GB T 18340.4-2001 地质样品有机地化测试 重馏分石油芳香烃族组成质谱分析方法.pdf

《GB T 18340.4-2001 地质样品有机地化测试 重馏分石油芳香烃族组成质谱分析方法.pdf》由会员分享，可在线阅读，更多相关《GB T 18340.4-2001 地质样品有机地化测试 重馏分石油芳香烃族组成质谱分析方法.pdf（28页珍藏版）》请在麦多课文档分享上搜索。

1、ICS 73- 010 D 04 道昌中华人民共和国国家标准GB/ T 18340.4-2001 地质样品有机地化测试重馆分石油芳香怪族组成质谱分析方法Organic geochemical measurements of geological samples Method for aromatic types analysis of gas-oils aromatic fractions by mass spectrometry 2001- 03 -19发布2001-10-01实施国家质量技术监督局发布GB/T .18340. 4- 2001 前本标准是在地质矿产石油地质海洋地质局企业标准重

2、锢分油芳香怪族组成质谱分析方法的基础上，参考石油化工科学研究院质谱法测定重馆分油芳香怪的烧类组成分析方法以及国内外有关资料，按照先进性、实用性与可靠性相结合的原则，并经过对比测试的实践检验后制定的。本标准确定了重锢分石油芳香炬族组成的质谱分析方法、定量计算过程和分析精密度要求。;$:标准的附录A是标准的附录。本标准由国土资源部提出。本标准由全国地质矿产标准化技术委员会岩矿测试标准物质和方法技术委员会归口。本标准起草单位:中国新星石油公司实验地质研究院、上海医药工业研究院、上海染料研究所。本标准主要起草人:周锦南、王惠敏、刘德晨。 1 范围中华人民共和回国家标准地质样晶有机地化测试重馆分石油芳香

3、短族组成质谱分析方法Organic geochemical measurements of geoJogical samples Method for aromatic types analysis or gas-oils aromatic rractions by mass spectrometry G/T 18340.4- 2001 本标准规定了重馆分石油芳香短族组成的质谱分析方法、定量计算过程和分析精密度等。本标准适用于重偏分石油沸程范围300C540C)芳香娃馆分中的18种芳香短及3种芳香嚓盼的怪族组成的定量分析。试样中必须不含烯炬，总硫量必须低于总重量的1%，-非芳怪含量应20 ;:

4、;5 4 GB/T 18340. 4- 2001 附录A标准的附录)计算重锢分石油芳香短族组成用的FORTRAN源程序A1 主程序ATAPROGRAM ATA C C 本程序是计算重馆分石油芳香怪族组成用的主程序C 程序中的H(M)代表质量M的多同位素峰高C 程序中的HDI(M)代表质量M的单同位素峰高C DIMENSION MS(8) ,HT(8) CHARACTER赞32FIL，TIT铸64，COAT铃11，CTIM铸8COMMON H(lOOO) ,HDI(lOOO) C 质谱的读人、平均、相减和储存CALL RMS (FIL , TIT ,COA T ,CTIM) C 校正重同位素影响

5、CALL OEISO C 计算和报告芳香怪族组成CALL AROMTC (FIL , TIT ,COA T ,CTIM) STOP ENO A2 子程序DEISOSUBROUTINE OEISO C C 本子程序计算单同位素峰强，假设所有离子的Z值在+2至一11之间C 分子通式为C(N)H(2N+Z).C COMMON H (lOOO) ,HDI (1000) DIMENSION NCARB(lOOO) ,NHYO(lOOO) 00101= 12,758 NCARB(I)= 0 10 NHYO(I)=0 0020 K=12,758 NCARB(K)= (K+11 ) / 14 NHYO(K)=

