Figure 5-- BATES grain

A BATES grain configuration is shown in Figure 5. This usually consists of two or more propellant segments, inhibited on the outer surfaces. This configuration is typically used when a nearly neutral Kn profile is desired (red curve in Figure 6). Kn slope rises to a maximum value then decays. The shape of the curve is determined by the Lo/D and do/D ratios. Judicious choice of segment length and core diameter is necessary, or else the Kn profile may instead have only a progressive profile (green curve) or regressive profile (blue curve). Ideally, Knmax should occur at a surface regression of one-half the web thickness, to produce a symmetric profile (initial Kn = final Kn).

Figure 6-- BATES Kn profiles

The instantaneous grain burning surface area is given by:

Figure 7-- BATES grain illustrating regression of the web

The initial and final burning surface areas are given by:
A_{b initial} = N [½ p (D² - do²) + p Lo do]
A_{b final} = N p D (Lo - 2t)      where t = ½ (D - do)

The value of x when the burning surface area reaches maximum is important as this determines maximum chamber pressure. This value of x may be found by setting the derivative (represented by the slope of the Kn v.s. web regression curve) to zero (i.e. dA_b/dx = 0), then solving for x.

As such, the value of x is found to be:

x = 1/6 (Lo - 2do)        at A_{b max}

Note that the Kn profile is progressive if the calculation gives   x > do. In this case, A_{b max} = A_{b final}
The Kn profile is regressive if the calculation gives   x < 0. In this case, A_{b max} = A_{b initial}

When designing a rocket motor, the dimension D is usually limited by factors such as casing or fuselage size. The choice of core diameter, do, is usually based upon desired web thickness (which determines burn time) and erosive burning considerations. Thus, segment length, Lo, is the parameter that may be available to control the Kn profile. The value of Lo may be found which gives a symmetric Kn profile (initial and final Kn are equal), if D and do are specified:

Lo = ½ (3D + do)          for symmetric profile

Figure 8-- Rod & Tube grain

The burning surface area for a Rod & Tube, as shown in Figure 8, is completely neutral (does not change during the burn) and may be calculated as follows:

A_b =p (d + D_r) L

However, the diameter of the Rod Grain needs to be equal to twice the web thickness of the Tube Grain in order for this to be strictly true. This ensures that burnout of the Rod Grain occurs simultaneously to that of the Tube Grain.

Dr = (Dg - d)/ 2

As well, the following surfaces are inhibited from burning:

• Tube Grain outer surface
• Tube Grain ends (both)
• Rod Grain ends (both)

The gap width should be chosen such that the initial port area (Ap) is a minimum of 2 times the nozzle throat cross-sectional area (At).

Ap > 2 At    or

p /4 (d² - Dr²) > 2p /4 Dt²     (where Dt is throat diameter). or,

(d² - Dr²) > 2 Dt²

Example 1

Determine the initial, maximum, and final steady-state chamber pressure for an unrestricted hollow cylindrical grain of KNDX with the following dimensions:

• Outer diameter 2.25 inches
• Core diameter 1.00 inches
• Grain length 10.50 inches

The nozzle throat diameter is 0.650 inches.

Solution:
A_{b max} = A_{b initial} = ½ p (D ² - d ²) + p L (D + d)
This gives        A_{b max} = A_{b initial} = ½ p (2.25 ² - 1.00 ²) + p 10.50 (2.25 + 1.00) = 114 in²

A_{b final} = ½ p (D + d) (L - t)       where t = ½ (D - d)
Therefore        t = ½ (2.25 - 1.00) = 0.625 in.
and       A_{b final} = p (2.25 + 1.00) (10.50 - 0.625) = 101 in²

The nozzle throat cross-sectional area is:    At = ¼ p (0.650)² = 0.332 in²
This gives an initial, and maximum, Kn = 114 / 0.332 = 343. The final Kn = 101 / 0.332 = 304.

From Figure 2, the initial and maximum steady-state chamber pressure is found to be 1080 psi. The final steady-state chamber pressure is 950 psi.

Example 2

Determine the initial, maximum, and final steady-state chamber pressure for a BATES grain configuration of KNSB with the following dimensions:

• Outer diameter 75 mm.
• Core diameter 22 mm.
• Segment length 100 mm
• 3 segments

The nozzle throat diameter is 13 mm.

Solution:

The value of web regression, x, at the point of maximum chamber pressure is found from the following expression:

x = 1/6 (Lo - 2do)
Therefore, x = 1/6 [100 - 2(22)] = 9.33 mm
Substitute the value of x into the following equations:
d = do + 2x       and       L = Lo - 2x

The initial and final burning surface areas are given by:
A_{b initial} = N [½ p (D² - do²) + p Lo do]
A_{b final} = N p D (Lo - 2t)      where t = ½ (D - do)

This gives:
A_{b initial} = 3 [½ p (75² - 22²) + p (100) 22] = 44 960 mm²
Initial web thickness is      t = ½ (75 - 22) = 26.5 mm
And       A_{b final} = 3 p 75 [100 - 2(26.5)] = 33 220 mm²
The nozzle throat cross-sectional area is: At = ¼ p (13)² = 133 mm²
The initial, maximum and final Kn can now be calculated:
Kn_initial = 44 960 / 133 = 338
Kn_max = 49 890 / 133= 375
Kn_final = 33 220 / 133= 250

From Figure 2, the maximum steady-state chamber pressure is found to be 6.3 MPa. The initial and final steady-state chamber pressure is 5.0 MPa and 3.1 MPa, respectively.

• Outer diameter 50 mm.
• Core diameter 18 mm.
• 4 segments

Solution:

Lo = ½ (3D + do)
This gives   Lo = ½ [3(50) + 18) = 84 mm.

Verify that initial and final surface areas are identical:
A_{b initial} = 4 [½ p (50² - 18²) + p (84) 18] = 32 673 mm²
A_{b final} = 4 p 50 [84 - 2(16)] = 32 673 mm²      where t = ½ (50 - 18) = 16 mm

Originally posted  May 18, 2001
Last updated October 19, 2022