Paradoxical as it may seem at first glance, a convex curvature of a pulley causes a flat belt to keep track and not jump off!

To understand this paradoxical behavior, let’s take a closer look at the belt on a crowned pulley. For this purpose, the figure below shows a flat belt attached to a cylindrical pulley (driving pulley) and to a crowned pulley (driven pulley). The curvature is greatly exaggerated for the sake of clarity. In practice, the curvature on pulleys is hardly visible and is often only a fraction of a millimeter.

Due to the curvature, the belt is tensioned differently on the two sides. The side of the belt closer to the center is tensioned more than the other side. This difference in tension means that the belt no longer wraps symmetrically around the pulley, but curves. This effect can be demonstrated relatively easily with a rubber band or even an elastic cloth. If you pull the rubber band apart on one side, the result is not a straight line, but the band curves outwards and forms an arc!

This effect of the bulge on the more tensioned side is also seen on the belt when it wraps around the crowned pulley. The belt is, so to speak, superelevated. This effect is all the greater the greater the differences in the tensions of the two sides of the belt and thus the more the pulley is curved.

Because the pulley and the belt itself are curved, the side of the belt with the bulge hits the pulley first.

This point of the belt, which runs onto the pulley first, is also under high tension and thus sticks very strongly to the pulley. This point, however, is tempted to run along a circular path around the pulley (line drawn in white with contact points marked in red). In this way, the belt thus pulls itself up towards the center of the pulley.

As long as the belt does not lie symmetrically in the center of the pulley, the more tensioned side will continue to bulge and thus always pull the belt further upwards. This effect only disappears when the belt lies symmetrically in the center of the pulley and both sides of the belt are equally tensioned. In this case, there is no bulge and the belt does not pull in any direction. This leads to the self-centering effect of the crowned pulleys and to the fact that the belt does not jump off the pulley.

Note that such a self-centering effect is only present with flat belts as long as the belt width is significantly greater than the belt height. With round belts or belts with a square cross-section, this self-centering effect no longer works with curved pulleys.

Flow process work

As explained in detail in the article on flow process work, the flow process work W_p is made up of the sum of flow work W_f and pressure-volume work W_v:

\begin{align}
\label{5715}
&W_\text{p} = W_\text{v} + W_\text{f} \\[5px]
\label{4461} 
&\boxed{W_\text{p} = – \int\limits_{V_1}^{V_2}p(V)~\text{d}V + p_2~V_2 – p_1~V_1 }
\end{align}

The individual terms were explained in more detail using the example of a continuously operating pump or compressor. In the following, however, the individual terms in equation (\ref{4461}) will be discussed in more detail on the basis of a discontinuously operating piston compressor. The piston compressor offers the didactic advantage that the individual thermodynamic processes (suction, compression and discharge) take place one after the other and can be clearly understood.

How a piston compressor works

The piston compressor basically consists of a cylinder provided with an inlet and outlet valve. Inside the cylinder, an electrically driven piston moves up and down. First, air flows into the cylinder with the piston moving down and the inlet valve open. Then the air is compressed with the valves closed. After the compression, the outlet valve is opened. This discharge valve connects the cylinder to a large pressure tank in which the compressed air is stored.

The pressure in the tank is kept constant by the compressor starting immediately when compressed air is taken and the pressure is about to drop. Ideally, this means that there is no pressure drop in the vessel. The compressor’s discharge valve opens precisely when the pressure in the cylinder is the same as the pressure in the pressure tank, so that the compressed air is pushed out of the cylinder at constant pressure.

Thermodynamic processes

In the following, we will take a closer look at the intake, compression and discharge of the air. However, we will not consider the intake process and discharge process from the point of view of the air sucked in or discharged, but from the point of view of the surroundings (where the air is sucked in) and the pressure vessel (where the air is discharged).

Intake process

First, the compressor sucks in air from that room in which the compressor is located – for example, from a basement room whose windows and doors are closed. The environment thus forms a closed system (in principle, the location of the compressor is irrelevant, as the entire earth could also be considered a closed system). If the piston of the compressor now retracts, the volume of the surroundings (i.e. the room) is increased by the volume of cylinder V₁.

This expansion process of the ambient air can be considered isobaric, since the pressure in the room will not change during this process. The ambient air in the closed room thus expands at constant pressure p₁, whereby the volume of the surroundings increases by ΔV = V₁. Consequently, the pressure-volume work W₁=-p₁⋅V₁ is performed by the surroundings (this work is to be counted mathematically negative, because the work is done by the ambient air):

\begin{align}
\label{8780}
W_1= -p_1~V_1
\end{align}

Discharge process

Now we consider the discharge process, i.e. when the compressed air is pushed out into the pressure tank. In this case, the tank is again regarded as a closed system. The compressed air already in the container can be thought of as being in a “bubble”. If the (compressed) volume V₂ coming from the compressor is now pushed into the pressure tank, the air bubble imagined therein is compressed by this volume. Thus, the volume of the air bubble decreases by the volume V₂.

Here again, this compression process can be assumed to be isobaric, if only the pressure tank is large enough. In this case, the small inflow will not significantly change the pressure in the tank. The air already in the tank is thus compressed at constant pressure p₂ by the volume ΔV = V2. Thus, the pressure-volume work W₂=p₂⋅V₂ is performed on the air in the pressure tank (this work is to be counted mathematically positive, because work is done on the compressed air in the tank):

\begin{align}
\label{eq:8844}
W_2 = p_2~V_2
\end{align}

Change in energy of the air in the surroundings and in the pressure tank

The energy content of the air in the surroundings therefore changes by the pressure-volume work W₁, while the energy content of the air in the pressure tank changes by W₂. The cause of these changes is obviously due the open system. The change in energy content is due to the air being pushed in and out and thus corresponds to the flow work W_f of the open system:

\begin{align}
\label{6852}
W_\text{f} = W_2+W_1=p_2~V_2-p_1~V_1
\end{align}

This example shows that the flow work of an open system can be interpreted as pressure-volume work of closed systems which are connected to the open system! The open system (compressor) transfers energy from one closed system (environment) to another closed system (pressure tank), so to speak.

Change in energy of the air inside the compressor

However, the compressor does not only change the energy content of the two closed systems, but also the energy content of the sucked-in air, i.e. the air that is pushed through the open system. For a complete energy balance, therefore, the compression process of the sucked in and compressed air from volume V₁ to V₂ must also be taken into account.

This volume change leads to a third work transfer W_v, which, however, no longer takes place isobarically. Therefore, the integral -∫p(V)⋅dV must be determined to calculate the pressure-volume work in the open system. The sum of all three work transfers is to be done by the compressor and finally corresponds to the total flow process work W_p:

\begin{align}
\label{3568}
W_\text{p} = W_\text{f} + W_2 + W_1=-\int\limits_{V_1}^{V_2}p(V)~\text{d}V + p_2~V_2-p_1~V_1
\end{align}

Note that the compression process of the air from the volume V₁ to V₂ takes place in the compressor with closed valves. Thus, this work transfer can also be considered as pressure-volume work W_v of a closed system! Therefore, it can be seen that the individual terms in the formula (\ref{4461}) for the calculation of the flow process work can always be traced back to processes of closed systems!

Change in temperature during compression

The isobaric process of the surrounding during the intake process (work transfer W₁) is in principle accompanied by a temperature change. The temperature change depends on how much the volume of the ambient air changes. For ideal gases, the law of Gay-Lussac applies. It states that in an isobaric process, the temperature changes to the same extent as the volume changes. However, the increase in volume of the surrounding air due to the retracting piston can be neglected and so can the temperature change. The temperature change of the compressed air in the pressure tank during the isobaric discharge process (work transfer W₂) can also be neglected in practice for the same reason.

But nevertheless, in reality, one will notice a significant increase in temperature between the air sucked in and the air pushed out of the compressor! However, this temperature change does not result from the intake or discharge process, but from the compression in the compressor itself (work transfer W_v)! As a rule, not only the pressure but also the temperature of the gas increases during compression. Depending on the type of compressor, temperatures of more than 100 °C can occur. For this reason, compressors are usually cooled.

To determine the temperature change during compression, the first law of thermodynamics for closed systems can be applied. This is because, as already explained, the compression takes place with valves closed and thus in a closed system. Even if the compressor operates without classical valves (e.g. rotary-screw compressor), this process can still always be traced back to a closed system. For example, by simply imagining a bubble around the sucked-in air mass, which represents a closed system.

