Aircraft Propulsion

Aircraft Propulsion 2E Solution Manual

1

Chapter 11: Aircraft Engine Component Matching and Off-Design

Problem 11.1 The corrected and physical mass flow rates at the engine face are related by:

2 2 2 2/cm m = 2 = Tt2/Tref = 268/288.2 = 0.9299 2 = pt2/pref = (0.99)112/101.33 = 1.0942 Therefore, 2 110.2 /cm kg s

22

cNN

=6400 rpm/ 0.9299 6,637 rpm

22 f

fc

mm

2.37 kg/s

0/FFc

0 0 /t refp p =112/101.33 = 1.1053 Therefore the corrected thrust is Fc= 145 kN/(1.1053) = 131.2 kN

0TSFCTSFCc

First, we calculate the TSFC to be: TSFC=(2.5 kg/s)/145 kN = 17.24 mg/s/N TSFCc = (17.24 mg/s/N)/(0.9299)0.5 17.88 mg/s/N

Problem 11.2 The corrected mass flow rate per unit area (at the engine face) is

),,(2

11/ 2)1(2)1(

22222 cczz

cz

ref

ref

c

cc RMfMMT

pR

Am cc

=

+=

+

We collect the constants on one side: [(180 kg/s)/(1 m2)][287 J/kg.K/1.4]0.5[288.2 K]0.5/[101.33 x1000 N/m2]=0.4331776 The function of axial Mach number is:


2

( )( 1)

2( 1) 32 22 2 2 2

11 1 0.22

c

ccz z z zM M M M

+ + = +

= 0.4331776

We use an Excel spreadsheet to get Mz2 to be Mz2 0.50 From the relationship between the corrected and physical mass flow rates:

2 2 2 2/cm m = We need to calculate 2 and 2 to arrive at the physical mass flow rate: Tt2=Tt0= (250 K) [1+0.2(0.85)2]= 286.125 K pt2=pt0. d = 30 kPa [1+0.2(0.85)2]3.5 (0.995) = 47.874 kPa 2 =286.125/288.2 = 0.9928 2 = 47.874/101.33 = 0.47246 Therefore the physical (air) mass flow rate at the engine face is:

2m = 180 kg/s [0.47246]/[0.9928]0.5 85.35 kg/s

From the continuity equation, we calculate the capture area, A0

00 ( 1)

2( 1)200 0

0

112

c

cc t c

c t

mAp M M

R T

+

= +

0.758 m2

Problem 11.3 Using the chain rule: pt4/pt2 = (pt4/pt3)(pt3/pt2)= (0.95)(10) = 9.5 The compressor exit total temperature is: Tt3=Tt2.c(-1)/ec= 300 K(10)0.4/1.4/0.9 623 K Energy balance across the burner gives Tt4 Tt4=[cpcTt3 +fQRb]/(1+f) cpt 1684 K Therefore Tt4/Tt2=1684/300 5.613


3

The area ratio A4/A2 is related to corrected mass flow rates, pressure and temperature ratios according to:

2

422

442

4 //)1(

t

t

c

c

cb TTAm

Amf

AA

+=

The corrected mass flow rate per unit area at the compressor face is:

( 1)2( 1)2 2

2 2 2 21/ 1 180.86 / /

2

c

crefc cc z z

c ref

pm A M M kg s m

R T

+ = + =

For a choked throat at the turbine inlet, M4=1.0, we have:

( 1)2( 1) 2

4 41/ 237.06 / /

2

t

treft tc

t ref

pm A kg s m

R T

++ = =

Therefore, A4/A2= 0.195745, which gives: A4=0.1957 m2

4 4237.06cm A= Therefore, 4 46.4 /cm kg s= The physical mass flow rate is the sum of air and fuel flow rates, i.e.,

4 180 6 186 /m kg s= + =

Problem 11.4 From equation 11.19, we have:

t

t

c

c

mm

=5

4

Also, turbine pressure and temperature ratios are related to the turbine efficiency according to:

( )ttttt /)1(11 = or

111t

tt

tt

=

=0.3441

t

t

c

c

mm

=5

4

=0.344/(0.8)0.5 = 0.3847


4

Similarly, the ratio of corrected mass flow rates between stations 3 and 4 is

4

3

(1 ) bcc b

m fm

= +

=(1+0.03)(1.8)0.5/0.95 = 1.455

Problem 11.5

22

cNN

Since, Tt2=Tref, 2=1, therefore Nc2=N=8,000 rpm

44

cNN

Since 4=Tt4/Tref=Tt4/Tt2=6, then Nc4=8,000 rpm/(6)0.5 3,266 rpm Compressor temperature ratio is c=Tt3/Tt2=872 K/288.2 K = 3.0257 From polytropic efficiency and temperature ratio, we get

