Chemical Engineering Chemical Engineering Thermodynamics I
Thermodynamics IApplication of Thermodynamics to Flow process
Application of Thermodynamics to Flow process2105370 2105370
2105370 2105370Apinan Soottitantawat Apinan
[email protected] 7 Chapter 72Energy
Balance: Differentiate forms Energy Balance: Differentiate
formsS.ApinanEnergy Balance Energy Balance = + + + + + +insys
sysSout dtU m dW Q gzuH m gzuH m ) ()2( )2( 2 2 0 ) ( = + + SW d Q
d gdz udu dH m Systemmsysu1m1P1A1u2m2P2A2H1H2UsysQWS3Objectives
Objectives Use the mass, energy and entropy balance with common
Unit Operation as throttling process, compressors, turbine, pump,
nozzle, pipe and duct Develop the general relations for
compressible flows encountered when gases flow at high speeds.
Derive the effects of area changes for one-dimensional isentropic
subsonic and supersonic flows. Solve problems of isentropic flow
through converging and convergingdiverging nozzles. Solve the
problems for compressible fluid flow in adiabatic and isothermal
pipeS.Apinan4SOME STEADY SOME STEADY- -FLOW ENGINEERING DEVICES
FLOW ENGINEERING DEVICESA modern land-based gas turbine used for
electric power production. This is a General Electric LM5000
turbine. It has a length of 6.2 m, it weighs 12.5 tons, and
produces 55.2 MW at 3600 rpm with steam injection.Many engineering
devices operate essentially under the same conditionsfor long
periods of time. The components of a steam power plant
(turbines,compressors, heat exchangers, and pumps), for example,
operate nonstop formonths before the system is shut down for
maintenance. Therefore, these devices can be conveniently analyzed
as steady-flow devices.At very high velocities, even small changes
in velocities can cause significant changes in the kinetic energy
of the fluid.S.Apinan5S.ApinanMass Balance Mass
BalanceSystemmsysu1V1P1A1u2V2P2A2dtdmm m sysout in = dtdmA u A u
sys= 2 2 2 1 1 1 Continuity equation Continuity
equation6S.ApinanEnergy Balance Energy
BalanceSystemmsysu1m1P1A1u2m2P2A2H1H2UsysQWS = + + + + + +insys
sysSout dtU m dW Q gzuH m gzuH m ) ()2( )2( 2 2 11st stlaw
Thermodynamics law Thermodynamics 7S.ApinanEntropy Balance Entropy
BalanceSystemmsysu1m1P1A1u2m2P2A2H1H2UsysQWSdtS m dSTQS m S m sys
systotal Gsur) (, 2 2 1 1 = + + 8S.ApinanEntropy balance + Energy
balance Entropy balance + Energy balanceSystemmsysu1m1P1A1u2m2P2A2
H1H2UsysQWSV1V2From Entropy Balance From Entropy BalancedtS m dSTQS
m S m sys systotal Gsur) (, 2 2 1 1 = + + dtS m dTWTQS m S m S T W
sys syssurLSsurtotal G surr LS ) (2 2 1 1 , = + + = dt S m T dW S m
S m T Q sys sys surrLS surr ) () ( 1 1 2 2 + = 9S.Apinan = + + + +
+ +insys sysSout dtU m dW Q gzuH m gzuH m ) ()2( )2( 2 2 From
Energy Balance From Energy BalanceFrom Energy and Entropy Balance,
at steady state From Energy and Entropy Balance, at steady state0 )
( )2( )2( 1 1 2 2 2222 2 1211 1 = + + + + + + S LS surr W W S m S m
T gzuH m gzuH m For only For only 1 1 input input 1 1 out put out
put0 ) ( )2( 2= + A + A A A mWmWS T z guH S LSsurrVdP TdS dH + =And
from0 )2( 2= + A A A mWmWz guP V S LSMechanical Energy Balance
Mechanical Energy Balance10S.ApinanMechanical Energy Balance
Mechanical Energy BalanceSystemmsysu1m1P1A1u2m2P2A2H1H2UsysQWSV1V2
= + + + + + + + +insys sysSout dtU m dW Q gzuPV U m gzuPV U m )
()2( )2( 2 2 o oEnergy Balance Energy Balance11S.ApinanMechanical
Energy Balance Mechanical Energy
BalanceSystemmsysu1m1P1A1u2m2P2A2H1H2UsysQWSV1V2| | = + + + + + +
+insys sysS out inout dtU m dW U U m Q gzuPV m gzuPV m ) () ( )2(
)2( 2 2 o oQ TdS Q S m S m T W surr LS = ) ( 1 1 2 PdV dU TdS PdV
TdS dU + = =| | dtU m dW U m Q V P m z guP V m sys sysS ) ()2( 2= +
A + A + A +A+ A o12S.ApinanMechanical Energy Balance Mechanical
Energy Balance| | dtU m dW U m Q V P m z guP V m sys sysS ) ()2( 2=
+ A + A + A +A+ A oFriction Heating Friction Heating Term Term Q
PdV dU WLS + =dtU m dW W z guP V m sys sysS LS ) ()2( 2= + A +A+ A
of LS gh m W = ||.|
\||.|
\|= =guDLfgDL fuhf242 2 22Re5 . 14269 . 0 ln(Re185 . 2269 . 0 ln
737 . 1 )`((
+ =D Df c cRe16= fRe < Re < 2000 2000Re > Re > 2000
2000f: Fanning friction factor13S.ApinanMechanical Energy Balance
Mechanical Energy BalanceFor Incompressible fluid For
Incompressible fluidf p h hguzgPguzgP+ + + = + +o o 2
222222111Balance Equation in Balance Equation in Differentiate
form: in the Differentiate form: in the steady flow steady
flow2105370 2105370 2105370
[email protected] Balance: Differentiate
form Entropy Balance: Differentiate
formSystemmsysu1m1P1A1u2m2P2A2H1H2UsysQWS0, = total Gsur S dTQ ddS
m 16S.ApinanME MEBB Balance: Differentiate form Balance:
Differentiate formSystemmsysu1m1P1A1u2m2P2A2H1H2UsysQWS0 ) ( = + +
+ S LS W d W d gdz udu VdP m 17Equation for solving engineering
problem Equation for solving engineering problemS.ApinanMass
Balance Mass BalancedtdmA u A u sys= 2 2 2 1 1 1 Entropy Balance
Entropy BalancedtS m dSTQS m S m sys systotal Gsur) (, 2 2 1 1 = +
+ Mechanical Energy Balance Mechanical Energy Balance = + + + + +
+insys sysS LSout dtU m dW W gzuPV m gzuPV m ) ()2( )2( 2 2 o
oEnergy Balance Energy Balance = + + + + + +insys sysSout dtU m dW
Q gzuH m gzuH m ) ()2( )2( 2 2 total G surr LS S T W , =There are
There are 4 4 mains equation for solving the problems in addition
to EOS. mains equation for solving the problems in addition to EOS.
