大气污染控制工程第三版课后习题答案18章全.docx
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大气污染控制工程第三版课后习题答案18章全
大气污染控制工程第三版课后习题答案(18章全)
大气污染控制丄程课后作业习题解答第一章概论
1.1解:
按lmol干空气计算,空气中各组分摩尔比即体积比,故n=0.781mol,
n二0.209mol,N202n二0.00934mol,n二0.00033mol。
质量白分数为ArC02
0.781,28.010.209,32.00,;N%,,100%,75.51%^,,
100%,23.08%2228.97,128.97,1
0.00934,39.940.00033,44.01,。
Ar%,,100%,1.29%C0%,,
100%,0.05%228.97,128.97,1
1.2解:
山我国《环境空气质量标准》二级标准查得三种污染物日平均浓度限值如下:
3333S02:
0.15mg/m,N02:
0.12mg/m,CO:
4.00mg/mo按标准状态下lm干空气计算,其
31,10摩尔数为。
故三种污染物体积百分数分别为:
44.643mol22.4
3,30.15,100.12,10S0:
N0:
0.052ppm,0.058ppm2264,44.64346,
44.643
34.00,10C0:
o,3.20ppm28,44.643
1.3解:
41.50,10,154331)(g/m),,,1.031g/mNN,322.4,10
41.50,10,333c(mol/m)o,,6.70,lOmol/mNN,322.4,10
32)每天流经管道的CC1质量为1.031X10X3600X24X10kg=891kg4
1.4解:
6每小时沉积量200X(500X15X60X10)X0.12=10.8,g,g
1.5解:
由《大气污染控制工程》P14(1,1),取M二210
4pC0Hb2.2,10,,M,,210,,0.2369,20Hbpl9.5,10202
COHb/OHbCOHbO.23692C0Hb饱和度,,,,,19.15%C0C0Hb,OHbl,COHb/OHbl,
0.236922
1.6解:
4800,20含氧总量为。
不同CO白分含量对应CO的量为:
960mL100
9609602%:
7%:
2%,19.59mL,7%,72.26mL93%98%
72.261)最初CO水平为0%时;t,,172.Omin,434.2,10,10
72.26,19.592)最初CO水平为2%时。
t,,125.4min,434.2,10,10
1.7解:
由《大气污染控制工程》P18(1,2),最大能见度为
2.6d2.6,1400,1.4pp。
L,,,11581.8mv,K2.2,0.2
第二章燃烧与大气污染
2.1解:
lkg燃油含:
重量(g)摩尔数(g)需氧数(g)
C85571.2571.25
H113,2.555.2527.625S100.31250.3125
H022.51.2502
N元素忽略。
1)理论需氧量71.25+27.625+0.3125=99.1875mol/kg设干空气0:
N体积比为
1:
3.78,则理论空气量99.1875X4.78=474.12mol/kg重油。
223即
474.12X22.4/1000=10.62m/kg重油。
N
烟气组成为C071.25mol,HO55.25+11.25二56.50mol,SOO.1325mol,N3.78X222299.1875=374.93mol。
理论烟气量71.25+56.50+0.3125+374.93=502.99mol/kg重油。
即
502.99X22.4/1000二11.27
3m/kg重油。
X
2)干烟气量为502.99,56.50二446.49mol/kg重油。
0.3125S0百分比浓度为,,100%,0.07%2446.49
71.25空气燃烧时C0存在最大浓度。
,100%,15.96%2446.4933)过剩空气为
10%时,所需空气量为1.1X10.62=11.68m/kg重油,N3产生烟气量为
11.267+0.1X10.62=12.33m/kg重油。
N
2.2解:
相对于碳元素作如下计算:
%(质量)mol/100g煤mol/mol碳
C65.75.4751H3.23.20.584S1.70.0530.01002.30.0720.013灰分18.13.306g/mol碳
水分9.01.644g/mol碳
故煤的组成为CHSO,0.5840.0100.013
100燃料的摩尔质量(包括灰分和水分)为。
燃烧方程式为,18.26g/molC5.475
CHSO,n(0,3.78N),C0,0.292H0,0.010S0,
3.78nN0.5840.0100.013222222n二1+0.584/4+0.010,0.013/2=1.1495
