Chapter 4 : HYDROLOGY¶
Example 4.1 pg : 113¶
In [1]:
#Given
p = [78.8, 90.2, 98.6, 102.4, 70.4]; #rain guage readings at respective stations
s = 0.;
# Calculations and Results
for i in range(5):
s = s+p[i];
pavg = s/5;
u = 0;
for i in range(5):
u = u+(p[i]-pavg)**2;
sx = (u/4)**0.5;
Cv = sx*100/pavg;
N = (Cv/6)**2;
N = round(N*100)/100;
print "mean rainfall = %.2f cm."%(pavg);
print "total stations needed = %.2f."%(N);
#taking N = 7
N = 7;
n = N-5;
print "additional guages needed = %i."%(n);
Example 4.2 pg : 115¶
In [2]:
#Given
pB = 48.; #precipitation at B
pC = 51.; #precpitation at C
pD = 45.; #precipitation at D
# Calculations
pA = (pB+pC+pD)/3;
# Results
print "precipitation at A = %i mm."%(pA);
Example 4.3 pg : 115¶
In [3]:
#Given
pA = 6.6; #precipitation at A
pB = 4.8; #precpitation at B
pC = 3.7; #precipitation at C
nA = 72.6; #normal precipitation at A
nB = 51.8; #normal precipitation at B
nC = 38.2; #normal precipitation at C
nX = 65.6; #normal precipitation at X
# Calculations
pX = (nX*pA/nA+nX*pB/nB+nX*pC/nC)/3;
pX = round(pX*100)/100;
# Results
print "precipitation at x = %.2f cm."%(pX);
Example 4.4 pg : 115¶
In [4]:
#Given
pB = 74.; #precipitation at B
pC = 88.; #precpitation at C
pD = 71.; #precipitation at D
pE = 80.; #precipitation at E
Bx = 9.;
By = 6.;
Cx = 12.;
Cy = -9.;
Dx = -11.;
Dy = -6.;
Ex = -7.;
Ey = 7.;
Ax = 0;
Ay = 0;
# Calculations
Db = (Bx**2+By**2);
Dc = (Cx**2+Cy**2);
Dd = (Dx**2+Dy**2);
De = (Ex**2+Ey**2);
Wb = 1/Db;
Wc = 1/Dc;
Wd = 1/Dd;
We = 1/De;
s = pB*Wb+pC*Wc+pD*Wd+pE*We;
pA = s/(Wb+Wc+Wd+We);
pA = round(pA*10)/10;
# Results
print "precipitation at A = %.2f mm."%(pA);
Example 4.5 pg : 119¶
In [5]:
import math
from numpy import zeros
#Given
p = [58., 61., 69., 56., 84., 86., 69., 79., 71.]; #values of precipitation
s = 0;
# Calculations and Results
for i in range(9):
s = s+p[i];
ar = s/9;
ar = round(ar*10)/10;
print "umath.sing arithmatic average method:"
print "Average rainfall = %.2f cm."%(ar);
I = [86., 85., 80., 75., 70., 65., 60., 55., 50.]; #isphytes
A = [0.43, 5.20, 4.0, 5.04, 5.85, 4.53, 4.09, 1.27]; #area between isohytes
a = zeros(9)
for i in range(8):
a[i] = (I[i]+I[i+1])/2;
P = zeros(8)
for i in range(8):
P[i] = A[i]*a[i];
s = 0;
for i in range(8):
s = s+P[i];
t = 0;
for i in range(8):
t = t+A[i];
ar = s/t;
ar = round(ar*10)/10;
print "isohytel method:"
print "Average rainfall = %.2f cm."%(ar);
A = [3.26, 0.39, 1.61, 2.04, 2.46, 0.84, 3.91, 5.09, 0.41, 3.94, 2.06, 4.40]; #thiessen area
p = [58., 63., 71., 69., 86., 81., 84., 56., 53., 69., 61., 79.]; #observed precipitation
P = zeros(12)
for i in range(12):
P[i] = A[i]*p[i];
s = 0;
for i in range(12):
s = s+P[i];
t = 0;
for i in range(12):
t = t+A[i];
ar = s/t;
ar = round(ar*10)/10;
print "thiesson polygon method:"
print "Average rainfall = %.2f cm."%(ar);
#mean rainfall obtained by thiesson polygon method is different from book as product(A*P) is round offed in book.
Example 4.6 pg : 127¶
In [3]:
%matplotlib inline
import math
from matplotlib.pyplot import plot
from numpy import zeros
#Given
X = [69., 55., 62., 67., 87., 70., 65., 75., 90., 100., 90., 95., 85., 90., 75., 95.]; #annual rainfall at X
Y = [77., 62., 67., 68., 86., 90., 65., 75, 70, 70, 70 ,75 ,65 ,70, 55, 75]; #average rainfall at 10 base stations
cx = zeros(16)
cx[0] = 69; #accumulated annual values at station X
for i in range(1,16):
cx[i] = cx[i-1]+X[i];
cy = zeros(16)
cy[0] = 77;
for i in range(1,16):
cy[i] = cy[i-1]+Y[i]; #accumulated annual values at ten stations
#since curve is not having unform slope
print "Record at X is not consistent.";
print "From the curve regime is observed in the year 1978."
Q = [1970., 1971., 1972., 1973., 1974., 1975., 1976., 1977.];
O = [95., 75., 90., 85., 95., 90., 100., 90.];
A = zeros(8)
for i in range(8):
A[i] = 0.7051*O[i];
print "Year Observed rainfall Adjusted rainfall";
for i in range(8):
print "%i %i %i"%(Q[i],O[i],A[i]);
#graph is plotted between cx and cy
plot(cy,cx,cy,cx,"ro")
Out[3]:
Example 4.7 pg : 130¶
In [5]:
%matplotlib inline
from matplotlib.pylab import bar,xlabel,ylabel
from numpy import zeros,linspace
#Given
c = [0, 12.4, 22.1, 35.1, 52.7, 63.7, 81.9, 109.2, 123.5, 132.6, 143.3, 146., 146.]; #cumulative rainfall
T = linspace(0,13,len(c)) #Time
t = 15./60; #time interval
r = zeros(13)
I = zeros(13)
r[0] = 0;
print "Rainfall intensity:";
I[0] = 0;
for i in range(1,13):
r[i] = c[i]-c[i-1];
I[i] = r[i]/t; #Rainfall intensity
print "%.2f"%(I[i]);
#graph is plotted between I and T
bar(T,I)
xlabel("Time hr")
ylabel("Rain fall insentity")
Out[5]:
Example 4.8 pg : 131¶
In [7]:
%matplotlib inline
import math
from numpy import zeros_like,array
from matplotlib.pylab import plot,xlabel,ylabel
#Given
CR = array([0, 12.4, 22.1 ,35.1, 52.7, 63.7, 81.9, 109.2, 123.5, 132.6, 143.3, 146.0, 146.0]); #cumulative rainfall
c15 = zeros_like(CR)
c30 = zeros_like(CR)
c45 = zeros_like(CR)
c60 = zeros_like(CR)
c90 = zeros_like(CR)
c120 = zeros_like(CR)
# Calculations and Results
c15[1] = 12.4;
c30[2] = 22.1;
c45[3] = 35.1;
c60[4] = 52.7;
c90[6] = 81.9;
c120[8] = 123.5;
for i in range(2,13):
c15[i] = CR[i]-CR[i-1];
for i in range(3,13):
c30[i] = CR[i]-CR[i-2];
for i in range(4,13):
c45[i] = CR[i]-CR[i-3];
for i in range(5,13):
c60[i] = CR[i]-CR[i-4];
for i in range(7,13):
c90[i] = CR[i]-CR[i-6];
for i in range(9,13):
c120[i] = CR[i]-CR[i-8];
print "15min 30min 45min 60min 90min 120min";
for i in range(13):
print "%.2f %.2f %.2f %.2f %.2f %.2f"%(c15[i],c30[i],c45[i],c60[i],c90[i],c120[i]);
I = [109.2, 91, 79.7, 74.1, 67.6, 61.75]; #maximum intensity at respective durations
D = [15 ,30 ,45 ,60 ,90 ,120]; #durations
#greph is plotted between I and D
plot(D,I,D,I,"ro")
xlabel("Duration")
ylabel("max rain fall intensity")
Out[7]:
Example 4.9 pg : 132¶
In [21]:
#Given
p = [475, 377, 731, 1066, 361, 305, 926, 628, 409, 236, 337, 853]; #precipitation value
N = 12.; #total number of years
T = 6; #recurrence interval
# Calculations
m = N/T;
# Results
print "Ranking of storm = %i."%(m);
#hence pick 2nd severest storm
print "preciptation value which has recurrence period of 6 years = %i mm."%(p[6]);
Example 4.10 pg : 132¶
In [8]:
%matplotlib inline
import math
from numpy import arange,array,zeros_like
#Given
I = linspace(25,16,10) #isohytes
a = array([407 ,1008, 1522, 1909, 2216, 2460, 2651, 2782, 2910, 2936]); #enclosed area
ia = zeros_like(a)
ia[0] = 407.;
# Calculations
for i in range(1,10):
ia[i] = a[i]-a[i-1];
r = linspace(25.5,16.5,10)
rv = r*ia
cv = zeros_like(rv)
cv[0] = 10378;
for i in range(1,10):
cv[i] = cv[i-1]+rv[i];
eud = cv/a
print "From depth area curve we obtain average depth of precipitation = 24.1 mm for an area of 1800 sq. km.";
#graph is plotted between eud and a.
