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Does Altitude the correction factor of specific heat?

48K views 83 replies 15 participants last post by  E Silberstein  
#1 ·
When checking CFM using a gas furnace
Delta T X 1.08 Specific Heat Divide this number against BTUH Output of the furnace = CFM
Does altitude change the number 1.08? Is there a chart for different altitudes?
 
#2 ·
Yes altitude affects the specific heat number. In fact quite a bit. Example, at 70deg sea level SH of 1 at 5000' sh .83.
NEEB's book HVAC Testing'Adjusting, and Ballancing Manual has all the conversions on a chart. The range is .3 - .4 per 1000'. So at 70degF 1000' .95, 2000' .92 etc.
Most of the time altitudes below 2000' are disregarded for T&B purposes.
 
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#3 ·
This brings up an interesting point

If you think altitude should be corrected for, what about measuring the air quantities after the air is heated.
The TAB scientific approach is to correct for anything to bring it back to SCFMs Standard CFM @ 70 degrees
 
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#4 ·
Altitude and Temperature will both change the 1.08 constant.

As temperature or altitude increases, the density of the air decreases.

Standard air = 70°f and weighs .075 pounds per cubic foot at sea level.

If I don’t have a chart, I use 3.5% (.035) decrease for each 1000 feet in altitude above sea level, and/or a 1.5% (.015) decrease for each 10 degrees of temp increase above 70 degrees, to get a new constant value.

Altitude Example for BTUH Correction:

For an altitude of 4000 feet: 3.5% x 4 = 14% (.14)

1.00 - .14 = .86 correction factor

1.08 x .86 = .93 new constant for 4000 feet

.93 x delta T x CFM = BTUH

To determine the actual amount of air being delivered, multiply the airflow reading by the correction factor.

EXAMPLE:

1200 CFM of 70 deg. air was measured at an elevation of 7000 feet above sea level, what is the actual amount of standard air in the duct?

1200 CFM x 0.76 correction factor = 912 CFM of Standard Air.


To correct airflow for elevation, divide the amount of standard air needed by the correction factor.

EXAMPLE:

1600 CFM of standard air is needed for the system to operate properly at an elevation of 5000 feet above sea level. How much air will be required for the system to operate with 1600 CFM of 70 deg air.?

1600 CFM of Standard Air needed / 0.83 correction factor = 1928 CFM of 5000 foot air required.
 
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#16 ·
Just a little more clarification



Rundawg,

Where can I find charts that will show the correction factors I need to use? I have searched the internet everywhere and I am not having any luck. Just so I can be clear, would you be so kind as to give an example of what the BTU output would be at 5000 ft at 80 degrees CFM is 1000 with a 50 Delta T? Thanks for your help.
 
#5 · (Edited)
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#7 ·
WOW

are techs suppose to be this scientific? You're going to give the biz a GOOD Name if this keeps up!
Impressive response
 
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#8 ·
Hey Rundawg

EXAMPLE:

1200 CFM of 70 deg. air was measured at an elevation of 7000 feet above sea level, what is the actual amount of standard air in the duct?

1200 CFM x 0.76 correction factor = 912 CFM of Standard Air.

If you did this why not just apply 1.08 times the standard air? Just off the cuff, I haven't thought it through yet
 
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#11 ·
If you did this why not just apply 1.08 times the standard air? Just off the cuff, I haven't thought it through yet
I'm not exactly sure what you are asking.

Specific Heat x Specific Density x 60 min./hr = Standard Air Constant

.24 (Specific Heat of standard air) x .075 lbs/cuft (specific density of standard air) x 60 (min/hr) = 1.08

The 1.08 constant, is a constant for "standard air" in the airflow equation.

1.08 x delta T x CFM = BTUH

Does this even come close to answering your question?
 
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#12 · (Edited)
ok, ive got a pretty good handle on this but still unclear on a few things

i understand how on different points on a psych chart the specific density of air will change, which then changes the constant to other than 1.08.

my question is, does the specific heat of air ( 0.24 ) change at various points of the psych chart??

rundawg, davidr, any TAB guys??????

oh one other question...

i recently read a users maunal for the new Amprobe thermal anomometer which combines a hot wire with a k-type tc and RH sensor. the manual states to punch in the current barometric pressure for accuracy.

this begs the question. if on the coast,sea level. 70 deg, 50% rh and in denver 70 deg, 50% rh, will my ADCF be differnent if both locations are experiencing say a barometer reading 29.92" hg??. after all that is a measure of atmospheric pressure?

im so confused:gah:
 
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#15 ·
If the ocean front were to experience the same conditions as Denver airport:
Current conditions 25°F 71%RH
Oceanfront 29.855"Hg (mean sea level pressure)
Denver airport @ 5431ft elevation would be 24.343"Hg. (actual true pressure)
This is due to elevation difference of 5431ft.
 
