化工原理習題

下冊主觀試題答案

j06c15001221.

1.(1)所需填料層高度Z：

Z=HOGXNoG=V(Yl-Y2)/(KyaQAYm)

已知Y|=0.02/(l-0.02)=0.0204

Y2=Y(l-90%)=0.00204

X=0X=?Y=0Y=0(m=0)

貝IJNoG=l/(l-mV/L)Xln[(l-mV/L)(Y-mX)/(Y-mX)+mV/L]

=ln(Y/Y)=ln10=2.30

AZ=HOGXNOG=2.30X0.5=1.15(m)

(2)￥1=0.02/(1-0.02)=0.0204

Y2=Y(l-99%)=0.000204

N0G=ln(Y/Y)=ln100=4.605

Z=H°GXNOG=2.03(m)

(3)￥)=0.04/(1-0.04)=0.0416

Y2=Y(l-90%)=0.00416

N0G=ln(Y/Y)=ln10=2.30

，Z—1.15(m)

2.由m=0知，全部傳質(zhì)阻力集中于氣相，氣膜阻力占100%,

由HOG=V/(KyaQ)=0.5

V/Q=5800/(29X3600)=0.0556(Kmol/s.m)

Kya=G/(PHOG)=0.0556/(1X0.5)=0.112Kmol/m.s.atm

3.由于m=0,可知溶劑S中A含量并不降低推動力，所以可采取溶液循環(huán)操作來加

大液量，保證設(shè)計效率。

j06c2000229

①已知yi=0.05y2=0.00263

V=2000(l-0.05)=1900m惰/h=1900/22.4=84.82(Kmol惰/h)

Yi=0.05/(1-0.05)=0.05263[KmolA/KmolB]

Y2=0.00263/(1-0.00263)=0.0026369[kmolA/kmolB]

每小時回收的A的量G

G=V(Y-Y)

=84.82(0.05263-0.0026369)=4.24[kmolA/h]

②XF61.2/58/((1000-61.2)/18)=0.02023[kmolA/kmol溶劑]

X2=0

Aym=((Y-Y)-(Y-Y)/In(Y-Y)/(Y-Y)

YI-YF0.05263-2X0.02023=0.01217

Y2-Y2=o.0026369-0

=0.0026369

△Ym=(0.01217-0.0026369)/In(0.01217/0.0026369)=0.0062333

NoG=(Y-Y)/AYm=(0.05263-0.0026369)/0.0062333=8.02

H=HgXN℃=0.7X8.02=5.61(m)

③1.02G=V(Y-Y')

Y'=Y-1.02G/V

=0.05263-1.02X4.24/84.82=0.001642

X'=61.2X1.02/58/((1000-61.2X1.02)/18)=0.02066

Y'-Y'=0.05263-2X0.02066=0.01131

Y'-Y'=0.001642-0=0.001642

△Ym'=(0.01131-0.001642)/ln(0.01131/0.001642)=0.0050099

N'=(Y'-Y')/AYm'

=(0.05263-0.001642)/0.0050099=10.2

H'=HOGXNOG'

=0.7X10.2=7.14(m)

AH=7.14-5.61=1.53(m)

j06cl50037

解：Z=HOGXNOGZ'=HOGXNOG'

N=l/(l-mV/L)Xln[(l-mV/L)(YI-MX2)/(Yi-mX2)+mV/L]

=1/(1-1.2/1.5)Xln[(l-0.8)Yi/((1-0.9)Y,)+0.8]=5.148

N'=l/(l-0.8)Xln[0.2XYi/(Yi(1-0.95))+0.8]=7.843

H^const.

...Z'=NOG'/N?；xZ=(7.843/5.148)5=7.618?=7.62m

AZ—2.62m

j06bl50048

解：MCHOH=132M110=18

以lOOKg溶液為基準

mH0=85KgmCH0H=15Kg

(1)摩爾分率=15/32/(15/32+85/18)=0.0903

(2)摩爾比=15/32/(85/18)=0.0993

(3)質(zhì)量比=15/85=0.176

(4)質(zhì)量濃度:V=m/p=100/970=0.1031(m)

質(zhì)量濃度=01。10H/V=15/0.1031=145.5(Kg/m)

(5)摩爾濃度=15/32/0.1031=4.55(kmol/m)

j06c2000516

解：低濃度y、x,近似以Y、X代替，則

(1)Yi=l.5X1=1.5X0.008=0.012

AY尸0.02-0.012=0.008,△Y2=0.004

AYm=(0.008-0.004)ln(0.008/0.004)=0.00577

NOG=(0.02-0.004)/0.00577=2.77

/.Hoc=4/2.77=1.44m

(2)L/V=(O.02-0.004)/0.008=2(L/V)min=(0.02-0.004)/(O.02/1.5)=1.2

AL/V/(L/V)min=2/1.2=1.67

(3)X1=(Y-Y)/(L/V)=(O.02-0.002)/2=0.009

Yi=l.5X=0.0135

△YFO.0065,AYFO.002

AYm=O.00382

NOG=4.71

Z=IIOGXNOG=1.44X4.71=6.80[m]

