# Chapter 15 : IRRIGATION CHANNEL 2 DESIGN PROCEDURE¶

## Example 15.1 pg : 739¶

In [1]:
import math
#design the distributory umath.sing Laecey theory

#Given
f = 0.85;           				#silt factor
AR = 3600.;          				#area for rabi
AK = 1400.;          				#area for kharif
delta_r = 0.135;    				#kor depth for rabi
delta_k = 0.19;     				#kor depth for kharif
tr = 4.;             				#kor period for rabi
tk = 2.5;           				#kor period for kharif

# Calculations
Du_r = 8.64*tr*7/delta_r;     				#duty for rabi
Du_k = 8.64*tk*7/delta_k;    				#duty for kharif
q_r = AR/Du_r;               				#discharge for rabi
q_k = AK/Du_k;               				#discharge for kharif
Q = q_r;                   				#math.since q_r>q_k
V = (Q*f**2/144)**(1./6);
A = Q/V;
P = 4.75*(Q)**0.5;
D = (P-(P**2-6.944*A)**0.5)/3.472;
S = f**(5./3)/(3340*Q**(1./6));
P = round(P*100)/100;
D = round(D*100)/100;

# Results
print "Bed slope = %.2e."%(S);
print "Perimeter of channel section = %.2f m."%(P);
print "Depth of channel section = %.2f m."%(D);

Bed slope = 2.03e-04.
Perimeter of channel section = 6.73 m.
Depth of channel section = 0.81 m.


## Example 15.2 pg : 740¶

In [2]:
import math
#design an irrigation channel in alluvial soil by Laecy's theory

#Given
Q = 15.;              				#Full supply discharge
f = 1.;               				#silt factor
s = 1./2;             				#side slope of channel

# Calculations
#from Laecey regime channel (Fig.15.4(b)) B and D is obtained as;
B = 15.1;
D = 1.38;
#also from Fig.15.5 we get slope as
S = 0.19/1000;

# Results
print "Width of channel section = %.2f m."%(B);
print "Depth of channel section = %.2f m."%(D);
print "Bed slope = %.2e."%(S);

Width of channel section = 15.10 m.
Depth of channel section = 1.38 m.
Bed slope = 1.90e-04.


## Example 15.3 pg : 740¶

In [3]:
import math
from numpy import array,zeros,linspace,float64

#design and prepare the longitudnal section;schedule of area statistics and channel dimension of irrigation channel

#Given
dl = 157.7;               				#datum level
fsl = 157.;               				#full supply level of parent channel
bl = 156.;                				#bed level of parent channel
kor_r = 4.;               				#kor period of rabi
kor_k = 2.5;             				#kor period of kharif
kord_r = 13.4;           				#kor depth of rabi
kord_k = 19.;             				#kor depth of kharif
s = 0.5;                 				#side slope
m = 1.;                   				#critical velocity ratio
N = 0.0225;              				#Kutter n
qo_r = 8.64*7*kor_r*100/kord_r;  				#outlet discharge for rabi(calculation is wrong in book)
qo_k = 8.64*7*kor_k*100/kord_k;  				#outlet discharge for kharif(calculation is wrong in  book)
ca = 16000.;               				#culturable commanded area
Ir = 0.3;                				#intensity of irrigation in rabi
Ik = 0.125;              				#intensity of irrigation in rabi

# Calculations and Results
Ar = Ir*ca;              				#area under rabi
Ak = ca*Ik;              				#area under kharif
q_r = Ar/qo_r;
q_k = Ak/qo_k;
q_r = round(q_r*100)/100;
q_k = round(q_k*100)/100;
print "discharge neede for rabi crop = %.2f cumecs."%(q_r);
print "discharge neede for kharif crop = %.2f cumecs."%(q_k);
print "outlet discharge factor adopted = %i hectares per cumecs."%(qo_r);
#at km 5
ca = 8000;          				#culturable area
Ar = Ir*ca;              				#area under rabi
q_r = Ar/qo_r;
l = 0.5               				#total loss after 5 km
q = q_r+l;             				#total discharge
dq = 1.1*q;            				#desigm discharge
S = 1./4000;           				#slope
B = array([5.5, 4.9, 4.55]);   				#Bed width
D = array([0.73, 0.79, 0.84]); 				#water depth
Vo = array([0.448, 0.472, 0.488]);  				#critical velocity
A = zeros(3)
V = zeros(3)
m = zeros(3)
print "Bed width     water depth      area        velocity      critical velocity    C.V.R";
for i in range(3):
A[i] = B[i]*D[i]+D[i]**2/2;
V[i] = dq/A[i];
m[i] = V[i]/Vo[i];
A[i] = round(A[i]*100)/100;
V[i] = round(V[i]*1000)/1000;
m[i] = round(m[i]*100)/100;
print "%.2f       %.2f      %.2f      %.2f       %.2f           %.2f"%(B[i],D[i],A[i],V[i],Vo[i],m[i]);

