# Chapter 3 : WATER REQUIREMENTS OF CROPS¶

## Example 3.1 pg : 44¶

In [1]:
#Given
Na = 24.;                                         				#concentration of sodium ion
Ca = 3.6;                                        				#concentration of calcium ion
Mg = 2.;                                         				#concentration of magnesium ion
EC = 180.;                                       				#electrical conductivity

# Calculations
SAR = Na/(((Ca+Mg)/2)**(0.5));                   				#Sodium absorption ratio
SAR = round(SAR*100)/100;

# Results
print "SAR = %.2f."%(SAR);
print "Water falls under S2 class.";    				#from table 3.2
print "For EC = 180";
print "water falls under C1 class.";    				#from table 3.1
print "Water is medium sodium and low saline water.";

SAR = 14.34.
Water falls under S2 class.
For EC = 180
water falls under C1 class.
Water is medium sodium and low saline water.


## Example 3.2 pg : 52¶

In [2]:
#Given
gammad = 15.;        				#dry weigth of soil
gammaw = 9.81;      				#unit weigth of water
Fc = 0.3;           				#field capacity
pwp = 0.08;         				#permanent wilting point
d = 0.8;            				#root zone depth

# Calculations
d1 = gammad*Fc*1000/gammaw;
d2 = gammad*pwp*1000/gammaw;
d3 = gammad*d*(Fc-pwp)*1000/gammaw;

# Results
print "Depth of moisture in root zone at field capacity = %.f mm/m."%(d1);
print "Depth of moisture in root zone at permanent wilting point = %.f mm/m."%(d2);
print "Depth of moisture available in root zone = %.f mm/m."%(d3);

Depth of moisture in root zone at field capacity = 459 mm/m.
Depth of moisture in root zone at permanent wilting point = 122 mm/m.
Depth of moisture available in root zone = 269 mm/m.


## Example 3.3 pg : 52¶

In [3]:
#Given
gammad = 15.3;      				#dry weigth of soil
gammaw = 9.81;      				#unit weigth of water
Fc = 0.15;          				#field capacity
Mc = 0.08;          				#moisture content before irrigation
D = 60.;             				#Depth of water applied

# Calculations
d = (gammaw*D)/(gammad*(Fc-Mc));

# Results
print "Depth upto which soil profile is wetted = %.f mm."%(d);

Depth upto which soil profile is wetted = 550 mm.


## Example 3.4 pg :53¶

In [4]:
#Given
Sg = 1.6;    				#Apparent specific gravity
Fc = 0.2;        				#Field capacity
M1 = 150.;        				#mass of sample soil
M2 = 136.;        				#mass of sample after drying
d = 0.9;         				#depth of soil to be irrigated

# Calculations
Mc = (M1-M2)/M2;
D = Sg*d*1000*(Fc-Mc);

# Results
print "Depth of water required to irrigate the soil = %.f mm."%(D);

Depth of water required to irrigate the soil = 140 mm.


## Example 3.5 pg : 53¶

In [5]:


#Given
d = 2.;            				#root zone depth
Wc = 0.05;        				#existing water content
gammad = 15;      				#dry density of soil
gammaw = 9.81;    				#unit weigth of water
Vw = 500.          				#water applied to the soil
Wl = 0.1;         				#water loss
A = 1000.;         				#area of plot

# Calculations
Vu = Vw*0.9;      				#volume of water used in soil
Wu = Vu*gammaw;   				#weigth of water used in soil
Ws = A*d*gammad;  				#total dry weigth of soil
Wa = Wu*100/Ws;   				#percent water added
Fc = Wc*100+Wa;
Fc = round(Fc*100)/100;

# Results
print "The Field Capacity of soil is = %.2f percent."%(Fc);

The Field Capacity of soil is = 19.72 percent.


## Example 3.6 pg : 53¶

In [6]:

#Given
Fc = 0.22;                                             				#Field Capacity
wc = 0.1;                                             				#wilting coefficient
gammad = 15.;                                          				#dry unit weigth of soil
gammaw = 9.81;                                        				#unit wiegth of water
d = 0.7;                                              				#root zone depth
w = 0.14;                                            				#falled moisture content
E = 0.75;                                           				#water application efficiency

# Calculations
SC = gammad*d*(Fc-wc)*100/gammaw;
D = gammad*d*(Fc-w)*1000/gammaw;
FIR = D/E;                                         				#Field irrigation requirement
SC = round(SC*10)/10;
D = round(D);
FIR = round(FIR)+1;