1c c

c

e

c c

= = 32.7

Since pt2=pref and Tt2=Tref, the corrected and physical mass flow rates at the engine face are equal, therefore,

2 2 360 /cm m kg s= = Also the compressor power is

2 3 2( )c pc t tm c T T = 211.0 MW The power balance across the burner yields, f, but first lets get Tt4=6Tt2=1729 K:

4 3

4

1156(1729.2) 1004(872) 0.028442000(1000)(0.99) 1156(1729.2)

pt t pc t

R b pt t

c T c Tf

Q c T

= =

The compressor-turbine power balance gives

6

4 50

211.01 10(1 ) ( )360 /

cm pt t t

x Wf c T Tm kg s

+ = =

Therefore Tt5 1236 K And Tt5/Tt4 = 1236/1729 = 0.71487


5

pt5/pt4 = 111

t

tt

tt

=

0.1926

5 5 4 3

2 4 3 2

t t t t

t t t t

p p p pp p p p

= = (0.1926)(0.94)(32.7) = 5.922

5 5 4

2 4 2

t t t

t t t

T T TT T T

= = (0.71487)(6) = 4.289

Problem 11.6 Using the same calculation steps as in Example 11.1 and an Excel spreadsheet, as follow, we calculate and graph the gas generator pumping characteristics. Tt4/Tt2 Nc2 (rpm) pi-c m-c2 (kg/s) eta-c tau-c fQeta/cpT tau-t

6.5 9000 27.5 180 0.84 2.878 3.621747 0.755 6.5 8500 22.5 143 0.84 2.707 3.792742 0.777 6.5 8000 16 116 0.87 2.389 4.111288 0.819 6.5 7500 14 85 0.845 2.332 4.168024 0.826

Nc4 (rpm) m-c4 (kg/s) m-c4.Nc4 eta-t pi-t pt5/pt2 Tt5/Tt2

3530 18.093 63870 0.88 0.269 7.019 4.908 3334 17.568 58571 0.88 0.309 6.600 5.053 3138 20.040 62884 0.88 0.395 6.007 5.323 2942 16.783 49370 0.88 0.4123 5.484 5.371

Nc2 tau-c-1 Nc2/Nc2)2 tuac-1/tauc-1

9000 1.878 1 1 8500 1.707 0.891975 0.908961 8000 1.389 0.790123 0.739363 7500 1.332 0.694444 0.709157

The pumping characteristics are summarized in the following table: Nc2%des pt5/pt2 Tt5/Tt2 mc2%des fQeta/cpT

0.833333 5.483834 5.371213 0.4722222 4.168024 0.888889 6.006888 5.323132 0.6444444 4.111288 0.944444 6.59916 5.053179 0.7944444 3.792742

1 7.018628 4.908269 1 3.621747


6

Problem 11.7 The flow area A4 can be calculated from the given corrected flow rate

skgmc /804 = And the Mach number, M4=1.0 as well as gas properties, t=1.33 and cpt=1,156 J/kg.K.

( 1)2( 1) 2

4 41/ 237.1 / /

2

t

treft tc

t ref

pm A kg s m

R T

++ =

Therefore, A4 = 0.3375 m2

Since, we have t

t

c

c

mm

=5

4

and t=0.8 with t=0.85, we calculate

3.5 4

4.5 5

5.5 6

6.5 7

0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1

Nc2 % Design

Gas Generator Pumping Characteristics

0.4 0.5 0.6 0.7 0.8 0.9

1

0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1

Nc2 % Design

Corrected flow rate at compressor face (% design)


7

111t

tt

tt

=

= 0.3392 and thus

5cm 211 kg/s

( 1)

2( 1)25 5 5 5

1/ 12

t

treft tc z z

t ref

pm A M M

R T

+ = +

= 176.2 kg/s/m2

Therefore, A5 = 1.197 m2 Energy balance across the afterburner yields

7 5

7 5 7 5

7 7

(1 ) (1 )

