Lect-31...Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay 20 Lect-31 Problem 4 • Air enters a converging –diverging nozzle, shown in the Figure, at 1.0

1

Lect-31

Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay2

Lect-31

In this lecture...

• Tutorial on intakes and nozzles


Lect-31

Problem 1• Consider a turbofan engine operating at a

Mach number of 0.9 at an altitude where the ambient temperature and pressure are -56.5oC and 22.632 kPa, respectively. The mass ingested into the engine is 235 kg/s through an inlet area of 3 m2. If the diffuser efficiency is 0.9 and the Mach number at the fan entry is 0.45, calculate: (a) the capture area, (b) the static pressures at the inlet and fan face, (c) velocities at inlet and the fan face, (d) the diffuser pressure recovery.


Lect-31

Solution: Problem 1• The ambient temperature, Ta=216.5 K• The flight speed is u=M√(γRT)=271.6 m/s• The freestream density is

operation. cruise of typical area, inlet the thansmaller is area capture The

m486.2u/mAisareacapturethe,Therefore

m/kg3479.0RT/P

2

3aa

==

==

∞ ρ

ρ


Lect-31

Solution: Problem 1• From gas tables, for a Mach number of

0.9, the area ratio, A/A*=1.00886.• Therefore, A*=2.486/1.00886=2.465 m2

• Now, A1/A*=3/2.465=1.217 m2

• From the gas tables, the corresponding Mach number is M1=0.577.

• We can also determine the temperature and pressure ratios for this Mach number from the gas tables.


Lect-31

Solution: Problem 1• P1/P01=0.798, T1/T01=0.93757• Since, T01=T0a and P01=P0a,• P1=30.547 kPa and T1=246.9 K• And, u1=M1√ (γRT1)=181.7 m/s• Since Mach number at the fan is M2=0.45,• For M2=0.45,P2/P02=0.87027 • From the definition of diffuser efficiency,

)1/(2

da

02 M2

11PP −

−

+=γγγη


Lect-31

Solution: Problem 1

952.0278.38/442.36P/PiserycovrepressureThe

s/m5.143RTMu,Therefore

K1.253M2

11/TTand

kPa714.31M2

11/PP,Since

kPa442.36P,ngSubstituti

M2

11PP

a002

222

22022

)1/(22022

02

)1/(2

da

02

==

==

=

−

+=

=

−

+=

=

−

+=

−

−

γ

γ

γ

γη

γγ

γγ


Lect-31

Solution: Problem 1

)1(2/)1(2

1

101

011

0

2

)1/(2

0

M2

11

MRTPAm

,1stationFor

T

)M2

11(

RM

211

MPRTMRTPu

Am

:equation continuity From:way different a in solved be also can problem This

−+

−

−

+

=

−+

−

+

===

γγ

γγ

γγ

γγ

γγρ


Lect-31

Solution: Problem 1

)1/(2

da02,02

)1/(21

a01

21

a011

1

M2

11P/PPfindTo

M2

11

PP

M2

11

T Tby, determined be can T

y.iterativel solved be can Thisunknown. is M equation, above the In

−

−

−

+=

−

+

=

−+

=

γγ

γγ

γη

γ

γ


Lect-31

Solution: Problem 1

222

)1/(22

022

22

a02

RTMu,Therefore

M2

11

PPandM

211

T TNow,

recovery. pressure the determine now can We

γ

γγ γγ

=

−

+

=−

+= −


Lect-31

Problem 2• Consider the mixed compression two-

dimensional supersonic intake as shown in the figure. The free stream Mach number is 3.0. The intake has a three shock system as shown. Determine the overall total pressure ratio and the overall static pressure ratio.


Lect-31

Problem 2

M1=3.0

1

2

3

4

5o

15o


Lect-31

Solution: Problem 2

• The first oblique shock has an upstream Mach number of 3.0 and δ1=15o.

• From the shock tables, the shock angle is β1=32.25o.

• With β1=32.25o and δ1=15o

• M1n=M1sinβ1= 3.0sin32.25 =1.60,• From the normal shock tables, we can

find M2n.• M2 = M2n/sin(β1-δ1)=2.25


Lect-31

Solution: Problem 2

• From the normal shock tables, P02/P01=0.8935, P2/P1=2.82

• For region 2, the deflection angle, δ2=15+5=20o

• For M2=2.25 and δ1 =20o, β2= 46.95o

• We find M3 in the same way as we calculated M2.

• M3=1.444 and P03/P02=0.878, P3/P2=2.992


Lect-31

Solution: Problem 2

• Similarly, M4=0.7219 (from the normal shock tables)

• P04/P03=0.9465 and P4/P3=2.333• The overall pressure ratios:• P04/P01=P04/P03xP03/P02xP02/P01

=0.9465x0.878x0.8935=0.7865• P4/P1=P4/P3xP3/P2xP2/P1

=2.333x2.992x2.82=19.691


Lect-31

Problem 3

• Air enters a converging duct with varying flow area at T1 =400 K, P1=100 kPa, and M1=0.3. Assuming steady isentropic flow, determine T2, P2, and M2 at a location where the flow area has been reduced by 20 percent.