18Equation for solving engineering problem: Steady Equation for
solving engineering problem: Steady flow flowS.ApinanMass Balance
Mass BalanceEntropy Balance Entropy BalanceMechanical Energy
Balance Mechanical Energy BalanceEnergy Balance Energy Balancetotal
G surr LS S d T W d , =There are There are 4 4 mains equation for
solving the problems in addition to EOS. mains equation for solving
the problems in addition to EOS. 0 = AdAuduVdV0 ) ( = + + SW d Q d
gdz udu dH m 0, = total GsurS dTQ ddS m 0 ) ( = + + + S LS W d W d
gdz udu VdP m Flow in various type of Units Flow in various type of
Units2105370 2105370 2105370
[email protected] Throattling Process
ProcessS.Apinan 21Throttling Process Throttling ProcessS.Apinan
Throttling Throttling Orifice 111HuP222HuP1 21 21 2H H u u P
P==~22Throttling Process: Orifice Throttling Process:
OrificeS.Apinan111HuP222HuP1 21 21 2z z u u P P==
56Compressor: Isentropic Compressor: Isentropic S.ApinanMass
Balance Mass Balance2 2 2 1 1 1 A u A u =Entropy Balance Entropy
Balance02 2 1 1 = A = S S m S m Energy Balance Energy Balancemin ,
1 2 ) ( sW H H m = Because of reversible process, the using
equations are 3 equations. VdP WS =min ,Mechanical Energy balance
Mechanical Energy balanceCompressor CompressorAdaibatic Adaibatic
compression process compression processS.Apinan5758S.Apinan12112
||.|
\|=||.|
\| VVTT 11212||.|
\|=||.|
\|PPTT ||.|
\|=||.|
\|2112VVPPVPCC= From isentropic Ideal gas process From
isentropic Ideal gas process 2 2 1 1 V P V P =o o2 2 1 1 V P V P =#
# Polytropic Polytropic process processEntropy Balance Entropy
Balance0 = dSCompressible fluids flow: Isentropic flow of the
Compressible fluids flow: Isentropic flow of the ideal gas fluid
ideal gas fluidQExample Example 77..9 9: Compressor : CompressorIf
methane (assumed to be an ideal gas) is compressed adiabatically
from 20 C and 140 kPa to 560 kPa, estimated the requirement and the
discharge temperature of the methane. The compressor efficiency is
0.75.S.Apinan592 6 3 ,10 64 . 2 10 081 . 9 702 . 1 T TRC methane P
+ =dPPRdTTCdS P =dT C dH P=Example 7.9: Compressor Example 7.9:
CompressorS.Apinan60) ( 1 2 min , 1 2 T T C W H H P s = =
11212||.|
\|=||.|
\|PPTT) 1 ( 1121 min , ||.|
\|= PPT C W P sFor CPis constant Example Example 77..9 9:
Compressor : CompressorS.Apinan61Assume CPof methane constant at
38.5 J mol-1 K-1 =1.271 -mol J 3866 ) 1 ( 1121 min , = ||.|
\|= PPT C W P sHHWW isentropicsisentropic sAA= = ,q1mol J 7 .
5154 =SWK 42 . 393, 2 =isenTK 89 . 4262 = TProblem: Compressor
Problem: CompressorS.Apinan62) ( 1 2 min , 1 2 T T C W H H P s = =
11212||.|
\|=||.|
\|PPTT) 1 ( 1121 min , ||.|
\|= PPT C W P sProvevpCCk = = R C C v p + = : Gas IdealExample
Example 77..8 8: Compressor : CompressorSaturated vapor steam at
100 kPa (tsat = 99.63 C) is compressed adiabatically to 300 kPa. If
the compressor efficiency is 0.75 what is the work required and
what are the properties of the discharge
stream?S.Apinan6364MINIMIZING THE COMPRESSOR WORK MINIMIZING THE
COMPRESSOR WORKObviously one way of minimizing the compressor work
is to approximatean internally reversible process as much as
possible by minimizing theirreversibilities such as friction,
turbulence, and nonquasi-equilibriumcompression. The extent to
which this can be accomplished is limited byeconomic
considerations. A second (and more practical) way ofreducing the
compressor work is to keep the specific volume of the gasas small
as possible during the compression process. This is done
bymaintaining the temperature of the gas as low as possible
duringcompression since the specific volume of a gas is
proportional totemperature. Therefore, reducing the work input to a
compressor requiresthat the gas be cooled as it is compressed.One
common way of cooling the gas during compression is to use cooling
jackets around the casing of the compressors.VdP WS =min
,65MINIMIZING THE COMPRESSOR WORK (ideal gas) MINIMIZING THE
COMPRESSOR WORK (ideal gas)Isentropic (Pvk= constant) No
cooling:Polytropic (Pvn= constant): Some coolingIsothermal (Pv =
constant): Maximum cooling P-v diagrams of isentropic, polytropic,
and isothermal compression processes between the same pressure
limits. the area to the left of the process curve is the integral
of v dP.When kinetic and potential energies are negligibleThe
adiabatic compression (Pvk= constant) requires the maximum work and
the isothermal compression (T = constant) requires the minimum.