# Results
plot(a,eud,a,eud,"ro")
xlabel("Area")
ylabel("mean precipitation depth")
Out[8]:
Example 4.11 pg :133¶
In [10]:
%matplotlib inline
import math
from numpy import array,float64
#24h max. rainfall with return period of 8,15 and 25.
#24h max rainfall with 40%,24% and 8% probability.
#probabilty of rainfall of magnitude equal to or exceeding 100 mm.
#Given
N = 20.;
r = array([142, 126, 116, 108, 102, 95, 92, 88, 86, 82, 80, 78, 76, 73, 71, 69, 68, 66, 65, 64],dtype=float64); #rainfall in respective years
m = linspace(1,20,20) #ranking of storm
p = m*100/(N+1)
T = 100/p
# Calculations and Results
#from frequency curve obtained we get
#Part (a)
T1 = array([8, 15, 25]);
r1 = array([119, 134, 149]);
print "Tyears Rainfallmm";
for i in range(3):
print "%i %i"%(T1[i],r1[i]);
#Part (b)
p1 = [40 ,24, 8];
r2 = [87, 101, 130];
print "probabilitypercent Rainfallmm";
for i in range(3):
print "%i %i"%(p1[i],r2[i]);
print "For rainfall = 100 m.T = 4 years.Probability = 25 percent.";
Example 4.12 pg : 135¶
In [14]:
%matplotlib inline
import math
from matplotlib.pylab import plot,xlabel,ylabel
from numpy import array,float64
#plot IDF curve for return period of 10,2 and 1 years umath.sing california formula
#Given
t = array([5, 10, 20, 30, 60, 90, 120],dtype=float64); #duration
#value of P for respective return period is
p10 = array([10.6, 14.7, 19.3, 20.8, 25.5, 29, 34.7]); #rainfall for T = 10 years
p2 = array([8.2, 10.3, 13.2, 14.2, 16.6 ,19.4, 21.4]); #rainfall for T = 2 years
p1 = array([3.5, 6.2, 8.9, 10, 13.2, 15, 16.5]); #rainfall for T = 1 year
# Calculations
i1 = p10*60/t; #intensity of rainfall with return period of 10 years
i2 = p2*60/t; #intensity of rainfall with return period of 2 years
i3 = p1*60/t; #intensity of rainfall with return period of 1 year
# Results
#graph is plotted between #t and i1 #t and i2 #t and i3
plot(t,i1)
plot(t,i2)
plot(t,i3)
xlabel("Duration")
ylabel("Intensity")
Out[14]:
Example 4.13 pg : 137¶
In [15]:
%matplotlib inline
import math
from numpy import linspace,array
from matplotlib.pylab import subplot,plot,xlabel,ylabel
#Given
N = 20.;
m = linspace(1,20,20) #rank number
rd = array([82, 78, 75, 72, 70, 68, 65, 63, 61, 58, 56, 54, 52, 50, 46, 40, 36, 34, 32, 30]); #rainfall in decremath.sing order
# Calculations
ri = rd[::-1]
T = N/(m-0.5);
# Results
#from the curves
print "maximum rainfall = 79cm for T = 15 years.";
print "minimum rainfall = 31 cm for T = 15 years.";
#graph is plotted between rd and T;ri and T
subplot(121)
plot(T,rd)
xlabel("Reccurance interval")
ylabel("rainfall cm")
subplot(122)
plot(T,ri)
xlabel("Reccurance interval")
ylabel("rainfall cm")
Out[15]:
Example 4.14 pg : 144¶
In [45]:
import math
from numpy import zeros
#average evaporation for one week
#Given
w = [12, 5, 2, -3, 1, 6, 11]; #water added or taken out
r = [0, 6, 8, 12, 9, 5, 0]; #rainfall
pan = zeros(7)
le = zeros(7)
for i in range(7):
pan[i] = w[i]+r[i]; #Pan evaporation
le[i] = 0.8*pan[i]; #lake evaporation
s = sum(le)
print "daily lake evaporationmm:";
for i in range(7):
print "%.2f"%(le[i]);
av = s/7;
av = round(av*100)/100;
print "average evaporation for one week = %.2f mm."%(av);
Example 4.15 pg : 145¶
In [46]:
import math
#total depth and volume of evaporation loss
#Given
Rh = 0.4; #relative humidity
A = 4.8; #average surface spread of reservior
v3 = 18.; #wind velocity at 3m above ground
es = 31.81; #saturated vapour pressure
Km = 0.36; #for large deep waters
# Calculations and Results
#umath.sing Meyer's formula
ea = es*Rh;
v9 = v3*(9./3)**(1./7);
E = Km*(es-ea)*(1+v9/16);
d = 7*E;
v = d*A*100/1000;
E = round(E*10)/10;
d = round(d*10)/10;
v = round(v*100)/100;
print "umath.sing Meyers formula:";
print "average evaporation loss from reservior = %.2f mm/day."%(E);
print "total depth = %.2f mm"%(d);
print "total volume = %.2f hectare-m."%(v);
#umath.sing Rohwer's formula
Pa = 760.;
vdash = (0.6/2)**(1./7)*18;
E = 0.771*(1.465-0.000732*Pa)*(0.44+0.0733*vdash)*(es-ea);
d = 7*E;
v = d*A*100/1000;
E = round(E*10)/10;
d = round(d*10)/10;
v = round(v*10)/10;
print "umath.sing Rohwers formula:";
print "average evaporation loss from reservior = %.2f mm/day."%(E);
print "total depth = %.2f mm"%(d);
print "total volume = %.2f hectare-m."%(v);
Example 4.16 pg : 149¶
In [17]:
%matplotlib inline
import math
from numpy import array,zeros_like
from matplotlib.pylab import plot,xlabel,ylabel
#plot infiltration capacity curve