A加高6.80-4=2.8[m]

(4)如圖。

j06bl500615

已知：D=0.8mLs=1200Kg/hG=1000Nn)7h

1

yi=O.01341=0.995x2=0ye=0.75xKya=0.OSSkmol/m.s.atm

Ls=1200/(3600X18)=0.0185kmol/s

G=1000/(22.4X3600)=0.0124kmol/s

為低濃度吸收A(L/G)=O.0185/0.0124=1.49

x)=(G/L)(y-y)+x=(0.013-0.000065)/1.49=0.00868

y?=y(1-巾)=0.013(1-0.995)=0.000065

VZ=HOGXN?；而N?；=(y-y)/Aym

Ayi=y-y=O.013-0.75X0.00818=0.006490

△yz=y-y=O.000065-0=0.000065

△ym=(△yL△y2)/ln(△yi/△y2)

=(0.006490-0.000065)/In(0.006490/0.000065)=0.001396

Noc=(0.013-0.000065)/0.001396=9.266

HkG'/(KyaQ)=0.0124/(0.785XO.8XO.8X0.055=0.449(m)

AZ=9.266X0.449=4.2m

j06bl500715

解：P=0.15X10x

x=CMs/p

P=0.15X10CMs/P

(1)C=PP/(MsXO.15X10)

=0.054X1000/(18X0.15X10)=0.002=C0

氣液兩相達到平衡

(2)C=0.001mol/l<C

A由氣相轉(zhuǎn)移到液相，發(fā)生吸收過程。

(3)C=0.003>C

A由液相轉(zhuǎn)入汽相，發(fā)生解吸過程。

(4)x=yeP/E=0.051X5/(0.15X10)=0.00018

C=0.003

x=CMs/P=0.003X18/1000=0.000054

x<x

A由氣相轉(zhuǎn)向液相。

j06cl500814

解：(1)P=yAP=XAPA°

yA=PA-/PXXA=380/760XXA=0.5X

Xi'=yi/O.5=0.015/0.5=0.03

(2)6=G(y-y)/(Gy)(低濃度)

y2=(l-<t>)y=(l-0.85)X0.015=0.00225

(L/G)min=(y-y2)/(xi*-x2)

=(0.015-0.00225)/(0.03-0.002)=0.4554

(3)h=H(>cXNOG

NOG=1/(1-1/A)Xln[(l-1/A)(y-mx)/(y-mx)+l/A]

L=3Lmin

1/A=m/(L/G)=0.5/(3X0.4554)=0.3660

N^l/d-O.3660)ln[(1-0.3660)(0.015-0.5X0.002)/(0.00225-0.5X

0.002+0.366)

=3.171

h=1.2X3.171=3.805m

j06c200098

解：(1)L/G=(yb-ya)/(xb-xa)=(0.03-0.002)/(0.015-0)=1.866

"=Xb/xb=0.015/(0.03/1.4)=0.7

(2)mG/L=l.4/1.866=0.75

NOG=1/(1-O.75)Xln[(l-0.75)(0.03-0)/(0.002-0)+0.75]=6

HOG=ho/NOG=9/6=1.5m

(3)Nm'=l/(l-0.75)Xln[(1-0.75)(0.03/0.0015)+0.75]=7

ho/=1.5X7=10.5m

△h<)=10.5-9=1.5m

j06bl501021

解：1.求吸收劑用量

Yi=yi/(l-yi)=O.13/(l-0.13)=0.149

Y尸Yi(l-<b)=O.149(1-0.90)=0.0149

Xi=0.2/44/(1000/18)=8.18X10

Xz=0

V=TV,/(TiX22.4)

=1000(1-0.13)X273/(22.4X293)=36.2kmol/h

L=V(YT)/(XLX2)

=36.2(0.149-0.0149)/(8.18X10)=59345kmol/h

=1.07X10KgHO/h

2.求填料高度

H”=L/(KxaQ)=59345/(10695Xn/4X1.33)=3.99M

Noi=(Xi-X2)/AXm

X產(chǎn)YMm=0.149/1420=1.05X10?

