Ôn tập bài tập giữa kỳ - Cơ học và tính năng tàu bay | Học viện Hàng Không Việt Nam

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34 17 lượt tải Tải xuống
Chia sẻ Báo cáo tài liệu

Môn: Cơ học và tính năng tàu bay (ĐHKL01)

Trường: Học viện Hàng Không Việt Nam

Thông tin:

12 trang 3 tháng trước

Tác giả:

Profile Mai Nguyệt
Ôn tập bài tập giữa kỳ cơ học và tính năng tàu bay
Bài tập Turning
Bài tập bay bằng
T =
W
CL CD/
V
stall
=
√
2 mg
ρ x S x CLmax
V
TAS
=
√
2 mg
ρ x S x CL
=
√
2mg
ρ
0 x S x CL
x
1
√
σ
= V
EAS
x
1
√
σ
L =
1
2
x ρ x V
2
x S x
C
L
=
1
2
x
ρ
0
x V
2
EAS
x S x
C
L
D =
C
D
=
1
2
x
ρ
0
x V
2
EAS
x S x
C
D
(T
min
) khi (D ) khi
min
(
CD
CL
)
min
C
D
= C + kC C
d L
2
Dmd
= 2C
d
C
Lmd
=
√
Cd
k
V
EASmd
=
√
2mg
ρ
0 x S
x
(
k
Cd
)
1
4
(P
min
) khi
(
CD
C L
3
2
)
min
C
D
= C + kC C
d L
2
Dmp
= 4C
d
C
Lmp
=
√
3 Cd
k
=
√
3
C
Lmd
V
EASmp
=
√
2mg
ρ
0 x S
x
(
k
3 Cd
)
1
4
=
VEASmd
3
1
4
Exercise 1: An airplane weighing 150000 N has a wing area of 210 m , and thrust of engine is 65.4 kN.
2
At the altitude of 9 km (σ =0.38), its cruise speed is 170 m/s.
(a) What are the values of C and C of the airplane?
L D
(b) If the maximum lift coefficient that the airplane can achieve when fluing at the altitude of 9 km is
1.85, determine the "stall" speed of the airplane.
W = 150000 N
S = 210 m
2
T = 65400 N
σ =0.38
V
eas
= 170 m/s
a/ C ? C ?
L D
L = W = mg = 150000
ρ = ρ x σ = 1.225 x 0.38 = 0,4655
0
L =
1
2
x ρ x V
2
x S x
C
L
C
L =
2 L
ρ x V
2
x S
=
2 x 150000
0.4655
x 170 210
2
x
= 0.106
T = D = 65400 N
D =
C
D
C
D
=
2 D
ρ x V
2
x S
=
2 x 65400
0.4655
x 170 210
2
x
= 0.046
b/ C = 1.85; V
Lmax stalll
?
V
stall
=
√
2 mg
ρ x S x CLmax
=
√
2 x 150000
0.4655 x 210 x 1.85
= 40,72 m/s
Exercise 2: The Messerschmidt 109 was produced in huge quantities during World War 2 and was a
very able and dependable aircraft. Over 35,000 were produced between 1936 and 1945. The M109
has a wing area of 16m and a mass of 3000 kg.
2
(a) At a height of 6000 m, where ρ = 0.55ρ0 the M109 has a cruise speed of V =160m/s. Calculate the
corresponding lift coefficient, CL.
(b) If CLmax = 2.9, calculate the airspeed at stall at sea level.
S = 16m
2
m = 3000 kg
a/ ρ = 0.55 ρ = 0.55 x 1.225 = 0.67375; V =160m/s; C ?
0 eas L
L = W = mg = 3000 x 9,8 = 29400
L =
1
2
x ρ x V
2
x S x
C
L
C
L =
2 L
ρ x V
2
x S
=
2 x 29400
0.67375
x 160 16
2
x
= 0.213
b/ C = 2.9; V
Lmax stall
at sea level?
V
stall at sea level
=
√
2 mg
ρ0 x S x CLmax
=
√
2 x 29400
1.225 x 16 2.9x
= 32.16 m/s
Exercise 3: A Spitfire aircraft has wing area of 22.48 m , a mass of 3000 kg and a cruise speed of V =
2
139 m/s at an altitude of 4570 m, where the air density ρ ≈ 0.7 kg/m .
3
(a) Calculate the lift coefficient at the cruising speed.
