Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Sm A Mathematics Boot Camp for Science and Engineering Students, 1st Edit, Exams of Engineering

Sm A Mathematics Boot Camp for Science and Engineering Students, 1st Edit

Typology: Exams

2024/2025

Available from 05/05/2025

QuizBank
QuizBank 🇺🇸

201 documents

1 / 21

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
1
Solutions to problems for
Aircraft Performance: An Engineering Approach, Mohammad Sadraey, 2nd ed., CRC, 2023
Chapter 1-10
Ch. 1
The software package Mathcad is used to solve problems.
1.1. Determine the temperature, pressure and air density at 5,000 m and ISA condition.
There are two methods:
a. Using appendix:
From Appendix A:
- Temperature: 255.69 K
- Pressure: 54,048 Pa
- Air density: 0.7364 kg/m3
b. Calculations:
Same results.
Sea level:
5000 m:
(Equ 1.6)
(Equ 1.16)
(Equ 1.23)
h5000 m
ISA
L1 6.5 K
1000 m

R1 287 J
kg K

Po101325 Pa
To15 273( ) K288 K
T5ToL1 h 255.5K
P5Po
T5
To
5.256
54000.3Pa
5
P5
R1T5
0.736 kg
m3

pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15

Partial preview of the text

Download Sm A Mathematics Boot Camp for Science and Engineering Students, 1st Edit and more Exams Engineering in PDF only on Docsity!

Solutions to problems for

Aircraft Performance: An Engineering Approach, Mohammad Sadraey, 2

nd ed., CRC, 2023

Chapter 1-

Ch. 1

The software package Mathcad is used to solve problems.

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

There are two methods:

a. Using appendix:

From Appendix A:

  • Temperature: 255.69 K
  • Pressure: 54,048 Pa
  • Air density: 0.7364 kg/m

3

b. Calculations:

Same results.

Sea level:

5000 m: (Equ 1.6)

(Equ 1.16)

(Equ 1.23)

h 5000 m ISA L1^ 6.^

K

1000 m

  R1 287

J

kg K

  P

o

101325 Pa

T

o

( 15  273 ) K 288K

T

T

o

  L1 h 255.5K

P

P

o

T

T

o

  54000.3Pa

P

R1 T

kg

m

3

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

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

Sea level:

5000 m: (Equ 1.6)

(Equ 1.16)

Sea level:

20000 ft: (Equ 1.6)

(Equ 1.16)

(Equ 1.23)

h 5000 m ISA  10 L1 6.

K

1000 m

  R1 287

J

kg K

  P

o

101325 Pa

T

o

( 15  273  10 ) K 278K

T

T

o

  L1 h 245.5K

P

P

o

T

T

o

  52714.2Pa

h 20000 ft ISA  15 L1 2

K

1000 ft

  R1 287

J

kg K

  P

o

101325 Pa

T

o

[( 15  273 )  15 ] K 303K T

o

545.4 R

T

T

o

  L1 h 263K T 20

473.4 R

P

P

o

T

T

o

  48143.9Pa P 20

lbf

ft

2

P

R1 T

kg

m

3

slug

ft

3

1.5. Determine relative density () at ISA-20 condition and 80,000 ft altitude.

1.6. Determine the temperature at 70,000 ft and ISA condition.

Sea level:

Second layer:

(Equ 1.16)

(Equ 1.23)

(Equ 1.25)

Sea level:

Second layer:

h 80000 ft ISA  20 R1 287

J

kg K

  P

o

101325 Pa  o

slug

ft

3

T

o

[( 15  273 )  20 ] K 268K T

o

482.4 R T

o

5.15 °C

T

( 56  273  20 ) K T

354.6 R T

197K

P

P

o

T

T

o

  20098.1Pa P 80

lbf

ft

2

P

R1 T

kg

m

3

slug

ft

3

o

h 70000 ft ISA

T

o

( 15  273 ) K 288K T

o

518.4 R

T

( 56  273 ) K T

390.6 R T

217K

1.7. An aircraft is flying at an altitude at which its temperature and pressure are 255 K and 4.72 × 10

4

Pa.

Calculate:

a. pressure altitude

From Appendix A, for T = 255 K; h = 5,000 m

b. temperature Altitude

From Appendix A, for P = 4.72 × 10

4

Pa; h = 6,000 m

c. density altitude

Interpolation:

From Appendix A, for  = 0.645 kg/m

3

; h = 6,215 m

(Equ 1.23)

T1 255 K P1^ 4.72 10

4   Pa R1 287

J

kg K

P

R1 T1

kg

m

3

given h  6500

7000 h

find (h ) 6215.

