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.
