M 2 ≈ 3.4 A2 = 3.664 A1 = δ 42.9° 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 halfdiamond airfoil. Region 1 is the free stream condition.
pt ,2' ≈ 53.4 kPa pt ,3' ≈ 6.02 kPa
pt ,4' ≈ 32.8 kPa Problem 2.8
cl = 1.732 cd cm , LE = 0.3694 Problem 2.9
u2 ≈ 255.2 m / s at ≈ 456.5 m / s h2 ≈ 488.4 kJ / kg Problem 2.10
aT .S . = 253.6 m / s Fglass = 1.744 kN Problem 2.11
cd ≈ 0.977 Problem 2.12
θ= 32.15° wall V2 = 1.264 V1
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x ≈ 0.373 c (t / c )max
p p2 = 0.14478 , 3 ≈ 2.389 p∞ p∞
pt ,2 = 157.59 kPa Problem 2.15
m ≈ 11.88 kg / s Lmax ≈ 5.35 m pt 2 ≈ 0.746 ; i.e. 25.4% total pressure loss. pt1 Problem 2.16
q* = 121.49 kJ / kg Problem 2.17
p2 ≈ 0.468 and static pressure drop is ~53.23% p1 Problem 2.18
Tt1 941 K p2 = 5.667 . Therefore, static pressure rise is 466.7% p1 Problem 2.19
Tmax ≈ 1.0285 T* p@Tmax ≈ 1.2073 p* Problem 2.20
ρ 2 = 0.5 kg / m3 Problem 2.21
ρ 2 ≈ 0.5 kg / m3 p2 ≈ 90 kPa
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L = 7.712 m Problem 2.23
L* ≈ 2.54 m ~ 40.7% total pressure loss Problem 2.24
M2=1.0 * p= p= 33.52 psia 2 * p= p= 63.46 psia t2 t
T2 = T * ≈ 1439 ° R
q1'* = 300 BTU / lbm Problem 2.25
M 1 ≈ 0.24 Tt 2 ≈ 2, 733 ° R Problem 2.26
M 4 ≈ 0.36 Problem 2.27
L* ≈ 53.455 D M 1' ≈ 0.48 Problem 2.28
xs ≈ 15.235 D pty = 0.6205 , i.e. 37.95% loss of total pressure pt1 I 2* − I1 = −0.1093 or ~11%. I1 Problem 2.29
M 1 ≈ 0.66 or 2.0 ~ 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
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T2 = 4.806 T1 pt 2 / pt1 = 0.845 , therefore 15.5% total pressure loss Problem 2.31
m ≈ 242.7 kg / s M 2 ≈ 0.52 CPR ≈ 0.4643
Problem 2.34 2.1 MJ/kg, which divides into700 kJ/kg per duct (for three ducts) Problem 2.35 Mcone ≈ 2.4
C p ,cone ≈ 0.1887
C p ,cone = C D ,cone Problem 2.36
m ≈ 81.8 kg / s A2 = 0.1973 m 2
p2 ≈ 161 kPa I1 = 90.75 kN I 2 ≅ 42.82 kN
M 2 ≈ 1.83 p2 ≈ 0.9706 p0 Fx ,con− wall ≅ 0.185 pt1 Ath 5
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Fx ,div − wall pt1 Ath Fx ,nozzle
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M 2 ≈ 0.34 CPR = 0.611 Fx
qmin=796.8 kJ/kg T2≈1911 K Fuel-to-air ratio=0.00664 or 0.664% Problem 2.42
q1* = 0.484 c pTt1
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. 6
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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
Delta pt Loss (%) 8.50 Problem 2.48 M-2 (from Table) 1.4 Tt2 (K) 450 pt2 (kPa) 129.3
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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
