Fundamentals of Rocket Propulsion

Chapter 5 Thermodynamic Performance Analysis of Chemical Rocket Engine

Lectured by Prof. Liang Guozhu

Department of Rocket Propulsion, School of Astronautics,

Beijing University of Aero. & Astro.(BUAA) September 2011

1

5 Thermodynamic Performance Analysis of Chemical Rocket Engine

5.1 Introduction

In Chapter 2 & 4 the one-dimensional ideal performance relations were developed. They require a

knowledge of the properties of the combustion products, such as their combustion temperature , average

molecular weight

fT

rM , gas constant R , and specific heat ratio etc. The present chapter discusses a

theoretical approach to determine the properties of the propellant reaction products for a given composition of

propellant (knowledge of the fuel and oxidizer chemicals and their relative proportions), chamber pressure, and

nozzle exit pressure.

k

The analysis is usually divided into two somewhat separate sets of calculations：The combustion process

and the nozzle gas expansion process.

5.2 Basic Principles for Thermodynamic Analysis for combustion chamber process

5.2.1 The principle of the conservation of mass

The mass of any of the atomic species present in the reactants before the chemical reaction must be equal

to the mass of the same species in the products. For example of hydrogen-oxygen combustion,

2 2 22 2 2 2 2H O H O H O O H OH O H O O H OHa b n n n n n n H+ → + + + + +

the mass balance would be:

for hydrogen: 2 2

2 2 2H O H H Oa n n n n= + + + H

for oxygen: 2 2

2 2H O O O Ob n n n n= + + + H

In more general form, the mass for any given element must be the same before and after the reaction, i.e.

(5.1) 1 1

L N

ij j ij jj jpropellants products

a n a n= =

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦∑ ∑

Here L gives the number of species in the reactants, the gives the number of gas species in the

combustion products. The atomic coefficient are the number of kilogram atoms of element per kg-mol

of species . is expressed as kg-mol for a particular species per kg of mixture.

N

ija i

j jn j

The effective average molecular mass rM of a gas mixture is then

r

1r

1

N

j jj

N

jj

n MM

n

=

=

=∑

∑ (5.2)

2

For example of hydrogen-oxygen reaction,

2 2 2

2 2 2

r

18 2 32 16 17H O H O O H O

H O H O O H OH

n n n n n nM

n n n n n nH+ + + + +

=+ + + + +

5.2.2 The principle of the energy balance (the conservation of energy)

The energy balance can be thought of as a two-step process. The chemical reaction occurs instantaneously

but isothermally at the reference temperature, and the resulting energy release then heats the gases from this

reference temperature to the final combustion temperature.

For the first step, the heat of reaction need to determined which is defined as the energy

released or absorbed when products are formed from its reactants at the standard reference conditions

(298.15K or 25ºC and 1atm). The heat of reaction can be negative or positive, depending on whether the

reaction is exothermic or endothermic. In general the heat of reaction can be determined from sums of the

heats of formation of the products and the reactants, namely

0r H∆

0 0 0r f fproducts reactants

1 1

N L

j j j jj j

H n H n H= =

⎡ ⎤ ⎡ ⎤∆ = ∆ − ∆⎣ ⎦ ⎣ ⎦∑ ∑ (5.3)

where the heat of formation is defined as the energy released (or absorbed), or the value of enthalpy

change, when 1 mole of chemical compound is formed from its constituent atoms or elements at the standard

reference conditions. By convention, the heat of formation of the gaseous elements (e.g., H

0f H∆

2, O2, Ar, Xe, etc.)

is set to zero at the standard reference conditions. When heat is released in the formation of a product, then

has a negative value. 0r H∆

For the second step, the physical heating process can be described by the following equation:

f

ref ref

0,

1 1

dN NT T

r j p j jT Tj j

H n c T n h= =

−∆ = = ∆∑ ∑∫ %%f

j (5.4)

here is the increase in enthalpy per molar species , and is the molar specific heat at constant pressure.

jh∆ % j ,p jc%

3

5.2.3 The principle of the chemical equilibrium

Definition of chemical equilibrium is that, if the concentrations of products and reactants remain constant,

the two reactants “forward and backward” are in equilibrium with each other. We define the state of chemical

equilibrium as the state in which there is no spontaneous tendency of the system to change its internal

composition, either by diffusion or by chemical reactions.

