# Chapter 12 : DIVERSION HEADWORKS¶

## Example 12.1 pg : 576¶

In [3]:
import math

#uplift presuures and thickness of floor at 6m, 12m and 18m from u/s

#Given
rho = 2.24;             				#relative density of material
gamma_w = 9.81;         				#unit weigth of water
L = 22.;                 				#total length
lc = (2.*6)+L+(2*8);     				#length of creep
print "avearge hydraulic gradient = %.2f."%(hg);
#at 6 m from u/s
x = 6.;
lg = (6.*2)+x;
up = gamma_w*h1;
t = 4.*h1/(3*(rho-1));
up = round(up*100)/100;
t = round(t*100)/100;
print "uplift at 6 m from u/s = %.2f kN/square metre."%(up);
print "thickness at 6 m from u/s = %.2f m."%(t);

#at 12 m from u/s
x = 12.;
lg = (6.*2)+x;
up = gamma_w*h1;
t = 4.*h1/(3*(rho-1));
up = round(up*100)/100;
t = round(t*100)/100;
print "uplift at 12 m from u/s = %.2f kN/square metre."%(up);
print "thickness at 12 m from u/s = %.2f m."%(t);

#at 18m from u/s
x = 18.;
lg = (6.*2)+x;
up = gamma_w*h1;
t = 4*h1/(3*(rho-1));
up = round(up*10)/10;
t = round(t*100)/100;
print "uplift at 18 m from u/s = %.2f kN/square metre."%(up);
print "thickness at 18 m from u/s = %.2f m."%(t);
uplift at 6 m from u/s = 25.11 kN/square metre.
thickness at 6 m from u/s = 2.75 m.
uplift at 12 m from u/s = 20.40 kN/square metre.
thickness at 12 m from u/s = 2.24 m.
uplift at 18 m from u/s = 15.70 kN/square metre.
thickness at 18 m from u/s = 1.72 m.

## Example 12.2 pg : 589¶

In [4]:
import math

#check whether section is safe against overturning and piping

#Given
b = 54.;            				#width of section
D1D2 = 16;         				#dismath.tance between points D1 and D2
D2D3 = 37;         				#dismath.tance between points D2 and D3

#first pipe line
#taking data from figure
d = 105-97;
b1 = 0.5;
alpha = b/d;
#from the curves we get
fic1 = 0.665;
fid1 = 0.76;
fie1 = 1;
t = 105-104;                      				#floor thickness
corec = (fid1-fic1)*100*t/d;          				#correction for floor thickness
#for pile no. 2
D = 104-97;
d = 104-97;
bdash = 16;
C = 19*(D/bdash)**0.5*(d+D)/b;    				#correction for pile no. 2
fic1 = fic1*100+corec+C;         				#corrected pressures

#intermedite pipe line
d = 105-97;
b1 = 16.5;
alpha = b/d;
r = b1/b;               				#ratio b1/b
#from the curves we get
fic2 = 0.52;
fie2 = 0.725;
fid2 = 0.615;
corec_c1 = (fid2-fic2)*100*t/d;
corec_e1 = (fie2-fid2)*100/d;

#for pile no. 1
C1 = C;
d = 104-97;
bdash = 37;
D = 104-95;
C2 = 19*(D/bdash)**0.5*(d+D)/b;
#correction due to slope
corec_e2 = 3.3;                				#from table 12.4
#correction is negative due to upwrd slope
l = 4;               				#horizontal length of slope
corec_c2 = corec_e2*l/bdash;

fie2 = fie2*100-corec_e1-corec_e2;
fic2 = fic2*100+corec_c1+C2-corec_c2;

#pile no. 3 at d/s end
d = 103.5-95;
alpha_ = d/b;
#for curves
fie3 = 0.35;fid3 = 0.242;
corec_t = (fie3-fid3)*100*(103.5-102)/d;

#correction for interference at pile no. 2
d = 102-95;
D = 102-97;
C3 = 19*(D/bdash)**0.5*(d+D)/b;
fie3 = fie3*100-corec_t-C3;

point = ['C1', 'C2' ,'E2' ,'E3'];            				#Point
P = [fic1 ,fic2 ,fie2 ,fie3];                				#pressure percent
P_ = [3.55 ,2.78, 3.39, 1.58];               				#pressure head
print "Points         Pressure percent        Pressure head";

for i in range(4):
P[i] = round(P[i]*10)/10;
print "%s                %.2f               %.2f"%(point[i],P[i],P_[i]);

#check for floor thickness
Pa = P_[1]-((P_[1]-P_[3])*6.5/37);
Pb = P_[1]-((P_[1]-P_[3])*24/37);
Pc = P_[1]-((P_[1]-P_[3])*30/37);
rho = 2.24;                              				#specific gravity of concrete
ta = Pa/(rho-1);
tb = Pb/(rho-1);
tc = Pc/(rho-1);
ta = round(ta*100)/100;
tb = round(tb*100)/100;
tc = round(tc*100)/100;
print "Thickness required at A = %.2f m."%(ta);
print "Thickness required at B = %.2f m."%(tb);
print "Thickness required at C = %.2f m."%(tc);
t = 103.5-102;
print "Thickness provided = %.2f m."%(t);
print "Floor thickness at B and C are adequate";

d = 103.5-95;                 				#depth cut-off
alpha = 6.35;
lambda1 = (1+(1+alpha**2)**0.5)/2;
Ge = H/(d*math.pi*lambda1**0.5);
print " it is less than permissible exit gradient  < 1/6Hence safe..";
C1                67.70               3.55
C2                52.80               2.78
E2                67.80               3.39
E3                33.10               1.58
Thickness required at A = 2.07 m.
Thickness required at B = 1.61 m.
Thickness required at C = 1.46 m.
Thickness provided = 1.50 m.
Floor thickness at B and C are adequate
it is less than permissible exit gradient  < 1/6Hence safe..

