16.522, Space Propulsion Lecture 6
Prof. Manuel Martinez-Sanchez Page 1 of 6
16.522, Space Propulsion
Prof. Manuel Martinez-Sanchez
Lecture 6: Hydrazine Decomposition: Performance Estimates
(3) Electrothermal Augmentation Concept
Geostationary satellites are most of the time exposed to the sun, but they still are
subject to eclipse periods around the two vernal points (March 21, September 21);
when the intersection of the orbital plane (Equator) and the ecliptic plane points to
the sun. The maximum eclipse length is
-1 E
ecl.
GEO
R24
t=2 sin =1.16 hr
2R
??
??
??
π
(once per day in the “eclipse
season”)
If the satellite is to remain active
during these occultations, enough
battery capacity must be
incorporated.
But obviously, these batteries have ample time to recharge even in eclipse season,
and are idle most of the time.
Starting from the payload requirements, the solar array capacity at End of Life (EOL)
is typically dimensioned with a 15% margin to allow for battery charging losses. A
further 15% is then added (depending on mission duration and altitude) to allow for
array degradation from Beginning of Life (BOL) to EOL. An example is the military
satellite DSCS III:
Array output (BOL/EOL) 977 watt/837 watt (28V ±1%)
Batteries 750 watt
1968 watt-hr (100% DoD)
[depth of discharge]
1180 watt-hr (60% DoD)
Payload requirements 723 watt
Thus, there is the possibility of using occasionally both the batteries and the arrays
to provide power for overheating the gas generated by a
2 4
NH decomposition
16.522, Space Propulsion Lecture 6
Prof. Manuel Martinez-Sanchez Page 2 of 6
chamber and increase the performance. In the case of the DSCS III, the power
available for electrothermal augmentation (ETA) is
If the firing is for 1 hr, the batteries provide 750 watt-hr, which is 38% depth of
discharge only, and should not compromise battery life.
To calculate the performance achievable, we can use the enthalpy equation again;
this time, since the temperature is going to be high and the residence time in the
heater long, we can assume equilibrium is reached, which essentially means x=1 (all
NH
3
decomposed)
22
4
2
NH N +2H→
23
(+0.5% H O+ 0.5% NH )~
Thus
()( )()( )
22
gas
h at T Kc per 32 grams = -2.83+7.75 +0.183 +2 -1.967+6.6 +0.367 θ θθθ
()
gaso
p
d h
1
c T, cal/g C = = +0.6547+0.0573
32 d
θ
θ
o
vp
1.987
c (T, cal/g C) = c - = 0.4685+0.0573
10.67
θ
pv
(T) = c cγ = 2 r =1.319θ→
=1.4 r =1.335θ→
28+2x2
M = =10.67 g/mole
3
Given the available electrical energy (E) per Kg of gas (or power/mass flow rate), we
can now solve for θby setting
4180J/Kcal
0.032Kg/mol
()
gas
Joule
E = h Kcal/32g. -12.0 x130,800
Kg
??
??
??
??
??
liquid N
2
H
4
at 298°K
We can also calculate the performance in the same way as done before (from γ , M,
P
e
/P
0
). However, since c
p
does vary with T, let us do it a bit better this time.
Assuming the nozzle expansion is ideal,
BOL EOL
977 w array 837 w array
+750 w batteries +750 w batteries
-723 w payload -723 w payload
1004 w for ETA
864 w for ETA
16.522, Space Propulsion Lecture 6
Prof. Manuel Martinez-Sanchez Page 3 of 6
1
ds = 0 dh = dp→
ρ
g
p
RT
cdT= dp
P
p
g
c (T)
d ln T = d ln P
R
0
0
T
p
gT
c(T)
P
=exp d ln T
PR
??
??
??
∫
Notice if c
p
= constant, this gives
0
-1
0
PT
=,
PT
γ
γ
??
??
??
as usual.
Now, since
p
c=a+bT,
()
00
TT
p 0
0TT
aT
c d ln T = +b dT = a ln +b T - T
TT
??
??
??
∫∫
With our values, this gives
()
g
g
a
b
T-TR
0
R
00
PT
=e;
PT
??
??
??
g
o
o
cal
0.6547
a gC
= = 3.515, etc
1.986 calR
10.67 gC
??
??
??
3.515
T-T
0
0.3075
1000
00
PT
=e
PT
??
??
??
The nozzle is likely to be specified by a given area ratio rather than a given pressure
ratio. To relate these two, let us first calculate the throat conditions. In general
() ()
()
0
gg
PT PT
m
=u= u= 2h-h
ARTRT
ρ
i
enthalpy per Kg
At the throat A is minimum, so, setting
d ln A
=0,
dT
()
*
*
T
* *
T
0
dH
dT
1d ln P
-+ - =0
dTT
2h-h
??
