Let's solve the problem step by step. The **given data** is: - Dry bulb temperature, \( T_{db} = 26^\circ \mathrm{C} \) - Relative humidity, \( \phi = 70\% = 0.70 \) - Atmospheric pressure, \( P = 1\,\text{atm} = 101.325\,\text{kPa} \) We will use standard psychrometric equations and charts. If you have a psychrometric chart, you can follow these steps visually, but I'll show you how to calculate each step without one. --- ## (a) **Specific Humidity (\( \omega \))** **Step 1: Find the saturation pressure of water vapor at \( T_{db} = 26^\circ \mathrm{C} \).** From steam tables: - \( P_{ws} \) at \( 26^\circ \mathrm{C} \) ≈ **3.363 kPa** **Step 2: Calculate the partial pressure of water vapor (\( P_w \))** \[ P_w = \phi \cdot P_{ws} = 0.70 \times 3.363 = 2.354\,\text{kPa} \] **Step 3: Calculate the specific humidity (\( \omega \)):** \[ \omega = 0.622 \frac{P_w}{P - P_w} \] \[ \omega = 0.622 \frac{2.354}{101.325 - 2.354} = 0.622 \frac{2.354}{98.971} \] \[ \omega \approx 0.622 \times 0.0238 \approx 0.0148\,\text{kg H}_2\text{O/kg dry air} \] --- ## (b) **Enthalpy (\( h \), in kJ/kg dry air)** \[ h = 1.006\,T_{db} + \omega \left(2501 + 1.805\,T_{db}\right) \] \[ h = 1.006 \times 26 + 0.0148 \left(2501 + 1.805 \times 26\right) \] \[ = 26.156 + 0.0148 \times (2501 + 46.93) \] \[ = 26.156 + 0.0148 \times 2547.93 \] \[ = 26.156 + 37.73 \] \[ = 63.89\,\text{kJ/kg dry air} \] --- ## (c) **Wet-bulb Temperature (\( T_{wb} \))** Normally, this is read from a psychrometric chart, but we can estimate using formulas or interpolation. - At 26°C, 70% RH, **\( T_{wb} \approx 21.3^\circ \mathrm{C} \)** (from chart or calculation). --- ## (d) **Dew-Point Temperature (\( T_{dp} \))** Find the temperature at which the air would be saturated with the current vapor pressure (\( P_w \)). - \( P_w = 2.354\,\text{kPa} \) From steam tables, the saturation temperature at this pressure is **\( T_{dp} \approx 19.5^\circ \mathrm{C} \)**. --- ## (e) **Specific Volume (\( v \), in m³/kg dry air)** \[ v = \frac{R_{a} T}{P_{da}} \] - \( R_a = 0.287\,\text{kJ/kg K} \) - \( T = 26 + 273.15 = 299.15\,\text{K} \) - \( P_{da} = P - P_w = 101.325 - 2.354 = 98.971\,\text{kPa} \) \[ v = \frac{0.287 \times 299.15}{98.971} \] \[ v = \frac{85.061}{98.971} \] \[ v \approx 0.86\,\text{m}^3/\text{kg dry air} \] --- ## **Final Answers** \[ \boxed{ \begin{aligned} \text{(a) Specific humidity, } \omega &: 0.0148\,\text{kg H}_2\text{O/kg dry air} \\ \text{(b) Enthalpy, } h &: 63.89\,\text{kJ/kg dry air} \\ \text{(c) Wet-bulb temperature, } T_{wb} &: 21.3^\circ \mathrm{C} \\ \text{(d) Dew-point temperature, } T_{dp} &: 19.5^\circ \mathrm{C} \\ \text{(e) Specific volume, } v &: 0.86\,\text{m}^3/\text{kg dry air} \\ \end{aligned} } \] --- **If you need any part explained in more detail, let me know!**