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+//Transport Processes and Seperation Process Principles

+//Chapter 3

+//Example 3.3-2

+//Principles of Momentum Transfer and Applications

+//given data

+Ps=741.7;//suction pressure in mm hg

+Pd=769.6;//discharge pressure in mm hg

+Patm=760;//atmospheric pressure in mm Hg

+rho1=28.97*(1/22.414)*(273.2/366.3)*(Ps/Patm);//air density at suction: mol wt= 28.97 kg air/kg mol for 22.414 m3/kg mol at 101.3 kPa and 273.2 K

+rho2=rho1*(Pd/Ps);

+rhoavg=(rho1+rho2)/2

+V=28.32;//volumetric flow rate in m3/s

+Ts=294.1;//temp at suction

+mdot=V*(1/60)*(1/22.414)**(273.2/Ts)*28.97;//mass flow rate of gas

+Patm=760;//atm pressure in mm Hg

+Hp=((Pd-Ps)/Patm)*(101325/rhoavg);//pressure head in J/kg

+v1=0;//air is stationary

+v2=45.7;//discharge velocity in m/s

+vd=(((v2^2)-(v1^2))/2);//developed velocity

+z1=0;

+sumF=0;

+Ws=Hp+vd;//substituting and solving for Ws by mechanical energy balance equation for a closed system in J/kg

+n=60/100;//efficiency given is 60%

+bkW= (Ws*mdot)/(n*1000);// brake kW

+mprintf(" brake kW= %f hP",bkW)