A nozzle of diameter 20 mm is fitted to a large tank which contains air at 20°C. The air flows from the tank into atmosphere. For adiabatic flow, find the mass rate of flow of air through the nozzle when pressure of air in tank is (i) 5.886 N/cm² (gauge) and (ii) 29.43 N/cm² (gauge). Take k = 1.4 and R = 287 J/kg K and atmospheric pressure = 9.81 N/cm².

$$ P_1 = 5.886 + 9.81 = 15.696 \, \text{N/cm}^2 = 156960 \, \text{Pa} \text{ (abs)} $$ $$ P_2 = P_{atm} = 9.81 \, \text{N/cm}^2 = 98100 \, \text{Pa} \text{ (abs)} $$ $$ \text{Pressure Ratio, } \frac{P_2}{P_1} = \frac{9.81}{15.696} \approx 0.625 $$ $$ \text{Critical Pressure Ratio} = \left(\frac{2}{k+1}\right)^{\frac{k}{k-1}} = \left(\frac{2}{2.4}\right)^{3.5} \approx 0.528 $$

$$ \rho_1 = \frac{P_1}{RT_1} = \frac{156960}{287 \times 293.15} \approx 1.865 \, \text{kg/m}^3 $$ $$ V_2 = \sqrt{2 \times \frac{k}{k-1} \times \frac{P_1}{\rho_1} \left[1 - \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}}\right]} $$ $$ V_2 = \sqrt{2 \times 3.5 \times (287 \times 293.15) \left[1 - (0.625)^{0.2857}\right]} \approx 268 \, \text{m/s} $$ $$ T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} = 293.15 \times (0.625)^{0.2857} \approx 262.6 \, \text{K} $$ $$ \rho_2 = \frac{P_2}{RT_2} = \frac{98100}{287 \times 262.6} \approx 1.30 \, \text{kg/m}^3 $$ $$ \dot{m} = \rho_2 A_2 V_2 = 1.30 \times 0.00031416 \times 268 \approx 0.1095 \, \text{kg/s} $$

$$ P_1 = 29.43 + 9.81 = 39.24 \, \text{N/cm}^2 = 392400 \, \text{Pa} \text{ (abs)} $$ $$ P_2 = P_{atm} = 9.81 \, \text{N/cm}^2 = 98100 \, \text{Pa} \text{ (abs)} $$ $$ \text{Pressure Ratio, } \frac{P_2}{P_1} = \frac{9.81}{39.24} \approx 0.25 $$

$$ \dot{m} = A_2 \frac{P_1}{\sqrt{T_1}} \sqrt{\frac{k}{R} \left(\frac{2}{k+1}\right)^{\frac{k+1}{k-1}}} $$ $$ \dot{m} = 0.00031416 \times \frac{392400}{\sqrt{293.15}} \sqrt{\frac{1.4}{287} \left(\frac{2}{2.4}\right)^{\frac{2.4}{0.4}}} $$ $$ \dot{m} = 7.19 \times \sqrt{0.004878 \times (0.8333)^6} \approx 7.19 \times \sqrt{0.001605} \approx 0.288 \, \text{kg/s} $$