5. 522, Space Propulsic 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) hen the intersection of the orbital plane(Equator) and the ecliptic plane points to the sun. The maximum eclipse length is te.=2 RE =1.16hr Conce per day in the eclipse 2πtecl season If the satellite is to remain active sin ga during these occultations, enough battery capacity must b 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 provide power for overheating the gas generated by a n,H4 decomposition 16. 522, Space Propulsion Lecture 6 Prof. Manuel martinez-Sanchez Page 1 of 6
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 24 R t = 2 sin = 1.16 hr 2 R 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 N H2 4 decomposition
chamber and increase the performance. In the case of the dscs ill, the power available for electrothermal augmentation(ETA)is 977 w array 837 w array +750w batteries +750w batteries 723 w payload 723 w payload 1004w for eta 864 w for ETA 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=l(all NH3 decomposed) N2H4→N2+2H (+0.5%H2O+~0.5%NH Thus hs( at T)(Koper32 grams)=(-283+7756+0.1830)+2(-1.967+6.60+036702) C(T, cal/go c)=i dhas 32d=+0.6547+0.05730 c(c/9p10.67 =0.4685+0.05730 Y(T)=cp/cy e=2→r=1.319 e=1.4→r=1.335 +2X2 10.67 g/mole Given the available electrical energy(E) per Kg of gas(or power/mass flow rate),we can now solve for e by setting 0. 032Kg/mol Joule 「hea(Kca/32g)-120×130,800 liquid N2H4at298°K We can also calculate the performance in the same way as done before(from y, M Pe/Po). However, since cp does vary with T, let us do it a bit better this time Assuming the nozzle expansion is ideal, 16. 522, Space Propulsion Lecture 6 Prof. Manuel martinez-Sanchez Page 2 of 6
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 NH3 decomposed) N H N + 2H 2 2 4 → 2 2 3 (+0.5% H O+ 0.5% NH ) ∼ Thus ( )( ) ( ) ( ) 2 2 h at T Kc per 32 grams = -2.83 + 7.75 + 0.183 + 2 -1.967 + 6.6 + 0.367 gas θ θ θθ ( ) o gas p 1 d h c T, cal/g C = = +0.6547 + 0.0573 32 d θ θ o v p 1.987 c (T, cal/g C) = c - = 0.4685 + 0.0573 10.67 θ p v γ(T) = c c θ → = 2 r = 1.319 θ → = 1.4 r = 1.335 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 θ by setting 4180J/Kcal 0.032Kg/mol ( ) gas Joule E = h Kcal/32g. - 12.0 x130,800 Kg liquid N2H4 at 298°K We can also calculate the performance in the same way as done before (from γ , M, Pe/P0). However, since cp 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
ds=0→dh=dr SdT= R,T d in t= d In p P P= exp d In t Notice if Cp= constant, this gives P/ P T Now, since a+ bt spdInT=T+b dT=alnT+b(T-To) With our values, this gives 0.6547 °。(}"eb g。C cals3. 515, etc 3.515 Po To 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 m P R R At the throat a is minimum, so, setting d in a 1, dIn P 0 d t 2(-h) 16. 522, Space Propulsion Lecture 6 Prof. Manuel martinez-Sanchez Page 3 of 6
16.522, Space Propulsion Lecture 6 Prof. Manuel Martinez-Sanchez Page 3 of 6 1 ds = 0 dh = dp → ρ g p R T c dT = dp P p g c (T) d ln T = d ln P R 0 0 T p T g P c (T) = exp d ln T P R ∫ Notice if cp= constant, this gives 0 -1 0 P T = , P T γ γ as usual. Now, since p c = a+ bT, ( ) 0 0 T T p 0 T T 0 a T c d ln T = + b dT = a ln + b T - T T T ∫ ∫ With our values, this gives g ( ) g a b R T-T R 0 0 0 P T = e; P T g o o cal 0.6547 a g C = = 3.515, etc R 1.986 cal 10.67 g C 3.515 T-T0 0.3075 1000 0 0 P T = e P T 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 g g m PT PT = u= u= 2 h -h A RT RT ρ i enthalpy per Kg At the throat A is minimum, so, setting d ln A = 0, dT ( ) * * T * * T 0 dH 1 d ln P dT - + - =0 T dT 2 h -h
Cp 0 R。T2(h-h) C-R。_c/2 ho-h(T) R RT M Not (h-hi)_( RT R u=hRT (speed of sound at throat) (T=temperature at throat. Using the known functions h (T),Y(D), this equation can be solved for T If y =constant and h=C T, this would give the usual To y+1 The area ratio now follows from the a"pression A。_TPh-h ho -h Given A/Aand To, this equation can be solved for T. and hence Pe/Po can be calculated Finally, the usual performance parameters cF=FPo A and c=o.can be found mue+PA_ PuA+PAe_Pe Ae(2(ho-he) PA P RT 16. 522, Space Propulsion Lecture 6 Prof. Manuel martinez-Sanchez Page 4 of 6