??
????
??
??
16.522, Space Propulsion Lecture 6
Prof. Manuel Martinez-Sanchez Page 4 of 6
() ( )
()
**
pp
**
*
g
0
cT cT
1
-+ - =0
TRT
2h-h
*
v
c
**
pg p
**
g0
c-R c 2
=
RT h -h
()
*
0
*
*
g
h-hT
1
=(T)
2RT
γ
g
R
R=
M
??
??
??
Note:
( ) ( )
γ
*
*2
0
*
**
gg
2h-h u
==
RT RT
So
γ
***
g
u= RT (speed of sound at throat)
(
*
T = temperature at throat.)
Using the known functions h(T), (T),γ this equation can be solved for
*
T.
If γ =constant and
p
h=c T,this would give the usual
*
0
T2
=
T+1γ
.
The area ratio now follows from the
m
A
i
expression:
ee 0
e0e
**
**
AT h-hP
=
Ph-hAT
Given
e
*
AAand T
0
, this equation can be solved for
e
T , and hence P
e
/P
0
can be
calculated.
Finally, the usual performance parameters
F0
*
c=FPA and
*
* 0
PA
c=
m
i
can be found:
()
2
0eeee eeeee ee
F
0ge00
***
2h-hmu +P A ρ uA +PA P A
c= = = +1
PRTPA PA A
??
??
??
i
16.522, Space Propulsion Lecture 6
Prof. Manuel Martinez-Sanchez Page 5 of 6
()
*
*
g*
00
*
*
0
RT
PA P
c= =
P
2h-h
m
i
From these,
* *
0 F
sp *
0
PA ccFF 1
I= = =
ggPA
mg m
ii
, as usual.
Using an assumed area ratio
e
*
AA 50,= these calculations give the following results:
(Notice the weak variation of c
F
and P
e
/P
0
, due to the shifting γ . Also (but not
shown), calculation using constant γ (using the
?
γ for instance), gives very nearly
the same results).
T
0
( K
null
)(superheated gas temp.)
1200
1400
1600
1800
2000
2200
2400
T
*
( K
null
) (temperature at throat)
1021
1194
1369
1544
1720
1897
2075
γ
*
(at throat)
1.3536
1.3470
1.3407
1.3345
1.3286
1.3228
1.3172
h
0
(Kcal/mole N
2
H
4
)
19.693
24.36
29.1
33.913
38.8
43.76
48.793
P
*
/P
0
0.5360
0.5370
0.5381
0.5392
0.5402
0.5412
0.5415
P
e
/P
0
0.000758
0.000768
0.000778
0.000787
0.000796
0.000810
0.000821
T
e
( K
null
)
170.0
202.1
235.3
269.4
304.5
341.7
379.5
c
F
(thrust coefficient)
1.746
1.748
1.750
1.750
1.751
1.759
1.762
c
*
(
m
sec
), (characteristic velocity)
1431
1548
1659
1766
1868
1954
2046
I
sp
(sec)
255.0
267.3
296.2
315.3
333.7
350.9
367.8
E=P
elec.
/m
i
(MJ/Kg, augmentation
power)
1.005
1.614
2.234
2.862
3.501
4.149
4.806
Performance of Electrothermally Augmented Hydrazine Microthrusters, as a Function
of Ideal Augmentation Power. (For A
e
/A
*
=50, vacuum operation).
16.522, Space Propulsion Lecture 6
Prof. Manuel Martinez-Sanchez Page 6 of 6
As we saw before, the thrust needed for NSSK depends on satellite mass, firing time
and firing frequency
c
bNSK
Mi
F= v
t η
;
0.9
i=
N
null
??
??
??
c
v = circular velocity in GEO
N = no. of firings per year
From the engine viewpoint,
sp
F=mgI
i
and if we have an electric power P
el.
and use it with an efficiency η (the rest are heat
losses), the specific ETA energy is
el el el
sp
PPP
E = m = F = g I
EE
m
η η η
→→
i
i
where I
sp
depends upon E (table).
For the DSCSIII satellite, assume =0.75η and BOL power (1004 watts). Suppose
one can achieve a superheated gas temperature of 1800°K (mostly limited by
materials).
Then, from the table,
Isp = 315.3 sec,
6
E = 2.862 10 J/Kg×
So, the thrust is
6
0.75 1004
F = 9.8 315.3 = 0.8130 Nt
2.862 10
×
×
×
This implies, for a DSCSIII mass of 1043 Kg
-4b
NSSK
c
Ft 0.8130 3600
i = = 0.9971 = 9.111 10 rad = 0.0522
Mv 1043 3071
×
η×
×
null
and so
0.9
N = = 17.2 per year
0.0522
(every 21.2 days)