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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+
+//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);