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1 Concepts and Notations Thermodynamics belongs to the theoretical foundations of engineering. One of its tasks is to describe the dependence between the properties of substances (temperature, pressure, density, energy content etc.) and to develop the structure of such dependences. In addition, thermodynamics deals with the transformation of energy and of substance (separation of substances from mixtures and chemical reactions), and with the direction and velocity of the corresponding processes. It therefore provides the basis for the application of such processes in technical devices such as steam engines, gas turbines, internal combustion engines, air-conditioners, and many more. In our effort to understand the structure of thermodynamics we have to introduce a clear nomenclature. We therefore start with a few concepts and notations. Other concepts, notations and symbols will follow in the treatment of the subjects. In some cases, terms of everyday speech will be used in a more restricted or modified meaning. In these cases, we will stick to their specific physical meaning; this is especially true if the everyday meaning can lead to misconceptions (like it is the case for terms like “heat” or “heat capacity”). Fortunately, many terms originate from Latin or ancient Greek, so their meaning can hardly be mistaken.

1.1 System, Property of State, State 1.1.1 System We shall call the object of our consideration a “system“. The quantity of substance in the system may vary by an exchange (addition or removal of substance) across the border of the system.

Figure 1.1: Examples of a system 1.1.2 Property of State Properties of state describe properties of a system. Some properties of state are will already be known to the student, others will be introduced and defined in this lecture. Pressure is the ratio of a force that acts perpendicularly onto an area, and the size of this area:

p = lim

A!0

F"A

The physical unit of pressure is 1 Pa = 1 N/m2 (Pa: Pascal). 105 Pa = 1 bar. Also, “at” (atmosphere) was used as a unit for pressure (now outdated, 1 at ! 1.013 bar). Volume is a measure of the amount of space occupied by a system and has the unit m3.

a.

b

air in a cylinder

border of the system

air in a bottle and inflowing ambient air

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Mass m has unit kg. The definition of temperature is somewhat more complex. In everyday life or in technical applications, temperature is mostly used as read from a thermometer. Thermometers are based on temperature-dependent properties, e.g., the length of a rod, the volume of a liquid, the electrical resistance of a wire, and so on. In thermodynamics, a more precise definition of temperature is introduced (see below). Properties of state can be categorized into intensive and extensive properties: Intensive properties of state are equalized when systems are united: temperature: uniting means removing of barriers to heat transfer. pressure: uniting means removing of mechanical barriers. Extensive properties are those which are added together – combined – when systems are united: mass, energy, volume, entropy. Specific properties are extensive properties per unit mass (in recent literature they are

sometimes called “massic” properties). We use “mass” instead of “quantity of substance” because all velocities in our considerations are very small compared to the velocity of light. Other important quantities of state are the specific volume v and the density ρ. There holds

v = lim

m→0

Vm

specific volume [v]=m3/kg

ρ = lim

V→0

mV

density [ρ]=kg/m3

v = 1/ρ 1.1.3 State Collective of the values of the properties of state at a given moment e. g.: air at ambient conditions: . Sometimes it is required to specify more quantities of state in order to characterize the state (e.g., fractional amount of mass of the different substances [like nitrogen, oxygen, etc] in air). 1.1.4 Change of State Change of state of one system from an initial state “1“ to a final state “2“. Differences of the properties are e.g.:

Δp = p2 − p1 ;ΔT = T2 −T1;ΔV = V2 −V1,... , Example: Compression of air in a pump with a closed outlet. 1.1.5 Process Change of state of a system combined with all interactions with neighboring systems and the changes in those systems. Interactions: transfer of heat, of work, of substance.

Or sequence of changes of state of one substance (e. g. gas turbine process; or cyclic sequence as cyclic process).

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1.2 Limitations to Thermodynamic Considerations 1.2.1 Macroscopic Systems Technical thermodynamics is phenomenological thermodynamics. Phenomena in our sense are macroscopic. Technical thermodynamics, therefore, considers macroscopic systems. Microscopic considerations are used merely for additional illustration.

microscopic: concerning a single molecule or a few molecules

macroscopic: concerning many molecules (“many” in the sense of statistics) Example: Five molecules of a gas are contained in one half of the volume of a container. The other half of the container is empty (vacuum). The wall which separates the two halves is removed. The five molecules “expand” into the whole volume. The probability that the five molecules all are back again in the first half at a given moment is 2-5=1/32. For 100 molecules, this probability would be 2-100≈10-30, i.e., practically zero. (For comparison: The age of the universe is about 5·1017 seconds.) Consequently, the expansion of the 100 molecules into the vacuum is irreversible. This corresponds to our experience with macroscopic systems. The system of five molecules, however, would be microscopic – its expansion could be reversed. 1.2.2 System of coordinates fixed to the substance The thermodynamic state is considered in a system of coordinates which is fixed to the substance of the system. This system of coordinates is moved with the substance. Properties of state therefore are e.g. pressure and temperature measured with sensors which are moved with the substance, density and energy content. The velocity of the center of gravity, the corresponding kinetic energy and the potential energy associated with the altitude are not thermodynamic properties of state. They do not appear in the system of coordinates which is moved with the substance.

1.2.3 Homogeneous substance We shall consider homogeneous substances only (an exception will be the case of humid air treated in thermodynamics II).

• Homogeneous substance: Substance in which the composition of chemical compounds is locally and temporally constant e.g.: − air in the lecture-hall (neglecting small deviations) − water in a container (liquid at the bottom, vapour at the top)

• non-homogeneous substance: − brandy in a bottle (in the lower part much liquid water, less alcohol, little dissolved

air; in the upper part much air, little gaseous water, little gaseous alcohol) − O2, H2, H2O during a chemical reaction.

Homogeneous substances, therefore, are all pure substances and all well stirred, chemically non-reacting, one-phase mixtures.

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1.2.4 Internal Equilibrium The thermodynamic state will be considered only for systems which are in internal equilibrium. There is internal equilibrium when the state is not changed within a defined period of time. Internal equilibrium approximated e. g., by air in a pressure container or water (vapor and

liquid) in a container, both at constant temperature. Internal non-equilibrium such as in a system with local temperature differences or in a flow

with friction (pressure variation).

1.3 Equation of State Not all properties of a system are independent from each other. Therefore, certain variables can be dependent or independent. Example: If in a container with nitrogen the specific volume v volume and temperature T are given, then a certain pressure p will result, with p = f(T, v). Alternatively, if pressure and volume are given, a certain temperature will result: T = g(p, v). It is therefore important to determine which variables are independent and which ones are dependent in a given situation. If Y is a property of state and x1,…,xn are independent quantities of state, then

Y = f(x1, x2, …, xn) describes a function of state. Since equations of state are path-independent, changes in Y can be described by a total differential dY:

Example: p = f(T, v)

.

From path-independence, there follows for all quantities of state:

Often, an implicit representation of the equation of state is used: F(Y, x1, x2, …, xn)

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1.4 Absolute Temperature, Model Systems, Real Gas Some thermodynamic state variables (p,V,…) have a immediately obvious meaning and are already familiar from mechanics

,

What is „temperature” and how can it be defined?

Zeroth law of thermodynamics: Two systems which are in thermodynamical equilibrium with another system, are also in thermodynamical equilibrium with each other.

Basis of temperature measurement: A in thermal equilibrium with C A in thermal equilibrium with B, and B in thermal equilibrium with C equal temperature

→ System C serves as a measurement instrument → thermometer

Figure 1.2: Gas thermometer

Figure 1.3: Example of a -diagram

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The product is dependent on and leads to a limit for . The temperature is defined by this very limit:

limp→0

p ⋅υ = const ⋅T

One fixed point is sufficient to determine this constant.

depends on the type of gas used!

Fixed point: triple point of water (273,16 K) Individual gas constant:

The absolute temperature is defined by means of the behaviour of real gases (gas thermometer, cf. equ. 1.11): Absolute temperature (1.1)

The absolute zero T = 0, therefore, is defined by .

The unit “Kelvin”, K, is fixed on the temperature scale by the definition: T = 273.16 K at the triple point of water. (All three phases of a pure substance coexist at the triple point.). With these definitions, the values of the individual gas constants R as well as the number of the universal gas constant, i. e., the gas constant per mole, per number of molecules,

, because ; with as the mass of a mole of substance, the molar mass.

Examples for individual gas constants; units: He : 2.08 Ar : 0.208 Xe : 0.063 H2 : 4.12 N2 : 0.297 O2 : 0.260 CH4 : 0.518 C2H6 : 0.277 C3H8 : 0.186 H2O : 0.461 CO2 : 0.189 Air : 0.287

Reason for using the triple point of water at 273,16 K as a fixed point? → precisely 100 K between freezing point and boiling point of water The value 273.16 K at the triple point of water along with the absolute zero is by definition exact. The values of all other fixed points on the temperature scale have to be negotiated internationally for the definition of the International Temperature Scale. At the moment there holds the scale of 1990, the “ITS-90”. Examples of fixed points of the ITS 90:

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Triple points: equilibrium hydrogen 13.80 K oxygen 54.36 K water 273.16 K

Freezing points: zinc 692.68 K silver 1234.93 K gold 1337.33 K The boiling point of water (about 373.13 K, i. e. 99.98 °C) is no longer included in the fixed points of the ITS. Gas thermometers are too complicated for application in science and technology. Their sensors are too big for many measurements. The platinum resistance thermometer, therefore, is applied for the realization of the ITS-90 from 13.80 K up to 1234.93 K. Platinum resistance thermometers which are calibrated according to the conventions of the ITS-90 can be used to calibrate other thermometers. Various methods of temperature measurement: Contact thermometer

• gas thermometer (evaluation of pressure measurement) • liquid thermometer (alcohol, quicksilver) • bimetallic strip thermometer (different coefficient of expansion of two metal plates attached to

each other) • resistance thermometer (resistance of a wire increases by approx. 0.4% per °C) • thermocouples (thermoelectric effect, electric potential of the junction of two metals depends

on the temperature) • vapour pressure thermometer (dependence of the vapor pressure of liquids on the temperature) • liquid crystal thermometer (thermochromic properties of liquid crystals) • Galileo-thermometer

Non-contact thermometer

• pyrometers (based on thermal radiation) • spectroscopic measurements

Measurement accuracy: Precision measurements: error less than 1 mK Measurements in flames: Error about ± 40 K (often much more) 1.4.1 Model Systems The ideal gas consists of molecules without interaction, i.e. mass points without attraction (“mass points“ means: no volume of the molecules and, therefore, no repulsion at v>0 – that means at distances larger than zero; cf. chapter 9.1). Equivalent forms of the equation of state (cf. equ. 1.10 and 1.11) (1.2) (1.3) = molar volume (1.4) n = number of moles (1.5) The molecules of the ideal gas do not interact in the whole range of density, i.e. at all distances. This seams to contradict the assumption of a maximum probability for the

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distribution of energy to the molecules according to an equilibrium temperature. Such a distribution could be reached via interaction of the molecules with a wall. Some special cases of changes of state:

Isothermal: T = const. law of Boyle and Mariotte (1.6)

Isobaric: p = const. law of Gay-Lussac (1.7)

Isochoric: v = const. law of Gay-Lussac (1.8)

"Incompressible" (isopycnic) fluid:

compressibility and

thermal expansion , so that with v = v(p, T), there results

. (1.9)

1.4.2 Example for an Equation of State of a Real Gas Virial equation (virial series)

first kind (1.10)

second kind (1.11) “Virial“, from Latin vis (= force), means: describing forces of molecular interaction. R individual gas constant

second virial coefficient third virial coefficient

Note that the influence of the molecular interaction – of the virial coefficients – upon p⋅v vanishes with v→∞ and p→0: .

