3 files changed, 181 insertions, 0 deletions
diff --git a/497/CH4/EX4.1/Chap4_Ex1.sce b/497/CH4/EX4.1/Chap4_Ex1.sce
new file mode 100755
index 000000000..ae7ee1021
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+++ b/497/CH4/EX4.1/Chap4_Ex1.sce
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+//Kunii D., Levenspiel O., 1991. Fluidization Engineering(II Edition). Butterworth-Heinemann, MA, pp 491
+
+//Chapter-4, Example 1, Page 106
+//Title: Design of a Perforated Plate Distributor
+//==========================================================================================================
+clear
+clc
+
+//INPUT
+dt=4;//Vessel diameter in m
+Lmf=2;//Length of the bed in m
+ephsilonmf=0.48;//Void fraction of bed
+rhos=1500;//Density of solid in kg/m^3
+rhog=3.6;//Density of gas in kg/m^3
+myu=2E-5;//Viscosity of gas in kg/m s
+po=3;//Pressure of inlet gas in bar
+uo=0.4;//Superficial velocity of gas in m/s
+uorm=40;//Maximum allowable jet velocity from holes in m/s
+g=9.80;//Acceleration due to gravity in m/s^2
+gc=1;
+pi=3.1428;
+
+//CALCULATION
+//Computation of minimum allowable pressure drop through the distributor
+deltapb={(1-ephsilonmf)*(rhos-rhog)*g*Lmf}/gc;//Calculation of pressure drop in bed using Eqn.(3.17)
+deltapd=0.3*deltapb;//Calculation of pressure drop in distributor using Eqn.(3)
+
+//Computation of orifice coefficient
+Ret=(dt*uo*rhog)/myu;
+if    Ret>=3000 then Cd=0.60;
+elseif    Ret>=2000 then Cd=0.61;
+elseif    Ret>=1000 then Cd=0.64;
+elseif    Ret>=500  then Cd=0.68;
+elseif    Ret>=300  then Cd=0.70;
+elseif    Ret>=100  then Cd=0.68;
+end
+
+//Computation of gas velocity through orifice
+uor=Cd*((2*deltapd)/rhog)^0.5;//Calculation of gas velocity through orifice by using Eqn.(12)
+f=(uo/uor)*100;//Calculation of fraction of open area in the perforated plate 
+
+
+//Computation of number of orifices per unit area of distributor
+dor=[0.001;0.002;0.004];//Different orifice diameters in m
+n=length(dor);
+i=1;
+while i<=n
+    Nor(i)=(uo*4)/(pi*uor*(dor(i))^2);//Calculation of number of orifices by using Eqn.(13)
+    i=i+1;
+end
+    
+//OUTPUT
+mprintf('\nThe pressure drop in bed:%fPa',deltapb);
+mprintf('\nThe minimum allowable pressure drop in distributor:%fPa',deltapd);
+if uor<uorm then mprintf('\nThe gas veleocity of %fm/s is satisfactory',uor);
+    else mprintf('\nThe gas veleocity of %fm/s is not satisfactory',uor);
+end
+if f<10 then mprintf('\nThe fraction of open area of %f percent is allowable',f);
+    else mprintf('\nThe fraction of open area of %f percent is not allowable',f);
+end   
+printf('\nDiameter of orifice(m)');
+printf('\tNumber of orifices per unit area(per sq.m)');
+j=1;
+while j<=n
+    mprintf('\n%f',dor(j));
+    mprintf('\t\t%f',Nor(j));
+    j=j+1;
+end
+printf('\nThis number can be rounded off.');
+printf('\nSince orifices that are too small are liable to clog and those that are too large cause uneven distribution of gas, we choose orifice of diameter %fm',dor(2));
+
+//====================================END OF PROGRAM ======================================================
+\ No newline at end of file
diff --git a/497/CH4/EX4.2/Chap4_Ex2.sce b/497/CH4/EX4.2/Chap4_Ex2.sce
new file mode 100755
index 000000000..53e00c02a
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+++ b/497/CH4/EX4.2/Chap4_Ex2.sce
@@ -0,0 +1,50 @@
+//Kunii D., Levenspiel O., 1991. Fluidization Engineering(II Edition). Butterworth-Heinemann, MA, pp 491
+
+//Chapter-4, Example 2, Page 108
+//Title: Design of a Tuyere Distributor
+//==========================================================================================================
+clear
+clc
+
+//INPUT
+lor=0.1;//Minimum allowable tuyere spacing in m
+uorm=30;//Maximum allowable jet velocity from the tuyere in m/s
+uo=0.4;//Superficial velocity of gas in m/s
+uor=30.2;//Gas velocity through orifice,from Exa 1, in m/s
+Cd=0.6;//Dicharge coefficient from Exa 1
+rhog=3.6//Density of gas in kg/m^3
+pi=3.1428;
+
+//CALCULATION
+Nor=1/(lor^2);//Calculation of number of orifices per unit area by assuming minimum spacing for tuyeres
