diff options
Diffstat (limited to '2087/CH18/EX18.3/example18_3.sce')
| -rwxr-xr-x | 2087/CH18/EX18.3/example18_3.sce | 214 |
1 files changed, 214 insertions, 0 deletions
diff --git a/2087/CH18/EX18.3/example18_3.sce b/2087/CH18/EX18.3/example18_3.sce new file mode 100755 index 000000000..151954226 --- /dev/null +++ b/2087/CH18/EX18.3/example18_3.sce @@ -0,0 +1,214 @@ +
+//example 18.3
+//design a cross -regulator and head regulatorfor a distributory channel
+clc;funcprot(0);
+//givrn
+Q=100; //discharge of parent channel
+Qd=15; //discharge ofdistributory
+fsl_u=218.1; //F.S.L of upstream parent channel
+fsl_d=217.9; //F.S.L of downstream of parent channel
+bw_u=42; //bed width of parent channel upstream
+bw_d=38; //bed width of parent channel downstream
+hw=2.5; //depth of water in parent channel
+fsl_dis=217.1; //F.S.L of distributory
+hw_dis=1.5; //depth of water in distributory
+Ge=1/5; //permissible exit gradient
+
+//design of cross regulator
+mprintf("DESIGN OF CROSS-REGULATOR::");
+//design of crest and waterway
+mprintf("\n\ndesign of crest and waterway:");
+cl=fsl_u-hw;
+h=fsl_u-fsl_d;
+d=fsl_d-cl;
+C1=0.557;C2=0.8;
+L=Q/(2*C1*(2*9.81)^0.5*h^1.5/3+C2*d*(2*9.81*h)^0.5);
+L=round(L*10)/10;
+mprintf("\ncrest level=%f m.",cl);
+mprintf("\nlength of crest=%f m.",L);
+mprintf("\nprovide 4 bays of 7 m each with a clear water-way.");
+tw=28+4.5;
+mprintf("\nprovide 3 piers of 1.5 m width each.\ntotal width of cross regulator=%f m.",tw);
+//design of d/s floor
+L=28;
+q=Q/L;
+Hl=fsl_u-fsl_d;
+Ef2=1.89; //from blench curve
+fl_d=fsl_d-Ef2;
+mprintf("\n\ndesign of d/s floor:");
+mprintf("\nd/s floor level=%f m.; which is higher than d/s bed level.\nadopt floor level =d/s bed level=215.40 m.",fl_d);
+Ef1=Ef2+Hl;
+//from specific energy curve
+D1=0.7;D2=1.65;
+cil=5*(D2-D1); //cistern length
+tl=2*16/3;
+tl=round(tl*10)/10;
+mprintf("\ncistern length =%f m.\nlength of d/s floor=%f m.",cil,tl);
+//design of impervious floor
+d1=hw/3+0.6; //depth of u/s cut-off
+w=0.5; //width of cut-off
+d2=hw/2+0.6; //deth of d/s cut-off
+d2=2; //keep
+Hs=fsl_u-(fsl_d-hw); //maximum static head
+n=Ge*d2/Hs; //n=1/%pi*(lambda)^0.5;
+//from exit gradient curves we get
+alpha=8;n=0.148;
+b=alpha*d2;
+mprintf("\n\ndesign of impervious floor:");
+mprintf("\ntotal length of impervious floor=%i m.;which is divided as-",b);
+mprintf("\nd/s floor length=10.6 m.\nd/s glacis length with 2:1 slope=0.4 m.\nbalance to be provided upstream=5 m.");
+d1=1.5;b=16;
+alpha_=d1/b;
+//hence
+fic1=100-28;
+fid1=100-19;
+t=0.5;
+fic1=fic1+(fid1-fic1)*t/d1;
+mprintf("\n\npressure calculation:\nupstream cut-off:\npressure =%f percent.",fic1);
+d2=2;b=16;
+alpha_=d2/b;
+//hence
+t=0.6;
+fie2=31;fid2=22;
