diff options
Diffstat (limited to '2087/CH18')
| -rwxr-xr-x | 2087/CH18/EX18.1/example18_1.sce | 172 | ||||
| -rwxr-xr-x | 2087/CH18/EX18.2/example18_2.sce | 112 | ||||
| -rwxr-xr-x | 2087/CH18/EX18.3/example18_3.sce | 214 |
3 files changed, 498 insertions, 0 deletions
diff --git a/2087/CH18/EX18.1/example18_1.sce b/2087/CH18/EX18.1/example18_1.sce new file mode 100755 index 000000000..43b6afb7c --- /dev/null +++ b/2087/CH18/EX18.1/example18_1.sce @@ -0,0 +1,172 @@ +
+
+//example 18.1
+//design Sarda type fall
+clc;funcprot(0);
+//given
+Q=40; //full supply discharge
+sl_u=218.3; //supply level at upstream
+sl_d=216.8; //supply level at downstream
+D=1.8; //suplly depth
+L=26; //bed width
+bl_u=216.5; //bed level upstream
+bl_d=215; //bed level downstream
+drop=1.5;
+
+//from the eqauation; Q=1.99LH^1.5*(H/B)^(1/6);
+//B=0.55*(H+d)^0.5;
+//H+d=drop+D;
+//we get
+H=(0.774)^0.6;
+d=3.3-H;
+Hc=D-H;
+d=round(d*100)/100;
+H=round(H*100)/100;
+Hc=round(Hc*100)/100;
+mprintf("H=%f m.\nd=%f m.",H,d);
+mprintf("\ncrest height above bed=%f m.",Hc);
+
+//adopt trapezoidal crest
+B=1; //top width
+mprintf("\n\nD/S batter=1:3; U/S batter=1:8.");
+Va=Q/((27+D)*D);
+vh=Va^2/(2*9.81);
+tel_up=sl_u+vh;
+crest=sl_u-H;
+E=sl_u-crest;
+mprintf("\nR.L of crest=%f m.",crest);
+mprintf('\nE=%f m.',E);
+//design of cistern
+x=(E*drop)^(2/3)/4; //depth of cistern
+lc=5*(E*drop)^0.5; //length of cistern
+cb=bl_d-x;
+x=round(x*100)/100;
+cb=round(cb*1000)/1000;
+lc=round(lc*10)/10;
+mprintf("\n\ndepth of cistern=%f m.",x);
+mprintf("\nlength of cistern=%f m.",lc);
+mprintf("\nR.L of bed of cistern=%f m.",cb);
+mprintf("\nkeep cistern at R.L 214.69.");
+//design of impervious floor
+Hs=2.44; //seepage head
+c=8; //Bligh's coefficient
+li=Hs*c;
+d1=1;d2=1.6;
+vl=2*(d1+d2);
+lh=li-vl;
+mprintf("\n\ndesign of impervious floor:");
+mprintf("\nprovide upstream cut-off=%i m.; downstream cut-off=%f m.",d1,d2);
+mprintf("\nlength of horizontal impervious floor=%f m.",lh);
+mprintf("\nprovide 15 m length impervious floor.");
+ld=2*(D+1.2)+drop;
+mprintf("\nminimum length of impervious floor to the d/s of toe of crest wall=%f m.",ld);
+mprintf("\nprovide ld=8 m.");
+bl=15-8;
+mprintf("\nthe balance of the length %i m is to be provided under and u/s of the crest.",bl);
+
+tcl=15+2*(1+16);
+mprintf("\n\nuplift pressure is counter balanced by weigth of water.\n hence provide thickness of 0.4 m.");
+rho=2.24;
+static=2.44*(1-0.446)+x;
+t=static/(rho-1);
+t=round(t*100)/100;
+mprintf("\nfor other points; thickness required =%f m.",t);
+mprintf("\nprovide thickness of 1.40 m.");
+mprintf("\nat downstream end of floor provide thickness of 0.6 m overlaid by 0.2 m brick pitching.");
+
+n=d2/(Hs*5); //n=1/%pi*(lambda)^0.5
+//from khosla exit curve we get
+alpha=10.5;
+lambda=(1/(%pi*n))^2;
+alpha=((2*lambda-1)^2-1)^0.5;
+b=alpha*d2;
+b=round(b*100)/100;
+mprintf("\n\nchecking of floor thickness by khosla theory:");
+mprintf("\nlength of floor provided=%f m. > length by Bligh theory.",b);
+b=15;
+d2=1.8;
+alpha=b/d2;
+n=0.145;
+Ge=Hs*n/d2;
+Ge=round(Ge*10)/10;
+mprintf("\nexit gradient after increase in depth cut-off=%f. which is in permissible limit",Ge);
+mprintf('\nprovide depth cut-off to 1.8 m.');
+//calculation of pressure
+mprintf("\n\ncalculation of pressure:");
+mprintf("\nU/S cut-off:");