The first law for a closed system states that the transfer of (pressure-volume) work W_v and heat Q result in a change of the internal energy ΔU of the fluid:

\begin{align}
\label{7560}
\underbrace{-\int\limits_{V_1}^{V_2}p(V)~\text{d}V}_{=W_\text{v}} + Q = \Delta U
\end{align}

For an ideal gas, the change in internal energy ΔU is due only to the temperature change ΔT:

\begin{align}
\label{8373}
\Delta U = c_\text{v}~m~\Delta T
\end{align}

Finally, for an ideal gas, the resulting temperature change ΔT can be determined from the supplied work W_V>0 and the dissipated heat Q<0 due to cooling. Or vice versa, the temperature change can then be used to determine the pressure-volume work.

Shaft work in an open system

In the article Shaft work in open systems it was shown that the shaft work W_s is generally composed of the flow process work (W_p), the gravitational work to lift the fluid (W_g) and the work to accelerate the fluid (W_a) as it passes through the open system (as well as the frictional work W_diss, which, however, shall be neglected):

\begin{align}
\label{8918}
&\boxed{W_\text{s} = W_\text{p}+W_\text{g}+W_\text{a}~~~(+W_\text{diss})} \\[5px]
&~~~W_\text{p} = W_\text{v} + W_\text{f} \\[5px]
&~~~W_\text{f} = \Delta (pV) \\[5px]
&~~~W_\text{g} = mg \cdot \Delta h \\[5px]
&~~~W_\text{a} = \tfrac{1}{2}m \cdot \Delta c^2 \\[5px]
\end{align}

The flow process work W_p in turn can be expressed as the sum of flow work W_f and pressure-volume work W_v. Thus, the shaft work W_s for the frictionless case (W_diss=0) can also be expressed as follows:

\begin{align}
\label{9318}
&W_\text{s} = \underbrace{W_\text{v} + W_\text{f}}_{W_\text{p}} + W_\text{g} + W_\text{a} \\[5px]
\end{align}

Flowing fluid element as a closed system

The pressure-volume work W_v in equation (\ref{9318}) is responsible for the fact that the volume of a considered fluid element, which just flows through the open system, changes from V₁ to V₂. Such a change in volume of a fluid element can be considered to take place in a closed system (see also the article flow process work).

Just imagine a bubble around a fluid volume of mass m with volume V₁. This bubble is now sucked in and compressed to the volume V₂ (in pumps) or expanded (in turbines). The envelope of the imaginary bubble represents the system boundary across which there is no mass transfer with the surroundings or vice versa. Thus, the volume change of the fluid element in the open system (then also called the control volume) can be considered as taking place in a closed system.

The laws of closed systems can therefore be used as a basis for the calculation of the pressure-volume work, in particular the first law of thermodynamics for a substance in a closed system. This states that for the frictionless case, the supply of pressure-volume work W_v and heat Q leads to a change in the internal energy ΔU of the substance:

\begin{align}
\label{6682}
W_\text{v} + Q = \Delta U \\[5px]
\end{align}

Note: In principle, a change in the kinetic and gravitational energy of the considered fluid element would have to be taken into account for moving closed systems. In the present approach, however, one moves with the center of gravity of the fluid element while it flows through the control volume. Regarding this center of gravity, however, then neither the gravitational energy nor the kinetic energy changes! When viewed from a fixed coordinate system (external observer), the change of gravitational energy and kinetic energy is already accounted for by equation (\ref{9318}) anyway.

Equation (\ref{6682}) explains why the air sometimes has a temperature of over 100 °C when it is pushed out of a compressor. This is because during the compression process, pressure-volume work is done on the gas (W_v>0), which without cooling (Q=0) leads directly to an increase in internal energy (W_V=ΔU). Since for ideal gases the internal energy is directly linked to the temperature, the temperature increases. To avoid such, mostly undesirable temperature increases, compressors are usually cooled (Q<0)!

First law of thermodynamics for open systems

The pressure-volume work W_v can thus also be determined according to equation (\ref{6682}) on the basis of the change in internal energy ΔU and the heat transfer Q:

\begin{align}
\label{1957}
W_\text{v} = \Delta U – Q \\[5px]
\end{align}

Equation (\ref{1957}) can now be put into equation (\ref{9318}). Then the terms are sorted according to process quantities and state quantities. Finally, the first law of thermodynamics is obtained for a fluid moving through an open system:

\begin{align}
\label{7962}
&W_\text{s} = \overbrace{\Delta U – Q}^{=W_\text{v}} + W_\text{f} + W_\text{g} + W_\text{a} \\[5px]
\label{6700}
&\underbrace{W_\text{s} + Q}_{\text{process quantities}} = \underbrace{\Delta U + W_\text{f} + W_\text{g} + W_\text{a}}_{\text{changes of state quantities}} \\[5px]
\label{6671}
&\boxed{W_\text{s} + Q = \Delta U + \Delta \left(p~V \right) + m~g~\Delta z + \tfrac{1}{2}~m~\Delta c^2} \\[5px]
&\text{First law for open systems}\\[5px]
\end{align}

To the left of the equals sign are only process quantities which are transferred to/from the flowing fluid. On the one hand, this is the shaft work W_s, which is usually transferred via shafts, e.g. by the pump shaft or the turbine shaft. On the other hand, heat Q can be transferred, as for example in turbines by the burning of fuel (heat supply) or in the case of compressors by cooling fins (heat dissipation). On the right side of the equals sign, however, there are only state variables which are changed accordingly by this external energy transfer.

Note that the flow work W_f can also be interpreted as a change of state variables. The flow work expresses that a fluid element with initial pressure p₁ and volume V₁ has been changed to final pressure p₂ and volume V₂ ⇒ Δ(pV). The flow work is in the thermodynamic sense therefore no process quantity as one could think on the basis of the term work!

The first law can also be stated in specific form (i.e. related to the fluid mass m) and is thus independent of the actual mass flowing through the open system:

\begin{align}
&\boxed{w_\text{s} + q = \Delta u + \Delta \left(p~v \right) + g~\Delta z + \tfrac{1}{2}~\Delta c^2} \\[5px]
\end{align}

Interpretation of the first law of thermodynamics for open systems

Thus, the first law for open systems can be interpreted according to equation (\ref{6671}) as follows: If shaft work W_s and heat Q are transferred across the system boundary to/from a fluid moving through an open system, the fluid will generally undergo the following changes:

• ΔU: the internal energy of the fluid will change. This usually results in a change in temperature.
• Δ(pV): the energy with which the fluid is pushed through the open system due to the acting (static) pressure at the system boundary changes. This is associated with a change in volume and pressure.
• Δz: the gravitational energy of the moving fluid changes, resulting in a change in the position of the center of gravity.
• Δc²: the kinetic energy changes, resulting in altered flow velocities.

In equation (\ref{7962}) the term which accounts for dissipation (W_diss) was neglected. If dissipation energy is taken into account, the term W_diss should also appear in equation (\ref{6671}). However, the dissipation energy transferred will sooner or later be dissipated into internal energy anyway, where it will lead to the change in pressure p and volume V . Thus, the dissipation energy is already considered in the terms ΔU and Δ(pV)!

Remark: Equation (\ref{6671}) includes, among other things, the change in internal energy ΔU and the change in pressure and volume Δ(pV). These are typical thermodynamic state quantities. With the concept of enthalpy, these two terms can be combined to form the so-called enthalpy change.

Excursus: Bernoulli equation

The first law of thermodynamics for an open system explained in the previous section can also be applied to pure flow processes of incompressible fluids such as liquids. For this purpose, a frictionless water flow through a pipe with a variable cross-section is considered.

