Fall 2006

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ME 440 Aerospace Engineering Fundamentals

Compressible Flow Examples

Example: Compressible Flow Nozzle Analysis Consider our simple turbojet example of our propulsion unit. We found that entering the nozzle we had T5 = 1492 K P5 = 674 kPa For an ambient pressure of 69.2 kPa and an aircraft velocity of 300 m/s, determine the specific thrust of the engine accounting for compressible flow. Solution: We begin by calculating our critical pressure for this converging nozzle. We have P* = 0.528P5 = (0.528)(674) = 356 kPa Since this is above the ambient pressure we will not have complete expansion in our nozzle. Then P6 = 356 kPa The temperature at the exhaust is the critical temperature

K 124314.1

21492)(

12

T TT 5*

6 =+

=+γ

==

The velocity at the exhaust is the sonic velocity s/m 707)1243)(1005)(14.1( Tc)1( V 6P6 =−=−γ= The specific thrust is then ft = V6-Vac = 707-300 = 407 Nt/(kg/s) When we ignored the affect of compressible flow we found a specific thrust of 897 Nt/(kg/s), which means that by not having complete expansion we loose over half of our thrust. Example: Required Area Ratio for Complete Expansion Determine the area ratio required in a converging-diverging nozzle to have complete expansion in the previous example. Solution: In this case we will follow our 4 step approach.

1. Assume full expansion, Pout = 69.2 kPa

ME 440 Aerospace Engineering Fundamentals Fall 2006

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2. Calculate Tout from the isentropic relationship

K 779 673.969.2

(1492) P

PT T

4.1/)14.1(/)1(

in

outinout =

=

=−γ−γ

3. Calculate Vout from the 1st law

( ) m/s 1197 779)-922(1005)(14 = T-T2c = V outinPout = 4. Calculate area ratio by applying continuity

in

P

outout

outin1/2-1)-/(

t

out

T)1(c

VP

TP

12

A

A −γ

+γ=

γγ

1.90 )1429(

)14.1)(1005()1197)(2.69()779)(9.673(

14.12

A

A 1/2-1)-4.1.4/(1

t

out =−

+=

Example: Normal Shock Wave Air at 270 K, 50 kPa, and a Mach number of 2.4 undergoes a normal shock. Determine the velocity, temperature and pressure downstream of the shock wave. Solution: We begin by calculating M2, T2, and P2.

0.274 1)4.2()}14.1/()4.1(2{

)14.1/(2)4.2(

1M)}1/(2{

)1/(2M M 2

2

21

212

2 =−−

−+=

−−γγ−γ+

=

M2 = (0.274)(1/2) = 0.523

3.13 1)4.2(1-.41

1.4)(2)4.2(

21-.41

1)4.2(1).41(

1)-.412(

1M1-

2M

21-

1M1)(

1)-2(

TT

2222

21

212

12

1

2

=

−

++

=

−γ

γ

γ

++γ

γ=

T2 = (3.13)(270) = 846 K

55.614.11-.41

- 2.4)(14.1

1.4)(2

11-

- M1

2

P

P 221

1

2 =++

=+γ

γ+γγ

=

P2 = (6.55)(50) = 328 kPa Then s/m305)846)(287)(4.1()523.0(RTMV 222 ==γ=

ME 440 Aerospace Engineering Fundamentals Fall 2006

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Example: Oblique Shock Wave Consider a supersonic flow with M1 = 1.8 P1 = 90 kPa T1 = 268 K that is turned to produce a shock wave. Determine the wall angle and resulting shock angle for a sonic downstream condition. Also determine the downstream temperature and pressure. Solution: Reading from Figure 1 for M1 = 1.8 and M2 = 1, we find θ = 18º β = 61º Then 1.574 )1(1.8)sin(6 )sin(M M 11N =°=β= 0.682)18-1(1.0)sin(6 )sin(M M 22N =°°=θ−β= Using the normal shock relations we find

1.3694 1)574.1(1-.41

1.4)(2)574.1(

21-.41

1)4.2(1).41(

1)-.412(

1M1-

2M

21-

1M1)(

1)-2(

TT

2222

21

212

12

1

2

=

−

++

=

−γ

γ

γ

++γ

γ=

T2 = (1.3694)(270) = 367 K

722.214.11-.41

- )574.1(14.1

1.4)(2

11-

- M1

2

P

P 221

1

2 =++

=+γ

γ+γγ

=

P2 = (2.722)(90) = 245 kPa