# Solution: Moist Air Heating & Humidification Problem Let’s break down the problem step by step. ## **Given Data** - **Inlet air conditions:** - Dry-bulb temperature (\(T_1\)): 15°C - Relative humidity (\(\phi_1\)): 70% - Volumetric flow rate (\(V_1\)): 1.5 m³/s - **Heating section:** - Heat added (\(Q\)): 18 kW = 18,000 W - **Humidifier outlet:** - Dry-bulb temperature (\(T_3\)): 31°C - Relative humidity (\(\phi_3\)): 60% - **Atmospheric pressure (\(P\)):** Assume 101.3 kPa --- ## **Step 1: Find the Inlet Air Properties (State 1)** ### a. Find partial pressure of water vapor (\(p_{w,1}\)) Use saturation pressure tables for water at 15°C: - \(P_{ws,1}\) (saturation pressure at 15°C) ≈ 1.705 kPa \[ p_{w,1} = \phi_1 \times P_{ws,1} = .7 \times 1.705 = 1.194 \text{ kPa} \] ### b. Find humidity ratio (\(w_1\)) \[ w_1 = .622 \frac{p_{w,1}}{P - p_{w,1}} = .622 \frac{1.194}{101.3 - 1.194} = .622 \times \frac{1.194}{100.106} \] \[ w_1 \approx .622 \times .01193 = .00742\ \text{kg water/kg dry air} \] ### c. Find specific volume (\(v_1\)) \[ v_1 = \frac{R_{air} T_1 (1 + 1.6078 w_1)}{P} \] \[ \text{where } R_{air} = .287 \text{ kJ/kg}\cdot\text{K} \] \[ T_1 = 15 + 273.15 = 288.15\text{ K} \] \[ v_1 = \frac{.287 \times 288.15 \times (1 + 1.6078 \times .00742)}{101.3} \] \[ v_1 \approx \frac{.287 \times 288.15 \times 1.01193}{101.3} \] \[ v_1 \approx \frac{83.01 \times 1.01193}{101.3} \approx \frac{84.01}{101.3} \approx .829 \text{ m}^3/\text{kg dry air} \] ### d. Find mass flow rate of dry air (\(\dot{m}_a\)) \[ \dot{m}_a = \frac{V_1}{v_1} = \frac{1.5}{.829} \approx 1.81\ \text{kg/s} \] --- ## **Step 2: Find State 2 (after heating, before humidification)** ### a. Let temperature after heating = \(T_2\), humidity ratio unchanged (\(w_2 = w_1\)) ### b. Find enthalpy at state 1 (\(h_1\)) \[ h_1 = 1.005 T_1 + w_1 (2501 + 1.88 T_1) \] \[ h_1 = 1.005 \times 15 + .00742 \times (2501 + 1.88 \times 15) \] \[ = 15.075 + .00742 \times (2501 + 28.2) \] \[ = 15.075 + .00742 \times 2529.2 \] \[ = 15.075 + 18.77 = 33.85\ \text{kJ/kg dry air} \] ### c. Find enthalpy at state 2 (\(h_2\)), using heat added \[ \text{Total heat added: } Q = \dot{m}_a (h_2 - h_1) \] \[ 18,000 = 1.81 (h_2 - 33.85) \] \[ h_2 = \frac{18,000}{1.81} + 33.85 = 9,944.75 + 33.85 = 9,978.6\ \text{kJ/kg dry air} \] This value is unreasonably high (should be ~50 kJ/kg). Let's check units: Actually: \[ h_2 - h_1 = \frac{Q}{\dot{m}_a} = \frac{18,000}{1.81} \approx 9,944\ \text{J/kg} = 9.94\ \text{kJ/kg} \] \[ h_2 = 33.85 + 9.94 = 43.79\ \text{kJ/kg} \] ### d. Find temperature at state 2 (\(T_2\)), since \(w_2 = w_1\): \[ h_2 = 1.005 T_2 + w_2 (2501 + 1.88 T_2) \] Plug in \(w_2 = .00742\): \[ 43.79 = 1.005 T_2 + .00742 \times (2501 + 1.88 T_2) \] \[ 43.79 = 1.005 T_2 + .00742 \times 2501 + .00742 \times 1.88 