Chapter 5.5, Problem 184RP

A pressure cooker is a pot that cooks food much faster than ordinary pots by maintaining a higher pressure and temperature during cooking. The pressure inside the pot is controlled by a pressure regulator (the petcock) that keeps the pressure at a constant level by periodically allowing some steam to escape, thus preventing any excess pressure buildup. Pressure cookers, in general, maintain a gage pressure of 2 atm (or 3 atm absolute) inside. Therefore, pressure cookers cook at a temperature of about 133°C instead of 100°C, cutting the cooking time by as much as 70 percent while minimizing the loss of nutrients. The newer pressure cookers use a spring valve with several pressure settings rather than a weight on the cover.

A certain pressure cooker has a volume of 6 L and an operating pressure of 75 kPa gage. Initially, it contains 1 kg of water. Heat is supplied to the pressure cooker at a rate of 500 W for 30 min after the operating pressure is reached. Assuming an atmospheric pressure of 100 kPa, determine (a) the temperature at which cooking takes place and (b) the amount of water left in the pressure cooker at the end of the process. Answers: (a) 116.04°C, (b) 0.6 kg

To determine

The temperature at which cooking takes place.

Answer to Problem 184RP

The temperature at which cooking takes place is $116.04°\text{C}$.

Explanation of Solution

Write the formula for absolute pressure build up in the cooker.

$P_{\text{abs}}=P_{\text{gage}}+P_{\text{atm}}$ (I)

Here, the absolute pressure is $P_{\text{abs}}$, the gage pressure is $P_{\text{gage}}$, and the atmospheric pressure is $P_{\text{atm}}$.

When the pressure cooker reaches it maximum pressure, the petcock valve allows the steam to exit the cooker, the condition of exit steam is saturated vapor.

Hence, the cooking temperature $\left(T_2\right)$ in the pressure cooker is equal to the saturation temperature corresponding to the absolute pressure $\left(P_{\text{abs}}\right)$.

$T_2=T_{\text{sat @}{P}_{\text{abs}}}$

Conclusion:

Substitute $75\text{kPa}$ for $P_{\text{gage}}$ and $100\text{kPa}$ for $P_{\text{atm}}$ in Equation (I).

$\begin{array}{c}{P}_{\text{abs}}=75\text{kPa}+100\text{kPa}\\ =175\text{kPa}\end{array}$

Refer Table A-5, “Saturated water-Temperature table”.

The saturation temperature corresponding to the pressure of $175\text{kPa}$ is $116.04°\text{C}$.

$T_2=T_{\text{sat}\text{}\text{@ 175}\text{kPa}}=116.04°\text{C}$

Thus, the temperature at which cooking takes place is $116.04°\text{C}$.

To determine

The amount water left in the pressure cooker at the end of the process.

Answer to Problem 184RP

The amount water left in the pressure cooker at the end of the process is $\text{0}\text{.6}\text{kg}$.

Explanation of Solution

Write the equation of mass balance.

$m_{\text{in}}-m_{\text{e}}=\Delta m_{\text{system}}$ (II)

Here, the inlet mass is $m_{\text{in}}$, the exit mass is $m_{\text{e}}$ and the change in mass of the system is $\Delta m_{\text{system}}$.

The change in mass of the system for the control volume is expressed as,

$\Delta m_{\text{system}}={\left(m_2-m_1\right)}_{\text{cv}}$

Here, the suffixes 1 and 2 indicates the initial and final states of the system.

Consider the given pressure cooker as the control volume.

Initially the cooker is filled with liquid and vapor and further no other mass is allowed to enter the cooker. Hence, the inlet mass is neglected i.e. $m_{\text{in}}=0$. The pressure relief cork is released when the cooker reaches it maximum pressure that is the exit mass.

Rewrite the Equation (I) as follows.

$\begin{array}{l}0-m_{\text{e}}={\left(m_2-m_1\right)}_{\text{cv}}\\ m_{\text{e}}=m_1-m_2\end{array}$ (III)

At initial state:

There is saturated mixture of water present in the pressure cooker.

Write the formula of initial specific volume $\left(v_1\right)$.

$v_1=\frac{\nu }{m_1}$ (IV)

Here, the volume of pressure cooker is $\nu$ and the initial mass is $m_1$.

Write the formula for quality of mixture at initial state.

$x_1=\frac{v_1-v_{f,1}}{v_{fg,1}}$ (V)

Write the formula for internal energy of steam at initial state.

$u_1=u_{f,1}+x_1{u}_{fg,1}$ (VI)

Here, the specific volume is $v$, the internal energy is $u$, the subscript $f$ indicates the liquid phase, the subscript $fg$ indicates the vaporization phase.