1.10. An aircraft is flying at 20,000 ft altitude with speed of 400 km/hr and ISA condition. What is

aircraft Mach number?

1.11. Calculate air density at sea level and ISA condition, when humidity is 100%.

Sea level:

(Equ 1.6)

(Equ 1.35)

Sea level:

Table 1.

(Equ 1.29)

(Equ 1.30)(Equ 1.23)

h 20000 ft ISA L1^2

K

1000 ft

  V1 400

km

hr

  V1 111.

m

s

 1.4 R1 287

J

kg K

T

o

( 15 273.15) K 288.15K T

o

518.67 R

T

T

o

  L1 h 248.15K T 20

446.67 R

a   R1T 20

m

s

M

V

a

R1 287

J

kg K

   100% P

o

101325 Pa

T

o

( 15 273.15) K 288.15K P

g

1.7051 kPa

P

v

P

g

P

v

P

g

3    Pa

P

a

P

o

P

v

4    Pa

P

a

R1 T

o

kg

m

3

1.12. Determine the speed of sound at sea level and

a. ISA condition

b. ISA+

c. ISA-

Sea level:

(Equ 1.34)

(Equ 1.34)

(Equ 1.34)

ISA  1.4 R1 287

J

kg K

T

o

( 15 273.15) K 288.15K T

o

518.67 R

a   R1T o

m

s

T 12 K T

b

T

o

  T 300.15K

a   R1T b

m

s

T  18 K T

c

T

o

  T 270.15K

a  R1 T c

m

s

1.15. Fighter aircraft Mig-31 is able to fly with Mach 2.2 at an altitude of 60,000 ft altitude. What is the

dynamic pressure if flying at ISA+20 condition?

Atmospheric pressure is not a function of temperature, so ISA+20 has no effect and is not used.

1.16. The temperature at the summit of a mountain at ISA condition is -

o

C. What is the height of this

summit from sea level?

From Appendix A, the altitude is almost 3000 m.

1.17. The highest peak of Mount Everest has the elevation of 29,035 ft. Calculate temperature, pressure

and air density at this peak.

(Equ 1.21)

(Equ 1.36)

Sea level:

(Equ 1.6)

App B: Interpolation:

(Equ 1.23)

P

o

 101325 Pa h  60000 ft M1 2.

P

60

P

o  0.2234 e

36089 ft h

20807 ft

  7173.267 Pa

q 0.7 P 60

 M

2

  24303Pa q 24.303kPa

T1 ( 5 273.15) K 268.15K

h 29035 ft R1 287

J

kg K

  L1 2

K

1000 ft

T

o

( 15 273.15) K 288.15K T

o

15°C

T1 T

o

  L1 h 230.08K T1 43.07 °C

h  29035 ft

P1  2000

629.7 P

30000 h Find P1( ) 657.

P1 657.

lbf

ft

2

P1 3.149 10

4   Pa

P

R1 T1

kg

m

3

slug

ft

3

1.18. The fighter aircraft F-15C is capable of flying at the altitude of 12,000 m with the speed of 2443

km/hr. What is its speed in terms of Mach number at ISA condition?

Second layer:

(Equ 1.35)

h 12000 m ISA V1 2443

km

hr

  V1 678.

m

s

  1.4 R1 287

J

kg K

T

( 56  273 ) K T

390.6 R T

217K

a   R1T 12

m

s

M

V

a

b. ISA+20 condition

c. ISA-20 condition.

1.21. Determine the air viscosity for an altitude of 10,000 m and ISA condition.

(Equ 1.6)

(Equ 1.34)

(Equ 1.35)

(Equ 1.6)

(Equ 1.34)

(Equ 1.35)

Sea level:

(Equ 1.6)

(Equ 1.26)

T 20 K T

b

T

o

  T 308.15K

T

T

b

  L1 h 268.15K T 20

482.67 R

a   R1T 20

m

s

M

V

a

V1 M1 a 590.

m

s

  V1 1321.

mile

hr

T  20 K T

c

T

o

  T 268.15K

T

T

c

  L1 h 228.15K T 20

410.67 R

a   R1T 20

m

s

M

V

a

V1 M1 a 545

m

s

  V1 1219.

mile

hr

h 10000 m ISA L1 6.

K

1000 m

T

o

( 15 273.15) K 288.15K T

o

518.67 R

T

T

o

  L1 h 223.15K T 10

401.67 R

a 1.485 10

 6 

kg

m s K

1

2 

  b 110.4 K

a T 10

b

T

 5 

kg

m s

1.22. The Earth is rotating around itself once a day. Calculate its velocity at a city at Equator in terms

of Mach number, if speed of sound is assumed to be 340 m/sec. The average radius of Earth is about

6,400 km.

