For an adiabatic combustion(no heat loss, no heating), we must have(starting from liquid N2H4 at 298K) h(,298)4(1×)hm(T)+31+2×)h2(T)+2xh(T)→ Equation for T The respective molar enthalpies can be fitted by h(T)=16.83+12.350+0.983020=7 mole 1000°K h2(T)=-2.83+7756+01830 h4(T)=1967+660+03670(300≤T≤4000K (Notice c,(cal/mole C))is o, with h in Kcal/mole) From the table given, hm((, 298 K)=12 Kcal/mole. We can now solve for T at various arbitrary values of x: √39.3-20.75×+845×2-(9.525+095×) 1.372-0.455X x(fraction of NH3 decomposed) 0 0.2040.60.81 T(K), adiabatic temperature 16591502134311821023863 16. 522, Space Propulsion Lecture 5 Prof. Manuel martinez-Sanchez age 7 of 1216.522, Space Propulsion Lecture 5 Prof. Manuel Martinez-Sanchez Page 7 of 12 For an adiabatic combustion (no heat loss, no heating), we must have (starting from liquid N2H4 at 298 KD ) ( ) ( ) () ( ) () () N H NH N H 24 3 2 2 4 1 h ,298 1-x h T 1+2x h T 2xh T Equation for T 3 3 A D = + +→ The respective molar enthalpies can be fitted by ( ) 3 2 cal NH K T h T = -16.83 +12.35 + 0.983 = mole 1000 K   θ θθ    D ( ) 2 2 h T = -2.83 + 7.75 + 0.183 N θ θ ( ) 2 2 h T = -1.967 + 6.6 + 0.367 H θ θ (300 T 4000 K ≤ ≤ ) D (Notice c cal mole C p ( ) D ) is dh dθ , with h in Kcal/mole) From the table given, ( ) N H2 4 h ,298 K 12 Kcal/mole = D A . We can now solve for T at various arbitrary values of x: ( ) 2 139.3 - 20.75x + 8.45x - 9.525 + 0.95x = 1.372 - 0.455x θ x (fraction of NH3 decomposed) 0 0.2 0.4 0.6 0.8 1 T( K), D adiabatic temperature 1659 1502 1343 1182 1023 863