### Enthalpy Table and Calculations**f. Construct and fill out the enthalpy table. Define variables for unknown enthalpies.**| Species/Phase | \(\dot{m}_{\text{in}}\) (kg/s) | \(\hat{H}_{\text{in}}\) (kJ/mol) | \(\dot{m}_{\text{out}}\) (kg/s) | \(\hat{H}_{\text{out}}\) (kJ/mol) ||---------------|-------------------|------------------|--------------------|--------------------|| H\(_2\)O (v) | | | | || Air (g) | | | | || H\(_2\)O (l) | | | | |**g. Determine unknown enthalpies.****h. Determine \(\dot{Q}\). Please give your answer in kW.**---For each species/phase, fill in the mass flow rates \(\dot{m}_{\text{in}}\) and \(\dot{m}_{\text{out}}\), as well as the specific enthalpies \(\hat{H}_{\text{in}}\) and \(\hat{H}_{\text{out}}\) where possible. Use these variables to help solve for any unknown enthalpies and calculate the energy transfer rate, \(\dot{Q}\), in kilowatts. **Problem Statement:**Follow the standard energy balance steps to determine the water condensed and the heat removed from a humid air stream. The feed stream contains 12 wt% water (balance is air), has a temperature of 250°C and a pressure of 1 atm. The exit gas stream is at 1 atm and 25°C. The exit vapor composition should be determined from the psychrometric chart (saturated). The heat of vaporization of water at 25°C is 44.0 kJ/mol.**Tasks:**a. Draw a picture and label appropriately (Basis: 1 kg/s feed). b. Perform a degree of freedom analysis.c. Write appropriate equations.d. Complete all material balances.e. Specify reference states.

A stream of humid air is fed to a condenser.(a) Labelled process flow chart is shown below: (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 kgs27.68 kgkmol= 0.0361 kmols × 1000 mol1kmolMolar flow rate of feed = 36.1 mols Convert the composition of feed from mass fraction to mole fraction water = 0.12180.1218 + 0.8829= 0.18 mol H2OvmolAir = 0.82 mol air gmol Apply overall material balance 36.1 = n1 + n2 .......... (1) Apply air balance 36.1×0.82 = 1-y×n1 ............ (2)(d) It is given that the exit gas stream is saturated. It means the partial pressure of water becomes equal to vapor pressure of water. From the steam table, vapor pressure of water at 25 0C = 3.1698 kPa Total pressure of the stream = 1 atm or 101.325 kPa The mole fraction of water and air in gas stream becomes, y = 3.1698101.325y= 0.031 mol H2Ovmol1-y = 0.969 mol air gmol Plugin the value of y in equation (2) 36.1×0.82 = 1-0.031 × n1n1= 30.55 mols Therefore, n2 = 36.1 - 30.55n2= 5.55 mol H2O ls(e) Reference states: Water = H2O (l), 25 0C, 1 atm Air = air (g), 25 0C, 1 atm (f) The enthalpy table becomes,(g) Enthalpy of each stream is calculated as follows: H1 = ∫25100cp,ldT + ∆Hv + ∫100250cp,vdT∫25100cp,ldT= ∫2510075.4×10-3dT = 75×75.4×10-3= 5.655 kJmol∆Hv = 44 kJmol ∫25100cp,vdT = ∫10025033.46×10-3+0.6880×10-5T+0.7604×10-8T2-3.593×10-12T3dT= 33.46×10-3150+0.6880×10-522502-1002+0.7604×10-832503-1003-3.593×10-1242504-1004=5.019 +0.1806+0.03707-0.00341=5.233 kJmolThus,H1 = 54.89 kJmolThe enthalpy of air entering the condenser is calculated as follows: H2 = ∫25250cpdT= ∫2525028.94×10-3 + 0.4147×10-5T + 0.3191×10-8T2-1.965×10-12T3 dT= 28.94×10-3 225 + 0.4147×10-522502-252 + 0.3191×10-832503-253 - 1.965×10-1242504-254= 6.5115 + 0.12829 + 0.0166 -0.001918H2=6.6545 kJmol H3 = H2Ol, 25°C →∆HvH2Ov, 25°CThus,H3 = 44 kJmol (h) Apply the energy balance Hout - Hin = Q0.94705 mols×44 kJmol - 6.498 mols×54.89kJmol + 29.602 mols×6.6545 kJmol= QQ = -511.99 kJs × 1 kW1 kJsQ= -511.99 ≃-512 kW