Assignment #4 Solutions
MAE 150/250R, S16, Prof. M. Spearrin
Problem 1.
= latitude
RE = 6380 km
v = ground speed
E = 0.000721 deg/s
E RE = 0.46 km/s
v=E d = E RE cos ( ) = 0.46 cos ( )
a) Equator, = 0, v = 0.46 km/s
b) Cape Carnavel, = 28o 27 ', v = 0.40
Assignment #1 Solutions
MAE 150/250R, S16, Prof. M. Spearrin
Problem 1.
(A) To determine work per unit mass, proceed as follows:
For an adiabatic process, heat transfer from the system to its surrounding is zero,
dQ = 0
ui2 u 2f
W = m dh
2 2
Write th
Assignment #5 Solutions
MAE 150/250R, S16, Prof. M. Spearrin
Problem 1.
Mo
=
u I sp g=
I sp g e ln
e ln (R )
ML + Ms
Structural mass of the solid propellant
( M S )solid = ( M o M L ) =
=
=
( M S )liquid 1.32
( 44055)
0.090 ( 500000 10500 ) = 44055 k
Assignment #7 Solutions
MAE 150/250R, S16, Prof. M. Spearrin
Problem 1.
el =+
(100 0.1Va )1.60 1019 J
Va
(g I )
=
2
e sp
2q m
0.5m ( g e I sp )
=
2
2
0.2 ( g e I sp )
1.60 1019
0.5m ( g e I sp ) + 100 +
2q m
q
Xenon: m =
2.18 1025 kg,
=7.34 105 C/kg
m
Assignment #3 Solutions
MAE 150/250R, S16, Prof. M. Spearrin
Problem 1.
Gas
Nitrogen
Hydrogen
Helium
Sum
mi
12
16
8
36
M
28
2
4
N
i
0.4286 0.0411
8
0.7671
2
0.1918
10.4286
1.0
i
i i 1
i ( i i 1)
3.5
3.5
2.5
0.1439
2.6849
0.4795
3.3083
1.4
1.4
1.67
Mixtu
Assignment #2 Solutions
MAE 150/250R, S16, Prof. M. Spearrin
Problem 1:
3
L
0.004 =
Po 1.5 MPa =
To 1000 K
= = 10 C
=
f
(A)
D 0.3
= 1.4 M = 29 M 1 < 1 M 2 = 1
4C f L*
= 4=
( 0.004 )10 0.160
D
From Fig 3-5: M 1 = 0.72
1
+1
P1
1 2 1 2 2 Po1 1 2 1 2 2( 1)
Assignment #6 Solutions
MAE 150/250R, S16, Prof. M. Spearrin
Problem 1
Initial steady buring (with o < p )
1
1 n
A
a p
b1
=
Po1 =
5 MPa
+1
A*
2 1
RTo + 1
Sudden change A b1 A b2
New steady pressure level:
1
A 1 n
Po 2 = Po1 b 2
Ab1
For n=0.86 an