X2=Y2/m=0.0149/1420=1.05X10

AXm=((X；-X,)-(X2*-X2))/ln((Xj-X)/(X4X2))

=((1.05X10-8.18X10)-1.05X10ln(1.05X10-8.18X10)/(1.05

X10)=1.60X10

NoL=8.18X10/(I.60X10)=5.113

Z=3.99X5.113=20.4m

j06bl00117

體系服從亨利定律，貝小

y=E/PXx

現(xiàn)己知P增大一倍，即Pz=2Pi,x、E不變，

則y『=E/PzXx=l/2(E/P,Xx)=yJ/2

分壓P2=P2y2=2Pl(l/2)yi-Py=Pi

平衡氣相濃度y*減小一-倍，分壓不變

故選擇(Oo

w06bl50121如圖

j06al00133

證：設(shè)進口氣體中A的摩爾比為Y,出口氣體中A的摩爾比為Y,而X=0,

則(L/V)min=(Y,-Y2)/(Y/m)n=(Y1-Y2)/Y1

/.(L/V)min=mH?

j06c!501412

圖1的操作線方程為：

Y=(L'/V)X+(Y-(L'/V)X')

或Y=(L'/V)X+(Y-(L'/V)X)

圖2的操作線方程為：

Y=((L'+R)/V)X+(Y-(U/V)X/-(R/V)Xi)

或Y=((L'+R)/V)X+(Y「((L'+R)/V)X)

對圖1：L'(XLX2')=V(YT2)

如圖中的AB線為操作線，A,B點為兩端面的狀態(tài)點，斜率為L'/V；

對圖2：V(Y>-Y2)=L(XI-X2)

L=L'+RLX=L'X'+RX

圖中AC為操作線，A,C為其端面狀態(tài)點，斜率為(L'+R)/V,

且有BC/CD=R/L'。

j06c150151如圖。

w06bl00165

解：在解吸塔中，氣體入口濃度低，而液體入口濃度高。

mG/L>l時，塔足夠高，兩相在塔底趨于平衡。

X可達較低濃度，但Xz>Xz*,而氣相出口濃度Y也較低；

mG/L<1時，塔足夠高，兩相在塔項趨于平衡，氣相出口濃度Y可達較高濃度,

但YiVYj,而液相出口濃度X較高。

w06al001715

答：根據(jù)吸收速度方程，氣相體積吸收總系數(shù)Kya為

Kya=G(Y|-Y2)/(ZQAYm)

GB為空氣摩爾流量，需由空氣體積流量算出，故應測定空氣流量。

GB=V/22.4=VP/22.4X(T/(TP)[kmoVh](空氣流量)

Vo——標準狀態(tài)下空氣流量[m/h]

V1--流量計操作狀態(tài)下空氣流量[m/h]

而V尸V指、((T|P指)/(P】T指))

V指---轉(zhuǎn)子流量計指示的流量|m/h|

To,Po-標準狀態(tài)下空氣溫度、壓力

T指,P指-轉(zhuǎn)子流量計標定的壓力和溫度

PL流量計操作狀態(tài)下空氣溫度、壓力

，

且有△Ym=((丫「丫|*)-(丫2-丫2*))/ln(Y1-Y1*)/(Y2-Y2)

Yj=Mxi

XI=GB(YI-Y2)/L而Y尸yi/(l-yjY2=y2/(l-y，)

因此，為求出Kya,還應測量T,P,y和塔內(nèi)操作條件下的溫度和壓力。

w06bl00181如圖

j06bl00191如圖。

j06b1502014

解：y尸15/17(1000/22.4)-0.01977[kmolNH^/kmolB+NL]

y?=3.954X10-4[kmolNH3/kmolB+NH3]

Y^y^d-yt)=0.01977/(1-0.01977)=0.02017[kmolNH3/kmolB]

Y2=y2/(l-y2)=3.9555X10[kmolNM/kmolB]

V=2500/22.4X(1-0.01977)=109.4[kmolB/h]

1=3.6X1000/18=200[kmolB/h]

全塔物料衡算

L(X-X2)=V(YI-Y2)

200(X1-0)=109.4(0.02017-3.9555X10)得X尸0.01072

△Yi=Yi-Yi=0.02017-1.2X0.01072=0.0072

AY2=Y2-Y2=0.0004

△Ym=(0.0072-0.0004)/In(0.0072/0.0004)=0.00235

NoG=(Y-Y)/△Ym=(0.02-0.0004)/0.00235=8.34

HoG=NoGXNoG=8.34X0.7=5.84m

j06d2002126

解：(1)y1=1.5%/(l-1.5%)=0.0152

y2=yi(l-99%)=0.000152

X2=0

(L/V)min=(yi-y2)/((yi/m)-0)

/.(I7V)min=mEA(mV/L)=l/EA

原工況下：mV/L=1/1.5X(mV/L)min=l/(0.99X1.5)

NoG=l/(l-mV/L)XIn[(1-mV/L)y/y+mV/L]

=l/(l-l/(0.99X1.5))Xln[(M/(0.99X1.5))0.0152/0.000152+l/(0.99X1.5)

=10.74

在新工況下：H不變，V1.2V,Kya'=1.2(Kya)

=1.16KyaH=IIoGXNoG

=V/(KyaQ)X10.74=V7(Kya')XN'