Chemical equilibrium is usually described by either of two equivalent formulations, i.e. equilibrium

constants or minimization of free energy. The comparisons between the two formulations show that, if a

generalized method of solution is used, the two formulations reduce to the same number of iteration equations.

However several disadvantages of the equilibrium constant method are found. Briefly these disadvantages are

numerical difficulties with use of components, more difficulty in testing for presence of some condensed

species, and more difficulty in extending the generalized method for nominal equations of state. For these

reasons, the free-energy minimization formulation is used. The condition for equilibrium may be stated in

terms of any of several thermodynamic functions such as the minimization of the Gibbs free energy or

Helmholtz free energy or the maximization of the entropy. If one wishes to use temperature and pressure to

characterize a thermodynamic state, the Gibbs free energy is most easily minimized in as much as temperature

and pressure are its natural variables. Similarly, the Helmholtz free energy is most easily minimized if the

thermodynamic state characterized by temperature and volume (or density).

Here the approach is the minimization of the Gibbs free energy is applied. The Gibbs free energy (often

called chemical potential) for mixtures of gas as well as an individual gas species is defined as g

g e pv Ts h Ts= + − = − (5.5)

where is the internal energy, the pressure, the specific volume, the temperature, the

entropy, and is the enthalpy. The free energy is a function of temperature and pressure. It is another

property of a material, just like enthalpy or density. The free energy may be thought of as the tendency or

driving force for a chemical material to enter into a chemical (or physical ) change.

e p v T s

h

For a system including different species the mixture free energy is g

(5.6) 1

N

j jj

g g=

= ∑ n

where is the number of species and is the mole number for species. N jn jThe change in free energy for reactions as the basic thermodynamic relation of a chemical system is

(5.7) 1

d d d dN

j jj

g s T v p g n=

= − + +∑

The change in the Gibbs free energy function of a chemical system is zero at equilibrium state. That is

0dg = (5.8)

When the system is under the condition of constant pressure and constant temperature, for example in rocket engine thrust chambers, then

1

dN

j jj

g n=

0=∑ (5.9)

The equation (5.9) is the chemical equilibrium equation for a chemical system with constant pressure and temperature. The equation (5.9) also can be expressed with equilibrium constants. For an arbitrary reversible chemical

4

reaction

aA bB cC dD+ ⇔ + (5.10)

where A and B are reactants and and are products. C D The chemical equilibrium equation can be written as

( ) ( )a b c dc d

C Dpa b

A B

n n pKn n n

+ − +⎛ ⎞= ⎜ ⎟⎝ ⎠

(5.11)

where is the total mole number of all the species and is the equilibrium constant. is only a function of temperature and can be calculated as follows:

n pK pK

c dC D

p a bA B

p pKp p

= (5.12)

Let

c dC D

n a b

n nKn n

=A B

(5.13)

which is the equilibrium constant described by molar fraction. Then

( ) (a b c d

n ppK Kn

)+ − +⎛ ⎞= ⎜ ⎟⎝ ⎠

(5.14)

Note that is not only a function of temperature but also a function of pressure. When =0,

then is only a function of temperature. nK a b c d( + )-( + )

nK

5.2.4 Solving method

The approach used commonly today for solving for the gas composition, combustion temperature, and gas

properties is relied on the minimization of the Gibbs free energy and on mass balance and energy balance

equations. As was explained in Eq.(5.8), the change in the Gibbs free energy function is zero at equilibrium. To

assist in solving this equation a Lagrangian multiplier or a factor of the degree of the completion of the

reaction is often used. An alternative method is to use the mass balance and energy balance equations together

with the equilibrium relationships (Eq.(5.14)).

After assuming a chamber pressure and setting up the energy balance, mass balance, and equilibrium

relations, one method of solving all equations is to estimate a serial of combustion temperatures and solve

mass balance and equilibrium relations for the various values of until there is a convergence where a

energy balance is achieved between the absolute value of heat of reaction and the heat absorbed by

the gases, that is , to go from the reference temperature to the final combustion temperature.

jn

0r H−∆

00TH H− 0

5

In the combustion chamber a constant pressure process is assumed. In the nozzle an isentropic expansion

with the entropy being constant is used.