1.1495,(1,3.78),3331)理论空气量;,1000,22.4,
lOm/kg,6.74m/kgl8.26
0.010SO在湿烟气中的浓度为,100%,0.174%21.6441,0.292,0.010,3.78,
10002)产生灰分的量为18.1,,80%,144.8g/kg100,33烟气量
(1+0.292+0.010+3.78X1.1495+1.644/18)X1000/18.26X22.4X10=6.826m/kg
144.83343灰分浓度为mg/m二2.12X10mg./m,106.826
1000,1.7%,1.7,4032.003)需石灰石/t煤,103.21kg35%
2.3解:
按燃烧lkg煤讣算
重量(g)摩尔数(mol)需氧数(mol)
C79566.2566.25H31.12515.56257.78S60.18750.1875H052.875
2.9402
设干空气中\:
0体积比为3.78:
1,22
所需理论空气量为4.78X(66.25+7.78+0.1875)二354.76mol/kg煤。
理论烟气量C0266.25mol,S020.1875mol,H2015.5625+2.94=18.50mol
3.78,354.76N2,280.54mol4.78
总计66.25+'8.50+0.1875+280.54=365.48mol/kg煤
0.1875实际烟气量365.48+0.2X354.76=436.43mol/kg煤,SO浓度为。
100%,0.043%2436.43
2.4解:
取lmol煤气计算
HS0.002mol耗氧量0.003mol2
CO0.05mol02
CO0.285mol0.143mol
H(0.13-0.004)mol0.063mol2
共需00.003+0.143+0.063+0.014=0.223molo设干空气中N:
O体积比为
3.78:
1,则222理论干空气量为0.223X(3.78+l)=1.066niol。
取,则实际干空气
1.2X,,1.21.066mol=l.279mol<>
333空气含湿量为12g/m,即含H00.67mol/m,14.94L/m。
故H0体积分数为
1.493%oN2NN2
1.279故实际空气量为。
1.298moll,1.493%
烟气量S0:
0.002mol,C0:
0.285+0.007+0.05=0.342mol,
N:
0.223X3.78+0.524=1.367mol,
222H00.002+0.126+0.014+1.298X1.493%+0.004=0.201mol2
故实际烟气量0.002+0.342+1.367+0.201+0.2X1.066=2.125mol
2.5解:
1)N%=1,11%,8%,2%,0.012%二78.99%2
由《大气污染控制工程》P46⑵11)
8,0.5,2,100%,50.5%空气过剩0.264,78.99,(8,0.5,2)
2)在测定状态下,气体的摩尔体积为
PVT101325,22.4,443112;V,,,,39.46L/mol2TP273,700,133.32212
363取lm烟气进行计算,则S0120X10m,排放浓度为2
612010,3o,,,,(18%)640.179/gm,339.4610,
22.433)。
,,,,m5663.37(18%)2957/minN39.46
39.4634)o30.0,,52.85g/mN22.4
2.6解:
按lkg煤进行计算
重量(g)摩尔数(mol)需氧数(mol)C75863.1763.17
S160.50.5
HO83.254.62502
需氧63.17+10.19+0.5=73.86mol设干空气中N:
0体积比为3.78:
1,则干空气
量为73.86X4.78X1.2二423.66mol,22
含水423.66X0.0116=4.91molo
烟气中:
CO63.17mol;S00.5mol;H04.91+4.625+20.375=29.91mol;222
N:
73.86X3.78=279.19mol;过剩干空气0.2X73.86X4.78=70.61molo2
实际烟气量为63.17+0.5+29.91+279.19+70.61=443.38mol
63.170.5其中CO;SO;22,100%,14.25%,100%,0.11%443.38443.38
279.19,0.79,70.6129.91H0;N。
22,100%,6.74%,
100%,75.55%443.38443.38
70.61,0.2090o2,100%,3.33%443.38
2.7解:
SO含量为0.11%,估计约1/60的SO转化为SO,则SO含量2233
1,,55,即P二1.83X10,lgP=-4.737o0.11%,,1.83,10H2S04H2S0460
查图2,7得煤烟气酸露点约为134摄氏度。
2.8解:
以lkg油燃烧计算,
C860g71.67mol;