#Given
D = 30.; #diameter of inside ring of infiltrometer
A = math.pi*D**2/4;
V = array([0, 200, 470, 840, 1405, 1840, 2245, 2510, 2745, 2885, 2990, 3130, 3270],dtype=float64); #cumulative volume;
t = array([0, 2, 5, 10, 20, 30, 45, 60, 80, 100, 120, 150, 180],dtype=float64); #Time(minutes)
# Calculations and Results
dt = zeros_like(t)
for i in range(1,13):
dt[i] = (t[i]-t[i-1])/60;
F = V/A;
Fd = zeros_like(F)
Fd[0] = F[0];
for i in range(1,13):
Fd[i] = F[i]-F[i-1];
ft = Fd/dt #infirltration rate
#from the graph
print "constant rate of infiltration = 0.40 cm/hr.";
avg10 = F[3]*60/10;
avg30 = F[5]*60/30;
avg10 = round(avg10*100)/100;
avg30 = round(avg30*100)/100;
print "average rate of infiltration for first 10 min = %.2f cm/hr."%(avg10);
print "average rate of infiltration for first 30 min = %.2f cm/hr."%(avg30);
#graph is plotted between ft and t
plot(t,ft)
xlabel("time in mins")
ylabel("Infiltrtion rate")
Out[17]:
Example 4.17 pg : 162¶
In [50]:
#drainage desity
#form factor
#channel slope
#average overland flow length
#Given
A = 82.; #area of watershed
d = 12.6; #dismath.tance between outlet and farther most point
l = 440.; #total length of channel
e = 656.; #elevation differnce between outlet and further most point
# Calculations
Dd = l/A;
ff = A/d**2;
cs = e/(d*1000);
lo = 1000/(2*Dd);
Dd = round(Dd*100)/100;
ff = round(ff*1000)/1000;
# Results
print "drainage desity = %.2f km/square.km."%(Dd);
print "form factor = %.2f."%(ff);
print "channel slope = %.2f."%(cs);
print "average overland flow length = %i m."%(lo);
Example 4.18 pg : 163¶
In [51]:
import math
#compute fi and W index
#Given
R = 3.6; #surface runoff
r = [0, 1.3, 2.8, 4.1, 3.9, 2.8, 2.0, 1.8, 0.9]; #rainfall at respective time
t = 4.; #total time
s = sum(r[2:]);
# Calculations and Results
fi = (s-R*2)/6;
#math.since fi >1.3 and <1.8
print "fi index = %.2f cm."%(fi);
print "computations are correct.";
s = sum(r);
P = s/2;
Sr = 0.;
W = (P-R-Sr)/t;
print "W index = %.2f cm/hr."%(W);
Example 4.19 pg : 163¶
In [3]:
import math
from numpy import linspace,array
#Given
T = linspace(1,9,9) #time from start
r = array([0.7, 1.4, 2.4, 3.7, 2.9, 2.6, 1.7, 0.8, 0.5]); #increamental rainfall
R = 9.3; #total run-off
s = sum(r)
ti = s-R;
#first trial
tr = 9.; #assumed
fi1 = ti/tr;
#this makes 1st,8th and 9th hour ineffective
#second trial
tr = 6.;
ti = s-R-r[0]-r[7]-r[8];
fi = ti/tr;
P = zeros_like(r)
for i in range(9):
P[i] = r[i]-fi;
if (P[i]<0):
P[i] = 0;
print "Timeh rainfall excess.";
for i in range(9):
print "%.2f %.2f"%(T[i],P[i]);
print "fi index = %.2f cm/hr."%(fi);
print "time of rainfall excess = %i hours.."%(tr);
Example 4.20 pg : 164¶
In [4]:
from numpy import array
import math
#Given
P = array([72.2, 70.1, 73.3, 42.5, 81.3, 50.6, 52.9, 59.4, 60.3, 64.3, 68.8, 56.7, 77.2, 40.5, 44.1, 65.5]); #Precipitation
R = array([24.1, 22.7, 25.6, 11.3, 28.4, 12.7, 13.4, 15.7, 16.2, 17.7, 19.2, 14.9, 25.4, 10.6, 11.7, 17.9]); #runoff
# Calculations
Ps = P**2
Rs = R**2;
PR = P*R;
s = sum(Ps)
t = sum(Rs)
u = sum(PR)
q = sum(P)
w = sum(R)
N = 16.;
a = (N*u-q*w)/(N*s-q**2);
b = (w-a*q)/N;
b = round(b*1000)/1000;
a = round(a*10000)/10000;
# Results
print "Equation is:%.4f P %.3f."%(a,b);
Example 4.21 pg : 165¶
In [19]:
%matplotlib inline
from numpy import linspace,array
import math
from matplotlib.pylab import plot,xlabel,ylabel
#Given
A = 8.6; #catchment area
T = linspace(0,4,9) #time
r = array([0, 0.4, 1.1, 2.3, 3.8, 4.8, 5.6, 6.2, 6.7]); #accumulated rainfall
fi = 0.4; #fi index
dt = 0.5; #time interval
# Calculations and Results
d = zeros_like(r)
for i in range(1,9):
d[i] = r[i]-r[i-1]; #accumulated rainfall
print "Intensity of effective Rainfall:";
I = zeros_like(r)
p = zeros_like(r)
s = 0;
for i in range(1,9):
p[i] = d[i]-fi; #effective rainfall
I[i] = p[i]/dt; #Intensity of effective Rainfall
s = s+I[i];
print "%.2f"%(I[i]);
#graph is plotted between I and T
run = s*dt;
V = run*A*10000;
print "Volume of direct run-off = %.2f cubic metre."%(V);
plot(T,I)
Out[19]:
Example 4.22 pg : 166¶
In [20]:
%matplotlib inline
from numpy import array,linspace
#total rainfall
#total rainfall excess
#W index
#Given
r = array([3.5, 6.5, 8.5 ,7.8, 6.4, 4, 4, 6]); #rainfall intensity
T = linspace(0,240,8) #time
dt = 30.; #time interval
# Calculations
s = sum(r);
P = s*dt/60;
Pe = ((6.5-4.5)+(8.5-4.5)+(7.8-4.5)+(6.4-4.5)+(6-4.5))*dt/60; #area of graph above r = 4.5.
