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

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 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 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

\begin{aligned}&m=\frac{\Delta V_{\text {liquid }}}{v_{f}}=\frac{0.6 L }{0.001053 m ^{3} / kg }\left(\frac{1 m ^{3}}{1000 L }\right)=0.570 kg \\&\dot{m}=\frac{m}{\Delta t}=\frac{0.570 kg }{40 min }=0.0142 kg / min =2.37 \times 10^{-4} kg / s \\&V=\frac{\dot{m}}{\rho_{g} A_{c}}=\frac{\dot{m} v_{g}}{A_{c}}=\frac{\left(2.37 \times 10^{-4} kg / s \right)\left(1.1594 m ^{3} / kg \right)}{8 \times 10^{-6} m ^{2}}=34.3 m / s\end{aligned}

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

\begin{aligned} e_{\text {flow }} &=P \vee=h-u=2693.1-2519.2=173.9 kJ / kg \\ \theta &=h+ ke + pe \cong h=2693.1 kJ / kg \end{aligned}

Note 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 the product of the mass flow rate and the total energy of the exiting steam per unit mass,

\dot{E}_{\text {mass }}=\dot{m} \theta=\left(2.37 \times 10^{-4} kg / s \right)(2693.1 kJ / kg )=0.638 kJ / s = 0 . 6 3 8 kW

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 h_{f g} ) since it relates directly to the amount of energy supplied to the cooker.