problem answers

Aircraft Propulsion 2E Answers to All Problems

1

Answers to All Problems in Aircraft Propulsion, 2nd Edition

Chapter 1 Problem 1.1 Thermal efficiency is 85.6%

Problem 1.2 Thermal efficiency is 66.7%

Problem 1.3 66.7%th Braytonη − ≈

70.8%th Humphreyη − ≈

Problem 1.4

N=3,250,000 rpm

Problem 1.5 5, 467

AB OffFuel lbm

−=

18,375AB On

Fuel lbm−

=

Chapter 2 Problem 2.1 1 592 /u m s *

2 0.615M ≈

Problem 2.2

2

1

1.458pp

≈

3

1

2.06pp

≈

0.036dC ≈

Problem 2.3 ,3 683tp kPa≈

Problem 2.4 M1 = 2.6 2 2.19M ≈

3 1.70M ≈

Problem 2.5 2 21.54p kPa≈


2

3 9.50p kPa≈ 4 12.59p kPa=

Problem 2.6 2 3.4M ≈

2

1

3.664AA

=

42.9δ = °

Problem 2.7

First, we state that regions 2 and 3 are on the top and region 4 is on the bottom surface of the half-diamond airfoil. Region 1 is the free stream condition.

',253.4

tp kPa≈

',36.02

tp kPa≈

',432.8

tp kPa≈

Problem 2.8

1.732l

d

cc

=

, 0.3694m LEc =

Problem 2.9 * 416.7 /a m s≈ 2 255.2 /u m s≈ 456.5 /ta m s≈ 2 488.4 /h kJ kg≈

Problem 2.10 . . 253.6 /T Sa m s= 1.744glassF kN=

Problem 2.11 0.977dc ≈

Problem 2.12 32.15wallθ = °

2

1

1.264VV

=

Problem 2.13


3

( )max/

0.373t c

xc

≈

9.06δ ≈ °

p1/p∞=1.0, 2 0.14478pp∞

= , 389.23 ≈∞p

p

Problem 2.14 ,2 157.59tp kPa=

Problem 2.15 11.88 /m kg s≈ max 5.35L m≈

2

1

0.746t

t

pp

≈ ; i.e. 25.4% total pressure loss.

Problem 2.16 * 121.49 /q kJ kg=

Problem 2.17 0.00267fC ≈

2

1

0.468pp

≈ and static pressure drop is ~53.23%

Problem 2.18 1 941tT K

2

1

5.667pp

= . Therefore, static pressure rise is 466.7%

Problem 2.19

max* 1.0285T

T≈

max@* 1.2073Tp

p≈

Problem 2.20 2 47.9p kPa=

32 0.5 /kg mρ =

Problem 2.21 3

2 0.5 /kg mρ ≈ 2 90p kPa≈


4

Problem 2.22 7.712L m=

Problem 2.23 * 2.54L m≈ ~ 40.7% total pressure loss

Problem 2.24 M2=1.0 *

2 33.52p p psia= =

*2 63.46t tp p psia= =

*2 1439T T R= ≈ °

*1' 300 /q BTU lbm=

Problem 2.25 1 0.24M ≈ 2 2,733tT R≈ °

Problem 2.26 4 0.36M ≈

Problem 2.27

*

53.455LD≈

1' 0.48M ≈

Problem 2.28

15.235sxD≈

6205.01

=t

ty

pp

, i.e. 37.95% loss of total pressure

*

2 1

1

0.1093I II−

= − or ~11%.

Problem 2.29 1 0.66 2.0M or≈ ~ 37.2% static pressure drop ~ 145% static pressure rise ~ 6.35% loss of fluid impulse (subsonic) and ~10.93% loss of fluid impulse (supersonic)

Problem 2.30 2 0.70M ≈


5

2

1

4.806TT

=

2 1/ 0.845t tp p = , therefore 15.5% total pressure loss

Problem 2.31 242.7 /m kg s≈ 2 0.52M ≈ 0.4643PRC ≈

20.26x wallF kN≈ −

Problem 2.32 ( ) 25x vane

F N=

( ) 25y vaneF N= −

Problem 2.33

1 1

0.2317x nozzleF

p A≈

Problem 2.34 2.1 MJ/kg, which divides into700 kJ/kg per duct (for three ducts)

Problem 2.35 Mcone ≈2.4 , 0.1887p coneC ≈

coneDconep CC ,, =

Problem 2.36 81.8 /m kg s≈

22 0.1973A m=

2 161p kPa≈ 1 90.75I kN= 2 42.82I kN≅

47.9x wallF kN≅

Problem 2.37 2 1.83M ≈

2

0

0.9706pp

≈

185.01

, ≅−

tht

wallconx

ApF


6

1229.01

, −≅−

tht

walldivx

ApF

,

1

0.062x nozzle

t th

Fp A

≅

Problem 2.38 2 0.34M ≈ 0.611PRC =

1 1

0.507x wallsFp A

= −

Problem 2.39

1 1

1.1675x wallsFp A

≈ −

Problem 2.40

1 1

0.244x wallsFp A

=

Problem 2.41 qmin=796.8 kJ/kg

T2≈1911 K Fuel-to-air ratio=0.00664 or 0.664%

Problem 2.42

*

1

1

0.484p t

qc T

=

1

3.99 399%p orp∆

≅

Problem 2.43 Mass flow rate=0.07258 kg/s Dh=0.02 m M2=0.66 (from Table) Δpt (%)=44.64

Problem 2.44 q1

*≈1511 kJ/kg f=1.275%

Problem 2.45 Fx (duct) ≈ 69 kN Delta Power=-147.2 MW

Since the flow was adiabatic and the exit power is 147.2 MW less than the power of the flow in, therefore, there has been shaft power extraction, which means that the component is a turbine.


7

Problem 2.46 T_t1 (K)

1170

p-t1(kPa) 156.5

M_2 (from table)

1.42

%Delta p_t/p_t1 33.3

L_1*/D

15.25

Delta(s)/R 0.4044

Problem 2.47

pt1 (kPa) 118.6

u1 (m/s) 245.5

M2 (~) 0.8

Delta pt Loss (%) 8.50

Problem 2.48

M-2 (from Table) 1.4

Tt2 (K)

450

pt2 (kPa) 129.3


8

Problem 2.49

q-1 (kPa) 14.75

p-2 (kPa)

96.7

C-PR 0.84

Problem 2.50

Lx/D=13.57 L/D≈16.36

~40.8% loss in total pressure due to wall friction and the NS

Problem 2.51 q1

*= 155.5 kJ/kg Tt2

*=1164 K M1’ ≈0.44 Mass flow rate drops by about 20%.

Problem 2.52

A1≈1.60 m2 A2≈1.13 m2

Fx)nozzle≈31 kN

Problem 2.53 L-x/D

8.447

L/D 18.855

L-x/L

0.448

(pt1-pt2)/pt1*100 40.8

Problem 2.54

T-t2 (K) 1981

M_2 (from Table)

1.2


9

p_2 (kPa) 69.4

Del (p_t)/p_t1 (%)

57.8

q* (kJ/kg) 616

Problem 2.55 q0=75.6 kPa A1=2.117 m2

�̇�0 ≈ 352.6 𝑘𝑔/𝑠

Chapter 3 Problem 3.1

( )max / 0.0144fA B

n

wf

w= =

( )max / 0.054fA B

n

wf

w= =

Maximum (A/B on): 2.5

2

3.67pp

= ; 3

2.5

3.0926pp

= and 5

4

0.2279pp

=

Military: 2.5

2

3.67pp

= ; 3

2.5

3.0926pp

= and 5

4

0.2279pp

=

Maximum (A/B on): 9 2,841 /V ft s= Military: 9 1,974 /V ft s= Maximum (A/B on): 0.168thη ≈ or ~16.8% Military: 0.295thη ≈ or 29.5%

Maximum (A/B on): 2.103 / /f

n

wTSFC lbm hr lbf

F= =

Military: 0.8353 / /f

n

wTSFC lbm hr lbf

F= =

Maximum (A/B on): 0.827carnotη ≈ or 82.7% Military: 0.744carnotη ≈ or 74.4%

57%Percent thrust increase ≈ and 295%Percent fuel flow rate increase ≈

Problem 3.2

359.1===core

fan

ww

BPR

α


10

0.01113f = 7,813 /fw lbm hr=

) 17,727n staticF lbs=

0.3658thη = Thermal efficiency of this engine is higher than the engine in Problem 3.1. This difference can be attributed to the thermally-inefficient AB-on engine. Also, JT3D has a higher pressure ratio than J57; therefore its thermal efficiency is higher.

