1
銆佹皵鍖栧伐娈?/p>
1.1
寰峰＋鍙ゆ按鐓ゆ祮鍔犲帇姘斿寲鍘熺悊
寰峰＋鍙ゆ按鐓ゆ祮鍔犲帇姘斿寲灞炴皵娴佸簥锛?/p>
姘ф皵鍜屾按鐓ゆ祮閫氳繃寰峰＋鍙ょ儳鍢?/p>
娣峰悎鍚庡柗灏勮繘鍏ユ皵鍖栫倝锛屽湪
0.1
绉掑唴鐓ゆ祮涓按琚苯鍖栵紝鐓ょ矑瀛愯姘斾綋
闅斿紑锛?/p>
鍚勭叅绮掔嫭绔嬬殑鍦ㄧ叅鐏扮啍鐐逛互涓婃俯搴﹀彂鐢熷弽搴旓紝
鍥轰綋鍦ㄦ皵鍖栫倝鍐?/p>
鍋滅暀
3
锝?/p>
5
绉掗挓锛?/p>
鍙嶅簲鐢熸垚鐨勭矖鍚堟垚姘旂敳鐑峰惈閲忓緢灏戯紝
涓�鑸粎涓?/p>
0.1%
浠ヤ笅锛岀⒊杞寲鐜囬珮杈?/p>
98%
锛岀敱浜庡弽搴旀俯搴﹂珮锛屼笉鐢熸垚娓ｆ补銆侀厷鍙婇珮
绾х儍绛夊彲鍑濊仛鐨勫壇浜х墿銆?/p>
1.1.1
姘斿寲杩囩▼鐨勫寲瀛﹀弽搴?/p>
姘斿寲鐐夊唴鍙戠敓鐨勫寲瀛﹀弽搴旀槸寰堝鐨勶紝
鍦ㄧ伀鐒颁腑鍙兘鍚屾椂杩涜浠ヤ笅
鍖栧鍙嶅簲锛?/p>
CmHn+(m+n/4)O
2
=mCO
2
+n/2H
2
O
鈶?/p>
2C+O
2
=2CO
+246.6KJ
鈶?/p>
C+O
2
=CO
2
+409.1KJ
鈶?/p>
2CO+O
2
=2CO
2
+573.2KJ
鈶?/p>
H
2
+1/2O
2
=H
2
O
鈶?/p>
C+H
2
O=CO+H
2
锛?/p>
122.7KJ
鈶?/p>
CO2+H2=CO+H
2
O
锛?/p>
42.3KJ
鈶?/p>
CH
4
+H
2
O=CO+3H
2
锛?/p>
412.8KJ
鈶?/p>
C+CO
2
=2CO
锛?/p>
165.0KJ
鈶?/p>
CH
4
=C+2H
2
鈶?/p>
姝ゆ椂杩樺彲鑳藉彂鐢熷涓嬩粯鍙嶅簲锛?/p>