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+// Example 8_5

+clc;funcprot(0);

+// Given data

+V=1.00;// m^3

+m=2.00;// kg

+T_1=20.0;// °C

+T_2=95.0;// °C

+T_b=100.0;// °C

+

+// Calculation

+// (a)

+v_1=V/m;// m^3/kg

+v_2=v_1;// m^3/kg

+// From Table C.1b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that

+// At 20.0°C

+v_f1=0.001002;// m^3/kg

+v_g1=57.79;// m^3/kg

+v_fg1=v_g1-v_f1;// m^3/kg

+u_f1=83.9;// kJ/kg

+u_g1=2402.9;// kJ/kg

+u_fg1=u_g1-u_f1;// kJ/kg

+// At 95.0°C

+v_f2=0.00104;// m^3/kg

+v_g2=1.982;// m^3/kg

+v_fg2=v_g2-v_f2;// m^3/kg

+u_f2=397.9;// kJ/kg

+u_g2=2500.6;// kJ/kg

+u_fg2=u_g2-u_f2;// kJ/kg

+x_1=(v_1-v_f1)/v_fg1;// The quality in the container when the contents are at 20.0°C

+x_1p=x_1*100;// %

+// (b)

+x_2=(v_2-v_f2)/v_fg2;// The quality in the container when the contents are at 95.0°C.

+x_2p=x_2*100;// %

+// (c)

+u_1=u_f1+(x_1*u_fg1);// kJ/kg

+u_2=u_f2+(x_2*u_fg2);// kJ/kg

+Q_12=m*(u_2-u_1);// kJ

+// (d)

+s_f1=0.2965;// kJ/kg.K

+s_fg1=8.3715;// kJ/kg.K

+s_f2=1.2503;// kJ/kg.K

+s_fg2=6.1664;// kJ/kg.K

+s_1=s_f1+((x_1)*s_fg1);// kJ/kg.K

+s_2=s_f2+((x_2)*s_fg2);// kJ/kg.K

+S_p_12=((m*(s_2-s_1))-(Q_12/(T_b+273.15)))*1000;// J/K

+T_b_minimum=Q_12/(m*(s_2-s_1));// K

+T_b_minimum=T_b_minimum-273.15;// °C

+printf('\n(a)The quality in the container when the contents are at 20.0°C,x_1=%0.3f percentage \n(b)The quality in the container when the contents are at 95.0°C,x_2=%2.1f percentage \n(c)The heat transport of energy required to raise the temperature of the contents from 20.0 to 95.0°C,Q_12=%4.0f kJ/kg \n(d)The entropy production,S_P=%3.0f J/K \n   The minimum boundary temperature,(T_b)minimum=%2.1f°C',x_1p,x_2p,Q_12,S_p_12,T_b_minimum);