CHAPTER 10 SOLID PROPELLANT ROCKETS
10.1 INTRODUCTION The sketch below shows a section view of a typical solid propellant rocket.
motor case nozzle
propellant grain port igniter propellant grain
A
1
*
2
Figure 10.1 Solid rocket rocket cross cross section
There are basically two types of propellant grains. 1) Homogeneous or double base propellants - Here fuel and oxidizer are contained within the same molecule which decomposes during combustion. Typical examples are Nitroglycerine and Nitrocellulose 2) Composite propellants - heterogeneous mixtures of oxidizing crystals in an organic plasticlike fuel binder typically synthetic rubber. Sometimes metal powders such as Aluminum are added to the propellant to increase the energy of the combustion process as well as fuel density. Typically these may be 12 to 22% of propellant mass although in the space shuttle booster Aluminum is the primary fuel.
10.2 COMBUSTION
CHAMBER PRESSURE
The combustion proceeds from the surface of the propellant grain. The rate at which combustion gases are generated is expressed in terms of the regression speed of the grain as indicated in the figure below.
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10.1
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Combustion chamber pressure
combustion products flame front
r ˙
Figure 10.2 Surface regression and gas generation
The gas generation rate integrated over the port surface area is m˙ g = ρ p A b r ˙
(10.1)
where
ρ p = solid propellant density A b = area of the burning surface
(10.2)
r ˙ = surface regression speed m˙ g = rate of gas generation at the propellant surface
The phase transition and combustion physics underlying the surface regression speed is extremely complex. In general r ˙ depends on the propellant initial temperature, the chamber pressure and weakly on the velocity of the combustion gases in the port. In general, K n r ˙ = -------------------- ( P t2 ) T 1 – T p
(10.3)
where P t2 = combustion chamber pressure K = impirical constant for a given propellant T 1 = impirical detonation temperature
(10.4)
n = impirical exponent, approximately independent of temperature
In general 0.4
< n < 0.7 and T 1 » T b .
Let M g be the mass of gas in the combustion chamber at a given instant, ρ g is the gas density and V is the chamber volume.
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10.2
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Combustion chamber pressure
d ρ g d M g d dV ----------------- = ( ρ V ) = ρ g + V --------- . dt g dt dt dt
(10.5)
The chamber volume changes as the propellant is converted from solid to gas. dV ------- = r ˙ A b . dt
(10.6)
To a good approximation the chamber stagnation temperature, T t 2 , is determined by the propellant energy density and tends to be independent of P t2 . From the ideal gas law, P t 2 = ρ g RT t 2 , d ρ g 1 d P t2 ---------------------- . --------- = RT t 2 dt dt
(10.7)
The mass flow out of the nozzle is
m˙ n =
( γ + 1 ) * – --------------------γ + 1 2 ( γ – 1 ) γ P t2 A
------------- 2
------------------- . γ RT t 2
(10.8)
The mass generated at the propellant surface is divided between the mass flow exiting the nozzle and the time dependent mass accumulation in the combustion chamber volume. d M g m˙ g = ----------- + m ˙n. dt
(10.9)
Fill in the various terms in (10.9).
( γ + 1 ) * – --------------------γ + 1 2 ( γ – 1 ) γ P t 2 A
d ρ g ρ p A b r ˙ = ρ g r ˙ A b + V --------- + ------------- dt
2
------------------γ RT t2
(10.10)
or
( γ + 1 ) * – --------------------γ + 1 2 ( γ – 1 ) γ P t2 A
K ( ρ p – ρ g ) A b n V d P t2 ------------------------------------( P t 2 ) = ------------ ----------- + ------------- 2 T 1 – T p RT t2 dt
------------------- . γ RT t 2
(10.11)
Rearrange (10.11) to read
( γ + 1 ) * – --------------------P A γ 2 1 ( – ) γ γ + 1 t2
V d P t2 ------------ ----------- + ------------- 2 RT t2 dt
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K ( ρ p – ρ g ) A b n ------------------- – ------------------------------------( P t 2 ) = 0 . T 1 – T p γ RT t2
10.3
(10.12)
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Dynamic analysis
After a startup transient, during which P t2 changes rapidly with time, the pressure reaches a quasi-steady state where the time derivative term in (10.12) can be regarded as small compared to the other terms. To a good approximation,
( γ + 1 ) * – --------------------γ + 1 2 ( γ – 1 ) γ P t 2 A
K ( ρ p – ρ g ) A b n ------------------- = ------------------------------------( P t2 ) . T 1 – T p γ RT t2
------------- 2
(10.13)
Solve for the chamber pressure,
P t2 =
( γ + 1 ) --------------------2 γ + 1 ( γ – 1 ) K
------------- 2
( ρ p – ρ g ) A b ----------------------------- ------ γ ( T 1 – T p ) A *
1 -----------1–n
γ RT t2
.
(10.14)
This formula can be used as long as A b is a slow function of time. All the quantities in (10.14) are apriori data with the exception of T t 2 which must be estimated or calculated from a propellant chemistry model. Note that there is a tendency for the chamber pressure to increase as the burning area increases.
