Quick Freeze

What if we apply $V$ at $280.73\,Hz$ directly?  Where $V=\Delta Tsin(\omega t)+V_{min}-\Delta T$ is sinusoidal.

And $V$ tracks $V_{min}$ as temperature $T$ falls.  $V_{min}$ changes with $T$ when the gas departs from an ideal gas.  And $\Delta T$ is adjusted such that,

$\Delta T\omega=-2A\cfrac{P^2}{T^3}\cfrac{dT}{dt}=-B\cfrac{1}{T}\cfrac{dT}{dt}$

where $A=\cfrac{3.4354^2}{2}\cfrac{m_i^3}{qZ^2k^2}$  and

$B=2A\cfrac{P^2}{T^2}=2V_{min}$

Strictly,

$\Delta T\omega=\left.-2\cfrac{V_{min}}{T}\cfrac{dT}{dt}\right|_{max}$

given an unchanging rate of removing $T^{-}$ from the system.  $\Delta T\le \small{\cfrac{1}{2}}V_{min}$, which provides for very high rate of cooling.  At $\Delta T=\small{\cfrac{1}{2}}V_{min}$,

$\left.\cfrac{dT}{dt}\right|_{max}=-\cfrac{1}{4}\omega{T}$

In this case, the inertia in $P$ and $T$ cannot be too slow to be $280.73\,Hz$.  $V_{min}$ can be fixed given $P$ and $T$ but $\cfrac{dT}{dt}$ is very high.

Have a nice day.

Note:  $T$ is not the temperature of the room being cooled.  It is the operating temperature of the cooling system.  $\small{\cfrac{dT}{dt}}$ removes $T^{-}$ from the cooling system and dumps them into the room being cooled.

In a refrigerator, the radiation coils and compressor are both operating at a higher temperature than the freezer compartment.