thermodynamics v0thermofluids 1 school ofmechanical engineering university of leeds

7/30/2019 Thermodynamics v0Thermofluids 1 School ofMechanical Engineering University of Leeds

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MECH 1215 Thermofluids 1 1

MECH1215: Thermofluids 1School of Mechanical Engineering

University of LeedsCourse Structure

MECH1215 is a 20 credit, level 1 module. It consists of two related subjects thermodynamics and fluid mechanics. There are natural links between the twosubjects. The course is taught in two strands (thermo- and fluids). The course isassessed using a mixture of final examination (60%) and course-work (40%). Thecourse work element is based on 4 practical sessions. Full details, including acomplete syllabus, can be found on the VLE.

Course Tutor and Module Leader

Dr Nik Kapur will present the thermo- part of the thermofluids course. He is also themodule leader. He can be contacted by email at [email protected].

Prof Phil Gaskell will present the fluid- part of the course. He can be contacted byemail at [email protected].

Book List

There is no set text for this course, however it is often useful to have access tosupport books since they offer a different view-point on a subject. Books that haveproved to be useful to students in the past include the following:

G.J.Van Wylen, R.E.Sonntag and C.Borgnakke Fundamentals ofthermodynamics 5th ed. John Wiley.

.A.Cengel and M.A.Boles Thermodynamics: An Engineering Approach McGraw Hill.

P.B.Whalley Basic engineering thermodynamics Oxford SciencePublications.

C.Borgnakke and R.E. Sonntag Thermodynamic and transport properties Wiley.(includes computer disc).

Mechanics of Fluids. 7th edition, B.S. Massey, Stanley Thornes Publishing,1998.

Fluid Mechanics. 3rd edition, Douglas, Gasiorek and Swafield, Longman,

1994. Fluid Mechanics. Foundation Engineering Series, Widden, McMillan Press,1996.

Thermofluids. Sherwin and Horsley, Chapman and Hall, 1996. Introduction to Fluid mechanics. 4th edition, Fox and McDonald, John Wiley

and Son (also for use in level 2)

These are available in good supply in the University library. If you choose to buy abook spend some time looking at the examples above to find one that suits yourlearning style.


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Contents

1. Introduction to thermodynamics 3

2. State and Fluid

propertiesPhase diagrams 8P, , T

Ideal gas behaviour P, , T 10Tabulated and graphical data

The P diagram forsteam 12Steam tables 14Tabulated data forrefrigerants 18

Internal energy

and Cv 19Enthalpy and Cp 21Relationship between Cp,Cv, R 23

3. Path Behaviour 24End state ofprocesses 26Displacement

work 27

4. First law 28Cyclic systems 29Closed (non-flow) systems

Displacement work 34Internalenergy 37

Open (flow)systems 39

Enthalpy 40

5. Air standard Cycles 42


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MECH 1215

1 INTRODUCTION TO

1.1 What is Thermodyn

From the Greek words ther

and entropy.

The science of heat and wrelation to heat and work

Thermofluids 1

HERMODYNAMICS

mics?

me (heat) and dynamis (force). It is the

Orork and those properties of a substanc

3

Science of energy

that that bear a


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For your chosen device (on the left) identify the following:

1. Importance to society ?

2. Impact on the environment ?

3. Key parameters within the system (why would you select one over another) ?

For the corresponding component (on the right) identify:

1. Heat inputs

2. Heat outputs

3. Mass inputs

4. Mass outputs

5. Input conditions

6. Output conditions

7. What conversions are taking place (between the inputs and the outputs)


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1.1 The Importance of Thermodynamics and Designing Systems

The understanding of thermodynamics is an important tool in the design ofequipment. Understanding thermodynamics allows design decisions to be made, andto understand the interplay between factors such as the operating temperatures,

pressures and efficiency..

Table 1.1 shows some of the engineering considerations for a variety ofthermodynamic systems. The two right-hand columns show the specific power andthe efficiency. The specific power is the amount on power produced per kg of plantequipment. The efficiency is a measure of useful output / energy input. The lowest

value of specific power output is for the steam turbine, this also has the greatestefficiency. To put it another way, a lot of engineering hardware and services (e.g.cooling water) can give a thermodynamic process that is relatively efficient. Theother extreme is the jet engine. Here the design criterion will be very different theengine needs to be lightweight and powerful, so having a high specific power. This is

at the expense of the overall efficiency. Were the engine to be more efficient, theadditional weight would be prohibitive and the plane would not leave the ground.

Thermodynamics therefore allows us to balance conflicting criteria within the overalldesign cycle. For example with a car engine, allowing a higher compression of thegases before ignition may give a greater efficiency, but the elevated pressures andtemperatures will require additional material to withstand the pressure, enhancedcooling or perhaps even exotic alloys to deal with the high temperature. The weightand size will also affect the design. The thermodynamic framework will allow theseconditions to be evaluatedbeforebuilding the engine.

From an environmental perspective, and as fuel becomes a more expensivecommodity then design criterion (and thus decisions) will change. For example itmay make more sense to invest more on the capital equipment (the initial plant maycost more) and have a greater fuel efficiency (the running cost will be less). So theeconomic cost of the designs will also form part of the design decisions. Some of this

will be speculative and projections of future fuel prices (and availability) will need tobe built into any engineering design.

Figure 1.1 shows an example of the array of choices available for the design of athermodynamic system; here this shows the relationship between a primary energysource and electrical power generation. Along each of these routes will be a vast

array of design choices and thermodynamic considerations. Thermodynamics helpsestablish a framework on which these choices can rationally be made.


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Table 1.1: Order of Magnitude Performance Characteristics of a number of PowerGeneration Devices. The specific power is a measure of the amount of energy output

per kilogram of plant.

Please refer to web page http://www.bpamoco.com/worldenergy for a statistical

review of world energy.

Device Energy Source UnitCapacity

SpecificPower(kW/kg)

Efficiency%

Heat Engines

Steam Turbine Power Plant

Combined Cycle Plant (CCGT)

Regenerative Gas Turbine

Jet Engine

Diesel Engine

Automobile Engine

Model Aircraft Engine

Fission Nuclear Plant

heat

natural gas

fuel oil

kerosene

fuel oil

petrol

petrol

-

660 MW/Turbine

500 MW

10 MW

10 MW

200 kW

200 kW

75 W

660 MW/reactor

0.05

-

0.45

3.50

0.55

0.83

1.10

-

40

50

23

20

35

25

15

10

Direct Energy Conversion Devices

Fuel Cell

Solar Cell

Radiation Cell

Thermoelectric

Thermionic

Magnetohydrodynamic (M.H.D.)