+fie2=fie2-(fie2-fid2)*t/d2;
+mprintf("\ndownstream cut-off:\npressure=%f percent.",fie2);
+t=10.6;
+p=fie2+(fic1-fie2)*t/b;
+p=round(p*10)/10;
+mprintf("\ntoe of glacis:\npressure=%f percent.",p);
+mprintf("\n\nthickness of floor:\nminimu thickness for u/s floor=0.5 m.");
+rho=2.24;
+t=fie2*2.7/(100*(rho-1));
+t=round(t*100)/100;
+mprintf("\nthickness of floor near d/s cut-off=%f m.\nprovide 0.7 m thick floor for last 2.1 m length.",t);
+t=1.6/(rho-1);
+t=round(t*100)/100;
+mprintf("\nthickness of floor at toe of glacis=%f m.",t);
+t=6.6;
+p=fie2+(fic1-fie2)*t/b;
+t=p*2.7/(100*(rho-1));
+t=round(t*100)/100;
+mprintf("\nthickness of floor at 4 m from toe of glais=%f m.\nprovide 1.1 m thick floor for next 2 m length",t);
+t=4.6;
+p=fie2+(fic1-fie2)*t/b;
+t=p*2.7/(100*(rho-1));
+t=round(t*100)/100;
+mprintf("\nthickness of floor at 6 m from toe of glais=%f m.\nprovide 0.9 m thick floor for next 2.5 m length",t);
+
+//design of u/s protection
+d1=hw/3+0.6;
+v=d1;
+v=round(v*100)/100;
+mprintf("\n\ndesign of u/s protection:\nvolume of block protection=%f cubic metre/metre.",v);
+mprintf("\nkeep thickness of protection=1 m.\nprovide 0.8mx0.8mx0.6m thick concret blocks over 0.4 m thick apron in length of 0.6 m.");
+cu=2.25*d1;
+cu=round(cu*100)/100;
+mprintf("\ncubic content of launching apron=%f cubic metre/metre.\nprovide 1 m thick and 3.5 m long launching apron.",cu);
+//design of d/s protection
+d2=hw/2+0.6;
+v=d2;
+v=round(v*100)/100;
+mprintf("\n\ndesign of d/s protection:\nvolume of inverted filter=%f cubic metre/metre.",v);
+mprintf("\nkeep thickness of concrete block=0.6 m.\nprovide 2 rows of 0.8mx0.8mx0.6m thick concret blocks over 0.6 m graded filter for length of 1.6 m.");
+cu=2.25*d2;
+cu=round(cu*100)/100;
+mprintf("\nlaunching apron volume=%f cubic metre/metre.\nprovide 1 m thick launching apron for length of 4.5 m.\nprovide a toe wall 0.4 m wide and 1.5 m deep between filter and launching apron.",cu);
+
+//design of head regulator
+mprintf("\n\n\nDESIGN OF DISTRIBUTORY HEAD REGULATOR::");
+//design of crest and waterway
+mprintf("\n\ndesign of crest and waterway:");
+cl=fsl_u-hw+0.5;
+h=fsl_u-fsl_dis;
+d=fsl_dis-cl;
+C1=0.557;C2=0.8;
+L=Qd/(2*C1*(2*9.81)^0.5*h^1.5/3+C2*d*(2*9.81*h)^0.5);
+L=round(L*100)/100;
+mprintf("\ncrest level=%f m.",cl);
+mprintf("\nlength of crest=%f m.",L);
+mprintf("\nprovide 2 bays of 3.5 m each with a 1 m thick pier in between.");
+tw=8;
+mprintf("\ntotal width of cross regulator=%f m.",tw);
+//design of d/s floor
+L=7.5;
+q=Q/L;
+Hl=fsl_u-fsl_dis;
+Ef2=1.58; //from blench curve
+fl_d=fsl_dis-Ef2;
+mprintf("\n\ndesign of d/s floor:");
+mprintf("\nd/s floor level=%f m.;\nkeepR.L of d/s floor=215.50 m.",fl_d);
