Use the chart!!! -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 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.

Humid air is fed to condenser to get the condensed water.(a) Process flow chart is shown below: (b) Degree of freedom analysis: Number of unknows = 4 (n1, n2, y, and Q) Number of independent equations = 2 (H2O, dry air balance) Number of independent energy balance equation = 1 Number of other relations = 1 (exit stream is saturated) Therefore, the DOF becomes, DOF = 4-2-1-1 DOF = 0(c) Apply overall material balance 1 = n1 + n2 ............ (1) Apply dry air balance 0.88 ×1 = n1 ×1-y .............. (2) Apply energy balance Q = ∑outniHi - ∑inniHi ............ (3)(d) It I given that the exit stream is saturated. It means the % RH humidity of the exit stream is 100 %. From the given psychrometric chart, the absolute humidity of saturated air-water mixture at 25 0C is 0.02 kg H2O/kg dry air. Therefore, the mass fraction of water and dry air in exit gas stream becomes, y = 0.020.02 + 1y = 0.0196 kg H2Ovkg Plugin the value of y in equation (2) 0.88 = n1×1-0.0196n1 = 0.8976 kgsFrom overall material balance1 = 0.8976 + n2n2 = 0.1024 kg H2Ols (e) Reference states: For water: H2O (l), 25 0C, 1 atm For air: air at 250C, 1 arm