Example 3.4 A converging-diverging nozzle (Fig. 3.7a) has a throat area of 0.002 m 2 and an exit area 0.008 m 2. Air stagnation conditions are Compute.

Example 3.4 A converging-diverging nozzle (Fig. 3.7a) has a throat

area of 0.002 m2 and an exit area 0.008 m2 . Air stagnation conditions are Compute (a) the design exit pressure and mass flux and the exit pressure and mass flux if (b) . Assume γ = 1.4 Solution

Interpolating into supersonic Isentropic Table for we get Mass Flow Parameter (MFP)

Prof. Dr. MOHSEN OSMAN 1

.500&1000 KTkPaP oo

kPaPckPaP bb 900)(&300 4

004.0

008.0*

A

A

A

A e

t

e

4*

A

A

1.34&95.2 e

oe P

PM kPa

PP o

e 3.291.34

1000

1.34

[email protected] MMFP

If No supersonic flow.(b) We are definitely far below the subsonic isentropic condi-tion C in Fig. 3.7b but we may even be below condition F with a normal shock in the exit plane. It is condition g , where because no shock has yet occurred. To find out, compute condition F by assuming an exit normal shock with , that is the design Mach number just upstream of the shock. From Normal – Shock Table


eb PP

skgmxx

m

R

MFP

AP

Tm o

/61.3040416.0002.0101000

500

0404.0

max3max

*0

max

;300kPaPb

kPaPP designee 3.29,

95.21 M

kPaPP

P293)3.29(99.9&99.9 2

1

2

Since this is less than the given there is a normal shock just upstream of the exit plane (condition E). The exit flow is subsonic and equals the back pressure Also (c) we compute for condition (C) as a comparison . Again For this condition, with a subsonic estimated from subsonic Isentropic Table. We get

The given back pressure of 900 kPa is less than this value (of 985) corresponding roughly to condition (D) in Fig. 3.7b. Thus, for this case, there is a normal shock just downstream of the throat Prof. Dr. MOHSEN OSMAN 3

,300kPaPb

kPaPP be 300skgmm /61.3max

,900kPaPb eeandPM

.4t

e

A

A

eM

kPaPkPaP

P

PM

be

o

ee

900985

985.0&147.0

and the throat is chocked For this large exit-area ratio the exit pressure would have to be larger than 985 kPa to cause a subsonic flow in the throat and a mass flux less than maximum. Governing Relations of the Normal-Shock Wave

. x y .

. . Normal-Shock Wave

Energy Equation 1

“adiabatic, no work, steady”

Continuity Equation 2


skgmmkPaPP be /61.3&900 max

yx VV .............

oy

yx

x hV

hV

h 22

22

yyxx VVA

m

Momentum Equation 3 Equation of state h = h (s & ρ) 4 5 What is meant by Fanno Line ?

Combination of Fanno Flow andNormal Shock


22

)()(

yyyxxx

xyyx

VPVP

VVmAPP

),( Pss

If conditions of x are fixed, if we are required to calculate conditions at y, Equations (1) and (2) and (4) define a locus of states passing through x → this is called Fanno Line.

i.e.,

● Choose

● Calculate from 2



● repeat for other values of

Note: Fanno Line is a line of constant

Frictional effects are required to pass along Fanno Line.


yV

y

yhys

yV

ohA

m&

What is meant by Rayleigh Line?● Similar to Fanno, but using equations (2), (3) & (5)It is a line of constant (friction coefficient)


FA

m&

Fanno & Raleigh Lines

Since Normal Shock must satisfy equations (1), (2), (3),(4) and (5)

Shock must lies on both Fanno and Rayleigh Lines !! The points of maximum entropy s on both Fanno & Rayleigh Lines are of sonic velocity.