## Example 12.3 pg : 605¶

In [6]:
import math
from numpy import roots,ceil

#design a vertical drop weir on Bligh's theory
#test floor by Khosla's theory

#Given
Q = 2800;                    				#maximum flood discharge
hfl = 285;                   				#H.F.L before construction
hw = 278;                    				#minimum water level
fsl = 284;                   				#F.S.L of canal
c = 12;                      				#coefficient of creep
flux = 1;                    				#allowable afflux
Ge = 1./6;                    				#permissible exit gradient
rho = 2.24;                  				#specific gravity of concrete

#Hydraulic calculation
L = 4.75*Q**0.5;
q = Q/L;
q = round(q*10)/10;
print "Hydraulic calculation:";
print "discharge per unit width of river = %.2f cumecs."%(q);
f = 1;
R = 1.35*(q**2/f)**(1./3);
R = round(R*100)/100;
print "regime scour depth = %.2f m."%(R);
V = q/R;                   				#regime velocity
l_down = hfl+vh;
l_up = l_down+flux;
hfl_up = l_up-vh;
hfl_down = hfl-0.5;
hfl_down = round(hfl_down*100)/100;
print "actual d/s H.F.L allowing 0.5 m for retrogation = %.2f m."%(hfl_down);
K = (q/1.7)**(2./3);
cl = l_up-K;               				#crest level
cl = round(cl*100)/100;
print "crest level = %.2f m."%(cl);
pl = fsl+0.5;              				#pond level
s = hfl_down-cl;                				#heigth of shutter
print "heigth of shutter = %.2f m."%(s);
rl_up_pile = hfl_up-1.5*R;    				#R.L of bottom u/s pile
d_up_cut = hw-276;           				#depth of upstream cut-off
print "depth of upstream cut-off = %.2f m."%(d_up_cut);
print " provide concrete cut off 2 m depth.";
rl_bot_ds = hfl_down-2*R;
Hc = cl-hw;                  				#heigth of crest
print "R.L of gates crest = %.2f m."%(Hs);
print "Heigth of crest = %.2f m."%(Hc);

#design of weir wall
d = hfl_up-cl;
a = d/(rho)**0.5;
a = 3*d/(2*rho);            				#from sliding consideration
a = s+1;                    				#from practical consieration
a = a+1;
print "design of weir wall:"
print "provide top width of %i m."%(a);
Mo = 9.81*Hs**3/6;                				#overtirning moment
#equating the moment of resismath.tance to overturning moment and putting the values we get
#y = poly([-1.084,0.020,0.039],'x','c');
y = [0.039,0.020,-1.084]
b = roots(y)[1];
#we get b =  - 5.5347261 and 5.0219056
#taking
b = 5;
#when weir is submerged
C = 0.58;
d = (q**2/((2*C/3)**2*2*9.81))**(1./3);
Mo = 9.81*d*Hc**2/2;
#from equation of moment of resistence we get
y = [1,3,-77.55]
b = ceil(roots(y)[1]);       #we get b =  - 10.433085 and 7.4330846
print "bottom width = %i m."%(b);

#design of impervious and pervious aprons
C = 12;
L = C*Hs;
print "design of impervious and pervious aprons:";
print "total creep length = %i m."%(L);
l1 = 2.21*C*(Hs/13)**0.5;
l1_ = l1+1;
print "length of downstream impervious apron = %i m."%(l1_);
d1 = hw-276;
d2 = hw-271;
l2 = L-l1-(b+2*d1+2*d2);
print "length of upstream impervious apron = %i m."%(l2);
l3 = 18*C*(Hs*q/975)**0.5;
print "total length of d/s apron = %i m."%(l3);                				#calculation is wrong in book
l = l3-l1;
le = l/2;
le = round(le*100)/100;
print 'provide filter of length %.2f m. and launching apron of length %.2f m.'%(le,le);
t = d2*10**0.5/le;
print "thickness of launching apron in horizontal position = %.2f m."%(t);
print "provide launching apron of thickness 1.5 m.";
T = 2*d1;
V = d1*10**0.5;
ta = V/T;
ta = round(ta*10)/10;
print "thickness of apron in horizontal position = %.2f m."%(ta);
Hr = Hs-Hs*(4+33+8)/L;
t = 4*Hr/(3*(rho-1));
t = round(t*10)/10;
print 'provide thickness of %.2f m from d/s of weir wall to point 6 m from it.'%(t);
Hr = Hs-Hs*(4+33+8+6)/L;
t = 4*Hr/(3*(rho-1));
t = round(t*10)/10;
print "provide thickness of %.2f m from 6 m to 12 m from d/s end of weir wall."%(t);
Hr = Hs-Hs*(4+33+8+12)/L;
t = 4*Hr/(3*(rho-1));
t = round(t*10)/10;
print "provide thickness of %.2f m for rest of length of weir floor."%(t);

