发布时间 : 星期二 文章第二章习题答案更新完毕开始阅读711ddf0b76c66137ee0619f7
第二章作业
1. 已知煤的空气干燥基成分：Cad=60.5% ，Had=4.2%，Sad=0.8%，Aad=25.5%，Mad=2.1%和风干水分=3.5%，试计算上述各种成分的收到基含量。 （Car=58.38%，Har=4.05%，Sar=0.77%，Aar=24.61%，Mar=5.53%)
f100?Mar100?3.5?3.5?2.1??5.53% 解：Mar?M?Mad100100farK?100?Mar100?5.53??0.965
100?Mad100?2.1Car?KCad?0.965?60.5?58.38% Har?KHad?0.965?4.2?4.05% Sar?KSad?0.965?0.8?0.77% Aar?KAad?0.965?25.5?24.61%
2, 已知煤的空气干燥基成分：Cad=68.6%，Had=3.66%，Sad=4.84%，Oad=3.22%，
Nad=0.83%，Aad=17.35%，Mad=1.5%，Vad=8.75%，空气干燥基发热量Qnet,ad=27528kJ/kg和收到基水分Mar=2.67%，煤的焦渣特性为3类，求煤的收到基其他成分,干燥无灰基挥发物及收到基低位发热量，并用门捷列夫经验公式进行校核。
（Car=67.79%，Har=3.62%，Sar=4.78%，Oar=3.18%，Nar=0.82%，Aar=17.14%，Vdaf=10.78%，Qnet，ar=27172kJ/kg;按门捷列夫经验公式Qnet,ar=26825kJ/kg) 解：从空气干燥基转换为收到基的换算系数K?100?Mar100?2.67??0.9881
100?Mad100?1.5Car?KCad?0.9881?68.6?67.79%
Har?KHad?0.9881?3.66?3.62% Sar?KSad?0.9881?4.84?4.78% Oar?KOad?0.9881?3.22?3.18% Nar?KNad?0.9881?0.83?0.82% Aar?KAad?0.9881?17.35?17.14% 从空气干燥基转换为干燥无灰基的换算系数
100100K???1.2323 100?Mad?Aad100?1.5?17.35Vdaf?KVad?1.2323?8.75?10.78%
Qnet,ar?(Qnet,ad?25Mad)?100?Mar100?2.67?25Mar?(27528?25?1.5)??25?2.67?27172kJ/kg
100?Mad100?1.5门捷列夫公式
Qnet,ar?339Car?1030Har?109(Oar?Sar)?25Mar?339?67.79?1030?3.62?109?(3.18?4.78)?25?2.67?26817.06kJ/kg
4，某工厂贮存有收到基水分Mar1=11.34%及收到基低位发热量Qnet,ar1=20097kJ/kg的煤100t，由于存放时间较长，收到基水分减少到Mar2=7.18%，问这100t煤的质量变为多少？煤的收到基低位发热量将变为多大？
100113.?4%?x?00.718解：设减少的水分为x（t），，所以x=4.48t，
100?x100t煤变为100-4.48=95.52t， 由收到基转为干燥基：
100100Qnet,d1?(Qnet,ar1?25Mar1)??(20097?25?11.34)??22987kJ/kg
100?Mar1100?11.34由干燥基转为收到基：
100?Mar2100?7.18Qnet,ar2?Qnet,d1??25Mar2?22987??25?7.18?21157kJ/kg
100100
7，一台4t/h的链条炉，运行中用奥氏烟气分析仪测得炉膛出口处RO2=13.8%， O2=5.9%，CO=0；省煤器出口处RO2=10.0%，O2=9.8%，CO=0。如燃料特性系数β=0.1，试校核烟气分析结果是否准确？炉膛和省煤器出口处的过量空气系数及这一段烟道的漏风系数有多大？
21（该题计算中学生很多用简化公式计算！?l?）
21?O221?O221?5.9解：因为CO=0，所以炉膛出口处RO2???13.7%，省煤器出口处
1??1?0.121?O221?9.8RO2???10.2%，分析准确。
1??1?0.111\??1.38 炉膛出口处：?l?O25.91?3.761?3.76100?(13.8?5.9)100?(RO2?O2)11\??1.85 省煤器出口处：??O29.81?3.761?3.76100?(10?9.8)100?(RO2?O2)漏风系数：????\??l\?1.85?1.38?0.47
8、SZL10-1.3-WⅡ型锅炉所用燃料成分为Car?59.6%，Har?2.0%，
