Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa (Fig. 5-17 ). It is observed that the amount of liquid in the cooker has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of the exit opening is 8mm^2 Determine (a) the mass flow rate of the steam and the exit velocity, (b) the total and flow energies of the steam per unit mass, and (c)the rate at which energy leaves the cooker by steam.

Question

Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa (Fig. 5-17 ). It is observed that the amount of liquid in the cooker has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of the exit opening is 8mm^2 Determine (a) the mass flow rate of the steam and the exit velocity, (b) the total and flow energies of the steam per unit mass, and (c)the rate at which energy leaves the cooker by steam.

Answer 1

To solve this problem, we need to use the principles of fluid mechanics and the ideal gas law.(a) Mass flow rate of the steam and the exit velocity:Given information:- Pressure cooker volume: 4 L- Operating pressure: 150 kPa- Decrease in liquid volume: 0.6 L in 40 min- Cross-sectional area of the exit opening: 8 mm²To find the mass flow rate of the steam, we can use the equation:Mass flow rate = (Change in liquid volume) / (Time taken)Substituting the given values:Mass flow rate = 0.6 L / 40 min = 0.015 L/minTo find the exit velocity, we can use the equation:Exit velocity = (Mass flow rate) / (Cross-sectional area of the exit opening)Substituting the given values:Exit velocity = (0.015 L/min) / (8 mm²) = 0.001875 m/s(b) Total and flow energies of the steam per unit mass:To find the total energy of the steam per unit mass, we can use the equation:Total energy = (Internal energy + Kinetic energy) per unit massThe internal energy of the steam can be calculated using the ideal gas law:Internal energy = (Pressure * Volume) / (Gas constant * Temperature)The kinetic energy of the steam can be calculated using the equation:Kinetic energy = (1/2) * (Velocity)² * (Mass flow rate)Substituting the given values:Total energy = (150 kPa * 4 L) / (0.287 kJ/kg·K * 373 K) + (1/2) * (0.001875 m/s)² * (0.015 L/min / 60 s/min) = 1.01 kJ/kgThe flow energy of the steam per unit mass can be calculated using the equation:Flow energy = (Pressure * Velocity) / 2Substituting the given values:Flow energy = (150 kPa * 0.001875 m/s) / 2 = 0.141 kJ/kg(c) Rate at which energy leaves the cooker by steam:The rate at which energy leaves the cooker by steam can be calculated using the equation:Rate of energy leaving = Total energy * Mass flow rateSubstituting the given values:Rate of energy leaving = 1.01 kJ/kg * 0.015 L/min / 60 s/min = 0.002525 kW

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