But the influence upon the derivatives does not vanish:

There is, e. g.: .

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2 First Law of Thermodynamics (Conservation of Energy) We accept the conservation of energy because we got acquainted very early with the concept of energy and learned that we have to pay for useful energy. We, therefore, forget easily that the concept of energy and the law of its conservation are not given by nature. They are introduced as abstractions to generalize manifold experiences and to make predictions about processes. The first law is the conservation of energy expressed with the forms of energy considered in thermodynamics.

2.1 Verbal Formulation The energy content of a system is changed by the sum of energies which are transferred across the border of the system.

2.2 Forms of Energy 2.2.1 Forms of Energy Content Examples: Internal energy U (per mass u) is a thermodynamic property of state. It is thermal energy because it is the sum of the energy of single molecules: kinetic energy, energy of rotation and oscillation, energy of molecular interaction. The kinetic energy of a molecule results from its velocity relative to the center of inertia of the system. It is assumed in this explanation that there is no macroscopic motion within the system relative to the center of inertia. Such a motion of partial systems exists as the motion of expansion or compression or as turbulent motion. Kinetic energy from the motion of the total system and potential energy from its altitude are not thermodynamic properties. They do not appear in the thermodynamic system of coordinates which is moved with the system. They are mechanical energy, the product of macroscopic forces and macroscopic paths. 2.2.2 Forms in which Energy is Transferred Across the Border of a System which is

Closed for Mass Flows Work W, per mass w

Definition: A system delivers work when the only change in its surroundings can be reduced to the lifting of a weight. This definition is not reversible. It does not hold: A system accepts work when the only change in its surroundings can be reduced to the lowering of a weight. There are many different types of work:

• mechanical work • work associated with volume change • shaft work • electrical work • surface work • magnetic work

A generalization of the concept of work becomes possible, if the terms “force” and “displacement” are generalized. Heat Q, per mass q

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Heat is defined by the first law. It is the transferred energy which is no work (no mass flows across the border of the system).

The definition is not very satisfying. This is especially true because we will have to distinguish carefully between heat and work. But this will not be difficult. At this point it is enough to understand that work is transferred mechanical energy and heat is transferred thermal energy; “thermal” as energy of single molecules in the sense of the internal energy. Transferred electric energy is work, transferred radiation is heat.

adiabatic system: perfectly heat insulated system

adiabatic change of state: change of state with Q = 0

2.3 First Law for a Closed and Contained System closed: no mass flows across the border and therefore no energy transport as con- tent of mass takes place across the border (in contrast to the open system of chapter 2.6). contained: resting relative to the observer, i.e. observed in the thermodynamic system of

coordinates which is fixed to the system, in which only the thermodynamic state appears.

Heat and work which are transferred into the system are positive. Heat and work which are transferred out of the system are negative.

"d" for a property of state like U or u (for an exact differential of a property of state) "δ" for Q and W, which are quantities associatd with a process, not with a state.

2.4 Calculation of the work for “reversible” changes of state The change of state has to go through a continuous sequence of equilibrium states if pA should be set equal to p = p(V, T). Such a change of state is not forced to go in the direction from an unequilibrium state to an equilibrium state – it is reversible. And it is infinitely slow, because the process speed follows from the deviation from equilibrium. Consequently, all processes which take place in finite time are irreversible! And – strictly speaking – nothing in our world can be in equilibrium because everything is in change (

, all is flowing; Heraklitos, Greek philosopher, Ephesos about 500 B.C.). In spite of that we shall take changes of state which run through a continuous sequence of equilibrium states – which are reversible – as idealizations of real changes of state when we have to apply an equation of state like p = p(v, T). We shall avoid the term “quasistatic”. Instead of that we shall say: “it should be assumed that the change of state goes through a continuous sequence of equilibrium states” or, which is the same: “the change of state should be considered as reversible”. This reminds us of the irreversibility of every real change of state.

per mass per mass

First law in differential notation:

First law

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Figure 2.1: Work, transferred by a piston upon a gas (with a constant amount of gas) The following equations hold for reversible changes of state:

(2.7)

(2.8)

The equations 2.8 with the equations 2.2 and 2.4 yield:

(2.9) and

. (2.10)

Conditions for the application of the equations 2.7 and 2.8: a) an equation of state p = p(v, T) for the considered substance must be given b) the change of state must be considered as reversible, to allow us to apply the equation of state c) a condition in form of a path of the change of state must be given to reduce p(v, T) to p(v) for the integration of the equations 2.7 and 2.8.

The work for volume change can nicely be depicted in p,v diagrams (see fig. 2.2), since it corresponds to areas under the curves representing the change of state. Fig. 2.2 shows three ways to get from state a to state b.

1. isobaric + isochoric chage of state 2. isothermal change of state 3. isochoric + isobaric change of state

The areas under the curves are different, and therefore also the works done are different.

pA pressure at the surface of the piston

x dx

A

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The hatched areas

represent the work which is transferred on the different paths from 'a' to 'b'.

Figure 2.2: Compression of a gas on different paths

We see that work is dependent on the path of the change of state. It belongs to the process, not to the state. Heat also belongs only to the process because 'u = q+w and 'u (= u2-u1) is

independent of the path. The path p(v) in also contains implicitly a statement

about the transferred heat. For irreversible changes of state, heat or work can only be calculated from 'u = w+q.

Examples:

1. Air is to be compressed reversibly (condition b) and isothermally (cond. c) from the specific volume v1 to v2. Air should be considered as an ideal gas (cond. a). Ideal gas: ; isothermally: T = const.

wrev = ! p(v,T) "dvv1

v2

# = !R "T dvvv1

v2

# = !R "T " ln v2

v1

(2.11)

2. Air is to be compressed reversibly from v1 to v2. It should hold for the change of state: const. with n = const.

The three conditions a, b, and c are fulfilled with const.

With

(2.12)

it follows that

wrev = ! p1 "v1n " dv

vnv1

v2

# = ! p1 "v1n

1! nv2

1!n ! v11!n( ) (2.13)

and wrev = p1 !v1

n "1v1

v2

#$%

&'(

n"1

"1#

$%

&

'( (2.14)

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2.5 First Law for a system which is moved through a control volume and closed for mass flows

Moved through a control volume: moved through control surfaces into and out of the volume; the change of state takes place within the control volume. We consider the individual system, not a mass flow and not the control volume! At this point, we are not interested whether the content of the control volume changes with time — whether the content of the control volume is stationary or unstationary. The state of the system which is moved through the control volume can change. We assume that there are no forces of inertia acting upon the control volume and upon the ducts leading to and from the control volume (as they are acting, e.g., upon an equipment in an accelerated vehicle).

Control volume: Turbine, compressor, nozzle, tube, etc. The system is moved relative to the control volume (machine). The thermodynamic state is always considered in a system of coordinates which is attached to the moving system. The change of energy is the sum of transferred heat and work

Q + W = energy (outlet) – energy (inlet)

Energy content to be considered in addition to the content of the contained system: mechanical energy content according to the motion p·v: specific energy of displacement c2/2: specific kinetic energy c: velocity g·z: specific potential energy g: gravitational acceleration z: altitude

Figure 2.3: Element of the system, moved through a control volume In addition to the internal energy, now also forms of mechanical energy have to be considered. These are the energy of displacement, kinetic energy and potential energy. The energy of displacement is the mechanical energy which is transported by the system through the control surfaces into (p1•V1) and out of (p2•V2) the control volume. The system is pushed against the pressure at the control surfaces. The kinetic energy is given by m·c2/2.

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The potential energy results from changes of height z in a gravitational field and is given by m·g·z with g the gravitational acceleration. There follows the energy balance, the first law, for a system which is moved through a control volume:

(2.15)

Change of the energy content of the system: (a) change of the content of thermal energy (b) change of the content of mechanical energy which corresponds to the motion.

per mass: (2.16)

differential form: δq + δw = du + d(p·v) + c·dc + g·dz There holds for a reversible change of state of the system inside the control volume – after entering and before leaving the control volume – i.e. for the change of state of the system “contained” in the control volume (for the thermodynamic process, separated from the mechanical process): δqrev - p·dv = du (2.17) Insertion into 2.16 yields: δqrev + δwrev = δqrev – p·dv + d(p·v) + c·dc + g·dz (2.18) With d(p·v) = p·dv + v·dp there follows: δwrev = v·dp + c·dc + g·dz (2.19) or after integration:

(2.20)

Illustration of the integral for a piston compressor:

Changes of kinetic and potential energy are not considered. Sum of all works (delivered by the compressor, accepted by the air):

Changes of kinetic or potential energy could be considered additionally with equation 2.20. The term could in this case be considered as work of displacement delivered by the compressor. But the energy of displacement generally can also be changed by changes of the

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kinetic or potential energy (equ. 2.16) or by the transfer of heat (simple case: isobaric heating of a gas with 'V > 0 yields > 0 or cooling with 'V < 0 yields < 0).

(p1·V1 work of the atmosphere for

pushing the air into the compressor (energy of displacement accepted by the compressor)

(!p·dV work for the change of

volume

p2·V2 work of the compressor for

pushing the air into a pressure container (energy of displacement accepted by the content of the pressure container)

!V·dp effective work done by the

compressor to the air flowing through the control volume (cylinder)

Figure 2.4: Loading, compression and exhaust in a piston compressor

2.6 First Law for a control volume of changing or unchanging content Content of the control volume: open system, open for mass flows. Till now, we observed an element of one moving system. We considered the entering of the element into the control volume, the change of state within the control volume and the leaving of the control volume step by step. We were interested in the fate of an individual element: the change of its energy content from entering to leaving the control volume. We observed the energy transfer within the control volume (analogy: loading and unloading of a vehicle driving through in a building). The content of the control volume could be unchanging or changing. It e.g. would be changing if a heat source within the control volume delivered heat to the system or a spring delivered work.In the following, we shall consider the unchanging or changing content of the control volume. We are interested in the change of its energy content from time to time. This change is caused by energy flows across the border of the control volume: heat flows, flows of work, and energy flows as content of mass flows. The temporal change of the energy content of the control volume at a moment is the sum of all energy flows across the border at this moment. We therefore do not consider anymore consecutive processes at one moving element but simultaneous flows across the border and the change of the energy content of the control volume.

