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Compressible Fluid Flow
.
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OutlineOutline
OutlineOutline Concept of compressible fluid
Mach number a criteria for compressible fluid flowcharacterization.
Processes of compressible fluid flow:
Adiabatic flow with friction
Isothermal flow with friction
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Compressibility, Z
A measure of the change in density that will beproduced in the fluid by a specified change in
.
Gases highly compressible.
Liquid very low compressibility.
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In fluid flow there are usuall occur chan esin pressure associated with changes of otherparameters of the flow
For example, changes in the velocity in the.
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ese pressure c anges w , n genera ,
cause density changes which will
the fluid involved will have an
influence on the flow.
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en ens y c anges are mpor an ,
temperature change in the flow that may arise
also influence on the flow
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In other words
,
the temperature changes in the flow
.
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Although the density changes in the flow fieldcan be very important, there exist many situations
of great practical importance in which the effects
of these density and temperature changesare negligible.
Example: Flow of incompressible fluid
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Incompressible fluid flow
The pressure and kinetic energy changes are so small
temperature changes in the fluid flow
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There are however a number of flows that are of reat
practical importance in which this assumption is notadequate.
The density and temperature changes being so large .
In such cases it is necessar to stud thethermodynamics of the flow simultaneously
with its dynamics.
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The study of these flows in which
e c anges n ens y an empera ure are mpor an
is known as
compress e u ow or gas ynam cs.
s c ap er w ocus on gas ows w erecompressibility effects are important.
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ApplicationsApplicationsApplicationsApplications
Although most obvious applications ofcompressible fluid flow theory are
in the design of high speed aircraft,
a knowledge of compressible fluid flow theoryis required in the design and operation
of many devices commonly encountered in
engineering practice.
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Gas and steam turbines e ow n e a ng an nozz es s rea e as compress e.
Reciprocating engines the flow of the gases through the valves and in the intake and
exhaust systems.
Natural gas transmission lines compressibility effects are important in calculating the flow through
such problems.
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Processes of Compressible FlowProcesses of Compressible FlowProcesses of Compressible FlowProcesses of Compressible Flow
Conver ent Diver ent
FlowReservoir Receiver
Thermal insulation
Isentropic flow
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, ,
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Isentropic Friction section
Flow
Thermal insulation
Adiabatic flow with friction
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IsentropicFriction section
ow
Isothermal flow with friction
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Speed of Sound, cSpeed of Sound, cSpeed of Sound, cSpeed of Sound, c
Small ressure disturbance that move throu h a
continuous medium
wave (sound wave) propagates through a fluid)
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Sound
A series of small air-pressure disturbances oscillation insinusoidal fashion in the frequency range from
, cyc es per secon .
More rigid material, speed of sound greater.
Note: Other text use the notation a for speed of sound.
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A property of a material / compound.
..(1)kRTkPPc22
S
= =
=
Where:k = specific heat ratio, Cp/CvP = absolute pressure of the fluid (kPa, psi or equivalent)
= 3
R = specific gas constant (kJ/kgK or equivalent)
T = absolute temperature of the fluid (K oroR)
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Mach Number, MaMach Number, MaMach Number, MaMach Number, Ma
For incompressible fluid, Reynolds number, Re is.
In compressible fluid, Mach number, Ma is useful in.
Mach number, Ma a o o u ve oc y an spee o soun .
Dimensionless.
c
VMa =
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BornFebruary 18, 1838
Brno, Austrian Empire
Februar 19, 1916 1916-02-19 a ed 78
Munich, German Empire
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Characteristic of compressible fluid flow:
a < : u son c ow
Ma = 1 : Sonic flow
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Concorde, Ma = 2
(supersonic aircraft)Boeing 747, Ma = 0.85 0.95
(high speed, subsonic aircraft)
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.Sukhoi 30 MKM Ma 2.3
MZA@UTPChemEFluidMechF 18 Ma 1.8 Hawk Ma 0.84
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Analysis of compressible fluid flowAnalysis of compressible fluid flowAnalysis of compressible fluid flowAnalysis of compressible fluid flow
The gas (compressible fluid) is transferred from a very
temperature, TR
pressure, PR
velocity, VR = 0
roper es o u a reservo r are
called reservoir (stagnation)
conditions.
RESERVOIR Receiver
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Stagnation conditions are those that would exist if the
brought to rest (velocity = 0)
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Assumptions:
Flow is one-dimensional
Velocity gradients within a cross section areneglected
Friction is restricted to wall shear
Shaft work is zero
Gravitational effects are negligible
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Steady, frictionless, adiabaticSteady, frictionless, adiabaticSteady, frictionless, adiabaticSteady, frictionless, adiabatic
flowflowflowflow
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Convergent Divergent
FlowReservoir Receiver
Thermal insulationThroat
Provided the reservoir conditions (TR and PR) how
interest?