Sar?0.5%，Oar?0.8%，Nar?0.8%，Aar?26.3%，Mar?10.0%，Vdaf?8.2%，
0、理论烟气量Vy0以及在过Qnet,ar?22190kJ/kg。求燃料的理论空气量VK量空气系数分别为1.45和1.55时的实际烟气量Vy，并计算??1.45时300℃及400℃烟气的焓和??1.55时200℃及400℃烟气的焓。
解：
0VK?CSHO1159.60.520.8(1.866ar?0.7ar?5.55ar?0.7ar)?(1.866?0.7?5.55?0.7)?5.81m3/kg 0.211001001001000.21100100100100 CarSarHarMarNar00V?1.866?0.7?(11.1?1.24?0.016VK)?(0.8?0.79VK)
100100100100100
0y?1.866?59.60.52100.8?0.7??(11.1??1.24??0.016?5.81)?(0.8??0.79?5.81)100100100100100
0当??1.45时，Vy?Vy0?1.0161(??1)VK?6.15?1.0161?(1.45?1)?5.81?8.81m3/kg
0当??1.55时，Vy?Vy0?1.0161(??1)VK?6.15?1.0161?(1.55?1)?5.81?9.40m3/kg
?6.15m3/kg
当??1.45时300℃的烟气的焓：
000Iy?VRO2(c?)RO?VN(c?)?V(c?)HO H2N2O222?(1.866CarSNHM00?0.7ar)(c?)RO?(0.8ar?0.79VK)(c?)N?(11.1ar?1.24ar?0.016VK)(c?)HO222100100100100100
59.60.50.8210?0.7)?559?(0.8?0.79?5.81)?392?(11.1?1.24?0.016?5.81)?463100100100100100 =2628.63kJ/kg
?(1.86600IK?VK(c?)K?5.81?403?2341.43kJ/kg
00Iy?Iy?(??1)IK?2628.63?(1.45?1)?2341.43?3682.27kJ/kg 当??1.45时400℃的烟气的焓： 000Iy?VRO(c?)?VN(c?)?VH O(c?)2RO22N22H2O?(1.866CarSNHM00?0.7ar)(c?)RO?(0.8ar?0.79VK)(c?)N?(11.1ar?1.24ar?0.016VK)(c?)HO222100100100100100
59.60.50.8210?0.7)?772?(0.8?0.79?5.81)?527?(11.1?1.24?0.016?5.81)?626100100100100100 =3505.94kJ/kg
?(1.86600IK?VK(c?)K?5.81?542?3149.02kJ/kg
00Iy?Iy?(??1)IK?3505.94?(1.45?1)?3149.02?4923.00kJ/kg 当??1.55时200℃的烟气的焓： 000Iy?VRO(c?)?VN(c?)?VH O(c?)2RO22N22H2O?(1.866CarSNHM00?0.7ar)(c?)RO?(0.8ar?0.79VK)(c?)N?(11.1ar?1.24ar?0.016VK)(c?)HO222100100100100100
59.60.50.8210?0.7)?357?(0.8?0.79?5.81)?260?(11.1?1.24?0.016?5.81)?304100100100100100 =1726.76kJ/kg
?(1.86600IK?VK(c?)K?5.81?266?1545.46kJ/kg
00Iy?Iy?(??1)IK?1726.76?(1.55?1)?1545.46?2576.76kJ/kg 当??1.55时300℃的烟气的焓： 000Iy?VRO2(c?)RO?VN(c?)?V(c?)HO H2N22O22?(1.866CarSNHM00?0.7ar)(c?)RO?(0.8ar?0.79VK)(c?)N?(11.1ar?1.24ar?0.016VK)(c?)HO222100100100100100
59.60.50.8210?0.7)?559?(0.8?0.79?5.81)?392?(11.1?1.24?0.016?5.81)?463100100100100100 =2628.63kJ/kg
?(1.86600IK?VK(c?)K?5.81?403?2341.43kJ/kg
00Iy?Iy?(??1)IK?2628.63?(1.55?1)?2341.43?3916.42kJ/kg
（该题学生偷懒耍花招，只给出结果，根本没有过程）