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#17 ·
If you know the RH or wb° then it's easy...
Use the attached 5000' altitude psychrometric chart and find the specific volume. The inverse of SV is Density.

Here is the formula you will use.

((specific volume)-Âą x 60 x 0.24) x 50 x 1000 = btuh
density x time x specific heat of air x ΔT x cfm = btuh

So, at 80°db 50%RH 50°Δ 1000cfm

.0598 x 60 x 0.24 x 50 x 1000 = ~ 43113.77 btuh
 
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#19 ·
One more fun fact. You should use the average density to get a more accurate calculation.
example: (return temp 70° 50%RH, supply temp 120° 18%RH)
(return density 0.06135 + supply density 0.05615)/2= 0.0587 giving a density of 0.0587, output of ~ 42293 btuh
 
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#27 · (Edited)
First, let me explain that this thread is a discussion on basic heat transfer calculations. I emphasize the word BASIC. They are a means to achieve a close estimate. The formula's are not exact, they are in simple math. We should all know that a firm grasp on calculus and physics would be necessary to use the proper formulas. With that said....

We know that temperature, pressure, and humidity are relative to density, no? A psychrometric chart will show this.
As temperature rises, and/or pressure falls, air expands and takes up more space, and there is a rise in specific volume ftÂł/lb. As temperature falls, and/or pressure rises, the air compresses and there is a fall in specific volume.
Density is the inverse being lb/ftÂł. So, as temperature rises and pressure falls, density will fall. This is due to molecules having more energy bouncing off one another, or less compression; this provides more space for the molecules to move in. As we rise in altitude air loses density and gains space between molecules due to loss in pressure. So, the air is less efficient at containing heat, because there are less molecules per cubic foot. You might ask, why is there less heat at higher altitude? Because our air is heated by the Earth through convection. The Sun heats the Earth by radiation.
As humidity rises, density will fall due to hydrogen having less molecular mass than oxygen and nitrogen. Remember density is mass/Volume, or m/V or lb/ftÂł.

davidr
If we are calculating heat transfer, we use temperature rise in the formula. In order to identify the amount of heat transfer, we need to know the amount of air we are dealing with. This is measured in cfm. However, the same 1000cfm at a specific volume of 13.5 ftÂł/lb, will not be equal molecular mass to air at 17.5 ftÂł/lb. Thus not having equal heat content. So, in order to know the heat content we need to know the density.
Now, if there is a temperature or pressure change, then density will change.

If we were only reading heat content in the air, then one density measurement would be fine. However, we are measuring heat gain from return to supply, not just heat in the supply.


I found a handy derivation heat calc formula sheet. I've attached it to this post, and it does mention to average the two.
In a discussion on the Testo 435, Jim Bergmann explained the function of this tool. I would not quote him, because I can't find the thread, but I believe he explained this meter, equipped with two real time probes one in the return and one in the supply, would account for and compensate for the density change.
 

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#31 ·
First, let me explain that this thread is a discussion on basic heat transfer calculations. I emphasize the word BASIC. They are a means to achieve a close estimate. The formula's are not exact, they are in simple math. We should all know that a firm grasp on calculus and physics would be necessary to use the proper formulas. With that said....

We know that temperature, pressure, and humidity are relative to density, no? A psychrometric chart will show this.
As temperature rises, and/or pressure falls, air expands and takes up more space, and there is a rise in specific volume ftÂł/lb. As temperature falls, and/or pressure rises, the air compresses and there is a fall in specific volume.
Density is the inverse being lb/ftÂł. So, as temperature rises and pressure falls, density will fall. Thuis is due to molecules having more energy bouncing off one another, or less compression; this provides more space for the molecules to move in. As we rise in altitude air loses density and gains space between molecules due to loss in pressure. So, the air is less efficient at containing heat, because there are less molecules per cubic foot. You might ask, why is there less heat at higher altitude? Because our air is heated by the Earth through convection. The Sun heats the Earth by radiation.
As humidity rises, density will fall due to hydrogen having less molecular mass than oxygen and nitrogen. Remember density is mass/Volume, or m/V or lb/ftÂł.

davidr
If we are calculating heat transfer, we use temperature rise in the formula. In order to identify the amount of heat transfer, we need to know the amount of air we are dealing with. This is measured in cfm. However, the same 1000cfm at a specific volume of 13.5 ftÂł/lb, will not be equal molecular mass to air at 17.5 ftÂł/lb. Thus not having equal heat content. So, in order to know the heat content we need to know the density.
Now, if there is a temperature or pressure change, then density will change.

If we were only reading heat content in the air, then one density measurement would be fine. However, we are measuring heat gain from return to supply, not just heat in the supply.