#check by khosla's theory
b = 33+8+19;            				#total horizontal length of impervious floor
d = 7;                  				#depth of downstream pile
alpha = b/d;
n = 0.14;                				#n = 1/math.pi*(lambda)**0.5;
Ge = Hs*n/d;
print "check by Khosla theory:";
print "exit gradient = %.2f. < 1/6 hence safe"%(Ge);
alpha_ = d/b;
fic1 = 0.83;fid1 = 0.88;
corec_c1 = (fid1-fic1)*100/2;
bdash = b;
d = 2;D = 7;
C1 = 19*(D/bdash)**0.5*(d+D)/b;
fic1 = fic1*100+corec_c1+C1;
Pc = Hs*fic1/100;                        				#pressure head at C
alpha_ = d/b;
fie2 = 0.31;fid2 = 0.21;
corec_e1 = (fie2-fid2)*1.7*100/7;
bdash = b;
d = 7;D = 2;
C1 = 19*(D/bdash)**0.5*(d+D)/b;
fie2 = fie2*100-corec_e1-C1;             				#in book 3.53 value is wrong
Pe = Hs*fie2/100;                         				#pressue head at E
#assuming linear variation of pressure for intermediate points
Pa = Pc-(Pc-Pe)*(33+8)/b;
t = Pa/1.24;
Pa = round(Pa*100)/100;
t = round(t*100)/100;
print "pressure at d/s of weir wall = %.2f m."%(Pa);
print "thickness at d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe."%(t);
Pb = Pc-(Pc-Pe)*(33+8+6)/b;
t = Pb/1.24;
Pa = round(Pa*100)/100;
t = round(t*100)/100;
print "pressure at 6 m from d/s of weir wall = %.2f m."%(Pb);
print "thickness at 6m from d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe."%(t);
Pc = Pc-(Pc-Pe)*(33+8+12)/b;
t = Pc/1.24;
Pa = round(Pa*100)/100;
t = round(t*100)/100;
print "pressure at 12 m from d/s of weir wall = %.2f m."%(Pc);
print "thickness at 12m from d/s of weir wall = %.2f m. > thickness by Bligh theory;hence unsafe."%(t);
print "hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.";
Hydraulic calculation:
discharge per unit width of river = 11.10 cumecs.
regime scour depth = 6.72 m.
actual d/s H.F.L allowing 0.5 m for retrogation = 284.50 m.
crest level = 282.65 m.
heigth of shutter = 1.85 m.
depth of upstream cut-off = 2.00 m.
provide concrete cut off 2 m depth.
R.L of gates crest = 6.50 m.
Heigth of crest = 4.65 m.
design of weir wall:
provide top width of 3 m.
bottom width = 8 m.
design of impervious and pervious aprons:
total creep length = 78 m.
length of downstream impervious apron = 19 m.
length of upstream impervious apron = 33 m.
total length of d/s apron = 58 m.
provide filter of length 20.00 m. and launching apron of length 20.00 m.
thickness of launching apron in horizontal position = 1.11 m.
provide launching apron of thickness 1.5 m.
thickness of apron in horizontal position = 1.60 m.
provide thickness of 3.00 m from d/s of weir wall to point 6 m from it.
provide thickness of 2.40 m from 6 m to 12 m from d/s end of weir wall.
provide thickness of 1.90 m for rest of length of weir floor.
check by Khosla theory:
exit gradient = 0.13. < 1/6 hence safe
pressure at d/s of weir wall = 3.03 m.
thickness at d/s of weir wall = 2.44 m. < thickness by Bligh theory;hence safe.
pressure at 6 m from d/s of weir wall = 2.66 m.
thickness at 6m from d/s of weir wall = 2.14 m. < thickness by Bligh theory;hence safe.
pressure at 12 m from d/s of weir wall = 2.29 m.
thickness at 12m from d/s of weir wall = 1.85 m. > thickness by Bligh theory;hence unsafe.
hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.

## Example 12.4 pg : 617¶

In [7]:
import math

#design a slopeing glacis

#Given
q = 10;                     				#maximum discharge intensity on weir crest
hfl = 255;                  				#H.F.L before construction of weir
rb = 249.5;                 				#R.L of river bed
pl = 254;                   				#pond level
s = 1;                      				#heigth of crest shutter
dhw = 251.5;                				#anticipated downstream water level in river when water is dischrging with pond level upstream
br = 0.5;                   				#bed retrogression
f = 0.9;                    				#Laecey silt factor
Ge = 1./7;                   				#permissible exit gradient
flux = 1;                   				#permissible afflux

cl = pl-s;                   				#crest level
print "crest level = %.2f m."%(cl);
K = (q/1.7)**(2./3);
tel_up = cl+K;
tel_up = round(tel_up*100)/100;
print "elevation of u/s T.E.L = %.2f m."%(tel_up);
R = 1.35*(q**2/f)**(1./3);
R = round(R*10)/10;
print "regime scour depth = %.2f m."%(R);
V = q/R;                     				#regime velocity
hfl_up = tel_up-vh;
tel_down = hfl+vh;
flux = hfl_up-hfl;
flux = round(flux*100)/100;
print "afflux = %.2f. which is near to permissible"%(flux);
hfl_down = hfl-br;             				#downstream H.F.L after retrogression
tel_down = tel_down-br;        				#downstream T.F.L after retrogression
Hl = tel_up-tel_down;          				#loss of head in flood
Hl = round(Hl*100)/100;
print "loss of head in at high flood = %.2f m."%(Hl);
K = pl-cl;              				#head over crest
q_ = 1.7*(K)**1.5;
Hl_ = pl-dhw;             				#loss of head
print "loss of head = %.2f m."%(Hl_);
Ef2 = 4.3;
Ef2_ = 1.7;              				#from Blench curve
jump = tel_down-Ef2;
jump_ = 251.5-Ef2_;      				#level at which jump will form
Ef1 = Ef2+Hl;
Ef1_ = Ef2_+Hl_;
D1 = 1.03;
D1_ = 0.15;
D2 = 3.96;D2_ = 1.68;
hj = D2-D1;
hj_ = D2_-D1_;            				#heigth of jump
concrete = 5*hj;
concrete_ = 5*hj_;         				#length of concrete floor
print "Hydraulic jump calculation:";
print "heigth of jump for high flood condition = %.2f m."%(hj);
print "length of concrete floor for high flood condition = %.2f m."%(concrete);
print "heigth of jump for pond level condition = %.2f m."%(hj_);
print "length of concrete floor for high pond level condition = %.2f m."%(concrete_);