Need for relation between reservoir condition and
MZA@UTPChemEFluidMech
point of interest.
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From THERMODYNAMICS, open-system energy balancefrom reservoir (R) to point/state of interest (1):
22 VV
1R2
gz2
gz
++=
++
, R ,
.. 1RP1R1 ==
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For ideal gas: CRC vP +=
Ck
v
P=
( )1kCP
=
Then Eq. (2) will become;
Rk2
( ) ( )TT2Chh2V 1RP1R2
1 ==
( )= 1k 1R1
MZA@UTPChemEFluidMech
( )
= T1k
1
11
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Taking the term RkT1 to the left hand side:
( )1
TT
1k2
RkTV
1
R
1
2
1
=
From Eq. (1), for condition at point 1 11 kRTc =21
21
11c =
2
kRTcS
=
=
=
Then; .(3)( ) 1T1kc 1R
2
1
1 =
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V
= From Eq (3), substitute Ma and rearrange:
c
..(4)( ) 121kMa
TT
2
1R +=
,temperature and Mach numberat point of interestfor a given fluid.
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For compressible fluid, pressure and density change
The isentropic (frictionless, adiabatic) relation is givenb :
(a)T
P
1k
k
RR
=11
1
(b)TT 1k
1
R
1
R
=
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Substituting Eq. (4) into (a) and (b):
..(5)( )1
1kMa
P
1kk
2
1R
+
=2P1
12 ..
12
1
1
R
+
=
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FlowReservoir Receiver
Steady, frictionless, adiabatic flow can be achieved iffluid flow in a variable cross sectional area nozzle.
Then, there is a need to find relation between areaperpendicular to the flow with respect to the area of
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.
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FlowReservoir Receiver
In any flow, mass is conserved.
From continuit e uation: AVAV =
11R VA =
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RR1
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RR
11
1
R
VA =
Using equation above is not practical since: A is ver lar e
VR = 0From previous assumptions
Therefore other reference point is needed apart fromthe reservoir
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In any such flow there will be a state where Ma = 1.
s s ca e e cr ca s a e.
Properties related to critical state is called critical* .
(Dont confuse yourself with definition of Pcror Tcr)
Then the relation is written as:
1 VA =
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11
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Relation between area perpendicular to theow an area a cr ca s a e:
( )( )1k2
1k2
+
)
1
1k
12
Ma
1
A
A
+=
2
At subsonic
to get the fluid go faster, one must reduce thecross sectional area perpendicular to the flow.
At supersonic to get the fluid go faster, one must increase
MZA@UTPChemEFluidMech
the cross sectional area perpendicular to the flow.
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Similarly, mass flow rate of fluid can be determinedfrom continuity equation with respect to the critical
2
RRT
kPm
( )( )( )1k2
1k
1
1kA
+
+
=
For air with k = 1.4: RT6847.0m R
R
=
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When the gas is blown down, there will be a decrease.
=,
Mass flow rate can also be determined usin :
VAVAm ==
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Example 1Example 1Example 1Example 1
Air flowing through an insulated, frictionless nozzle is, .
Determine: The Mach number Ma
The temperature, T
The density, The air velocity, V
at a location in the duct where the pressure is 430 kPa.
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SolutionSolutionSolutionSolution
Assumption:
r c on ess, nsu a e uc a a a c
Air is an ideal gas
. , p . ,R = 0.287 kJ/kgK
Given: TR = 400 K, PR = 500 kPa
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SolutionSolutionSolutionSolution
For isentropic gas flow:k
( ) 121kMa
PP
1k
k2
1
1
R
+=
T
T
P
P 1k
1
R
1
R
=
1P2
Mak
1k
R
1
=
P
PTT
k
R
1R1
=
5002 4.114.1
( )
500
430400
4.1
14.1
=
( )
469.0
43014.1
=
=K383 =
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SolutionSolutionSolutionSolution
Density, 11
RT
P= The speed of sound
( )( )383287.0430
=K
mkPa383287.04.1
213
=
3m
kg3.91 =
mkPa
Kkg
2
13
Fluid velocity, V1m
kg.
=
=
12.410.469
ca1
=
=s
.
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s82.5 =
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Adiabatic Flow with FrictionAdiabatic Flow with FrictionAdiabatic Flow with FrictionAdiabatic Flow with Friction
Isentropic Friction section
Thermal insulation
Occurs when a gas flows through a length of pipe at
high velocity.
If pipe is insulated or flow is fast, heat transfer is
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considered negligible adiabatic.