0.4407 / /TSFC lbm hr lbf= 1.113SPF =

0.721carnotη =

3

2

13.6t

t

pp

=

13 0.858M =

Problem 3.3

slbwww

f fcore

f /199.2=⇒=

) 14,284n exhaustF lbs≈

0.3232thη ≈ 0.7621carnotη ≈ 0.554 / /TSFC lbm hr lbf≈

85.152

3 =t

t

pp

1.1===core

fan

ww

BPR

α

0.7621carnotη ≈ 9 0.87M ≈

2.5

2

4.082t

t

pp

≈

3

2.5

3.88t

t

pp

≈

Problem 3.4

5.212

3 =t

t

pp

) 34,300g fanF lbs=

0.017f =


11

) 9,284g coreF lbf=

) 43,584g totalF lbf=

0.335thη =

0.351 / /TSFC lbm hr lbf=

06.5===core

fan

ww

BPR

α

0.786carnotη =

18.22

3 =t

t

pp

84.9

3

4 =t

t

pp

19 0.788M =

9 0.70M ≈

Problem 3.5 14.971ramD kN≈ 27.040gF kN≈

) 12.1n un installedF kN

−≈

/0.2237 kg hrTSFCN

≈

0.303thη ≈ 0.726pη ≈

0.22oveallη ≈

Problem 3.6 , 1,000 1s prop kW MW℘ = =

9 200 /V m s= 0.9pη =

Problem 3.7

0

1

10.0AA

=

1

0

0.9505pp

=

1 0

0.0405addDA p

≈

Problem 3.8


12

175F kN= 356.8sI s=

Problem 3.9 19.1ramD kN≈ 7.5gcF kN≈

33.4gfF kN≈

21.8netF kN≈ 0.663pη ≈

Problem 3.10 5.99ramD kN

) 15.5g nozzleF kN≈

9.5netF kN≈ 0.583pη =

Problem 3.11 30ramD kN≈

) 61.5g nozzleF kN≈

31.5netF kN≈ 0.133thη ≈ 0.677pη ≈

860Range km≈ No!

Problem 3.13 13.986s MW℘ ≈

Problem 3.14 3.924F MN= 9 3,924 /V m s= MNFrocket 924.3=

Problem 3.15 318.4pF kN≈

Problem 3.16 2

2 0.508A m= 1 182.61I kN= kNI 26.1572 =


13

25.35x wallsF kN=

Problem 3.17 2

7 0.67A m≈ 7 38.1I kN≈ 8 31.1I kN≈

) 7.1x nozzleF kN≈

Problem 3.18 Ftotal=Fcore+Fprop=50 kN 10thrust MW℘ = 11.765shaft MW℘ =

Problem 3.19 9 1200 /V m s≈

Problem 3.20 1.59 /fm kg s≈

2,569sI s= 0.0371f ≈

Problem 3.21

D_r (kN) 68.0

F_g (kN) 137.0

F_n (kN)

69.0

TSFC (mg/NS) 60.86

Eta-th Eta-p D_r (lbf) F_g (lbf)

0.340 0.174 15294 30808 TSFC (lbm/hr/lbf)

2.148

Problem 3.22 D_r (kN) F_g (kN)

TSFC (mg/NS)

26.0 97.4

35.0

Eta-th Eta-p D_r (lbf) F_g (lbf)

TSFC (lbm/hr/lbf) 0.401 0.433 5845 21891

1.24


14

Chapter 4

Problem 4.1

0.80dπ ≈ 0.889dη ≈

0.223sR∆

≈

Problem 4.2 3 661.6tT K= 0.8484cη = 42.5c MW℘ =

Problem 4.3 0.02f ≈ 4 1, 485tT K≈ 4 3.104tp MPa≈

Problem 4.4 5 976.2tT K= 0.6573tτ = 0.813te ≈ 5 1,102tp kPa= 60t MW℘ ≅

Problem 4.5 , 1,185t mixed outT K− ≈

Problem 4.6 1,643.60tgT K=

900.4tcT K= , 2.09t mixed outp MPa− =

1.2687sR∆

= −

Problem 4.7 0.818nπ ≈ A9/A8≈2.506 9 2.15M∴ ≈

Problem 4.8


15

0.0952pη ≈

0.5455pη ≈

Problem 4.9 02141.0=f 9 1.830M ≈

0

781.26 / /F N kg sm

≈

0.3157thη ≈ 0.6147pη ≈

Problem 4.10 0 631.5 /V m s=

20 1.128A m=

0 78.2tp kPa= 4 66.9tp kPa= 4 1,368.7tT K= 9 61.5tp kPa= 9 1.59M ≈

9 967 /V m s≈

29 1.965A m≈

109.4gF kN= 9 1,062 /

efffV m s≈

0.3231thη = 0.7654pη =

Problem 4.11 16.46rD kN≈ Tt2=438.9 K pt2=90.9 kPa pt3=1363.3 kPa Tt3= 1036.8 K Tt13 ≈ 499.15 K pt15=pt13= 136.3 kPa 0.01586f ≈ Tt5 ≈925.7 K pt5=136.3 kPa α = 2.043 0 8.213 /m kg s≈ 16.787 /fanm kg s≈ Tt6M≈640.8 K, pt6M=133 kPa


16

M9≈2.272 and V9≈808.5 m/s 20.32gF kN≈

0 0

0.5158(1 )

nFm aα

≅+

33.76 / /TSFC mg s N≅ 50.13%thη ≅ 91%pη ≈ and ηo ≈ 45.6%.

Problem 4.12 25.8ramD kN≈ 39.57c MW℘ = 0.02f = 4 1,508tT K≈ 0.7437tτ = 39.57t MW℘ = 8.522ABλτ = 0.0272ABf = Fg≈154 kN TSFC=36.92 mg/s/N Fn≈128 kN 0.477thη = ηp≈0.341

Problem 4.13 3 722tp kPa≈ 13 331tT K≈

0.0215f ≈

5 935tT K≈ 19 1.24M ≈

9 3.1M ≈ 9 1,108 /V m s≈ 19 396 /V m s≈ 0 271 /V m s≈

0.826fan

core

FF

≈

Problem 4.14 6 522.7t MT K= 9 1.00M ≈ 9 502 /V m s≈


17

5

1,506 / /gFN kg s

m≈

Problem 4.15

8

8

1.821on

off

AA

−

−

≈

Problem 4.16 5 15 370.4p p kPa= = 6 404.3Mp kPa≈

Problem 4.17 3 3,009tp kPa= 3 848tT K≈ 4 1,729tT K≈ 0.663tτ ≈

23.0c off designπ

−≈

Problem 4.18 0.0232f = 0.705tτ = 18.6c Off Designπ − − ≈

Problem 4.20

02 ( 1)

