diff options
Diffstat (limited to '1052/CH26')
| -rwxr-xr-x | 1052/CH26/EX26.10/2610.sce | 23 | ||||
| -rwxr-xr-x | 1052/CH26/EX26.11/2611.sce | 20 | ||||
| -rwxr-xr-x | 1052/CH26/EX26.2/262.sce | 31 | ||||
| -rwxr-xr-x | 1052/CH26/EX26.3/263.sce | 37 | ||||
| -rwxr-xr-x | 1052/CH26/EX26.4/264.sce | 25 | ||||
| -rwxr-xr-x | 1052/CH26/EX26.5/265.sce | 15 | ||||
| -rwxr-xr-x | 1052/CH26/EX26.6/266.sce | 13 | ||||
| -rwxr-xr-x | 1052/CH26/EX26.7/267.sce | 24 | ||||
| -rwxr-xr-x | 1052/CH26/EX26.8/268.sce | 23 | ||||
| -rwxr-xr-x | 1052/CH26/EX26.9/269.sce | 16 |
10 files changed, 227 insertions, 0 deletions
diff --git a/1052/CH26/EX26.10/2610.sce b/1052/CH26/EX26.10/2610.sce new file mode 100755 index 000000000..15c512dca --- /dev/null +++ b/1052/CH26/EX26.10/2610.sce @@ -0,0 +1,23 @@ +clc;
+//Example 26.10
+//page no 395
+printf("Example 26.10 page no 395\n\n");
+//a bed of 200 mesh particles is fluidized with air
+d_b=0.2//diameter of bed,m
+d_p=7.4e-5//particle diameter
+L_mf=0.3//bed height at minimum fludization
+e_mf=0.45//bed porosity at min. fluidization
+L_o=L_mf*(1-e_mf)//the zero porosity bed height
+printf("\n zero porosity bed height L_o=%f m",L_o);
+rho_s=2200//density of particles
+rho_f=1.2//density of fluid
+g=9.807//grav. acc
+meu_f=1.89e-5//viscosity of fluid
+//assuming laminar flow ,use equation 26.9
+v_mf =(e_mf^3)*(g*(rho_s-rho_f)*(d_p^2))/(150*(1-e_mf)*meu_f)//velocity at minimum fluidization
+printf("\n velocity at min. fluidization v_mf=%f m/s",v_mf);
+v_t=0.35//terminal velocity from example 26.3
+e=0.91//value of e porosity from eq26.9
+L_f=L_o/(1-e)//expanded bed height L_f
+m=rho_s*%pi*d_b^2*L_o//bed inventory
+printf("\n expanded bed height L_f=%f m\n bed inventory m=%f kg",L_f,m);
diff --git a/1052/CH26/EX26.11/2611.sce b/1052/CH26/EX26.11/2611.sce new file mode 100755 index 000000000..be8f33712 --- /dev/null +++ b/1052/CH26/EX26.11/2611.sce @@ -0,0 +1,20 @@ +clc;
+//Example 26.11
+//page no 396
+printf("\n Example 26.11 page no 396\n\n");
+//refer to illustrative example 26.9
+d_p=7.4e-5//particle diameter
+L_mf=0.3//bed height at minimum fludization
+e_mf=0.45//bed porosity at min. fluidization
+L_o=L_mf*(1-e_mf)//the zero porosity bed height
+printf("\n zero porosity bed height L_o=%f m",L_o);
+rho_s=2200//density of particles
+rho_f=1.2//density of fluid
+g=9.807//grav. acc
+meu_f=1.89e-5//viscosity of fluid
+//assuming laminar flow ,use equation 26.9
+v_mf =(e_mf^3)*(g*(rho_s-rho_f)*(d_p^2))/(150*(1-e_mf)*meu_f)//velocity at minimum fluidization
+printf("\n velocity at min. fluidization v_mf=%f m/s",v_mf);
+F_mf=v_mf^2/(g*d_p)//fluidization mode
+printf("\n fluidization mode F_mf=%f ",F_mf);
+//from value of F_mf ,fluidization is smoth,F_mf =0.66<0.13
diff --git a/1052/CH26/EX26.2/262.sce b/1052/CH26/EX26.2/262.sce new file mode 100755 index 000000000..0b0e47d26 --- /dev/null +++ b/1052/CH26/EX26.2/262.sce @@ -0,0 +1,31 @@ +clc;
+//Example 26.2
+//page no 382
+printf("Example 26.2 page no 384\n\n");
+//a water softner unit consists of a large diameter tank ,the bottom of tank is connected to a vertical ion exchange pipe
+h_f=1.25//total fluid height
+h_l=h_f
+g=32.174//grav. acc
+e=0.25// bed porosity
+d_p=0.00417//ion exchange resin particle diameter,ft
+L=1//pipe length ,ft
+//assume turbulent flow ,apply burke purmer equation
+v_s=sqrt(g*h_f*e^3*d_p/(1.75*(1-e)*L))//superficial velocity
+printf("\n superficial velocity v_s=%f ft/s",v_s);
