[Narrator] If you're in the market for a refrigeration system to cool produce on your farm,
be aware that manufacturers guidelines for sizing a system based on the area of a cooler are for retail,
not for produce coming in from the field.
In retail, the goods going into a cooler are already at a lower temperature,
so the system will have far less heat to remove.
This video we'll walk you through how to size a refrigeration system for farming operations.
[Sanford] You put a lot of work into growing the product, and so you need to make sure
that you have enough cooling capacity to get the product cold.
So that it has a long shelf life and maintains quality for market.
In this video, we're going to talk about a number of things so we're going to calculate.
There's math involved-- we're going to take it step-by-step and the object is that
we're taking it step-by-step and using tables and stuff, so the average person can do these calculations.
You don't need to be a mathematician!
So the first thing we're going to talk about is the types of heat that we have to remove,
and that's going to dictate the size of our cooler.
You harvest the crops and you bring it into the cooler.
We have to get it from the temperature it comes out of the field to the temperature of the cooler, and that's called Field Heat.
Once we get the produce cold and in the cooler it's going to continue to generate heat,
because the produce is converting sugar to energy and that's called Respiration Heat.
It's a smaller amount but it's still an amount.
Next is Conduction Heat.
So, we have a box we're putting this stuff into, we're trying to keep the temperature in that box at a certain temperature,
so we have to either add heat or refrigeration to maintain that, so that's Conduction Heat.
The next is Air Exchange. Every time we open the door to put produce in or pull it out,
we're going to lose cold air and it's going to be replaced by warm air that we have to cool.
And then we're going to call that Infiltration Heat or Air Exchange.
The next is equipment inside the cooler,
we have fans to move the air around, and run air across the refrigerator coils.
We also have lights, most likely, we may have a fork truck in there.
So that's going to be Equipment Heat that we have to account for.
The next is people. Occasionally, some farmers will do sorting and stuff inside the cooler
for a product that's very sensitive to temperature.
So if you have people in a cooler that's going to go to require some heat removal.
Now, today, to try to help us to understand this we're going to use an example.
So the example I have today, is a 10 by 12 by 8 foot high cooler.
It's made of prefab white panels,
the west end is exposed to the sun, the rest of it is under a building or within a building,
it has a 3 by 7 foot door,
we have r-values of the panels that this is made of.
We have to have a design temperature and this is typically worst-case temperature,
and we're going to use 90 degrees so the
maximum temperature outside the cooler is going to be 90 degrees and inside the
cooler we're to try to maintain 32 degrees or just above it.
And that's basically because the crops we're going to look at today are broccoli, cabbage, and sweet corn,
and that's the temperature that's best to keep them at.
Field Heat is the mass times specific heat times a temperature difference.
So the mass is just going to be the pounds of the product that we're putting into the cooler in a 24-hr period.
Specific heat is going to be the amount of energy
that you have to remove from the product to reduce it by one degree Fahrenheit.
So, it's in units of BTUs per pound degree Fahrenheit,
and the temperature difference is the temperature that it's coming into the
cooler at and what we want to cool it to with in 24 hours.
This tends to be the biggest load in a cooler.
Okay, now we're going to take our formula and put it into an example.
So, we've got the mass times the specific heat times temperature difference,
and we're going to do that for each of our crops here from our daily harvest,
because that's what we have to get the field heat out of.
So we've got 200 pounds of broccoli, 400 pounds of cabbage, and 600 pounds of sweet corn.
so we take that mass times the specific heat,
and these you can look up in a table.
Typically, if you want to use the averages it's about 0.9.
It's going to come in from the field at 80 degrees and we're going to cool it to 32 degrees,
to get our total amount of heat we have to remove.
We're going to take each crop and sum them.
And then we're going to sum all of them.
So, in this example, we have to remove about 52,000 BTUs from the crop,
in order to get it down to the cooler temperature.
And this is step one. In step two, we're going to look at how fast we have to do that,
because refrigeration is rated in BTUs per hour not BTUs.
So the question is: How fast do we need to remove it?
The rate we're going to have to do is going to depend on the produce.