Why?66Multistage Compression with Multistage Compression with
Intercooling IntercoolingIt is clear from these arguments that
cooling a gas as it is compressed isdesirable since this reduces
the required work input to the compressor. However,often it is not
possible to have adequate cooling through the casing of
thecompressor, and it becomes necessary to use other techniques to
achieveeffective cooling. One such technique is multistage
compression withintercooling, where the gas is compressed in stages
and cooled between eachstage by passing it through a heat exchanger
called an intercooler.67Multistage Compression with Multistage
Compression with Intercooling IntercoolingThe gas is compressed in
stages and cooled between each stage by passing it through a heat
exchanger called an intercooler.P-v and T-s diagrams for a
two-stage steady-flow compression process.68Multistage Compression
with Multistage Compression with Intercooling IntercoolingTo
minimize compression work during two-stage compression, the
pressure ratio across each stage of the compressor must be the
same. When this condition is satisfied, the compression work at
each stage becomes identical, that is, wcomp I,in= wcomp II,in.The
only variable in this equation is Px. The Px value that minimizes
the total work is determined by differentiating this expression
with respect to Px and setting the resulting expression equal to
zero. It yields69Staged compression Staged compressionIn the case
of centrifugal compressors, commercial designs currently do
notexceed a compression ratio of more than a 3.5 to 10 in any one
stage (for atypical gas). Since compression generates heat, the
compressed gas is to becooled between stages making the compression
less adiabatic and moreisothermal. The inter-stage coolers
typically result in some partial condensationthat is removed in
vapor-liquid separators.In the case of small reciprocating
compressors, the compressor flywheel maydrive a cooling fan that
directs ambient air across the intercooler of a two or morestage
compressor.Because rotary screw compressors can make use of cooling
lubricant to removethe heat of compression, they very often exceed
a 9 to 10 compression ratio. Forinstance, in a typical diving
compressor the air is compressed in three stages. Ifeach stage has
a compression ratio of 7 to 1, the compressor can output 343times
atmospheric pressure (7 x 7 x 7 = 343 atmospheres). (343 atm/34.8
MPa;5.04 ksi) NPPr / 1stage first in,stage last out,(((
=70Example: Multistage Compression Example: Multistage
Compression71Example: Multistage Compression Example: Multistage
Compression72Example: Multistage Compression Example: Multistage
Compression73Example: Multistage Compression Example: Multistage
Compression74Turbines and Compressors Turbines and
CompressorsTurbine drives the electric generator In steam, gas, or
hydroelectric power plants.As the fluid passes through the turbine,
work is done against the blades, which are attached to the shaft.
As a result, the shaft rotates, and the turbine produces
work.Compressors, as well as pumps and fans, are devices used to
increase the pressure of a fluid. Work is supplied to these devices
from an external source through a rotating shaft.A fan increases
the pressure of a gas slightlyand is mainly used to mobilize a gas.
A compressor is capable of compressing the gas to very high
pressures. Pumps work very much like compressors except that they
handle liquids instead of gases.Energy balance for the compressor
in this figure:S.Apinan75Example: Power Generation by a Steam
Example: Power Generation by a Steam Turbine TurbineS.ApinanThe
power output of an adiabatic steam turbine is 5 MW, and the inlet
andthe exit conditions of the steam are as indicated in Figure.(a)
Compare the magnitudes of h, ke, and pe.(b) Determine the work done
per unit mass of the steam flowing throughthe turbine.(c) Calculate
the mass flow rate of the steam.76Solution : Solution : Power
Generation by a Steam Turbine Power Generation by a Steam
TurbineS.Apinan77Solution : Solution : Power Generation by a Steam
Turbine Power Generation by a Steam TurbineS.Apinan78Solution :
Solution : Power Generation by a Steam Turbine Power Generation by
a Steam TurbineS.Apinan79Pump PumpS.ApinanQHHWW
isentropicsisentropic sAA= = ,q1 2 P P >80Pump: Isentropic Pump:
Isentropic S.ApinanMass Balance Mass Balance2 2 2 1 1 1 A u A u
=Entropy Balance Entropy Balance02 2 1 1 = A = S S m S m Energy
Balance Energy Balancemin , 1 2 sW H H = Because of reversible
process, the using equations are 3 equations. VdP WS =min
,Mechanical Energy balance Mechanical Energy balance81Equation for
solving engineering problem: Steady Equation for solving
engineering problem: Steady flow flowS.ApinanMass Balance Mass
BalanceEntropy Balance Entropy BalanceMechanical Energy Balance
Mechanical Energy BalanceEnergy Balance Energy Balancetotal G surr
LS S d T W d , =There are 4 mains equation for solving the problems
in addition to EOS. There are 4 mains equation for solving the
problems in addition to EOS. 0 = AdAuduVdV0 ) ( = + + SW d Q d gdz
udu dH m 0, = total GsurS dTQ ddS m 0 ) ( = + + + S LS W d W d gdz
udu VdP m 82Pump PumpS.ApinanQHHWW isentropicsisentropic sAA= = ,q)
( 1 2 P P V Hisentropic = AdP T V dT C dH P ) 1 ( | + =VdP dTTCdS
P| =83Isentropic Efficiencies of Compressors and Pumps Isentropic
Efficiencies of Compressors and PumpsThe h-s diagram of the actual
and isentropic processes of an adiabatic compressor.Compressors are
sometimes intentionally cooled to minimize the work
input.Isothermal efficiencyFor a pumpWhen kinetic and potential
energies are negligibleCan you use isentropic efficiency for a
non-adiabatic compressor? Can you use isothermal efficiency for an
adiabatic compressor?84Example Example 77..10 10: Pump :
PumpS.ApinanWater at 45 C and 10 kPa enters an adiabatic pump and
is discharged at a pressure of 8600 kPa. Assume pump efficiency to
be 0.75. Calculate the work of pump, the temperature change of
water and the entropy change of water. Properties saturated water
at 45 C: V = 1010 cm3kg-1, | = 425 x10-6K-1, Cp = 4.178 kJ
kg-1K-1)85Example : Pump VS Compressor Example : Pump VS
CompressorS.Apinan86Example : Pump VS Compressor Example : Pump VS
CompressorS.Apinan87Example : Pump VS Compressor Example : Pump VS
CompressorS.Apinan88Example : Pump VS Compressor Example : Pump VS
CompressorS.Apinan89Mixing chambers Mixing chambersIn engineering
applications, the section where the mixing process takes place is
commonly referred to as a mixing chamber.The T-elbow of an ordinary
shower serves as the mixing chamber for the hot- and the cold-water
streams.Energy balance for the adiabatic mixing chamber in the
figure is:10C60C43C140 kPaS.Apinan90Heat exchangers Heat
exchangersHeat exchangers are devices where two moving fluid
streams exchange heat without mixing. Heat exchangers are widely
used in various industries, and they come in various designs.A heat
exchanger can be as simple as two concentric pipes.Mass and energy
balances for the adiabatic heat exchanger in the figure is:The heat
transfer associated with a heat exchanger may be zero or nonzero
depending on how the control volume is selected.S.Apinan91Pipe and
duct Pipe and duct flow flowThe transport of liquids or gases in
pipes and ducts is of great importance in many engineering
applications. Flow through a pipe or a duct usually satisfies the
steady-flow conditions.Heat losses from a hot fluid flowing through
an uninsulatedpipe or duct to the cooler environment may be very
significant.Pipe or duct flow may involve more than one form of
work at the same time.Energy balance for the pipe flow shown in the
figure isS.ApinanMore focus in the More focus in the Compressible
fluid flow: Compressible fluid flow: Steady flow Steady flow2105370
2105370 2105370 [email protected] of
compressible flow Processes of compressible flowS.ApinanThere are 3
main processes of compressible flowa. Isentropic flow through
nozzlesb. Adiabatic friction flowc. Isothermal friction flowUsing 4
equations to solve the problemUsing 4 equations to solve the
problemUsing 4 equations to solve the problem94Compressible fluids
flow: Compressible fluids flow:S.Apinan-What is the relation
between pressure and distance ?-What is the velocity different?
