# Chapter 16 : WATERLOGGING AND CANAL LINING¶

## Example 16.1 pg : 764¶

In :
import math
from numpy import roots
from sympy import acot

#design a trapezoidal concrete lined channel

#Given
Q = 100.;               				#discharge
S = 25./100000;           				#bed slope
N = 0.016;             				#rogsity coefficient
s = 1.5;               				#side slope
V = 1.5;               				#limiting velocity

# Calculations
#umath.sing manning's equation  V = (R**2/3*S**1/2)/N;
R = (V*N/(S**0.5))**(1.5);      				#hydraulic mean depth

#for s = 1.5;
theta = acot(1.5);
A = Q/V;
P = A/R;
#umath.sing equation of area and perimeter of trapezium
#perimeter of trapezium = b+2d(theta+cot(theta));
#area of trapezium = bd+d**2(theta+cot(theta));
#we get
y = [1,-17.1,31.9]
d = roots(y);
#we get D = 14.968917 and 2.1310826.
#taking d = 2.1310826;
b = P-4.18*d;
b = round(b*100)/100;
d = round(d*100)/100;
print "required bed width = %.2f m."%(b);
print "required bed depth = %.2f m"%(d);

required bed width = 26.74 m.
required bed depth = 2.13 m


## Example 16.2 pg : 765¶

In :
import math

#design a trapezoidal concrete lined channel

#Given
Q = 100.;               				#discharge
S = 25./100000;           				#bed slope
N = 0.016;             				#rogsity coefficient
s = 1.5;               				#side slope
r = 8.;                 				#b/d ratio

# Calculations
#umath.sing manning equation V = (R**2/3*S**1/2)/N;
#Perimeter = A/R
#V = Q/A and on simplification we get
d = ((101/10.09)*(12.18/10.09)**(2./3))**(3./8);
b = r*d;
b = round(b);
d = round(d*100)/100;

# Results
print "required bed width = %.2f m."%(b);
print "required bed depth = %.2f m"%(d);

required bed width = 20.00 m.
required bed depth = 2.49 m


## Example 16.3 pg : 766¶

In :
import math
from sympy import acot

#design a concrete lined channel

#Given
Q = 45.;              				#discharge
S = 1./10000;         				#bed slope
s = 5./4;             				#side slope
N = 0.018;           				#rogosity coefficient(manning N)

#channel is assumed to be of triangular section
theta = acot(s);
#umath.sing manning equation V = (R**2/3*S**1/2)/N;
#V = Q/A;
#perimeter of trapezium = b+2d(theta+cot(theta));
#area of trapezium = bd+d**2(theta+cot(theta));
#we get
d = (Q*2.86/1.925)**(3./8);
d = round(d*100)/100;

# Results
print "required depth of triangular channel = %.2f m"%(d);

required depth of triangular channel = 4.84 m


## Example 16.4 pg : 767¶

In :
import math

#design a concrete lined channel of trapezoidal section

#Given
Q = 250.;             				#discharge
S = 1./6000;          				#bed slope
s = 1.5;             				#side slope
d = 3.;               				#limiting depth
N = 0.015;           				#rogosity coefficient

#umath.sing Perimeter = A/R;
#perimeter of trapezium = b+2d(theta+cot(theta));
#area of trapezium = bd+d**2(theta+cot(theta));
#Q = A*V; and on simplification
#we get
#(3b+18.81)**5/3/(b+12.54)**2/3 = 290.47;
#solving it by trial and error method we get
b = 44.6;

# Results
print "required bed width = %.2f m."%(b);
print "required bed depth = %i m"%(d);

required bed width = 44.60 m.
required bed depth = 3 m


## Example 16.5 pg : 767¶

In :
import math

#Given
H = 10.;             				#depth of impervious stratum from top soil
D = 1.8;           				#position of drain below top soil surface
Hw = 1.5;          				#depth of highest point of water
k = 1.e-4;         				#permeability consmath.tant

# Calculations
#math.since water has to be removed in 24 hours
b = H-Hw;
a = H-D;
L = (4*k*(b**2-a**2)*100*24*3600/0.8)**0.5;

# Results
print "drains should be spaced at %i m c/c."%(L);

drains should be spaced at 147 m c/c.


