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+clc;
+warning("off");
+printf("\n\n example7.12 - pg301");
+// given
+Q=50/(7.48*60); //[ft/sec] - volumetric flow rate of water
+d1=1; //[inch] - diameter of pipe
+deltaz=-5; //[ft] - distance between end of pipe and tank
+g=32.1; //[ft/sec] - acceleration due to gravity
+Cp=1; //[Btu/lb*F] - heat capacity of water
+p=62.4; //[lb/ft^3] - density of water
+S1=(%pi/4)*(d1/12)^2;
+U1=Q/S1;
+w=p*Q;
+U2=0;
+gc=32.174;
+// using the formula deltaH=(w/2)*((U2)^2-(U1)^2)+w*g*deltaz
+deltaH=-(w/(2*gc))*((U2)^2-(U1)^2)-w*(g/gc)*deltaz;
+disp(deltaH);
+deltaH=deltaH/778; // converting from ftlb/sec to Btu/sec
+deltaT=deltaH/(w*Cp);
+printf("\n\n The rise in temperature is %fdegF",deltaT);
+
+
+
+