# Results
print "Maximum storage capacity of soil = %.2f cm."%(SC);
print "Water depth required to be applied = %.2f mm"%(D);
print "Field Irrigation Requirement = %.2f mm"%(FIR);

Maximum storage capacity of soil = 12.80 cm.
Water depth required to be applied = 86.00 mm
Field Irrigation Requirement = 115.00 mm


## Example 3.7 pg : 55¶

In [7]:

#Given
Fc = 0.27;           				#Field capacity
pwp = 0.14;         				#permanent wilting point
gammad = 15.;        				#dry density of soil
gammaw = 9.81;      				#unit weigth of water
d = 0.75;          				#effective depth of root zone
Du = 11.;          				#daily consumptive use of water

# Calculations
Am = Fc-pwp;      				#Available moisture
#let readily available moisture be 80 percent of available moisture
RAm = 0.8*Am;
Mo = Fc-RAm;
D = gammad*d*(Fc-Mo)*100/gammaw;
WF = D*10/Du;

# Results
print "Watering Frequency = %i days."%(WF);

Watering Frequency = 10 days.


## Example 3.8 pg : 56¶

In [8]:

#Given
Fc = 0.22;     				#Field capacity
Sg = 1.56;     				#Apparent specific gravity
d = 0.6;       				#root zone depth
#irrigation is started when 70 percent of moisture is used
l = 250.;        				#length of field
b = 40.;         				#width of field
q = 20.;         				#Discharge

# Calculations
m = (1-0.7)*Fc;
D = Sg*d*(Fc-m)*1000;
A = l*b;
t = A*D/(q*3600);
D = round(D);
t = round(t);

# Results
print "Net depth of irrigation water required = %.2f mm."%(D);
print "Time required to irrigate field = %.2f hours."%(t);

Net depth of irrigation water required = 144.00 mm.
Time required to irrigate field = 20.00 hours.


## Example 3.9 pg : 64¶

In [9]:

#Given
B = 110.;     				#Base period
D = 1400.;    				#Duty of water

# Calculations
delta = 8.64*B*100/D;
delta = round(delta);

# Results
print "Delta for crop is = %.2f cm."%(delta);

Delta for crop is = 68.00 cm.


## Example 3.10 pg : 64¶

In [10]:

#Given
B = 120.;      				#Base period
delta = 92.;   				#total depth requirement of crop

# Calculations
D = 8.64*B*100/delta;

# Results
print "Duty of water = %.2f hectares/cumec."%(D);

Duty of water = 1126.96 hectares/cumec.


## Example 3.11 pg : 65¶

In [11]:

#Given
Cr = 2.;            				#crop ratio
A = 80000.;         				#Area of field
CI = 85.;           				#percent field culturable irrigable
IK = 30.;           				#irrigation intensity during kharif season
IR = 60.;           				#irrigation intensity for rabi season
DuK = 800.;           				#Duty of water for kharif season
DuR = 1700.;          				#Duty of water for rabi season

# Calculations
CIA = A*CI/100;    				#Culturable irrigable area
AK = CIA*IK/100;   				#Area under kharif season
AR = CIA*IR/100;   				#Area under rabi season
DK = AK/DuK;
DR = AR/DuR;

# Results
print "Dischage required at head of canal during Kharif season = %.2f cumecs."%(DK);
print "Dischage required at head of canal during Rabi season = %.2f cumecs."%(DR);
print "Water requirement during kharif is greater than during rabi season";
print "Hence,canal should be designed to carry discharge of %.2f cumecs."%(DK);

Dischage required at head of canal during Kharif season = 25.50 cumecs.
Dischage required at head of canal during Rabi season = 24.00 cumecs.
Water requirement during kharif is greater than during rabi season
Hence,canal should be designed to carry discharge of 25.50 cumecs.


## Example 3.12 pg : 65¶

In [12]:
#Given
CA = 2600.;       				#culturable area
IS = 20.;         				#irrigation intensity for sugarcane
IR = 40.;         				#irrigation intensity for rice
DuS = 750.;         				#Duty of water for sugarcane
DuR = 1800.;        				#Duty of water for rice
PK = 1.2;          				#Peak demand

# Calculations
AS = CA*IS/100;     				#Area under sugarcane
AR = CA*IR/100;     				#Area under rice
DS = AS/DuS;
DR = AR/DuR;
DT = DS+DR;
DD = PK*DT-0.005333+0.01;
DR = round(DR*1000)/1000;
DT = round(DT*1000)/1000;

# Results
print "Water required for Rice = %.2f cumecs."%(DR);
print " Sugarcane is a perennial crop.";
print "Hence,Water required for Sugarcane = %.2f cumecs."%(DT);
print "Design dischage to meet the peak demand = %.2f cumecs."%(DD);

Water required for Rice = 0.58 cumecs.
Sugarcane is a perennial crop.
Hence,Water required for Sugarcane = 1.27 cumecs.
Design dischage to meet the peak demand = 1.53 cumecs.