6、K-12铃NCARB(K)IFCNHYO(K). LT. 0) NHYO(K)=O 20 CONTINUE 00 30 K=14,758 HOI(K)=H(K)-HOI(K-1)铃(.010811铃FLOAT(NCARB(K一1)+.00015祷祷FLOAT (NHYO (K - 1) ) ) 5 GB/ T 18340.4- 2001 HDI(K)=HDI(K)+HDI(K-2)铃(.00005844铃FLOAT(NCARB(K-2)祷铃(1-NCARB(K一2)+.1125E一7赞FLOAT(NHYO(K-2)铃(1-NHYO(K一2) 养一.162165E-5铃FLOAT(NCARB(K

7、-2)铃NHYD(K- 2) IF(HDI(K). LT. 0. 0) HDI(K) = 0.0 30 CONTINUE RETURN END A3 子程序AROMTC6 SUBROUTINE AROMTC(W ,TITLE ,CDAT1 ,CTIM1) C C 本子程序给出了石油重芳娃馆分计算过程C 详细的计算过程描述见AnalChem 41 ,1548-54(1969). C C COMMON H (1 000) ,HDIOOOO) DIMENSION AIN (7, 7) ,BA(7) ,BB(7) ,SR(1 000) CHARACTER ANS赞1，W祷32，OEV铃32，F2501

8、祷8，TITLE祷64，CDAT1铃11，CTIM1铃8DATA AIN/ + 1. 8094，一.1952，+.0124，一.0027，一.0015，一.0011，一.0028，铃一.1601，+2.0479，一.2806，一.0401，+.0082，+.0012，+.0000，铃一.0943，一.2287，+2.3024，一.4936，.0601，一.0155，一.0089，祷.0292，+.0033，一.0580，+1.9404，一.1337，一.0117，一.0043，铃.0022，一.0033，一.0026，一.0195，+1.9773，一.1823，+.0123，铃一.0420，+

9、.0026，一.0018，一.0151，-.0584，+2.0616，-.4193，铃一.2346，-.1069，一.0267，一.0019，一.0057，一.0904 , + 1. 9904/ NC1 =LEN T RIM (W) NC2 = LEN T RIM (T1TLE) C 开方数组初始化C DO 2132 1= 12,750 2132 SR(I)= O. O C C 加和芳怪馆分相关谱峰以用于7X7矩阵处理C A6，A7，A8.的值先是代表Z值为6汀，8.的峰强加和C 然后在A6，A7，A8.中又加入了相应于分子离子峰一1系列的奇数Z值峰强C A6=A6+A7,A8 =A8+ A9

10、. . . C A6=0. 0 00 2106 M = 78,750 ,14 2106 A6=A6+H(M) A7=0.0 GB/T 18340. 4- 2001 DO 2107 M=91 ,750 ,14 2107 A7=A7+HDI(M) A6=A6+A7 IF(A6. LT. 1. 0) A6= 1. 0 A8=0.0 DO 2108 M =104.750,14 2108 A8=A8+H (M ) A9=0.0 DO 2109 M=117 ,75 0,14 2109 A9=A9+ HDI (M) A8= A8+A9 IF(A8. LT. 1. 0) A8=1. 0 A10=0. 0 DO

11、 2110 M=130,750.14 2110 A10=A10+H(M) A11 =O.O DO 2111 M=129 ,750.14 2111 All = A11 +HDICM) A10=A10+A11 IF(A10. LT. 1. 0) A1 0=1. 0 A12=0. 0 DO 211 2 M = 128 ,750 ,14 2112 A12=A12+ H(M) A13=0.0 DO 2113 M=141 ,750.14 2113 A13=A13+HDI (M) A12=A12十A13IF(A12. LT. 1. 0) A12 = 1. 0 A14= 0.0 DO 2114 M=154,7

