# Step-by-Step Solution: Inflated Paddleboard Thermodynamics ## Given Data - **Length (L)** = 2.7 m - **Width (W)** = 80 cm = .80 m - **Thickness (H)** = 15 cm = .15 m - **Ambient temperature \( T \)** = 30°C = 303.15 K - **Ambient pressure \( P_{\text{amb}} \)** = 1 bar = 100,000 Pa - **Specific gas constant \( R_s \)** = 287 J/kg·K - **Specific heat at constant pressure \( c_p \)** = 1. kJ/kg·K = 100 J/kg·K - **Overpressure after inflation** = 1.2 bar (gauge) - **Compressor Power** = 110 W - **Pumping Time** = 10 min = 600 s --- ## a) Air Volume of the Inflated Board The volume \( V \) is: \[ V = L \times W \times H = 2.7 \, \text{m} \times .80 \, \text{m} \times .15 \, \text{m} = .324 \, \text{m}^3 \] **Final Answer:** \[ \boxed{V = .324 \text{ m}^3} \] --- ## b) Mass of Air in the Board after Inflation **Absolute pressure in the board:** \[ P_{\text{abs}} = P_{\text{amb}} + P_{\text{over}} = 1\,\text{bar} + 1.2\,\text{bar} = 2.2\,\text{bar} = 220,000\,\text{Pa} \] **Using the ideal gas law:** \[ m = \frac{P V}{R_s T} \] Plug in the numbers: \[ m = \frac{220,000 \times .324}{287 \times 303.15} \] \[ m = \frac{71,280}{87,015.05} \approx .819 \, \text{kg} \] **Final Answer:** \[ \boxed{m \approx .82\,\text{kg}} \] --- ## c) Minimum Absolute Work Required to Inflate the Board (Isothermal Process) For isothermal compression: \[ W = m R_s T \ln\left(\frac{P_2}{P_1}\right) \] Or, since \( m = \frac{P_2 V}{R_s T} \) (after filling): \[ W = P_2 V \ln\left(\frac{P_2}{P_1}\right) \] Where: - \( P_1 = 1. \) bar = 100,000 Pa (ambient) - \( P_2 = 2.2 \) bar = 220,000 Pa So, \[ W = 220,000 \times .324 \times \ln\left(\frac{220,000}{100,000}\right) \] \[ = 71,280 \times \ln(2.2) \] \[ \ln(2.2) \approx .7885 \] \[ W = 71,280 \times .7885 \approx 56,224.68 \text{ J} = 56.2 \text{ kJ} \] **Final Answer:** \[ \boxed{W = 56.2\,\text{kJ}} \] --- ## d) Change in Entropy at Beginning (Before Inflation, \( P_{\text{in}} = P_{\text{amb}} \)) **At the start, the pressure inside the board equals ambient pressure, and the air inside and outside is at the same temperature and pressure.** - There is **no entropy difference** between the air inside the board and the surroundings at this point. \[ \Delta S_{\text{board}} = \] \[ \Delta S_{\text{ambient}} = \] \[ \Delta S_{\text{total}} = \] **Final Answer:** \[ \boxed{ \Delta S_{\text{board}} = \text{ J/kg} \\ \Delta S_{\text{ambient}} = \text{ J/kg} \\ \Delta S_{\text{total}} = \text{ J/kg} } \] --- ## e) Isothermal Efficiency of the Compressor **Isothermal efficiency:** \[ \eta = \frac{\text{Minimum isothermal work}}{\text{Actual work input}} \] - **Minimum work (from c):** \( W_{\text{min}} = 56,200 \) J - **Actual work input:** \( W_{\text{actual}} = \text{Power} \times \text{Time} = 110\,\text{W} \times 600\,\text{s} = 66,000\,\text{J} \) \[ \eta = \frac{56,200}{66,000} = .8515 \] \[ \eta = 85.2\% \] **Final Answer:** \[ \boxed{\eta = 85.2\%} \] --- ## Summary Table | Subtask | Quantity | Value | |---------|----------|-------| | a | Air volume | .324 m³ | | b | Mass of air | .82 kg | | c | Minimum work | 56.2 kJ | | d | Entropy changes | J/kg (all) | | e | Isothermal efficiency | 85.2% | --- **Note:** - All calculations use SI units. - Overpressure is always "gauge"; absolute pressure is used for calculations.