//(12.14) A stream consisting of 142 m3/min of moist air at a temperature of 5C and a humidity ratio of 0.002 kg(vapor)kg(dry air) is mixed adiabatically with a second stream consisting of 425 m3/min of moist air at 24C and 50% relative humidity. The pressure is constant throughout at 1 bar. Using the psychrometric chart, determine (a) the humidity ratio and (b) the temperature of the exiting mixed stream, in C. //solution //variable initialization AV1 = 142 //in m^3/min T1 = 5 //in degree celcius omega1 = .002 AV2 = 425 //in m^3/min T2 = 24 //in degree celcius psi2 = .5 p = 1 //in bar //part(a) //from the psychrometric chart, Fig. A-9. va1 = .79 //in m^3/kg va2 = .855 //in m^3/kg omega2 = .0094 ma1dot = AV1/va1 //in kg/min ma2dot = AV2 /va2 //in kg/min omega3 = (omega1*ma1dot+omega2*ma2dot)/(ma1dot + ma2dot) printf('the humidity ratio is: %f',omega3) //part(b) //Reduction of the energy rate balance gives //(ha + omega*hv)3 = [ma1dot*(ha + omega*hv)1 + ma2dot*(ha + omega*hv)2]/(ma1dot+ma2dot) //with (ha + omega*hv)1 = 10kj/kg and (ha + omega*hv)2 = 47.8kj/kg from figure A-9 LHS = (ma1dot*10+ma2dot*47.8)/(ma1dot + ma2dot) //This value for the enthalpy of the moist air at the exit, together with the previously determined value for omega3, fixes the state of the exiting moist air. From inspection of Fig. A-9, T3 = 19 //in degree celcius printf('\n\nthe temperature of the exiting mixed stream in degree celcius is: %f',T3)