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p_2 (kPa) 69.4 Del (p_t)/p_t1 (%) 57.8 q* (kJ/kg) 616
q0=75.6 kPa A1=2.117 m2
𝑚̇0 ≈ 352.6 𝑘𝑔/𝑠 Chapter 3
wf = 0.0144 wn wf = 0.054 f( max A= / B) wn p p p Maximum (A/B on): 2.5 = 3.67 ; 3 = 3.0926 and 5 = 0.2279 p2.5 p4 p2 p p p Military: 2.5 = 3.67 ; 3 = 3.0926 and 5 = 0.2279 p2 p4 p2.5 Maximum (A/B on): V9 = 2,841 ft / s Military: V9 = 1,974 ft / s Maximum (A/B on): ηth ≈ 0.168 or ~16.8% Military: ηth ≈ 0.295 or 29.5% wf Maximum (A/B on): TSFC = = 2.103 lbm / hr / lbf Fn wf Military: TSFC = = 0.8353 lbm / hr / lbf Fn Maximum (A/B on): ηcarnot ≈ 0.827 or 82.7% Military: ηcarnot ≈ 0.744 or 74.4% Percent thrust increase ≈ 57% and Percent fuel flow rate increase ≈ 295% f( max A= / B)
w fan = 1.359 w core 9
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f = 0.01113 w f = 7,813 lbm / hr
Fn ) static = 17, 727 lbs
ηth = 0.3658 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. TSFC = 0.4407 lbm / hr / lbf FSP = 1.113 ηcarnot = 0.721 pt 3 = 13.6 pt 2 M 13 = 0.858 Problem 3.3
w f ⇒ w f = 2.199 lb / s w core
Fn )exhaust ≈ 14, 284 lbs
ηth ≈ 0.3232 ηcarnot ≈ 0.7621 TSFC ≈ 0.554 lbm / hr / lbf pt 3 = 15.85 pt 2 w BPR = α = fan = 1.1 w core
ηcarnot ≈ 0.7621 M 9 ≈ 0.87 pt 2.5 ≈ 4.082 pt 2 pt 3 ≈ 3.88 pt 2.5 Problem 3.4
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ηth = 0.335 TSFC = 0.351 lbm / hr / lbf w BPR = α = fan = 5.06 w core
ηcarnot = 0.786 pt 3 = 2.18 pt 2 pt 4 = 9.84 pt 3
M 19 = 0.788 M 9 ≈ 0.70 Problem 3.5
Dram ≈ 14.971 kN Fg ≈ 27.040 kN
Fn )un −installed ≈ 12.1 kN TSFC ≈ 0.2237
ηth ≈ 0.303 η p ≈ 0.726
ηoveall ≈ 0.22 Problem 3.6
℘s , = 1, 000 = kW 1 MW prop
V9 = 200 m / s η p = 0.9 Problem 3.7
A0 = 10.0 A1 p1 = 0.9505 p0 Dadd ≈ 0.0405 A1 p0 Problem 3.8
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I s = 356.8 s Problem 3.9
Dram ≈ 19.1 kN Fgc ≈ 7.5 kN Fgf ≈ 33.4 kN
Fnet ≈ 21.8 kN η p ≈ 0.663 Problem 3.10
Fnet ≈ 9.5 kN η p = 0.583 Problem 3.11
Fnet ≈ 31.5 kN ηth ≈ 0.133 η p ≈ 0.677 Range ≈ 860 km No! Problem 3.13
℘s ≈ 13.986 MW Problem 3.14
F = 3.924 MN V9 = 3,924 m / s Frocket = 3.924 MN Problem 3.15
Fp ≈ 318.4 kN Problem 3.16
I1 = 182.61 kN I 2 = 157.26 kN 12
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I 7 ≈ 38.1 kN I 8 ≈ 31.1 kN
Fx )nozzle ≈ 7.1 kN Problem 3.18
Ftotal=Fcore+Fprop=50 kN ℘thrust = 10 MW ℘shaft = 11.765 MW Problem 3.19
V9 ≈ 1200 m / s Problem 3.20
I s = 2,569 s f ≈ 0.0371 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
0.340 TSFC (lbm/hr/lbf) 2.148 Problem 3.22 D_r (kN) 26.0 Eta-th 0.401
35.0 F_g (lbf) 21891
TSFC (lbm/hr/lbf) 1.24 13
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Chapter 4 Problem 4.1
