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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /249/CH6 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '249/CH6')
| -rwxr-xr-x | 249/CH6/EX6.1/6_01.sce | 12 | ||||
| -rwxr-xr-x | 249/CH6/EX6.2/6_02.sce | 20 | ||||
| -rwxr-xr-x | 249/CH6/EX6.3/6_03.sce | 62 |
3 files changed, 94 insertions, 0 deletions
diff --git a/249/CH6/EX6.1/6_01.sce b/249/CH6/EX6.1/6_01.sce new file mode 100755 index 000000000..7ff0fea6e --- /dev/null +++ b/249/CH6/EX6.1/6_01.sce @@ -0,0 +1,12 @@ +clear
+clc
+V1=50;V2=30;V3=40;
+//Branch D consists of 2 reactor in series,can be considered a single reactor of volume
+VD=V1+V2;
+VE=V3;
+//For Reactor in parallel,V/F must be same for same conversion
+//FE/FD=VE/VD;FD/F=1/(1+VE/VD)
+m=VE/VD
+fr_D=1/(1+m);
+printf("\n Fraction of feed going to branch D is %f \n",fr_D)
+
diff --git a/249/CH6/EX6.2/6_02.sce b/249/CH6/EX6.2/6_02.sce new file mode 100755 index 000000000..d201f8fc5 --- /dev/null +++ b/249/CH6/EX6.2/6_02.sce @@ -0,0 +1,20 @@ +clear
+clc
+//For single reactor and 90% Conversion
+//From fig 6.6
+kCot=90;
+//For 2 reactor space time is doubled and from fig
+kCot=180;
+//From graph X=97.4%
+X=97.4;
+printf("\n Part a")
+printf("\n The conversion in percentage is %f \n",X)
+//For 90% Conversion & N=2.from graph
+kCot=27.5;
+//Comparing the reaction rate group for N=1 and N=2,We get
+//(V2/v2)/(V1/v1)=27.5/90
+//V2=2V1
+//Ratio of flow rates
+ratio=90*2/27.5;
+printf("\n Part b")
+printf("\n Treatment rate can be increased by %f \n",ratio)
diff --git a/249/CH6/EX6.3/6_03.sce b/249/CH6/EX6.3/6_03.sce new file mode 100755 index 000000000..b938d3502 --- /dev/null +++ b/249/CH6/EX6.3/6_03.sce @@ -0,0 +1,62 @@ +clear
+clc
+CAo=[2;5;6;6;11;14;16;24];//mmol/m3
+CA=[0.5;3;1;2;6;10;8;4];//mmol/m3
+t=[30;1;50;8;4;20;20;4];//min
+vo=0.1;//m3/min
+for i=1:8
+ inv_rA(i)=t(i)/(CAo(i)-CA(i));
+end
+//Sorting CA and accordingly changing -1/rA for plotting graph between CA and -1/rA
+for i=1:8
+ for j=i:8
+ if CA(i)>CA(j)
+ temp=CA(i);
+ CA(i)=CA(j);
+ CA(j)=temp;
+ temp1=inv_rA(i);
+ inv_rA(i)=inv_rA(j);
+ inv_rA(j)=temp1;
+ end
+ end
+end
+plot(CA,inv_rA)
+xlabel('CA(mmol/m3)');ylabel('-1/rA(m3.min/m mol)');
+disp('From the graph,we can see that we should use plug flow with recycle')
+//From fig
+CAin=6.6;//mmol/m3
+R=(10-6.6)/(6.6-1);
+//V=t*vo=area*vo
+V=(10-1)*1.2*vo;
+vr=vo*R;
+printf("\n Part a")
+printf("\n The vol of reactor is %f",V)
+printf("m3 \n The recycle flow rate is %f",vr)
+printf("m3/min")
+//Part b,from fig
+t=(10-1)*10;
+t1=(10-2.6)*0.8;
+t2=(2.6-1)*10;
+//For 1 tank
+V=t*vo;
+//For 2 tank
+V1=t1*vo;
+V2=t2*vo;
+Vt=V1+V2;
+printf("\n Part b")
+printf("\n For 1 tank volume is %f",V)
+printf("m3 \n For 2 tank the volume is %f",Vt)
+printf("m3")
+printf("\n Part c")
+disp('We should use mixed flow followed by plug flow')
+//For MFR
+tm=(10-4)*0.2;
+Vm=tm*vo;
+//For PFR
+tp=5.8;//by graphical integration
+Vp=tp*vo;
+Vtotal=Vp+Vm;
+printf("\n For MFR volume(m3) is %f",Vm)
+printf("\n For PFR volume(m3) is %f",Vp)
+printf("\n Total volume is %f",Vtotal)
+printf("m3")
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