No energy in the form of heat or work is to be transferred to/from the water. The water therefore just flows through the pipe. Thus, both shaft work W_s and heat Q in equation (\ref{6671}) are zero. A change in temperature can therefore also be neglected, so that the internal energy does not change ΔU=0. For this case, the first law then results as follows:

\begin{align}
\label{6570}
\underbrace{W_\text{s}}_{=0} + \underbrace{Q}_{=0} &= \underbrace{\Delta U}_{=0} + \Delta \left(p~V \right) + m~g~\Delta z + \tfrac{1}{2}~m~\Delta c^2 \\[5px]
\label{eq:4700}
0 &= \Delta \left(p~V \right) + m~g~\Delta z + \tfrac{1}{2}~m~\Delta c^2 \\[5px]
\end{align}

If the pressure at the inlet of the pipe is denoted by p₁, the volume by V₁, the flow velocity by c₁ and the inlet height by z₁, and the corresponding quantities at the outlet are given the index “2”, the following relationship applies:

\begin{align}
\label{7577}
&0 = \underbrace{p_2~V_2-p_1~V_1}_{\Delta \left(p~V \right)} + \underbrace{m~g~z_2-m~g~z_1}_{m~g~\Delta z} + \underbrace{\tfrac{1}{2}~m~ c_2^2-\tfrac{1}{2}~m~ c_1^2}_{\tfrac{1}{2}~m~\Delta c^2} \\[5px]
\end{align}

Note that for incompressible fluids there is basically no need to distinguish between V₁ and V₂, since for such cases both volumes are identical. The volume is therefore simply denoted by V (without index). Thus, the following relationship applies between the different quantities at the inlet and outlet:

\begin{align}
\label{8183}
&V~p_1 + m~g~z_1 + \tfrac{1}{2}~m~c_1^2 = V~p_2 + m~g~z_2 + \tfrac{1}{2}~m~c_2^2 \\[5px]
\end{align}

Dividing this equation by the volume V of the considered fluid element yields the so-called Bernoulli equation for inviscid flows of incompressible fluids:

\begin{align}
\label{4404}
&\boxed{p_1 + \rho~g~z_1 + \tfrac{1}{2}~\rho~c_1^2 = p_2 + \rho~g~z_2 + \tfrac{1}{2}~\rho~c_2^2} ~~~\text{Bernoulli equation} \\[5px]
&~~~\text{where: } \rho = \frac{m}{V} \\[5px]
\end{align}

Flow process work

In the article on flow process work, the discussion of the work performed in open systems (e.g. in compressors or pumps) was only limited to the flow process work. It was assumed that the open system only has to do work to push the fluid through the system (flow work) and to change the volume of the fluid (pressure-volume work). This work of an open system, called flow process work W_p, can be determined by the following integral:

\begin{align}
\label{7410}
W_\text{p} = \int\limits_{p_1}^{p_2}V(p)~\text{d}p
\end{align}

In this integral, V(p) denotes the volume of a fluid element as a function of the pressure when it flows through the open system. Besides the flow process work, however, other energetic processes generally take place in an open system. The following example of a water pump will be used to illustrate this.

Work to accelerate the fluid

We consider a pump where the inlet cross-section is larger than the outlet cross-section. Nevertheless, in the steady-state operation, the same amount of mass flows through the smaller outlet cross-section as through the larger inlet cross-section within a certain time. This fact is due to the conservation of mass (continuity equation), because in steady-state operation, no mass can accumulate or be annihilated in the pump. Thus, the same amount of water will enter the pump through the inlet cross-section as exits at the outlet cross-section at the same moment.

However, since the outlet cross-section is smaller, the water mass there must flow through at a higher velocity. Only in this way can the same mass of water flow through as at the inlet in the same time. This phenomenon is also familiar from everyday experience: if you squeeze the front end of an open garden hose and thus reduce the cross-section, the water flows out at a higher speed.

Accordingly, a fluid element of mass m must be accelerated from the inflow velocity c₁ to the outflow velocity c₂ when flowing through the pump. Thus, in addition to the flow process work, the pump also has to do work W_a on the fluid to accelerate it:

\begin{align}
\label{6723}
\boxed{W_\text{a} = \tfrac{1}{2}~m~\Delta c^2} ~~~\text{where}~~~ \Delta c^2 = c_2^2 – c_1^2
\end{align}

Note: With identical inlet and outlet cross sections, no work needs to be done to accelerate the fluid for incompressible fluids such as liquids, since the velocities do not change. For compressible fluids such as gases, however, the flow velocity will change even with identical inlet and outlet cross sections, since the gas is pushed out with a significantly different volume, which in turn requires different velocities in order to keep the mass flow constant.

Gravitational work to lift the fluid

In the previous considerations, the inlet of the pump was at the same height as the outlet. However, this does not necessarily have to be the case! In general, the inlet and outlet are at different heights. In centrifugal pumps, the pump outlet is usually higher than the pump inlet. To overcome this height difference, the pump must do work on the fluid to lift it.

The water must now no longer simply be pushed through the pump against the acting pressure difference between inlet and outlet, but the fluid mass m flowing through must be lifted against gravity by the height Δz=z₂-z₁. This requires the gravitational work W_g:

\begin{align}
\label{2735}
\boxed{W_\text{g} = m~g~\Delta z} ~~~\text{where}~~~ \Delta z = z_2-z_1
\end{align}

Shaft work

The flow of the fluid through the pump will not be frictionless in reality due to the viscosity of the fluid. And also the friction in the bearings must be taken into account. As a result, the pump must also do work to compensate for the friction, which is commonly referred to as dissipation work W_diss.

All the above mentioned energetic processes, i.e. the sum of

• flow process work W_p (= flow work + volume-pressure work),
• acceleration work W_a,
• lifting work W_g and
• dissipation work W_diss (frictional work),

must be done by the pump. An open system need not always be a pump, which is why this type of work is commonly referred to as shaft work W_s. Shaft work is the total work that an open system must do on the fluid (for pumps) or that is done by the fluid (for turbines). Since this work is usually transferred by shafts, it is referred to as shaft work.

\begin{align}
\label{8918}
&W_\text{s} = W_\text{p} +W_\text{a} +W_\text{g} +W_\text{diss} \\[5px]
\label{e8761}
&\boxed{W_\text{s} = \int\limits_{p_1}^{p_2}V(p)~\text{d}p + \tfrac{1}{2}~m~\Delta c^2 + m~g~\Delta z + W_\text{diss}} ~~~~~\text{shaft work} \\[5px]
\end{align}

or as specific shaft work w_s (shaft work per unit mass):

\begin{align}
\label{9228}
\boxed{w_\text{s} = \int\limits_{p_1}^{p_2}v(p)~\text{d}p + \tfrac{1}{2}~\Delta c^2 + g~\Delta z + w_\text{diss}} ~~~~~\text{specific shaft work}
\end{align}

Finally, the shaft power P_s can be calculated from the specific shaft work w_s multiplied by the mass flow rate m’ which is transported through the open system (for the derivation of this formula see article Flow work in open systems):

\begin{align}
\label{4794}
\boxed{P_\text{s} = w_\text{s}~\dot{m}}
\end{align}

Note: So far, only pumps or compressors have been considered as open systems. These systems are transferring shaft work to the fluid, i.e. work is done on the fluid by the open system. However, such thermodynamic processes can also take place in reverse, i.e. the fluid does work on the open systems and transfers shaft work to the surroundings. This is the case, for example, in water turbines or steam turbines. These transfer mechanical work to the surroundings by the turbine shaft. The equations considered so far can in principle be applied to any open system.

Choice of system boundaries

Note that the flow velocities c heights z and pressures p given in the upper equations are defined at the boundaries of the open systems, i.e. when the fluid flows into or out of the open system (also called control volume).

Depending on the choice of the system boundary, this can result in completely different values for the respective quantities. It would also be possible to include the entire piping system and the water tank in the consideration as part of the open system. The system boundaries would then be at the surface of the water tank and at the end of the pipe. This approach would be based on completely different pressures, volumes, flow velocities and heights at the boundaries!

Energetic processes at the boundaries of an open system (flow work)

The article on flow work used the example of a water pump to explain that flow work is the work that must be done to push a fluid through an open system. This is due to the pressure difference between entering the system (e.g. pump inlet with low pressure) and leaving the system (e.g. pump outlet with high pressure). In pumps, the fluid is pushed in by the surroundings with lower energy W₁ and pushed out by the pump with higher energy W₂.

The energy with which the fluid ist pushed in or out result from the product of the fluid volume V, which is transferred across the system boundary, and the pressure p acting there. The difference between these energies is the work done by the system, e.g. the work of the pump, and is generally called flow work W_f:

\begin{align}
&W_\text{f} = W_2 – W_1 \\[5px]
&W_\text{f} = p_2 \cdot V_2 – p_1 \cdot V_1 \\[5px]
\end{align}

The knowledge gained from the example of the water pump can be applied to any open system, because in every open system, a certain fluid mass m is pushed into the system with volume V₁ and pressure p₁, and is pushed out of the system again with a changed volume V₂ and a changed pressure p₂.