=1.2V/(1.16Kya)XN,

???NoG'=10.74X1.16/1.2=10.382

mV/L^VX1.2/L=1.2/(0.99X1.5)=0.808

10.382=l/(l-0.808)Xln[l-0.808]yl/(yl(l-EA))+0.808)

得EA=97.06%o

(2)在新工況下，回收率仍99%時

NoG"=1/(1-0.808)Xln[(l-0.808)y/(y(l-EA))+0.808]

將EA=99%代入上式，得

Nn=15.61

H,=HoG,XNoGn

=V7(KyaQ)X15.61=1.2VX15.61/(1.16KyaQ)

H=V/(KyaQ)X10.74

???H'/H=1.2X15.61/(1.16X10.74)=1.504倍。

j06d2002336解

①1)求NoG;已知X“=0Y=l.2X,n=0.99

(Ls/GB)min=(YT)/(Xt*-XJ=(Y「YJ/(Y/m)=mn

=1.2X0.99=1.188

Ls/GB=1.5(Ls/GB)min=1.5X1.188=1.782

解法1:mGB/Ls=1.2/1.782=0.673Yb/￥a=l/(l-n)=100

NoG=l/(l-S)Xln[(l-S)Y/Y+S]

-1/(1-0.673)Xln[(1-0.673)X100+0.673]=10.7

解法2:Yu=Y(l-n)=0.0134(1-0.99)=1.34X10

X?=GB/LsX(Yb-Yu)

=1/1.782X(0.0134-1.34X10-4)=0.00744

AYm=((Yb-mXb)-Ya/In(Y.-mXb)/Ya

=(0.00447-0.000134)/In(0.00447/0.000134)=0.00124

NoG=(Yb-Y?)/AYm=(0.0134-0.000134)/0.00124=10.7

2)求H;已知G=5000Nm/h,KYa=200kmol/m.h

GB=G/22.4X(l-Yb/d+Yb))

=5000/22.4X(1-0.0134/1.0134)=220.3kmol/h

IIoG=GB/(KYaQ)=220.3/(200X1.4)=0.787m

3)求h；

h=HoGXNoG=0.787X10.7=8.42m

②進塔氣量增至G'=5500Nm/h,增加10%

GB'=242.3kmol/h

n=0.99,h不變，KYa℃U℃GB

HoG=GB/(KYaQ)8GB

HoG'=HoGX1.1=0.787X1.1=0.81m

NoG'=h/H'oG=8.42/0.81=10.4

NoG'=(Y-Y)/AYm'設(shè)AY=0.00464

△Ym'=((Yb-mX；)-￥,,)/In(Y「mXJ)/Y

=(0.00464-0.000134)/ln(0.00464/0.000134)=0.001271

，

(Yb-Ya)/AYm=(0.0134-0.000134)/0.001271=10.42

與NoG'值相近

.?.Xb'=(Y『△Y；)/m=(0.0134-0.00464)/1.2=0.0073

則Ls'=GB'(YD/X；=242.3X(0.0134-0.000134)/0.0073

=440.3kmol/h

即Ls=1.782GB=1.782X220.3=392.6kmol/h

吸收劑用量增加：

Ls'/Ls=440.3/392.6=1.122

j06d2002417

Mm=29X0.94+17X0.06=28.28G,=580/28.28=20.51kmol/m2.h

G=20.51(1-0.06)=19.28kmol/m2.hL=770/18=42.78kmol/m2.h

yi=0.06Yi=yi/(1-yJ=0.0638yz4Y2as丫2(1-0.99)=6.38X10

G(Y,-Y2)=L(XI-X2)X2=0AX^O.0285X=0.0277

出口氨水濃度XLO.0277

ye=0.755xye；=0.755X0.0277=0.0209

Aym=((0.06-0.0209)-(0.000638-0))/ln(0.0391/0.000638)=0.00935

NoG=(YB-Y,J/△ym=(0.06-0.000638)/0.00935=6.35

HoG=h/NoG=3/6.35=0.472

當壓強增加,?倍，水質(zhì)量速度不變時，x增加一倍，但m=E/P

故ye/=mx維持不變，即N不變，對氣膜控制Ky弋k

HoG=G/(Kya)=G/(KGPa)

0808

故HoG,/H0g2=(G./AG,-Pa)/(G2/AG2Pa)

=(G1/G2)(P2/P1)

=(1/2)02(2/1)=1.74

故H0G,=H0G2/1.74

hoG=NoGXHoG=6.35X0.472/1.74=1.72m

j06cl502510

解：系統(tǒng)服從拉烏爾定律

(1)Y=(P°/P)X=685/760X=0.9X

X=Y,/m=0.01/0.9=0.011

(2)(L/G)min=(Y1-Y2)/X*=(y!-y2)/(yi/m)=mn=0.9X0.9=0.81

(3)(L/G)=2(Ls/G)min=2X0.81=1.62

S=mG/L=0.9/1.62=0.556

Y/Y評1/(1-n)=10

NoG=l/(l-S)Xln[(l-S)y>/y2+S]

