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CP502 Advanced Fluid Mechanics. Compressible Flow. Lectures 1 & 2 Steady, quasi one-dimensional, isothermal, compressible flow of an ideal gas in a constant area duct with wall friction. Speed of the flow ( u ). M ≡. Speed of sound ( c ) in the fluid at the flow temperature.

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cp502 advanced fluid mechanics

CP502 Advanced Fluid Mechanics

Compressible Flow

Lectures 1 & 2

Steady, quasi one-dimensional, isothermal,

compressible flow of an ideal gas in a

constant area duct with wall friction

slide2

Speed of the flow (u)

M≡

Speed of sound (c) in the fluid

at the flow temperature

Incompressible flow assumption is not valid if Mach number > 0.3

What is a Mach number?

Definition of Mach number (M):

For an ideal gas,

specific heat ratio

specific gas constant (in J/kg.K)

absolute temperature of the flow at the point concerned (in K)

slide3

For an ideal gas,

u

u

=

M=

c

Unit of u = m/s

Unit of c = [(J/kg.K)(K)]0.5

= [J/kg]0.5

= (N.m/kg)0.5

= [kg.(m/s2).m/kg]0.5

= [m2/s2]0.5

= m/s

slide4

constant area duct

quasi one-dimensional flow

compressible flow

steady flow

isothermal flow

ideal gas

wall friction

is a constant

Diameter (D)

speed (u)

u varies only in x-direction

x

Density (ρ) is NOT a constant

Mass flow rate is a constant

Temperature (T) is a constant

Obeys the Ideal Gas equation

is the shear stress acting on the wall

where is the average Fanning friction factor

slide5

Friction factor:

  • For laminar flow in circular pipes:
  • where Re is the Reynolds number of the flow defined as follows:
  • For lamina flow in a square channel:
  • For the turbulent flow regime:

Quasi one-dimensional flow is closer to turbulent velocity profile than to laminar velocity profile.

slide6

Ideal Gas equation of state:

temperature

pressure

specific gas constant

(not universal gas constant)

volume

mass

Ideal Gas equation of state can be rearranged to give

K

Pa = N/m2

kg/m3

J/(kg.K)

slide7

Problem 1 from Problem Set 1 in Compressible Fluid Flow:

Starting from the mass and momentum balances, show that the differential equation describing the quasi one-dimensional, compressible, isothermal, steady flow of an ideal gas through a constant area pipe of diameter D and average Fanning friction factor shall be written as follows:

where p, ρ and u are the respective pressure, density and velocity at distance x from the entrance of the pipe.

(1.1)

slide8

p

p+dp

D

u

u+du

dx

x

Write the momentum balance over the differential volume chosen.

(1)

steady mass flow rate

cross-sectional area

shear stress acting on the wall

is the wetted area on which shear is acting

slide9

p

p+dp

D

u

u+du

dx

x

Equation (1) can be reduced to

Substituting

Since , and , we get

(1.1)

slide10

Problem 2 from Problem Set 1 in Compressible Fluid Flow:

Show that the differential equation of Problem (1) can be converted into

which in turn can be integrated to yield the following design equation:

where p is the pressure at the entrance of the pipe, pL is the pressure at length L from the entrance of the pipe, R is the gas constant, T is the temperature of the gas, is the mass flow rate of the gas flowing through the pipe, and A is the cross-sectional area of the pipe.

(1.2)

(1.3)

slide11

The differential equation of problem (1) is

in which the variables ρ and umust be replaced by the variable p.

(1.1)

Let us use the mass flow rate equation and the ideal gas equation to obtain the following:

and

and therefore

It is a constant for steady, isothermal flow in a constant area duct

slide12

,

Using

and

in

(1.1)

we get

(1.2)

slide13

p

pL

L

Integrating (1.2) from 0 to L, we get

which becomes

(1.3)

slide14

Problem 3 from Problem Set 1 in Compressible Fluid Flow:

Show that the design equation of Problem (2) is equivalent to

where M is the Mach number at the entry and ML is the Mach number

at length L from the entry.

(1.4)

slide15

Design equation of Problem (2) is

which should be shown to be equivalent to

where p and M are the pressure and Mach number at the entry and pL

and ML are the pressure and Mach number at length L from the entry.

(1.3)

(1.4)

We need to relate p to M!

slide16

We need to relate p to M!

which gives

= constant for steady, isothermal flow in a

constant area duct

Substituting the above in (1.3), we get

(1.4)

slide17

Summary

Design equations for steady, quasi one-dimensional, isothermal,compressible flow of an ideal gas in a constant area duct with wall friction

(1.1)

(1.2)

(1.3)

(1.4)

slide18

Problem 4 from Problem Set 1 in Compressible Fluid Flow:

Nitrogen (γ = 1.4; molecular mass = 28) is to be fed through a 15 mm-id commercial steel pipe 11.5 m long to a synthetic ammonia plant. Calculate the downstream pressure in the line for a flow rate of 1.5 mol/s, an upstream pressure of 600 kPa, and a temperature of 27oC throughout. The average Fanning friction factor may be taken as 0.0066.

γ = 1.4; molecular mass = 28;

= 1.5 mol/s;

= 0.0066

pL = ?

D = 15 mm

p = 600 kPa

T = 300 K

L = 11.5 m

slide19

γ = 1.4; molecular mass = 28;

= 1.5 mol/s;

= 0.0066

pL = ?