## Example 16.6 pg : 768¶

In :
import math

#Given
L = 30.;         				#spacing between drans
Q = 4.e-6;       				#discharge
a = 8.;
b = 8.3;

# Calculations
k = 1000000*Q*L/(4*(b**2-a**2));
k = round(k*100)/100;

# Results
print "permeability coefficient = %.2fD-6 m/sec."%(k);

permeability coefficient = 6.13D-6 m/sec.


## Example 16.7 pg : 768¶

In :
import math

#Given
L = 50.;         				#spacing between drains
k = 1.e-5;       				#permeability coefficient
a = 10.;
b = 10.3;

# Calculations
Q = 4*k*(b**2-a**2)/L;
Pav = Q*24*3600*100*100/L;

# Results
print "annual average rainfall = %i cm"%(Pav);

annual average rainfall = 84 cm


## Example 16.8 pg : 768¶

In :
import math

#Given
r1 = 2;        				#ka/kb
r2 = 1./1.5;    				#La/Lb
r3 = 5./6;      				#(b**2-a**2)a/((b**2-a**2)b)

# Calculations
Rq = r1*r3/r2;
Rp = Rq/r2;

# Results
print "ratio of discharge at A and B = %.2f."%(Rq);
print "ratio of average rainfall at A and B = %.2f."%(Rp);

ratio of discharge at A and B = 2.50.
ratio of average rainfall at A and B = 3.75.


## Example 16.9 pg : 775¶

In :
import math

#decide whether it is economically feasible to provide canal lining

#Given
li = 2.5;               				#seepage loss for lined channel
p1 = 25.;                				#wetted perimeter for lined channel
t = 12.;                 				#thickness of concrete lining
lf = 0.02;              				#seepage loss for unlined channel
p2 = 20.;                				#wetted perimeter for unlined channel

#assume 1 km length of canal
#annual benifit

#(1).seepage
A1 = p1*1000;             				#area of wetted perimeter
li = li*p1/1000;          				#seepage loss
A2 = p2*1000;            				#area of wetted perimmeter for unlined channel
lf = p2*lf/1000;        				#seepage loss for unlined channel
s = li-lf;               				#saving in water loss
a1 = s*p1*100000;        				#annual revenue saved

#(2)maintainence
a2 = 0.4*25000;         				#saving in maintainance math.cost
ts = a1+a2;             				#total annual benifit

#annual math.cost
A1 = p2*1000;           				#area of lining for unlinrd canal
C = 100*A1;             				#math.cost of lining
#interest rate is 6%
i = 0.06;
N = 50;
a = (C*i*(i+1)**N)/((1+i)**N-1);  				#annual math.cost of lining or capital recovery factor
bcr = ts/a;                 				#benifit math.cost ratio
bcr = round(bcr*1000)/1000;
print "Benifit math.cost ratio = %.2f."%(bcr);
#as bcr>1
print " ;Since it is more than 1.Hence, it is economically justifiable. ";

Benifit math.cost ratio = 1.30.
;Since it is more than 1.Hence, it is economically justifiable.


## Example 16.10 pg : 776¶

In :
import math

#Given
Ecd = 20;           				#electrical conductivity of drainage water
Eci = 1.5;        				#m mho/cm
Dc = 55.5;        				#consumptive use

# Calculations
Lr = Eci/Ecd;
D = Dc/(1-Lr);

# Results
print "required depth of water to be applied = %i mm."%(D);

required depth of water to be applied = 60 mm.


## Example 16.11 pg : 776¶

In :
import math

#Given
Eci = 1.4;        				# m mho/cm
Ece = 11.;         				#saturated extract of soil
Dc = 85.;          				#consumptive use requirement of crop

# Calculations
#let us assume Ecd = 2Ece
Lr = Eci/(2*Ece);
Di = Dc/(1-Lr);
Di = round(Di*10)/10;

# Results
print "required depth of water to be applied = %.2f mm."%(Di);

required depth of water to be applied = 90.80 mm.