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Figure 2.6: Control volume with energy flows across its border, the content is an “open system“ The dot characterizes flows, e.g. mass flows or flows of heat or flows of work . τ designates the time. First law for a control volume of changing or unchanging content:

(2.22)

(a) Constant or non-constant flows across the border of the control volume, in: positive

out: negative

(b) Change of the energy content of the open system ( = content of the control volume). The open system consists of n partial systems. , , are average specific energy contents of the partial systems.

Remark: Energies of displacement which are transferred between partial systems within the control volume cancel each other. Therefore, the term (b) does not contain energy of displacement.

2.7 Introduction of the Enthalpy For moving systems or for the control volume, the energy of displacement had to be considered. To simplify notation, internal energy and energy of displacement are often summarized and the result is defined as a new quantity, the enthalpy. This is simply a definition. Definition: = enthalpy (2.23) = specific enthalpy (2.24) (2.25) Equation 2.10 with Equation 2.23 and Equation 2.19 yields:

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(2.26)

(2.27)

With the enthalpy as the sum of thermal energy U and mechanical energy , which are both thermodynamic properties of state, there result the following conservation laws:

First law for one system which is cosed against mass flows and moved through a control volume (equation 2.15 and 2.16)

(2.28)

(2.29)

First law for a control volume (equ. 2.22)

(2.30)

We see that the enthalpy is the energy which is transported through a control surface as energy content of a moving system as far as this energy content is a thermodynamic property of state. Equation 2.29 with equation 2.27 would yield equation 2.20. And with equation 2.19 we would get again equation 2.17. About the concept of enthalpy: We saw that the enthalpy H consists of thermal energy U and mechanical energy . The product enters an energy balance only if the system is moved through a control surface. We then can interpret of the moving system as work of expansion of a contained system out of which the moving system is pushed through the control surface and vice versa as work of compression of a contained system into which the moving system is pushed through the control surface. The moving system, therefore, transfers work from one contained system to the other, from one side of the control surface to the other. The energy of displacement of the moving system is like a mechanical coupling of the two contained systems. The enthalpy has also been termed „heat content“ in the past. This term is misleading in a number of ways: heat is only defined as a form of transferred energy. It is defined only as belonging to a process, not to a state. Heat, therefore, cannot be content. The change of enthalpy, furthermore, is only in special cases equal to the heat which is transferred to a contained or a moving system. Contained system:

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1. Heating of a gas at constant volume (W = 0):

Balance: With and there follows

.

In this case, the rise of enthalpy is larger than the transferred heat. It is, therefore, larger than the totally transferred energy and consequently larger than the change of energy content.

2. Heating of a gas at constant pressure:

Balance with yields

In this case, the enthalpy increase is equal to the transferred heat. The rise of the energy content is smaller than the transferred heat by the delivered work. The product , therefore, rises for the equivalent of an energy which is no more contained in the system, which is delivered as work.

Moving system:

In this case, the change of enthalpy is equal to the transferred heat ('h = q) only if in eq. 2.29 (q + w = 'h + 'c2/2 + g·'z) there is w = 'c2/2 + g·'z. The product p·V formally can be considered as energy content. But we can encounter contradictions if we formulate energy balances with the comfortable abbreviation „enthalpy“ without being aware of the meaning of the two parts U and p·V.

Remarks on the first law: We came relatively fast to the first law for a control volume. The equations 2.22 and 2.30 can be interpreted as a comprehensive form of the first law. To illustrate this, we shall now apply equation 2.30 to some particular cases:

(2.30)

1. Contained system, closed against mass flows:

Q

W

Q

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From the assumptions: one heat flow, one flow of work, no mass flow across the border, one system of the constant mass m, , , , there follows:

(2.31)

(2.32)

These are again the equations 2.1 through 2.4, which we previously formulated using our intuition. 2.1 Stationary mass flow through a stationary control volume (control volume of

stationnary content):

(2.33)

With there comes

(2.34)

(2.35)

The equations 2.34 and 2.35 are equal to the equations 2.15 and 2.16. We there conceived to travel with a certain element and saw the transfer of energy to or from the element in a control volume of stationary or unstationary content. Here we assumed a stationary control volume with an unchanging content and the sum of the constant flows across the border, therefore, had to be zero at every moment.

Equation 2.30 with and with one heat

flow and one flow of work yields:

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2.8 Changes of the Internal Energy and the Enthalpy The first law yields a balance of energy. For instance, it describes the change of energy of a system, if work is done (e.g., compression of air in a cylinder by a piston). If we are interested in the consequences for temperature, pressure or specific volume, we need to know the dependence of internal energy on temperature and specific volume. From the fact that internal energy and enthalpy are state variables, it follows

With u = u(T,v) there comes . (2.36)

with there comes . (2.37)

= “specific (massic) heat capacity“ at constant volume

= “specific (massic) heat capacity" at constant pressure

The differences 'u and 'h are generally calculated with the equations 2.36 and 2.37, i.e. with cv and cp, also if v or p is not constant and also when no heat is transferred. The expression “heat capacity“, therefore, is misleading. It is additionally misleading because it suggests that there exists a capacity of the substances to contain heat. But heat never is content. The reversible or irreversible change of state is substituted by special reversible paths when the equations 2.36 or 2.37 are applied. For example:

'u calculated for an isochoric–isothermal change of state

(2.38)

calculated for an isobaric – isothermal change of state:

(2.39)

Ideal gas: p·v = R·T yields

and (deduction in chapter 3.4)

T

T

v

p

2

2

1

1

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There follows and and

With and , there follows

Perfect gas: p·v = R·T and cp = const. (cv = const.)

It is obvious that the equations 2.40, 2.41, 2.43, 2.44 hold if v or p are constant or not constant, and they are independent of the form of the transferred energy from which 'u or 'h results.

Example for the application of the first law: Air flows into an initially evacuated bottle. Which temperature will be reached in the bottle? Air should be considered as perfect gas. Heat transfer and changes of potential energy should be neglected.

Figure 2.6: Bottle with inflowing air

Changes of potential energy neglected, only one flow and one substance:

The consideration starts at )1 before the inflowing of the air and at )2 after the inflowing. There are no more macroscopic movements after the inflowing.

ambience:

in the valve

(2.42)

(2.43)

(2.44)

(2.40)

(2.41)

22

The content of the bottle is m2 with u2; and m1 = 0 (the bottle was evacuated initially). The inflowing air is min:

There holds for the acceleration from the rest state in the ambience into the valve:

This yields With and comes

We can see immediately that u2 = ha if we understand that the atmosphere delivers the work pa·va to the unit mass of inflowing air (Figure 1.1b). This yields u2−ua = pa·va, and therefore u2 = ha. The problem can also immediately be solved if one recognizes that the air transports through the valve:

internal energy uin energy of displacement (p·v)in and kinetic energy (c2/2)in

and that the air in the bottle has u2. With , it follows again: .

The resting system air which flows into the bottle – the border of the system always surrounding also the inner volume of the bottle – is considered in the first short solution of the problem. The system moving through the valve and its following rest state are considered in the second short solution of the problem. This solution corresponds to equation 2.30, specialized to the given problem and is much more elegant than the lengthy application of equation 2.30 shown above. But it demands a good understanding of the first law.

2.9 Equations of Euler and Bernoulli Consider a reversible flow of a system through a control volume.

23

qrev + wrev = Δh + Δ c2/2 + g·Δz (2.45)

In differential form:

δqrev + δwrev = dh + c·dc + g·dz (2.46)

If the system is considered without mechanical energy, there follows du = dqrev + dwrev = δqrev − p·dv dh = du + v·dp + p·dv = δqrev + v·dp, and it follows

δwrev = v·dp + c·dc + g·dz (2.47)

Consider now systems without work ⇒ δwrev = 0 With 1/ρ = v, we obtain Euler’s equation

0 = dp + ρ·c·dc + ρ·g·dz (2.48)

For incompressible fluids (ρ = const.), there results Bernoulli’s equation

0 = Δp + ρ·Δc2/2 + ρ·g·Δz (2.49)

or: p + ρ·c2/2 + ρ·g·z = const. (2.50)

By limiting the consideration to reversible flows of incompressible fluids, the first law (eq. 2.35 and 2.16) decomposes into a Balance of thermal energy

qrev = Δu (eq. 2.10 with v = const.) (2.51)

and a balance of (fluid-)mechanical energy (after integration of 2.47):

wrev = v·Δp + Δc2/2 + g·Δz (eq. 2.32)

Thermal and mechanical energy are decoupled, because

“reversible” means that no mechanical energy is transformed into thermal energy by friction, and

“v = const” means that no mechanical energy is transformed into thermal energy by compression (then Δv would be < 0) and no thermal energy is transformed into mechanical energy by expansion (Δv would be > 0).

This decoupling allows to limit the quest for reversibility to a quest for absence of friction. The temperature has no influence onto the flow, and therefore a heat transfer across temperature differences has no influence. It causes an irreversibility that does not affect the flow.

Examples for consequences of Bernoulli’s equation:

1. Horizontal flow: Δz = 0

Δp = -ρ·Δc2/2

24

* 'c2/2 > 0 (accelerated flow) * 'p < 0

* 'c2/2 < 0 (retarded flow) * 'p > 0 } physical foundation for airfoils!

2. resting (contained) system 'c = 0

'p = -"·g·'z

For water there is "·g ! 103 kg/m3·10 m/s = 104 N/m3 = 0.1 bar/m. When diving, the pressure increases by about 1 bar per meter of depth in water.

3. Leaking water container

Same pressure (air pressure). "/2·'c2 + "·g·'z = 0 c1 ! 0 (with small leakage: c1 can be neglected) c2

2/2 + g·'z = 0 * c2 = Compare this to a freely falling body.

c2

25

3 Entropy and Second Law of Thermodynamics

3.1 Introduction of the Concept of Entropy We expressed the non-exact differential δW of the process quantity work under the assumption of internal equilibrium by p·dV: δWrev = − p·dV. This is the product of the intensive property of state p and the differential of the extensive property of state V. In analogy to this, we now express δQrev by the intensive property of state T and the differential of an extensive property of state. This property of state is the entropy S which we introduce now: δQrev = T·dS (3.1) We get, with δQrev = dU + p·dV, the differential of the entropy S (constant mass):

(3.2a)

and of the specific entropy s:

this corresponds to the function s = s(u, v) (3.2b)

With , it follows that

and

We introduced an integrating factor 1/T by which the non-exact differential δqrev = du + p·dv was transformed into an exact one, namely ds. There results for the surface s = s(u, v)

(law of Schwarz) (3.3)

which we shall use in chapter 3.4. (We do not explain the mathematical procedure to find an integrating factor.) Equation 3.2b yields , or with du = dh − v·dp − p·dv: dh = T·ds + v·dp (3.4)

T, s and p, v are couples of energy conjugate variables. T, p being intensive properties of state are contact variables.