-What is the density/ specific volume change with distance?
222HuP1A 2A0 = Q111HuP95Equation for solving engineering problem:
Steady Equation for solving engineering problem: Steady flow
flowS.ApinanMass Balance Mass BalanceEntropy Balance Entropy
BalanceMechanical Energy Balance Mechanical Energy BalanceEnergy
Balance Energy Balancetotal G surr LS S d T W d , =There are There
are 4 4 mains equation for solving the problems in addition to EOS.
mains equation for solving the problems in addition to EOS. 0 =
AdAuduVdV0 ) ( = + + SW d Q d gdz udu dH m 0, = total GsurS dTQ ddS
m 0 ) ( = + + + S LS W d W d gdz udu VdP m Compressible fluid flow:
Compressible fluid flow: adiabatic steady flow adiabatic steady
flow2105370 2105370 2105370
[email protected] fluids flow: adiabatic
Compressible fluids flow: adiabaticS.Apinan222HuP1A 2A0 =
Q111HuPMass Balance Mass BalanceEnergy Balance Energy Balance 0 =
AdAuduVdVudu dH =2 1 z z =98S.ApinanWhy sonic velocity ?Why this
value relate to the gas flow?How it relate to the gas flow
????Compressible fluids flow Compressible fluids flowIn
incompressible flow, an increase in velocity is associated with
adecrease in the cross-sectional area of the ductHowever, because
of the compressibility, the flow behavior ofcompressible fluid is
different from incompressible one. It could beexplained with the
sonic velocity. Therefore, in the flow of compressiblefluid (mainly
in gas), the velocity of fluids will be normally reportedthem
comparing to the sonic velocity (mach number).99S.ApinanFor the
compressible fluid, let consider the specific volume of fluid in
the function of entropy and pressure: V=V(S,P)dPPVdSSVdVS P |.|
\|cc+|.|
\|cc=Therefore Therefore????????Compressible fluids flow
Compressible fluids flow100S.ApinanP P P STTVSV|.|
\|cc|.|
\|cc=|.|
\|ccFrom FromPTVV ((
cc= 1|TCTS PP =((
ccAND ANDP P CT VSV |=|.|
\|ccTherefore Therefore(Volume expansivity)Compressible fluids
flow Compressible fluids flow101S.ApinanSVPV c |.|
\|cc = 2 2From physics: the speed of sound, From physics: the
speed of sound, cc, in a fluid could be calculated from , in a
fluid could be calculated from 22cVPVS =|.|
\|ccTherefore ThereforeCompressible fluids flow Compressible
fluids flow102S.ApinanConsider the specific volume of fluid in the
function of entropy and pressure: V=V(S,P)From FromdPcVdSCT VdVP 22
= |Therefore, Therefore, dPcVdSCTVdVP 2 = |Compressible fluids flow
Compressible fluids flow103Compressible fluids flow: adiabatic
Compressible fluids flow: adiabaticS.Apinan222HuP1A 2A0 =
Q111HuPMass Balance Mass BalanceEnergy Balance Energy Balance0 =
AdAuduVdVudu dH =2 1 z z =Compressible fluid Compressible fluid
dPcVdSCTVdVP 2 = |104Compressible fluids flow: adiabatic
Compressible fluids flow: adiabaticS.Apinan222HuP1A 2A0 =
Q111HuPThe system is the adiabatic, The system is the adiabatic,
there are the loss (irreversibility in the processSG,Total=02 1 z z
=The entropy balance will not be used at this step. The entropy
balance will not be used at this step. However, the fundamental
relation will be used. However, the fundamental relation will be
used. VdP TdS dH + =105S.ApinanThere are 4 equations with 6
differentials (dH, du, dV, dA, dSand dP). If we treats dS and dA as
dependent. From Fromand and dPcVdSCTVdVP 2 = |0 = AdAuduVdVudu dH
=VdP TdS dH + =Compressible fluids flow: adiabatic Compressible
fluids flow: adiabatic Mass BalanceEnergy BalanceCompressible
fluidFundamental Relation106S.ApinanTherefore Therefore VdP TdS udu
+ = 02 = AdAududPcVdSCTP|02 2 2 = + + AdAdPuVdSuTdPcVdSCTP|0 ) ( )
( 2 2 2 = + +AdAdPcVuVdSCTuTP|0 ) 1 ( ) 1 ( 222 2= + +
dAAuVdPcuTdSCuP|Compressible fluids flow: adiabatic Compressible
fluids flow: adiabaticudu dH =VdP TdS dH + =107S.ApinanFrom From0 )
1 ( ) 1 ( 222 2= + + dAAuVdPcuTdSCuP|When WhencuM =Mach Number Mach
NumberTherefore Therefore0 ) 1 ( ) 1 ( 222= + + dAAuVdP M
TdSCuP|Compressible fluids flow: adiabatic Compressible fluids
flow: adiabatic108Compressible fluids flow: adiabatic Compressible
fluids flow: adiabaticS.ApinanFrom Fromand andTherefore Therefore0
) 1 ( ) 1 ( 222= + + dAAuVdP M TdSCuP|VdP TdS udu + = 0 ) )( 1 ( )
1 ( 222= + + dAAuTdS udu M TdSCuP|0) 1 ( 1 22222= ||||.|
\|+dAM A uudu TdSMCuMP|109Compressible fluids flow: adiabatic
Compressible fluids flow: adiabaticS.Apinan0) 1 ( 1 22222=
||||.|
\|+dxdAM A udxduudxdSTMCuMP|0 ) 1 ( ) 1 ( 222= + +dxdAAudxdPV
MdxdSTCuP|Therefore, divide by Therefore, divide by dx dx0 > dx
dSFor spontaneous process:(with irreversibility)110Compressible
fluids flow: Pipe flow Compressible fluids flow: Pipe
flowS.ApinandxdSMCuMTdxduu P||||.|
\|+= 2221|dxdSMCuVTdxdP P||||.|
\|+ = 2211 |For flow in pipe: For flow in pipe: dA dA//dx dx = =
0 00 > dx dSFor spontaneous process (with
irreversibility):Subsonic flow: M1P2< P1u2> u1P2> P1u2<
u1However, for the supersonic flow is unstable in constant area
pipe, the compression shock occurs results in increase in pressure
and decrease in velocity to a subsonic value111SHOCK WAVES AND
EXPANSION WAVES SHOCK WAVES AND EXPANSION WAVESFor some back
pressure values, abrupt changes in fluid properties occur in a very
