From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 2087/CH2/EX2.5/example2_5.sce | 48 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 48 insertions(+)
 create mode 100755 2087/CH2/EX2.5/example2_5.sce

(limited to '2087/CH2/EX2.5')

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);
-- 
cgit