T_2 \] \[ 43.79 = 1.005 T_2 + 18.57 + .01395 T_2 \] \[ 43.79 - 18.57 = (1.005 + .01395) T_2 \] \[ 25.22 = 1.01895 T_2 \] \[ T_2 = \frac{25.22}{1.01895} \approx 24.76\ ^\circ\text{C} \] ### e. Find relative humidity at state 2 (\(\phi_2\)): First, find \(p_{ws,2}\) at \(T_2 = 24.76^\circ\)C From steam tables, \(P_{ws,2} \approx 3.13\) kPa \[ p_{w,2} = \frac{w_2 P}{.622 + w_2} = \frac{.00742 \times 101.3}{.622 + .00742} \approx \frac{.752}{.6294} \approx 1.195\ \text{kPa} \] \[ \phi_2 = \frac{p_{w,2}}{P_{ws,2}} = \frac{1.195}{3.13} \approx .382 = 38.2\% \] --- ## **Step 3: State 3 (after humidification: \(T_3 = 31^\circ\)C, \(\phi_3 = 60\%\))** ### a. Find \(P_{ws,3}\) at 31°C From tables: \(P_{ws,3} \approx 4.45\) kPa \[ p_{w,3} = \phi_3 \times P_{ws,3} = .6 \times 4.45 = 2.67\ \text{kPa} \] \[ w_3 = .622 \frac{p_{w,3}}{P - p_{w,3}} = .622 \frac{2.67}{101.3 - 2.67} = .622 \times \frac{2.67}{98.63} \] \[ = .622 \times .02709 = .01687\ \text{kg/kg dry air} \] --- ## **Step 4: Moisture Added by Humidifier** \[ \Delta w = w_3 - w_2 = .01687 - .00742 = .00945\ \text{kg/kg dry air} \] \[ \text{Amount of moisture added per second: } \dot{m}_a \times \Delta w = 1.81 \times .00945 = .0171\ \text{kg/s} \] \[ \text{Per minute: } .0171 \times 60 = 1.03\ \text{kg/min} \] --- ## **Step 5: Dew Point of Air Leaving Humidifier (State 3)** - \(p_{w,3} = 2.67\) kPa - Dew point is the temperature where saturation pressure equals \(2.67\) kPa. From steam tables: - At 22°C, \(P_{ws} = 2.64\) kPa - At 23°C, \(P_{ws} = 2.81\) kPa Interpolate: \[ T_{dp} = 22 + \frac{2.67 - 2.64}{2.81 - 2.64} \times (23 - 22) \] \[ = 22 + \frac{.03}{.17} \approx 22.18^\circ\text{C} \] --- ## **Summary Table** | State | Temp (°C) | RH (%) | Humidity Ratio (kg/kg) | Enthalpy (kJ/kg) | |---|---|---|---|---| | 1 | 15 | 70 | .00742 | 33.85 | | 2 | 24.8 | 38.2 | .00742 | 43.79 | | 3 | 31 | 60 | .01687 | *(not calculated but can be)* | --- ## **Final Answers** 1. **Temperature of air leaving heating section (\(T_2\))**: **24.8°C (dry-bulb)** 2. **Relative humidity of air leaving heating section (\(\phi_2\))**: **38.2%** 3. **Amount of moisture added by humidifier**: **1.03 kg/min** 4. **Dew-point temperature of air leaving humidifier**: **22.2°C** --- ## **Psychrometric Chart** - Plot the three states: - **State 1**: 15°C, 70% RH (initial point) - **State 2**: 24.8°C, 38.2% RH (after heating, horizontal to the right) - **State 3**: 31°C, 60% RH (after humidification, vertical upwards to higher RH) *Note: For actual plotting, use a psychrometric chart and plot these points accordingly.* --- ## **Key Formulas Used** - \(w = .622 \frac{p_w}{P - p_w}\) - \(h = 1.005 T + w (2501 + 1.88T)\) - \(\dot{m}_a = V / v\) - \(\phi = p_w / p_{ws}\) --- **If you need a psychrometric chart sketch, let me know!**