So, strawberries --they spoil very quickly-- we're going to want to get that those cool in an hour.
Something like carrots, they're not as sensitive, we might be able to go 12 hours.
And that's going to reflect on the capacity of refrigeration we're going to need.
For our example today, we're going to use this line here. We're going to cool it in four hours.
So that's the first one of our numbers we're going to need here.
So, this is going to be in BTUs per day.
Next, we're going to talk about Respiration Heat.
So, even though this carrot has been harvested the cells in this are still active.
And they're taking the sugar, stored sugars, and converting them to heat.
As the temperature is reduced, the metabolic rate is reduced, and the heat generated is reduced,
and the shelf life of the product is extended.
The formula we use for this is the mass in pounds, divided by 2,000 pounds per ton,
and then we're going to go by the respiration rate, which you're going to look up on a table.
After the product is been cooled down and it's being maintained in a cooler this
is the rate of respiration or the rate of heat generation that's going to be put out by the crop.
Now we're gonna do an example to show.
So, we're going to look down here in our example chart, we're going to pull this information.
So this is the amount of a produce we're going to store in the cooler at any one time.
So we're going to take that, so the 400 divided by 2,000 to get our tons,
and this is the 4,400 off-the-chart. So if you look on the chart, you go over to
32 and drop down to broccoli and you're going to see that's 4,400 and that's
BTUs per ton per day and you're going to do the same for the cabbage, and the sweet corn.
One thing I want you to notice is these are all 32 degrees in the cooler.
You notice that the cabbage and the sweet corn there's nine times
more heat generated from the sweet corn than the cabbage.
So there's quite a bit of difference between different types of produce,
so you have to be aware of what your storing is as to that number.
So we get down here we have about 6900 BTUs per day that we have to remove,
so that's going to go up here as 6900 BTUs per day that we have to remove.
So the next step we're going to look at is a conduction heat, so we can do that
by looking at each component and then adding them together,
so we've got the walls, the doors, the ceiling, and the floors.
The heat gains of any of these you use the same formula.
It's the area of the surface, times the temperature difference between the outside and the inside,
divided by the R-value, times 24 hours because we want it in hours or BTUs per day.
The r-value it can be looked up on a chart,
and it's going to be in units of square feet per degree Fahrenheit per hour per BTU.
So over here we've got the r-value of the walls is 25,
and we're going to need the r-value of the doors which is 15.
So to do the heat loss on the walls, we've got two walls that are 12 by 8 and have a
temperature difference of 58 degrees, we divide it by the R-value which is 25,
The next wall has a door in it so the area is my 10 by 8 minus the door area, which
is 21 square feet, times the temperature difference, or 58 degrees, divided by the
r-value of 25. The next wall is the west wall, it's exposed to sun so it's 8 by 10
times the temperature difference, and here we have 94 degrees, so we've got 62
degrees of temperature difference here because it's exposed to sun, divided by
the r-value of 25, and then we have to add the door back in so the door is 21
square feet with 58 degree temperature difference divided by 15, it's a little
different than the walls. We sum all this up and multiply the whole thing by 24
hours per day because these are all in BTUs per hour and then we get to sum
total for just the walls of 20,686 BTUs per day.
So we have the wall, we got the heat from the ceiling, and from the floor, and we're going to
summarize them and we're going to get about 31,000 BTUs per day.
So that's going to be up here, so we got 31,000 BTUs per day, and that's going
to be added to our list for our Conduction Heat.
Next, we're going to talk about Infiltration or Air Exchange, and this
has to do with opening and closing of a door for the cooler.
So the first thing is a volume we've got 12 ft by 10 ft by 8 ft which comes out to 960 cubic feet for our cooler.
We've got the air exchanges, which we're going to look up on a table,
if you go look up on the table, up there, you're not going to find 960 but you'll find a 1000 which is close to it.
I'm going to assume a normal opening and closing rate, so you have a choice of light, normal, or heavy,
I'm assuming normal. So that's 17.5 air exchanges per day and then we go to another table,
which is specific heat of the air. And this is going to be base-- you're going to go across the top of the table
and you're going to find 90 and then you go down the table until you
find (it doesn't have 32 it has 35 and 30), it's going to be about 2.48 BTUs per cubic foot out of that table.