Lect-31

Problem 3

T1=400KP1=100 kPaM1=0.3A1

T2=?P2=?M2=?A2 = 0.8A1


Lect-31

Solution: Problem 3

• From the isentropic tables, for a Mach number of 0.3,

• A1/A*=2.0351, T1/T0=0.9823, P1/P0=0.9395

• With a 20% area reduction, A2=0.8A1• A2/A*= A2/A1 x A1/A* =0.8 x 2.0351

=1.6281• For this value of area ratio, from the

isentropic tables, T2/T0=0.9701, P2/P0=0.8993 and therefore M2=0.391


Lect-31

Solution: Problem 3

number. Mach the in increase an is Therenozzle. converging a through flow in drops etemperatur and etemperatur static The

kPa7.95P9395.08993.0100

P/PP/PPP

P/PP/P

PP

K395T9823.09701.0400

T/TT/TTT

T/TT/T

TT

2

01

0212

01

02

1

2

2

01

0212

01

02

1

2

=

=

=→=

=

=

=→=


Lect-31

Problem 4

• Air enters a converging–diverging nozzle, shown in the Figure, at 1.0 MPa and 800 K with a negligible velocity. For an exit Mach number of M=2 and a throat area of 20 cm2, determine (a) the throat conditions, (b) the exit plane conditions, including the exit area, and (c) the mass flow rate through the nozzle.


Lect-31

Problem 4

T0 = 800 KP0 = 1 MPaV = 0

Throat area = 20cm2

Me=2.0


Lect-31

Solution: Problem 4• The nozzle exit Mach number is given as

2.0. Therefore the throat Mach number must be 1.0.

• Since the inlet velocity is negligible, the stagnation pressure and stagnation temperature are the same as the inlet temperature and pressure, P0=1.0 MPaand T0=800 K.

3000 m/kg355.4RT/P ==∴ ρ


Lect-31

Solution: Problem 4

s/m5.517*RTV,Therefore

m/kg761.20.6339*K6.666T8333.0*TMPa5283.00.5283PP*

6339.0*,8333.0T*T,5283.0

P*P

tables, isentropic the From 1.M throat, the At (a)

*

30

0

0

000

==

==

==

==

===

=

γ

ρρ

ρρ


Lect-31

Solution: Problem 4

2e

30e

0e

0e

e

0

e

0

e

0

e

cm75.33*A6875.1Am/kg002.12300.0

K5.444T5556.0TMPa1278.0P1278.0P

,Therefore

6875.1*A

A,6330.1*M

,2300.0,5556.0TT,1278.0

PPtables,

isentropic the From 2.M exit, nozzle the At (b)

==

==

==

==

==

===

=

ρρ

ρρ


Lect-31

Solution: Problem 4

rate. flow mass choking :nozzle the throughpossible flow mass highest the to scorrespond This

s/kg86.25.517x0002.0x761.2*V*A*m

choked. is flow the since throat, the at properties theon based calculated be can rate flow mass The)c(

s/m2.8455.4442874.12RTMV from

determined be can velocity exit nozzle The

eee

===

=

××==

ρ

γ


Lect-31

Exercise Problem # 1

• A turbofan engine ingests air at 500 kg/s through an inlet area of 3.0 m2. If the ambient conditions are 288 K and 100 kPa, calculate the Mach number when the capture area will be equal to the inlet are.

• Ans: 0.405


Lect-31

Exercise Problem # 2• An aircraft flies at a Mach number of 2.4 at

an altitude where the ambient conditions are 70 kPa and 260 K. The aircraft has a two-dimensional intake with a wedge of half-angle 10°. If the axis of the intake and hence the wedge is tilted 25° with respect to the upstream airflow, determine the downstream Mach number, pressure, and temperature above the wedge.

• Ans: 3.105, 23.8 kPa, 191 K


Lect-31


• Air enters a nozzle at 0.2 MPa, 350 K, and a velocity of 150 m/s. Assuming isentropic flow, determine the pressure and temperature of air at a location where the air velocity equals the speed of sound. What is the ratio of the area at this location to the entrance area?

• Ans: 0.118 MPa, 301 K, 0.629


Lect-31


• Air enters a converging–diverging nozzle at 0.5 MPa with a negligible velocity. Assuming the flow to be isentropic, determine the back pressure that will result in an exit Mach number of 1.8.

• Ans: 0.087 MPa


Lect-31


• Air enters a converging–diverging nozzle of a supersonic wind tunnel at 1.5 MPa and 350 K with a low velocity. If a normal shock wave occurs at the exit plane of the nozzle at Me=2, determine the pressure, temperature, Mach number, velocity, and stagnation pressure after the shock wave.

• Ans: 0.863 MPa, 328 K, 0.577, 210 m/s, 1.081 MPa

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Lect-31...Prof. Bhaskar Roy, Prof. A M Pradeep, Department of Aerospace, IIT Bombay 20 Lect-31 Problem 4 • Air enters a converging –diverging nozzle, shown in the Figure, at 1.0