mech101 thermodynamics lecturer: dr m. w. johnson module ...em22/thermo1/lectures.pdf · 1 x 2 hour...

MODULE SPECIFICATION

Full details in your Year book.

MECH101 is a 15 credit (1 unit) module which is 50%Fluid Mechanics and 50% Thermodynamics. 1 x 2 hour lecture (using the interactive teaching system)in each topic with 1 x 1 hour problem class per week. Assessment is through 4 labs, 1 full report, 1 oralpresentation and a 3 hour exam.

The module webpage is www.liv.ac.uk/~em22/Thermo1/

TEXTBOOKSEssential steam tablesRogers, J.F.C. and Mayhew, Y.A., ‘Thermodynamicsand Transport Properties of Fluids’, Blackwell Recommended textsCengel, Y.A. and Boles, M.A., ‘Thermodynamics – AnEngineering Approach’, McGraw HillalsoRogers, J.F.C. and Mayhew, Y.A., ‘EngineeringThermodynamics - Work and Heat Transfer (SI units)’,Longman McGovern, J.A., ‘The Essence of Thermodynamics’,Prentice Hall

MECH101 ThermodynamicsLecturer: Dr M. W. Johnson

1

INTRODUCTIONWhat is Thermodynamics?

Thermodynamics

Heat - Thermal Energy Mechanical Energy

Thermodynamics is concerned with the conversion ofheat into mechanical energy or vice versa.

Heat usually generated through combustion of fossil fuelor nuclear reaction.

Examples:-Heat to Mechanical Energy Petrol, Diesel engines - cars, trucks, boats, trains.Steam turbines - electricity power generation, ships.Gas turbines - ships, electric power generation, aircraft.

Mechanical Energy to HeatRefrigerationHeat Pumps

Stored chemical Energy to HeatCombustionChemical reactionsPhase changes in metals


2

CONCEPTS AND DEFINITIONSSystemA system (often called a closed system) is any collectionof matter within a prescribed boundary, but no mattermay enter or leave the system.

Note1. That the boundary may be real or imaginary and theshape of the boundary can changeE.g.


3

2. Different parts of the system may be at differenttemperatures, pressures and velocities and may containdifferent sorts of matter.

3. If a fluid is passing steadily through a reservoir so thatthe mass in the reservoir is constant, we still can notconsider the reservoir as a system. Matter is crossingthe boundary.

NOT a system

Actually a controlvolume – later


4

4. A system only has two kinds of interaction with itssurroundings. Work and Heat.

Positive WORK is done by a system, during a givenoperation, when the sole effect external to the systemcould be reduced to the rise of a weight.


5

Wording‘Sole effect’ to exclude cases where HEAT leaves thesystem and enters an engine which raises a weight. Weshall see later that as well as raising the weight, eitherthe engine rejects heat or the engine gets hotter.

‘External to’ to exclude for e.g.

‘could be reduced to’. We are allowed to assume nofriction, no electrical resistance etc, but we must notchange the interaction across the system boundary.

WORKDisplacement Work


6

Work done by system on piston = Force x Distance= p A x

= p x Volume swept by piston.

Note that in general p is not constant and hence

Work = where 1 and 2 are the start and end2

1�pbdVb

states.= pressure on the boundary.pb

= Volume swept by boundary.dVb

Work done in cycle = ´pdVe.g. Petrol engine Indicator diagram


7

Area represents work done.

Indicated Mean Effective Pressure (IMEP)

IMEP =(Area P - Area Q)

Swept volume

Also Brake Mean Effective Pressure (BMEP), which ispressure which would result in the work done at theflywheel if there were no frictional losses between thecylinder and flywheel.

e.g. Jaguar V12.The measured power output on an engine dynamometeris 272 b.h.p. (203 kW) @ 5850 r.p.m.Four stroke, 12 cylinders 90.0 mm bore 70.0 mm stroke

2 revs to complete 1 cycle.

585060 � 2cycles/sec = 48.75 cycles/sec

Work done by 1 cylinder in one cycle = 203 � 103

12 � 48.75= 347 J

Swept volume of one cylinder = � 902

4 � 70

= 4.4532 x 105 mm3

= 445.32 cm3

(Total capacity = 12 x 445.32 = 5344 cm3)

BMEP = = 0.779 x 106 N/m2 (=7.79 atm)347445.32 x 10−6

(Note 9.73 atm at 3600 r.p.m.)


8

TEMPERATURETwo systems are equal in temperature when no

change in any property occurs when they are broughtinto communication.