26

The equation u = u(s,v) is a fundamental equation (Gibbs potential) out of which the thermodynamic behavior can be completely derived:

“thermal” equation of

state * “caloric” equation of state

The linear differential form, equation 3.2b or 3.3 or 3.4 is the basis of the general structure of equilibrium thermodynamics (chapter 3.4). This equation and the formalisms derived from it, therefore, describe only systems which are in thermodynamic equilibrium. They describe, instead of irreversible changes of state, always reversible substitutes (e.g. equations 3.8, 3.14, 3.18).

Equation 3.1 yields (3.5)

Consequently is equivalent to . This means: the entropy content of a system is constant during a reversibly adiabatic change of state. This change of state is called “isentropic”. Reversibly adiabatic is isentropic, a change of state at constant entropy.

3.2 Temperature-Entropy Diagram The p,v diagram which was previously introduced allows to determine the work transferred during reversible changes of state. In analogy, a T,S diagram helps in determining the transferred heat.

=

Figure 3.1: Heat represented by the area under the curve of the reversible change of state in the T, S–diagram.

The heat which is transferred during a reversible change of state is represented by the volume under the curve in the T, S–diagram which describes the reversible change of state. (Irreversible changes of state cannot be plotted into the T,S-diagram! No volume can be bounded by non-existing curves!).

T

S

1

2 2 1

!S !S

27

3.3 Temperature-Entropy Diagram of a Perfect Gas With the internal energy u:

There is valid generally: (3.1b)

ideal gas: and u = u(T) (cf. section 3.4),

(3.6)

slope of the isochors. (3.7)

The integration from T1, v1 to T, v (variable) yields for the perfect gas ( and = const.):

(3.8)

(3.9)

s = const. (3.10)

v = const. (3.11) Isochors of a perfect gas are exponential functions in a T,s-diagram. The subtangent of the isochors in the T,s-diagram is . With the enthalpy h:

There is valid generally: (3.5)

ideal gas: and (cf. part 3.4)

⇒ (3.12)

⇒ slope of the isobars. (3.13)

The integration from T1, p1 to T, p (variable) yields for a perfect gas ( and

= const.):

28

(3.14)

(3.15)

s = const. (3.16)

p = const. (3.17)

Isobars of a perfect gas are exponential functions in a T,s-diagram. The subtangent of the isobars is .

Figure 3.2: The specific heat capacities as subtangents in the T,s-diagram

Now that isochoric, isobaric and isothermal processes can be represented in p,v and T,sdiagrams, the question occurs what an isentropic change of state looks like in a p,v diagram. Dependence between entropy, pressure and volume:

(cf. eq. 3.14)

T

s cv cp

29

with

* with eq. 2.42 (3.18)

* (3.19)

isentropic exponent. (3.20)

With s = const, there follow the relations for the isentrope of a perfect gas:

(3.21)

(3.22) and with + in eq. 3.16

(3.23)

and in eq. 3.10

. (3.24)

Introduction of the polytrope: The polytrope is defined in analogy to the isentrope with n = polytropic exponent.

3.4 General Relations between the Properties of State (general structure of equilibrium thermodynamics)

30

Internal energy, enthalpy, entropy depend on the intensive variables (p, v, T). In the following, the quantitative description of this dependency will be demonstrated. The basis are the total differentials for the specific entropy and for the specific internal energy.

(3.2b)

By substitution, one obtains a formulation of the total differential of s(T, v). This yields with eq. 3.2b

(3.25)

with there results:

(3.26)

(This is valid generally (cf. eq.3.7), not for the perfect gas only) (3.27)

The terms and are not yet known explicitly.

Law of Schwarz: with eq. 3.25 (3.28)

(3.29)

Example: For the ideal gas there is and

i.e.: For v > 0 there is no interaction between particles of of the model gas “ideal gas”, and therefore no change of potential energy due to a variation of distance between particles, i.e., a variation of the volume. The ideal gas is a mass point gas, there is no repulsion of particles, no intrinsic “volume” of the particles, without attraction of particles.

31

We can also obtain now:

with eq. 3.29

(3.30)

All results together yield

There follows analogously from: (3.4)

with

(this is valid generally, cf.eq.3.13) (3.31)

(3.32)

(3.33)

Equation 3.25 with equation 3.29 yields: (3.34)

There follows analogously: (3.35)

The equations 3.34 and 3.35 belong to the Maxwell relations. Difference cp – cv Change of the entropy

32

(3.36)

It follows that, if the pressure is held constant during this change of the entropy,

(3.37)

and with the equations 3.26 and 3.31

(3.38)

and with equation 3.34

(3.39)

It follows from : .

For , this becomes

. (3.40)

This yields with equation 3.39

(3.41)

A system is stable only (can exist only) if

. (3.42)

This yields . (3.43)

Remark: This also holds for water between 0 °C and +4 °C with and, therefore,

positive work during an isobaric heating (i.e., accepted compression work).

33

3.5 Second Law of Thermodynamics The second law can be formulated also verbally and without the concept of entropy. Any experience about the direction of real processes can be used for such a formulation: “Water always flows downhill” (strictly: to the level of lower potential energy). “Gases mix when a separating wall is removed“. “Heat flows to lower temperature“. These statements are too special, and the concepts used are not precise enough to suggest a general statement about the direction of real processes. We therefore shall quantify the experiences which are contained in such sentences by means of the entropy concept and shall generalize the results to the second law.

During the treatment of the first law we could assume the concept of energy and its conservation as being accepted. The concept of entropy had to be introduced. It therefore, seems to be more abstract than the concept of energy with which we were already acquainted. On the other hand, our experiences with the direction of processes in our surroundings which are summed up and generalized in the second law by means of the entropy are so obvious that we hardly realize them. We therefore first have to get used to the entropy concept and to become conscious of our experiences to be able to comprehend the second law: Experiences: Direction of all processes (real processes in finite time) Question: Which is the consequence for such irreversible processes concerning the

entropy? Aim: Abstraction of the experiences by means of entropy and generalization of

the abstraction as the second law.

3.5.1 Examples of Experiences which will be generalized as the Second Law 3.5.1.1 Adiabatic systems, closed for mass flows Example 1: Expansion of a gas into a vacuum

Figure 3.3: Spontaneous expansion of a gas into a vacuum

p1 ,V1 p2 ,V2 vacuum

34

Experience: V2 > V1. A gas never contracts itself to a smaller volume under the given conditions.

Consequence: S2 > S1; 'S > 0 (with T > 0 and p > 0) The entropy rises without a transfer of entropy, i. e. there is entropy produced. Reversible limiting case: V2 = V1, 'S = 0 (not a real process) The time was introduced implicitly with the expansion, i.e. with the process which runs with the time to a higher volume. If we introduce the proceeding time as ) explicitly, the consequence can be written as:

(3.44)

Example 2: Heat conduction in a heat insulated system consisting of two partial systems a and b of the temperatures Ta and Tb (e.g. copper blocks). There is a heat resistor between a and b which does not change its state.

Figure 3.4: Heat conduction in an adiabatic system

The entropy change of the system is with

Experience: if . Heat never flows in the direction of rising temperature.

"Q a

Ta

Tb b

heat resistor

heat insulation

35

Consequence: and for n partial systems and

The entropy of the total system rises without a transfer of entropy, i.e. there is entropy produced. The entropy change with time is for n partial systems:

(3.45)

Reversible limiting case: and (The heat flows

infinitely slowly, i.e. the process is not real.)

3.5.1.2 Non-adiabatic system, closed for mass flows.

Example 3: Entropy transferred to the system (coupled with heat) and internal heat conduction. The system is simplified in analogy to example 2. Partial systems: • a surface at TA with no mass, in internal equilibrium • a heat resistor which is a model for all heat resistors of

the system, does not change its state • the substance of the system at , in internal

equilibrium

Figure 3.5: Heat transfer to a system and heat conduction within the system

The transferred entropy is

The entropy change of the system is

"Q #$ surface

heat resistor

heat insulation

substance of the system

36

Experience: if (corresponds to the experience from example 2)

Consequence: ;

The increase of the entropy content of the system is larger than the transferred entropy, i. e. there is entropy produced. For m heat flows and n partial systems:

(3.46)

Reversible limiting case: and

(The heat flows infinitely slowly, i.e. the process is not real.) 3.5.1.3 Adiabatic system, open for mass flows (content of a control volume)

Example 4: Flow through an adiabatic control volume of constant content. Simple case: fluid of in a horizontal duct of constant cross-section; no transfer of work.

Figure 3.6: Fluid flowing in a duct

with

Experience: . A fluid never flows in the direction of rising pressure under the given conditions. The pressure drop caused by friction is p1 – p2.

Consequence: (with T > 0 and v > 0) Reversible limiting case: (The ideal case of the frictionless flow is not a real process.) There holds with the entering mass flow as positive and :

More entropy flows out of the control volume than into the control volume. Generalized for k mass flows across the border of the control volume, constant content of the control volume:

!

37

(3.47)

The entropy which is produced within the adiabatic control volume of constant content is delivered out of the control volume as a part of the content of mass flows. 3.5.2 Second Law Formulated with Entropy

Sum of the consequences for the considered examples and generalization as the Second Law of Thermodynamics: 3.5.2.1 System closed for mass flows

Adiabatic system

consisting of one partial system (example 1, eq. 3.44):

(3.48)

consisting of n partial systems (example 2, eq. 3.45):

(3.49)

Non-adiabatic system consisting of n partial systems (example 3, eq. 3.46):

(3.50)

3.5.2.2 Non-adiabatic system, open for mass flows (content of a control volume)

There results for n partial systems as content of the control volume and m heat flows and k mass flows across the border of the control volume (examples 3 and 4, eq. 3.46 and 3.47):

is the sum of m entropy flows which are transferred across the border of the system. The entropy flows are coupled with the heat flows which are transferred at the temperatures of the surfaces.

