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)