ZEROTH LAW OF THERMODYNAMICSTwo systems which are equal in temperature to a

third system are equal in temperature to each other.

HEAT is the interaction between systems which occursby virtue of their temperature difference when theycommunicate.

Symbol for Heat is Q. Positive when heat enters asystem.

If heat transfer is zero during a process it is anADIABATIC process.


9

HEAT TRANSFERThree mechanisms:1) Conduction - occurs in solids and stationary fluids(liquids or gases).

2) Convection - occurs in fluids. Change in the buoyancyof the fluid due to its change in temperature results inmovement of the fluid.

3) Radiation - an electromagnetic waveform whichtherefore doesn’t require any medium. Can occur acrossa vacuum.

CONDUCTIONFourier’s law governs steady heat conduction. ‘Steady’means that Thermal Equilibrium has been achievedsuch that T at any point does not change with time. Forone-dimensional heat conduction


10

Observations show that Q` A

Q` − �T Q` 1

�xand so

as Q` −AdT

dx �x G 0

or Q= −kAdT

dx

where k is a property called the COEFFICIENT OFTHERMAL CONDUCTIVITY

ANALOGY BETWEEN HEAT AND MASS TRANSFER The conduction of heat is analogous to mass transfer(diffusion) where the temperature gradient is replaced bya concentration gradient.


11

CONVECTION

For many fluids !` 1T

As the fluid adjacent to the heat source moves! < !a

upwards due to the buoyancy force. Heat transfer to airadjacent to the heat source is by conduction, but as airmoves away a high is maintained at the heat source. �TThis circulation assisted conduction is called FREE orNATURAL CONVECTION. If the circulation is assistedby mechanical means (e.g. a fan) then it is calledFORCED CONVECTION.

Q= −hA�T

where and h is the heat transfer coefficient.�T = Ts − Ta

h depends on the geometry and orientation of thesurface and the properties of the fluid, but forengineering calculations tables of values are generallyused.


12

RADIATION

An energy balance gives .! + � + $ = 1

Most solid surfaces are opaque to radiation (even glassonly transmits a naarow range of frequencies) and so

. $ = 0

A material with absorbs all the radiant energy� = 1whatever the frequency and is known as a BLACKBODY.


13

A black body emits more radiant energy per unit area ata given T then any other body and this is proportional to

.T4

STEFAN BOLTZMANN LAW

qb = "T4

where is the Stefan Boltzmann constant " 56.7nW/m2K4

Radiant energy is random as regards direction andfrequency. Analysis for real ‘grey’ bodies is difficult,therefore use an empirical emissivity �

� = q qb T

where T is the same for the grey and black bodies. Forthermal equilibrium, (KIRCHHOFF’S LAW). � = �

Q=

q A = � qb A = �"AT4

All bodies radiate some energy, but because of the T4

law this can usually be neglected for bodies attemperatures close to ambient.


14

COMPOSITE WALLS

Many situations in engineering involve severalheat transfer processes. Consider a wall made up oflayers of three layers which separates two fluids.

Heat transfer will be due to conduction within thewall, but due to convection between the surfaces ofthe wall and the fluids. The heat goes through eachof these transfer processes in turn and hence is

Q

the same for each process.

�

QA = −h12(T2 − T1 ) = −k23

x23(T3 − T2 ) = −k34

x34(T4 − T3 )

= −k45x45

(T5 − T4 ) = −h56(T6 − T5 )


15

The overall temperature difference is equal to thesum of the temperature differences fo each of theprocesses. i.e.

T6 − T1 = (T6 − T5 ) + (T5 − T4 ) + (T4 − T3 ) + (T3 − T2 ) + (T2 − T1 )

Therefore substituting for each of the temperaturedifferences from the first equation.

T6 − T1 = − 1h56

+ x45

k45+ x34

k34+ x23

k23+ 1

h12

�QA

so rearranging

�Q = −A(T6 − T1 )

1h56

+ x45

k45+ x34

k34+ x23

k23+ 1

h12

The result for other situations can be deduced. Eachconvection process results in a term and each1

hconduction process by a term.x

k

RADIAL HEAT CONDUCTIONFor the composite wall considered, the area for

each of the heat transfer processes is the same.This is not always the case e.g. in a cylindrical pipe.


16

Consider a small element of thickness δδr of thecylinder.