Energetic processes inside the open system (pressure-volume work)

In an open system, the flow work is due solely to the fact that the fluid must be displaced against the pressure difference between system inlet and system outlet. Obviously, only the energetic processes at the system boundaries are relevant. However, this consideration does not take into account the internal processes inside the open system itself when the fluid flows through the system. So far, the open system has only been considered as a “black box”. Therefore, in the following we will take a closer look at the energetic processes inside an open system.

For an overall energy balance, not only the energetic processes at the system boundary must be considered (flow work), but also the processes in the system itself must be examined more closely. For example, a change in pressure occurring inside an open system generally also causes a change in volume of the fluid. Such an internal process of volume change has not been discussed so far. This volume change did not have to be taken into account in the example of the water pump, since water is an (almost) incompressible fluid. With such incompressible fluids, there is no volume change.

In the case of compressible fluids such as gases, however, a pressure-induced change in volume requires pressure-volume work, because a gas does not change its volume by itself, but only by performing work! Thus, in addition to the flow work which is performed at the boundaries of the open system, the pressure-volume work which is performed inside the open system must also be taken into account, especially in the case of gases.

Control volume

The processes taking place in an open system can be imagined more clearly if the entering fluid is regarded as a closed system. One thinks of a “bubble” around a considered fluid element with mass m. This closed system – i.e. the mass inside it – is now observed as it flows through the open system. In such an approach, the pump is then often no longer referred to as an open system, but as a so-called control volume.

In such an approach, the different energy transfers become clear once again. The flow work refers only to the pushing through of the considered fluid mass due to the acting pressure difference at the control volume boundaries. The pressure-volume work, on the other hand, refers to the work done inside the control volume, i.e. to the work required to compress the considered fluid mass as it flows through the control volume.

Both energy transfers – flow work and pressure-volume work – are to be provided by the open system (e.g. by the motor of a compressor). Since both the pressure-volume work W_v and the flow work W_f are due to the process of pressure change inside the open system, the sum of both is also referred to as the flow process work W_p:

\begin{align}
\label{5715}
&W_\text{p} = W_\text{V} + W_\text{S} \\[5px]
\label{4461}
&\boxed{W_\text{p} =~ – \int\limits_{V_1}^{V_2}p(V)~\text{d}V + p_2~V_2 – p_1~V_1 }
\end{align}

or as specific flow process work w_p (flow process work per unit mass):

\begin{align}
\label{8858}
&\boxed{w_\text{p} =~ – \int\limits_{v_1}^{v_2}p(v)~\text{d}v + p_2~v_2 – p_1~v_1 }
\end{align}

The fact that an open system also performs pressure-volume work in addition to flow work can be clearly demonstrated using a compressor. If the inlet and outlet of a compressor are closed for a short time, the compressor no longer has to perform flow work for this time, but the gas inside is still compressed. So there is still pressure-volume work to be done on the gas. The open system effectively becomes a closed system in which only pressure-volume work occurs.

Flow process work refers to the work of an open system to transfer a fluid through the system boundaries and change the volume.

In the simplest case, the flow process work corresponds to the total amount of work done in an open system. This flow process work must be provided by the driving motor in the case of a pump (for liquids) or in the case of a compressor (for gases). In this frictionless consideration, changes in kinetic and potential energy are neglected. This will be discussed in more detail in the section on shaft work.

Flow process work for incompressible fluids

Note that in an open system, the flow work and the pressure-volume work always occur together in general, namely as flow process work. After all, a pressure change is generally always associated with a volume change. Only in the theoretical case of a perfectly incompressible fluid would only the flow work occur in the open system. Because without a change in volume there is no pressure-volume work (dV=0). Therefore, only flow work must be done to push the incompressible fluid through the open system.

For such an incompressible fluid, the flow process work W_p when the volume V=V₁=V₂ flows through the open system results only from pressure change Δp generated by the open system:

\begin{align}
\label{2394}
&W_\text{p} =~ – \int\limits_{V_1}^{V_2}p(V)~\underbrace{\text{d}V}_{=0} + p_2~\underbrace{V_2}_{=V} – p_1~\underbrace{V_1}_{=V}=p_2~V-p_1~V = V~(p_2-p_1)= V \cdot \Delta p \\[5px]
\label{9679}
&\boxed{W_\text{p} = V\cdot\Delta p}~~~\text{applies only to incompressible fluids}
\end{align}

or as specific flow process work w_p (flow process work per unit mass):

\begin{align}
\label{3376}
&\boxed{w_\text{p} = v\cdot\Delta p}~~~\text{applies only to incompressible fluids}
\end{align}

Even if such an incompressible case practically does not exist, it can still be considered as such for liquids in a very good approximation, since these are considered to be almost incompressible. For all other cases, especially for compressible fluids such as gases, the flow process work must be determined according to equation (\ref{4461}) or equation (\ref{8858}). However, this equation can be represented in a much simpler form. This becomes clear if one visualizes the process of an open system in a volume-pressure diagram. This will be discussed in more detail in the next section.

Flow process work represented in a volume-pressure diagram

In the following, a much more simpler formula for calculating the flow process work will be given. For this purpose, a gas is considered which is initially sucked in by a compressor with the volume V₁ at a pressure p₁. The gas is then compressed in the system to a volume V₂. The process should run according to the curve shown in the p(V) diagram. After compression, the fluid volume V₂ is pushed out of the compressor at a final pressure p₂.

Equation (\ref{5715}) can be used to determine the flow process work. In this equation, the individual terms can be represented as areas in the p(V) diagram. The term of the pressure-volume work W_v = -∫p(V)⋅dV can be represented as the area under the curve (note that pressure-volume work is transferred to the gas during compression and this amount of work is thus mathematically positive)! According to equation (\ref{5715}) the term p₂⋅V₂ is added now, which can be represented as a rectangular area from the point (V₂|p₂) to the origin (0|0) (“push-out energy”). According to equation (ref{5715}), the rectangular area extending from the point (V₁|p₁) to the origin (0|0) is subtracted from this total area (“push-in energy”).

So, what is left is therefore only the area left of p(V)-curve up to pressure axis. This area left of the curve can therefore be interpreted as the flow process work!

Mathematically, this area can be determined as the integral W_p=-∫V(p)⋅dp within the boundaries between p₁ and p₂. This becomes clear if one simply imagines the coordinate system rotated by 90° and mirrored at the same time. The horizontal axis then corresponds to the pressure and the vertical axis to the volume. In this case, the result is a V(p) function whose area under the curve corresponds to the flow process work W_p:

\begin{align}
\label{8449}
&\boxed{W_\text{p} = \int\limits_{p_1}^{p_2}V(p)~dp}~~~\text{applies to an open system}
\end{align}

Since open systems are usually described by specific quantities, it makes sense to represent the thermodynamic process also in specific diagram form. Therefore, the specific volume is usually plotted instead of the total volume. This means that the volume shown refers to a mass of 1 kg.

\begin{align}
\label{1956}
&\boxed{w_\text{p} = \int\limits_{p_1}^{p_2}v(p)~dp}~~~\text{applies to an open system}
\end{align}

Note: These equations show that for the calculation of the (specific) flow process work, the function p(V) must be known. In contrast to the flow work, the flow process work is thus dependent on the path along which the process takes place and is therefore a process quantity!

Introduction

Many thermodynamic processes take place in open systems. In contrast to closed systems, in open systems there is not only a transfer of energy as heat or work but also a mass exchange with the surroundings. This is the case, for example, with pumps, compressors, aircraft engines or gas turbines. In these systems, mass enters the system from one side and leaves on the other.

To push this mass through an open system, energy is required. This will be illustrated in the following using the example of a water pump. First, we consider the case without a pump, where water flows from a tank through a horizontal pipe. In this case, the water exits the pipeline at a relatively slow speed.

Now a pump is installed in the pipeline. Similar to a ship’s propeller, the pump contains a rotating impeller (see also the article on centrifugal pumps). This impeller puts the water in the front part of the pump or pipeline under high pressure. Thus, the water flows faster through the pipeline and increases the mass flow, so that more water can be pumped in the same time than without a pump.