38

(3.51)

(3.52)

(a) change with time of the entropy content of the system which is the content of the control volume

(b) entropy flows coupled with heat flows transferred across the border of the control volume

(c) entropy flows as content of mass flows transferred across the border of the control volume

(d) entropy production per time

We now generalize the equations 3.51 and 3.52 as the SECOND LAW OF THERMODYNAMICS, saying that they are valid everywhere and always. This is a step which surpasses our examples of experiences and all experiences which we can make.

The second law says that entropy can be produced but not destroyed. There is entropy produced in all real processes. The entropy is constant in the limiting cases of the reversible processes. We, therefore, can call the second law a “half” conservation law. The structure of equation 3.51 corresponds to that of equation 2.40. The equation 2.40 is longer than equation 3.51 because there are different forms of energy but no different forms of entropy.

The mathematical formulations of the second law (eq. 3.51 and 3.52) correspond to the following - historic - verbal formulations:

1. “A machine of constant content which only accepts heat and transforms this heat into work is impossible.” It would be a perpetuum mobile of second order, an entropy destroying machine because it had to destroy the entropy which it accepts coupled with the heat. (The perpetuum mobile of first order produces energy out of nothing.)

2. “Heat cannot flow in the direction of rising temperature”.

This corresponds to the experience of the examples 2 and 3. The entropy flow coupled with a heat flow would decrease in the direction of increasing temperature. This would make possible a perpetuum mobile of second order.

39

3.6 Cycle Processes 3.6.1 Clausius’ Inequality

Figure 3.7 Irreversible cycle process in a T,S-diagram

There is valid S2 S1 for a substance which goes through a cycle process, or expressed otherwise .

If the cycle process is realized in a stationary machine, the produced entropy has to be delivered to the surroundings coupled with heat (cf. eq. 3.50):

This is Clausius’ inequality. (3.53)

TA = temperature at the surface where & Q is transferred. Equality would hold for a machine performing a reversible function. 3.6.2 Reversible Cycle Processes and Areas in the T,s- and p,v-diagram

Figure 3.8: Reversible cycle process There is and , and ,

There follow with the first law , because .

and

3.6.3 Heat Engine

T

S

S = entropy of the system which undergoes the cycle process

(not specific entropy)

2 1

T p

V S

1 1

effectively accepted heat = effectively delivered work

(3.54)

2 2

40

A heat engine (HE) in effect accepts heat and delivers work: . The average temperature level at which the heat Q1 is supplied to the engine is higher than that at which the heat Q0 is rejected from the engine. The cycle process of the HE runs clockwise in the T,S-diagram and in the p,V-diagram. Figure 3.9: Heat engine, scheme and cycle process

Thermal efficiency:

It follows that

and with

* (3.55)

3.6.4 Refrigerating Engine, Heat Pump

41

Refrigerating engines (RE) and heat pumps (HP) in effect accept work and deliver heat: . The average temperature at which the heat Q0 is transferred to the engine

is lower than that at which the heat Q1 is transferred from the engine. The cyclic process runs counterclockwise in the T,S-diagram and in the p,V-diagram.

Figure 3.10: Refrigerating engine or heat pump, scheme and cycle process

Coefficient of performance:

expenditure: work, effectively transferred to the engine .

Refrigerating engine: yield = heat Q0 , accepted at low temperature (e.g. from a cold-storage)

(3.56)

Heat pump: yield = heat Q1, delivered at higher temperature (e.g. for heating in a building)

(3.57)

3.6.5 Carnot Cycle

Reversible cycle process, consisting of 2 isentrops and 2 isotherms:

42

Figure 3.11: Carnot cycle

Heat engine:

(3.58)

Refrigerating engine

(3.59)

Heat pump:

(3.60)

3.7 Exergy

The thermal efficiency of a heat engine can never be „one“, even in the case of ideal performance (reversible processes). We have seen that the heat (the expenditure) can never be

T

S !S

T1

T0

43

transformed completely into work (the yield). On the other hand, the energetic efficiency of a mechanical engine can reach „one“ in the case of ideal performance (no friction). This led to the desire to introduce a quantity for a heat engine, equivalent to the work input to a mechanical engine. Using this quantity one can express the efficiency of a heat engine as “one” in the ideal case. This quantity is the “exergy”. It is a kind of potential to gain mechanical energy. It combines the first law and the second law.

Verbal formulation: Exergy is mechanical energy which can be gained when a system is brought reversibly from its state to ambient conditions and heat can be exchanged at no cost only at ambient temperature Ta (e.g. from or to ambient air, water etc.). The exergy is assigned to the state “1”. Abbreviations: exergy: E; specific exergy: ε 3.7.1 Closed and contained System

Here mechanical energy is work.

The process can be represented easily in a T,s-diagram. Initial state: state 1 Final state: state a Additional conditions: and reversible How can we understand this in reality and in a Ts-, -diagram? For ideal gases:

1→1‘ 1‘→2 because Idea: during 1→1‘ „mechanical work“ is retrieved from during 1‘→2 higher pressure is again used for work

44

First law: (3.61) Conditions from the verbal formulation:

"reversibly" : (3.62)

"q only at Ta" : (3.63)

"change of st. to amb. st. " : with i.e. from “1” to “a” and (3.64) "gained work": (3.65)

(3.66) specific exergy generally: (3.67)

The first law was introduced as equation 3.61. The restrictions from the second law were: "reversibly": no entropy production "q only at ": heat from the ambience cannot flow to higher temperature than . Hint: it is also possible, that :

45

How can exergy be explained in a T,s-diagram? Two parts: ,

The first part is easy to describe. For the second part, we build an alternative route in the T,s-diagram: It applies: and for an isochoric process:

For ideal gases u depends only on the temperature, i.e. independent of s all states have the same internal energy for T = Ta . Thus as an alternative route a process is used, that begins at Ta and leads along an isochor to state 1. Illustration of the specific exergy for the ideal gas:

and ideal gas

Figure 3.12: Specific exergy as an area in the T,s-diagram (for the ideal gas) Exergy of a cold stone and a warm stone: 1. Warm stone:

Ta

s

a

T

46

2. Cold stone:

Remark about the exergy of a cold stone or a warm stone (volume of the stone considered as constant):

1

47

Figure 3.13: Exergy of a cold stone or a warm stone as area in a T,S-diagram The stone has at T1 Ta always a positive exergy . Exergy is a potential to transform thermal energy into mechanical energy. Thermal energy can be taken from the ambience for nothing. Exergy, therefore, is not necessarily contained in a system as energy.

Until now we have considered only contained (resting) systems where mechanical energy is work and exergy is the potential to gain work under the introduced conditions, so called availability.

3.7.2 Moving System

For a moving system one has to consider not only the work but also the mechanical energy which is contained in the system:

• energy of displacement • kinetic energy (macroscopic) • potential energy (macroscopic)

Formulation of the specific technical exergy based on the formulation of the specific exergy:

First law:

(corresponds to equ. 2.16) (3.68)

Interpretation in the sense of exergy: (a) expended thermal energy (b) mechanical energy gained from thermal energy (c) yielded work (d) rise of the content of mechanical energy

In this case exergy is generally mechanical energy which can be gained from thermal energy. There follows under the conditions introduced for the exergy:

(cf. equ. 3.68) (3.69)

3.7.3 Technical Exergy There follows with that

T

s

Ta

Tc

Tw

a

% warm > 0

% cold > 0 1

48

(3.70)

The sum is called specific technical exergy at the state "1". The specific technical exergy is abbreviated with : (3.71) The technical exergy is no longer mechanical energy which can be gained from thermal energy because h contains also the mechanical energy . Illustration of the specific technical exergy for the ideal gas:

Figure 3.14: Specific technical exergy as an area in the T,s-diagram (for the ideal gas) 3.7.4 "Exergy" of a Heat

The so called exergy of a heat which is available at T1 is the work which is delivered by a reversibly working engine of constant content into which this heat is transferred and which delivers heat only at Ta into the ambience.

(3.72)

T

s

Ta

T

1'' 1'

1

49

Figure 3.15: "Exergy" of a heat as an area in the T,S-diagram

3.7.5 Loss of Exergy El

First way Verbal explanation: The entropy Sprod, produced in an engine of constant content, has to be delivered into the ambience at Ta

coupled with additional heat Qad. This heat results from additional expended work or from lost work Wad (transformation of work into heat by irreversibilities: friction, heat transfer across temperature differences etc. “additional” and “lost” compared with a reversibly working engine). Wad is the loss of exergy.

First law: .

with

yields (3.73) Second way Balance of exergy:

(3.74)

(a) transferred exergy (work and exergy of heat) (b) change of the exergy content Work is transferred exergy. TA is the temperature of the border of the system where Q is transferred.

(3.75)

T

T1

Ta T1

Ta

S S HE: E > 0 RE: E < 0

Q

E

50

: Change of entropy of the system during the irreversible change of state.

: transferred entropy.

(3.76)

(3.77) Example: The loss of exergy and the loss of work during an irreversibly adiabatic expansion of a gas from 1 to :

Figure 3.16: Specific loss of work and specific loss of exergy as areas in the T,s-diagram

s sprod

1

2

2*

T1

Ta

%l = specific loss of exergy

wl = h2*& h2 specific loss of work

work which could be gained by means of a reversibly working engine which causes the gas to change its state from to with the ambience as heat sink.

T

51

4 Engine Processes

4.1 Stirling Engine Idealized process:

2 isochors 2 isotherms

considered as reversible. Working fluid considered as ideal gas. The absolute quantities of heat and are equal. They are transferred at the same variable temperature. They can be stored in a porous regenerator ( ) and can be delivered ( ) from the regenerator. That means, they can be transferred in internal heat transfer. They do not appear outside the engine. The reversible isochors, therefore, are equivalent to reversible adiabatics, to isentrops, seen from outside the engine. The process, therefore, is equivalent to a Carnot cycle:

Heat engine: (4.1)

Refrigerating engine: (4.2)

Heat pump: (4.3)

Remark to the Stirling motor: The Stirling process is a closed process, a real cycle process. The heats have to be transferred through walls into the engine and out of the engine. The load of Stirling engines, therefore, can only be changed slowly. This is an essential disadvantage of a Stirling automobile motor compared with a motor with internal combustion (gasoline or Diesel engine, gas turbine). An advantage of the Stirling engine is, that it does not demand special fuels. But, another disadvantage of the Stirling engine is the high temperature at the part of the engine where the heat is supplied. This causes considerable problems with materials when a high thermal efficiency is desired with a high temperature T2. This, above all, is the reason why it still is not possible to develop a competitive Stirling motor. The Stirling process has been applied very successfully for decades in refrigeration engineering, e.g. Stirling engines of Philips Company for liquifying air (temperature of condensation: ! 80 K).