(1 )( )AB pAB t pt t AB RAB ab

pAB t pt t pAB t pt tAB

RAB ab pAB t RAB ab pAB t

f f c T f c T f Qf c T c T c T c T

fQ c T Q c T

+ + + =

+ =

First we calculate Tt5 and Tt7 to be: Tt5= Tt4.t = 1760 K (0.8) = 1408 K Tt7=2.Tt5 = 2816 K Therefore fAB 0.0514 Application of the continuity equation at the nozzle throat produces the throat area, A8, but first

4 44

4c

mm

8 8 4 88

8 8

(1 )ABc

m f mm

+

The ratio of the two equations gives

8 8 8 44

4 4 8 4 8

c t t

c t t

m T pm T p

= =

For the dry case Tt8=Tt5 = 1408 K and

8 78 5

7 5

t tt t

t t

p pp pp p

=

pt5 = t . pt4 = 0.3392 (2000 kPa) = 678.4 kPa


8

pt8 = (0.98)(0.96)(678.4 kPa) = 638 kPa Therefore the corrected mass flow rate at the nozzle throat is

81408 2000(80 / ) 224.2 /1760 638.2c

m kg s kg s =

And for a choked throat, we have

( 1)2( 1) 2

8 81/ 237.1 / /

2

t

treft tc

t ref

pm A kg s m

R T

++ =

Hence the throat area is A8-dry 0.946 m2

8 8 4

4 4 8

c wet t wet t

c t t wet

m T pm T p

=

Tt8-wet = 2 Tt5 = 2816 K

8 78 5

7 5

t wet tt wet t

t t wet

p pp pp p

=

=(0.95)(0.90)(678.4 kPa)

pt8-wet 580 kPa

82816 2000(80 / )1760 580c wet

m kg s

348.94 kg/s

( 1)2( 1) 2

8 81/ 235.16 / /

2

AB

ABrefAB ABc wet wet

AB ref

pm A kg s m

R T

+

+ = =

Therefore A8-wet 1.4838 m2 Before we calculate A9/A8, let us first calculate M9-dry pt9 = (pt9/pt8) pt8 = (0.99)(638.2 kPa) = 631.8 kPa

1

99

9

2 11

t

tt

dryt

pMp

=

1.875

The physical mass flow rate through the nozzle is

8 88

8

cdry

dry

mm

=

= (224.2 kg/s) (638.2/101.33)/(1408/288.2)0.5 = 638.8 kg/s


9

Continuity demands: 9

9

12( 1)29 9 9

9 9 9 99 9

112

t

t

pm A M MR T

+

= +

A9-dry 1.4939 m2 A9-dry/A8-dry 1.579 In wet-mode, we calculate the nozzle exit Mach number for perfect expansion, i.e., M9-wet

1

99

9

2 11

AB

ABt wet

wetAB

pMp

=

pt9-wet = (0.95)(580 kPa) = 551 kPa therefore, M9-wet 1.794

9 8(1 )wet AB drym f m + =1.0514(638.85 kg/s) = 671.7 kg/s Now using continuity equation at the nozzle exit, we calculate the exit area, A9-wet A9-wet 2.424 m2 (A9/A8)wet = 2.424/1.4838 = 1.634 V9-dry = M9-dry a9-dry T9-dry is calculated from total temperature and Mach number to be 891.2 and thus the speed of sound, a9 is 583.1m/s V9-dry = 1093 m/s Fg-dry = (638.85 kg/s)(1093 m/s) = 698.372 kN T9-wet is similarly calculated to be ~1900 K and speed of sound is a9-wet= 842 m/s V9-wet = (1.7938)(842 m/s) = 1510 m/s Fg-wet = (671.7 kg/s)(1510 m/s) 1014 kN

Problem 11.8 p-t-0 (kPa) T-t-0 (K) Tau-r Pi-r a-0 (m/s) V-0 (m/s) p-t-2 (kPa) p-t-3 (kPa) 101.33 288.2 1.0000 1.0000 340 0.0000 99.34 2483


10

Tau-c T-t-3 (K) T-t-4 (K) p-t-4 (kPa) f Tau-t 2.7784 801 1729 2358 0.0232 0.7044

T-t-5 (K) Pi-t p-t-5 (kPa) 1218 0.2363 557 p-t-9 (kPa) M-9 T-9 (K) a-9 (m/s) V-9 (m/s) Fn/m-a0 TSFC (mg/s/N) 540 1.75 755 551 964 2.9 23.5 At off design conditions, we keep t constant and calculate c-O-D from