+d1=1;
+b=15;
+alpha_=d1/b;
+fic1=100-24;
+fid1=100-17;
+t=0.4;
+fic1=fic1+(fid1-fic1)*t/d1;
+mprintf("\ncorrected fic1=%f percent.",fic1);
+mprintf("\nD/S cut-off wall:");
+d2=1.8;
+b=15;
+alpha_=d1/b;
+fie2=31;
+fid2=21.5;
+t=0.6;
+fie2=fie2-(fie2-fid2)*t/1.8;
+fie2=round(fie2*10)/10;
+mprintf("\ncorrecte fie2=%f percent.",fie2);
+//calculation of thickness
+mprintf("\n\nprovide a minimum thickness of 0.4 m for u/s floor.");
+pre=fie2+(fic1-fie2)*8/b;
+static=pre*Hs/100+x;
+t=static/(rho-1);
+t=round(t*100)/100;
+mprintf("\nthickness at d/s toe of crest=%f m.",t);
+mprintf("\nprovide thickness of 1.4 m thick concrete overlaid by 0.2 m brick pitching.");
+pre=fie2+(fic1-fie2)*5/b;
+static=pre*Hs/100+x;
+t=static/(rho-1);
+t=round(t*100)/100;
+mprintf("\nthickness at 3 m from d/s toe of crest=%f m.",t);
+mprintf("\nprovide thickness of 1.2 m thick concrete overlaid by 0.2 m brick pitching.");
+pre=fie2+(fic1-fie2)*2/b;
+static=pre*Hs/100; //calculation is wrong in book
+t=static/(rho-1);
+t=round(t*100)/100;
+mprintf("\nthickness at 6m from d/s toe of crest=%f m.",t);
+mprintf("\nprovide thickness of 0.7 m thick concrete overlaid by 0.2 m brick pitching.");
+//design of downstream wings
+wing=6*(E*drop)^0.5;
+hw=D+0.5;
+mprintf("\n\nheigth of top of downstream wings above the bed=%f m.",hw);
+projec=hw*3;
+mprintf("\nlength of warped wing measured along centre line of canal=%f m.",projec);
+//downstream pitching
+l=9+2*1.5;
+mprintf("\n\nlength of bed pitching=%f m.",l);
+mprintf("\nlength of sloping pitching=7 m.\nlength of horizontal pitching=6 m.");
+mprintf("\nprovide one toe wall of 1 m depth and 0.4 m width.");
+mprintf("\nside pitching is curtailed at 45 degree from the end of bed pitching in plan.\nsupprot the side pitching on toe wall 0.4 m thick and 1 m deep. ");
+//energy dissipators
+q=Q/L;
+dc=(q^2/9.81)^(1/3);
+mprintf("\n\nsize and position of friction blocks:");
+L=2*dc;
+w=dc;
+h=dc;
+di=1.5*dc;
+L=round(L*10)/10;
+w=round(w*10)/10;
+h=round(h*10)/10;
+di=round(di);
+mprintf("\nlength of block=%f m.\nwidth of block=%f m.\nheight of block=%f m.\ndistance from toe of crest=%f m.",L,w,h,di);
+mprintf("\nprovide two rows staggered ata distance of 1 m from toe of crest.");
+mprintf("\nsize and position of cube blocks:");
+L=D/10;
+w=D/10;
+h=w;
+L=round(L*10)/10;
+w=round(w*10)/10;
+h=round(h*10)/10;
+mprintf("\nlength of block=%f m.\nwidth of block=%f m.\nheight of block=%f m.",L,w,h);
+mprintf("\nprovide two rows staggered at the end of impervious floor.");
+//u/s approach
+r=6*H;
+mprintf("\n\nprovide wing wall segmental with 5 m radius subtending angle of 60 degree at the centre.");
diff --git a/2087/CH18/EX18.2/example18_2.sce b/2087/CH18/EX18.2/example18_2.sce new file mode 100755 index 000000000..bf0cd6722 --- /dev/null +++ b/2087/CH18/EX18.2/example18_2.sce @@ -0,0 +1,112 @@ +
+
+//example 18.2
+//design an unflumed straight glacis non-meter fall
+clc;funcprot(0);
+//given
+Q=40; //full supply discharge
+sl_u=218.3; //supply level at upstream
+sl_d=216.8; //supply level at downstream
+D=1.8; //suplly depth
+L=26; //bed width
+bl_u=216.5; //bed level upstream
+bl_d=215; //bed level downstream
+drop=1.5;
+Ge=1/6; //permissible exit gradient
+
+//design of crest
+mprintf("design of crest:");
+E=(Q/(1.84*L))^(2/3);