## Example 3.13 pg : 66¶

In [13]:

#Given
ql = 20.;          				#discharge in left branch
Al = 20000.;       				#culturable area in left branch
Bl = 120.;         				#Base period in left branch
Il = 0.8;         				#intensity of rabi in left branch
qr = 8.;           				#discharge in rigth branch
Ar = 12000.;       				#culturable area in rigth branch
Br = 120.;         				#Base period in rigth branch
Ir = 0.5;         				#intensity of rabi in rigth branch

# Calculations and Results
#for left canal
ARl = Al*Il;
Dl = ARl/ql;
print "Duty for left canal is = %i hectares/cumecs."%(Dl);

#for rigth canal
ARr = Ar*Ir;
Dr = ARr/qr;
print "Duty for left canal is = %i hectares/cumecs."%(Dr);
print "Since,left canal has higher duty,it is more efficient";

Duty for left canal is = 800 hectares/cumecs.
Duty for left canal is = 750 hectares/cumecs.
Since,left canal has higher duty,it is more efficient


## Example 3.14 pg : 66¶

In [14]:
#Given
CA = 1200.;        				#culturable area
IA = 0.4;          				#intensity of irrigation of crop A
IB = 0.35;         				#intensity of irrigation of crop B
bA = 20.;           				#kor period of crop A
bB = 15.;           				#kor period of crop B
deltaA = 0.1;          				#kor depth of crop A
deltaB = 0.16;          				#kor depth of crop B

# Calculations and Results
#crop A
A = CA*IA;
Du = 8.64*bA/deltaA;
qA = A/Du;
qA = round(qA*1000)/1000;
print "Discharge required for crop A = %.2f cumec."%(qA);

#crop B
A = CA*IB;
Du = 8.64*bB/deltaB;
qB = A/Du;
qB = round(qB*1000)/1000;
print "Discharge required for crop B = %.2f cumec."%(qB);
D = qA+qB;
D = round(D*10)/10;
print "Design discharge of water course = %.2f cumec."%(D);

Discharge required for crop A = 0.28 cumec.
Discharge required for crop B = 0.52 cumec.
Design discharge of water course = 0.80 cumec.


## Example 3.15 pg : 67¶

In [15]:

#Given
B = 12.;         				#transplantaion period
D = 0.5;        				#total depth of water required by the crop
R = 0.1;        				#rain falling on field
L = 0.2;        				#loss of water
A = 600.;        				#irrigated area
I = 0.6;        				#intensity of irrigation
delta = D-R;
Dui = 8.64*B/delta;

# Calculations and Results
#math.since water loss is 20 percent
Du = (1-L)*Dui;
print "Duty of water required = %.2f hectares/cumec."%(Du);

TA = I*A;
q = TA/Du;
q = round(q*100)/100;
print "Discharge at head of water course = %.2f cumecs."%(q);

Duty of water required = 207.36 hectares/cumec.
Discharge at head of water course = 1.74 cumecs.


## Example 3.16 pg : 67¶

In [16]:
import math

#Given
CF = 0.8;                        				#Capacity factory
Tf = 13./20;                      				#time factor
A = [850., 120., 600., 500., 360.];
#Given values of area
B = [320., 90., 120., 120., 120.];
#Given values of Base period
D = [580., 580., 1600. ,2000., 600.];
#Given values of duty at head canal

# Calculations and Results
DS = A[0]/D[0];                 				#discharge for sugarcane
DOS = A[1]/D[1];                				#discharge for overlap sugarcane
DW = A[2]/D[2];                 				#discharge for wheat
DB = A[3]/D[3];                 				#discharge for bajri
DV = A[4]/D[4];                 				#discharge for vegetables
DR = DS+DW;
DM = DS+DB;
DH = DS+DOS+DV;
print "Maximum demand is in hot weather";
q = DH/Tf;
D = q/CF;
q = round(1000*q)/1000;
D = round(100*D)/100;
print "Full supply discharge at head = %.2f cumecs"%(q);
print "Design discharge = %.2f cumecs."%(D);

Maximum demand is in hot weather
Full supply discharge at head = 3.50 cumecs
Design discharge = 4.37 cumecs.