=l/(l-0.556)Xln[(l-0.556)XI0+0.556]=3.62

(4)h=HoGXNoG=0.5X3.62=1.81m

jo6d20036

解：由題意：X2=0

Y(=0.05/(1-0.05)=0.0526

Y2=0.005/(1-0.005)=0.005025

m=0.0177/(1-0.0177)/((18.89/17)/(1000/18))=0.9

已知吸收因數(shù)為1,所以液氣比L/V=m=0.9

(1)由物料衡算

V(YLY2)=L(X「X2)

:.X2=(V/L)(Y1-Y2)=(1/0.9)(0.0526-0.005025)=0.05286

(2)HoG=H/NoG

NoG=(Y,-Y2)/AYm/

△Y尸Y「YJ=Y「mY尸0.0526-0.9X0.05286=0.005025

同理：得：AY2=0.005025

即各處推動力相同(因操作線平行于平衡線)，所以AYm=0.005025

NoG=(0.0526-0.005025)/0.005025=9.468

HoG=3/9.468=0.317(m)

(3)VA=1.-.NT=NOG,等板高度HETP=HoG=0.317m

j06b1502710

解：L=3000X(1/18)X(1/3600)=0.0463kmol/m2.s

氨分壓p=PXy

由題圖x屋=0.050

貝ljxb=0.050X70%=0.035

ya=yb-L(xb-xb)/G=0.042-0.0463(0.035-0)/0.04=0.00149

吸收率<|>?(yb-ya)/y=(0.042-0.00149)/0.042=96.5%

y；=0;由題圖%*=0.027

△ym=((0.042-0.027)-0.00149)/ln(0.042-0.027)/0.00149=0.00585

h=G(yb-ya)/(Ky△ym)

=0.04(0.042-0.00149)/(0.03X0.00585)=9.23m

jo6c15028

解：(1)y2=yi(l-)=0.03(1-0.98)=0.0006

l/A=mG/L=1.05/2=0.525

NoG=l/(l-l/A)X]n[(l-l/A)(yi-mx)/(y-mx2)+l/A]

=1/(1-0.525)Xln[(l-0.525)(0.03-1.05X0.0001)/(0.0006-1.05X

0.0001+0.525)

=7.104

(2)填料塔無限高時:X1與月平衡此時液氣比變?yōu)?.04,m>L7V)

xei=y!/m=0.03/1.05=0.02857

作物料衡算求出y：

L2x2+Glyi=L|X|+G2y2

?..是低濃氣體吸收，可認為L、G不發(fā)生變化。

(L/G)X2+yi=(L/G)X|+y2

?"-yz'=yi+(L/G)(x2-x1)

=0.03+1.04(1X10-0.02857)=3.9X10

此時極限回收率為

，

獷=(yi-y2)/y=(0.03-3.9X10)/0.03=98.7%

j06cl502912

解：y?=yi(1-6)=0.05(1-0.95)=0.0025

1.Xi*=y/2.0=0.025

xi=x「X70%=0.0175

L=G(yl-y2)/(xi-x2)=80(0.05-0.0025)/(0.0175-0)=17kmol/h=3.91m/h

2.Aym=((y「yi*)-(y2-yz*))/ln((yi-yi*)/(y2-y2,)

(0.05-2.0X0.0175)-(0.0025-2.0X0))/In((0.05-0.0350)/0.0025)

=0.00698

G'=80/(n/4X0.8)=159kmol/m.h

h=(G/Kya)(yi-y2)/Aym=(159/150)(0.05-0.0025)/0.006698=7.21m

3.L/Lmin=(G(y=yz)/(x「X2))/(G(yi-yz)/(Xi*-X2)

=X,7x>=0.025/0.0175=1.43

j06c2003014(1)m=2090/760=2.75

合股后yb=(0.015X0.1+0.0I5X0.04)/0.03=0.07

ya?=0.07/(1-0.07)X(l-0.85)=0.0113

xb=0.07/2.75=0.02545

Lmin=0.03X(0.07-0.0113)/0.02545=0.0692kmol/s

(2)G=0.5/(22.4X293/273)=0.0208kmol/m.s

L/G=1.2X0.0692/0.03=2.768;x=0.02545/l.2=0.0212

或:xb=(0.07-0.0113)/2.768=0.0212

Ayb=yb-yb*=0.07-0.0212X2.75=0.0117

△ya=ya-ya*=0.0U3

Aym=(0.0117-0.0113)/ln(0.0117/0.0113)=0.0115

ho=0.0208/(8X10-3)X(0.07-0.0113)/0.0115=13.27m

(3)LminI,hI

進料口應在塔內(nèi)相應濃度處。

jO6bl5O3119

解：n=90%,yb=0.05,Yb=yb/(l-yb)=O.05/(1-0.05)=0.05263

Y,=Yb(l-n)=0.005266

G=940/22.4=41.96kmol/h

GB=G(l-yb)=41.96(1-0.05)=39.87kmol/h

(Ls/GB)min=(YT,)/(Y/m)

