CE309 Quiz #5
Name A fuel tank for a rocket in space under a zero-g environment is rotated to keep the fuel in one end of the tank. The system is rotated at 3 rev/min. The end of the tank (point A) is 1.5 m from the axis of rotation, and the fuel level is

IN UNSTEADY ONE DIMENSIONAL GAS DYNAMICS CHAPTER
14 Oblique-Shock 14.1 Preface to Oblique Shock = 0 Fig. 14.1: A
view of a straight normal shock as a limited case for oblique shock In
Chapter (5), discussion on a normal shock was presented. A normal
shock

temperature ratio in equation (5.58) and the rest of the righthand side
show clearly that Msx has four possible solutions (fourthorder
polynomial Msx has four solutions). Only one real solution is possible.
The solution to equation (5.58) can be obtained

repeat previous stage until the solution is obtained. M1 M2 4fL D up
4fL D down Mx My 3.0000 1.0000 0.22019 0.57981 1.9899 0.57910
(c) The way of the numerical procedure for solving this problem is by
finding 4fL D up that will produce M1 = 3. In the proc

flow must be parallel to the wall. For the first shock, the upstream Mach
number is known together with deflection angle. Utilizing the table or
the PottoGDC, the following can be obtained: 258 CHAPTER 14.
OBLIQUE-SHOCK Mx Mys Myw s w P0y P0x 4.0000 0.461

reduces the upstream flow velocity and therefore the shock does not
exist at close proximity to the wall. In larger distance from the wall, the
shock becomes possible. 14.4.8 Maximum Value of Oblique shock The
maximum values are summarized in the followin

P0y P0x 3.0000 0.47641 2.8482 88.9476 21.5990 3.0000 0.99879 For
the shock BC the results are Mx Mys Myw s w P0y P0x 2.8482
0.48610 2.7049 88.8912 22.7080 3.0000 0.99894 And the isentropic
relationships for M = 2.7049, 2.7008 are M T T0 0 A A? P P0 AP
AP0

wedge with zero angle of attack. The discussion so far was about the
straight infinite long wedgea which is a pure 2D configuration.
Clearly, for any finite length of the wedge, the analysis needs to account
for edge effects. The end of the wedge must hav

equation, the derivative isnt zero. Hence, the mass equation is reduced
to Ur + (U) = 0 (15.13) Equation (15.13) can be rearranged as
transformed into 1 U Ur + U = 1 (15.14) The
momentum equations now obtain the form of U r Ur U 2 r = 0
U Ur U = 0 (15.15)

increase of Mach angle (weak shock only) M1 > M2 = 1 < 2. The
Mach number decreases after every shock. Therefore, the maximum
deflection angle decreases with a decrease the Mach number.
Additionally, due to the symmetry a slip plane angle can be guessed t

1.18192 To reduce the pressure ratio the deflection angle has to be
reduced (remember that at weak weak shock almost no pressure change).
Thus, the pressure at zone 3 has to be reduced. To reduce the pressure
the angle of slip plane has to increase from 1

explanation for the combination of diverging-converging nuzzle with
tube for Fanno flow first appeared in Shapiros book? Meta End
expanding model by others Isothermal Flow The earliest reference to
isothermal flow was found in Shapiros Book. The model If

requires a larger energy to change the flow direction. Once, the
inclination angle reaches the maximum potential energy, a change in
the flow direction is no longer possible. In the alternative view, the fluid
sees the disturbance (in this case, the wedge

shock occurs at the same time. In this chapter, the stability issue will be
examined in greater detail. 14.3 Oblique Shock Comparsion Line - !#"
$ %'& ( ) * + ,-/. + 0 - 1#243 Fig. 14.3: A typical oblique shock
schematic The shock occurs in reality in sit

the tube. The only change will be at tube surroundings which are
irrelevant to this discussion. If the feeding nozzle is a converging
diverging then it has to be differentiated between two cases; One case is
where the 4fL D is short or equal to the critic

