5 files changed, 105 insertions, 0 deletions
diff --git a/2762/CH14/EX14.2.1/14_2_1.sce b/2762/CH14/EX14.2.1/14_2_1.sce
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+++ b/2762/CH14/EX14.2.1/14_2_1.sce
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+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.2-1
+//Mechanical-Physical Seperation Processes
+//given data
+t=[4.4 9.5 16.3 24.6 34.7 46.1 59 73.6 89.4 107.3];
+V=(1/1000)*[0.498 1 1.501 2 2.498 3.002 3.506 4.004 4.502 5.009]
+tbyVd=[];
+for(i=1:10)
+    tbyVd(i)=(t(1,i)/V(1,i))
+end
+tbyV=(1/1000)*tbyVd
+plot2d((V*1000),tbyV);
+xtitle("Determination of the constants","Vx1000","(t/V)x(10^-3)")
+//as seen from the graph,
+B=6400;//intercept in s/m3
+//as it is seen the given graph resembles a straight line we can find Kp/2
+Kpby2=[tbyV(6)-tbyV(5)]*1000/(V(6)-V(5));//slope of st line in s/m6
+Kp=Kpby2*2;
+mu=8.937e-4;
+A=0.0439;
+delP=338*1000;
+//Now Kp=mu*alpha*cs/(A*A*delP)
+Cs=23.47;
+alpha=Kp*(A*A*delP)/(mu*Cs);
+//also B=(mu*Rm)/(A*A*(-delP))
+Rm=B*(A*(delP))/mu
+mprintf("alpha= %f m/kg",alpha)
+mprintf(" Rm= %f m-1",Rm)
+//As the slope has been calculated in numerical value deviations have been found.Kpby2=6*10^6 according to the example but the value calculated here is 5.815*10^6
diff --git a/2762/CH14/EX14.2.2/14_2_2.sce b/2762/CH14/EX14.2.2/14_2_2.sce
new file mode 100755
index 000000000..f7e02995e
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+++ b/2762/CH14/EX14.2.2/14_2_2.sce
@@ -0,0 +1,27 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.2-2
+//Mechanical-Physical Seperation Processes
+//given data
+t=[4.4 9.5 16.3 24.6 34.7 46.1 59 73.6 89.4 107.3];
+V=(1/1000)*[0.498 1 1.501 2 2.498 3.002 3.506 4.004 4.502 5.009]
+tbyVd=[];
+for(i=1:10)
+    tbyVd(i)=(t(1,i)/V(1,i))
+end
+tbyV=(1/1000)*tbyVd
+plot2d((V*1000),tbyV);
+xtitle("Determination of the constants","Vx1000","(t/V)x(10^-3)")
+//as seen from the graph,
+B=6400;//intercept in s/m3
+//as it is seen the given graph resembles a straight line we can find Kp/2
+Kpby2=[tbyV(6)-tbyV(5)]*1000/(V(1,6)-V(1,5));//slope of st line in s/m6
+Kp=Kpby2*2;
+A=20*0.837;//total area= (no. of plates)*(area of one plate)
+Aold=0.0439;
+Kpn=Kp*((Aold/A)^2);
+Bn=B*((Aold/A));
+V=3.37;//amt6 of filtrate to be recovered inm3
+t=((Kpn/2)*V*V)+(Bn*V);
+mprintf("the time taken to recover the filtrate= %f s",t)
+//As the slope has been calculated in numerical value deviations have been found.Kpby2=6*10^6 according to the example but the value calculated here is 5.815*10^6
diff --git a/2762/CH14/EX14.3.2/14_3_2.sce b/2762/CH14/EX14.3.2/14_3_2.sce
new file mode 100755
index 000000000..614b2d6dd
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+++ b/2762/CH14/EX14.3.2/14_3_2.sce
@@ -0,0 +1,24 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.3-2
+//Mechanical-Physical Seperation Processes
+//given data
+rhowater=998;
+muwater=1.005e-3;
+rhop=2467;
+Dp=1.554e-4;
+solid=60;//solid weight percent
+liquid=100-solid;//liquid weight percent
+E=(liquid/rhowater)/((liquid/rhowater)+(solid/rhop));//volume fraction
+rhom=E*rhowater+(1-E)*rhop;
+ship=(1/(10^(1.82*(1-E))));
+g=9.807;//gravity accelaration
+vt=((g*Dp*Dp)*(rhop-rhowater)*(E*E*ship))/(18*muwater);
+Re=(Dp*vt*rhom)/((muwater/ship)*E)
+if (Re<2100)
+    disp(Re)
+    disp('Settling is in laminar range')
+else
+    disp('Settling is not in laminar range')
+end
+//end
diff --git a/2762/CH14/EX14.4.3/14_4_3.sce b/2762/CH14/EX14.4.3/14_4_3.sce
new file mode 100755
index 000000000..7574e312c
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+++ b/2762/CH14/EX14.4.3/14_4_3.sce
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+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.4-3
+//Mechanical-Physical Seperation Processes
+//given data
+rhol=919.5;//density of oil in kg/m3
+rhoh=980.3;//density of the aqueous phase
+rin=10.16;
+rout=10.414;
+r2=sqrt((rhoh*(rout^2)-rhol*(rin^2))/(rhoh-rhol));
+mprintf("the location of centrifuge,r= %f mm",r2)
+//end
diff --git a/2762/CH14/EX14.5.1/14_5_1.sce b/2762/CH14/EX14.5.1/14_5_1.sce
new file mode 100755
index 000000000..297001bab
--- /dev/null
+++ b/2762/CH14/EX14.5.1/14_5_1.sce
@@ -0,0 +1,12 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.5-1
+//Mechanical-Physical Seperation Processes
+//given data in english units
+Df=3/12;//feed size in feet
+Dp=(1/8)/12;//productr size in ft
+T=10/60;//feed rate in ton/min
+//Bonds Equation(English units):(P/T)=1.46E1((1/Dp^0.5)-(1/Df^0.5))
+E1=12.68;
+P=T*(1.46*E1*((1/sqrt(Dp))-(1/sqrt(Df))));
+mprintf("gross power reqd= %f hp",P)