cw = 2;                    				#crets width
us = 2;                    				#upstream slope
ds = 3;                    				#downstream slope
l = 15;
print " upstream slope of glacis = %i:1."%(us);
print "downstream slope of glacis = %i:1."%(ds);
print "horizontal length of floor beyond the toe = %i m.."%(l);

R = 6.5;
sh_up = hfl_up-1.5*R;
sh_down = hfl_down-2*R;
sh_up = round(sh_up*100)/100;
print "R.L of bottom of upstream sheet pile = %.2f m."%(sh_up);
print "R.L of downstream sheet pile = %.2f m."%(sh_down);
print "provide intermediate sheet pile at d/s toe of glacis.";
Hs = pl-249.6;                       				#maximum percolation head
d = 249.6-sh_down;                   				#depth of d/s cut-off
n = Ge*d/Hs;                          				#n = 1/(math.pi*lambda**0.5);
alpha = 1.5;
b = alpha*d;
print "length of impervious floor = %.2f m."%(b);
fl = (2*(253-249.5))+2+(3*(253-249.6))+15;
us = 36-fl;
print "length of floor already provide = %.2f m."%(fl);
print "which is more than required from permissible exit gradient.no upstream floor is required.";
print "provide %.2f m upstream floor so that total length becomes 36 m."%(us);
alpha_1 = 0.089;
alpha_2 = 0.225;            				#alpha_ = 1/alpha
b1 = 21;
alpha = 4.44;
print "Pressure percent at points:";
point = ['C1', 'D1' ,'C2' ,'E2' ,'D2' ,'D3' ,'E3'];
bc = [72 ,82 ,31.5 ,45.5 ,58.5 ,29 ,44];
crt = [3.1, 0, 3.5, 0, -3.2, 0, 0, -3.6];
crs = [0 ,0, 0, 0, 2.3, 0, 0, 0];
cri = [3.7, 0, 6.4, 0, -2.4, 0, -6.4];
after = [0,0,0,0,0,0,0]
print "Points       Before correction            After correction";
for i in range(7):
after[i] = bc[i]+crt[i]+crs[i]+cri[i];
print "%s                   %i                       %.2f"%(point[i],bc[i],after[i]);

Hs = 254-249.6;               				#no flow condition
Hs_ = 256.13-254.5;           				#high flood condition
Hs__ = 254-251.5;             				#flow at pond level
print "elevation of subsoil H.G above datum:";
print "no flow condition:";
fie1 = 1*Hs;
fid1 = 0.82*Hs;
fic1 = 0.788*Hs;
fie2 = 0.552*Hs;
fid2 = 0.455*Hs;
fic2 = 0.414*Hs;
fie3 = 0.34*Hs;
fid3 = 0.29*Hs;
fic3 = 0;
fie1 = round(fie1*100)/100;fid1 = round(fid1*100)/100;fic1 = round(fic1*100)/100;
fie2 = round(fie2*100)/100;fid2 = round(fid2*100)/100;fic2 = round(fic2*100)/100;
fie3 = round(fie3*100)/100;fid3 = round(fid3*100)/100;fic3 = round(fic3*100)/100;
print "fie1 = %.2f.;fid1 = %.2f.;fic1 = %.2f.fie2 = %.2f.;fid2 = %.2f.;fic2 = %.2f.fie3 = %.2f.;\
fid3 = %.2f.;fic3 = %.2f."%(fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);
print "high flood condition:";
fie1 = 1*Hs_;
fid1 = 0.82*Hs_;
fic1 = 0.788*Hs_;
fie2 = 0.552*Hs_;
fid2 = 0.455*Hs_;
fic2 = 0.414*Hs_;
fie3 = 0.34*Hs_;
fid3 = 0.29*Hs_;
fic3 = 0;
fie1 = round(fie1*100)/100;fid1 = round(fid1*100)/100;fic1 = round(fic1*100)/100;
fie2 = round(fie2*100)/100;fid2 = round(fid2*100)/100;fic2 = round(fic2*100)/100;
fie3 = round(fie3*100)/100;fid3 = round(fid3*100)/100;fic3 = round(fic3*100)/100;
print "fie1 = %.2f.;fid1 = %.2f.;fic1 = %.2f.fie2 = %.2f.;fid2 = %.2f.;fic2 = %.2f.fie3 = %.2f.;\
fid3 = %.2f.;fic3 = %.2f."%(fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);
print "flow at pond level:";
fie1 = 1*Hs__;
fid1 = 0.82*Hs__;
fic1 = 0.788*Hs__;
fie2 = 0.552*Hs__;
fid2 = 0.455*Hs__;
fic2 = 0.414*Hs__;
fie3 = 0.34*Hs__;
fid3 = 0.29*Hs__;
fic3 = 0;
fie1 = round(fie1*100)/100;fid1 = round(fid1*100)/100;fic1 = round(fic1*100)/100;
fie2 = round(fie2*100)/100;fid2 = round(fid2*100)/100;fic2 = round(fic2*100)/100;
fie3 = round(fie3*100)/100;fid3 = round(fid3*100)/100;fic3 = round(fic3*100)/100;
print "fie1 = %.2f.;fid1 = %.2f.;fic1 = %.2f.fie2 = %.2f.;fid2 = %.2f.;fic2 = %.2f.fie3 = %.2f.;\
fid3 = %.2f.;fic3 = %.2f."%(fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);