10.3 DYNAMIC
ANALYSIS
Rearrange (10.12) to read,
1 ⁄ 2 * ( γ RT t 2 ) A d P t2 K ( ρ p – ρ g ) A b RT t2 n ----------- + ------------------------------------------ P t2 – ------------------------------------ ------------ ( P t 2 ) V ( γ + 1 ) T 1 – T p dt -------------------- γ + 1 2 ( γ – 1 ) ------------- V 2
= 0.
(10.15)
This is a nonlinear first order ordinary differential equation for the chamber pressure of the form, d P t2 1 n ----------- + --- P t2 – β ( P t2 ) = 0 τ dt
(10.16)
where the characteristic time is
( γ + 1 ) --------------------γ + 1 2 ( γ – 1 )
------------- 2
V
τ = -------------------------------------- ------ . ( γ RT t 2 ) 1 ⁄ 2 A *
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10.4
(10.17)
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Dynamic analysis
This time is proportional to the time required for an acoustic wave to travel the length of the combustion chamber multiplied by the internal area ratio of the nozzle. The system has the character of a Helmholtz resonator and the inverse of (10.17) is the natural “Coke bottle” frequency of the rocket motor. The constant in the nonlinear term is,
β =
K ( ρ p – ρ g ) A b RT t2 ------------ . ----------------------------------- T T V – 1 p
(10.18)
Let’s look at the linear behavior of (10.16) near a steady state operating point. Let, P t2 ( t ) = P t 2 + p t 2 ( t )
(10.19)
where p t 2 is a small deviation in the pressure from the steady state. Substitute into (10.16) and expand the nonlinear term in a binomial series. With higher order terms in the series neglected, the result is, d p t2 n n–1 1 1 ----------- + --- P t 2 + --- p t2 – β ( P t2 ) – β n ( P t2 ) p t2 = 0 . τ τ dt
(10.20)
The steady state terms satisfy
1--- P – β ( P ) n t2 τ t 2
= 0
(10.21)
and the dynamical equation becomes, d pt 2 n–1 1 p = 0 . ----------- + --- – β n ( P t2 ) τ t2 dt
(10.22)
Note that from (10.21)
β ( P t2 )
n–1
=
1--- τ
(10.23)
and so d p t2 1–n ----------- + ------------ p t 2 = 0 . τ dt
(10.24)
The solution of (10.24) is
n – 1
t p t2 -----------τ ------------- = e . p t2
(10.25)
0
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Dynamic analysis
< 1 a small deviation in pressure will be restored to the equilibrium value (the extra nozzle flow exceeds the extra gas generation from the propellant surface). But if n˙ > 1 the gas generIf n
ation rate exceeds the nozzle exhaust mass flow and the chamber pressure will increase exponentially - the vehicle will explode! If the fluid velocity over the surface becomes very large, enhanced heat transfer can lead to a situation called erosive burning. In this case the burning rate can vary considerably along the port and excessive gas generation can lead to a failure. In the case of very low chamber pressure, the combustion process can become unsteady or cease altogether this defines the combustion limit of a particular propellant. There is also an upper pressure limit above which combustion again becomes erratic or unpredictable. For most propellants this is above 5000psi. 10.3.1
EXACT
SOLUTION
The chamber pressure is governed by the equation d Pt 2 1 n ----------- + --- P t2 – β ( P t 2 ) = 0 . τ dt
(10.26)
Let’s determine the exact integral of this equation and compare the behavior of the system with the linearized solution for both n < 1 and n > 1 . It is virtually always best to work in terms of dimensionless variables. The steady state solution of (10.26) for which the time derivative term is zero is
Pt 2
= steady state
( τβ )
1 -----------1–n
(10.27)
Let t – t 0 η = -------------
P t2 H = ---------------------------------Pt 2
τ
(10.28)
steady state
In terms of new variables (10.26) becomes n dH -------- = H – H d η
(10.29)
Equation (10.29) is rearranged as follows dH ------------------ = d η n H – H
(10.30)
which integrates to
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10.6
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Dynamic analysis
1–n
1 – H – ( 1 – n ) η = Log ------------------------1–n 1 – H 0
(10.31)
where H 0 is the initial value of P t2 ⁄ P t2 and the initial value of η is taken to be steady state zero. Now solve for H
H =
( 1– ( 1–
1–n
H0
)e
1 -----------------–( 1 – n ) η ( 1 – n )
)
(10.32)
Several cases are shown below. 6
Pt 2 ---------------------------------Pt 2
steady state
5
4
3
2
n = 1.2
n = 0.6
1
n = 1.2 2
4
6
8
10
t ⁄ τ Figure 10.3 Chamber pressure response of a solid rocket.
The exact solution is consistent with the linear analysis and shows that if n actual steady state, the chamber pressure either decays to zero or blows up.
