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4. The First Law of Thermodynamics for Closed Systems

## 4.4 The first law of thermodynamics for closed systems

The first law of thermodynamics is essentially an energy conservation law. Both heat and work are energy transfer mechanisms. They play an important role in the first law of thermodynamics. Table 4.4.1 summarizes the main differences between heat and work, and internal energy.

The first law of thermodynamics states that t he change in the total energy stored in a system equals the net energy transferred to the system in the fo rm of heat and work.

[latex]\Delta \rm{energy = + in - out}[/latex]

The change in the total energy of a system during a process from states 1 to 2 can be expressed as

[latex]\Delta E = E_2 - E_1 ={}_{1}Q_{2} - {}_{1}W_{2}[/latex]

If the changes in the kinetic and potential energies of the system are negligible, i.e., [latex]\Delta KE = \Delta PE = 0[/latex], then the first law of thermodynamics for a closed system can be simplified as

[latex]\Delta U = U_2-U_1 = {}_{1}Q_{2} - {}_{1}W_{2}[/latex]

[latex]E[/latex]: total energy stored in a system

[latex]U[/latex]: internal energy of a system

[latex]Q[/latex]: heat transfer in a process. A common sign convention: positive sign (+) for the heat transfer into a system, and negative sign (-) for the heat transfer out of a system. In short, the sign for heat transfer: in (+), out (-). See Figure 4.4.1 .

[latex]W[/latex]: work done by or to a system. A common sign convention: positive sign (+) for the work output (work done by a system to its surroundings), and negative sign (-) for the work input (work done by the surroundings to the system). In short, the sign for work: in (-), out (+). See Figure 4.4.1 .

Subscripts 1 and 2 refer to the initial and final states of a process.

The following procedure may be followed when solving problems with the first law of thermodynamics.

- Sketch the physical system described in the problem and show its main components.
- Set up an appropriate closed system by drawing the system boundary. How a system is set up may determine if a means of energy transfer can be regarded as heat or work.
- Indicate the heat and work transferred into or out of the system and their signs, see Figure 4.4.1 .
- Identify the type of the processes (e.g., isobaric, isothermal, isochoric, polytropic, or isentropic). Show the processes on the [latex]P-v[/latex] and [latex]T-v[/latex] diagrams if possible, and list all of the known and unknown states.
- Apply the first law of thermodynamics to the closed system, eliminating the terms that are not applicable to the system.
- Solve for the unknowns by combining the first law of thermodynamics with the ideal gas law, thermodynamic tables, and other physical laws as appropriate.

The following examples demonstrate how to apply the first law of thermodynamics to closed systems.

Consider the vapour compression refrigeration cycle consisting of a compressor, condenser, expansion device, and evaporator as shown. The compressor must consume work, [latex]W_{in}[/latex], from an external energy source such as electricity. The evaporator and condenser absorb and reject heat, [latex]Q_H[/latex] and [latex]Q_L[/latex], respectively. What is the relation between [latex]W_{in}[/latex], [latex]Q_H[/latex], and [latex]Q_L[/latex]?

The vapour compression refrigeration cycle can be regarded as a closed system with the initial and final states being identical; therefore, [latex]\Delta U = 0[/latex].

[latex]\because \Delta U = 0=Q-W[/latex]

[latex]\therefore Q_L-Q_H-(-W_{in})=0[/latex]

[latex]\therefore Q_H=Q_L+W_{in}[/latex]

Note the sign convention for heat: in (+), out (-) and for work: in (-), out (+). This relation can be interpreted as the total energy transferred out of the cycle remains the same as the total energy transferred into the cycle.

A rigid tank has two rooms, both containing R134a at the following initial states.

Room A: m = 2 kg, P =200 kPa, v = 0.132 m 3 /kg

Room B: m = 3 kg, P =500 kPa, T =100°C

A crack is developed in the partition between the two rooms, which allows R134a in the tank to mix. Assume the mixing takes place slowly until R134a in the whole tank reaches a uniform state at 50°C. Find the heat transfer during this process. The process can be treated as a quasi-equilibrium process.

First, set the whole rigid tank as the closed system.

Because the volume of the tank remains constant, the boundary work during the mixing process is zero; therefore, from the first law of thermodynamics,

[latex]\because \Delta U = Q - W[/latex] and [latex]W = 0[/latex]

[latex]\therefore \Delta U = Q[/latex]

The heat transfer during the process depends on the internal energies of the initial and final states.

[latex]\begin{align*} \Delta U &= U_3 - (U_1 + U_2) \\&= (m_1 + m_2)u_{3} - (m_{1}u_{1} + m_{2}u_{2}) \end{align*}[/latex]

where the subscripts 1, 2, and 3 represent the initial states of R134a in rooms A and B, and the final state of R134a in the whole tank, respectively.

Second, find the specific internal energies, u 1 , u 2 , and u 3.

Room A at the initial state: P =200 kPa, v = 0.132 m 3 /kg

From Table C1 , at T = -10°C, P sat = 200.6 kPa, v g = 0.09959 m 3 /kg. Since v > v g , R134a at this state is a superheated vapour.

From Table C2 , at P = 200 kPa, T = 60°C, v = 0.132057 m 3 /kg ≈ 0.132 m 3 /kg

[latex]u_1 = 427.51 \ \rm{kJ/kg}[/latex]

[latex]\mathbb{V}_A = m_{1}v_{1} = 2 \times 0.132 = 0.264 \ \rm{m^3}[/latex]

Room B at the initial state: P =500 kPa, T = 100°C

From Table C1 , at T = 100°C, P sa t = 3972.38 kPa. Since P < P sat , R134a at this state is a superheated vapour.

From Table C2 , at P =500 kPa, T = 100 °C

[latex]u_2 = 459.65 \ \rm{kJ/kg}[/latex]

[latex]v_2 = 0.058054 \ \rm{m^3/kg}[/latex]

[latex]\mathbb{V}_B = m_{2}v_{2} = 3 \times 0.058054 = 0.1742 \ \rm{m^3}[/latex]

The final state of R134a in the whole tank: T = 50°C

[latex]v_3 = \dfrac{\mathbb{V}_{tot}}{m_{tot}} = \dfrac{\mathbb{V}_A + \mathbb{V}_B}{m_1 + m_2} =\dfrac{0.264 + 0.1742}{2+3} = 0.0876 \ \rm{m^3/kg}[/latex]

From Table C1 , at T = 50°C, v g = 0.015089 m 3 /kg. Since v 3 > v g , R134a at the final state is a superheated vapour.

From Table C2 ,

At P = 200 kPa, T = 50°C, v = 0.127663 m 3 /kg, u = 419.29 kJ/kg

At P = 300 kPa, T = 50°C, v = 0.083723 m 3 /kg, u = 418.19 kJ/kg

Use linear interpolation,

[latex]\because\dfrac{P_{3}-200}{300-200} =\dfrac{0.0876-0.127663}{0.083723-0.127663} =\dfrac{u_{3} - 419.29 }{418.19-419.29}[/latex]

[latex]\therefore P_3 = 291.2 \ \rm{kPa}[/latex] and [latex]u_3 = 418.287 \ \rm{kJ/kg}[/latex]

Last, substitute u 1 , u 2 and u 3 into the simplified first law,

[latex]\begin{align*} Q &= \Delta U = (m_1 + m_2)u_{3} - (m_{1}u_{1} + m_{2}u_{2}) \\& = 5 \times 418.287 - (2 \times 427.51 + 3 \times 459.65) \\&= -142.535 \ \rm{kJ}\end{align*}[/latex]

During the mixing process, the heat is transferred from the tank to the surroundings; therefore, the sign for the heat transfer is negative.

Consider 0.5 kg of ammonia in a piston-cylinder device initially at P 1 =100 kPa, T 1 =0 o C. The ammonia is compressed until its pressure reaches P 2 =150 kPa in a polytropic process with n =1.25. Calculate the heat transfer in this process.

