initial commit / add all books

7 files changed, 202 insertions, 0 deletions

diff --git a/2087/CH2/EX2.1/example2_1.sce b/2087/CH2/EX2.1/example2_1.sce
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+++ b/2087/CH2/EX2.1/example2_1.sce
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+

+

+//example 2.1

+//calculate time required to cover 0.1 hectare area by tubewell

+clc;

+//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

+t=(60*2.303*y*log10(Q/(Q-I*A)))/I,

+t=round(t);

+mprintf("Time required to cover given area=%f minutes.",t);

diff --git a/2087/CH2/EX2.2/example2_2.sce b/2087/CH2/EX2.2/example2_2.sce
new file mode 100755
index 000000000..1f046003c
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+++ b/2087/CH2/EX2.2/example2_2.sce
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+

+

+//example 2.2

+//calculate maximum area that can be irrigated

+clc;

+//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

+Amax=Q/I;

+mprintf("Maximum area that can be irrigated =%f hectare.",Amax);

diff --git a/2087/CH2/EX2.3/example2_3.sce b/2087/CH2/EX2.3/example2_3.sce
new file mode 100755
index 000000000..fcb14146a
--- /dev/null
+++ b/2087/CH2/EX2.3/example2_3.sce
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+//example 2.3            

+

+//calculate 

+//time of water application

+//optimum length of each border strip

+//dischrge for each border strip

+

+clc;

+//Given

+d=0.05;//depth of root zone

+I=1.25D-5;//average infiltration rate

+s=0.0035//slope of border strip

+t=d/(I*3600);

+t=round(t*1000)/1000;

+mprintf("Time of water application=%f hours.",t);

+

+//Part (a)

+q=2D-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-%e^(-d/y));

+L=round(10*L)/10;

+mprintf("\nPart (a):");

+mprintf("\nOptimum length of each border strip=%f m.",L);

+

+//Part (b)

+Lgiven=150//given value of length

+//First Trial

+q=3D-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-%e^(-d/y));

+//second trial

+q=3.15D-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-%e^(-d/y));

+q=9*Lgiven*q*1000/L;

+q=round(q*10)/10;

+mprintf("\nPart (b):");

+mprintf("\nDischarge for each border strip=%f lps.",q);

diff --git a/2087/CH2/EX2.4/example2_4.sce b/2087/CH2/EX2.4/example2_4.sce
new file mode 100755
index 000000000..ab903ce5c
--- /dev/null
+++ b/2087/CH2/EX2.4/example2_4.sce
@@ -0,0 +1,51 @@
+

+

+//example 2.4

+//calculate 

+//deep percolation loss

+//water application efficiency and time of irrigation.

+

+clc;

+//Given

+B=12;//breadth of basin

+L=36//length of basin

+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

+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;

+mprintf("Part (a):")

+mprintf("\nDeep percolation loss= %f percent.",p);

+mprintf("\nWater application efficiency= %f percent.",eita);

+mprintf("\nTime of irrigation= %f 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;

+mprintf("\nPart (b):")

+mprintf("\nDeep percolation loss= %f percent.",p);

+mprintf("\nWater application efficiency= %f percent.",eita);

+mprintf("\nTime of irrigation= %f minutes.",Tdash);

diff --git a/2087/CH2/EX2.5/example2_5.sce b/2087/CH2/EX2.5/example2_5.sce
new file mode 100755
index 000000000..5a5257f91
--- /dev/null
+++ b/2087/CH2/EX2.5/example2_5.sce
@@ -0,0 +1,48 @@
+//example 2.5

+//calculate 

+//size of cut-back stream.

+//time required for putting 37.5 mm depth of water

+//average depth of water applied

+

+clc;

+//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

+

+for i=1:5//loop to find respective values of time of ponding

+    B(i)=143-A(i);

+end

+

+

+for j=1:5//loop to find respective furrow infiltration

+    C(j)=B(j)^(n)*k;

+end

+

+

+for K=1:4//loop to find respective average infiltration

+   

+   D(K)=(C(K)+C(K+1))/2;

+end

+

+E(1)=D(1);

+for l=2:4//loop to determine cumulative infiltration

+    E(l)=D(l)+E(l-1);

+end

+I=E(4);

+

+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;

+mprintf("Maximum size of cut-back stream=%f lpm.",I);

+mprintf("\nMinimum size of cut-back stream=%f lpm.",q);

+mprintf("\nTime required for putting 37.5mm depth of water=%f minutes.",T);

+mprintf("\nAverage depth of water required=%f mm.",dav);

diff --git a/2087/CH2/EX2.6/example2_6.sce b/2087/CH2/EX2.6/example2_6.sce
new file mode 100755
index 000000000..de2918a11
--- /dev/null
+++ b/2087/CH2/EX2.6/example2_6.sce
@@ -0,0 +1,14 @@
+

+//example 2.6

+//calculate Average depth of water applied

+clc;

+//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

+qm=0.6/s;

+q=qm*0.4;

+dav=((q*t2*60)+(2*t1*60))/100;

+mprintf("Average depth of water applied=%f mm.",dav);

diff --git a/2087/CH2/EX2.7/example2_7.sce b/2087/CH2/EX2.7/example2_7.sce
new file mode 100755
index 000000000..5d4169136
--- /dev/null
+++ b/2087/CH2/EX2.7/example2_7.sce
@@ -0,0 +1,17 @@
+

+

+//example 2.7

+//calculate 

+//time required to irrigate

+//maximum area that can be irrigated

+clc;

+//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

+t=(2.303*y*60/I)*log10(Q/(Q-I*A));

+Amax=Q/I;

+t=round(t*100)/100;

+mprintf("Time required to irrigate=%f minutes.",t);

+mprintf("\nMaximum area that can be irrigated=%f ha.",Amax);