Solving the set of equations to determine the gas compositions and the thermodynamic properties of the

combustion products will obtain those important gas parameters, such as fT , , , etc. eT k

5.3 Equilibrium constant method for calculating the gas compositions for given pressure and

temperature

Example 5-1. The liquid monopropellant nitromethane (CH3NO2) can be decomposed into gaseous

reaction products with the assumption of no dissociation and no O2. Determine the values of products of

combustion temperature ，average molecular weight fT rM ，specific heat ratio ，characteristic velocity

，thrust coefficient ，and specific impulse

k

*c FC sI using the water-gas equilibrium conditions.

SOLUTION. Neglect minor products, the chemical reaction for 1 mol of reactant can be described as

2 2 23 2 CO CO 2 H 2 H O 2 N1.0CH NO CO CO H H O Nn n n n n→ + + + +2 2

The mass balances are obtained for each atomic element:

C: 1= + COn2COn

H: 3=2 +2 2Hn

2H On

O: 2= +2 + COn2COn

2H On

N: 1=2 2Nn

From the last one of the above equations, can be solved directly as =0.5. 2Nn

2Nn

The reaction commonly known as the water-gas reaction is

2 2 2H CO H O CO+ ⇔ +

Its equilibrium constant , expressed as molar concentration, is only a function of temperature (see also

(5.14)), i.e.

nK

2

2 2

H O CO

H COn

n nK

n n=

The for equation above have five unknowns: namely, the four molar concentrations and the equilibrium

constant , which is a function of temperature. can be obtained from a table of the water-gas reaction

as a temperature. Try =2500K and =6.440, then the four molar concentrations can be solved as

nK nK

fT nK

2Hn =0.664, =0.836, =0.164, =0.836. 2H On

2COn COn

The heats of formation for various specifies are listed in the Table 5- 1 below from the JANAF

thermochemical tables. By definition, the heat of formation of H

0f H∆

2 or N2 is zero. The enthalpy change of the

gases going from the reference conditions to the combustion temperature can also be obtained from the

JANAF thermochemical tables and is again listed in Table 5- 1. The heat of reaction is obtained from Eq.(5.3),

6

5 1

0 0 0r f fproducts reactants

1 1

0.836( 241.826) 0.164( 393.522) 0.836( 110.53) 1.0( 113.1)246kJ / mol

j j j jj j

H n H n H= =

⎡ ⎤ ⎡ ⎤∆ = ∆ − ∆⎣ ⎦ ⎣ ⎦

= − + − + − − −= −

∑ ∑

Table 5- 1 The calculation data at 2500K

Species 0f H∆ f

ref

T

j Th∆ % rM jn

N2 0 74.296 28 0.500

H2O -241.826 99.108 18 0.836

H2 0 70.498 2 0.664

CO -110.53 74.985 28 0.836

CO2 -393.522 121.917 44 0.164

CH3NO2 -113.1 61 1.000

The gas enthalpy change of the hot gas in the combustion chamber should be equal to the negative value

of the heat of reaction from the view of energy banlance. Using data from the above table,

f

ref

52500298

1

249.5kJ/molT

j j Tj

H n h=

∆ = ∆ =∑ %

This is not identical to the 246kJ/mol obtained previously, and therefore a lower temperature is to be tried.

After one or two iterations the final combustion temperature of 2470K and the new molar composition of

species will be found where the heat of reaction balances the enthalpy rise of the products. The molecular

weight can then be obtained from Eq.(5.2): 5

r1

r 5

1

20.3j j

j

jj

n MM

n

=

=

= =∑

∑

The specific heat varies with temperature, and average specific heat values pc can be obtained from each

species by integrating the following formula from the JANAF thermochemical tables: 2470

