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Eng3901 - Thermodynamics I 1

1 Introduction

1.1 Thermodynamics

• Thermodynamics is the study of the relationships between heat transfer, work inter-actions, kinetic and potential energies, and the properties of systems.

• The basic principles involved are:

1. conservation of mass;

2. conservation of energy (1st Law of Thermodynamics); and

3. the 2nd Law of Thermodynamics.

• The traditional emphasis of thermodynamics is on turbines, pumps, engines (i.e. me-chanical devices), but it can also be applied to electric devices, or any system thattransforms energy from one form to another.

• Much of the historical difficulty in thermodynamics was due to imprecise definitions.The power and utility of thermodynamics centers around the fact that there arerelatively few variables:

Properties: Process variables:

pressure workvolume heattemperatureinternal energyentropy

and these variables can be applied to a wide variety of systems. The following sub-sections define terms that arise in thermodynamics.

1.2 Systems

• A system is a region enclosed by an imaginary boundary that may be rigid or flexible(control volume).

• The system in thermodynamics is similar to the free body diagram in mechanics.

• Closed System (or Control Mass)

– No mass crosses the system boundary,however, energy (in the form of heatand/or work) may cross the boundary.

– e.g. piston-cylinder


• Open System (or Control Volume)

– permits the transfer of both mass and energy across the system boundary

– a region in space through which mass and energy pass (control volume)

– e.g. pump

• Isolated System

– neither mass nor energy may cross the system boundary

– idealization

• common to all systems

– surroundings (everything external to the system)

– system boundary

– system volume

• e.g. An oxy-acetylene torch is used to boil water in a cylinder, thus creating steamand raising a weight. Is the system open, closed or isolated?

• The answer depends on where the system boundary is placed.

– A frequent error is to assume a system is closed when in fact it is open.

– Note: a system may consist of a single substance or many substances.


1.3 Property

• A property is any measurable characteristic of a system, or any quantity that dependson the state of matter in the system.

• e.g. T , p, V , ρ, µ, k, R, etc.

• Properties are independent of the past history of matter, they are conditions at anequilibrium state. (Note: a system is in equilibrium if its properties do not changewith time.)

• Intensive Property

– independent of the amount of mass present

– e.g. T , p, ρ, v (specific volume)

• Extensive Property

– dependent on the amount of mass present

– e.g. V (volume), Ek (kinetic energy), U (internal energy)

• Note:

– extensive property divided by mass → intensive property

– convention - lower case → intensive, upper case → extensive, but there are ex-ceptions (T , m)

1.4 State

• The properties of a system describe the state of the substance within the system, i.e.a phase may exist at different states (p,T ).

• A relationship among properties is an equation of state.

1.5 Process

• The transformation of a system, resulting in a change of properties, from one state toanother is a process.

• Consider a frictionless piston-cylinder assembly in equilibrium. If a weight is removed,there will be a change in the system properties, i.e. p1, V1, T1 → p2, V2, T2.


• The path of the succession of states through which the system passes is called aprocess.

• In thermodynamics, processes are not considered to occur in a continuous manner,but to move through a series of equilibrium steps which are infinitely small in size.A continuous process gives rise to conceptual difficulties (e.g. kinetic energy of thepiston, friction in the cylinder).

• Equilibrium

– mechanics - balance maintained by equality of opposing forces

– thermodynamics (deals with forms of energy)

– mechanical - balance of forces

– thermal - no change in temperature

– phase - balance between phases

– chemical - no further chemical reactions

– Remember - properties are conditions at equilibrium

• What if heat is added to the same piston-cylinder?

Note: p1 = p2 since the load on the piston has not changed, but V2 > V1, and T2 > T1.


• The change in properties can be plotted on process diagrams, which are very usefulin the visualization of how a change in state occurs during a process.

• An inherent assumption in the use of process diagrams is that the system passesthrough a series of equilibrium states (since properties only have meaning at equilib-rium). This gives rise to what is called a quasi-equilibrium or internally reversibleprocess.

• An internally reversible process could be reversed, and the system and surround-ings would retrace the same series of equilibrium steps. This requires an infinitesimalprocess. Real processes are not reversible (finite rate and losses due to friction andheat transfer). Internally reversible (ideal) processes are often used in the analysis ofsystems (e.g. calculation of maximum efficiencies).

• Common types of processes:

– isothermal - constant temperature

– adiabatic - no heat transfer

– isobaric - constant pressure

– isochoric - constant volume

1.6 Point function

• A property is a point function; the value of any property of any system at any stateis independent of the path or process used to reach that state.

• The integral of the differential change of a property between two states, 1 and 2, is:∫ 2

1dV = V2 − V1 (1)

• d is used to represent the exact differential of a point function.


1.7 Path function

• The value of a path function is dependent on the path followed during a process, andthe path must be specified before a path function can be evaluated (e.g. heat transferand work).

• The total integral of the work done between two states is:∫ 2

1δW = W12 (2)

• Note: ∫ 2

1δY 6= Y2 − Y1

• δ is used to represent the inexact differential of a path function.

1.8 Cycle

• a process or series of processes whose initial and final states are identical

• The change in a property during a cycle is zero.∮dX = 0 (3)

• The cyclic integral of a path function is not necessarily zero, as it depends on the pathtaken. ∮

δY 6= 0 (4)

e.g. δW = pdV , and W12 6= W21, therefore,∮δW 6= 0


1.9 Pressure (p)

• normal force per unit area acting on the surface of the system

• intensive property

• Thermodynamic Pressure

– results from the cumulative effect of individual molecules striking the walls of acontainer

p = limδA→δA′

δFnδA

(5)

– where δFn is the normal force component acting on a small area δA, and δA′ isthe smallest area over which the substance can be considered a continuum

• Dynamic Pressure (ρu2/2)

– pressure due to the motion of a fluid

• Absolute Pressure (pabs)

– absolute force per unit area on a surface (including atmospheric pressure)

• Gauge Pressure (pg)

– pressure read on a gauge (e.g. tire pressure)

– pg < 0→ vacuum

pabs = pg + patm (6)

• units of measure

Pa = 1 N/m2

1 kPa = 1000 Pa1 bar = 105 Pa1 atm = 101,325 Pa1 atm = 760mm Hg1 atm = 14.7 psi

• measured using gauges, transducers and manometers

1.10 Continuum

• Pressure and other properties imply the use of a continuum (or macroscopic view).

• Classical thermodynamics studies the macroscopic behaviour of matter, and is, there-fore, concerned with the time-averaged behaviour of a large collection of atoms (con-tinuum).

• Statistical thermodynamics takes a microscopic view, and studies the behaviour ofindividual atoms.


• To use the concept of a continuum, dimensions must be much larger than the meanfree path of the molecules under study (i.e. the mean distance between molecules),otherwise the behaviour of individual molecules would be important.

• In the definition of pressure, δA′ is chosen large enough to ensure that the influenceof individual molecules is negligible.

• e.g. ρ = m/V or ρ = ∆m/∆V :

• As ∆V is decreased, the curve becomes irregular since the volume sample has beenmade so small that the number of molecules in the sample would vary with time.

1.11 Temperature (T )

• Temperature is a measure of the kinetic energy of the molecules in a substance.

• With a temperature increase the molecular activity and velocity increase, therefore,Ek increases, and if temperature decreases, Ek decreases.

• Temperature is not defined, rather the equality of temperatures is defined.

• Two bodies are equal in temperature if they do not interchange heat when placed incontact, i.e. they are in thermal equilibrium. Temperature is used as a measure ofthermal equilibrium.

• Zeroth Law of Thermodynamics

When two bodies have equality of temperature with a third body, they haveequality of temperature with each other.

The Zeroth Law allows the use of a thermometer to measure temperature. It does notdefine temperature.

• A temperature scale is defined with respect to some references (e.g. freezing, boiling,triple points).

• units of measure

SI: ◦C CelsiusK = ◦C + 273.15◦ Kelvin

Imperial: ◦F Fahrenheit◦R = ◦F + 460◦ Rankine

Conversion: ◦C = (◦F - 32)∗5/9


1.12 Heat Transfer (Q)

• Heat transfer is the energy transfer across a system boundary due to a temperaturedifference between the system and its surroundings.

• There are three modes of heat transfer:

1. conduction - solids and fluids

2. convection - fluids

3. radiation - solids, fluids, and vacuum

• Heat transfer always occurs from high to low temperatures, and increases for largertemperature differences.

• The concept of thermal resistance is used when solving heat transfer problems. Ther-mal resistances are used in methods similar to those applied to electric circuits, there-fore, the concepts of good thermal conductors and insulators arise.

• If no heat transfer occurs during a process (due to large thermal resistances, or in-finitesimal temperature difference) the process is adiabatic. Note the difference be-tween adiabatic (no heat transfer) and isothermal (constant temperature) processes.

• Heat transfer is directional and the standard convention is:

Q < 0 for heat transfer from the system to its surroundings; and

Q > 0 for heat transfer to the system from its surroundings.

• units of measure:

SI: Q Jq J/kg

Q J/s or W (heat transfer rate)Imperial: Q Btu (1 Btu = 1.0551 kJ)

Q Btu/h (1 Btu/h = 0.293 W)


• A Btu (British thermal unit) is the amount of energy required to raise the temperatureof 1 lbm of water at 68◦F by 1◦F.

• Heat transfer is a path function, therefore, the amount of heat transfer in a processbetween states 1 and 2 is: ∫ 2

1δQ = Q12 (7)

1.13 Work Interactions (W )

• A work interaction is an energy transfer across a system boundary that is equivalentto a force acting through a distance.

• The sign convention used for work is:

W < 0 when work is done on the system by its surroundings; and

W > 0 when work is done by the system on its surroundings.

Note: this convention is opposite to that used for heat transfer. It was chosen sothat work produced by a steam engine would be positive.

• Work is a path function.

• Equations used to evaluate work only give the magnitude of the work, the sign mustbe inferred from the direction of the force and the displacement with respect to thesystem.

• e.g. force acting through a distance

W12 =

∫ 2

1

~F · d~s

• e.g. electric work (electric force is used to displace electrons)

W12 =

∫ 2

1EI dt


• e.g. rotating shaft work (e.g. pumps, compressors, turbines)

W12 =

∫ 2

1T dθ

• e.g. pdV work (e.g. piston-cylinder)

– If the piston is displaced by a distance ds, what is the work done during a quasi-equilibrium process?

δW = pAds = pdV

therefore

W12 =

∫ 2

1p dV (8)

– The pdV work done during a process is equal to the area under the p -V curvedrawn for the process.


– If dV > 0 (expansion) work is done by the system on its surroundings (i.e.W > 0).

– If dV < 0 (compression)work is done on the system by its surroundings (i.e.W < 0).

– Often, the functional relationship between p and V can be written in the formof a polytropic equation:

pV n = c (a constant) (9)

such a process is called a polytropic process. If the exponent n is known, onecan easily integrate to determine the pdV work:

W12 =

∫ 2

1p dV

but p = c/V n, therefore:

W12 =

∫ 2

1cdV

V n(10)

where the integration will depend upon the value of the exponent n (see e.g. 2.1of Moran and Shapiro).

– e.g. if n = 1 (an isothermal process of an ideal gas)

W12 = c ln

∣∣∣∣V2V1∣∣∣∣ (11)

where c = p1V1 = p2V2.

• Units of measure:

SI: W J or N·mW W or J/s (power)

Imperial: W ft·lbf or Btu (1 Btu = 778.17 ft·lbf = 1.0551 kJ)

W hp or Btu/h (1 hp = 2545 Btu/h = 0.7457 kW)

1.14 Energy

• Forms of energy:

– kinetic, potential, chemical, internal, surface tension, magnetic . . .

– Microscopic forms of energy are those related to molecules and the interactionof molecules.

– Macroscopic forms of energy are those related to the gross characteristics ofa substance, i.e. they are based on dimensions much larger than the mean freepath of the molecules (continuum), e.g. a mass acting at a center of gravity.

