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+// Example 17_5

+clc;funcprot(0);

+//Given data

+T_s=56;// Temperature of steam entering the condenser in °C

+T_a=46;// Temperature at the air pump suction in °C

+P_b=76;// The barometer reading in cm of Hg

+Q=90;// The discharge of dry air pump in m^3/min

+R=287;// J/kg.k

+

+//Calculation

+//(a)

+//From steam tables,at saturation temperature of 56°C

+p_s=0.1684;//Pressure of steam in bar

+p_s=p_s/0.01359// cm of Hg

+p_a=0;// Partial pressureair at the inlet of condenser in cm of Hg

+p_t=p_s+p_a;

+p_v=P_b-p_t;//Vacuum in the condenser in cm of Hg

+

+//(b)

+//From steam tables,at saturation temperature of 46°C

+p_s1=0.1028;// bar

+v_s=14.56;// m^3/kg

+p_a1=(p_t*0.01359)-p_s1;// bar

+m_a=(p_a1*10^5*Q*60)/(R*(T_a+273));//The air leakage in the condenser per hour in kg/hr

+//(c)

+Ls=(Q*60)/v_s;// Loss of condensate in kg/hr

+printf('\n(a)The vacuum in the condenser=%0.1f cm of Hg \n(b)The air leakage in the condenser=%0.1f kg/hr \n(c)Loss of condensate=%0.0f kg/hr',p_v,m_a,Ls);

+// The answer vary due to round off error