1M λτγ

= − −

Problem 4.21

( )11

2 3,0 −

−= λτγoptM

Problem 4.24 7.588λτ = 0.033f =

0 0

2.1324nFm a

=

0.68pη =

0.415thη ≈

Problem 4.25


18

2.304c

f

℘=

℘

0.0181f ≈ V19/V9=0.6758

2.537n fan

n core

FF

−

−

≅

ηth ≈39.8 % ηp ≈ 75.0%

18.47 / /TSFC mg s N≈ 5,523sI s≈

Problem 4.26 0.901nη =

72.30

9 =aV

0913.0)(

09

909 =−

amApp

Problem 4.28 0.044f = 9 2.363M =

0 0

2.588nFm a

=

Problem 4.29 0.313thη ≈ 0.504thη ≈

Problem 4.30 9.096t MW℘ = 8.96

coregF kN=

88

8

1.00VMa

= =

4.648corenF kN=

0.935nη = 0.900LPTη =

Problem 4.31

0.45

15 =mm


19

35

6 =A

A M

kNMp MM 106.164)1( 266 =+γ

40.05

6 =t

Mt

TT

Problem 4.32 7.15t MW℘ =

7.01prop MW℘ =

56.1propF kN

Problem 4.34 M0,opt≈2.40 The effect of fuel heating value on optimum flight Mach number is negligibly small

Problem 4.35 M9≈3.81

0 1

50.4rDp A

≈

0 1

64.0gFp A

≈

Is≈2,841 s

Problem 4.36 Percent change in fuel-to-air ratio = +78.95%

Percent change in TSFC = +30.1% Percent change in nozzle exit temperature = +72.8% Percent change in thrust = +37.6% Percent change in exhaust speed = +27.5% Percent change in thermal efficiency = - 5.64%

Problem 4.37 Dr=3.5546 kN

ηc=0.8482 HPT℘ =0.9562 MW

f℘ =0.4387 MW Fg,c≈1.718 kN Fn≈2.444 kN TSFC≈20.155 mg/s/N thη ≈ 49% pη ≈ 60%

Problem 4.38

T-t3 (K) f Tau-Lamb p-t4 (kPa) h-t-4.5 T-t-4.5 (K) Tau-t-HPT Pi-t-HPT 883 0.0256 7.338 1552 1317110 1143 0.6929 0.1575


20

p-t-4.5 h--t5 T-t5 Tau-LPT e-t (LPT) p-t5 P-LPT(MW) F-prop (kN)

244.5 964340 837 0.73216 0.8590 56.6 18.09 57.4

T-9i (K) T-9 from eff. V-9 from eff. a-9 from eff M-9 (exact) pt-9 from eff Pi-n 715.0 721 517 523 0.987 54.7 0.966

F-core (kN) F-total (kN) P-prop (MW) P-core (MW)

PSFC (mg/s/kW)

TSFC (mg/s/N)

Etha-thermal

Etha-propuls

13.3 70.7 17.8 5.1 55.7 18.1 0.427 0.814

Problem 4.39

.th regenη ≈ 70.3%

th Braytonη ≈ 48.2%

Problem 4.40

Tau-r Pi-r a-0 (m/s)

V-0 (m/s)

p-t0 (kPa) T-t0 (K)

p-t2 (kPa)

p-t13 (kPa)

2.25 17.0859 309 773 427 535.5 363 545

Tau-f T-t13 (K)

p-t3 (kPa) Tau-c T-t3 (K) p-t4 (kPa) f

p-t15 (kPa)

1.137371 609 4357 2.20 1179 4095 0.02233 539

Pi-t Tau-t T-t5 (K) T-5 (K) a-5 (m/s)

p-t5 (kPa)

p-5 (kPa) Alpha

0.131649 0.6687 1204 1156 663 539 458 0.293

h-t6M (J/kg) M-15 T-15 (K) a-15 (m/s)

p-15 (kPa) A15/A5 c-p6M gama-6M

1213921.4 0.488 581 483 458.1 0.203261 1119.03 1.343423

C1 T-t6M (K) C2 C M-6M p-6M p-t6M-i (kPa) Pi-M (ideal)

1.332673 1084.8 0.51551 6.683 0.511 451.8 536.5 0.995 p-t6M (kPa) Pi-M

p-t7 (kPa) f-AB

p-t9 (kPa) M-9 T-9 (K)

a-9 (m/s)

V-9 (m/s)

509.7 0.9453 468.9 0.0387 445.4 1.93 1411 725 1400 V-9eff F-n/(1+Alfa)m.a0 TSFC (mg/s/N) Eta-therm Eta-prop Eta-overall f+f-AB

1577 2.842 53.7 0.497 0.677 0.336 0.061

Problem 4.41

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) Tau-c

107 439 1.968 10.6927 299.3 658.4 94.10 753 1.935

T-t-3 (K) T-t-4 (K)

p-t-4 (kPa) f Tau-t

T-t-5 (K) Pi-t p-t-5 (kPa)


21

849.2 1449.5 692.5 0.01518 0.7064 1024.0 0.2627 182

p-t-7 (kPa) T-t-7 (K) f-AB p-t-9 (kPa) M-9 T-9 (K) a-9 (m/s) V-9 (m/s) 164 2007 0.0264 155.6 2.440 916.2 606.6 1480.0

Fn/m-a0

TSFC (mg/s/N) Eta-th

Eta-p (exact)

Eta-p (approx) Eta-thermal

2.95 47.1 0.52854 0.6293 0.6158 0.5285

D-ram (kN)

Power-comp (MW) F-g (kN) 6.58 4.120 15.42

A9 ≈ 0.203 m2

Problem 4.42

M_0 1.992

M_9 1.67

V_9 (m/s) 1088

Problem 4.43 A_9 (m^2) F_g (N) V_9eff F_TO (kN) I_s (s)

0.233 115137 1112.4 115.1 3357

Problem 4.44 0.0275f ≈ pt5≈ 507 kPa fAB≈ 0.0359 9 1.61M ≈

Problem 4.45 Power_i (MW) p_t5 T_t5

26.83 30 676.5

Power-LPT (MW) T_9 (K) V_9 (m/s) 22.31 619 363

Problem 4.46 Eta-propulsive

0.935

Problem 4.47


22

f Tau-t T-t-5 (K) Pi-t m-dot-f 0.02683 0.7299 1285 0.2248 2.683

p-t-9 (kPa) M-9 T-9 (K) D-ram (kN) F-net (kN) 525.0 2.62 604 60.5 68.3

Problem 4.48 V-9 (m/s)

634

V-9 (m/s) 750

Problem 4.49

D-r (kN) T-t2 (K) p-t0 (kPa) p-t2 (kPa) T-t3 (K) 11.045 240 43.6 42.70 638.7

T-t4 (K) Pi-c p-t3 (kPa) p-t4 (kPa) T-t4.5 (K)

1509 21.74 928 882.0 1169

p-t4.5 T-t5 (K) Pi-LPT p-t5 (kPa) F-prop( kN) p-t9 (kPa) Power-LPT (MW) M-9

263 826.0 0.203541 53.52 74.24 52.99 20.41 0.960

F-c (kN)

TSFC (mg/s/N) 14.73

15.74

Problem 4.50 T-t3

e_c

686

0.900

Problem 4.51

V_0

f 627

0.0347

V_9

F_g/D_r

1477

2.354

Problem 4.52 D-r (kN)

F-g (fn) kN

127.4

142.1

F-n (kN) Eta-th Eta-p TSFC (mg/s/N) 47.2 0.34 0.82 22.2


23

Problem 4.53 c-p6M (J/kg.K) gamma-6M T-t6M (K) M-15

1042 1.38 605.4 0.390 A-15/A-5

2.200

Problem 4.54 f

0.021

T-t-5 (K) 1170

p-t-5 (kPa)