�

Q = −kAdTdr

where and so A = 2�rl

�Q = −k2�rdT

dr

�Q

2

1� dr

r = −k2�l2

1� dT

�Q = −k2�l(T2 − T1 )

lnr2r1


17

JOULE’S MECHANICAL EQUIVALENT OF HEAT

The water temperature rises by 1oC in both cases.

1 Joule = 1 Newton Metre = 1 Watt Second

FIRST LAW OF THERMODYNAMICS

For a system ´dW = ´dQ

or Net Work Done = Net Heat Added

Definitions

PROPERTY is any observable characteristic of asystem. e.g. p, V, T

STATE of a system is defined by the values of theproperties.

A PROCESS is that which brings about a change ofstate of a system.


18

A CYCLIC PROCESS is one where the value of allproperties is the same at the end of the process as itwas at the beginning.

Note1. Since a property is observable the change in thevalue of the property depends only on the start and endstates and not on the path of the process.

2. If a quantity related to a system changes during aprocess by an amount which depends only on the startand end states and not on the path then that quantity isa PROPERTY of the system.

[HEAT and WORK are NOT PROPERTIES]

INTERNAL ENERGY

For a system Q = W +�E

where is the change in E, the stored energy in the�Esystem.

E has various forms

1. Kinetic Energy

�E12 = 12mc2

2 − 12mc1

2


19

2. Potential Energy

Work done on system =mg(z2 − z1 )

k�E12 = mg(z2 − z1 )

3. Other forms of energy are due to random movementof molecules, chemical energy etc. and are combined inU, the internal energy.


20

Consider three processes A, B and C between states 1and 2 which are defined by two properties x and y.

Consider the cyclic process A + C. From the first law

´(dQ − dW) = 0

1A

2� (dQ − dW) +2C

1� (dQ − dW) = 0

and for the cyclic process B + C

1B

2� (dQ − dW) +2C

1� (dQ − dW) = 0

Subtracting

1A

2� (dQ − dW) =1B

2� (dQ − dW)

So is the same for any process path1

2� (dQ − dW)

between states 1 and 2.

But and and so1

2� dQ = Q121

2� dW = W12


21

is the same for any path and therefore isQ12 − W12 �E12

the same for any path and hence E is a property.

SPECIFIC PROPERTIES

Properties such as volume V and internal energy U canalso be expressed per unit mass

Specific volume v = Vm m3/kg

note that where is density v = 1! !

Specific Internal Energy u = Um kJ/kg


22

CONTROL VOLUMES (or Open Systems)

The boundary of a control volume is called a CONTROLSURFACE. A control volume can in general changeshape like a system, but often it does not changeposition or shape. Matter can cross a control surface.

EXTERNAL WORK is all the work transfer acrossWx

the control surface other than that due to fluid flow. Wx

is normally shaft work or electrical work.

For this system Q = W +�E

Q = [Wx + p2(m2v2 ) − p1(m1v1 )] + [(Ecf + E2 ) − (Eci + E1 )]

But E1 = m1u1 +m1c1

2

2 + m1gz1

E2 = m2u2 +m2c2

2

2 + m2gz2


23

write h = u + pv

Q − Wx = m2 h2 +c2

2

2 + gz2 − m1 h1 +c1

2

2 + gz1 + Ecf − Eci

This equation is the NON-STEADY FLOW ENERGYEQUATION

STEADY FLOW ENERGY EQUATIONIf there is steady flow, nothing changes with time

1. The properties of the fluid at entry and exit areconstant

2. The properties of the fluid at any point in the controlvolume are constant

3. and both are constantdm1

dt = dm2

dt

4. Q and Wx are constant

and k Ecf = Eci m1 = m2

Q − Wx = h2 +c2

2

2 + gz2 − h1 +c1

2

2 + gz1

where Q and Wx are per unit mass in kJ/kg. Thisequation can also be written as a rate equation

Q −

Wx =

m h2 +c2

2

2 + gz2 − h1 +c1

2

2 + gz1


24

where is the mass flow rate [kg/s] and and are m

Q

Wx

the heat transfer rate and rate of doing work [kW].

ENTHALPY

is a combination of properties and hence musth = u + pvalso be a property.

H is the ENTHALPY in kJ

h is the SPECIFIC ENTHALPY in kJ/kg


25

PURE SUBSTANCESA pure substance is homogeneous and invariable

in chemical composition. i.e. The same proportions ofchemical elements combined in the same waythroughout the system and not undergoing chemicalchange.

PHASEEach physical homogeneous type (solid, liquid orvapour) is a phase.