2982470

298

d

d

pp

c Tc

T= ∫∫

=2040kJ/kg.K

The specific heat ratio is

1.258314.420.3

p

p

ck

c= =

−

With rM , , and now determined, the ideal performance of a nitromethane rocket engine can be

established on the basis of Chapter 错误！未找到引用源。 for

k fT

cp =69atm and ep =1.0atm. The results are

7

f

r* 1525m / s

T RM

c = =Γ

o

1

es ef f

r c

2 1 2394m / s1

kkpk RI u T

k M p

−⎡ ⎤⎛ ⎞⎢ ⎥= = − =⎜ ⎟⎢ ⎥− ⎝ ⎠⎢ ⎥⎣ ⎦

o

s* 1.57F

ICc

= =

Example 5-2. For a given solid propellant formulation chemical

formula:C11.2738H40.8672O25.8632N5.7692Cl5.6773 S2.4177Al1.8532 (the subscripts represent atomic molar quantities of

the propellant with the molecular weight 1003.2606kg), the requirements are to calculate the equilibrium

combustion composition by the method of equilibrium constants under the combustion temperature = 2800

K, the combustion pressure (

cT

cp = 70atm), and the reaction products being considered in this system are: Al2O3

(condensed)、CO2、H2O、CO、H2、N2、HCl、SO2、O2、OH、NO、H、Cl (13 species).

Calculation procedure

According to the assumed combustion composition, 13 unknown species require 13 equations to solve

them. Then we can start by mass balance as follows:

(1) For carbon:

2CO CO 11.2738n n+ = (5.15)

(2) For hydrogen:

(5.16) 2 2

2 3 2 2 2 2

2

H O H HCl OH H2 2 40.8672n n n n n+ + + + =

(3) For oxygen:

(5.17) Al O CO SO O H O CO OH NO3 2 2 2 25.8632n n n n n n n n+ + + + + + + =

(4) For nitrogen:

N NO2 5.7692n n+ = (5.18)

(5) For chlorine:

HCl Cl 5.7692n n+ = (5.19)

(6) For sulfur:

8

9

2 SO 2.4177n = (5.20)

(7) For aluminum:

2 3Al O2 1.8532n = (5.21)

The equilibrium constants may be applied when chemical compounds are formed from their elements.

According to the 13 species, the following 6 independent reversible chemical reactions can be considered:

(1) 2 21H O H O2

⇔ + 2 and it’s chemical equilibrium equation is:

2 2

2

11/ 2 2

H O,H O

H O gp

n n pKn n

2

−⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

(5.22)

(2) the water gas reaction: and it’s equilibrium equation is: 2 2 2CO H CO H O+ ⇔ +

2

0

CO H O

2 2

pCO H g

p

n n pKn n n

⎛ ⎞K= =⎜ ⎟⎜ ⎟

⎝ ⎠ (5.23)

(3) 21H O OH H2

⇔ + 2 and it’s chemical equilibrium equation is:

2

2

11/ 2 2

OH H,H O

H O g

'p

n n pKn n

2

−⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

(5.24)

(4) and it’s chemical equilibrium equation is: 2H 2⇔ H

2

12H

,HH g

pn

2

pKn n

−⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

(5.25)

(5) and it’s chemical equilibrium equation is: HCl H Cl⇔ +

1

H Cl,HCl

HCl gp

n n pKn n

−⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

(5.26)

(6) and it’s chemical equilibrium equation is: 2 2N O 2NO+ ⇔

2 2

2 2 2 2 2 2 Cn n n n n n n n n n n n n= + + + + + + + + + + +

2NO

,NON O

pn K

n n= (5.27)

The equilibrium constants, listed in Table 5- 2, are depending on combustion temperature. As mentioned

before, the number of equations now is 13 but a new unknown variable appeared which increase the

number of unknowns to 14. represents the total number of moles of the products considered for this

system, but only for the gaseous products because, as an assumption, any condensed phases add a negligible

amount to the total pressure. Then the number of moles of (Al

gn

gn

2O3) will be not added to the total amount

because it will be at the liquid state (solidification temperature of Al2O3 ≤ 2327K) and the following equation

can be considered:

l (5.28) g CO H O CO H N HCl SO O OH NO H

Table 5- 2 The equilibrium constants at 2800K

T equilibrium constants 2800K

OHpK2, 0.2233×10－1

Kp 7.0080

2,'p H OK 0.2091×10－1

2,HpK 0.6649×10－2

HClp,K 4.5983×10－3

NOpK , 0.8786×10－2

The above mentioned system of equations were used to calculate the average molecular weight of the

produced gases ( gM ) and equilibrium combustion composition for propellant formulations based on (C, H, O,

N, Cl, S and Al) for given combustion temperature and pressure using an iterative technique till convergence.

