Commercial Heat Load Question

[superdave44]

I have recently purchased wrightsoft universal including commercial. Does anyone have any sample loads for a commercial kitchen that would simulate a30 X 48 kitchen and an 80 X 80 area for 500 people. The 500 people area may also be used for dancing as well as other activities. How does this software handle exhaust and makeup air calculations? What information will the software calculate and what do I need to supply? I understand this is a vague question - just looking to be pointed the wright direction. My initial stab at this is calling for 15 tons in the kitchen and a little over 100 tons in the lodge area. Just seems a little high to me. Any thoughts appreciated...

[beenthere]

Don't know how that program handles things.
Sounds like it is calculating the fresh air for the 80 x 80 room as a constant load.

Not knowing your area, or the design of the builing, I wouldn't guess more then 50 to 60 tons for that room.

[BigJon3475]

Don't know how that program handles things.
Sounds like it is calculating the fresh air for the 80 x 80 room as a constant load.

Not knowing your area, or the design of the builing, I wouldn't guess more then 50 to 60 tons for that room.

Do you ever sleep :p (joking)

[beenthere]

Do you ever sleep :p (joking)
I can't believe you sleep in this late. :)

[Carnak]

It really depends on your location and how humid the outdoor air is.

I am in the tropics, your kitchen load sounds like what I could expect down here. For kitchens it is mainly the heat generated inside, and the effect of make up air. For my worst case here I try to keep kitchens under 80F and 60% RH.

Your dining and dancing area load sounds high. Your software should let you model this area per the time of day. I do not think you will have a room full of people dancing up a storm at 2 PM in the afternoon when the sun is blaring down, and the outside air temperature is at a max. This kind of load will peak after the sun has set. You have a lot of fresh air for people as well, a code like the ICC may want a high amount of fresh air, something like the current ASHRAE 62.1 may require 4500 CFM.

500 people dancing is a lot of heat load, but I do not think you could fit them all on the dance floor at once. Maybe its 100 dancing, 100 just finished dancing and are sitting down so 200 people sweating, 300 people just sitting there at rest.

So depending on your fresh air moisture content and people, maybe vent/people is a 50 ton load maximum. Not a lot of lighting when they are dancing, sun is set.

So I do not think you will have 50 tons of heat radiating off of the bands PA system or conducting in through the walls and roof at night. 100 tons sounds high to me for the dining/dancing area. You run this at 30+ CFM a person for smoking?

May want to hire an engineer on this one. Computer programs are notorious for GIGO.

[beenthere]

carnak.

We generally allow for the hot food for those 500 people also. It usually comes out to 9 people per ton.

So we'd be looking at 55 tons. 60 depending on which township we were in.

[Carnak]

yes hot food also

I just looked at it like dancing, some babes are hotter than any tamale :)

so after dinner and the speeches is when the real heat begins

But it really depends on that ventialtion load, some places yes it could be 55 tons for everything, other places 50 tons just for the people and the ventilation.

[dan sw fl]

Does anyone have any sample loads for a commercial kitchen that would simulate a30 X 48 kitchen and an 80 X 80 area for 500 people.

How does this software handle exhaust and makeup air calculations?

What information will the software calculate and what do I need to supply?

My initial stab at this is calling for 15 tons in the kitchen and a little over 100 tons in the lodge area. Just seems a little high to me. Any thoughts appreciated...

Pretty difficult starting with the applications
that are among the most complex in the A/C world.

ASHRAE guidelines and

http://www.energy.ca.gov/reports/2003-06-13_500-03-034F.PDF


Nothing is Simple in the kitchen, otherwise,
we would all be cooking

K.I.S.S.

[superdave44]

Great suggestions and comments from everyone. Thank you! I was brought in to this project after the first building constructed like the one I am calculating did not handle the cooling load. This building is in the central part of Mississippi where the humidity is very high. I have calculated this building on a constant load and on maximum occupancy. This building will be rented out for various activities so there is noway I can determine when the load may exist. For most buildings they have outlawed smoking and want to crucify anyone who smokes so I have not considered any smoking provisions in the load. Personally, I am a smoker but thats a story for another day. You are exactly correct in that 500 people will probably not all be on the floor at the same time, But if they do they will come to me saying "It wont cool". This might be what happened to the first building. Not sure. Wrightsoft has the ICC information already loaded into the program and its calculating the fresh air on a per person basis. I know vent hoods come in different variaties and not sure how to add this element into the mix. Will the vent hood in the kitchen make up the same air being exhaused. According to research a 10X6 hood will require 9000 CFM of air. At 400 CFM per ton thats a little over 20 tons if the hood were only exhausting the air. How should the vent hood be handled in the calculation if it also brings in makeup air too. I foresee a possible pressurization problem for the kitchen...

Hi Dan, just read your link ... Great information on the subject!!!!

Thanks again for all your comments and suggestions!

