} \) is the volumetric flow rate. In our exercise, after determining the density using the ideal gas law, we calculate the mass flow rate at the exit. For instance, with an exit density of 1.267 kg/m³ and a volumetric flow rate of 0.35 m³/s, the mass flow rate comes out to be approximately 0.444 kg/s. This parameter is crucial because it remains constant for steady-state flow, simplifying calculations for other parts of the system.

volumetric flow rate

Volumetric flow rate (\( \dot{V} \)) is a measure of the volume of fluid passing through a cross-section per unit time, usually expressed in m³/s. In the context of our problem, we know the exit flow rate (0.35 m³/s) but need to find the inlet flow rate. Using the continuity equation, \( \dot{m} = \rho_1 \dot{V}_1 = \rho_2 \dot{V}_2 \), we solve for the inlet flow rate. This calculation shows how the volumetric flow rate changes due to varying densities at different points in the system. Solving for \( \dot{V}_1 \) yields 0.366 m³/s, illustrating how volumetric flow adapts to maintain a constant mass flow rate.

continuity equation

The continuity equation is a fundamental principle in fluid mechanics that states the mass flow rate must remain constant from one cross-section of a pipe to another in a closed system. For incompressible fluids, it simplifies as \( A_1 v_1 = A_2 v_2 \), where \( A \) is the cross-sectional area and \( v \) is the fluid velocity. For compressible fluids like air, it couples with density, forming \( \dot{m} = \rho A v \). By using \( \dot{m} = \rho_1 \dot{V}_1 \) and \( \rho_2 \dot{V}_2 \), we instinctively maintain the principle that the amount of mass entering a system equals the amount exiting. This is deeply tied into ensuring the derived flow rates and velocities are accurate for both the inlet and exit points.

velocity calculation

Calculating velocity is crucial for understanding how fast air moves through the fan's cross-section. Utilizing the formula \( \dot{V} = A \cdot v \), where \( A \) is the area and \( v \) is the velocity, we can determine how changes in flow rate affect speeds. Given the fan diameter (0.6 m), we first calculate the area: \( A = \frac{\pi D^2}{4} = 0.2827 \text{ m}^2 \). To find the velocities, we reframe the formula for speed: \( v = \frac{\dot{V}}{A} \). Applying this formula, the inlet velocity (\( v_1 \)) calculates to approximately 1.295 m/s and the exit velocity (\( v_2 \)) roughly 1.239 m/s. This step underscores how areas and flow rates inform velocity, critical for performance analysis of such systems.