The water pump is a typical open system into which mass flows at the pump inlet at relatively low pressure and is pushed out again at higher pressure at the pump outlet. Under simplified, frictionless conditions, the question of what work the pump has to do in order to deliver a certain mass of water will now be discussed. For this purpose, the energetic processes at the inlet and outlet of the pump will be examined in more detail.

Energy with which the fluid is pushed in

Just before entering the pump, the water is under a certain pressure p₁. In this case, this pressure is caused on the one hand by the hydrostatic pressure of the water column in the tank and on the other hand by the ambient pressure of the air acting on the water surface. Both pressures push the water into the pump on the so-called suction side.

Note: Not the pump sucks the water by a negative pressure, as one might think by the term suction side, but the ambient pressure and the hydrostatic pressure push the water into the pump! Because even if the pump would create a vacuum at the inlet, no water could be pushed into the pump without ambient pressure and hydrostatic pressure (see also the article How does a drinking straw work?).

Based on the acting pressure p₁ at the inlet and the cross-sectional area A₁, the force F₁ with which the water is pushed into the pump by the surroundings can be determined:

\begin{align}
\label{7481}
&F_1 = p_1 \cdot A_1
\end{align}

With this force F₁, a certain amount of water located directly in front of the inlet is pushed into the pump within a certain time. One can also imagine a piston that pushes the water into the pump with the force F₁ along the distance s₁. From the product of force and distance, the energy W₁ can be determined, with which the surroundings (“piston”) pushes the water mass into the pump. This energy corresponds to the work done by the surroundings on the water mass:

\begin{align}
&W_1 = F_1 \cdot s_1 = p_1 \cdot \underbrace{A_1 \cdot s_1}_{=V_1} = p_1~V_1 \\[5px]
\label{5792}
&\boxed{W_1 = p_1 \cdot V_1} ~~~~~\text{pushed-in energy}
\end{align}

The product of displacement s₁ and cross-sectional area A₁ in the above equation equals the water volume V₁ that is pushed into the system.

The energy with which a fluid is pushed through a cross-section results from the product of pressure and volume!

Numerical example

If the ambient pressure is assumed to be 1 bar and the height of the water column is assumed to be 1 m, a (hydrostatic) pressure of 0.1 bar acts in addition to the ambient pressure of 1 bar. In total, the water is thus pushed into the pump with a total pressure of p₁=1.1 bar. Within a time of t = 4 s, m = 2 kg of water flows through the pump inlet. With a water density of 999.25 kg/m³, the volume of this water mass is V₁ =2.0015 liters. This volume of water is pushed into the pump by the surroundings with an energy of W₁ = 220 J:

\begin{align}
&\underline{W_1} = p_1 \cdot V_1 = 1.1 \cdot 10^5\frac{\text{N}}{\text{m²}} \cdot 2.0015 \cdot 10^{-3} \text{ m³} = \underline{220 \text{ J}}
\end{align}

Energy with which the fluid is pushed out

After the energetic process at the pump inlet has been clarified, the energetic process at the pump outlet, the so-called pressure side, will now be considered in more detail. Basically, the same mass of water is pushed out at the outlet as flows in at the inlet within a certain time, because in the steady state no water can accumulate in the pump nor be annihilated (law of conservation of mass). The conservation of mass applies not only to incompressible fluids such as liquids, but also to compressible substances such as gases!

In the previous section, it was calculated on the basis of the numerical example that the surroundings push the water with an energy of 220 J into the pump. If the pump were switched off, the water would be pushed through the pump with this energy and would only leave the pump again with this energy (flow and friction losses neglected). However, the pump now causes an increase in pressure by the rotating impeller. The water is therefore pushed out of the pump at a higher pressure than it was pushed into the pump.

The water therefore exits the pump with a greater force, which results in an increase in the energy with which the water flows out. This energy at the pump outlet can be derived analogously to equation (\ref{5792}). The water is pushed out with the force F₂=p₂⋅A₂ along the distance s₂ through the cross-section A₂ at the pump outlet. This leads to the following energy W₂ with which the water ist pushed out:

\begin{align}
&W_2 = F_2 \cdot s_2 = p_2 \cdot \underbrace{A_2 \cdot s_2}_{=V_2} = p_2 \cdot V_2 \\[5px]
\label{3018}
&\boxed{W_2 = p_2 \cdot V_2} ~~~~~\text{pushed-out energy}
\end{align}

Numerical example

If, for example, the pump generates a pressure increase of Δp = 20 bar, the water pressure increases from p₁ = 1.1 bar at the pump inlet to a total of p₂ = 21.1 bar at the pump outlet. The increased pressure in turn affects the displaced volume. Due to the conservation of mass, the masses pushed in and out within a certain time are identical, but the respective volumes differ due to the different pressures. The water mass of 2 kg no longer has a volume of V₁ = 2.0015 liters at the pump outlet, but a slightly lower volume of V₂ = 1.9996 liters due to the greater pressure.

At a pressure of p₂=21.1 bar and a volume of V₂ =1.9996 liters, the energy with which the water mass of 2 kilograms is pushed out is W₂ = 4220 J:

\begin{align}
&\underline{W_2} = p_2 \cdot V_2 = 21.1 \cdot 10^5 \frac{\text{N}}{\text{m²}} \cdot 1.9996 \cdot 10^{-3} \text{ m³} = \underline{4220 \text{ J}}
\end{align}

If one compares the energy of W₁ = 220 J, with which the water ist pushed in, with the energy of W₂ = 4220 J, with which the water is pushed out, it becomes clear that the pump obviously ensures that the entering water mass is pushed out with greater energy than it was pushed in by the surroundings. Thus, the water flows out at the end of the pipe with greater power compared to the case without pump.

Note: Due to the incompressibility of liquids, the small volume changes caused between the inlet and outlet of an open system can often be neglected. In the case of compressible liquids such as gases, however, the volume changes can be very large and thus significant, since further energy is then required to compress the volume (pressure-volume work must be done on the gas). This will be discussed in more detail in the article Flow process work in open systems.

Energy to push through (flow energy)

The 4000 J higher energy with which the water flows out of the pump compared to the energy at the pump input can of course only be due to the pump itself. The energy of 4000 J corresponds to the work that the pump must perform on the flowing liquid. It is supplied to the pump in the form of electrical energy and transferred to the water by the impeller. This consideration is highly simplified and will be discussed in more detail in the article Flow process work in open systems and Shaft work in open systems.

However, this simplified approach shows that the work of the pump is calculated from the difference between the push-out energy W₂ and the push-in energy W₁. In an open system, this work is generally referred to as flow work (W_f):

\begin{align}
\label{9372}
&W_\text{f} = W_2 – W_1 \\[5px]
\label{7076}
&\boxed{W_\text{f} = p_2~V_2 – p_1~V_1 } = \Delta \left(p~V \right) ~~~~~\text{flow work}
\end{align}

The work which is required to maintain the flow against the different static pressures between inlet and outlet of an open system is referred to as flow work!

Equation (\ref{7076}) shows that the flow work of an open system is determined only by the state of a flowing mass just before it enters the system (p₁, V₁) and just after it leaves the system (p₂, V₂). It does not matter how exactly the flowing mass gets from the initial state 1 to the final state 2. The thermodynamic process inside the open system is therefore completely irrelevant for the determination of the flow work. Thus, the flow work is not a process quantity, as one might mistakenly think on the basis of the term work, but a state variable!

Note: Sometimes, only the individual products of pressure and volume are referred to as flow work. This is because the product p₁⋅V₁ is due to the pushing into the open system and the product p₂⋅V₂ is due to the pushing out of the open system. For didactic reasons, however, in the following the term flow work shall always mean the difference of both products according to equation (\ref{7076}), i.e. the work required to push the mass through the open system (not only in or out).

Specific flow work

The example of the pump explained in the previous sections showed that the pump has to perform a flow work of 4000 J for pumping a water mass of 2 kg. If the pump is to deliver 10 times the water mass of 20 kg, then the pump must perform a 10-fold work of 40,000 J in total.