DcDOr

DrDOc )1()/(

)/(1, +

= 1+(1/6)(2.7784-1)/(1.1445/6.5) 2.6833

In the above, we substituted the ram temperature ratio and limit parameter at off design according to their simple relations on flight Mach number and given -O-D

1ce

c OD c OD

= 22.4

( )( )

4 22,

2, 4 2

/ 1456.5 / 295.3 1.111729.2 / 288.2/

t tc O D O D

c D t t D

T TNN T T

= =

( )( )

4 22, ,

2, , 4 2

/ 22.4 1 0.803725 1.11/

t tc O D c O D D

c D c D t t O D

T Tmm T T

And thus, 2 0.8037(80 / ) 64.3 /c ODm kg s kg s = From corrected mass flow rate and the axial Mach number at the engine face at the design point, we calculate A2

( 1)2( 1)2 2

2 2 2 21/ 1 180.1 / /

2

c

crefc cc z z

c ref

pm A M M kg s m

R T

+ = +

The compressor face flow area is thus A2 = 0.4443 m2 From the off-design corrected mass flow rate and the physical flow area at 2,

( 1)2( 1)2 2

2 2 2 21 64.3/ 1 144.71 / /

2 0.4443

c

crefc cc OD z OD z OD

c ref

pm A M M kg s m

R T

+

= + = =


11

We arrive at one equation, one unknown in Mz2-OD, using an Excel spreadsheet, we get Mz2-OD 0.378 We calculated the TSFC at design in our cycle analysis (spreadsheet). For the off-design analysis, we repeat our cycle analysis with the off-design compressor pressure ratio and the parameters that are specified in the problem. The spreadsheet is shown below: M0 p0 (kPa) T0 ( K ) gam-c cp-c (J/kg.K) Pi-d Pi-c Design-Pt 0 101.33 288.2 1.4 1004 0.98 25 Off-Design 0.85 20 258 1.4 1004 0.98 22.4 p-t-0 (kPa) T-t-0 (K) Tau-r Pi-r a-0 (m/s) V-0 (m/s) p-t-2 (kPa) Design-Pt 101.33 288.2 1.0000 1.0000 340 0.0000 99.3 Off-Design 32.08 295.3 1.1445 1.6038 322 274 31.4 p-t-3 (kPa) Tau-c T-t-3 (K) T-t-4 (K) p-t-4 (kPa) f 2482 2.7784 801 1729 2358 0.02318 704 2.6832 792 1456 669 0.02206 Tau-t T-t-5 (K) Pi-t p-t-5 (kPa) 0.7044 1218 0.2363 557.2 0.7086 1032 0.1839 123.0 p-t-9 (kPa) M-9 T-9 (K) a-9 (m/s) V-9 (m/s) Fn/m-a0 TSFC (mg/s/N) 540.5 1.75 755 551 964 2.9001 23.5 113.3 1.80 671 506 913 2.0502 33.4

Problem 11.9 We first calculate the design values for r, f, cH and :

1)0(2.01 =+=r

0.2857 / 0.90 0.31746(1.8) 1.2051f f = = =

0.2857 / 0.90 0.31746(14) 2.3113cH cH = = =

4

0

1146 1600 273 7.421004 15.2 273

pt t

pc

c Tc T

+ = = +

Now, we can calculate the first constant, C1

11.2051( 1) (1.3113) 0.2130

7.42r f

cHC

= =

Constants C2 and C3 are:


12

2 (1 )( 1) 6(0.2051)(1/ 7.42) 0.1659rfC

= + =

3

. 1.2051. 5(14) 28.2147.42

r fcHC

= =

Let us calculate r, and for the off-design operation:

2, 1 0.2(0.90) 1.162r OD = + =

4

,0

1146 1300 273 7.0971004 273 20

pt tOD

pc

c Tc T

+ = =

Now, let us substitute all the parameters in equation 10.66

21

11.

3 1)1(

)1()(

CCC

C

cHr

cHr

re

cH

rcH

=

+

3.15

1.162 28.214 1.162 0.213(7.097) 1 0.16597.097 1.162( 1)0.213(7.097)7.097

1.162( 1)cH cH

cH

+ =

The solution to this equation is found using an Excel spreadsheet to be: cH2.135 Therefore the high-pressure compressor pressure ratio at off design is cH=cHe/(-1)=(2.135)3.1510.90 The fan pressure ratio at off-design is calculated from:

)1(1

=

cHrf

C

=1.1463

f=f3.15=1.14633.15 1.537 The off-design bypass ratio is calculated from

3..

