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The inlet ducting to a jet engine forms a diffuser that steadily decelerates the entering air to zero velocity relative to the engine before the air enters the compressor. Consider a jet airplane flying at \(1000 \mathrm{~km} / \mathrm{h}\) where the local atmospheric pressure is \(0.6\) bar and the air temperature is \(8^{\circ} \mathrm{C}\). Assuming ideal gas behavior and neglecting heat transfer and potential energy effects, determine the temperature, in \({ }^{\circ} \mathrm{C}\), of the air entering the compressor.

Short Answer

Expert verified
The temperature of the air entering the compressor is approximately 46.33 \text{°C}.

Step by step solution

01

Convert Speed to m/s

The airplane's speed is given as 1000 km/h. First, convert this speed to meters per second (m/s). Use the conversion factor: \text{1 km/h} = \frac{1}{3.6} \text{ m/s} Calculate the speed in m/s:\[ 1000 \text{ km/h} = 1000 \times \frac{1}{3.6} \text{ m/s} = 277.78 \text{ m/s} \]
02

Calculate the Stagnation Temperature

To find the temperature of the air entering the compressor, we need the stagnation temperature. The formula for stagnation temperature in terms of static temperature and Mach number is used.Stagnation Temperature (\text{T}_0) is calculated using:\[ \text{T}_0 = \text{T} \times \bigg( 1 + \frac{\text{k} - 1}{2} \times \text{M}^2 \bigg) \]where:\text{T} is the static temperature in Kelvin,\text{k} is the ratio of specific heats (for air, \text{k} \text{≈ 1.4}),\text{M} is the Mach number which can be calculated as the ratio of velocity to speed of sound.Convert the static air temperature \text{8}^{\text{°C}} to Kelvin:\[T = 8 + 273.15 = 281.15 \text{ K} \]
03

Calculate the Speed of Sound

Use the formula for the speed of sound in air:\[ \text{a} = \text{√} (\text{k} \times \text{R} \times \text{T}) \]where:\text{a} is the speed of sound,\text{R} is the specific gas constant for air = 287 J/(kg·K),\text{T} is the static temperature.Now calculate the speed of sound:\[ \text{a} = \text{√} (1.4 \times 287 \times 281.15) \text{ m/s} = 336.43 \text{ m/s} \]
04

Calculate the Mach Number

Mach number (\text{M}) is calculated as:\[ \text{M} = \frac{\text{velocity}}{\text{speed of sound}} = \frac{277.78 \text{ m/s}}{336.43 \text{ m/s}} = 0.826 \]
05

Find Stagnation Temperature

The stagnation temperature can now be calculated:Using the formula from Step 2\[ \text{T}_0 = 281.15 \times (1 + \frac{1.4 - 1}{2} \times 0.826^2) \] \[ \text{T}_0 = 281.15 \times (1 + 0.2 \times 0.682) \] \[ \text{T}_0 = 281.15 \times (1 + 0.1364) \] \[ \text{T}_0 = 281.15 \times 1.1364 = 319.48 \text{K} \]
06

Convert Stagnation Temperature to Celsius

Finally, convert the stagnation temperature back to \text{°C}:\[ \text{T}_0 = 319.48 - 273.15 = 46.33 \text{°C} \]

