Cooling Load Calculation

[UDP10]

I'm working in a partition wall space (small, easily calculated cooling load) within a larger building. The users will be using a process in which they heat a material to 400 degrees.

Assuming I can determine the surface area of the heated material, how do I calculate the cooling required to offset the 400 degree material?

 

[gepman]

You don't mention how large a process this is. If it is a large process that is difficult to model then I would take the input energy of the device that is heating the material to 400 degrees subtract the energy that is removed by any exhaust system and add the energy required to condition the makeup air. If the material stays in the room and cools down then what I gave is sufficient. If the 400 deg. F material is moved out of the room then subtract the amount of energy added to the material (mCpdeltaT). Energy inputs are usually well known from the manufacturer's data sheets. You will probably need to multiply it by some type of usage factor or duty cycle (i.e. 50% if used 1 minute out of each two).

 

[GMcD]

And don't forget the massive radiant heat off the hot equipment/stuff that is in the room, which must be added to the usual air temperature load calculations. You can blow all the cold air you want on a person, but there is a substantial radiant heat gain that must be dealt with which the cold air cannot overcome. I would start with trying to reduce the heat gains in the space by reviewing the equipment and process to see how much heat can be removed directly before it impacts the people and air in the room.

 

[UDP10]

I should clarify. I am clear with the engergy ins and outs (makeups, exhausts, etc).
Forget the process for now. A magical, constant temperature 400 degree black box of known surface area exists in the room. How is the convective heat transferred into the space calculated? How is the radiated heat calculated?

 

[gepman]

What I was trying to say in my reply was that it is very difficult to model convection and radiation heat transfer on some unusual shaped object. Therefore if you know the energy inflow and the energy outflow then you can calculate the net heating effect without going through all the heat transfer equations. GMcD rightly pointed out that even with the calculation that I described another surface could get hot up due to the heat transfer by radiation.

Now if you want to try to model all of convective heat transfer (which will not be as accurate as the calculation that I described, unless it is a simple shape and you have the airflows etc. across the body) then you will have to calculate the Nusselt number, the Prandtl number, and the Reynolds number, find a good relationship for you shaped object and air velocity direction.

The heat transfer by radiation is a little simpler. If you assume that the heat transfer from the rest of the room to your hot material is negligible then q/A=esT4 where q is the heat flux, A is the area of your body that is radiating, e is the emissivity of the object, s is the Stefan Boltzman constant, and T4 is the temperature of the object in absolute terms to the fourth power.

If I were you I would calculate the total heat gain as I originally described and then calculate the radiative heat gain by the equation that I gave to see if radiation is a significant contributor. Depending on the color of the object at 400 deg. F, I suspect that the radiation will not be a major contributor. If it did glow cherry red (like an IR heater element (which is usually higher than 400 deg. F) then it will be significant.

If you do calculate the convective and radiative heat transfer please post your calculations and sources of equations since I would be interested in seeing them. I advocate using the energy balance method instead of using the rate (heat transfer) equations.

 

[IRstuff]

Your answer is rather glib "I am clear with the engergy ins and outs".

If you already know the energy in, then the cooling load is just that. Your "black box" was heated through those energy "ins". Unless you've got perpetual motion...

TTFN

FAQ731-376

 

[omarbakr]

cooling a space with a material at 400 degrees being cooled in it is what I understand from your posting.
it depends the material is in what form! Is it spread over a large surface or in a vessel or divided into parts, all these aspects differ from each other because it will account for the heat rate you are supposed to remove from the closed space. If the heated material is in a vessel then you will do lump system analysis and this will give you the least heat rate removal of all probabilites, but if they are a large number of blocks and spread over a big space then this will give you the most heat removal rate. In alll cases most of the heat is by natural convection, except if you apply forced air to the product.
you have a lack of information in what you are posting, surface area andtemp is not enough but there are other aspects to the problem.

 

[UDP10]

Apologies. My intent was not to be glib. This is the exact scenario:
A local manufacturing company is expanding one of their processes. The current line is located in a little shanty adjacent to their factory. It has NO ventilation or exhaust. Coils of a material I'm not to divulge is heated (outside the room) to 400 degrees and wheeled in where they go through a stamping/pressing type process and then are wheeled out. I'm sure they are not exactly 400 when wheeled out but for the sake of conversation I'll assume no loss. At any given time there are 6 coils in the room.

They are relocating/expanding their process and I'm designing the HVAC system. Though it doesn't have to be an exact science, I'm trying to establish a cooling load for the new room and I don't want to oversize too much as it's DX. The existing room is really hot and the only sources of load are the lights, people, and these hot coils (middle of blizzard winter north Ohio). I can't figure how to calc the coils.

Again, I didn't mean to be glib, I'm just getting frustrated because it seems like it should be relatively easy and I should know how to do it but I don't.

 

[gepman]

Actually it isn't relatively easy. Natural (more complicated) and forced convection heat transfer solutions are very dependent on geometry. Closed solutions have been determined for horizontal and vertical plates, horizontal and vertical tubes, etc.

Now if this is a coil of sheet metal that is unrolled and then stamped you will have a high surface to volume ratio which will allow it to cool off faster.

Again if you can I would suggest the energy balance approach instead of the rate (heat transfer) approach.

Determine the mass of each coil. This gives you "m". The manufacturer should know this information. Determine the heat capacity of the material. This gives you "Cp". This should be available from some type of materials handbook. You know the temperature going in (T1)is 400 deg. F. Now if as you state the temperature going out (T2) is 400 deg. F then you don't have to worry because there is no heat gain into the room which of course isn't very likely. All you have to do find the temperature going out of the room. If the manufacturer does not know this information then you can do one of the following a) try to calculate the heat loss of the material using some natural or forced convection closed solution or numerical solution which unless you know what you are doing will not be very accurate and either will take some time or b) Measure the temperature of the material as it leaves the existing process with an IR temperature probe (not that accurate), a thermocouple or RTD surface probe, or some of the sticks which melt at different temperatures, all of which are available from Omega (or other sources). Now that you have all of this information you can calculate the heat gain in the room by Q=mCp(T1-T2).