. CconstfrcH =

5.972

Off-Design

cH10.90

Off-Design

f 1.537

Off-Design

5.97


13

Based on these off-design parameters, we may proceed to calculate other engine off-design performance such as thrust specific fuel consumption or specific thrust, thermal and propulsive efficiencies.

Problem 11.10 The turbojet cycle analysis at design point follows the steps developed in Chapter 4. The spreadsheet of input data is shown below: M0 p0 (kPa) T0 ( K ) gam-c cp-c (J/kg.K) Pi-d Pi-c e-c Design-pt. 0 101.33 288.2 1.4 1004 0.95 30 0.9

Tau-Lamb Q-R

(kJ/kg) Eta-b Pi-b gam-t cp-t (J/kg.K) e-t Eta-m 7.882377 42600 0.98 0.95 1.33 1156 0.85 0.98

Pi-n p9/p0 0.9 1

We solve the above design point cycle to arrive at the following cycle parameters: p-t-0 (kPa) T-t-0 (K) Tau-r Pi-r a-0 (m/s) V-0 (m/s) p-t-2 (kPa) Design-pt. 101.33 288.2 1.0000 1.0000 340 0.0000 96.3 p-t-3 (kPa) Tau-c T-t-3 (K) T-t-4 (K) p-t-4 (kPa) f Tau-t 2888 2.9439 848 1973. 2744 0.03621 0.7571

T-t-5 (K) Pi-t p-t-5 (kPa) p-t-7 (kPa) T-t-7 (K) 1494 0.2674 734 734 1494 p-t-9 (kPa) M-9 T-9 (K) a-9 (m/s) V-9 (m/s) Fn/m-a0 TSFC (mg/s/N) 660 1.894 938 598 1133 3.452 30.8 Eta-th

0.4314 Now, we use the design-point turbine expansion parameter, t=0.7571 at off design to calculate the compressor pressure ratio at off design. First, the ram and limit temperature ratios at off design: r,O-D = 1.128 ,O-D = 8.577

DcDOr

DrDOc )1()/(

)/(1, +

= 1+[(1/7.882)/(1.128/8.577)](2.9439-1) = 2.875

c,O-D= c,O-D

ec/ (-1) 27.85


14

M0 p0 (kPa) T0 ( K ) gam-c cp-c (J/kg.K) Pi-d Pi-c Design-pt. 0 101.33 288.2 1.4 1004 0.95 30 Off-design 0.8 20 238 1.4 1004 0.95 27.85 The off-design engine analysis is now based on the above parameters. The results are: p-t-0 (kPa) T-t-0 (K) Tau-r Pi-r a-0 (m/s) V-0 (m/s) p-t-2 (kPa) Design-pt. 101.3300 288.2 1.0000 1.0000 340.2 0.0000 96 Off-design 30.4868 268.5 1.1280 1.5243 309.2 247.3 29 p-t-3 (kPa) Tau-c T-t-3 (K) T-t-4 (K) p-t-4 (kPa) f 2888 2.9439 848 1973. 2743.5 0.0362 807 2.8752 772 1773. 766.3 0.0321 Tau-t T-t-5 (K) Pi-t p-t-5 (kPa) p-t-7 (kPa) T-t-7 (K) 0.7571 1494 0.2674 733.5 733.5 1494 0.7599 1347 0.2593 198.7 198.7 1347 p-t-9 (kPa) M-9 T-9 (K) a-9 (m/s) V-9 (m/s) Fn/m-a0 660 1.89 938 598 1133 3.4518 179 2.09 782 546 1143 3.0153 TSFC (mg/s/N) Eta-th

30.8 0.4314 34.4 0.4704

p 2/(1+V9/V0)=0.355 at cruise.

Problem 11.11 Tt4/Tt2 Nc2 (rpm) pi-c m-c2 (kg/s) eta-c tau-c m-c3/m-c2 m-c3 (kg/s) pi-b

