( )( )
( )
( )( )
( )
=
=
== s
m
661
lbm
kg
2.2
1
lbm25.0
s
mkg
800,242
lbm
kg
2.2
1
lbm25.0
kJ8.242
22
2
m
W
Vp
This is over twice the speed of sound! To get a more realistic answer let’s look at the ideal gas
equation of state. If the projectile doesn’t move, the volume behind it is
( )
( )
( )
34
3
3
3
2
2
m106.4
in
m
0254.0in4in
4
3
V−
=
==
Al
i
The volume of the exhaust pipe when the potato is at the end is twice this value. The initial
pressure is calculated from the ideal gas equation of state then using this volume
V
VTNR
pTNRpu
u=→=
( )
( ) ( )
( )
kPa280,15
m106.4
K500,1
Kkgmol
kJ
314.8kgmol1036.56
34
5
=
=−
−
p
( ) ( )
−
−
−
+=
+
=
l
c
l
cdxlx
l
pA
dx
l
lx
pAlmV
00
22
2
1
l
l
0
0
1
2
−
−
−
Inserting the limits of integration we obtain
( ) ( )
−
−
−
−
−
2
2
11
1
1
2ll
ll
pA
c
pcair V–VV =
(3)
333 in677in483in1160V=−=
air
(4)
The number of moles of air then can be found from the ideal gas equation of state as follows
airu
air
air TR
p
NV
=
(5)
2,605.5 Btu/lbmol. The tables in the book are in kJ/kgmol so we have some unit conversions to
do.
( ) ( ) ( )
=
=kgmol
kJ
046,6
kgmol
lbmol
2.2
Btu
kJ
055.1
lbmol
Btu
5.605,2
9286 ONHC
0
f
h
(11)
We can change this over to internal energy (because there was no flow work by using the
definition of enthalpy, thus
( ) ( ) ( )
=
−
=kgmol
kJ
568,3K298
Kkgmol
kJ
314.8
kgmol
kJ
046,6
R
U
We can put equation (9) on a per mol of propellant basis by dividing by 0.115. We then have
( ) ( ) ( ) ( ) ( ) ( )
sC496.2N007.1CO504.3O4HN00687.0O0018.0ONHC 222229286 +++→++
(12)
To get everything in gmol we would multiply by 52.27 but we will hold off on this for now.
2.6 Solid Propellant Combustion
Problem 11 – M1 propellant is measured in a closed bomb. Its adiabatic flame temperature is
3906 F. Its molar mass is 22.065 lbm/lbmol, what is the effective mean force constant in ft-
lbf/lbm?
lbf–ft
Solution: This is just a straight use of the definition of the force constant from the notes.
( ) ( )
R4603906
R–lbmol
lbf–ft
1545
lbm
lbmol
065.22
1
0+
== TnRu
=lbm
lbf–ft
709,305
Problem 12 – M15 propellant was tested in a strand burner to determine the linear burning rate.
The average pressure evolved was 10,000 psi. If the burning exponent,
was known to be 0.693
and the pressure coefficient,
was known to be 0.00330 in/s/psi0.693, determine the average
linear burning rate, B in in/s.
( ) ( ) ( )
=
693.0
2
693.0
0.693
2
in
lbf
000,10
in
lbf
s
in
00330.0pB
=lbm
709,305
( )
=s
in
952.1pB
Problem 13 – . Please derive the functional form of
in terms of f for a flake propellant.
Assume cylindrical geometry.
Hint Flake propellant consists of grains that have thicknesses muck smaller than any other
characteristic dimension.
Answer
ft −= 1)(
4
4
where, D is the diameter of the grain and
is the thickness (length). Then the initial weight of
the propellant grain is
4
2
D
gc graininit =
(3)
the geometry is not
circular for that matter) is
insignificant with respect
to the loss of volume due
to the thickness change.