H140g70mol,耗氧35molo
设生成COxmol,耗氧0.5xmol,则生成CO(71.67,x)mol,耗氧
(71.67,x)molo2
1.5%x烟气中0量。
2,6600,10
1.5%x总氧量,干空气中N:
0体积比22,0.5x,(71.67,x),35,106.67,
24.5x,6600,10
为3.78:
1,则含N3.78X(106.67+24.ox)<>根据干烟气量可列出如下方程:
21.5%xx,解得x二0.306,71.67,3.78(106.67,24.5x),,6,6600,10600,1071.67,0.306故CO%:
;,100%,13.99%20.306
6600,10
3.78(24.5,0.306,106.67)N%:
100%,84.62%20.306
6600,10
由《大气污染控制工程》P46(2,11)
1.5,0.5,0.06空气过剩系数,,1,,1.070.264,84.62,(1.5,0.5,0.06)第三章大气污染气象学3.1解:
由气体静力学方程式,大气中气压随高度的变化可用下式描述:
(1)dPgdZ,,,,
将空气视为理想气体,即有
mmPM可写为
(2),,,PVRT,MVRT
将
(2)式带入
(1),并整理,得到以下方程:
dPgM,,dZPRT
假定在一定范围内温度T的变化很小,可以忽略。
对上式进行积分得:
gMPgM2即(3)InPZC,,,ln(),,,ZZ21RTPRT1
o假设山脚下的气温为10C,带入(3)式得:
5009.80.029,In,,,Z10008.314283,
得,,Zkm5.7
即登山运动员从山脚向上爬了约5.7kmo
3.2解:
T297.8,298,不稳定,,,,,,2.35K/100m,,1.5,10d,zlO,1.5
T297.5,297.8,不稳定,,,,,,1.5K/100m,,10,30d,z30,10
T297.3,297.5,不稳定,,,,,,1.OK/100m,,30,50d,zoO,30
T297.5,298,不稳定,,,,,,1.75K/100m,,1.5,30d,z30,1.5
T297.3,298,不稳定。
,,,,,,1.44K/100m,,1.5,50d,z50,1.5
3.3解:
TPO.28811,,()TPOO
P6000.2880.2881T,T(),230(),258.49K10P4000
3.4解:
uZZm由《大气污染控制工程》P80(3,23),,取对数
得,muu(),lglg()IZZul11
uZ设,,由实测数据得lg,ylg(),xZull
x0.3010.4770.6020.699y0.06690.11390.14610.1761由excel进行直线拟合,取截距为0,直线方程为:
y=0.2442x故m,0.2442。
3.5解:
Z50Z1000.070.070.070.0712,
uums,,,,()2()2.24/uums,,,,()2()2.35/1020Z10Z1000
Z200Z3000.070.070.070.0734,
uums,,,,()2()2.47/uums,,,,()2()2.54/3040Z10Z1000
Z4000.070.075ouums,,,,()2()2.59/50Z100
稳定度D,m二0.15
ZZ501000.150.150.150.1512,u,u(),2,(),2.55m/su,u(),2,
(),2.82m/sl020Z10Z1000
ZZ2003000.150.150.150.1534,u,u(),2,(),3.13m/su,u(),2,(),3.33m/s3040Z10Z1000
Z4000.150.155ou,u(),2,(),3.48m/s50Z100
稳定度F,m=0.25
ZZ501000.250.250.250.2512,u,u(),2,(),2.99m/su,u(),2,
(),3.56m/sl020Z10Z1000
ZZ2003000.250.250.250.2534,u,u(),2,(),4.23m/su,u(),2,
(),4.68m/s3040Z10Z1000
Z4000.250.255u,u(),2,(),5.03m/s50Z100
风速廓线图略。
3.6解:
dPgMl)根据《AirPollutionControlEngineering》可得高度与压强的关系
为,,dzPRT
dP2将g=9.81m/s>M二0.029kg、R=8.31J/(mol.K)代入上式得。
dz,,29.21TP。
o当t二11.0C,气压为1023hPa;当t二9.8C,气压为1012hPa,
o故P二(1023+1012)/2二1018P&,T二(11.0+9.8)/2二10.4C二283.4K,dP二1012-
1023=,llPao
11因此,z二119m。
dz,,29.21283.4m,89ml018
同理可计算其他测定位置高度,结果列表如下:
测定位置2345678910
