Chapter 19 : CROSS DRAINAGE WORKS¶

Example 19.1 pg : 857¶

In :
import math
from numpy import linspace,zeros

#design an expansion transition for canal by Mitra's method

#Given
Lf = 16.;            				#length of flume
Bf = 9.;            				#width of throat
Bo = 15.;           				#width of canal

#width at any dismath.tance x from flumed section is given by
#Bx = Bo*Bf*Lf/(Lf*Bo-(Bo-Bf)x)
#on solving we get
#Bx = 2160/(240-6x)

x = linspace(2,16,8)       				#dismath.tance
print "width at any dismath.tance x from flumed section:";
Bx = zeros(8)
for i in range(8):
Bx[i] = 2160./(240-6*x[i]);
Bx[i] = round(Bx[i]*100)/100;
print '%.2f'%(Bx[i]);
width at any dismath.tance x from flumed section:
9.47
10.00
10.59
11.25
12.00
12.86
13.85
15.00

Example 19.2 pg : 857¶

In :
import math
from numpy import linspace,zeros

#design an expansion transition for canal by Chaturvedi's method

#Given;
Lf = 16.;            				#length of flume
Bf = 9.;            				#width of throat
Bo = 15.;           				#width of canal

x = linspace(2,16,8);     				#dismath.tance

#dismath.tance x is related as x = Lf*Bo**(2/3)(1-(Bf/Bx)**1.5)/(Bo**1.5-Bf**1.5)
#on solving we get
#(Bf/Bx)**1.5 = 1-(x/29.893)   (relation is misprinted in book)
#let (Bf/Bx)**1.5 = r
r = zeros(8)
R = zeros(8)
Bx = zeros(8)

print "width at any dismath.tance x from flumed section:";
for i in range(8):
r[i] = 1-(x[i]/29.893);       				#Bf/Bx**(1.5)
R[i] = r[i]**(2./3);            				#Bf/Bx
Bx[i] = Bf/R[i];
Bx[i] = round(Bx[i]*100)/100;
print "%.2f."%(Bx[i]);
width at any dismath.tance x from flumed section:
9.43.
9.90.
10.45.
11.08.
11.81.
12.67.
13.71.
15.00.

Example 19.3 pg : 860¶

In :
import math
from numpy import linspace,zeros,zeros_like

#design a syphon aqueduct

#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

#demath.sing of drainage water-way
P = 4.75*(hfq)**0.5;     				#laecey P-Q formula
print "design of drainage water-way:wetted perimeter of river = %i m.provide 13 spans \
of 6 m each,separated by 12 piers each of 1.25 m thick."%(P);
t = 78.+15;
print "total 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;
print "height of barrels = %.2f m.provide recmath.tangular barrels 6 m wide and 2.5 m high.actual \
velocity through barrels = %.2f m/sec."%(hb,ac);

#design of canal waterway
print "design of canal waterway:Type 3 aqueduct is adopted.";
l1 = B-10;
l2 = (20-10)*3/2;
print "providing a splay 2:1 in expansion,length of contraction transition = %i m.providing\
a splay of 3:1 in expansion,length of expansion transition = %i m."%(l1,l2);
print 'In transition side slopes are warped from original slope of 1.5:1 to vertical.';

#design of levels of different sectionn
print "design of levels of different sectionn:at section 4-4:";
A = (B+1.5*D);        				#area
V = Q/A;              				#velocity of flow
ws = gl+D;            				#R.L of water surface
tel = ws+vh;
tel = round(tel*1000)/1000;
print "R.L of T.E.L = %.2f m. at section 3-3:"%(tel);
A = 10*D;                    				#area of trough
V = Q/A;                    				#velocity
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;
print "elevation of T.E.L = %.2f m.R.L of bed to maintain consmath.tant water depth = %.2f 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
tel = tel+hl;
rlw = tel-vh1;
rlb = rlw-D;
tel = round(tel*1000)/1000;
rlb = round(rlb*1000)/1000;
print "at section 2-2:R.L of T.E.L = %.2f m.R.L of bed to maintain consmath.tant water depth = %.2f 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;
print "at section 1-1:R.L of T.E.L = %.2f m.R.L of bed to maintain consmath.tant water depth = %.2f 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 = linspace(10,20,11)
print "design of contraction transition on the basis of chaturvedi formula:\nBx            x";
x = zeros_like(Bx)
for i in range(11):
x[i] = 15.45*(1-(10/Bx[i])**1.5);
x[i] = round(x[i]*100)/100;
print "%i         %.2f"%(Bx[i],x[i]);

#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);
print "design of expansion transition on the basis of chaturvedi formula:\nBx            x";
for i in range(11):
x[i] = 23.15*(1-(10/Bx[i])**1.5);
x[i] = round(x[i]*100)/100;
print "%i         %.2f"%(Bx[i],x[i]);

