4 files changed, 83 insertions, 0 deletions
diff --git a/1052/CH16/EX16.2/162.sce b/1052/CH16/EX16.2/162.sce
new file mode 100755
index 000000000..8ad5a6f30
--- /dev/null
+++ b/1052/CH16/EX16.2/162.sce
@@ -0,0 +1,21 @@
+clc;
+//Example 16.2
+//page no 183
+printf(" Example 16.2 page no 183\n\n");
+//cal. pressure drop if the flow for both phases is turbulent
+//a. since the flow is tt and 1<X<10 ,apply equatuion 16.16b to obtain Y_g
+X=1.66
+Y_g=5.80+6.7143*X+6.9643*X^2-0.75*X^3
+printf("\n Y_g=%f ",Y_g);
+//the value of Y_g is an excellent agreement with the values provided by lockhart and Martinelli
+//then pressure drop is
+P_drop_g=2.71
+P_drop_t=Y_g*P_drop_g
+printf("\n P_drop_t=%f psf/100 ft",P_drop_t);
+//b. applying eq. 16.17b to generate Y_l
+Y_l=18.219*X^-.8192
+printf("\n Y_l =%f ",Y_l);
+//pressure drop from eq. 16.2
+P_drop_l=7.50
+P_drop=Y_l*P_drop_l
+printf("\n P_drop=%f psf/100ft",P_drop); 
diff --git a/1052/CH16/EX16.3/163.sce b/1052/CH16/EX16.3/163.sce
new file mode 100755
index 000000000..2df5e2d96
--- /dev/null
+++ b/1052/CH16/EX16.3/163.sce
@@ -0,0 +1,20 @@
+clc;
+//Example 16.3
+//page no 185
+printf(" Example 16.3 page no 185\n\n");
+//if the flow for the gas phase is turbulent and the liquid phase is viscous
+//cal. pressure drop total
+X=1.66//from ex. 16.1
+Y_G_tv=20-21.81*X+16.357*X^2-1.8333*X^3
+printf("\n Y_G_tv=%f ",Y_G_tv);
+//pressure drop from eq 16.1
+P_drop_g=2.71
+P_drop_a=Y_G_tv*P_drop_g
+printf("\n pressure drop P_drop_a=%f psf/100 ft",P_drop_a);
+//b. applying eq 16.20b to generate Y_l
+Y_l_tv=11.702*X^-0.7334
+printf("\n Y_l_tv=%f ",Y_l_tv);
+//pressure drop from equation 16.2
+P_drop_l=7.50
+P_drop_b=Y_l_tv*P_drop_l
+printf("\n P_drop_b=%f psf/100 f",P_drop_b);
diff --git a/1052/CH16/EX16.4/164.sce b/1052/CH16/EX16.4/164.sce
new file mode 100755
index 000000000..67e214f25
--- /dev/null
+++ b/1052/CH16/EX16.4/164.sce
@@ -0,0 +1,21 @@
+clc;
+//Example 16.4
+//page no 187
+printf("Example 16.4 page no 187\n\n");
+//if flow for both phases is laminar then cal pressure drop total
+//a. apply eq. 16.22b to obtain Y_G
+X=1.66
+Y_G=10-10.405*X+8.6786*X^2-0.9167*X^3
+printf("\n Y_G=%f ",Y_G);
+//pressure drop from eq 16.1
+P_drop_g=2.71
+P_drop=Y_G*P_drop_g
+printf("\n pressure drop P_drop=%f psf/100 ft",P_drop);
+//b. apply eq 16.23b to generate Y_l
+Y_l=6.4699*X^-0.556
+printf("\n Y_l =%f ",Y_l);
+//pressure drop from eq. 16.2
+P_drop_l=7.50
+P_drop_b=Y_l*P_drop_l
+printf("\n pressure drop P_drop_b=%f psf/100 ft",P_drop_b);
+
diff --git a/1052/CH16/EX16.6/166.sce b/1052/CH16/EX16.6/166.sce
new file mode 100755
index 000000000..ff1bf49cc
--- /dev/null
+++ b/1052/CH16/EX16.6/166.sce
@@ -0,0 +1,21 @@
+clc;
+//Example 16.6
+//page no 191
+printf("\n Example 16.6 page no 191\n\n");
+//a mixture of air(a) and kerosene(k) are flowing in a horizontal pipe 
+rho_a=0.075//density of airlb/ft^3
+meu_a=1.24e-5//viscosity of air ,lb/ft.s
+q_a=5.3125//flow rate ft^3/s
+rho_k=52.1//density of kerosene,lb/ft^3
+meu_k=0.00168//viscosity lof kerosene,lb/ft.s
+q_k=1.790//flow rate ft^3/s
+D=.19167//diameter of pipe ,ft
+S=(%pi/4)*D^2//cross sectional area,ft^2
+printf("\n S=%f ",S);
+//superficial velocity of each phase can be obtained by applying either eq, 16.7 and 16.8
+v_a=q_a/(S*60)//for air
+v_k=q_k/(S*60)//for kerosene
+printf("\n velocity v_a =%f ft/s\n velocity v_k=%f ft/s",v_a,v_k);
+R_e_a=D*rho_a*v_a/meu_a//reynolds no. of Air
+R_e_k=D*rho_k*v_k/meu_k//reynolds no. of kerosene
+printf("\n R_e_a=%f\nR_e_k=%f ",R_e_a,R_e_k);