Results

The combustion composition calculated, total number of moles of the produced gases for 1kg combustion

products and average molecular weight of them are listed below in Table 5- 3.

Table 5- 3 Results of computationsEquilibrium Species Values/mol/kg

2 3Al O (c)n 0.9266

2COn 0.8060

2H On 6.1378

COn 10.4678

2Hn 11.3743

2Nn 2.8839

HCln 5.6064

10

2SOn 2.4177

2On 8.2975×10-5

OHn 2.8767×10-2

NOn 1.4500×10-3

Hn 0.2079

Cln 7.0867×10-2

gn 40.0029

gM 22.6365

5.4 Basic Principles for Thermodynamic Analysis for Nozzle Flow Process

The nozzle gas expansion process constitutes the second set of calculations. The fully reacted, equilibrated

gas combustion products enter the nozzle and undergo an adiabatic expansion in the nozzle. For the nozzle

expansion thermodynamic calculation the isentropic equation need to be adoped to determine the gas

temperature at certain location along nozzle axis. This is the different characteristic from combustion chamber

where the energy balance equation is applied. The entropy remains constant during a reversible (isentropic)

nozzle expansion for an ideal rocket engine.

The entropy is another thermodynamic property of matter. In the anlysis of isentropic nozzle flow, it is

assumed that the entropy remains constant. It is defined as

d d d dd p

e p v T ps cT T T p

= + = − R (5.29)

and the corresponding intergral is

00 0

ln lnpT ps s c RT p

− = − (5.30)

where the subscript zero applies to the reference state. For a mixture the entropy is

(5.31) 1

N

j jj

s s=

= ∑ n

Here entropy is in . The entropy for each gaseous species is J/(kmol K)⋅

0,

g

ln lnjj T j

ns s R R

n= − − p (5.32)

For solid and liquid species the last two terms are zero. Here refers to the standard pressure temperature

.

0Ts

T

11

5.5 Free-energy minimization method for calculating the gas compositions for given pressure and

temperature

5.5.1 Equation of state

We assume all gases to be ideal and that interactions among phases may be neglected. The equation of

state for the mixture is

pv nR T= o (5.33)

where p is pressure (N/m2), v specific volume (m3/kg), n moles (kg-mole/kg), T temperature (K), and Ro is

the universal gas constant ( Ro R=8314.4 J/(kg-mole.K) if the international system of units (SI units) are used.

Equation (5.33) is assumed to be correct, even when small amounts of condensed species up to several

percent by weight are present. In this event the condensed species are assumed to occupy a negligible volume

and exert a negligible pressure compared to the gaseous species. In this variables volume, moles, and density

the volume and moles refer to gases only while the mass is for the entire mixture including condensed species.

The molecular weight of the mixture M is then defined to be

nM /1= (5.34)

Where

(5.35) ∑=

=m

jjnn

1

Where nj is the number of kilogram-moles of species j per kilogram of mixture. An equivalent expression for

molecular weight is

∑

∑

=

== m

jj

n

jjj

n

MnM

1

1 (5.36)

Where Mj is the molecular weight of species j. As implied in equation (5.36), among the n possible species

which may be considered, gases are indexed from 1 to m and condensed species from m+1 to n.

5.5.2 Minimization of Gibbs Free Energy

The Gibbs free energy per kilogram of mixture for n mixture species is defined as:

12

13

n (5.37) 1

n

j jj

g µ=

=∑

where the chemical potential per kilogram of species is defined as:

( ), , i i j

jj T p n

gn

µ≠

⎛ ⎞∂= ⎜ ⎟⎜ ⎟∂⎝ ⎠

(5.38)

The condition for chemical equilibrium is the minimization of free energy. This minimization is usually

subjected to certain constraints such as the following mass balance constraints:

(5.39) 0

10, ( 1,.... )

n

ij j ij

a n b i L=

− = =∑

or

(5.40) 0 0, ( 1,.. )i ib b i L− = =

where

),...1(,1

Linabn

jjiji == ∑

=

where the stoichiometric coefficients aij are the number of kilogram-atoms of element i per kilogram-mole of

species j, is the assigned number of kilogram-atoms of element i per kilogram of total reactants, and

is the number of kilogram-atoms of element i per kilogram of mixture.

