Errata Farokhi 5z646h

Errata Sheet for Aircraft Propulsion, 2nd Ed., 2014

Page

Line Number

27 35 36 43 44

3rd line in 3rd Paragraph Equation 2.56 Equation 2.57 Below Eq. 2.79, 3rd line 15 from top

51 52 57

75

Figure

Table Change mote (u+du) -udρ LHS is the pressure… The entropy remains constant, hence

10 from top 13 from bottom 5 from bottom

top example

Change to more (u-du) +udρ RHS is the pressure…

Equation 2.61a Equation 2.101 angle θ

p

t1

/ pt *    pt 2 / pt * 

p

t1

/ pt

p

t1



[delete this part of the sentence] Equation 2.63 Equation 2.102a angle θ [note the turning angle is called “deflection angle” and is labeled δ in FIGURE 2.23]

/ pt *    pt 2 / pt * 

p

t1

/ pt * 

106

last line of Problem 2.20

p2 and p2

p2 and ρ2

106 126 176 177 196

2nd line Problem 2.21 Equation 3.39 2 lines from bottom 2 lines from top Last sentence

1  100 kPa

p1  100 kPa

Internal-Thrust ,mixed-out=1594 1.594 kJ/kgK=1154.7 K …into Equation 4.90

Internal Thrust ,mixed-out=1094 1.094 kJ/kgK=1682.7 K …into Equation 4.91

205

Equation 4.120 box

(ideal) thrust

specific work

206 206

4.38 caption 4 line from bottom

thrust πd(78.24 kPa)

specific work πd=(78.24 kPa)

206 207 207 244 244

Last equation 2nd line 3rd line line 2 from 2nd paragraph Equation 4.202)

𝜋𝑐 ptTt4 tλ …rate of kinetic ...

𝜋𝑐 tTt4 τλ rate of residual kinetic …

244 244

th

𝛾𝑐 𝑒𝑐 ∗(𝛾𝑐 −1)

𝑉2

𝑉2

𝑉2

𝛾𝑐 −1 𝑒𝑐 ∗(𝛾𝑐 )

(𝑉 −𝑉 )2

𝑉2

13th line from bottom 9 from bottom of last paragraph

𝑚̇𝑝 ( 1 + 𝜃1 − 0 ) 2 2 2 Aquations axial kinetic energy

𝑚̇𝑝 1 0 + 𝑚̇𝑝 𝜃1 2 2 Equations residual axial kinetic energy

244

Equation 4.206

𝑚̇𝑝 (

𝑚̇𝑝

245

Equation 4.210

𝜌0 𝑉𝑝 𝐴𝑝 (

245

Equation 4.211

𝑉12 2

−

𝑉02

2 𝑉12

)

−

𝑉02

3 𝑉1 2

2 2 1 𝑉1

2 𝑉0

2 𝑉0

)

( ) − ( )−

(𝑉1 −𝑉0 )2

2 (𝑉1 −𝑉0 )2

𝜌0 𝑉𝑝 𝐴𝑝 3

1 𝑉1 2

2 1 𝑉1

1

2

2 𝑉0

2 𝑉0

2

( ) − ( )−

1

246

Example 4.18

3 𝑉1 2

1 𝑉

3

1 𝑉1 2

1 𝑉

1

2 𝑉0

2 𝑉0

2

2 𝑉0

2 𝑉0

2

( ) − ( 1) −

in first equation

( ) − ( 1) −

V1/V0=1.127 V1/V0=1.225 V1=112.7 m/s V1=122.5 m/s Vp=(100+112.7)/2=106.4 m/s Vp=(100+122.5)/2=111.25 m/s 𝐹𝑝𝑟𝑜𝑝 ≈ 𝜌0 𝑉𝑝 𝐴𝑝 (𝑉1 − 𝑉0 ) ≈ 5.821 𝑘𝑁 (This is about 1,310 lbf) 𝐹𝑝𝑟𝑜𝑝 ≈ 𝜌0 𝑉𝑝 𝐴𝑝 (𝑉1 − 𝑉0 ) ≈ 10.79 𝑘𝑁 (This is about 2426.4 lbf) 𝑚̇0 = 𝑚̇𝑝 ≈ 𝜌0 𝑉𝑝 𝐴𝑝 = 458.3 𝑘𝑔/𝑠 𝑚̇0 = 𝑚̇𝑝 ≈ 𝜌0 𝑉𝑝 𝐴𝑝 = 479.7 𝑘𝑔/𝑠 FpropV0=0.5821 MW FpropV0=1.08 MW ηp ≈ (2V0)/(V0+V1)=94% ηp ≈ (2V0)/(V0+V1)=89.9% 𝐹𝑝𝑟𝑜𝑝 𝑉0 𝐹 𝑉 𝜂𝑝𝑟𝑜𝑝 ≡ ≈ 48.5% 𝜂𝑝𝑟𝑜𝑝 ≡ 𝑝𝑟𝑜𝑝 0 ≈ 89.9%

