John Sayers' Design Forum

John Sayers' Recording Studio Design Forum

A World of Experience
Click Here for Information on John's Services
It is currently Fri Nov 24, 2017 2:45 am

All times are UTC + 10 hours




Post new topic Reply to topic  [ 4 posts ] 
Author Message
PostPosted: Wed Aug 31, 2016 12:50 am 
Offline

Joined: Tue Nov 17, 2015 1:13 am
Posts: 26
Location: Austin, TX, USA
Hi guys. Thanks in advance for any help you can give me. FYI, I have a related and parallel discussion going on HERE about air flow into and out of my vocal booth.

I’m building a machine room in which all gear with fans will be located in a tall server rack. This rack will contain two UPS’s, a console power supply, a power amp for monitors and a computer plus lighting, etc. I haven’t finished figuring out what the total wattage is but, for the sake of discussion, let’s say that it’s an even 1,000 watts for now.

The room is going to be approximately 50 sq. ft. with 8 foot ceilings for a total volume of approximately 400 cu. ft. My intent is to use an intake fan and/or exhaust fan to exchange air with the adjacent live room which is already air conditioned. I’ll be using silencer boxes at each penetration to keep fan/air noise from entering into the adjacent live room. My questions revolve around figuring out the amount of air flow into and out of the machine room which is necessary to keep it sufficiently cooled.

I’ve been having difficulty finding any concrete guidance on how to get from watts (or btu’s or tons) to volumetric air flow (cfm). I need to be able to figure out how much air flow I need so that I can appropriately size my fans, duct work and silencer boxes for the machine room.

Thus far I have the following:

1,000 watts x 3.4129 btu/watt = 3,754.19 btu/hr
3,754.19 btu/hr / 12,000 btu/ton = 0.312849 ton/hr

Is everything above correct? Now this is the part where I’m stuck. I’ve managed to read that a good rule of thumb is 400 cfm of air flow per ton. I don’t know the origin of that number but I read it on an hvac site. I’m not even sure if I’m applying it in the right context. However, if that rule of thumb is correct, then I would have the following:

0.312849 ton x 400 cfm/ton = 125.1397 cfm

This leaves me with the following questions:

1. Does 125 cfm into and out of my machine room seem reasonable based on a 1,000 watt heat load?
2. Is this 400 cfm per ton conversion factor even correct?
3. If this 400 cfm per ton conversion factor is not correct, then what IS the correct conversion factor to use and how do I use it?
4. Is there some better or more appropriate way to figure out how much air flow I need based on the total heat load of my machine room?
5. Being that the silencer boxes for the machine room are ONLY for preventing fan/air noise from escaping into the adjacent room, would one layer of 5/8" plywood be sufficient or should I still double up with two layers on all sides of the box?


Report this post
Top
 Profile  
 
PostPosted: Mon Sep 12, 2016 2:14 pm 
Offline
Site Admin
User avatar

Joined: Thu Aug 21, 2008 10:17 am
Posts: 10278
Location: Santiago, Chile
Quote:
I’ve been having difficulty finding any concrete guidance on how to get from watts (or btu’s or tons) to volumetric air flow (cfm).
The reason you can't find an info on that relationship is because there isn't any relationship! It's that simple. You are trying to convert between unrelated things. It's like saying "I'm trying to convert miles per hour into kilograms". They measure different things, so there is no conversion factor.

Heat energy flow is not related to air flow volume. The amount of heat that air can remove depends on the temperature difference, not how fast the air moves, or what volume of air is moving. If the air is at the same temperature as the equipment, then no cooling at all takes place: if the air is at 35°C and the equipment is at 35°C then it does not matter how much air you move thorough the equipment, it will never get any cooler.

In other words, you are asking for a conversion that makes no sense.

You are looking at this from the wrong point of view: You don't need to move a certain number of cubic feet per minute to cool the equipment: you need to provide a certain amount of heat removal (cooling). The amount of air that you need to circulate through there depends on the conditions of the air itself. The temperature and humidity. If the air is very cold, you don't need to move so much of it, but if it is warm then you need to a lot more. You need to look at the cooling capacity of the air itself in order to figure out how much you need.

