CP502

Advanced Fluid Mechanics

Compressible Flow

Part 01_Set 03:

Steady, quasi one-dimensional, isothermal, compressible flow of an ideal gas in a constant area duct with wall friction (continued)

Problem 10

from Problem Set 1 in Compressible Fluid Flow: Determine the isothermal mass flow rate of air in a pipe of 10-mm i.d. and 1m long with upstream condition of 1 MPa and 300 K with a exit pressure low enough to choke the flow in the pipe assuming an average Fanning friction factor of 0.0075. Determine also the exit pressure.

Given

μ

= 2.17 x 10 -5 kg/m.s, calculate the Reynolds number of the flow to check if the given flow were turbulent.

p

= 1 MPa = ?; Air:

γ =

1.4; molecular mass = 29;

D

= 10 mm

T

= 300 K

f

= 0.0075

flow is choked

L

= 1 m At choking condition,

p L = p*, M L =

R. Shanthini 16 Feb 2012

M

*  1 /  and

L = L max

p

= 1 MPa = ?; Air:

γ =

1.4; molecular mass = 29;

D

= 10 mm

T

= 300 K

f

= 0.0075

M

*  1 / 

L

=

L

max = 1 m 

A



u

 

D

2 4

p RT M

= 4 

RT

 

D

2 4

pM

(

π

) (10/1000 m) 2 (1000,000 Pa)

M

( 

RT

1.4

(8314/29)(300) J/kg 0.5

) = 0.317

M M

at the entrance could be determined using (1.9) R. Shanthini 16 Feb 2012

p

= 1 MPa = ?; Air:

γ =

1.4; molecular mass = 29;

D

= 10 mm

T

= 300 K

f

= 0.0075

M

*  1 /  Use 4

f L

max

D

 1   

M M

2 2  ln

L

=

L

max = 1 m (part of 1.9) 4 ( 0 .

0075 )( 1 m ) ( 10 / 1000 m )  1  1 .

4

M

2 1 .

4

M

2  ln  1 .

4

M

2  Solving the nonlinear equation above gives

M

= 0.352 at the entrance = 0.317

M

= 0.317 x 0.352 = 0.1116 kg/s R. Shanthini 16 Feb 2012

p

= 1 MPa = ?; Air:

γ =

1.4; molecular mass = 29;

D

= 10 mm

T

= 300 K

f

= 0.0075

M

*  1 / 

L

=

L

max = 1 m Determine the exit pressure.

Since 

A



u

 

D

2 4

p RT M



RT

 

D

2 4

pM



RT

(

pM

) entrance = (

pM

) exit (1 MPa) (0.352) =

p

exit ( 1 /  )

p

exit = (1 MPa) (0.352) (1.4) 0.5

R. Shanthini 16 Feb 2012 = 0.417 MPa

p

= 1 MPa = ?; Air:

γ =

1.4; molecular mass = 29;

D

= 10 mm

T

= 300 K

f

= 0.0075

M

*  1 / 

L

=

L

max = 1 m Reynolds Number: Re  

uD

 

A D

  4 

D

2

D

  4

m



D

 = 4 (0.1116 kg/s)

π

(10/1000 m) (2.17 x 10 -5 kg/m.s) = 6.5 x 10 5 Therefore, flow is turbulent R. Shanthini 16 Feb 2012

Problem 11

from Problem Set 1 in Compressible Fluid Flow: Air flows at a mass flow rate of 9.0 kg/s isothermally at 300 K through a straight rough duct of constant cross-sectional area 1.5 x 10 -3 m 2 . At one end A the pressure is 6.5 bar and at the other end B the pressure is 8.5 bar. Determine the following: (i) Velocities

u

A and

u

B (ii) Force acting on the duct wall (iii) Rate of heat transfer through the duct wall In which direction is the gas flowing?

R. Shanthini 16 Feb 2012

Air:

γ =

1.4; molecular mass = 29; = 9.0 kg/s;

p

A bar = 6.5

A

= 1.5x10

-3 m 2

T

= 300 K

p

B = 8.5 bar (i) Velocities

u

A and

u

B = ?

u A



A



A



RT Ap A

= (9 kg/s) (8314/29 J/kg.K) (300 K) (1.5x10

-3 m 2 ) (6.5 bar) (100,000 Pa/bar)

u A p A

 

RT A



u B p B

= 794 m/s

u B



p A u A p B

= 6.5 bar 8.5 bar 794 m/s R. Shanthini 16 Feb 2012 = 607 m/s

p

A bar = 6.5 Air:

γ =

1.4; molecular mass = 29;

A

= 1.5x10

-3 m 2 = 9.0 kg/s;

T

= 300 K

p

B = 8.5 bar (ii) Force acting on the duct wall = ?