7 10000 13.5 90 0.85 2.298292 0.1122972 10.106746 0.95 7 9500 11.2 72 0.89 2.117119 0.1299136 9.3537796 0.953 7 9000 9.3 58 0.9 1.990087 0.1516886 8.7979406 0.956 7 8500 8 46 0.88 1.922099 0.1732997 7.9717881 0.958

fQeta/cpT tau-t Nc4 (rpm) m-c4 (kg/s) m-c4.Nc4 eta-t pi-t

4.701708 0.842823 3779.645 19.123676 72280.7 0.85 0.438654 4.882881 0.864756 3590.662 18.440688 66214.29 0.85 0.497365 5.009913 0.880135 3401.68 17.889891 60855.69 0.85 0.541923 5.077901 0.888366 3212.698 16.494171 52990.79 0.85 0.566968

pt5/pt2 Tt5/Tt2 Nc2 tau-c-1 Nc2/Nc2)2 tuac-1/tauc-1

5.625733 5.899758 10000 1.298292 1.234568 0.691223 5.30868 6.053294 9500 1.117119 1.114198 0.594765

4.818129 6.160948 9000 0.990087 1 0.527132 4.345246 6.218564 8500 0.922099 0.891975 0.490935


15

Nc2%des pt5/pt2 Tt5/Tt2 mc2%des fQeta/cpT 0.85 4.345246 6.218564 0.5111111 5.077901 0.9 4.818129 6.160948 0.6444444 5.009913

0.95 5.30868 6.053294 0.8 4.882881 1 5.625733 5.899758 1 4.701708

Nc2%des mc2%des

0.85 0.511111 0.9 0.644444

0.95 0.8 1 1

Problem 11.12 We start with the definition of corrected mass flow rates at the nozzle exit and the engine face. Then, we take the ratio of the two expressions.

4 4.5

5 5.5

6 6.5

0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1

Nc2 % Design

Gas Generator Pumping Characteristics

0.4 0.5 0.6 0.7 0.8 0.9

1

0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1

Nc2 % Design

Corrected flow rate at the compressor face (% design)


16

( )

9 99

9

2 22

2

9 2 9 2 5 29 9

2 2 9 2 9 2 5 2

/ / /(1 ) (1 )

/ / /

c

c

t t t tc

c t t n t t

mm

mm

T T T Tm m f fm m p p p p

= = + = +

Nozzle total pressure ratio as a function of engine corrected mass flow rate is graphically specified. We insert a column of n values in our table. Then, we insert a column of nozzle exit corrected mass flow rates according to the above formula. Since the nozzle exit is choked, we can calculate the area, A8, from the corrected mass flow rates. This becomes a new column in our spreadsheet. Finally, we graph the exit area variation as a percent of design value and graph it using Excel, as follows: Tt4/Tt2 Nc2 (rpm) pi-c m-c2 (kg/s) eta-c tau-c m-c3/m-c2 m-c3 (kg/s) pi-b pi-n

7 10000 13.5 90 0.85 2.298292 0.1122972 10.106746 0.95 0.97 7 9500 11.2 72 0.89 2.117119 0.1299136 9.3537796 0.953 0.98 7 9000 9.3 58 0.9 1.990087 0.1516886 8.7979406 0.956 0.99 7 8500 8 46 0.88 1.922099 0.1732997 7.9717881 0.958 0.99

pt5/pt2 Tt5/Tt2 m-c9 (kg/s) A-8 (m2)

5.625733 5.899758 41.261591 0.197619 5.30868 6.053294 35.071391 0.167971

4.818129 6.160948 31.086752 0.148887 4.345246 6.218564 27.465691 0.131544

Nc2%des mc2%des A-8%des

0.85 0.511111 0.665647 0.9 0.644444 0.753405

0.95 0.8 0.849975 1 1 1

0.6 0.7 0.8 0.9

1

0.85 0.86 0.87 0.88 0.89 0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1

Nc2 % Design

Nozzle exit area, A8 (% design)


17

Problem 11.13 We first calculate the design values for r, f, cH and :

1)0(2.01 =+=r

0.2857 / 0.90 0.31746(1.65) 1.1723f f = = =

0.2857 / 0.90 0.31746(20) 2.5884cH cH = = =

4

0

1146 1650 273 7.621004 15.0 273

pt t

pc

c Tc T

+ = = +

Now, we can calculate the first constant, C1

11.1723( 1) (1.5884) 0.2444

7.62r f

cHC

= =

Constants C2 and C3 are:

2 (1 )( 1) 8(0.1723)(1/ 7.62) 0.18086rfC

= + =

3

. 1.1723. 7(20) 54.9077.62

r fcHC

= =

Let us calculate r, and for the off-design operation:

2, 1 0.2(0.85) 1.1445r OD = + =

4

,0

1146 1500 273 7.8441004 273 15

pt tOD

pc

c Tc T

+ = =

Now, let us substitute all the parameters in equation 11.66

21

11.