+V=Q/((L+D)*D);
+vh=V^2/(2*9.81);
+tel_up=sl_u+vh;
+cl=tel_up-E;
+w=2*E/3;
+w=round(w*10)/10;
+mprintf("\nlength of crest=%f m.",L);
+mprintf("\nwidth of crest=%f m.",w);
+//design of cistern
+q=Q/L;
+Hl=1.5;
+//from blench curve
+Ef2=1.44;
+cistern=sl_d+0.03-1.25*Ef2;
+mprintf("\n\nR.L of cistern=%f m. > d/s bed level.",cistern);
+mprintf("\nkeep R.L of cistern at 214.5 m.");
+l=6*Ef2;
+mprintf("\nlength of cistern=%f m.",l);
+mprintf("\nprovide cistern of 9 m length ");
+d=bl_d-214.5;
+mprintf("\ndepth of cistern=%f m.",d);
+
+//design of impervious floor
+d1=D/3;
+mprintf("\n\ndesign of impervious floor:");
+mprintf("\nprovide 0.4 m wide and 1 m deep curtain wall at u/s.");
+d2=D/2;
+mprintf("\nprovide 0.4 m wide and 1 m deep curtain wall at d/s.\nthe curtain wall will project the above the d/s bed by 0.18 m.");
+Hs=cl-bl_d;
+d2=1;
+n=d2*Ge/Hs; //n=1/(%pi*(lambda)^0.5)
+//from khosla exit curves we get
+alpha=40;
+lambda=(1/(%pi*n))^2;
+alpha=((2*lambda-1)^2-1)^0.5;
+b=alpha*d2;
+//since length is to excessive
+d2=2;
+n=d2*Ge/Hs; //n=1/(%pi*(lambda)^0.5)
+//from khosla exit curves we get
+alpha=10;
+lambda=(1/(%pi*n))^2;
+alpha=((2*lambda-1)^2-1)^0.5;
+b=alpha*d2+1;
+mprintf("\ntotal length=%i m.\nlength of cistern=9 m.\nlength of d/s glacis=5.88 m.\nwidth of crest=0.6 m.\nlength of u/s glacis=0.47 m.\nbalance to be provided to u/s of the u/s glacis=4.05 m.",b);
+
+//pressure calculations
+mprintf("\n\npressure calculations:");
+mprintf("\nupstream curtain wall:");
+d1=1;b=20;
+alpha_=d1/b;
+t=0.3;
+fic1=100-22;
+fid1=100-15;
+corec=(fid1-fic1)*t/d1
+fic1=fic1+corec;
+mprintf("\ncorrected fi_c1=%f percent.",fic1);
+mprintf("\ndownstream curtain wall:");
+d2=2;b=20;
+alpha_=d2/b;
+t=0.5;
+fie=29;
+fid=21;
+corec=(fie-fid)*t/d2
+fie=fie-corec;
+mprintf("\ncorrected fi_e=%f percent.",fie);
+mprintf("\ntoe of glacis:");
+//assuming linear variation of pressure
+p=fie+(80-fie)*9/20;
+mprintf("\npressure at downstream of the glacis=%f percent.",p);
+
+//floor thickness
+rho=2.24;
+mprintf("\n\nfloor thickness:\nprovide minimum thickness of 0.3 m at the u/s floor.");
+static=p*2.44/100+(bl_d-214.5);
+t=static/(rho-1);
+t=round(t*100)/100;
+mprintf("\nfloor thickness required at toe of glacis=%f m.\nprovide 1.5 m thick floor for length of 3 m.",t);
+p=fie+(80-fie)*6/20;
+static=p*2.44/100+(bl_d-214.5);
+t=static/(rho-1);
+t=round(t*100)/100;
+mprintf("\nfloor thickness required at 3m from toe of glacis=%f m.\nprovide 1.3 m thick floor from 3 m to 6.5 m from toe of glacis.",t);
+t=0.27*2.44/(rho-1);
+t=round(t*100)/100;
+mprintf("\nthickness of d/s end of cistern=%f m.\nprovide thickness of 0.6 m at d/s end of floor.",t);
+
+//design of d/s protection
+mprintf("\n\nno bed protection is needed as deflector wall is provided.");
+sp=3*D;
+mprintf("\nlength of side protection=%f m.\nprovide 5.5 m length of 20 cm thick brick pitching beyond impervious floor.\npitching will rest on toe wall 0.4 m wide and 0.9 m deep.\nprovide 0.4 m wide profile at the end of pitching",sp);
+//design of u/s approach
+mprintf("\n\nu/s wing wall is splayed at 45 degree from u/s end of impervious floor.\nextend 1 m into earthen banks from line of F.S.L.");
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);
+
+
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