16.522, Space Propulsion Lecture 6 Prof. Manuel Martinez-Sanchez Page 4 of 6 ( ) ( ) ( ) * * p p * * * g 0 1 cT cT - + - =0 T RT 2 h -h * v c * * pg p * * g 0 c -R c 2 = R T h -h ( ) * 0 * * g h -h T 1 = (T ) R T 2 γ g R R = M Note: ( ) ( ) γ * * 2 * 0 * * g g 2 h -h u = = RT RT So γ * ** u = RTg (speed of sound at throat) ( * T = temperature at throat.) Using the known functions h(T), γ(T), this equation can be solved for * T . If γ =constant and h = c T, p this would give the usual * 0 T 2 = T +1 γ . The area ratio now follows from the m A i expression: ee 0 e 0e * * * * A T h -h P = A T P h -h Given e * A A and T0, this equation can be solved for Te , and hence Pe/P0 can be calculated. Finally, the usual performance parameters F 0 * c =F P A and * * P A0 c = m i can be found: ( ) 2 e e e ee e e e e e 0 e F 0 0 0 ge * ** mu +P A ρ u A +P A P A 2 h -h c = = = +1 PA PA A P RT i
RT ≈P√2(n-h) From these Ff PA 1 C as usual mg PA Using an assumed area ratio Ae/A'=50, these calculations give the following results (Notice the weak variation of c and Pe/Po, due to the shifting y. Also(but not shown), calculation using constant y(using the y'for instance), gives very nearly the same results) To(K)(superheated gas temp. 1200 1400 1600 1800 2000 2200 2400 T(K)(temperature at throat) 1021 1194 1369 1544 1897 2075 y(at throat) 135361.3470 1.3407 1.3345 1.3286 1.3228 1.3172 (Kcal/mole N2H4) 19.693 244.36 33.913 38.8 43.76 48.793 53600.53700.5381 053920.5402 0.5412 0.5415 Pe/Pc 0.0007580.0007680.0007780.0007870.0007960.0008100.000821 T(°K) 170.0 202.1 235.3 269,4 304.5 341.7 379.5 C thrust coefficient) 1.746 1.748 1.750 1.750 1.751 1.759 1.762 c(,(characteristic velocity) 1431 1868 1954 2046 Isp(sec) 255.0 267.3 296.2 315.3 333.7 350.9 367.8 m(Mo/Kg, augmentation 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 Ae/A=50, vacuum operation) 16. 522, Space Propulsion Lecture 6 Prof. Manuel martinez-Sanchez Page 5 of 6
16.522, Space Propulsion Lecture 6 Prof. Manuel Martinez-Sanchez Page 5 of 6 ( ) * * * 0 0 g * * 0 PA P R T c= = m P 2 h -h i From these, * * 0 F sp * 0 FF 1 P A c c I= = = P A g g mg m i i , as usual. Using an assumed area ratio e * A A 50, = these calculations give the following results: (Notice the weak variation of cF and Pe/P0 , due to the shifting γ . Also (but not shown), calculation using constant γ (using the ∗ γ for instance), gives very nearly the same results). T0( K D )(superheated gas temp.) 1200 1400 1600 1800 2000 2200 2400 T* ( K D ) (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 h0 (Kcal/mole N2H4) 19.693 24.36 29.1 33.913 38.8 43.76 48.793 P* /P0 0.5360 0.5370 0.5381 0.5392 0.5402 0.5412 0.5415 Pe/P0 0.000758 0.000768 0.000778 0.000787 0.000796 0.000810 0.000821 Te( K D ) 170.0 202.1 235.3 269.4 304.5 341.7 379.5 cF(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 Isp(sec) 255.0 267.3 296.2 315.3 333.7 350.9 367.8 E=Pelec./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 Ae/A* =50, vacuum operation)
As we saw before, the thrust needed for NSsK depends on satellite mass, firing time and firing frequency M 0.9 v=circular velocity in GEO N N=no of firings per year From the engine view point F=mgIs and if we have an electric power Pel and use it with an efficiency n(the rest are heat losses), the specific ETA energy is E=卫→m=吗→F=吧PgLn where Isp depends upon E(table) For the DSCSIlIl satellite, assume n=0.75 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=3153 sec, E=2862×105/Kg So, the thrust 0.75×1004 2.862×10° 9.8×315.3=0.8130Nt This implies, for a DSCSIII mass of 1043 Kg 0.8130×3600 NSSK 0.9971 9.111×10-rad=0.0522° 1043×3071 N=-=17.2 per year 0.0522 (every 21.2 days) 16.522, Space Propulsion Lecture 6 Prof. Manuel martinez-Sa Page 6 of 6
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 b NSSK M i F= v t η ; 0.9 i = N D c v = circular velocity in GEO N = no. of firings per year From the engine viewpoint, F = mgIsp i and if we have an electric power Pel. and use it with an efficiency η (the rest are heat losses), the specific ETA energy is el el el sp P PP E = m = F = g I E E m η η η → → i i where Isp depends upon E (table). For the DSCSIII satellite, assume η = 0.75 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 b -4 NSSK c Ft 0.8130 3600 i = = 0.9971 = 9.111 10 rad = 0.0522 Mv 1043 3071 × η × × D and so 0.9 N = = 17.2 per year 0.0522 (every 21.2 days)