– Total energy is a sum of the macroscopic and microscopic forms of energy:

E = Emicro + Emacro

– Thermodynamics studies changes in total energy, however, there are usuallysignificant changes in only a few forms of energy (all other changes are negligible).


– This course will be concerned with chemically non-reacting systems, and, in mostcases, magnetic, electrical, and surface tension energy levels will be neglected.

– The most frequent mode of work encountered will be either pdV or shaft work,therefore, most examples will be concerned with simple compressible sub-stances.

• Kinetic Energy (Ek, ek)

– physical property

Ek = mV 2

2(12)

– or intensive kinetic energy, ek:

ek =V 2

2(13)

• Potential Energy (Ep, ep)

– physical propertyEp = mgz (14)

– where z is the measured elevation of mass, m, above some reference datum in agravity field ~g

– intensive potential energy, ep:ep = gz (15)

• Internal Energy (U , u)

– thermodynamic property

– microscopic form of energy due to:

– molecular level energies – kinetic, rotational, vibrational

– intermolecular forces

– All substances possess some level of internal energy.

– temperature increase → increase u

– change from solid to liquid, or liquid to vapour → increase u

– Internal energy cannot be measured directly, but changes in u are related toother properties (e.g. T , p, v)

• Ignoring chemical, electrical, surface tension, and magnetic energies, the energy pos-sessed by a substance is:

E = Ek + Ep + U

= mV 2

2+mgz +mu (16)

therefore, the intensive energy is:

e =V 2

2+ gz + u (17)

The most significant changes in E are often due to changes in U , therefore, the kineticand potential energies (Ek and Ep) are often neglected.


1.15 Enthalpy

• Enthalpy is a defined thermodynamic property that is obtained from other properties.

• Often, the sum U + pV arises in thermodynamic analyses. To make life easy, athermodynamic property called enthalpy is defined:

H ≡ U + pV (18)

h ≡ u+ pv (19)

1.16 Conservation of Energy for a Closed System

• The principle of the conservation of energy provides the foundation of thermodynam-ics, as it supplies the basic framework required to study the relationships among work,heat transfer, and the various forms of energy.

• The principle of the conservation of energy can be written in the following form:

Energy is a conserved property. It can neither be created nor destroyed; only itsform can be altered from one form of energy to another.

• The conservation of energy equation, or energy balance, for a closed system will haveto account for the following energies and energy transfers:

1. Energy transfer due to differences in pressure and temperature (heat transferand work), which are only identifiable at a system boundary (e.g. a system cando work on its surroundings).

2. Energy related to the mass of the substance(s) within the system (i.e. U , Ek,Ep).


• The conservation of energy equation may be written in the following form: Total rate at whichenergy enters a cv

across its boundary

− Total rate at which

energy leaves a cvacross its boundary

=

Net rate of increaseof energy within the

control volume

(20)

i.e. if an amount of energy enters a control volume, then the same amount of energymust leave the control volume, or it will produce a net change in the energy withinthe control volume.

• The first two terms in Eq. (20) represent heat transfer and work interactions.

Q− W =dEsysdt

(21)

orδQ

δt− δW

δt=dEsysdt

orδQ− δW = dEsys

• In a process from state 1 to state 2 during time interval ∆t = t2 − t1 then:∫ 2

1δQ−

∫ 2

1δW =

∫ 2

1dEsys

orQ12 −W12 = (E2 − E1)sys (22)

and for a closed system undergoing a cycle:∮δQ−

∮δW =

∮dEsys = 0 (23)


2 Properties of Pure Substances

2.1 State Postulate

• The state postulate defines how many properties are required to specify a state.

The number of independent intensive thermodynamic properties required to com-pletely and uniquely specify the thermodynamic state of a homogeneous sub-stance is one more than the number of relevant, reversible modes of work.

Note:

1. Not all properties are independent (e.g. v = 1/ρ, and T and p at the boilingpoint of a substance).

2. The state postulate is for homogeneous substances, i.e. substances that are uni-form in physical and chemical structure.

• Most mechanical systems use simple compressible substances, for which pdV work isthe only significant work mode, therefore, only two independent intensive propertiesare required to specify a state.

– e.g. If T and v for water are known the state of the water may be determinedand other properties evaluated.

• The state postulate is used as a basis for relationships between difficult to measureproperties (e.g. internal energy, u) and those that are easily measured (e.g. T , p, V ).

2.2 Equilibrium Diagrams

• There is a relationship between the properties which define the state of any substance.

• Equilibrium diagrams are used to obtain a qualitative feel for the behaviour of sub-stances in thermal equilibrium.

• Consider a closed system (e.g. frictionless piston cylinder) containing 1 kg of water at0.1 MPa (i.e. 1 bar) and 20◦C. Heat is added to the system in an isobaric process.


• As heat is added to the the liquid water (state 1), the temperature of the water willincrease to 99.63◦C, and the volume will increase to 0.0010432 m3 (state 2).

• If more heat is added, the temperature will remain at 99.63◦C, but the volume willincrease as vapour forms (i.e. boiling or vaporization), (e.g. V = 1 m3, state 3).

• The temperature will remain at 99.63◦C until all of the water is vaporized, and thevolume has increased to 1.694 m3 (state 4).

• Continued heat addition will increase both the temperature and volume (e.g. T =500◦C, V = 3.565 m3, state 5).

• This process can be plotted on a T -v diagram for a constant pressure of p = 0.1 MPa.

• State 1 – Subcooled or Compressed Liquid

– any addition of heat will increase temperature and volume (slightly)

• State 2 – Saturated Liquid

– liquid at its boiling point

– any addition of heat vaporizes the water but does not change its temperature (ifpressure is constant)

– T = 99.63◦C, v = 0.0010432 m3/kg


• State 3 – Liquid-Vapour Saturation Region

– T = 99.63◦C, 0.0010432 m3/kg ≤ v ≤ 1.694 m3/kg

• State 4 – Saturated Vapour

– vapour (only) at its condensation point

– any addition of heat will increase temperature and volume

– T = 99.63◦C, v = 1.694 m3/kg

• State 5 – Superheated Vapour

– e.g. T = 500◦C, v = 3.565 m3/kg

• The temperature and pressure at which boiling and condensation occur are the satu-ration temperature, Tsat, and pressure, psat, respectively.

• State 1 is “subcooled”, because its temperature is below Tsat for this pressure (psat =0.1 MPa → Tsat = 99.63◦C). Similarly, it is “compressed”, because its pressure ishigher than the psat that corresponds to the given temperature (Tsat = 20◦C →psat = 2.339 kPa).

• The amount of water vapour present varies from 0% at the start of boiling to 100%at the start of superheating. The fraction of water in the vapour form is called thequality, x, (steam quality for water).

x ≡ mg

mf +mg(24)

where, mg is the mass of vapour, and mf is the mass of liquid.


• Similar experiments can be performed at different pressures, and the result is:

• As pressure increases, the length of the saturated liquid-vapour line decreases untilit reaches a point (the critical point or state), where the liquid and vapour statescoincide.

• At pressures above the critical pressure, there is no clear distinction between super-heated vapour and compressed liquid phases, and a phase change cannot occur at atemperature or pressure higher than the critical state values, therefore, the substanceis called a fluid.

• For water: Tcrit = 647.3 K, pcrit = 22.09 MPa (Table A-1, Moran and Shapiro).

• For air: Tcrit = 133 K, pcrit = 3.77 MPa (Table A-1, Moran and Shapiro), i.e. atstandard atmosphere conditions air is a superheated vapour.


• A similar experiment can be performed, but starting with the solid phase. The resultswill depend on whether the substance contracts on freezing (Fig. 3.2, Moran andShapiro) or expands on freezing (Fig. 3.1, Moran and Shapiro). For a substance thatcontracts on freezing:

• Note: each line separates two phases, except the triple line, which separates threephases. This occurs for all substances. The triple point for water is Tt = 0.01◦C,pt = 0.6113 kPa.

• The equilibrium values of p-v-T can be plotted on a 3-D plot (as in Figs.3.1 and 3.2 of Moran and Shapiro). It is obvious from these plots that a single equationof state cannot be derived → tables are used to evaluate properties.

• It is possible to project the p-v-T diagram to the p-T plane (see Figs. 3.1 and 3.2 ofMoran and Shapiro).


• The triple line becomes a point representing the infinite number of states at whichthe solid, liquid, and vapour states can co-exist.

• Note : dry ice sublimes at 1 atm, because the pressure is below the triple point value.(CO2: Tcrit = 216.6 K, pcrit = 516.6 kPa)

2.3 Evaluation of the Properties of Pure Substances

• The tables to be used in the evaluation of the properties of pure substances areincluded in Appendix A of Moran and Shapiro. Tables are presented for water (TablesA-2 to A-6), Refrigerant 22, R-22, (Tables A-7 to A-9), Refrigerant 134a, R-134a, (A-10 to A-12), Ammonia (Tables A-13 to A-15), and Propane (Tables A-16 to A-18).

• Tables are supplied for the following phases of water: saturated water (liquid andvapour), Tables A-2 and A-3; superheated water vapour, Table A-4; compressed liquidwater, Table A-5; and saturated water (solid and vapour), Table A-6.

• Note: The saturated liquid specific volume, vf , is multiplied by 103 in Tables A-2, 3,5, 6, 7, 8, 10, 11, 13, 14, 16 and 17. So, vf = 0.0010018 m3/kg for saturated liquidwater at 20◦C (Table A-2).

• The examples in the notes are for water, but the methods employed apply to anysubstance.

• The first step is to identify the phase(s) present → which table is to be used.

• Assume a pure substance (e.g. water) at a given temperature, pressure, and specificvolume, T1, p1, and v1, respectively.


• Superheated Vapour:

– if T1 > Tsat, assuming p1 is psat

– if p1 < psat, assuming T1 is Tsat

– if v1 > vg|T1 or v1 > vg|p1 (this test also applies to u, h, and s)

– if T1 > Tcrit

– Any two independent intensive thermodynamic properties are sufficient to fullydescribe the state of the substance.

• Compressed (or Subcooled) Liquid

– if T1 < Tsat, assuming p1 is psat (subcooled)

– if p1 > psat, assuming T1 is Tsat (compressed)

– if v1 < vf |T1 or v1 < vf |p1 (this test also applies to u, h, and s)

– if T1 < Tcrit when p1 = pcrit


• Liquid-Vapour Saturation Region

– psat and Tsat are not independent, therefore, another property is required (e.g.v, u, h, s or x)

– Note: vf ≤ v ≤ vg in the saturation region. This is also true for u, h and s.

– Obviously, the properties of mixtures with different qualities will not be equiv-alent, but the tables only supply properties for the saturated liquid and thesaturated vapour states.

– e.g. Determine the volume of a mixture of saturated liquid and vapour.

V = Vf + Vg (25)

= mfvf +mgvg (26)

Defining the total mass: m = mf +mg, and the quality x = mg/m:

V = m(1− x)vf +mxvg (27)

= m(vf + xvfg) (28)

orv = vf + xvfg (29)

where vfg = vg − vf .

– Note: this expression also applies for u, h, and s.


• e.g. Assuming water is the working fluid, determine the phase(s) present, and thedesired properties for the following states:

1. p = 100 kPa, T = 500◦C, v =?


2. T = 400◦C, v = 4.434 m3/kg, p =?

3. p = 1.0 MPa, T = 560◦C, h =?

4. T = 100◦C, p = 105 kPa, u =?

5. T = 120◦C, h = 1000 kJ/kg, u =?

6. p = 160 kPa, u = 1500 kJ/kg, T =?, x =?

2.4 Specific Heats and Latent Heats

• Using the state postulate (i.e. two independent intensive properties define the stateof a simple compressible substance) the internal energy can be written as a functionof temperature and specific volume (two easily measured quantities).

u = u(T, v)

The change in u can then be written as:

du =

(∂u

∂T

)v

dT +

(∂u

∂v

)T

dv

• The first term is defined to be the constant volume specific heat.

cv ≡(∂u

∂T

)v

(30)

then

du = cvdT +

(∂u

∂v

)T

dv (31)

and for an isochoric (constant volume) process:

du = cvdT (32)

which facilitates the evaluation of changes in u.

