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±ùµã½µÎÂ: ½ð±Ò+2, лл»ØÌû£¬ÐÁ¿àÁË£¡ 2015-12-01 20:17:37
steve.jobs(ÇàÈô´ú·¢): ½ð±Ò+5, лл»ØÌû 2015-12-22 21:09:08
£¨1£©¸ù¾ÝÌâ¸ÉÏÈÐè¾Ý¿Ë-¿Ë·½³Ì¼ÆËãAÔÚ320.2KʱµÄp*(A)=50.44kPa
p=p(A)+p(B)=101=0.1p*(A)+0.9p*(B)=5.044+0.9p*(B)
p*(B)=106.62kPa
(2)x(A)=0.4
p=p(A)+p(B)=0.4p*(A)+0.6p*(B)=84.15kPa
y(A)=p(A)/p=0.24   y(B)=1-0.24=0.76
(3)y(A)=0.4 y(B)=0.6
y(A)/y(B)=p(A)/p(B)=x(A)P*(A)/(1-x(A))P*(B)=0.4/0.6=2/3
X(A)=0.58  X(B)=0.41
p=p(A)+p(B)=0.58p*(A)+0.42p*(B)=74.04kPa
2Â¥2015-12-01 19:25:12
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steve.jobs

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ÒýÓûØÌû:
2Â¥: Originally posted by ¹ÈÖÐÃ÷ at 2015-12-01 19:25:12
£¨1£©¸ù¾ÝÌâ¸ÉÏÈÐè¾Ý¿Ë-¿Ë·½³Ì¼ÆËãAÔÚ320.2KʱµÄp*(A)=50.44kPa
p=p(A)+p(B)=101=0.1p*(A)+0.9p*(B)=5.044+0.9p*(B)
p*(B)=106.62kPa
(2)x(A)=0.4
p=p(A)+p(B)=0.4p*(A)+0.6p*(B)=84.15kPa
y(A)=p(A)/p=0.24   ...

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3Â¥: Originally posted by steve.jobs at 2015-12-01 20:14:30
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steve.jobs

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ÒýÓûØÌû:
4Â¥: Originally posted by ¹ÈÖÐÃ÷ at 2015-12-01 20:25:00
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