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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+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);