H2 - O2CH4

sunlight

Sr - Y

heat

heat

heat

15 kW2 MW

250 W

10-8W

5 kW

300 W

20 MW

--

0.1

-

0.45

2.0

-

9055-60

14

14

10

13

60

Alternative Power Generators

Tidal Power Plant

Windmills

Hydraulic Turbine

Waterwheel

Geothermal

Focused Solar Energy Plant

-

-

reservoir

river/millpond

-

240 MW

100-500 kW

550 MW/turbine

- 150 kW

100 - 500 MW

40 kW

-

-

-

-

-

-

18

(average)

40

90

20 -60

30


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Figure 1.1: Relationship between primary energy source and electrical generation.


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MECH 1215

2. STATE AND FLUID P

State properties are the prtime. In thermodynamics,and T) are commonly e

design, for example in dedesign. There are additionand S) that also help define

2.1 The Phase Diagram

The phase diagrams, showrange of conditions. Here tand pressure on the y-axisput inside a closed cylindetemperature (regulated by

controlled by moving the p

Figure 2.1 (left) a typicconditions (oxyge

Figure 2.2 A possible ex

Thermofluids 1

ROPERTIES

operties that define a substance at a paroperties such as pressure, volume andcountered. These quantities feed into

ining the materials or the mechanicall properties, internal energy, enthalpy athe thermodynamic state.

in figure 2.1, describe the behavior of ae phase diagram is shown with tempera

. To construct such a diagram the matewhich, at one end, has a movable pisto

its surroundings) the pressure within t

ston in or out (figure 2.2).

al phase diagram for a material that is an in this case); (right) a phase diagram f

eriment to relate pressure, temperature

and phase information.

8

ticular instant inemperature (P, Vthe engineering

properties of thend entropy (U, H

material under ature on the x-axisial in question isn. At a particulare cylinder can be

as under roomr water.

, specific volume


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The phase diagrams in figure 2.1 show the behaviour of the material as theseconditions are varied. Three states exist solid, liquid and gas (plasma is oftenregarded as a fourth state but dont worry about this just now). The left hand figure(for oxygen) will be used for the following discussion. Point A represents the solidphase. Here the molecules are close together, with relatively strong intermolecular

(that is forces between the molecules) bonds that form the structure. The moleculesvibrate about a fixed point. As the temperature increases the molecular spacingremains relatively constant, but now the kinetic energy of the molecules is such thatthe intermolecular forces no longer dominate, and molecules can switch places withtheir neighbours. This is the liquid state, point B. In this experiment, thetemperature is further increased until the kinetic energy dominates and there isrelatively little intermolecular attraction between the molecules. This is the gas state,point C, with the molecules filling the container they are within. The spacing betweenthe molecules will be at least 10 times the radius of the molecule. As the temperatureis further increased (noting that the piston will need to be moved out to maintain thepressure at a constant value) the spacing between the molecules will continue to

increase.

There are two additional points marked on the diagram.

1. The triple point, where (in the closed piston arrangement of this experiment)solid, liquid and gas co-exist. One use of water at the triple point is for thecalibration of thermometers, since this is a unique state point.

2. The critical point. Above the critical temperature, it is impossible to compressthe gas and turn it into liquid. The kinetic energy of the molecules is so highthat it dominates and the intermolecular forces are relatively weak. Even at

high pressures the intermolecular forces never become significant enough toform a liquid phase. An alternate way to interpret this in that when thepressure is greater than the critical pressure it is impossible to cool the gasand turn it into a liquid.

There is a subtle difference between water and many other materials commonlyencountered in thermodynamic processes (e.g. oxygen, nitrogen constituents ofair) since hydrogen bonding is an additional force that plays a role in the solid-liquidinteraction, see the line between the solid and liquid region in figure 2.1. Note thatthe gradient of the line between the solid and liquid phase has been exaggerated inreality it is much closer to vertical than this.

Figure 2.3 shows the typical engineering space in which most thermodynamicprocesses operate. The boundaries that define this are entirely practical and include:

The design of equipment to withstand pressure. Materials (mostly metals) to withstand the temperatures. Operating fluids within the thermodynamic system (e.g. steam within power

stations).

An observation is that within many thermodynamic processes, operating conditions

are close to the critical point of water, but far above the critical temperature ofoxygen (taken as representative of air). This forms an important point in the


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MECH 1215

discussion of when a matetreated as non-ideal.

Figure 2.3 the enginee

2.2 Ideal Gases press

Ideal gases are gases that ffirst observed experimentathe theory also has originsis of interest here.

Equation 2.1 shows the idevolume (V) and temperatu

A consistent set of units an

P=PressureV=Volumen=Number moles gasR=Gas constantT=Temperature

In engineering, it is oftenworking fluid within a gas.equations, where m is the

1 The volume of a gas is inversel

i.e. (V 1/P) when T is constant.

2 The volume of a gas increasedproviding the pressure remains c

Thermofluids 1

rial can be treated as an ideal gas and

ing workspace in relation to the three phpoint of air and water.

re, volume and temperature

llow a certain pattern of behavior. Ideallly, for example by Boyle (1660)1 or Chain molecular theory. It is the results of t

l gas law, that relates three state propere (T).

value of R are given by:

kPaL

8.314472 kPaLK1mol1

K

more convenient to work with the maThe working below shows the relationsolecular mass.

y proportional to its pressure when the temperat

by 1/273 part of its volume at 0C for each 1Constant.

Commonthermodynamic

system workspace

TRnPV

10

when it must be

ase and critical

gas behavior wasles (~1780)2, butis hard work that

ties, pressure (P),

(2.1)

ss of a particularip between these

re is kept constant.

rise of temperature,


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(2.2)

P=Pressure kPav=Specific volume m3/kgR=Gas constant (8.314 / molecular mass) kJK1kg1

T=Temperature K M=Molecular mass g/mol

Air: 0.287 kJK1kg1

Water: 0.462 kJK1kg1

Note: v is the specific volume, that is the volume, V, taken by 1 kg of gas and is equalto 1/density. Specific values are generally given the lower case symbol do notconfuse the two! A useful tip is, to convert density to specific volume, as it helps youmake a judgement if this has a sensible value.

2.3 Assumptions Behind an Ideal Gas

There are three main conditions that need to exist for a gas to be treated as ideal:

Randomly-moving, elastic collisions. Non-interacting (ie no forces between the particles).

Point particles.

So when, for the real gases, will these conditions be obeyed (ie the ideal gas law canbe used) and when will these conditions be violated (ie some other means will needto be used to establish the relationship between P,V and T) ?

Under conditions of high pressure the molecules will be forced to be closetogether, and the assumption that the molecules are point particles (that isthey take no volume) will be violated.

Under conditions close to the liquid-gas transition the forces between theparticles are significant.

On figure 2.3, when using air as the working fluid in a thermodynamic cycle it is arelatively good assumption to treat it as an ideal gas as the working temperature will

be significantly greater than the critical temperature and for most processes thepressure will be less than the critical pressure.

For water, however, most practical thermodynamic processes take place close to thepoint where the liquid-gas transition takes place. Under these conditionsexperimental measurements must be used to give the relationships between the state

properties.

