Moisture content of air - STHE intercooler 1

[patrickraj]

Dear friends,

An air compressor second stage intercooler data sheet indicates that the air moisture content is 1513 Kg/hr. But as per my calculation, it is only 1415.6 Kg/hr. Following are the calculation details :
Dry air flow First stage =61600.0Kg/Hr
First stage outlet air temperature=42°C (567.6 °R)
First stage operating pressure =2.45 kscA (34.84 psiA)
LN vapour pressure =(0.001786-1/567.6)x9015
LN (VP) =0.2181
VP =1.244 PSIA
Partial pressure of water vapour=34.84 - 1.244 =33.6 psiA

Moisture Content =61600/29 x 1.244 / 33.6 x 18= 1415.6 Kg/hr

% Error=1513 - (1415.6 / 1513) x 100 = 6.43 %

Someone could clarify the reason for the difference/ What is wrong in my calculation.

Thanks to all in advance.

 

[TD2K]

Your calculation looks right to me. Your water vp agrees with the steam tables as close as I can read them off to your temperature.

I took your dry air flow, converted that to kgmole/hr, upped that to the total flow given the partial pressure of the dry air and the water vapor and then worked out the water vapor flow rate, essentially the same rate as you.

One thing to remember is that water vapor pressure is very sensitive to temperature. A very slight change in your temperature of 42C leaving the cooler could easily affect the vapor pressure by the 6%. Simply changes in cooling water temperatures to the upstream cooler during operation could swing your outlet temperature of the air by more than this.

Basically, you need to consider how significant is this error and what is the effect on the rest of your process.

 

[Mrepp]

LN vapour pressure =(0.001786-1/567.6)x9015
Could you explain the elements of this equation

 

[TD2K]

I sort of wondered that myself but the vapor pressure agrees with the steam tables and unlike school, I don't have to give marks for his work

 

[Montemayor]

patrickraj:

You request us to clarify the differences you get and ask what is wrong in your calculation. As a professional engineer, I'm going to assume you mean what you ask and I'll take it from there.

1) Reason for the difference:
This could be, as TD2K indicates, that you trying to dot the i's and cross the t's on accuracy when there is no justification to do so. Your data sheet was filled in by another engineer that may have been using other basic data. Additionally, assuming you are basing yourself on John Dalton's famous Law, you assume that the partial pressure fraction is equal to the mole fraction of a component in a mixture (water, in this case). Water vapor is regarded as and ideal gas for the purpose of Dalton's theory, i.e. water molecules act independently from each other in the air mixture. In reality there is a small interaction between molecules which leads to a small increase of saturation vapor in air. As the system pressure increases, the effect of the compressibility factor also increases. Other factors to consider are found below, in 2).

2) Contrary to what TD2K states, I find your calculations flawed and poorly documented - resulting, as usually is the case in such work, in errors. Assuming you'll accept well-intentioned, constructive criticism, lets look at the flaws:
a) You state that you are checking the air moisture content in the air compressor SECOND stage intercooler; yet, you take basic data related to the FIRST stage. Obviously, something is wrong here. I'll assume you just got your stages crossed-up for now.
b) You cite something called "LN (VP)" without identifying what it is or where you obtained (or based) the relationship. This is bad engineering. No responsible supervisor will accept this type of documentation. Sorry, but that is the basic law in engineering: document, document.
c) Your vapor pressure value for water at 42 oC is off by 4.5% (1.1907 psia) according to the NIST website information - which is my documented source. NIST = National Institute of Standards & Technology (USA).
d) You say that the partial pressure of water vapor = 33.6 psia; I believe this is wrong. The partial pressure of the water vapor is the same as its vapor pressure - 1.1907 psia. I think the partial pressure of the air is 33.6 psia
e) I believe Dalton's Law states that the partial pressures in a mixture are cummulative to the total system pressure. That means the partial pressure fraction of water = 1.1907/34.84 = 0.03417 and not 0.03702 as you note. The important point here is applying Dalton's Law and differentiating between the different values and why.
f) If you really wanted accuracy (since you note the answer down to the first decimal place), you would use the molecular weights of air and water as 29.964 and 18.015. Otherwise, you'll get an estimated answer -- which is OK.

On a more practical and less theoretical level, your answer is close enough for engineering purposes - in my opinion. What I object to is the wrong employment of Dalton's great work and not documenting your calculations the way they should be. Engineering answers are worthless if they are not backed by credible references or sources. I don't care a hoot what the calculator or the computer spit out. The answer is only as valid as the logic and engineering prowess of the engineer. And if this isn't clearly denoted in the calculations, I can't tell if it exists. I didn't want to go into details, but you asked what was wrong with your calculation and I did the best I could.

Regards,



Art Montemayor
Spring, TX

 

[TD2K]

Art, not trying to nitpick here, but the source of your MW for air is from where?

I've always used 28.96. I checked the CRC Handbook of Tables for Applied Engineering Science, they list the MW of air as 28.966. Ingersoll-Rand's Compressed air and data book use 28.97.

 

[Montemayor]

TD2K:

You caught my typo error; it should, of course, be 28.964. Thanks for the correction. My source is the GPSA Engineering Data Book; 9th Ed.; 1972; Section 16; Physical Properties Table, Fig. 16-1. Although I have the latest electronic version, I frequently resort to the old-time hard copy because it's handy and I know exactly where to look. The GPSA MW values were significant to the 3rd decimal place.