The weight of the
unburnt propellant grain at time t is given by
f
D
gtc grain 4
)(
2
=
(4)
As stated earlier
)(tccc initgas −=
(5)
Substitution of equations (3) and (4) into (5) yields
( )
f
D
gf
D
gc graingraingas −=−= 1
44
22
(6)
We now introduce the fraction of gas generated
. Initially
= 0 and
= 1 at “all burnt”. This
function is defined as
init
gas
c
c
t=)(
(7)
Substitution of equations (3) and (6) into (7) yields the desired relation
ft −= 1)(
(8)
We need to note that this formulation will work for any geometry of flake propellant where the
thickness is much smaller than any other dimension.
Problem 14 – An M60 projectile is to be fired from a 105mm M204 Howitzer. The propellant
used in this semi-fixed piece of ammunition is 5.5 lbm of M1 propellant. M1 propellant consists
of single perforated grains (
= 0) with a web thickness of 0.0165 inches. If the average pressure
(over the launch of this projectile) developed in the weapon is 20,455 psi. Calculate the average
burning rate coefficient in in3/lbf-s if the burn rate is (we use a negative sign in the burn rate to
make the form come out right later)
1–
s9.185−=
dt
df
( ) ( )
s–lbf
in
lbf
20455
2
Problem 15
is actually a function of pressure and temperature (it is really given in tables at
25F at this value). For simplification (and illustration) we will assume it is constant. Given this
assumption, calculate the functional form of the web fraction, f from problem P 2.6.4 , above.
p
t = 0. Performing this task and substituting the constant back into the above equation yields the
( ) ( ) ( )
tmRTtmtp ggB
== 0
V
(SP-23)
( )
( )
tm
tp
g
BV
=
Since all of the material is presumably gas at the end of the combustion we can simply insert
values into this equation
( ) ( )
( )
=
grains
lbm
000,7
1
grains003189.0
in
ft
12
1
in5.0
in
lbf
3.706 3
2
=lbm
lbf–ft
000,339
A note here about why we used the gage pressure. This is because the bomb is usually evacuated
to obtain a true measure – if there was air in there we would have accounted for that too. The
burn rate coefficient can be approximated from equation (SP-38)
c
D
tt
D
c
BB
V
350
V
2
700 =→=
(SP-38)
ct
D
B
V
350
Inserting our numbers we have
( ) ( )
( )
( )
( )
( ) ( ) ( )
grains003189.0
ft
in
12
lbm
lbfft
000,339s063.0
lbm
grains
000,7in5.0in0.01
350
3
−
1.) Ideal gas behavior – even though in the burn portion the Noble-Able equation of state
2.) Combustion products are generated at the adiabatic flame temperature of the
3.) No heat transfer occurred to the walls of the container – adiabatic behavior – There is
4.) The combustion of the propellant grain neglected end effects – so the propellant will
2.7 Fluid Mechanics
Problem 18 – The principle behind a muzzle brake on a gun is to utilize some of the forward
momentum of the propelling gases to reduce the recoil on the carriage. In the simple model
below, the brake is assumed to be a flat plate with the jet of gases impinging upon it. If the jet
diameter is 105mm and the velocity and density of the gas (assume air) are 750 m/s and 0.457
kg/m3, find the force on the weapon in Newtons assuming the gases are directed 90 to the tube
and the flow is steady.
F
( ) ( ) ( )
N9.2225m
4
105.0
s
m
750
m
kg
0.457 2
2
2
2
2
3−=
−=
x
F
Problem 19 – Some engineer gets the idea that if deflecting the muzzle gases to the side is a good
idea, then deflecting it rearward would be better (until of course an angry gun crew gets hold of
him!). If the jet diameter is again 105mm and the velocity and density of the gas (again assume
air) are 750 m/s and 0.457 kg/m3, find the force on the weapon in Newtons assuming the gases
are directed 150 to the tube and the flow is steady.
F
30
(typ.)