So, the formula we're using putting all the numbers in we got 960 times 17.5 air exchanges per day,
times 2.48 BTUs per cubic foot, which was going to give us about 42,000 BTUs per day,
then we're going to have to account for or remove from the refrigeration system is going to have to remove that
from the cooler to maintain your temperature and worst-case scenario.
42,000 BTUs per day.
Next, we're going to talk about the energy gain from equipment that's in the cooler.
This can be lights, fans, it could be a fork truck in the cooler.
So today we're going to talk a lot about lights and fan motors.
So we're assuming that the cooler has 3, 4-ft LED lights, that are 22 Watts each
So the formula for that is the lamp rating times the hours of operation,
times the conversion factor 3.4 BTUs per Watt.
So, in my example here, I'm going to pull down 22 Watts, times 3 fixtures,
times 24 hours a day-- We're going to assume that the lights are running 24 hours a day,
because they're in worst case scenario--- times a conversion factor.
So we have 5,400 BTUs per day.
The next is we have evaporation motor,
we're assuming we have two of them, they're a 1/4 horsepower fans, and they run 24/7.
So, a formula for that is for the motor is going to be the horsepower, times the
heat gain factor which we're going to pull out of a table, times the hours,
go to the table into that range that includes quarter horsepower and we're
going to pull out pull out the 4,250. This is BTUs per horsepower-hour, times 24 hours,
which is going to give us 51,000 BTUs per day.
Now all you just need to do is add them together, so we have 5400 BTUs plus 51,000 BTUs,
would give us 55,000 BTUs per day.
So that's our next number up here on the chart, so this is going to be 55,000 BTUs per day.
Okay, now we're going to talk about the people heat load.
Now this we only want to include if people are working in the cooler for a prolonged period of time,
so maybe sorting, or packaging, or something where they're in
there for more than just walking in and and putting their crates in and coming back out.
So this calculation it's going to be the number of people,
times the heat equivalent, which we're going to pull off a table, times 24 hours.
Yes, they're not going to be there in for 24 hours but we're trying to get everything on a worst-case,
24-hour a day, per day basis.
So, we're going to assume that we have two people working in the cooler at 32 degrees,
we're going to the table which gives me about 45,000 BTUs per day.
Okay, now we're going to take our final summary and calculate our refrigeration size.
So, we've got all these components, the total down here is 491,000.
You can see that Field Heat is the biggest component of that, and dwarfs everything else.
So that's important --that's the most important-- calculation to get right,
because that's going to affect the refrigeration sizing the most,
and if it isn't large enough you're not going to get the product cool enough, fast enough to retain that quality.
So, we're going to add these up as we did over here and
then we're going to divide it by either 16 or 18 hours basically as a safety factor.
We don't want the refrigeration system to run 24 hours a day,
because if we're doing that we're right on the verge and if our numbers aren't quite right,
or you actually put in a little more produce than you thought,
you're not going to get it cool fast enough.
So we're gonna take it and we're going to divide the 491,000 by 18 hrs, and we get about 27,000.
If we do it by 16 hours, we get about 30,500.
So that's the range when you go to shop for a refrigeration system that you would need for this example.
We also have a publication here that can help you,
it covers what we've covered in this presentation in more detail.
This can be gotten from the Learning Store at the University of Wisconsin.
We've learned that we need to understand the heat loads for the refrigeration system,
we have to the size it for all the loads occurring at once,
and the other thing I want to point out is that
some manufacturers provide a table based on your cooler size of the ideal refrigeration needs,
and those are aimed at the retail sector,
and it will undersize your cooler for what your trying to do with cooling produce.
So, an example, this clear size: the 10 by 12 by 8 ft,
they would recommend a 10,000 to 12,000 BTU per hour cooler.
We calculate out a 27,000 to 30,000 BTU cooler!
So you're going to be almost 1/3 less than than what's required.
So it may work for everything but removing field heat.
And the other thing is it's safer to oversize the cooler slightly than undersize.
You've got a lot of investment in what you're putting in there you don't want to lose it.