First, set ammonia in the piston-cylinder as the closed system. From the first law of thermodynamics,

[latex]\because \Delta U = Q - W[/latex]

[latex]\therefore Q = W + \Delta U = W + m(u_2 - u_1)[/latex]

Second, consider the boundary work in a polytropic process. The specific volumes are unknowns

[latex]W = \dfrac{P_{2}\mathbb{V}_{2} - P_{1}\mathbb{V}_{1}}{1-n} = \dfrac{m\left({P_{2}{v}_{2} - P_{1}{v}_{1}}\right)}{1-n}[/latex]

Third, find the specific volumes and specific internal energies at both initial and final states.

At the initial state: P 1 = 100 kPa, T 1 = 0°C. From Table B1 , P sat = 429.39 kPa at 0°C. Since P 1 < P sa t , ammonia is a superheated vapour.

From Table B2 ,

v 1 = 1.31365 m 3 /kg, u 1 = 1504.29 kJ/kg.

At the final state P 2 = 150 kPa. The process is polytropic with n =1.25.

[latex]\because P_1v_{1}^n = P_2v_{2}^n[/latex]

[latex]\therefore v_2=v_1\left(\dfrac{P_1}{P_2}\right)^{1/n}=1.31365\times\left(\dfrac{100}{150}\right)^{1/1.25} = 0.94974 \ \rm{m^3/kg}[/latex]

From Table B1 : at T = -25 °C and P = 151.47 kPa ≈ 150 kPa, v g = 0.771672 m 3 /kg. Since v 2 > v g , ammonia at the final state is a superheated vapour.

At P = 150 kPa, T = 20°C, v = 0.938100 m 3 /kg, u = 1535.05 kJ/kg

At P =150 kPa, T = 30°C, v = 0.972207 m 3 /kg, u = 1551.95 kJ/kg

[latex]\because\dfrac{T_{2}-20}{30-20} =\dfrac{0.94974-0.938100}{0.972207-0.938100} =\dfrac{u_{2} - 1535.05 }{1551.95-1535.05}[/latex]

[latex]\therefore T_2 = 23.4 \ ^ \rm{{\circ}} \rm{C}[/latex] and [latex]u_2 = 1540.82 \ \rm{kJ/kg}[/latex]

Last, the heat transfer in this process can now be found from

[latex]\begin{align*} Q &= W + \Delta U = m\dfrac{P_{2}{v}_{2} - P_{1}{v}_{1}}{1-n} + m(u_2 - u_1) \\&= 0.5 \left[ \dfrac{150 \times 0.94974 - 100 \times 1.31365}{1-1.25} + (1540.82 - 1504.29)\right] \\&= - 3.928 \ \rm{kJ}\end{align*}[/latex]

During this process heat is rejected to the surroundings; therefore, the sign for heat transfer is negative.

A piston-cylinder device contains steam initially at 200 o C and 200 kPa. The steam is first cooled isobarically to saturated liquid, then isochorically until its pressure reaches 25 kPa.

- Sketch the whole process on the [latex]P - v[/latex] and [latex]T - v[/latex] diagrams
- Calculate the specific heat transfer in the whole process

1. [latex]P - v[/latex] and [latex]T - v[/latex] diagrams

2. Calculate the specific heat transfer

First, set the steam in the piston-cylinder device as a closed system. From the first law of thermodynamics,

[latex]\because\Delta u = q - w[/latex]

[latex]\therefore q= \Delta u + w = (u_3 - u_1) + w[/latex]

Second, analyze the processes.

The process is isobaric from state 1 to state 2, then isochoric from state 2 to state 3. The specific work is the shaded area of the rectangle shown in the [latex]P - v[/latex] diagram; therefore,

[latex]w = P_1(v_3 - v_1)[/latex] and [latex]v_2 = v_3[/latex]

Third, determine the specific volumes and specific internal energies at states 1 and 3.

At state 1, P 1 = 200 kPa and T 1 = 200 o C.

From Table A2 ,

v 1 = 1.08048 m 3 /kg, u 1 = 2654.63 kJ/kg

State 2 is saturated liquid water at P 2 = 200 kPa.

From Table A1 ,

At T = 120 o C, P = 198.67 kPa, v f = 0.001060 m 3 /kg

At T = 125 o C, P = 232.24 kPa, v f = 0.001065 m 3 /kg

[latex]\because \dfrac{v_{2}-0.001060}{0.001065-0.001060} =\dfrac{T_{2}-120}{125-120} =\dfrac{200-198.67}{232.24-198.67}[/latex]

[latex]\therefore v_2 = 0.0010602 \ \rm{m^3/kg}[/latex] and [latex]T_2 = 120.2 \ ^ \rm{{\circ}} \rm{C}[/latex]

At state 3, P 3 = 25 kPa. v 3 = v 2 = 0.0010602 m 3 /kg.

From Table A1 , v f < v 3 < v g ; therefore, state 3 is a two phase mixture of saturated liquid and saturated vapour with T 3 = T sat ≈ 65 o C.

v f = 0.001020 m 3 /kg, v g = 6.19354 m 3 /kg

u f = 272.09 kJ/kg, u g = 2462.42 kJ/kg

The quality and specific internal energy of the two phase mixture are

[latex]x = \dfrac{v_3 - v_f}{v_g - v_f} = \dfrac{0.0010602 - 0.001020}{6.19354 - 0.001020} = 6.5 \times 10^{-6}[/latex]

[latex]\begin{align*} u_3 &= u_f + x(u_g - u_f) \\&= 272.09 + 6.5 \times 10^{-6}(0.0010602 - 1.08048) \\&=272.10 \ \rm{kJ/kg} \end{align*}[/latex]

Note that state 3 is almost a saturated liquid with very small quality; therefore, u 3 ≈ u g.

Last, calculate the specific boundary work and specific heat transfer in this whole process

[latex]\begin{align*} w &= P_1(v_3 - v_1) \\&= 200\times (0.0010602 - 1.08048) \\&= -215.884 \ \rm{kJ/kg} \end{align*}[/latex]

[latex]\begin{align*} q &=( u_3 - u_1) + w \\&= (272.10 - 2654.63) + (-215.884) \\&= -2598.4 \ \rm{kJ/kg}\end{align*}[/latex]

In this cooling process, the volume decreases, resulting in a negative specific boundary work. The temperature and the internal energy decrease too. As a result, the specific heat transfer is negative indicating a heat loss from the system to its surroundings.

Practice Problems

## Media Attributions

- Vapor compression refrigeration cycle © WGisol is licensed under a CC BY-SA (Attribution ShareAlike) license

Introduction to Engineering Thermodynamics by Claire Yu Yan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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We consider the First Law of Thermodynamics applied to stationary closed systems as a conservation of energy principle. Thus energy is transferred between the system and the surroundings in the form of heat and work, resulting in a change of internal energy of the system. Internal energy change can be considered as a measure of molecular activity associated with change of phase or temperature of the system and the energy equation is represented as follows:

Energy transferred across the boundary of a system in the form of heat always results from a difference in temperature between the system and its immediate surroundings. We will not consider the mode of heat transfer, whether by conduction, convection or radiation, thus the quantity of heat transferred during any process will either be specified or evaluated as the unknown of the energy equation. By convention, positive heat is that transferred from the surroundings to the system, resulting in an increase in internal energy of the system.

In this course we consider three modes of work transfer across the boundary of a system, as shown in the following diagram:

In this course we are primarily concerned with Boundary Work due to compression or expansion of a system in a piston-cylinder device as shown above. In all cases we assume a perfect seal (no mass flow in or out of the system), no loss due to friction, and quasi-equilibrium processes in that for each incremental movement of the piston equilibrium conditions are maintained. By convention positive work is that done by the system on the surroundings, and negative work is that done by the surroundings on the system, Thus since negative work results in an increase in internal energy of the system, this explains the negative sign in the above energy equation.