394

f-AB 0.026

TSFC (mg/s/N) Eta-th Eta-p (exact) Eta-p (approx) Eta-thermal 53.42 0.493 0.666 0.648 0.493

D-ram (kN) Power-comp (MW) F-g (kN)

37.41 18.75 81.64

Problem 4.55

m-dot-f (kg/s) f P-c (kW) P-t (kW) P-f (kW) Alpha T-t6M (K) M-8 0.62 0.0248 9849 18965 9116 7.844 404 0.691

T-8 (K) a-8 (m/s) V-8 (m/s) F-n (kN) 369 385 266 59

Problem 4.56 Delta-h_t (kJ/kg) Delta-T-t (K) P-prop (MW) F-prop (kN)

441.9

383.6

4.44

18.87

Problem 4.57 Shaft Power (MW)

A-2 (m^2)

121.0

1.125

Rho-3 (kg/m^3)

Delta (s)/R 13.18

0.40

Problem 4.58

Tau-r Pi-r

T-t4 (K) Tau-Lamb 1.8 7.82

1885.0 10.19


24

Press. Thrust/mom. Thrust

V-9eff (m/s) 0.26

1236

Eta-p

TSFC (mg/s/N)

0.679

72.5

Problem 4.59

p-t-13 (kPa)

M-19 p-19 (kPa)

T-t-13 (K) 73.09

1 37.45

315.7

a-19=V-19 (m/s) V-19-eff (m/s) Since p-19>p-0, then M-19 is 1.

325.0 371.2

Problem 4.60

Power-LPT/m-dot (kJ/kg) 147

V-9 (m/s)

114.9

Problem 4.61 p-t-5 (kPa) Shaft Power (MW) Eta-t

171.3 90.5 0.89

Problem 4.62

gamma-6-M

c-p-6M (J/kg.K)

T-t-6M (K) 1.386

1030

463

Problem 4.63 Tau_r Pi_r

T_t4 Tau_L

1.8 7.82

1885.0 10.19 V_9 Pressure thrust/Nozzle momentum thrust

978 0.264 V_9eff Eta_p TSFC (mg/s/N)

1236 0.680 72.50

Problem 4.64

D-ram (kN)

p-t3 (kPa) 24.3

1084

Power-c (MW)

Eta-d

20.97

0.866


25

Problem 4.65

q_0 9.856

T-t3 (K)

854.4

f 0.026

T-t-4.5 (K)

1157

p-t-4.5 205

T-t5

780.3

p-t5 33.6

M-9 (exact)

0.80

V-9 from eff. 415.3

F-prop (kN)

49.15

Problem 4.66

Tau-Lamb

f 9.30

0.0394

V-9 (m/s)

F-n/(m-dot)(a0)

1351

2.576

Eta-p Eta-th

0.663 0.440

Problem 4.67

D_ram (kN)

A_18 (m^2) F_g(fn) (kN) V_18_eff 28.54

0.588 35.57 356

Tau_fan

A_8 (m^2) F_g(cn) (kN) V_8_eff 1.161

0.163 11.93 586


26

Pi_fan

1.60

TSFC (mg/s/N) Eta_th (%)

Eta_p (%) 19.40 40.8

70.2

Problem 4.68

D-r (kN)

p-t4 (kPa)

T-t4 (K) 74.5

313

2003

M-9

V-9 (m/s) F-g (kN)

TSFC (mg/s/N) 2.27

1472 154.6

62.46

Eta-th 0.410

Eta-p

0.694


BHP = IHP*ηmech = 481.70 HP IHP = 535.2 HP FHP =53.5 HP BMEP = 144.00 psi Compression Ratio = 7.54

Problem 5.2

BHP = 171.27 HP

Problem 5.3 BSFC = 0.43 lbs/hp/hr

ηthermal = 29.69%

Problem 5.4

BMEP = 115.6 psi IMEP = 128.44 psi


27

Problem 5.5 245o

Problem 5.6

373o

Chapter 6 Problem 6.1 0.968dπ ≈ 0.928dη ≈

Problem 6.2 Pi-d

0.948 A2/A0 1.385

p2/p0 1.2814

Problem 6.3 Cp,stag=1.170 M1≈0.66 A1/Ath=1.062 A2/Ath=1.317

0 1

0.0061addDp A

≈

Problem 6.4

≈10 Ap

Dadd 0.01246

Problem 6.5 τr=1.8

Δs/R≈0.1056 ηd≈0.933

Problem 6.6 M2≈0.485 CpR≈0.473

Problem 6.7 M1≈0.65

Dadd≈935 N


28

skgm /5.2540 ≈ 68.3rD kN≈

A2≈3.58 m2

Problem 6.8 A0/A2≈2.557

Problem 6.9 Dadd≈5376 N

Problem 6.10 Mth≈0.53

Below critical throat Mach number of 0.75, therefore, there is no adverse shock boundary layer interaction due to the convex throat geometry.

Problem 6.11 AM/A1≈4.16

Problem 6.12

)948.0(2

21, Vpp avgareat

+=−ρ

)9545.0(2

21, Vpp avgmasst

+=−ρ

Problem 6.13 M1 ≈0.52 M2≈0.36 p1/p0≈1.2677 p2/p1≈1.0994

Problem 6.14

−+

−

−

= 12112

0

12

00

1

1

1, p

pMV

VA

AAAC spil

McowlD γ

Problem 6.15 % opening is =(Ath’-Ath)/Ath ≈ 621%

Problem 6.16 pt2/pt0=0.900 A2/Ath=1.2378

Problem 6.17 M1≈0.56 πd ≈0.850

Problem 6.18


29

M-Des A1/A-th (geo) M0 sub. (choked) M0>M0 sub A0/Ath sub % spillage (sub)

1.6 1.250 0.56 0.7 1.0944 12.47 M-y (NS)des Ay/A* % spill.(sup) My-overspeed A/A*-oversp M-th-oversp

0.668 1.1192 10.48 2.108 1.850 2.1

Problem 6.19 Mover-speed≈2.92

q0≈ 119.37 kPa

Problem 6.20 M-Design A-1/A* A-1/A-th M-y (NS) A-y/A* A-1/A-th' % opening

3 4.2345 4.234 0.475 1.390 1.390 204.5

Problem 6.21 % opening is ≈369.6% Mth

’≈3.1

Problem 6.22 A1/Ath (design) M-th (started) M-y (NS-th) pt-recovery(best)

1.2165 1.55 0.684 0.913

Problem 6.23 A1/Ath≈1.1446 Mth

’≈1.45 πd ≈0.945

Problem 6.24 A1/Ath (design) M-th (started) M-y (NS-th) pt-recovery(best)

1.4145 1.75 0.628 0.835

Problem 6.25 Mx≈1.85 % gain in total pressure recovery≈13.3%

Problem 6.26 M-Design A-1/A* A-1/A-th M-y (NS) A-y/A* A-1/A-th'

1.6 1.250 1.250 0.668 1.119 1.119 % opening A-th'/A* M-th' pt-recov (best)

11.71 1.117 1.4 0.958

Problem 6.27 Mover-speed≈1.784


30

Problem 6.28

M-design M-y (NS) A1/Ath (design) M-th (started) M-y (NS-th) pt-recovery(best)

2 0.577 1.216 1.55 0.684 0.913

Problem 6.29 Loss~8.5%

Problem 6.30 M-Design A-1/A* A-1/A-th M-y (NS) A-y/A* A-1/A-th'

2.6 2.896 2.896 0.504 1.332 1.332 % opening A-th'/A* M-th' pt-recov (best)

117.3 2.17 2.25 0.606

Problem 6.31 % opening is =38.72%

Problem 6.32 Mx, shock=2

py=211.3 kPa pty=264.8 kPa A2/A1=0.8694

Problem 6.33 Pty/ptx=0.785 πd ≈0.697 πd ≈0.7464 q0=72.65 kPa

Problem 6.34 pt3/pt0=0.813

Normal shock at Mach 2.5 gives πd=49.9%, therefore two external ramps improve the inlet recovery significantly. Compare 81.3% to 49.9%.

