room deminsions 10'*10'*10'
u factor for walls, floor, ceiling = .25
desired temp =75deg(f)
design temp =105deg(f)
SOLVE FOR BTU INCREASE
(NO DOORS OR WINDOWS)
You forget to mention time in the question. Are we to wait until the indoor temp matches the outdoor temp? Or are we running at a steady state and you just want to know the btus that need to be removed to maintain indoor temp?
Read, read, read!
I was just wanting to know the way to manually calculate the heat load of a room. I know my example room could not possibly exist in reality but if you assume a .25 U factor (R4) for all 6 sides of the room, how do you calculate the heat load?
Remember that the desired indoor temperature is 75 and the design temperature is 105.
It may come as a surprise but none of my competitors seem to posses this knowledge either. When I ask, the excuses I get are, "I've got that information in a book at home." Or "I use a computer program that calculates all that for me." Or, "Oh, It's been so long since I've done that, I'd have to look back on some old notes to give you that information." (Please don't ask me where I am from because it would impose we're all idiots here. Well, ok the facts speek for themselves I guess.) Fact is, most of my competitors never are involved in building houses and are never (required) to perform a manual J calculation.They simply replace what's there, use a rule of thumb and of course (guess) on heat loads.
To be totally honest, what got me started on this (quest) is my fishing buddy has asked me to calculate how much ice it would take to keep his shad alive. He plans to use a new type oxygenator that doesn't introduce (hot)outside air.
What I am supposed to do for him is calculate the heat gain through how much and how thick and what type insulation. Determine how many pounds of ice per day is needed to maintain the desired temperature on the hottest of days. (ie,,, 105)
I realized I couldn't even properly calculate a single room load using the information we are all supposed to possess.
using the formula you gave me:
lets exchange the letters for real numbers.
Q = A * U * TD
AREA = 10X10= 100 sq ft.
TD = 105-75 = 30 ????
U = .25 ????
IF I HAVE THIS RIGHT:
q = 100 * .25 * 30
q = 750
There are six sides of equal proportion and u value so:
750 * 6 = 4500
so, BTU = 4500
????????? Did I do it right????
I assume this is 4500 btu/hr. or, appx 150 cfm (7 inch duct)
If this how it is done, I should easily discover the thickness and type of insulation required to manitain my friends shad tank. Once the btu increase is calculated, I could then determine the lbs of ice required to maintain his desired temperature.
Since 1 lb of ice requires 3.41 btu to turn into water, I will know how many lbs of ice to use per hour (then per day).
Of couse, I am assuming a 10x10x10 room and a 2x3x4 tank (actual size still not known) full of water that are of equal U values would permit or transfer heat at the same rate.
Please don't pay much attention to the temperatures here. I'm sure my friend will need 55 or 60 deg water. I will (plug) real numbers in later. What is important is that I have the formula and method of calculation right. I have U factors for a multitude of materials (insulation) he might could use.
each side is done seperately and added up to a total sansible heat content. and Q is equal to btu/hr stop using that damn program. buy a trane manual and rip out the load calc sheet, then use the tables to practice.
the floor of the space is a different situation all together. not calculated the same as walls. or roof.
it's not my formula. ASHRAE (ASHVE), willis carrier god knows who else came up with that one.
much respect to the older and wiser, thanks Sal.
i just about let the whole thing go and then saw your bit about the ice. put the brakes on. first psychrometrics are going to get you here. unless your freind worked it out already. i hope you are not planning to count evaporative cooling here. and there are more like 144 btu per pound of ice melting. now you know why everyone raves about latent thermal storage systems.if you wanna post more details i will look for them.
and there are different u values or heat tranfer coefficients for different materials, wether they be homogenous or non-homogenous, a simple insulation book may not help with all. i have more formula avail.
[Edited by hvac3901 on 06-25-2004 at 01:23 AM]
Here are some R Values I have:
BLOWN CELLULOSE 3.70/INCH
EXPANDED POLY STYRENE 4.00/INCH
EXTRUDED POLYSTYRENE 5.00/INCH
(I was thinking more on the lines of that expanding insulation you can buy in a spray can at home depot called "Great Stuff". Maybe it gives the R value/inch on the can somewhere and I can then invert it for the U factor.)
I am hoping to be able to use one of these materials to construct the insulation for the tank.
Thanks for the correction on BTU/lb ice = 144. I was thinking that didn't sound right when I wrote it.
As far as the room is concerned. It is purely hypothetical. I was just looking for the math. Did I have that part right?
The program......... O'brians was purchased about 8 years ago and I never trusted it for a real job. I called and talked with the programer of the program and found out I had many of the default U factors set wrong but even he suggested I get the latest windows version. (They apparently made a lot of improvements...... I still am trying to find the time to evaluate the trial version before [time] runs out.)
The trial version this forum recommends, "HVAC CALC" is so limited in any of the settings, I found it impossible to evaluate.....I'm sure its a fine program just going by the advice of the people talking here about it.
A pen and a piece of paper is much slower but I don't need such calculations that often anyway. I'd probably put it in a spread sheet if I got the math right. I like doing things (my way) most of the time but even I realize sometime I shouldn't (always) try re-inventing the wheel. A better mouse trap???,,,,, ALWAYS!
Thanks for letting me (pick) your brain!
The more I learn about things the more I realize how ignorant I am. That's why I quit going to school. (got tired of realizing how ignorant I was! Now I realize if I don't use my brain cells, they will die off one by one even if they are left empty. Life just ain't fair!)
those values for materials would all be concidered homogenous materials, and calculated by thickness and coresponding ht transfer coefficients.
the comments about using the program were dirrected more toward sarge, with no offence intended. said formula lacked time factor. but if the heat transfer coefficient is known and the delta t is known then the time for the temp to equalize is also known.
you do have the jist of it good luck
[Edited by hvac3901 on 06-25-2004 at 09:55 AM]