Force balance on the entire duct gives the following:

p A



A = p B



B

+ Force acting on the duct wall Force acting on the duct wall

=

(

p A – p B

) (

u A – u B

)

=

(6.5

–

8.5) bar x 100,000 Pa/bar x 1.5 x 10 -3 m 2

+

(9.0 kg/s) (794

–

607) m/s

=

-300 Pa.m

2 + 1683 kg.m/s 2

=

-300 N + 1683 N R. Shanthini 16 Feb 2012

=

1383 N

p

A bar = 6.5 Air:

γ =

1.4; molecular mass = 29;

A

= 1.5x10

-3 m 2 = 9.0 kg/s;

T

= 300 K

p

B = 8.5 bar (iii) Rate of heat transfer through the duct wall = ?

Energy balance on the entire duct gives the following: Rate of heat transfer through the duct wall from the surroundings +

h A

+

u A

2

/

2

= h B

+

u B

2 /2 Enthalpy at A Enthalpy at B Kinetic energy at A R. Shanthini 16 Feb 2012 Kinetic energy at B

Air:

γ =

1.4; molecular mass = 29; = 9.0 kg/s;

p

A bar = 6.5

A

= 1.5x10

-3 m 2

T

= 300 K

p

B = 8.5 bar Rate of heat transfer through the duct wall from the surroundings

=

(

h B

–

h A

) + (

u B

2 –

u A

2 )/2 Since (

h B

–

h A

) =

c p

(

T B

–

T A

) = 0 for isothermal flow of an ideal gas Rate of heat transfer through the duct wall from the surroundings

=

(

u B

2 –

=

(607

u A

2 )/2 2 – 794 2 )/2 m 2 /s 2

=

(-130993.5 m 2 /s 2 )

=

(-130993.5 J/kg)

=

(-130993.5 J/kg) (9.0 kg/s) 16 Feb 2012

=

-1178942 J/s

=

-1179 kW

Heat is lost to the surroundings

Air:

γ =

1.4; molecular mass = 29; = 9.0 kg/s;

p

A bar = 6.5

A

= 1.5x10

-3 m 2

T

= 300 K Direction of the gas flow: Determine first the limiting pressure as follows:

p

B = 8.5 bar

p

*



( /

A

)

RT

= (9.0/1.5x10

-3 ) kg/m 2 .s (8314*300/29 J/kg) 0.5

= (9.0/1.5x10

-3 ) kg/m 2 .s (8314*300/29 J/kg) 0.5

= 17.6 bar Since

p A p

and

p B

are lower than the limiting pressure, increases along the flow direction (see Problem 6). 16 Feb 2012

Summarizing the results of Problem 11:

= 9.0 kg/s;

p

A = 6.5 bar = 794 m/s

T

= 300 K

p

B

u

B = 8.5 bar = 607 m/s Pressure increases in the flow direction and therefore velocity decreases according to the following equation:

u A p A

 

RT A



u B p B

Force acting on the entire duct wall is 1383 N Velocity decreases and therefore kinetic energy is lost across the duct. The lost energy is transferred from the duct to the surroundings through the duct wall.

R. Shanthini 16 Feb 2012

Problem 12

from Problem Set 1 in Compressible Fluid Flow: Gas produced in a coal gasification plant (molecular weight = 0.013 kg/mol,

μ

= 10 -5 kg/m.s,

γ

= 1.36) is sent to neighbouring industrial users through a bare 15-cm-i.d. commercial steel pipe 100 m long. The pressure gauge at one end of the pipe reads 1 MPa absolute. At the other end it reads 500 kPa. The temperature is 87 o C. Estimate the flow rate of coal gas through the pipe? Additional data:

ε

= 0.046 mm for commercial steel. For fully developed turbulent flow in rough pipes, the average Fanning friction factor can be found by use of the following:

f

 1 /[ 4 log( 3 .

7

D

/  )] R. Shanthini 16 Feb 2012

Properties of gas produced: Molecular weight = 0.013 kg/mol;

μ

= 10 -5 kg/m.s;

γ

= 1.36

p

= 1 MPa

D

= 15 cm

T

= (273+87) K

L

= 100 m What is the flow rate through the pipe?

p L

= 500 kPa R. Shanthini 16 Feb 2012

Design equation to be used: 4

f L D



RT

(

p

2 /

A

) 2   1 

p L

2

p

2    ln  

p L

2

p

2   (1.3)

f f

 1 /[ 4 log( 3 .

