Chapter IV Compressible Duct Flow with Friction Chapter 3 showed the effect of area change on a...

Chapter IVCompressible Duct Flow with Friction

Chapter 3 showed the effect of area change on a compressible flow while neglecting friction and heat transfer. We could now add friction and heat transfer to the area change and consider coupled effect. Instead, as an elementary introduction, this chapter treats only the effect of friction, neglecting area change and heat transfer. The basic assumptions are :

1 – Steady one-dimensional adiabatic flow.

2 – Perfect gas with constant specific heats.

3 – Constant – area straight duct.Prof. Dr. MOHSEN OSMAN 1

4 – Neglecting shaft-work and potential – energy changes.5 – Wall shear stress correlated by a Darcy friction factor.

In effect, we are studying a Moody-type pipe-friction problem but with large changes in kinetic energy, enthalpy, and pressure in the flow.

Consider the elemental duct control volume of area A and length dx in Fig. 4.1. The area is constant, but other flow properties (P, ρ, T, h, V) may vary with x. τw πD dx

V V+dV

P P+dP

ρ → ρ+dρ

T T+dT

h h+dh

x x+dx Fig.4.1 Elemental control volume for flow in a constant area duct with friction. Prof. Dr. MOHSEN OSMAN 2

Application of the three conservation laws to this control volume gives three differential equations : Continuity : (4.1a) or x – momentum :

(4.1b)

Energy : (4.1c)

Prof. Dr. MOHSEN OSMAN 3

tconsGA

mV tan

atedifferenticonstV .....).........ln(lnln

0.0V

dVd

0.04

....

4

...

])[()(

2

VdVD

dxdP

dVA

AV

D

DdxdP

VdVVmDdzAdPPPA

w

w

w

0.0.........

tan2

1 2

VdVdTCOR

tconsVTCTCh

p

popo

Since these three equations have five unknowns, P, ρ, T, V and τw , we need two additional relations. One is the perfect–gas law

(4.2)

To eliminate τw as an unknown, it is assumed that wall shear is correlated by a local Darcy factor f

(4.3)

Equations (4.1) and (4.2) are first-order differential equations and can be integrated, using friction – factor data, from any inlet section (1), where etc., are known, to determine - , etc., along the duct. It is practically impossible to

, Prof. Dr. MOHSEN OSMAN 4

T

dTd

P

dPOR

T

dTzero

d

P

dP

TRP

RTP

...........

lnlnlnln

22

8

1

8

1PMfVfw

,,, 111 VTP

)(),( xTxP

eliminate all but one variable to give, say, a single differential equation for P(x), but all equations can be written in terms of the Mach number M(x) and the friction factor, using the definition of Mach number.

(4.4)

Prove for extra class-work bonus marks ! By eliminating variables between equations (4.1) to (4.4), we obtain the working relations:

(4.5a)

(4.5b)


T

dT

M

dM

V

dVOR

TRMV

RTMV

22.......

ln)ln(ln2ln2

22

D

dxf

M

Md

D

dxf

M

MM

P

dP

)1(2

))1(2

)1(1(

2

2

2

22

Due Date : next week lecture All these except have the factor (1 - M2 ) in the denominator, so that, like the area–change formula, subsonic and supersonic flow have opposite effects.


D

dxf

M

MM

M

dM

D

dxf

M

M

T

dT

D

dxfM

d

P

dP

o

o

o

o

))1(2

)1(2

11

(

)1(2

)1(

2

1

2

2

22

2

2

4

2

)5.4(

)5.4(

)5.4(

e

d

c

o

o

P

dP

Property Subsonic Supersonic

P Decreases Increases

ρ Decreases Increases

V Increases Decreases

…

We have added to the list above that entropy must increase along the duct for either subsonic or supersonic flow as a consequence of the second law of thermodynamics for adiabatic flow. For the same reason, stagnation pressure and density (Po , ρo ) must both decrease. The key parameter above is the Mach number whether the inlet flow is subsonic or supersonic, the Mach number always tends downstream toward M=1 because this is the path along which the entropy increases. If the pressure and density are computed from Eqs. (4.5a) and (4.5b) and the entropy from the perfect–gas relation. Prof. Dr. MOHSEN OSMAN 7

Property Subsonic Supersonic

Po , ρo Decreases Decreases

T Decreases Increases

M Increases Decreases

S Increases Increases

, the result can be plotted in Fig 4.2versus Mach number for γ = 1.4. The maximum entropy occurs at M=1, so that the second law requires that the duct–flow properties continually approach the sonic point.

Fig. (4.2) Adiabatic Frictional Flow in a constant–area duct always approaches M=1 to satisfy the second law of thermodynamics.


])(ln[2

1

1

212

P

P

C

SS

V

Example 4.1 Air flows subsonically in an adiabatic 1-inch-diameter duct.

