Thermodynamics

Chapter 5 | 229EXAMPLE 5–3Energy Transport by MassSteam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa(Fig. 5–16). It is observed that the amount of liquid in the cooker hasSteamdecreased by 0.6 L in 40 min after the steady operating conditions areestablished, and the cross-sectional area of the exit opening is 8 mm 2 .Determine (a) the mass flow rate of the steam and the exit velocity, (b) thetotal and flow energies of the steam per unit mass, and (c) the rate at whichenergy leaves the cooker by steam.m ¢V liquid 0.6 L 1 m3av f 0.001053 m 3 >kg 1000 L b 0.570 kg 150 kPaSolution Steam leaves a pressure cooker at a specified pressure. The velocity,flow rate, the total and flow energies, and the rate of energy transfer bymass are to be determined.PressureCookerAssumptions 1 The flow is steady, and the initial start-up period is disregarded.2 The kinetic and potential energies are negligible, and thus they arenot considered. 3 Saturation conditions exist within the cooker at all timesso that steam leaves the cooker as a saturated vapor at the cooker pressure.Properties The properties of saturated liquid water and water vapor at 150kPa are v f 0.001053 m 3 /kg, v g 1.1594 m 3 /kg, u g 2519.2 kJ/kg,and h g 2693.1 kJ/kg (Table A–5).Analysis (a) Saturation conditions exist in a pressure cooker at all times FIGURE 5–16after the steady operating conditions are established. Therefore, the liquidhas the properties of saturated liquid and the exiting steam has the propertiesSchematic for Example 5–3.of saturated vapor at the operating pressure. The amount of liquid that hasevaporated, the mass flow rate of the exiting steam, and the exit velocity arem # m ¢t0.570 kg40 min 0.0142 kg>min 2.37 104 kg/sV m# m# v g 12.37 104 kg>s211.1594 m 3 >kg2 34.3 m/sr g A c A c8 10 6 m 2(b) Noting that h u Pv and that the kinetic and potential energies aredisregarded, the flow and total energies of the exiting steam aree flow Pv h u 2693.1 2519.2 173.9 kJ/kgu h ke pe h 2693.1 kJ/kgNote that the kinetic energy in this case is ke V 2 /2 (34.3 m/s) 2 /2 588 m 2 /s 2 0.588 kJ/kg, which is small compared to enthalpy.(c) The rate at which energy is leaving the cooker by mass is simply theproduct of the mass flow rate and the total energy of the exiting steam perunit mass,E # mass m # u 12.37 10 4 kg/s2 12693.1 kJ/kg2 0.638 kJ/s 0.638 kWDiscussion The numerical value of the energy leaving the cooker with steamalone does not mean much since this value depends on the reference pointselected for enthalpy (it could even be negative). The significant quantity isthe difference between the enthalpies of the exiting vapor and the liquidinside (which is h fg ) since it relates directly to the amount of energy suppliedto the cooker.