3.2 Lagrange Gradient
Problem 1 – You are asked to analyze the pressure of a charge zero (igniter) firing in an M31
boom for a 120mm mortar projectile. You decide to examine it as a closed bomb first. Assume
we have 59 g of M48 propellant (properties given below). The volume of the closed bomb is
5.822 in3. The propellant grains are ball (roughly spherical) with a diameter (web) of 0.049
inches.
M48 propellant properties
Density
=3
in
lbm
056.0
Ratio of specific heats
21.1=
Covolume
=lbm
in
72.26
3
b
Isochoric flame temperature
F3720
0
=T
Burn rate exponent
9145.0=
Average burn rate coefficient
−
=psis
in
0095.0
Burn rate
=s
in
341.40
dt
df
D
Force constant
=lbm
lbf–ft
000,391
a.) Come up with the equation for the web fraction, f, as a function of time.
Answer
( )
%29.8231tf −=
b.) For a sphere the fraction of propellant burnt has the functional form
3
1f−=
, write
this in terms of time and
.
Answer
32 251,031,558419,033,2470,2 ttt +−=
c.) Determine how long it will take the propellant to burn halfway through and all the
way through.
Answer Time to burn through halfway is 0.6 milliseconds
d.) Using the Noble-Abel equation of state, determine the pressure in the vessel when
half of the propellant is burnt and when all of the propellant is burnt. Note that this
can not usually occur as the propellant is a charge zero firing is vented into the main
ullage volume behind the mortar bomb (significantly greater volume.
lbf
in0.049
dt
Here the negative sign is required since f must decrease with time.
( )
%29.823 Ctf +−=
or 0.6 milliseconds
If the problem statement said that we have used up half the propellant, that would indicate that
s00061.0
29.823
5.01 =
−
=t
s00025.0
29.823
794.01 =
−
=t
or 0.25 milliseconds
Problem 2 – If we use the Lagrange approximation in examination of a 155mm projectile launch,
what is the average pressure in the volume behind a 102 lbm projectile if the breech pressure is
55,000 psi? The propelling charge weighs 28 lbm.
lbf
We need to calculate the effective length of the chamber for use in our calculations so we
assume the chamber has the same cross-sectional area as the bore of the weapon. We
( )
( ) ( )
( )
( )
( )( )
( )( )
+
+
+
=
+
+
+
=
4.503
25.12
1
4.502
25.12
1
in82.1848in53.17
41.0
lbm
in–lbf
000,044,4lbm25.12
3
1
2
1
2
w
c
w
c
lxA
c
pB
=2
in
lbf
990,17
B
p
This is not too far off the mark.
Problem 4 – The Paris gun was a monstrous 210mm weapon designed by Germany during the
First World War to bombard Paris from some 70 miles away. It was unique in that it fired the
first exo-atmospheric projectile ever designed. The weapon had a chamber volume of 15,866
in3. Very little of the projectile protrudes into the chamber after it seats (so ignore the volume
the base occupies). The length of travel for the projectile from shot start to shot exit is 1,182 in.
The projectile weighs 234 lbs. The propelling charge weighs 430.2 lbs. The propellant used was
specially designed and was similar to U.S. M26 propellant. It consisted of 64-68% NC, 25-29%
NG with 7% Centralite (symmetrical diethyl diphenylurea C17H20N2O) and some other additives.
The propellant was single perforated with a web thickness described below. Assume the
propellant has the following properties (Note that these are the Authors guesses – a better
estimate of the properties can be found in reference [6]).
Adiabatic flame temperature T0 = 2881 K
Specific heat ratio γ = 1.237
Co-volume b = 1.06 cm3/g
Density of solid propellant
= 1.62 g/cm3
Propellant burn rate coeff.
= 0.0707 (cm/s)/(MPa)
Web thickness D = 0.217 in
Propellant force
= 1,019 J/g
a.) Using the above data determine the projectile base pressure in psi, velocity in ft/s and
distance down the bore of the weapon in inches for peak pressure
lbf
max =
p
b.) Determine the pressure in psi at a point 3” behind the projectile base when the charge
−2
3in
x
c.) Assuming the gas behaves according to the Noble-Abel equation of state, determine the
muzzle velocity of the projectile in ft/s
ft
Solution:
The central ballistic parameter is found directly by plugging in the given values from the
problem statement. First we must calculate the bore area. Since no other information was given
1== ww
lbm2.430=c
We also need to get the burn rate coefficient into consistent units as follows:
( )
=
=
22 in
lbf
s
in
000192.0
in
lbf
MPa
145
1
cm
in
54.2
1
MPa
s
cm
0707.0
Then the central ballistic parameter is:
+
+
=2
1
1
2
1
22
2
1
3
1
w
c
w
c
cw
DA
M
( )
( )
( )
( )( )
( )
( )( )
( )
( )
( ) ( )
−
−
−
+
+
=
4
2
2
2
2
2
2
2
2
4
2
in
lbf
s
in
000192.0
ftlbm
slbf
2.32
1
lbm
lbfft
597,341lbm2.430lbm234
2342
2.430
1
2343
2.430
1
in217.0in69.53
M
510.1=M
0.