ib 0ib

We define a term G as:

(5.41) ( 0

1ii

L

ii bbgG −+= ∑

=

λ )

=

n

where λi are Lagrangian multipliers, the condition for equilibrium becomes:

(5.42) ( )0

1 1 1

0n L L

j i ij i i i ij i i

G a n b bδ µ λ δ δλ= = =

⎛ ⎞= + + −⎜ ⎟⎝ ⎠

∑ ∑ ∑

By treating the variations δnj and δλi as independent gives

(5.43) 1

0, ( 1,..., )L

j i iji

a jµ λ=

+ = =∑

and also the mass balance equation (5.40).

Based on the assumptions in the section 5.5.1 the chemical potential may be written as:

0

0

ln ln { 1,..., ( )}

{ 1,..., ( )}

jj j

j

j

nR T R T p for j m gas phase

n

for j m n condensed phase

µµ

µ

⎧ ⎛ ⎞+ + =⎪ ⎜ ⎟

= ⎝ ⎠⎨⎪ = +⎩

o o

(5.44)

Where 0jµ for gases (j=1 to m) and for condensed species (j > m) is the chemical potential in the standard

state.

For a gas, the standard state is the hypothetical ideal gas. For a pure solid or liquid, the standard state is the

substance in the condensed phase under a pressure of one atmosphere (101325 Pa). The numerical value of 0jµ is depending partly on a term involving units of atmospheres. Therefore, the unit of pressure in equation

(5.44) should be consistent with the units of pressure in the thermodynamic data being used.

Equations (5.40) and (5.43) permit the determination of equilibrium compositions for thermodynamic

states specified by an assigned temperature and pressure 0T 0p that is, in addition to equations (5.40) and

(5.43), we have the pair of trivial equations

0T T= ， 0p p= (5.45)

However, the thermodynamic state can be specified by assigning, any two states function. For example the

thermodynamic state corresponding to constant pressure combustion is specified, instead of equation (5.45) by:

0h h= ， opp = (5.46)

where h is the enthalpy of the mixture and is a constant equal to the enthalpy of the reactants. The

enthalpy h is defined by this equation:

0h

(5.47) ( )1

no

j T jj

h n H=

= ∑

where ( )oT j

H is the standard state enthalpy for species j.

For assigned entropy and pressure (such as for an isentropic compression or expansion to a specified

pressure), the thermodynamic state is specified instead of equation (5.46) by:

0s s= ， opp = (5.48)

where s is the entropy of the mixture and is the assigned entropy, or entropy of the total reactant. The

entropy s is defined by this equation:

0s

14

15

s (5.49) 1

n

j jj

s n=

= ∑

where

( )

( )

0

0

ln ln ( 1,..., )

( 1,..., )

jT jj

j

T j

ns R R p j

nss j m

⎛ ⎞⎛ ⎞− − =⎜ ⎟⎜ ⎟

⎝ ⎠= ⎜ ⎟⎜ ⎟= +⎜ ⎟⎝ ⎠

o o m

n (5.50)

where is the standard state entropy for species j. Equation (5.50) is similar to equation (5.44), and

same discussion concerning standard state pressure that applied to equation (5.44) also applied to equation

(5.50).

( )0T j

s

5.5.3 Gibbs Iteration Equations

The equations required to obtain composition are not all linear in the composition variables and therefore

an iteration procedure is generally required. In the iteration procedure to be described it will be convenient to

treat n as an independent variable. A descent Newton-Raphson method is used to solve for corrections to initial

estimates of compositions , Lagrangian multipliers jn iλ , moles of gaseous species n, and (when required)

temperature T. This method involves a Taylor series expansion of the appropriate equations with all terms

truncated that contain derivatives higher than the first. The correction variables used are