• Enthalpy may be written as a function of pressure and temperature:

h = h(p, T )

and the change in enthalpy is:

dh =

(∂h

∂T

)p

dT +

(∂h

∂p

)T

dp

• The first term is defined as the constant pressure specific heat:

cp ≡(∂h

∂T

)p

(33)

then

dh = cpdT +

(∂h

∂p

)T

dp (34)

and for an isobaric (constant pressure) process:

dh = cpdT (35)

which facilitates the evaluation of changes in h.


• The units of the specific heats are kJ/kg·K.

• The values of cp and cv are more strongly dependent upon temperature than pressure.

• The ratio of the specific heats is often used:

k ≡ cpcv

(36)

and it becomes of particular importance for processes involving ideal gases.

• Consider a closed system receiving heat in a constant volume process:

The 1st Law for this closed system is:

du = δq − δw

but δw = pdv = 0, therefore, du = δq or δq = du = cvdT .

• Consider a closed system receiving heat in a constant pressure process:

The work during the constant pressure process can be expressed as: δw = pdv = d(pv)

Therefore, the 1st Law can be written as:

du = δq − d(pv)

and rearranged to give:

δq = du+ d(pv)

= d(u+ pv) = dh

= cpdT


• Note: The amount of heat gained (or lost) by a system undergoing constant volumeor constant pressure processes can easily be evaluated using specific heats.

• If the internal energy and enthalpy of a substance can be defined as functions oftemperature only (i.e. u = u(T ), h = h(T )) then the changes in internal energy andenthalpy for that substance can be calculated using specific heats for any process (i.e.du = cvdT , dh = cpdT ). This becomes of particular significance for ideal gases.

• Latent heat is the amount of heat that must be added to or removed from a substanceduring a phase change at a constant temperature.

hfg – latent heat of vaporization

hig – latent heat of sublimation

hif – latent heat of fusion

• Sensible heat is used to change the temperature of a substance.

2.5 Ideal Gases

• It has been found from experiment that the p-v-T behaviour of gases can be approx-imated by the following equation (under certain conditions):

pv = RT (37)

where, p is the absolute pressure (kPa),

v is the specific volume (m3/kg),

R is the gas constant (kJ/kg·K),

T is the absolute temperature (K).

Note: When using the ideal gas law be careful of the units, and use absolute tem-perature and pressure.

• The ideal gas law can also be written on a mass basis (m in kg):

pV = mRT (38)

• It can also be written on a molal basis:

pv = RT (39)

where, v = V/n is the molal volume (m3/kmol),

R is the universal gas constant=8.314 kJ/kmol·K,

n = m/M is the number of moles,

M is the molar mass (kg/kmol), (Table A-1, Moran and Shapiro).

Multiplying both sides of Eq. (39) by n gives:

pvn =m

MRT

pV = mRT

so the gas constant is R = R/M .


• The ideal gas law is very simple to use (convenient):

e.g. Constant mass processp1V1T1

=p2V2T2

(40)

e.g. Constant pressure, constant mass process

v1T1

=v2T2

(41)

• The ideal gas law is an idealization of the behaviour of real gases at low pressures.The accuracy of this approximation increases for high temperatures and low molecularweight.

• The ideal gas law neglects intermolecular forces. Since the mean free path increasesfor increases in temperature, the intermolecular forces will decrease, and the approx-imation improves at high temperatures (a similar process happens at low pressures).

• A gas is more nearly ideal if p < pcrit and T > Tcrit. One check for ideal gas behaviouris to compare the given state to the critical state using the values in Table A-1, Moranand Shapiro. Air at 1 atm and 300 K is highly superheated, and the assumption ofideal gas behaviour is acceptable. Note: air at 300K can be assumed an ideal gas upto pressures of 10 MPa.

• Strictly, to verify ideal gas behaviour, one should check the compressibility, Z:

Z ≡ pv

RT(42)

Z ∼ 1 for ideal gas behaviour.

• Nonideal gas behaviour will result if:

1. The temperature is decreased (or pressure increased) the molecules become moredensely packed and attractive forces between the molecules become significant.

2. The molecules become too close, the electron clouds will interact, producingrepulsive forces.

3. The pressure is very high, then the space occupied by the molecules becomes afactor, and they can no longer be treated as point masses.

• Ideal gas behaviour can be assumed in the vapour phase of a pure compressible sub-stance, when the pressure and temperature are not near the critical values.

– At low pressures ideal gas behaviour can be assumed irrespective of temperature.

– Ideal gas behaviour can be assumed at temperatures double Tcrit.

• e.g. Can the following substances be considered ideal gases?

1. N2 at 30◦C, 3 MPa

2. CO2 at 30◦C, 3 MPa

3. H2O at 1300◦C, 3 MPa


4. H2O at 50◦C, 10 kPa

5. H2O at 30◦C, 10 kPa

Critical States (Table A-1, Moran and Shapiro):

N2: Tcrit = 126 K, pcrit = 3.39 MPa

CO2: Tcrit = 304 K, pcrit = 7.39 MPa

H2O: Tcrit = 647.3 K, pcrit = 22.09 MPa

1. T >> Tcrit, OK (highly superheated)

2. T ∼ Tcrit, but p << pcrit, OK

3. T >> Tcrit, p << pcrit, OK

4. Table A-2: Tsat = 50◦C, psat = 12.35 kPa, → superheated vapour, and p <<pcrit, OK (Note: be careful near the saturated vapour line)

5. Table A-2: Tsat = 30◦C, psat = 4.246 kPa, → compressed liquid, not OK

• For an ideal gas, it can be shown that u = u(T ), therefore, du = cvdT for any processof an ideal gas. Changes in internal energy can be evaluated as follows:

u2 − u1 =

∫ 2

1cv dT (43)

• Since h = u + pv = u + RT , then h = h(T ) for an ideal gas, and dh = cpdT for anyprocess of an ideal gas:

h2 − h1 =

∫ 2

1cp dT (44)

• Note: since h = u+RT , then:dh

dT− du

dT= R

substituting the expressions for the specific heats, Eqs. (30) and (33):

cp − cv = R (45)

Since k = cp/cv then

cp =k

k − 1R (46)

cv =R

k − 1(47)

Eqs. (45) through (47) are valid for any ideal gas. An ideal gas that has a constant kis called a perfect gas.

• If cp and cv are known, problems related to the evaluation of properties can be signif-icantly reduced.

• In general, cp and cv are functions of temperature:

– for a small ∆T , cp and cv may be assumed constant


– for a large ∆T , the temperature variation must be taken into account → molarzero-pressure specific heats (cp and cv) or tables

cp

R= α+ βT + γT 2 + δT 3 + εT 4 (48)

where values of the constants α, β, γ, δ, and ε for common gases are given inTable A-21 for the temperature range 300 to 1000 K

• The constant pressure specific heat is found from:

cp =cp

R

R

M(49)

• Equations (48) and (49) would then be substituted into Eq. (44) to determine thechange in enthalpy.

• To save time, use Tables A-22 and A-23, Moran and Shapiro, to evaluate changes inh, u and s for air, N2, O2, H2O, CO, CO2.

• These tables allow the evaluation of enthalpy through the following equation:

h(T ) =

∫ T

Tref

cp(T ) dT + h(Tref ) (50)

where Tref is an arbitrary reference temperature, and h(Tref ) is the enthalpy at thereference temperature. Tables A-22 and A-23 use a reference value of h(0 K) = 0kJ/kg. Therefore, the change in enthalpy in air as it changes temperature from 300K to 600 K is h(600 K) − h(300 K) where the enthalpy values are determined fromTable A-22 (i.e. 607.02− 300.19 = 306.83 kJ/kg).

• To obtain h and u from the h and u values given in Table A-23, divide h and u byM , the molar mass of the gas.

• The changes in enthapy and internal energy for gases with linearly varying specificheats (or gases undergoing processes with a small temperature difference) can beevaluated as follows:

h2 − h1 = cpav(T2 − T1) (51)

u2 − u1 = cvav(T2 − T1) (52)

where cpav and cvav are appropriate mean values of cp and cv determined from TableA-20.

• Table A-20 may also be used to determine the specific heats at a specified temperature(use linear interpolation if necessary).

• In summary, changes in enthalpy and internal energy should be evaluated using:

– Eqs. (73) and (52) when changes in temperature are small (i.e. ∆T < 200 K);

– Tables A-22 and A-23 for processes with ∆T > 200 K. Note: these tables arevalid for all processes, and the enthalpy and internal energy values are referencedto 0 kJ/kg at 0 K.


• Note: for monotonic gases (Ar, Ne, He, Xe) the specific heats are constants:

cp =5

2R (53)

cv =3

2R (54)

2.6 Polytropic Processes of Ideal Gases

• Using a p-v diagram, pressure may be written as a function of specific volume:

p = p(v)

and in some cases a polytropic equation may be written for the process:

pvn = c (55)

Equation (55) can be used to determine the pdv work of a process (see e.g. 2.1 inMoran and Shapiro):

w12 =

∫ 2

1p dv = p1v1 ln

v2v1

= p2v2 lnv2v1

(n = 1) (56)

w12 =

∫ 2

1p dv =

p2v2 − p1v11− n

(n > 1) (57)

Note: these expressions give specific work since specific volume was used (W12 =mw12).

• Polytropic equations are not restricted to ideal gases, however, further relations canbe derived using the polytropic equation and the ideal gas law.

• If pvn = c and pv = RT , then:

p

(T

p

)n= c′

or p = c′Tn

n−1 , therefore

p2p1

=

(T2T1

) nn−1

(58)

alsoT

vvn = c′′

or Tvn−1 = c′′, therefore:

v2v1

=

(T1T2

) 1n−1

(59)

• So, for an ideal gas undergoing a polytropic process:

pvn = c (60)

p/Tn

n−1 = c′ (61)

Tvn−1 = c′′ (62)

the value of the exponent n defines specific processes for an ideal gas:


n = 0 constant pressure1 constant temperature∞ constant volumek reversible and adiabatic (i.e. isentropic)

2.7 Incompressible Substances

• A substance with constant density (or specific volume) regardless of what happens tothe other properties is an incompressible substance.

• Many substances are assumed incompressibe, e.g. liquid water (Tables A-2 and A-5):

T (◦C) p (MPa) v (m3/kg) h (kJ/kg) u (kJ/kg)

100 0.1014 0.0010435 419.04 418.94100 5 0.0010410 422.72 417.52100 30 0.0010290 441.66 410.78

• For incompressible substances ∆V = 0, therefore:

du = cvdT +

(∂u

∂v

)T

dv

= cvdT (63)

and

u2 − u1 =

∫ 2

1cv dT (64)

Also

dh = d(u+ pv)

= du+ vdp+ pdv

= cvdT + vdp (65)

andh2 − h1 = u2 − u1 + vf |T1(p2 − p1) (66)

• To evaluate the properties of a compressed liquid at low pressures, use the the satu-rated liquid values at the given temperature(s), and use Eq. (66) to evaluate changesin enthalpy for large changes in pressure.

• Note: for liquid water v ∼ 10−3 m3/kg, therefore, for a 1 kJ/kg change in enthalpyin Eq. (66) a pressure difference of 1 MPa is required. The compressed liquid watertables in Moran and Shapiro begin at 2.5 MPa.

• For incompressible substances only one specific heat is defined c = cp = cv. Values ofc for some liquids and solids are given in Table A-19, Moran and Shapiro.


3 Conservation of Mass

3.1 Principle of the Conservation of Mass

• The principle of the conservation of mass can be written as follows:

Mass is a conserved property. It can neither be created nor destroyed; only itscomposition can be altered from one form to another.

• Or, the mass of a system must always be accounted for or conserved. Note: changesin chemical composition occur during chemical reactions, however, the mass of theconstituents and the final products must be the same.

3.2 Conservation of Mass for a Control Volume

• The conservation of mass statement can be written in the following form for a controlvolume (cv): Total rate at which

mass enters a cvacross its boundary


mass leaves a cvacross its boundary

=

Net rate of increaseof mass within the

control volume

or, in mathematical form: ∑

i

mi −∑e

me =dmcv

dt(67)

• This is the general statement of the conservation of mass equation.