w = (P-Pe)/4;
# Results
print "total rainfall = %.2f cm."%(P);
print "total rainfall excess = %.2f cm."%(Pe);
print "W index = %.2f cm/hr."%(w);
plot(T,r)
Out[20]:
Example 4.23 pg : 166¶
In [66]:
import math
from numpy import array,zeros_like
#Given
r = array([0, 8, 22, 74, 92, 105, 114, 120],dtype=float64); #raccumulated rainfall
T = array([0, 2, 4, 6, 8, 10, 12, 14],dtype=float64); #time for start of rainfall
V = 2e6; #volume of run-off
A = 40.; #catchment area
tr = 14.; #duration of rainfall
# Calculations
d = V*1000/(40*1000000);
l = r[7]-d;
W = l/tr;
I = zeros_like(r)
for i in range(1,8):
I[i] = r[i]-r[i-1]; #incremental rainfall
#rainfall excess is available in 4 time intervals of 2 hrs
tre = 8.;
fi = (l-I[1]-I[6]-I[7])/tre;
fi = round(fi*100)/100;
# Results
print "fi index = %.2f mm/hr."%(fi);
Example 4.24 pg : 167¶
In [68]:
import math
from numpy import array
#Given
r = array([2.0 ,2.5, 7.6, 3.8, 10.6, 5.0, 7.0, 10.0, 6.4, 3.8, 1.4, 1.4]); #rainfall depths
R = 25.5;
# Calculations
s = sum(r)
tf = s-R;
af = tf/12;
#rainfall is less than average infiltration in1st,2nd,11th and 12th hours
f = (tf-r[0]-r[1]-r[10]-r[11])/8;
f = round(f*10)/10;
# Results
print "average infiltration index = %d cm/hour."%(f);
Example 4.25 pg : 167¶
In [69]:
import math
from numpy import zeros,zeros_like,array
#Given
r = array([2.0, 2.5, 7.6, 3.8, 10.6, 5.0, 7.0, 10.0, 6.4, 3.8, 1.4, 1.4]); #rainfall depths
A1 = 20.;
A2 = 40.;
A3 = 60.;
# Calculations and Results
A = A1+A2+A3;
fi1 = 7.6;
fi2 = 3.8;
fi3 = 1.0;
R1 = zeros_like(r)
R2 = zeros_like(r)
R3 = zeros_like(r)
for i in range(12):
R1[i] = r[i]-fi1; #rainfall excess
R2[i] = r[i]-fi2;
R3[i] = r[i]-fi3;
if (R1[i]<0):
R1[i] = 0;
if (R2[i]<0):
R2[i] = 0;
if (R3[i]<0):
R3[i] = 0;
print "average depth of hourly rainfall excesscm/hr";
a1 = zeros(12)
a2 = zeros(12)
a3 = zeros(12)
T = zeros(12)
for i in range(12):
a1[i] = R1[i]*A1/A; #average rainfall excess
a2[i] = R2[i]*A2/A;
a3[i] = R3[i]*A3/A;
T[i] = a1[i]+a2[i]+a3[i]; #total hourly rainfall excess
T[i] = round(T[i]*100)/100;
print "%.2f"%(T[i]);
Example 4.26 pg : 185¶
In [7]:
from numpy import array,linspace
#derive the unit hydrograph
#Given
A = 92.; #area of drainage bamath.sin
t = array([6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 2, 4, 6, 8, 10, 12, 14, 16],dtype=float64); #time
r = array([10.6, 9.7, 107.8, 175.6, 193.9, 150.3, 126.2, 106.9, 90, 72.8, 58.2, 48, 36.2, 28.4, 20.2, 14, 10.2, 10.4]); #total run-off
B = array([10.6, 9.7, 9.73, 9.77, 9.8, 9.83, 9.87, 9.9, 9.93, 9.97, 10, 10.03, 10.07, 10.10, 10.13, 10.16, 10.20, 10.40]); #base flow
s = 0;
# Calculations and Results
d = r - B
s = sum(d)
n = 0.36*s*2/A;
print "ordinates of unit hydrograph:";
u = zeros(18)
for i in range(18):
u[i] = d[i]/n; #ordinates of unit hydrograph
u[i] = round(u[i]*100)/100;
print "%.2f"%(u[i]);
print "Hydograph is 4-hr unit hydrograph";
Example 4.27 pg : 186¶
In [74]:
import math
from numpy import zeros
#Given
A = 316.; #drainage area
B = 17.; #base flow
t = 6.;
O = [17.0, 113.2, 254.5, 198.0, 150.0, 113.2, 87.7, 67.9, 53.8, 42.5, 31.1, 22.6, 17.0]; #ordinates of storm hydrograph
Or = zeros(13)
Oh = zeros(13)
for i in range(13):
Or[i] = O[i]-B; #ordinates of direct run-off
Oh[i] = Or[i]/6.477; #ordinates of unit hydrograph
s = sum(Or);
re = s*60*60*t/(A*10000);
re = round(re*1000)/1000;
# Results
print "rainfall excess = %.2f cm."%(re);
Example 4.28 pg : 188¶
In [75]:
import math
from numpy import zeros
#Given
fi = 2.5; #infiltration index
B = 10; #Base flow
O = [0, 110, 365, 500, 390, 310, 250, 235, 175, 130, 95, 65, 40, 22, 10, 0, 0, 0]; #ordinates of unit hydrograph
R1 = 2;R2 = 6.75;R3 = 3.75;
r1 = (R1*10-(fi*3)-5)/10; #rainfall excess in first three hour
r2 = (R2*10-(fi*3))/10; #rainfall excess in second three hour
r3 = (R3*10-(fi*3))/10; #rainfall excess in third three hour
s1 = zeros(18)
for i in range(18):
s1[i] = r1*O[i];
s2 = zeros(18)
for i in range(1,18):
s2[i] = r2*O[i-1];
s3 = zeros(18)
for i in range(2,18):
s3[i] = r3*O[i-2];
#surface run-off from rainfall excess during succesive unit periods
print "ordinates of storm hydrograph";
T = zeros(18)
t = zeros(18)
for i in range(18):
T[i] = s1[i]+s2[i]+s3[i];
t[i] = T[i]+B;
t[i] = round(t[i]*10)/10;
print "%.2f"%(t[i]);
Example 4.29 pg : 189¶
In [21]:
%matplotlib inline
from numpy import array,linspace,zeros
from matplotlib.pylab import plot,xlabel,ylabel
#derive and plot 6 hr unit hydrograph
#Given
A = 103.4; #area of bamath.sin
t = linspace(0,36,9); #time
q = [0, 21, 80, 82, 189, 123, 184, 87, 55.5, 25.25, 9, 6, 0]; #flow
print "ordinates of unit hydrograph are:";
u = zeros(9)
u[0] = 0;
u[1] = q[1]/2.;
u[2] = (q[2]-4*u[0])/2;
u[3] = (q[3]-4*u[1])/2;
for i in range(4,9):
u[i] = (q[i]-3*u[i-4]-4*u[i-2])/2; #ordinates of unit hydrograph
for i in range(9):
print "%.2f"%(u[i]);
print "The succesive unit hydrograph will have same ordinates but will be shiftedlaterally by 6 hrs.";
#graph is plotted between u and t.
