MAE 3253 Exam 2
Name:________________________
Equation List
General Relations
P = TV∞
m
m
q∞ = (1/2) ρV∞ 2
c = γRT
σ=
Cm x c / 4 − x cg
=
CL
c
dV∞
= T cosε − D − W sin θ
dt
m
V∞2
= L cos φ + T sin ε cos φ − W cos θ
rp
V2
m ∞ = L sin φ + T sin ε sin φ
rb
2 2 C1/L 2
R=
c t ρS CD
E=
VL / Dmax =
2
1
W
ρ πeARCDo S
Drag Equations
CDo
1− M
2
∞
2
R = V∞
ω=
g n 2 −1
V∞ g n 2 −1
=
R
V∞
φ = cos−1 (1/n)
(W− W)
i
f
1 L Wi
ln
ct D W f
VC 1/ 2 /C |
L
D max
= 1.32VL / Dmax
VC 3 / 2 /C |
L
⎛ W ⎞2 1 S
D = q∞ SCDo + ⎜ ⎟
⎝ S ⎠ πeAR q∞
for M∞ < 1
T = m˙ (Ve − V∞ ) + ( pe − po )Ae
Constants
Standard Atmosphere (English)
hG [ft]
T [oR]
p [lb/ft2]
0
519
2116
10,000
483
1455
20,000
447
973
30,000
412
630
35,000
394
499
40,000
390
393
50,000
390
243
60,000
390
151
l S
Vvt = vt vt
bS
dV∞
= T − D − μr (W − L)
dt
CL2
πeAR
CD = CDo + CDi = CDo +
CD =
l S
Vht = ht ht
cS
D max
= 0.76VL / Dmax
Viscosity
ρ [slugs/ft3]
.002377
.001756
.001267
.000891
.000738
.000587
.000364
.000226
μ = 1.7894⋅ 10 −5 kg/(m⋅ s)
= 3.7372⋅ 10 −7 slug/(ft⋅ s)
Gas Constants
γ = 1.4 R = 1,716 ft - lbf/slug - o R
Conversions
p : 1 lb/ft 2 = 47.9 Pa [N/m2 ]
ρ : 1 slugs/ft 3 = 517.5 kg/m3
V∞ : 1 knot = 1.69 fps = 0.514 m/s
P : 1 hp = 550 ft - lbf/s
R : 1 nm = 6,076 ft
0