D = 15 mm

p = 600 kPa

T = 300 K

L = 11.5 m

Design equation:

(1.3)

= 0.0066;

= 20.240

L = 11.5 m;

D = 15 mm = 0.015 m;

unit?

slide20

γ = 1.4; molecular mass = 28;

= 1.5 mol/s;

= 0.0066

pL = ?

D = 15 mm

p = 600 kPa

T = 300 K

L = 11.5 m

Design equation:

(1.3)

p = 600 kPa = 600,000 Pa;

T = 300 K;

R = 8.314 kJ/kmol.K = 8.314/28 kJ/kg.K = 8314/28 J/kg.K;

= 1.5 mol/s = 1.5 x 28 g/s = 1.5 x 28/1000 kg/s;

A = πD2/4 = π(15 mm)2/4 = π(0.015 m)2/4;

= 71.544

unit?

slide21

γ = 1.4; molecular mass = 28;

= 1.5 mol/s;

= 0.0066

pL = ?

D = 15 mm

p = 600 kPa

T = 300 K

L = 11.5 m

Design equation:

(1.3)

20.240

= 71.544

p = 600 kPa = 600,000 Pa

Solve the nonlinear equation above to determine pL

pL = ?

slide22

20.240

= 71.544

p = 600 kPa = 600,000 Pa

Determine the approximate solution by ignoring the ln-term:

pL = p (1-20.240/71.544)0.5

= 508.1 kPa

Check the value of the ln-term using pL= 508.1 kPa:

ln[(pL /p)2] = ln[(508.1 /600)2]

= -0.3325

This value is small when compared to 20.240. And therefore pL= 508.1 kPa is a good first approximation.

slide23

Now, solve the nonlinear equation for pL values close to 508.1 kPa:

20.240

= 71.544

p = 600 kPa = 600,000 Pa

slide24

Problem 4 continued:

Rework the problem in terms of Mach number and determine ML.

γ = 1.4; molecular mass = 28;

= 1.5 mol/s;

= 0.0066

ML = ?

D = 15 mm

p = 600 kPa

T = 300 K

L = 11.5 m

Design equation:

(1.4)

= 20.240 (already calculated in Problem 4)

M = ?

slide25

γ = 1.4; molecular mass = 28;

= 1.5 mol/s;

= 0.0066

ML = ?

D = 15 mm

p = 600 kPa

T = 300 K

L = 11.5 m

u

u

1

1

RT

M =

=

=

=

c

A

p

(

)

0.5

4 (1.5x 28/1000 kg/s)

(8314/28)(300) J/kg

=

1.4

π

(15/1000 m)2 (600,000 Pa)

= 0.1

slide26

γ = 1.4; molecular mass = 28;

= 1.5 mol/s;

= 0.0066

ML = ?

D = 15 mm

p = 600 kPa

T = 300 K

L = 11.5 m

Design equation:

(1.4)

20.240

Solve the nonlinear equation above to determine ML

ML = ?

slide27

20.240

Determine the approximate solution by ignoring the ln-term:

ML = 0.1 / (1-20.240 x 1.4 x 0.12)0.5

= 0.118

Check the value of the ln-term using ML= 0.118:

ln[(0.1/ML)2] = ln[(0.1 /0.118)2]

= -0.3310

This value is small when compared to 20.240. And therefore ML= 0.118 is a good first approximation.

slide29

Problem 5 from Problem Set 1 in Compressible Fluid Flow:

Explain why the design equations of Problems (1), (2) and (3)

are valid only for fully turbulent flow and not for laminar flow.

slide30

Problem 6 from Problem Set 1 in Compressible Fluid Flow:

Starting from the differential equation of Problem (2), or otherwise,

prove that p, the pressure, in a quasi one-dimensional, compressible, isothermal, steady flow of an ideal gas in a pipe with wall friction should always satisfies the following condition:

(1.5)

in flows where p decreases along the flow direction, and

(1.6)

in flows where p increases along the flow direction.

slide31

Differential equation of Problem 2:

(1.2)

can be rearranged to give

In flows where p decreases along the flow direction

(1.5)

slide32

Differential equation of Problem 2:

(1.2)

can be rearranged to give

In flows where p increases along the flow direction

(1.6)

slide33

Problem 7 from Problem Set 1 in Compressible Fluid Flow:

Air enters a horizontal constant-area pipe at 40 atm and 97oC with a velocity of 500 m/s. What is the limiting pressure for isothermal flow?

It can be observed that in the above case pressure increases in the direction of flow. Is such flow physically realizable?

If yes, explain how the flow is driven along the pipe.

Air: γ = 1.4; molecular mass = 29;

40 atm

97oC

500 m/s

p*=?

L

slide34

Air: γ = 1.4; molecular mass = 29;

40 atm

97oC

500 m/s

p*=?

L

Limiting pressure:

slide35

Air: γ = 1.4; molecular mass = 29;

40 atm

97oC

500 m/s

p*=?

L

(40 atm) (500 m/s)

=

= 61.4 atm

0.5

[(8314/29)(273+97) J/kg]

slide36

Air: γ = 1.4; molecular mass = 29;

40 atm

97oC

500 m/s

p*=61.4 atm

L

Pressure increases in the direction of flow. Is such flow physically realizable?

YES

If yes, explain how the flow is driven along the pipe.

Use the momentum balance over a differential element of the flow (given below) to explain.