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The kinetic energy of a fluid increases as it is accelerated in an adiabatic nozzle. Where does this energy come from?

Short Answer

Expert verified
Answer: The energy that increases the kinetic energy of the fluid comes from the internal energy of the fluid, which is converted into work as the fluid does work on the surrounding environment due to the adiabatic process.

Step by step solution

01

Understanding Adiabatic Process

First, let's understand what an adiabatic process is. It is a process in which the system (in this case, the fluid) does not exchange heat with its surroundings. In simpler terms, no heat is gained or lost by the fluid during the process.
02

Recall the First Law of Thermodynamics

The first law of thermodynamics states that the change in internal energy (∆U) of a system is equal to the heat exchanged (Q) with the surroundings, minus the work done (W) by the system on the surroundings. Mathematically, this can be represented as: ∆U = Q - W Since the process is adiabatic, there is no heat exchange (Q = 0). Therefore, the equation becomes: ∆U = -W
03

Identify the Source of Energy

Now we know that the change in internal energy of the fluid is equal to the negative of the work done on the surroundings by the fluid. This means that the fluid's internal energy is being converted into work, which manifests in the form of increased kinetic energy as the fluid accelerates through the nozzle. In summary, the energy that increases the kinetic energy of the fluid comes from the internal energy of the fluid as it does work on the surrounding environment due to the adiabatic process.

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Key Concepts

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

Adiabatic Process
Imagine you have a balloon that you don't want to cool down or heat up—you want its temperature to stay the same as you carry it through a carnival full of hot grills and cold ice cream stands. Translating this idea to physics, an adiabatic process is somewhat similar. It is a thermodynamic process in which a fluid, like our hypothetical balloon, doesn't exchange heat with its environment. This means that the energy content of the fluid is all kept 'inside'—no heat enters or leaves the system. In practice, such a scenario is ideal and assumes a perfectly insulated system, which is hard to come by in reality, but this concept is crucial for understanding various phenomena in thermodynamics, from how a gas-powered engine cycles to how natural atmospheric processes occur.

An important takeaway is that during an adiabatic process, any changes in the energy of the fluid must come from work done on or by the system rather than from heat transfer, setting the stage for a deeper look into the first law of thermodynamics.
First Law of Thermodynamics
The first law of thermodynamics is like a universal energy accountant—it makes sure that all energy in a system is accounted for. It tells us that the energy of the universe is constant; energy can be transformed from one form to another but can't be created or destroyed. When looking at a specific system, such as a fluid in an adiabatic nozzle, this law translates to a simple yet powerful formula: \[\Delta U = Q - W\] This equation states that any change (\(\Delta\)) in internal energy (\(U\)) of our system is due to the heat (\(Q\)) added to it minus the work (\(W\)) it does on its surroundings. In an adiabatic process, since \(Q = 0\), the equation becomes \[\Delta U = -W\] This makes sense because if the fluid can't borrow energy as heat from elsewhere, it has to use what it already has—its internal energy—to do work, such as moving through a nozzle.
Internal Energy
What is this 'internal energy' that's so important? In the same way that a battery contains juice for your electronics, the internal energy (\(U\)) is the total energy stored within a system. It's the sum of all the kinetic energy of the molecules zooming around inside the system, as well as the potential energy from their interactions with each other. When a fluid's internal energy decreases in an adiabatic process, it's because that energy isn't disappearing—it's changing forms. For the fluid moving through a nozzle, the decrease in internal energy powers the increase in the speed of the fluid particles, a form of work. This internal energy can also be thought of as a measure of the temperature of the system, with higher internal energy correlating with higher temperatures. Hence, when a gas expands adiabatically and does work on its surroundings, it often cools down, because its internal energy decreases.
Kinetic Energy
Now let's dive into kinetic energy. It's the energy of motion. Think of a skateboard zipping down a hill—that's kinetic energy in action. Back to our fluid in the nozzle: as it speeds up, its kinetic energy increases. But energy doesn't come from nowhere. The kinetic energy that the fluid gains as it accelerates comes from its internal energy. To put it in everyday terms, the fluid is using its 'savings' of energy to 'pay for' its motion through the nozzle. This kinetic energy is precisely what accelerates the fluid, causing it to move faster and faster until it exits the nozzle. The magical conversion from stored (internal) energy to motion (kinetic) energy, without any heat exchanged, is the heart of what happens in an adiabatic nozzle.

Understanding these conversions is crucial for fields like aerospace engineering, where predicting the behavior of gases through nozzles determines everything from the efficiency of jet engines to the success of rocket launches.

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Most popular questions from this chapter

A 3-ft' rigid tank initially contains saturated water vapor at \(300^{\circ} \mathrm{F}\). The tank is connected by a valve to a supply line that carries steam at 200 psia and \(400^{\circ} \mathrm{F}\). Now the valve is opened, and steam is allowed to enter the tank. Heat transfer takes place with the surroundings such that the temperature in the tank remains constant at \(300^{\circ} \mathrm{F}\) at all times. The valve is closed when it is observed that one-half of the volume of the tank is occupied by liquid water. Find (a) the final pressure in the tank, ( \(b\) ) the amount of steam that has entered the \(\tan \mathrm{k},\) and \((c)\) the amount of heat transfer.

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.

In a heating system, cold outdoor air at \(7^{\circ} \mathrm{C}\) flowing at a rate of \(4 \mathrm{kg} / \mathrm{min}\) is mixed adiabatically with heated air at \(70^{\circ} \mathrm{C}\) flowing at a rate of \(3 \mathrm{kg} / \mathrm{min} .\) The exit temperature of the mixture is \((a) 34^{\circ} \mathrm{C}\) (b) \(39^{\circ} \mathrm{C}\) \((c) 45^{\circ} \mathrm{C}\) \((d) 63^{\circ} \mathrm{C}\) \((e) 77^{\circ} \mathrm{C}\)

Reconsider Prob. \(5-168 .\) Using EES (or other) software, investigate the effect of the inlet temperature of cold water on the energy saved by using the lowflow shower head. Let the inlet temperature vary from \(10^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\). Plot the electric energy savings against the water inlet temperature, and discuss the results.

\(\begin{array}{lllllll} & \text { A } & \text { long } & \text { roll } & \text { of } & \text { 2-m-wide } & \text { and } & \text { 0.5-cm-thick }\end{array}\) 1-Mn manganese steel plate \(\left(\rho=7854 \mathrm{kg} / \mathrm{m}^{3} \text { and } c_{p}=\right.\) \(\left.0.434 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) coming off a furnace at \(820^{\circ} \mathrm{C}\) is to be quenched in an oil bath at \(45^{\circ} \mathrm{C}\) to a temperature of \(51.1^{\circ} \mathrm{C}\) If the metal sheet is moving at a steady velocity of \(10 \mathrm{m} / \mathrm{min}\) determine the required rate of heat removal from the oil to keep its temperature constant at \(45^{\circ} \mathrm{C}\).

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