RTP

Tm

R

n

VP

TRnPV

.


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2.4 Steam Tables, Steam Charts and Mollier Diagrams

The properties of steam are represented in many ways including tabular, graphical ofvia computer software3.

2.4.1 Steam chart P- v diagram

To develop understanding around the behavior of steam (as an example of a materialclose to the gas-liquid transition) the focus will be on a diagram describing the

behavior of steam in terms of pressure and specific volume. Experiments would becarried out in a similar apparatus to that shown in figure 2.2, where (for example)the temperature is set by regulating the external environment, and the pressure set

by moving the piston. The specific volume would be given by the cylinder volumedivided by the mass of the gas contained within. This data would then be plottedappropriately (P on the abscissa, v on the ordinate).

Figure 2.4 shows the behaviour of the water plotted in this way. The solid lines aretemperature isotherms at any point on a given isotherm, the temperature isconstant. The dotted line represents two things on the left side of the cricical pointit is where liquid is just at its boiling point, on the right of the critical point itrepresents the point where the water has just boiled and turned to vapour. At anypoint in this diagram it is possible to read off the pressure, the specific volume fromthe axes, and the temperature from the isotherms. Additional information can bederived when in the liquid/vapour region, described later.

Figure 2.4 P-v diagram for water

Figure 2.5 is used to describe the behaviour of the water at points 1-4 shown in figure2.4. In figure 2.5, the arrows to the right show, for each stage of the process thechange in temperature and specific volume.

3 E.g. http://www.spiraxsarco.com/resources/steam-tables.asp

Temperature of

isotherms increasing

L+V

V

G

L

tpTT

CT

CTT

CTT

P

v1g

v1v1f

v

1

triple line

sat.

vapour

sat.

liquid

Critical Point

2 3 4A

S

OLI

D


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MECH 1215

Figure 2.5(a) showsis added and the liqis called sensible he

Point 2 represents tits boiling point (f

cylinder will be usevapour phase terNote at any point ifperfectly insulated)

Ultimately though,(point 3) where the

vapour. Just at this Now as further he

molecules and the gsaturated vapour linmolecular interactiassumption of zero

Note the difference betweheat (point 1-2 and point 3the addition of latent heaovercome the intermolecul

vapour phase. The temperthe vapour phase.

Figure 2.5 State of water.

Water can also exist in a so2.4. This is a simplificatio

boundaries.

Thermofluids 1

the change in the water in moving fromid water expands. Since no phase chantaddition.

e saturated liquid line. This is where tr a given pressure). Any additional h

in freeing the molecules from this liqued latent heat addition. This is show

heat addition were to stop (and assumihe liquid and vapour phase would remais more and more heat is added, a pointliquid phase no longer exists all thoint, the vapour is described as saturate

at is added, more kinetic energy isas starts to expand (point 4). The behae wont follow the ideal gas laws as thereons (and the molecules are relative

olecular volume is not valid).

n heat and temperature. During the ad-4) the energy addition causes a temper

a phase change takes place, and ther bonds within the liquid, and liberate

ture will not rise until all the liquid has

he numbers beneath each of the diagrathose in figure 2.3

lid form (ice). This is shown on the left h, in reality there will be solid-liquid an

13

1 to 2. Here heate takes place this

e liquid is just ateat added to the

id state to form ain figure 2.5(b).

ng the cylinder isn in equilibrium.

ould be reachedwater would be

d vapour.

imparted to theiour close to the

will be still somey close, so the

dition of sensibleture rise. Duringnergy is used toolecules into theeen converted to

s correspond to

nd edge of figured solid-gas phase


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2.4.2 Steam Tables

In the Tables, shown schematically in figure 2.6 and found in full in appendix A,data for water/steam are set out in a format corresponding to the P-v diagram infigure 2.4. In the main table the data are given in increments of pressure (MPa) with

the corresponding value of saturation temperature (boiling temperature at thatpressure) given in brackets. The temperature symbol tis used to indicate units of 0C(instead of the usual symbol T used for absolute temperature, K).

The other symbols within this table are:v Specific volume m3/kgu Specific internal energy kJ/(kg K)h Specific enthalpy kJ/(kg K)s Specific entropy kJ/(kg K)

The symbols (f) and (g) represent saturated liquid and saturated gas, respectivel.

To interpret the data, first specify the pressure (column 3). To the right of each valueof pressure (e.g. atmospheric, 1.0133 bar or 0.10133 MPa with ts=100C ); are giventhe values of specific volume (and other properties) for a range of temperaturesshown across the top of the table. When on the left of the stepped double line the

water is in its liquid state. At the saturation temperature (given by ts in the pressurecolumn) the water will either be in a liquid state or a gas state (or some mix of thetwo). The double line hence corresponds to the two saturation lines on the P-vdiagram. The data at this point is given as properties of saturated liquid (column 1)or properties of saturated gas (column 2). To the right of the stepped line represents

water in its gas form.


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MECH 1215

Figure 2.6. Rela

SaturatedConditions P

(MN/m2)

f g ts (C)

v 0.00104 1.673 0.10133u 421 2508h 419.1 2675.8 100s 1.307 7.355

For cases where the tempervapour is given, Table (B)pressure at saturation. Thissame data as the Pressure

but equally spaced in temperThe data for 100C exists in bdata from table A (above) wit

Thermofluids 1

ionship between P-v diagram and steam

50 (C) 100 200 300

v 0.00101 2 1.67325072676

7.355

2.14526592876

7.828

2.60428113075

8.209

u 209.3h 209.4s 0.7037

ts (C)P

(MN/m2)

100 v

u

0.1013 h

s

ture of saturated liquid /can be used to find thetable merely contains theolumn in the main table,ture rather than pressure.

oth columns, compare thethat from table B (right).

A

15

tables.

400 500 600

3.062 3.519 3.9752968 3131 33023278 3488 3705

8.537 8.828 9.091

aturated Conditions

f g

0.001044 1.673

418.9 2506

419 2676

1.307 7.355


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(i) Subcooled Liquid

This is water below its saturation temperature. Within this region, the specificvolume of the liquid varies little with either temperature and pressure (i.e. water isalmost incompressible). Hence in the liquid region of the P-v diagram, isotherms are

(in reality) almost vertical and very close together; in working diagrams thesedifferences are usually exaggerated for clarity.

(ii) Superheated/Supercritical Vapour

To the right of the stepped double lines (saturation lines) are given values ofspecific volume for steam at temperatures above the saturated (boiling) temperatureat that pressure - note that the values are very much larger than those for the liquid!

At temperatures below the critical temperature (374.15oC) the vapour is usuallytermed superheated, at temperatures above the critical temperature the gas istermed supercritical.

(iii) Saturated Steam

The values of specific volume for saturated liquid (vf) and saturated vapour (v ) [aswell as other properties] corresponding to conditions at the exact saturationtemperature (often falling between two values in the main table) are given in thetabulation to the left at each pressure.