Normal Shock in a Perfect GasFor a perfect gas use Mach number relations to find the Normal Shock downstream conditions of all fluid properties , if you know that .What is meant by supersonic Pitot tube ?


yyy TPM ,,

yyoyy andsAP ,,, * ooyox TTT

Where :

Which is known as Rayleigh Supersonic Pitot-tube FormulaNotes: You can use Normal-Shock Tables ….. to get .Homework Assignment and submit next lecture in a report Plot (a) Air (γ=1.4)(b) Helium (γ=1.66) Prof. Dr. MOHSEN OSMAN 10

x

y

y

oy

x

oy

P

P

P

P

P

P*

)

)1

1

1

2(

)2

1(

...............................

)1(2

2)1(......

)2

11(&

1

1

1

2

1

1

2

12

2

22

12

2

x

x

x

oy

x

xy

yoy

xx

y

M

M

P

PyAccordingl

M

MMwhere

MPy

P

MP

P

xM

:............ forMvsP

Px

x

oy

Example 4.1A pitot tube is inserted into a supersonic air stream, and records a pressure of 0.7 bar. The static pressure upstream of the tube is0.15 bar, and the static temperature is 350 K.Calculate the flow velocity upstream of the tube.

Solution

Interpolating into Normal-Shock Table

we get


667.415.0

7.0

1

02 P

P

P

P

x

oy

8.1xM

smV

TMCMV

x

xxxxx

/675)350046.20(8.1

)046.20(

Supersonic Wind TunnelsMajor Types:(a) Blowdown Wind Tunnels (open)

(b) Continuous-Flow Wind Tunnels (closed)

which consists of:

1) Convergent – Divergent (Supersonic) Nozzle

2) Test Section

3) Convergent-Divergent (Supersonic) Diffuser

4) Heat Exchanger

5) Compressor


(a) Continuous Wind Tunnel (no diffuser)


(b) Continuous Wind Tunnel with diffuser


Wind Tunnel Start-Up Sequence To provide a test section with supersonic flow requires a converging–

diverging nozzle. To operate economically, the nozzle–test-section combination must be followed by a diffusing section which also must be converging–diverging. This configuration presents some interesting problems in flow analysis. Starting up such a wind tunnel is another example of nozzle operation at pressure ratios above the second critical point. Figure 3.8 shows a typical tunnel in its most unfavorable operating condition, which occurs at startup. A brief analysis of the situation follows.

As the exhauster is started, this reduces the pressure and produces flow through the tunnel. At first the flow is subsonic throughout, but at increased power settings the exhauster reduces pressures still further and causes increased flow rates until the nozzle throat (section 2) becomes choked. At this point the nozzle is operating at its first critical condition.


As power is increased further, a normal shock is formed just downstream of the throat, and if the tunnel pressure is decreased continuously, the shock will move down the diverging portion of the nozzle and pass rapidly through the test section and into the diffuser. Figure 3.9 shows this general running condition, which is called the most favorable condition. We return to Figure 3.8, which shows the shock located in the test section. The variation of Mach number throughout the flow system is also shown for this case. This is called the most unfavorable condition because the shock occurs at the highest possible Mach number and thus the losses are greatest. We might also point out that the diffuser throat (section 5) must be sized for this condition.


Figure 3.8 Supersonic tunnel at startup (with associated Mach number variation).

Due to the shock losses (and other friction losses), we know that pt 5 < pt 2 , and therefore A5 must be greater than A2 . Knowing the test-section-design Mach number fixes the shock strength in this unfavorable condition and A5 is easily determined from equation (pt 2 A2 = pt 5 A5 ). Keep in mind that this represents a minimum area for the diffuser throat. If it is made any smaller than this, the tunnel could never be started (i.e., we could never get the shock into and through the test section). In fact, if A5 is made too small, the flow will choke first in this throat and never get a chance to reach sonic conditions in section 2.


Once the shock has passed into the diffuser throat, knowing that A5 > A2 we realize that the tunnel can never run with sonic velocity at section 5. Thus, to operate as a diffuser, there must be a shock at this point, as shown in Figure 3.9. We have also shown the pressure variation through the tunnel for this running condition.