12、50 ,14 211 4 A14=A14+H (M) A1 5=0.O DO 211 5 M = 167 ,750 ,14 2115 A15=A15+HDICM) A14= A14+A15 IF(A1 4. LT. 1. 0) A1 4=1. 0 A16 =0.0 DO 2116 M =166,750 ,14 2116 A16=A16+H(M) A17=0. 0 DO 2117孔1=179，750，142117 A17=A17+HDICM ) A16 = A16+A17 7 8 C GB / T 18340. 4- 2001 IF(AI6. LT. 1. 0) A16=1. 0 A18=0.O

13、 DO 2118 M = 178 ,750 ,14 2118 A18=A18十H(M)A19=0.0 DO 2119 M=191 ,750 ,14 211 9 A19=A19十HDI(M)A18=A18+A19 IF(A18. LT. 1. 0) A18= 1. 0 C 对质量数175，176.189，190，200，213处的不恰当的离子峰强作校正C C 校正HDI(175)C C CD1175 =HDI(161)一(HDIC161)-HDI(203)/3.0IF(HDI(175). GE. CDI1 75) GO TO 1046 CDI175=HDI(175) C 校正H(176)C C

14、1046 CH176=H(l62) - (H(162) - H (204) )/3.0 IF(H(176). GE. CH1 76) GO TO 1048 CH176= H (176) C 校正HDI(189)C C 1048 CD1189=CDI175一CCDl175-HDl(203 ) ) /2.0 IFCHDI(189). GE. CDl189) GO TO 1049 CDI1 89 =HDl(189) C 校正H(190)C 1049 CH190=CH176一(CH176-H(204)/2.0IF(H (190). GE. CH190) GO TO 2101 CH190=H(190)

15、C C 校正H(200)C C 2101 CH200= (H(186)十H(214) / 2. 0 IF(H (200). GE. CH200) GO TO 2102 CH200 =H(200) C 校正HDl(213)C C GB/ T 18340.4-:-2001 2102 CDI213=(HDI(l99)+HDI(227)/2. 0 IF(HDI(213). GE. CDI213) GO TO 2103 CDI213=HDI(213) C 下述语句校正A6，AI0加和2103 A6=A6一(HDI(175)十HDI(189)+H(176)+H(190) 铃+CDIl75+CDI189+C

16、H176+CH190A10=A10一(H(200) + HDI (213) +CH200+CDI213 C C 把A6，A8.加和值重定义为数组元素变量，然后以芳娃逆矩阵AIN(l，)乘之C C BA(l)=A6 BA(2) =A8 BA(3)=A10 BA(4)=A12 BA(5)=A14 A(6)=A16 BA(7)=A18 DO 2125 J=1 , 7 BB() =0. 0 DO 21241=1 ,7 2124 BB()=BB(J)+BA(I)祷AIN(,1) 2125 CONTINUE DO 2127 J = 1,7 IF(BB() 2126 ,2127 ,2127 2126 BB(

17、) =0.0 2127 CONTINUE AA6=BB(1) AA8=BB(2) AA10=BB(3) AA12=BB(4) AA14 = BB(5) AA16=BB(6) AA18= BB(7) SUMAA=O.O DO 2128 J=1 ,7 2128 SUMAA=SUMAA+BB(J) C 计算出的AA6，AA8.即相应于Z=-6，-8.等类芳怪的分量大小C SUMAA是AA值的加和，它代表所计算芳煌的总量C 下面用基线法求解Z=-6重叠型C 9 10 C GB/ T 18340. 4- 2001 A 7=A7- HDI (175) - HDI (189)+CDI175+CD1189 H

18、DI(175)=CD1175 HDI (189) =CD1189 002130 M= 105.750.14 IFCHDICM) 2130 ,2131.2130 2130 CONTINUE 2131 MM=M一14* SLOPE= ( (0.72铃HDI(105)铃铃O.5)一(HDI(MM)棒祷0.5)/(90. 71一(1000.0/FLOAT(如1M)赞祷2)B= (0.72怜HDI(105)善赞0.5-90.71赞SLOPE00 2133 M = 147.MM.14 REALM=M 2133 SR(M) =SLOPE铃(1000.0/REALM)怜铃2+BC 以上是2=-6重叠型，皖基苯

