initial commit / add all books

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+clc;

+clear;

+printf("\n Example 8.6\n");

+

+P1=101.3e3;

+Q_watr=0.01;

+printf("\n Given:\n Flow rate of Water = %.2f m^3/s",Q_watr);

+depth=100;

+printf("\n Depth of well = %d m",depth);

+d=100e-3;

+printf("\n Diameter of pipe = %d mm",d*1e3);

+depth_watr=40;

+printf("\n Level of water below water = %d m",depth_watr);

+Q_air=0.1;

+printf("\n Flow rate of Air = %.2f m^3/s",Q_air);

+P2=800e3;

+Gamma=1.4;

+//V1=Q_air;

+G_watr=Q_watr*1000;//Mass flow of water

+W=G_watr*depth_watr*9.81;

+//The energy needed to compress 0. 1 m^3/s of air is given by:

+E=P1*Q_air*(1.4/0.4)*((P2/P1)^(0.4/1.4)-1);//     equation 8.37

+printf("\n\n Calculations:\n The power required for this compression is = %d W",E);

+effi=W/E*100;

+printf("\n Efficiency = %.1f per cent",effi);

+//the mean pressure

+P=345e3;

+printf("\n The mean pressure = %d kN/m^2",P);

+v1=8314*273/(29*P);

+printf("\n The specific volume v of air at 273 K and given pressure is = %.3f m^3/kg",v1);

+v2=8314*273/(29*P1);

+printf("\n The specific volume v of air at 273 K and 101.3 kN/m^2 is = %.3f m^3/kg",v2);

+G_air=Q_air/v2; //mass flowrate of the air is:

+Q_mean=G_air*v1;//Mean volumetric flowrate of air

+Q_tot=Q_watr+Q_mean;//Total volumetric flowrate

+A=%pi/4*d^2;//Area of pipe

+v_mean=Q_tot/A;

+printf("\n Mean velocity of the mixture = %.2f m/s",v_mean);

+

+

+