plot(t,u)
xlabel("Time in hours")
ylabel("Discharge")
Out[21]:
Example 4.30 pg : 190¶
In [79]:
import math
from numpy import zeros
#derive ordinates of 6 hrs unit hydrograph
#Given
R = [0, 1, 2.7, 5, 8, 9.8, 9, 7.5, 6.3, 5, 4, 2.9, 2.1, 1.3 ,0.5, 0, 0, 0, 0, 0]; #2hrs unit hydrograph
print "ordinates of 6 hrs unit hydrograph";
O1 = zeros(20)
for i in range(18):
O1[i+2] = R[i];
O2 = zeros(20)
for i in range(16):
O2[i+4] = R[i];
S = zeros(20)
f = zeros(20)#offset unit hydrograph
for i in range(20):
S[i] = O1[i]+O2[i]+R[i]; #sum
f[i] = S[i]/3; #ordinates of 6 hrs unit hydrograph
f[i] = round(f[i]*10)/10;
print "%.2f"%(f[i]);
Example 4.31 pg : 190¶
In [10]:
import math
from numpy import array,linspace,zeros
#Given
t = linspace(0,45,16) #time
O = [0 ,9, 20, 35, 49, 43, 35, 28, 22, 17, 12, 9, 6, 3, 0, 0]; #ordinate of 2 hr unit hydrograph
# Calculations and Results
of = zeros(16)
for i in range(2,16):
of[i] = O[i-2]+of[i-2]; #offset ordinate
s = zeros(16)
for i in range(16):
s[i] = O[i]+of[i]; #ordinate of s-curve
of1 = zeros(16)
for i in range(3,16):
of1[i] = s[i-3]; #offset of s-curve
print "ordinates of 9 hrs unit hydrograph:";
y = zeros(16)
u = zeros(16)
for i in range(16):
y[i] = s[i]-of1[i];
u[i] = 2*y[i]/3; #ordinate of 9 hrs unit hydrograph
u[i] = round(u[i]*10)/10;
print "%.2f"%(u[i]);
Example 4.32 pg : 210¶
In [2]:
import math
from numpy import zeros,linspace
#california method
#Hazens method
#gumbels method
#Given
q = [9200, 7800, 6600, 5800, 5260, 4980, 4525, 3810, 3630, 3250, 3110, 3090, 2380, 2390, 1723]; #Discharge arranged in decreamath.sing order
N = 15;
C = 0.3;
m = linspace(1,15,15)
C = [0.3, 0.44, 0.52, 0.57, 0.61, 0.66, 0.7, 0.74, 0.78, 0.82, 0.86, 0.88, 0.94, 0.96, 1]; #from table 4.25
print "California Hazen Gumbel";
Ca = zeros(15)
H = zeros(15)
G = zeros(15)
Ca = zeros(15)
G = zeros(15)
for i in range(15):
Ca[i] = N/m[i];
H[i] = 2*N/(2*m[i]-1);
G[i] = N/(m[i]+C[i]-1);
Ca[i] = round(Ca[i]*100)/100;
G[i] = round(G[i]*100)/100;
H[i] = round(H[i]*100)/100;
print "%.2f %.2f %.2f"%(Ca[i],H[i],G[i]);
Example 4.33 pg : 211¶
In [4]:
import math
#Given
T1 = 40.;
T2 = 80.; #Return period
F1 = 27000.;
F2 = 31000.; #Peak flood
# Calculations
y80 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));
y40 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));
y = (F2-F1)/(y80-y40);
T = 240.;
y240 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));
x240 = F2+(y240-y80)*y;
# Results
print "flood magnitude with return period of 240 years = %i cumec."%(x240);
Example 4.34 pg : 212¶
In [8]:
import math
#Given
N = 40;
Sn = 1.1413;
yn = 0.5436; #from table 4.21 (a) and(b)
q = [1330, 1095, 1030, 980, 975, 950, 945, 940, 925, 855, 853, 840, 835, 825, 810, 795, 756, 710, 708, 705, 700, 670, 625, 620, 610, 605, 595, 585, 570, 550, 530, 505, 500, 495, 485, 465, 460, 420, 390, 380]; #discharge
s = sum(q)
xavg = s/N;
w = 0;
# Calculations
t = zeros(40)
for i in range(40):
t[i] = (q[i]-xavg)**2;
w = w+t[i];
sigma = (w/(N-1))**0.5;
N = 10.;
y10 = -(2.303*math.log10(2.303*math.log10(N/(N-1))));
K10 = (y10-yn)/Sn;
x10 = xavg+K10*sigma;
N = 20.;
y20 = -(2.303*math.log10(2.303*math.log10(N/(N-1))));
K20 = (y20-yn)/Sn;
x20 = xavg+K20*sigma;
N = 5.;
y5 = -(2.303*math.log10(2.303*math.log10(N/(N-1))));
K5 = (y5-yn)/Sn;
x5 = xavg+K5*sigma;
T = 100.;
y100 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));
K100 = (y100-yn)/Sn;
x100 = xavg+K100*sigma;
T = 200.;
y200 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));
K200 = (y200-yn)/Sn;
x200 = xavg+K200*sigma;
x100 = round(x100);
# Results
print "For T = 100 years:flood discharge = %.2f cumecs.\
\nFor T = 200 years:flood discharge = %.f cumecs."%(x100,x200);
Example 4.35 pg : 214¶
In [9]:
import math
from numpy.linalg import solve
#Given
sigma = 1.1413; #smath.radians(numpy.arcmath.tan(ard deviation
yn = 0.5436;
T = 50.;
# Calculations
y50 = -2.303*math.log10(2.303*math.log10(T/(T-1)));
K50 = (y50-yn)/sigma;
T = 100.;
y100 = -2.303*math.log10(2.303*math.log10(T/(T-1)));
K100 = (y100-yn)/sigma;
x50 = 878; x100 = 970;
#Given peak flood
A = [[K50, 1],[K100, 1]];
B = [x50,x100];
C = solve(A,B)#A\B;
xavg = C[1];
sigmad = C[0];
T = 200.;
y200 = -2.303*math.log10(2.303*math.log10(T/(T-1)));
K200 = (y200-yn)/sigma;
x200 = xavg+K200*sigmad;
x200 = round(x200);
# Results
print "200 year flood for stream = %.2f cumecs."%(x200);
Example 4.36 pg : 214¶
In [85]:
import math
#risk of failure of cofferdam
#return period
#Given
T = 30.; #deign for period
n = 6.; #period of construction
# Calculations
R = (1-(1-(1/T))**n)*100;
R1 = 0.1; #reduced risk
T1 = 1./(1-(1-R1)**(1./6));
R = round(R*10)/10;
T1 = round(T1*100)/100;
# Results
print "risk of failure of cofferdam = %.2f percent."%(R);
print "return period = %.2f years."%(T1);
Example 4.37 pg : 215¶
In [87]:
import math
#probability of excedence
#probability of flood magnitude occuring at:
#at least once in 10 years
#two times in 10 succesive years
#once in 10 succesive years
#Given
T = 40.; #return period
P = 1./T;
n = 10;
Rsk = 1.-(1-P)**n;
s = 1.;
t = 1.;
for i in range(1,n+1):
s = s*i;
for i in range(1,n-1):
t = t*i;
P2n = s*P**2*(1-P)**8/(t*2);
P1n = n*P*(1-P)**(n-1);
Rsk = round(Rsk*1000)/1000;
P2n = round(P2n*10000)/10000;
P1n = round(P1n*1000)/1000;