Boundary work is evaluated by integrating the force F multiplied by the incremental distance moved dx between an initial state (1) to a final state (2). We normally deal with a piston-cylinder device, thus the force can be replaced by the piston area A multiplied by the pressure P, allowing us to replace Adx by the change in volume dV, as follows:

This is shown in the following schematic diagram, where we recall that integration can be represented by the area under the curve.

Note that work done is a Path Function and not a property, thus it is dependent on the process path between the initial and final states. Recall in Chapter 1 that we introduced some typical process paths of interest:

- Isothermal (constant temperature process)
- Isochoric or Isometric (constant volume process)
- Isobaric (constant pressure process)
- Adiabatic (no heat flow to or from the system during the process)

It is sometimes convenient to evaluate the specific work done which can be represented by a P-v diagram thus if the mass of the system is m [kg] we have finally:

where: P is pressure [kPa], V is volume [m 3 ]

We note that work done by the system on the surroundings (expansion process) is positive, and that done on the system by the surroundings (compression process) is negative.

Finally for a closed system Shaft Work (due to a paddle wheel) and Electrical Work (due to a voltage applied to an electrical resistor or motor driving a paddle wheel) will always be negative (work done on the system). Positive forms of shaft work, such as that due to a turbine, will be considered in Chapter 4 when we discuss open systems.

## Internal Energy [u]

The third component of our Closed System Energy Equation is the change of internal energy resulting from the transfer of heat or work. Since specific internal energy is a property of the system, it is usually presented in the Property Tables such as in the Steam Tables .

## Enthalpy [h]

In the case studies that follow we find that one of the major applications of the closed system energy equation is in heat engine processes in which the system is approximated by an ideal gas, thus we will develop relations to determine the internal energy for an ideal gas. We will find also that a new property called Enthalpy will be useful both for Closed Systems and in particular for Open Systems, such as the components of steam power plants or refrigeration systems. Enthalpy is not a fundamental property, however is a combination of properties and is defined as follows:

As an example of its usage in closed systems, consider the following constant pressure process:

However, since the pressure is constant throughout the process:

Substituting in the energy equation and simplifying:

Values for specific internal energy (u) and specific enthalpy (h) are available from the Steam Tables , however for ideal gasses it is necessary to develop equations for Δu and Δh in terms of Specific Heat Capacities. We develop these equations in terms of the differential form of the energy equation in the next section of this chapter (Specific Heat Capacities of an Ideal Gas).

## Solved Example

Two kilograms of water at 25°C are placed in a piston cylinder device under 3.2 MPa pressure as shown in the diagram (State (1)). Heat is added to the water at constant pressure until the temperature of the steam reaches 350°C (State (2)). Determine the work done by the fluid (W) and heat transferred to the fluid (Q) during this process using the energy equations, then compare this value of Q to that obtained from the change in enthalpy of the system.

## Solution Approach:

We first draw the diagram of the process including all the relevant data as follows:

Since work involves the integral of Pdv we find it convenient to sketch the P-v diagram of the problem as follows:

Notice on the P-v diagram how we determine the specific work done as the area under the process curve. We also notice that in the Compressed Liquid region the constant temperature line is essentially vertical. Thus all the property values at State (1) (compressed liquid at 25°C) can be determined from the saturated liquid table values at 25°C.

Process Energy

## Specific Heat Capacities of an Ideal Gas

For a simple system, internal energy (u) is a function of two independent variables, thus we assume it to be a function of temperature T and specific volume v, hence:

Substituting equation (2) in the energy equation (1) and simplifying, we obtain:

Now for a constant volume process (dv = 0):

where: C v is the specific constant volume heat capacity

That is, the specific constant volume heat capacity of a system is a function only of its internal energy and temperature. Now in his classic experiment of 1843 Joule showed that the internal energy of an ideal gas is a function of temperature only, and not of pressure or specific volume. Thus for an ideal gas the partial derivatives can be replaced by ordinary derivatives, and the change in internal energy can be expressed as:

Consider now the enthalpy. By definition h = u + Pv, thus differentiating we obtain:

Again for a simple system, enthalpy (h) is a function of two independent variables, thus we assume it to be a function of temperature T and pressure P, hence:

Substituting equation (6) in the energy equation (5), and simplifying:

Hence for a constant pressure process, since dP = 0:

where: C p is the specific constant pressure heat capacity

That is, the specific constant pressure heat capacity of a system is a function only of its enthalpy and temperature. Now by definition:

Now since for an ideal gas Joule showed that internal energy is a function of temperature only, it follows from the above equation that enthalpy is a function of temperature only. Thus for an ideal gas the partial derivatives can be replaced by ordinary derivatives, and the differential changes in enthalpy can be expressed as:

Finally, from the definition of enthalpy for an ideal gas we have:

Values of R, C P , C v and k for ideal gases are presented (at 300K) in the table on Properties of Various Ideal Gases . Note that the values of C P , C v and k are constant with temperature only for mon-atomic gases such as helium and argon. For all other gases their temperature dependence can be considerable and needs to be considered. We find it convenient to express this dependence in tabular form and have provided a table of Specific Heat Capacities of Air .

## The Stirling Cycle Engine

Conceptually the Stirling engine is the simplest of all heat engines. It has no valves, and includes an externally heated space and an externally cooled space. It was invented by Robert Stirling, and an interesting website by Bob Sier includes a photograph of Robert Stirling, his original patent drawing of 1816, and an animated model of Stirling’s original engine.

In its original single cylinder form the working gas (typically air or helium) is sealed within its cylinders by the piston and shuttled between the hot and cold spaces by a displacer. The linkage driving the piston and displacer will move them such that the gas will compress while it is mainly in the cool compression space and expand while in the hot expansion space. This is clearly illustrated in the adjacent animation which was produced by Richard Wheeler ( Zephyris ) of Wikipedia .

Refer also to the animation produced by Matt Keveney in his Stirling engine animation website. Since the gas is at a higher temperature, and therefore pressure, during its expansion than during its compression, more power is produced during expansion than is reabsorbed during compression, and this net excess power is the useful output of the engine. Note that there are no valves or intermittent combustion, which is the major source of noise in an internal combustion engine. The same working gas is used over and over again, making the Stirling engine a sealed, closed cycle system. All that is added to the system is steady high temperature heat, and all that is removed from the system is low temperature (waste) heat and mechanical power.

Athens, Ohio, is a hotbed of Stirling cycle machine activity, both engines and coolers, and includes R&D and manufacturing companies as well as internationally recognized consultants in the area of Stirling cycle computer analysis. The parent company of this activity is Sunpower, Inc. It was formed by William Beale in the early 1970’s, mainly based on his invention of the free-piston Stirling engine which we describe below. Update (Jan. 2013) : Sunpower was recently acquired by AMETEK, Inc. in Pensylvania, however continues doing Stirling cycle machine development in Athens, Ohio. Update (Nov. 2013) : Sunpower has recently introduced a 1 kW Stirling Developers Kit based on a free piston Stirling engine fired by Propane or natural gas.

Some examples of single cylinder Stirling engines: Stirling Technology Inc. is a spinoff of Sunpower, and was formed in order to continue the development and manufacture of the 5 kW ST-5 Air engine. This large single cylinder engine burns biomass fuel (such as sawdust pellets or rice husks) and can function as a cogeneration unit in rural areas. It is not a free-piston engine, and uses a bell crank mechanism to obtain the correct displacer phasing. Another important early Stirling engine is Lehmann’s machine on which Gusav Schmidt did the first reasonable analysis of Stirling engines in 1871. Andy Ross of Columbus, Ohio built a small working replica of the Lehmann machine , as well as a model air engine .