19、峰强的开方储存于数组SR(D中C 以下是解2=-8重叠型C C C C C 002134 M = 215 ,750.14 IF(HDI(M) 2134;2135,2134 2134 CONTINUE 2135 MN=M-14 * SLOPE= ( (0.66铃HDI(73)铸铃0.5)一(HDI(MN)赞祷0.5)/(34. 12一OOOO.O/FLOAY(MN)祷祷2)B= (0.66铃HDI(173)铃赞0.5-34.12铃SLOPE00 2136 M=21 5,MN , 14 REALM = M 2136 SR(M)=SLOPE祷(1000.0/REALM)铃铃2+B以下是解2=-10重

20、叠型Al1= All-HDI(213) + CDI213 HDI(213)二CDI213002137 M =241.750.14 IFCHDI(M ) 2137.2138 , 21 37 2137 CONTINUE 2138 MO= M-14 SLOPE = (HDI(185)特祷O.5一(HOIMO)祷祷0.5)/* (29. 22一(1000.0/FLOAT(MO)铃铃2)B=HOI(185)铃铃0.5-29.22铃SLOPE002139 M =241 ,MO , 14 REALM= M 2139 SR(M)=SLOPE祷(1000.0/REALM)祷赘2+BGB/ T 18340.4-

21、2001 C 以下是解2=-12重叠型C DO 2140 M=197 . 750.14 IF(HDI(岛1)2140.2141.2140 2140 CONTINUE 2141 MP=M-14 SLOPE=(0.25铃HDI(183)铃铃0.5)一(HDI(MP)赞铃0.5)/铃(29.86一(1000.0/FLOAT(MP)铃铃2)B= (0.25铃HDI(183)铃铃0.5-29.86铃SLOPEDO 2142 M = 197.MP. 14 REALM= M 2142 SR(M)=SLOPE祷(1000.0/REALM)赞铃2+BC C 以下是解2=-14重叠型C C DO 2143 M =

22、 265.750.14 IF(HDI(肌1)2143.2144.2143 2143 CONTINUE 2144 MQ=M-14 * SLOPE=(0.64赞HDI(251)费铸0.5)一(HDI(MQ)铃铃0.5)/(15.87一(1000.0/FLOAT(MQ)铃提2)B= (0.64铃HDI(251)棒赞0.5-15.87祷SLOPEDO 2145 M =265.MQ.14 REALM = M 2145 SR(M)=SLOPE怜(1000.0/REALM)祷怜2+BC 以下是解2=-16重叠型C C DO 2146 M = 291.750.14 IF(HDICM) 2146. 2147 .

23、2146 2146 CONTINUE 2147 MR=M-14 SLOPE=(0.7铃HDI(277)养铃0.5)一(HDI(MR)祷铃0.5)/祷(13.03一(1000.0/FLOAT(MR)铃祷2)B= (0.7祷HDI(277)铃铃0.5-13.03铃SLOPEDO 2148 M =291.MR.14 REALM= M 21 48 SR(M) =SLOPE铃(1000.0/REALM)铃铃2+BC 以下是解2=一18重叠型C DO 2149 M =247.750.14 IF(HDI(M) 2149.2150. 2149 11 12 C GB/ T 18340. 4-2001 2149

24、CONTINUE 2150 MS=M - 14 * SLOPE= ( (0.58祷HDI(233)铸铃0.5)一(HDI(MS)铃铃0.5)/(18.42一(1000.0/FLOAT(MS)提提2)B= (0.58祷HDI(233)铸协0.5-18.42祷SLOPEDO 2151 M=247 ,MS,14 REALM=M 2151 SR(M)=SLOPE铃(1000.O/REAL岛1)祷祷2+BC 至此已计算了开方数组C 对某些谱在零峰高区仍得到斜率和截距。如发生这一情况，将会把错误带入SR数组C 以下语句使HDI=O.O时SR也为零C DO 2153 1= 12 ,750 lF(HDI (1