Problem 6.35 θ1≈12.2o θ2 ≈13.87o

Problem 6.36 A9/A-th (spec)

M9 (isen) p9/pt-9

Neu-9 (deg) Plume angle (deg)-given

Neu-10 (deg)

2 2.15 0.101 30.42 15 45.42 M10 p10/pt-10 p0/pt9 NPR(perfect) NPR(actual)

2.75 0.0398 0.0398 9.89 25.1 In the over-expanded mode: A9/A-th (spec) M9 (isen) p9/pt-9 NPR (perfect) NPR (NS at the lip)

2 2.15 0.101 9.89 1.892


31

Problem 6.37 p0= 46.4 kPa

Ty=708 K 96.36≅m kg/s

Problem 6.38 Altitude is ~8 kft Altitude is ~26 kft Altitude is ~ 1400 ft.

Problem 6.39 A9/A8=8.17 M9=3.7 V9=1715 m/s

Problem 6.40 A9/A-th (spec) M9 (isen) p9/pt-9

2.4 2.35 0.0739 plume angle (deg) Beta (deg) M9.sin(beta) p-y/p-x NPR

15.95 40 1.510 2.495 5.418

Problem 6.41 M10=4 θ≈7.26o

μ≈14.48o

Problem 6.42 A9/A-th (spec) M9 (isen) p9/pt-9

3.5 2.79 0.0374 plume angle (deg) Beta (deg) M9.sin(beta) p-y/p-x NPR

15 33.5 1.54 2.60 10.3 pt7 = 102.8 kPa

Problem 6.43 P9=224.6 kPa T9=817 K V9=556 m/s ηn≈0.97 V9s=564 m/s % gross thrust gain=2.43 CD8=0.98

Problem 6.44 CA(conical)=(1+cosα)/2=0.953


32

CA(2D)=sinα/α=0.968 2D rectangular nozzle produces ~1.6% higher gross thrust than the conical nozzle

Problem 6.45 Tm/Tc=1.444 % Fg improvement ≈3.4%

Problem 6.46

a) p0=p9=2.656 kPa

b) p0=39.7 kPa c) A9=3.95 m2 d) Fg= 44 kN e) Fg= 24 kN f) Fg=46.6 kN

Problem 6.47 Aj=27.98 cm2 A=111.92 cm2 A1=83.94 cm2

Percent increase in mass flow rate = 147.1% Percent drop in exit temperature = 40.4% Gross thrust with ejector = 896.2 N Gross thrust of the jet = 784.6 N

Problem 6.48

9M ≈ 1.56

nη ≈ 0.975

Problem 6.49 pt/p0=61.5 NPR=19.8 NPR (actual)=255.2 NPR (perfect)=61.5

Problem 6.50

NPR ~ 654

Problem 6.51

.

1.0918g CD

g Conv

FF

−

−

≈

Gross thrust is enhanced by ~9.2% in de Laval vs. convergent

Problem 6.52


33

Mth≈1.54 Mx≈1.8 Two-Shock Total Pressure Recovery ≈ 0.756

Problem 6.53 M_9 (from isentropic table) p_9 (kPa)

Pressure thrust/momentum thrust

2.52 7.71

0.0425

Problem 6.54 Pi-KD (NS table)

Pi-NS (NS table)

% gain in Pi

M-th (table)

0.936

0.856

9.3841

~ 1.48

Problem 6.55 A-spillage/A-1 Pi-d

0.061 0.930

Problem 6.56 q-1 (kPa)

14.75 p-2 (kPa) C-PR

96.7 0.839

Problem 6.57

I0=21.25 kN I1=23.73 kN Dadd=2.48 kN Dr≈10 kN

Problem 6.58 q_1 (kPa) p_2 (kPa) p_t2 (kPa)

A_2/A_1

11.2 108.4 110.31

3.745

Problem 6.59 M1 p-1/p0

V-1/V0

0.64 1.230

0.767

Problem 6.60 Pi_shock M_3

0.939 0.786

Problem 6.61 % Opening

621

Problem 6.62 % spillage = 7.93


34

Problem 6.63

M8=1.0 p8/p0=1.60

Problem 6.64

M9=2.82 p9/pt7=0.033 py/px=1.502

Problem 6.65

M-8 p-8 (kPa)

V-8 (m/s) 1 40.9

441

Problem 6.66

p-t-2 (kPa) p-th (kPa) q-th (kPa) 73.6 54.1 18.55

Problem 6.67

M8=0.78 V8=373 m/s

Problem 6.68

M1=0.62 M2-0.5

Problem 6.69

Cfg=0.90 % Increase in Fg=3.96

Problem 6.70

πd=0.732

Problem 6.71 M9=2.4 A9/A*=2.653

Problem 6.72

Dr=22.5 kN πd=0.75

Problem 6.73

Dr=25.75 kN A0=0.9585 p1=34.43 kPa T1=268.3 K A1=1.05 m2 Dadd=228.3 kN

Problem 6.74


35

M9=2.52 p9=11.7 kPa β=34.3o

Problem 6.75

M1=0.62 Mth=0.68 M2=0.52 CPR=0.648

Problem 6.76

M9=2.52 p0=77.6 kPa

Problem 6.77

A1/Ath=1.09 pt2/pt1=0.976 Mth=1.32

Problem 6.78

A0=0.930 m2 �̇� = 112.2 𝑘𝑔/𝑠

Problem 6.79 M1=1.6 pt2/pt0=0.895 p3=50.6 kPa

Problem 6.80

Dadd/p0A1=0.0229

Problem 6.81 MHL≈0.6

Problem 6.82

Dadd/p0A1=0.0352 for A0/A1=0.82 Dadd/p0A1=0.0578 for A0/A1=0.76

Problem 6.83

p1=50 kPa, p2=30 kPa and p3=25 kPa D’=6.75 kN/m M0≈1.7

Problem 6.84

pt2/pt1=0.979 per shock 𝜋𝑠ℎ𝑜𝑐𝑘 𝑠𝑦𝑠𝑡𝑒𝑚 = 0.920 𝜋𝑖𝑛𝑙𝑒𝑡 = 0.915 𝜋4−𝑠ℎ𝑜𝑐𝑘 𝑜𝑝𝑡𝑖𝑚𝑢𝑚 = 0.920

M0≈2.6


36


21COn kmol=

2

4Nn kmol=

2

1On kmol=

2

16Oχ =

2

0.167COp bar=

20.667Np bar=

2

0.167Op bar=

31.3 /mMW kg kmole=

612 =

m

CO

VV

322 =

m

N

VV

612 =

m

O

VV

Problem 7.2

fstoich=0.029

Problem 7.3

2 0.22O

m

nn

= ; 2 0.33N

m

nn

= ; 2 0.45CO

m

nn

=

36.08 /mMW kg kmole≈

Problem 7.4

fstoich=0.0684

Problem 7.5 44,879 /LHV kJ kg≈ 48,350 /HHV kJ kg≈

Problem 7.6

Since N2 is assumed to have no chemical reaction, the LHV and HHV are the same as problem 7.5.