Solar Heat and Power Cogeneration: With the current energy and global warming crises, there is renewed interest in renewable energy systems, such as wind and solar energy, and distributed heat and power cogeneration systems. Cool Energy, Inc. of Boulder, Colorado, is currently in advanced stages of developing a complete solar heat and power cogeneration system for home usage incorporating Stirling engine technology for electricity generation. This unique application includes evacuated tube solar thermal collectors, thermal storage, hot water and space heaters, and a Stirling engine/generator.

Ideal Analysis: Please note that the following analysis of Stirling cycle engines is ideal, and is intended only as an example of First Law Analysis of closed systems. In the real world we cannot expect actual machines to perform any better than 40 – 50% of the ideal machine. The analysis of actual Stirling cycle machines is extremely complex and requires sophisticated computer analysis (see for example the course notes on: Stirling Cycle Machine Analysis .)

The free-piston Stirling engine developed by Sunpower, Inc is unique in that there is no mechanical connection between the piston and the displacer, thus the correct phasing between them occurs by use of gas pressure and spring forces. Electrical power is removed from the engine by permanent magnets attached to the piston driving a linear alternator. Basically the ideal Stirling engine undergoes 4 distinct processes, each one of which can be separately analyzed, as shown in the P-V diagram below. We consider first the work done during all four processes.

- Process 1-2 is the compression process in which the gas is compressed by the piston while the displacer is at the top of the cylinder. Thus during this process the gas is cooled in order to maintain a constant temperature T C . Work W 1-2 required to compress the gas is shown as the area under the P-V curve, and is evaluated as follows.

- Process 2-3 is a constant volume displacement process in which the gas is displaced from the cold space to the hot expansion space. No work is done, however as we shall see below, a significant amount of heat Q R is absorbed by the gas from the regenerator matrix.
- Process 3-4 is the isothermal expansion process. Work W 3-4 is done by the system and is shown as the area under the P-V diagram, while heat Q 3-4 is added to the system from the heat source, maintaining the gas at a constant temperature T H .

- Finally, process 4-1 is a constant volume displacement process which completes the cycle. Once again we will see below that heat Q R is rejected by the working gas to the regenerator matrix.

The net work, W net , done over the cycle is given by: W net = (W 3-4 + W 1-2 ), where the compression work W 1-2 is negative (work done on the system).

We now consider the heat transferred during all four processes, which will allow us to evaluate the thermal efficiency of the ideal Stirling engine. Recall from the previous section that in order to do a First Law analysis of an ideal gas to determine the heat transferred we needed to develop equations to determine the internal energy change Δu in terms of the Specific Heat Capacities of an Ideal Gas .

The two constant volume processes are formed by holding the piston in a fixed position, and shuttling the gas between the hot and cold spaces by means of the displacer. During process 4-1 the hot gas gives up its heat Q R by passing through a regenerator matrix, which is subsequently completely recovered during the process 2-3.

Now from the First Law for a cycle:

Thus thermal efficiency:

We will find in Chapter 5 that this is the maximum theoretical efficiency that is achievable from a heat engine, and usually referred to as the Carnot efficiency.

Note that the practical Stirling cycle has many losses associated with it and does not really involve isothermal processes, nor ideal regeneration. Furthermore since the Free-Piston Stirling cycle machines involve sinusoidal motion, the P-V diagram has an oval shape, rather than the sharp edges defined in the above diagrams. Nevertheless we use the ideal Stirling cycle to get an initial understanding and appreciation of the cycle performance.

## The Stirling Cycle Cooler

One important aspect of Stirling cycle machines that we need to consider is that the cycle can be reversed – if we put net work into the cycle then it can be used to pump heat from a low temperature source to a high temperature sink. Sunpower, Inc. has been actively involved in the development of Stirling cycle refrigeration systems and produces Stirling cycle cryogenic coolers for liquefying oxygen. In 1984 Sunpower developed a free piston Duplex Stirling Machine having only three moving parts including one piston and two displacers, in which a gas fired Stirling cycle engine powered a Stirling cycle cooler. Global Cooling, Inc. was established in 1995 as a spinoff of Sunpower, and was formed mainly in order to develop free-piston Stirling cycle coolers for home refrigerator applications. These systems, apart from being significantly more efficient than regular vapor-compression refrigerators, have the added advantage of being compact, portable units using helium as the working fluid (and not the HFC refrigerants such as R134a, having a Global Warming Potential of 1,300). More recently Global Cooling decided to concentrate their development efforts on systems in which there are virtually no competitive systems – cooling between -40°C and -80°C, and they established a new company name: Stirling Ultracold .

We are fortunate to have obtained two original M100B coolers from Global Cooling. The one is used as a demonstrator unit, and is shown in operation in the following photograph. The second unit is set up as a ME Senior Lab project in which we evaluate the actual performance of the machine under various specified loads and temperatures.

## The Air-Standard Diesel Cycle (Compression-Ignition) Engine

The Air Standard Diesel cycle is the ideal cycle for Compression-Ignition (CI) reciprocating engines, first proposed by Rudolph Diesel over 100 years ago. The following link by the Kruse Technology Partnership describes the four-stroke diesel cycle operation including a short history of Rudolf Diesel. The four-stroke diesel engine is usually used in motor vehicle systems, whereas larger marine systems usually use the two-stroke diesel cycle . Once again we have an excellent animation produced by Matt Keveney presenting the operation of the four-stroke diesel cycle .

The actual CI cycle is extremely complex, thus in initial analysis we use an ideal “air-standard” assumption, in which the working fluid is a fixed mass of air undergoing the complete cycle which is treated throughout as an ideal gas. All processes are ideal, combustion is replaced by heat addition to the air, and exhaust is replaced by a heat rejection process which restores the air to the initial state.

The ideal air-standard diesel engine undergoes 4 distinct processes, each one of which can be separately analyzed, as shown in the P-V diagrams below. Two of the four processes of the cycle are adiabatic processes (adiabatic = no transfer of heat), thus before we can continue we need to develop equations for an ideal gas adiabatic process as follows:

The Adiabatic Process of an Ideal Gas (Q=0)

The analysis results in the following three general forms representing an adiabatic process:

where k is the ratio of heat capacities and has a nominal value of 1.4 at 300K for air.

Process 1-2 is the adiabatic compression process. Thus the temperature of the air increases during the compression process, and with a large compression ratio (usually > 16:1) it will reach the ignition temperature of the injected fuel. Thus given the conditions at state 1 and the compression ratio of the engine, in order to determine the pressure and temperature at state 2 (at the end of the adiabatic compression process) we have:

Work W 1-2 required to compress the gas is shown as the area under the P-V curve, and is evaluated as follows.

An alternative approach using the energy equation takes advantage of the adiabatic process (Q 1-2 = 0) results in a much simpler process:

During process 2-3 the fuel is injected and combusted and this is represented by a constant pressure expansion process. At state 3 (“fuel cutoff”) the expansion process continues adiabatically with the temperature decreasing until the expansion is complete.

Process 3-4 is thus the adiabatic expansion process. The total expansion work is W exp = (W 2-3 + W 3-4 ) and is shown as the area under the P-V diagram and is analyzed as follows:

Finally, process 4-1 represents the constant volume heat rejection process. In an actual Diesel engine the gas is simply exhausted from the cylinder and a fresh charge of air is introduced.

The net work W net done over the cycle is given by: W net = (W exp + W 1-2 ), whereas before the compression work W 1-2 is negative (work done on the system).

In the Air-Standard Diesel cycle engine the heat input Q in occurs by combusting the fuel which is injected in a controlled manner, ideally resulting in a constant pressure expansion process 2-3 as shown below. At maximum volume (bottom dead center) the burnt gasses are simply exhausted and replaced by a fresh charge of air. This is represented by the equivalent constant volume heat rejection process Q out = -Q 4-1 . Both processes are analyzed as follows:

At this stage we can conveniently determine the engine efficiency in terms of the heat flow as follows:

Again from the First Law for a cycle:

## The Air-Standard Otto Cycle (Spark-Ignition) Engine

The Air Standard Otto cycle is the ideal cycle for Spark-Ignition (SI) internal combustion engines, first proposed by Nikolaus Otto over 130 years ago, and which is currently used most motor vehicles. The following link by the Kruse Technology Partnership presents a description of the four-stroke Otto cycle operation including a short history of Nikolaus Otto. Once again we have excellent animations produced by Matt Keveney presenting both the four-stroke and the two-stroke spark-ignition internal combustion engine operation.