25、) 2152 ,2152 ,2153 2152 SR(I) =0.0 2153 CONTINUE C C SR数组平方，得到各个Z值的未校准标称型峰高C DO 2154 1=12,750 2154 SR(1) =SR(I)提提2C C 对某些非线性开方关系的SR(l)值进行校准C SR(147) =SR(147)铃1.44 SR (197) = SR (197)祷3.10 SR (211 ) = SR ( 211 )祷2.52SR(225)=SR(225)铃2.07 SR(239)=SR(239)赞1.83 SR(253)=SR(253)祷1.59 SR(267)=SR(267)祷1.39 S

26、R(281)=SR(281)提1.28 SR(295) =SR(295)铃1.26 SR(309)=SR(309)铃1.14 SR(323)=SR(323)保1.06 SR(265)=SR(265)祷1.42 SR(279)=SR(279)祷1.24 SR (293) =SR (293)赞1.12 SR(307) =SR(307)铃1.06 SR(291) =SR(291)铃1.24 SR(305) =SR(305)铃1.15 SR(319) =SR(319)骨1.07 飞、电C G/ T 18340. 4-2001 SR(333) =SR(333)铃1.0.6 SR(347)=SR(347)

27、升1.0. 5 SR(361) =SR(361)铃1.0.3 SR(24=SR(247)蕃1.61 SR(261) =SR(261)铃1.50. SR(275) =SR(275)头1.44 SR(289)=SR(289)铃1.37 SR(3D3) =SR(3D3)赞1.28 SR(317) =SR(317)赞1.28 SR(331) =SR(331)祷1.21 SR(345)=SR(345)祷1.10. SR(359)=SR(359)铃1.0. 9 SR (373) =SR (373)铃1.0.7 SR(387) =SR(387)铃1.0. 5 C 如发现有SR(M)值超过对应的HDI(M)则

28、进行校正C DO 2156 M=12,75D IF(SR(M)-HDI(M) 2156 ,21 56,2155 2155 SR(M)=HDI(M) 2156 CONTINUE C C 计算A7的各个部分即A6A，A1DS，A2DA各个类型C C A6A= D. D DO 2157 M=91 ,133 ,14 2157 A6A=A6A十HDI(M)DO 2158 M=147,MM,14 2158 A6A=A6A+SR(M) A1DS=D.D DO 21 59 M=147 ,189,14 2159 A1DS=A1DS+HDIM) - SR(M) A1DS = A1DS/D.75 A2DA=A7-A

29、6A- A1DS IF(A2DA) 216 0., 2161 , 2161 216 0. A2DA= D.D A1o.S=A7- A6A 2161 CONTINUE C 计算分量A6A，A1DS和A2DAC TRASH=(A6-AA6赞D.5579)祷(A7/A6)IF(TRASH. LT. D. 0.) TRASH= D. 0. 13 14 C A7=A7-TRASH A6A=A6A-TRASH GB/ T 18340. 4- 2001 IF(A6A. LT. o. 0) A6A=0. 0 IF(A6A. EQ. o. 0) A7 = A10S+A20A IFCA7. EQ. o. 0) T

30、HEN A6A = 0.0 AI0S=0.0 A20A=0.0 ELSE A6A=(A6A/A7)铃AA6AIOS= CA10S/ A7)铸AA6A20A = CA20A / A 7)祷AA6ENOIF C 计算A9的各个部分即A8A，A22A和A36AC C A8A=0.0 00 2162 M = 11 7 , 201 , 14 2162 A8A=A8A+HDI(M) 00 2163 M =215 ,MN , 14 21 63 A8A=A8A+SRCM) A22A=0. 0 002164 M = 215 ,257 ,14 2164 A22A=A22A+HDICM) - SRCM) A22A

