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Air at 13 psia and \(65^{\circ} \mathrm{F}\) enters an adiabatic diffuser steadily with a velocity of \(750 \mathrm{ft} / \mathrm{s}\) and leaves with a low velocity at a pressure of 14.5 psia. The exit area of the diffuser is 3 times the inlet area. Determine \((a)\) the exit temperature and \((b)\) the exit velocity of the air.

Short Answer

Expert verified
Question: Determine the exit temperature and velocity of air through an adiabatic diffuser given the following parameters: inlet pressure of \(13 \, \mathrm{psia}\), inlet temperature of \(65^{\circ} \mathrm{F}\), inlet velocity of \(750 \, \mathrm{ft/s}\), exit pressure of \(14.5 \, \mathrm{psia}\), and the exit area being three times the inlet area. Answer: (a) The exit temperature of the air is approximately \(537.98 \, \mathrm{R}\) and (b) the exit velocity of the air is approximately \(227.75 \, \mathrm{ft/s}\).

Step by step solution

01

Convert temperature and pressure to absolute units

Since we're dealing with an ideal gas, we need the absolute temperature and pressure values. Convert the given Fahrenheit temperature to Rankine: \(T_1=65+459.67=524.67\, \mathrm{R}\), and convert psi to psia (absolute pressure): \(P_1=13 \, \mathrm{psia}\) and \(P_2=14.5 \, \mathrm{psia}\).
02

Find the values of constant pressure specific heat and the ratio of specific heats

From the air properties table, we can find the constant-pressure specific heat, \(c_p = 0.240 \, \mathrm{Btu/(lb \cdot R)}\), and the ratio of specific heats, \(\gamma = 1.4\).
03

Calculate the exit temperature using the isentropic relation

Use the isentropic relation: \(T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{(\gamma - 1)/\gamma}=524.67\left(\frac{14.5}{13}\right)^{0.4/1.4}\). Calculate \(T_2\) to find exit temperature: \(T_2\approx 537.98\, \mathrm{R}\).
04

Use the ideal gas law to find the inlet and outlet densities

For air, \(R=53.35\, \mathrm{ft \cdot lb/(lb_m \cdot R)}\). Apply the ideal gas relation for both inlet and outlet: \(\rho_1=\frac{P_1}{RT_1}=\frac{13*144}{53.35*524.67}\approx 0.00262 \, \mathrm{lb_m/ft^3}\), \(\rho_2=\frac{P_2}{RT_2}=\frac{14.5*144}{53.35*537.98}\approx 0.00287 \, \mathrm{lb_m/ft^3}\).
05

Apply the conservation of mass equation to find the exit velocity

Using the conservation of mass and area ratio: \(\rho_1A_1V_1 = \rho_2A_2V_2\), where \(A_2=3A_1\). Rearrange for the unknown, \(V_2\): \(V_2=\frac{\rho_1A_1V_1}{\rho_2A_2}=\frac{0.00262A_1(750)}{0.00287(3A_1)}\). Calculate \(V_2\): \(V_2\approx 227.75 \, \mathrm{ft/s}\). Answer: (a) The exit temperature of the air: \(T_2 \approx 537.98 \, \mathrm{R}\), (b) The exit velocity of the air: \(V_2 \approx 227.75 \, \mathrm{ft/s}\).