The same data are set out, transposed in (smaller) steps of temperature, in thesupplementary table (B). Pressure and temperature are not independentvariables atthe saturated condition, requiring an additional property value to define the

thermodynamic state.

In the subcooled region, since property values vary little with temperature, values ofsaturated liquid at a given temperature can generally be used with reasonableaccuracy regardless of the pressure.

(iv) Two Phase (Liquid-Vapour) Steam

Consider the state indicated by Point (A) in Fig. 2.4 / 2.5. This apparently indicatessome condition intermediate between liquid and vapour phases. This is of course notso, the point is representative of a mixture of saturated liquid and saturated vapour.

This mixture may exist in the form of a mist of liquid droplets in vapour or separatedas shown in Fig. 2.7.

Figure 2.7. Wet steam: The two cases areequivalent, since the liquid droplets in (a)could be gathered together as one mass ofliquid as shown in (b).

vapour

droplet

(a) (b)


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A way of characterising how much material exists as a liquid (or droplets) and howmuch material is as vapour (ie gas) is through the term quality fraction. A qualityfraction of 1 means that all of the water is (saturated) gas, a quality fraction of 0means that all of the water is (saturated) liquid. Another term that is sometimes usedis dryness fraction. This is exactly the same as quality fraction. For example, if you

were running a turbine, the rotational speed of the turbines can be very high, thepresence of droplets can cause erosion so quality steam means that with a qualityfraction of 1 ie without droplets.

(2.3)

Quality factor is defined as the ratio of the mass of the vapour to the overall mass ofwater in the system. For saturated liquid x=0, whilst for saturated vapour x=1. In

the 2 phase region , where P and T are not independent, x effectively replaces one ofthese quantities in defining the thermodynamic state.

The total volume (V) is equal to the volume of the vapour (Vg) plus the volume of theliquid (Vf), where these volumes are each given by their masses multiplied by their(specific) volumes per unit mass:

(2.4)

Combining Eqs. 2.3 and 2.4. the specific volume of the mixture is given by:

,1g

g

gv

m

mv

m

mv

(2.5)4

Note any specific property value of a two phase mixture (e.g. specific entropy,specific internal energy, specific enthalpy) is calculated in a similar way:

(2.6)

4 Other convenient forms of Eq. 2.5 can be generated by setting v f= v vf, thusfgf vvxv

m

mx

g

gfg mmmmmm or,

ffgg vmvmmv fg VVV

fg vxvxv 1

fgx )propertyspecific(1)propertyspecific(propertySpecific


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(v) Interpolation

For thermodynamic states falling between tabulated values it is usual to interpolate,by assuming property values vary linearly between given values. This is not alwaysquite correct, variation can be non-linear in some regions and the steps in property

values in the illustrative tabulations provided are rather large. However, linearinterpolation will be adequate for this course. Commercially available tables adoptfiner meshes, but often present information in more compacted forms which canconfuse those not used to them. In current textbooks, the data are often availablethrough software.

2.4.3 Other Fluids

There are a number of materials that have a phase change in typical engineeringworkspace (figure 2.3). Refrigerants (for example HCFC, ammonia) are a commonexample, but equally processes such as gas liquefaction will require understanding of

the state relationships. Tables and charts are available for these materials.

In general, however, most fluids have behaviour similar to water/steam, albeit atrather different absolute temperature and pressure - i.e. having different criticalpoints. data in tabular, graphical and computer disk formats are available for mostcommon fluids. At a given reduced pressure (P/Pcrit) and reduced temperature(T/Tcrit) values of reduced specific volume (v/vcrit) are very similar for most fluids.This can be useful for determining property values for fluids where property data areunavailable - as long as the critical values are known for that fluid.

The term reduced in this context simply means the actual value divided by the

critical value.


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MECH 1215

2.5 Internal Energy & C

Internal energy is the enermolecules) and the potemolecules. For a rigid sea

shows that the internal enthe fluid. Figure 2.8 showsenergy with the temperatexperiment shown in figuconstant specific volume. Tto the rise in internal energa range of masses of a gas cfamily of curves (each for apoint is termed the specific

From the experimental meis observed to be relatively

Figure 2.8 I

At high specific volumes (lgases and superheated vaponly. (This is sometimes kdemonstrated this behavio

be written as.

and thus the change in integiven by:

Thermofluids 1

onstant Volume Specific Heat

y associated with the kinetic energy (e.gntial energy (e.g. intermolecular forcled container of fluid, the first law of

rgy can be changed by adding heat orthe typical results of experiments to cor

ure. One way to collect this data woure 2.2. Fixing the piston at a given pohe first law would then states the heat a

y (section 4). The curves show the obserontained within a given volume thesedifferent specific volume). The gradientheat at constant volume, Cv.

asurements, at a given temperature theconstant and independent of the specific

ternal energy as a function of temperat

ow densities) the constant volume linesurs, tend to be co-incident - i.e. a functiown as Joules Law, after James Presco

ur experimentally). Under these condi

dTcdudT

duc vv or

rnal energy between State (1) and some

vv dT

duc

1

001 dTcuu v

19

. vibrations of thees) of the fluidthermodynamics

y doing work onelate the internalld be to use thesition will give aded will be equal

ved behaviour forre expressed as a

of these lines at a

(2.7)

gradient (dU/dT)volume (v).

re

on a diagram, forn of temperaturet Joule, who firstions, Eq. 2.7 can

atum State (0) is


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MECH 1215

To solve equation 2.7, Cvthat the right-hand side ctemperature is written as

tables.

c

At conditions of high spapproximates to a straight

Figure 2.9: A gas which hthe more generally applica

For a perfect gas Cv is conrange, Cv is taken to be con

Since in thermodynamics,giving rise to the followingconditions that Cv is taken

Tabulated data for the spethe same form as for spsuperheated and compresregion is given by a relatio

5It is possible to define datums i

T=0K, for real gases this is taken as

Thermofluids 1

eeds to be written as a function of teman be integrated. In practice the dep

a polynominal, with values for many

etcTaTaav

2

210

ecific volume (low density) the plotine, i.e. Cv is constant as shown in Fig 2.

s a constant volume specific heat in adle ideal gas rule (Pv = RT) is termed a

stant, and for many real gases over a listant.

it is the change in internal energy thequation for perfect gases or when tho be constant.

cific internal energy are available for pecific volume, i.e. saturated liquid aned liquid states. The internal energysimilar to those met previously:

fg uxuu 1

e a value at which the internal energy is zero. For pu=0 at T=25C.

1212 TTcuu v

20

(2.7)

perature, T, suchndence of Cv onases available in

f u T often.

dition to obeyingerfect gas.

ited temperature

at is important5,real gas is under

(2.8)

re substances ind vapour states,in the two phase

erfect gases u=0 when


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MECH 1215

2.6 Enthalpy and Const

Since ,Pvuh it ienthalpy values from ta

Alternatively, enthalpy valflow heating experiments.