So that's the issue. You are looking for something that cannot be found.


- Stuart -

_________________
I want this studio to amaze people. "That'll do" doesn't amaze people.


Report this post
Top
 Profile  
 
PostPosted: Mon Sep 12, 2016 5:38 pm 
Offline

Joined: Tue Nov 17, 2015 1:13 am
Posts: 26
Location: Austin, TX, USA
Soundman2020 wrote:
Quote:
I’ve been having difficulty finding any concrete guidance on how to get from watts (or btu’s or tons) to volumetric air flow (cfm).
The reason you can't find an info on that relationship is because there isn't any relationship! It's that simple. You are trying to convert between unrelated things. It's like saying "I'm trying to convert miles per hour into kilograms". They measure different things, so there is no conversion factor.

Heat energy flow is not related to air flow volume. The amount of heat that air can remove depends on the temperature difference, not how fast the air moves, or what volume of air is moving. If the air is at the same temperature as the equipment, then no cooling at all takes place: if the air is at 35°C and the equipment is at 35°C then it does not matter how much air you move thorough the equipment, it will never get any cooler.

In other words, you are asking for a conversion that makes no sense.

You are looking at this from the wrong point of view: You don't need to move a certain number of cubic feet per minute to cool the equipment: you need to provide a certain amount of heat removal (cooling). The amount of air that you need to circulate through there depends on the conditions of the air itself. The temperature and humidity. If the air is very cold, you don't need to move so much of it, but if it is warm then you need to a lot more. You need to look at the cooling capacity of the air itself in order to figure out how much you need.

So that's the issue. You are looking for something that cannot be found.


- Stuart -


I understand what you're saying but then HOW would I go about figuring the cfm I need? "Conversion" was probably a poor choice of words, as I get that there isn't an actual direct conversion factor. I simply meant to ask how to get from what I know (watts) to what I don't know (cfm) based on my stated conditions and needs? As you pointed out, cfm may not be directly related to heat but somehow the cool air still needs to get in there and the warm air needs to get out so a calculation of cfm must be made at SOME point, right? There has to be some way to calculate the volumetric flow rate of air I need to move through there to achieve the desired heat removal.

I've finished figuring up the total wattage of the equipment going in there and it's going to be right around 4,000 watts. The adjacent live room will be kept at around a constant 75 degrees (f) and I'd like to keep the machine room at around that same temperature or a little warmer, say up to 85 degrees. Let's say that the relative humidity inside the adjacent live room is at 50%, as it has already been cooled and removed of some of it's moisture by the ac system in there. I need to size my silencer boxes, ducts and fans to deal with this. How would you calculate this? I have Rod's book and I don't see anywhere how to do something like this?


Report this post
Top
 Profile  
 
PostPosted: Sun Oct 09, 2016 12:03 pm 
Offline
Site Admin
User avatar

Joined: Thu Aug 21, 2008 10:17 am
Posts: 10278
Location: Santiago, Chile
This is very much an indirect method of figuring it, but look at the flow rate of a typical mini{-split system that can handle that heat load.

4000 watts heat output is roughly 13,000 BTU/hr. A typical 12,000 BTU mini-split will move about 250 to 350 CFM, very roughly. It will likely need to run at max capacity to to provide the full 12,000 BTU cooling, so I would assume that you'll need to move at least 350 CFM, probably more like 400, to provide sufficient airflow for cooling 4 KW of gear.

Like I said, this is a very mangled, indirect-method of figuring it, but I'm betting it wont be far off the mark.


- Stuart -

_________________
I want this studio to amaze people. "That'll do" doesn't amaze people.


Report this post
Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 4 posts ] 

All times are UTC + 10 hours


Who is online

Users browsing this forum: No registered users and 19 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Jump to:  
Powered by phpBB® Forum Software © phpBB Group