4.2 Otto Engine (gasoline engine, considered as a four stroke engine, strongly idealized process)

Figure 4.1: Stirling engine as heat engine

T

s

1

2 3

4

52

1 2 compression of the gasoline air mixture (first stroke) considered as isentropic 2 3 ignition and combustion of the gasoline, considered as isochoric heating 3 4 expansion of the flue gas (second stroke, delivering work) considered as isentropic 4 1 exhaust of the flue gas (third stroke), intake of fresh mixture of gasoline and air (forth

stroke); not shown in the diagrams. Working fluid considered as air, i.e. gasoline and the products of the combustion are neglected. Air considered as a perfect gas.

Figure 4.2: Otto process

. (4.4)

With and

with and (isentrop)

and with and

it follows that and therefore

(4.5)

Example: ratio of volumes

p T

v s

2

3

4

1 1

2

3

4

53

Real process (friction, heat transfer between gas and walls, throttling of the air in the intake etc.): The Otto process (and the Seiliger process, the Diesel process and the process of an open gas turbine) ( strictly speaking ( can only be considered as cycle processes, if the exhaust and intake are considered as cooling and reverse of the internal combustion.

4.3 Diesel Engine (considered as four stroke engine, idealized process) 1 2 compression of the air (first stroke) considered as isentropic 2 4 expansion (second stroke)

2 3 injection and combustion of the fuel, considered as reversible and isobaric heating

3 4 continuation of the expansion, considered as isentropic 4 1 exhaust and intake (third and fourth stroke) Working fluid considered as air. Air considered as perfect gas.

Figure 4.3: Diesel process

(4.6)

The Diesel engine has the important advantage compared with the Otto engine that its load can be changed by changing the amount of injected fuel and not by throttling of the air in the intake – with its entropy production. There are gasoline engines developed in the moment with direct fuel injection which do not need the throttling of the air in the intake. Comparison of the efficiency of constant-volume and constant-pressure process for the same compression ratio.

p T

s v

1 1

2

2 3 3

4 4

T

54

# For the same compression ratio a constant-volume process has a greater efficiency.

4.4 Seiliger Process (Model of the Otto process and the Diesel process) 1 2 compression of the gasoline air mixture (first stroke) 2 3 combustion, considered as isochoric 3 4 continuation of the combustion, considered as isobaric, first part of the expansion 4 5 second part of the expansion, considered as isentropic (3#5: second stroke) 5 1 exhaust and intake (third and fourth stroke)

Figure 4.4: Seiliger process

4.5 Gas Turbine

Idealized process (Joule process): 1 2 compression, considered as isentropic

are of the same size for the same work

and

additional expenses

v = const. p = const.

same compression ratio

S

p T

v s

1

2

3 4

5

1

2

3

4

5

55

2 3 heating, considered as reversibly isobaric open cycle: combustion of the injected fuel in the compressed air closed cycle: heat supplied to the compressed gas

3 4 expansion, considered as isentropic 4 1 cooling, considered as reversibly isobaric

open cycle: exhaust of flue gas and intake of fresh air closed cycle: heat rejected

Working fluid, considered as a perfect gas:

open cycle: working fluid, considered as air, fuel and products of combustion neglected closed cycle: different gases, e.g. helium.

The pressure level of the closed cycle is independent of the ambi- ent pressure.

Figure 4.5: Gas turbine process open cycle: 4 1 closed cycle: 4 1 perfect gas, ,

applies for 1#2 and 4#3

Consequence: choose as high as possible.

p

v s

T 2 3

1 4 1

2

3

4

56

this yields analogously to the Otto process

(4.7)

Example: open cycle gas turbine, pressure ratio yields

and with T1 = 300 K T2 = 793 K

Real process (friction, heat transfer between gas and walls, consumption of compressed air for the cooling of the turbine blades etc.)

4.6 Jet Engine

Idealized process: gas turbine with a diffuser before the compressor and a nozzle behind the turbine; no net work yielded because the turbine drives – roughly speaking – only the compressor.

1 2 deceleration (compression) in the diffuser from the velocity c1 (velocity of the airplane) to the velocity at the entrance of the turbo-compressor

2 3 compression in the turbo-compressor 3 4 heating in the burning chamber (considered as isobaric) 4 5 expansion in the turbine 5 6 expansion in the nozzle to the exit velocity c6 Compression and expansion considered as isentropic. Working fluid considered as air. Air considered as a perfect gas.

Figure 4.6: Jet engine

s

57

First law for a system moved through control volume: Δc2/2 + g·Δz + Δh = w + q

for a perfect gas

⇒ (4.8)

Efficiency of the propulsion:

Thrust: (4.9) air flow through the engine (mass flow rate)

(4.10)

The efficiency of the propulsion is the higher the lower the difference (c6 - c1) between the velocity c6 of the jet at the exit and the flight velocity c1 is, i.e. at given thrust, the higher the mass flow through the engine is. (The lower the difference c6 - c1 is, the lower will be the noise caused by the jet in the atmosphere.) This leads to the jet engine with ducted fan (the bypass engine) or to the gas turbine with propeller (the turbo-prop engine).

4.7 Bypass Engine (jet engine with ducted fan or gas turbine with propeller)

Thermodynamic process in the core of the engine:

58

Figure 4.7: Thermodynamic process in the core of the bypass engine 1 4 as in the jet engine without bypass 4 5 expansion in the turbine to drive the compressor 5 5' expansion in the turbine to drive the fan (in the bypass of the core engine) or the propeller 5' 6 expansion in the nozzle Thermodynamic process: subscript t Fan or propeller: subscript p Increase of the kinetic energy of the total jet (from the thermodynamic process in the core engine and from the fan or propeller):

(4.11)

Simplifying assumption: , perfect gas

= velocity behind the fan or propeller (relative to the engine)

Thermal efficiency:

(4.12)

There follows with the equations 4.8 and 4.14:

(4.13)

Efficiency of the propulsion:

}used for thrust 5’

59

(4.14)

The equations 4.16 and 4.17 correspond to the equations 4.11 and 4.13; but, at a given thrust the difference c6-c1 is diminished and the efficiency of the propulsion is enlarged, if

.

4.8 Ramjet The ramjet is a jet engine without fan, without turbo compressor and without turbine. It, therefore, consists only of the diffuser, the burning chamber and the nozzle.

Figure 4.8: Ramjet (numbers analogous to figure 4.7) The thermal efficiency is again (cf. equ. 4.16)

(4.15)

It holds for an isentropic compression in the diffuser with w = 0, q = 0, !z = 0 and c3 = velocity behind the diffuser:

(4.16)

The equations 4.18 and 4.19 yield

(4.17)

We see that the thermal efficiency of the ramjet rises with the flight velocity c1. It is zero at c1

= 0 (and c3 = 0), i.e. a ramjet cannot be used to start an airplane. It is mainly used to propel missiles which reach a starting velocity by other means of propulsion. Examples for the thermal efficiency of a ramjet calculated with equation 4.20. Assumptions: T1 = 300 K; is negligible compared to .

100 300 900

1.6 13 57

60

The efficiency of the propulsion is again calculated with equation 4.10.

4.9 Sound speed, nozzle flow Sound waves are longitudinal waves. They result from disturbances of the velocity- and pressure field. Consider a planar disturbance, which propagates in a medium with the sound speed c*:

From mass conservation there follows: "·c = const. For a reversible disturbance (small disturbance) there holds Euler’s equation (2.48), which reads for negligible changes of potential energy: dp + "·c·dc = 0 Combination of the equations yields: dp = c*2 d" or c*2 = dp/d" Since we assumed an isentropic change of state, there holds:

Therefore, a sound wave propagates in a fluid with the speed c*. For a ideal gas, there holds: cp·dT = dh = T·ds + v·dp p = "·R·T dp = R·T·d" + R·"·dT For an isentropic change of state, there follows: cp·dT = v·dpcp(dp " R·T·d") = R·"·v·dp (cp(R)·dp = cp·R·T·d" cv·dp = cp·R·T·d"

*

* the sound speed increases with increasing temperature * the sound speed is larger for gases with smaller molar mass * the sound speed is larger for gases with smaller cv.

61

Flow in a nozzle:

Figure 4.9: Nozzle; first convergent, divergent afterwards Initial state, (“rest state” at c0 = 0): . One is inclined to assume that the velocity increases in the convergent part and decreases in the divergent part. Is that correct? From mass conservation, there follows:

(A: cross section) * A·d("·c) + "·c·dA = 0

*

'·c: Mass flow density * Convergent cross section * mass flow density increases : A$ "·c % Divergent cross section * mass flow density decreases: A% "·c $ For incompressible media (" = const), it is easy to determine the change of velocity: A$ c% A% c$ For gases, density and velocity are coupled by energy conservation: dh + dc2/2 + g·dz = &q+&w Approximations: potential energy negligible dz = 0 adiabatic &q = 0 no work &w = 0 * dh + c·dc = 0 additional approximation: reversible * ds = 0 * dh = v·dp * v·dp + c·dc = 0 Consequence: c% * p$, the pressure decreases with increasing velocity

It is still not clear if c increases or decreases. For isentropic changes of state, there holds: dp/p = + d"/"

62

and it follows:

The consequences are:

a) Density decreases with increasing velocity b) For c2 < κ·R·T the relative change of velocity is larger then the relative change of density c) For c2 > κ·R·T the other way round d) For c2 = κ·R·T both are equal Remember that for an incompressible medium, there can be no change of density. Introducing the speed of sound c*, one obtains:

An increase of velocity leads to a decrease of density and vice versa. For the relative change of velocity there holds with the Mach number M = c/c*

M < 1 subsonic ⇒

M = 1 sonic ⇒

M > 1 supersonic ⇒

This means: In the subsonic regime, the relative density change is smaller than the relative velocity change, in the supersonic regime it is larger. With the conservation of mass

one obtains

M < 1 ⇒ A↓ ⇒ c↑, A↑ ⇒ c ↓

M > 1 ⇒ A↑ ⇒ c↑, A↓ ⇒ c ↓

Reason: For M > 1 there is

63

In the supersonic regime, the velocity increases due to a decreasing cross sectional area. Reason: If A increases, "·c has to decrease. Since " decreases faster then c increases, c has to increase. What is the temperature when the speed of sound is reached? Assumption: c0 # 0 'h + c2/2 = 0 Assumption: Perfect gas cp·(T*(T0) + c*2/2 = 0 cp·(T(T0) + +·R·T*/2 = 0 (cp + +·R/2)·T* = cp·T0

Since + > 1, there follows: T* < T0 * the temperature decreases. What is the pressure when the speed of sound is reached?

