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+
+//example 12.12
+//calculate
+//number of gates required for the barrage
+//head regulator if each gate has 10 m clear span(neglect end contractions and approach velocity)
+//length and R.L of basin floor if silting basin is provided downstream of barrage
+clc;funcprot(0);
+//given
+Lmax=212; //maximum reservior level
+Lp=211; //pond level
+hfl=210; //downstream high flood level in the river
+Qmax=3500; //maximum design flood discharge
+Lcrest=207; //crest level of the barrage
+Lcrest_r=208; //crest level of head regulator
+Cd=2.1; //coefficient of discharge for barrage
+Cd_r=1.5; //coefficient of discharge for head regulator
+rbl=205; //river bed level
+Q=500; //design discharge of main canal
+
+//design of water way for barrage during flood
+H=Lmax-Lcrest;
+L=Qmax/(Cd*H^1.5);
+//which gives L=149.07.
+//provide 15 bays of 10m clear span
+mprintf("nunmber of gates for the barrage=15.");
+
+//design of waterway for canal head regulator
+H=Lp-Lcrest_r;
+L1=Q/(Cd_r*H^1.5);
+//which gives L=64.2
+//hence provide 7 bays of 10 m each
+mprintf("\n\nnunmber of gates for the head regulator=7.");
+
+//design of stilling basin
+Hl=Lmax-hfl;
+q=Qmax/L;
+yc=(q^2/9.81)^(1/3);
+Z=Hl/yc;
+//since Z<1
+Y=1+0.93556*Z^0.368;
+y2=Y*yc;
+Lc=5*y2;
+Lc=round(Lc*10)/10;
+mprintf("\n\nLength of cistern=%f m.",Lc);
+Ef2=yc*(Y+1/(2*Y^2));
+j=hfl-Ef2;
+j=round(j*10)/10;
+mprintf("\nR.L of cistern=%f m.",j);
+
+