Energy Transport by Mass Steam 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 has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of the exit

Question

Energy Transport by Mass

Steam 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 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 8 mm[latex]^{2}[/latex]. 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.

Accepted Answer

Steam leaves a pressure cooker at a specified pressure. The velocity, flow rate, the total and flow energies, and the rate of energy transfer by mass are to be determined.
Assumptions 1 The flow is steady, and the initial start-up period is disregarded. 2 The kinetic and potential energies are negligible, and thus they are not considered. 3 Saturation conditions exist within the cooker at all times so that steam leaves the cooker as a saturated vapor at the cooker pressure. Properties The properties of saturated liquid water and water vapor at 150 kPa are [latex]v_{f}=0.001053 \mathrm{~m}^{3} / \mathrm{kg}, v_{g}=1.1594 \mathrm{~m}^{3} / \mathrm{kg}, u_{g}=2519.2[/latex] [latex]\mathrm{kJ} / \mathrm{kg}[/latex], and [latex]h_{\mathrm{g}}=2693.1 \mathrm{~kJ} / \mathrm{kg}[/latex] (Table A-5).

Analysis (a) Saturation conditions exist in a pressure cooker at all times after the steady operating conditions are established. Therefore, the liquid has the properties of saturated liquid and the exiting steam has the properties of saturated vapor at the operating pressure. The amount of liquid that has evaporated, the mass flow rate of the exiting steam, and the exit velocity are

[latex]m=\frac{\Delta V_{ ext {liquid }}}{V_{f}}=\frac{0.6 \mathrm{~L}}{0.001053 \mathrm{~m}^{3} / \mathrm{kg}}\left(\frac{1 \mathrm{~m}^{3}}{1000 \mathrm{~L}} ight)=0.570 \mathrm{~kg} [/latex]

[latex]\dot{m}=\frac{m}{\Delta t}=\frac{0.570 \mathrm{~kg}}{40 \mathrm{~min}}=0.0142 \mathrm{~kg} / \mathrm{min}=2.37 imes 10^{-4} \mathrm{~kg} / \mathrm{s} [/latex]

[latex]V=\frac{\dot{m}}{ ho_{e} A_{c}}=\frac{\dot{m} v_{g}}{A_{c}}=\frac{\left(2.37 imes 10^{-4} \mathrm{~kg} / \mathrm{s} ight)\left(1.1594 \mathrm{~m}^{3} / \mathrm{kg} ight)}{8 imes 10^{-6} \mathrm{~m}^{2}}=34.3 \mathrm{~m} / \mathrm{s}[/latex]

(b) Noting that h=u+P v and that the kinetic and potential energies are disregarded, the flow and total energies of the exiting steam are

[latex]e_{ ext {flow }}=P V=h-u=2693.1-2519.2=173.9 \mathrm{~kJ} / \mathrm{kg} [/latex]

[latex] heta=h+\mathrm{ke}+\mathrm{pe} \cong h=2693.1 \mathrm{~kJ} / \mathrm{kg}[/latex]

Note that the kinetic energy in this case is [latex]k e=V^{2} / 2=(34.3 \mathrm{~m} / \mathrm{s})^{2} / 2=[/latex] [latex]588 \mathrm{~m}^{2} / \mathrm{s}^{2}=0.588 \mathrm{~kJ} / \mathrm{kg}[/latex], which is small compared to enthalpy.
(c) The rate at which energy is leaving the cooker by mass is simply the product of the mass flow rate and the total energy of the exiting steam per unit mass,

[latex]\dot{E}_{ ext {mass }}=\dot{m} heta=\left(2.37 imes 10^{-4} \mathrm{~kg} / \mathrm{s} ight)(2693.1 \mathrm{~kJ} / \mathrm{kg})=0.638 \mathrm{~kJ} / \mathrm{s}=0.638 \mathrm{~kW}[/latex]

Discussion The numerical value of the energy leaving the cooker with steam alone does not mean much since this value depends on the reference point selected for enthalpy (it could even be negative). The significant quantity is the difference between the enthalpies of the exiting vapor and the liquid inside (which is [latex]h_{f g}[/latex] ) since it relates directly to the amount of energy supplied to the cooker.

Question 5.3: Energy Transport by Mass Steam is leaving a 4-L pressure coo...

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