If you really want to calculate the natural and/or forced convection and radiation then I suggest that you post your question in the "Heat Transfer and Thermodynamics" forum under Mechanical Engineering.

 

[IRstuff]

You can swag it a number of different ways. But, you need to specify how long these coils stay in the room, and some idea of how much they cool. If they don't cool at all, then there is no cooling load to contend with, right? So, you have to assume some sort of heat loss, otherwise, there's no heat transfer problem to be solved.

So, a simple swag is to assume that they cool by, say, 100º (F or C?) and that they are in the room for some dwell time. Then the specific heat multiplied by the change in temperature divided by the dwell time is an estimate of the heat load. Since this stamping process requires mechanical contact with some sort of metal structure, this might well constitute the biggest heat transfer component of all, hence the guess of 100º. In any case, you should be able ask the company what the exit temperature of the coils is.

To that, you'd have to add the cooling load of the press operation itself.


TTFN

FAQ731-376

 

[omarbakr]

Well gepman and ir stuff are saying and talking right, it's as simple as Q=m*Cp(dT), but there is something missing is that this depends on the coil diameter which means the coil wire diameter and if the coils are wrapped around something. if the coil wire has a large diameter compared to the time it will stay in the room then only surface heat will convect and the center of the coil will stay hot and the heat load will be small and in this case you should use lumped system analysis, but if the coil wires are small in diameter so you should use a constant temp distribution and assume that the whole coil has the same temp and use q=m*Cp*(dT)

 

[gepman]

The effect that omarbakr is describing can be determined by calculating the Biot number which is equal to hc (convective heat transfer coefficient) * l (characteristic length [which depends on shape])/k (thermal conductivity of the solid). If the Biot number is < 0.1 there is negligible internal resistance (uniform temperature in the mass). If the Biot number is >40 then there is negligible surface resistance and you can you an infinite series solution for which there are charts available. If the Biot number is >0.1 and <40 then you an use Heisler charts. I used to do this a lot when I modeled food sterilization in cans. Current practice is to use numerical methods. All of this is time consuming. You could just dunk the whole coil into a container of water and measure the water temperature increase. From this information (including the mass of water) you determine the amount of energy the heated mass had released to the environment.

 

[IRstuff]

I'm not sure if that was what I was saying at all.

The process exists, therefore, there should be some data that either exists or can be trivially measured with an IR thermometer.

You then only need to know the material and its specific heat, from which you can determine the total energy loss during the process. That energy loss is your cooling load, regardless of how it leaves the coil.

Doing the heat transfer from fundamental principles seems to be overly complicated and rife with assumptions that cannot be readily verified.

You already have stuff that loses heat, so that can be measured and confirmed. Likewise, your coil material should be sufficiently well known to have a specific heat that could also be verified, if need be, with a simply lab experiment.

TTFN

FAQ731-376

 

[whisperinghill]

I believe the easist method of determining the heat load is to measure the temperature before you start your process and then after you complete your process, will give you the temperature rise you need to deal with.

Then you can go to this website

http://www.heatershop.com/btu_calculator.htm

and put in your temperature rise, room size and use tight walls and see what the BTU and wattage is calculated to, just for the metal heat.

Then download my Heat load calculator, (attached) and go to the requirement tab. Plug in your real time data and use the metal load wattage as other load.

Thgis will give you a final tonnage and some ways to calculate air flow requirements, etc

 

[sterl]

Presuming there is some control on the process, its very likely that the entire extent of the coil, side to die adn through and through, is at 400 Deg. when it enters the shed...
But if its shaped and reconditioned by machines and tools it is not going to be 400-deg, and neither is it going to be any singular surface temperature on its way out...especially if the duration is short compared to the (heat content of total volume, divided by exposed area) which is much more significant in short term processes on odd shapes, than is the Biot number...
Chances are a lot of the heat from One Part was spread around in contact with the machinery in the room to a larger volume and surface area of machine, so your Heat Transfer will get pretty ugly in a hurry....

But never mind: The most realistic way of doing it was suggested by somebody up there; I'll add my twist.

Get a Whole Bunch of ice; a little water; put both in a "sieved" container like a deep frier, such that the ice is in the basket. Lift the basket out, weigh it and the ice. Throw a hot part in for a good healthy time; if its weight compared to the weight of water is such that steam will be generated: Slap a lid on all of it. When the whole thing dies down: Fish the part out, make usre its as cold as the ice water; then weigh the ice and the basket. The ice that melted, thus is not in the basket, took on the heat of the part and cooled it all the way to 32-F. from 400 deg F.


Do the whole thing over again with a typical Outbound Part. This time you will have a lower total heat in the part; thus you will melt less ice. The difference between the wieght of the lost ice in Round 1 vs Round 2 is the heat that was lost from the typical part on the way through the Shed, in BTU or Calories; to get the cooling load for the many parts: multiply heat lost from one part times the rate of production.
RE: Workign tools and machines: Those points of contact, at 3-400 FPM velocities, become the focus of your air flow if you are looking for dimensional stability or safety in terms of cool surfaces of the machines...Otherwise: Its a matter of where the hottest pieces stay still long enough to have a local radiant effect on the opposing surfaces...Its those surfaces that need some air movement to focus the cooling and ensure stability and comfort.