ROCKET PROPELLANT TANK DESIGN
MAE 4262: Rockets and Mission Analysis
Introduction
One area that is important to the design of rocket propulsion systems is the design and
sizing of the propellant tanks. This structural issue is not covered in Hill & Peterson,
which focuses primarily on modeling the fluid and thermodynamic aspects of rocket
propulsion. This is appropriate because it is these features that set rockets apart from
most other devices. On the other hand it is not possible to understand the characteristics
and limitations of rocket systems without at least a rudimentary understanding of their
structural characteristics, which determine mass, durability, etc. This document develops
a simplified model for one such aspect: the sizing and scaling of the propellant tanks.
Model Development
Consider the “pill-shaped” propellant tank shown in the figure below:
D
L
Figure 1: Simplified Propellant Tank Representation
The tank is made of a cylinder of length, L, and diameter, D. The tank is capped with two
spherical end-caps, each also of diameter, D. The tank has a wall thickness that is
everywhere equal and defined using the variable, t. There are two primary stresses that
act on this tank, which are in the circumferential, y, and axial, x, directions.
Consideration of the stress states through the application of a free body diagram shows
that the circumferential stress is the limiting stress, and it is for this case that we must
design. The circumferential stress is given in Equation 1.
1
y 
pc D
2t
1
In this equation, pc, is the pressure inside the tank, D is the diameter, and t is the wall
thickness. We also make the assumption that the thickness of the tank is much less than
the diameter of the tank, t<<D. The mass of the case alone is:
mcase  mcylinder  mcaps
2.1
mcase  Vcase  case
2.2
2.3
2

 D  

mcase  DLt  4   t  case

 2  

mcase  DL  D 2 t case
L p

mcase  D 3 1   c  case
 D  2 y
2.4
2.5
In equation 2.5, we have substituted the circumferential stress for the case thickness,
Equation 1 into Equation 2.4. The mass of the propellant is found as:
m prop  V prop  prop
3.1
 D 2 L 4  D  3 
m prop  
     prop
 4
3  2  

 D 2 L D 3 
 prop
m prop  

6 
 4
3.2
1 1 L 
m prop  D 3  
 prop
6 4 D
3.4
3.3
Here we have used, , to denoted the fraction of the case volume filled by propellant.
Next, we can express the ratio of the mass of the case divided by the mass of the
propellant as:
mcase
m prop
L p

31   c  case
 D  2 y

 3 L
1 
 prop
 2 D
4
Next, examine some typical numbers.
Example 1: Solid Rocket:
Consider a propellant tank with chamber pressure of 50 atm., which is about 750 psi. A
typical value of the circumferential stress is around y~150,000 psi, and the density of
2
steel for the case material is case~0.3 ln/in3, and the density of a solid propellant,
prop~0.06 lb/in3, with a propellant fill factor of =0.8. Putting these representative
numbers in equation 4 we find that:
mcase
m prop
L  750

0.3
31  
D  2150000


 3 L
1 
0.80.06
 2 D
L
1
mcase
D
 0.93
3 L
m prop
1
2D
5.1
5.2
For an L/D of 3, the ratio of the case to propellant mass is 0.034, or 3.4 percent. The next
question we might ask is, is this a reasonable estimate? Here is some data from Sutton
and Ross on the Minuteman missile:
Table 1: Minuteman Missile Data
mcase
mprop
mcase
m prop
2557 lb
45831 lb
0.056
It seems that our estimate is reasonable, especially given that we have left out all other
fittings, attachments, tank connectors, hoses, pipes, etc.
Example 2: Pressure-Fed N2H4, N2O4
For this example, take the chamber pressure, pc=20 atm., which is about 300 psi, and the
density of the propellant to be 62 lb/ft3, and using a stainless steel tank, case=500 lb/ft3,
and the circumferential stress to be around 50,000 psi. Again using equation 5.1, we can
show that this results in:
mcase
m prop
L
D
 0.097
2 L

3 D
1
6
This value turns out to be around 10 percent, which is pretty large considering that we are
trying to save as much weight as possible. Also keep in mind that the mass of the tank
does not contribute at all to the propulsion of the vehicle. We would be very interested in
trading tank mass for extra propellant mass.
Example 3: Design for Thrust Load
The force is equal to the pressure times the area, or as expressed below:
3
 D 2 

F  p
 4 
7
Let us also suppose that the acceleration of the vehicle is twice that of gravity, or 2 g’s,
then we have:
 D 2 

F  2m prop  p
 4 
m prop
p 8
D 2
1 1 L 
 propD 3  

6 4 D

p 8
D 2
8.1
8.2
8.3
We can again express the ratio of the tank mass to the propellant mass combined with
equation 8.3 to yield:
2  prop D 
mcase

L
 2 case
1  
m prop
 prop

D

mcase 4 D case 
L

1  
m prop
  D
9.1
9.2
Using the same numbers from example 2, we can show:
mcase 40.3
D  L 

m prop 50000
10
For a reasonable sized vehicle, say L+D=100 ft, this gives a ratio of 0.029 or 3 percent.
Note that the scaling gets worse as the vehicle gets larger.
4