B = 4.55;
D = 0.84;
print "hence take B = %.2f .; D = %.2f m."%(B,D);
#at km 4
q = round(q*100)/100;
print "discharge at 5 km = %.2f cumecs."%(q);
ca = 10000;          				#culturable area
Ar = Ir*ca;              				#area under rabi
q_r = Ar/qo_r;
l = 0.5               				#total loss below 5 km
P = B+D*5**0.5;        				#wetted perimeter
l1 = P*1000*2/1000000;  				#loss between 5 km and 4km
l2 = l1+l;
q = q_r+l2;
dq = 1.1*q;
q = round(q*1000)/1000;
print "discharge at 4 km  = %.2f cumecs"%(q);
print "other discharge are calculated and are tabulated as:";
x = linspace(1,5,6)
A1 = array([4800, 4200, 3600, 3300, 3000, 2400],dtype=float64);
A2 = array([2000, 1750, 1500, 1375, 1250, 1000],dtype=float64);
S = array([22.5, 22.5, 22.5, 24, 24, 25]);
B = array([5.5, 5.2, 4.85, 4.7, 4.55, 4.55]);
D = array([1.04, 1.007, 0.975, 0.945, 0.915, 0.840]);
dq = array([3.56, 3.17, 2.8, 2.6, 2.4, 2.02]);
V = array([0.570, 0.555, 0.538, 0.530, 0.521, 0.484]);
m = array([1.015, 1, 1, 1, 1, 0.992]);
print "Below km     area to irrigate rabi   area to irrigate kharif  bed slope  bed width water depth design discharge  velocity C.V.R";
for i in range(6):
print "%8i         %i       %i        %.2f           %.2f            %.2f\
%.2f           %.2f        %.2f"%(x[i],A1[i],A2[i],S[i],B[i],D[i],dq[i],V[i],m[i]);

discharge neede for rabi crop = 2.66 cumecs.
discharge neede for kharif crop = 2.51 cumecs.
outlet discharge factor adopted = 1805 hectares per cumecs.
Bed width     water depth      area        velocity      critical velocity    C.V.R
5.50       0.73      4.28      0.47       0.45           1.05
4.90       0.79      4.18      0.48       0.47           1.02
4.55       0.84      4.17      0.48       0.49           0.99
hence take B = 4.55 .; D = 0.84 m.
discharge at 5 km = 1.83 cumecs.
discharge at 4 km  = 2.17 cumecs
other discharge are calculated and are tabulated as:
Below km     area to irrigate rabi   area to irrigate kharif  bed slope  bed width water depth design discharge  velocity C.V.R
1         4800       2000        22.50           5.50            1.04                3.56           0.57        1.01
1         4200       1750        22.50           5.20            1.01                3.17           0.56        1.00
2         3600       1500        22.50           4.85            0.97                2.80           0.54        1.00
3         3300       1375        24.00           4.70            0.94                2.60           0.53        1.00
4         3000       1250        24.00           4.55            0.92                2.40           0.52        1.00
5         2400       1000        25.00           4.55            0.84                2.02           0.48        0.99


## Example 15.4 pg : 744¶

In [4]:
import math
from numpy import roots

#Given
B = 5.;                  				#bed width
t = 2.;                  				#top width of banks
h = 2.92;               				#heigth of banks from bed
n = 1.5;

#sectional area of digging = sectional area of two banks
#By+zy**2 = 2(h-y)+2n(h-y)**2
#substituting the values and on simplificatio we get
s = [1,-13.26,18.59]
y = roots(s)[1];
#from this we get y = 11.666556 and 1.5934436.
#taking  y = 1.5934436;
y = round(y*10)/10;

# Results
print "economical depth of cutting = %.2f m."%(y);

economical depth of cutting = 1.60 m.