Chapter 2 : METHODS OF IRRIGATION

Example 2.1 pg : 21

In [1]:
import math 


#Given
Q = 0.0108   #discharge through well
y = 0.075    #average depth of flow
I = 0.05     #average infiltration rate
A = 0.1      #area to cover

# Calculations
t = (60*2.303*y*math.log10(Q/(Q-I*A)))/I

# Results
print "Time required to cover given area = %.f minutes."%(t)

Time required to cover given area = 56 minutes.

Example 2.2 pg : 21

In [2]:
#Given
Q = 0.0108   #discharge through well
y = 0.075   #average depth of flow
I = 0.05   #average infiltration rate
A = 0.1   #area to cover

# Calculations
Amax = Q/I;

# Results
print "Maximum area that can be irrigated  = %.2f hectare."%(Amax);

Maximum area that can be irrigated  = 0.22 hectare.

Example 2.3 pg : 22

In [8]:
import math 



#time of water application
#optimum length of each border strip
#dischrge for each border strip

#Given
d = 0.05;								#depth of root zone
I = 1.25E-5;								#average infiltration rate
s = 0.0035								#slope of border strip
t = d/(I*3600);

# Calculations and Results
t = round(t*1000)/1000;
print "Time of water application = %.2f hours."%(t);

#Part (a)
q = 2E-3;								#discharge entering water source
qdash = q*100**2*60;
n = 0.55425-(0.0001386*qdash);
yo = (n*q/(s**0.5))**0.6;
y = 0.665*yo;
L = (q/I)*(1-math.e**(-d/y));
L = round(10*L)/10;
print "Part a:";
print "Optimum length of each border strip = %.2f m."%(L);

#Part (b)
Lgiven = 150								#given value of length
#First Trial
q = 3E-3;
qdash = q*100**2*60;
n = 0.55425-(0.0001386*qdash);
yo = (n*q/(s**0.5))**0.6;
y = 0.665*yo;
L = (q/I)*(1-math.e**(-d/y));
#second trial
q = 3.15E-3;
qdash = q*100**2*60;
n = 0.55425-(0.0001386*qdash);
yo = (n*q/(s**0.5))**0.6;
y = 0.665*yo;
L = (q/I)*(1-math.e**(-d/y));
q = 9*Lgiven*q*1000/L;
q = round(q*10)/10;
print "Part b:";
print "Discharge for each border strip = %.2f lps."%(q);

Time of water application = 1.11 hours.
Part a:
Optimum length of each border strip = 101.90 m.
Part b:
Discharge for each border strip = 28.20 lps.

Example 2.4 pg : 26

In [9]:
import math 




#deep percolation loss
#water application efficiency and time of irrigation.

#Given
B = 12.;				#breadth of bamath.sin
L = 36.				#length of bamath.sin
d = 70.				#depth of irrigation
Ic = 70.				#cumulative infiltration
kdash = 9;
ndash = 0.42;
#Part (a) 
a = 5;
b = 0.6;
q = 1.5;				#stream size

# Calculations and Results
Q = (q*B)/1000;
tl = (L/a)**(1/b);
td = (Ic/kdash)**(1/ndash);
T = tl+td;
p = (1-(td/T)**(ndash))*100;
eita = (1-p/100)*100;
Tdash = (d*L*B)/(10*eita*Q*60);
p = round(p*100)/100;
eita = round(eita*100)/100;
Tdash = round(Tdash*10)/10;
print "Part a:"
print "Deep percolation loss =  %.2f percent."%(p);
print "Water application efficiency =  %.2f percent."%(eita);
print "Time of irrigation =  %.2f minutes."%(Tdash);
#part (b)
a = 8;
b = 0.6;
q = 3;
Q = (q*B)/1000;
tl = (L/a)**(1/b);
td = (Ic/kdash)**(1/ndash);
T = tl+td;
p = (1-(td/T)**(ndash))*100;
eita = (1-p/100)*100;
Tdash = (d*L*B)/(10*eita*Q*60);
p = round(p*100)/100;
eita = round(eita*100)/100;
Tdash = round(Tdash*10)/10;

print "Part b:"
print "Deep percolation loss =  %.2f percent."%(p);
print "Water application efficiency =  %.2f percent."%(eita);
print "Time of irrigation =  %.2f minutes."%(Tdash);

Part a:
Deep percolation loss =  7.47 percent.
Water application efficiency =  92.53 percent.
Time of irrigation =  30.30 minutes.
Part b:
Deep percolation loss =  3.66 percent.
Water application efficiency =  96.34 percent.
Time of irrigation =  14.50 minutes.

Example 2.5 pg : 31

In [3]:
import math 
from numpy import zeros


#given
d = 37.5				#crop water requirement
W = 1.				#furrow spacing
L = 120.				#length of furrow
n = -0.49;
k = 38.;
Ttotal = 143.;				#Total time of irrigation
A = [0 ,23, 52 ,88, 127]				#given values of time of advance
B = zeros(5)
C = zeros(5)
D = zeros(5)
E = zeros(5)

# Calculations
for i in range(5):				#loop to find respective values of time of ponding
    B[i] = 143-A[i] 

for j in range(5):				#loop to find respective furrow infiltration
    C[j] = B[j]**(n)*k;

for K in range(4):				#loop to find respective average infiltration
    D[K] = (C[K]+C[K+1])/2;


E[0] = D[0];
for l in range(1,4):				#loop to determine cumulative infiltration
    E[l] = D[l]+E[l-1];

I = E[3];

T = (30*d*W*(n+1)/k)**(1/(n+1));
dav = ((24.5*Ttotal)+(I*(T-Ttotal)))/L;
q = ((120*37.5)-(24.5*143))/62;
T = round(T);
dav = round(dav*10)/10;
q = round(q*100)/100;
I = round(I*100)/100;

# Results
print "Maximum size of cut-back stream = %.2f lpm."%(I);
print "Minimum size of cut-back stream = %.2f lpm."%(q);
print "Time required for putting 37.5mm depth of water = %.2f minutes."%(T);
print "Average depth of water required = %.2f mm."%(dav);

Maximum size of cut-back stream = 19.69 lpm.
Minimum size of cut-back stream = 16.07 lpm.
Time required for putting 37.5mm depth of water = 205.00 minutes.
Average depth of water required = 39.40 mm.

Example 2.6 pg : 32

In [4]:
#Given
L = 100.;				#length of furrow
W = 1.;				#furrow spacing
s = 0.3				#longitudnal slope of furrow
t1 = 80.				#initial time flow of  stream
t2 = 35.				#final time flow of stream

# Calculations
qm = 0.6/s;
q = qm*0.4;
dav = ((q*t2*60)+(2*t1*60))/100;

# Results
print "Average depth of water applied = %.2f mm."%(dav);

Average depth of water applied = 112.80 mm.

Example 2.7 pg : 40

In [5]:
import math 


#Given
Q = 0.0072;				#discharge through well
y = 0.1;				#average depth of flow
I = 0.05				#infiltration capacity of soil
A = 0.04				#area of land

# Calculations
t = (2.303*y*60/I)*math.log10(Q/(Q-I*A));
Amax = Q/I;
t = round(t*100)/100;

# Results
print "Time required to irrigate = %.2f minutes."%(t);
print "Maximum area that can be irrigated = %.2f ha."%(Amax);

Time required to irrigate = 39.06 minutes.
Maximum area that can be irrigated = 0.14 ha.