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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clc;
+warning("off");
+printf("\n\n example7.13 - pg303");
+// given
+deltaz=30; //[ft] - distance between process and the holding tank
+Q=100; //[gpm] - volumetric flow rate of water
+p1=100; //[psig]
+p2=0; //[psig]
+g=32.1; //[ft/sec] - acceleration due to gravity
+sv=0.0161; //[ft^3/lb] - specific volume of water
+p=1/sv; //[lb/ft^3] - density of water
+e=0.77; // efficiency of centrifugal pump
+deltap=(p1-p2)*(144); //[lbf/ft^2]
+gc=32.174;
+// using the equation deltap/p+g*(deltaz)+Ws=0;
+Wst=-deltap/p-(g/gc)*(deltaz);
+// using the formula for efficiency e=Ws(theoritical)/Ws(actual)
+// therefore
+Wsa=Wst/e;
+// the calulated shaft work is for a unit mass flow rate of water,therfore for given flow rate multiply it by the flow rate
+w=(Q*p)/(7.48*60);
+Wsactual=Wsa*w;
+power=-Wsactual/(778*0.7070);
+printf("\n\n the required horsepower is %fhp",power);