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+//Fluid Systems - By Shiv Kumar
+//Chapter 16- Hydraulic Power and Its Transmissions
+//Example 16.10
+//To Find   (i)The Weight of Loaded Cylinder and energy stored by the Cylinder   (ii)Ther Power supplied by the Accumulator     (iii)The Diameter of ram of an ordinary Accumulator.
+      clc
+      clear
+
+//Given Data:-
+         D=180;        // mm
+        d=150;      //mm
+        L=1.25;      //Stroke length, m
+         p=100;      //Pressure of Water, bar
+
+//Computations:-
+        D=D/1000;      //m
+        d=d/1000;      //m
+        p=p*10^5;      //N/m^2
+
+        A=(%pi/4)*(D^2-d^2);       //Annular area of Ram, m^2
+    //(i)         
+          W=p*A/1000;        //Weight of Loaded Cylinder, kN
+          Energy=W*L;         //Energy stored in the Accumulator, kNm
+   //(ii)
+           t=90;      //Time taken by ram to complete the stroke, seconds                                   
+           P=W*L/t;       //kW
+   //(iii)
+            D=(W*1000/(p*%pi/4))^(1/2)*1000;        //mm
+
+//Results:-
+         printf("(i)Weight of Loaded Cylinder, W=%.2f kN\n",W)       //The answer vary due to round off error
+             printf("    Energy stored in the Accumulator=%.3f kNm\n",Energy)       //The answer vary due to round off error
+         printf("(ii)Power Supplied by the Accumulator=%.3f kW\n",P)       //The answer vary due to round off error
+         printf("(iii)Ram Diameter (In case of Ordinary Accumulator) = %.2f mm\n",D)       //The answer vary due to round off error