石油大学化学工程原理课后习题答案

-3\r= 1.687 x l O\r,6.777 x l O -3\r^ 1 . 3 2 1 x 1 0 -\rY - Y _ 0.99 X 2.041 X 10-2\r{ 2\r= 11.98\r1.687 x l O -3\rA @\rV 0.015x(1-0.02)\r" O G = = 0.98m\rK Y a ; Q 0.015\rZ = HOGNOG = 0.98 X 11.98 = 11.74/71\r~ @ M D N K Y\rm V 0.36\rS = = 0.6711\rL 0.5364.\r1 j h °lX 4 mX 2\rN OG ~ In ( 1 - 5 ) - ------- - + Sp\r1 -S L % - ¤\r1 ,  ˜¥l2.041x 10-2 40.36x2x10-4 0 [\rIn (1-0.6711)-------------- ----------------------7 + 0.6711\r1-0.6711 2.041 x IO- 4 -0 .3 6 x 2 x l0 -4\r= 11.97\rZ = H COCO/CCr N C()/CU = 0.98 x 11.98 = 11.74m