Let's solve each part step by step: --- # **3(a) Solution** **Given:** - Dry bulb temperature (\(T_{db}\)) = **130°F** - Relative humidity (\(RH\)) = **35%** We are asked for: - Humidity (absolute) - Wet bulb temperature (\(T_{wb}\)) (approximate) - Dew point --- ## **Step 1: Convert 130°F to °C and K** \[ T_{db} (°C) = \frac{5}{9}(130 - 32) = \frac{5}{9}(98) = 54.44°C \] \[ T_{db} (K) = 54.44 + 273.15 = 327.59\,K \] --- ## **Step 2: Find Saturation Vapour Pressure at \(T_{db}\)** We need the **saturation vapor pressure (\(P_{ws}\))** at 54.44°C. Using steam tables or the Antoine equation (simplified here): At 54.44°C, \[ P_{ws} \approx .134\, \text{bar} = 13.4\, \text{kPa} = 13,400\, \text{Pa} \] --- ## **Step 3: Find Actual Vapour Pressure** \[ P_w = RH \times P_{ws} = .35 \times 13.4 = 4.69\, \text{kPa} \] --- ## **Step 4: Calculate Humidity Ratio (\(Y\))** Assume atmospheric pressure \(P = 101\, \text{kPa}\). \[ Y = .622 \times \frac{P_w}{P - P_w} \] \[ Y = .622 \times \frac{4.69}{101 - 4.69} = .622 \times \frac{4.69}{96.31} = .622 \times .0487 = .0303\, \text{kg H}_2\text{O}/\text{kg dry air} \] --- ## **Step 5: Dew Point Temperature** Dew point is the temperature at which air becomes saturated for the actual vapor pressure (\(P_w = 4.69\, \text{kPa}\)). From steam tables, \(P_{ws} = 4.69\, \text{kPa}\) at about **32°C**. **Dew Point ≈ 32°C** --- ## **Step 6: Wet Bulb Temperature (Approximation)** Wet bulb temperature can be estimated by psychrometric charts or approximate formula: \[ T_{wb} \approx T_{db} - \left(\frac{T_{db} - T_{dp}}{3}\right) \] \[ T_{wb} \approx 54.44 - \left(\frac{54.44 - 32}{3}\right) = 54.44 - 7.48 = 46.96°C \] --- ## **Summary for 3(a):** - **Humidity (Humidity ratio):** \(Y = .0303\, \text{kg H}_2\text{O}/\text{kg dry air}\) - **Wet Bulb Temperature:** \(\approx 46.96°C\) - **Dew Point Temperature:** \(\approx 32°C\) --- # **3(b) Solution** **Given:** - Gas flow = **1.1 m³/s** - Dew point changes: **280 K → 260 K** - Vapour pressure at 280 K = **2.4 kN/m²** - Vapour pressure at 260 K = **.75 kN/m²** **Find:** - Water removed - Volume of gas after drying --- ## **Step 1: Find Humidity Ratio at Each Dew Point** Assume atmospheric pressure \(P = 101\, \text{kPa} = 101,000\, \text{N/m}^2\) ### At 280 K (\(P_{ws1} = 2.4\, \text{kN/m}^2 = 240\, \text{N/m}^2\)): \[ Y_1 = .622 \frac{240}{101000 - 240} = .622 \frac{240}{98600} = .622 \times .02435 = .01514\, \text{kg/kg} \] ### At 260 K (\(P_{ws2} = .75\, \text{kN/m}^2 = 750\, \text{N/m}^2\)): \[ Y_2 = .622 \frac{750}{101000 - 750} = .622 \frac{750}{100250} = .622 \times .007484 = .004655\, \text{kg/kg} \] --- ## **Step 2: Find Mass Flow Rate of Dry Air** For 1.1 m³/s at 280 K (assume ideal gas, dry air): \[ P V = nRT \implies n = \frac{PV}{RT} \] - \(P = 101,000\, \text{N/m}^2\) - \(V = 1.1\, \text{m}^3/\text{s}\) - \(R = 287\, \text{J/kg K}\) (for air) - \(T = 280\, \text{K}\) \[ \text{Mass flow of dry air, } \dot{m}_a = \frac{PV}{RT} \] \[ \dot{m}_a = \frac{101000 \times 1.1}{287 \times 280} = \frac{111100}{80360} = 1.382\, \text{kg/s} \] --- ## **Step 3: Water Removed per Second** \[ \dot{m}_{\text{water}} = \dot{m}_a (Y_1 - Y_2) \] \[ \dot{m}_{\text{water}} = 1.382 \times (.01514 - .004655) = 1.382 \times .010485 = .01449\, \text{kg/s} \] --- ## **Step 4: Volume of Gas After Drying** After drying, the mass of water vapor decreases, so total mass of air + vapor: \[ \text{Final mass flow} = \dot{m}_a (1 + Y_2) \] Final temperature = 280 K (if unchanged), or else use 260 K. Assume temperature is unchanged (280 K): \[ \text{Final volume flow} = \frac{\text{Final mass flow} \times R \times T}{P} \] \[ \text{Final mass flow} = 1.382 \times (1 + .004655) = 1.382 \times 1.004655 = 1.388\, \text{kg/s} \] \[ V_{\text{final}} = \frac{1.388 \times 287 \times 280}{101000} = \frac{1.388 \times 80360}{101000} = \frac{111,508.7}{101000} = 1.104\, \text{m}^3/\text{s} \] --- ## **Summary for 3(b):** - **Water removed per second:** \(\boxed{.0145\, \text{kg/s}}\) - **Volume of gas after drying:** \(\boxed{1.104\, \text{m}^3/\text{s}}\) --- ## **Final Answers** ### **3(a)** - **Humidity (Y):** .0303 kg H₂O/kg dry air - **Wet Bulb Temp:** 46.96°C - **Dew Point:** 32°C ### **3(b)** - **Water removed per second:** .0145 kg/s - **Final volume of gas:** 1.104 m³/s Let me know if you need further explanation or calculations!