• The total mass within a cv is defined as follows:

mcv =

∫Vρ dV (68)


• Since the velocity and density can vary over an inlet (or exit), the mass flow rate atan inlet (or exit) can be expressed as:

m =

∫AρVn dA (69)

where, A is the area of the inlet (or exit) and Vn is the velocity normal to the inlet(or exit), i.e. Vn = ~V · ~n, and ~n is the normal to the cv boundary.

• The mass conservation equation can then be rewritten as:∑i

(∫AρVn dA

)i

−∑e

(∫AρVn dA

)e

=d

dt

∫Vρ dV (70)

3.3 Forms of the Mass Conservation Equation

• Equation (70) is a general form of the mass conservation equation, but it requiresinformation regarding the variation of ρ within the cv and the variation of ρ and Vn atall inlets and exits. Often this information is unavailable, and simplfying assumptionsare made.

3.3.1 One-Dimensional (or Uniform) Flow

• Assuming all intensive properties are uniform with position (i.e. bulk averaged values)over each inlet and exit, and the given velocity is normal to the cv boundary, then:

m = ρAV (71)

and Eq. (70) can be written as:∑i

(ρAV )i −∑e

(ρAV )e =dmcv

dt(72)

• The assumption of uniform flow is appropriate in typical engineering applicationswhere flows are turbulent, because the velocity profile is full, and the properties arerelatively uniform over a cross-section.


3.3.2 Steady State

• When all properties of a control volume are independent of time the control volumeor system is said to be at steady state (e.g. a turbine or pump a suitably long timeafter start up).

• For steady state conditions, Eq. (67) can be reduced to:∑i

mi =∑e

me (73)

i.e. (mass in) = (mass out). If the flow is incompressible (ρ =const) mass conservationgives (volume in) = (volume out).

• The steady state assumption is a realistic and useful assumption for mechanical de-vices.

• Note: if all properties are independent of time, then dV/dt = 0, and there can be nopdV work in a device operating under steady state conditions.

3.3.3 Closed Systems

• By definition, a closed system does not have mass crossing its boundary, therefore,the mass conservation equation, Eq. (67), reduces to:

mcv = const (74)


3.3.4 Transient Analysis

• When a system operates under transient conditions (e.g. start up, shutdown, varyinginlet or exit conditions, varying environment) the mass within the cv can change.

• Usually, the change in conditions over a time interval ∆t = t2 − t1 are of interest.Integrating Eq. (67) over this time interval:

∑i

∫ t2

t1

mi dt−∑e

∫ t2

t1

me dt =

∫ t2

t1

dmcv

dtdt

gives: ∑i

mi −∑e

me = (m2 −m1)cv (75)

i.e. the net change in the mass in the cv over a time interval ∆t is equal to thedifference between the total mass entering and the total mass leaving the cv during∆t.


4 Conservation of Energy (1st Law of Thermodynamics)

4.1 Principle of the Conservation of Energy

• The principle of the conservation of energy provides the foundation of thermodynam-ics, as it supplies the basic framework required to study the relationships among work,heat transfer, and the various forms of energy.

• The principle of the conservation of energy can be written in the following form:

Energy is a conserved property. It can neither be created nor destroyed; only itsform can be altered from one form of energy to another.

• This principle is so fundamental to the study of thermodynamics that it is called the1st Law of Thermodynamics.

• The general form of the 1st Law will contain the following energies and energy trans-fers:

1. Energy transfer due to differences in pressure and temperature (i.e. work andheat transfer), which are only identifiable at a system boundary (e.g. a systemcan do work on its surroundings).

2. Energy related to the mass of a substance (i.e. U , Ek, Ep). This would includeenergy due to the mass within a control volume and that due to the mass beingtransported across the control volume boundary at inlets and exits.

• The energy equation (or 1st Law) may be written in the following form: Total rate at whichenergy enters a cv

across its boundary


energy leaves a cvacross its boundary

=

Net rate of increaseof energy within the

control volume

(76)

i.e. if an amount of energy enters a control volume, then the same amount of energymust leave the control volume, or it will produce a net change in the energy withinthe control volume.

• The first two terms in Eq. (76) represent heat transfer, work interactions, and energytransport across the control volume boundary due to mass flow.


4.2 A Mathematical Statement of the 1st Law of Thermodynamics

• To obtain a mathematical statement of Eq. (76), consider the following control vol-ume:

• The rate at which mass flows across an elemental area dA is ρVndA, therefore, therate at which energy is transported across the elemental area dA by this mass flow iseρVndA. This expression can be integrated over all inlets to give: Total rate of energy

transport due tomass entering the cv

=∑i

∫Ai

e(ρVn) dA (77)

Similary, the energy transport by mass flow out of the control volume can be writtenas: Total rate of energy

transport due tomass leaving the cv

=∑e

∫Ae

e(ρVn) dA (78)

• The mass within a control volume is∫V ρ dV , therefore, the total energy contained

within a control volume is∫V eρ dV . We are interested in the time rate of change of

the total energy within a control volume:[Net rate of increase of the totalenergy within a control volume

]=

d

dt

∫Veρ dV (79)

• Using Eqs. (77), (78) and (79), Eq. (76) can be written as follows:

Qcv − W +∑i

∫Ai

e(ρVn) dA−∑e

∫Ae

e(ρVn) dA =d

dt

∫Veρ dV (80)

where

Qcv = δQ/δt =

{Net rate of heat transfer to thecontrol volume from the surroundings


W = δW/δt =

Net rate of work leaving the controlvolume due to all work interactions(e.g. pdV , shaft, electric, flow work)

Note: both the work and heat transfer are assumed in the positive directions.

• Flow work is the work that must be performed to cause mass to flow across controlvolume boundaries. Due to pressure differentials, mass must be pushed into a controlvolume at inlets, and pushed out of the control volume at exits. This work is usuallyseparated from other forms of work.

• Consider the element of fluid, shown below, that is being pushed into a control volume.

• The work done pushing this element of fluid into the control volume is ~F · ~s, and therate at which this work is done can be written as:

~F · d~sdt

= ~F · ~V = FnVn = pVndA

since Fn = pdA. With this definition of flow work, the net rate at which flow work isdone by the control volume can be defined:[

Net rate offlow work

]= −

∑i

∫Ai

pv(ρVn) dA+∑e

∫Ae

pv(ρVn) dA

The flow work at the inlets has a negative sign, because work is being done on thesystem at an inlet. Note: each term has been multiplied by ρv (= 1).

• Defining:W = Wcv + Wflow

where Wcv represents all reversible and irreversible forms of work except flow work,allows the energy equation to be rewritten in the following form:

Qcv − Wcv +∑i

∫Ai

(e+ pv)(ρVn) dA−∑e

∫Ae

(e+ pv)(ρVn) dA =d

dt

∫Veρ dV

Bute+ pv = u+ pv + ek + ep = h+ ek + ep


therefore, the energy equation, or 1st Law of thermodynamics, can be written asfollows:

Qcv−Wcv+∑i

∫Ai

(h+ek+ep)(ρVn) dA−∑e

∫Ae

(h+ek+ep)(ρVn) dA =d

dt

∫Veρ dV

(81)for a thermodynamic system, or control volume, consisting of a simple compressiblesubstance.

4.3 The 1st Law of Thermodynamics for a Closed System

• By definition, no mass crosses the boundary of a closed system, therefore conservationof mass gives:

msys = const (82)

and the 1st Law (or conservation of energy) gives:

Q− W =dEsysdt

(83)

orδQ

δt− δW

δt=dEsysdt

orδQ− δW = dEsys

• In a process from state 1 to state 2 during time interval ∆t = t2 − t1 then:

Q12 −W12 = (E2 − E1)sys (84)

and for a closed system undergoing a cycle:∮δQ−

∮δW =

∮dEsys = 0 (85)

4.4 Forms of the 1st Law for a Control Volume (or an Open System)

• Since an open system, or control volume, can have mass and energy crossing its bound-ary the general form of the 1st Law, Eq. (81), must be used. But some simplifyingassumptions are usually employed to ease the use of this equation.

4.4.1 One-dimensional (or Uniform) Flow

• If one-dimensional flow is assumed, then the fluid properties (ρ, Vn, h, ek, and ep) areuniform at each inlet and exit, therefore:∫

Ai

(h+ ek + ep)(ρVn) dA = mi(h+ ek + ep)i

and the 1st Law, Eq. (81) becomes:

Qcv − Wcv +∑i

mi(h+ ek + ep)i −∑e

me(h+ ek + ep)e =dEcvdt

(86)

where (h+ ek + ep) are evaluated at each inlet and exit and multiplied by the appro-priate mass flow rate.


4.4.2 One-Dimensional, Steady State Flow

• Under steady state conditions all properties are independent of time, therefore, Eq.(86) reduces to:

Qcv − Wcv +∑i

mi(h+ ek + ep)i −∑e

me(h+ ek + ep)e = 0 (87)

and there will be a balance between the total rate of transport of energy into and outof the control volume.

• Equation (87) is the form of the 1st Law most often used in this course.

• Since dV/dt = 0 there will be no pdV work. Typically, mechanical engineers dealwith rotating machinery, and under steady state conditions the only significant modeof work is shaft work. This is why the work in steady state control volume (or opensystem) analyses is usually termed shaft work.

4.4.3 Transient Analysis of Open Systems (or Control Volumes)

• During the transient operation of a device (e.g. start up, shutdown, varying load ona motor, etc.) the time rate of change of properties becomes significant. Usually, thechange over a time interval ∆t = t2 − t1 is of interest, therefore, the energy equationis integrated over this time interval (as for the mass conservation equation).

• Since Qcv = δQcv/δt, Wcv = δWcv/δt, and m = dm/dt, these three terms can besubstituted into the 1st law for uniform flow, Eq. (86), and the resulting equation canbe integrated over the time interval ∆t = t2 − t1:∫ t2

t1

δQcvδt

dt−∫ t2

t1

δWcv

δtdt +

∑i

∫ t2

t1

dmi

dt(h+ ek + ep)idt

−∑e

∫ t2

t1

dme

dt(h+ ek + ep)edt =

∫ t2

t1

dEcvdt

dt

which can be rewritten as:

Q12−W12 +∑i

∫ t2

t1

(h+ek+ep)i dmi−∑e

∫ t2

t1

(h+ek+ep)e dme = (E2−E1)cv (88)

• Note:

1. The fluid flow entering and/or leaving the control volume is usually assumeduniform (1D), and the flow rate is often assumed constant with time.

2. Perfect mixing (i.e. the state of the substance throughout the control volume isthe same (except at inlets and exits)) is often assumed. This is not the same asthe assumption used for uniform flow control volume analyses, where the stateis uniform at a location, but can vary from location to location.


4.5 Applications of the 1st Law of Thermodynamics

4.5.1 Potential and Kinetic Energies

• Potential energy is one of the forms of energy included in the 1st Law, but it is oftenneglected.

• The change in potential energy is defined as ∆ep = g∆z. To give a change inpotential energy of ∆ep = 1 kJ/kg will require a change in elevation of ∆z =(1000J/kg)/(9.81m/s2) = 102 m.

• For a change in enthalpy of ∆h = 1 kJ/kg for an ideal gas (∆h = cp∆T ) or anincompressible substance (∆h = cav∆T ):

Substance T (K) cp (kJ/kg· K) Required ∆T (◦C)

Air 300 1.005 0.9951000 1.142 0.876

Liquid 273 4.217 0.237H2O 373 4.218 0.237

Superheated 500 1.955 0.512H2O 1500 2.614 0.383

• Since ∆z is often less than 10 m, but ∆T is significant, then ∆h >> ∆ep, and ∆epis neglected. Note: this would not be true for hydroelectric plants, or when pumpingfluids between reservoirs, where ∆ep is the most significant change in energy.