## Example 3.17 pg : 68¶

In [17]:
import math
from numpy import zeros

#Given
CL = 0.2;                   				#Canal loss
RL = 0.12;                  				#Reservior loss
A = [4800., 5600., 2400., 3200., 1400];
#Given values of area under crop
D = [1800., 800., 1400., 900., 700];
#Given values of duty at field
B = [120., 360., 200., 120., 120];
#Given values of base period

# Calculations and Results
#(a) Wheat
d = A[0]/D[0];
V1 = d*B[0];
#(b) Sugarcane
d = A[1]/D[1];
V2 = d*B[1];
#(c) Cotton
d = A[2]/D[2];
V3 = round(d*B[2]);
#(d) Rice
d = A[3]/D[3];
V4 = round(d*B[3]);
#(e) vegetables
d = A[4]/D[4];
V5 = d*B[4];

Vd = (V1+V2+V3+V4+V5)*8.64;
SC = Vd/((1-CL)*(1-RL));
print "Reservior capacity = %.2f hectare-metres."%(SC);

#Alternative method
delta = zeros(5)
for i in range(5):
delta[i] = 8.64*B[i]/D[i];
V = zeros(5)
for j in range(5):
V[j] = A[j]*delta[j];

s = 0;
for k in range(5):
s = s+V[k];

SC = s/((1-CL)*(1-RL));

print " By Alternative method.Storage capacity = %.f hectare-metres."%(SC);

Reservior capacity = 47250.00 hectare-metres.
By Alternative method.Storage capacity = 47244 hectare-metres.


## Example 3.18 pg : 82¶

In [18]:
import math
from numpy import zeros

#Given
eita = 0.7;        				#water application efficiency
k = 0.75;          				#crop factor
T = [19., 16., 12.,.5, 13.];
#Given values of temperature
p = [7.19 ,7.15, 7.30, 7.03];  				#daytime hours of the year
RD = 1.2;              				#rainfall in december
RJ = 0.8;              				#rainfall in january

f = zeros(4)
for i in range(4):
f[i] = p[i]*(1.8*T[i]+32)/40;

s = 0;
for i in range(4):
s = s+f[i];

C = k*s;
R = RD+RJ;
CIR = C-R;
FIR = CIR/eita;
C = round(10*C)/10;
CIR = round(CIR*10)/10;
FIR = round(FIR*10)/10;

# Results
print "Consumptive use = %.2f cm."%(C);
print "consumptive irrigatin requirement = %.2f cm."%(CIR);
print "field irrigatio reqiurement = %.2f cm."%(FIR);

Consumptive use = 28.70 cm.
consumptive irrigatin requirement = 26.70 cm.
field irrigatio reqiurement = 38.20 cm.


## Example 3.19 pg : 83¶

In [19]:
import math

#Given
L = 20.;             				#latitude of place(degree North)
T = 15.;             				#mean monthly temperature(degree celcius)
RH = 0.5;            				#relative humidity
E = 250.;            				#elevation of area
V = 25.;             				#wind velocity at 2 m heigth

#from table 3.10
VP = 12.79;         				#saturation vapour pressure
s = 0.8;            				#slope of curve between vapur pressure and temperature
#from table 3.11
R = 10.8;
#from table 3.12
N = 11.1;
#from table 3.9
n = 7.74;

# Calculations
p = n/N;
e = VP*RH;
Ea = 0.002187*(160+V)*(VP-e);
r = 0.2;
alpha = 0.49;
sigma = 2.01E-9;
Ta = 293;
H = R*(1-r)*(0.29*math.cos(math.pi/9)+0.55*p)-sigma*Ta**4*(0.56-0.092*e**0.5)*(0.10+0.9*p);
Et = (s*H+alpha*Ea)*31/(s+alpha);
Et = round(Et*10)/10;

# Results
print "consumptive use of rice in january = %.2f mm of water."%(Et);

consumptive use of rice in january = 71.60 mm of water.