# Results
print "probability of excedence = %.2f."%(P);
print "probability of flood magnitude occuring at least once in 10 years = %.2f"%(Rsk);
print "probability of flood magnitude occuring at two times in 10 succesive years = %.2f"%(P2n);
print "probability of flood magnitude occuring at once in 10 succesive years = %.2f"%(P1n);
Example 4.38 pg : 215¶
In [10]:
import math
#Given
C1 = 0.22;C2 = 0.12;C3 = 0.32; #run-off coefficient
A1 = 3.2;A2 = 4.8;A3 = 1.8;
L = 2.4; #length of water course
H = 30; #fall
T = 30; #frequency
# Calculations
t = 60*0.000323*(L*1000)**0.77*(H/(L*1000))**(-0.385);
i = 78*T**0.22/(t+12)**0.45;
q = 2.778*i*(C1*A1+C2*A2+C3*A3);
q = round(q*10)/10;
# Results
print "peak rate of run off = %.2f cumecs."%(q);
Example 4.39 pg : 216¶
In [11]:
import math
#Given
T = 30; #return period
A = 2.4; #area of watershed
s = 1./200; #slope oof catchment
L = 1.8; #length of travel of water
C = 0.25; #average run-off coefficient
r = [2.5, 3.8, 4.8, 5.9, 6.7, 7.4, 8.4, 8.7, 9.2]; #rmath.sinfall depth
# Calculations
t = 60*0.000323*(L*1000)**0.77*(s)**(-0.385);
rmax = r[6]+(r[7]-r[6])*7.84/10;
i = rmax*60/t;
q = 2.778*C*A*i;
q = round(q*100)/100;
# Results
print "peak flow rate = %.2f cumecs."%(q);
Example 4.40 pg : 216¶
In [12]:
import math
#Given
pA = 75.; #precipitation at A
pB = 58; #precpitation at B
pC = 47; #precipitation at C
nA = 826; #normal precipitation at A
nB = 618; #normal precipitation at B
nC = 482; #normal precipitation at C
nX = 757; #normal precipitation at X
# Calculations
pX = (nX*pA/nA+nX*pB/nB+nX*pC/nC)/3.;
pX = round(pX*10)/10;
# Results
print "precipitation at x = %.2f cm."%(pX);
Example 4.41 pg : 216¶
In [13]:
import math
#Given
p = [41, 51, 32, 55, 50, 68]; #rain guage readings at respective stations
s = sum(p)
pavg = s/6;
u = 0;
for i in range(5):
u = u+(p[i]-pavg)**2;
# Calculations
sx = (u/5)**0.5;
Cv = sx*100/pavg;
N = (Cv/8)**2;
N = round(N*100)/100;
# Results
print "mean rainfall = %.2f cm."%(pavg);
print "total stations needed = %.2f."%(N);
Example 4.42 pg : 217¶
In [14]:
import math
#Given
a = 4; #dimension of plot sides
P1 = 4.8;P2 = 13;P3 = 8;P4 = 5.4;P5 = 3.2;P6 = 9.4; #precipitaion at respective stations
# Calculations
A1 = a**2/8+a**2/(4*1.73);
A2 = a**2/8;
A3 = A2;A4 = A1;
A5 = a**2/(4*1.73);
A6 = a**2/2;
A = A1+A2+A3+A4+A5+A6;
Pavg = (P1*A1+P2*A2+P3*A3+P4*A4+P5*A5+P6*A6)/A;
# Results
print "Mean precipitaion = %.2f cm."%(Pavg);
Example 4.43 pg : 218¶
In [15]:
import math
from numpy import linspace
#Given
A = [90, 140, 125, 140, 85, 40, 20]; #area of isohytes
I = linspace(13,1,7) #average isohytel interval
s = 0;t = 0;
for i in range(7):
s = s+A[i]*I[i];
t = t+A[i];
Pavg = s/t;
Pavg = round(Pavg*10)/10;
# Results
print " average depth of precipitation = %.2f cm."%(Pavg);
Example 4.44 pg : 218¶
In [16]:
import math
#Given
p = [120, 95, 96, 60, 65, 70, 45, 21]; #rain guage readings at respective stations
# Calculations and Results
s = sum(p)
pavg = s/8;
u = 0;
for i in range(8):
u = u+(p[i]-pavg)**2;
sx = (u/7)**0.5;
Cv = sx*100/pavg;
N = (Cv/13.99)**2;
N = round(N*100)/100;
print "mean rainfall = %.2f cm."%(pavg);
print "total stations needed = %.2f."%(N);
#taking N = 10
N = 10;
n = N-8;
print "additional guages needed = %i."%(n);
Example 4.45 pg : 219¶
In [22]:
%matplotlib inline
from matplotlib.pylab import plot
from numpy import zeros,linspace
import math
#compute maximum rainfall intensities for 5,10,15,20,25,30,35,40,45,50 minutes
#plot intensity duration graph
#Given
CR = [0, 1.02, 2.08, 3.30, 4.72, 5.58, 6.40, 7.16, 7.88, 8.54, 9.14]; #cumulative rainfall
c5 = zeros(11)
c10 = zeros(11)
c15 = zeros(11)
c20 = zeros(11)
c25 = zeros(11)
c30 = zeros(11)
c35 = zeros(11)
c40 = zeros(11)
c45 = zeros(11)
c50 = zeros(11)
c5[1] = CR[1];
c10[2] = CR[2];
c15[3] = CR[3];
c20[4] = CR[4];
c25[5] = CR[5];
c30[6] = CR[6];
c35[7] = CR[7];
c40[8] = CR[8];
c45[9] = CR[9];
c50[10] = CR[10];
for i in range(2,11):
c5[i] = CR[i]-CR[i-1];
for i in range(3,11):
c10[i] = CR[i]-CR[i-2];
for i in range(4,11):
c15[i] = CR[i]-CR[i-3];
for i in range(5,11):
c20[i] = CR[i]-CR[i-4];
for i in range(6,11):
c25[i] = CR[i]-CR[i-5];
for i in range(7,11):
c30[i] = CR[i]-CR[i-6];
for i in range(8,11):
c35[i] = CR[i]-CR[i-7];
for i in range(9,11):
c40[i] = CR[i]-CR[i-8];
for i in range(10,11):
c45[i] = CR[i]-CR[i-9];
#rainfall in any possible time interval
print "5min 10min 15min 20min 25min 30min 35min 40min 45min 50min";
for i in range(11):
print "%4.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f"%(c5[i],c10[i],c15[i],c20[i],c25[i],c30[i],c35[i],c40[i],c45[i],c50[i]);
I = [17.04, 15.84, 14.80, 14.16, 13.39, 12.80, 12.27, 11.82, 11.39, 10.97]; #maximum intensity at respective durations
D = linspace(5,50,len(I)) #durations
#graph is plotted between I and D
plot(I,D)
Out[22]:
Example 4.46 pg : 221¶
In [23]:
%matplotlib inline
import math
from matplotlib.pylab import plot
#draw storm hyetograph and intensity duration curve
#Given
p = [0, 5, 7.5, 8.5, 9]; #accumulated precipitation
t = [0, 30, 60, 90, 120]; #time
r = zeros(5)
I = zeros(5)
print "Rainfall intensity:";
for i in range(1,5):
r[i] = p[i]-p[i-1]; #rainfall in succesive 30 min interval
I[i] = r[i]*60/30; #rainfall intensity
print "%.2f"%(I[i]);
#graph is plotted between I and t.