> 1 there is in no
Whereas if n < 1 then the chamber pressure will return to the steady state value even in the face of a large deviation away from steady state. The motor is stable in the face of finite disturbances. 10.3.2
CHAMBER
PRESSURE HISTORY
The analysis in the last section is useful for determining the transitory behavior of the motor during transients such as start-up and shut down where the chamber pressure responds on a very short time scale measured by τ . As the burning area of a circular port increases over the course
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10.7
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Dynamic analysis
of the burn the chamber pressure changes on a much longer time scale and we can use the steady state balance (10.14) together with the regression rate law (10.3) to determine the port radius as a function of time. Rewrite (10.14) as
Pt 2 =
1 -----------1–n
α --r -- r i
(10.33)
where
α =
( γ + 1 ) --------------------γ + 1 2 ( γ – 1 ) K
( ρ p – ρ g ) 2 π r i L ----------------------------- --------------- γ RT t2 γ ( T 1 – T p ) A *
------------- 2
(10.34)
and L is the length of the port assumed to be constant. Now solve n -----------1–n
K r dr ----- = ------------------------- α ---- ( T 1 – T p ) r i dt
(10.35)
for the radius of the port as a function of time. d ( r ⁄ r i ) --------------------------- =
( r ⁄ r i )
n -----------1–n
n -----------1–n
K ( α ) ----------------------------- ( T 1 – T p ) r i
dt
(10.36)
Integrating (10.36) leads to
r ---- = r i
1 – 2n K ------------------------------ ( α ) 1 + --------------- 1 – n ( T 1 – T p ) r i
n -----------1–n
1–n ---------------1 – 2n
t
n
≠ 0.5 (10.37)
n
r ---- = Exp r i
----------- K ( α ) 1 – n t ----------------------------- ( T 1 – T p ) r i
n = 0.5
This defines a much longer time scale
τ burn =
( T 1 – T p ) r i ----------------------------- n ------------ K ( α ) 1 – n
(10.38)
where the negative sign applies if n > 0.5 . This time scale characterizes the change in chamber pressure during the burn. The burntime is determined by the outer radies of the motor.
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10.8
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Problems
t burnout =
1 – 2n --------------- r f 1 – n -r ---- i
1 – n – 1 ---------------- τ burn 1 – 2n
r f t burnout = Log ----- τ burn r i
n ≠ 0.5
(10.39)
n = 0.5
10.4 PROBLEMS Problem 1 - It is a beautiful summer day at the cape and a space shuttle astronaut on her second
mission finds that the g forces during launch are noticeably larger than during her first mission that previous December. Can you offer a plausible explanation for this? Problem 2
- A solid propellant rocket operates in a vacuum with a 10 cm diameter nozzle throat and a nozzle area ratio of 100. The motor has a cylindrical port 300 cm long. At the beginning of the burn the port is 20 cm in diameter and the propellant recession velocity is 1 cm/sec. The port diameter at the end of the burn is 80 cm. The regression rate law is r ˙ =
α P 0.5 t2
(10.40)
3
The solid propellant density is 2 grams/cm and the combustion gas has γ = 1.2 and molecular weight equal to 20. The combustion chamber temperature is 2500K . Determine the thrust versus time history of the motor. Problem 3 - One of the simplest types of solid rocket designs utilizes an end burning propellant
grain as shown below.
m˙
The motor diameter is 100 cm and the grain length at the beginning of the burn is 200 cm. The 3
solid propellant density is 2 grams/cm and the combustion gas has
γ
= 1.2 and molecular
weight equal to 20. The combustion chamber temperature is 2500K and, at the beginning of
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10.9
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Problems
5
2
the burn, the pressure is P t2 = 5x10 N ⁄ M . The motor exhausts to vacuum through a 30 cm diameter nozzle throat and a nozzle area ratio of 10. Sketch the thrust-time history of the motor and determine the total impulse t b
=I
∫ (
Thrus ) d tt
(10.41)
0
in units of Kg-M/Sec. Problem 4
- The thrust versus time history of a solid rocket with a circular port is shown
below.
The regression rate of the propellant surface follows a law of the form r ˙ =
α P nt 2
(42)
where the exponent n is in the range of 0.4 to 07. Briefly show why the thrust tends to increase over the course of the burn.
Problem 5
- A solid propellant upper stage rocket operates in space. The motor has a 0.2 m
diameter nozzle throat and a cylindrical port 4.2 m long. At the end of the burn the port is 0.8 m in diameter. The regression rate law is r ˙ = 3.8 × 10
– 6 0.5 P t 2 m/sec
2
(43) 3
where the pressure is expressed in N/m . The solid propellant density is 2000 kg/ m and the combustion gas has γ = 1.2 and molecular weight equal to 32. The combustion chamber temperature is 3000K . The quasi-equilibrium chamber pressure at the end of the startup transient is 2
P t2 = 3.0 × 10 6 N/m .
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10.10
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Problems
1)Determine the characteristic time τ for the start-up transient. 2)Determine the propellant mass expended during the startup transient. Take the start-up time to be 8 τ . 3) Determine the mass flow and quasi-equilibrium chamber pressure P t 2 at the end of the burn. 4) Once the propellant is all burned the remaining gas in the chamber is expelled through the nozzle and the pressure in the chamber drops to zero. Calculate the time required for the pressure to drop to 10% of its value at the end of the burn. 5) Sketch the pressure-time history of the motor.
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Problems
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10.12
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