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+// Example 22_18

+clc;funcprot(0);

+//Given data

+T_1=500;// °C

+p_1=40;// bar

+p_2=10;// bar

+p_3=0.04;// bar

+m_b=50;// The boiler generation rate in tons/hour

+n_m=85/100;// Mechanical efficiency

+n_g=95/100;// Electrical generation efficiency

+

+// Calculation

+// From h-s chart:

+h_1=3400;// kJ/kg

+h_2=3050;// kJ/kg

+h_3=2150;// kJ/kg

+// From steam tables

+h_f4=121.4;// kJ/kg(at 0.04 bar)

+h_f5=762.6;// kJ/kg(at 10 bar)

+h_f6=h_f5;// kJ/kg

+//Assume m_1=y(1);h_fm=y(2)

+function[X]=mass(y)

+    X(1)=((y(1)*h_f6)+((1-y(1))*h_f4))-(y(1)*y(2));

+    X(2)=(y(1)*(h_2-h_f5))-(1*(h_f5-y(2)));

+endfunction

+y=[0.1 100];

+z=fsolve(y,mass);

+m=z(1);// kg/kg of steam generated

+h_fm=z(2);// kJ/kg

+W=(h_1-h_2)+((1-m)*(h_2-h_3));// kJ/kg 

+m_b=m*100;// Bled steam in %

+Q_s=h_1-h_f5;// Heat supplied per kg of steam in kJ/kg

+n=(W/Q_s)*100;// Efficiency in %

+P=(((m_b*10^3)*W*n_m*n_g)/3600)/1000;// Power developed in MW

+printf('\nThe percentage of bled steam=%0.0f percentage \nThe thermal efficiency of the plant=%0.1f percentage \nThe generating capacity of the plant=%0.1f MW',m_b,n,P);

+// The answer provided in the textbook is wrong

+