, ln ( 1,..., )jn j m∆ = ( 1,..., )jn j km n∆ = + , ln n∆ , ii R T

λ−Π =

o, and lnT∆ . As Zeleznik and Gordon

(1968)[错误！未定义书签。] pointed out, it is no restriction to start each iteration with the estimate for the Lagrangian

multipliers equal to zero inasmuch as they appear linearly in equation (5.43). After making dimensionless these

equations containing thermodynamic functions, the Newton-Raphson equations obtained from equations (5.39),

(5.43), (5.46), and (5.48) are:

( )

1ln ln ln 1,...,

oL T j j

j ij ii

Hn a n T j

R T R Tµ

=

⎡ ⎤⎢ ⎥∆ − Π −∆ − ∆ = − =⎢ ⎥⎣ ⎦

∑ o om (5.51)

( )

1ln 1,...,

oL T j j

ij ii

Ha T j m

R T R Tµ

=

⎡ ⎤⎢ ⎥− Π − ∆ = − = +⎢ ⎥⎣ ⎦

∑ o on (5.52)

(5.53) 1 1

ln 1,...,m n

okj j j kj j k k

j j ma n n a n b b k l

= = +

∆ + ∆ = − =∑ ∑

16

j∑ (5.54) 1 1

ln lnm m

j jj j

n n n n n n= =

∆ − ∆ = −∑

( ) ( ) ( )

1 1 1ln ln

o o om n nj T T j pj j j o

j jj j m j m

n H H n C h hn n TR T R T R= = + = +

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −⎢ ⎥ ⎢ ⎥ ⎢ ⎥∆ + ∆ + ∆ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∑ ∑ ∑o o o R To (5.55)

( ) ( )0

1 1 1ln ln

om n nT j pj j j j o

j jj j m j m

s n Cn s s sn n T nR R R R= = + = +

⎡ ⎤ ⎡ ⎤⎡ ⎤ −⎢ ⎥ ⎢ ⎥∆ + ∆ + ∆ = + −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦∑ ∑ ∑o o o o

1

m

jj

n=∑ (5.56)

Where is the standard state specific heat at constant pressure for species j at temperature T. ( )0p j

C

5.5.4 Reduced Gibbs Iteration Equations

For problems with assigned thermodynamic states TP, HP, or SP various combinations of equations (5.51)

to (5.56) could be used to obtain corrections to estimates. However, for chemical systems containing many

species, it would be necessary to solve a large number of simultaneous equations. This large number of

equations can be reduced quite simply to a much smaller number by algebraic substitution, this is

accomplished by substituting the expression for ln jn∆ obtained from equation (5.51) in to equation (5.53) to

(5.56). Including equation (5.52) written with signs reversed, the resulting reduced number of equations are:

1 1 1 1 1 1

( )ln ln ( )

oL m n m m mkj j T j kj j jo

kj ij i kj j kj j k ki j j m j j j

a n H a na a a n a n n T b b

R T R Tµ

= = = + = = =

⎡ ⎤Π + ∆ + ∆ + ∆ = − +⎢ ⎥

⎢ ⎥⎣ ⎦∑∑ ∑ ∑ ∑ ∑o o

(5.57)

( 1,... )k L=

1

( )ln ( 1,..., )

oLT j j

ij ii

Ha T j m

R T R Tµ

=

⎡ ⎤Π + ∆ = = +⎢ ⎥

⎢ ⎥⎣ ⎦∑ o o

n (5.58)

∑∑∑∑∑∑===== =

+−=∆⎥⎥⎦

⎤

⎢⎢⎣

⎡+∆⎟⎟

⎠

⎞⎜⎜⎝

⎛−+Π

m

j

jjm

jj

m

j

joTj

m

jj

L

i

m

jijij RT

nnnT

RTHn

nnnna11111 1

ln)(

lnµ

(5.59)

1 1 1 1

2

2 21 1 1

( ) ( ) ( )ln

( ) ( ) ( )ln

( ) ( )

o o oL m Tn mij j T j T j j T j

i ji j m j

o o on m mj p j j T j j T j jo

j j j

a n H H n Hn n

R T R T R T

n C n H n Hh hTR R T R T R T

µ

= = + =

= = =

⎡ ⎤ ⎡ ⎤ ⎡Π + ∆ + ∆⎢ ⎥ ⎢ ⎥ ⎢

⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣⎡ ⎤ −

+ + ∆ = +⎢ ⎥⎢ ⎥⎣ ⎦

∑ ∑ ∑ ∑

∑ ∑ ∑

o o o

o o o o

⎤⎥⎥⎦

(5.60)