thin section of a pipe under supersonic flow conditions, creating a
shock wave. .Normal ShocksNormal shock waves: The shock waves that
occur in a plane normal to the direction of flow. The flow process
through the shock wave is highly irreversible.Schlieren image of a
normal shock in a Laval nozzle. The Mach number in the nozzle just
upstream (to the left) of the shock wave is about 1.3. Boundary
layers distort the shape of the normal shock near the walls and
lead to flow separation beneath the shock.S.Apinan112Example
Example 77..1 1S.ApinanConsider the steady state, adiabatic,
irreversible flow of an incompressible fluid in a horizontal pipe
of constant cross-sectional area. Show thata. The velocity is
constantb. The temperature increases in the direction of flowc. The
pressure decreases in the direction of flow 0 = AdAuduVdV0 ) ( = +
+ SW d Q d gdz udu dH m 0, = total GsurS dTQ ddS m 0 ) ( = + + + S
LS W d W d gdz udu VdP m Compressible fluid flow: Compressible
fluid flow: Isentropic steady flow Isentropic steady flowcovergent
covergent/divergent pipe /divergent pipe2105370 2105370 2105370
[email protected] NozzleS.Apinan111HuP222HuP1A
2ANozzle Process Nozzle Process--Change internal energy to kinetic
energy Change internal energy to kinetic energy--usually operated
usually operated isentropically isentropically1 21 21 21 2S S z z u
u P P==> 1) can be accomplished only by attaching a diverging
flow section to the subsonic nozzle at the throat (a
convergingdiverging nozzle), which is standard equipment in
supersonic aircraft and rocket propulsion.Convergingdiverging
nozzles are commonly used in rocket engines to provide high
thrust.S.Apinanu = cM=1 (Asterisk
condition)123S.Apinan124Compressible fluids flow: Isentropic flow
of the Compressible fluids flow: Isentropic flow of the ideal gas
fluid ideal gas fluidS.ApinanMass Balance Mass Balance 0 = W 0 = Q
0 = AZ 0 = AS 222HuP111HuP0 = AdAuduVdVudu dH =Energy Balance
Energy BalanceEntropy Balance Entropy Balance0 = dSMechanical
Energy Balance Mechanical Energy Balanceudu VdP = VdP TdS dH + =We
can use the fundamental relation instead of MEB125S.Apinan12112
||.|
\|=||.|
\| VVTT 11212||.|
\|=||.|
\|PPTT ||.|
\|=||.|
\|2112VVPPVPCC= From isentropic Ideal gas process From
isentropic Ideal gas process 2 2 1 1 V P V P =o o2 2 1 1 V P V P =#
# Polytropic Polytropic process processEntropy Balance Entropy
Balance0 = dSCompressible fluids flow: Isentropic flow of the
Compressible fluids flow: Isentropic flow of the ideal gas fluid
ideal gas fluid126S.ApinanCompressible fluid : Compressible fluid :
They will be used when the mach number is used to They will be used
when the mach number is used to present the velocity of fluid
present the velocity of fluid0 = W 0 = Q 0 = AZ 0 = AS
222HuP111HuPdPcVVdV2 = cuM =Ideal gas and isentropic Ideal gas and
isentropicSVPV c |.|
\|cc = 2 22 22V P c =VPVPS =|.|
\|cc 2 2 1 1 V P V P =Compressible fluids flow: Isentropic flow
of the Compressible fluids flow: Isentropic flow of the ideal gas
fluid ideal gas fluid22RT c =127For an ideal gasMach numberSpeed of
soundThe Mach number can be different at different temperatures
even if the velocity is the same.The speed of sound changes with
temperature and varies with the fluid.Ma = 1 Sonic flowMa < 1
Subsonic flowMa > 1 Supersonic flowMa >> 1 Hypersonic
flowMa ~ 1 Transonic flowS.ApinanTVPc |.|
\|cc = 2VRTP =PV RT c = =2128S.Apinan} = 2122122PPVdP u uP1
u1V1P2 u2V2} = 211 1112122 2PPdPP V Pu u ( )(((
||.|
\|= 112 1 12122 112PP V Pu uFrom Mechanical Energy Balance From
Mechanical Energy BalanceCompressible fluids flow: Isentropic flow
of the Compressible fluids flow: Isentropic flow of the ideal gas
fluid ideal gas fluidudu VdP =129S.ApinanP1 u1V1P2 u2V2( )(((
||.|
\|= 112 1 12122 112PP V Pu uWhen u When u22reaches the speed of
sound reaches the speed of soundSVPV c u |.|
\|cc = = 2 2 22 2 222 V P u =VPVPS =|.|
\|ccCompressible fluids flow: Isentropic flow of the
Compressible fluids flow: Isentropic flow of the ideal gas fluid
ideal gas fluid130S.ApinanP1 u1V1P2 u2V2) 1 (1212 ||.|
\|+= PP2 22 22 2 1 1 , 0 V P c u M u = = = =Compressible fluids
flow: Isentropic flow of the Compressible fluids flow: Isentropic
flow of the ideal gas fluid ideal gas fluid131Example: Isentropic
of compressible fluid flow Example: Isentropic of compressible
fluid flow S.ApinanAir enters a convergen-divergent nozzle at a
temperature of 555.6 K and a pressure of 20 atm. The throat area is
one-half that of the discharge of divergent section.a. Asterisk
condition at the throat b. At M=0.8 in the throat what is pressure,
temperature, velocity, specific volume ?Assume: Isentropic flow
with air as the ideal gas with Mw = 29 g/gmol and = 1.420 atm555.6
K132Example: Isentropic of compressible fluid flow Example:
Isentropic of compressible fluid flow S.Apinana. Asterisk condition
at the throat 20 atm555.6 Ku1=0 m/sP2?? atmT2?? KM = 1) 1 (1212
||.|
\|+= PP 2 2 1 1 V P V P =2 222 V P u =133Example: Isentropic of
compressible fluid flow Example: Isentropic of compressible fluid
flow S.Apinanb. At M=0.8 in the throat what is pressure,
temperature, velocity, specific volume ?20 atm555.6 Ku1=0 m/sP2??
