design equation prob

[dogma]

Ok guys I'm back with a new dilema.

I have 85kg of peaches at 18C and I wish to freeze them at -25C. I'm working on about 12hrs pull down time. My ambient is 36 and wish to condense at 45C.

Now as far as my calcs go I'm looking at 824Watts to cool in 12 hrs.


AM I RIGHT TO THIS POINT?



Now If I wish to size a system how do I go about it?


If I'm using R404a am I right by going straight to a capillary selection table and selecting 2m at 1.1mm dia?


Then I sized the compressor to 825watts, A Maneurope LTZ22JE.

Stop me guys if i'm off in the wrong direction.

How do I size my Evap and Condenser or should I firstly size my evaporator, then compressor, condenser, then suplimentary parts? And do I need to take into account the size of my cabinet as well as the "K" factor of my insulated walls?


How much info do I need to be able to build a prototype?


I would like to be able to size all components firstly without looking to manufacturers specs.

Is there a handbook which contains the equations I need?



This is just a theoredical exercise, getting ready for TAFE.








Thanks for the insight guys.

[Peter_1]

If I take water to freeze (85 x 15) +(85 x 8) + (85 x 25), divided by 12 hours, then I get +/- 950 W.
So 824 will be OK I think.
Add also heat disipated by the fans and the heat coming through the walls.
So, let's take +/- 1,000 W.
You want -25°C, so you need to evaporate at -30°C to -32°C.
An LTZ 22 gives at -35°C 1,000 Watt--> OK
Gives 1,400 W at -30°C
Select no an evaporator which gives at least 1,400 W at ambient -25°C and 5 K.

With heaters of course, heated drain, if possible stronger fans, make sure that air flow around the product is sufficient...

You have more capacity then needed, so it will freeze faster.

LTZ 22, max evaporating temperature = -20°C. So, install a crankcase pressure regulator or a MOP TEV.

[NoNickName]

Nice one by Peter: MOP is important because you will start with room at ambient temperature before it cools. So the evap temp will be way out of design until room temp settles.

[dogma]

Cool, so am I getting my head around these equations, and refrigeration concepts correctly???

Am I doing things in the right order.



You CAN be brutally honest.

Oh, trying not to look too stupid here, but I'm still unfamiliar with MOP. I know the very very basic concept of CPR and EPR, but not an MOP.


Cheers guys.

[Renato RR]

Maximum operating pressure for TEV.Insted of thet you can use crancase presure regulating valve but it isnt nessesary for this yob.

Best regards,
Renato

[Julian]

Hi Dogma

Your heat load calcultion is pretty close i make 834Watts
using
Q=m x c x delta T/time in seconds This is the load before freezing =145.8w

Q=m x h / time in seconds. This is the latent heat of fusion @ freezing = 586.34w

Q=m x c x delta T/time in seconds This is the load after freezing =92.6w

Heat of resperation= m x rl = 9.35w

This is only the product load you need to do heat load calculation for the room itself Q=U x A x delta T

& another for tha air change load
Air change load = Infiltration Rate x Heat Removal Factor

Then one for miscellanious load i.e fork lifts lighting, number of peaple ect.

Once you have your total heat load add a 10% safety margin
Then it will be possible to select an evaporator (dont forget to add evap fan load) then compressor (try to acheive a system balance) then one you no your total THR a condenser if yourusing a remote condenser.

Hope this helps

Julian

[Bob45]

Hi Dogma

Your heat load calcultion is pretty close i make 834Watts
using
Q=m x c x delta T/time in seconds This is the load before freezing =145.8w

Q=m x h / time in seconds. This is the latent heat of fusion @ freezing = 586.34w

Q=m x c x delta T/time in seconds This is the load after freezing =92.6w

Heat of resperation= m x rl = 9.35w

This is only the product load you need to do heat load calculation for the room itself Q=U x A x delta T

& another for tha air change load
Air change load = Infiltration Rate x Heat Removal Factor

Then one for miscellanious load i.e fork lifts lighting, number of peaple ect.

Once you have your total heat load add a 10% safety margin
Then it will be possible to select an evaporator (dont forget to add evap fan load) then compressor (try to acheive a system balance) then one you no your total THR a condenser if yourusing a remote condenser.

Hope this helps

Julian

Hi Julian,

I'm just woundering why are you taking the heat of respiration. The product will be frozen. I normaly don't take it in account if I'm freezing within 12 hours because heat of respiration varies with the product temperature.

Here in Canada I use Keeprite book for load calculation, work very well. Then you need a PH chart and tables to calculate your condenser and compressor load.

[Peter_1]

If vegetables or fruit are first have to be cooled before freezing, then you have to take in account this additional heat load (which is very small when you want to freeze it afterwards)
Both are living products which produces additional heat.

Reason why I take mostly water is because heat load through the walls, fan load, opening of the doors, some safety margin,.. are then also included in my estimation.

