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+
=
w
c
p
pmuzzle
S
3
1
(36)
Inserting the numbers yields
( )
( )( )
=
+
=22 in
lbf
904,10
in
lbf
485,13
8.313
1
672,11
muzzle
S
p
(37)
To plot the pressure behind the projectile at both peak pressure and muzzle exit we must use
equation (LG-59)
w
c
x
x
ppp g
ss 2
12
2
−+=
(LG-59)
Where xg is the position behind the projectile varying between 0 and x. For peak pressure this
variation is from 0 to xm and for muzzle exit this variation id from 0 to L. plotting the result for
the full charge we have
050 100 150
35000
36000
37000
38000
39000
pmz( )
z
0200 400 600
10800
11000
11200
11400
11600
11800
pmuzm z( )
z
( )
( )( )( )
408.0
1.02658.1
11.0658.1
2
1=
+
−+
=
+
−+
=
M
M
fm
(42)
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
485,12
35.235
1
ftlbm
slbf
2.32
1
lbm485,1
in
lbf
s
in
000298.0
408.01in165.0in9.153
2
2
2
max
p
V
(52)
=s
ft
014,1
max
p
V
(53)
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
485,12
35.235
1
ftlbm
slbf
2.32
1
lbm485,1
in
lbf
s
in
000298.0
01in165.0in9.153
2
2
2
c
V
(54)
=s
ft
712,1
c
V
(55)
( )
−
+
+
−
=
−
1
991.8627.335
991.868.540
23.11
223.0
(56)
758.0=
(57)
( )
( )( )
=
+
=22 in
lbf
248,9
in
lbf
485,13
35.235
1
737,9
muzzle
S
p
(62)
( )
( )
3
3
3in721,14
in
lbm
059.0
lbm90.156
in380,17V=
−=
i
(64)
( )
( )
( )
( )( )
( )
( )( )
( )
( )
( ) ( )
−
−
−
+
+
=
4
2
2
2
2
2
2
2
2
4
2
in
lbf
s
in
000298.0
ftlbm
slbf
2.32
1
lbm
lbfft
000,365lbm90.156lbm485,1
485,12
90.156
1
485,13
90.156
1
in165.0in9.153
M
(65)
569.2=M
(66)
( )
( )( )( )
603.0
1.02569.2
11.0569.2
2
1=
+
−+
=
+
−+
=
M
M
fm
(67)
( ) ( ) ( )
( )
( )
( )( )
( )
( )( )
( ) ( )
( )( )( )
+
++
+
+
−
=+
+
2
1.0
569.2
1
1.0
569.2
2
3
1.02569.2
1.0569.21.01
485,13
90.156
1
485,12
90.156
1
in14,721
lbm90.156
ft
in
12
lbm
lbfft
000,365
max
B
p
(68)
=2
in
lbf
773,7
max
B
p
(69)
At this point I checked the distance to charge burnout and got a very large number so the
1
1
−
+
=
M
L
f
(71)
in629.95
in9.153
in721,14
3
3
==l
(72)
218.0
1.0
1
629.95
629.958.540
1.0
1.01
569.2
1.0 =−
+
+
=
L
f
(73)
Now we insert this into equation (LG-133) to determine the breech pressure when the projectile
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
485,12
90.156
1
ftlbm
slbf
2.32
1
lbm485,1
in
lbf
s
in
000298.0
603.01in165.0in9.153
2
2
2
max
p
V
(79)
=s
ft
697
max
p
V
(80)
( )
( )
( )
( ) ( )
( )
( )( )
+
−
−
−
=
485,12
90.156
1
ftlbm
slbf
2.32
1
lbm485,1
in
lbf
s
in
000298.0
218.01in165.0in9.153
2
2
2
muzzle
V
(81)
=s
ft
373,1
muzzle
V
(82)
050 100 150 200
7300
7400
7500
7600
7700
7800
pmz( )
z
3.3 Lagrange Gradient for Spherical and Cubic Grains
Problem 17 - Below are the data for the 3 different powders used in the 8-inch rifled muzzle loader.
Initial data was provided by Kent Crawford. Some data was provided in the literature brought to the
attention of the authors by Byron Angel [9].
Assuming the following
Projectile weight 180 lbm
Propellant weight See table
Projectile and bore diameter 8 inches (in actuality the projectile diameter was
7.92 inches)
Adiabatic flame temperature 2600 K
Co-volume 5.02-5.68 in3/lbm
Specific Heat Ratio 1.24
Burn Rate Coefficient See table
Propellant Force See table
Chamber Length See table
Projectile Travel See table
0200 400 600
5400
5500
5600
5700
5800
pmuzm z( )
Pebble
0.625”
cube [9]
35
1.8
15.3
0.5
1,391
RLG
0.14”
diameter
sphere
[9]
30
1.77
n/a
n/a
1,330
Plot for Pebble Powder
010 20 30 40 50 60 70 80 90 100
0
3500
7000
10500
14000
17500
21000
24500
28000
31500
35000
Pressure (psi)
31788
0
p.B f( )
930x f( )
27000
31500
36000
40500
45000
42047