π d ≈ 0.80 ηd ≈ 0.889 ∆s ≈ 0.223 R
Tt 3 = 661.6 K ηc = 0.8484 ℘c =42.5 MW Problem 4.3
f ≈ 0.02 Tt 4 ≈ 1, 485 K
pt 4 ≈ 3.104 MPa Problem 4.4
Tt 5 = 976.2 K τ t = 0.6573 et ≈ 0.813 pt 5 = 1,102 kPa ℘t ≅ 60 MW Problem 4.5
Tt ,mixed −out ≈ 1,185 K Problem 4.6
Ttc = 900.4 K pt ,mixed −out = 2.09 MPa ∆s = −1.2687 R Problem 4.7
∴ M 9 ≈ 2.15 Problem 4.8
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η p ≈ 0.5455 Problem 4.9
f = 0.02141 M 9 ≈ 1.830 F ≈ 781.26 N / kg / s m 0
ηth ≈ 0.3157 η p ≈ 0.6147 Problem 4.10
V0 = 631.5 m / s A0 = 1.128 m 2
pt 0 = 78.2 kPa pt 4 = 66.9 kPa Tt 4 = 1,368.7 K pt 9 = 61.5 kPa M 9 ≈ 1.59 V9 ≈ 967 m / s A9 ≈ 1.965 m 2 Fg = 109.4 kN
V9efff ≈ 1, 062 m / s
ηth = 0.3231 η p = 0.7654 Problem 4.11
Dr ≈ 16.46 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 f ≈ 0.01586 Tt5 ≈925.7 K pt5=136.3 kPa α = 2.043
m 0 ≈ 8.213 kg / s m fan ≈ 16.787 kg / s Tt6M≈640.8 K, pt6M=133 kPa 15
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M9≈2.272 and V9≈808.5 m/s
Fn ≅ 0.5158 m 0 (1 + α )a0 TSFC ≅ 33.76 mg / s / N ηth ≅ 50.13%
η p ≈ 91% and ηo ≈ 45.6%. Problem 4.12
Dram ≈ 25.8 kN ℘c =39.57 MW f = 0.02 Tt 4 ≈ 1,508 K τ t = 0.7437 ℘t =39.57 MW τ λ AB = 8.522 f AB = 0.0272 Fg≈154 kN TSFC=36.92 mg/s/N Fn≈128 kN ηth = 0.477 ηp≈0.341 Problem 4.13
pt 3 ≈ 722 kPa Tt13 ≈ 331 K f ≈ 0.0215 Tt 5 ≈ 935 K M 19 ≈ 1.24 M 9 ≈ 3.1 V9 ≈ 1,108 m / s V19 ≈ 396 m / s V0 ≈ 271 m / s Ffan ≈ 0.826 Fcore Problem 4.14
Tt 6 M = 522.7 K M 9 ≈ 1.00 V9 ≈ 502 m / s
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Fg ≈ 1,506 N / kg / s m 5 Problem 4.15
A8−on ≈ 1.821 A8−off Problem 4.16
= p5 p= 370.4 kPa 15 p6 M ≈ 404.3 kPa Problem 4.17
pt 3 = 3, 009 kPa Tt 3 ≈ 848 K Tt 4 ≈ 1, 729 K τ t ≈ 0.663
π c off − design ≈ 23.0 Problem 4.18
f = 0.0232 τ t = 0.705
π c −Off − Design ≈ 18.6 Problem 4.20
2 (τ λ − 1) γ −1
f = 0.033 Fn = 2.1324 m 0 a0 η p = 0.68
ηth ≈ 0.415 Problem 4.25
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℘c = 2.304 ℘f f ≈ 0.0181 V19/V9=0.6758 Fn − fan ≅ 2.537 Fn −core ηth ≈39.8 % ηp ≈ 75.0%
TSFC ≈ 18.47 mg / s / N I s ≈ 5,523 s
ηn = 0.901 V9 = 3.72 a0 ( p9 − p0 ) A9 = 0.0913 m 9 a0
f = 0.044 M 9 = 2.363 Fn = 2.588 m 0 a0
ηth ≈ 0.313 ηth ≈ 0.504
℘t =9.096 MW Fgcore = 8.96 kN M= 8 Fncore
V8 = 1.00 a8 = 4.648 kN
ηn = 0.935 η LPT = 0.900 Problem 4.31
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A6 M =3 A5 p6 M (1 + γM 62M ) = 164.106 kN
Tt 6 M = 0.40 Tt 5 Problem 4.32
℘prop = 7.01 MW Fprop 56.1 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
Dr ≈ 50.4 p0 A1 Fg ≈ 64.0 p0 A1
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 ηth ≈ 49%
η p ≈ 60% Problem 4.38 T-t3 (K) 883
