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diff --git a/2087/CH19/EX19.3/example19_3.sce b/2087/CH19/EX19.3/example19_3.sce new file mode 100755 index 000000000..b8b30dcd8 --- /dev/null +++ b/2087/CH19/EX19.3/example19_3.sce @@ -0,0 +1,166 @@ +
+
+//example 19.3
+//design a syphon aqueduct
+clc;funcprot(0);
+//given
+Q=25; //design discharge of canal
+B=20; //bed width of canal
+D=1.5; //depth of water in canal
+bl=160; //bed level of canal
+hfq=400; //high flood discharge of drainage
+hfl=160.5; //high flood level of drainage
+bl_drain=158; //bed level of drainage
+gl=160; //general ground level
+
+//desing of drainage water-way
+P=4.75*(hfq)^0.5; //laecey P-Q formula
+mprintf("design of drainage water-way:\nwetted perimeter of river=%i m.\nprovide 13 spans of 6 m each,separated by 12 piers each of 1.25 m thick.",P);
+t=78+15;
+mprintf("\ntotal length of water-way=%i m.",t);
+v=2; //velocity through syphon
+hb=hfq/(78*v);
+ac=hfq/(6*2.5*1.3); //calculation is wrong in book
+hb=round(hb*100)/100;
+ac=round(ac*100)/100;
+mprintf("\nheight of barrels=%f m.\nprovide rectangular barrels 6 m wide and 2.5 m high.\nactual velocity through barrels=%f m/sec.",hb,ac);
+
+//design of canal waterway
+mprintf("\n\ndesign of canal waterway:\nType 3 aqueduct is adopted.");
+l1=B-10;
+l2=(20-10)*3/2;
+mprintf("\nproviding a splay 2:1 in expansion,length of contraction transition=%i m.\nproviding a splay of 3:1 in expansion,length of expansion transition=%i m.",l1,l2);
+mprintf('\nIn transition side slopes are warped from original slope of 1.5:1 to vertical.');
+
+//design of levels of different sectionn
+mprintf("\n\ndesign of levels of different sectionn:\nat section 4-4:");
+A=(B+1.5*D); //area
+V=Q/A; //velocity of flow
+vh=V^2/(2*9.81); //velocity head
+ws=gl+D; //R.L of water surface
+tel=ws+vh;
+tel=round(tel*1000)/1000;
+mprintf("\nR.L of T.E.L=%f m.\n at section 3-3:",tel);
+A=10*D; //area of trough
+V=Q/A; //velocity
+vh1=V^2/(2*9.81); //velocity head
+le=0.3*(vh1-vh); //loss of head in expansion from section 3-3 to 4-4
+tel=tel+le;
+rlw=tel-vh1;
+rlb=rlw-D;
+tel=round(tel*1000)/1000;
+rlb=round(rlb*1000)/1000;
+mprintf("\nelevation of T.E.L=%f m.\nR.L of bed to maintain constant water depth=%f m.",tel,rlb);
+
+//at section 2-2
+R=A/P;
+N=0.016;
+S=V^2*N^2/R^(4/3); //from manning's formula
+L=93; //length of trough
+hl=L*S; //head loss
+tel=tel+hl;
+rlw=tel-vh1;
+rlb=rlw-D;
+tel=round(tel*1000)/1000;
+rlb=round(rlb*1000)/1000;
+mprintf("\nat section 2-2:\nR.L of T.E.L=%f m.\nR.L of bed to maintain constant water depth=%f m.",tel,rlb);
+
+//at section 1-1
+hl=0.2*(vh1-vh); //loss of hed in contraction transition
+tel=tel+hl;
+rlw=tel-vh;
+rlb=tel-D;
+tel=round(tel*1000)/1000;
+rlb=round(rlb*1000)/1000;
+mprintf("\nat section 1-1:\nR.L of T.E.L=%f m.\nR.L of bed to maintain constant water depth=%f m.",tel,rlb);
+
+//design of contraction transition
+//it is designed on the basis of chaturvedi's formula
+Bo=20;
+Bf=10;
+L=10;
+//from chaturvedi formula we get relation between x and Bx as: x=15.45(1-(10/Bx)^1.5);
+Bx=[10:1:20];
+mprintf("\n\ndesign of contraction transition on the basis of chaturvedi formula:\nBx x");
+for i=1:11
+ x(i)=15.45*(1-(10/Bx(i))^1.5);
+ x(i)=round(x(i)*100)/100;
+ mprintf("\n%i %f",Bx(i),x(i));
+end
+
+//design of expansion transition on the basis of chaturvedi formula
+L=15;
+Bf=10;Bo=20;