This example shows that the flow work depends on the mass flowing through an open system. For this reason, it makes more sense in open systems not to relate the amount of energy converted to an arbitrary mass of 2 kg or 20 kg, but always to 1 kg. Thus, the water pump must deliver an energy of 2000 J per kilogram of water flowing through it. Such a quantity related to the mass is also referred to as a specific quantity. In this case, it is called specific flow work and is 2000 J/kg. 

The specific flow work w_f is generally determined as the quotient of the flow work W_f and the associated mass m:

\begin{align}
\label{3475}
&\boxed{w_\text{f} = {W_\text{f} \over m}} ~~~[w_\text{f}] = \frac{\text{J}}{\text{kg}} ~~~~~\text{specific flow work} \\[5px]
\end{align}

With the definition of the flow work as the difference of the products of pressure p and volume V, it can be seen that the specific flow work can also be calculated by the specific volumes v:

\begin{align}
&w_\text{f} =\frac{W_\text{f}}{m} = \frac{p_2 \cdot V_2 – p_1 \cdot V_1}{m} = {p_2 \cdot \underbrace{\frac{V_2}{m}}_{=v_2} – p_1 \cdot \underbrace{\frac{V_1}{m}}_{=v_1}} = p_1 \cdot v_1 – p_2 \cdot v_2 \\[10px]
\label{2070}
&\boxed{w_\text{f} = p_1 \cdot v_1 – p_2 \cdot v_2} ~~~ \text{where}~~~ \boxed{v={V \over m}} ~~~ [v]=\frac{\text{m³} }{\text{kg}} ~~~\text{specific volume} \\[10px]
\end{align}

The specific volume indicates the volume occupied by a substance with a mass of 1 kg. The specific volume v thus corresponds to the reciprocal of the mass density ρ (specific mass):

\begin{align}
\label{9579}
&\boxed{v = {1 \over \rho}} \\[5px]
\end{align}

Thus, the specific flow work can also be determined by the density of the fluid at the inlet and outlet of an open system:

\begin{align}
\label{2791}
 &\boxed{w_\text{f} = \frac{p_2}{\rho_2} – \frac{p_1}{\rho_1} }~~~\text{where}~~~ \boxed{\rho= {m \over V}} ~~~ [\rho]={\text{kg} \over \text{m³}} ~~~ \text{density} \\[5px]
\end{align}

The advantage of determining the specific flow work according to equation (\ref{2791}) or (\ref{2070}) is that these equations can be applied independently of the actual mass flowing through the open system. It is completely irrelevant whether 2 kg or 20 kg of water flow through the pump.

Numerical example

In the present example, the water density at the pump inlet is ρ₁ = 999.25 kg/m³ (at p₁ = 1.1 bar) and at the pump outlet ρ₂ = 1000.19 kg/m³ (at p₂ = 21.1 bar). From these quantities, the specific flow work can be determined without having to know in advance how much mass will actually flow through the pump:

\begin{align}
& \underline{w_\text{f} } = \frac{p_2}{\rho_2} – \frac{p_1}{\rho_1} = \frac{12.1 \cdot 10^5 \frac{\text{N}}{\text{m²}} }{1000.19 \frac{\text{kg}}{\text{m³}}} – \frac{1.1 \cdot 10^5 \frac{\text{N}}{\text{m²}} }{999.25 \frac{\text{kg}}{\text{m³}}} = \underline{2000 \frac{\text{J}}{\text{kg}}} \\[5px]
\end{align}

Flow power (power of the pump)

Especially in open systems it is relatively easy to calculate the transferred power by using specific quantities. Using the example of the water pump already considered, we will assume that it delivers a water mass of m = 2 kg within a time of t = 4 s at. Thus, 0.5 kg of water flows through the pump within one second. This transferred mass per unit time is also referred to as mass flow rate. The mass flow rate is often denoted by a dot above the symbol (\(\dot{m}\)) and is generally calculated as the quotient of the mass m delivered and the respective time t:

\begin{align}
\label{7914}
\boxed{\dot{m} = \frac{m}{t}} ~~~[\dot{m}]=\frac{\text{kg}}{\text{s}} ~~~~~\text{mass flow rate} \\[5px]
\end{align}

With this mass flow rate, the power of the pump (flow power) can be determined relatively easily. According to the example, the pump is to deliver a water mass of m = 2 kg within t = 4 s. With a specific flow work of the pump of w_f = 2000 J/kg, the pump then performs a total flow work of W_f = 4000 J on the 2 kg water mass; and this within a time period of t = 4 s. The quotient of transferred work W_f and time t finally results in a flow power of the pump of P_f = 1000 J/s (1000 W). In the ideal case, this would correspond to the electrical power supplied to the pump (neglecting flow or friction losses and assuming incompressibility of the water). Such a (flow) power P_f can be determined quite generally by the mass flow rate and the specific (flow) work:

\begin{align}
&P_\text{f}=\dot{W_\text{f}}= \frac{W_\text{f}}{t} = \frac{w_\text{f} \cdot m} {t} = w_\text{f}~\underbrace{\frac{m}{t}}_{=\dot{m}} \\[5px]
\label{6526}
&\boxed{P_\text{f} = w_\text{f} \cdot \dot m}
\end{align}

Equation (\ref{6526}) is not limited to the flow power, but can be generalized:

\begin{align}
\label{3679}
&\boxed{\text{Power = specific energy x mass flow rate}}
\end{align}

Outlook

As already indicated several times in this article, for incompressible fluids the flow work (or flow power) is the main work to be done by/on an open system. However, if compressible fluids such as gases are involved, then a further energy transfer takes place in an open system, since the volume must also be compressed in addition to just pushing the volume through (pressure-volume work). This is discussed in more detail in the article Flow process work in open systems.

Introduction

The free adiabatic expansion of an ideal gas in a vacuum showed that this thermodynamic process cannot be described by the isentropic equations, although these equations were derived for an adiabatic system. The adiabatic expansion against a vacuum is an isothermal process in which the temperature remains constant. Therefore, the question arises under which condition a thermodynamic process in an adiabatic system can actually be described with the isentropic equations.

Reversibility of thermodynamic processes

For this purpose, we will take a closer look at different thermodynamic processes. In doing so, we will see that, from an energetic point of view, all these processes can run in one direction as well as, in principle, in the other direction. For example, in the article Concept of pressure-volume work, we looked at a thermodynamic process in which a gas expands in a cylinder by adding heat and finally lifts a weight via a rack and pinion. From an energetic point of view, this process can be reversed in principle.

To do so, one can imagine the original process as being recorded with a video camera, which is then played backwards. In such a reverse case, the weight lowers by removal of heat (cooling). Here, work is no longer done by the gas on the weight, but conversely work is done on the gas by the weight – the gas is compressed by the weight, so to speak. At the same time, cooling lowers the temperature and with it the internal energy. The energy flows are therefore reversed in such a reversal process, without any contradiction arising from experience.

Isochoric process

Such a reversal process can also be found in an isochoric heating of a gas cylinder. The heat supplied causes an increase in internal energy, i.e. the temperature rises. If this process is considered energetically reversed, one would obtain an isochoric cooling. In such a reversal case, the internal energy would decrease due to the dissipated heat and the temperature would decrease.

Isobaric process

The process explained in the article on the isobaric process, in which a weight is lifted when heat is supplied, can also be reversed in terms of energy. In the original case, the gas performs work on the weight (lifting) by adding heat, while the internal energy increases and the temperature rises. In the reverse case, an isobaric cooling would be obtained, in which the gas is compressed by the weight (lowering), while at the same time heat is dissipated. The internal energy and thus the temperature are reduced.

Isothermal process

The compression of air in an air pump explained in the article on the isothermal process also shows such an energetic reversibility. If the gas in an air pump is compressed very slowly with the valve closed, the temperature (nearly) remains constant due to the heat transfer to the surroundings. Conversely, the gas can also expand isothermally from this compressed state. In this case, the expansion must again be very slow so that the surroundings can transfer enough heat to the expanding gas to prevent a drop in temperature.

Isentropic process

Also, the rapid compression of air in an air pump presented in the article on the isentropic process can be reversed from an energetic point of view. In the original process, a gas is compressed rapidly (work is done on the gas) so that heat cannot be transferred in such a short time (adiabatic system), which leads to an increase in internal energy and causes the temperature to rise. In the reverse case, the gas expands very rapidly starting from the compressed state. In this isentropic reversal process, work is then done by the gas at the expense of the internal energy and the temperature consequently drops.