( )
( )( )( )
338.0
02510.1
10510.1
2
1=
+
−+
=
+
−+
=
M
M
fm
0=
c
f
( )
( )
( ) ( )
3
3
3
3
3
3
base0 in500,80
in
cm
54.2
g
lbm
1000
2.2
cm
g
1.62
lbm2.430
in866,15VVV =−
−=−−=
c
i
( )
( )
fM
i
Bef
w
c
w
c
c
p−−
−
+
+
=1
1
11
3
1
2
1
V
At peak pressure the pressure at the breech is
( ) ( ) ( )
( )
( )
( )( )
( )
( )( )
( )
( )( )
338.015.1
3338.01
2343
2.430
1
2342
2.430
1
in8,500
lbm2.430
ft
in
12
lbm
lbfft
597,341
max
−−
−
+
+
−
=epB
=2
in
lbf
547,60
max
B
p
This is a breech pressure value so we need to see what the value is at the base of the projectile so
we use the Lagrange gradient
w
c
p
pB
s
2
1+
=
So
( )
( )( )
=
+
=2
2
in
lbf
548,31
2342
2.430
1
in
lbf
547,60
max
s
p
Now let’s look at the velocity
( )
+
−
=
1
12
1
1
w
c
w
fAD
V
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
2342
2.430
1
ftlbm
slbf
2.32
1
lbm234
in
lbf
s
in
000192.0
338.01in217.0in69.53
2
2
2
max
p
V
=s
ft
880,2
max
p
V
Keep in mind that this number is the velocity at peak pressure NOT the muzzle velocity.
Now we need to determine where the projectile is in the bore at maximum pressure. We first
i
Bef
w
c
+
1
3
1
V
( ) ( ) ( )
( )
( )
( )( )
( )
( )( )
( )
( )( )
0151.1
301
2343
2.430
1
2342
2.430
1
in8,500
lbm2.430
ft
in
12
lbm
lbfft
597,341
−−
−
+
+
−
=ep c
B
=2
in
lbf
539,54
c
B
p
This is a breech pressure value but we need to see what the value is at the base of the projectile
so we use the Lagrange gradient again
w
c
p
pB
s
2
1+
=
So
( )
( )( )
=
+
=2
2
in
lbf
417,28
2342
2.430
1
in
lbf
539,54
c
s
p
0. Thus, at burnout:
( )
llexfM −= −1
( )
( )( )
in26.157in26.157 0151.1 −= −
exc
in6.554=
c
x
Since we have both the breech pressure and the base pressure we can use this and knowledge of
the Lagrange gradient to determine the pressure at any point behind the base of the projectile.
We know the breech pressure at burnout was
=2
in
lbf
539,54
c
B
p
We also know that the base pressure at charge burnout was
=2
in
lbf
417,28
c
s
p
We write
w
c
x
x
ppp SS 2
12
2
3
−+=
Inserting values we get
( )
( ) ( )
( )( )
=
+
+
−+
=
−2
2
2
2
3in
lbf
637,28
2342
2.430
26.1576.554
26.1576.551
11
in
lbf
417,28
x
p
We now assume that heat transfer to the barrel is negligible and that means that our pressure
( )
+
−
=
1
12
1
1
w
c
w
fAD
V
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
2342
2.430
1
ftlbm
slbf
2.32
1
lbm234
in
lbf
s
in
000192.0
01in217.0in69.53
2
2
2
c
V
=s
ft
351,4
c
V
Then we have
−
+
+
−
+
+
=−
−
1
1
3
21
1
22
lx
lx
lx
c
w
pA
VV
c
cc
c
( ) ( )( )
( )
( )
( )
−
+
+
−
+
−
−
+
=
−
−
1
26.1576.554
26.157182,1
in
237.11
26.1576.554
ftlbm
slbf
2.32
1
3
lbm2.430
lbm234
in
ft
12
1
in
lbf
832,45in69.532
s
ft
351,4
237.11
2
2
2
2
2
2
V
=s
ft
791,5V
Problem 5 – A British 14 inch Mark VII gun has a chamber volume of 22,000 in3. 5” of the
projectile protrude into the chamber after it seats. The length of travel for the projectile from
shot start to shot exit is 515.68 in. The weapon has a uniform twist of 1 in 30. The projectile
weighs 1590 lbs. The propelling charge weighs 338.25 lbs. The propellant used is called “SC”
and consists of 49.5% NC (12.2% nitrated), 41.5% NG with 9% Centralite (symmetrical diethyl
diphenylurea C17H20N2O). Assume SC propellant has the following properties:
Adiabatic flame temperature T0 = 3090 K
Specific heat ratio γ = 1.248
Co-volume b = 26.5 in3/lbm
Density of solid propellant
= 0.0567 lbm/in3
Propellant burn rate
= 0.000331 (in/s)/(psi)
Web thickness D = 0.25 in
Specific molecular weight n = 0.04262 lb-mol/lbm
a.) Determine the force constant,
in ft-lbf/lbm
b.) Determine the central ballistic parameter for this gun/projectile combination.