1 1 1 1

2 21 1 1 1

ln

( ) ( )ln

( ) ( )

L m Tn mij j j j j j

i ji j m j

o on m m mj p j j T j j j j jo

jj j j j

a n s s n sn n

R R R

n C n H s n ss sT n nR R T R R T

µ

= = + =

= = = =

⎡ ⎤ ⎡ ⎤⎛ ⎞Π + ∆ + ∆⎢ ⎥ ⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤ −

+ + ∆ = + − +⎢ ⎥⎢ ⎥⎣ ⎦

∑ ∑ ∑ ∑

∑ ∑ ∑ ∑

o o o

o o o o

(5.61)

After obtaining the previous correction variables, the corrections for gases species

are then obtained from equation (5.51). Of course convergence on the size of

corrections before they are applied to obtain improved estimates should be controlled.

ln ( 1,..., )jn j m∆ =

5.6 Results of Thermodynamic Calculations

1. Rocket engine performance

(1) Frozen equilibrium rocket performance

(2) Shifting equilibrium rocket performance

Actual thermochemical rocket performances fall between the results determined on the basis of

shifting and frozen equilibria.

2. Any intermediate rate cannot easily be treated analytically, depending on reaction kinetics.

3. In general, high specific impulse or high characteristic velocity can be obtained if the average

molecular weight of the reaction products is low (usually this implies a formulation rich in hydrogen)

or if the available chemical energy (heat of reaction) is large, which means high combustion

temperatures.

4. Actual values of specific impulse obtained from firing actual propellants in rocket units are, in

general, 3 to 12% lower than those calculated by the method explained in this chapter because of

some inefficiencies in the combustion chamber and the nozzle, only a portion of perhaps 1 to 4% is

due to combustion inefficiencies.

5. For the liquid propellant combination, liquid oxygen-RP-1:

For maximum specific impulse the following figures show an optimum mixture ratio of approximately

2.3 for frozen equilibrium expansion and 2.5 for shifting equilibrium expansion. The maximum values of

are at slightly different mixture ratios. It should be noted that this optimum mixture ratio is not the value for

highest temperature, which is usually fairly close to the stoichiometric value.

*c

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Fig.5- 1 Calculated performance analysis of liquid oxygen and hydrocarbon fuel as a function of mixture

ratio

Fig.5- 2 Calculated chamber gas composition for liquid oxygen and hydrocarbon fuel as a function of

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mixture ratio. Aggressive gases, such as O2, O or OH can cause oxidation of the wall materials in the chamber and the nozzle.

Fig.5- 3 Calculated nozzle exit gas composition for shifting equilibrium conditions as a function of mixture

ratio. Breakdown into O, OH or H and free O2 occurs only at the higher mixture ratios.

Fig.5- 4 Variation of theoretical specific impulse with mixture ratio and pressure ratio, calculated for frozen

equilibrium.

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Fig.5- 5 Variation of calculated parameters with pressure ratio for liquid oxygen hydrocarbon propellant at a mixture ratio 2.20. An increase in pressure ratio is due to an increase in chamber pressure, a decrease of nozzle exit pressure (larger area ratio and higher altitude), or both.

Fig.5- 6 Variation of exhaust gas composition at nozzle exit with pressure ratio at a fixed mixture ratio and for shifting equilibrium. For frozen equilibrium the composition would be the same as in the chamber, as shown in Fig.5- 2.

5.7 ASSIGNMENTS

For hydrazine monopropellant N2H4, it decomposes into hot gas composition including NH3, N2, and H2,

i.e., N2H4→NH3+N2+H2. Among the gas composition, there is an equilibrium dissociation reaction

2NH3 ⇔ 3H2+N2. Finish a computer program (C/Fortran/Matlab…) to calculate equilibrium combustion

compositions (molar numbers for 1kg combustion products) and temperature for combustion chamber pressure

20

cp =10atm by using the method of equilibrium constants (refer to the JANAF thermochemical tables and other

chemical literatures). Write an analysis report about this assignment. Submit the electronic Word version of the

report and the computer source code.

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