Reversibility

The processes considered in the previous sections make it clear that they are all reversible from an energetic point of view. The reversibility of thermodynamic processes presupposes, among other things, that there is no friction or, more generally, no dissipation of energy (e.g. when a piston slides in a cylinder).

In thermodynamics, reversible processes are processes which are reversible from an energetic point of view without violating everyday experience!

Furthermore, the gas states must always be in equilibrium, so that there is no temperature difference within the gas itself. Both the assumption that no energy is dissipated by friction and the assumption of the always existing equilibrium exist, of course, only in the ideal conception. Reversible processes therefore do not exist in reality. However, these ideal conditions have been assumed for all thermodynamic processes considered so far.

Adiabatic expansion against a vacuum as a non-reversible process

As far as reversibility is concerned, however, the situation is different for the free adiabatic expansion of an ideal against a vacuum. This thermodynamic process is not reversible from everyday experience! In adiabatic expansion against a vacuum, the gas expands at constant temperature without transfer of work or heat. For the reverse case, this would have to mean that the gas compresses itself without help.

However, this contradicts all experience, because in order to compress a gas in a cylinder, work must be done on the gas – a gas will not do this by itself. The temperature would then also no longer remain constant, but it would increase due to the work transferred to the gas. Consequently, an adiabatic expansion against a vacuum is a non-reversible process from an energetic point of view. Such non-reversible processes are also called irreversible processes.

Basically all equations presented in the respective articles – in particular those of the isochoric, isobaric, isothermal and isentropic process – are valid only for reversible processes. This now explains why the equations of the isentropic process cannot be applied to the adiabatic expansion against a vacuum: the process is irreversible! The equations of the isentropic process may only be used if it is a reversible process of an adiabatic system. Thus, an isentropic process be defined as follows:

An isentropic process is a reversible thermodynamic process of an adiabatic system!

The reversibility of thermodynamic processes is described by the quantity entropy. For a reversible process of an adiabatic system, the entropy remains constant. Therefore, such a process is called an isentropic process (process at constant entropy).

In the article Free expansion of an ideal gas in a vacuum it was explained that the free expansion of a gas against a vacuum taking place in an (ideal) adiabatic system is an isothermal process. For this purpose, a gas-filled cylinder was considered, which is closed with a massless and frictionless sliding piston.

The temperature of the ideal gas remains constant during such an adiabatic expansion. However, this applies only to ideal gases. If, on the other hand, such an experiment is carried out with a real gas, then a slight decrease in temperature would be observed in reality (even though the piston can still be considered to be massless). This phenomenon is due to the attractive forces acting between the gas molecules, which are neglected in the case of ideal gases.

Because when a real gas expands, the distances between the molecules increase due to the increasing volume. The gas must therefore increase its volume against the attractive forces acting between the molecules. This process obviously requires energy. This energy must be delivered by the gas itself and can only be provided at the expense of the kinetic energy of the gas molecules. This means a decrease in temperature, because the temperature is directly linked to the kinetic energy. This phenomenon does not only occur during the expansion against a vacuum, but ultimately whenever a real gas expands against a lower pressure. This effect of temperature decrease is also called Joule-Thomson effect.

The Joule-Thomson effect is the phenomenon of a decrease in temperature of real gases when they expand against a lower pressure!

Note that in many cases the gas temperature decreases during an expansion. This generally has nothing to do with the Joule-Thomson effect. The Joule-Thomson effect only describes the stronger decrease in temperature for real gases than would be the case for ideal gases, since binding energies must be taken into account for real gases. And as already explained, these binding energies cause an additional amount of work during an expansion, which is at the expense of the kinetic energies, which additionally decreases the temperature. Only this additional temperature decrease is called the Joule-Thomson effect.

Note also that even with an adiabatic (!) expansion of a real gas against a vacuum, the internal energy of the gas remains constant. In a real gas, the internal energy is not only composed of the kinetic energy of the molecules as in an ideal gas, but also binding energies are part of the internal energy. In real gases, there is only an “internal” redistribution of energies, namely at the expense of the kinetic energy and in favor of the binding energies. The work done by the gas to increase the molecular distances is not transferred across the system boundary, but only redistributed internally!

The first law of thermodynamics only balances energies that are tranferred accross the system boundary! In an adiabatic expansion against a vacuum, neither such boundary work nor heat is transferred (W=0, Q=0)! Thus, even for real gases, there is no change in the internal energy ΔU:

\begin{align}
\underbrace{W}_{=0} + \underbrace{Q}_{=0} = \Delta U = 0
\end{align}

In the article on the isentropic process it was already indicated that the often used term “adiabatic process” can be misleading in some situations and is generally not correct. To clarify this, the expansion of an ideal gas against a vacuum is considered in the following.

For this purpose, imagine a gas in a cylinder which is closed by a piston with negligible mass. The piston slides frictionless inside the cylinder. The piston is initially held in position by force. The cylinder is located in an evacuated chamber so that there is a vacuum on the outside of the piston. Cylinder and piston are assumed to be perfectly thermally insulated. It is therefore an adiabatic system. Now the piston is released and the gas expands against the vacuum.

However, just because the cylinder and the piston form an adiabatic system, this does not mean that the derived laws of the supposed “adiabatic process” can be applied! According to these laws, the temperature should decrease during an expansion with increasing volume (T⋅V^κ-1=constant). However, this is not observed in the case considered, where the gas expands against the vacuum! The surprising result of this process shows up, if one takes a look at the first law of thermodynamics:

\begin{align}
\underbrace{W}_{=0} + \underbrace{Q}_{=0} = \Delta U = 0
\end{align}

Since it is an adiabatic system, by definition no heat is transferred (Q=0). Furthermore, the gas does not perform any work (W=0), because the gas expands against a vacuum. The gas does not have to exert any force to move the (massless!) piston. There is, so to speak, no counterpressure against which the gas has to push. Thus, the gas does not have to perform any work for the expansion of its volume. However, if no energy is transferred, neither as work nor as heat, the internal energy consequently remains constant (ΔU=0). Since the internal energy of an ideal gas is directly linked to the temperature (and this does not change), the free expansion of an ideal gas against a vacuum is therefore an isothermal process!

When an ideal gas expands in an adiabatic system against a vacuum, the temperature remains constant (isothermal process)!

Note: The fact that the temperature remains constant during expansion against the vacuum applies only to ideal gases. For real gases, on the other hand, a slight decrease in temperature will be observed. This phenomenon is referred to Joule-Thomson effect.

One can explain the isothermal expansion against a vaccum quite clearly. Imagine you are in a vacuum and you hold wildly flying particles in your hand. If you open your hand, the particles fly into the vacuum and are not slowed down by any other particles (since the piston is considered massless in the above example, you can also remove the piston and let the particles fly freely into the vacuum – this is the same situation). So the gas molecules keep their kinetic energy. Since the kinetic energy of the molecules is directly linked to the temperature, this means a constant temperature.

The adiabatic expansion of an ideal gas into a vacuum represents an isothermal process!

This example makes clear that thermodynamic processes in adiabatic systems cannot necessarily be described with the laws of isentropic processes. Therefore, in connection with the isentropic equations derived for this purpose, one should not speak of “adiabatic” processes. Because this suggests falsely that for all adiabatic systems these equations could be used. However, the expansion against a vacuum disproves exactly this.

Of course, the question arises under which condition a thermodynamic process of an adiabatic system can actually be described by the derived isentropic equations. The answer to this question can be found in the article Reversibility of thermodynamic processes (entropy).