(b) The stall speed is 32.6 m/s at sea level. Calculate CL,max at stall and also calculate the stall speed
at 4570m
S = 22.48 m
2
M = 3000 kg
V
eas
= 139 m/s
ρ ≈ 0.7 kg/m
3
a/ C ?
L
L = W = mg = 3000 x 9,8 = 29400 N
L =
1
2
x ρ x V
2
x S x
C
L
C
L =
2 L
ρ x V
2
x S
=
2 x 29400
0.7
x 139 22.48
2
x
= 0.1933
b/ V
stall at sea level Lmax at stall stall, 4570m
= 32.6 m/s; C ? V ?
V
stall at sea level
=
√
2 mg
ρ0 x S x CLmax
C
Lmax at stall
=
2 mg
ρ
0 x S x V
2
=
2 x 29400
1.225 22.48
x x 32.6
2
= 2.009
V
stall, 4570m
=
√
2 mg
ρ x S x CLmax
=
√
2 x 29400
0.7 x 22.48 x 2.009
= 43.12 m/s
Exercise 4: An airplane weighing 50000 kg has a wing area of 210 m , and thrust of engine is 23.4 kN.
2
At the altitude of 5.5 km (σ =0.75), its cruise speed is 100 m/s.
(a) What are the values of C and C of the airplane?
L D
(b) Determine the stall speed of the airplane at this height if CLmax = 1.42
m = 50000 kg
S = 210 m
2
T = 23400 N
σ =0.75
V
eas
=100 m/s
a/ C ? C ?
L D
L = W = mg = 50000 x 9.8 = 490000 N
ρ = ρ x σ = 1.225 x 0.75 = 0.91875
0
L =
1
2
x ρ x V
2
x S x
C
L
C
L =
2 L
ρ x V
2
x S
=
2 x 490000
0.91875
x 100 210
2
x
= 0.5079
T = D = 23400 N
D =
C
D
C
D
=
2 D
ρ x V
2
x S
=
2 x 23400
0.91875
x 100 210
2
x
= 0.02425
b/ C = 1.42; V
Lmax stalll
?
V
stall
=
√
2 mg
ρ x S x CLmax
=
√
2 x 490000
0.91875 x 210 x 1.42
= 59.8 m/s (cái này sai đáp số so với slide)
Exercise 6: An aircraft weighs 56,000 lbs and has a 900 ft wing area. Its drag polar is given by: C =
2
D
0.01575 + 0.03334 CL
2
.
(a)Find the minimum thrust required for cruising flight and the corresponding airspeeds at sea-level
(true air speed) and at 30,000 ft (0.000889 slugs/ft³)
(b) Find the minimum power required and the corresponding true airspeeds for cruising flight at sea-
level and at 30,000 ft
W = 56000 lbs
S = 900 ft
2
C
D
= 0.01575 + 0.03334 CL
2
C = 0.01575; k = 0.03334
d
a/ T ? V
min EAS at sea level md, 30000ft
? V ?
(T
min
) khi (D ) khi
min
(
CD
CL
)
min
C
Dmd
= 2C
d
= 2 x 0.01575 = 0.0315
C
Lmd
=
√
Cd
k
=
√
0.01575
0.03334
= 0.687
C
Lmd
/C =
Dmd
0.687
0.0315
= 21.80
Tmin =
W
CLmd /CDmd
=
56000
21.8
= 2568.80 N
ρ
0= 1.225 kg/m = 1.225 x 0.002377 slugs/ft .
3
0.00194 =
3
V
EAS at sea level
=
√
2 mg
ρ0 x S x CL
=
√
2 x 56000
0.002377 x 900 x 0.687
= 276.054 ft/s
ρ30,000= 0.458 kg/m³ = 0.000889 slugs/ft³
σ = ρ
0
x ρ
30,000
=
0.000889
0.002377
= 0.374
V
md, 30000ft =
V
TAS EAS
= V
x
1
√
σ
= 276.054 x
1
√
0.374
= 451.39 ft/s
b/ P ? V
min EASmp
? V
mp, 30000ft
?