print "Prejump profile:";
print "high flood condition:";
dist = [3 ,6, 8.4];                 				#dismath.tance
glacis = [252, 251, 250.32];        				#R.L of glacis
D1 = [1.3 ,1.15, 1.03];
Ef1 = [0,0,0]
print "Ef1              D1";
for i in range(3):
Ef1[i] = 256.25-glacis[i];
print "%.2f         %.2f"%(Ef1[i],D1[i]);

print "pond level flow:";
dist = [3, 6, 9, 9.6];             				#dismath.tance
glacis = [252, 251, 250, 249.9];       				#R.Lof glacis
D1 = [0.31, 0.23, 0.16, 0.15];
Ef1 = [0,0,0,0]
print "Ef1              D1";
for i in range(4):
Ef1[i] = 254-glacis[i];
print "%.2f         %.2f"%(Ef1[i],D1[i]);

rho = 2.24;
Uf = 4;                           				#unbalanced head for high flood condtion
Us = 2.56;                        				#unbalanced static head
Hf = 2*Uf/3;
t = Hf/(rho-1);
t = round(t*10)/10;
print "floor thickness at the point of formation of hydraulic jump = %.2f m."%(t);
Uf = 2.9;                           				#unbalanced head for high flood condtion
Us = 2.2;                        				#unbalanced static head
Hf = 2*Uf/3;
t = Us/(rho-1);
t = round(t*10)/10;
print "floor thickness at the point of formation of hydraulic jump at the pond level condition = %.2f m."%(t);
P = 1.5;                        				#pressure head at d/s end of floor
t = P/(rho-1);
t = round(t*10)/10;
print "floor thickness at downstream side of sloping glacis = %.2f m."%(t);
D = rb-sh_up;                 				#depth of u/s scour hole above bed level
a = 1.5*D;
a = round(a*10)/10;
print "minimum length of upstream launching apron = %.2f m."%(a);
print "provide 1.5 m thick apron for length of 5 m.";
D = 249.6-241.5;
a = 1.5*D;
print "minimum length of downstream launching apron = %.2f m."%(a);
print "provide 1.5 m thick apron for length of 12 m.";
crest level = 253.00 m.
elevation of u/s T.E.L = 256.26 m.
regime scour depth = 6.50 m.
afflux = 1.14. which is near to permissible
loss of head in at high flood = 1.64 m.
loss of head = 2.50 m.
Hydraulic jump calculation:
heigth of jump for high flood condition = 2.93 m.
length of concrete floor for high flood condition = 14.65 m.
heigth of jump for pond level condition = 1.53 m.
length of concrete floor for high pond level condition = 7.65 m.
upstream slope of glacis = 2:1.
downstream slope of glacis = 3:1.
horizontal length of floor beyond the toe = 15 m..
R.L of bottom of upstream sheet pile = 246.39 m.
R.L of downstream sheet pile = 241.50 m.
provide intermediate sheet pile at d/s toe of glacis.
length of impervious floor = 12.15 m.
length of floor already provide = 34.20 m.
which is more than required from permissible exit gradient.no upstream floor is required.
provide 1.80 m upstream floor so that total length becomes 36 m.
Pressure percent at points:
Points       Before correction            After correction
C1                   72                       78.80
D1                   82                       82.00
C2                   31                       41.40
E2                   45                       45.50
D2                   58                       55.20
D3                   29                       29.00
E3                   44                       37.60
elevation of subsoil H.G above datum:
no flow condition:
fie1 = 4.40.;fid1 = 3.61.;fic1 = 3.47.fie2 = 2.43.;fid2 = 2.00.;fic2 = 1.82.fie3 = 1.50.;fid3 = 1.28.;fic3 = 0.00.
high flood condition:
fie1 = 1.63.;fid1 = 1.34.;fic1 = 1.28.fie2 = 0.90.;fid2 = 0.74.;fic2 = 0.67.fie3 = 0.55.;fid3 = 0.47.;fic3 = 0.00.
flow at pond level:
fie1 = 2.50.;fid1 = 2.05.;fic1 = 1.97.fie2 = 1.38.;fid2 = 1.14.;fic2 = 1.03.fie3 = 0.85.;fid3 = 0.73.;fic3 = 0.00.
Prejump profile:
high flood condition:
Ef1              D1
4.25         1.30
5.25         1.15
5.93         1.03
pond level flow:
Ef1              D1
2.00         0.31
3.00         0.23
4.00         0.16
4.10         0.15
floor thickness at the point of formation of hydraulic jump = 1.60 m.
floor thickness at the point of formation of hydraulic jump at the pond level condition = 1.80 m.
floor thickness at downstream side of sloping glacis = 1.20 m.
minimum length of upstream launching apron = 4.70 m.
provide 1.5 m thick apron for length of 5 m.
minimum length of downstream launching apron = 12.15 m.
provide 1.5 m thick apron for length of 12 m.

