Follow the standard energy balance steps to determine the water condensed and the heat removedfrom a humid air stream. The feed stream contains 12 wt% water (balance is air), has atemperature of 250°C and a pressure of 1 atm. The exit gas stream is at 1 atm and 25°C. Theexit vapor composition should be determined from the psychrometric chart (saturated). The heatof vaporization of water at 25°C is 44.0 kJ/mol.Draw a picture and label appropriately (Basis: 1 kg/s feed).a.b. Perform a degree of freedom analysis.c. Write appropriate equations.d. Complete all material balances.e. Specify reference states. -5.-1015 20-5825"0Wet Bulb or Saturation Temperature, °C5+0.20+0.4+0.6+0.810+1.045velalafate15Enthalpy At Saturation, kJ/kg Dry/VOL5909SSos-0.05no10-275 80 85 90 95 100 105202515Dry Bulb Temperature, °C3090%80%70%60%3540%30%50% Relative Humidity20%4010%450.9050-0.031-0.030€0.0290.0280.027€0.026€0.0250.024-0.023-0.0220.021€0.0200.0190.0180.0170.95 0.0160.015€0.0140.0130.012€0.0110.010-0.009-0.008E-0.007-0.0060.0050.0040.003€0.0020.0010.00100.750.800.85Humid Volume, m³/kg Dry AirBelow 0°C Properties and Enthalpy Deviation Lines Are For IceFIGURE 8.4-1 Psychrometric chart-SI units. Reference states: H₂O (liquid, 0°C, 1 atm), dry air (0°C, 1 atm). (Data obtained from Carrier Corporation.)55Moisture Content, kg/kg Dry Air

a) b) Degree of freedom analysis Number of unknowns = 4 (Q, n1, n2, y) Number of independent material balance equations = 2 (water, air) Number of other relations = 2 (exit gas stream is saturated) Number of energy balance equation = 1 Therefore, DOF = 4-2-1-1 DOF = 0 c) Convert the mass flow rate of feed inot molar flow rate as follows: The average molar mass of the feed stream Mavg = 0.12×18 + 0.88×29= 27.68 kgkmolThus,molar flow rate of feed = 1 Kg/s27.68 Kg/Kmol = 36.1 mol/sConvert the composition of feed from mass fraction to mole fractionwater = 0.12180.1218 + 0.8829 = 0.18 mol H2Ohence Air = 0.88 mol air(g) / mol36.1 = n1 + n2 ----- 1 Apply air balance 36.1×0.82 = (1−y)×n1 ----- 1d) It is given that the exit gas stream is saturated. It means the partial pressure of water becomes equal to the vapor pressure of water. From the steam table, the vapor pressure of water at 25 0C = 3.1698 kPa The total pressure of the stream = 1 atm or 101.325 kPa The mole fraction of water and air in the gas stream becomes, e) Reference states: Water = H2O (l), 25 0C, 1 atm Air = air (g), 25 0C, 1 atm