Two or three phases may coexist. e.g. Ice and water at0 oC. Mixtures of gases (e.g. Air) can only be treated as puresubstances if phase changes are excluded. This isbecause phase changes of constituents occur atdifferent temperatures. Oxygen evaporates at -183 oC,Nitrogen at -194 oC.

TWO PROPERTY RULE The state of a pure substance of given mass is

defined by specifying two appropriate propertiesprovided that

1. The system is in equilibrium2. Gravity, motion, electricity, magnetism, capillarity etc.have a negligible effect.


26

MIXTURE OF PHASESPhases are usually identifiable because of a

difference in density which causes gravitationalseparation. e.g. Ice floats on water, steam rises.

Phase diagram for H2O

DefinitionsSATURATION TEMPERATURE - Vaporisationtemperature at a given pressure

SATURATION PRESSURE - Vaporisation pressure at agiven temperature

SATURATED LIQUID (subscript f) - Liquid alone atsaturated p and T

SUBCOOLED LIQUID - Liquid at T<Tsat for given p.


27

COMPRESSED LIQUID - Liquid at p>psat for given T.

(DRY) SATURATED VAPOUR (subscript g) - Vapouralone at saturated p and T.

SUPERHEATED VAPOUR - Vapour at T>Tsat at given p.

DEGREE OF SUPERHEAT - Difference betweencurrent temperature and Tsat at given p.


28

On vapourisation (fusion and sublimation) curveschanges in properties occur at constant p and T.p and T are no longer independent properties andtherefore an additional extensive property (e.g. u, v, h)needs to be stated to satisfy the two property rule.

DRYNESS FRACTION (QUALITY)

x =Mass of Vapour

Mass of Vapour and Liquid

for mixtures of vapour and liquid.

Therefore for a saturated vapour x=1 (i.e. its ‘dry’) andfor a saturated liquid x=0 (i.e. its ‘wet’).

For extensive properties, the value is obtained bysumming the proportions of each phase.

u = xug + (1 − x)u f

h = xhg + (1 − x)h f

v = xvg + (1 − x)v f

As it is frequently taken as zero.v f << vg


29

PHASE DIAGRAMS FOR METALSSome pure substances, for example, metallic alloys canhave a number of different cystalline structures whichcan co-exist in equilibrium. In addition, because thedifferent constituents of the alloy in their pure form meltat different temperatures phase regions exist wheresome constituents are in their liquid phase whereasothers are in their solid phase. A simple example of thisis a copper nickel alloy.


30

PERFECT GASESAVOGADRO’S HYPOTHESIS

Equal volumes of all gases contain the samenumber of molecules when the volumes are measuredat the same p and T.

Therefore for a given volume at the same p and T,with the same number of molecules, the amount (ormass) of the gas will be directly proportional to theRELATIVE ATOMIC or MOLECULAR MASS orMOLECULAR MASS RATIO M.

MOLAR MASS = Mass of one mole = M g/mol

EQUATION OF STATE FOR A PERFECT GAS

pvT = R =Constant (different for each gas)

R is in kJ/kg K.

Note vapours close to the saturation line may NOT betreated as perfect gases. (i.e. Steam is NOT a perfectgas). R can be determined for a perfect gas

R = R0

M

R0 is the universal gas constant = 8.3143 kJ/kmol K.

Values of M are H2 - 2, CO2 - 44, N2 - 28, O2 - 32.

For Air - and so R = 0.287 kJ/kg KM M 29


31

SPECIFIC HEATS AT CONSTANT VOLUME Cv ANDCONSTANT PRESSURE Cp

1. Definitions and Cv = �u�T v

Cp = �h�T p

For a perfect gas onlyu = u(T)Also onlyh = u + pv = u(T) + RT = h(T)

So and Cv = dudT Cp = dh

dT

2. h = u + pv = u + RT

k dh = du + dT

CpdT = CvdT + RdT

for a perfect gas onlyCp − Cv = R

3. For an adiabatic process dQ = 0 and for the fullyresisted case

dW = pdv

and so the first law gives −pdv = du

For a process where pvn = Constant

npvn−1dv + vndp = 0npdv + vdp = 0

n =−vdppdv = dh

du =Cp

Cv= �


32

So for an adiabatic process for a perfect gas

pv� = constant

where is the ratio of specific heats. �


33

CYCLIC HEAT POWER PLANT (CHPP)A heat engine is a thermodynamic system with

only heat and work transfers across the boundary (i.e.no mass flow). The engine must either operatecontinuously or cyclically so that after a cyclic period thesystem is returned to its initial state.

e.g. Steam Power Plant

Note that no mass transfer occurs across the systemboundary as cooling water and fuel/air and combustionproducts remain outside the system.