To keep the losses during running at a minimum, the shock in the diffuser should occur at the lowest possible Mach number, which means a small throat. However, we have seen that it is necessary to have a large diffuser throat in order to start the tunnel. A solution to this dilemma would be to construct a diffuser with a variable- area throat. After startup, A5 could be decreased, with a corresponding decrease in shock strength and operating power. However, the power required for any installation must always be computed on the basis of the unfavorable startup condition.

Although the supersonic wind tunnel is used primarily for aeronautically oriented work, its operation serves to solidify many of the important concepts of variable-area flow, normal shocks, and their associated flow losses. Equally important is the fact that it begins to focus our attention on some practical design applications.Prof. Dr. MOHSEN OSMAN 18

Figure 3.9 Supersonic tunnel in running condition (with

associated pressure variation).Prof. Dr. MOHSEN OSMAN 19

Wind Tunnel Start–Up SequenceDraw the following sequence of Wind–Tunnel start-up !

(a) Initial Start-Up

(b) First Throat Sonic

(c) Shock in Diverging Section

(d) Shock in Test Section Entrance

(e) Shock Swallowed

(f) Shock-Free Deceleration with

Variable-Area Diffuser Throat


Example 4.2A continuous–duty supersonic wind tunnel is to be designed. The test section specifications are a Mach number of 2, a static pressure of 40 kPa, a static temperature of 250 K&area of 0.5 m2 . Determine the stagnation conditions required upstream of the test section, the mass flow rate required, and the throat

area required. During the startup process, what is the maximum stagnation pressure loss across the shock system? Also, calculate the pressure loss during the steady-state operating conditions. If a constant–area diffuser is used, what is the minimum diffuser throat area? Sketch the startup process on a T–S diagram. Solution: Consider Figure 3.9 for fixed diffuser steady-state operating conditions.


A continuous–duty supersonic wind tunnel with: From Isentropic Table @ M=MT = 2.0, we get

occurs when shock takes place at maximum Mach number; i.e. at MT


25.0.......&.......250..,.........40.,.........0.2 mAKTkPaPM TTTT

max

max

max

2

/1767.068473.0450287

3132963.0

1@*

2963.06875.1

5.0*

45055556.0

250....,.........313

1278.0

40

6875.1*

...,.........55556.0...,.........1278.0

o

o

o

oo

T

o

T

o

T

P

skgxx

xmm

R

MMFPx

T

PAmm

mA

KTkPaP

A

A

T

T

P

P

From Normal-Shock Wave Tables @ M=MT = 2.0, we get

Question 1: How do we calculate during the steady–state operating conditions) ?Question 2: Show start–up on T–s using Fanno–line !


2*

*

*

max

max

411.072087.0

2963.0............................................

72087.0.......sin

37.87)72087.01(313

)1(

72087.0

mAthen

P

P

A

Ace

kPaP

P

PPPPP

P

P

y

ox

oy

y

x

o

ox

oyoxoyoxo

ox

oy

oo PeiP .,.(min

Solution to Question 1For steady–state operating conditions, shock occurs at the diffuser throat; i.e.at

from Isentropic Table @ & @ from Normal–Shock Table

If a variable throat diffuser is used, the throat must initially posses at least an area of 0.411 m2 in order to swallow the normal shock wave. The power to the compressor would be


7509.1

3871077.12963.0

411.0*

x

x

M

A

A

3871.1*

xA

A

7509.1xM

kPaP

P

PPPPP

P

P

o

ox

oyoxoyoxo

ox

oy

9.51)8342.01(313

)1(

8342.0

min

min

increased until the shock wave was swallowed. Once the shock wave is swallowed, the variable–area throat would be reduced to the area of the first throat, 0.2963 m2 . Simultaneously, the back pressure would be increased to place the shock at the diffuser throat. Ideally, the limit for a variable throat diffuser would be shock–free flow as illustrated in Figure (C). Solution to Question 2

T – S Diagram of wind tunnel start – up using Fanno – Line


Example 3.4 A converging-diverging nozzle (Fig. 3.7a) has a throat area of 0.002 m 2 and an exit area 0.008 m 2. Air stagnation conditions are Compute.

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