atmT2?? KM = 0.8Determine u Determine u22SVPV c |.|
\|cc = 2 22 22V P c =c u M / = Mc u =2From From134Example:
Isentropic of compressible fluid flow Example: Isentropic of
compressible fluid flow S.ApinanFrom From01 = uDetermine: P
Determine: P22Then, Then, ( )(((
||.|
\|= 112 1 12122 112PP V Pu u) 1 (2122 ) 1 ( 2 ||.|
\|+ = M PP 2 2 1 1 V P V P =2 22V P c = Mc u =2How to derive
this equation !!135Example Example 77. .3 3: Isentropic of
compressible fluid : Isentropic of compressible fluid flow flow
S.ApinanConsider the nozzle with the steam behave as an ideal gas
flow calculatea. The critical pressure ratio and the velocity at
the throatb. The discharge pressure for M = 2 at the nozzle
exhaust700 kPa300 Cu1= 30 m/sGiven: = 1.4, Mw = 18P2kPaT2 Cu2= ??
m/sM =1 P3kPaT3 Cu3= ?? m/sM = 2136S.Apinan12112 ||.|
\|=||.|
\| VVTT 11212||.|
\|=||.|
\|PPTT ||.|
\|=||.|
\|2112VVPPVPCC= From isentropic Ideal gas process From
isentropic Ideal gas process 2 2 1 1 V P V P =o o2 2 1 1 V P V P =#
# Polytropic Polytropic process processEntropy Balance Entropy
Balance0 = dSCompressible fluids flow: Isentropic flow of the
Compressible fluids flow: Isentropic flow of the ideal gas fluid
ideal gas fluid137Example Example 77. .3 3: Isentropic of
compressible fluid : Isentropic of compressible fluid flow flow
S.Apinan700 kPa300 Cu1= 30 m/sP2kPaT2 Cu2= ?? m/sM =1 P3kPaT3 Cu3=
?? m/sM = 212 ( )(((
||.|
\|= 112 1 12122 112PP V Pu uRT PV =) 1 (1212 ||.|
\|+= PP138Example Example 77. .3 3: Isentropic of compressible
fluid : Isentropic of compressible fluid flow flow S.Apinan23 (
)(((
||.|
\|= 112 1 12122 112PP V Pu uRT PV =2 22V P c =2 22 2 2 22 V P M
c M u = =Compressible fluid flow: Compressible fluid flow:
Isothermal steady flow Isothermal steady flow2105370 2105370
2105370 [email protected] fluids flow:
Isothermal Compressible fluids flow: IsothermalS.ApinanTemperature
constant along the pipe Temperature constant along the pipeAs the
similar way, using the balance equation to solve the problem. As
the similar way, using the balance equation to solve the problem.
Using the Re and moody chart to calculate the lost work. Using the
Re and moody chart to calculate the lost work. 141Equation for
solving engineering problem: Steady Equation for solving
engineering problem: Steady flow flowS.ApinanMass Balance Mass
BalanceEntropy Balance Entropy BalanceMechanical Energy Balance
Mechanical Energy BalanceEnergy Balance Energy Balancetotal G surr
LS S d T W d , =There are There are 5 5 mains equation for solving
the problems in addition to EOS. mains equation for solving the
problems in addition to EOS. 0 = AdAuduVdV0 ) ( = + + SW d Q d gdz
udu dH m 0, = total GsurS dTQ ddS m 0 ) ( = + + + S LS W d W d gdz
udu VdP m Compressible fluid Compressible fluiddPcVdSCTVdVP 2 =
|They will be used when the velocity is in the form of mach
number142Problem HW: Isothermal friction flow Problem HW:
Isothermal friction flowS.ApinanAir at 1.7 atm gauge and 15 C enter
a horizontal 75 mm steel pipe that is 70 m. long. The flow rate of
the entering air q is 0.265 m3/s. Assuming isothermal flow, what is
the pressure at the discharge end of the pipe and rate of heat
transfer?Properties We take the properties of air to be = 1.4, Cp=
1.005 kJ/kg K, and R = 0.287 kJ/kg K. =0.0174 cP, Assume as the
ideal gas1.7 atm15 Cu1P2?? atmT215 Cu2 = ??143Equation for solving
engineering problem: Steady Equation for solving engineering
problem: Steady flow flowS.ApinanMass Balance Mass BalanceEntropy
Balance Entropy BalanceMechanical Energy Balance Mechanical Energy
BalanceEnergy Balance Energy Balancetotal G surr LS S d T W d , =0
= AdAuduVdVmQ dudu dH= +0, = total GsurS dTQ ddS m 0 = + +mW dudu
VdP LS2211VuVu= 0 ) ( = + + + S LS W d W d gdz udu VdP m
144Equation for solving engineering problem: Steady Equation for
solving engineering problem: Steady flow flowS.Apinanf LS gh m W
=||.|
\||.|
\|= =guDLfgDL fuhf242 2 2|.|
\|||.|
\|= =DdL uf gdhmW dfLS24 2Therefore02 2= + + dLDfuudu VdPFrom
mass balanceconstant2211= = = =VuAmGVA uVA u 145Equation for
solving engineering problem: Steady Equation for solving
engineering problem: Steady flow flowS.Apinanso02 2 2= + + dLDV
fGudu VdPAnd From Ideal gas lawconstant2211= = = =VuAmGVA uVA u 02
22 = + + dLDfGVuduVdP02 22 2 = + + dLDfGG uududPRTP02 22= + +
dLDfGuduG dPRTP146Equation for solving engineering problem: Steady
Equation for solving engineering problem: Steady flow
flowS.Apinan02 22= + + dLDfGuduG dPRTP02) ln( ) (21 2122 2122 = + +
D L fGuuG P PRTAnd From Ideal gas law and mass balance2211VuVu= 2 2
1 1 V P V P =and211 2PPu u =0044 . 0 , 00061 . 0 / , 10 8.56 Re S,
kg/m 5 . 198 , cP 0174105 . 0 5 2= = = = = f D k G Re is constant
???? Prove !!147Equation for solving engineering problem: Steady
Equation for solving engineering problem: Steady flow
flowS.Apinan||.|
\|+ = ) ln(22212 21 2PPDfLRT G P P0044 . 0 , 00061 . 0 / , 10
8.56 Re S, kg/m 5 . 198 , cP 0174105 . 0 5 2= = = = = f D k G K
kJ/kg. 0.287 R K, 288 T m, 70 , kg/m.s 273578 kPa 578 . 273 atm 7 .