Properties change around the cycle, but at any instant oftime properties at a particular point are the same.

From the first law

´dQ = ´dWQ1 + Q2 = W1 + W2


34

Note that W2 and Q2 are -ve but are smaller than W1 andQ1 and so there is net heat in and net work out.

REVERSED HEAT ENGINES - REFRIGERATOR ORHEAT PUMPThe throttle valve is used to drop the pressure (and

temperature) between the condenser and evaporator. Aturbine would provide a more efficient CHPP as aturbine would produce some of the work required todrive the pump, however the capital cost of a turbine isusually too high to be viable. Again

Q1 + Q2 = W1 + W2

but here there is net work in and net heat out. If the coldspace is required this CHPP is called a refrigerator, butif the heat from the condenser is required it is called aheat pump.


35

PERFORMANCE PARAMETERSPerformance parameters are defined as the ratio

of the energy transfer required to the necessary energyexpenditure. (i.e. Ratio of what you want and what youpay for).

Heat Engine

EFFICIENCY

� = WQ1

=Q1 + Q2

Q1

Refrigerator

COEFFICIENT OF PERFORMANCE

CoP = −Q2

W

Heat Pump

CoP =Q1

W

Note that Q1 and W are -ve.


36

SECOND LAW OF THERMODYNAMICS KELVIN PLANCK STATEMENT

It is impossible to construct asystem which will operate in acycle and produce no othereffects than the raising of aweight (i.e. Work) and heattransfer with a single reservoir.

This is what is known as aperpetual motion machine ofthe second kind (PMM2).

There must be a second heat transfer to another thermalreservoir at a temperature T2 < T1.

It is also impossible to construct a PMM1 which wouldcontravene the frirst law.

CLASIUS STATEMENT

It is impossible to construct asystem which operates in acycle and produces no othereffects than the transfer of heatfrom a cooler body to a hotterone.

CoP =Q1

W G �


37

The two statements can be shown to be equivalentStarting from the Kelvin Planck statement the Clasiusstatement can be reached.

And vice versa.


38

REVERSIBILITYReversibility has a special meaning in

Thermodynamics. It is quite possible to perform anirreversible process backwards.

A REVERSIBLE process is one whose effects on thesystem and on the surroundings can be effaced.

A CHPP is reversible if when it is exchanging heat withtwo reservoirs each having a fixed and uniformtemperature the sign of all the work an heat exchangescan be reversed and their magnitudes can remainunchanged. It is often easier to check instead that eachindividual stage of the cycle is itself reversible.


39

Example

Electric fire

Effacer is a PMM2 andso the electric fire isirreversible.

Examples of irreversible processes

Movement with friction - All real engines have bearingsetc.

Unrestrained expansion - Results in kinetic energy whichis then dissipated by viscous friction.

Heat transfer over a non-infinitesimal temperaturedifference - All real heat transfer processes.

Spontaneous chemical reactions - Combustion.

Mixing of different gases.


40

Examples of reversible processes

Frictionless motion

Fully resisted expansion - Slow so there is no significantkinetic energy

Heat transfer due to infinitesimal temperature difference

Mixing of same substance in same state

Reversible processes are therefore an ideal which wecan strive for. e.g. Reducing bearing friction. In practicewe can never achieve reversibility. Irreversibility can alsobe thought of as increasing chaos. e.g. If we mix twogases together we increase chaos and it is much moredifficult to unmix the gases i.e. reduce chaos.


41

DEDUCTIONS FROM THE SECOND LAW

1. It is impossible to construct an engine operatingbetween two reservoirs which has a higher efficiencythan a reversible engine operating between the sametwo reservoirs.

2. All reversible engines operating between the sametwo reservoirs will have the same efficiency.

3. A scale of temperature can be defined which isindependent of any thermodynamic substance andwhich provides an absolute zero of temperature.

Deduction 2 implies that the efficiency only depends onthe temperature of the reservoirs and is thereforeindependent of the particular properties of the workingfluid and of the quantity of heat Q0 supplied to theengine.


42

But

� =Q0 − Q

Q0= 1 −

QQ0

If T0 is assigned as a convenient reference temperature(e.g. the melting point at a fixed pressure) then we candefine the temperature T of any other reservoir by

T = T0QQ0

If Q is measured for different reservoirs and differenttemperatures, a scale of T varying linearly with Q isobtained. This is the THERMODYNAMICTEMPERATURE scale as it depends only on the laws ofthermodynamics and not on the properties of anyparticular substance.

