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+// Example 17_8
+clc;funcprot(0);
+//Given data
+T_s=38;// The temperature of the steam entering the condenser in °C
+T_a=34;// The temperature of the air  entering the air pump in °C
+T_c=36;// The temperature of the air of the condensate in °C
+m_a=3;// kg/hr
+m_c=8000;//The condensate removed in kg/hr
+R=287;// J/kg.k
+
+//Calculation
+//(a)
+//From steam table, a saturation temperature at 38°C
+p_s1=0.0676;// bar
+p_a1=0.0;// bar
+p_t=p_a1+p_s1;// bar
+//From steam table, a saturation temperature at 34°C
+v_s1=26.5;// kg/hr
+p_s=0.0542;// bar
+p_a=p_t-p_s;// Partial pressure of air at the entry of air pump in bar
+V_1=(m_a*R*(T_a+273))/(p_a*10^5);// m^2/hr
+
+//(b) 
+// From steam table, a saturation temperature at 36°C
+v_s2=24;// kg/hr
+p_s=0.0606;// bar
+p_a=p_t-p_s;// bar
+V_2=(m_a*R*(T_c+273))/(p_a*10^5);// m^2/hr
+V=m_c*0.001006;// m^3/hr
+Tv=V_2+V;// Total volume removed by wet air pump in m^3/hr
+Pi_apc=((Tv-V_1)/V_1)*100;// Percentage increase in air-pump capacity in %
+m_wd=(V_1/v_s1);// Mass of water vapour carried with air when dry air-pump is used to remove the air in kg/hr
+m_ww=(Tv/v_s2);// Mass of water vapour carried with air when wet air-pump is used to remove the air in kg/hr
+Pi_lwv=((m_ww-m_wd)/m_wd)*100;// Percentage increase in loss of water vapour
+printf('\n(a)The Capacity of the air pump=%0.0f m^3/hr \n(b)Percentage increase in air-pump capacity=%0.0f percentage \n   Percentage increase in air-pump capacity=%0.1f percentage',Tv,Pi_apc,Pi_lwv);
+// The answer vary due to round off error