Heat is transferred to a pressure cooker at a specified rate for a specified time period. The cooking temperature and the water remaining in the cooker are to be determined.
Assumptions 1 This process can be analyzed as a uniform-flow process since the properties of the steam leaving the control volume remain constant during the entire cooking process. 2 The kinetic and potential energies of the streams are negligible, ke ≅ pe ≅ 0. 3 The pressure cooker is stationary and thus its kinetic and potential energy changes are zero; that is, \Delta KE =\Delta PE =0 and \Delta E_{\text {system }}=\Delta U_{\text {system }} . 4 The pressure (and thus temperature) in the pressure
cooker remains constant. 5 Steam leaves as a saturated vapor at the cooker pressure. 6 There are no boundary, electrical, or shaft work interactions involved. 7 Heat is transferred to the cooker at a constant rate.
Analysis We take the pressure cooker as the system (Fig. 5–49). This is a control volume since mass crosses the system boundary during the process.
We observe that this is an unsteady-flow process since changes occur within the control volume. Also, there is one exit and no inlets for mass flow.
(a) The absolute pressure within the cooker is
P_{ abs }=P_{ gage }+P_{ atm }=75+100=175 kPa
Since saturation conditions exist in the cooker at all times (Fig. 5–50), the cooking temperature must be the saturation temperature corresponding to this pressure. From Table A–5, it is
T=T_{\text {sat } @ 175 kPa }=116.04^{\circ} C
which is about 16°C higher than the ordinary cooking temperature.
(b) Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as
Mass balance:
m_{\text {in }}-m_{\text {out }}=\Delta m_{\text {system }} \rightarrow-m_{e}=\left(m_{2}-m_{1}\right)_{ CV } \quad \text { or } \quad m_{e}=\left(m_{1}-m_{2}\right)_{ CV }
Energy balance: \underbrace{E_{\text {in }}-E_{\text {out }}}_{\begin{array}{c} \text { Net energy transfer } \\ \text { by heat, work, and mass } \end{array}}=\underbrace{\Delta E_{\text {system }}}_{\begin{array}{c} \text { Change in internal, kinetic, } \\ \text { potential, etc., energies } \end{array}}
Q_{\text {in }}-m_{e} h_{e}=\left(m_{2} u_{2}-m_{1} u_{1}\right)_{ CV } \quad(\text { since } W=0, \text { ke } \cong \text { pe } \cong 0)
Combining the mass and energy balances gives
Q_{\text {in }}=\left(m_{1}-m_{2}\right) h_{e}+\left(m_{2} u_{2}-m_{1} u_{1}\right)_{ CV }
The amount of heat transfer during this process is found from
Q_{\text {in }}=\dot{Q}_{\text {in }} \Delta t=(0.5 kJ / s )(30 \times 60 s )=900 kJ
Steam leaves the pressure cooker as saturated vapor at 175 kPa at all times (Fig. 5–51). Thus,
h_{e}=h_{g @ 175 kPa }=2700.2 kJ / kg
The initial internal energy is found after the quality is determined:
\begin{aligned} &v_{1}=\frac{v}{m_{1}}=\frac{0.006 m ^{3}}{1 kg }=0.006 m ^{3} / kg \\ &x_{1}=\frac{v_{1}-v_{f}}{v_{f g}}=\frac{0.006-0.001}{1.004-0.001}=0.00499 \end{aligned}
Thus,
u_{1}=u_{f}+x_{1} u_{f g}=486.82+(0.00499)(2037.7) kJ / kg =497 kJ / kg
and
U_{1}=m_{1} u_{1}=(1 kg )(497 kJ / kg )=497 kJ
The mass of the system at the final state is m_{2}=V / v_{2} . Substituting this into the energy equation yields
Q_{\text {in }}=\left(m_{1}-\frac{V}{v_{2}}\right) h_{e}+\left(\frac{V}{v_{2}} u_{2}-m_{1} u_{1}\right)
There are two unknowns in this equation, u_{2} \text { and } v_{2} . Thus we need to relate them to a single unknown before we can determine these unknowns. Assuming there is still some liquid water left in the cooker at the final state (i.e., saturation conditions exist), v_{2} \text { and } u_{2} can be expressed as
\begin{aligned} &v_{2}=v_{f}+x_{2} v_{f g}=0.001+x_{2}(1.004-0.001) m ^{3} / kg \\ &u_{2}=u_{f}+x_{2} u_{f g}=486.82+x_{2}(2037.7) kJ / kg \end{aligned}
Recall that during a boiling process at constant pressure, the properties of each phase remain constant (only the amounts change). When these expressions are substituted into the above energy equation, x_{2} becomes the only unknown, and it is determined to be
x_{2}=0.009
Thus,
v_{2}=0.001+(0.009)(1.004-0.001) m ^{3} / kg =0.010 m ^{3} / kg
and
m_2=\frac{V}{v_2} =\frac{0.006 m^3}{0.01m^3/kg}=0.6kg
Therefore, after 30 min there is 0.6 kg water (liquid + vapor) left in the pressure cooker.
Discussion Note that almost half of the water in the pressure cooker has evaporated during cooking.