The analysis of the Otto cycle is very similar to that of the Diesel cycle which we analyzed in the previous section. We will use the ideal “air-standard” assumption in our analysis. Thus the working fluid is a fixed mass of air undergoing the complete cycle which is treated throughout as an ideal gas. All processes are ideal, combustion is replaced by heat addition to the air, and exhaust is replaced by a heat rejection process which restores the air to the initial state.

The most significant difference between the ideal Otto cycle and the ideal Diesel cycle is the method of igniting the fuel-air mixture. Recall that in the ideal Diesel cycle the extremely high compression ratio (around 18:1) allows the air to reach the ignition temperature of the fuel. The fuel is then injected such that the ignition process occurs at a constant pressure. In the ideal Otto cycle the fuel-air mixture is introduced during the induction stroke and compressed to a much lower compression ratio (around 8:1) and is then ignited by a spark. The combustion results in a sudden jump in pressure while the volume remains essentially constant. The continuation of the cycle including the expansion and exhaust processes are essentially identical to that of the ideal Diesel cycle.

A schematic diagram followed by an animated schematic of the cooler (both courtesy of Global Cooling ) are shown below:

Conceptually the cooler is an extremely simple device, consisting essentially of only two moving parts – a piston and a displacer. The displacer shuttles the working gas (helium) between the compression and expansion spaces. The phasing between the piston and displacer is such that when the most of the gas is in the ambient compression space then the piston compresses the gas while rejecting heat to the ambient. The displacer then displaces the gas through the regenerator to the cold expansion space, and then both displacer and piston allow the gas to expand in this space while absorbing heat at a low temperature.

Thermodynamics Copyright © by Diana Bairaktarova (Adapted from Engineering Thermodynamics - A Graphical Approach by Israel Urieli and Licensed CC BY NC-SA 3.0) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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## Understanding Closed System Thermodynamics

In the vastness of engineering studies, you may come across the principle of Closed System Thermodynamics quite often. This integral concept of thermodynamics revolves around systems in which no mass may enter or leave.

## The Meaning of Closed Systems in Thermodynamics

Technically, a closed system refers to a type of Thermodynamic System where energy can be transferred in or out, but not mass. This is one of the most fundamental concepts encountered in thermodynamic studies.

A closed system allows the transfer of energy ( Heat and Work ), but it prohibits the transfer of mass. The boundary that encases this system may change its shape or volume, but always remains impermeable to matter.

The key equation that governs the behaviour of closed thermodynamics systems is the First Law of Thermodynamics, mathematically represented by:

Understanding closed system thermodynamics is pivotal in conceptualising conservation of energy and matter principles which happen to be the foundation stones of all physical sciences.

## The Role of Closed Systems in Thermodynamics

Closed systems play a pivotal role in thermodynamics. Such systems help to explain the energy balance of a system and can be utilised to examine a wide range of phenomena.

## Examples of Closed Systems in Thermodynamics

There are several practical instances of closed systems in the realm of thermodynamics. A few examples include water in a boiler or the refrigerant in an air conditioning system.

- A gas in a piston-cylinder assembly
- A steam in a steam generator
- Refrigerant in a refrigerator coil
- A mass of fluid being transported by a pump or fan

## Everyday Examples of Closed Systems in Thermodynamics

You are in continual interaction with closed systems in everyday life. For instance, a pressured aerosol spray can is an excellent example of a closed system. The can's content doesn't change by mass, but when sprayed, it undergoes a pressure-volume change - an Isochoric Process typical to closed systems.

Another everyday example is a sealed pot of boiling water. Here, Heat energy is transferred into the pot, altering the state of water from a liquid to a gas (steam). However, since the pot is sealed, no mass escapes the system, illustrating a classic example of a closed system in thermodynamics.

## Applications of Closed Systems in Thermodynamics

In the realm of thermodynamics, the concept of a closed system is of central importance. Understanding its applications can help you, as an aspiring engineer, devise solutions that improve the efficiency of various technologies and industrial processes.

## Practical Applications of Closed Systems in Thermodynamics

Applying the principles of closed systems in thermodynamics spans numerous sectors, including but not limited to, power generation, refrigeration, automotive engineering, and countless industrial processes.

Power Generation : In power plants, closed system thermodynamics is used to harness energy. For example, in a Rankine cycle - commonly used in power plants, the working fluid is recycled continuously in a closed loop. The system absorbs heat energy and converts it into mechanical work, ultimately producing electricity.

You can use this equation to evaluate the performance and efficiency of the power plants.

Refrigeration and Air Conditioning : The operations of refrigeration and air conditioning systems are textbook examples of closed systems. The refrigerant flows in a closed cycle, where it undergoes phase changes to remove heat from the designated area.

Automotive Engineering : An internal combustion engine is a classic closed system example. The explosion of fuel in the cylinder, captured in a closed space, generates power that drives your vehicles.

The operation of this engine can be broken down into a sequence of thermodynamic processes – primarily isochoric (constant volume) and adiabatic expansions and compressions.

The applications above give you a glimpse of how pervasive and crucial closed system thermodynamics is in the realms of modern engineering and industrial processes.

## Advanced Applications of Closed Systems in Thermodynamics

Some advanced applications of closed systems in Thermodynamics include the designing of spacecraft propulsion systems, supercritical power plants, and innovative energy storage systems.

Spacecraft Propulsion : In spacecraft propulsion, the propulsion system acts as a closed system where the exhaust gases, by expanding and pushing against the walls of the engine, produce the necessary thrust. These processes can be examined in-depth using the principles of closed system thermodynamics.

Supercritical Power Plants : Supercritical power plants operate under high pressures and temperatures, transitioning from liquid to gas states without a definitive boiling point – a situation that presents unique modelling challenges. Thermodynamics of the closed system significantly aids in designing and optimising such power plants.

Energy Storage Systems : Closed system thermodynamics is also vital for designing energy storage systems such as pumped hydro storage and compressed air energy storage, where energy conservation principles are critical.

In all these applications, using the principles of closed system thermodynamics , you can analyse and optimise the energy transferral processes, thereby holding the potential for enhancing efficiencies and advancements in the technology.

In conclusion, understanding and applying closed system thermodynamics has significant implications in enhancing the efficiency, performance, and overall feasibility of various systems across numerous industries.

## Closed Systems in Thermodynamics – The Formula

In thermodynamics, the concept of closed systems is governed by a specific formula, originating from the First Law of Thermodynamics. This law is essentially a statement of energy conservation which, for a closed system, can be expressed as:

## Interpreting the Closed System Thermodynamics Formula

To understand the closed system thermodynamics formula, it is imperative to break down its constituents, each representing a particular energy transaction.

- \(\Delta U\) illustrates the internal energy change of the system.
- \(Q\) represents the net heat transfer - heat absorbed by the system from its surroundings.
- \(W\) symbolises the work done by the system on its surroundings.

The formula essentially encapsulates how energy in the form of heat (\(Q\)) and work (\(W\)) alters the system's internal energy (\(\Delta U\)).

For a closed system, \(\Delta U = Q - W\) is the fundamental equation. This equation states that the net heat supplied to the system minus the work done by the system is equal to the change in internal energy.

Let's consider a scenario where a system is subjected to an amount of heat \(Q\) while it does work \(W\) on its surroundings.