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Adiabatic Flow with FrictionAdiabatic Flow with FrictionAdiabatic Flow with FrictionAdiabatic Flow with Friction
Isentropic Friction section
Thermal insulation
Effect of friction due to the flow will cause the entropy
o ow ng gas o ncrease en ropy s no cons an Therefore isentropic relation cannot be applied in the
MZA@UTPChemEFluidMech
analysis.
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Ffriction
T
V
+T + dT
V + dV
dx
Applying the momentum balance:
Net pressure force Force due to wall shear stress= Mass flow rate x (Velocity out Velocity in)
( )[ ] ( )[ ] frictionFPdPPAVdVVm ++=+
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Ffriction
T
V
+T + dT
V + dV
dx
Applying the continuity equation:
AV = constantV = constant (since A is constant) V = ( + d)(V + dV)
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F
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Ffriction
T
V
+T + dT
V + dV
dx
Applying the energy balance:
( )dVVV22
+= 22
P + dP = ( + d) R (T + dT)
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F
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Ffriction
T
V
+T + dT
V + dV
dx
Also from Mach numberdefinition:
VV
2
2
( )dVVkRTc
22 +=
dTTkR +
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F
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Ffriction
T
V
+T + dT
V + dV
dx
The equations represents a set of equations withunknown dP, dT, d, dV and dMa
Have to be solved accordingly to obtain appropriateex ressions.
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F
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Ffriction
T
V
+T + dT
V + dV
dx
In momentum balance, there exist the term wallshear stress,
wall.
In pipeline system, this is expressed as dimensionlessvalue friction factor, f
Most compressible gas flows in duct involve turbulentflow
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Solving from the equation (for circular pipe):
( )
+++ =
2
2
2
21
22
Ma1k2
1Maln1k111x4 f
( )
+ 21221 Ma1k2
1
The equation describe the change of Ma over a givenlength.
,condition Ma 1.
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When friction is involved, flows tend to reach soniccondition Ma 1. (Ma2 approaching 1)
By setting Ma2 = 1, the length of duct required to givee va ue o a1 s o a ne as max mum eng , max
(or critical length, L*)
++ 22 M11M1*
( ) +
+
=2
1
21 Ma1k
2112
n2kMakD
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ExampleExample
ExampleExample
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ExampleExampleExampleExample
Air flows in a 5 cm diameter pipe. The air entersa a = . an s o eave a a = . . e erm ne
the length of pipe required. What would be the
Assume f = 0.002 and adiabatic flow.
MZA@UTPChemEFluidMech
SolutionSolution
SolutionSolution
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SolutionSolutionSolutionSolution
Flow is adiabatic.
( )Ma1k2
1
1Maln1k111x4
2
2
2
21
22
f
+++ =
( )Ma1k2
1 21221
+
( )( )
2.111
1
.2
1.5
2.5ln
2(1.4)
141
1.5
1
2.5
1
1.4
1
050
L00204
22
2
22
.
..
.
.
+
+
++
=
m1850L2
.=
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Maximum pipe length, L*:
( )1Ma1kln
2k1k
MkMa1
DL4
2
12
2
1*f +++ =a
2
22
1
*
+
( ) ( ) ( )( )2.51412
112
.ln1.42
.2.541.
050
22
.
.
...
+
+
=
m72L .* =
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Pressure relation: ( )
+ 22Ma
2
1k1
MaP
( )
+
=2112 Ma2
1k1MaP
Temperature relation:
( )
+ 22Ma
1k1
( ) +=
2
12 Ma
21k1T
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Density relation:
( ) 2Ma1k1 +
( ) 22
1
2
1
2
2
1
2
1
Ma
1k
1
Ma
a
TP
+
==
MZA@UTPChemEFluidMech
Isothermal Flow with FrictionIsothermal Flow with Friction
Isothermal Flow with FrictionIsothermal Flow with Friction
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Isothermal Flow with FrictionIsothermal Flow with FrictionIsothermal Flow with FrictionIsothermal Flow with Friction
IsentropicFriction section
FlowReservoir Receiver
Occurs in long, small, uninsulated pipe in contact withenvironment transmit sufficient heat to kee the flow
isothermal. E.g.: flow of natural gas through long distance pipelines.