plot(I,t)
Out[23]:
Example 4.47 pg : 222¶
In [24]:
%matplotlib inline
import math
from numpy import zeros,linspace
#Given
I = linspace(21,12,10) #isohytes
a = [543, 1345, 2030, 2545, 2955, 3280, 3535, 3710, 3880, 3915]; #enclosed area
ia = zeros(10)
ia[0] = 543;
for i in range(1,10):
ia[i] = a[i]-a[i-1]; #net incremental area between isohytes
rv = zeros(10)
r = linspace(21.5,12.5,10)
for i in range(10):
rv[i] = r[i]*ia[i]; #rainfall volume
cv = zeros(10)
cv[0] = 11675;
for i in range(10):
cv[i] = cv[i-1]+rv[i]; #cumulative volume
eud = zeros(10)
for i in range(10):
eud[i] = cv[i]/a[i]; #depth(mm)
print "From depth area curve we obtain average depth of precipitation = 20.15 mm for anarea of 2400 sq. km.";
#graph is plotted between eud and a
plot(eud,a)
Out[24]:
Example 4.48 pg : 222¶
In [21]:
import math
#Given
h1 = 7.75; #initial depth of water
r = 3.80; #rainfall during the week
hr = 2.50; #depth of water removed
C = 0.7; #pan coefficient
# Calculations
ha = r-hr;
hl = ha+h1;
h2 = 8.32;
ev = hl-h2;
evs = ev*C;
evs = round(evs*100)/100;
# Results
print "evaporation from reservior surface during the week = %.2f cm."%(evs);
Example 4.49 pg : 223¶
In [22]:
import math
from numpy import linspace
#Given
T = linspace(1,12,12) #time from start
r = [1.8, 2.6, 7.8, 3.9, 10.6, 5.4, 7.8, 9.2, 6.5, 4.4, 1.8, 1.6]; #increamental rainfall
R = 24.4; #total run-off
s = sum(r)
ti = s-R;
#first trial
tr = 7; #assumed
ti = s-R-r[0]-r[1]-r[3]-r[10]-r[11];
fi = ti/tr;
P = zeros(12)
for i in range(12):
P[i] = r[i]-fi;
if (P[i]<0):
P[i] = 0;
print "Timeh rainfall excess.";
for i in range(12):
print "%.2f %.2f"%(T[i],P[i]);
print "fi index = %.2f cm/hr."%(fi);
Example 4.50 pg : 224¶
In [23]:
import math
from numpy import linspace
#Given
r = [0.6, 1.35, 2.25, 3.45, 2.7, 2.4, 1.5, 0.75]; #incremental rainfall
T = linspace(1,8,8) #time from start of rainfal
t = 8.;
P = 15.; #total rainfall
R = 8.7; #direct run-off
# Calculations
W = (P-R)/t;
#math.since fi wil be more than W
tre = 6;
fi = ((P-R)-r[0]-r[7])/tre;
# Results
print "fi index = %.2f cm/hr."%(fi);
Example 4.51 pg : 224¶
In [25]:
import math
from scipy.integrate import quad
#Given
I = 10; #total infiltration rate
fI = 5; #final infiltration rate
k = 0.95; #rate of decay of difference between final and initial infiltration rate
# Calculations
def f8(t):
return fI+(I-fI)*math.e**(-k*t)
q = quad(f8,0,6)[0]
q = round(q*100)/100;
# Results
print "total infiltration depth = %.2f mm."%(q);
Example 4.52 pg : 224¶
In [26]:
import math
from numpy import linspace
#find the equation of infiltration capacity
#Given
fc = 1; #consmath.tant infiltration rate
ft = [10.4, 5.6, 3.2, 2.1, 1.5, 1.2, 1.1, 1, 1]; #infiltration capacity
f = ft[0]-fc;
t = linspace(0,2,9)
r = zeros(9)
for i in range(9):
r[i] = ft[i]-fc;
h = zeros(7)
for i in range(7):
h[i] = math.log10(r[i]);
s = 0.775; #from graph
k = 1/(math.log10(math.e)*s);
k = round(k*100)/100;
# Results
print "Equation is:ft = fc+%.2fe**-%.2ft)"%(f,k);
Example 4.53 pg : 226¶
In [25]:
%matplotlib inline
from matplotlib.pylab import bar
import math
#total rainfall
#net run-off
#W index
#Given
r = [2, 2, 8, 7, 1.25, 1.25, 4.5]; #rainfall intensity
T = [15, 30, 45, 60, 70, 90, 105]; #time
dt = 15.; #time interval
fi = 3.; #fi index
#graph is plotted between r and T
bar(T,r)
s = sum(r)
P = s*dt/60;
Pe = ((8-3)+(7-3)+(4.5-3))*dt/60; #area of graph above r = 3.0.
w = (P-Pe)/(105./60);
w = round(w*1000)/1000;
# Results
print "total rainfall = %.2f cm."%(P);
print "net run-off = %.2f cm."%(Pe);
print "W index = %.2f cm/hr."%(w);
Example 4.54 pg : 226¶
In [31]:
import math
#run-off by rainfall of 3.3cm in 3hrs
#Given
A = [36, 18, 66]; #area of catchment
fi = [0.9 ,1.1, 0.5]; #fi index
r1 = [0.6 ,0.9, 1.0]; #rainfall in first hour
r2 = [2.4 ,2.1, 2.0]; #rainfall in second hour
r3 = [1.3, 1.5, 0.9]; #rainfall in third hour
# Calculations and Results
t36 = r1[0]+r2[0]+r3[0];
t18 = r1[1]+r2[1]+r3[1];
t66 = r1[2]+r2[2]+r3[2];
p = (t36*A[0]+t18*A[1]+t66*A[2])/(A[0]+A[1]+A[2]);
print "Total rainfall in catchment = %.2f cm."%(p);
ro1 = [0 ,0, 0.5];
ro2 = [1.5 ,1.0, 1.5];
ro3 = [0.4, 0.4, 0.4]; #rainfall-fi
t1 = ro1[0]+ro2[0]+ro3[0];
t2 = ro1[1]+ro2[1]+ro3[1];
t3 = ro1[2]+ro2[2]+ro3[2];
run = (A[0]*t1+A[1]*t2+A[2]*t3)/(A[0]+A[1]+A[2]); #run-off from entire catchment
print "run-off by rainfall of 3.3cm in 3hrs = %.2f cm."%(run);
fia = (fi[0]*A[0]+fi[1]*A[1]+fi[2]*A[2])/(A[0]+A[1]+A[2]);
tr = (1.1-fia)*3;
print "Total run-off = %.2f cm."%(tr);
Example 4.55 pg : 227¶
In [33]:
import math
from numpy import array
#Given
P = array([4, 22, 28, 15, 12, 8, 4, 15, 10, 5]); #Precipitation
R = array([0.2, 7.1, 10.9, 4.0, 3.0, 1.3, 0.4, 4.1, 2.0, 0.3]); #runoff
# Calculations
Ps = P**2
Rs = R**2
PR = P*R
s = sum(Ps)
t = sum(Rs)
u = sum(PR)
q = sum(P)
w = sum(R)
N = 10.;
a = (N*u-q*w)/(N*s-q**2);
b = (w-a*q)/N;
a = round(a*10000)/10000;
b = round(b*10000)/10000;
# Results
print "Equation is:%.2f P %.2f."%(a,b);
Example 4.56 pg : 228¶
In [117]:
import math
#Given
Q = 470; #peak discharge of flood hydrograph
B = 15; #base flow
l = 0.25; #infiltration loss
Qr = Q-B;
d = 8; #average depth of rainfall
# Calculations
re = d-l*6; #rainfall excess
q = Qr/re;
# Results
print "peak discharge of 6 hrs unit hydrograph = %i cumecs."%(q);
Example 4.57 pg : 228¶
In [34]:
import math
from numpy import array,zeros
#Given
fi = 0.25; #infiltration index
B = 20; #Base flow
O = array([0, 20, 60, 150, 120, 90, 70, 50, 30, 20, 10, 0, 0, 0]); #ordinates of unit hydrograph
R1 = 5;
R2 = 0.8;
R3 = 3;
r1 = R1-(fi*4); #rainfall excess in first four hour
r2 = R2-(fi*4); #rainfall excess in second four hour
r3 = R3-(fi*4); #rainfall excess in third four hour
if r2<0 :
r2 = 0;
# Calculations and Results
s1 = r1*O
s2 = zeros(14)
for i in range(1,14):
s2[i] = r2*O[i-1];
s3 = zeros(14)
for i in range(2,14):
s3[i] = r3*O[i-2];
#surface run-off from rainfall excess during succesive unit periods
print "ordinates of storm hydrograph";
T = zeros(14)
t = zeros(14)
for i in range(14):
T[i] = s1[i]+s2[i]+s3[i]; #sub-total
t[i] = T[i]+B; #ordinate of flood hydrograph
print "%i"%(t[i]);
Example 4.58 pg : 229¶
In [35]:
import math
from numpy import array