1.23. A Boeing 767 is flying at the altitude of 40,000 ft. How much air pressure must be increased by

its air pressure system in order to provide pressurized air of 0.8 atm for the passengers inside cabin

and cockpit?

At an altitude of 40,000 ft, from Appendix B, the pressure is:

To provide pressurized air of 0.8 atm for the passengers inside cabin:

Total Distance traveled; Circumference:

(Equ 1.35)

a 340

m

s

  R

E

6400 km

t 24 hr t^ 8.64^10

4   s

X 2  R

E

  40212.4 km

V

E

X

t

m

s

M

V

E

a

P

lbf

ft

2

  P

18821.7Pa P 40

0.186atm

1.26. The humidity is 80% at a room with the volume of 160 m

3

in a summer day. Determine the water

vapor content (in kg) of this room at ISA+30 condition, if located at 2,000 ft altitude.

T T T Lh T   C K

o

ISA o

15 0. 002 2000 30 41 314. 15 2000

         

P Pa

T

T

T

T

P

P

o o

ISA

o

ISA

101325 0. 9356 94800

  1. 9356

15 273. 15 30

  1. 15

  2. 256

  3. 256

2000

  1. 256

  

 

 

 

 

The saturation pressure Pg at 41

o

C can be found from table 1.2 as 7.8 kPa.

     7. 8  0. 8  6. 24 v g

g

v P P

P

P

P P P kPa a v

   94. 8  6. 24  88. 56

kg dry air

kgHO

P

P

a

v (^) 2

  1. 0438

  2. 56

  3. 24

 0. 622  0. 622  

  kg

RT

PV

m m

a

v a

  1. 88

  2. 287 314. 15

  3. 56 160

  4. 0438   

 

  

1.27. A transport aircraft is flying at an altitude of 25,000 ft, ISA condition. The pilot observes that the

ice has been formed on the leading edge of the wing. How much deicing system must increase the

temperature of the leading edge in order to melt the ice.

As long as the temperature of the ice is above 0

o

C, it will be melted.

Sea level:

(Equ 1.6)

h 25000 ft L1^2

K

1000 ft

T

o

( 15 273.15) K 288.15K T

o

15 °C

T

T

o

  L1 h 238.15K T 12

 35 °C

T

LE

o

C

1.28. The temperatures of a city in a summer day and a winter day are 115

o

F and 15

o

F respectively.

Find the ratio between air densities at a summer day to that of the winter day.

The ratio between air densities is 0.826.

1.29. The humidity of a city in a summer day (ISA+20) is 100%, and in a winter day (ISA-20) is 10%.

Find the ratio between air densities at a summer day to that of the winter day.

Sea level:

(Equ 1.23)

(Equ 1.23)

Sea level:

Summer:

Table 1.

(Equ 1.29) Winter:Table 1.2 (Equ 1.30)(Equ 1.23)(Equ 1.29)(Equ 1.30)(Equ 1.23)

T1 115°F T2 15°F R1^287

J

kg K

  P

o

110325 Pa

P

o

R1 T1

kg

m

3

slug

ft

3

P

o

R1 T2

kg

m

3

slug

ft

3

R1 287

J

kg K

  P

o

101325 Pa

T  20  100% T

o

( 15  273.15T) K 308.15K T

o

35°C

P

g

5.628 kPa

P

v

P

g

P

v

P

g

3 P    Pa a

P

o

P

v

4    Pa

s

P

a

R1 T

o

kg

m

3

T  20  10%

T

o

( 15  273.15T) K 268.15K T

o

P  5 °C

g

0.4 kPa

P

v

P

g

P

v

P

g

P   40Pa a

P

o

P

v

5    Pa

w

P

a

R1 T

o

kg

m

3

s

w

1.31. What is the dynamic pressure when an aircraft is cruising at an altitude of 14,500 m and Mach

number of 0.6, ISA-12 flight condition?

1.32 An aircraft is cruising at an altitude of 40,000 m and a Mach number of 0.93, in ISA +

flight condition. Determine the dynamic pressure.

1.33 Determine the air viscosity for an altitude of 30,000 ft and ISA+20 condition.

(Equ 1.20)

(Equ 1.36)

P

o  101325 Pa h  14500 m M1 0.

P1 P

o  0.2234 e

11000 m h

6342 m

  13035.357 Pa

q 0.7  P1 M

2   3284.9 Pa q 3.285kPa

1.34 Determine temperature, pressure, and air relative density in ISA + 30 condition and 50,000 ft

altitude.

Sea level:

(Equ 1.6)

(Equ 1.26)

50000 ft: (Equ 1.7)

(Equ 1.21)

(Equ 1.23)