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FfrictiondQ
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V
+V + dV
+ d
dx
Applying the continuity equation:
AV = constantV = constant (since A is constant)V = ( + d)(V + dV)
=
ddV
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FfrictiondQ
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V
+V + dV
+ d
dx
Applying the momentum balance:
Net pressure force Force due to wall shear stress= Mass flow rate x (Velocity out Velocity in)
( )[ ] ( )[ ] frictionFPdPPAVdVVm ++=+
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FfrictiondQ
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V
+V + dV
+ d
dx
Applying the energy balance:
+
VdVV
22
= 22 PP
=
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FfrictiondQ
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V
+V + dV
+ d
dx
From ideal gas EOS (P = RT)P + dP = ( + d) RT
=
ddP
P
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FfrictiondQ
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V
+V + dV
+ d
dx
Also from Mach numberdefinition:
V
Ma=(since T constant, c constant)dVV
dMaMa
c
+=+
dV
M
dMa =
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FfrictiondQ
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V
+V + dV
+ d
dx
The equations represents a set of equations withunknown dP, dT, d, dV and dMa
expressions.
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Isothermal Flow with FrictionIsothermal Flow with Friction
Isothermal Flow with FrictionIsothermal Flow with Friction
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Q
1
V1
1
2
V2
2
x
1 2
1
2
2
1
2
1
1
2
Ma
Ma
P
P
V
V==
=
= 2
Mal
111L2f
MZA@UTPChemEFluidMech
121 MaMaMa2kD
Q
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V11Ma
V22Ma
x
For isothermal flow with friction, Ma tend to reachk
1
1 When Ma ~ , dq ~ infinity.
An infinite amount of heat must be transferred or removed to kee
k
the temperature of the gas constant.
Limiting value for Ma =
k1
MZA@UTPChemEFluidMech
Isothermal Flow with FrictionIsothermal Flow with Friction
Isothermal Flow with FrictionIsothermal Flow with Friction
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By setting Ma2 = , maximum length Lmax or L*1
2kMa1L4 *12
1
ankMaD
+
=
MZA@UTPChemEFluidMech
ExampleExample
ExampleExample
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pppp
Air flows through a 5 cm diameter pipeline. The flow, .
pressure of 900 kPa, and exit at Ma = 0.5. Determine: Len th of the i e.
Maximum pipe length and the correspondingpressure.
Mass flow rate of the air.
e spec c ea ra o or a r an e mean r c onfactor may be taken as 1.4 and 0.004 respectively.
MZA@UTPChemEFluidMech
SolutionSolution
SolutionSolution
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Length of pipe:
( )( ) 0.1
0.5
ln0.5
1
0.1
1
1.42
1
050
L0040222.
.
=
m2204L .=
( ) ( )( )0.11.41L00404 22
*.
( )( )m43233L
..
0.11.4050
2
.*.
=
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Pressure at the maximum length:
MaPFrom 12 =
1Maand*PPL*,at 2 ==
kPa51069000.1
P*
.
.==
1.4
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At T = 293 K,c = = 10.85 m/s
V1 = Ma1 c = (0.1)(10.85) = 1.85 m/skRT
( ) kg710kPa900P 3311 ===.
K293Kkg
2870
.
851050710
m
2
11
=
=
...
kg0.0389 =
MZA@UTPChemEFluidMech
ExampleExample
ExampleExample
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Natural gas flows through a 0.075 m diameter pipeline.
assumed to be isothermal with a temperature of 15
o
C.The Mach number and pressure at the inlet are 0.09and 900 kPa, respectively. If the mean friction factorfor the flow is 0.002, determine the Mach number at
.of the pipe and the exit pressure with this length of pipe.
Assume the flow is steady and the specific heat ratiofor the natural gas is 1.3.
MZA@UTPChemEFluidMech
SolutionSolution
SolutionSolution
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Assumption: Isothermal, steady flow, ideal gas
= Maln111L22
22f
( )( )
=Ma
ln11175000202 2
121
.
=Ma
l111
4
...
2
2.
0.09Ma0.00812.6
2
MZA@UTPChemEFluidMech
Unknowns Ma2 solve through trial-and-error
Ma2, uess LHS RHS Error
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0.1 40 8.916 -31.084
0.2 40 37.068 -2.932
0.25 40 40.3 0.300
o grap a vs rror
y = 64.76x - 15.88
0
0.5
,Ma, x = 0.245
-1.5
-1
-0.5 0.2 0.25
Ma
-3.5
-3-2.5
-2
MZA@UTPChemEFluidMech
Error
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MaP 21
1
2 =
kPa6330900
2450
P2 .
.
.==
Maximum possible length:
2kMaln
kMaD
2
12
1
1 +
=
( )( ) ( )( )( )0.091.3ln0.091.30.091.31
0750
L00204 22 +
=.
*.
MZA@UTPChemEFluidMech
m838.2L =*
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MaPFrom 12 =
1Maand*PPL*,at 2
21
==
kPa4299000.09
P*
.
.==
1.3
MZA@UTPChemEFluidMech
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