=(0.05263-0.00526)/(0.05263/0.14)=0.126

Ls=1.5Lsmin=1.5X0.126X39.87=7.536kmol/h

=7.536X170Kg/h=1281Kg/h

Xb=(Yb-Ya)/(1.5(Ls/GB)min)

=(0.05263-0.00526)/(1.5X0.126)=0.251

Xb=Xb/(l+Xb)=0.2006

HoG=GB/(KYaQ)

=(39.87/3600)/(0.035X0.785X0.6)=1.12m

N=(Y「Yj/AYm

△Yb=Yb-Yb*=0.05263-0.14X0.251=0.01749

AY=Ya-Y>0.00526

AYm=(0.01749-0.00526)/In(0.01749/0.00526)=0.01018

NoG=(0.05263-0.00526)/0.01018=4.65

Z=HoGXNoG=l.12X4.65=5.2m

j06c2003227

解：(1)由已知條件可知：m=0.14X2=0

進入塔內(nèi)隋性組分摩爾流量為：

V=1000/22.4X273/(273+25)X(780/760)X(1-0.05)

=39.88(kmol/h)

進塔氣體苯的濃度：

YFO.05/(1-0.05)=0.05263

則出塔氣體中苯濃度為：

Yz=Yi(1-4>)=0.05263X(1-0.95)=0.00263

由吸收劑最少用量公式Lmin=V(Y,-Y2)/(Yl/m-X2)

則L=1.5Lmin=1.5V(Y-Y2)/(Y/m-X2)

=1.5X39.88X(0.05263-0.00263)/(0.05263/0.14)

=7.956(kmol/h)

(2)由全塔物料衡算式，X1=X2+V/L(Y1-Y2)

=0+41.12/8.22X(0.05263-0.00263)=0.25

(3)mV/L=0.14X41.12/8.22=0.7

Z=HoGXNoG

HoG=(V/Q)/(Kya)=39.88/(n/4X0.62)/125=1.13(m)

NoG=l/(l-mV/L)Xln[(l-mV/L)(Y-mX)/(Y-mX)+mV/L]

=l/(l-0.7)Xin[(1-0,7)(0.05263/0.0263)+0.7]=6.34

AZ=6.34X1.13=7.16(m)

(4)吸收苯的質(zhì)量為：

m=MV(YI-Y2)=39.88X178X(0.05263-0.00263)=155.5(Kg/h)

(5)由回收率公式6=l-Y〃Yi,要使t須Yl

在逆流吸收塔操作線圖中可以看到Y(jié)I,相當于增大操作線斜率，即L/Vt(操

作液氣比增大)可以提高回收率，增加塔高，提高操作壓力也可提高回收率。

j06c1503316

由(I)塔：L/G=(yB-yA)/(xB-xA)

=(0.02-0.004)/(0.008-0)=2

s=m/(L/G)=l.5/2=0.75

兩個塔的HoG相等，由h=HoGXNoG得

h(II)/h(I)=NoG(II)/NoG(I)

=(l/(l-s)Xln[(1-s)(yis,-mx；)/(y；-mx，)+s])/(1/(1-s)X

ln[(l-s)(yB-mx；/(y；mxj+s))

=ln[(l-O.75)(0.004-0)/(0.002-0)+0.75]/

In[(1-0.75)(0.02-0)/(O.004-0)+0.75]=0.322

/.h(II)=O.322Xh(I)=0.322X8=2.58m

(2)同理h'/h=ln[(1-0.75)(0.02-0)/(0.002-0)+0.75]/

In[(1-0.75)(0.02-0)/(O.004-0)+0.75]=1.70

h'=1.7X8=13.6m

顯然h'>h(I)+h(II)

j06c1503411

解：1.n=(y「y2)/yi(yiWlO%)

y2=yi(l-H)=0.015(1-0.85)=0.225%

由平衡關(guān)系知：出口的礦物油中最大溶質(zhì)濃度為x「=y/O.5=3/

2.(L/G)min=(yi-y2)/(xf-xz)=(0.015-0.00225)/O.3=0.425

3.(L/G)=3(L/G)min->(yi-y2)/xi=3(yi-yz)/x>

Xi=xM3=l%

而NoG=(yi-y2)/Aym

Aym=(Ayi-Ay2)/ln(Ayi/Ay2)

yi-y2=l.275%Ayi=yi-mxi=l.5%-0.5X1%=1%

Ay2=y2；.Aym=(l%-0.225%)/ln(l%/0.225%)=0.52%

，NoG=(l.5%-0.225%)/O.52%=2.45

4.ho=HoGXNoG=lX2.45=2.45(m)

j06b1503520

連續(xù)操作的解吸塔：L'=40T/hX,=8X10

X2=2X10Y2=0.001V=20Vmin

E=1.6X1OKN/mP=100KN/mKxa=800kmol/m.h

m=E/P即m=1600

解：液相按純水計算，即L'=4X10000/18=2222kmol/h

(1)Vmin=2222X(8X10-2X10)/(1600X8X10-0.001)

