化工原理第五章.docx
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化工原理第五章
五蒸馏习题解答
1解:
(1)作x-y图及t-x(y)图,作图依据如下:
■/XA=(p-pB0)/(pA°-pB0);yA=pA0g/p
以t=90C为例,xa=(760-208.4”(1008-208.4)=0.6898
yA=1008>0.6898/760=0.9150
计算结果汇总
tc
80.02
90
100
110
120
130
131.8
x
1
0.6898
0.4483
0.2672
0.1287
0.0195
0
y
1
0.9150
0.7875
0.6118
0.3777
0.0724
0
4.612x/(1+
3.612x)
1
0.9112
0.7894
0.6271
0.4052
0.0840
0
(2)用相对挥发度计算x-y值:
y=ax/[1+(-1)x]式中a=a=1/2(aa2)
•/a=A°/pB°
a1=760/144.8=5.249;2=3020/760=3.974
二aM=1/2(1+02)=1/2(5.249+3.974)=4.612
y=4.612x/(1+3.612x)
由此计算x-y值亦列于计算表中,y-x图,t-x(y)图如下:
1题附图
2解:
(1)求泡点:
在泡点下两组分的蒸汽分压之和等于总压P,即:
pA+pB=pA°XA+XB°XB=p求泡点要用试差法,先
设泡点为87C
lgpA°=6.89740-1206.350/(87+220.237)=2.971
pA0=102.971=935.41[mmHg]
lgpB0=6.95334-1343.943/(87+219.337)=2.566
pB0=102.566=368.13[mmHg]
935.41X0.4+368.13X0.6=595〜600mmHg
•••泡点为87C,气相平衡组成为
y=pA/p=pA0xA/P=935.410X.4/600=0.624
(2)求露点:
露点时,液滴中参与甲苯组成应符合下列关系:
xA+xB=1或pA/pA0+pB/pB0=1
式中pA=0.4X760=304[mmHg];pB=0.6X760=456[mmHg]
求露点亦要用试差法,先设露点为103C,则:
lgpA°=6.8974-120.635/
(103+220.237)=3.165
•pA0=1462.2[mmHg]
lgpB0=6.95334-1343.943/(103+219.337)=2.784
•pB0=608.14[mmHg]
于是:
304/1462.2+456/608.14=0.96<1
再设露点为102C,同时求得Pa°=1380.4;pB0=588.84
304/1380.4+456/588.84=0.995~1
故露点为102C,平衡液相组成为
xA=pA/pA0=304/1380.4=0.22
3解:
(1)XA=(p总-pB0)/(pA°-pB0)
0.4=(p总-40)/(106.7-40)
•p总=66.7KPa
yA=xApA0/p=0.4X06.7/66.7=0.64
(2)a=p/pB°=106.7/40=2.67
4解:
(1)yD=?
aD=(y/x)A/(y/x)B
=(yD/0.95)/((1-yD)/0.05)=2
yD=0.974
(2)L/VD=?