Figure 2.10 Enthalpy

The curves show the obsergradient of these lines at a

Similar to internal energy,between h and T ie therthe Joule - Thompson Laexperimentally. The suff

dropped.

and thus the change in integiven by:

To solve equation 2.10, Cthat the right-hand side c

Thermofluids 1

ant Pressure Specific Heat

possible to calculate (and hence si

ulated experimental uTvP and,, da

es can be determined by applying the Fypical behaviour is shown in figure 2.10.

as a function of temperature for a range

ed behaviour of the enthalpy for a rangeoint is termed the specific heat at const

for gases at low pressures there is a uis no pressure dependence. This is so

after the workers who demonstratedx p from the right hand side of equ

dTcdhdT

dhc pp or

rnal energy between State (1) and some

needs to be written as a function of tean be integrated. In practice the dep

p

pdT

dhc

1

001 dTchh p

21

milarly tabulate)

a at any state.

rst Law to steady

f pressures.

of pressures. Thent pressure, Cp.

(2.9)

ique dependenceetimes known asthe relationshiption 2.9 can be

atum State (0) is

(2.10)

perature, T, suchndence of C on


22/45

MECH 1215

temperature is written astables.

cp

At conditions of high spapproximates to a straight

Figure 2.11 Entha

Since in thermodynamics,to the following equation fthat C is taken to be const

Tabulated data for the spsame form as for specific vand compressed liquid starelation similar to those m

6Again, it is possible to define dat

T=0K, for real gases this is taken as

Thermofluids 1

a polynominal, with values for many

etcTaTaa 2210

ecific volume (low density) the plotine, i.e. Cv is constant as shown in Fig 2.

py as a function of temperature for a pre

it is the change in enthalpy that is impor perfect gases or when the real gas isnt.

cific enthalpy are available for purelume, i.e. saturated liquid and vapour st

tes. The enthalpy in the two phase regt previously:

ms ie a value at which the enthalpy is zero. For ph=0 at T=25C.

1212 TTchh p

fg hxhh 1

22

ases available in

f h T often1.

fect gas.

rtant6, giving riseunder conditions

(2.11)

ubstances in theates, superheatedion is given by a

erfect gases h=0 when


23/45


2.7 Relationship Between C , Cv and R for an Ideal Gas

The relationship between vp cc and follows from the definition of enthalpy:

pvuh (2.12)

,Pvddudh where for an ideal gas RTPv

i.e. RdTdudh

or (2.13)

The ratio of specific heats is defined by the relation,

v

p

c

c (2.14)

or substituting Eq. 2.14 into 2.13 gives

(2.15)

Since is a function only of properties, it must also be a property. For a perfect gas

its value is constant, e.g. for air .4.1 For real gases, as for ,and vp cc is notconstant and varies (slightly) with temperature:

Gas Air N2 O2 H2 CO CO2 ArC012@ 1.4 1.4 1.4 1.4 1.4 1.3 1.67C0900@ 1.32 1.32 1.3 1.36 1.32 1.17 1.67

Note that is approximately the same for all diatomic gases.

RccRcceiRdT

du

dT

dhvpvp or,..,

1,

1

Rc

Rc pv


24/45

MECH 1215

3 PROCESS PATH

Although property values atwo states defines the wor

process. Section 3.1: Where

information regardiproperties for that sat state 2 (note notof the path from s

variables to be estab Section 3.2: As sho

given by 12wsubstituted into this e

stage further contex

For gases and superheathermodynamic processes

where n is called the indeare called polytropic proirreversible processes.

Note for the case wherefollowing mathematical de

Thermofluids 1

re independent of the path taken, the rok done and the heat transfer in a give

nly one property is known at one end sg the path would enable the determin

ate. For example, if P,,T are known atenough information to determine P or Tate 1 will allow a second (and then

lished.wn in section 4.2.1 the reversible disp

2

Pd . To solve this, the P needs to b

quation (you can treat this as a mathemati

will be given later).

Fig. 3.1 Process Path

ed vapours it has been found expay often be represented by a relationsh

of expansion (or compression). Procesesses. The relation may be applied to b

=, for processes taking place at consivations hold, with replacing n.

P v n = constant

24

te taken betweenthermodynamic

tate of a process,tion of the otherstate 1 but only) then knowledgeubsequent) state

acement work is

related to and

al problem at this

erimentally thatp of the form,

3.1

ses of this natureth reversible and

ant entropy, the


25/45

MECH 1215

The value n defines the sh3.2 shows the behaviourthe temperature is constan

Figure 3.2. Various end p

Thermofluids 1

ape of the curve between the start andor different values of n, all starting at p

(isothermal case), since P v = constant

ints of a polytropic compression processthe value of n.

25

end point. Figureoint 1. When n=1

RT.

as a function of


26/45


3.1 End State of Processes

Since two states on a polytropic process (Fig. 3.1) are related by Eq. 3.1, nn vPvP 2211 ,or:

n

2

1

1

2

P

P

3.2

thus if State 1 is defined and the final pressure (or specific volume) known, then Eq.3.2 enables one to define the final state by determining a second property.

For an ideal gas (Pv = RT), it is similarly possible to relate endstate temperatures to pressures or volumes, since

:/,/ 222111 hencePRTvandPRTv

n

2

22

n

1

11

P

RTP

P

RTP

or

1n

2

1n

1n

1

2

1

2

P

P

T

T

3.3

Noteyou should be able to derive equation 3.2 and 3.3 from first principles.

3.2 Displacement Work in a Reversible Polytropic Process

If a polytropic process occurs reversibly (frictionless - a better definition will come

later) one may combine the displacement work done (w=Pdv) and the polytropicexpression (Pvn - constant) to evaluate any displacement work:

2

112 (say)kconstantPwhere,Pdw

2

1 n12

dkw

2

1

n)(1

1k

n)(1

)(11)(12 kk.n)(1

1

where ,PP 2211 k hence

1111122212 .P.Pn)(1

1w

n

n1

PP 1122

3.3


27/45


This work (the integral of Pd for the process) is represented by the area under theprocess path line from State 1 to State 2, Fig. 3.1.

Noteyou should be able to derive equation 3.3 this from first principles.

For an ideal gas, the reversible work done expression given byEq. 3.3 can be re-arranged in terms of end state temperatures,since (from the equation of state for an ideal gas) Pv = RT, hence:

1212 TTn)(1

Rw

3.4

For an ideal gas isothermal process n = 1 and so both the

numerator and the denominator in Eq. 3.4 become zero!. Thisdifficulty may be avoided by returning to first principles,

where Pv = RT = constant (for an isothermal process)

)/v(v1nRTv

dvRTw 12

2

112

3.5

ote that the relationships have used the relationship Pv = RT -hence they apply to ideal gases

3.3 Summary

Things you should know and be able to do at the end of this section:

1. The path of many real fluid and ideal gas thermodynamic processes canbe described by Pvn = const.

2. The polytropic expression can often be used to determine a

thermodynamic end state.3. How to combine the polytropic and (reversible) displacement workdone expression for a closed system (w=Pdv).