[dogma]

............... "for the room itself Q=U x A x delta T"

I'm lost here mate. Can you elaborate further on this equation?

............"& another for tha air change load
Air change load = Infiltration Rate x Heat Removal Factor"

How is this calculated? I'll hazard a guess that flow rates are required and TD over the coil is taken into account no?



With fresh produce, how do you determine the correct Humidity? I was told that in comfort AC the coil temp is about 12C to give a room temp of 22C and rel hum of 50%. Increasincing TD will give higher rel. hum no?




Thanks again guys. I'm needing all the imput you can give. Just got to Bris today. Go to TAFE on monday for a seven week stint.





Cheers.......

[Bob45]

If vegetables or fruit are first have to be cooled before freezing, then you have to take in account this additional heat load (which is very small when you want to freeze it afterwards)
Both are living products which produces additional heat.

Reason why I take mostly water is because heat load through the walls, fan load, opening of the doors, some safety margin,.. are then also included in my estimation.

Hi Peter,

Heat of respiration or evolution is a chemical reaction if you keep a fruit or veg at a certain temperature over 0°C. If you are cooling it down to freeze it, first it will not really have time for the chemical reaction to occur and the load in consideration is so low not worth taking it into account.

My point of view.

[Bob45]

Hi Dogma,

What you need is a course my friend.

............... "for the room itself Q=U x A x delta T"

I'm lost here mate. Can you elaborate further on this equation

Q= Btu/hrs or kw

U= 1/R where R is your isulation factor
A= surface of wall, ceiling whatever you are mesuring
Delta T= difference between you temperature inside and outside

............... "& another for tha air change load
Air change load = Infiltration Rate x Heat Removal Factor"

How is this calculated? I'll hazard a guess that flow rates are required and TD over the coil is taken into account no?

Air change load = Q = kw or btu/hrs
Infiltration = cubic meter/sec of air coming into the room or cfm.

Heat removal factor is a bit more complicated.

Enthalpy of the outdoor air - enthalpy of the inside air which you can get of a psycometric chart, dry bulb wet bulb or relative humidity.

........With fresh produce, how do you determine the correct Humidity? I was told that in comfort AC the coil temp is about 12C to give a room temp of 22C and rel hum of 50%. Increasincing TD will give higher rel. hum no?

No, a bigger TD will give you less humidity, output temperature of your coil will be lower which will make more condensation.





Thanks again guys. I'm needing all the imput you can give. Just got to Bris today. Go to TAFE on monday for a seven week stint.





Cheers.......[/QUOTE]

[dogma]

Thanks BOB45.

I'm making an attempt to get learn in front of my courses.
I know its probably a pain in the butt for you engineers.
All the Q's.


Thanks guys for the help.





Cheers.

[Bob45]

Thanks BOB45.

I'm making an attempt to get learn in front of my courses.
I know its probably a pain in the butt for you engineers.
All the Q's.


Thanks guys for the help.





Cheers.

No problem Dogma, I help you and you helped me on an other one

[dogma]

U= 1/R where R is your isulation factor


Hi, just a question to clarify.

Is "R" the same as the "K" factor?????

[Andy]

Hi Dogma.

A U value is the value for the insulation one figuire for that insulation, no units of thinkness are discussed.

An R value is the resistance of a component that makes up the insulation value of a structure we are interested in.

In K values we are interested in units of thickness to determine the R or U value.

Hope this is correct, most calculations now rely on a U value that is given, so I wouldn't use the above much.

Kind Regards. Andy

[dogma]

So am I right in saying R*K = U ?

Is "U" a value I would get from knowing the properties of the insulating materials by taking into consideration the blowing agent and density of the compond?


I'm confused.

sorry guys

[Bob45]

So am I right in saying R*K = U ?

Is "U" a value I would get from knowing the properties of the insulating materials by taking into consideration the blowing agent and density of the compond?


I'm confused.

sorry guys

No Dogma you're wrong.

Let say you have a wall of 6" thick. If your K value is 1 per inche. Then your R value is 6" x 1 = R6

Then your U = 1/R where R is the resistance and U the conductance.

So U = 1/6

Hope it's clear.

[dogma]

Thank you bob. It is much clearer now




cheers.

[Lc_shi]

Let say you have a wall of 6" thick. If your K value is 1 per inche. Then your R value is 6" x 1 = R6

the k is 1/λ? if yes,you're right

[dogma]

what is lambda Ic shi?

[Lc_shi]

lambda is conductivity

[dogma]

Ok. thanks.

If you don't mind me asking Ic shi, Is there a handbook available out there which you know of which contains all the equations required for the refrigeration and HVAC industry?

And if so do you know where in Australaisa I may be able to obtain one, ie refigeration company.











Question guys, does anyone use a pitot and manometer in there line of work, and is it a nessasarry tool of the trade????//???