If the system absorbs heat, \(Q\) is positive. Conversely, if the system loses heat or work is done on the system, \(W\) is negative. These conventions follow the physics sign conventions – work done by the system and heat added to the system are positive, and vice versa.

Therefore, the change in internal energy, \(\Delta U\), can be increased by adding heat to the system or doing work on it, and it can be reduced by removing heat from the system or letting the system do work.

Understanding this formula and the science it embodies forms the bedrock of thermodynamic analysis and is crucial to guide experimental procedures accurately.

## Application of the Closed System Thermodynamics Formula

The closed system thermodynamic formula is instrumental in numerous applications. Power generation, cooling systems, engines, and industrial processes - all use this formula for optimisation and efficiency improvements.

In power generation , the formula can be used to calculate the thermal efficiency of a power plant by quantifying the internal energy change and measuring the work output.

For cooling systems like refrigerators and air conditioners , the formula can be utilised to calculate the coefficient of performance (COP), thereby evaluating the system's effectiveness.

The internal combustion engines that power most cars apply the thermodynamic formula during each cycle of operation. By altering the mechanics of this operation, improvements to fuel efficiency and power output can be obtained.

Lastly, various industrial processes apply the formula to enhance the efficiency of chemical reactions. A fundamental understanding of the formula allows engineers to leverage thermodynamic principles to optimise heat exchange, carry out accurate mass balance analyses, and much more.

To summarise, the formula for closed system thermodynamics, \(\Delta U = Q - W\), informs much of the day-to-day applications in the field of engineering and beyond. Whether you are an engineer, applied physicist, or environmental scientist, this formula will always be your guide in understanding and manipulating the invisible energy flows around us.

## Exploring Open and Closed Systems in Thermodynamics

In the universe of thermodynamics, the concepts of open and closed systems stand as two pillars that shape our understanding of how energy and matter interact and transform. Getting to grips with these structures can be really advantageous for students like you in mastering thermodynamics.

## Contrasting Open and Closed Systems in Thermodynamics

Open systems and closed systems are foundational elements in studying thermodynamics. These are theoretical constructs that help you comprehend the behaviour of systems under various conditions. They are defined based on the exchange of matter and energy with the surrounding environment.

An open system refers to a system where both matter and energy can transfer between the system and its surroundings. This exchange of energy can occur in the forms of heat, work, or both. On the other hand, mass can enter or leave the open system. As an everyday example, imagine a cup of hot coffee left on a table. The system (coffee) loses heat to the surroundings, and it can receive dust particles from the environment.

A closed system refers to a system which allows the transfer of energy ( Heat and Work ) but does not allow the transfer of mass with the outside environment. To visualise this, think of a sealed thermos with hot liquid. The content can transfer heat through the thermos walls but does not allow any transfer of matter in and out of it.

Using these definitions as the premise, thermodynamic analyses can be classified based on whether they deal with open or closed systems. This forms the basis of different branches of thermodynamics: Classical Thermodynamics primarily focuses on closed systems, whereas Fluid Mechanics and Heat Transfer primarily deal with open systems.

There is also a third category known as an Isolated System which does not allow either matter or energy to exchange with its surroundings. Such systems are purely theoretical as perfectly isolated systems do not exist in nature.

## Applications and Examples of Open vs Closed Systems in Thermodynamics

There are a multitude of real-life examples and applications for both open and closed systems. Here, we will delve into them in more detail to better understand their practical relevance.

Open Systems : A boiling pot on a stove is an example of an open system. The system (water in the pot) gains heat from the burner (surroundings) and also allows mass (steam) to escape into the surrounding environment. Another magnificent example is a car radiator, an open system where coolant absorbs heat from the engine and exchanges heat (usually loses) with the surrounding air.

A commercial aircraft cruising at high altitude can also be described as an open system. As it consumes fuel (a decrease in mass), it simultaneously produces thrust by expelling hot gases (also a change in mass) while also exchanging heat with its environment.

Closed Systems : On the other hand, let's take a pressure cooker as an example of a closed system. Once sealed, the pressure cooker does not allow mass to cross its boundaries, but it does allow for the transfer of heat from the stove's flame.

Another example is the internal combustion engine of a car. During combustion, mass is not exchanged with the surroundings, but heat is transferred by the working fluid. As the fuel and air inside the engine are ignited, they release energy that is transferred to the piston in the form of work, which then helps to move the car.

To conclude, the concepts of open and closed systems are elemental to understanding thermodynamics. By contrasting their properties and examining their applications, you can gain a profound understanding of their roles in various physical, chemical, and engineering phenomena. As you further explore thermodynamics, these concepts will serve as vital tools in your scientific analysis and problem-solving toolkit.

## Closed System Thermodynamics Problems and Solutions

If you're studying engineering, physics, or a related discipline, you're likely to encounter numerous problems related to closed system thermodynamics. These problems require not only a thorough understanding of the underlying principles, but also an ability to practically apply these principles to real-world scenarios.

## Case Studies on Closed System Thermodynamics

There is a myriad of case studies stemming from the closed system thermodynamics domain. These cases illustrate variations in heat exchange, work transfer, and changes in internal energy. They serve to examine thermodynamics in practical contexts, providing valuable insight into resolving complex closed system scenarios.

Here are a couple of illustrative examples:

Case Study 1: Consider a tank containing a certain amount of inert gas, and the tank is being heated. The tank, in this case, acts as a closed system. As heat is supplied, the internal energy of the gas (\(\Delta U\)) increases – this increase must be calculated. The first problem here is balancing the heat exchange. Assuming the tank is perfectly insulated (i.e. no heat loss), the entire heat supplied will contribute to the increase in internal energy and can be calculated using the formula \(Q = nC\Delta T\), where \(n\) is the number of moles, \(C\) is the specific heat at constant volume, and \(\Delta T\) is the change in temperature.

Case Study 2: Picture a closed system of a gas inside a piston-cylinder device that is expanding against the atmospheric pressure. As the gas expands, it performs work on its surroundings by pushing the piston outward. This is an example of a closed system that does work on its surroundings. The work done can be calculated using the generalized work equation \(W = P\Delta V\), where \(P\) stands for pressure and \(\Delta V\) is the change in volume.

These case studies elucidate how fundamental thermodynamic principles apply to real-world situations. Working through these problems helps you better comprehend the principles of thermodynamics and their applications.

## Practical Solutions to Closed System Thermodynamics Problems

Having looked at some common problems relating to closed systems in thermodynamics, let's now discuss their solutions in more detail.

Solution for Case Study 1: Here, the heat supplied to the gas causes the internal energy (\(\Delta U\)) to increase. This increase in internal energy can be calculated by rearranging the first law of thermodynamics formula \(\Delta U = Q - W\). Since there's no work being done (the gas neither expands nor compresses), \(W=0\). Hence, the entire heat supplied \(Q\) contributes to the increase in internal energy. We then can compute \(Q\) using the formula \(Q = nC\Delta T\), and this value gives us the increase in internal energy \(\Delta U\). Here, \(n\) represents the number of moles of gas, \(C\) is the specific heat capacity of the gas, and \(\Delta T\) is the change in the gas's temperature.

Solution for Case Study 2: In this example, the gas's expansion performs work on the surroundings by moving the piston.The work done by the system is given by the formula \(W = P\Delta V\). Since the pressure \(P\) is atmospheric pressure, and \(\Delta V\) is the change in volume, \(W\) can be calculated readily. However, this work done comes at the cost of the internal energy (\(\Delta U\)). Hence, if there is no exchange of heat with the surroundings (an adiabatic process), this work done is exactly equal to the decrease in internal energy, according to the first law of thermodynamics (\(\Delta U = Q - W\), but \(Q=0\)).

Arriving at these solutions involves comprehending the problem, identifying the relevant thermodynamics principles, and then applying the appropriate formulas or algorithms. Mastery of this process not only provides deeper insights into the principles of thermodynamics but also cultivates robust problem-solving skills that are highly applicable across scientific and engineering disciplines.