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Key Concepts

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

Isentropic Relation
Understanding isentropic processes is crucial in thermodynamics, as they describe ideal transformations typically occurring in nozzles and diffusers, where heat exchange with the surroundings doesn't occur, making the process adiabetic (no heat transfer), and entropy remains constant (isentropic). The relationship pertains to the ratio of final to initial temperature and pressures of a gas during such a thermodynamic process, typically expressed as
\( T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{(\gamma - 1)/\gamma} \),
where \(\gamma\) is the ratio of specific heats at constant pressure to constant volume. This formula hinges on the ideal-gas assumption and represents the connection between pressure and temperature changes without the influence of external heat. In the solved problem where an adiabatic diffuser is used, the isentropic relation helped us to find the exit temperature of the air, demonstrating the change solely due to pressure variations.
Ideal Gas Law
The ideal gas law, a cornerstone of gas-related calculations in thermodynamics, relates the pressure, volume, and temperature of an ideal gas through the equation
\( PV = nRT \),
where \(P\) is the pressure, \(V\) is the volume, \(n\) is the mole count, \(R\) is the universal gas constant, and \(T\) is the temperature. However, in engineering, we often use the specific gas constant in terms of mass instead of moles, modifying the equation to \( P = \rho RT \), with \(\rho\) symbolizing density. This formula is pivotal for finding gas densities at different points in the flow, both in aerodynamics and the operation of machinery such as diffusers or turbines. When we apply this law to our adiabatic diffuser problem, it allows us to compute the air density at both inlet and exit points using the corresponding specific gas constant for air.
Conservation of Mass
The principle of conservation of mass, also known as mass continuity, asserts that mass cannot be created or destroyed within a closed system. Applied to fluid dynamics, the principle gives birth to the continuity equation, which translates to \( \rho_1A_1V_1 = \rho_2A_2V_2 \) for steady-flow processes, where \(\rho\) is the fluid density, \(A\) is the cross-sectional area, and \(V\) is the velocity at any two points along the flow. In the context of our exercise, this principle is indispensable in finding the exit velocity of the air through the adiabatic diffuser by correlating the changes in areas and densities.
Specific Heat Capacity
Specific heat capacity, symbolized by \(c\), is a measure of the amount of heat required to raise the temperature of a given mass of a material by one degree. In gases, this property varies with whether the process is at constant volume \(c_v\) or constant pressure \(c_p\). The distinction is pivotal as it relates closely to the work done by or upon the gas. The ratio of these specific heats, \(\gamma = \frac{c_p}{c_v}\), is essential in thermodynamics, especially when dealing with isentropic processes as it helps in describing how gases respond to pressure and temperature changes. In our problem, we utilize the specific heat capacity at constant pressure, \(c_p\), to assess the heat capacity of air for changes in temperature at a constant pressure during the adiabatic process.
Pressure-Temperature Relationship
The pressure-temperature relationship is a key aspect of thermodynamics and fluid mechanics, describing how a change in temperature reflects in a change in pressure for a given mass of gas and vice versa, particularly under constant volume. For example, Gay-Lussac's law, a derivative of the ideal gas law, states that pressure of a gas is directly proportional to its temperature when volume remains unchanged. In practical applications like refrigeration, heating systems, and aerodynamics, understanding this relationship is crucial for controlling processes. In the context of the diffuser problem, where the volume changes as air flows through, the underlying principle assists in analyzing the relationship between pressure increases and the corresponding temperature changes, leading to finding the exit temperature.

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Most popular questions from this chapter

A vertical piston-cylinder device initially contains \(0.2 \mathrm{m}^{3}\) of air at \(20^{\circ} \mathrm{C}\). The mass of the piston is such that it maintains a constant pressure of \(300 \mathrm{kPa}\) inside. Now a valve connected to the cylinder is opened, and air is allowed to escape until the volume inside the cylinder is decreased by one-half. Heat transfer takes place during the process so that the temperature of the air in the cylinder remains constant. Determine \((a)\) the amount of air that has left the cylinder and (b) the amount of heat transfer.

Water is heated in an insulated, constant-diameter tube by a \(7-\mathrm{kW}\) electric resistance heater. If the water enters the heater steadily at \(20^{\circ} \mathrm{C}\) and leaves at \(75^{\circ} \mathrm{C}\), determine the mass flow rate of water.

In large gas-turbine power plants, air is preheated by the exhaust gases in a heat exchanger called the regenerator before it enters the combustion chamber. Air enters the regenerator at \(1 \mathrm{MPa}\) and \(550 \mathrm{K}\) at a mass flow rate of \(800 \mathrm{kg} / \mathrm{min}\). Heat is transferred to the air at a rate of \(3200 \mathrm{kJ} / \mathrm{s}\). Exhaust gases enter the regenerator at \(140 \mathrm{kPa}\) and \(800 \mathrm{K}\) and leave at \(130 \mathrm{kPa}\) and \(600 \mathrm{K}\). Treating the exhaust gases as air, determine ( \(a\) ) the exit temperature of the air and \((b)\) the mass flow rate of exhaust gases.

Two mass streams of the same ideal gas are mixed in a steady-flow chamber while receiving energy by heat transfer from the surroundings. The mixing process takes place at constant pressure with no work and negligible changes in kinetic and potential energies. Assume the gas has constant specific heats. (a) Determine the expression for the final temperature of the mixture in terms of the rate of heat transfer to the mixing chamber and the inlet and exit mass flow rates. (b) Obtain an expression for the volume flow rate at the exit of the mixing chamber in terms of the volume flow rates of the two inlet streams and the rate of heat transfer to the mixing chamber. (c) For the special case of adiabetic mixing, show that the exit volume flow rate is the sum of the two inlet volume flow rates.

An \(\quad\) insulated, vertical piston-cylinder device initially contains \(10 \mathrm{kg}\) of water, \(6 \mathrm{kg}\) of which is in the vapor phase. The mass of the piston is such that it maintains a constant pressure of \(200 \mathrm{kPa}\) inside the cylinder. Now steam at \(0.5 \mathrm{MPa}\) and \(350^{\circ} \mathrm{C}\) is allowed to enter the cylinder from a supply line until all the liquid in the cylinder has vaporized. Determine ( \(a\) ) the final temperature in the cylinder and \((b)\) the mass of the steam that has entered.

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