#Given
fi = 2.5; #fi index
t = 24.;
A = 200.; #area of catchment
R1 = 7.5;
R2 = 2.0;
R3 = 5.; #rainfall
r1 = R1-fi;
r2 = R2-fi;
r3 = R3-fi;
r2 = 0;
r = [5, 0, 2.5]; #excess rainfall
D = array([5 ,15, 40, 25, 10, 5, 0, 0, 0]); #distribution
d1 = D*r[0]/100
d2 = zeros(9)
for i in range(8):
d2[i+1] = D[i]*r[1]/100;
d3 = zeros(9)
for i in range(7):
d3[i+2] = D[i]*r[2]/100;
#distribution run-off for rainfall excess
tr1 = zeros(9)
tr2 = zeros(9)
for i in range(9):
tr1[i] = d1[i]+d2[i]+d3[i]; #total run-off as depth
tr2[i] = 23.148*tr1[i]; #total run-off as discharge
tr2[i] = round(tr2[i]*1000)/1000;
s = sum(tr2)
print "Total run-off:";
print "as depth as discharge";
for i in range(9):
print "%.2f %.2f"%(tr1[i],tr2[i]);
r = 0.36*s*t/A;
r = round(r*10)/10;
print "total run-off = %.2f cm."%(r);
Example 4.59 pg : 230¶
In [36]:
import math
from numpy import zeros
#Given
O = [10, 30, 90, 220, 280, 220, 166, 126, 92, 62, 40, 20, 10]; #ordinates of 6 hr flood hydrograph
B = 10; #Base flow
r = zeros(13)
for i in range(13):
r[i] = O[i]-B; #ordinates of direct run-off
print "Ordinates of 6 hr unit hydrograph";
u = zeros(13)
for i in range(1,13):
u[i] = r[i]-u[i-1]; #ordinates of 6 hrs unit hydrograph
for i in range(13):
print "%i"%(u[i]);
Example 4.60 pg : 231¶
In [37]:
import math
from numpy import zeros
#determine the ordinates of 1 cm-6 hour hydrograph
#Given
t = 6;
A = 450; #catchment area
O = [5, 15, 40, 80, 60, 50, 25, 15, 5]; #ordinates of flood hydrograph
B = 5; #base flow assumed
s = 0;
r = zeros(9)
for i in range(9):
r[i] = O[i]-B; #ordinates of direct run-off
s = s+r[i];
n = s*0.36*12/A;
print "ordinates of unit hydrograph";
for i in range(9):
u[i] = r[i]/n;
u[i] = round(u[i]*100)/100;
print "%.2f"%(u[i]);
Example 4.61 pg : 232¶
In [38]:
import math
from numpy import zeros
#obtain ordinates 24 hr unit hydrograph
#Given
O = [0, 5.5, 13.5, 26.5, 45, 82, 162, 240, 231, 165, 112, 79, 57, 42, 31, 22, 14, 9.5, 6.6, 4, 2, 1, 0, 0, 0, 0, 0]; #ordinates of 1st 8 hrs unit hydrograph
o1 = zeros(27)
o2 = zeros(29)
for i in range(25):
o1[i+2] = O[i]; #ordinates of 2nd 8 hrs unit hydrograph
o2[i+4] = O[i]; #ordinates of 3rd 8 hrs unit hydrograph
o3 = zeros(27)
t = zeros(27)
print "ordinates 24 hr unit hydrograph:";
for i in range(27):
o3[i] = o1[i]+o2[i]+O[i]; #total 24 hr hydrograph of 3 cm run-off
t[i] = o3[i]/3;
t[i] = round(t[i]*10)/10;
print "%.2f"%(t[i]);
Example 4.62 pg : 233¶
In [26]:
%matplotlib inline
import math
from numpy import zeros,linspace
#ordinates of 1 hr unit hydrograph
#Given
t = linspace(0,12,13) #time
O = [0, 0, 54, 0, 175, 0, 127, 0, 58, 0, 25, 0, 0, 0]; #ordinate of 2 hr unit hydrograph
of = zeros(13)
for i in range(2,13):
if (i%2) == 0:
of[i] = 0
else:
of[i] = O[i-2]+of[i-2];
s = [0, 25, 54, 120, 229, 300, 356, 390, 414, 430, 439, 439, 439]; #Ordinates of S-curve
of1 = zeros(13)
for i in range(1,13):
of1[i] = s[i-1];
y = zeros(13)
u = zeros(13)
print "ordinates of 1 hr unit hydrograph:";
for i in range(13):
y[i] = s[i]-of1[i];
u[i] = y[i]*2;
print "%i"%(u[i]);
#graph is plotted between u and t
plot(t,u)
# graph in book is wrong. Please check.
Out[26]:
Example 4.63 pg : 234¶
In [43]:
import math
#Given
xavg = 1200; #sample mean
n = 50.; #assurance year
A = 0.95; #assurance percent
Rsk = 1-A;
sigma = 650; #smath.radians(numpy.arcmath.tan(ard deviation
yn = 0.53622; #mean of reduced variate
sigma30 = 1.11238; #smath.radians(numpy.arcmath.tan(ard deviation of reduced variate
# Calculations
T = 1/(1-(1-Rsk)**(1/n));
yt = -2.303*math.log10(2.303*math.log10(T/(T-1)));
K = (yt-yn)/sigma30;
xt = xavg+K*sigma;
# Results
print " design disharge = %i cumecs."%(xt);
Example 4.64 pg : 235¶
In [44]:
import math
#Given
T1 = 50.;
T2 = 100.; #Return period
F1 = 20600.;
F2 = 22150; #Peak flood
# Calculations
y100 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));
y50 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));
y = (F2-F1)/(y100-y50);
T = 500.;
y500 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));
x500 = F2+(y500-y100)*y;
x500 = round(x500);
# Results
print "flood magnitude with return period of 240 years = %.2f cumec."%(x500);
Example 4.65 pg : 235¶
In [45]:
import math
#Given
xavg = 1.65; #mean of data
sigma = 0.45; #smath.radians(numpy.arcmath.tan(ard deviation
x = 3;
# Calculations
y = 1.2825*(x-xavg)/sigma+0.577;
l = math.e**(math.e**(-y));
T = l/(l-1);
T = round(T*10)/10;
# Results
print "recurrence interval of 10 minutes storm = %.2f years."%(T);
Example 4.66 pg : 236¶
In [46]:
import math
#Given
T1 = 50.;
T2 = 100.; #Return period
F1 = 30800.;
F2 = 36300.; #Peak flood
# Calculations
y100 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));
y50 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));
y = (F2-F1)/(y100-y50);
T = 200.;
y200 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));
x200 = F2+(y200-y100)*y;
x200 = round(x200);
# Results
print "flood magnitude with return period of 240 years = %.2f cumecs."%(x200);
Example 4.67 pg : 236¶
In [49]:
import math
#Given
xavg = 4200.; #mean
sigma = 1705.; #smath.radians(numpy.arcmath.tan(ard deviation
xt = 9550.; #flood value
# Calculations
K = (xt-xavg)/sigma;
yt = 1.2825*K+0.577;
l = math.e**(math.e**(-yt));
T = l/(l-1);
# Results
print "Return period of flood of 9950 cumec/s = %.2f years."%(T);
Example 4.68 pg : 237¶
In [50]:
import math
#Given
T1 = 100.;
T2 = 50.; #Return period
F1 = 485.;
F2 = 445.; #Peak flood
# Calculations
y50 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));
y100 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));
y = (F2-F1)/(y50-y100);
T = 1000.;
y1000 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));
x1000 = F2+(y1000-y50)*y;
x1000 = round(x1000*10)/10;
# Results
print "flood magnitude with return period of 240 years = %.2f cumecs."%(x1000);
Example 4.69 pg : 238¶
In [51]:
import math
#probability of exceedence
#probability of occurence in next 12 years
#Given
T = 25.; #return period
n = 12.;
# Calculations
P = 1/T;
Rsk = 1-(1-P)**n;
P = round(P*100)/100;
Rsk = round(Rsk*10000)/10000;
# Results
print "probability of exceedence = %.2f."%(P);
print "probability of occurence in next 12 years = %.2f."%(Rsk);