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Combustion gases with a specific heat ratio of 1.3 enter an adiabatic nozzle steadily at \(800^{\circ} \mathrm{C}\) and \(800 \mathrm{kPa}\) with a low velocity, and exit at a pressure of 85 kPa. The lowest possible temperature of combustion gases at the nozzle exit is \((a) 43^{\circ} \mathrm{C}\) \((b) 237^{\circ} \mathrm{C}\) \((c) 367^{\circ} \mathrm{C}\) \((d) 477^{\circ} \mathrm{C}\) \((e) 640^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: (b) 237°C

Step by step solution

01

Adiabatic equation and parameters

The adiabatic equation is represented by the formula (assuming ideal gas behavior): \(p_1^{(\gamma - 1)/\gamma} \cdot T_1^{1 - 1/\gamma} = p_2^{(\gamma - 1)/\gamma} \cdot T_2^{1 - 1/\gamma}\), where: \(p_1\) = initial pressure (800 kPa) \(T_1\) = initial temperature (800 + 273.15 K) \(T_2\) = final temperature \(p_2\) = final pressure (85 kPa) \(\gamma\) = specific heat ratio (1.3)
02

Rearrange the adiabatic equation for final temperature

In order to solve for \(T_2\), we need to rearrange the adiabatic equation to isolate \(T_2\). The rearranged equation is given by: \(T_2 = \frac{p_2^{(\gamma - 1)/\gamma} \cdot T_1^{1 - 1/\gamma}}{p_1^{(\gamma - 1)/\gamma}}\) Now, we can plug in the given values to determine the final temperature.
03

Calculate the final temperature

Plugging in the given values into the rearranged adiabatic equation, we obtain: \(T_2 = \frac{(85 \times 10^3)^{(1.3 - 1)/1.3} \cdot (800 + 273.15)^{1 - 1/1.3}}{(800 \times 10^3)^{(1.3 - 1)/1.3}}\) After calculating this expression, we find that the temperature \(T_2\) is approximately 510.15 Kelvin.
04

Convert the final temperature to Celsius

Since the given answer choices are in Celsius, we need to convert the final temperature to Celsius: \(T_2^{\circ} \mathrm{C} = T_2 - 273.15\) \(T_2^{\circ} \mathrm{C} = 510.15 - 273.15 = 237^{\circ} \mathrm{C}\) The lowest possible temperature of combustion gases at the nozzle exit is \((b) 237^{\circ} \mathrm{C}\).

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Key Concepts

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

Specific Heat Ratio
The specific heat ratio, often denoted by the Greek letter \(\gamma\), is a dimensionless number that represents the ratio of the specific heat capacity at constant pressure (\(C_p\)) to the specific heat capacity at constant volume (\(C_v\)). It is a crucial parameter in the study of thermodynamics, particularly when dealing with gases, as it affects how the temperature of a gas changes as it expands or compresses.

For ideal gases, \(\gamma\) helps determine the nature of several thermodynamic processes. In an adiabatic process, for instance, where there is no heat exchange with the surroundings, the pressure and volume of the gas change in a way that is described by the specific heat ratio. Since the value of \(\gamma\) varies depending on the gas, knowing this ratio is essential for accurately predicting the behavior of a gas during processes such as expansion or compression in a nozzle, as seen in the exercise.

For monatomic gases, such as helium, \(\gamma\) typically has a value around 1.67, while for diatomic gases like nitrogen or oxygen, it's around 1.4. The specific heat ratio provided in the original exercise, 1.3, suggests we're likely dealing with a polyatomic gas, where the molecules have more complex structures and thus, different energetic properties.
Adiabatic Process
An adiabatic process is a thermodynamic process in which a system (usually a gas) does not exchange heat with its surroundings. This means that all the energy in the system is conserved, and any work done by the system will cause a change in its internal energy, resulting in a change in temperature.