Work performed in adiabatic systems

Many thermodynamic processes take place within very short times, such as the expansion of the burned fuel-air mixture in internal combustion engines during the power stroke (approx. 10 ms) or the expansion of hot gases in turbines. Within such short times, a significant heat transfer between gas and surroundings can be neglected in a first approximation (Q=0). The thermodynamic processes are then considered to take place in an adiabatic system. According to the first law of thermodynamics, the work done W by/on the gas is given directly by the change in internal energy ΔU of the gas:

\begin{align}
&\boxed{W+Q=\Delta U} &&~\text{first law of thermodynamics} \\[5px]
&W = \Delta U &&~\text{only applies to adiabatic systems (}Q=0 \text{)}\\[5px]
\end{align}

Since for ideal gases the change in internal energy ΔU is independent of the thermodynamic process and only results from the temperature change, …

\begin{align}
&\Delta U = c_\text{v} \cdot m \cdot \Delta T ~~~~~\text{where}~~~~~\Delta T = T_2-T_1\\[5px]
\end{align}

… the work done by/on the adiabatic system can be determined as follows:

\begin{align}
\label{9067}
&\boxed{W = c_\text{v}~m~\Delta T} ~~~\text{only applies to adiabatic systems (}Q=0 \text{)} \\[5px]
\end{align}

Note that the work denoted by W does not necessarily correspond to the pressure-volume work W_v of the gas. This would only be the case for a non-dissipative process where no friction occurs. Therefore, the work denoted by W describes quite generally the work transferred across the system boundary (boundary work) and thus also includes dissipative processes! More information about this can be found in the article Dissipation of energy in closed systems.

Equation (\ref{9067}) clearly shows that during the expansion of a gas in an adiabatic system, the greater the difference between the initial and final temperature, the more work is transferred from the system to the surroundings. For example, during the power stroke inside a diesel engine, the temperature in the cylinder drops from about 2000 °C to about 500 °C during expansion. However, at a given initial temperature T₁, dissipative processes with friction always lead to higher final temperatures T₂ compared to frictionless processes due to dissipation of energy that benefits the internal energy. The temperature obviously does not decrease as much during the expansion. Therefore, according to equation (\ref{9067}), the work done by the system ist lower!

Dissipative processes in the volume-pressure diagram

The volume-pressure diagram below shows an expansion of a gas in a cylinder which has been closed with a frictionless moving piston (blue curve). In comparison, an expansion is shown in which a constant frictional force between cylinder and piston was assumed over the duration of the expansion (red curve). It was assumed that the entire friction energy was dissipated in internal energy.

Due to the larger temperature difference, the greatest possible work is done by the gas during non-dissipative expansion → see equation (\ref{9067}). For such reversible processes, the laws of the isentropic process (n=κ) apply in adiabatic systems. Dissipative processes, on the other hand, run at a higher pressure level due to the higher temperatures. Such processes can then be approximated by a polytropic process, where in the case of an expansion the polytropic index n has a lower value than the isentropic index κ (n<κ):

\begin{align}
\label{3475}
&\boxed{p_1~V_1^n=p_2~V_2^n} \\[5px]
\label{2070}
&\boxed{T_1~V_1^{n-1}=T_2~V_2^{n-1}} \\[5px]
\label{9579}
&\boxed{T_1^n~p_1^{1-n}=T_2^n~p_2^{1-n}} \\[5px]
\text{ where: }~~ &n=\kappa ~~~\text{for non-dissipative (isentropic) processes} \nonumber \\
&n < \kappa ~~~\text{for dissipative (polytropic) expansion processes} \nonumber \\
&n > \kappa ~~~\text{for dissipative (polytropic) compression processes} \nonumber \\
\end{align}

With the use of these polytropic equations, the work W done by the system at a given initial temperature T₁ can also be determined by the pressure or volume ratio:

\begin{align}
\label{1}
&W=c_\text{v}~m~\left[T_2-T_1 \right] = c_\text{v}~m~T_1~\left[{T_2 \over T_1}-1 \right] \\[5px]
&\boxed{W=c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{n-1}}-1 \right]}~~~\text{only applies to adiabatic systems} \\[5px]
\label{3}
&\boxed{W=c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-n} \over n}}-1 \right]}~~~\text{only applies to adiabatic systems} \\[5px]
\text{ where: }~~ &n=\kappa ~~~\text{for non-dissipative (isentropic) processes} \nonumber \\
&n < \kappa ~~~\text{for dissipative (polytropic) processes} \nonumber \\
\end{align}

At this point, it should be explicitly mentioned that one should refrain from using the term “adiabatic” process. This is because both the isentropic process (n=κ) and the polytropic process (n<κ) take place in an adiabatic system. In this respect, all these processes are “adiabatic”, but they differ, of course, due to the different polytropic index n used to describe the respective processes (see also the article Free expansion of an ideal gas in a vacuum)!

Calculation of the polytropic index for dissipative processes

Which polytropic index n best describes a given dissipative process can be determined on the basis of the initial and final states. In practice, of course, this requires an experimental determination of the respective values. Then either equation (\ref{3475}), equation (\ref{2070}), or equation (\ref{9579}) must be solved for the polytropic index n:

\begin{align}
\label{6526}
\boxed{n
=\frac{\ln{\left(\frac{p_2}{p_1}\right)}} {\ln{\left(\frac{V_1}{V_2}\right)}}
={\frac{\ln{\left(\frac{T_2}{T_1}\right)}} {\ln{\left(\frac{V_1}{V_2}\right)}} +1}
=\frac{\ln{\left(\frac{p_2}{p_1}\right)}} {\ln{\left(\frac{T_1}{T_2}\right)} + \ln{\left(\frac{p_2}{p_1}\right)} }
} \\[5px]
\end{align}

Based on the given initial and final state, a polytropic index of n=1.1689 seems to best reflect the dissipative process shown in the diagram (green curve). The comparison in the diagram below between the approximated polytropic process and the actual process (red curve) shows very good agreement at the beginning and towards the end of the process; however, there are minor deviations in the area in between.

Calculation of dissipated energy

In the article Dissipation of energy in closed systems it has already been explained in detail that the work W transferred across the system boundary (boundary work) is the sum of the pressure-volume work W_v of the gas and the dissipated energy W_diss due to friction:

\begin{align}
\label{ums}
&\boxed{W = W_\text{v} + W_\text{diss}} ~~~\text{work transferred across the system boundary}
\end{align}

In the article Polytropic process in a closed system, it was shown that the pressure-volume work W_v for a polytropic process can be determined using the following formula:

\begin{align}
\label{wv}
&\boxed{W_\text{v} = \left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~\left(T_2-T_1 \right)}
\end{align}

Solving equation (\ref{ums}) for the dissipated W_diss energy and putting in equation (\ref{9067}) and equation (\ref{wv}), gives the dissipated energy:

\begin{alignat}{2}
\label{3440}
&W_\text{diss}= W – W_\text{v} = \underbrace{c_\text{v}~m~(T_2-T_1)}_{W} – \underbrace{\left[\frac{\kappa-1}{n-1}\right]~c_\text{v}~m~(T_2-T_1)}_{W_\text{v}} \\[5px]
\label{5715}
&\boxed{W_\text{diss}= \left[\frac{n-\kappa}{n-1}\right]~c_\text{v}~m~(T_2-T_1)} = \left[\frac{n-\kappa}{n-1}\right]~c_\text{v}~m~T_1~\left[{T_2 \over T_1}-1 \right] \\[5px]
\label{4461}
&\boxed{W_\text{diss}=\left[\frac{n-\kappa}{n-1}\right]~c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{n-1}}-1 \right]} \\[5px]
\label{2394}
&\boxed{W_\text{diss}=\left[\frac{n-\kappa}{n-1}\right]~c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-n} \over n}}-1 \right]} \\[5px]
\end{alignat}

A closer look at these equations shows that they are the same equations that describe the transferred heat in a (non-dissipative) polytropic process (see article Polytropic process in a closed system). However, for the dissipative process at hand, they can no longer be interpreted as transferred heat, since the process takes place in an adiabatic system. However, the identical equations can still be explained from logic, because from an energetic point of view the dissipated energy has the same effect as a heat transfer, namely an increase of the internal energy!

Note

Note that the work W done by/on a closed system according to equations (\ref{1}) to (\ref{3}) depends only on the initial and final state. These equations therefore reflect the actual boundary work, even if the polytropic equations do not accurately describe the actual process (see diagram below). This is due to the fact that the initial and final values agree with reality, provided that the polytropic index was selected according to equation (\ref{6526})!

However, the situation is different for the dissipated work W_diss. For its determination, according to equation (\ref{3440}), the pressure-volume work W_v is relevant, which results as the area under curve in the volume-pressure diagram. However, as can be seen in the diagram, there are deviations between the approximated polytropic process and the actual process. Therefore, equations (\ref{5715}) to (\ref{2394}) will generally not accurately account for the dissipated energy. In the present case, the area under the polytropic curve is larger than for the actual process and thus suggests a somewhat larger dissipated energy than will be the case in reality.