The unit of Thermodynamictemperature is the KELVIN.

All other temperature scalesdepend on the method ofmeasurement.


43

CLASIUS INEQUALITYTo avoid contravening the 2nd law

´(dWR + dW) > 0

1st law gives´ dWR = ´ dQ0 −´ dQ

Third deduction from the 2nd law

gives dQ0

T0=

dQT

k ´ dWR = ´ T0

T − 1 dQ

Also by the first law ´ dW = ´ dQ

´(dWR + dW) = ´ T0

T − 1 dQ + dQ = ´ T0dQT > 0

Since T0 is positive for a particular reservoir

´ dQT > 0

This is the Clasius Inequality. Note that for a reversible

cycle and for irreversible cycles.´ dQT = 0 ´ dQ

T < 0


44

ENTROPY

Consider the reversible cycles A + C and B + C

´ dQT = 0 =

1A

2� dQT +

2C

1� dQT =

1B

2� dQT +

2C

1� dQT

k1A

2� dQT 1B

=2� dQ

T

So is the same for any reversible path between1

2� dQT

states 1 and 2 and hence only depends on the statesand so must be a property

is the change in a property called entropy S or if 1

2� dQrev

T is for a unit mass the specific entropy s.dQrev


45

ENTROPY CHANGE FOR A PERFECT GAS

ds =dQrev

Tk Tds = dQrev

From the first lawdQrev = dW + du = pdv + du

k Tds = pdv + du

For a perfect gas and du = CvdT pv = RT

k Tds = RTv dv + CvdT

� ds = R � dvv + Cv � dT

T

s2 − s1 = R lnv2v1

+ Cv lnT2

T1

Using to get , alternativepv = RTp2p1

v2v1

= T2

T1expressions can be obtained

s2 − s1 = Cp lnT2

T1− R ln

p2p1

s2 − s1 = Cp lnv2v1

+ Cv lnp2p1


46

GIBBS FUNCTIONDefinition g = h − Ts

It can be shown that (see textbooks for the proof) whena mixture of different phases of a substance existtogether in equilibrium then each phase has the samevalue for the Gibbs function. The Gibbs function istherefore only a function of the chemical composition ofa substance and its pressure and temperature.

Gibbs function is particularly useful when consideringchemical reactions. A chemical reaction will alwaysproceed in the direction which reduces g and hence willcease when g reaches a minumum. The energyreleased in the reaction can also be computed.


47

THERMODYNAMICS APPLIED TO MATERIALSMost of this module is concerned with the

application of thermodynamics to engines or enginecomponents. The thermodynamic processes aretherefore concerned with fluids. Thermodynamics canalso be applied to solids. Energy methods will be used inthe Year 2 Mechanics of Solids module.

The TF1 laboratory considers theThermodynamics involved in stretching rubber. Whenwork is done on a substance the first law means thatthere will be a change in internal energy of thesubstance and also in its entropy (as from the�Q = Tdssecond law). Internal energy is stored in a substanceeither due to an increase in the molecular kinetic energy(an increase in temperature) or due to stretching of themolecular bonds (an increase in pressure or stress). Inthe case of rubber the molecular bonds are notstretched and hence if the rubber is stretched atconstant temperature the internal energy remainsconstant. The entropy of the rubber will decreasethough. The work done on the rubber can therefore bedirectly related to the change in the entropy. A fullerexplanation is given in the laboratory handout.


48

POWER PLANT CARNOT CYCLE

The ideal steam power plant is the Carnot cycle.

1 - 2. Heat added at constant temperature and pressureto convert saturated liquid to saturated vapour.

Q12 = h2 − h1


49

2 - 3. Ideal expansion at constant entropy (Isentropic,reversible).

W23 = −(h3 − h2 )

3 - 4. Heat extracted at constant T and p.

Q34 = h4 − h3

4 - 1. Ideal compression (Isentropic, reversible).

W41 = −(h1 − h4 )

In practice the Carnot cycle can not be used in practice

1. Irreversibilities2. Steam can not easily be partly condensed - it is allcondensed.3. Compression of vapour and liquid is difficult.4. Wet steam in the turbine causes blade erosion.


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mech101 thermodynamics lecturer: dr m. w. johnson module ...em22/thermo1/lectures.pdf · 1 x 2 hour...

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