C
D
= C + kC C
d L
2
Dmp
= 4C
d
= 4 x
0.01575
= 0.063
C
Lmp
=
√
3 Cd
k
=
√
3 x 0.01575
0.03334
= 1.188
(P
min
) khi
(
CD
C L
3
2
)
min =
0.063
1.188
3
2
= 0.048
V
EASmp
=
√
2mg
ρ
0 x S
x
(
k
3 Cd
)
1
4
=
√
2 x 56000
0.002377
x 900
x
(
0.03334
3 x 0.01575
)
1
4
= 209.7 ft/s
V
EASmp, 30000ft = TAS, 30000ft
V = V
EAS
x
1
√
σ
= 209.7 x
1
√
0.374
= 342.9 ft/s
P
min
= T x V
min EASmp
= 2568.80 x 209.7 = 538677.36 ft-lb/s
P
min, 30000ft EASmp, 30000ft
= T
min
x V = 2568.80 x 342.9 = 880841.52 ft-lb/s
Exercise 7: An aircraft has a mass of 14000 kg and a wing area of 18 m2, sải cánh =13.41 m, its drag
polar is given by: C =0.015+ 1.2*CL2/(πAR). Find;
D
(a) Equivalent airspeeds if the thrust of the engine at mean sea level is 11000 N..
(b) CD,CL at minimum drag flight condition
(c) CD,CL at minimum power flight condition
m = 14000 kg
S
= 18 m
2
Wingspan, =13.41 𝑏
T = 11,000 N
C
D
=0.015+ 1.2*C
L
2
/(πAR)
AR =
b
2
S
=
13.41
2
18
=9.99
Cd = 0.015; k = 0.038
a/ V ?
eas
L = W = mg = 14000 x 9.8 = 137200 N
T = D = 11000 N
D =
1
2
x ρ 0 x Veas
2
x S x
C
D
C
D
=
2 D
ρ
0 x V
2
x S
=
2 x 11000
1.225
x Veas
2
x 18
(1)
C
L =
2 L
ρ
0 x V
2
x S
=
2 x 137200
1.225
x Veas
2
x 18
Ta có:
C
D
=
0.015
+1.2 x
C L
2
πAR
=
0.015 0.038+
x C = 0.015 + 0.038 x
L
2
(
2 x 137200
1.225
x Veas
2
x 18
)
2
(2)
Từ (1) và (2) ta có:
2 x 11000
1.225
x Veas
2
x 18
=0.015 +0.038 x
(
2 x 137200
1.225
x Veas
2
x 18
)
2
Veas = 80.988 m/s
b/ C
Dmd
,C
Lmd
?
C
D
= C + kC C
d L
2
Dmd
= 2C
d
= 2 x 0.015 = 0.03
C
Lmd
=
√
Cd
k
=
√
0.015
0.038
= 0.628
c/ C ?
Dmp
,C
Lmp
C
D
= C + kC C
d L
2
Dmp
= 4C
d
= 4 x
0.01 5
= 0.06
C
Lmp
=
√
3 Cd
k
=
√
3 x 0.015
0.038
= 1.08
| 1/12
| | Tải xuống

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Ôn tập bài tập giữa kỳ cơ học và tính năng tàu bay Bài tập Turning Bài tập bay bằng W T = CL/CD Vstall = √ 2mg ρ x S x CLmax 1 1
VTAS = √ 2mg =√ 2mg x = VEAS x ρ x S x CL ρ0 x S x CL √σ √σ 1 1 L = 2
x ρ x V x S xCL = x ρ 2 2 0 x V2EASx S x CL 1 1 D = 2
x ρ x V x S xCD = x ρ0 x V2EASx S x CD 2 2 (T ( CD ) min) khi (Dmin) khi min CL C 2 D = Cd + kCL C  Dmd = 2Cd CLmd = √Cd k
VEASmd = √ 2mg x( k )14 ρ0 x S Cd ( CD ) (Pmin) khi 3 min C L2 C 2 D = Cd + kCL C  Dmp = 4Cd
CLmp = √3Cd = √3CLmd k VEASmd
VEASmp = √ 2mg x( k )14 = 1 ρ0 x S 3 Cd 3 4
Exercise 1: An airplane weighing 150000 N has a wing area of 210 m2, and thrust of engine is 65.4 kN.
At the altitude of 9 km (σ =0.38), its cruise speed is 170 m/s.