I found a handy derivation heat calc formula sheet. I've attached it to this post, and it does mention to average the two.
In a discussion on the Testo 435, Jim Bergmann explained the function of this tool. I would not quote him, because I can't find the thread, but I believe he explained this meter, equipped with two real time probes one in the return and one in the supply, would account for and compensate for the density change.
Let's simplify this. For comfort heating with no change in humidity ratio (grains) through the system, the sensible capacity (btu/hr) formula can be expressed in these terms.

(cfm_2 x 60 x dry air density_2 x h2) - (cfm_1 x 60 x dry air density_1 x h1)

The left hand side represents the total quantity of btus passing into the furnace per hour and the right hand side represents the total quantity of btus exiting the furnace per hour. The capacity is simply the difference between these to btu/hr rates. The affect of moisture content is accounted for in the values h1 and h2.

Note that this formula can only be simplified to the more familiar

cfm x 4.5 x H by assuming constant cfm(in vs out) constant density(in vs out) and standard air.

Both the change in cfm and the change in density are factors in the full equation. However, the cfm is inversely proportional to the density, and thus any correction factors for these would cancel out. IOW, the averaging of inlet and outlet densities that you mentioned will provide an incorrect result.

Don't forget to figure in blower motor heat as well.
 
#28 ·
Mgenius33,

Who is Jim Bergmann? your help has been greatly appreciated. We work for a utility and they have a program that calculates Btuh a little different. I think your formula is more accurate and I am just trying to reference something more than just the hvac-talk forum :).
 
#29 ·
If you Google, Jim Bergmann, you'll find some links related to him. He frequents this site and has made a few helpful instructional videos and pdf manuals on topics such as combustion analysis and airflow.
There are quite a few instructors that use this site, and many high quality technicians.
 
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#41 ·
There is a change in pressure. This is shown by using a total pressure tip. This is due to the weight of a fixed cubic ft of air being less at 120° than the weight of a fixed cubic ft of air at 60°. The averaging of inlet and outlet densities are due to this differential mass.
 
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#45 · (Edited)
The heat wasn't added to 1 lb of air going in only to be contained by a mere 0.9 lbs of air coming out. Where did the other 0.1 lb of air and the heat that was added to it go? Didn't the same mass come out that went in? Or is conservation of mass just a fiction?

The total mass is constant, but it is taking up more space coming out due to its thermal expansion. That's exactly why the discharge cfm is higher than the return cfm.

The constant pressure comment was supposed to be followed by some other stuff, but the edit button disappeared before I got around to correcing the post. I was going to say that the formulas posted here assume constant pressure. The specific heat of 0.24 is for standard air pressure. In reality the specific heat is dependent on pressure. Also in relality there is a slight difference in pressures, in vs out, and even between these and room pressure or ambient outdoor pressure. The pressures will depend upon the exact points of measurement. This can be adjusted for using the "equivalent altitude" method if extreme accuracy is required, but for our purposes we can consider the error to be too small to be relevant.
 
#42 ·
This is one reason it is handy to use pounds per hour, that is the only thing the blower wheel is concerned with regardless of changes in air properties that occur on the discharge of the fan.
 
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#43 ·
For capacity and temp diff you are right



But Dave we can't forget that the fan will deliver the same CFMs at sea level and Denver CO but with the difference in temp because as you point pout, the pounds are different
 
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#47 ·
The equation 1.08 x cfm x ΔT is not taking in the amount of Work done by the thermal isobaric process, if the pressure of a gas is held constant. The Work would be the expansion of the gas. This is calculated by using:
W = P x ΔV = P ( V2-V1)
W= work
P= pressure constant
V= volume

The first law of thermodynamics states ΔU = U2-U1 = Q - W
Where:
U= internal energy
Q= heat
W= work

If you calculate the total of heat transferred - work done by form of expansion, you will find the difference is near the simple compensation of averaging the two specific volumes.
Problem is, it's very difficult to take into account all of the changes occuring in a thermal exchange system. There could be instances where these simple calculations are off due to outside changes unaccounted for. I'm just trying to get as close as possible without being a rocket scientist.
 
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#48 ·
The equation 1.08 x cfm x ΔT is not taking in the amount of Work done by the thermal isobaric process, if the pressure of a gas is held constant. The Work would be the expansion of the gas. This is calculated by using:
W = P x ΔV = P ( V2-V1)
W= work
P= pressure constant
V= volume

The first law of thermodynamics states ΔU = U2-U1 = Q - W
Where:
U= internal energy
Q= heat
W= work

If you calculate the total of heat transferred - work done by form of expansion, you will find the difference is near the simple compensation of averaging the two tyspecific volumes.
Problem is, it's very difficult to take into account all of the changes occuring in a thermal exchange system. There could be instances where these simple calculations are off due to outside changes unaccounted for. I'm just trying to get as close as possible without being a rocket scientist.
Unfortunately it IS rocket science. That's why we use all of these dumbed down equations.
;)

What is the expansion doing work on in this case?
 
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