4. How to combine the expressions Pvn = const. and Pv = RT, for idealgases.

5. That constant pressure (n = o) and constant volume (n = ) processesare special cases of the polytropic process and that, for ideal gases only,the isothermal process (n = 1) is a further special case.

2

112 Pdvw


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MECH 1215

4 FIRST LAW OF THE

The First Law of Thermconservation applied to thforms, according to the typ

First Law is that applied todifferent manifestations oconcept today (implicit indebate in the 19th century,

were laid down.

Energy can be trans

The change in interheat supplied to the sy

s

An important note on s

Within thermodynamics it(and other properties). The

Heat into a system Heat out of a syste Work into a syste Work out of a syst

This can be summarised onnumber) brought to youLawes):

Another example is a refrig

Heat into the engine in (burning fuel) will be psign, heat lost (e.g. via th

will be negative in sign. Wthe engine will be positive.

Thermofluids 1

MODYNAMICS

odynamics is a statement of the prirmodynamic systems. The First Law t

e of system under consideration. The si

cyclic systems. It effectively states thatf the same thing energy. This is aur definitions!) but was the subject of fiwhen the foundations of the science of

ormed from one form to another,created nor destroyed

al energy of a system is equal to ttem minus the amount of work pe

ystem on its surroundings

gn convention:

is important to define a sign conventionsign convention used on this course is o

is positiveis negative

is negativem is positive

these slightly cheesy diagram (not evenby the founding fathers of thermodyn

erator

from thesitive in

e funnel)rk out of

heat is coming out owill be negative in sElectrical energycompressor, so thecompressor is beingsystem, so will be n(e.g. -5J).

28

nciple of energykes a number ofplest form of the

eat and work arereadily acceptedrce, acrimoniousthermodynamics

ut cannot be

e amount offormed by the

for heat and worke where

orthy of a figuremics (no, not Dr

f the fridge, sogn (e.g. -10J).

drives thework of thedone on the

gative in sign


29/45


4.1 Cyclic Systems

Consider a Steam Turbine Generator Plant as shown in Fig. 4.1.

Fig. 4.1 Steam Turbine

Applying the energy conservation principle (gain = in out) to the complete system,boundary (a), noting that as the system is cyclic there can be no energy build upwithin the system.

then, energy in = energy out, i.e. :

01 cextcpfuelEEWEWE (4.1)

The energy sources in are the fuel, the work in the pump and the coolant in.The energy sources out are the output work, the exhaust gases and the coolant out.

Both mass and energy cross boundary (a) the outer dotted line. The next sectionswill look at how this can be simplified to just consider heat and work.


30/45


4.1.1 Heat Engines

As thermodynamics is principally concerned with the relationships between heat andwork the term fuel energy (Efuel) in Eq. (4.1) is inconvenient. It is therefore usual toredraw the system boundary, (b) in Fig. 1.1, with the plant as a heat engine as shown

in Fig. 4.2.

Fig. 4.2 Heat Engine

Heat Engine A heat engine is any continuously operating (cyclic) thermodynamic

system across the boundaries of which flow only heat and work. It is assumed thatthe chemical energy in the fuel is converted into heat in the boiler, which can beregarded as a hot reservoir, at temperature TH.

Reservoir A system at a fixed and uniform temperature.

Similarly, the condenser may be regarded as a low temperature reservoir at TL, towhich the engine rejects heat. It is convenient to consider that the turbine suppliesthe necessary pump work, such that the engine has a single net work output Wn;

where Wn = (Wt Wp).

The First Law for such a device is then, simply:

LHn QQW (4.2)

The more general form that includes unsteady cycles, such as spark ignition anddiesel engines is nothing more than mathematically stating the heat and work roundeach part of the cyclic process add up to zero:

OWQ (4.3)


31/45


Note that it is impossible to obtain work from a heat engine without supplying energyto it ie perpetual motion is not possible. Such a device, which inventors devisefrom time to time, is known as a erpetual motion machine of the first kind as itdefies the First Law. Remember, we obey the Laws of Thermodynamics within theSchool of Mechanical Engineering.

The thermal efficiency - () can be defined as this gives information about the useof energy in a process, and in general can be considering:

""

""

forpaywewhat

wantwewhat (4.4)

For the example shown in figure 4.2, the useful part of the cycle is the net workoutput, the cost is the heat input giving:

H

n

Q

W

inputheat

outputworknet (4.5)

or, from Eqs. (4.2) & (4.5),

H

L

H

LH

thQ

Q

Q

QQ

1 (4.6)

Note that in Eqs. (4.1) through (4.6) the usual heat transfer sign convention has beenignored in that QL (e.g. we might say the heat rejected is 10 kW not -10 kW).


32/45


4.1.2 Reversed Heat Engines (refrigerators/heat pumps)

Thermodynamics is not just concerned with power generation devicestransforming heat into work but also with machines that accomplish the reversee.g. refrigerators in which work is transformed into heat as shown in Fig. 4.3.

The detailed operation of refrigerators and heat pumps are discussed later in yourstudies. For the present analysis, it is sufficient to recognise that each operates byexpanding and compressing a working fluid such that heat can be absorbed from oneenvironment (e.g. inside the refrigerator) and rejected elsewhere (e.g. into yourkitchen behind the fridge). For the compression process, a compressor (=work) isrequired.

The First Law applies equally to such devices:

Figure 4.3 Reversed heat engine (note convention or representing a reversed heat

engine by , rather than E.

Again Wn = QH - QL

For a refrigerator the thermal efficiency, or performance criterion, is called theCoefficient of Performance, CR:

)(""

""

LH

L

n

L

R QQ

Q

W

Q

forpaywewhat

wantwewhatC

(4.7)

Such reversed engines may also be used for heating, taking heat from the atmosphere(or other suitable reservoir) and pumping it into (for example) a room. In this modethe device is called a heat pump and the corresponding Coefficient of Performance,CHP, is given by:

)(""

""

LH

H

n

H

HPQQ

Q

W

Q

forpaywewhat

wantwewhatC

(4.8)

Note that CHP CR = 1


33/45

MECH 1215

4.1.3 Combined heat e

Consider a heat engine ruengine is coupled to a heatemperature Tc and the te

in figure 4.4. Note how cardistinguish which heat flo

Figure 4.4:

The overall efficiency of tenergy flow, to the input ouseful energy flow is QH2there is energy entering th it is drawn from the surr

For this case, the efficiencyof the heat engine multiplishould work from first prin

QH1

For heat en

Thermofluids 1

gines / heat pumps

ning between limits of THE and TC. Thpump, used to warm a building, runniperature of the room Troom. This is sho

e has been taken to label the heatflowsis which.

combined heat engine / heat pump cyc

e system will be given by considering tf energy. In this case (since the room ishilst the energy input to the engine is Qheat pump (QL1) but there is no cost as

undings. This gives the definition:

of the combined system can be expressed by the coefficient of performance of thciples in deriving such agreements.