•/V=Vd+L
(V/VD)=1+(L/VD)
V0.96=VD0.974+L0.95
(V/VD)0.96=0.974+(L/VD)0.95
(1+L/VD)0.96=0.974+(L/VD)0.95
(L/VD)=1.4
5解:
简单蒸馏计算:
x1dx
InWi/W2=x2yx
W2=(1-1/3)Wi=2/3Wi;y=0.46x+0.549,x1=0.6,代入上式积分解得
釜液组成:
X2=O.498,
馏出液组成:
Wdxd=W1x1-W2x2
(1/3W1)xd=W1X0.6-(2/3W1)>0.498
/•xd=0.804
6解:
FxF=Vy+Lx/•0.4=0.5y+0.5x
(1)
y=ax/(1+(-1)a)=3x/(1+2x)
(2)
(1),
(2)联立求解,得y=0.528,x=0.272
回收率=(Vy)/(Fxf)=0.50528/0.4=66%
7.解:
f=d+w
Fxf=Dxd+Wxw
已知Xf=0.24,xd=0.95,xw=0.03,解得:
D/F=(xf-xw)/(xd-xw)=(0.24-0.03)/(0.95-0.03)=0.228
回收率Dxd/Fxf=0.228>95/0.24=90.4%
残液量求取:
W/D=F/D-1=1/0.228-1=3.38
•••W=3.38D=3.38(V-L)=3.38(850-670)=608.6[kmol/h]
8解:
(1)求D及W,全凝量V
f=d+w
Fxf=Dxd+Wxw
xf=0.1,xd=0.95,xw=0.01(均为质量分率)
F=100[Kg/h],代入上两式解得:
D=9.57[Kg/h];W=90.43[Kg/h]
由恒摩尔流得知:
F(0.1/78+0.9/92)=V(0.95/78+0.05/92)
[注意:
如用质量百分数表示组成,平均分子量Mm=1/(aA/MA+aB/Mb)]
解得V=87[Kg/h]由于塔顶为全凝器,故上升蒸汽量V即为冷凝量,
(2)求回流比R
V=D+L•L=V-D=87-9.57=77.43[Kg/h]
R=L/D=77.43/9.57=8.09(因为L与D的组成相同,故8.09亦即为摩尔比)
(3)操作线方程.
因塔只有精馏段,故精馏段操作线方程为
yn+1=Rxn/(R+1)+xd/(R+1)
式中Xd应为摩尔分率
xD=(xD/MA)/[xD/MA+(1-xD)/MB]=(0.95/78)/(0.95/78+0.05/92)=0.961
/•yn+i=8.09xn/9.09+0.961/9.09=0.89xn+0.106操作线方程为:
yn+1=0.89xn+0.106
9解:
y=[R/(R+1)]x+xD/(R+1)
(1)R/(R+1)=0.75R=0.75R+0.75R=0.75/0.25=3
(2)xD/(R+1)=0.2075xD/(3+1)=0.2079xD=0.83
(3)q/(q-1)=-0.5q=-0.5q+0.5q=0.5/1.5=0.333
(4)0.75x+0.2075=-0.5x+1.5xf0.75xq'+0.2075=-0.5xq'+1.50^4
I.25xq'=1.50.44-0.2075=0.4425xq'=0.362
(5)010解:
(1)求精馏段上升蒸汽量V和下降的液体量L,提馏段上升蒸汽量V'和下降的液体量L'.
进料平均分子量:
Mm=0.4478+0.6492=86.4
F=1000/86.4=11.6[Kmol/h]
FxF=DxD+WxW
F=D+W
II.640.4=D40.97+(11.6-D)0.02
•••D=4.64[Kmol/h]
W=6.96[Kmol/h]
R=L/D,•L=3.744.64=17.17[Kmol/h]V=(R+1)D=4.744.64=21.8[Kmol/h]
平均气化潜热r=3080740.4+3332040.6=32313.6[KJ/Kmol]
从手册中查得xf=0.4时泡点为95C,则:
q=[r+cp(95-20)]/r=(32313.6+159.275)/324313.6=1.37
•L'=L+qF=17.17+1.37141.6=33.1[Kmol/h]
V'=V-(1-q)F=21.8+0.37114.6=26.1[Kmol/h]
(2)求塔顶全凝器热负荷及每小时耗水量.
Qc=Vr
•r=0.97430804+3332040.03=30879.5[KJ/Kmol]
•Qc=21.8430879.5=673172.7[KJ/h]
耗水量Gc=673172.7/4.18(50-20)=5368.2[Kg/h]
(3)求再沸器热负荷及蒸汽耗量.