• The change in kinetic energy is ∆ek = (V 22 − V 2

1 )/2. The required change in velocityto give ∆ek = 1 kJ/kg depends on the magnitudes of the two velocities (i.e. large Vrequires small ∆V , and vice versa). For example, if V1 = 10 m/s → V2 = 46 m/s,and if V1 = 100 m/s → V2 = 110 m/s. So whether kinetic energy should be neglectedrelative to changes in enthalpy is problem dependent. Often, ek is neglected due toinsufficient information, or to obtain a quick solution (i.e. no need to determine flowareas, and velocity profiles).

4.5.2 Turbines

• Turbines are used to convert the energy of a working fluid (e.g. steam, liquid water,combustion gases) to rotating shaft work.

• In a steam turbine, the decrease in enthalpy of the steam as it flows through theturbine is used to increase the kinetic energy of the steam, and rotate a shaft.

• In a hydroelectric plant, the decrease in potential energy of the water is used toincrease the kinetic energy of the water, and rotate a shaft.


• Steam turbine:

– Consider steady state operation, and one-dimensional (uniform) flow in a steamturbine with one inlet and one outlet.

– Conservation of mass will give:

mi = me = m (89)

– The 1st Law of thermodynamics applied to the working fluid (steam) gives:

Qcv − WT + mi(h+ ek + ep)i − me(h+ ek + ep)e = 0 (90)

– In a real steam turbine, ∆z is small (∆eP → 0), and the turbine casing is wellinsulated (Qcv → 0), therefore, the 1st Law can be written as:

WT = mi

(hi +

V 2i

2

)− me

(he +

V 2e

2

)or

WT = m (hi − he) + m

(V 2i

2− V 2

e

2

)(91)

– If the inlet and exit velocities are known, they should be included for an accuratecalculation, although in most cases ∆ek << ∆h, and over 90% of ∆h is convertedto useable power. If the velocities are unknown or neglected, the 1st Law reducesto:

WT = m (hi − he) (92)

– To evaluate changes in h, use the tables or ideal gas relations (if the workingfluid is a superheated vapour, including the relations for adiabatic processes ofan ideal gas (when Qcv is neglected)).

– Ideally, the working fluid should not condense within a turbine, as the resultingliquid droplets can cause severe blade erosion.

– Turbines in real steam plants often have more than one exit. Steam that isbled off may be used to preheat water going to a boiler, or in another process(cogeneration).


– Consider steady state, one-dimensional flow in the turbine shown below:

– Conservation of mass gives:m1 = m2 + m3 (93)

– Assuming Q = ∆ep = ∆ek = 0, the 1st Law will reduce to:

WT = m1h1 − m2h2 − m3h3 (94)


– Multi-stage turbines are often used, as this allows for more efficient blade designas the pressure drops through the turbine. The turbines at Holyrood have threestages. Consider the three stage turbine with reheat as shown below:

– In this turbine, steam is bled form the high pressure (HP) stage to preheat thewater as it returns to the boiler. Before entering the intermediate pressure (IP)stage, the steam exiting the HP stage is reheated in the boiler to increase its Tat constant p to give a higher h4 → the potential for a larger ∆h in the IP stage.Note: T4 will not be as high as T1, because steam, not liquid water, will be theworking fluid.

– Conservation of mass for this turbine will give:

m1 = m2 + m3 (95)

m3 = m4 (96)

m4 = m5 (97)

m5 = m6 (98)

– Assuming Qcv = ∆ep = ∆ek = 0 for each stage of the turbine gives:

WHP = m1h1 − m2h2 − m3h3 (99)

WIP = m4h4 − m5h5 (100)

WLP = m5h5 − m6h6 (101)

WT = WHP + WIP + WLP (102)

– Or, using a cv boundary around the three turbine stages and the reheater:

m1 = m2 + m6 (103)

WT = m1h1 − m2h2 − m6h6 + Q (104)


4.5.3 Pumps and Compressors

• Pumps and compressors use a work input to increase the pressure of liquids and gases,respectively.

• Positive displacement compressors include reciprocating (piston-cylinder), rotary screw,and Roots types.

• Axial flow compressors are used in gas turbines (may be multi-staged).

• Centrifugal pumps inject liquid at the center and it is accelerated by vanes.

• Assume one-dimensional flow and steady state operation of the pump and compressorshown below.

• Conservation of mass gives:mi = me (105)

• The 1st Law of thermodynamics applied to the working fluid is:

Qcv − WP + mi(h+ ek + ep)i − me(h+ ek + ep)e = 0 (106)

• Note: in this equation the work and heat transfer are assumed in the positive direc-tions. I prefer to write the 1st Law in this form and include the signs in Qcv and WP

(i.e. I should expect Qcv and WP to be less than 0).

• In a real compressor ∆z is small, therefore, ∆ep is negligible.

• When a gas is compressed there can be a significant increase in temperature (fromthe ideal gas law), therefore, unless the compressor is insulated Qcv can be significant.When the pressure of a liquid is increased, the temperature increase is usually small,and Qcv is also small. Another reason Qcv may be small is due to the small surfacearea of a pump (thermal resistance ∝ 1/A). So Qcv is often neglected.


• The 1st Law then becomes:

WP = mi

(hi +

V 2i

2

)− me

(he +

V 2e

2

)= m

(hi − he +

V 2i

2− V 2

e

2

)(107)

• Often Vi is very small (e.g. pumping from a reservoir) and it is neglected. Also, ∆ekis often small relative to ∆h (problem dependent), or there is insufficient informationto calculate ∆ek, and, therefore, it is neglected.

• The evaluation of ∆h is substance dependent, i.e. tables or ideal gas relations may beused.

4.5.4 Nozzles and Diffusers

• Nozzles and diffusers are used to change the flow characteristics of a fluid (i.e. velocity).They are converging and diverging ducts, and are commonly used in jet engines, windtunnels, etc.

• Note: supersonic flows through converging and diverging ducts demonstrate oppositebehaviour to subsonic flows.

• Consider the steady one-dimensional flow through an insulated diffuser. The 1st Law:

Qcv − Wcv + mi(h+ ek + ep)i − me(h+ ek + ep)e = 0 (108)

reduces to:

hi +V 2i

2= he +

V 2e

2(109)


• A pressurized reservoir is used in a supersonic wind tunnel.

• The 1st Law for this situation becomes:

hi − he =V 2e

2(110)

4.5.5 Throttling

• If a fluid at a certain pressure in a pipe is required at a lower pressure downstream, therequired change in pressure may be achieved by using a flow restriction (e.g. a valve).This is called throttling, and the change in pressure is a measure of the throttlingprocess.

• Consider steady state one-dimensional flow through a valve:

• The 1st Law:

Qcv − Wcv + mi(h+ ek + ep)i − me(h+ ek + ep)e = 0 (111)

can be reduced to:hi = he (112)

since ∆ep = 0, ∆ek = 0, Wcv = 0 (no shaft work), and Qcv = 0 (small surface area).

• A throttling process is isenthalpic. For an ideal gas ∆h = cp∆T , therefore, thethrottling process for an ideal gas is also isothermal.


• For a real gas, however, ∆T 6= 0.

• The inversion curve is the point at which µJ = (∂T/∂p)h = 0 (i.e. ideal gas behaviour),and µJ is called the Joule-Thomson coefficient.

• A throttling process occurs in the direction of decreasing pressure, therefore, if a gasis throttled to the left of the inversion curve it is cooled, and if it is throttled to theright of the inversion curve it is heated.

• Gases can be liquefied by throttling to the left of the inversion curve.

• Throttling is used in refrigeration and air conditioning cycles to cool the refrigerant.


4.5.6 Heat Exchangers

• A heat exchanger is used to provide efficient heat transfer between two fluids.

• Three common types of heat exchangers are a finned tube (e.g. car radiator), a doublepipe exchanger, and a shell and tube exchanger (e.g. water-water exchanger used forthe heating system in this building).

• Consider steady state one-dimensional flow of the counterflow double pipe heat ex-changer show above. Conservation of mass gives:

m1,i = m1,e = m1 (113)

m2,i = m2,e = m2 (114)

• In a real heat exchanger ∆ep = ∆ek = Wcv = 0. The 1st Law applied to each fluidgives:


Q1 = m1(he − hi)1 (115)

Q2 = m2(he − hi)2 (116)

Note: if all of the heat from fluid 1 is assumed to be absorbed by fluid 2, and there isno heat loss from the external shell of the exchanger Q2 = −Q1.

• e.g. Given the mass flow rates and the inlet and exit conditions for fluid 1 and theinlet conditions for fluid 2, find T2,e?

– Using the two cvs above, and Eqs. (154) and (155):

1. Determine h1,e and h1,i → Q1.

2. Determine h2,i and use it with Q1 → h2,e, then use h2,e and another propertyto identify state (2, e) → T2,e.

i.e. a two step solution procedure.

– For an alternate (quicker) solution, place a cv boundary around the whole ex-changer, and the 1st Law becomes:

m1(hi − he)1 + m2(hi − he)2 = 0 (117)

and solve directly for h2,e.

– This is an example of how the correct choice of a control volume can speed upthe solution process.


4.6 Applications of the 1st Law to Practical Thermodynamic Cycles

• The 1st Law can be applied to the individual components of a cycle, or to the cycle asa whole. This section will study the steam power cycle and the vapour-compressionrefrigeration cycle.

4.6.1 Steam Power Cycle

• The simplest form of steam power cycle (or steam plant) consists of four devices: (1) aboiler, where the working fluid is vaporized and superheated; (2) a turbine, where thechange in enthalpy of the working fluid is used to produce shaft work; (3) a condenser,where the working fluid is condensed from a vapour to a liquid; and (4) a pump whichis used to increase the pressure of the condensate before it enters the boiler.

• The 1st law can be applied to each of these components (open systems) or to thewhole power plant or cycle (closed system). For the whole cycle:∮

δQ =

∮δW (118)

orQB + QC = WT + WP (119)

with appropriate signs (i.e. QC < 0 and WP < 0). The powers and heat transfer ratescan be obtained from 1st Law analyses of the individual components.

• The thermal efficiency is used as a measure of the efficiency of the steam plant (andall power cycles):

ηth =Wnet

Qin(120)

i.e. the net rate of work done by the system on the surroundings divided by the totalrate of heat transferred to the system.


• For the simple steam plant shown above:

ηth =WT + WP

QB(121)

with appropriate signs (i.e. WP < 0).

• Applying the 1st Law to the whole cycle:

QB + QC = WT + WP = Wnet

then

ηth =QB + QC

QB= 1 +

QC

QB< 1 (122)

since QC < 0.

• Note: a simple way of remembering efficiency is:

Efficiency =Desired Effect

Cost(123)

• To improve the efficiency of a steam plant one could use feedwater heaters, multi-staging, reheat, and of course efficient pumps, condensers, burners, blades...

4.6.2 Vapour-Compression Refrigeration Cycle

• The vapour-compression refrigeration cycle is used in refrigerators, air conditioners,and heat pumps. The simplest cycle consists of a condenser, a throttling valve, anevaporator, and a compressor.

• The refrigerant leaves the compressor as a superheated vapour.


• The superheated vapour is condensed to a liquid in the condenser (QC to the sur-roundings, e.g. the grid on back of a refrigerator).

• The flow exiting the condenser and entering the throttling valve is a compressed liquid.In the throttling valve, the pressure is reduced, but the process is isenthalpic. Therefrigerant leaves the throttling valve as a low pressure, low temperature mixture ofsaturated liquid and vapour.

• In the evaporator, heat (QE) is removed from the region to be cooled, and used toevaporate the refrigerant, which then enters the compressor. In the compressor, therefrigerant is further superheated (i.e. compression of a gas leads to an increase intemperature).

• The refrigeration cycle is similar to a steam plant run in reverse, where the throttlingvalve replaces the pump, and the compressor replaces the turbine.

– A power plant produces work at the expense of heat transfer to the working fluid.

– A refrigeration cycle produces heat transfer from a cool region to a warmer regionat the expense of work done on the fluid.