c.) Using the above data determine the projectile base pressure, velocity and distance down
the bore of the weapon for both peak pressure and charge burnout assuming the grain is
cylindrical propellant (θ = 1)
Answers
−
=lbm
lbfft
246,366
933.1=M
=2
in
lbf
200,41
max
B
p
=2
in
lbf
080,25
c
B
p
=2
in
lbf
240,37
max
s
p
=2
in
lbf
670,22
c
s
p
=s
ft
082,1
max
p
V
=s
ft
128,2
c
V
in5.79
max =
p
x
in7.293=
c
x
Solution:
0
TNRu
=
Looking at this you might be scared that I gave you n in the problem statement and here you see
N. Don’t worry, I asked for the units in ft-lbf/lbm. Remember that n is nothing more than N per
unit mass. In this field you need to be careful that the units come out correct because the
we shall estimate
( )
22
2
2
in94.153in
4
14
4===
d
A
( )
( )
( )
( )( )
( )
( )( )
( )
( )
( ) ( )
−
−
−
+
+
=
4
2
2
2
2
2
2
2
2
4
2
in
lbf
s
in
000605.0
ftlbm
slbf
2.32
1
lbm
lbfft
246,366lbm25.338lbm1590
15902
25.338
1
15903
25.338
1
in25.0in94.153
M
579.0=M
=
2
in
lbf
919,71
max
B
p
And at “all burnt” the pressure at the breech is
( ) ( ) ( )
( )
( )
( )( )
( )
( )( )
( ) ( )( )( ) ( )( )( )
( )
1
579.0
311
011
01101
15903
25.338
1
15902
25.338
1
in15,264
lbm25.338
ft
in
12
lbm
lbfft
246,366
+
+
+−
+
+
−
=
c
B
p
=2
in
lbf
355,67
c
B
p
These were breech pressure values we need to see what the values are at the base of the projectile
so we use the Lagrange gradient
w
c
p
pB
s
2
1+
=
So
( )
( )( )
=
+
=2
2
in
lbf
005,65
15902
25.338
1
in
lbf
919,71
max
s
p
( )
( )( )
=
+
=2
2
in
lbf
879,60
15902
25.338
1
in
lbf
355,67
c
s
p
Now let’s look at velocities
( )
+
−
=
1
12
1
1
w
c
w
fAD
V
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
15902
25.338
1
ftlbm
slbf
2.32
1
lbm1590
in
lbf
s
in
000605.0
224.01in25.0in94.153
2
2
2
max
p
V
=s
ft
6.903
max
p
V
and for the velocity at “all burnt” we have
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
15902
25.338
1
ftlbm
slbf
2.32
1
lbm1590
in
lbf
s
in
000605.0
01in25.0in94.153
2
2
2
c
V
=s
ft
164,1
c
V
Keep in mind that this number is the velocity at charge burnout NOT the muzzle velocity.
f
+
1
( )
( )( )
−
+
+
=1
224.011
11
in16.99
1
579.0
max
p
x
in61.32
max =
p
x
This is 1/16 of the way down the bore. Thus we have a pretty spikey p-t curve
( )
( )( )
−
+
+
=1
011
11
in16.99
1
579.0
c
x
in97.48=
c
x
This means that the propellant burns out relatively quickly and the only thing pushing on the
projectile after this point is the expanding (and cooling) gas.
This would wrap up our problem as stated but we may not be quite satisfied with the answer.
We burned out all of the propellant but the projectile only moved about 50 inches. This is
typical of propellants with very low central ballistic parameter. They have a very “spikey”
pressure time curve wherein the pressure rises very quickly and then drops down fast.
So what do we do now? Well, we can assume that heat transfer to the barrel is negligible
+
+
−
−
+
+
=−
−
1
1
22 1
1
3
2
lx
lx
lx
c
w
Ap
VV
c
cc
c
If we used the above we get
( ) ( )( )
( )
( )
( )
+
+
−
−
+
−
−
+
=
−
−248.11
2
2
2
2
2
2
16.9997.48
16.9968.515
1in
1248.1
16.9997.48
ftlbm
slbf
2.32
1
3
lbm25.338
lbm1590
in
ft
12
1
in
lbf
355,67in94.1532
s
ft
164,1V
=s
ft
676,2V
Or we could numerically integrate. To do this later step we would need to solve the equation of
motion for the acceleration
m
Ap
m
F
amaFii
avgi==→=