[frank]

Question guys, does anyone use a pitot and manometer in there line of work, and is it a nessasarry tool of the trade????//???

If you are working with the movement of air in ducted systems then yes it is a necessary tool.

If you are just installing a unit cooler then no it isn't.
Here's the one I use http://rswww.com/cgi-bin/bv/rswww/se...o?cacheID=ukie The Digitron

[Lc_shi]

ASHRAE handbook is very good for you to know all about HVAC＆refrigeration.

rgds
LC

[Dan]

Increasincing TD will give higher rel. hum no?

No. If you consider that "TD" means having a lower evaporator temperature than the air temperature, you will remove more moisture because of the lower dew point of the coil.

[Dan]

If you are working with the movement of air in ducted systems then yes it is a necessary tool.

If you are just installing a unit cooler then no it isn't.

True Frank. But an anemometer or a velometer are useful tools with all refrigeration that employs fans. I wish they were cheaper and hardier devices.

[Lc_shi]

Hi Frank，
Your avatar makes my eyes brightwhere is it

rgards
LC

[drk_in]

hai dogma,
i had read the question. if the product freezing temp. is above the freezing temp. then u have to take the respiration load in to account. other wise u shouldn't. u can get the freezing temp. of the product from any refrigeration book like dossat or from ASHRAE HANDBOOK.
remaing are the same as told by our frnds.
bye

[dogma]

Hi guys. Thanks again for the help and insight.

Lc shi, drk in, I've tried to get dossat and the ashrae handbooks from the university bookstores but they don't have them any more for some reason.

Frank, that manometer is a little more hi tech then the one at tafe. We are learning with the old school inclinded manometer. lol.

[Ireland]

I've just been reading down through this and it brings me back.

It's been a while since I've had to sit down and work out much of it.

I'm very impressed by the level of knowledge on here and how helpful people are.

[Ireland]

Oh, and for Lc_shi, here's another avtar to brighten your day.

See, I even put it on twice for you .... lol

[Ireland]

By the way, has the Dossat book gone off the market completely?

That was our bible years ago.

Just curious at it was a great help to us all back in our apprentice days.

[Peter_1]

There's one for sale at Ebay.com for the moment and cheap.

[dogma]

Hi guys, Just made a web page today thought you guys might like to take a look.


www.refrigerationportal.bravehost.com


It's under construction but there are some usefull links.

[Julian]

............... "for the room itself Q=U x A x delta T"

I'm lost here mate. Can you elaborate further on this equation?

............"& another for tha air change load
Air change load = Infiltration Rate x Heat Removal Factor"

How is this calculated? I'll hazard a guess that flow rates are required and TD over the coil is taken into account no?



With fresh produce, how do you determine the correct Humidity? I was told that in comfort AC the coil temp is about 12C to give a room temp of 22C and rel hum of 50%. Increasincing TD will give higher rel. hum no?




Thanks again guys. I'm needing all the imput you can give. Just got to Bris today. Go to TAFE on monday for a seven week stint.





Cheers.......

This is the standard equation for calculating the quantity of heat transmited through the walls of a refrigerated space
Heat Flow Rate Q = U x A x ∆T

U= To the overhaul coefficient of heat transfer for a given product. you can find U factors for brick, polystyrene, polyurothane most materails from the following eqation.
U=K/X
where K is the thermal conductivity of a materail in W/mK
(watts per metre per kelvin) you can get this from tables or books like dossat or ASHRAE manuals
X = thickness of the materail in metres
If more than one product is used you have to use
1= X1 + X2
U K1 + K2
So in your case you have to know the insulating materail & the thicknes of it to work this out including the floor (i.e concrete with insulation on top).

A = The total surface area of area of the room
say for a room 5 x 4 x3
=2(5x4) + 2( 5 x3)+ 2(4x3) = 47m2

delta T = the temperature differance across the walls i.e. Outside temp - room temp

Air Change load is the heat gain into a refrigerated space mainly due to door openings & can be calculated by the following.
Q = m(Ho - Hi)
Where Q = Air change load in Kw
M = mass of air entering the space
Ho = Enthalpy of outside air
Hi = Enthalpy of inside air

Although it all sounds quite daunting Dossat provides tables with all this data @ given conditions.

Relavity Humidity can be found from variuos sources i.e
Dossat or ASHRAE manuals heatcaft & Actrol should also be able to supply you with this information.


TD has a direct affect on room humidity the greater the TD the lower the humidity & the lower the TD the greater the humidity.

[dogma]

Thanks Julian. It's all starting to come together.

[mcgru-]

So, dogma, what is your decision?
Concerning the first 3-4 messages here, do you intend to use 850-950 W at Tev=-32C?

Why you don't take into account, that it is a deep freeze from high temperaures? You can save money, if you'll take it into account

how?