。
气温/C9.812.014.015.013.013.012.61.60.8气压/hPa1012
1000988969909878850725700高度差/m89991011635362902711299
281高度/m119218319482101813071578287731582)图略
T11,9.81,23),不稳定;,,,,,,1.35K/100m,,1,2d,z,891,2
T9.8,122,3,逆温;,,,,,,,2.22K/100m,02,3,z,992,3
T12,143,4,逆温;,,,,,,,1.98K/100m,03,4,z,1013,4
T14,154,5,逆温;,,,,,,,0.61K/100m,04,5,z,1634,5
T15,135,6,稳定;,,,,,,0.37K/100m,,5,6d,z,5365,6
T,13,136,7,,,,,,06,7z,,2906,7
T13,12.67,8,稳定;,,,,,,0.15K/100m,,7,8d,z,2717,8
T12.6,1.68,9,稳定;,,,,,,0.85K/100m,,8,9d,z,12998,9
Tl.6,0.89,10,稳定。
,,,,,0.28K/100m,,9,lOd,z,2819,10
3.7解:
T26.7,21.11,故,逆温;,,,G,0G,,,1.22K/100m,0111,z4581
T15.6,21.12,故,稳定;,,,G,0.72K/100m,,G,,,,0.72K/100m222d,z7632
T8.9,15.63,故,不稳定;,,,G,1.16K/100m,,G,,,,1.16K/100m333d,z5803
T5.0,25.04,故,不稳定;,,,G,lK/lOOm,,G,,,,lK/100m444d,z20004
T20.0,30.05,故,不稳定;,,,G,2K/100m,,G,,,,2K/100m555d,z5005
T28.0,25.06,故逆温。
,,G,0G,,,0.43K/100m,0666,z7006
3.8解:
以第一组数据为例进行计算:
假设地面大气压强为1013hPa,则山习题3.1推导得到的公式
PgM2,代入已知数据(温度T取两高度处的平均值)即ln(),,,ZZ21PRTl
P9.8,0.0292,由此解得P=961hPaoIn,,,458210138.314,297
III《大气污染控制工程》P72(3,15)可分别讣算地面处位温和给定高度处位温:
100010000.2880.288,,,T(),294.1(),293K地面地面P1013地面
100010000.2880.288,,,T0,299.7(),303.16K11P9611
293,303故位温梯度二,2.18K/100m0,458
同理可计算得到其他数据的位温梯度,结果列表如下:
测定编号123456
。
地面温度/C21.121.115.625.030.025.0高度/m4587635802000
500700
。
相应温度/C26.715.68.95.020.028.0位温梯度/
K/100m2.220.27,0.17,0.02,1.021.42
3.9解:
PgM2以第一组数据为例进行计•算,山习题3.1推导得到的公式,设地面ln(),,,ZZ21PRTl
9709.8,0.029压强为P,代入数据得到:
解得P=1023hPa«因此In,,,
45811P8.314,2971
100010000.2880.288,,T(),294.1(),292.2K地面地面P1023地面
同理可讣算得到其他数据的地面位温,结果列表如下:
测定编号123456
。
地面温度/C21.121.115.625.030.025.0高度/m4587635802000
500700
。
相应温度/C26.715.68.95.020.028.0地面压强/hPa102310121002
104010061007
。
地面位温/C292.2293.1288.4294.7302.5297.4
3.10略。
笫四章大气扩散浓度估算模式
4.1解:
吹南风时以风向为x轴,y轴指向悄壁,原点为点源在地面上的投影。
若不存在悄壁,则有
222()()Qyz,Hz,H,,(,,,)exp(){ex讥]exp[]}xyzH,,,,,222,,,222,,,2uyzzyz
现存在悄壁,可考虑为实源与虚源在所关心点贡献之和。
222()()Qyz,Hz,H,实源exp(){exp[]exp[]},,,,,1222,,,222,,,2uyzzyz
222
(2)()()QL,yz,Hz,H,虚源
exp[]{exp[]exp[]},2222,,,222,,,2uyzzyz
222()()Qyz,Hz,H,因此+exp(){exp[]exp,222,,,222,,,2uyzzyz
222
(2)()()QL,yz,Hz,Hexp[]{exp[]exp[]},,,,222,,,222,,,2uyzzyz2