## Example 12.5 pg : 631¶

In [8]:
import math

#Given
b = 16;       				#total length of floor
d = 5;        				#depth of downstream pile
D = 4;        				#depth of upstream pile
H = 2.5;      				#head created by weir

#pressure at E
alpha = b/d;
lambda1 = (1+(1+alpha**2)**0.5)/2;
fie = math.acos((lambda1-2)/lambda1)/math.pi;
C = 19*(D/b)**0.5*((d+D)/b);
fie = fie*100-C;
P = H*fie/100;
P = round(P*1000)/1000;
print "Pressure at E = %.2f m."%(P);

#pressure at C1
alpha = b/D;
lambda1 = (1+(1+alpha**2)**0.5)/2;
fie = math.acos((lambda1-2)/lambda1)/math.pi;
fic = 1-fie;            				#by principle reversibility of flow
C = 19*(d/b)**0.5*((d+D)/b);
fic = fic*100+C;
P = fic*H/100;
P = round(P*1000)/1000;
print " Pressure at C = %.2f m."%(P);
Pressure at E = 1.22 m.
Pressure at C = 1.43 m.

## Example 12.6 pg : 632¶

In [9]:
import math

#Given
b = 13;         				#length of floor
d = 2;          				#depth of downstream wall
D = 1.5;       				#depth of upstream cut-off
rho = 2.24;     				#relative density
H = 1.5;

#at junction of d/s cut-off with floor
alpha = b/d;
lambda1 = (1+(1+alpha**2)**0.5)/2;
fie = math.acos((lambda1-2)/lambda1)/math.pi;
C = 19*(D/b)**0.5*((d+D)/b);
fie = fie*100-C;
P = H*fie/100;
t = P/(rho-1);
t = round(t*10)/10;
print "floor thickness at junction of d/s cut-off with floor = %.2f m."%(t);

#at junction of u/s cut-off with floor
alpha = b/D;
lambda11 = (1+(1+alpha**2)**0.5)/2;
fie = math.acos((lambda11-2)/lambda11)/math.pi;
fic = 1-fie;            				#by principle reversibility of flow
C = 19*(D/b)**0.5*((d+D)/b);
fiec = fic*100+C;
P = fiec*H/100;
t = 0.3;              				#this the uplift will be counter balanced by downward weigth of impounded water
print "floor thickness at junction of u/s cut-off with floor = %.2f m."%(t);

#at mid-length
P = (1.08+0.489)/2;            				#assuming linear variation
t = P/(rho-1);
t = round(t*100)/100;
print "floor thickness at mid-length = %.2f m."%(t);

G = H/(d*math.pi*(lambda1)**0.5);
G = round(G*1000)/1000;
#math.since G<0.18
print " G = %.2f. <0.18./nfloor is safe against failure by piping."%(G);
floor thickness at junction of d/s cut-off with floor = 0.40 m.
floor thickness at junction of u/s cut-off with floor = 0.30 m.
floor thickness at mid-length = 0.63 m.
G = 0.13. <0.18./nfloor is safe against failure by piping.

## Example 12.7 pg : 634¶

In [10]:
import math

#Given
B = 30;            				#stream width
D = 3;             				#stream depth
V = 1.25;          				#mean velocity
Cd = 0.95;         				#discharge coefficient
Q = B*D*V;

# Calculations
C = 2*Cd*(2*9.81)**0.5/3;
x = 4-(Q/(C*B))**(2./3);
x = round(x*1000)/1000;

# Results
print "heigth of weir to be built = %.2f m."%(x);
heigth of weir to be built = 2.79 m.

## Example 12.8 pg : 635¶

In [11]:
import math

#Given
b = 50.;       				#length of floor
d = 8.;        				#depth of downstream pile
D = 8.;        				#depth of upstream pile
tu = 1.;        				#floor thickness at upstream
td = 2.;        				#floor thickness at downstream

# Calculations and Results
#downstream cut-off
alpha = b/d;
lambda1 = (1+(1+alpha**2)**0.5)/2;
fie = math.acos((lambda1-2)/lambda1)/math.pi;
fid = math.acos((lambda1-1)/lambda1)/math.pi;
Ct = (fie-fid)*td/d;
C = 19*(D/b)**0.5*((d+D)/b);
fie = fie*100-C-Ct*100;
P = H*fie/100;
P = round(P*100)/100;
print "Pressure at downstream cut-off = %.2f m."%(P);

#upstream cut-off
fie = math.acos((lambda1-2)/lambda1)/math.pi;
fid = math.acos((lambda1-1)/lambda1)/math.pi;
fic1 = 1-fie;
fid1 = 1-fid;
Ct = (fic1-fid1)*td/d;
C = -19*(D/b)**0.5*((d+D)/b);
fic1 = fic1*100-C-Ct*100;
P = H*fic1/100;
P = round(P*100)/100;
print "Pressure at upstream cut-off = %.2f m."%(P);
G = H/(d*math.pi*(lambda1)**0.5);
Pressure at downstream cut-off = 1.49 m.
Pressure at upstream cut-off = 3.51 m.

