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+//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 ======================================================
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