3 1)1(

)1()(

CCC

C

cHr

cHr

re

cH

rcH

=

+

3.15

1.1445 54.907 1.1445 0.2444(7.844) 1 0.180867.844 1.1445( 1)0.2444(7.844)7.844

1.1445( 1)cH cH

cH

+ =

The solution to this equation is found using an Excel spreadsheet to be:


18

cH 2.468 Therefore the high-pressure compressor pressure ratio at off design is cH=cHe/(-1)=(2.468)3.1517.21 The fan pressure ratio at off-design is calculated from:

)1(1

=

cHrf

C

=1.1409

f=f3.15=1.14093.15 1.515 The off-design bypass ratio is calculated from

3..

. CconstfrcH =

7.818 Based on these off-design parameters, we may proceed to calculate engine off-design performance such as thrust specific fuel consumption or specific thrust, thermal and propulsive efficiencies.

Problem 11.14 M0 p0 (kPa) T0 ( K ) gam-c cp-c (J/kg.K) Pi-d Pi-c Design 0 100 288 1.4 1004 0.98 15

e-c Tau-Lamb Q-R

(kJ/kg) Eta-b Pi-b gam-t cp-t (J/kg.K) 0.9 7.028303 42800 0.98 0.97 1.33 1156

e-t Eta-m Pi-n p9/p0 0.8 0.995 0.97 1

Based on the above input values, we calculate t at design point: p-t-0 (kPa) T-t-0 (K) Tau-r Pi-r a-0 (m/s) V-0 (m/s) p-t-2 (kPa) Design 100 288 1 1 340 0.0000 98 p-t-3 (kPa) Tau-c T-t-3 (K) T-t-4 (K) p-t-4 (kPa) f Tau-t 1470 2.3624 680 1758 1426 0.034 0.8115 Based on the following equation, we calculate compressor temperature ratio at off design:

DcDOr

DrDOc )1()/(

)/(1, +

The off design values for ram and limit temperature ratios are:

Off-Design

cH17.2

Off-Design

f 1.515

Off-Design

7.818


19

r-OD = 1.8 -OD = 7.80 Therefore, c-OD = 1.8401, which gives the compressor pressure ratio at off design as c-OD 6.83

( )( )

4 22, ,

2, , 4 2

/0.5797

/t tc O D c O D D

c D c D t t O D

T Tmm T T

2, 13.9 /c O Dm kg s

( )( )

4 22,

2, 4 2

/0.785

/t tc O D O D

c D t t D

T TNN T T

=

Therefore, Nc2,O-D=4,710 rpm From the design values of compressor corrected flow and axial Mach number, we get

( 1)2( 1)2 2

2 2 2 21/ 1 180.06 / /

2

c

crefc cc z z

c ref

pm A M M kg s m

R T

+ = + =

Therefore A2 = 0.13329 m2

( 1)2( 1)2 2

2 2 2 21 13.9124/ 1 104.38 / /

2 0.13329

c

crefc cc OD z OD z OD

c ref

pm A M M kg s m

R T

+

= + = =

The only unknown in the above equation is the axial Mach number at the engine face at off design. We use Excel to calculate Mz2,OD. Mz2,OD 0.261

Problem 11.15 We start our cycle analysis at design point and proceed to calculate turbine expansion parameter t, which we keep constant at off-design operation as well. The spreadsheet is shown below. The turbine expansion parameter is calculated to be t = 0.7936 Power balance between the turbine and compressor yields:


20

DcDOr

DrDOc )1()/(

)/(1, +

Which gives c,OD=1.90 and the corresponding c,OD = 7.564 M0 p0 (kPa) T0 ( K ) gam-c cp-c (J/kg.K) Pi-d Pi-c Design 0 101.33 288.2 1.4 1004 0.95 18 Off-Design 2.5 15 223 1.4 1004 0.82 7.56

e-c m-c2 (kg/s) N-c2 (rpm) Tau-Lamb Q-R

(kJ/kg) Eta-b 0.9 67 7120 7.083353 42800 0.98 0.9 36.37 5512 9.551927 42800 0.97

Pi-b gam-t cp-t (J/kg.K) e-t Eta-m Tau-L-AB 0.97 1.33 1156 0.8 0.995 9.66 0.98 1.33 1156 0.8 0.995 13.60