2
outoutinxinx
It is sometimes helpful to write this equation as
( )( ) ( ) ( )
+= 150cos
2
20cos out
outininx
V
mVmF
Now the answer is
( ) ( ) ( )
2
2
2
2
2
3m
4
105.0
s
m
750
m
kg
0.457–
=
x
F
4
105.0
s
m
m
kg
2
2
2
2
3−=−
Problem 20 – Consider a shock tube that is 6 feet long with a diaphragm at the center. Air is
contained in both sections (γ = 1.4). The pressure in the high pressure region is 2,000 psi. The
pressure in the low pressure region is 14.7 psi. The temperature in both sections is initially 68°
F. When the diaphragm is burst
Determine
a) The velocity that the shock wave propagates into the low pressure region.
Answer 2,798 ft/s
b) The induced velocity behind the wave.
Answer 1,946 ft/s
c) The velocity of a wave reflected normally off the wall (relative to the laboratory).
Answer 1,232 ft/s
d) The temperature behind the incident wave.
Answer 657F
e) Draw an x-t diagram of the event. Include the path of a particle located 2 feet from the
diaphragm.
RTaa
== 41
( )( ) ( )
=+
−
−
== s
ft
126,1R46068
Rslug
lbfft
716,14.1
41 aa
We need to iterate to determine the static pressure behind the wave. Since p2 has to lie between
p1 and p4 we shall use 20 psi as a starting point. The equation is:
( )
( )
1
2
2
1
2
4
1
4
1
2
1
4
4
4
11
1
−
−
−++
−
−
−=
p
p
p
a
a
p
p
p
p
Now we need to see how this would reflect off of the rear wall. We will first find the incident
wave Mach number
94.101422.1
2=→=−− RRR MMM
Thus
==+=T
+
+
+
+
7.14
14.1
1
1
p
ft
( ) ( )
( ) ( )
( )
( )
703.0
47.2
1
4.1147.2
14.1
14.12
1
147.2
47.2
12
2
222 =
+−
+
−
+
−
=
−
R
M
M
115.2
7.14
793.102
14.1
14.1
1
7.14
14.1
7.14
793.102
1
1
1
1
1
2
1
1
2
1
2=
−
+
+
+
−
=
−
+
+
−
=
p
p
p
p
p
T
T
Determine a.) The static pressure behind the wave (assume γ = 1.4 and since
we are far away from the effects of the explosion assume a1/a4 ≈
0.5)
b.) The velocity that the wave propagates in still air
c.) The induced velocity that a building would see after the wave passes
d.) The velocity of a wave reflected normally off a building
( )( ) ( )
=+
−
−
=s
ft
1126R46068
Rslug
lbfft
17164.1
1
a
We can now calculate the speed of propagation of the wave through still air as
This is a pretty respectable speed. We can now calculate the induced velocity behind the wave.
This high of a velocity would certainly make itself felt. Now we need to see how this would
reflect off of a building. We will first find the incident wave Mach number
38.20196.1
2=→=−− RRR MMM
Thus
( ) ( )
( )( )
=+
−
+
=+
−
+
=s
ft
529411
7.14
6.376
4.12
14.1
s
ft
112611
2
1
1
2
1p
p
aU
( )
( )
( )
=
+
+
−+
−
=
+
+
−
+
−= s
ft
4212
7.14
6.376
14.1
14.1
14.1
4.12
1
7.14
6.376
4.1
s
ft
1126
1
1
1
2
1
1
2
1
21
p
p
p
pa
up
ft
( ) ( )
( ) ( )
( )
( )
510.0
69.4
1
4.1169.4
14.1
14.12
1
169.4
69.4
12
2
222 =
+−
+
−
+
−
=
−
R
M
M
24.5
7.14
6.376
14.1
14.1
1
6.376
14.1
7.14
1
1
1
1
1
2
2
1
1
−
+
+
+
+
=
−
+
+
+
+
=
p
p
p
p
T
==+=T