DIN EN ISO 13443-2007 Natural gas - Standard reference conditions (ISO 13443 1996 including Corrigendum 1 1997) English version of DIN EN ISO 13443 2007-09《天然气 标准参比条件(ISO 13443 199.pdf

《DIN EN ISO 13443-2007 Natural gas - Standard reference conditions (ISO 13443 1996 including Corrigendum 1 1997) English version of DIN EN ISO 13443 2007-09《天然气 标准参比条件(ISO 13443 199.pdf》由会员分享，可在线阅读，更多相关《DIN EN ISO 13443-2007 Natural gas - Standard reference conditions (ISO 13443 1996 including Corrigendum 1 1997) English version of DIN EN ISO 13443 2007-09《天然气 标准参比条件(ISO 13443 199.pdf（16页珍藏版）》请在麦多课文档分享上搜索。

1、September 2007DEUTSCHE NORM English price group 10No part of this standard may be reproduced without prior permission ofDIN Deutsches Institut fr Normung e. V., Berlin. Beuth Verlag GmbH, 10772 Berlin, Germany,has the exclusive right of sale for German Standards (DIN-Normen).ICS 75.060!$I as constan

2、ts, the respective values being determined by trial calculations for a wide range of natural gases. The values used are (az/ag), = - 0,000 020/kPa (azia711, = + 0,000 025/K Zair/a, = +O,OOO 011/K I I 77,0(aH$aT) = - 0,000 I O/K I / R;(aR;/ ar) = - 0,000 011 The values of I+ and a% are taken to be ze

3、ro. Despite the simplicity of these approximations, it is expected that the accuracy of conversion will usually stil within the limits quoted in annex A. The equations may not be used to increase the number of digits given in conversion factors given in annex A. be the Note that if the reference con

4、ditions for the source data are not given in K and kPa (for example in “C or 0 F I and in atm, mbar, psia or psig, respectively) then these must be converted appropriately before the equations may be applied (see BS 350:Part 1 :I 974131). Ideal volume VO Vo(lS0) = Vo (T2, p2) x 288,15 p2/1 01,325 T2

5、 . . . (B.1) Ideal density po po(lSO) = po (72, 2) x 101,325 T2/288,15 p2 . . . iB 2) . Ideal relative density do d”(lSO) = d”(T2, 2) . . . 63) Compression factor 2 z(lS0) = Z(T2, p2) x I + 0,000 020 (p2 - 101,325)/1 + 0,000 025 (T2 - 288,15)1 . . . (B.4) Real volume V v(lS0) = V(T2,p2) x 288,15p2/1

6、01,325 T2x I + 0,000 020 (p2 - 101,325)/1 + 0,000 025 (T2 - 288,15) . . . (B.5) 8 EN ISO 13443:2005 (E) Real density p ISO) = p(T2, 2) x I 01,325 T2/288,15p$ x I + 0,000 025 (ISO) = HP (T1,pl) x I + 0,000 01 ITI - 288,15) Mass-basis ideal superior calorific value fig ii; (ISO) = I?; (T, pl) x I + 0,

7、000 10 (Tl - 288,15)1 Mass-basis ideal inferior calorific value fiy iiF (ISO) = iiF (Tl,pl) x I + 0,000 01 (T1 - 288,15) Molar-basis real superior calorific value HS 77, (ISO) = p, (T1, pl) x I + 0,000 10 (Tl - 288,15)1 Molar-basis real inferior calorific value q q (ISO) = q (T1, pl) x I + 0,000 01

8、(Tl - 288,15) Mass-basis real superior calorific value IiS ii, (ISO) = fis (Tl, pl) x I + 0,000 10 (T, - 288,15) Mass-basis real inferior calorific value I?, fi, (ISO) = ii, (T1, pl) x I + 000 01 (Tl - 288,15) Volume-basis ideal superior calorific value fig (ISO) = fi,o (Tj, T2, PI, 2) x I 01,325 T2

9、/288,?5 21 x I + 0,000 10 (T1 - 288,15) Volume-basis ideal inferior calorific value I? (ISO) = gp (TI, T2, 1, 2) x I 01,325 T2/288,l5 21 x I + 0,000 01 (TJ - 288,15) ideal Wobbe index W” W” (ISO) = W” ( TI, T2, 1, 2) x I 01,325 T2/288,l5 p2 x I + 0,000 10 (TI - 288,15) Volume-basis real superior cal

10、orific value es fi, (ISO) = fi, (TI, T2, PI, 2) x I 01,325 T2/288,l5 21 x I + 0,000 10 (TI - 288,15)1 x I + 0,000 025 (T2 - 288,15)/1 + 0,000 020 (p2 - 101,325) Volume-basis real inferior calorific value fi, fi, (ISO) = g, (TJ, T2,pl,p2) x 101,325 T2/288,15p2 x I + 0;OOO 01 (TI - 288,15) x I + 0,000

11、 025 (T2 - 288,15)/1 + 0,000 020 (p2 - 101,325)1 Real Wobbe index W W(lS0) = W (TJ, T2, 1, 2) x I 01,325 T2/288,l5 p2 x I + 0,000 10 (T1 - 288,15) x (I + 0,000 020 (p2 - 101,325)/1 + 0,000 036 (T2 - 288,15)1)-l/2 . . . u3.8) . . . (B.9) . . . (B.lO) . . . (B.1 I) . . . (B.12) . . . (B.13) . . . (B.1

12、4) . . . (B.15) . . . (B.16) . . . (B.17) . . . (B.18) . . . (B.19) . . . (B.20) . . . (B.21) 9 EN ISO 13443:2005 (E) Annex C (normative) Symbols Symbol Meaning d HI 4 4 % fr, 4 P T t V W z P relative density molar-basis inferior calorific value mass-basis inferior calorific value volume-basis infer