• To measure the efficiency (Desired Effect/Cost) of a refrigeration cycle (e.g. refriger-ator or air conditioner) one uses the Coefficient of Performance (COP):

β = − QE

WCP

(124)

where QE is the desired effect (rate of cooling), and WCP is the cost (power suppliedto the compressor, < 0).

• Note: unlike ηth, β can be > 1.

• Applying the 1st Law to the whole cycle:

QE + QC = WCP (with appropriate signs) (125)

so

β =−QE

QE + QC= − 1

(QC/QE) + 1(126)

• A heat pump provides heat transfer to a high temperature region at the expense of awork input, therefore:

γ =QC

WCP

=QC

QE + QC=

1

(QE/QC) + 1> 1 (127)

i.e. (The heat transfer rate to the warm area from the condenser (< 0))/ (The powerrequired by the compressor (< 0))


5 The Second Law of Thermodynamics

5.1 Introduction

• Conservation of mass and the 1st Law of thermodynamics may be used to providemass and energy balances for processes or cycles, but they cannot be used to determineif a process is possible.

• For example, by application of conservation of mass and energy, it is possible to provethat a hot cup of coffee can be further heated by placing the cup in room air. Toprove that this is impossible requires use of the 2nd Law of thermodynamics.

• The 2nd Law of thermodynamics can be used to:

1. Determine if a process is possible.

2. Determine the direction of a process.

3. Define maximum efficiencies and coefficients of performance, and help identifywhy these cannot be attained.

4. Define a thermodynamic temperature scale that is independent of the substanceused to measure temperature.

5. Define the property entropy.

5.2 Reversible and Irreversible Processes

• A process is internally reversible if a system proceeds through a series of equilibrumstates, and if the direction of the process is reversed at any point in the process, thesystem can be returned to its initial equilibrium state without leaving any permanentchange in the system. (Idealized process that never occurs in nature.)

• Consider the slow expansion of a gas in a frictionless piston-cylinder:

– The process runs infinitely slowly, therefore, there are no ∆ek losses.


– During the expansion and (compression):

du = δq − δw

and there is sufficient time for heat to flow in to maintain the temperature of thesystem and its surroundings.

– If the process is reversed, the same amount of work is done, and the same amountof heat can flow out, therefore, the process will retrace the same series of p-v-Tsteps, and it is reversible.

• Consider the rapid expansion of a gas in a piston-cylinder.

– When the pin is removed the piston will rise rapidly, and slam into the stops(dissipate ek).

– Note: W12 = patm∆V and W23 =∫ 32 p dV , but p > patm, therefore, |W12| < |W23|

– Since du = δq − δw, and δw is different for expansion and compression, then, δqwill be different for both processes, and the process is irreversible.

• Reversibilty requires no dissipative effects (e.g. friction, drag, finite heat transfer),and a slow (or infinitesimal process).

• A totally reversible process requires an internally reversible process and interactionsbetween the system and its surroundings must also be reversible.

• Factors that make a process irreversible are:

1. Friction (e.g. fluid dynamic drag)

– A body is heated by friction, therefore, to return it to its initial conditionsrequires heat transfer to the surroundings, i.e. not totally reversible.


2. Unrestrained expansion

– Consider, a pressurized gas and a vacuum separated by a membrane. Whenthe membrane is broken, the gas will rapidly expand. To return the gas toits initial conditions requires that it be compressed (work), and cooled (heattransfer to the surroundings), i.e. not totally reversible.

3. Heat transfer through a finite temperature difference

– Consider heat transfer between a high temperature body and a low temper-ature body.

– To restore the bodies to their initial temperature will require a refrigerator(work from, and heat transfer to the surroundings), i.e. irreversible.

– A reversible heat transfer process can be defined. Heat transfer will approachreversibilty as ∆T approaches zero, therefore, a reversible heat transfer pro-cess (quasi-equilibrium heat transfer) is one in which heat is transferredthrough an infinitesimal ∆T and will require infinite time and area, i.e. notpractical.

4. Mixing of different substances

– If O2 and N2 are separated by a membrane, and the membrane is broken,the gases expand and mix. To return the system to its initial state requirescompression, cooling, and work to separate the molecules.

5. i2R, hysterisis, combustion . . .

• A thermal energy reservoir is a body that remains at the same temperature re-gardless of how much heat is transferred to it (e.g. the ocean and atmosphere). Thedefinition of a reservoir is relative, i.e. the magnitude of the heat transfer is smallrelative to the thermal mass (mcv) of the body (e.g. cooling lead shot in a bucket ofoil). This leads to the concept of the heat source and sink.

5.3 Statements of the 2nd Law of Thermodynamics

• The 2nd Law is a generalization of the consequences of the principle that heat flowsfrom a high temperature to a low temperature. The initial development was throughthe analysis of the efficiency of heat engines by Carnot.

• The two statements of the 2nd Law are due to Clausius, and Kelvin and Planck:

It is impossible to construct a device that will operate in a cycle and produce noother effect on the surroundings than:

the transfer of heat from a low temperature body to a high temperaturebody. Clausius

the raising of a weight and the interchange of heat with a single thermalreservoir. Kelvin-Planck

the conversion of heat transfer to an equivalent amount of net positive work.Kelvin-Planck

• Both the Clausius and Kelvin-Planck statements of the 2nd Law are equivalent, andboth are negative statements, therefore, they cannot be proved.


• The Clausius statement of the 2nd Law applies to refrigeration, and heat pumpcycles.

– A refrigerator is not a violation of the 2nd Law as work is required in the com-pressor, and the heat transfer in the evaporator and condenser always occursfrom high to low temperatures.

– The Clausius statement implies that β <∞ and γ <∞.

• The Kelvin-Planck statement of the 2nd Law applies to heat engines.

– A heat engine is a device that operates in a cycle and produces net positive workdue to heat transfer to the device (e.g. steam power plant).

– A temperature differential is required to transfer heat, therefore, a heat enginerequires both high (TH) and low (TC) temperature reservoirs (i.e. heat transferto the engine from TH and heat transfer from the engine to TC).


– It would be impossible to build a heat engine with ηth = 1 since all of the heattransferred to the device would be converted to work→ violation of the 2nd Law(e.g. the heat rejected in the condenser of a steam power plant).

– If an ideal gas in a frictionless piston-cylinder is heated to raise a piston isother-mally, then Q12 = W12 (i.e. work is done with the exchange of heat from onethermal reservoir). But this process is not a cycle, however, and to lower the pis-ton (complete the cycle) heat has to be removed to a low temperature reservoir.

• The Kelvin-Planck and Clausius statements of the 2nd Law are equivalent. Thiscan be proven by proposing violations of the two 2nd Law statements. For example,propose a violation of the Kelvin-Planck statement (i.e. ηth = 1):

• This proposal can be converted to a cyclic device having no effect on the surroundingsother than the transfer of heat from low to high temperature reservoirs (i.e. a violationof Clausius).


• Propose a violation of the Clausius statement (i.e. β =∞):

• Since there is no net heat transfer to the low temperature reservoir, a cyclic devicethat produces net positive work with exchange of heat with a single thermal reservoirhas been created (i.e. violation of Kelvin-Planck).

• A violation of one statement of the 2nd Law leads to a violation of the other statement,therefore, the two statements are equivalent.

• Both the Clausius and Kelvin-Planck statements of the 2nd Law show the impossibilityof creating a perpetual motion machine of the second kind.

– A perpetual motion machine of the first kind would create work from nothing orcreate mass energy (i.e. violation of the 1st Law).

– A perpetual motion machine of the second kind would violate the 2nd Law.

– A perpetual motion machine of the third kind would have no friction, therefore,it would run indefinitely, but produce no work.


5.4 Carnot Cycle

• The Carnot cycle is a hypothetical reversible heat engine developed by Carnot. TheCarnot cycle operates between two thermal reservoirs, and each process in the cycleis reversible, and, therefore, the whole cycle may be reversed. It is an ideal cycle thatis used to determine maximum efficiencies and to derive corollaries to the 2nd Law.

• A Carnot heat engine consists of the following reversible processes:

– 1→2: A reversible isothermal process in which heat flows from a high temperaturesource to a working fluid (at a constant temperature infinitesimally less than TH).

– 2→3: A reversible adiabatic process during which the working fluid expandsthrough a turbine to produce positive work output.

– 3→4: A reversible isothermal process in which heat flows from the working fluidto a low temperature reservoir. The temperatures of the fluid and sink areconstant and of infinitesimal difference.

– 4→1: A reversible adiabatic process during which the temperature of the workingfluid is raised back to TH , and work is supplied to the fluid to compress it backto state 1.


5.5 Corollaries of the 2nd Law

5.5.1 Corollary 1

The thermal efficiency of all reversible heat engines are the same if the enginesoperate between the same two thermal energy reservoirs. (Carnot)

• To prove this corollary, consider two reversible heat engines operating between thesame thermal reservoirs.

• Propose a contradiction of this corollary, e.g. WA = WB, but QH,A > QH,B therefore|QC,A| > |QC,B|.

• Then

ηth,A < ηth,BW

QH,A<

W

QH,B

• Heat engine A can be reversed to give a refrigerator that can be powered by heatengine B.

• This can be rearranged to create a cyclic device which causes heat transfer from lowto high temperatures with no other effect on the surroundings, i.e. a violation of theClausius statement of the 2nd Law. Therefore, the thermal efficiencies of all reversibleheat engines operating between the same two thermal reservoirs are equivalent.


5.5.2 Corollary 2

The thermal efficiency of a reversible heat engine is greater than that of an irre-versible heat engine, when both engines operate between the same two thermalenergy reservoirs. (Carnot)

• Consider the reversible (R) and irreversible (I) heat engines operating between thesame thermal energy reservoirs. Propose a violation of the 2nd corollary, i.e. ηth,I >ηth,R. (Both engines receive QH from TH , therefore |QC,I | < |QC,R|.)

• The result is a cyclic device that produces net work while exchanging heat with onlyone reservoir, a violation of the Kelvin-Planck statement of the 2nd Law.

• Maximum Efficiencies

– The Kelvin-Planck statement limits ηth to less than one, but what is the maxi-mum efficiency?

– The maximum efficiency would be obtained with a reversible heat engine (Corol-lary 2), and it would be the same for any reversible heat engine operating betweenthe same thermal energy reservoirs (Corollary 1).

– Therefore ηth,R = f(TH , TC).

– Since,

ηth =Wnet

QH=

QH +QCQH

= 1 +QCQH

(QC < 0)

then

ηth,R = 1 +QCQH

= f(TC , TH)(QCQH

)R

= g(TC , TH)


Lord Kelvin proposed: (QCQH

)R

= −TCTH

then ηth,R, the Carnot efficiency, is:

ηth,R = 1− TCTH

(128)

where the temperatures are absolute. The Carnot efficiency cannot be attaineddue to finite rate processes, finite temperature differences, friction, and otherdissipative losses.

– e.g. for a steam plant operating between high and low temperature reservoirs at1000◦C and 5◦C, respectively, ηth,R = 0.78.

– The maximum coefficients of performance are:

βR = − 1

(QH/QC)R + 1= − 1

1− TH/TC=

TCTH − TC

(129)

γR =1

1 + (QC/QH)R=

1

1− TC/TH=

THTH − TC

(130)

5.5.3 Corollary 3

A temperature scale can be defined that is independent of the properties of thesubstance used for the measurement of temperature.

• The proof of this corollary uses Corollary 1 (i.e. all reversible heat engines, no matterwhat the working fluid, produce the same thermal efficiency when operating betweenthe same thermal reservoirs).

• From the derivation of maximum efficiencies:(QCQH

)R

= −TCTH

(QC < 0) (131)

This expression is universally accepted as the definition of the thermodynamic tem-perature scale.

• One reason for the choice of this linear function is because temperature defined inthis manner is fortuitously the same as the (absolute) ideal gas temperature, andtemperatures measured using this scale are identical to those measured with a constantvolume ideal gas thermometer.

• The triple point of water was defined as Tt = 273.16 K (1954), so

T = −273.16Q

Qt

where Qt is the finite amount of heat exchanged with a reservoir at 273.16 K, and Qis a finite amount of heat exchanged with a reservoir at T .