## Example 12.9 pg : 636¶

In [12]:
import math

#Given
Q = 1000.;        				#discharge of river
L = 256.;         				#crest length of diversion
f = 1.1;         				#silt factor
seg = 1./6;       				#safe exit gradient
hfl = 103;       				#high flood level
cf = 100;        				#reduced level of downstream concrete floor
H = 2.4;         				#maximum static head of weir
b = 40;          				#length of concrete floor

# Calculations and Results
q = Q/L;
R = 1.35*(q**2/f)**(1./3);
rld = hfl-1.5*R;
d = cf-rld;
d = round(d*100)/100;
print "depth of downstream cut-off = %.2f m."%(d);

alpha = b/d;
lambda1 = (1+(1+alpha**2)**0.5)/2;
G = H/(d*math.pi*(lambda1)**0.5);
#math.since G<seg
print " G = %.2f. <1/6./nfloor is safe against failure by piping."%(G);
depth of downstream cut-off = 1.87 m.
G = 0.12. <1/6./nfloor is safe against failure by piping.

## Example 12.10 pg : 636¶

In [13]:
import math

#Given
b = 60;       				#length of floor
H = 6;        				#static head of weir
d = 6;        				#downstream depth of pile
n = 0.3;      				#porousity of soil particles
G = 2.7;      				#relative density of soil particles

alpha = b/d;
lambda1 = (1+(1+alpha**2)**0.5)/2;
Ge = H/(d*math.pi*(lambda1)**0.5);
e = n/(1-n);
chg = (G-1)/(1+e);
f = chg/Ge;
f = round(f*100)/100;
print "critical exit gradient = %.2f.\
\nfactor of safety of system = %.2f."%(chg,f);
factor of safety of system = 8.79.

## Example 12.11 pg : 637¶

In [14]:
import math
from numpy import roots,ceil

#design a vertical drop weir on Bligh's theory
#test floor by Khosla's theory

#Given
Q = 2800.;                    				#maximum flood discharge
hfl = 285.;                   				#H.F.L before construction
hw = 278.;                    				#minimum water level
fsl = 284.;                   				#F.S.L of canal
c = 12.;                      				#coefficient of creep
flux = 1.;                    				#allowable afflux
Ge = 1./6;                    				#permissible exit gradient
rho = 2.24;                  				#specific gravity of concrete

#Hydraulic calculation
L = 4.75*Q**0.5;
q = Q/L;
q = round(q*10)/10;
print "Hydraulic calculation:";
print "discharge per unit width of river = %.2f cumecs."%(q);
f = 1;
R = 1.35*(q**2/f)**(1./3);
R = round(R*100)/100;
print "regime scour depth = %.2f m."%(R);
V = q/R;                   				#regime velocity
l_down = hfl+vh;
l_up = l_down+flux;
hfl_up = l_up-vh;
hfl_down = hfl-0.5;
hfl_down = round(hfl_down*100)/100;
print "actual d/s H.F.L allowing 0.5 m for retrogation = %.2f m."%(hfl_down);
K = (q/1.7)**(2./3);
cl = l_up-K;               				#crest level
cl = round(cl*100)/100;
print "crest level = %.2f m."%(cl);
pl = fsl+0.5;              				#pond level
s = hfl_down-cl;                				#heigth of shutter
print "heigth of shutter = %.2f m."%(s);
rl_up_pile = hfl_up-1.5*R;    				#R.L of bottom u/s pile
d_up_cut = hw-276;           				#depth of upstream cut-off
print "depth of upstream cut-off = %.2f m."%(d_up_cut);
print " provide concrete cut off 2 m depth.";
rl_bot_ds = hfl_down-2*R;
Hc = cl-hw;                  				#heigth of crest
print "R.L of gates crest = %.2f m."%(Hs);
print "Heigth of crest = %.2f m."%(Hc);

#design of weir wall
d = hfl_up-cl;
a = d/(rho)**0.5;
a = 3*d/(2*rho);            				#from sliding consideration
a = s+1;                    				#from practical consieration
a = a+1;
print "design of weir wall:"
print "provide top width of %i m."%(a);
Mo = 9.81*Hs**3/6;                				#overtirning moment
#equating the moment of resismath.tance to overturning moment and putting the values we get
y = [0.039,0.020,-1.084]
b = int(roots(y)[1]);
#we get b =  - 5.5347261 and 5.0219056
#taking
#b = 5;
#when weir is submerged
C = 0.58;
d = (q**2/((2*C/3)**2*2*9.81))**(1./3);
Mo = 9.81*d*Hc**2/2;
#from equation of moment of resistence we get
y = [1,3,-77.55]
b = ceil(roots(y)[1]);
#we get b =  - 10.433085 and 7.4330846
#taking
#b = 8;
print "bottom width = %i m."%(b);

#design of impervious and pervious aprons
C = 12;
L = C*Hs;
print "design of impervious and pervious aprons:";
print "total creep length = %i m."%(L);
l1 = 2.21*C*(Hs/13)**0.5;
l1_ = l1+1;
print "length of downstream impervious apron = %i m."%(l1_);
d1 = hw-276;
d2 = hw-271;
l2 = L-l1-(b+2*d1+2*d2);
print "length of upstream impervious apron = %i m."%(l2);
l3 = 18*C*(Hs*q/975)**0.5;
print "total length of d/s apron = %i m."%(l3);                				#calculation is wrong in book
l = l3-l1;
le = l/2;
le = round(le*100)/100;
print 'provide filter of length %.2f m. and launching apron of length %.2f m.'%(le,le);
t = d2*10**0.5/le;
print "thickness of launching apron in horizontal position = %.2f m."%(t);
print "provide launching apron of thickness 1.5 m.";
T = 2*d1;
V = d1*10**0.5;
ta = V/T;
ta = round(ta*10)/10;
print "thickness of apron in horizontal position = %.2f m."%(ta);
Hr = Hs-Hs*(4+33+8)/L;
t = 4*Hr/(3*(rho-1));
t = round(t*10)/10;
print 'provide thickness of %.2f m from d/s of weir wall to point 6 m from it.'%(t);
Hr = Hs-Hs*(4+33+8+6)/L;
t = 4*Hr/(3*(rho-1));
t = round(t*10)/10;
print "provide thickness of %.2f m from 6 m to 12 m from d/s end of weir wall."%(t);
Hr = Hs-Hs*(4+33+8+12)/L;
t = 4*Hr/(3*(rho-1));
t = round(t*10)/10;
print "provide thickness of %.2f m for rest of length of weir floor."%(t);

