//ques22 //Energy and Cost Savings by Fixing Air Leaks clear clc //The work needed to compress a unit mass of air at 20°C from the atmospheric pressure of 101 kPa to 700+101=801 kPa is R=0.287;//gas constant for water P2=801;//final pressure in kPa P1=101;//initial pressure in kPa n=1.4; nc=0.8; T1=293;//initial temperature in K w=n*R*T1/(nc*(n-1))*((P2/P1)^(1-1/n)-1);//work done in kJ/kg D=3*10^-3;//diameter in metre A=%pi*D^2/4;//area in m^2 //Line conditions are 297 K and 801 kPa, the mass flow rate of the air leaking through the hole is determined as Cdis=0.65; k=1.4;//k=n R=0.287;//gas constant for water Tline= 297;//temperature of line in K Pline=801;//pressure of line=P2 ms=Cdis*(2/(k+1))^(1/(k-1))*Pline/(R*Tline)*A*sqrt(1000*k*R*Tline*2/(k+1));//mass flow in kg/s pw=ms*w//power wasted in kW Esaving=pw*4200/0.92//in kWh/yr Energy saving=Power saved*operating hr/efficiency of motor printf('Energy saving = %.0f kWh/yr \n',Esaving); Csaving=Esaving *0.078;//Cost saving in $/yr = Energy saving * unit cost printf(' Cost saving = $ %.0f/yr \n',Csaving);