Isothermal flow with friction
GDJP Anna University
IntroductionØ The Fanno flow of an ideal gas through a constant area duct
under adiabatic condition is achievable in practice when theduct is not very long.
Ø Avoiding heat transfer to the environment is not convenientwhen a gas line is taken over long lengths.
Ø The supply of natural or by-product gases over long pipelines from an industrial area to a consuming city.
The pipe is exposed to the atmosphere and heattransfer through the pipe wall is a reality. The pipe gasattains the environment temperature by heat exchange.
Ø The study of isothermal flow of an ideal gas through aconstant area pipe is applicable to long pipe lines. Friction isaccounted for.
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Cont..
GDJP Anna University
Ø The Reynolds number may be written in the form
where, G - mass velocity or mass flow densityD - duct diameterμ - viscosity of the fluid
For a constant temperature the viscosity of the flowingfluid μ is a constant. Since G is constant, the Reynoldsnumber is constant at all points in the flow.
Friction coefficient for a given pipe surface is a functionof the Reynolds number alone, it follows that in isothermalflow friction coefficient is invariant along the pipe.
µµρ GDVD
==Re
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Isothermal flow process- T-s diagram
GDJP Anna University
Isothermal line
Stagnation temperature line
γ
1>M
γ
1=M
γ
1<M
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Isothermal flow – Assumptions
GDJP Anna University
The following assumptions are used to derive
equations for T=constant flow :
1. Perfect gas
2. Constant diameter duct
3. Absence of body forces
4. Steady, one dimensional flow
5. Simple Diabatic, frictional flow at T=cons.
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Isothermal flow – Fundamental equations
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To describe isothermal flow process, the following
governing relationships are generally used :
Ø Continuity equation
Ø Equation of state
Ø Energy equation
Ø Momentum equation
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Cont..
GDJP Anna University
From continuity G =ρc = constantc
dcd−=
ρρ
From first law of thermodynamics 21 ; ohohodhQ ≠=δ
From equation of state p = ρRTρρd
pdp
=
From momentum equation
Ddxf
M
Mp
dp 4 212
2
−
−=γ
γ
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Direction of isothermal flow process
GDJP Anna University
Ddxf
M
Mp
dp 4 212
2
−
−=γ
γ
Case (a) veM +=− 21 γ
γ
γ1M
12M
<∴
<givesThis
γ
γ
γ
112
2M-1 (b) Case
>∴
>
−=
M
MgivesThis
ve
γ
1*M
Case Limitingt
=
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Variation of flow parameters- Isothermal flow
GDJP Anna University
Parameter M < 1/√γ M > 1/√γ
Pressure
Density
Velocity
Mach number
Stagnation Temperature
Stagnation pressure
Decreases
Decreases
Increases
Increases
Increases
Decreases
Increases
Increases
Decreases
Decreases
Decreases
12M
12M
+>
+<
γ
γ
forDecreases
forIncreases
(Cooling)(Heating)
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Property Variations- Isothermal Flow
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0.845 0.912
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Isothermal Flow – Property ratios
GDJP Anna University
Velocity and Density
12
21
12
MM
CC
==ρρ
Stagnation Pressure
−+
−+
=222
11
212
11
0201
M
M
TT
γ
γ
Stagnation Temperature
1
22
211
21
211
12
02P01P
−
−+
−+
=
γγ
γ
γ
M
M
MM Impulse Function
221
211
12
21
M
MMM
FF
γ
γ
+
+×=
Change of Entropy
1M2M lnR
2p1p ln 12
=
=− RSS
Duct Length
21
max4max44MD
LfMD
LfDLf
−
=
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Limiting Conditions
GDJP Anna University
Choking velocityØ As the gas flows along the pipe, its static pressure
decreases. Eventually a limiting condition is reached wherethe pipe cannot be increased in length without altering theupstream conditions; that is the flow has become choked.
Ø Hence the limiting or choking velocity for the isothermalflow of a perfect gas in a constant-area duct in thepresence of wall friction alone is
LT1T 1 L 1
1
==⇒=
=
Qγγ
γ
aorLaCLaLc
LM
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Isothermal flow- Problem
GDJP Anna University
Air flows in a long pipe (diameter=0.150 m) under isothermal conditions. At the pipe inlet, the static temperature and pressure are 300 K and 3.5 bar, respectively, and the velocity is 175 m/s. The friction coefficient is 0.005
Calculate (a) the length of pipe required to choke the flow, (b) the limiting velocity and pressure, and (c) the length of pipe at the station where the Mach number is 0.60
Solution:(a) The inlet speed of sound and Mach number are
50.02.347
175
111
sec/ 2.3473002874.11*
1
===
=××====
acM
mRTaaa γ
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Cont..
GDJP Anna University
21
max4max44MD
LfMD
LfDLf
−
=
We know that
mf
DD
Lat
05.6005.04
150.0807.04
807.0maxL 807.0maxf4
table flow isothermalrefer 50.01M
=×
×=
×=⇒=
=
(b) Limiting velocity and pressure
barMMpp
MMthatknowWe
sm
aMFor
07.284515.0
50.05.3211
*2p
21
1p2p
/ 44.2932 . 347 84515.0
***c 84515.01*M flow isothermal
=×=×==⇒=
=×=
=⇒==γ
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Cont..
GDJP Anna University
(c) the length of pipe at the station where the Mach no. is 0.60
21
max4max44MD
LfMD
LfDLf
−
=
mL
LDLat
DLat
81.3005.04
150.0508.0
508.0299.0807.0150.0
4(0.005)
299.0f4 table isothermal from 60.02M
and 807.0f4 table isothermal from 50.01M
=×
×=∴
=−=
==
==
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