## Closed System Thermodynamics - Key takeaways

- Definition of Closed Systems in Thermodynamics: A closed system refers to a system which permits the transfer of energy (heat and work) but does not allow the transfer of mass with the outside environment.
- Examples of closed systems: Pressurized aerosol can, internal combustion engine, and a sealed pot of boiling water.
- Applications of closed system thermodynamics: Power plants such as Rankine cycle, refrigeration and air conditioning, internal combustion engines and numerous industrial processes including chemical reactions in closed containers and distillation.
- Closed System Thermodynamics Formula: \(\Delta U = Q - W\), which represents the First Law of Thermodynamics. \(\Delta U\) represents the change in internal energy of the system, \(Q\) represents the net heat transfer - heat absorbed by the system from its surroundings and \(W\) represents the work done by the system.
- Contrasting Open and Closed Systems: An open system allows the transfer of both matter and energy while a closed system allows transfer of energy but not matter. An Isolated System doesn't permit either matter or energy to exchange with its surroundings.

## Frequently Asked Questions about Closed System Thermodynamics

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What is a closed system in thermodynamics?

Show answer

A closed system in thermodynamics refers to a type of system where energy can be transferred in or out, but mass cannot. The boundary that encases the system can change its shape or volume but remains impermeable to matter.

Show question

What is the key equation that governs the behaviour of closed thermodynamics systems?

The key equation governing the behaviour of closed thermodynamics systems is the First Law of Thermodynamics, represented by: ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

What are some practical examples of closed systems in thermodynamics?

Some practical examples of closed systems in thermodynamics include water in a boiler, a gas in a piston-cylinder assembly, a steam in a steam generator, refrigerant in a refrigerator coil, and a mass of fluid being transported by a pump or fan.

What is the principle behind power generation in a closed system thermodynamics?

In power plants, closed system thermodynamics is used to harness energy through a Rankine cycle where the working fluid is recycled in a closed loop. The system absorbs heat energy and converts it into mechanical work to produce electricity.

How is closed system thermodynamics applied in the design of spacecraft propulsion systems?

The propulsion system in spacecraft acts as a closed system where the exhaust gases, by expanding and pushing against the engine walls, produce the necessary thrust. Closed system thermodynamics principles can be used to examine these processes in-depth.

What role does closed system thermodynamics play in the operation of refrigeration and air conditioning systems?

Refrigeration and air conditioning systems are illustrations of closed systems, where the refrigerant flows in a closed cycle, undergoing phase changes that remove heat from a designated area.

What does the ΔU = Q - W formula represent in thermodynamics?

In thermodynamics, the formula ΔU = Q - W represents the energy transactions in a closed system: ΔU is the change in internal energy, Q is the net heat absorbed by the system, and W is the work done by the system.

How does the ΔU = Q - W formula help in real-world applications?

The formula ΔU = Q - W is used in power generation to calculate thermal efficiency, in refrigeration and air conditioning to determine performance, in internal combustion engines to increase fuel efficiency and power, and in industrial processes to enhance efficiency of chemical reactions.

What do the variables Q and W represent in the ΔU = Q - W formula in closed system thermodynamics?

In the ΔU = Q - W thermodynamic formula, Q represents the net heat absorbed by the system from its surroundings, and W symbolises the work done by the system on its surroundings.

What is an open system in the context of thermodynamics?

An open system in thermodynamics is one where both matter and energy can transfer between the system and its surroundings. Examples include a cup of hot coffee that loses heat to the surroundings or a boiling pot on a stove.

In the field of thermodynamics, how is a closed system defined?

In thermodynamics, a closed system refers to one which allows the transfer of energy (like heat and work) but does not allow the transfer of mass with the outside environment. A sealed thermos with hot liquid serves as a good example.

What is the difference between open and closed system in the sphere of thermodynamics?

In thermodynamics, an open system allows both matter and energy to transfer between the system and its surroundings, while a closed system only permits the transfer of energy but not matter.

What is the principle of a closed system in thermodynamics?

In thermodynamics, a closed system is one in which there's no exchange of matter with its surroundings, but there is an exchange of energy. This energy exchange is governed by principles like heat exchange, work transfer, and changes in internal energy.

How can we calculate the work done by a gas in a closed system like a piston-cylinder device?

The work done by a gas in a closed system like a piston-cylinder device can be calculated using the generalized work equation \(W = P\Delta V\), where \(P\) is the pressure, and \(\Delta V\) is the change in volume.

How can you calculate the increase in internal energy of a gas heated in a closed system?

The increase in internal energy of a gas heated in a closed system can be calculated by using the formula \(\Delta U = Q - W\), where \(Q\) is the heat supplied, and \(W\) is the work done. In a perfectly insulated system with no work done, \(\Delta U = Q\), and \(Q\) can be calculated as \(Q = nC\Delta T\).

What does the Sackur Tetrode equation express?

The Sackur Tetrode equation expresses the entropy of an ideal gas in terms of the number of particles, volume, and energy.

Who developed the Sackur Tetrode equation and when?

The Sackur Tetrode equation was developed independently by Hugo Martin Sackur and Otto H. Tetrode around 1912.

What does the Sackur Tetrode equation imply in terms of the behaviour of an ideal gas?

The equation implies that in an isolated system with fixed number of particles and energy, but changeable volume, the entropy would increase if the volume increases. It suggests that gases expand to fill their containers to reach a state of maximum disorder, hence highest entropy.

What are the practical implications of the Sackur Tetrode equation?

The Sackur Tetrode equation provides insights into the nature of entropy and behaviour of ideal gases, helping to analyse situations where entropy, volume, or energy changes such as the expansion of a gas in a cylinder or the dispersion of helium atoms in a room.

Why is the Sackur Tetrode equation suitable for monatomic ideal gases?

For monatomic ideal gases like helium, neon, and argon, the kinetic molecular theory is a very close approximation of reality. The Sackur Tetrode equation, which accounts for the phase space that gas particles can occupy, provides an accurate measure of the entropy of such gases.

What is phase space in the context of the Sackur Tetrode equation?

In the Sackur Tetrode equation, phase space refers to the mathematical space representing all possible states of a system, here, all possible positions and momenta of monatomic ideal gas particles.

What does the Sackur Tetrode equation explain in the context of engineering thermodynamics?

The Sackur Tetrode equation provides detailed insights into how factors like volume, quantity of matter, and total energy dictate the entropy of an ideal gas. It is particularly useful in engineering thermodynamics, helping engineers understand and predict the behaviour of gases in different conditions.

What kind of calculations does the Sackur Tetrode equation assist with?

The Sackur Tetrode equation aids in numerous calculations relating to entropy changes. It simplifies entropy calculation into a precise formula, helping to understand entropy changes during processes like adiabatic expansion, asymptotic cooling, and varying pressures.

Can the Sackur Tetrode equation be used for diatomic gases?

Yes, with a few modifications to account for the extra degrees of freedom and the different statistical and quantum mechanical properties of diatomic gases, the Sackur Tetrode equation can indeed be applied to these gases as well.

What is the Sackur Tetrode equation and its significance in studying ideal gases?

The Sackur Tetrode equation is derived from principles of statistical mechanics, thermodynamics, and quantum mechanics. It helps us understand the behaviour and entropy of ideal gases at a deeper level, by factoring in quantum notions and statistical considerations of indistinguishable particles.

What are the essential components of the Sackur Tetrode equation derivation process?

The derivations of the Sackur Tetrode equation involve a few critical components: Boltzmann's definition of entropy, Quantum Mechanical Principle of Indistinguishability, Quantum-Grained Phase Space, and Stirling's Approximation for factorials.

What are some of the challenges when deriving the Sackur Tetrode equation and how can you overcome them?

Deriving the Sackur Tetrode equation includes handling large numbers and factorials, grasping quantum concepts, and maintaining consistency with macroscopic thermodynamic laws. Overcome these challenges using approximations like Stirling's, visualising quantum principles, and cross-verifying steps with thermodynamic laws.