*

What velocity can be reached maximally? Assumption; T*#0 (actually this is not reasonable, since the fluid is then not a gas anymore, especially, not a perfect gas!).

cp(0(T0) + c2/2 = 0 * Table 4.1: Examples for gasdynamic characteristic constants and sound velocities and corresponding temperatures of the rest state, gases considered as perfect

sound velocity and corresponding temperature of the rest state

gas constant gasdynamic constants as properties of the substances

T* = 300 K T* = 400 K

H2 4.124 1.41 0.83 0.53 0.63 1320 361.5 1525 482

64

He 2.079 1.66 0.75 0.42 0.56 1017 399 1175 533 H2O 0.462 1.33 0.86 0.54 0.63 429 349.5 496 465 air 0.287 1.41 0.83 0.53 0.63 350 361.5 402 482

CO2 0.189 1.30 0.87 0.55 0.63 271 345 313 460

Properties of state during the expansion:

It is valid that

Therefore the following relationship holds for the relative change of pressure, density and temperature

and

Local gasdynamic quantities during the expansion in dependence on the actual thermodynamic state:

Flow velocity:

Sound velocity:

Mach number:

Flow density (mass flow relative to the cross-section of the flow):

Cross-section of the flow for a given mass flow rate :

Maximum flow velocity:

65

Figure 4.10 and table 4.2: Density, temperature, flow velocity, local sound velocity, local Mach number, flow density and cross-section of the flow in a Laval nozzle as a function of the pressure ratio p/p0 of an isentropic expansion. Example: air; considered as a perfect gas;

Assumptions:

Table 4.2: Data corresponding to figure 4.10

1 0.98 0.95 0.9 0.8 0.7 0.6 0.5 1 0.99 0.96 0.93 0.85 0.78 0.70 0.61

1 0.99 0.99 0.97 0.94 0.9 0.86 0.82

0 59 94 134 194 243 287 331

350 349 347 345 339 332 325 317

0 0.17 0.27 0.39 0.57 0.73 0.88 1.04 0 6.7 10.4 14.3 19.0 21.7 23.0 23.2

4.47 2.88 2.09 1.58 1.38 1.31 1.29

0.4 0.3 0.2 0.1 0.05 0.02 0.01 0 0.52 0.43 0.32 0.19 0.12 0.06 0.04 0

0.77 0.71 0.63 0.51 0.42 0.32 0.26 0

subsonic supersonic

66

374 421 473 541 590 638 665 775

306 294 277 251 227 199 180 0

1.22 1.43 1.71 2.16 2.6 3.21 3.70 22.4 20.6 17.3 12.1 8.1 4.6 2.9 0

1.34 1.46 1.73 2.48 3.71 6.58 10.3

Conclusions about the shape of a Laval nozzle (assumption: isentropic expansion): The cross-section is unambiguously related to the local thermodynamic state (and the local gasdynamic quantities) if the mass flow and the rest state are given (figure 4.10, table 4.2). The mass flow, the local thermodynamic state and the local gas dynamic quantities, on the other hand, are fixed if the rest state and the shape of the nozzle are given. The pressure behind the nozzle will generally be larger than zero. In the optimal case the nozzle will be shaped so that this pressure is reached. The expansion will continue in the jet behind the nozzle if the nozzle is too short (if the pressure behind the nozzle is too low). The expansion can be terminated by special gasdynamic processes in the divergent (supersonic) part of the nozzle if the nozzle is too long (if the pressure behind the nozzle is too high). Such a process can be a vertical shock wave which leads to higher pressure and from supersonic flow to subsonic flow. Then a further pressure rise can be caused by the deceleration of the flow in the divergent part of the nozzle which acts now as a diffuser for the subsonic flow.

67

5 Properties of Real Pure Substances Represented by Diagrams and Tables

Solid: Lattice of atoms, strong interactions between atoms, ions or molecules. Liquid: short-range order, small distances, strong interaction, frequent collisions Gas: “Disorder”, “molecular chaos”, particle distances much larger than

particlediameter, collisions are rare

T

1

2 3 4

5

1 2 3 4 5

liquid boiling liquid

saturated vapour

liquid vapour

superheated vapour = gas

p p

p

p p

saturated vapour

boiling

first vapour bubble

saturated

last liquid drop

liquid saturated

boiling liquid

68

diagram (complete, including solid):

5.1 Pressure-Temperature Diagram

The boiling temperature is reached in an isobaric heating process when vapour bubbles develop and coexist with the liquid. The pressure of the liquid is then equal to the pressure in the bubbles. (A higher pressure in the bubbles is caused by the surface tension. This will not be discussed at this point.)

T

solid

liquid

liquid

liquid + vapour

vapour „normal“ substance:

T

ice

liquid + ice

liquid

liquid + vapour

vapour

water:

69

Boiling temperature of water at ambient pressure as an example:

Figure 5.1: Pressure-temperature diagram of a „normal“ substance (The slope of the melting pressure curve is positive.) The p,T-diagram shows that pressure and temperature of a two-phase system are unambiguously connected with each other. The two-phase system of a pure substance has only one degree of freedom for its intensive variables p and T. (A one-phase system has two degrees of freedom upon the surface of states p = p(v,T). This corresponds to the phase rule of Gibbs:

(5.1)

F = number of the degrees of freedom C = number of the components of a mixture (1 for a pure substance) P = number of phases e.g. triple point of a pure substance: , point between solid, fluid and gaseous phase in the p,T-diagram. Critical point (c.p.):If the pressure and the temperature rise in the two-phase system liquid-vapour, the density of the liquid will decrease in spite of the rising pressure and the density of the vapour will increase in spite of the rising temperature. Hence the difference between the densities decreases until it vanishes. The differences between all properties of the vapour and the liquid vanish together with the difference of the densities. This means the end of the vapour pressure curve – the critical point. (The meniscus between the liquid and the vapour vanishes at the critical point.)

curve of sublimation pressure desublimation

sublimation

triple point gaseous

liquid

evaporating condensing

freezing

melting

solid

critical point

cycle with only one phase tran-sition

1

4

2

3

70

5.2 Pressure-Volume Diagram Repetition: ideal gases

What changes for real gases? „Experiment“: constant temperature, continuous increase of pressure

p

71

The temperature coordinate is now the direction of the projection in a p,v,T-space.

1. 2. 3. 4.

5. 6. 7.

8.

p2 p3

gas

p4

p5

p6

p7

p8

first drop is forming

p1

p

liquid last gas bubble

compression of the liquid, „force to evade“

first crystals are forming last liquid

compression of the solid

1

2 3 4

5

6 7

8

p

72

Figure 5.2: Pressure-volume diagram of a „normal“ substance (increase of volume during the melting) 1 boiling liquid, left border of the two phase area liquid-vapour, superscript ,. 2 saturated vapour, right border of the two phase areas liquid-vapour and solid-vapour,

superscript -. 3 superheated vapour 4 liquid 5 two-phase area solid-liquid (melting area) Table 5.1: Data for the boiling liquid and for saturated vapour of water (H.-D. Baehr,

"Thermodynamik", 9. Auflage, Springer-Verlag, 1996)

80 0.04736 1.029 3.409 334.92 2643.0 1.0753 7.6132 100 0.10133 1.037 1.673 419.06 2676.0 1.3096 7.3554 120 0.19854 1.061 0.892 503.72 2706.0 1.5276 7.1293

Table 5.2: Examples of critical data

v

c. p.

liquid-vapour co-existence

solid and vapour

1

2

3

triple line

vapour

5

4

73

Hg 107.7 1460 0.2 H2O 22.0 374 3.11 C2H5OH (alcohol) 6.4 243 3.62 C2H6 5.0 35 4.93 CO2 7.5 31 2.15 C2H4 4.7 -82 6.17 O2 5.1 -119 2.34 N2 3.4 -147 3.18 H2 1.3 -240 31.65 He 0.23 !5.3 K 14.37

5.3 Description of a State in a Two-Phase Area Example: two-phase area liquid-vapour The state “1” will be described as the center of gravity of the mass fractions of

boiling liquid L and of saturated vapour V.

Figure 5.3: Representation of a state in the liquid-vapour area

(5.2)

With it follows that

and with

and with (because )

.

p

v v'' v'

1 V L

c. p.

v1

74

(5.3)

analogously:

(5.4)

(5.5)

(5.6)

Scale as an illustration of the mass averaged state 1

Figure 5.4: Equilibrium of torque L lies on the boiling-liquid line. V lies on the saturated-vapour line.

with yields and

.

5.4 Temperature-Entropy Diagram

There holds and with p = const.

75

In the liquid-vapour area (for the liquid-vapour equilibrium) it is valid with p = const. also T = const.

. This yields for a vaporization process at constant pressure: (5.7)

= enthalpy of vaporization (“heat“ of vaporization)

= enthalpy of melting (“heat“ of melting)

= enthalpy of sublimation (“heat“ of sublimation)

subscripts l: liquid on the right boundary line of the melting area s: solid substance on the left boundary line of the melting area and of the area of sublimation

Figure 5.5: Temperature-entropy diagram of a „normal“ substance (The melting pressure rises with the temperature.)

5.5 Clausius-Clapeyron Equation

T

s

solid

vaporiza- tion

triple line

gaseous

sublimation

c. p.

liquid

s'' s'

76

The Clausius-Clapeyron equation is a differential equation for the relation between pressure and temperature in a two-phase equilibrium, e.g. for the gradient of the vapour-pressure curve

.

We consider a differential cycle process in the p,v-diagram and the same cycle process in the T,s-diagram. (Reversibility is assumed a priori because phase equilibria are considered.)

work heat Figure 5.7: The same differential cycle process, shown in a p,v-diagram and in a T,s-diagram It is generally valid that: (

There holds for a reversible cycle (

And

Clausius-Clapeyron equation (5.8)

This yields with equation 5.7 (5.9)

Integration of the Clausius-Clapeyron equation for the vapour-pressure curve: Simplifying assumptions: 1. * is neglected.

2.

3. Consequences of the simplifying assumptions for the approach to the critical point:

becomes and is smaller than : The denominator is assumed to be larger than the real value.

becomes : The numerator is assumed to be larger than the real value. The errors compensate each other to a high extent.

The Clausius-Clapeyron equation becomes with the assumptions 1 and 2:

p

v v'' v' s' s'' s

c.p. c.p.

dp dT

T

77

This can be integrated with the assumption 3:

(5.10)

With this, vapour-pressure curves of different substances are represented by straight lines in a

-diagram.

Table 5.3: Phase-changes and elements of the Clausius-Clapeyron equation 5.9

process gradient of the h´´−h´ change during the

vaporization vapour-pressure curve enthalpy of vaporization

vaporization

melting melting-pressure curve enthalpy of melting melting sublimation sublimation-pressure curve enthalpy of

sublimation sublimation

The differential quotient has a discontinuity at the transition from the sublimation curve

to the vapour-pressure curve (i.e., at the triple point), because the enthalpy of sublimation at temperature TTr−ΔT is by the enthalpy of melting larger than the enthalpy of vaporization at TTr+ΔT. (ΔT is a very small positive temperature difference). The relative change of is smaller than the relative change of .

