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+// Example 15_5

+clc;funcprot(0);

+// Given data

+T_DB=90.0;// °F

+T_WB=75.0;// °F

+phi=50;// The relative humidity in %

+w=105*1/7000;// lbm H2O/lbm dry air

+M_da=28.97;// lbmdry air/lbmole dry air

+M_H2O=18.016;// lbmH2O/lbmoleH2O

+

+// Calculation

+w=w*(M_da/M_H2O);// lbmole H2O/lbmole dry air

+// From the balanced reaction equation of part a of Example 15.3, we find that the amount of dry air used per mole of fuel is

+a_da=21.9*(1+3.76);// moles

+a_w=w*a_da;// moles of water

+n_H2O=9.00+a_w;// moles per mole of fuel

+n_total=111.5;// moles per mole of fuel

+X_H2O=n_H2O/n_total;// The mole fraction of water vapor in the exhaust

+p_H20=X_H2O*14.7;// psia

+// Again, interpolating in Table C.1a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find

+T_DP=116.0;//°F

+T_DP=(T_DP-32)/1.8;// °C

+T_sat=T_DP;// °C

+printf("\n(a)The amount of water carried into the engine in the form of inlet humidity,w=%0.4f lbmole H2O/lbmole dry air \n(b)The new dew point temperature of the exhaust products,T_DP=%2.1f°C",w,T_DP);