91Ó°ÊÓ

At steady state, air at \(200 \mathrm{kPa}, 325 \mathrm{~K}\), and mass flow rate of \(0.5 \mathrm{~kg} / \mathrm{s}\) enters an insulated duct having differing inlet and exit cross-sectional areas The inlet cross-sectional area is \(6 \mathrm{~cm}^{2}\). At the duct exit, the pressure of the air is \(100 \mathrm{kPa}\) and the velocity is \(250 \mathrm{~m} / \mathrm{s}\). Neglecting potential energy effects and modeling air as an ideal gas with constant \(c_{p}=\) \(1.008 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the velocity of the air at the inlet, in \(\mathrm{m} / \mathrm{s}\). (b) the temperature of the air at the exit, in \(\mathrm{K}\). (c) the exit cross-sectional area, in \(\mathrm{cm}^{2}\).

Step by step solution

01

Identify and List Known Information

Identify the given data: - Inlet pressure: \(P_1 = 200 \mathrm{kPa}\)- Inlet temperature: \(T_1 = 325 \mathrm{K}\)- Mass flow rate: \(\dot{m} = 0.5 \mathrm{kg/s}\)- Inlet cross-sectional area: \(A_1 = 6 \mathrm{cm^2}\)- Exit pressure: \(P_2 = 100 \mathrm{kPa}\)- Exit velocity: \(V_2 = 250 \mathrm{m/s}\)- Heat capacity: \(c_p = 1.008 \mathrm{kJ/kg \,K}\)

02

Calculate the Inlet Velocity

Use the mass flow rate equation: \(\dot{m} = \rho_1 \cdot A_1 \cdot V_1\)Recall ideal gas law: \(\rho_1 = \frac{P_1} {R \cdot T_1}\), where the specific gas constant for air \(R = 0.287 \mathrm{kJ/kg\, K}\).Calculate air density at the inlet: \(\rho_1 = \frac{200 \mathrm{kPa}} {0.287 \mathrm{kJ/kg \cdot K} \cdot 325 \mathrm{K}}\)The mass flow rate equation becomes: \(0.5 \mathrm{kg/s}= \rho_1 \cdot 6 \mathrm{cm^2} \cdot V_1\)Convert cross-sectional area to square meters: \(A_1 = 6 \mathrm{cm^2} = 6 \times 10^{-4} \mathrm{m^2}\)Solve for inlet velocity \(V_1\): \(V_1 = \frac{\dot{m}} {\rho_1 \cdot A_1}\)

03

Calculate the Outlet Temperature

Use the steady flow energy equation assuming no heat transfer: \(\dot{m} \left( h_1 + \frac{V_1^2}{2} \right) = \dot{m} \left( h_2 + \frac{V_2^2}{2} \right)\) Neglecting potential energy changes, change in enthalpy is approximated by: \(h_2 - h_1 = c_p (T_2 - T_1)\)Write the steady flow energy equation as: \(c_p T_1 + \frac{V_1^2}{2} = c_p T_2 + \frac{V_2^2}{2}\)Solve for \(T_2\): \(T_2 = T_1 + \frac{V_1^2 - V_2^2}{2 c_p}\)

04

Calculate the Exit Cross-Sectional Area

Use the mass flow rate equation at the outlet: \(\dot{m} = \rho_2 \cdot A_2 \cdot V_2\)Calculate air density at the outlet using ideal gas law: \(\rho_2 = \frac{P_2} {R \cdot T_2}\)Rearrange to solve for the exit area \(A_2\): \(A_2 = \frac{\dot{m}} {\rho_2 \cdot V_2}\)Convert \(A_2\) from \(\mathrm{m^2}\) to \(\mathrm{cm^2}\) by multiplying by \(10^4\).

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Key Concepts

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

Ideal Gas Law

The ideal gas law is a fundamental principle used to relate the pressure, volume, and temperature of an ideal gas. This law is expressed as \[ PV = nRT \] where: \[ P \] is the pressure, \[ V \] is the volume, \[ n \] is the number of moles of gas, \[ R \] is the universal gas constant, and \[ T \] is the temperature in Kelvin. In many engineering problems, we use a version of the ideal gas law with density ( math \rho): \[ \rho = \frac{P} {R \times T} \] where \[ \rho \] is the density of the gas. This form of the equation is particularly useful when dealing with gases flowing through ducts, as it allows us to determine the density of the gas at different points along the duct.

Mass Flow Rate

Mass flow rate is a measure of the amount of mass moving through a cross-sectional area per unit of time. It is denoted by \[ \textmath{dot{m}} \] and is given by the equation: \[ \textmath{dot{m}} = \rho \times A \times V \] where: \[ \rho \] is the density of the gas, \[ A \] is the cross-sectional area, and \[ V \] is the velocity of the gas. The mass flow rate is crucial in determining how much gas is moving through a system, which affects the energy transfer and other properties. In the exercise, knowing the mass flow rate and the gas properties allows us to find other parameters like velocity and cross-sectional area at different points in the duct.

Steady Flow Energy Equation

The steady flow energy equation is used in thermodynamics to analyze the energy balance for a fluid flowing through a control volume in a steady-state process. The general form of the equation is: \[ \textmath{dot{m}} ( h_1 + \frac{V_1^2}{2} ) = \textmath{dot{m}} ( h_2 + \frac{V_2^2}{2} ) \] where: \[ h \] is the enthalpy of the fluid, \[ V \] is the velocity, \[ \textmath{dot{m}} \] is the mass flow rate. This equation assumes that there are no heat transfers, work interactions, or changes in potential energy. It balances the energy in terms of enthalpy and kinetic energy before and after the control volume. By rearranging this equation, we can solve for unknown parameters like the temperature at the exit of a duct.

Enthalpy

Enthalpy is a measure of the total energy of a thermodynamic system, including internal energy and the energy required to displace its environment to make room for it (pressure-volume work). The specific enthalpy ( math h) is given by: \[ h = c_p T \] where: \[ c_p \] is the specific heat capacity at constant pressure, and \[ T \] is the temperature. Enthalpy plays a key role when dealing with energy transformations in thermodynamic systems. It allows us to account for the energy carried by the fluid in different states within the system. In the context of the exercise, changes in enthalpy help us determine the temperature at different stages of the duct flow process, which is crucial for solving the problem.