## Example 3.20 pg : 89¶

In [20]:
import math

#Given
Fc = 0.27;               				#Field capacity
pwp = 0.13;              				#permanent wilting point
d = 80.;                 				#depth of soil(cm)
gammad = 1.5;           				#dry unit weigth of soil(g/cc)
gammaw = 1.;             				#unit weigth of water(g/cc)
M = 0.18;               				#avearge soil moisture
eita = 0.8;             				#field efficiency
FC = 0.15;              				#field channel

# Calculations
SC = gammad*d*(Fc-pwp)/gammaw;
D = gammad*d*(Fc-M)/gammaw;
FIR = D/eita;
W = FIR/(1-FC);
W = round(W*10)/10;

# Results
print "maximum storage capacity = %.2f cm"%(SC);
print "depth of irrigation water = %.2f cm"%(D);
print "field irrigation requirement = %.2f cm"%(FIR);
print "water required at canal outlet = %.2f cm"%(W);

maximum storage capacity = 16.80 cm
depth of irrigation water = 10.80 cm
field irrigation requirement = 13.50 cm
water required at canal outlet = 15.90 cm


## Example 3.21 pg : 89¶

In [22]:
import math
from numpy import zeros

#Given
W = 0.4;                     				#amount of water available from precipitation
Cl = 0.15;                   				#Channel loss
RL = 0.1;                    				#reservior loss
B = [120., 320., 120., 200., 100];   				#Base period
D = [1800., 800., 900., 1400., 1200];				#Duty at field
A = [500., 600., 300., 1200., 500];  				#Area under crop

# Calculations
for i in range(5):
delta[i] = 8.64*B[i]/D[i];

V = zeros(5)
for i in range(5):
V[i] = delta[i]*A[i];

s = 0;
for i in range(5):
s = s+V[i];

C = s*(1-W)/((1-Cl)*(1-RL));

# Results
print "Reservior capacity = %.f ha-m."%(C);

Reservior capacity = 3567 ha-m.


## Example 3.22 pg : 90¶

In [23]:

#Given
GCA = 10000.;                    				#gross commanded area
CCA = 0.75*GCA;                 				#Culturable commanded area
IR = 0.6;                       				#intensity of irrigation during rabi season
IK = 0.3;                       				#intensity of irrigation during kharif season
DuR = 2500.;                     				#duty during rabi season
DuK = 1000.;                     				#duty during kharif season

# Calculations
AR = IR*CCA;                    				#area under irrigation in rabi season
AK = IK*CCA;                    				#area under irrigation in kharif season
DR = AR/DuR;
DK = AK/DuK;

# Results
print "discharge required at head of distributory = %.2f cumecs."%(DK);

discharge required at head of distributory = 2.25 cumecs.


## Example 3.23 pg : 90¶

In [24]:
import math

#Given
Fc = 0.18;         				#field capacity
wc = 0.07;         				#wilting cofficient
Sg = 1.35;         				#bulk density of soil
d = 1.2;           				#root zone depth

# Calculations and Results
m = Fc-wc;
mo = wc+m/3;
dw = 100*Sg*d*(Fc-mo);
print "Depth of water required = %.2f cm"%(dw);
ev1 = 1.1;         				#average evapotranspiration rates in 1 NOV-30 NOV
ev2 = 1.7;         				#average evapotranspiration rates in 1 DEC-31 DEC
ev3 = 2.4;         				#average evapotranspiration rates in 1 JAN-31 JAN
ev4 = 1.5;         				#average evapotranspiration rates in 1 FEB-28 FEB
ev5 = 3.5;         				#average evapotranspiration rates in 1 MAR-25 MAR
#irrigation requirement from 1 NOV to 3 JAn
dev = (ev1*30+ev2*31+ev3*3)/10;
print "Water consumed by evapotranspiration = %.2f cm."%(dev);
print "No water is required during 1 NOV-3 JAN";

#irrigation requirement after 3rd JAN
ws = (ev3-1.5)*16/10;             				#water consumed from soil from 4 JAN-19 JAN
ts = ws+dev;                				#water withdrawn from soil from 1 NOV-19 JAN
s = (dw-ts)*10;
day = s/ev3;
depth = ts+(4*ev3)/10+(2*ev3)/10;
print "depth of water required in first irrigation = %.2f cm."%(depth);
#/irrigation requirement from 26 JAn to 25 MAR
w1 = ev3*6;
w2 = ev4*28;
w3 = ev5*25;
W = w1+w2+w3;
x = (dw*10-(14.4+42))/ev5;
print "Hence second irrigation is required after %.2f days i.e on 18th March."%(x);
depth1 = (W-(dw*10))/10;
print "required water depth = %.2f cm"%(depth1);
print "First Watering on 29 JAn and 30 JAN = %.2f cm.Second watering required on 18th March = %.2f cm."%(depth,depth1);

Depth of water required = 11.88 cm
Water consumed by evapotranspiration = 9.29 cm.
No water is required during 1 NOV-3 JAN
depth of water required in first irrigation = 12.17 cm.
Hence second irrigation is required after 17.83 days i.e on 18th March.
required water depth = 2.51 cm
First Watering on 29 JAn and 30 JAN = 12.17 cm.Second watering required on 18th March = 2.51 cm.


