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+clc;

+//Example 26.4

+//page no 387

+printf("\nExample 26.4 page no 387\n\n");

+//air is used to fluidize a bed of speherical particles

+D=0.2//bed diameter,m

+d_p=7.4e-5//diameter of 200 mesh particles from table 23.2,m

+rho_s=2200//ultimate solid density

+rho_f=1.2//density of air

+meu=1.89e-5//viscosity of air

+g=9.807//grav. constant

+e=0.45//bed porosity

+L_mf=0.3//length at minimum fluidization

+//assume laminar flow 

+//applying equation 26.29

+v_mf=(1-e)*g*rho_s*d_p^2/(150*e^3*meu)//minimum fluidizaton veloctiy 

+printf("\n min. fluidization velocity v_mf=%f m/s",v_mf);

+//check the flow regime

+R_e=v_mf*d_p/(meu*(1-e))

+printf("\n Reynolds no R_e=%f ",R_e);

+//since R_e= 1.79 <10,flow is laminar

+m_dot=%pi*v_mf*D^2*rho_f/4//mass flow rate

+printf("\n mass flow rate m_dot =%f kg/s",m_dot);

+P_fr=round((1-e)*rho_s*g*L_mf)//gas pressure drop across the bed

+printf("\n gas pressure drop P_fr=%f Pa",P_fr);