(a) What are the values of CL and C of the airplane? D
(b) If the maximum lift coefficient that the airplane can achieve when fluing at the altitude of 9 km is
1.85, determine the "stall" speed of the airplane. W = 150000 N S = 210 m2 T = 65400 N σ =0.38 Veas = 170 m/s a/ CL ? CD ? L = W = mg = 150000
ρ = ρ0 x σ = 1.225 x 0.38 = 0,4655 1 2 L 2 x 150000 L = 2 x ρ x V x S xC = = 0.106 2 L  CL = 2 2 ρ x V x S 0.4655 x 170 210 x T = D = 65400 N 1 2 D 2 x 65400 D = 2
x ρ x V x S xCD  CD = = = 0.046 2 2 2 ρ x V x S 0.4655 x 170 210 x b/ CLmax = 1.85; Vstalll ?
Vstall = √ 2mg = √ 2x150000 = 40,72 m/s ρ x S x CLmax
0.4655 x 210 x 1.85
Exercise 2: The Messerschmidt 109 was produced in huge quantities during World War 2 and was a
very able and dependable aircraft. Over 35,000 were produced between 1936 and 1945. The M109
has a wing area of 16m2 and a mass of 3000 kg.
(a) At a height of 6000 m, where ρ = 0.55ρ0 the M109 has a cruise speed of V =160m/s. Calculate the
corresponding lift coefficient, CL.
(b) If CLmax = 2.9, calculate the airspeed at stall at sea level. S = 16m2 m = 3000 kg
a/ ρ = 0.55 ρ0 = 0.55 x 1.225 = 0.67375; Veas =160m/s; C ? L
L = W = mg = 3000 x 9,8 = 29400 1 2 L 2 x 29400 L = 2 x ρ x V x S xC = = 0.213 2 L  CL = 2 2 ρ x V x S 0.67375 x 160 16 x
b/ CLmax = 2.9; Vstall at sea level ? Vstall at sea level = √ 2 mg
= √ 2x29400 = 32.16 m/s ρ0 x S x CLmax 1.225 x 16 2.9 x
Exercise 3: A Spitfire aircraft has wing area of 22.48 m2, a mass of 3000 kg and a cruise speed of V =
139 m/s at an altitude of 4570 m, where the air density ρ ≈ 0.7 kg/m3.
(a) Calculate the lift coefficient at the cruising speed.
(b) The stall speed is 32.6 m/s at sea level. Calculate CL,max at stall and also calculate the stall speed at 4570m S = 22.48 m2 M = 3000 kg Veas = 139 m/s ρ ≈ 0.7 kg/m3 a/ C ? L
L = W = mg = 3000 x 9,8 = 29400 N 1 2 L 2 x 29400 L = 2 x ρ x V x S xC = = 0.1933 2 L  CL = 2 2 ρ x V x S 0.7 x 139 22.48 x
b/ Vstall at sea level = 32.6 m/s; CLmax at stall ? Vstall, 4570m ? 2 mg 2 x 29400 Vstall at sea level = √ 2 mg  CLmax at stall = = = 2.009 2 2 ρ0 x S x CLmax ρ 0 x S x V 1.225 22.48 x x 32.6
Vstall, 4570m = √ 2mg = √ 2x29400 = 43.12 m/s ρ x S x CLmax
0.7 x 22.48 x 2.009
Exercise 4: An airplane weighing 50000 kg has a wing area of 210 m2, and thrust of engine is 23.4 kN.
At the altitude of 5.5 km (σ =0.75), its cruise speed is 100 m/s.
(a) What are the values of CL and C of the airplane? D
(b) Determine the stall speed of the airplane at this height if CLmax = 1.42 m = 50000 kg S = 210 m2 T = 23400 N σ =0.75 Veas =100 m/s a/ CL ? CD ?
L = W = mg = 50000 x 9.8 = 490000 N
ρ = ρ0 x σ = 1.225 x 0.75 = 0.91875 1 2 L 2 x 490000 L = 2 x ρ x V x S xC = = 0.5079 2 L  CL = 2 2 ρ x V x S 0.91875 x 100 210 x T = D = 23400 N 1 2 D 2 x 23400 D = 2
x ρ x V x S xCD  CD = = = 0.02425 2 2 2 ρ x V x S 0.91875 x 100 210 x b/ CLmax = 1.42; Vstalll ?
Vstall = √ 2mg = √ 2x490000 = 59.8 m/s (cái này sai đáp số so với slide) ρ x S x CLmax
0.91875 x 210 x 1.42
Exercise 6: An aircraft weighs 56,000 lbs and has a 900 ft2 wing area. Its drag polar is given by: CD = 0.01575 + 0.03334 CL2.