#check by khosla's theory
b = 33+8+19;            				#total horizontal length of impervious floor
d = 7;                  				#depth of downstream pile
alpha = b/d;
n = 0.14;                				#n = 1/math.pi*(lambda)**0.5;
Ge = Hs*n/d;
print "check by Khosla theory:";
print "exit gradient = %.2f. < 1/6 hence safe"%(Ge);
alpha_ = d/b;
fic1 = 0.83;fid1 = 0.88;
corec_c1 = (fid1-fic1)*100/2;
bdash = b;
d = 2.;D = 7;
C1 = 19*(D/bdash)**0.5*(d+D)/b;
fic1 = fic1*100+corec_c1+C1;
Pc = Hs*fic1/100;                        				#pressure head at C
alpha_ = d/b;
fie2 = 0.31;fid2 = 0.21;
corec_e1 = (fie2-fid2)*1.7*100/7;
bdash = b;
d = 7;D = 2;
C1 = 19*(D/bdash)**0.5*(d+D)/b;
fie2 = fie2*100-corec_e1-C1;             				#in book 3.53 value is wrong
Pe = Hs*fie2/100;                         				#pressue head at E
#assuming linear variation of pressure for intermediate points
Pa = Pc-(Pc-Pe)*(33+8)/b;
t = Pa/1.24;
Pa = round(Pa*100)/100;
t = round(t*100)/100;
print "pressure at d/s of weir wall = %.2f m."%(Pa);
print "thickness at d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe."%(t);
Pb = Pc-(Pc-Pe)*(33+8+6)/b;
t = Pb/1.24;
Pa = round(Pa*100)/100;
t = round(t*100)/100;
print "pressure at 6 m from d/s of weir wall = %.2f m."%(Pb);
print "thickness at 6m from d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe."%(t);
Pc = Pc-(Pc-Pe)*(33+8+12)/b;
t = Pc/1.24;
Pa = round(Pa*100)/100;
t = round(t*100)/100;
print "pressure at 12 m from d/s of weir wall = %.2f m."%(Pc);
print "thickness at 12m from d/s of weir wall = %.2f m. > thickness by Bligh theory;hence unsafe."%(t);
print "hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.";
Hydraulic calculation:
discharge per unit width of river = 11.10 cumecs.
regime scour depth = 6.72 m.
actual d/s H.F.L allowing 0.5 m for retrogation = 284.50 m.
crest level = 282.65 m.
heigth of shutter = 1.85 m.
depth of upstream cut-off = 2.00 m.
provide concrete cut off 2 m depth.
R.L of gates crest = 6.50 m.
Heigth of crest = 4.65 m.
design of weir wall:
provide top width of 3 m.
bottom width = 8 m.
design of impervious and pervious aprons:
total creep length = 78 m.
length of downstream impervious apron = 19 m.
length of upstream impervious apron = 33 m.
total length of d/s apron = 58 m.
provide filter of length 20.00 m. and launching apron of length 20.00 m.
thickness of launching apron in horizontal position = 1.11 m.
provide launching apron of thickness 1.5 m.
thickness of apron in horizontal position = 1.60 m.
provide thickness of 3.00 m from d/s of weir wall to point 6 m from it.
provide thickness of 2.40 m from 6 m to 12 m from d/s end of weir wall.
provide thickness of 1.90 m for rest of length of weir floor.
check by Khosla theory:
exit gradient = 0.13. < 1/6 hence safe
pressure at d/s of weir wall = 3.03 m.
thickness at d/s of weir wall = 2.44 m. < thickness by Bligh theory;hence safe.
pressure at 6 m from d/s of weir wall = 2.66 m.
thickness at 6m from d/s of weir wall = 2.14 m. < thickness by Bligh theory;hence safe.
pressure at 12 m from d/s of weir wall = 2.29 m.
thickness at 12m from d/s of weir wall = 1.85 m. > thickness by Bligh theory;hence unsafe.
hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.

## Example 12.12 pg : 637¶

In [15]:
import math

#number of gates required for the barrage
#length and R.L of bamath.sin floor if silting bamath.sin is provided downstream of barrage

#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.
print "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
print "nunmber of gates for the head regulator = 7.";

#design of stilling bamath.sin
Hl = Lmax-hfl;
q = Qmax/L;
yc = (q**2/9.81)**(1./3);
Z = Hl/yc;
#math.since Z<1
Y = 1+0.93556*Z**0.368;
y2 = Y*yc;
Lc = 5*y2;
Lc = round(Lc*10)/10;
print "Length of cistern = %.2f m."%(Lc);
Ef2 = yc*(Y+1/(2*Y**2));
j = hfl-Ef2;
j = round(j*10)/10;
print "R.L of cistern = %.2f m."%(j);
nunmber of gates for the barrage = 15.
nunmber of gates for the head regulator = 7.
Length of cistern = 33.30 m.
R.L of cistern = 202.70 m.