16.512, Rocket Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 13: Examples of Chemical Equilibrium
13.1 Numerical Iteration Procedure
For the Shuttle main engine, we take R = O/F = 6, and P = 210 atm. To get started,
we neglect
,
,
and
, so that Eqs. (12.2-3) and (12.2-4) reduce to
2
O
n
O
n
H
n
OH
n
+=
22
HO
H
8
2n
3
2
n1
=
which give
,
1/3 and (since all other
’s are taken to be zero, n = 4/3).
2
=
2
H
n
=
i
n
The enthalpy before reaction was (assuming very cold reactants, i.e.
),
0
i
T0
K
=
()
=+
41
HhO hO
32
=
-15,632 J (for the
4
3
moles of
,
2
H
1
2
mole of
).
2
O
For the products to have the same enthalpy, the temperature must be very
high. Using the table in Ref. 12.1 (pp. 692-693) we have the trial values tabulated
below:
T(K)
3400
4000
4200
2
h
(J/mole)
-92,973
-58,547
-46,924
2
h
(J/mole)
103,738
126,846
134,700
H
1
hh
3
+
-58.394
-16,265
-2,024
Since we must obtain an enthalpy of -15,632 J, these values indicate a
temperature very close to 4000 K. Linear interpolation between 4000 and 4200 K
gives
T = 4009 K
We can now use this temperature to calculate the four equilibrium constants
through
. These are given in Table 12 of Ref. 12.1. With some interpolation, we
obtain:
1
K
4
K
( )
=
1/2
1
K0
.
9
7
4
a
t
m
(
)
=
2
K2
.
6
1
a
t
m
=
3
K
0.589
atm
( )
=
4
K
2.286
atm
16.512, Rocket Propulsion
Lecture 13
Prof. Manuel Martinez-Sanchez
Page 1 of 5