nitrogen enthalpy 2

[MECE02]

I'm trying to calculate heat transfer from nitrogen to boiling water in a dryer system. N2 enters at a rate of 1400SCFM at 347K and 2psig and exits at 383K . I used the following link for N2 enthalpy data:

http://www.engineeringtoolbox.com/nitrogen-gas-properties-d_1252.html

enthalpy difference from chart = roughly 571kJ/mol from the table
delta Heat = 1009956.25 kJ/min.


I used the following for Cp data:

http://www.engineeringtoolbox.com/nitrogen-d_977.html

I took avg T and Cp and got

N = 1768.75 mol/min.
Qnitrogen = 24.7625 kg/min.
delta Heat = 928.676952 kJ/min.

Why don't the delta heat values correlate between the two approaches - especially because it states N2 gas specific heat? Which one is correct - I would normally use enthalpy here?

 

[Montemayor]

This is something I've been repeating throughout the years on these forums: for recognized and authoritative enthalpy values (as well as complete thermo properties) go to:

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

I don't know why your calculations don't match up since you don't furnish your detailed calculations. It could be a host of reasons - a typo; wrong data; wrong units; wrong equation(s); etc., etc.. The important thing is that you can easily obtain the accurate enthalpy data (or the accurate Specifi Heat) from the NIST thermo data base I identify above. Be sure to use the correct data input and units. You can obtain a convenient HTML table listing of all the values and copy & paste the table into an Excel spreadsheet for convenient future reference and calculations.

 

[MECE02]

Thanks Montemayor!

From the NIST website for N2 should I use Enthalpy (kJ/mol)or Cp (J/mol*K)for N2 vapor?

Cp seems right to me because:

1400SCFM*(28.3L/ft^3)*(mol/22.4L) = 1769 mol/min.

(1769mol/min)*(14g/mol)*(1kg/1000g) = 24.8 kg N2/min

(1769mol/min.)*(29.2J/mol*K)*(383K-347K)=1.85E6 J/min = 1850 kJ/min.

Great, so that is the rate of energy transferred to water at 155F. The water evaporates and becomes water vapor due to the heat transfer to the water. Should I use a humidity chart at this point? I need to determine the mass of water vapor leaving the vessel in the N2 stream.

Thanks!

 

[25362]

You have made two assumptions:

[•]The selected "standard" T,P conditions
[•]The Cp taken as a constant

Using your number of moles, I get a small difference with your estimate: 1860 kJ/min, using enthalpies.

 

[MECE02]

Correct, Cp doesn't change much with these T changes.

Do you have an approach to the water vapor flow rate?

Thanks!

 

[Montemayor]

MECE02:

You are confusing us (or me, at least) by not telling us ALL of the basic data.

Are you sparging dry, hot nitrogen into a water bath and, thereby, saturating it? In other words, I suspect you are NOT exchanging heat in a conventional heat heat exchanger apparatus, but that you are carrying out a simultaneous heat and mass transfer operation. Am I correct? If so, then you are saturating the flowing nitrogen at the system pressure and resulting temperature.

Correct me if I am wrong in my aasumptions because first you stated "I'm trying to calculate heat transfer from nitrogen to boiling water in a dryer system". Then you further state "The water evaporates and becomes water vapor due to the heat transfer to the water". If your water is boiling at 2 psig, how is it that you can take nitrogen at 370 oK (206 oF) and heat the hotter water with it???

Is this a school homework problem? If not, then gives us ALL of the basic data - the whole story, please.

 

[MECE02]

Sorry for any confusion. This is certainly not a school homework problem. I was trying to over simplify this plant scenario. At first I thought the water was near the boiling point, however, it isn't.

To further clarify:

Unreactive solids are added in the top of a steamer vessel. Steam is added to the steamer through a steam ring inside the vessel. Some steam condenses on contact with the solids and subcools. The slurry is transported to a dryer tank where hot nitrogen is added to fluidize the particulate bed and remove water. I am trying to calculate the amount of water vapor in the nitrogen stream.

I am not sure if the N2 stream is saturated.

I hope that helps and I look forward to your response.

 

[Montemayor]

MECE02:

OK. Thanks for coming through with the rest of the story. Now, I can visualize what is happening and I think I know exactly what you want to do.

You are going to have to assume that the nitrogen is saturated with water vapor as it exits your dryer. That way, you can calculate the amount of water vapor that exits with the saturated nitrogen. To do this calculation, you apply Dalton's Law of Partial Pressures.

I'm on my way out to go home for the weekend, but will try to get to the calculation if you don't know how to apply Dalton's Law to a gas (vapor) mixture. Yesterday was my birthday, so I may/may not get to this tonight since my real boss has arranged for a party at our house tonight.

 

[25362]

To Mr Montemayor : Feliz cumplea[ñ]os!

To MECEO2: It is clear I didn't understand the situation. Please explain how the nitrogen stream becomes heated from 347 to 383 K upon contacting the slurry at 155 F.

 

[MECE02]

Hi 25362!

Dryer N2 in = 383K

Dryer N2 + water vapor out = 165 F = 347K (type-o above, not 155 F) Therefore, the N2 is actually cooled from 383K to 347K.

Particulate + water (which are the heat sinks) entering dryer must be less than 347K. I am not sure of its value though.

Thanks!

 

[25362]

Therefore the process includes a transfer of heat from the nitrogen stream to the wet solids, and a transfer of mass by water evaporation to the nitrogen stream.

However, since part of the heat transferred may have gone to the "solids" and to liquid water as sensible heat, as you imply, the value of their entering temperature may be needed to better assess the degree of "gas humidification".

 

[MECE02]

So if I go with Dalton's law approach:

Ptotal = PN2 +PH20
Given:
PvapH20 @ 383K = 143.0 kPa
Pdryer = Ptotal = 2psi= 114.7 kPa

Calculated:
PN2 = Ptotal - PH2O = 28.2 kPa
yN2 = 0.246
yH2O = 0.753
QH2O = 1768.75 mol N/min. / yN2 = 7181 mol/min H2O

I don't think that is a reasonable amount of water vapor, it seems much to large. Thoughts?

 

[Montemayor]

MECE02:

Wow! I'm getting an excedrin headache from trying to understand your typo errors, & whether you mean gauge pressure or ABSOLUTE pressure (you really have to designate exactly what pressure you are talking about; there is no such pressure as "psi" or just kPa. It is either gauge or absolute, but we can't guess. Additionally, how do you get:

2 psi = 114.7 kPa
(2 psia = 13.7895 kPaA; 2 psig = 115.115 kPaA)

Also, you write: PvapH20 @ 383K = 143.0 kPa
But yet, you corrected yourself and said that the nitrogen+water vapor are at "165 F = 347 K". This gives a different vapor pressure for water.

How can you get:
PN2 = Ptotal - PH2O = 28.2 kPa
when Ptotal = 114.7 kPa and PH2O = 143 kPa????
I get a NEGATIVE value.

I know Dalton's Law works. It always has in the past. But in order for it to work, you've got to get the numbers correct and in the right form.

 

[25362]

MECEO2

[•] A rough check: if all the heat lost by the nitrogen stream were used only for water evaporation (which is not true) at 347 K, the maximum amount of water evaporated would be about 0.8 kg/min.

[•] Dalton's law is applicable, but then you must be sure the entering nitrogen stream is "dry" and not partly recirculated. If it is not dry, either you deduct the entering humidity or you carry a complete heat balance including the sensible heat gained by the wet solids.