Gas Dynamics-isothermal Flow

Isothermal flow with friction Introduction Ø The Fanno flow of an ideal gas through a constant area duct under adiabatic condition is achievable in practice when the duct is not very long. Ø Avoiding heat transfer to the environment is not convenient when a gas line is taken over long lengths. Ø The supply of natural or by-product gases over long pipe lines from an industrial area to a consuming city. The pipe is exposed to the atmosphere and heat transfer through the pipe wall is a reality. The pipe gas attains the environment temperature by heat exchange. Ø The study of isothermal flow of an ideal gas through a constant area pipe is applicable to long pipe lines. Friction is accounted for. GDJP PDF created with pdfFactory trial version www.pdffactory.com

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Cont.. Ø The Reynolds number may be written in the form

ρVD GD Re = = µ µ where, G - mass velocity or mass flow density D - duct diameter μ - viscosity of the fluid For a constant temperature the viscosity of the flowing fluid μ is a constant. Since G is constant, the Reynolds number is constant at all points in the flow. Friction coefficient for a given pipe surface is a function of the Reynolds number alone, it follows that in isothermal flow friction coefficient is invariant along the pipe. GDJP PDF created with pdfFactory trial version www.pdffactory.com

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Isothermal flow process- T-s diagram Stagnation temperature line

Isothermal line

M<

1 γ

M=

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1 γ

M>

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1 γ

Isothermal flow – Assumptions 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 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.. From continuity G =ρc = constant

dρ dc = − ρ c

From first law of thermodynamics

δQ = dho ; ho1 ≠ ho2

From equation of state p = ρRT From momentum equation dp dx γM 2 4 f = − p D 2  1 − γ M 2   

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dp dρ = p ρ

Direction of isothermal flow process 1 − γ M 2 = + ve

Case (a)

This

gives γ M 2 < 1

∴ M <

1 γ

Case (b) 1 - γ M 2 = − ve This gives γ M 2 > 1 ∴M >

γM 2

dp dx = − 4 f p D 2  1 − γ M 2   

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1 γ

Limiting Case t 1 * M = γ

Variation of flow parameters- Isothermal flow Parameter Pressure

M < 1/√γ

M > 1/√γ

(Heating)

(Cooling)

Decreases

Increases

Density

Decreases

Increases

Velocity

Increases

Decreases

Mach number

Increases

Decreases

Stagnation Temperature

Increases

Decreases

Decreases

Increases for M <

2 γ +1

Decreases for M >

2 γ +1

Stagnation pressure

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Property Variations- Isothermal Flow

0.845

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0.912

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Isothermal Flow – Property ratios Velocity and Density

Stagnation Temperature

C 2 ρ1 M 2 = = C1 ρ2 M 1

γ −1 2   1 M1  + T01  2 =  γ −1 2  T02  1+ M2   2

Stagnation Pressure P01 M 2 = P02 M1

γ  1 +   1 + γ 

−1 2  M1  2  −1 2  M2  2

γ γ −1

Impulse Function 1 + γ M 12 F1 M2 = × F2 M1 1 + γ M 22

Change of Entropy  p1   S 2 − S 1 = R ln   p 2    M2   = R ln    M1 

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Duct Length 4f

L L L     =  4 f max  −  4 f max  D  D M  D M 1 2

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Limiting Conditions Choking velocity Ø As the gas flows along the pipe, its static pressure decreases. Eventually a limiting condition is reached where the pipe cannot be increased in length without altering the upstream conditions; that is the flow has become choked. Ø Hence the limiting or choking velocity for the isothermal flow of a perfect gas in a constant-area duct in the presence of wall friction alone is ML =

1 γ

cL 1 = aL γ

a L or a1 ⇒ CL = γ

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Q T1 = TL

Isothermal flow- Problem 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

a1 = a = a* = γRT1 = 1.4 × 287× 300 = 347.2 m / sec c1 175 M 1= = = 0.50 a1 347.2 GDJP PDF created with pdfFactory trial version www.pdffactory.com

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Cont.. We know that 4f

L L    L  =  4 f max  −  4 f max  D  D M  D M 1 2

at M1 = 0.50 refer isothermal flow table Lmax 0.807× D 0.807× 0.150 4f = 0.807 ⇒ Lmax = = = 6.05 m D 4 × 0.005 4f

(b) Limiting velocity and pressure 1 = 0.84515 ⇒ c* = M*a* For isothermal flow M* = γ p2 M1 = We know that p1 M2

= 0.84515 × 347. 2 = 293.44 m / s 0.50 M1 * ⇒ p2 = p = p1× = 3.5 × = 2.07 bar M2 0.84515

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Cont.. (c) the length of pipe at the station where the Mach no. is 0.60 4f

L L    L  =  4 f max  −  4 f max  D  D M  D M 1 2

L at M1 = 0.50 from isothermal table 4f = 0.807 and D L at M2 = 0.60 from isothermal table 4f = 0.299 D L 4(0.005) = 0.807 − 0.299 = 0.508 0.150 0.508× 0.150 ∴L = = 3.81 m 4 × 0.005 GDJP PDF created with pdfFactory trial version www.pdffactory.com

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