During an adiabatic expansion, a gas does work on its surroundings and cools down, while in an adiabatic compression, the surroundings do work on the gas, causing it to heat up. This type of process is described by the adiabatic equation shown in the original exercise:\[p_1^{(\gamma - 1)/\gamma} \cdot T_1^{1 - 1/\gamma} = p_2^{(\gamma - 1)/\gamma} \cdot T_2^{1 - 1/\gamma}\]
Here, \(p\) stands for pressure, \(T\) for temperature, and subscripts 1 and 2 refer to the initial and final states of the gas, respectively. The absence of heat exchange makes the adiabatic process especially relevant in situations where the process happens too quickly for heat to be transferred, such as in the expansion or compression of gases in engines or nozzles.
Ideal Gas Behavior
The concept of ideal gas behavior is a simplified model that helps in understanding and predicting how gases respond to changes in pressure, volume, and temperature. According to the ideal gas law, the relationship between these variables is given by the equation \(PV = nRT\), where:\
  • \(P\) is the pressure of the gas,
  • \(V\) is the volume it occupies,
  • \(n\) is the amount of substance in moles,
  • \(R\) is the ideal gas constant, and
  • \(T\) is the absolute temperature of the gas.
  • \
Furthermore, an ideal gas is assumed to consist of perfectly elastic particles that are point-like, meaning they have no volume and do not attract or repel each other.

In practice, real gases only behave like an ideal gas under a certain range of temperatures and pressures, and deviations become notable under extreme conditions. However, for many engineering applications, such as the nozzle problem in our exercise, assuming ideal gas behavior simplifies calculations and still provides sufficiently accurate results.

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Most popular questions from this chapter

An inventor claims to have invented an adiabatic steady-flow device with a single inlet-outlet that produces \(230 \mathrm{kW}\) when expanding \(1 \mathrm{kg} / \mathrm{s}\) of air from \(1200 \mathrm{kPa}\) and \(300^{\circ} \mathrm{C}\) to \(100 \mathrm{kPa} .\) Is this claim valid?

A well-insulated heat exchanger is to heat water \(\left(c_{p}=4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) from 25 to \(60^{\circ} \mathrm{C}\) at a rate of \(0.50 \mathrm{kg} / \mathrm{s}\) The heating is to be accomplished by geothermal water \(\left(c_{p}=4.31 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) available at \(140^{\circ} \mathrm{C}\) at a mass flow rate of \(0.75 \mathrm{kg} / \mathrm{s} .\) Determine \((a)\) the rate of heat transfer and \((b)\) the rate of entropy generation in the heat exchanger.

Air enters the evaporator section of a window air conditioner at \(100 \mathrm{kPa}\) and \(27^{\circ} \mathrm{C}\) with a volume flow rate of \(6 \mathrm{m}^{3} / \mathrm{min} .\) The refrigerant- \(134 \mathrm{a}\) at \(120 \mathrm{kPa}\) with a quality of 0.3 enters the evaporator at a rate of \(2 \mathrm{kg} / \mathrm{min}\) and leaves as saturated vapor at the same pressure. Determine the exit temperature of the air and the rate of entropy generation for this process, assuming ( \(a\) ) the outer surfaces of the air conditioner are insulated and ( \(b\) ) heat is transferred to the evaporator of the air conditioner from the surrounding medium at \(32^{\circ} \mathrm{C}\) at a rate of \(30 \mathrm{kJ} / \mathrm{min}\).

Air is to be compressed steadily and isentropically from 1 atm to 16 atm by a two-stage compressor. To minimize the total compression work, the intermediate pressure between the two stages must be \((a) 3\mathrm{ atm}\) \((b) 4 \mathrm{atm}\) \((c) 8.5 \mathrm{atm}\) \((d) 9 \mathrm{atm}\) \((e) 12 \mathrm{atm}\)

The inner and outer surfaces of a \(2-m \times 2-m\) win dow glass in winter are \(10^{\circ} \mathrm{C}\) and \(3^{\circ} \mathrm{C}\), respectively. If the rate of heat loss through the window is \(3.2 \mathrm{kJ} / \mathrm{s}\), determine the amount of heat loss, in \(\mathrm{kJ},\) through the glass over a period of 5 h. Also, determine the rate of entropy generation during this process within the glass.

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