analysis of water injection into high-temperature mixture of ...

211/2( cy )∂y ∂∂c ∂y∂T ∂T ∂T∂T8 5 4 1/2 5 4= = y4 + D841/2 1/2( cy y )∂y ∂∂c ∂y∂T ∂T ∂T ∂T9 3 4 6 1/2 1/2 3 4= = y4 y6 + D94 + D961/2 1/2( cy y )∂y6∂T∂y ∂∂c ∂y ∂y∂T ∂T ∂T ∂T∂T10 7 4 3 1/2 1/2 7 3 4= = y4 y3 + D103 + D1041/2( cy )∂y∂11 6 3 1/2 ∂c6∂y = = y3 + D 3113∂T ∂T ∂T∂TEq.3-71Eq.3-72Eq.3-73Eq.3-74The independent set of derivatives is obtained by solution of matrix equationthat results from differentiating with respect to T (Temperature (K)) in matrix formas [ A][ ∂y/ ∂ T] + [ ∂f / ∂ T] = 0. The matrix [ A]is seen to be identical to that usedearlier in the Newton iteration from Eq.3-62 through Eq.3-65. To evaluate ∂f/ ∂ T ,which defined by inspection of Eq.3-48 through Eq.3-54 reveals that xiare functionsof y by i = 3, 4, 5 and 6. So,ix = y / c , x = y / c , x = y / c , x = y / c , x = y / c , x = y / c1 1 1 2 2 2 7 7 4 8 8 5 9 9 3 10 10 7and x11 = y11 / c8. Substitution of x iinto Eq.3-37 through Eq.3-40 followed bydifferentiation with respect to T , Eq.3-37 is expressed as below,y7 c5 y8 c3 y9 c7 y10 c611∂f1 ∂c1 ∂y1 ∂c2 ∂y2 ∂c ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂4∂y= + + + + + +∂T ∂T ∂c ∂T ∂c ∂T ∂c ∂T ∂c ∂T ∂c ∂T ∂c ∂T ∂c1 2 7 5 3 76Eq.3-75Where, ∂y1 ∂ c1 = y1 c1 ∂y2 ∂ c2 = y2 c2 ∂y7 ∂ c7 = y7 c7 ∂y8 ∂ c5 = y8 c5/ / , / / , / / , / / ,∂y9/ ∂ c3 = y9/ c3, ∂y10/ ∂ c7 = y10/ c7 and ∂y11/ ∂ c6 = y11/c6.Also, differential with respect to T expressed from Eq.3-37 through Eq.3-40 as below,∂f ∂c ∂c ∂c∂ c ∂ c ∂ c ∂ c∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂1 1 2 45 3 7 6= x1+ x2 + x7 + x8 + x9 + x10+11T T T T T T T T x∂f ∂c ∂c ∂ c ∂c∂ ∂ ∂ ∂ ∂2 2 4 31= 2 x2 + x7 + x9 −d11T T T T T x∂ f3 ∂c1 ∂c2 c5 c3 c7c12 x1 x ∂ 2x ∂ 8x ∂ ∂= + + +9+ x10 −d2x∂T ∂T ∂T ∂T ∂T ∂T ∂T∂f4∂c ∂c ∂c1= x + x −d x∂T ∂T ∂T ∂T7 610 11 3 11Eq.3-76Eq.3-77Eq.3-78Eq.3-79