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+// Example 17_4
+clc;funcprot(0);
+//Given data
+m_s=50000;// Steam condensed in kg/hr
+T_s=40;// Temperature of steam in a condenser in °C
+x=0.85;// Dryness of steam entering into condenser
+m_a=150;// The air leakage in the condenser in kg/hr
+T_c=35;// Temperature of the condensate in °C
+T_suction=32;//Temperature at the suction of the air pump in °C
+gradT=10;//The rise in cooling water temperature in °C
+R=287;// Gas constant in J/kg k
+p_b=1.013;// bar
+C_pw=4.2;// kJ/kg.k
+
+//Calculation
+//From steam tables
+//(a)
+//At 40°C saturation temperature 
+p_s=0.0752;// Pressure in bar
+v_s=19.5;// Specific volume in m^3/kg
+V=m_s*x*v_s;// Volume in m^3
+p_a=(m_a*R*(T_s+273))/(V*10^5);// The pressure of air in the condenser in bar
+p_t=p_a+p_s;// The total pressure in the condenser in bar
+P_v=(p_b-p_t)/0.013959;// Vacuum in condenser in cm of Hg
+
+//(b)
+// From steam tables ,At 32°C
+p_s1=0.0485;// Partial pressure of steam in bar
+p_a1=p_t-p_s1;// bar
+V_1=(m_a*R*(T_suction+273))/(p_a1*10^5);// Volume of air at 32°C in m^3/hr
+Apc=V_1;// Air pump capacity in m^3/hr
+
+//(c)
+v_s1=29.6;// Specific volume of steam at 32°C saturation temperature in m^3/hr
+Ls=V_1/v_s1;// Loss of steam in kg/hour
+
+//(d)
+// From steam tables ,At 40°C saturation temperature and 0.85 dry
+h_f1=168;// kJ/kg
+h_fg1=2414;// kJ/kg
+h_1=h_f1+(x*h_fg1);// kJ/kg
+h_f2=147;// kJ/kg
+m_w=(m_s*(h_1-h_f2))/(C_pw*gradT*1000);// Quality of cooling water passed through the condenser in tons/hr
+printf('\n(a)Vacuum in condenser=%0.2f cm of Hg \n(b)Capacity of dry air pump=%0.1f m^3/hr \n(c)Loss of steam in kg per hour=%0.1f kg/hr \n(d)Quality of cooling water passed through the condenser=%0.0f tons/hr',P_v,Apc,Ls,m_w);
+// The answer provided in the textbook is wrong