Write the energy balance equation.

$\begin{array}{l}{E}_{in}-E_{out}=\Delta E_{system}\\ \left\{\begin{array}{l}\left[Q_{\text{in}}+W_{\text{in}}+m_{\text{in}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{in}}\right]\\ -\left[Q_{\text{e}}+W_{\text{e}}+m_{\text{e}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{e}}\right]\end{array}\right\}={\left[\begin{array}{l}{m}_2{\left(u+\text{ke}+\text{pe}\right)}_2\\ -m_1{\left(u+\text{ke}+\text{pe}\right)}_1\end{array}\right]}_{\text{system}}\end{array}$ (VII)

Here, the heat transfer is $Q$, the work transfer is $W$, the enthalpy is $h$, the internal energy is $u$, the kinetic energy is $\text{ke}$, the potential energy is $\text{pe}$ and the change in net energy of the system is $\Delta E_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

Since the pressure cooker is not insulated, the heat transfer occurs through the pressure cooker wall. In control volume, there is no work transfer, i.e. $\left(W_{\text{in}}=W_{\text{e}}=0\right)$. Neglect the kinetic and potential energy changes i.e. $\left(\Delta \text{ke}=\Delta \text{pe}=0\right)$. There is no mass inlet for the pressure cooker, the inlet mass is neglected i.e. $\left(m_{\text{in}}=0\right)$.

The Equation (V) reduced as follows.

$\begin{array}{l}{Q}_{\text{in}}-m_{\text{e}}{h}_{\text{e}}=m_2{u}_2-m_1{u}_1\\ Q_{\text{in}}=m_2{u}_2-m_1{u}_1+m_{\text{e}}{h}_{\text{e}}\end{array}$ (VIII)

Write the formula for amount of heat supplied to the cooker.

$Q_{\text{in}}={\dot{Q}}_{\text{in}}\Delta t$ (IX)

Here, the rate of heat supply is ${\dot{Q}}_{\text{in}}$ and the time interval is $\Delta t$.

Consider, at final state the pressure cooker consist of saturated mixture.

Write the formula for specific volume $\left(v_2\right)$ of steam at final state.

$u_2=u_{f,2}+x_2{u}_{fg,2}$ (X)

Write the formula for internal energy $\left(u_2\right)$ of steam at final state.

$v_2=v_{f,2}+x_2{u}_{fg,2}$ (XI)

Write the formula for mass of steam $\left(m_2\right)$ at final state.

$m_2=\frac{\nu }{v_2}$ (XII)

Here, the subscript 2 indicates the final state.

The properties of the steam at both initial and final states are equal, the only variation occurs with their quality.

The pressure cooker consist of mixture of vapor $\left(g\right)$ and liquid $\left(f\right)$ at both initial and final state.

Refer Table A-5, “Saturated water-Pressure table”.

Obtain the following corresponding to the pressure of $175\text{kPa}$.

The initial fluid and gaseous specific volume is as follows.

$v_{f,1}=0.001057{\text{m}}^3\text{/kg}$

$v_{g,1}=1.0037{\text{m}}^3\text{/kg}$

The initial fluid and gaseous internal energy is as follows.

$u_{f,1}=486.82\text{kJ/kg}$

$u_{fg,1}=2037.7\text{kJ/kg}$

When the pressure cooker reaches it maximum pressure, the steam starts to exit the cooker, the condition of exit steam is saturated vapor.

Hence, the exit enthalpy $\left(h_g=h_{\text{e}}\right)$ corresponding to the pressure of $175\text{kPa}$ is,

$h_g=h_{\text{e}}=2700.2\text{kJ/kg}$

Conclusion:

Substitute $6\text{L}$ for $\nu$ and $1\text{kg}$ for $m_1$ in Equation (IV).

$\begin{array}{c}{v}_1=\frac{6\text{L}}{1\text{kg}}\\ =\frac{6\text{L}\times \frac{1{\text{m}}^3}{1000\text{}\text{L}}}{1\text{kg}}\\ =0.006{\text{m}}^3\text{/kg}\end{array}$

Substitute $0.006{\text{m}}^3\text{/kg}$ for $v_1$, $0.001057{\text{m}}^3\text{/kg}$ for $v_{f,1}$, and $1.0037{\text{m}}^3\text{/kg}$ for $v_{g,2}$ in Equation (V).

$\begin{array}{c}{x}_1=\frac{0.006{\text{m}}^3\text{/kg}-0.001057{\text{m}}^3\text{/kg}}{1.0037{\text{m}}^3\text{/kg}-0.001057{\text{m}}^3\text{/kg}}\\ =0.0049\end{array}$

Substitute $486.82\text{kJ/kg}$ for $u_{f,1}$, $0.0049$ for $x_1$, and $2037.7\text{kJ/kg}$ for $u_{fg,1}$ in

Equation (VI).