(a)Find the minimum thrust required for cruising flight and the corresponding airspeeds at sea-level
(true air speed) and at 30,000 ft (0.000889 slugs/ft³)
(b) Find the minimum power required and the corresponding true airspeeds for cruising flight at sea- level and at 30,000 ft W = 56000 lbs S = 900 ft2
CD = 0.01575 + 0.03334 CL2  Cd = 0.01575; k = 0.03334
a/ Tmin ? VEAS at sea level ? Vmd, 30000ft ? (T )
min) khi (Dmin) khi ( CD min CL
CDmd = 2Cd = 2 x 0.01575 = 0.0315
CLmd = √Cd = √0.01575 = 0.687 k 0.03334 0.687 CLmd /CDmd = = 21.80 0.0315 W 56000 Tmin = = = 2568.80 N CLmd /CDmd 21.8
ρ0= 1.225 kg/m3 = 1.225 x 0.00194 = 0.002377 slugs/ft3.
VEAS at sea level = √ 2mg = √ 2 x 56000 = 276.054 ft/s ρ0 x S x CL
0.002377 x 900 x 0.687
ρ30,000= 0.458 kg/m³ = 0.000889 slugs/ft³ 0.000889
σ = ρ0 x ρ30,000 = = 0.374 0.002377 1 1
Vmd, 30000ft = VTAS = VEAS x = 276.054 x = 451.39 ft/s √ σ √0.374
b/ Pmin ? VEASmp ? Vmp, 30000ft ? C 2 D = Cd + kCL C
 Dmp = 4Cd = 4 x 0.01575 = 0.063
CLmp = √3Cd =√3x0.01575 = 1.188 k 0.03334 0.063 (P ) min) khi ( CD3 min = 3 = 0.048 C L2 1.1882
VEASmp = √ 2mg x( k )14 = √ 2x56000 x( 0.03334 )14 = 209.7 ft/s ρ0 x S 3 Cd 0.002377 x 900 3 x 0.01575 1 1
VEASmp, 30000ft = VTAS, 30000ft = VEAS x = 209.7 x = 342.9 ft/s √ σ √0.374
Pmin = Tmin x VEASmp = 2568.80 x 209.7 = 538677.36 ft-lb/s
Pmin, 30000ft = Tmin x VEASmp, 30000ft = 2568.80 x 342.9 = 880841.52 ft-lb/s
Exercise 7: An aircraft has a mass of 14000 kg and a wing area of 18 m2, sải cánh =13.41 m, its drag
polar is given by: C =0.015+ 1.2*CL2/(πAR). Find; D
(a) Equivalent airspeeds if the thrust of the engine at mean sea level is 11000 N..
(b) CD,CL at minimum drag flight condition
(c) CD,CL at minimum power flight condition m = 14000 kg S = 18 m2 Wingspan, =13.41 𝑏 T = 11,000 N C 2 D=0.015+ 1.2*CL /(πAR) 2 2 13.41 AR = b = =9.99 S 18  Cd = 0.015; k = 0.038 a/ Veas ?
L = W = mg = 14000 x 9.8 = 137200 N T = D = 11000 N 1 2 D 2 x 11000 D = 2
x ρ 0 x Veas x S x C = (1) 2 D  CD = 2 2 ρ 0 x V x S 1.225 x Veas x 18 2 L 2 x 137200 CL = = 2 2 ρ 0 x V x S 1.225 x Veas x 18 Ta có: 2 C C L 2 D = 0.015 +1.2 x = 0.015 0.038 +
x CL = 0.015 + 0.038 x ( 2 x137200 )2(2) πAR 2 1.225 x Veas x 18 Từ (1) và (2) ta có: 2 x 11000
=0.015+0.038 x ( 2x137200 )2 2 2 1.225 x Veas x 18 1.225 x Veas x 18  Veas = 80.988 m/s b/ CDmd,CLmd ? C 2 D = Cd + kCL C
 Dmd = 2Cd = 2 x 0.015 = 0.03
CLmd = √Cd = √0.015 = 0.628 k 0.038 c/ CDmp,CLmp ? C 2 D = Cd + kCL C
 Dmp = 4Cd = 4 x 0.01 5 = 0.06
CLmp = √3Cd =√3x0.015 = 1.08 k 0.038