Rearrange and equate withoverall definition of efficiency

Equate in terms W

1HQ

W

W

W

QC HHP

2

HP

H

C

QW 2

overallHP

H

H CQ

Q

1

2

HP

H

HC

QQ 21

1

2

H

H

overall Q

Q

ine For heat pump

33

work out of thisg between a low

wn schematically

o it is possible to

le.

he ratio of usefulbeing heated) the

H1. Note also thatsociated with this

as the efficiencye heat pump. You


34/45


Summary

What you should know from this part of the course:

1. How to draw a system boundary for a heat engine.

2. That net work output from the system equals net heat transfer to the

systems, i.e. OWQQQW LH )(, 3. The various performance parameters for heat engines (normal or reversed)

- what we want/what we pay for, i.e.

n

H

HP

n

L

R

H

n

W

QC

W

QC

Q

W ,,

4. Be able to combine the above expressions and apply them to single heatengines or combinations of them.


35/45

MECH 1215

4.2 FIRST LAW OF TH

So far, the energy conservsystems. Because such sysinitial condition at the end

happens internally in suacross the system boundanon-cyclic systems (e.g. filanalysis of components osystems the closed system,

will be considered first.

4.2.1 Closed (Non-Flo

(a) Displacement Wo

Work transfers to and frowork - an energy transferpart of that boundary unexpansion of some mass (sectional area, A, as shown

Figure 4.4: work

Thermofluids 1

RMODYNAMICS: non-cyclic syst

tion principle (First Law) has been appems are cyclic, all states within the syst

of the cycle. Hence, one can, in energy t

ch systems and note only the energy trries. However thermodynamics is alsling gas bottles, humidifying, rockets ef cycles (e.g. condensers, turbines etc.in which no matter (mass) crosses the

) Systems

rk

closed systems are often in the formccurring at a system boundary by virtueder the action of a force. Consider,

) of fluid contained in a cylinder byin Fig. 4.4.

one in moving a piston over a small dist

35

ms

lied only to cyclicm return to theirrms, ignore what

ansfers occurringconcerned with

c.), and with the). Of non-cyclicystem boundary,

of displacementof a movement offor example, the

piston of cross-

ance


36/45


Consider moving the piston from a position of to a position + where thepressure difference between the inside and outside of the piston is at that instant intime.

The distance will be given by (final position initial position) =

The force will be given by : (the distance is so small that the pressureremains constant)

Since work = force distance then:

APW .

Note that we write since this represents the small amount of work over thedistance .

Now for a bit of simple manipulation. the (cross sectional area distance) is equal tothe volume that the piston moves. And the volume is equal to the (specific volume mass of gas), so it is possible to write the work expression as:

vPmVPW .

Now the total work done is given by the sum of the work for each of the little movesof a distance , so adding these up gives the total work.

area)wholetheoversummationindicates(W vPm

Accuracy is increased if smaller and smaller strips are considered:

)theovernintegratioindicates(dPW

0vlimW

movementwholevm

vPm

So that gives us our final expression for the work required to move the piston frompoint 1 to point 2 (remembering w is specific work (ie per kg) whilst W is actual workfor a given mass of fluid)

2

1

Pdvw (4.9)


37/45


To evaluate the expression shown in equation 4.9 (ie calculate the amount of workrequired to carry out the compression), the pressure P should be written in terms ofthe specific volume. This relationship will depend on how the compression is carriedout (e.g. isothermal constant temperature ; adiabatic no heat loss; isentropic contstant entropy). See section 3.2 and 3.3 for examples of how this can be used for

different compression ratios.

Note: The expression Pdvw only holds for displacement work and only if theprocess is reversible. A process may be considered reversible if it occurs slowly and

without friction - a more comprehensive definition will be developed later.

Compare for example the two expansion processes (a) and (b) shown in Fig. 4.5.

(a) Resisted frictionless expansion(reversible). The system does perform

work.:

Pdvw a 2

112

(b) Unresisted expansion system(irreversible). The system does not do

work when the diaphragm is ruptured;

2

112 PdvOw b

Note the convention of representing anirreversible process by a broken line onthe state diagram - this is because thesystem does not pass through a series ofdeterminable equilibrium states

Fig. 4.5 Resisted and Unresisted Expansion

The expression 2

112 Pdvw obeys the thermodynamic sign convention, i.e. 12w is

positive for work done by the system , and is negative for work done on the systemby its surroundings.


38/45


(b) Internal Energy

Consider unit mass of gas contained in a rigid (constant volume) vessel as shown inFig 4.6(a).

Figure 4.6: a (left) fixed mass of gas within a rigid box; b (right) change in pressureand temperature as heat is added.

As a result of heat transfer, energy is given to the system. Since the system mass andvolume (and so specific volume) are fixed, no displacement work is done.

.0 Pdv

What happens? Where does the energy go? Intuitively, one knows that temperatureand pressure increase as shown in Fig. 4.6(b). The gas molecules fly around faster(recall molecular theory of gases). Thus, the energy imparted by the heat transferremains internal to the system. The specific internal energy (u), which is the wholestock of energy per unit mass held within the fluid by its molecular forces, hasincreased in response to the applied heat transfer.

Since there is no work output, )0( dv , the First Law (gain = in - out) applied tothe above system reduces to:

Gain in energy = Energy transferred to system, i.e.

1212 quu (note special case of no volume change)

or for a more general system, as shown in Fig. 4.7, where work transfersimultaneously removes work from the system:

Gain in system energy = Energy transfer to system - Energy transfer from system.

121212 wquu 4.10


39/45


Figure 4.7: Simultaneous heat and work transfer to a system. The gain in internalenergy will be the difference between the heat in and the work out.

alternatively, re-arranging equation 4.10 in differential form gives:

wqdu (4.11)

Note again the sign convention that heat transfers to the system are positive (henceheat transfers from the system are negative) and work transfers from the system are

positive (and work transfers to the system are negative).

Note: Internal energy is a property, heat and work are only energy transfersoccurring at a system boundary.

Summary

What you should know from this part of the course:

1. That a closed system is one where no matter (mass) crosses thesystem boundary.

2. That the reversible specific displacement work associated with the

movement of a closed system boundary is given by Pdvw

3. That the energy per unit mass contained by the molecules making upthe fluid in the system is its specific internal energy (u) and that u,unlike specific heat and work transfers, is a property (having a unique

value at a given thermodynamic state) of the system.

4. How to generate and apply an energy conservation (gain = in - out)equation (First Law) for a closed system in specific, total ordifferential from.


40/45


4.3 Open (Flow) Systems Steady State

In open systems energy can enter or leave in association with mass transfers acrossthe system boundary, in addition to the heat and work transfers considered to date.To apply the First Law to such a system, it is necessary to know the energy associated

with the state (e.g. temperature, pressure) of a mass crossing the boundary.