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Key Concepts

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

Stagnation Temperature
Stagnation temperature is a crucial concept in jet engine thermodynamics. It represents the temperature an air stream would reach if it were decelerated to zero velocity isentropically (without energy loss or gain). This is essential in computing parameters for jet engines as it relates to the performance efficiency of compressors and turbines.
To calculate the stagnation temperature (\textrm{T}_0), we use the equation: \[ \textrm{T}_0 = \textrm{T} \times \bigg( 1 + \frac{\textrm{k} - 1}{2} \times \textrm{M}^2 \bigg) \] where:
  • \textrm{T} is the static air temperature in Kelvin
  • \textrm{k} is the ratio of specific heats, typically \textrm{k} ≈ 1.4 for air
  • \textrm{M} is the Mach number
Converting the final stagnation temperature back to degrees Celsius involves the conversion formula: \[ \textrm{Temperature in } ^{\textrm{°C}} = \textrm{T}_0 - 273.15 \] This calculation provides the stagnation temperature in a form usable for further thermodynamic analysis of air entering the compressor.
Mach Number
The Mach number is a dimensionless quantity representing the ratio of an object's speed to the speed of sound in the surrounding medium. It is essential for understanding the behavior of airflows in various regimes from subsonic to supersonic.
The Mach number (\textrm{M}) is calculated using: \[ \textrm{M} = \frac{\textrm{velocity}}{\textrm{speed of sound}} \] For our exercise, with an airplane speed of 277.78 m/s and speed of sound in air 336.43 m/s, the Mach number becomes: \[ \textrm{M} = \frac{277.78 \textrm{ m/s}}{336.43 \textrm{ m/s}} \text{ } \textrm{≈ 0.826} \] Knowing the Mach number helps in determining whether the flow is in subsonic, transonic, or supersonic regimes, influencing design and operational decisions in aerospace engineering.
Ideal Gas Behavior
In this exercise, we assumed ideal gas behavior to simplify calculations. Ideal gas behavior implies that the gas follows the ideal gas law: \[ \textrm{PV} = \textrm{nRT} \] where:
  • \textrm{P} is pressure
  • \textrm{V} is volume
  • \textrm{n} is the amount of substance (in moles)
  • \textrm{R} is the universal gas constant
  • \textrm{T} is temperature in Kelvin
For air, which is a mixture of gases behaving ideally under standard conditions, this assumption holds well, especially in high-speed aircraft applications. By assuming ideal gas behavior, we simplify the thermodynamic equations and can make accurate predictions about pressure, temperature, and volume changes within the jet engine.
Speed of Sound in Air
The speed of sound in air is a crucial parameter for calculating the Mach number. It depends on temperature and the properties of the air, and is calculated with: \[ \textrm{a} = \textrm{√} (\textrm{k} \times \textrm{R} \times \textrm{T}) \] where:
  • \textrm{a} is the speed of sound
  • \textrm{k} is the ratio of specific heats (1.4 for air)
  • \textrm{R} is the specific gas constant (287 J/(kg·K) for air)
  • \textrm{T} is the static temperature in Kelvin
For the given temperature of 281.15 K, the speed of sound is calculated to be 336.43 m/s.
The speed of sound is pivotal for understanding aerodynamic phenomena and designing efficient jet and piston engines, as well as for predicting sonic booms and other critical events.

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Most popular questions from this chapter

Air enters a one-inlet, one-exit control volume at 8 bar, \(600 \mathrm{~K}\), and \(40 \mathrm{~m} / \mathrm{s}\) through a flow area of \(20 \mathrm{~cm}^{2}\). At the exit, the pressure is 2 bar, the temperature is \(400 \mathrm{~K}\), and the velocity is \(350 \mathrm{~m} / \mathrm{s}\). The air behaves as an ideal gas. For steady-state operation, determine (a) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\). (b) the exit flow area, in \(\mathrm{cm}^{2}\).

Does \(\dot{Q}_{\mathrm{cv}}\) accounting for energy transfer by heat include heat transfer across inlets and exits? Under what circumstances might heat transfer across an inlet or exit be significant?

Would it be desirable for a coolant circulating inside the engine of an automobile to have a large or a small specific heat \(c_{p} ?\) Discuss.

Air is compressed at steady state from 1 bar, \(300 \mathrm{~K}\), to 6 bar with a mass flow rate of \(4 \mathrm{~kg} / \mathrm{s}\). Each unit of mass passing from inlet to exit undergoes a process described by \(p v^{1.27}=\) constant. Heat transfer occurs at a rate of \(46.95 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing to cooling water circulating in a water jacket enclosing the compressor. If kinetic and potential energy changes of the air from inlet to exit are negligible, calculate the compressor power, in \(\mathrm{kW}\).

A feedwater heater operates at steady state with liquid water entering at inlet 1 at 7 bar, \(42^{\circ} \mathrm{C}\), and a mass flow rate of \(70 \mathrm{~kg} / \mathrm{s}\). A separate stream of water enters at inlet 2 as a two-phase liquid-vapor mixture at 7 bar with a quality of \(98 \%\). Saturated liquid at 7 bar exits the feedwater heater at 3 . Ignoring heat transfer with the surroundings and neglecting kinetic and potential energy effects, determine the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), at inlet 2 .

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