31、= A22A/0.75 A36A=A9-A8A-A22A IFCA36A) 2165 ,21 66 ,2166 216 5 A36A =0. 0 A22A = A9 - A8A 2166 CONTINUE C 计算分量A8A，A22A和A36AC TRASH= CA8-AA8铃O.4997)提(A9/A8) IF(TRASH. LT. O. 0) TRASH=O. 0 A9= A9-TRASH A8A = A8A - TRASH IFCA8A. LT. O. 0) A8A=0. 0 IF(A8A. EQ. O. 0) A9=A22A+A36A IF(A9. EQ. O. 0) THEN A8A

32、=0.0 A22A =0.0 A36A=0.0 C GB/ T 18340.4- 2001 ELSE A8A= (A8A/ A9)头AA8A22A= (A22A/ A9)祷AA8A36A= (A36A/ A9)铃AA8ENDIF C 计算All的各个部分即A10A，A4A和A38AC A10A=0.0 DO 2167 M = 129 ,227 ,14 2167 A10A=A10A十HDI(M)DO 2168沁1=241，MO，142168 A10A=A10A十SR(M)A24A=0. 0 DO 2169 M=241 ,283 ,14 2169 A24A=A24A+HDI(M)-SR(M) A2

33、4A=A24A/0.75 A38A=All-A10A - A24A IF(A38A) 2170 , 2171,2171 2170 A38A=0.0 A24A=All-A10A 2171 CONTINUE C C 计算分量A10A，A24A和A38AC C TRASH=(A10-AA10祷。.4435)并(All/AIO) IF (TRASH. L T. 0, 0) TRASH = O. 0 All=All- TRASH A10A=A10A- TRASH IF(A10A. LT. O. 0) AIOA=O. 0 IF(A10A. EQ. O. 0) All=A24A十A38AIF(All. EQ

34、. O. 0) THEN A10A=0.0 A24A=0. 0 A38A=0. 0 ELSE A10A= CA10A / All)铃AA10A24A = (A24A/ All)必AA10A38A=(A38A/All)替AAIOEND IF C 计算A13的各个部分即A12A，A16S和A26AC 15 16 C GB/ T 18340. 4- 2001 A12A=0. 0 00 2172 M=141,183 ,14 2172 A1 2A=A12A+HDI(M) 002173 M=197 ,MR,14 2173 A12A=A12A+SR(M) A16S=0.O 002174 M=197, 225

35、 ,14 2174 A16S=A16S+HDI(M)-SR(M) A16S=A16S/ 0. 625 A26A=A13-A12A-A16S IF(A26A) 2175 ,2176 ,2176 2175 A26A = 0.0 A16S=A13-A12A 2176 CONTINUE C 计算分量A12A，A16S和A26AC C TRASH= (A12-AA12铃0.5192)赞(A13/A12) IF(TRASH. LT. O. 0) TRASH=O. 0 A13 = A13-TRASH A12A = A12A -TRASH IF(A12A. LT. O. 0) A12A = 0. 0 IF(A

36、12A. EQ. O. 0) A13=A16S+A26A IF(A13. EQ. O. 0) T HEN A1 2A=0. O A16S=0.O A26A=0.O ELSE A12A= (A12A/ A13)祷AA12A16S= (A16S/ A13)祷AA12A26A= (A26A/ A13)养AA12ENO IF C 计算A15的各个部分即A14A，A28A和A42AC A14A=0.0 DO 2177 M=167，2日，142177 A14A=A14A+HOI(M) DO 2178 M =265,MQ,14 2178 A14A=A14A十SR(M)A28A = O. 0 002179 M