The average friction factor is 0.024. What length of duct is necessary to accelerate the flow from M1 = 0.1 to M2 = 0.5 ? What

additional length will accelerate it to M3 = 1.0 ? Assume γ = 1.4

Solution Equation (4.8) applies, with values of computed from

Eqn. (4.7) or read from Fanno- line flow Table


D

Lf *

ftx

xLLengthDuct

L

D

Lf

D

Lf

D

Lf MM

229024.012

8525.651...

8525.65

12

1024.0

8525.650691.19216.66.............

)*

()*

( 5.01.0

For M3 = 1.0 Duct Length = L* @ M = 0.1

i.e., the additional length to go from M = 0.5 to 1.0 is taken directly from the Table Formulae for other flow properties along the duct can be derived from Equations (4.5). Equation (4.5e) can be used to eliminate from each of the other relations, giving, for example, as a function only of M and . For convenience in tabulating the results, each expression is then integrated all the way from (P,M) to the sonic point (P*, 1.0). The integral results are:


ftx

xL

D

Lf

M

M

7.3024.012

0691.11*

0691.1*

5.0

5.0

LftLL

M7.3*

5.0

D

fdx

P

dP2

2

M

dM

Eq. (4.5e)

Substitute into equation (4.5a)


2

24

2

2

2

22

2

)2

11(

1

12

11

dMMM

M

D

dxf

D

dxf

M

MM

M

dM

2

1......1)

2

1(..............

1.....

)1(1)2

11(

........min...........]

2

11

[

)2

11(2

)1(1

)2

11(

)1(

)1(2

)1(1

2

222

2

22

2

22

2

2

24

2

2

22

bbatermMoftCoefficien

atermAbsolute

MbMMa

fractionspartialofatorsnoequatingdMM

b

M

a

P

dP

dMMM

M

P

dP

dMMM

Mx

M

MM

P

dP

Accordingly,

Similarly:


)9.4......(..............................])1(2

1[

1

*

]2

11(

)1(

2[

*

)]2

11()

1

2ln[(

*ln

)2

11(

2

1

ln2

1*ln

)]2

11([ln

2

1*ln

]2

11ln([ln

2

1ln

]

2

11

2

11

[2

1

2

1

2

2

12

2

122

22

122

122*

2

2

1

2

*

aMMP

P

MMP

P

MMP

P

MMP

P

MMP

P

MMP

dMMMP

dP

M

M

P

P

M

P

P

)9.4(..................................................]1

)1(2[

1 2

12*

*b

M

MV

V

-- To derive working formulae, we first attack Eq. (4.5e) which relates

Mach number to friction. Separate the variables and integrate,


)9.4....(]1

)1(2[

1

)9.4.......(..........)1(2

1

)1(2

)1(2

**

22*

2

*

dM

MP

P

cMC

C

T

T

o

o

o

o

4

1

2

11

2

1

1

)2

1(.........0)

2

1(:........

1)2

1(:.......

1))(2

11(

]

2

11

[1

)2

11(

1

2

4

2

2422

1 12

24

22

24

2

0 2 2

*

d

b

a

bddbMoftCoefficien

baMoftCoefficien

MdMbMaM

dMM

d

M

bMadM

MM

M

D

dxf

M M

L

--


2

22

!

4

*

2

2

2

1

4

2

0

]

2

11

)2

1(

2

1

2

11

[1

)

2

11

4

1

2

11

(1

2

2

*

dMMMMD

Lf

dMMM

Mdx

D

f

M

M

L

)7.4(....................)1(2

)1(ln

2

11

]

2

11

2

1

ln[1

1{1

])2

11

ln(1

[1

)]2

11ln(

2

1ln

2

11[

1

2

2

2

2*

2

1

2

22

*

12

1

2

2

2

*

1222

*

2

2

M

M

M

M

D

Lf

M

MMD

Lf

M

M

MD

Lf

MMMD

Lf

M

M

All these ratios are also tabulated in the same Fanno Flow Table. For finding changes between points M1 and M2 which are not sonic, products of these ratios are used. For example

Example 4.2 For the duct flow of Example 4.1 assume that at M1 = 0.1 we have P1 = 100 psia and T1 = 600 oR. Compute at section 2 further downstream (M2 = 0.5) (a) P2 ; (b)

T2 ; (c) V2 ; (d) Po2 Solution: As preliminary information we can compute V1 and Po1 from the given information: V1 =M1 C1

= M1 x49 = 0.1x49 =120 ft/s From

Isentropic Table @ M=M1 =0.1 Prof. Dr. MOHSEN OSMAN 15

)10.4.........(..........)........./()(*1

*2

1

*

*2

1

2

P

P

P

P

P

Px

P

P

P

P

RT o1 600

99303.01

1 oP

P

Then Now enter Fanno Line Table, to find the following ratios

Use these ratios to compute all properties downstream:

Note the 77% reduction in stagnation pressure due to friction.Prof. Dr. MOHSEN OSMAN 16

psiaP

Po 7.10099303.0

100

99303.01

Section M P/P* T/T* V/V* Po /P*o

1 0.1 10.9435 1.1976 0.1094 6.8218

2 0.5 2.1381 1.1429 0.5345 1.3399

psiaP

Px

P

PxPP

sftV

Vx

V

VxVV

RT

Tx

T

TxTT

psiaP

Px

P

PxPP

o

o

o

ooo

o

2.23)8218.5

3399.1(7.100)(

/586)1094.0

5345.0(120)(

573)1976.1

1429.1(600)(

5.19)9435.10

1381.2(100)(

1

*

*2

12

1

*

*2

12

1

*

*2

12

1

*

*2

12

Choking Due To FrictionThe theory here predicts that for adiabatic frictional flow in a constant area duct, no matter what the inlet Mach number M1

is, the flow downstream tends toward the sonic point. There is a certain duct length L*(M1) for which the exit Mach number will be exactly unity. But what if the actual duct length L is greater than the predict-ed “maximum” length L* ? Then the flow condition must change and there are two classifications.

Subsonic Inlet If L ˃ L*(M1), the flow slows down until an inlet Mach number M2 is reached such that L = L*(M2). The exit flow is sonic and the mass flow has been reduced by frictional choking. Further increases in duct length will continue to decrease the inlet Mach number M and mass flux.


Supersonic Inlet From Table we see that friction has a very large effect on supersonic duct flow. Even an infinite inlet Mach number will be reduced to sonic conditions in only 41

diameters for @ M = 10 @ M = ∞

Fig. (4.3) Behavior of duct flow with a nominal supersonic inlet condition M = 3.0 (a) L/D ≤ 26, flow is supersonic throughout duct; (b) L/D = 40 ˃L*/D; normal shock at M = 2.0 with subsonic flow then accelerating to sonic exit point. Prof. Dr. MOHSEN OSMAN 18

02.0f38683.0

*

D

Lf

82.0*

D

Lf

410.....................82.002.0 *

*

LD

L

(c) L/D = 53, shock must now occur at M = 2.5; (d) L/D ˃ 63, flow must be entirely subsonic and choked at exit Some typical numerical values are shown in Fig. (4.3), assuming

an inlet M = 3.0 and . For this condition L* = 26 diam. If L is increased beyond 26 D, the flow will not choke but a normal shock will form at just the right place for the subsequent subsonic frictional flow to become sonic exactly at the exit. Figure (4.3) shows two examples, for As length increases, the required shock moves upstream until, for Fig.(4.3), the shock is at the inlet for Further increase in L causes the shock to move upstream of the inlet into the supersonic nozzle feeding the duct. Yet the mass flux is still the same as for the very short duct, because presumably the feed nozzle still has a sonic throat.


.02.0f

.53......40 andD

L

.63D

L

Eventually, a very long duct will cause the feed-nozzle throat to become choked, thus reducing the duct mass flux. Thus supersonic friction changes the flow pattern if L ˃L* but does

not choke the flow until L is much larger than L*.

Example 4.3 Air enters a 3-cm-diameter duct at and V1 = 100 m/s. The friction factor is 0.02. Compute (a) the maximum duct length for this condition, (b) the mass flux if the duct length is 15 m, and (c) the reduced mass flux if L = 30 m. Required: D = 3 cm & f = 0.02 (a) Lmax for given condition (b) for L = 15 m Po =200 kPa (c) for L = 30 m To =500 K

V1 = 100 m/s Prof. Dr. MOHSEN OSMAN 20

,500,.....200 KTkPaP oo

mm

Solution (a) Apply energy equation between stagnation and inlet states CP To = CP T1 + ½ V1

2

Interpolating into Fanno-Line Table across, from M=M1=0.225 we get


225.0445

100

/445495046.20046.20

49510052

)100(500

2

1

11

11

2

1

21

1

C

VM

smTC

Kx

KT

C

VTT

Po

mL

LD

Lf

5.16)11(02.0

03.0

1103.0

02.0

11

max

max

max

(b) The given L=15 m is less than Lmax , and so is not choked and the mass flux follows from inlet conditions;


iii

iiii

o

VART

PVAm

kPaP

P

PTableIsentropicM

06.193)9653.0(200

9653.0....225.0

1

11

(c) Since L = 30 m is greater than Lmax , the duct must choke

back until L = Lmax , corresponding to a lower inlet M1

Interpolation in Fanno-Line Table we find that this value of 20 corresponds to M1, choked = 0.174 (23% less) From Isentropic Table :


skgm

xx

m

/0961.0

)100()03.0(4495287.0

06.193 2

2003.0

3002.0max x

D

Lf

994.0,....9794.0 11 oo T

T

P

P

)%...22...(........../..0753.0

6.77)03.0(4497287.0

88.195

/6.77)446(174.0)(

/446497046.20

88.195)9794.0(200

497)994.0(500

2,11

1

1

11,1

,1

,1

,1

lessskgm

xxx

VART

Pm

smCMV

smC

kPaP

KT

new

newnew

newnew

newnew

new

new

new

Chapter IV Compressible Duct Flow with Friction Chapter 3 showed the effect of area change on a...

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