+Ef1=Ef2+Hl;
+//from specific energy curve
+D1=0.42;D2=2.55;
+cil=5*(D2-D1); //cistern length
+tl=2*14/3;
+mprintf("\ncistern length =%f m.",cil);
+
+//design of impervious floor
+d1=hw/3+0.6; //depth of u/s cut-off
+w=0.5; //width of cut-off
+d2=hw_dis/2+0.6; //deth of d/s cut-off
+d2=2; //keep
+Hs=fsl_u-215.5; //maximum static head
+n=Ge*d2/Hs; //n=1/%pi*(lambda)^0.5;
+//from exit gradient curves we get
+alpha=7;n=0.154;
+b=alpha*d2;
+mprintf("\n\ndesign of impervious floor:");
+mprintf("\ntotal length of impervious floor=%i m.;which is divided as-",b);
+mprintf("\nlength below the toe of glacis=10.5 m\nlength of d/s glacis at 2:1 slope=1.2 m.\nwidth of crest=1 m.\nlength of u/s glacis at 1:1 slope=0.5 m.\nu/s floor:balnce=0.8 m.");
+d1=1.5;b=16;
+alpha_=d1/b;
+//hence
+fic1=100-28;
+fid1=100-19;
+t=0.5;
+fic1=fic1+(fid1-fic1)*t/d1;
+mprintf("\n\npressure calculation:\nupstream cut-off:\npressure =%f percent.",fic1);
+d2=2;b=16;
+alpha_=d2/b;
+//hence
+t=0.6;
+fie2=31;fid2=22;
+fie2=fie2-(fie2-fid2)*t/d2;
+mprintf("\ndownstream cut-off:\npressure=%f percent.",fie2);
+t=10.6;
+p=fie2+(fic1-fie2)*t/b;
+p=round(p*100)/100;
+mprintf("\ntoe of glacis:\npressure=%f percent.",p);
+mprintf("\n\nthickness of floor:\nminimu thickness for u/s floor=0.5 m.");
+rho=2.24;
+t=p*2.6/(100*(rho-1));
+t=round(t*100)/100;
+mprintf("\nthickness under the crest=1 m.");
+mprintf("\nthickness of floor at toe of glacis=%f m.",t);
+t=9.5;
+p=fie2+(fic1-fie2)*t/b;
+t=p*2.7/(100*(rho-1));
+t=round(t*100)/100;
+mprintf("\nthickness of floor at 2 m from toe of glais=%f m.\nprovide 1.1 m thick floor for next 4 m length",t);
+t=4.5;
+p=fie2+(fic1-fie2)*t/b;
+t=p*2.7/(100*(rho-1));
+t=round(t*100)/100;
+mprintf("\nthickness of floor at 6 m from toe of glais=%f m.\nprovide 0.9 m thick floor for next 2.5 m length",t);
+t=2;
+p=fie2+(fic1-fie2)*t/b;
+t=p*2.7/(100*(rho-1));
+t=round(t*100)/100;
+mprintf("\nthickness of floor at 8.5 m from toe of glais=%f m.\nprovide 0.7 m thick floor for next 2 m length",t);
+
+//design of upstream protection
+d=hw/3+0.6;
+d=round(d*10)/10;
+mprintf("\n\ndesign of u/s protection:\nu/s scour depth=%f m.\nprovide same protection as in cross regulator",d);
+
+//design of d/s protection
+d2=hw_dis/2+0.6;
+v=d2;
+mprintf("\n\ndesign of d/s protection:\nvolume of inverted filter=%f cubic metre/metre.",v);
+mprintf("\nkeep thickness of concrete block=0.5 m.\nprovide 2 rows of 0.8mx0.8mx0.5m thick concret blocks over 0.5 m thick graded filter.");
+cu=2.25*d2;
+mprintf("\nlaunching apron volume=%f cubic metre/metre.\nprovide 1 m thick launching apron for length of 3.5 m.\nprovide a masonary toe wall 0.4 m wide and 1.2 m deep between filter and launching apron.",cu);
+
+
|