Problem 7.7 2700 K

2500 K


37

Problem 7.8 0.9091φ =

Problem 7.9

0.0554fuel

air

massmass

≈

0.0664fuel

air stoich

massmass

≈

0.834φ ≈

2 2032T K≈

Problem 7.10

20.254Oχ =

2

0.629Oχ =

Problem 7.11

20.927Hχ =

2

0.9921Hχ =

Problem 7.12 p=1.18 atm

Problem 7.14 Me=0.225 with pte/pti=0.982 (Dry Mode) Me=0.438 with pte/pti=0.916 (Wet Mode)

Problem 7.15 2 0.23M ≈

Problem 7.16

21,128 /LHV kJ kg≈ 23,877 /HHV kJ kg≈

Problem 7.17 46,454 /LHV kJ kg≈ 50,452 /HHV kJ kg≈

Problem 7.18 f=0.01457 1,522afT K≈

Problem 7.20

2

1

4AA

=


38

1

0.0796t

t

pp∆

=

2 0.0954M =

Problem 7.22

0.4uu∆

=

2, 24.5 /upperu m s= 2, 33.2 /loweru m s=

0.3uu∆

=

2, 14.14 /upperu m s=

2, 26.46 /loweru m s=

Problem 7.23 225.005.0 << f

Problem 7.24

0.969e

i

pp

=

0.234eM = 9693.0=

i

e

TT

0.9788te

ti

pp

=

Problem 7.25 0.3078eM =

0.9297e

i

pp

=

0.9603te

ti

pp

≈

0.3078eM =

Problem 7.26 ~ 97,200 kg

Problem 7.27 0.9819bη ≈ 0.95bη =


39

Problem 7.29 MW_air (hot) f

21.76 0.092

Problem 7.30 QR≈1906 kJ/kg

Problem 7.31 f=0.039 ϕ=0.667 Taf=1783 K

Problem 7.32 HHV=23.95 MJ/kg

Problem 7.33 Taf=2266 K

Problem 7.34 QR=114,868 kJ/kg LHV=114,868 kJ/kg Taf=4794 K

Problem 7.35 f f_stoich Phi 0.05281 0.06500 0.8125

Problem 7.36 f f_stoich Phi 0.041 0.067 0.62 T_adiabatic flame (K) 1728

LHV in kJ/kg (comb. in air) 44327

Problem 7.37 HHV=50.32 MJ/kg

Problem 7.38 T_adiabatic flame (K) 1879

Problem 7.39 A1=3.855 m2 Dfh=7.614 kN p2=29.6 kPa T2=670 K


40

Problem 7.40 a) Taf=2410 K b) Taf=2295 K

Problem 7.41 f=0.039 ϕ=0.5952 Taf=1529 K

Problem 7.42 Taf=2065 K LHV=27,772 kJ/kg HHV=30,640 kJ/kg

Problem 7.44 f=0.0685

Chapter 8 Problem 8.1 30.83 /cw kJ kg=

256.4 /m smτ=

Problem 8.2 ω=4074 rpm Cθ2m=228 m/s wcm=51.2 kJ/kg W2m=299.3 m/s τs=- τr=-120 m2/s ψm=0.78125 ϕm=0.605

Problem 8.3 1 179.6 /W m sθ =

2 81.6 /W m sθ = ↑ 1 250.8 /mW m s= 2 193.1 /mW m s= 0.393rmD =

24.9 /m m sΓ = ' 1,059.7 /L N m= 0.0337C =

20.58 /cmw kJ kg≈ 0.467mψ =

0.622omR =


41

Problem 8.4 2 1 2,min0.72 126 /W W W m s> → = 0.442pC ≈

Problem 8.5 1.1609sτ = 0.8932sη =

Problem 8.6 20.87 /cmw kJ kg=

3

1

1.072t

t

TT

=

2 0.49rM =

2

1

1.264t

t

pp

=

3

2

0.996t

t

pp

=

0.945sη =

0.5oR =

Problem 8.7 68.54 /cw kJ kg=

0.5oR =

Problem 8.8 2 0.682M = pt3=149.3 kPa M3=0.47 8,000s N mτ = − ⋅ 3 2 18.35p p kPa− = 3 2 12.75T T K− =

3 2 0.00478s s

R−

=

Problem 8.9 43.86 /cw kJ kg= 1.515sπ = 5.27cπ 4.387m MW℘ = 0.62rmD =


42

Problem 8.10 α1m=29o β1m=-35.8o α2m=35.8o β2m=-29o wc=6.9 kJ/kg

21.5 /r m smτ

=

Problem 8.11 0 224T K= 1 212.8T K= 5,585 rpmω ≈

Problem 8.12 0.34o

mR =

2

1

1.789t

t

pp

=

Problem 8.13 33.58 /cw kJ kg= 1mψ = 0.954mφ = 1 , 0.766r mM =

1.424sπ =

Problem 8.14 0.85cη = 57.6c MW℘ =

Tt3=870 K πstage=1.40

Problem 8.15 2.82cτ = 26.2cπ = 2 102.6 /cm kg s= 0.847cη =

Problem 8.16 ec≈0.90

Problem 8.17 2 1, 413 o

tT R=


43

0.878cη =

( )3 2 214.32 /c t tw h h BTU lbm= − =

21,432 /c BTU s℘ =

Problem 8.18 πs=1.495 ηs=0.915 τc=2.717

Problem 8.20 2 288.7mW fpsθ = 2 911.3mC fpsθ =

2911.3 /ft smτ=

1

0.347t

t

TT∆

=

2 1216a fps= 2 0.855mM = 1 12.8p psia= pt2=38.1 psia Cpm=0.864 ω=11,460 rpm

2911.3 /ft smτ=

518.4z bladeF lbf= −

323.5bladeF lbfθ = −

' 1.878'

LD

=

2 2109,360 /cw ft s=

Problem 8.21 6Re 2.46 10c = ×

Problem 8.22 2337.1 /ft sΓ = Ideal Lift=674 lbf/ft Actual Lift=671 lbf/ft % Loss of Ideal lift=0.445

Problem 8.23 188.5 /mU m s= 2 109.57 /mC m sθ =


44

2 ( ) 365.23 /C r r m sθ =

2

0.0261( ) 1oR rr

= −

Problem 8.24 M1r=0.544 Tt1=305 K wc=12.0 kJ/kg Tt2=317 K M2r=0.448 πs=1.135 oR=0.5

Problem 8.25 0 286.8T K= 0 87.45p kPa= 1 99.76tp kPa= 0 11.96q kPa= 1 284T K≈

1 84.291p kPa= 1 107.22t rp kPa= 2 0.4912rM = 2 106.59t rp kPa= p2 = 90.37 kPa pt2 ≈ 114.4 kPa

Problem 8.26 wc=1.414x106 ft.lbf/slug

Problem 8.27 32γ =

Problem 8.28 ρ3/ρ2=12.05 A3/A2=0.083

Problem 8.29 1 10.487 ; 0.244t hr m r m 8,504 rpmω ≈ 2 326 /mC m sθ ≈ 2 394t mT K≈ 2 281t mp kPa= 2 281t mp kPa=