2 21 = = = = = = L P||.|
\|+ = )273578ln( 213 . 8 5 . 198 288 287 2 27357822 22PPatm 26 .
12 = P148Example: Compressible flow, Rayleigh flow Example:
Compressible flow, Rayleigh flowS.ApinanAir enters a rectangular
duct at T1 = 300 K, P1 = 420 kPa, and Ma1 = 2. Heat is transferred
to the air in the amount of 55 kJ/kg as it flows through the duct.
Disregarding frictional losses, determine the temperature and Mach
numberat the duct exit. Assumption: associated with Rayleigh flow
(i.e., steady one-dimensional flow of an ideal gas with constant
properties through a constant cross-sectional area duct with
negligible frictional effects) , Properties We take the properties
of air to be = 1.4, Cp= 1.005 kJ/kg K, and R = 0.287 kJ/kg K.
149Equation for solving engineering problem: Steady Equation for
solving engineering problem: Steady flow flowS.ApinanMass Balance
Mass BalanceEntropy Balance Entropy BalanceMechanical Energy
Balance Mechanical Energy BalanceEnergy Balance Energy Balancetotal
G surr LS S d T W d , =There are 5 mains equation for solving the
problems in addition to EOS. There are 5 mains equation for solving
the problems in addition to EOS. 0 = AdAuduVdV0 ) ( = + + SW d Q d
gdz udu dH m 0, = total GsurS dTQ ddS m 0 ) ( = + + + S LS W d W d
gdz udu VdP m Compressible fluid Compressible fluiddPcVdSCTVdVP 2 =
|They will be used when the velocity is in the form of mach
number150Solution : Compressible flow, Rayleigh flow Solution :
Compressible flow, Rayleigh flowS.Apinanm/s 4 . 694 u m/s 2 . 347
12= = = = = Mc c RT PV c FromP1 = 420 kPaT1 = 300 KMa1 = 2T2, Ma2q
= 55 kJ/kgEnergy balance21 122 2 u H u HmQq + = =Mass balance22112
2 2 1 1 1 0VuVuA u A u = = Compressible fluid udu dHmQ d+ =0 =
AdAuduVdV151Solution : Compressible flow, Rayleigh flow Solution :
Compressible flow, Rayleigh flowS.ApinanEntropy balance:) ( 1 2 S S
mTQsur = EOS:22 211 1TV PTV P=Mechanical Energy Balance0, = total
GsurS dTQ ddS m 0 ) ( = + + + S LS W d W d gdz udu VdP m udu VdP
=152HW: Isothermal friction flow HW: Isothermal friction
flowS.ApinanAir at 25 C enter a section of 2-in. Schedule 40 steel
pipe at a gauge pressure of 310 kN/m2and flow rate of 1200 kg/h.
Assuming isothermal flow, what is the pressure drop in 60 m of pipe
?Compressible fluid flow: Compressible fluid flow: adiabatic steady
flow adiabatic steady flow2105370 2105370 2105370
[email protected] fluids flow:
Compressible fluids flow: Adaibatic AdaibaticS.ApinanTemperature
not constant along the pipe Temperature not constant along the
pipeAs the similar way, using the balance equation to solve the
problem. As the similar way, using the balance equation to solve
the problem. Using the Re and moody chart to calculate the lost
work. Using the Re and moody chart to calculate the lost work.
Because the temperature is not constant, the Re numbers is also not
Because the temperature is not constant, the Re numbers is also not
constant. However, it is not different so much along the pipe. The
average constant. However, it is not different so much along the
pipe. The average Re could be used. Furthermore, Re could be used.
Furthermore, for M< for M dx dSFor spontaneous process (with
irreversibility):Subsonic flow: M1P2< P1u2> u1P2> P1u2<
u1However, for the supersonic flow is unstable in constant area
pipe, the compression shock occurs results in increase in pressure
and decrease in velocity to a subsonic value157S.ApinanReference
Reference1. Thermodynamics an Engineering Approach, Y. A. engel and
M. A. Boles, 6thed., 20082. Introduction to Chemical Engineering
Thermodynamics, 6th Edition, J. M. Smith, H. C. Van Ness and M. M.
Abbott, McGraw-Hill, 20013. , , .. 25384. Chemical Engineering
Thermodynamics I: .. 5. Chemical Engineering Thermodynamics I: ..
6. Unit Operations of Chemical Engineering, 7thedition, W.L.
McCabe, J.C. Smith and P. Harriott158Stagnation Properties
Stagnation PropertiesStagnation (or total) enthalpyStatic enthalpy:
the ordinary enthalpy hSteady flow of a fluid through an adiabatic
duct.Energy balance (with no heat or work interaction, no change in
potential energy) S.ApinanDuring a stagnation process, the kinetic
energy of a fluid is converted to enthalpy, which results in an
increase in the fluid temperature and pressure. The properties of a
fluid at the stagnation state are called stagnation properties
(stagnation temperature, stagnation pressure, stagnation density,
etc.). The stagnation state is indicated by the subscript 0.159The
actual state, actual stagnation state, and isentropic stagnation
state of a fluid on an h-s diagram.Isentropic stagnation state:
When the stagnation process is reversible as well as adiabatic
(i.e., isentropic).The stagnation processes are often approximated
to be isentropic, and the isentropic stagnation properties are
simply referred to as stagnation properties. When the fluid is
approximated as an ideal gas with constant specific heatsT0is
called the stagnation (or total) temperature, and it represents the
temperature an ideal gas attains when it is brought to rest
adiabatically.The term V2/2cpcorresponds to the temperature rise
during such a process and is called the dynamic
temperature.S.Apinan160DUCT FLOW WITH HEAT TRANSFER AND DUCT FLOW
WITH HEAT TRANSFER AND NEGLIGIBLE FRICTION (RAYLEIGH FLOW)
NEGLIGIBLE FRICTION (RAYLEIGH FLOW)So far we have limited our
consideration mostly to isentropic flow (no heat transfer and no
irreversibilities such as friction). Many compressible flow
problems encountered in practice involve chemical reactions such as
combustion, nuclear reactions, evaporation, and condensation as
well as heat gain or heat loss through the duct wall. Such problems
are difficult to analyze exactly since they may involve significant
changes in chemical composition during flow, and the conversion of
latent, chemical, and nuclear energies to thermal energy. A
simplified model is Rayleigh flow.Rayleigh flows: Steady
one-dimensional flow of an ideal gas with constant specific heats
through a constant-area duct with heat transfer, but with
negligible friction. S.Apinan161Quiz III: Compressible fluid flow
Quiz III: Compressible fluid flowS.ApinanHelium expands in a nozzle
from 1 MPa, 500 K, and negligible velocity to 0.1 MPa. Calculate
the throat and exit areas for a mass flow rate of 0.25 kg/s,
assuming the nozzle is isentropic. Assume: He as the ideal gas with
the properties of R = 2.0769 kJ/kg.K, Cp = 5.1926 kJ/kg.K, = 1.667.