#design of trough
print "design of the trough:";
print "flumed water way of canal = 10 m.trough carrying canal will divide into two \
compartments each 5 m wide an dseparated by 0.3 m thick partiions.heigth of trough will be \
=  2 m.trough iss constructed umath.sing monolithic reinforced concrete.the outer and inner walls\
ca be kept 0.4 m thick.thus,outer width of trough  =  11.1 m.";

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;
print "head loss through syphon barrels = %.2f m.upstream H.F.L = %.2f 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;
print "uplift pressure on the roof = %.2f kN/square m.trough slab is 0.4 m thick and exert a downward load of 9.42 kN."%(up);
print "th ebalance of the uplift pressure has to be resisted by bending action of \
trough slab.so,reinforcement has to be provided at the top of the slab.";

#uplift on the floor of the barrel and its design
print "uplift on the floor of the barrel and its design: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;
print "static uplift on the floor = %.2f m."%(static);

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;
print "b) seepage head:total uplift = %.2f kN/square m.provide thickness of floor 0.8 m"%(tu);
bending = tu-17.58;
bending = round(bending*100)/100;
print "uplift to be resisted by bending action of floor = %.2f kN/square m."%(bending);

#design of cut-off and protection works for drainage floor
print "design 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);
print "R.L of bottom of u/s cut-off = %.2f m.R.L of bottom of d/s cut-off = %.2f m."%(bot_up,bot_down);
print "length of d/s protection consisting of 40 cm brick pritching = %.2f m.pitching is \
supported by toe wall 0.4 m wide and 1.5 m deep at its d/s end.length of d/s protection consisting\
of 0.4 cm brick pritching = %.2f m.pitching is supported by toe wall 0.4 m wide and\
1 m deep at its u/s end."%(l_down,l_down1);
design of drainage water-way:wetted perimeter of river = 95 m.provide 13 spans of 6 m each,separated by 12 piers each of 1.25 m thick.
total length of water-way = 93 m.
height of barrels = 2.56 m.provide recmath.tangular barrels 6 m wide and 2.5 m high.actual velocity through barrels = 20.51 m/sec.
design of canal waterway:Type 3 aqueduct is adopted.
providing a splay 2:1 in expansion,length of contraction transition = 10 m.providing a splay of 3:1 in expansion,length of expansion transition = 15 m.
In transition side slopes are warped from original slope of 1.5:1 to vertical.
design of levels of different sectionn:at section 4-4:
R.L of T.E.L = 161.56 m. at section 3-3:
elevation of T.E.L = 161.59 m.R.L of bed to maintain consmath.tant water depth = 159.95 m.
at section 2-2:R.L of T.E.L = 162.36 m.R.L of bed to maintain consmath.tant water depth = 160.72 m.
at section 1-1:R.L of T.E.L = 162.38 m.R.L of bed to maintain consmath.tant water depth = 160.88 m.
design of contraction transition on the basis of chaturvedi formula:
Bx            x
10         0.00
11         2.06
12         3.70
13         5.03
14         6.12
15         7.04
16         7.82
17         8.48
18         9.05
19         9.55
20         9.99
design of expansion transition on the basis of chaturvedi formula:
Bx            x
10         0.00
11         3.08
12         5.54
13         7.53
14         9.17
15         10.55
16         11.71
17         12.71
18         13.56
19         14.31
20         14.97
design of the trough:
flumed water way of canal = 10 m.trough carrying canal will divide into two compartments each 5 m wide an dseparated by 0.3 m thick partiions.heigth of trough will be  =  2 m.trough iss constructed umath.sing monolithic reinforced concrete.the outer and inner walls  ca be kept 0.4 m thick.thus,outer width of trough  =  11.1 m.
head loss through syphon barrels = 0.33 m.upstream H.F.L = 160.83 m.
uplift pressure on the roof = 11.01 kN/square m.trough slab is 0.4 m thick and exert a downward load of 9.42 kN.
th ebalance of the uplift pressure has to be resisted by bending action of trough slab.so,reinforcement has to be provided at the top of the slab.
uplift on the floor of the barrel and its design:a static head:
static uplift on the floor = 1.70 m.
b) seepage head:total uplift = 32.41 kN/square m.provide thickness of floor 0.8 m
uplift to be resisted by bending action of floor = 14.83 kN/square m.
design of cut-off and protection works for drainage floor:
R.L of bottom of u/s cut-off = 160.13 m.R.L of bottom of d/s cut-off = 159.79 m.
length of d/s protection consisting of 40 cm brick pritching = -4.00 m.pitching is supported by toe wall 0.4 m wide and 1.5 m deep at its d/s end.length of d/s protection consisting of 0.4 cm brick pritching = -4.00 m.pitching is supported by toe wall 0.4 m wide and  1 m deep at its u/s end.