How does cooling time (pull down time) relate to Q evaporator?

This feels like a dumb question, but here goes. (My understanding of refrigeration systems is super basic)

Is the cooling rate (pull down time) of the food or beverage in a refrigerated unit only dependent on the temperature inside the fridge or can you speed it up by have a a higher cooling capacity (Watts) on your system?

Normally, I would think that the heat transfer rate is solely determined by the temperature differential and that you cannot speed that up unless you increase the heat transfer in some way (higher flow, agitation, etc). But what confuses me is that a compressor will have a given cooling capacity for a known evaporator temperature. Since the cooling capacity is in Watts, which is Joules/second, then doesn't this mean that it will remove heat at that rate? Or is that the maximum that it can pull and it will be limited by the rate that the food can cool?

This is neither a dumb nor a simple question, you can tell by the number of replies!

Yes, cooling rate depends on the temperature but more precisely on the temperature difference between the food and the freezer like Newton’s Law of Cooling states (convection coolig) but it is not “SOLELY determined by the temperature differential” it also depends on the heat transfer coefficient of the food and exposed area.

Increasing the cooling capacity of the system will have the effect of lowering the temperature of the freezer thus increasing the rate of cooling of the food, but in REAL systems lowering the temperature of the freezer has the opposite effect of DECREASING the cooling capacity. So the net effect will depend on the temperature range and type of equipment used.

Increasing the heat transfer coefficient (higher flow and stirring) will also increase the rate of cooling but in REAL systems this has also the cost of increasing the energy rate you are transferring into your system which will tend to increase the freezer temperature thus having the opposite effect also.

The result (time of cooling) you will get in the end is a compromise between the rate at which you can extract energy from the cooling media and the rate your food is able to transfer its thermal energy to the cooling media.
The watt rating of the compressor states exactly that it will extract from your system as long as you have a perfect adiabatic (insulated) system with no energy loss. If you have losses you have to extract them from the compressor value and talk about the system’s “useful effect”.

If the watt rating of your compressor is not equal the rate your food loose energy plus losses then the system will tend to work in different conditions tending to equilibrium.

Going back to your original question, the pull down time does relate to the Qevaporator and several other variables like stirring, losses, controls cutting in before the system reaches equilibrium, heat transfer coefficient of your product, ambient temperature as it affects compressor watts and energy transfer rate into your freezer, losses, etc.

Excellent description by aramis.
There are many factors, which also this, including the product itself, a whole cow, chills different to a steak, even though they are the same thing. You also have to consider, if product losses ,moisture, what effect on defrost cycles. Packaging, breaking of boundary layers.

Dont want to get of topic to much . but how long would the compressor run in an hour at normal running temps no door opening .

On and off 4 times in an hour total of 20mins ? whats normal ?
I know there's alot of variables .

it depends on your design . for example :

your heat load from goods /h= 1000 w
your heat lose in room and.../h = 1000w
So your evaporator should be/h = 2000w
your compressor should cover 2000w plus more  for safety side=A w = 2200w
if you choose less than A w your design time goes up
if you choose more than A w your time comes down ( because your suction pressure comes down= your Evaporator temperature comes down ) also by coming suction pressure down your cooling capacity of compressor comes down
you asked you want to run your compressor 40 min. /h
so compressor capacity = A w* 60/ 40 = B w = 3300 w
by what system do you want to control it ? Suction pressure or timer
if you control by suction pressure your system comes on and goes off frequency an hour ( a lot of times ) may be more than 60 times / h and this matter damage your compressor . if you adjust it for lower pressure your designing system change
if you want to control by timer your frozen time (cooling time ) goes up so there is sense we put big size compressor and pay more money for that and then we turn it off
too low suction pressure has own attention we should care about that.( two stage system - oil - leaks -pipe size and more...)

Normally for frozen goods
run time = 8 h
defrosting time and loading and unloads goods to room = 2 h