+dor={(4/pi)*(uo/uor)*(1/Nor)}^0.5;//Calculation of diameter of inlet orifiec by using Eqn.(13)
+
+//Computation of diameter of hole for different number of holes per tuyere
+q=(lor^2)*uo;//Volumetric flow rate in m^3/s
+Nh=[8;6;4];//Different number of holes per tuyere
+n=length(Nh);
+i=1;
+while i<=n
+    dh(i)=((((q/Nh(i))*(4/pi))/uorm)^0.5);//Calculation of diameter of holes
+    i=i+1;
+end
+deltaph=(rhog/2)*((uor/Cd)^2);
+
+//OUTPUT
+printf('\nNumber of holes(number of holes/tuyeres)');
+printf('\tDiameter of hole(m)');
+j=1;
+while j<=n
+    mprintf('\n%f',Nh(j));
+    mprintf('\t\t\t\t\t%f',dh(j));
+    j=j+1;
+end
+printf('\nThe design chosen is as follows');
+printf('\n\tTuyeres are as shown in Fig.2(b),page 97');
+mprintf('\n\tNumber of holes = %f(Since rectangular pitch is chosen for tuyeres)',Nh(2));
+mprintf('\n\tDiameter of hole = %fm',dh(2));
+mprintf('\n\tDiameter of incoming high-pressure-drop orifice = %fm ID',dor);
+printf('\nChecking the pressure drop in tuyeres');
+mprintf('\nSince pressure drop of %fPa gives sufficiently high distributor pressure drop as seen in Exa.1, use of inlet orifice can be dispensed.',deltaph);
+
+//====================================END OF PROGRAM ======================================================
+\ No newline at end of file
diff --git a/497/CH4/EX4.3/Chap4_Ex3.sce b/497/CH4/EX4.3/Chap4_Ex3.sce
new file mode 100755
index 000000000..85ab5bbe9
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+++ b/497/CH4/EX4.3/Chap4_Ex3.sce
@@ -0,0 +1,59 @@
+//Kunii D., Levenspiel O., 1991. Fluidization Engineering(II Edition). Butterworth-Heinemann, MA, pp 491
+
+//Chapter-4, Example 3, Page 110
+//Title: Power Requirement for a Fluidized Coal Combustor(FBC)
+//==========================================================================================================
+clear
+clc
+
+//INPUT
+deltapd=[3;10]//Distributor pressure drop in kPa
+deltapd2=10;//Distributor pressure drop in kPa
+po=101;//Entering air pressure in kPa
+To=20;//Entering air temperature in degree C
+y=1.4;//Fugacity of air
+deltapb=10;//Pressure drop in bed in kPa
+p3=103;//Pressure at the bed exit in kPa
+F=8;//Feed rate of coal in tons/hr
+H=25;//Gross heatig value of coal in MJ/kg
+Fa=10;//Air required at standard condition in nm^3/kg
+etac=0.75;//Efficiency of compressor
+etap=36;//Efficiency of plant in %
+
+//CALCULATION
+//Calculation of volumetric flow rate of air
+vo=((F*1000)*Fa*((To+273)/273))/3600;
+
+//Case(a) Distributor Pressure drop = 3kPa and Case(b) Distributor Pressure drop = 10kPa
+n=length(deltapd);
+i=1;
+while i<=n
+    p2(i)=p3+deltapb;//Calculation of pressure at the entrance of the bed
+    p1(i)=p2(i)+deltapd(i);//Calculation of pressure before entering the bed
+    ws(i)=(y/(y-1))*po*vo*((p1(i)/po)^((y-1)/y)-1)*(1/etac);//Calculation of power required for the compressor by Eqn.(18) & Eqn.(20)
+    i=i+1;
+end
+
+//Case(c) 50% of the required bypassed to burn the volatile gases. Distributor Pressure drop = 3kPa
+//No change in pressure drop from case(a)
+v1=vo/2;//New volumetric flow rate of air
+ws1=ws(1)/2;//Power required for blower for primary air
+ws2=(y/(y-1))*po*v1*((p3/po)^((y-1)/y)-1)*(1/etac);//Power required for blower for bypassed air
+wst=ws1+ws2;//Total power required for the two blowers
+p=((ws(1)-wst)/ws(1))*100;//Saving in power when compared to case(a)
+
+//OUTPUT
+printf('\nCase(a)');
+mprintf('\n\tVolumetric flow rate of air = %f m^3/hr',vo);
+mprintf('\n\tPower required for compressor = %f kW',ws(1));
+printf('\nCase(b)');
+mprintf('\n\tVolumetric flow rate of air = %f m^3/hr',vo);
+mprintf('\n\tPower required for compressor = %f kW',ws(2));
+printf('\nCase(c)');
+mprintf('\n\tVolumetric flow rate of air = %f m^3/hr',v1);
+mprintf('\n\tPower required for compressor for primary air = %f kW',ws1);
+mprintf('\n\tPower required for blower for bypassed air = %f kW',ws2);
+mprintf('\n\tTotal power required for the two blowers = %f kW',wst);
+mprintf('\n\tPower saved compared to case(a) = %f percent',p);
+
+//====================================END OF PROGRAM ======================================================
+\ No newline at end of file