塔的热量衡算
QB+QF+QR=Qv+QW+QL
QB=Qv+QW+QL-QF-QR该式右边第一项是主要的,其它四项之总和通常只占很小比例,故通常有:
Qb~Qv=V・lv
Iv=(r+Cpt)=30879.5+159.28.24=43933.9[KJ/Kmol]
•••Qb=21.8H3933.9=957759.02[KJ/h]
2.5[KgF/cm2]下蒸汽潜热r=522Kcal/Kg=5224.18氷8=39275.3[KJ/Kmol]
•蒸汽需量为Gv
Gv=QB/r=957759.02/39275.3=24.4Kmol/h
=24.418=39.04[Kg/h]
(4)提馏段方程y=L'x/(L'-W)-WxW/(L'-W)=1.26x-0.005
11解:
提馏段:
ym+1'=1.25Mx'-0.0187
(1)
=L'xM'/V'-WxW/V',
L'=L+qF=RD+F
V'=(R+1)D
W=F-D,精馏段:
yn+1=Rxn/(R+1)+xD/(R+1)
=0.75xn+0.25xD
(2)
q线:
xf=0.50(3)
将(3)代入
(1)得出:
ym+1=1.250.5-0.0187=0.606,代入
(2)
0.606=0.750.5+0.25xD,
xD=0.924
12解:
(1)y1=xD=0.84,
0.84=0.45x1+0.55
x1=0.64,
yW=30.64/(3+1)+0.84/(3+1)=0.69,0.69=0.45xW+0.55,xW=0.311,
(2)D=100(0.4-0.311)/(0.84-0.311)=16.8(Kmol/h),
W=100-16.8=83.2(Kmol/h)
13解:
(1)求R,xD,xW
精馏段操作线斜率为R/(R+1)=0.723•R=2.61
提馏段方程y=L'x/(L'-W)-WxW/(L'-W)=1.25x-0.0187精馏段操作线截距为
xD/(R+1)=0.263•xD=0.95
提馏段操作线与对角线交点坐标为
y=x=xWxW=1.25xW-0.0187•xW=0.0748
(2)饱和蒸汽进料时,求取进料组成
将y=0.723x+0.263
y=1.25x-0.0187
联立求解,得x=0.535,y=0.65
因饱和蒸汽进料,q线为水平线,可得原料组成y=xF=0.65
14解:
⑴yi=XD=0.9,x1=09(4-30.0)=0.692,
(2)y2=1>0.692/(1+1)+0.9/2=0.796
(3)xD=xF=0.5,yD=0.5/2+0.9/2=0.7
15解:
(1)FxF=Vyq+Lxq
0.45=(1/3)yq+(2/3)xq
yq=2.5xq/(1+1.5xq)
•••xq=0.375yq=0.6
(2)Rmin=(xD-yq)/(yq-xq)
=(0.95-0.6)/(0.6-0.375)=1.56
R=1.5Rmin=2.34
D=0.95>0.45/0.95=0.45W=1-0.45=0.55
xW=(FxF-DxD)/W=(0.45-0.450.>95)/0.55=0.041
L=RD=2.34>0.45=1.053;V=(R+1)D=1.503
L'=L+qF=1.053+(2/3)1>=1.72;V'=V-(1-q)F=1.503-1/3=1.17
y'=(L'/V')x'-WxW/V'=1.72/1.17x'-0.550.04>1/1.17
=1.47x'-0.0193
16解:
精馏段操作线方程
yn+1=3/4xn+0.24
平衡线方程y=ax/[1+(-1)x]=2.5x/(1+1.5x)
提馏段操作线方程
y=1.256x-0.01278
其计算结果如下:
N0
x
y
1
0.906
0.96
2
0.821
0.92
3
0.707
0.86
4
0.573
0.77
5
0.462
0.70
6
0.344
0.567
7
0.224
0.419
8
0.128
0.268
9
0.065
0.148
10
0.029
0.069
由计算结果得知:
理论板为10块(包括釜),加料板位置在第五块;
17解:
D/F=(xF-xW)/(xD-xW)=(0.52-xW)/(0.8-xW)=0.5
解得:
xw=0.24
精馏段操作线方程:
yn+1=(R/(R+1))Xn+XD/(R+1)=0.75xn+0.2平衡线方程:
y=ax/(1+(-l)x)=3x/(1+2x)
或:
x=y/(-aa1)y)=y/(3-2y)
交替运用式
(1),
(2)逐板计算:
xd=y1=0.8.X1=0.571;
y2=0.628,x2=0.360;
y3=0.470,x3=0.228•••共需Nt=3块(包括釜).