37、 =265 , 307 ,14 2179 A28A=A28A十HDI(M)-SR(M ) C GB/ T .18340.4- 2001 A28A=A28A/D.75 A42A= A15-A14A - A28A IF(A42A) 2180.2181.2181 2180 A42A=0.0 A28A=A15-A14A 2181 CONTINUE C 计算分量A14A.A28A和A42AC C 。TRASH=(A14-AA14铃0.5075)祷(A15/A14) IF(TRASH. LT. o. 0) TRASH=O. 0 A15 = A15-TRASH A14A=A14A-TRASH IF(A14A

38、. LT. o. 0) A14A= 0. 0 IF(A14A. EQ. o. 0) A15=A28A+A42A IF(A15. EQ. o. 0) THEN A14A=0.0 A28A=0. 0 A42A=0.0 ELSE A14A = (A14A/ A15)铃AA14A28A= (A28A/ A15)铃AA14A42A= (A42A/ A15)铃AA14END IF C 计算A17的各个部分即A16A.A30A和A44AC A16A = 0.0 DO 2182 M=179.277.14 2182 A16A=A16A十HDI(M)DO 2183 M=291.MR.14 2183 A16A=A1

39、6ASR(M) A30A= 0. 0 DO 2184 M=291.333.14 2184 A30A=A30A+HDI(M) - SR(M) A30A=A30A/0.75 A44A = A17-A16A - A30A IF (A44A) 2185.2186.2186 2185 A44A= 0. 0 A30A=A17-A16A 2186 CONTINUE C C 计算分量A16A.A30A和A44A17 18 C C GB/ T 18340. 4-2001 TRASH= (A16-AA16怜0.4910)铃(A17/A16)IF(TRASH. LT. o. 0) TRASH=O. O A17=A1

40、7-TRASH A 16A = A 16A - TRASH IF(A16A. LT. o. 0) A1 6A =0. 0 IF(A16A. EQ. o. 0) A17=A30A- A44A IF(A17. EQ. 0.0) THEN A16A=0. 0 A30A=0.0 A44A=0. 0 ELSE A16A= (A16A/ Al7)祷AA16A30A=(A30A/A17)祷AA16A44A= (A4A/ A17)长AA16ENDIF C 计算A19的各个部分即A18A,A22S ,A32A C C A18A =0. 0 002187 M = 191,233 ,14 2187 A18A=A18

41、A+ HDI (M) 00 2188 M =247 ,MS,14 2188 A18A =A18A +SR (M) A22S= 0. 00 00 2189 M = 247 ,289 ,14 2189 A22S=A22S+ HDI (M) - SR(M ) A22S=A22S/0. 75 A32A = A19- A18A-A22S IF(A32A) 2190 ,219 1,2191 2190 A32A=0.0 A22S=A19- A1 8A 2191 CONTINUE C 计算分量A18A，A22S和A32AC TRASH = (A18-AA18祷。.5073)铃(A19/A18) IF (TRA

42、SH. L T . O. 0) TRASH = O. 0 A19= A1 9- TRASH A1 8A=A18A-TRASH IF(A18A. LT. o. 0) A18A = O.。IF(A1 8A. EQ. O. 0) A19 =A22S+A32A IF(A19. EQ. o. 0) THEN C C C C C A18A=O.0 A22S=0.0 A32A=0.0 ELSE GB/T 18340. 4- 2001 A18A= CA18A/ A19)铃AA18A22S= (A22S/ A19)祷AA18A32A= CA32A/ A19)祷AA18ENDIF 芳怪类型分离计算结束下面计算体积

43、百分数V6A=100.0祷A6A/SUMAAVI0S=100.0祷AI0S/SUMAAV20A=100.0铃A20A/SUMAAV8A=100.0铃A8A/SUMAAV22A=100.0禄A22A/SUMAAV36A=100.。祷A36A/SUMAAVI0A=100.0铃AI0A/SUMAAV24A=100.0铃A24A/SUMAAV38A=100.0铃A38A/SUMAAV12A=100.0铃A12A/SUMAAV16S=100.0铃A16S/SUMAAV26A=100.0铃A26A/SUMAAV14A=100.0铃A14A/SUMAAV28A=100.0椅A28A/SUMAAV42A=10