=1.365kmol/h

V=20Vmin=27.3kmol/h

Vh=27.3X22.4X(273+25)/273X(101.3/100)(1+0.001)=677m3/h

(2)Z=NoLXHoL

HoL=L/(KxaQ)=8000/(18X800)=0.556m

NoL=(X1-X2)/AXm

YFO.001+2222X(8X10-2X10)/27.3=7.35義10

△X2=X「Yi/m=8X10-7.35X10/1600=7.54X10

同理△X2=X2-Y〃m=l.375X10

△Xm=(AXi-AX2)/ln(AX1/AX2)

=7.403X10/4.004=1.848X10

于是NoL=(8X10-2X10)/(1.848X110)=4.22

Z=4.22X0.556=2.346m

j06c2003616

解：由于稀濃氣體，因此可進行近似計算

Yi^yiY2s?y2V=?GKya^kyaY^l.5X

(1)(L/V)min=(L/G)min=(0.05-0.005)/(0.05/1.5-0)=1.35

(L/G)=l.5X1.35=2.025仁2

Xi=V(0.05-0.005)/L=0.0225

(2)HoG=V/(KyaQ)=0.5/(0.85Xn/4X0.4)=1.17m

NoG=l/(l-l/A)Xln[(l-1/A)(Y-mX)/(Y-mX)+1/A]

l/A=mV/L=l.5/2=0.751-1/A=O.25

ANoG=l/0.25XIn[0.25X0.05/0.005+0.75]

=4.715七4.72m

H'=l.17X4.72=5.52m

(3)HoG'=5.52+2=7.52m

NoG'=H'/HoG=6.427

6.427=1/0.25Xln[(0.25)X0.05/Y,+0.75]

得=0.00295

EA=(YI-Y2))/YI=(0.05-0.00295)/0.05=94.1%

j06b100379

解：N=-CD(dyA/dZ)+yA(NA+NB)

又知NB=-2NA

，NA=-CD(dyA/dZ)+yA(NA-2NA)=-CDAB(dy/dZ)-yANA

NA=-l/(l+yA)XCD(dyA/dZ)很顯然在穩(wěn)態(tài)擴散時，N為常數(shù)，

CD為常數(shù)。

8y,6

NAX/odZ=-CDf(l/(l+y))dy=CDln(l/(l+y,6)

0y,0

NA=CDAB/8Xln(l/(l+y,8)

j06b1003812

氣相主體中CCh的分壓為

P=500X0.03=15KN/m2

與溶液成平衡的c02分壓為：

P*=Exi

對于稀溶液：C=1000/18=55.6kmol/m3

C。2的摩爾數(shù)n=370/(1000X44)=0.00841

n/C=0.00841/55.6=1.513X10~

，P'=1.73X10X1.513X10=26.16KN/m

P'>P

于是發(fā)生脫吸作用。

以分壓差表示的傳質(zhì)推動力為

AP=P*-P=11.16KN/m

j06d2003925

(1)(YI-Y2)/Y!=0.9Y2=0.IYI=0.005X2=0.01

G(Y,-Y2)=L(XI-X2)100(0.05-0.005)=60(X,-0.01)

AXFO.045X(100/60)+0.01=0.085

(X1(Y,)=(0.085,0.05)(X2,Y2)=(0.01,0.005)

Y「=0.08X0.085=0.0068Y2=0.08X0.01=0.0008

y「=0.0068/(1+0.0068)=0.00675

yj=0.0008/(1+0.0008)=?0.0008

yi=0.05/(l+0.05)=0.0476y2=0.005/(1+0.005)=^0.005

△ym=((0.0476-0.00675)-(0.005-0.0008))/ln(0.04085/0.0042)=0.0161

NoG=(0.0476-0.005)/0.0161=2.646

(2)近似取m=E/P,E=2X0.08atmE=C/HC?=PL/ML=1000/40=25

H=25/(2X0.08)=156.25kmol/m2.atm

HkL=156.25X6kmol/m'.atm.s

kc=O.03[kmol/m2.sKN/m]=3.04kmol/m'.s.atm

1/Ky=l/(K#)=(l/kc+l/Ik)/P

=(1/3.04+1/(156.25X6)/2=0.165

Ky=6.06kmol/nf.s

1/kc>>1/Hkl.?.氣膜控制

(3)按再循環(huán)操作

劑量32.55(1-0.085/(1+0.085))=30kmol/h

，30X0.085+60X0.01=90XX/

=0.035故再循環(huán)操作線端點為(Xi,YI)=(0.085,0.05)