CHAPTER 7 Nozzle Flow With External Forces This chapter is under
heavy construction. Please ignore. If you want to contribute and add any
results of experiments, to this chapter, please do so. You can help
especially if you have photos showing these effec

people have helped me with this book, directly or indirectly. I would like
to especially thank to my adviser, Dr. E. R. G. Eckert, whose work was
the inspiration for this book. I also would like to thank Amy Ross for her
advice ideas, and assistance. The

extreme case, only in several points (depending on the bumps) at the
leading edge can a very weak shock occur. Therefore, for the purpose of
an introductory class, no Mach wave at zero inclination should be
assumed. Furthermore, if it was assumed that no

DH = 4 Cross Section Area wetted perimeter (9.7) Or in other words A
= DH 2 4 (9.8) 3The equation of state is written again here so that all
the relevant equations can be found when this chapter is printed
separately. 9.3. NONDIMENSIONALIZATION OF THE
EQU

1 + k1 2 M2 (15.8) After integration of equation (15.8) becomes (M)
= r k + 1 k 1 tan1 r k 1 k + 1 (M2 1) + tan1 p (M2 1) +
constant (15.9) The constant can be chosen in a such a way that = 0 at
M = 1. 15.2.1 Alternative Approach to Governing Equations ba

Problem 3.13
Apply the grid method to each situation below. Unit cancellations are not shown in
these solutions.
a.)
Situation:
Pressure varies with elevation.
z = 8 ft.
Find:
Pressure change (Pa).
Properties:
air, = 1:2 kg= m3 .
Solution:
p= g z
p =
g z

3.72: PROBLEM DEFINITION
Situation:
A plug sits in a hole in the side of a tank.
z = 2 m, ro ring = 0:2 m, rplug = 0:25 m.
Find:
Horizontal and vertical forces on plug.
Properties:
Water, Table A.5:
= 9810 N= m3 .
SOLUTION
Hydrostatic force
Fh = pA
= zA
=

1.10: PROBLEM DEFINITION
Note: Student answers will vary. The CT process format (Issue/Reasoning/Conclusion)
should be used.
An example answer is provided here.
Issue:
A lift force on an airfoil is caused by air pressure on the bottom of the wing relative

Jennifer Berrios & Glenda Rodriguez
Fountain 1: Principle of Continuity:
m 1=m
2
A 1 V 1= A 2 V 2
t avg=
.33+.43+.32+.48+.34+.43+.33+ .50+.29+.50+.27+.43+.29+.46
=.375
14
t=0.38 s
y=35 2.92 ft
y=V 0 y t
( 12 ) g t
2
2.92 ft=V 20.38 s
( 12 )( 32.2s ft )(

NAME:
CE 309
Instructor: Dr. Patrick Lynett
EXAMINATION #1
1) [25 points] A thin 20-cm by 20-cm at plate is pulled at 1 this horizontally through a 3.6mm-
thick oil layer sandwiched between two walls. The top wall is xed (stationary) while the bottom

a suction section at M = 2.0, P = 1.0[bar], and T = 17C. Compare the
different conditions in the two 252 CHAPTER 14. OBLIQUE-SHOCK
different configurations. Assume that only a weak shock occurs. 7
/Duy / e/C/ 7 Normal shock
neglect the detached distance

examples will appear in the later versions. 7Those who are
mathematically inclined can include these kinds of questions but there
are no real world applications to isothermal model with shock. 148
CHAPTER 8. ISOTHERMAL FLOW Example 8.1: A tube of 0.25 [m]

then Uy = My cy = 0.75593 1.4 287 369.24 291.16[m/sec]
Example 6.6: A flow of gas is brought into a sudden stop. The mass flow
rate of the gas is 2 [kg/sec] and cross section A = 0.002[m3 ]. The
imaginary gas conditions are temperature is 350K and pressur

unstable but rather that the model is incorrect. There isnt any known
experimental evidence to show that flow is unstable for = 0.
CHAPTER 15 Prandtl-Meyer Function 15.1 Introduction positive angle
maximum angle Fig. 15.1: The definition of the angle for

kind of pressure ratio the perpendicular component has to be Mx My Ty
Tx y x Py Px P0y P0x 1.0008 0.99920 1.0005 1.0013 1.0019 1.00000
The shock angle, can be calculated from = sin1 1.0008/2.7049 =
21.715320879 The deflection angle for such shock angle wi