+meu=6.76e-4//absolute viscosity of water
+rho=62.4//density of water
+//check for turbulent flow
+R_e=d_p*v_s*rho/((1-e)*meu)
+printf("\n R_e=%f",R_e);
+//since reynold no is low the calculation is not valid
+//assume laminar flow and use Blake-Kozeny equation 26.9
+v_s_t=rho*g*h_f*e^3*d_p^2/(150*meu*((1-e)^2)*L)//superficial velocity
+printf("\n superficial velocity v_s_t=%f ft/s",v_s_t);
+ //check the porous medium reynolds no
+ R_e_t=v_s_t*d_p*rho/((1-e)*meu)
+ printf("\n reynolds no R_e_t=%f ",R_e_t);
+ //since reynolds no R_e < 10,the flow is therfor laminar
+ D=0.167//diameter of pipe
+ S=(%pi/4)*D^2//empty cross sectional area
+ q=v_s_t*S//volumetric flow rate
+ printf("\n vol. flow rate q=%f ft^3/s",q);
diff --git a/1052/CH26/EX26.3/263.sce b/1052/CH26/EX26.3/263.sce new file mode 100755 index 000000000..b2abdb6d9 --- /dev/null +++ b/1052/CH26/EX26.3/263.sce @@ -0,0 +1,37 @@ +clc;
+//Example 26.3
+//page no 384
+printf("Example 26.3 page no 384\n\n");
+//refer to Example 26.2
+//a water softner unit consists of a large diameter tank ,the bottom of tank is connected to a vertical ion exchange pipe
+h_f=1.25//total fluid height
+h_l=h_f
+g=32.174//grav. acc
+e=0.25// bed porosity
+d_p=0.00417//ion exchange resin particle diameter,ft
+L=1//pipe length ,ft
+//assume turbulent flow ,apply burke purmer equation
+v_s=sqrt(g*h_f*e^3*d_p/(1.75*(1-e)*L))//superficial velocity
+printf("\n superficial velocity v_s=%f ft/s",v_s);
+meu=6.76e-4//absolute viscosity of water
+rho=62.4//density of water
+//check for turbulent flow
+R_e=d_p*v_s*rho/((1-e)*meu)
+printf("\n R_e=%f",R_e);
+//since reynold no is low the calculation is not valid
+//assume laminar flow and use Blake-Kozeny equation 26.9
+v_s_t=rho*g*h_f*e^3*d_p^2/(150*meu*((1-e)^2)*L)//superficial velocity
+printf("\n superficial velocity v_s_t=%f ft/s",v_s_t);
+ //check the porous medium reynolds no
+ R_e_t=v_s_t*d_p*rho/((1-e)*meu)
+ printf("\n reynolds no R_e_t=%f ",R_e_t);
+ //since reynolds no R_e < 10,the flow is therfor laminar
+//calculation of the pressure drop due to friction and the pressure drop across the resin bed
+k=e^3*d_p^2/(150*(1-e)^2)//packed bed permeability
+P_drop_fr=rho*h_f//friction pressure drop across resin bed,psf
+printf("\n fricion pressure drop P_drop_fr=%f psf",P_drop_fr);
+z_d=-1//length from point 2 to 3,ft
+P_drop_r=rho*(z_d+h_f)//pressure drop across the resi bed
+printf("\n pressure drop across across the resin bed P_drop_r=%f psf",P_drop_r);
+
+
diff --git a/1052/CH26/EX26.4/264.sce b/1052/CH26/EX26.4/264.sce new file mode 100755 index 000000000..a04024cd2 --- /dev/null +++ b/1052/CH26/EX26.4/264.sce @@ -0,0 +1,25 @@ +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);
diff --git a/1052/CH26/EX26.5/265.sce b/1052/CH26/EX26.5/265.sce new file mode 100755 index 000000000..88e916814 --- /dev/null +++ b/1052/CH26/EX26.5/265.sce @@ -0,0 +1,15 @@ +clc;
+//Example 26.5
+//page no 389
+printf("Example 26.5 page no 389\n\n");
+//air flowing through a 10 ft packed bed
+V_o=4.65//superficial velocity,ft/s
+meu_g=1.3e-5//viscosity of air
+rho_g=0.67//density of air,lb/ft^3
+e=0.89//void volume
+g_c=32.2//grav. constant
+L=10//length of packed bed
+d_p=0.007815//effective particle diameter
+P_drop = [(150*V_o*meu_g/(g_c*d_p^2))*((1-e)^2/e^3) + (1.75*rho_g*V_o^2/(g_c*d_p))*((1-e)^2/e^3)]*L//pressure drop
+printf("\n pressure drop P_rop=%f lb/ft^2",P_drop);//calculation error in book
+