(V,Ya)=(0.035,0.005)

(4)由于操作線更靠近平衡線，故氣相總傳質(zhì)單元數(shù)將增加。

j06c1504012

解：(-)Z=HoGXNoG

HoG=l.11mNoG=0.9<b/AYm

AYm=((Yi-Mxi)-(Y2-mX2))/In(Yi-mXi)/Y2(X2=0)

L/V=l.5(L/V)min=1.5(Y「Yz)/YiXm=l.35m

XFV(Y-Y2)/L=(1/1.5)Y,

AYm=((1-1/1.5)Y-0.1Y1)/In(1-1/1.5)/0.1=0.194Y,

NoG=0.9Yi/0.194Yi=4.64

N另解：S=m/(L/V)=m/(1.35m)=l/l.35=0.7407

.-.NoG=l/(l-S)Xln[(l-S)Y/Y2+S]

=1/(1-0.7407)Xln[(l-0.7407)X1/0.1+0.7407]=4.64

Z=4.64X1.11=5.15m

(-)YtXt

j06bl504111解：

(1)Yi=3/97=0.03093X2=0

Yz=Y"l-*)=0.03093X(1-0.95)=0.00155

由最小溶劑用量公式

(Lmin/V)=(YLY?)/(Y/m-X?)=(Y「Yz)/(%/m)=m(Y>-Y2)/Y.

已知mV/L=0.8則L/V=(l/0.8)m=1.25m

(L/V)/(L/V)min=1.25/0.95=1.316

(2)由物料衡算式得：

X尸(Y1-Y2)/(L/V)=(0.03093-0.00155)/(1.25X1.013)=0.0228

(3)NoG=l/(l-mV/L)Xln[(l-mV/L)(YLmX2)/(Y2-mX2)+mV/U

=l/(l-0.8)Xln[(l-0.8)Y/(0.05Y1)+0.8]

=7.84

j06d2004219

解：1.n=(Yb-Ya)/YbYa^ya

Yb=yb/(l-yh)=0.04/(1-0.04)=0.0417

n=(0.0417-0.008)/0.0417=80.7%

.G=2240/22.4=100kmol/hGB=100(1-0.04)=96kmol/h

(L/G)min=(Y?-Ya)/(Xb*-0)=(0.0417-0.008)/(0.0417/0.0126-0)

~0.102

Lmin=0.102X96=9.792kmol/h

Ls=1.4Lmin=13.7kmol/h

3.Ls/GF(Y-Y)/(X-0)=13.7/96=0.143

Xb=(0.0417-0.008)/(L/G)=0.24

△Y產(chǎn)Y「Yb*=O.0417-0.126X0.24=0.001146

4.AYm=(Ay「Ay?)/In(Ay/△y,)=0.00346/ln(0.01146/0.008)

=0.00963

5.部分循環(huán)，入口液量為(L+L')，入塔濃度為Xa',在最大L'下，X'處在

最大。

(L/G)max=(0.0417-0.008)/(X「X：)

X；=Y,/0.126=0.008/0.126=0.0635

:.((L)+Ls')/G)max=(0.0417-0.008)/(0.24-0.0635)=0.191

L'smax'=0.191X96-13.7=4.64kmol/h

L'max'=Lsmax'/(1-X)=4.64/(1-0.24)=6.1kmol/h

j06b1504311

yb=O.02y“=0.02Xl%=0.0002

Xb,=yb/m=0.02/0.36=0.0556xa=0.0002

(L/G)min=(yb-ya)/(Xb*-xa)=(0.02-0.0002)/(0.0556-0.0002)=0.3577

L/G=l.5(L/G)min=1.5X0.3577=0.5366

xiFx?+(yb-y?)/(L/G)=0.0002+(0.02-0.0002)/0.5366=0.0371

Ayb=yb-mxb=0.02-0.36X0.0002=0.00664

△yFyjmxFO.0002-0.36X0.0002=0.000128

Aym=(0.00664-0.000128)/In(0.00664/0.000128)=0.00165

NoG=(yb-y?)/Aym=(0.02-0.0002)/0.00165=12

HoG=G/(Kya)=53/3600/(Jt/4X1.12)/0.015=1.033m

ho=HoGXNOG=1.033X12=12.4m

w06d100441如圖。

j06d2004537

解：A=lm2VH2S=5000標準米/hypO.03

逆流x2=0,6=90%,p=latm,t=27℃ye=2x

(1)低濃度=(y>-y2)/yFO.9即(0.03-y2)/0.03=0.9

.,.yz=0.003

(L/G)min=(yf)/(y/m-x?)=(0.03-0.003)/(0.03/2-0)=1.8

L/G=1.2(L/G)min=1.2X1.8