Solution Manual for Aircraft Performance: An Engineering Approach 2nd Edition by Mohammad H. Sadraey | Complete Chapters 1–10 | Verified Answers & Detailed Solutions

Solution Manual For Aircraft Performance, An Engineering
Approach 2nd Edition By Mohammad H. Sadraey | All
Chapters (1-10) | Latest Version A+ Guide




1

, Table of Contents
x x




1. Atmosphere.
x x




2. Equations of Motion.
x x x x




3. Drag Force and Drag Coefficient.
x x x x x x




4. Engine Performance.
x x x




5. Straight-Level Flight – Jet Aircraft.
x x x x x x




6. Straight-Level Flight: Propeller-Driven Aircraft.
x x x x x




7. Climb and Descent.
x x x x




8. Takeoff and Landing.
x x x x




9. Turn Performance and Flight Maneuvers.
x x x x x x




10. Aircraft Performance Analysis Using Numerical Methods and MATLAB(R)
x x x x x x x x x




2

, Ch. 1x




The software package Mathcad is used to solve problems.
x x x x x x x x




1.1. Determine the temperature, pressure and air density at 5,000 m and ISA condition.
x x x x x x x x x x x x




There are two methods:
x x x


a. Using appendix: x


From Appendix A:
x x




- Temperature: 255.69 K x x


- Pressure: 54,048 Pa x x


- Air density: 0.7364 kg/m3
x x x




b. Calculations:

K x x J
h x = ISA L1 x= x 6.5 xx
x
R1 x= x 287x Po x = x 101325Pa
x 5000m
1000 x

m
kgK

Sea level: x To x= x (15 x+ x 273)K x = x288


x 5000 m: x x K xT5 x= x To x − xL1h x = (Equ 1.6)x




x 255.5xK

5.256
 T5 
x x
P5 x= x Po  = x54000.3xPa (Equ 1.16)
x


 To  x x




P5 xkg

5 x = = x0.736 x (Equ 1.23)
x

R1T 3
x
m
5



Same results. x




3

, 1.2. Determine the pressure at 5,000 m and ISA-10 condition.
x x x x x x x x




x x K x x J
h x = ISA x − L1 x= x 6.5x R1 x= x 287x Po x = x 101325Pa
x 5000m x 10
1000m x



kgK

Sea level:
x To x= x (15 x+ x 273 x− x10)K x = x278


x 5000 m: x K xT5 x= x To x − xL1h x = x245.5xK
x (Equ 1.6)
x




5.256
 T5 x x
P5 x= x Po  = x52714.2xPa (Equ 1.16)
x


 To  x x




1.3. Calculate air density at 20,000 ft altitude and ISA+15 condition.
x x x x x x x x x




x x K x x J
h x = ISA x + L1 x= x 2x R1 x= x 287x Po x = x 101325Pa
x 20000ft x 15
1000ft x



kgK

Sea level:
x To x = x [(15 x + x 273) x + x 15]K x = To x = x 545.4R
x 303xK




20000 ft: x T20 x= x To x − x L1h x = T20 x = (Equ 1.6)
x


x 263xK x 473.4R




5.256
 T20 
x x
lbf
P20 x= x Po  = x48143.9xPa P20 x = (Equ 1.16)
x


 To 
x x x
x 1005.5
2
ftx

P20 xkg xslug

20 x= = x0.638 x 20 = x 0.001238x (Equ 1.23)
x

R1T2
x 3 3
m ftx
0




4