18解:
q=0,XD=0.9,xf=0.5,
xw=0.1,R=5,
精馏段操作线方程:
yn+1=Rxn/(R+1)+xd/(R+1)
=5xn/(5+1)+0.9/(5+1)
=0.833xn+0.15
图解:
得理论板数为11块(不包括釜),包括釜为12块
(1)
(2)
18题附图
19解:
(1)F=D+W
Fxf=Dxd+WxW
D=F(xF-xw)/(xD-XW)=100(0.3-0.015)/(0.95-0.015)=30.48Kmol/h=30.5Kmol/hW=F-D=69.50Kmol/h
(2)Nt及Nf=?
xd=0.95、xw=0.015、q=1、R=1.5;xd/(R+1)=0.38
作图得:
Nt=9-1=8(不含釜)
进料位置:
Nf=6
⑶L',VW/及xw-1
q=1,V'=V=(R+1)D
V'=30.5(1.5+1)=76.25Kmol/h
L'=L+qF=RD+F=1.5
X30.5+100=145.8Kmol/h
xw-1=0.03
由图读得:
yw=0.06,
(1)原料为汽液混合物,成平衡的汽液相组成为x,y
平衡线方程y=ax/[1+(-a)x]=4.6x/(1+3.6x)⑴
q线方程(q=2/(1+2)=2/3)则
y=[q/(q-1)]x-xF/(q-1)=-2x+1.35
(2)
联解
(1),
(2)两式,经整理得:
-2x+1.35=4.6x/(1+3.6x)7.2x2+1.740x-1.35=0
解知,x=0.329
y=0.693
(2)Rmin=(xD-ye)/(ye-xe)=(0.95-0.693)/(0.693-0.329)=0.706
21解:
因为饱和液体进料,q=1
ye=aX[1+(-a)Xe]=2.47区6心+1.470.6)=0.788
Rmin=(xD-ye)/(ye-xe)=(0.98-0.788)/(0.788-0.6)=1.02
R=1.5>Rmin=1.53
Nmin=lg[(xD/(1-xD))((1-xW)/xW)]/lga
=lg[(0.98/0.02)(0.95/0.05)]/lg2.47=7.56
x=(R-Rmin)/(R+1)=(1.53-1.02)/(1.53+1)=0.202
Y=(N-Nmin)/(N+1)Y=0.75(1-x0.567)
•••(N-7.56)/(N+1)=0.75(1-0.2020.567)解得N=14.5取15块理论板(包括釜)实际板数:
N=(15-1)/0.7+1=21(包括釜)
求加料板位置,先求最小精馏板数
(Nmin)精=lg[XD/(1-XD)(1-XF)/XF]/lga
=lg[0.98/0.020.4/0.6]/lg2.47=3.85
N精/N=(Nmin)精/Nmin
•N精=N(Nmin)精/Nmin=14.5885/7.56=7.4
则精馏段实际板数为7.4/0.7=10.6
取11块故实际加料板位置为第12块板上.
22解:
(1)由y=aX/[1+(-1a)X]=2.4X/(1+1.4X)作y-X图
由于精馏段有侧线产品抽出,故精馏段被分为上,下两段,抽出侧线以上的操作线方程式
yn+1=RXn/(R+1)+XD/(R+1)=2/3Xn+0.3
(1)
侧线下操作线方程推导如下:
以虚线范围作物料衡算V=L+D1+D2
Vys+1=LXs+D1XD1+D2XD2;
ys+1=LXs/V+(D1XD1+D2XD2)/V=LXs/(L+D1+D2)+(D1XD1+D2XD2)/(L+D1+D2);
L=L0-D2,则:
ys+i=(L0-D2)xs/(L0-D2+D1+D2)
+(Dixd1+D2XD2)/(L0-D2+D1+D2)=(R-D2/Di)Xs/(R+1)+(xdi
+D2XD2/Dl)/(R+1)(R=L0/D1)
将已知条件代入上式,得到:
ys+i=0.5x+0.416
22题附图
(2)用图解法,求得理论塔板数为(5-1)块,见附图.
23解:
根据所给平衡数据作x-y图.