162The effects of back pressure on the flow through a
convergingdiverging nozzle.When Pb= P0(case A), there will be no
flow through the nozzle. 1. When P0> Pb> PC, the flow remains
subsonic throughout the nozzle, and the mass flow is less than that
for choked flow. The fluid velocity increases in the first
(converging) section and reaches a maximum at the throat (but M
< 1). However, most of the gain in velocity is lost in the
second (diverging) section of the nozzle, which acts as a diffuser.
The pressure decreases in the converging section, reaches a minimum
at the throat, and increases at the expense of velocity in the
diverging section.Converging Converging Diverging Nozzles Diverging
NozzlesS.ApinanPPbb= Exit pressure = Exit pressure1632. When Pb=
PC, the throat pressure becomes P* and the fluid achieves sonic
velocity at the throat. But the diverging section of the nozzle
still acts as a diffuser, slowing the fluid to subsonic velocities.
The mass flow rate that was increasing with decreasing Pbalso
reaches its maximum value.3. When PC> Pb> PE, the fluid that
achieved a sonic velocity at the throat continues accelerating to
supersonic velocities in the diverging section as the pressure
decreases. This acceleration comes to a sudden stop, however, as a
normal shock develops at a section between the throat and the exit
plane, which causes a sudden drop in velocity to subsonic levels
and a sudden increase in pressure. The fluid then continues to
decelerate further in the remaining part of the convergingdiverging
nozzle. Converging Converging Diverging Nozzles Diverging
NozzlesS.Apinan1644. When PE> Pb > 0, the flow in the
diverging section is supersonic, and the fluid expands to PFat the
nozzle exit with no normal shock forming within the nozzle. Thus,
the flow through the nozzle can be approximated as isentropic. When
Pb= PF, no shocks occur within or outside the nozzle. When Pb<
PF, irreversible mixing and expansion waves occur downstream of the
exit plane of the nozzle. When Pb> PF, however, the pressure of
the fluid increases from PFto Pbirreversibly in the wake of the
nozzle exit, creating what are called oblique shocks.Converging
Converging Diverging Nozzles Diverging NozzlesS.Apinan165SHOCK
WAVES AND EXPANSION WAVES SHOCK WAVES AND EXPANSION WAVESFor some
back pressure values, abrupt changes in fluid properties occur in a
very thin section of a convergingdiverging nozzle under supersonic
flow conditions, creating a shock wave. We study the conditions
under which shock waves develop and how they affect the flow.Normal
ShocksNormal shock waves: The shock waves that occur in a plane
normal to the direction of flow. The flow process through the shock
wave is highly irreversible.Schlieren image of a normal shock in a
Laval nozzle. The Mach number in the nozzle just upstream (to the
left) of the shock wave is about 1.3. Boundary layers distort the
shape of the normal shock near the walls and lead to flow
separation beneath the shock.166Compressible fluids flow: Feature
Compressible fluids flow: FeatureCompressible flows can have
features that do not occur in low-speed flows. For example, shock
waves and expansion waves can occur in supersonic flows. Another
important phenomenon that can occur due to compressibility is
choking, where the mass flow rate through a duct system may be
limited as a result of the Mach number being equal to 1 at some
point in the flow. Another effect of compressibility is associated
with the acceleration of a gas flow through a duct. In
incompressible flow, an increase in velocity is associated with a
decrease in the cross-sectional area of the duct, this in fact
being true as long as M < 1. However, when M > 1, that is,
when the flow is supersonic, the opposite is true; that is, an
increase in the velocity is associated with an increase in the
cross-sectional area. Therefore, in order to accelerate a gas flow
from subsonic to supersonic velocities in a duct, it is necessary
first to decrease the area and then, once the Mach number has
reached 1, to increase the area, that is, to use a so-called
convergent-divergent nozzle. An example is the nozzle fitted to a
rocket engine. S.Apinan167Compressible fluids flow: Choke flow
Compressible fluids flow: Choke flowFluid flow through a restricted
area whose rate reaches a maximum when the fluid velocity reaches
the sonic velocity at some point along the flow path. The
phenomenon of choking exists only in compressible flow and can
occur in several flow situationsChoked flow can occur through a
convergent flow area or nozzle attached to a hugereservoir. Flow
exits the reservoir through the nozzle if the back pressure is less
than thereservoir pressure. When the back pressure is decreased
slightly below the reservoirpressure, a signal from beyond the
nozzle exit is transmitted at sonic speed to thereservoir. The
reservoir responds by sending fluid through the nozzle. Further,
themaximum velocity of the fluid exists at the nozzle throat where
the area is smallest.When the back pressure is further decreased,
fluid exits the reservoir more rapidly.Eventually, however, the
velocity at the throat reaches the sonic velocity. Then the
fluidvelocity at the throat is sonic, and the velocity of the
signal is also sonic. Therefore, furtherdecreases in back pressure
are not sensed by the reservoir, and correspondingly will notinduce
any greater flow to exit the reservoir. The nozzle is thus said to
be choked, and themass flow of fluid is a maximum. See also Mach
number; Nozzle; Sound; Supersonicdiffuser.Through varying Through
varying--area duct area ductS.Apinan168Compressible fluids flow:
Choke flow Compressible fluids flow: Choke flowChoked flow can also
occur through a long constant-area duct attached to areservoir. As
fluid flows through the duct, friction between the fluid and the
ductwall reduces the pressure acting on the fluid. As pressure is
reduced, other fluidproperties are affected, such as sonic
velocity, density, and temperature. Themaximum Mach number occurs
at the nozzle exit, and choked flow results whenthis Mach number
reaches.With friction With frictionS.ApinanWith heat addition With
heat additionA reservoir with a constant-area duct attached may
also be considered in the case that the flow through the duct is
assumed to be frictionless but heat is added to the system along
the duct wall.
