Question

A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out of liquid water for 75 min, determine the highest rate of heat transfer allowed.

Solution

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First the initial mass in the cooker is determined from the masses of the constituents. The specific volumes of the constituents are taken from A-5 for the given pressure:

m1=Vvapαvap+Vliqαliq=12V(1αvap+1αliq)=124103(11.0037+10.001057)kg=1.894kg\begin{align*} m_{1}&=\dfrac{V_{\text{vap}}}{\alpha_{\text{vap}}}+\dfrac{V_{\text{liq}}}{\alpha_{\text{liq}}}\\ &=\dfrac{1}{2}V\bigg(\dfrac{1}{\alpha_{\text{vap}}}+\dfrac{1}{\alpha_{\text{liq}}}\bigg)\\ &=\dfrac{1}{2}\cdot4\cdot10^{-3}\bigg(\dfrac{1}{1.0037}+\dfrac{1}{0.001057}\bigg)\:\text{kg}\\ &=1.894\:\text{kg} \end{align*}

The final mass is:

m2=Vαvap=41031.0037kg=0.003985kg\begin{align*} m_{2}&=\dfrac{V}{\alpha_{\text{vap}}}\\ &=\dfrac{4\cdot10^{-3}}{1.0037}\:\text{kg}\\ &=0.003985\:\text{kg} \end{align*}

The mass that has left the pressure cooker as vapor then is:

mout=m1m2=1.890015kg\begin{align*} m_{\text{out}}&=m_{1}-m_{2}\\ &=1.890015\:\text{kg} \end{align*}

The initial internal energy is determined from the masses and the internal energies of the constituents at the given pressure:

U1=m1u1=(mu)vap+(mu)liq=12V(uvapαvap+uliqαliq)=124103(2524.51.0037+486.820.001057)kJ=926.17kJ\begin{align*} U_{1}&=m_{1}u_{1}\\ &=(mu)_{\text{vap}}+(mu)_{\text{liq}}\\ &=\dfrac{1}{2}V\bigg(\dfrac{u_{\text{vap}}}{\alpha_{\text{vap}}}+\dfrac{u_{\text{liq}}}{\alpha_{\text{liq}}}\bigg)\\ &=\dfrac{1}{2}\cdot4\cdot10^{-3}\bigg(\dfrac{2524.5}{1.0037}+\dfrac{486.82}{0.001057}\bigg)\:\text{kJ}\\ &=926.17\:\text{kJ} \end{align*}

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