7

D

/  )] = 1/[4 log(3.7x15x10/0.046)] = 0.0613

= 0.0038

4

f L

= 4 x 0.0038 x 100 m / (15 cm) = 10.1333

 

D p L

2

p

2   = (500/1000) 2 = 0.25

Using the above in (1.3), we get

p

2

RT

( /

A

) 2 R. Shanthini 16 Feb 2012

=

10.1333 – ln(0.25) 1 – 0.25

= 15.3595

RT

(

p

2 /

A

) 2 = 15.3595

p

= 1 MPa = 1,000,000 Pa;

R

= 8.314 J/mol.K = 8.314/0.013 J/kg.K;

T

= 360 K;

A

=

πD

2 /4

= π

(15 cm) 2 /4 =

π

(0.15 m) 2 /4; Therefore, 

pA RT

15 1 .

3595 = 9.4 kg/s; Check the Reynolds number: Re = 

=

(9.4 kg/s)(15/100 m)/(10 -5 kg/m.s) R. Shanthini 16 Feb 2012 = 1.4x10

5 Therefore, flow is turbulent

Governing equation for incompressible flow:

Starting from the mass and momentum balances, obtain the differential equation describing the quasi one-dimensional,

incompressible

, isothermal, steady flow of an ideal gas through a constant area pipe of diameter

D

and average Fanning friction factor.

f

Incompressible flow Steady flow Density (

ρ

) is a constant 

A

 Constant area pipe

A

is a constant Therefore,

u

is a constant for a steady, quasi one-dimensional, compressible flow in a constant area pipe. R. Shanthini 16 Feb 2012



w p p+dp D u u+du x dx

Write the momentum balance over the differential volume chosen.

pA

 

u

 (

p



dp

)

A

  (

u



du

)  

w dA w Adp

 

du

 

w dA w

 0

A

 

D

2 / 4 

w



f



u

2

/ 2

dA w

 

Ddx dp

 R. Shanthini 

udu

16 Feb 2012  

u

2 2 4

f D dx

 0

Therefore, we get 4

f D dx

 2 

u

2

dp

 2

du u

 0 Rearranging (2) gives

dp dx

  4 2 /

f

/

D



u

2   2

f



u

2

D

 0

(2)

It means

p

decreases in the flow direction.

Since

ρ

and

u

are constants, integrating the above gives

p



p L

 2

f



u

2

L D

pressure at the exit 16 Feb 2012

Rework Problem 12 assuming incompressible flow:

Molecular weight = 0.013 kg/mol;

f

= 0.0038

p

= 1 MPa

D

= 15 cm

T

= (273+87) K

p L

= 500 kPa

L

= 100 m What is the flow rate through the pipe?

Design equation used for

compressible

flow 4

f L D



RT

(

p

2 /

A

) 2   1 

p p

2

L

2    ln  

p L

2

p

2   (1.3) Design equation to be used with

incompressible

flow

p

 R. Shanthini 16 Feb 2012

p L

 2

f



u

2

L D

Molecular weight = 0.013 kg/mol;

f

= 0.0038

p

= 1 MPa

D

= 15 cm

T

= (273+87) K

L

= 100 m

p



p L

 2

f



u

2

L D u



m A

 

p



p L

 2

f

(

D

 /

A

) 2

L p L

= 500 kPa 

A

(

p



p L

)

D

 2

f L

R. Shanthini 16 Feb 2012

Molecular weight = 0.013 kg/mol;

f

= 0.0038

p

= 1 MPa

D

= 15 cm

T

= (273+87) K

p L

= 500 kPa

L

= 100 m 

A

(

p



p L

)

D

 2

f L

 

D

2 4 What is

ρ

?

ρ =

(

ρ

entrance +

ρ

exit ) / 2 (

p



p L

)

D

 2

f L =

[(

p/RT

) entrance + (

p/RT

) exit )] / 2

=

(

p

entrance +

p

exit ) / 2

RT

Therefore, = (1,000,000 + 500,000) Pa / [2 x (8.314/0.013) x 300 J/kg] = 1.9545 kg/m 3 = 7.76 kg/s R. Shanthini 16 Feb 2012

Compare 7.76 kg/s with the 9.4 kg/s obtained considering the flow to be compressible.

Important Note:

Problems (13) and (14) from Problem Set 1 in Compressible Fluid Flow are assignments to be worked out by the students themselves in preparation to the mid-semester examination.

R. Shanthini 16 Feb 2012