$\begin{array}{c}{u}_1=486.82\text{kJ/kg}+0.0049\left(2037.7\text{kJ/kg}\right)\\ =486.82\text{kJ/kg}+9.9847\text{kJ/kg}\\ =496.8047\text{kJ}\\ \simeq 497\text{kJ/kg}\end{array}$

Substitute $500\text{W}$ for ${\dot{Q}}_{\text{in}}$ and $30\text{min}$ for $\Delta t$ in Equation (IX).

$\begin{array}{c}{Q}_{\text{in}}=\left(500\text{W}\right)\left(30\text{min}\right)\\ =\left(500\text{W}\times \frac{1\text{kJ/s}}{1000\text{W}}\right)\left(30\text{min}\times \frac{60\text{s}}{1\text{min}}\right)\\ =900\text{kJ}\end{array}$

Substitute $486.82\text{kJ/kg}$ for $u_{f,2}$ and $2037.7\text{kJ/kg}$ for $u_{fg,2}$ in

Equation (X).

$\begin{array}{c}{u}_2=486.82\text{kJ/kg}+x_2\left(2037.7\text{kJ/kg}\right)\\ =486.82+2037.7x_2\end{array}$

Substitute $0.001057{\text{m}}^3\text{/kg}$ for $v_{f,2}$, and $1.0037{\text{m}}^3\text{/kg}$ for $v_{g,2}$ in Equation (XI).

$\begin{array}{c}{v}_2=0.001057{\text{m}}^3\text{/kg}+x_2\left(1.0037{\text{m}}^3\text{/kg}\right)\\ =0.001057+1.0037x_2\end{array}$

Substitute $6\text{L}$ for $\nu$ and $0.001057+1.0037x_2$ for $v_2$ in Equation (XII).

$\begin{array}{c}{m}_2=\frac{6\text{L}}{0.001057+1.0037x_2}\\ =\frac{6\text{L}\times \frac{1{\text{m}}^3}{1000\text{}\text{L}}}{0.001057+1.0037x_2}\\ =\frac{0.006{\text{m}}^{\text{3}}}{0.001057+1.0037x_2}\\ =\frac{0.006}{0.001057+1.0037x_2}\end{array}$ (XIII)

Substitute $1\text{kg}$ for $m_1$ and $\frac{0.006}{0.001057+1.0037x_2}$ for $m_2$ in Equation (III).

$\begin{array}{c}{m}_e=1\text{kg}-\frac{0.006}{0.001057+1.0037x_2}\\ =1-\frac{0.006}{0.001057+1.0037x_2}\end{array}$

Substitute $900\text{kJ}$ for $Q_{\text{in}}$, $\frac{0.006}{0.001057+1.0037x_2}$ for $m_2$, $486.82+2037.7x_2$ for $u_2$, $1\text{kg}$ for $m_1$, $497\text{kJ/kg}$ for $u_1$, $1-\frac{0.006}{0.001057+1.0037x_2}$ for $m_{\text{e}}$, and $2700.2\text{kJ/kg}$ for $h_{\text{e}}$ in Equation (VIII).

$\begin{array}{c}900\text{kJ}=\left[\begin{array}{l}\left(\frac{0.006}{0.001057+1.0037x_2}\right)\left(486.82+2037.7x_2\right)\\ -\left(1\text{kg}\right)\left(497\text{kJ/kg}\right)\\ +\left(1-\frac{0.006}{0.001057+1.0037x_2}\right)\left(2700.2\text{kJ/kg}\right)\end{array}\right]\\ =\left[\begin{array}{c}\left(\frac{0.006}{0.001057+1.0037x_2}\right)\left(486.82+2037.7x_2\right)\\ -497\text{kJ}\\ +\left(1-\frac{0.006}{0.001057+1.0037x_2}\right)\left(2700.2\text{kJ/kg}\right)\end{array}\right]\end{array}$ (XIV)

Use Engineering Equation Solver (EES) or online calculator to solve the Equation (XIV) and obtain the value of $x_2$.

$x_2=0.00902$

Substitute $0.00902$ for $x_2$ in Equation (XIII).

$\begin{array}{c}{m}_2=\frac{0.006}{0.001057+1.0037\left(0.00902\right)}\\ =\frac{0.006}{0.01011}\\ =0.59345\text{kg}\\ \simeq \text{0}\text{.6}\text{kg}\end{array}$

Thus, the amount water left in the pressure cooker at the end of the process is $\text{0}\text{.6}\text{kg}$.