44、0.0祷A42A/SUMAAV16A=100.0势A16A/SUMAAV30A=100.0铃A30A/SUMAAV44A=100.0铃A44A/SUMAAV18A=100.。铃A18A/SUMAAV22S=100.。祷A22S/SUMAAV32A -:-100.。铃A32A/SUMAAAMONO=A6A+A8A+AI0A VMONO= V6A + V8A + VI0A ADI = A12A + A14A + A16A VDI = V12A + V14A + V16A ATRI = A18A + A20A VTRI=V18A+V20A A TETRA = A22A + A24A VTETRA=V

45、22A+V24A APENTA=A28A+A30A 19 20 G/ T 18340.4- 2001 VPENT A = V28A + V30A A THIO = A 10S+ A 16S+ A22S VTHIO = V1 OS + V1 6S+ V22S AUNID二A36A+A38A十A26A+ A42A + A44A + A32A VUNID=V36A+V38A十V26A十V42A十V44A十V32AC 储存和打印结果WRITE(铃，2300)2300 FORMAT(计算结果存盘否?Y /NJ ,) READ(铃，2310)ANS 2310 FORMAT(A1) IF(ANS.EQ. N

46、.OR.ANS.EQ. n) GOTO 2400 OPEN (UNIT = 2, FILE二AT5.LST , ACCESS = APPEND ,STATUS=UN KNOWN ) WRITE(2,2320) W(l :NC1) , TITLE(1 :NC2) 2320 FORMATC1 X ,A, ,A) WRITE(2,2330 ) VMONO , V6A , V8A , V10A ,VDI ,V12A, V14A, V16A , 养VTRI，V1 8A , V20A , VTETRA, V22A, V24A , 铸VPENTA，V28A , V30A , VTHI0, V10S, V16

47、S , V22S , VUNID, V36A , V38A , 铸V26A，V 42A , V 44A , V32A 23 30 FORMAT (lX, 27 (F5. 2, ) ,F5. 2) CLOSE (UNIT = 2) 2400 . WRITE归，2410)24 10 FORMATOX，选择打印设备或文件名隐含为打印机J:, ) READ(祷，(A)DEV lF(DEV. EQ.川)DEV=PRN WRITEC畴，2420)2420 FORMAT(确定打印份数o一的?, ) READ(祷，2440，ERR=3000)lTM 2440 FORMAT(Il) OPEN饵，FILE=DEV

48、, STATUS= NEW) DO 2460 K = l ,ITM WRITE(8,2500 ) C CALL GETDATCIYR,IMON ,IDAY) C CALL GETTIMCIHR,IMIN ,ISEC,IlOOTH) C WRITE(8 ,2510) W C1 :NC1) ,IYR ,IMON ,IDAY ,IHR ,IMIN ,ISEC WRITE(8 ,2510) W(l :NC1) ,CDAT1,CTIM1 IS = ( 66 - NC2 ) / 2 IS1 = IS/ 10 IS2 = IS -IS1讲10F2501 = ( / / CHAR (48十IS1)/ CHAR

49、(48+ IS2)/ / X,A) WRITE(8,F 2501) T ITLEO :NC2 ) WRITE(8, 2502) AMONO;VMONO,A6A ,V6A ,A8A ,V8A ,A10A ,V10A WRITEC8 , 2503) ADI , VDI , A1 2A , V12A ,A14A , V14A,A16A , V16A GB/ T 18340. 4-2001 WRITE(8,2504) ATRI, VTRI ,A18A, V18A ,A20A, V20A WRITE(8,2505) ATETRA,VTETRA ,A22A ,V22A ,A24A ,V24A WRITE(8,2506) APENTA,VPENTA,A28A,V28A,A3