32 1.3 /m kg mρ =


45

0.5otipR =

0.5ohubR =

0.99rmD = 0.3hr m=

Problem 8.30 1.4tσ = 3.5hσ =

0.463omR =

0.78rmD =

2m

1m

WDe-Haller Criterion= 0.56 0.72W

therefore not met= <

0.725mψ = 0.57mφ = 0.82tD =

0.463ohR =

tan

tanz

rC cons tC cons t

θ ==

Therefore, C2(r)≠constant

Problem 8.31 1 51.34mβ = −

2 0mβ =

0.5omR =

Problem 8.32 1 54.4h hγ β= =

11 1

7

418.9( ) ( ) tan

1500.1

h

rr rr r

γ β −

= =

−

Problem 8.33 9.54cπ ≈

Problem 8.34 3,083 Hz 3,417 Hz


46

Problem 8.35 21.26 /cw kJ kg=

238.9 /m smτ=

0.7786omR =

Problem 8.36 2 58.75 /mW m sθ = 2 ( ) 376.7 (1/ )C r r s r in metersθ =

( ) 0.64 .oR r const= =

Problem 8.37

160.517bending

pσ

=

Problem 8.38 ( ) 7,134rpmω <

Problem 8.39 0.376rmD =

( ) 0.698omR r =

Problem 8.40 Stall Margin =11.8%

Problem 8.41 ω=6939 rpm Cθ2=173 m/s 0Rm=0.667 Dr=0.378 τs=1.124 πs=1.445

Problem 8.42 M1r=0.93 W2=150 m/s T2=313.7 K M2=0.876 M3=0.400 τs=1.256 πs=2.07

Problem 8.43 De Haller=0.773 oR=0.824 wc=31.67 kJ/kg Dr=0.4272


47

Problem 8.44 Power=121 MW A2=1.125 m2 ρ3=13.18 kg/m3 Δs/R=0.395

Problem 8.45 β1=63.4o β2=53.4o Cθ2=98 m/s ψ=0.327

Problem 8.46 M1r=0.97 W2=193 m/s T2=306 K M2=0.74 M3=0.485 τs=1.18 πs=1.66

Problem 8.47 Tt3= 680.4 K M2= 0.481 M3= 0.31 ρ3/ρ2= 6.78 A3/A2= 0.1475

Problem 8.48 M1=0.523 M1r=0.80 p1=83 kPa Tt1=303.4 K Cθ2=192 m/s oRm=0.53 wc=38.45 kJ/kg T2=308.1 K M2=0.738 a3=362

M3=0.491

Problem 8.49 ω=5002 rpm wc=42.2 kJ/kg M2=0.563

Problem 8.50 Cθ2=155 m/s Tt2=325.8 K T2=302.7 K M2=0.62


48

M2r=0.62 q1r=67.1 kPa

pt1r=174.3 kPa p2=132.1 kPa pt2=170.9 kPa Dr=0.524

Problem 8.51 Cθ2=102.1 m/s Wθ1=-204.1 m/s Wθ2=-102.1 m/s De Haller=0.754 Dr=0.436

Problem 8.52 wc=39.48 kJ/kg ψ=0.4 ϕ=0.477 M1r,m=1.04 πs,m=1.51 Dr,m=0.458 De Haller=0.692<0.72, therefore not satisfied

Problem 8.53 Cθ2=106 m/s wc=20.8 kJ/kg Tt2m=303 K De Haller=0.709<0.72 not satisfied Dr=0.56

Problem 8.54 M1=0.50 Shaft Power= 39.415 MW A1=0.562 m2 M2=0.32 A2=0.082 m2

Chapter 9 Problem 9.1 Shaft Power ≈ 624 kW ΔTt ≈ 62 oC (or K)

Problem 9.2 r2≈19.48 cm

Problem 9.4 Mass flow rate=33.95 kg/s Cθ2=330 m/s wc≈121 kJ/kg Shaft Power=4109 kW


49

πc=2.713 M2=0.985 Cθ3≈257 m/s

Problem 9.5 1.279c MW℘ ≈ 22 θCrm ≈ 3.086 kN.m

Problem 9.6 Cr(r)=25/r where r is in meters an Cr is in m/s Cθ(r)=50/r where r is in meters an Cθ is in m/s

Problem 9.7 pt3/p1 ≈ 3.19 pt3/pt1 = 3 0.8895TSη ≈ 0.885TTη ≈ 0.9012ce ≈

Problem 9.8 Cθ1=1204.3 r in m/s where ris in meters 1

1( ) tan (12.043 )r rα −= The IGV exit blade angle is = δ+α

Where δ is the deviation angle and is specified to be constant and α(r) is described above. Therefore IGV twist is higher than the flow angle variation with span by a constant deviation angle.

Problem 9.9 M1r ≈ 1.414 CPR=0.714

Problem 9.10 A3/A2=r3/r2=4 Cr3/Cr2=1/4 Cθ3/Cθ2=r2/r3=1/4 CPR=1-1/AR2=1-1/16=0.937

Problem 9.12 3 1/tp p ≈7.816

3

1

t

t

TT

≈ 1.953

TSη = =0.7869

Problem 9.13 ε ≈ 0.9365 ΠM=1.076 wc =100.7 kJ/kg τs=1.348


50

sπ ≈ 2.53

Problem 9.14 ε =0.971 ΠM=1.212 wc = 177.8 kJ/kg τs=1.653 sπ ≈ 4.79

Problem 9.15 A2eff ≈ 157.08 cm2 A2eff ≈ 153.28 cm2 A2eff ≈ 137.95 cm2

Problem 9.16 U2=502.2 m/s wc=226.95 kJ/kg Tt2=514 K ω =19,183 rpm

Problem 9.17 NtbbrA −= 22 2π Aeff=A2- (N-1)θwaker2b B2=(N-1)θwaker2b/A2

Problem 9.18 M2=0.81 Cθ3=167 m/s (inviscid) Cθ3=158 m/s (viscous)

Problem 9.19 Mach Index=1.482 Cθ2=410.4 m/s M2=1.21 τc=1.79 πc=6.02

Problem 9.20 U2=555 m/s Cθ2=450.4 m/s wc= 250 kJ/kg M2=1.205 Cθ3=257.4 m/s

Cr3=114 m/s

Problem 9.21 Cθ2=356 m/s M2=1.02 Cr3=112.5 m/s Cθ3=267 m/s


51

Problem 9.22 Cθ2=349 m/s MT=1.14 a2=382 m/s

Problem 9.23 A1=0.0236 m2 Cz1=128 m/s ρ1=1.23 kg/m3 Mass flow rate=3.71 kg/s U2=366.5 m/s Cθ2=330.2 m/s Shaft Power=449.2 kW M2=0.986 Cθ3=220.2 m/s πc=3.1

Problem 9.24 A1=0.1114 m2 rh1=1.893 cm rt1=18.93 cm ω=1403 rad/s ≈13,400 rpm

Problem 9.25 pt3/pt1=3.79 p3/p1=4.08 M3=0.35 wc=156.6 kJ/kg

Chapter 10 Problem 10.1 C1 ≈ 412 m/s M2 = 1.287 2/ 318.8 /m m sτ ≈

Problem 10.2 α2 ≈ 58.6o

Problem 10.3 Cz2 ≈343 m/s Cz3 = 387 m/s Um ≈ 427.4 m/s oRm= 0.0795 wt = 269.4 kJ/kg


52

Problem 10.4 α2 = 67.4o α2 = 78.2o

wt= 360 kJ/kg wt=720 kJ/kg

Problem 10.5 wt =320 kJ/kg ψm= 2

Problem 10.6 wt=154.2 kJ/kg β3= -49.1o ( ) 0.070o

mR r ≈

Problem 10.7 T1 ≈ 1,936 K M1 = 0.4706 p1 = 1,736 kPa

Problem 10.8 α2 ≈ 61.1o

Problem 10.9 T2=1354 K Cz2 ≈ 268 m/s Cθ2 ≈ 735 m/s M1 ≈ 0.35 pt2 ≈ 2987 kPa p2 ≈ 1,450 kPa

Problem 10.10 M2 = 1.267 M2r= 0.829 0.087o R ≈ wt = 388.8 kJ/kg T3 ≈ 1464 K M3r = 0.907

Problem 10.11 U2 ≈ 425 m/s wt= 229.3 kJ/kg 0.366o R ≈ Γ = 26.94 m2/s M2 = 0.86 ρ2= 3.36 kg/m3 M3r = 0.767 pt3 = 1.6499 MPa ρ3= 2.901 kg/m3