• An interesting consequence of this temperature scale (defined through the 2nd Law)is that it is impossible to use a cyclic device to cool something to 0 K!

– Since Q = 0 when T = 0 K, then a heat engine operating between reservoirs at0 K and Tt would produce net work while rejecting zero heat to the low temper-ature reservoir. This is a violation of the Kelvin-Planck statement. Another wayof proving this is to use the expression for maximum efficiency.

ηth,R = 1− TCTH

< 1

Therefore, TC > 0, i.e. it cannot be 0 K (10−8K was reached in 1984).

5.5.4 Corollary 4 – The Clausius Inequality

Whenever a system undergoes a complete cycle, the cyclic integral of δQT around

the cycle is equal to or less than zero (i.e.∮ δQ

T ≤ 0).

• Consider a Carnot cycle heat engine operating between two constant temperaturereservoirs:

– Since there is positive work produced:∮δQ = QH +QC > 0 (QC < 0)

– Since TH and TC are constant:∮δQ

T=QHTH

+QCTC

(132)

where QH occurs at TH and QC occurs at TC . The thermodynamic temperaturescale:

QHQC

= −THTC

givesQHTH

= −QCTC


therefore, Eq. (132) becomes: ∮δQ

T= 0

for any reversible heat engine.

– For all reversible heat engines: ∮δQ ≥ 0 (133)∮δQ

T= 0 (134)

where∮δQ can be reduced to zero by making TH → TC , reversibly.

– If the heat engine is irreversible (ηth,I < ηth,R by the 2nd Corollary), and bothreversible and irreversible engines receive QH from TH then:

WI < WR

QH +QC,I < QH +QC,R (QC < 0)

|QC,I | > |QC,R|

So for an irreversible heat engine:∮δQ = QH +QC,I > 0∮δQ

T=QHTH

+QC,ITC

< 0

– If the engine is made more irreversible (i.e. increase |QC,I |, while QH , TH , andTC are held fixed) then the following is true for all irreversible heat engines:∮

δQ > 0 (135)∮δQ

T< 0 (136)

• Consider a Carnot cycle refrigerator operating between two constant temperaturethermal reservoirs.


– For a refrigerator QH < 0, W < 0, and QH +QC = W .

– For the reversible refrigerator:∮δQ = QH +QC < 0∮δQ

T=QHTH

+QCTC

= 0

– As for the reversible heat engine,∮δQ can be reduced to zero by making TH →

TC , then the following expressions that are true for all reversible refrigerators:∮δQ ≤ 0 (137)∮δQ

T= 0 (138)

– If the refrigerator (β = −QC/W ) is irreversible (βI < βR, by the 2nd Corollary),and both reversible and irreversible refrigerators receive QC from TC then:

|WI | > |WR||QH,I +QC | > |QH,R +QC | (QH < 0)

|QH,I | > |QH,R|

So for an irreversible refrigerator:∮δQ = QH,I +QC < 0∮δQ

T=QH,ITH

+QCTC

< 0

– If the refrigerator is made more irreversible (i.e. increase |QH,I |, while QC , TH ,and TC are held fixed) then the following is true for all irreversible refrigerators:∮

δQ < 0 (139)∮δQ

T< 0 (140)

• So, for all cycles: ∮δQ

T≤ 0 (141)

i.e. proof of Corollary 4 (or Clausius’ Inequality). The equality in this equation holdsfor internally reversible processes, and the inequality is true for irreversible processes.

• Moran and Shapiro write the Clausius Inequality as an equality:∮δQ

T= −σcycle (142)

where

σcycle = 0 reversible

σcycle > 0 irreversible

σcycle < 0 impossible


5.5.5 Corollary 5 – Entropy

A property of a closed system exists such that a change in its value for anyreversible process undergone by the system between state 1 and state 2 is equalto∫ 21δQT .

• From Corollary 4: (∮δQ

T

)int rev

= 0

and the definition of a property states:∮dH = 0

then a property (entropy) can be defined:

dS =

(δQ

T

)int rev

(143)

or

ds =

(δq

T

)int rev

(144)

• For an internally reversible process:

TdS = δQ

or

Q12,int rev =

∫ 2

1T dS (145)

i.e. Q12 is the area underneath a T -S curve.

• Entropy is defined in the context of reversible processes. Can it be used for irreversibleprocesses?


• Consider a cycle composed of one reversible and one irreversible process.

∮δQ

T=

∫ 2

1,A

(δQ

T

)I

+

∫ 1

2,B

(δQ

T

)R

< 0

since the cycle is irreversible. The reversible process can be reversed to give:∫ 2

1,A

(δQ

T

)I

−∫ 2

1,B

(δQ

T

)R

< 0

or ∫ 2

1,B

(δQ

T

)R

>

∫ 2

1,A

(δQ

T

)I

Using the definition of entropy:

S2 − S1 >∫ 2

1,A

(δQ

T

)I

or

dS >

(δQ

T

)I

(146)

• So, for any process:

dS ≥ δQ

T(147)

S2 − S1 ≥∫ 2

1

δQ

T(148)

where the equality holds for internally reversible processes, and the inequality holdsfor irreversible processes.

• Note: for an adiabatic process:dS ≥ 0 (149)

and dS = 0 for a reversible, adiabatic (i.e. isentropic) process.


5.6 Gibbs (Tds) Equations

• The Gibbs or Tds equations are derived from a combination of the 1st and 2nd Laws.They are relations between entropy and the other properties of a substance.

• Applying the 1st Law to a closed, stationary system undergoing an internally reversibleprocess:

δq − δw = du

orTds = du+ pdv (150)

Also, since h = u+ pv:Tds = dh− vdp (151)

• Equations (150) and (151) are the Gibbs or Tds equations. These equations are validfor any simple compressible substance, open and closed systems, and reversible andirreversible processes. The equations can be written as follows:

ds =du

T+p

Tdv (152)

ds =dh

T− v

Tdp (153)

5.7 Evaluation of Entropy

5.7.1 Ideal Gases

• For an ideal gas: du = cvdT and p/T = R/v, therefore, the first Gibbs equation, Eq.(152) can be written as:

ds =cvTdT +

R

vdv

or

s2 − s1 =

∫ 2

1cvdT

T+R ln

(v2v1

)(154)

Also, dh = cpdT and v/T = R/p for an ideal gas, and Gibbs second equation, Eq.(153), can be written as:

ds =cpTdT − R

pdp

or

s2 − s1 =

∫ 2

1cpdT

T−R ln

(p2p1

)(155)

• Note: changes in entropy are not dependent only on T , therefore, entropy cannot betabulated. What is tabulated is:

s◦(T ) =

∫ T

0cpdT

T(156)

which takes into account the temperature dependence of cp. The values of s◦ and s◦

are tabulated in Tables A-22 and 23 in Moran and Shapiro. Changes in entropy wouldbe evaluated as follows:

s2 − s1 = s◦2 − s◦1 −R ln

(p2p1

)(157)


• If cp is a linear function of T , or constant, or if ∆T < 200 K, the following simplifiedequations may be used to evaluate changes in entropy:

s2 − s1 = cv,av ln

(T2T1

)+R ln

(v2v1

)(158)

s2 − s1 = cp,av ln

(T2T1

)−R ln

(p2p1

)(159)

• Isentropic Processes of Ideal Gases

– Constant Specific Heats

– Equation (158) can be rearranged, for an isentropic process, to give:

−cv ln

(T2T1

)= R ln

(v2v1

)v2v1

=

(T1T2

)cv/Rbut cp − cv = R and k = cp/cv, therefore:

v2v1

=

(T1T2

)1/(k−1)(160)

Equation (159) can be rearranged to give:

cp ln

(T2T1

)= R ln

(p2p1

)p2p1

=

(T2T1

)k/(k−1)(161)

Finally

p2p1

=

(v1v2

)k(162)

This demonstrates that an isentropic process of an ideal gas with constantspecific heats is polytropic with n = k = cp/cv.

– Nonconstant Specific Heats

– For an isentropic process between states 1 and 2, Eq. (157) can be writtenas:

s◦2 = s◦1 +R ln

(p2p1

)where s◦2 = s◦(T2) and s◦1 = s◦(T1).

– If p2/p1 and T1 are known for an isentropic process, then s◦2 can be deter-mined, and used to find T2.


– If p1, T1, and T2 are known for an isentropic process, p2 can be determined:

p2 = p1 exp

(s◦2 − s◦1R

)or

p2p1

=exp(s◦2/R)

exp(s◦1/R)

Defining the relative pressure, pr:

pr = exp(s◦(T )/R) (163)

thenp2p1

=pr2pr1

(164)

where pr is a function of temperature only, and values for the relative pressurefor air are tabulated in Table A-22.

– For an ideal gas:

v =RT

p

thereforev2v1

=

(RT2p2

)(p1RT1

)and for an isentropic process between states 1 and 2:

v2v1

=

(RT2pr2

)(pr1RT1

)Defining the relative volume, vr:

vr =RT

pr(T )(165)

thenv2v1

=vr2vr1

(166)

where vr is tabulated in Table A-22 for air.

– The relative pressure and volume may be used to determine pressure andvolume or temperature changes in isentropic processes of ideal gases.

– e.g. Given p1, T1, and T2 for air undergoing a reversible adiabatic process, findp2 and v2/v1.

– Since the process is reversible and adiabatic it is, by definition, isentropic.

– Assuming constant specific heats:

p2 = p1

(T2T1

) kk−1

v2v1

=

(p1p2

) 1k


– Assuming nonconstant specific heats:

p2 = p1pr2pr1

v2v1

=vr2vr1

where pr2, pr1, vr2, and vr1 are evaluated using Table A-22 with the knowntemperatures T2 and T1.

5.7.2 Pure Substances

• For a pure substance (e.g. water, R-134a...) entropy is evaluated as for the otherproperties, with s = sf + x(sg − sf ) for a mixture of saturated liquid and vapour.

5.7.3 Incompressible Substances

• For an incompressible substance, Gibbs first equation, Eq. (152), reduces to:

ds =du

T

because dv = 0 for incompressible substances. Since cv = cp = c

ds = cdT

T(167)

or

s2 − s1 = cav ln

(T2T1

)(168)

And note that since ∆s is only dependent on ∆T , then for any incompressible sub-stance an isothermal processisentropic process.

5.8 Increase in Entropy Principle

• For an isolated system (i.e. a system and its surroundings) the total change in entropyis the sum of the net change in entropy of the system and the net change in entropyfor the surroundings.

dStot = dSisolated = dSsys + dSsur

and from the definition of entropy:

dS ≥ δQ

T

• Since there is no heat transfer across the boundary of an isolated system:

dStot ≥ 0 (169)

where the equality holds for reversible processes, and the inequality holds for irre-versible processes.

• Since all real processes are irreversible, the entropy of the universe is always increasing(otherwise there is a violation of the 2nd Law).


5.9 What is Entropy?

• Consider a crystal (e.g. ice) in its lowest possible energy state.

• The significance of S = 0 is that the material will be in the most ordered state possible.Consider the addition of heat to this crystal, and plot a q-T curve.

• The crystal becomes more disordered as heat is added. If temperature is consideredto be an index of disorder there is a difficulty at phase changes, since T is constant,but the state of disorder changes substantially as heat is added.

• If, however,∫ T0 δQ/T = S is plotted versus T , the difficulty during the phase changes

disappears.


• Entropy is a measure of the state of disorder of a substance.

• e.g. Liquid water is more disordered than ice, but by how much?

∆s =hifTt

=333.5kJ/kg

273.16K= 1.22kJ/kg ·K

• The following properties may be perceived as indices:

T – index of vibration

p – index of molecular momentum

v – index of space occupied

s – index of disorder (pattern is a property)

5.10 The 3rd Law of Thermodynamics

• The 3rd Law of Thermodynamics, due to Nernst, is:

The entropy of all pure substances in thermodynamic equilibrium approacheszero as the temperature of the substance approaches absolute zero.