## Example 3.24 pg : 91¶

In [25]:

#Given
Fc = 0.26;              				#Field capacity of soil
A = 3000.;               				#Area of field
OM = 0.12;              				#optimum moisture
pwp = 0.1;              				#permanent wilting point
d = 80.;                				#depth of root zone
RD = 1.4;               				#relative density of soil
f = 10.;                 				#frequency of irrigation
eita = 0.23;            				#overall efficiency

# Calculations
D = RD*d*(Fc-OM);
U = D*10/f;
Wr = A*D*100;
q = Wr/(f*24*3600);
q = round(q*100)/100;

# Results
print "daily consumptive = %.2f mm."%(U);
print "discharge in canal = %.2f q cumecs."%(q);

daily consumptive = 15.68 mm.
discharge in canal = 5.44 q cumecs.


## Example 3.25 pg : 92¶

In [26]:

#Given
C1 = 0.2;              				#consumptive requirement of crop for 1 to 15 days
C2 = 0.3;              				#consumptive requirement of crop for 16 to 40 days
C3 = 0.5;              				#consumptive requirement of crop for 41 to 50 days
C4 = 0.1;              				#consumptive requirement of crop for 51 to 55 days
A = 50.;                				#area of land
wr = 5.;                				#presowing water requirement
R = 3.5;               				#rainfall during 36th and 45th day

# Calculations
w1 = 15*C1*100;
w2 = 25*C2*100;
w3 = 10*C3*100;
w4 = 5*C4*100;
w5 = 5*100;
W = w1+w2+w3+w4+w5;
ER = 3.5*100;
q = (W-ER)*A;

# Results
print "total water to be delivered = %i cubic metre."%(q);

total water to be delivered = 87500 cubic metre.


## Example 3.26 pg : 93¶

In [27]:

#Given
Fc = 0.3;          				#field capacity
pwp = 0.11;        				#permanent wilting percent
gammad = 1300.;     				#density of soil
gammaw = 1000.;     				#density of water
d = 700.;           				#root zone depth
CW = 12.;           				#daily consumptive use of water

# Calculations
WHC = Fc-pwp;
mo = Fc-(0.75*WHC);
D = gammad*d*(Fc-mo)/gammaw;
I = D/CW;

# Results
print " watering interval = %i days"%(I);

 watering interval = 10 days


## Example 3.27 pg : 94¶

In [28]:

#Given
A = 1000.;                				#total area
AI = 0.7*A;              				#area under irrigation
B = 15.;                  				#Base period
d = 500.;                 				#depth of water required during transplantation
R = 120.;                 				#useful rain falling
Wl = 0.2;                 				#water loss

# Calculations
delta = d-R;
Du = 8.64*B*1000/delta;
DuH = Du*(1-Wl);
q = AI/DuH;
q = round(q*100)/100;

# Results
print "Duty of water = %i hec/cumec."%(Du);
print "discharge required in water course = %.2f cumecs."%(q);

Duty of water = 341 hec/cumec.
discharge required in water course = 2.57 cumecs.


## Example 3.28 pg : 94¶

In [29]:
from numpy import zeros

#Given
Ar = 4000.;                 				#culturable commanded area
CL = 0.25;                				#canal loss
RL = 0.15;                 				#reservior loss
B = [120., 360., 180., 120., 120.];  				#base period
D = [1800., 1700., 1400., 800., 700.];				#duty of water
I = [20., 20., 10., 15., 15.];       				#intensity of irrigation

A = zeros(5)
# Calculations
for i in range(5):
A[i] = Ar*I[i]/10;     				#area under crop

Q = zeros(5)
for i in range(5):
Q[i] = A[i]/D[i];        				#discharge required

for i in range(5):
V[i] = 8.64E4*Q[i]*B[i];     				#quantity of water

s = 0;
for i in range(5):
s = s+V[i];

SC = round(s/((1-CL)*(1-RL)*1000000));

# Results
print "Storage capacity = %iD+06 cubic metre."%(SC);

Storage capacity = 633D+06 cubic metre.