53

Problem 10.12 Ttc,r ≈ 661 K

Problem 10.13 tπ =0.18736 tη ≈ 0.8735 A5/A4=5.02

Problem 10.14 Cz ≈ 140.0 m/s wt= 90 kJ/kg ψ =1.0 φ = 0.466

T1-T2 ≈ 38.9 K T2-T3 ≈ 38.9 K

Problem 10.15 ρ5/ρ4 = 0.361 A5/A4 = 2.77

Problem 10.16 Tt2r=1918 K T2=1769 K awrT ≈1902 K

Problem 10.17 Δη/ηo ≈ 0.0299

Problem 10.18 z Zσ ψ ≈ 1.466 σz ≈ 1.833

Problem 10.19 Taw ≈1624 K for turbulent boundary layer Ttg ≈ 1639 K

Problem 10.20 Ttg=1654 K Taw=1628 for TBL Taw=1616 for LBL hg=125.34 W/m2K qw=53.65 kW/m2

Problem 10.21 hg=3641 W/m2K

Problem 10.22


54

/ ( )m span m ≈ 149 (kg/s)/m

Problem 10.23 Q =150.6 kW ,p mixedc =1237 J/kg.K

mixedT ≈ 1,796 K

Problem 10.24 M2r=0.74 β3=-48.8o α3=-11.1o oR=0.1881

Problem 10.25 Δη (knife edge)=0.0092 Δη (groove)=0.0103

Problem 10.26 Ttg=1890 Taw=1861 K % Error=1.53

Problem 10.27 5,t cT = ≈ 1,324.5 K pt5,c ≈ 227.4 kPa

Problem 10.29 Throat opening, o =1.757 cm

Problem 10.30 σc/ρblade =1.76827x106 ft2/s2

7100.1=blade

c

ρσ

ksi/slug/ft3

Single crystal super alloy

Problem 10.31 U2=346 m/s ΔTt=207 K a2=735 m/s a3=735 m/s M2=1.00 M3r=0.583

Problem 10.32 α2=64.9o wt=256 kJ/kg Tt2r=1348 K M2=1.01 a3=688 m/s


55

M3r=0.727

Problem 10.33 Cθ2=694.3 m/s M2=1.019 M3r=0.65 ψm=1.7

Problem 10.34 Mean torque=27.5 kN.m T2=1452 K

Problem 10.35 Cθ2=728 m/s M2=1.03 β2=51.1o wt=291.2 kJ/kg ψm=1.82 oR=0.09 M3r=0.641

Problem 10.36 M2=1.05 β2m=37.6o Tt2r=1485 K oR=0.154

Problem 10.37 Taw=1747 LBL Taw=1756 TBL Ttg=1776 K

Problem 10.38 ε≈0.043

Problem 10.40 No. of cycles to failure ~ 800 for conventionally cast Rene 80 ~ 4000 for DS Rene 150 ~ 104 for DS eutectic

Chapter 11 Problem 11.1 2 110.2 /cm kg s

2cN ≈ 6,637 rpm

fcm ≈ 2.37 kg/s Fc= 131.2 kN TSFCc ≈ 17.88 mg/s/N

Problem 11.2


56

Mz2 ≈ 0.50 A0≈0.758 m2

Problem 11.3 pt4/pt2 = 9.5 Tt4/Tt2 ≈ 5.613 A4=0.1957 m2 4 46.4 /cm kg s= 4 186 /m kg s=

Problem 11.4

4

5

c

c

mm

= 0.3847

4

3

c

c

mm

= 1.455

Problem 11.5 Nc2= 8,000 rpm Nc4≈ 3,266 rpm cπ = 32.7 c℘ =≈ 211 MW f ≈ 0.0284 Tt5/Tt4 = 0.71487 pt5/pt4 ≈ 0.1926

5

2

t

t

pp

= 5.922

5

2

t

t

TT

= 4.289

Problem 11.7 A4 = 0.3375 m2 A5 = 1.1973 m2 fAB ≈ 0.0514 A8-dry ≈ 0.946 m2 A8-wet ≈ 1.4838 m2 A9-dry/A8-dry ≈ 1.579 (A9/A8)wet = 1.634 Fg-dry = 698.4 kN Fg-wet ≈ 1014 kN

Problem 11.8 f=0.0232 τt=0.7044 c ODπ − ≈ 22.4

2,

2,

1.11c O D

c D

NN

− ≈


57

2,

2,

0.8037c O D

c D

mm

− ≈

Mz2-OD ≈ 0.378 TSFC)design=23.5 mg/Ns TSFC)OD=33.4 mg/Ns

Problem 11.9 πcH≈10.90 πf ≈1.537 α ≈5.97

Problem 11.10 V9-Des=1133 m/s ηth=0.4314 at Design Point TSFC)Des.=30.8 mg/Ns πc-OD=27.85 V9,OD=1143 m/s ηth=0.4704 at Off-Design Point (cruise) ηp=0.355 at cruise TSFC)cruise=34.4 mg/Ns

Problem 11.13 πf ≈1.515 πcH≈17.21

α ≈ 7.818

Problem 11.14 πc-OD ≈ 6.83 2, 13.9 /c O Dm kg s− ≈ Nc2,O-D=4,710 rpm Mz2,OD ≈ 0.261

Problem 11.15 πc-OD ≈7.564 2, 36.4 /c O Dm kg s− ≈

2, 57.2 /O Dm kg s− ≈ f ≈ 0.03 V9=1801 m/s TSFC (DP) = 47.3 mg/s/N TSFC (O-D) = 56.1 mg/s/N

Problem 11.16 A2=0.557 m2

Problem 11.17 πc-Des ≈34 πc-OD ≈12.9

Problem 11.18


58

Stall Margin≈10.25 % Mz2@stal≈0.465

Problem 11.19 C1=0.2130 C2=0.1659 C3=28.209 τcH-OD=2.192 τf-OD=1.111 πf-OD=1.394 πcH-OD=11.85 αOD=5.63 �̇�𝑐2−𝑂𝐷 = 475 𝑘𝑔/𝑠

Problem 11.20 �̇�𝑐0 = 199 𝑘𝑔

𝑠

A0=0.844 m2 �̇�0 = 65.75 𝑘𝑔

𝑠

Ath=0.905 Dr=16.56 kN

Chapter 12 Problem 12.1 D2 ≈ 5.598 ft

Problem 12.2 A2/Ath=36.871 c=3305 m/s F=7.908 MN Pressure Thrust=4924 kN

Problem 12.3 F=3.5 MN Is ≈ 357 s

Problem 12.4 CF≈1.5965 F=344,863 lbf M2≈3.19 A2/Ath=8.87

Problem 12.5

21.488On =

r = 1.707 MW=44.258 kg/kmol

Problem 12.7 ΔV ≈ 110,100 m/s neglecting gravity


59

ΔV ≈ 104,300 m/s with gravity

Problem 12.8 ΔV=8082 m/s

Problem 12.9

sin7,888 /

gle stageV m s

−∆ ≅

2

12,525 /stage

V m s−

∆ ≈

Problem 12.10 c = 3,283 m/s ηp ≈ 0.964 ηo ≈ 0.439

Problem 12.11 A0/Af=1.665 Vo=13 m/s Vf=26.6 m/s

Problem 12.12 r=0.037 m/s tburning≈96.5 s

Problem 12.14 Tg=2679 K Vg, th=1096 m/s hg=8.936 kW/m2K

Problem 12.15 A2/A1=35.2 at SL A2/A1=30,862 at Alt

Problem 12.16 Pressure Thrust=21.68 MN c=2168 m/s

Problem 12.17 Ratio of Specific Impulse=0.988

Problem 12.18 ΔL=0.015 m

Problem 12.20 T2/T0=3.49 T4=5731 K p4/p2=1 f=0.0621 M10=2.98 Dr/p0A1=50.4


60

Fg/p0A1=85.2 Is=2124 s A4/A2=6.76 A10/A4=7.1 ηth=0.531 ηp=0.817

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