+//from chaturvedi formula we get relation between x and Bx as: x=23.15(1-(10/Bx)^1.5);
+mprintf("\n\ndesign of expansion transition on the basis of chaturvedi formula:\nBx x");
+for i=1:11
+ x(i)=23.15*(1-(10/Bx(i))^1.5);
+ x(i)=round(x(i)*100)/100;
+ mprintf("\n%i %f",Bx(i),x(i));
+end
+
+//design of trough
+mprintf("\n\ndesign of the trough:");
+mprintf("\nflumed water way of canal=10 m.\ntrough carrying canal will divide into two compartments each 5 m wide an dseparated by 0.3 m thick partiions.\nheigth of trough will be = 2 m.\ntrough iss constructed using monolithic reinforced concrete.\nthe outer and inner walls ca be kept 0.4 m thick.\nthus,outer width of trough = 11.1 m.");
+
+//head loss through syphon barrels
+V=2.05; //velocity through barrels
+f1=0.505; //coefficient of loss of head at entry
+a=0.00316;b=0.030;
+R=(6*2.5)/(2*(6+2.5));
+f2=a(1+b/R);
+L=11.1; //length of barrel
+h=(1+f1+f2*L/R)*V^2/(2*9.81);
+hfl_up=hfl+h;
+h=round(h*1000)/1000;
+hfl_up=round(hfl_up*1000)/1000;
+mprintf("\n\nhead loss through syphon barrels=%f m.\nupstream H.F.L=%f m.",h,hfl_up)
+
+//uplift pressure on the roof
+bt=gl-0.4; //R.L of bottom of the trough
+hl=0.505*V^2/(2*9.81);
+u=hfl_up-hl-159.6;
+up=u*9.81;
+mprintf("\n\nuplift pressure on the roof=%f kN/square m.\ntrough slab is 0.4 m thick and exert a downward load of 9.42 kN.",up);
+mprintf("\nth ebalance of the uplift pressure has to be resisted by bending action of trough slab.\nso,reinforcement has to be provided at the top of the slab.");
+
+//uplift on the floor of the barrel and its design
+//(a) static head
+mprintf("\n\nuplift on the floor of the barrel and its design:\n(a) static head:");
+bf=bt-2.5; //R.L of barrel floor
+t=0.8; //tentative thickness of floor
+bot=bf-t;
+static=bl_drain-bot;
+static=round(static*100)/100;
+mprintf("\nstatic uplift on the floor=%f m.",static);
+
+//(b) seepage head
+L=10; //length of u/s transition
+bs=3; //half the barrel span
+df=11; //end drainage floor
+tcl=24; //total creep length
+tsh=161.5-bl_drain; //total seepage head
+rs=tsh*(1-13/tcl); //residual seepage at B
+tu=(static+rs)*9.81;
+tu=round(tu*100)/100;
+mprintf("\n(b) seepage head:\ntotal uplift=%f kN/square m.\nprovide thickness of floor 0.8 m",tu);
+bending=tu-17.58;
+bending=round(bending*100)/100;
+mprintf("\nuplift to be resisted by bending action of floor=%f kN/square m.",bending);
+
+//design of cut-off and protection works for drainage floor
+mprintf("\n\ndesign of cut-off and protection works for drainage floor:");
+Q=400;f=1;
+R=0.47*(Q/f)^(1/3);
+d_up=1.5*R; //depth of u/s cut-off
+bot_up=hfl_up-d_up; //R.L of bottom of u/s cut-off
+d_down=1.5*R; //depth of d/s cut-off
+bot_down=hfl-d_down; //R.L of bottom of d/s cut-off
+l_down=2.5*(bl_drain-bot_down);
+l_down1=2*(bl_drain-bot_up);
+bot_up=round(bot_up*100)/100;
+bot_down=round(bot_down*100)/100;
+l_down=round(l_down);
+l_down1=round(l_down1);
+mprintf("\nR.L of bottom of u/s cut-off=%f m.\nR.L of bottom of d/s cut-off=%f m.",bot_up,bot_down);
+mprintf("\nlength of d/s protection consisting of 40 cm brick pritching=%f m.\npitching is supported by toe wall 0.4 m wide and 1.5 m deep at its d/s end.\nlength of d/s protection consisting of 0.4 cm brick pritching=%f m.\npitching is supported by toe wall 0.4 m wide and 1 m deep at its u/s end.",l_down,l_down1);
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