精馏段操作线
yn+1=Rxn/(R+1)+xD/(R+1)
=1.5xn/(1.5+1)+0.95/(1.5+1)
=0.6xn+0.38
q线方程与q线:
料液平均分子量:
Mm=0.35X0.65>18=22.9
甲醇分子汽化潜热:
r=25213214.2=33868.8[KJ/Kmol]
水的分子汽化潜热:
r=552H8>4.2=41731.2[KL/Kmol]
料液的平均分子汽化潜热:
r=0.3533868.8+0.6541731.2=38979.4[KL/Kmol]
料液的平均分子比热
Cp=0.88>22.94.2=84.6[KL/KmolG]
q=[r+Cp(ts-tf)]/r=[38979.4+84.6(78-20)]/38979.4=1.13
q线斜率q/(q-1)=1/13/0.13=8.7
提馏段操作线方程与操作线:
由于塔釜用直接蒸汽加热,故提馏段操作线过横轴上(xw,0)一点汙是在x-y图上,作出三条线
用图解法所得理论板数为7.6块,可取8块(包括釜).
24解:
对全塔进行物料衡算:
F1+F2=D+W
(1)
F1XF1+F2XF2=DxD+WxW
100>0.6+2000.2=D10.8+W>0.02
(2)
100=0.8D+0.02W
由式
(1)W=F1+F2-D=100+200-D=300-D代入式
(2)得:
D=120.5Kmol/h
L=RD=2X120.5=241kmol/h
V=L+D=241+120.5=361.5Kmol/h在两进料间和塔顶进行物料衡算,并设其间液汽流率为L",V",塔板序号为s.
V''+F1=D+L''
V''ys+1"+F1xF1=L''xs''+DxDys+1=(L''/V'')xs''+(DxD-F1xF1)/V''
L''=L+qiFi=241+1x100=341Kmol/h
V''=V=361.5
ys+i"=(341/361.5)xs”+(120.50.&1000.6)/361.5ys+i"=0.943xs''+0.i
25解:
对于给定的最大V',V=(R+1)D,回流比R愈小,塔顶产品量D愈大,但R需满足产品的质量要求xD》0.98,故此题的关键是求得回流比R.
由题已知加料板为第14层,故精馏段实际板数为13层,精馏段板数为:
13.0.5=6.5
取苯-甲苯溶液相对挥发度为a=2.54
用捷算法求精馏段最小理论板数
(Nmin)精=ln[0.98/0.02-0.5/0.5]/In2.54=4.175
y=[N精馏段-(Nmin)精]/(N精馏段+1)=(6.5-4.175)/(6.5+1)
=1.31
由y=0.75(1-x0.567)x=(1-Y/0.75)(1/0.567)=0.392=(R-Rmin)/(R+1)
•••R=(0.392+Rmin)/(1-0.392)
Rmin=(xD-ye)/(ye-xe)
对泡点进料xe=xF=0.5
ye=ax/[1+(-1a)x]
=2.54.0.5/(1+1.540..5)=1.27/1.77=0.72
•Rmin=(0.98-0.72)/(0.72-0.5)=0.26/0.22=1.18
•R=(0.392+1.18)/(1-0.392)=1.572/0.608=2.59
•D=V/(R+L)=2.5/(2.59+1)=0.696[Kmol/h]故最大馏出量为0.696[Kmol/h]
26解:
求n板效率:
Emv=(yn-yn+1)/(yn*-yn+1),
因全回流操作,故有yn+1=xn,yn=xn-1
与xn成平衡的yn*=axn/[1+(-a1)xn]=2.430..285/(1+1.430.2.85)=0.492
于是:
Emv=(xn-1-xn)/(yn*-xn)=(0.43-0.285)/(0.492-0.285)=0.7
求n+1板板效率:
Emv=(yn+1-yn+2)/(yn+1*-yn+2)=(xn-Xn+)/(yn+1*-Xn+1)y'n+1=2.430.173/(1+1.430.173)=0.337
•••Emv=(0.285-0.173)/(0.337-0.173)=0.683
27解:
由图可知:
该板的板效率为