247

℘𝑝

𝜏𝑝𝑟𝑜𝑝

264

𝑉̅𝜃𝑝′ ≈25 m/s 𝑉̅𝜃1 ≈25.8 m/s ℘𝑝 = =≈ 11.5 𝑘𝑁. 𝑚 𝜔

℘𝑝

𝜏𝑝𝑟𝑜𝑝

𝑉̅𝜃𝑝′ ≈24 m/s 𝑉̅𝜃1 ≈25.1 m/s ℘𝑝 = =≈ 11.46 𝑘𝑁. 𝑚 𝜔

Problem 4.13

nozzles are convergent

442 479 480

nozzles are convergentdivergent and perfectly expanded 9th line from bottom Table 7.2 HO NO rd 3 from top Me Me Replace Figures 7.31 and 7.32 and their captions with the ones below

519

Problem 7.13

The last sentence goes at the end of the problem.

534

Equation 8.18

559

Line 8

W2 ≥ 0.62 W1

W2 ≥ 0.72 W1

559

10

Equation 8.20

Equation 8.70

571

1

𝑊𝜃1 = 3253

595

𝑊𝜃1 = 𝐶𝑧1 𝑡𝑎𝑛𝛽1

8.52

𝑚 𝑠

𝑊𝜃1 = 325.3

Rator

Rotor

𝑚 𝑠

623

13 from bottom

ruputure

rupture

657

7

graph Equation 9.7b

graph Equation 9.9b

663

First equation in solution

𝐶21

𝐶12

666

Equation 9.24

(9.24)

(9.24a)

667

Insert this sentence and equation before the first paragraph: Therefore, from continuity of incompressible fluid, we note that the radial velocity, Cr is inversely proportional to radius, i.e.,

2

rCr  const. or Cr 

σi-1

1 r

(9.24b)

σi=1

672

8

683 722

Figure P9.22 Ω ω The right-hand column of solution (Top Eq) /1157≈1889 K /(1157x2)≈1674 K The right-hand column of solution (Next Eq) /1157≈1865 K /(1157x2)≈1662 K 1st column, 9th line from bottom and 771 K and 704 K nd st 2 from bottom of 1 column σn2=0.2 σn2=2.0 Example 11.4 Item 9 in the 1st column Delete p8=p18=p0 Item 12 in the 2nd column πd=0.995 πd=0.98 Example 11.4 τcH=2.1184 τcH=2.24 πcH=10.64 πcH=12.68 τf=1.3304 τf=1.20 πf=2.458 πf=1.78 α=7.66 α=6.77 Problem 11.13 (7th line from bottom) πd=0.995 πd=0.98 Problem 11.19 Delete p8=p18=p0 Problem 12.9 Assume MR1=MR2 Clarification Note: the definitions used under U.S. Standard Atmosphere SI Units are intended for the table on Page 906 Equation (B.7) Insert “=” sign before the bracket (see equation 2.92) Value of T/T* at Mach 0.2 is 1.1905+E00 [not 1905E+00 that is in the table] Equation D.3 γ=1 γ+1

747 752 797

798

817 818 900 905 907 908 937

3

Afterburner total pressure loss (dry mode) 1 0.95 0.9

pte/pti

0.85 CD=

0.8 0.75 0.7

0.5 0.75 1.0 1.25

0.65 0.6 0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Mi FIGURE 7.31 Afterburner total pressure loss due to flameholder drag (γi=γe=1.33)

Afterburner exit Mach number (dry mode) 1 0.9

CD=

0.8

1.0 1.25

Me

0.7 0.6 0.5 0.4 CD=

0.3

0.5 0.75

0.2 0.1 0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Mi FIGURE 7.32 Frictional choking of afterburner caused by flameholder drag (γi=γe=1.33)

4