5.11 T -s Diagrams

• Similar to p-v diagrams, T -s diagrams can aid in the visualization of processes. Forexample, for the Carnot cycle heat engine:

• Since each process in the Carnot cycle is reversible, Tds = δq, and:

Q12 =

∫ 2

1T dS


Then, for the Carnot cycle above:

Q12 = TH(S2 − S1)Q34 = TC(S4 − S3) = TC(S1 − S2)Q23 = Q41 = 0

from the application of the 1st Law to the whole cycle (closed system):

Wnet = Q12 +Q34

and

ηth =Wnet

Q12=TH(S2 − S1)− TC(S2 − S1)

TH(S2 − S1)= 1− TC

TH

which is the maximum efficiency that was derived in the context of the 2nd Corollary.

5.12 Summary

• The following expressions must be true, otherwise there is a violation of the 2nd Law:

ηth < ηth,R (170)

β < βR (171)

γ < γR (172)∮δQ

T≤ 0 (173)

dS ≥ δQ

T(174)

S2 − S1 ≥∫ 2

1

δQ

T(175)

dStot ≥ 0 (176)

and only one of the expressions has to be checked (i.e. the most convenient for theproblem at hand).


6 Application of the Second Law to Thermodynamic Sys-tems

6.1 Entropy Balance for Closed Systems

• Consider a closed system undergoing a cycle composed of one reversible and oneirreversible process.

• The Clausius Inequality states:∫ 2

1

(δQ

T

)b

+

∫ 1

2

(δQ

T

)R

≤ 0

or as Moran and Shapiro write it:∫ 2

1

(δQ

T

)b

+

∫ 1

2

(δQ

T

)R

= −σ (177)

where, σ > 0 for irreversible processes, and σ = 0 for reversible processes. Thesubscript b indicates that the term is evaluated at the boundary of the system, andthe subscript R indicates an internally reversible process.

• When the definition of the property entropy:

S1 − S2 =

∫ 1

2

(δQ

T

)R

is substituted into Eq. (177) the following entropy balance equation is defined.

S2 − S1 =

∫ 2

1

(δQ

T

)b

+ σ (178)

• (S2 − S1) is the entropy change in a process, and it is only dependent on the initialand final states (since entropy is a property).


•∫ 21

(δQT

)b

is the entropy transport across the system boundary due to heat transfer.

The entropy transport is in the same direction as the heat transfer, therefore, thesame sign convention applies. The evaluation of this term depends on the process,since heat transfer is a path function.

• σ is the entropy production term, i.e. it represents the amount of entropy produceddue to irreversibilities (e.g. friction). Since the irreversibilities depend on the process,σ is a path function.

• Note: σ ≥ 0, but S2 − S1 can be equal to, less than, or greater than 0 for a system(but dStot ≥ 0).

• The usual goal when employing the entropy balance is to determine the entropy pro-duction term. This term can be evaluated if the initial and final states are known,and the heat transfer at the boundary can be determined.

• The entropy production term, σ, is indicative of the irreversibilities present in a pro-cess.

– It is possible to determine the entropy production term for each component in alarger device. This will indicate which components are most wasteful, and wherethe most useful gains in efficiency can be made.

– A comparison of σ’s for two components that perform the same task indicateswhich is the better component (thermodynamically speaking).

6.1.1 Forms of the Entropy Balance for Closed Systems

• If heat transfer occurs at several locations on the boundary of the system, the entropybalance is:

S2 − S1 =∑j

QjTj

+ σ (179)

where T is assumed uniform and constant at each location j.

• The entropy balance can be written in terms of a time rate equation:

dS

dt= S =

∑j

QjTj

+ σ (180)

where Tj is the instantaneous uniform temperature at location j.

• The entropy balance can also be written in a differential form:

dS =

(δQ

T

)b

+ δσ (181)


6.2 Entropy Balance for Open Systems (Control Volumes)

• Consider an open system:

• An entropy balance for this system (or control volume) can be written as follows:

[Net rate of increaseof entropy in the cv

]=

Net rate at whichentropy is transported

into the cv

+

Net rate of productionof entropy within

the cv

which can be written in the following algebraic form:

dScvdt

=∑j

QjTj

+∑i

misi −∑e

mese + σcv (182)

where the assumption of 1D (uniform) flow has been used. The misi and mese termsaccount for the transport of entropy into and out of the cv due to mass flow (similarto the transport of internal energy in the 1st Law).

• This equation can be written in the following more general form:

d

dt

∫Vρs dV =

∫A

(Q

T

)b

+∑i

(∫A

(sρVn dA

)i

−∑e

(∫A

(sρVn dA

)e

+ σcv

• For steady state, uniform flow conditions, the entropy balance for a control volumecan be written as:

0 =∑j

QjTj

+∑i

misi −∑e

mese + σcv (183)

• For only one inlet and one exit:

0 =∑j

QjTj

+ m(si − se) + σcv (184)


or

se − si =1

m

∑j

QjTj

+σcvm

(185)

and for adiabatic conditions:

se − si =σcvm

(186)

i.e. entropy must increase, since σcv > 0 for a steady state, adiabatic, irreversibleprocess.

6.3 Isentropic Efficiencies

• The isentropic efficiency is a comparison of the actual performance of a device to theperformance of an ideal device operating under the same inlet conditions, and exitpressure.

6.3.1 Turbines

• Consider the steady state operation of a turbine:

• Assuming ∆ek = ∆ep = Q = 0, the 1st Law for the turbine is:

Wcv = m(h1 − h2)

• The ideal turbine would be reversible, and since the turbine is assumed adiabatic theprocess would be isentropic. Or, a steady state entropy balance would give:

0 = m(s1 − s2) + σcv

But the process is reversible, therefore, σcv = 0, and s2 = s1, or the process isisentropic.

• The exit state can be defined since p2 and s2 are known.

• Defining h2s as the enthalpy at state 2 when s2 = s1 gives:(Wcv

m

)s

= h1 − h2s (187)

for the ideal turbine.


• Defining the isentropic turbine efficiency, ηt:

ηt =(Wcv/m)

(Wcv/m)s< 1 (188)

allows an evaluation of the quality of a turbine. Typically 0.7 < ηt < 0.9.

• The value of ηt < 1 since in reality s2 > s1, therefore, h2 > h2s, and (h1 − h2) <(h1 − h2s). Since p2 is held constant, then the exit flow from the real turbine is at ahigher temperature than that from the ideal turbine.

6.3.2 Nozzles

• The 1st Law for the steady state operation of a nozzle (W = Q = ∆ep = ~V1 = 0) is:

V 22

2= h1 − h2

• If the process within the nozzle is reversible and adiabatic, it will be isentropic, anda nozzle efficiency can be defined as:

ηn =(V 2

2 /2)

(V 22 /2)s

=h1 − h2h1 − h2s

(189)


• A typical nozzle efficiency is 0.95.

6.3.3 Pumps and Compressors

• Consider the steady state operation of a pump or a compressor:

• Assuming ∆ep = ∆ek = Q = 0 the 1st Law for both devices can be written in thefollowing form:

Wcv

m= h1 − h2 (Wcv < 0)

• For a reversible adiabatic process through the device, s1 = s2 and h2s < h2, andcompressor and pump efficiencies are defined as follows:

ηc = ηp =(Wcv/m)s

(Wcv/m)=h1 − h2sh1 − h2

(190)

• Typical compressor efficiencies are 0.75 to 0.85.


6.4 Second Law Efficiency (Effectiveness)

• Consider two irreversible heat engines, with ηth = 0.3, that reject heat to a lowtemperature reservoir at 300 K. Engine A receives heat from a high temperaturereservoir at 600 K, and Engine B receives heat from a reservoir at 1000 K. Which isthe “better” engine?

• Since both engines have the same thermal efficiency, they convert the same fractionof input heat to work, but the 1st Law cannot be used to identify the “better” engine.

• Using the 2nd Law, the maximum efficiency for a heat engine is:

ηR = 1− TCTH

So, for Engine A:

ηR,A = 1− 300

600= 0.5


and for Engine B:

ηR,B = 1− 300

1000= 0.7

– Since ηR,B > ηR,A, engine B has greater potential to do work, but it attains thesame thermal efficiency as Engine A, therefore, Engine B is performing poorlyrelative to Engine A.

– Introducing the effectiveness (ε):

ε = ηth/ηth,R (191)

= β/βR (192)

= γ/γR (193)

– Using the definition of effectiveness, εA = 0.6 and εB = 0.43, therefore, EngineA converts 60% of its available work potential into useful work, while Engine Bis performing at only 43% of its potential.

– Engine A is the “better” machine, as it is working closer to its potential.

6.5 A Combination of the 1st and 2nd Laws

• Consider the 1D, steady state flow through a control volume with one inlet and oneexit, and no irreversibilities. The 1st Law can be written as:(

Wcv

m

)R

=

(Qcvm

)R

+ (h1 − h2) +

(V 21 − V 2

2

2

)+ g(z1 − z2)

where the subscript R indicates an internally reversible process.

• From an entropy balance, or the definition of entropy:(Qcvm

)R

=

∫ 2

1T ds

therefore (Wcv

m

)R

=

∫ 2

1T ds+ (h1 − h2) +

(V 21 − V 2

2

2

)+ g(z1 − z2) (194)

• The assumption of an internally reversible process implies that the system proceedsthrough a series of equilibrium steps, therefore, property relations hold throughoutthe process, and Gibb’s second equation:

T ds = dh− v dp

integrated between states 1 and 2:∫ 2

1T ds = (h2 − h1)−

∫ 2

1v dp

can be used in Eq. (194) to give:(Wcv

m

)R

= −∫ 2

1v dp+

(V 21 − V 2

2

2

)+ g(z1 − z2) (195)


• Consider the case where Wcv = 0, e.g. frictionless flow in nozzles, diffusers, and pipes:∫ 2

1v dp+

(V 22 − V 2

1

2

)+ g(z2 − z1) = 0 (196)

which is a form of the Bernoulli equation (Fluids I).

• If the fluid is incompressible (v = const):

v(p2 − p1) +

(V 22 − V 2

1

2

)+ g(z2 − z1) = 0 (197)

or, dividing by g:p2 − p1ρg

+V 22 − V 2

1

2g+ (z2 − z1) = 0 (198)

Note: the units of this equation in the SI system are meters. When this equation isused one would speak of an energy head.

• When Eq. (195) is applied to devices in which ∆ek = ∆ep = 0, the following form ofthe 1st Law results: (

Wcv

m

)R

= −∫ 2

1v dp (199)

Consider a pump and a compressor used to subject liquid water and water vapour,respectively, to the same pressure rise. The pump will require less work input, becausethe specific volume of liquid water is much less than that of water vapour.

• If the fluid is incompressible, Eq. (199) reduces to:(Wcv

m

)R

= −v(p2 − p1) (200)

• If the process between states 1 and 2 is polytropic (pvn = c), Eq. (199) can be writtenas: (

Wcv

m

)R

= − n

n− 1(p2v2 − p1v1) (n 6= 1) (201)

= −p1v1 ln(p2/p1) (n = 1) (202)

• For a polytropic process of an ideal gas, Eq. (199) can be written as:(Wcv

m

)R

= − nR

n− 1(T2 − T1) (n 6= 1) (203)

= −RT1 ln(p2/p1) (n = 1) (204)


• Since the processes here are internally reversible, a p-v diagram may be used.

• Note: (W/m)R is the area to the left of the curve.

• Equation (199) can be used to determine minimum power requirements (pumps andcompressors) and maximum power outputs (turbines).


References

Black, W.Z. and Hartley, J.G., Thermodynamics, 2nd edition, HarperCollins Pub-lishers, New York, 1991.

Cengel, Y.A. and Boles, M.A., Thermodynamics An Engineering Approach,5th edition, McGraw-Hill, Toronto, 2005.

Moran, M.J. and Shapiro, H.N., Fundamentals of Engineering Thermodynam-ics, 6th edition, John Wiley and Sons, Inc., Toronto, 2008.

VanWylen, G.J. and Sonntag, R.E., Fundamentals of Classical Thermodynam-ics, 2nd edition, John Wiley and Sons, Inc., Toronto, 1978.

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