diff --git a/1052/CH26/EX26.6/266.sce b/1052/CH26/EX26.6/266.sce new file mode 100755 index 000000000..12d03b345 --- /dev/null +++ b/1052/CH26/EX26.6/266.sce @@ -0,0 +1,13 @@ +clc;
+//Example 26.6
+//page no 392
+printf("Example 26.6 page no 392\n\n");
+//a bed of pulverized is to be fluidized with liquid oil
+D=4//diameter of bed ,ft
+d_p=0.00137//particle diameter ,ft
+rho_s=84//coal particle density ,lb/ft^3
+rho_f=55//oil density,lb/ft^3
+e_mf=0.38//void fraction
+L_mf=8//bed height at minimum fluidization,ft
+P_drop=(rho_s-rho_f)*(1-e_mf)*L_mf +rho_f*L_mf
+printf("\npressure drop P_drop=%f psf",P_drop);
diff --git a/1052/CH26/EX26.7/267.sce b/1052/CH26/EX26.7/267.sce new file mode 100755 index 000000000..07161ed47 --- /dev/null +++ b/1052/CH26/EX26.7/267.sce @@ -0,0 +1,24 @@ +clc;
+//Example 26.7
+//page no 393
+printf("Example 26.7 page no 393\n\n");
+//refer to example 26.6
+D=4//diameter of bed ,ft
+d_p=0.00137//particle diameter ,ft
+rho_s=84//coal particle density ,lb/ft^3
+rho_f=55//oil density,lb/ft^3
+meu_f=3.13e-4//viscosity of oil
+e_mf=0.38//void fraction
+L_mf=8//bed height at minimum fluidization,ft
+L_f=10//bed height,ft
+e=1-L_mf*(1-e_mf)/L_f//bed voidage
+g=32.174//grav acc
+v_s=(d_p^2)*g*(e^3)*(rho_s-rho_f)/(150*meu_f*(1-e)) //superficial velocity
+printf("\n superficial velocity v_s=%f ft/s",v_s);
+q=(%pi/4)*D^2*v_s//volumetric flow rate
+printf("\n vol. floe rate q=%f ft^3/s",q);
+//check on the laminar flow assumption
+meu_f=0.01
+R_e=d_p*v_s*rho_f/(meu_f*(1-e))
+printf("\n reynolds no R_e=%f",R_e);
+printf("\n since R_e is less than 10 ,flow is laminar");
diff --git a/1052/CH26/EX26.8/268.sce b/1052/CH26/EX26.8/268.sce new file mode 100755 index 000000000..9374e0fc1 --- /dev/null +++ b/1052/CH26/EX26.8/268.sce @@ -0,0 +1,23 @@ +clc;
+//Example 26.8
+//page no 393
+printf(" Example 26.8 page no 393\n\n");
+//refer to example 25.6
+//obtain the porous medium friction factor usingthe burke -plummer equation
+///since the flow is turbulent ,eq.26.6 applies
+f_pm=1.75//porous medium friction facot
+v_s=2//superficial velocity
+e=.4//porosity
+L=50//length of bed
+d_p=0.0833//particle diameter
+g=32.174//grav. acc
+h_f=(f_pm)*(v_s^2)*(1-e)*L/(g*(e^3)*d_p)//head loss
+printf("\n head loss h_f=%f ft of propane ",h_f);
+//applying bernoulli eq. between the entrance and gas exit
+//neglect the dynamic head
+P2=4320//pressure at the bottom of the catalyst bed
+rho_f=0.0128//density of fluid
+z_d=-50//length from point 2 to 3,z2-z1
+P1 = P2 + rho_f*(z_d-h_f)// absolute pressure of the inlet gas
+printf("\n pressure P1=%f psf",P1);
+//since flow is turbulent , permeablity of the medium k can not be calculated
diff --git a/1052/CH26/EX26.9/269.sce b/1052/CH26/EX26.9/269.sce new file mode 100755 index 000000000..d0f4d0764 --- /dev/null +++ b/1052/CH26/EX26.9/269.sce @@ -0,0 +1,16 @@ +clc;
+//Example 26.9
+//page no 394
+printf("Example 26.9 page no 394\n\n");
+//turbulent flow of water through a carbon bed
+d_p=0.001//particle diameter
+meu=0.001//viscosity of water
+e=0.25//porosity
+R_e=1000//R_e is >1000 for turbulent flow,for minimum pressure drop
+rho=1000//density of water,kg/m^3
+v_s=R_e*meu*(1-e)/(d_p*rho)//superficial velocity
+printf("\n superficial velocity v_s=%f m/s",v_s);
+phi_s=1//spehercity
+L=0.5//length of bed,m
+P_drop = 1.75*rho*L*v_s^2*(1-e)/(phi_s*d_p*(e^3))//presssure drop
+printf("\npressure drop P_drop=%f Pa",P_drop);
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