How does the Sackur Tetrode equation change for monatomic and diatomic gases?

Monatomic gases have three degrees of freedom and their Sackur Tetrode equation depends on kinetic energy. Diatomic gases have additional degrees of freedom and their Sackur Tetrode equation is modified to account for rotational and vibrational modes of energy.

What factors lead to the modification of the Sackur Tetrode equation for different gases?

The modifications in the Sackur Tetrode equation are due to the extra degrees of freedom available to diatomic and polyatomic gases and the quantum mechanical nature of these gases.

How does the Sackur Tetrode equation adapt to complex gases like polyatomic gases?

The Sackur Tetrode equation alters to account for additional contributory factors like rotational, vibrational, and translational degrees of freedom. For larger polyatomic gases, it considers the vibrational spectrum of these gases.

What is the principle of Maximum Entropy in engineering thermodynamics?

The principle of Maximum Entropy in engineering thermodynamics is a statistical method that prioritizes the distribution set with the highest level of randomness or unpredictability, under certain constraints.

What does "Entropy" mean in the context of Maximum Entropy?

Entropy is a measure of the system's disorder, randomness, or unpredictability, as used in the Maximum Entropy principle.

How does the Maximum Entropy Principle apply to a coin toss scenario?

According to the Maximum Entropy Principle, a coin toss scenario would suggest a 50-50 chance for heads and tails, as this distribution possesses the highest entropy.

What is the principle of Maximum Entropy?

Maximum Entropy is the principle that suggests the decision of the statistical distribution with the highest entropy given certain constraints.

How is the principle of Maximum Entropy applied in real-world scenarios?

Maximum Entropy can be applied in various fields like audio recognition, image processing, traffic modelling and medical diagnosis. For instance, in traffic modelling, engineers apply Maximum Entropy to predict network flow, given constraints about traffic at various points.

What is the role of Maximum Entropy in Engineering Thermodynamics?

In Engineering Thermodynamics, Maximum Entropy refers to the state of thermal equilibrium, where the entropy of a thermodynamic system is at its peak. It has various applications, including optimizing heat engines and exploring thermal conduction paths.

What is one application of the Maximum Entropy concept within the field of engineering?

The Maximum Entropy concept is used in traffic modelling, where it aids in predicting network flow. Given certain constraints about traffic at different points, the traffic flow distribution that maximises the entropy tends to be the most reliable.

Where does the Maximum Entropy concept apply in the realm of research and academic practice?

Maximum Entropy is employed in imaging and image processing, helping in edge detection and resolution improvement. In econometrics, it's used to create models that align with observed values with minimal assumptions. Quantum Physics utilises it in density matrix models.

How is Maximum Entropy employed in the digital realm?

Maximum Entropy is used in speech and audio processing to differentiate sound types, in machine learning for flexible feature integration, and in data science for creating predictive models and incorporating newfound feature constraints.

Who is Edwin Thompson Jaynes and what did he contribute to the field of statistical mechanics and information theory?

Edwin T. Jaynes is a renowned physicist who proposed the Maximum Entropy Principle as an inference principle. He suggested its application across a variety of fields including image processing, linguistics, economics and machine learning, where predictions are made based on incomplete data.

What is Jaynes' Maximum Entropy Principle?

Jaynes' Maximum Entropy Principle states that given a set of constraints, one should choose the probability distribution with the maximum entropy. This principle suggests that the 'most probable' outcome is the one that preserves the most ignorance and aligns with the highest possible disorder or randomness.

How has Jaynes' perspective influenced the understanding and application of Maximum Entropy?

Jaynes' perspective has fundamentally changed the understanding and interpretation of Maximum Entropy. His outlook made Maximum Entropy a guiding principle for decision-making and probabilistic prediction. It has found use in diverse fields such as information theory, machine learning, physics, and engineering.

What is the Maximum Entropy Markov Model (MEMM) used for in machine learning?

The Maximum Entropy Markov Model (MEMM) is used to predict sequences of labels for sequences of observations, particularly in areas such as natural language processing, bioinformatics, and speech and handwriting recognition.

What unique feature does Bayesian Maximum Entropy (BME) add to the Maximum Entropy principle?

Bayesian Maximum Entropy (BME) uses the Maximum Entropy principle in conjunction with Bayesian inference, allowing the incorporation of subjective knowledge into the statistical model.

How does Maximum Entropy Markov Model (MEMM) differ from Bayesian Maximum Entropy (BME) in their applications?

MEMM predicts sequences of labels for sequences of observations, commonly used in areas like machine learning and natural language processing. On the other hand, BME predicts probabilistic events at specific locations and is used in spatial prediction.

What does a negative temperature in thermodynamics signify?

Negative temperature in thermodynamics refers to a state where the population of particles in higher energy states is greater than those in lower energy states. It doesn't mean a temperature colder than zero.

What factors influence the occurrence of negative temperature?

The occurrence of negative temperature is influenced by population inversion (more particles reside in a higher energy state than a lower one) and bounded energy levels (the system has a maximum energy level a particle can possess).

In what scenarios can the concept of negative temperature manifest?

The concept of negative temperature can manifest in scenarios like quantum systems, where population inversions are achievable, and real-world applications like lasers, which operate based on the principle of population inversion.

What are two real-world examples where negative temperatures can occur in Thermodynamics?

Two real-world examples of negative temperatures are Lasers and Magnetic Systems.

How is population inversion achieved in a laser for it to enter a negative temperature regime?

Population inversion in a laser is achieved by pumping energy into the laser, prompting electrons to 'jump' from lower-energy orbitals to higher-energy orbitals. This leads to more electrons in the high-energy state than the low-energy state, thus entering a negative temperature regime.

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MDME: MANUFACTURING, DESIGN, MECHANICAL ENGINEERING

## CLOSED SYSTEMS

A thermodynamic system can be defined as a boundary chosen for the analysis of a problem. It can be either a quantity of matter (closed system) or a region of space (open system).

## System Definitions

- Isolated system: No mass transfer and no energy transfer across the system boundary.
- Closed system: No mass transfer, but energy may transfer across the system boundary.
- Open system: Both mass and energy allowed to transfer across the system boundary.

## Internal Energy U

The kinetic energy of a molecule is more than just velocity. It can be a combination of all the degrees of freedom of the structure.

More Examples: Kinetic theory of gases: (Java) Perfect Gas Simulator (Java)

## Closed Systems

A solid is a closed system (no mass transfer across boundary)

Expanding or compressing a gas can also be a closed system if there is no flow or major leakages.

Non-flow Energy Equation (1st Law of Thermodynamics);

U 2 -U 1 = Q - W Where U = internal energy (J) Q = heat transfer (J) (+ = heat IN) W = work transfer (J) (+ = work OUT )

WARNING! The (Kinsky) book uses this definition U 2 -U 1 = Q - W. (i.e. Positive work is work done BY the system (not To the system which is more consistent with the concept of a Free Body Diagram) See Ref 1 Confused? Just remember... Expansion = Positive Work !

## Isolated Systems

Here is a bit more detail on the double negative in the outboard motor question.

The 35.1MJ of work we calculated is a negative work because it is being applied TO the system.

Now put this into the first law of thermo: U 2 -U 1 = Q - W = 0 - (-35.1) = +35.1 MJ

It says that the change of internal energy is found by the change in temperature.

## References and Notes

Relevant pages in mdme.

- thermo_energy.html
- Google search: "first law of thermodynamics".
- http://en.wikipedia.org/wiki/Energy_transformation Energy transformation
- http://www.edfenergy.com/powerup/keystage3/out/page4.html Energy
- http://en.wikipedia.org/wiki/First_law_of_thermodynamics First Law of Thermodynamics

## IMAGES

## VIDEO

## COMMENTS

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