[beenthere]

Make up air for the exhaust hood can be brought in low to the floor near the grille/cooking area. You don't have to condition it then. Bring in atleast 80% of your exhaust.
In a high humidity are, It may take 1 ton for every 200 CFM of fresh air.

[cxagent]

Rightsuite is going to calculate what you tell it to calculate. For the kitchen you have to tell it what pieces of equipment are in the kitchen and pick the heat load each give off. Don't just stick all the kitchen equipment in there and running full blast at the same time. You have to use some good judgement on how much and when each run.

Hoods will have about 80% of the exhaust as make up air. You only need to cover the 20% as outside air in the heat load calculation. But that 20% will be at 1 ton for each 200 to 250 cfm of outside air - not 400 cfm because of the latent load in 100% outside air. See beenthere's post.

If you design the dining room as 500 people all the time - you will have MAJOR humidity control problems. At 12 sq ft per person (using your numbers) that doesn't leave room for tables and chairs and aisles for the servers to deliver food. You better be planning on oudoor air preconditioning or at minimum demand controlled ventilation. Even with DCV I would recommend several stages of cooling with some kind of enhanced humidity control. What is enhanced humidity control? Chilled water, split face coils (NOT intermingled coils), deep evaporator coils (4 or more rows), hot gas reheat, etc.

I would STRONGLY recommend you find a GOOD engineer to design this. Those are rare as hen's teeth but there are a few. You don't find them by looking for the cheapest "bid" or the quickest turn around. You probably have to find one by looking for similar building that work and asking for those engineers. Usually the HVAC contractors, TAB contractors, and some of the building owners know them but don't always share their name with anyone.

[dan sw fl]

Great suggestions and comments from everyone. Thank you!

I was brought in to this project after the first building constructed like the one I am calculating did not handle the cooling load.

This building is in the central part of Mississippi where the humidity is very high.

You are exactly correct in that 500 people will probably not all be on the floor at the same time, But if they do they will come to me saying "It wont cool".

... and its calculating the fresh air on a per person basis.

I know vent hoods come in different variaties and not sure how to add this element into the mix.

Will the vent hood in the kitchen make up the same air being exhaused.
According to research a 10X6 hood will require 9000 CFM of air.

At 400 CFM per ton thats a little over 20 tons if the hood were only exhausting the air. How should the vent hood be handled in the calculation if it also brings in makeup air too.

Hi Dan, just read your link ... Great information on the subject!!!!


10x6 foot hood may have the air flow designed at ~ 4,000 CFM depending on Capture Air effectivness. ~ 80% can be make-up air through the hood.
One might need to provide ~1,000 CFM additional Outside Air.

However, with 500 people one is already designing for ~ 4,000 + CFM Outside Air.
With high enthalpy ( ~42) for the Outside Air, the total peak load would be ~ 30 Tons.

Enthalpy at 53'F WB =~ 22

Q = 4.5 * Enthalpy difference * CFM
Q total = 4.5 * 20 * 4000 = 360,000 BTUh

Q sensible = 1.08 * dT * CFM = 1.08 * 40 * 4000 = 173,000

Q latent is a little more than half of Q total.

Kitchen will obviously be a negative pressure relative to the rest of the building.

Use DCV too reduce operating costs.
Use ERV to reduce equipment size and operating costs.

[Carnak]

It will not be difficult controlling temperature at all. What will be difficult will be controlling huimidity especially with the 'your better safe than sorry' appraoch with 500 people dancing.

For 500 people dancing, that load program you have your blind faith in is probably saying that your supply air conditions are something like 55F dry bulb and 51F wet bulb. When supply dry bulbs and wet bulbs required are far apart, that is a big red flag. It means that you will not be able to control humidity without a reheat system.

Your program is also probably spitting out what appeears to be a rediculously low CFM needed for the tonnage. It is telling you this is not a normal type of job.

You would almost be better off going with a 100% outside air unit if in fact you were designing for that.

For 500 people, not overly active, even with an ERV doing the ventilation and feeding air to the return of your air conditioning system, you could still be challenged to control humidity.

I just did a Church Hall that you can fit up to to 500 Pentacostal parishoners in, it is doubling as a hurricane shelter. ERV knocked a lot of load off of the system but it still needed some supplemental dehumidification. Mind you it is more humid where I am compared to Mississippi or Florida and I am not stuck with excessive ICC vent rates, I only have to follow ASHRAE.

9000 CFM up the hood sounds excessive. Is this two short Island hoods back to back?

6 foot hood dimension sounds wrong

9000 sounds worse than a 9'x6'x150 FPM capture velocity dictatred by some code for a hood being built from scratch to go above some big wok. Talk to a hood manufacturer like Greenheck or CaptiveAire.

If the hood is mounted on a wall, you can supply a good chunk of the make up air like 70% through a back supply plenum and in the summer you would not even waste energy conditioning it.

You can bring in a little more of the make up air through the kitchen AC unit, and then allow air from the hall to transfer to the kitchen to make up the rest.

This sounds like a pretty big job to be trying to cut your teeth on with advise from an internet forum.