Many important open thermodynamic systems (e.g. compressors, turbines, nozzles)can be considered to be of steady flow (i.e. invariant with time). Obviously, for suchsystems to be steady there can be no build-up (or decline) of mass or energy withinthe system, and the influx and efflux of mass must be the same. Figure 4.8 shows aschematic of such a system.

The Isothermal Case

For the isothermal case, the internal energy of the fluid entering the system will beequal to that leaving the system.

Fluid entering or leaving the system will carry with it energy in one of three forms:pressure; kinetic and potential. In addition, heat may be added into the system (Q12or work may be taken out of the system W12).

o

2

0

0

12i

2

i

i

12 g/2cP

mW/2cP

mQ

4.12

This is an extension to Bernoullis equation, which you will see a proof of in the fluidspart of this module. The addition of heat in (left hand side) and work out (right hand

side) complete this for an isothermal case.

The non-Isothermal Case

In many thermodynamic processes, the system is more complex, in that the inlet andoutlet temperatures will be different. This implies that the internal energy of the fluiddiffers (in effect this is another place for energy to be stored within the fluid).Equation 4.12 is then written in its more general form:

o20012i2ii12 g/2chmWz/2chmQ 4.13


41/45

MECH 1215

Where the enthalpy, h, of t

Remember, the specific vopressure (displacement wSince internal energy, prenthalpy of the flow is also.

1.3.1 Enthalpy

Consider the mass flow (m

The system (that in the dotthis, the pressure outside tthat that has been pushedsurroundings impart energpushing the mass into thremembering the pressuresystem:

worflow

Thus the total energy assokinetic or potential energy

Pv). Both these forms of e

a system. Equally, since upoint) then the sum (u +the sum (u + Pv) is also a(property) a separate nameeach time, i.e.

Similarly, the energy assocplus, of course, any kinetic(by virtue of its state, e.g.of a thermodynamic syste

Thermofluids 1

e fluid is defined as

Pvuh

lume =1/. The stored energy within trk) or the internal energy is capturedssure and specific volume are all pro

into the thermodynamic system shown

ted line) will have some pressure, so to pe system must be greater. Imagine the sinto the system (at pressure, P) by the

y to the mass, this energy being the (fle system; where the flow work done

at the inlet will be constant since this

mPv=mvP.PV=PdV=

iated with the mass entering the systewill be the sum of internal energy andergy are always associated with the flo

P and v are all roperties (with fixed vv) will always have a fixed value at anyproperty of the fluid and it is convenien(enthalpy) and symbol (h) rather than t

Pvuh

ated with any mass leaving a system wiland potential energy. Remember, theressure and temperature) with any masis its enthalpy (not its internal energy).

41

4.14

he fluid as eitherby the enthalpy.perties, then the

elow:

ush material intoaded mass being

piston. Thus thew) work done inis given by (andis a steady state

(4.15)

(apart from anylow work, m (u +

of any mass into

alues at any statestate point. Thus

to give this sumadd up u and Pv

be its enthalpy -nergy associated

s flow into or out


42/45


Summary

What you should know from this part of the course

1. That, in an open system, energy can be transmitted across the system boundary

by mass transfers; in addition to heat and work transfers.

2. The energy associated with the thermodynamic state (i.e. as described by twoproperties, such as temperature and pressure) of a mass entering or leaving anopen system is its enthalpy.

3. Enthalpy h is the sum of internal energy u and flow work Pv , i.e. (h=u+Pv),and that enthalpy is a property (having a fixed value at any thermodynamicstate).

4. How to construct the First Law (energy conservation equation) for a steady

(invariant with time) open system; whatever the number of mass, heat and worktransfers.


43/45


5 AIR STANDARD CYCLES: THERMODYNAMICS IN PRACTICE

One of the ultimate goals of thermodynamics is to relate the parameters that definean engine to the conditions within in. Thermodynamics are often complicated takea combustion engine - within it the mix of chemistry (combustion), heat transfer, gas

(and liquid) flows, geometry all affect the behaviour of the engine. However, in thedesign process it is often helpful to simplify the physics to build a working model toat least get an approximation of the engine behaviour. One such tool withinthermodynamics is the air standard cycle.

The Otto Cycle an example Air Standard Cycle

Figure 5.1(left) shows a (real) petrol engine cycle. This engine cycle has been brokendown into 4 main parts. So the challenge is to design a simple model of this to allowthe performance to be better understood. The following simplifications apply:

1. The working fluid is a mixed mass air (i.e. it neglects the fuel or combustionproducts). This is assumed to be a perfect gas (i.e. having constant values of

PvPv cccc ,, and RTPv ).2. Rather than the combustion taking place internally, let the combustion take

place outside the engine, and add the equivalent heat into the cycle.3. Instead of exhausting hot gases at the end of the cycle remove heat to return it

back to its initial process.4. All processes are internally reversible (no fluid friction and the second law

applies - PdvwTdsq , for internal processes).

Figure 5.1(right) shows the idealised cycle on a P- diagram and how this relates tothe main points of the engine cycle.


44/45

MECH 1215

Figure 5.1: (left) the proces

Note that different enginchoosing the appropriate cycle (or Otto cycle), the crepresented by adiabatic cheat rejection (4-1) take pinternal processes that tak

by measuring how t

change with time Consideration of thfuel combusts (e.g.

ou will see examples ofworked question will illustbe used to analyse such pro

Thermofluids 1

ses within an engine and (right) the repran air standard cycle.

cycles will have different P- diagraath that represents the cycle. For exammpression stroke 1-2 and the power str

ompression and expansion. The heat alace at a constant volume. This best mplace within the engine itself determine

e pressure and the position of the pisto

physical processes e.g. the charactehether it burns very quickly or slowly)

ther cycles further on in your studies,ate how the understanding you have decesses.

44

esentation within

s, the skill is inle within a petroloke 3-4 are oftendition (2-3) and

atches the actual, for example

n (ie the volume)

istics of how the

but the followingeloped so far can


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A Worked Example

At the beginning of compression, P1=100kPa and T1=300K. The heat addition per unit mass of air is 1350 kJ/kg The compression ratio given to you in the lecture (alternatively use a

compression ratio of 9)

(a) Draw:a. The Pv diagram and TS diagram

(b) Determine:a. The net work, in kJ per kg of air

b. The thermal efficiency of the cyclec. The maximum pressure, in kPad. The maximum temperature in the cycle, in K

Since this is a first model of the process, assume that the compression and expansionprocesses are isentropic (ie take place reversibly and adiabatically). This immediately

means that the relationship constP can be used in calculations. The values for

have already been measured and tabulated.

SEE YOUR OWN NOTES FOR THE WORKING!(and the example on the VLE)

thermodynamics v0thermofluids 1 school ofmechanical engineering university of leeds

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