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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 /542/CH7/EX7.1/Example_7_1.sci | |
| 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 '542/CH7/EX7.1/Example_7_1.sci')
| -rwxr-xr-x | 542/CH7/EX7.1/Example_7_1.sci | 63 |
1 files changed, 63 insertions, 0 deletions
diff --git a/542/CH7/EX7.1/Example_7_1.sci b/542/CH7/EX7.1/Example_7_1.sci new file mode 100755 index 000000000..7ce893d09 --- /dev/null +++ b/542/CH7/EX7.1/Example_7_1.sci @@ -0,0 +1,63 @@ +clear;
+clc;
+printf("\n Example 7.1");
+//In the leaf filter filtration is at const pressure from the start
+//V^2 + 2ALV/v = 2(-deltaP)A^2t/(ruv)
+
+//In the filter press,a volume V1 of filtrate is obtained under const rate conditions in time t1,and filtration is then carried out at constant pressure.
+//V1^2 + 2ALV1/v = 2(-deltaP)A^2t1/(ruv)
+//and (V^2 − V1^2 ) + 2AL/υ(V − V1) = 2(−P)A^2/rμυ(t − t1)
+
+//for the leaf filter
+t2 = 300; //t2 is in secs
+V2 = 2.5*10^(-4); //V2 is in m^3
+t3 = 600; //t3 is in secs
+V3 = 4*10^(-4); //V3 is in m^3
+A = 0.05; //A is in m^2
+deltaP = -7.13*10^(4); //it is in N/m^2
+//putting these values in above eq
+
+a = [2*7.13*10^(4)*0.05^(2)*300 -2*0.05*2.5*10^(-4);2*7.13*10^(4)*0.05^(2)*600 -2*0.05*4*10^(-4)];
+b = [(2.5*10^(-4))^2;(4*10^(-4))^2];
+x = inv(a)*b;
+y = [1/x(1);x(2)];
+printf("\n L/υ=%f*10^(-3) and rμυ = %f*10^(11)",y(2)*10^3,y(1)*10^(-11));
+
+//for the filter press
+V1 = poly([0],'V1');
+s = roots(V1^2 + (2.16*y(2)*V1)-(4*10^(5)*2.16^2)/y(1)*180);
+printf("\n the value of V1 = %fm^3",s(2));
+
+//For a constant pressure period (t - t1)=900secs
+//Calculting the total volume of filtrate
+V = poly([0],'V');
+d = roots((V^2-3.33*10^(-4))+(1.512*10^(-2)*(V-1.825*10^(-2))-5.235*10^(-6)*900));
+printf("\n The value of V = %.3f m^3",d(2));
+
+f = (4*10^(5)*(2.16)^2)/(7.13*10^(11)*(6.15*10^(-2) + 2.16*3.5*10^(-3)));
+printf("\n The final rate of filtration is %.2f*10^(-5) m^3/sec",f*10^(5));
+
+// Assuming viscosity of the filtrate is the same as that of the wash-water
+rw_400 = (0.25)*f;
+printf("\n Rate of washing at 400 kN/m2 = %.1f*10^(-6) m^3/sec",rw_400*10^(6));
+
+rw_275 = rw_400*(275/400);
+printf("\n Rate of washing at 275 kN/m^2 = %.1f*10^(-6) m^3/sec",rw_275*10^6);
+printf("\n Thus the amount of wash-water passing in 600s = %.3f m^3",600*rw_275);
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