## Example 16.12 pg : 776¶

In :
import math
#percentage of earth work is saved in lined section

#Given
s = 1.5;        				#side slope
Q = 15.;         				#discharge
S = 1./4000;     				#bed slope
Nl = 0.014;      				#manning n for lined channel
Nu = 0.028;      				#manning n for ulined channel
fb = 0.75;         				#free board

#considering the perimeter of trapezoidal section
#taking minimum perimeter for given area
#i.e dP/dD = 0
#we get
#A = 2.1D**2; R = D/2; and P = 4.2D

#for linrd channel
#Q = AR**(2/3)*S**0.5
#substituting above values we get
D = (10.0396)**(3./8);
B = 0.6*D;
R = D/2;
tau = 9.81*R*S*1000;
tau = round(tau*1000)/1000;
print "for lined canal:";
print "average boundary shear stress = %.2f N/square m."%(tau);
Dc = D+fb;            				#total depth of cutting
A1 = (B+1.5*Dc)*Dc;

#for unlined channel
#Q = AR**(2/3)*S**0.5
#substituting above values we get
D = 3.08;
B = 0.6*D;
R = D/2;
tau = 9.81*R*S*1000;
tau = round(tau*100)/100;
print "for unlined canal:";
print "average boundary shear stress = %.2f N/square m."%(tau);
Dc = D+fb;            				#total depth of cutting
A2 = (B+1.5*Dc)*Dc;
per = (A2-A1)*100/A2;
per = round(per*100)/100;
print "percent saving of earth = %.2f percent."%(per);

for lined canal:
average boundary shear stress = 2.91 N/square m.
for unlined canal:
average boundary shear stress = 3.78 N/square m.
percent saving of earth = 34.32 percent.


## Example 16.13 pg : 778¶

In :
import math
from numpy import roots

#design a lined canal

#Given
Q = 100.;              				#discharge
S = 1./2500;           				#bed slope
V = 2.;                				#maximum permissible velocity
n = 0.013;            				#manning n
s = 1.25;             				#side slope

# Calculations
A = Q/V;
#from manning formula  V = (R**2/3*S**1/2)/N;
R = (V*n/S**0.5)**1.5;
P = A/R;

#now umath.sing the equation of area and perimeter of trapezoid
#area = D(B+2.5D)
#perimeter = B+3.2D;
#we get
y = [1.95,-33.73,50]
D = roots(y);
#we get D = 15.660087 and 1.6373489
#taking D = 1.6373489;
B = P-3.2*D;
B = round(B*10)/10;
D = round(D*100)/100;

# Results
print "required bed width = %.2f m."%(B);
print "required bed depth = %.2f m"%(D);

required bed width = 28.50 m.
required bed depth = 1.64 m


## Example 16.14 pg : 778¶

In :
import math

#Given
B = 5.;         				#bed width
D = 2.;         				#bed depth
S = 1./1600;    				#bed slope
n = 0.015;     				#manning n

# Calculations and Results
A = B+2*D;     				#area of lining
#let B1 and D1 be new  width and depth of bed
#for getting maximum discharge we diffrentiate Q and equating it to zero
#Q = S**0.5*B1D1**5/3/n
#we get
D1 = 45./16;
B1 = 9-2*D1;
Q1 = S**0.5*B1*D1**5/3/n;
D1 = round(D1*10000)/10000;
print "new width of bed = %.2f m."%(B1);
print "new depth of bed = %.2f m."%(D1);
print " maximum discharge = %.2f cumec."%(Q1);
R = D;
V = R**(2./3)*S**0.5/n;
F = V/(9.81*D)**0.5;       				#froud number
R = D1;
V = R**(2./3)*S**0.5/n;
F = V/(9.81*D1)**0.5;       				#froud number
print "Froud number is less than 1 in both case.Hence flow doesnot change from sub-critical to super critical.";

new width of bed = 3.38 m.
new depth of bed = 2.81 m.
maximum discharge = 329.96 cumec.
Froud number is less than 1 in both case.Hence flow doesnot change from sub-critical to super critical.


## Example 16.15 pg : 779¶

In :
import math
from sympy import acot

#area to be irrigated

#Given
B = 5.;          				#bed width
D = 2.5;        				#bed depth
s = 1.5;        				#side slope
S = 1./1000;     				#bed slope
n = 0.016;       				#manning n
k = 10.;         				#kor period
d = 150.;        				#field irrigation requirement

# Calculations
theta = acot(s);
A = B*D+D**2*(theta+1/math.tan(theta));
P = B+2*D*(theta+1/math.tan(theta));
R = A/P;
Q = A*R**(2./3)*S**0.5/n;
V = Q*k*24*3600;   				#volum of water supply by channel
A = V*10/(d*10000);
Q = round(Q*100)/100;
A = round(A)*100;

# Results
print "maximum carrying capacity of canal = %.2f cumec."%(Q);
print "Area to be irrigated = %.2f hectares."%(A);

maximum carrying capacity of canal = 70.65 cumec.
Area to be irrigated = 40700.00 hectares.