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+clear;

+clc;

+printf("\t\t\tProblem Number 3.12\n\n\n");

+// Chapter 3 : The First Law Of Thermodynamics

+// Problem 3.12 (page no. 112) 

+// Solution

+

+Z1=2; //Unit:m //Inlet position

+g=9.81 //Unit:m/s^2 //g=The local gravity

+V1=40; //Unit:m/s //Inlet velocity

+h1=3433.8; //Unit:kJ/kg //Inlet enthalpy

+q=1 //Unit:kJ/kg //Heat losses

+Z2=0; //Outlet position //unit:m

+V2=162; //Unit:m/s //Outlet velocity

+h2=2675.5; //Unit:kJ/kg //Outlet enthalpy

+

+//Energy equation is given by

+//((Z1*g)) + (V1^2/2) + h1 + q = ((Z2*g) + (V2^2/2) + h2 + w

+

+w= ((Z1*g)/1000) + ((V1^2/2)/1000) + h1 - q - ((Z2*g)/1000) - ((V2^2/2)/1000) - h2 ; //Unit:kJ/kg //Conersation: 1 kJ=1000 J

+printf("The work output per kilogram is %f kJ/kg\n",w);