5.6 Pressure-Enthalpy Diagram

78

Figure 5.8: Pressure-enthalpy diagram of a „normal“ substance (The melting tempera- ture rises with the pressure.) gradient and curvature of the isentrops in the p,h-diagram:

(5.10)

(5.11)

The first and the second derivative of the isentrops are positive in the p,h-diagram.

5.7 Enthalpy-Entropy Diagram

79

Figure 5.9: Enthalpy-entropy diagram of a „normal“ substance (The melting temperature rises

with the pressure.)

Gradient and curvature of the isobars in the h,s-diagram:

with p = const. follows

and (cf. equ. 3.31)

The first and second derivative of the isobars are positive in the h,s-diagram. The isobars have no discontinuities of slope in the h,s-diagram because T varies continuously along the lines of constant pressure. This is true also on the boundary lines of the two-phase areas. Remark (supplement to chapter 3.1)

h

s

sublimation

c.p.

vaporiza-tion

80

The function is a fundamental equation, a thermodynamic potential, a Gibbs potential. All thermodynamic properties can be determined by differentiation of this function:

corresponds to . This yields:

“thermal” equation of state

with “caloric” equation of state Other thermodynamic potentials:

specific Helmholtz-free-energy specific Gibbs-free-energy

81

82

6 Engine Processes - Continuation

Engine Processes Engine Processes Engine Processes Continuation Continuation Continuation Continuation Continuation Continuation Continuation Continuation ContinuationEngine Processes Engine Processes Engine Processes Engine Processes 6 Continuation Continuation ContinuationEngine Processes Engine Processes Engine Processes Engine Processes Engine Processes Engine Processes Engine Processes Continuation Continuation Continuation Continuation Continuation Continuation

“normal“ substance

water

Figure 5.11: Qualitative diagrams for a „normal“ substance and for water

Figure 5.12: Quantitative diagram for a substance with a large number of excited degrees of freedom of oscillation (toluene; cf. chapter 9.2, equ. 9.17)

83

6.1 Clausius-Rankine Process, Steam Power Cycle Preliminary considerations: Carnot-process in the liquid-vapour area

Problem: compression respectively expansion of liquid-vapour are technically problematic

Figure 6.1: Simplified scheme of a steam-power unit

Figure 6.2: Idealized steam power cycle in the T,s-diagram 1#2 pumping of the condensate by the feed pump, considered as isentropic

2

1

5

6

3 4

feed pump

turbine

steam generator

pre-heater boiler superheater

condenser

T

s v

p

84

2#3 pre-heating of the condensate to the boiling temperature 3#4 vaporization 4#5 superheating 2#5 considered as isobaric 5#6 expansion of the steam in the turbine, considered as isentropic 6#1 condensation of the steam in the condenser, considered as isobaric Thermal efficiency:

Changes of kinetic and potential energy are neglected.

*

Remark:

, because the (average) specific

volume of the steam in the turbine is much larger than the specific volume of the condensate in the feed pump. Reheat cycle:

+

85

Figure 6.3: Idealized steam power cycle with intermediate reheating

5#6 first partial expansion to the intermediate pressure 6#7 reheat, considered as isobaric 7#8 second partial expansion to the condensing pressure Advantages of the reheat cycle:

1. less condensate in the turbine flow ( > ) : less erosion of the blades in the low pressure turbine, less flow losses.

2. higher thermal efficiency because

Typical data of a modern steam power unit fired with fossil fuels:

boiling pressure ! 15 MPa maximum steam temperature ! 800 K condensing pressure ! 0.004 MPa condensing temperature ! 300 K thermal efficiency 0.40 … 0.42

For comparison: thermal efficiency of a nuclear power plant: ! 0.35

T

s

c.p.

1

2

3

5

+

7

+

8

6

6*

4

86

Figure 6.4: Idealized steam power cycle without reheat in different diagrams

6.2 Combined Gas Turbine and Steam Cycle The exhaust gas of a gas turbine can be used to heat the boiler (“waste heat boiler”) of a steam power plant. The thermal efficiency of such a combined unit can reach nearly 60 % (58.6 % in

isobar:

isentrope:

enthalpy change in the turbine

87

the Rheinhafen-Dampfkraftwerk Karlsruhe, Block 4S, EnBW AG). Such a unit can employ supplementary heating using fossil fuels (e.g. Heizkraftwerk 2, HKW 2, Kraftwerk Altbach-Deizisau, Neckarwerke Stuttgart AG).

Figure 6.5: Coupling of a gas turbine process with a steam power cycle by means of a waste heat boiler in a (qualitative) diagram Repetition gas turbine:

T

! 1500 K c

d

5

b

a

1 2

6

3 4

entropy flow

e

88

Idea: The heat of the gas turbine is used to heat the steam power cycle

entropy, produced by the heat transfer in the waste heat boiler per time. The two processes which are coupled by means of the waste heat boiler can be represented in one diagram because the entropy – in our considerations – contains an arbitrary constant of integration. Subscripts: gas turbine: “g”; steam power unit: “s” Heat, rejected by the exhaust gas of the gas turbine: = Heat, supplied to the water in the boiler of the steam power cycle: =

Balance:

Thermal efficiency:

Additionally heat can be used for district heating in a combined heat and power unit:

T

2 isentrops, 2 isobars is used instead of S, because in the

following, gas turbine and steam power cycle have different mass flows

89

( partially and

( at higher condensing temperature (! 120 °C) and correspondingly higher condensing pressure (turbine back-pressure) .

If the sum of heat and power is considered as yield, the efficiency should no more be called „thermal efficiency“ but, might be termed “energetic efficiency”.

It is assumed that the waste heat boiler is a counter flow heat exchanger. In the - diagram we cannot see immediately the temperature differences across which the heat is transferred from the exhaust gas to the water. But, we can couple a t,h-diagram of the exhaust gas with a t,h-diagram of the water by introducing the isobaric cooling of the exhaust gas into the t,h-diagram of water. In this diagram we can see the temperature differences across which the heat is transferred as vertical distances 'T between the isobar of the exhaust gas and the boiler isobar of the water (cf. figure 6.6). Scheme of a counterflow heat exchanger:

This diagram applies to water, i.e. if we want draw in TG , we must calculate how TG changes with hD! We have to consider the enthalpy balance for the two substances exhaust gas and water in the waste heat boiler to introduce the isobar of the exhaust gas into the t,h-diagram of the water.

T

hs 1

2

3 4

5

6

dh = cp dT + !h

!p"#$

%&' T

d p

Tlow

Thigh Tlow

Thigh

water from boiler feed pump

gas from gas turbine

90

In a counter flow heat exchanger, the enthalpy of the exhaust gas and the enthalpy of the water both rise in the same direction. It is valid for a differential step along the heat exchanger (thermally isolated from the ambience):

and with the mass flows

Therefore, a step on the enthalpy axis of the t,h-diagram of water corresponds to a

simultaneous step on the enthalpy axis of the t,h-diagram.

With , there follows the slope of the isobar of the exhaust gas in the t,h-

diagram of water

The mass flows and the temperatures of the water and of the exhaust gas have to be arranged so that ΔT is everywhere sufficient for the heat transfer.

91

We see also an isentropic partial expansion of the steam (2→3) and an isobaric re-heating (3→4) in the diagram of figure 6.6. If the reheating of the steam should be caused by heat from the exhaust gas, this heat transfer can be represented by carrying a line, parallel to the isobar 3→4, to the right until is reached. The heat supplied to the water is now plotted continuously above the h-axis without overlapping and corresponds to the continuous heat rejection from the exhaust gas. (The parallel transfer of the isobar only changes the constant of integration in the enthalpy of water for the re-heating. It does not influence the enthalpy differences which enter into the energy balance for the heat exchanger.)

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6.3 Refrigeration Cycle Compressor engine as refrigerating engine: cooling a cold storage, as heat pump: heating a building.

Figure 6.7: Simplified scheme of a refrigeration cycle The evaporator of a refrigerating engine which accepts heat at low temperature is placed e.g. in the interior of a refrigerator or a deep freezer while the condenser delivers heat to the environment. The evaporator of a heat pump accepts heat, e.g. from the environment and the condenser delivers heat at a higher temperature level e.g. for the heating of a building.

Figure 6.8: Simplified refrigeration cycle in a T,s diagram. 1#2 compression of the vapour, considered as isentropic 2#3 cooling and condensing of the vapour, considered as isobaric 3#4 throttling of the condensate from the condenser pressure to the evaporator pressure;

assumption: (throttling in a capillary tube in a household refrigerator). 4#1 evaporation, considered as isobaric. Changes of kinetic and potential energy are neglected.

condenser

evaporator

compressor throttling device

4

3 2

1

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Heat supplied to the evaporator:

corresponds to the “refrigeration power”.

Heat rejected from the condenser:

+ corresponds to the “heating power” of a heat pump.

Work supplied to the compressor:

Coefficients of performance:

Typical data of a household refrigerator. Working fluid: refrigerant "R600a" (isobutane): condenser pressure: 0.49 MPa

condenser temperature: 310 K evaporator pressure: 0.16 MPa

evaporator temperature: 273 K refrigerating capacity: 60 W coefficient of performance:

Figure 6.9: Simplified refrigeration cycle in different diagrams

+

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Figure 6.9: (continued)

6.4 The Effect of Irreversibilities in the Steam Power Cycle and in the Refrigeration Cycle

We only consider pressure drops caused by friction. The differences between the reversible cycles and the irreversible cycles are exaggerated for the reason of clearness. 6.4.1 Steam power cycle The isobars = const. and = const. and the highest temperature limit the irreversible cycle and a comparable reversible - ideal - cycle (reversible cycle characterized by asterisks).

Figure 6.10: Simplified irreversible steam power cycle compared to a reversible cycle

between the isobars = const. and = const. and with the highest temperature .

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Irreversible cycle: Supplied heat:

Rejected heat:

Thermal efficiencies:

(hatched areas cf. fig. 6.2)

If we describe of the reversible cycle as the ratio of the areas hatched in figure 6.2, we

see that the thermal efficiency is diminished by the irreversibilities:

<

6.4.2 Refrigeration cycle, heat pump cycle Limiting temperatures: Heat rejection (heating capacity):

Heat supply (refrigerating capacity):

Figure 6.11: Simplified irreversible refrigerating/heat pump cycle compared to a reversi- ble cycle with the same temperature span (reversible cycle characterized by asterisks).

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Irreversible cycle: Rejected heat:

Supplied heat:

Coefficient of performance of the refrigerating cycle:

(hatched areas cf. fig. 6.8)

The comparison shows that

and because it shows also that .

(