Art Montemayor
Spring, TX

 

[patrickraj]

Thanks to all.

1)Yes.There is an typographical error,it is PP of dry air, not PP water vapour.
2).My friends in my office use this formulae extensively,so, I started using. I do not know the source.
3).From first stage outlet air goes to 2nd stage,so I have taken first stage outlet temperature & pressure as the inlet conditions to the 2nd stage, to calculate moisture content.
4)567.6 is in RANKINE, of 42deg C
5)Per data sheet inlet air (to 2nd stage) molecular weight is 28.55,Since some moisture is knocked out in the first stage compression.First stage inlet air molecular weight is 28.128 per data sheet
6)Is it not possible to arrive at the same value as calculated by others.
7)You may be correct. (Vapor pressure value for water at 42 oC is off by 4.5%). Tomorrow I write on this.

Though I seems to be wrong, why I am not getting the data sheet moisture value.

Encore Thanks to all.

 

[Montemayor]

patrickraj:

As I was able to prove, typo errors are easy to make. I wouldn't worry about this part of engineering too much. Documenting and explaining your work is more important. It is becoming exceedingly important to always leave "audit trails" in your engineering calculations in these days of increasing litigation.

The formula your friends employ is an application of John Dalton's famous Law that says:

mol fraction of water vapor = mols water vapor/total gas mols = partial pressure water/total system pressure

You have calculated the water content in the 1st stage discharge where it is 42 oC - not in the 2nd stage discharge. Dalton's Law is applied to the discharge after it has been cooled to 42 oC (& is in equilibrium with condensed water). This discharge mixture is saturated with water vapor (since it is in equilibrium) and thus, the vapor pressure of water at its temperature is equal to the partial pressure of the same water. Hence, Dalton's Law.

It is very possible to arrive at the same value as calculated by others; all you have to do is use the same relationships and basic data. As I said previously, obviously you are not. But this is not critical to your application. Believe me, an error of 6% for this type of calculation where you will be designing the intercooler and condensate trap. Your engineering experience should tell you that you must allow for contingency in such calculations. For example, as I pointed out, you've applied Dalton's relationship as it theoretically exists. You have not allowed for compressibility and mutual attraction effects (empirical deviations from Dalton). This is where pure science is separated from real life, practical engineering.

I hope you are not discouraged because you are not getting the same, identical numerical answer.



Art Montemayor
Spring, TX

 

[patrickraj]

Dear Mr. Montmeyar,

Following equation is available in CE June 10,1985 issue.
(T/Tb) – 1 = Ax + B(x)^2 + C(x)^3 + D(x)^4
Where “x “ is Ln(P), P in AtmA
A = .0749586
B = .00636077
C = .000608182
D = .0000626433
Tb = 373.16 °K
Vapour press calculated to be 1.1893 PSIA compared to 1.24 PSIA obtained using the previous equation

With the above VP of 1.1893 PSIA, moisture content works out to 1358.9Kg/Hr. Now the VP is very close to table values but the error works out to higher at 11.343%.

 

[patrickraj]

Dear TD2K,
Is it (28.96) universaly value, do not this vary region to region(in coastal area MW will be low because of high moisture content).
In very cold regions, air MW will be high due to low moisture content.

 

[TD2K]

28.96 is the MW of air with no water vapor (dry). Obviously, the water vapor content will affect the average MW of the mixture.

 

[25362]

To patrickraj, the pressure read at the discharge of the 1st stage is higher than that prevailing after the intercooler, to cover for the friction drop in this part of the system. If so, the explanation for the difference with the data sheet may be that those writing this sheet took into account the actual inlet pressure to 2nd stage being lower than the one you are using by a few psia.

What do you think ?

 

[patrickraj]

Dear 25362,
You are absolutely correct that the actual O/L pressure read at the discharge of the 1st stage is higher than that prevailing after the intercooler. So, pressure will be marginally high, and the moiture content will be still low(the denominator will be high) only. For calculated low moisture content oly I could not find the reason or how to calculate?.

What is your opinion?

 

[patrickraj]

Dear Montemayor,

Please give website address of NIST = National Institute of Standards & Technology (USA).

Thanks in advance.

 

[Montemayor]

patrickraj (Chemical):

Go to

http://webbook.nist.gov/chemistry/fluid/

This website gives FREE thermodynamic data on 34 Fluid Systems. These data include the following:

Density
CP
CV
Enthalpy
Internal energy
Viscosity
Joule-Thomson coefficient
Specific volume
Entropy
Speed of Sound
Thermal conductivity
Surface tension (saturation curve only)

The substances defined with FREE information are:
1 Water
2 Nitrogen
3 Hydrogen
4 Parahydrogen
5 Deuterium
6 Oxygen
7 Fluorine
8 Carbon Monoxide
9 Carbon Dioxide
10 Methane
11 Ethane
12 Ethene
13 Propane
14 Propene
15 Butane
16 Iso-Butane
17 Pentane
18 Hexane
19 Heptane
20 Helium
21 Neon
22 Argon
23 Krypton
24 Xenon
25 Ammonia
26 Nitrogen Trifluoride
27 R-22
28 R-32
29 R-123
30 R-124
31 R-125
32 R-134A
33 R-143A
34 R-152A

Of course, the NIST has thermo data on just about every substance and compound known but it probably is kept in the databases that they sell to the public - and not at a bad price.

Hope this helps.



Art Montemayor
Spring, TX

 

[patrickraj]

Dear Montemayor,

Thanks a lot from heart.