Q-R-AB (kJ/kg) Eta-AB Pi-AB-On gam-AB-On cp-AB-On (J/kg.K) Pi-n

42800 0.98 0.95 1.3 1243 0.9 42800 0.98 0.95 1.3 1243 0.88

p9/p0

1 1

p-t-0 (kPa) T-t-0 (K) Tau-r Pi-r a-0 (m/s) V-0 (m/s) Design 101.3300 288.2 1.0000 1.0000 340.2075 0.0000 Off-Design 256.2891 501.7 2.2500 17.086 299.2604 748.1 p-t-2 (kPa) p-t-3 (kPa) m-2 (kg/s) Tau-c T-t-3 (K) 96.3 1733 63.6500 2.5032 721 210.2 1590 57.1639 1.9009 954 T-t-4 (K) p-t-4 (kPa) f Tau-t T-t-5 (K) Pi-t p-t-5 (kPa) 1773 1681 0.03322 0.7936 1407 0.3120 524.4 1850 1558 0.02999 0.7929 1467 0.3107 484.0 p-t-7 (kPa) T-t-7 (K) f-AB p-t-9 (kPa) M-9 498.1 2250. 0.0309 448.3 1.652 459.8 2450. 0.0357 404.6 2.756 T-9 (K) a-9 (m/s) V-9 (m/s) Fn/m-a0 TSFC (mg/s/N) Eta-th 1596 772 1275 3.9871 47.26 0.6050 1145 654 1801 3.9135 56.12 1.1211


21

Problem 11.16 m-dot-c2 (kg/s) M-z2 M-0 p-0 (kN) T-0 (K) R (J/kg.K) gamma 100

0.5 0 101 288 287 1.4

a-0 (m/s)

f(M-z2)

A-2 (m^2) 340

0.43192

0.557

Problem 11.17 M_4 M_8 M_0-D p_0-D T_0-D Tau_t T_t2-D 1 1 0 101 288 0.7 288

f-D Eta_m-D e_c-D gamma c_p T_t4-D

0.023 0.993 0.9 1.4 1004 1950

M_0-OD f-OD Eta_m-OD e_c-OD p_0-OD T_0-OD T_t4-OD 2 0.023 0.993 0.9 25 223 1650

Tau_c-D Pi_c-D T_t2-OD Tau_c-OD Pi_c-OD 3.063 34.0 401.4 2.253 12.9

Problem 11.18 Pi-c-des Pi-c-OD m-c-2,D (kg/s) m-c-2,OD (kg/s)

Stall Margin (%)

30 31.5 250

238.1 10.25

R T-ref p-ref

M-z2,D m-c2,D/mc2,St 287 288.2 101.33

0.5 1.050

Mz2 @ Stall

0.465


22

Problem 11.19

Design Point

M_0 p_0 T_0 gam_c c_pc Pi_d

m-dot_c2 M_z2

0 101.33 288.2 1.4 1004 0.98 600 0.5 Pi_f e_f Alpha Pi_cH e_cH T_t4 Q_R Eta_b 1.8 0.9 5 14 0.9 1873 42800 0.99 Pi_b e_tH Eta_mH e_tL Eta_mL gam_t c_pt Pi_nf 0.95 0.85 0.995 0.89 0.995 1.33 1146.4 0.98

Off-Design Parameters M_0-OD p_0-OD T_0-OD T_t4-OD 0.84 20 253 1573

Design Point Analaysis Tau_r Tau_f Tau_lam Tau_cH C_1 C_2 C_3

1 1.205 7.421 2.311 0.2130 0.1659 28.209

Off-Design Analysis Tau_r-OD

Tau_lam-OD

1.14

7.10

Tau_cH-OD Equation 11.67-C_2

Pi_cH-OD

2.192 0.000663

11.85

Tau_f-OD

Pi_f-OD

Alpha-OD

m-dot_c2-OD (kg/s)

1.111

1.394

5.63

475

Problem 11.20

From Appendix A for 12 km US Standard Atmosphere

M_0 gam R p_0 (kPa) T_0 (K) 0.85 1.4 287 19.39 218.7

Pi_d m-dot_c2

m-dot_c0 (kg/s) p_ref (kPa)

T_ref (K) A_0 (m^2) M_th

0.995 200 199 101 288 0.844 0.7


23

p_t0 (kPa) T_t0 (K)

m-dot_0 (kg/s) A_th (m^2)

a_0 (m/s) V_0 (m/s) D_r (kN)

31.11 250.3 65.75 0.905 296.4 252.0 16.56

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