13、ior calorific value molar-basis superior calorific value mass-basis superior calorific value volume-basis superior calorific value (absolute) pressure (absolute) temperature Celsius temperature = T - 273,15 volume Wobbe index compression factor density SI unit (or multiple) kJ mol-1 MJ kg-1 MJ m-3 k

14、J mol-1 MJ kg-1 MJ m-3 kPa K 0 C m3 MJ m-3 kg m-3 Subscripts for the combustion reference condition 2 for the volumetric or metering reference condition Superscript 0 for the ideal-gas state (no superscript indicates the real-gas state) 10 EN ISO 13443:2005 (E) Annex D (informative) Example calculat

15、ions The examples given below are all for real dry natural gases consisting preponderantly (greater than about 70 % molar) of methane. EXAMPLE 1 What is the compression factor at IS0 standard reference conditions of a natural gas which has a compression factor of 0,997 1 at a temperature of 273,15 K

16、 and a pressure of 101,325 kPa? From table A.1, line 4, column 3 (used inversely): Z(lS0) = 0,997 1 x (l/O,999 6) = 0,997 5 The same value (to four significant digits) is obtained from equation (B.4). EXAMPLE 2 What is the volume occupied at IS0 standard reference conditions of a natural gas which o

17、ccupies 1 000 m3 at normal conditions? The term “normal conditions” is taken to mean 273,15 K (0 “C) at 101,325 kPa. From table A.1, line 5, column 3 (used inversely): V(lS0) = 1 000 x (l/O,947 6) = 1 055,3 m3 The same value (to five significant digits) is obtained from equation (8.5). EXAMPLE 3 Wha

18、t is the mass-basis superior calorific value at IS0 standard reference conditions of a natural gas determined as having a calorific value of 54,21 MJ.kg-1 at 25 “C and a pressure of 100 kPa? Molar-basis and mass-basis calorific values are taken to be independent of pressure p1 within the pressure ra

19、nge given in annex B (which covers the range of normal atmospheric variation); the pressure assignment is therefore irrelevant in this case. From table A.1, line 14, column 2: & (ISO) = 54,21 x 1,001 0 = 54,26 MJ.kg-1 The same value (exactly) is obtained from equation (B.14). 11 EN ISO 13443:2005 (E

20、) EXAMPLE 4 What is the volume-basis superior calorific value at IS0 standard reference conditions of a natural gas determined as having a calorific value of 38,57 MJ.m-3 at 60 “F and 101,560 kPa? This example cannot be solved using table A.1 because that table is only valid for a pressure of 101,32

21、5 kPa and does not include the Fahrenheit temperature quoted. First convert 60 “F to the kelvin scale: 60 “F = (60 - 32) x 5/9 + 273,15 = 288,706 K Then, from equation (B.19): ii,” (ISO) = 38,57 x 101,325 x 288,706 x I + 0,000 10(288,706 - 288,15) x I + 0,000 025(288,706 - 288,15)/288,15 x 101,560 x

22、 I + 0,000 020(101,560 - 101,325) = 38,56 MJ.m-3 to four significant digits. EXAMPLE 5 What is the volume-basis inferior calorific value at IS0 standard reference conditions of a natural gas determined as having a calorific value of 37,35 MJ.m-3 when the volume of gas burned is measured at 0 OC, com

23、bustion takes place at 25 “C and the reference pressure in both cases is one standard atmosphere? The term “one standard atmosphere” is taken to mean 101,325 kPa. From table A.1, line 20, column 4: i, (ISO) = 37,35 x 0,947 7 = 35,40 MJ.m-3 The same value (to four significant digits) is obtained from

24、 equation (B.20). 12 EN ISO 13443:2005 (E) Annex E (informative) National usage of reference conditions For those countries where the use of non-metric (e.g. Imperial) measurement units and reference conditions are still in use alongside metric units and reference conditions, only the metric referen

25、ce conditions are listed. In table E.l below: t,/“C = combustion reference temperature t*/OC = volumetric or metering reference temperature The reference pressure is 101,325 kPa in all cases. Table E.l - National usage of reference conditions Argentina Australia Austria Belgium Brazil Canada China C

26、zechoslovakia Denmark Egypt Finland France Germany Hong Kong Hungary India tj/“c 15 25 25 15 20 25 25 0 25 t2/0C 15 Indonesia 15 ran 0 reland 0 taly 0 Japan 15 Netherlands 20 New Zealand 20 and 0 Norway 0 Pakistan 15 Romania 15 Russia 0 Spain 0 Sweden 15 United Kingdom 0 USA 0 Yugoslavia q/“C 15 25

27、0 25 25 25 0 15 15 0 t21C 0 15 15 0 0 0 15 15 15 15 and 0 20 and 0 0 0 15 15 0 13 EN ISO 13443:2005 (E) Annex F (informative) Bibliography I IS0 5024: 1976, Petroleum liquids and gases - Measurement - Standard reference conditions. Z IS0 7504:1984, Gas analysis - Vocabulary . 3 BS 350:Part 1 :I 974,

28、 C onversion factors and tables - Part 7: Basis of tables - Conversion factors. 4 lGU/G-64 (1964), Report of the committee on documentation and sundry questions - Report of the subcommittee on units, pp. 24-38. 5 lGU/G-73 (1973), Report of the committee on documentation and sundry questions - Report of the subcommittee on units, pp. 83-101. 6 lGU/G-76 (1976), Report of the committee on statistics, documentation and sundry questions - Report of the subcommittee on units, pp, 36-43. 14 EN ISO 13443:2005 (E)