bank of heat exchangers 1

[AO1958]

Hi there,

Hope somebody can help me in this topic.

I have to calculate three different heat exchagers - put in bank in the air flow direction - to cool water coming from a glass furnace.

Following is the topic: I have the fan curve of the producer given according to standard conditions, but inlet temperature of air in the first heat exchanger is 106F.
Threfore the point is that this air flow will pass through these three heat exchangers.

I think I need to fix the air flow.
My opinion is not to consider volumetric air flow, since we all know volumetric flow will change when passing through these heat exchangers.
A more reasonable solution would be:
1 - To calculate the volumetric air flow with given fan curve at standard conditions.
2 - To calculate the mass flow @conditions where the curve is given (mass flow = volumetric*density@standard conditions)
3 - To utilize this mass flow to calculate the capacity for the three heat exchangers.

Please, has anybody any different idea ?
Many thanks for your feedback, comments or opinions.

 

[ione]

The fan performance curve is usually given for standard conditions, which very hardly match your real actual conditions. Just convert actual conditions to standard conditions (conversion is based on the fact mass flow rate is kept constant) using the ideal gas law:

Flow standard = (Pactual/Pstandard)*Flow actual*(Tstandard/Tactual)

This will allow you to use the fan performance curve you have.

 

[AO1958]

Hello,

many thanks for your kind answer.

just to summarize.

Pactual = actual pressure ?
Pstandard = standard pressure ?
T Standard = absolute temperature standard
T actual = absolute temperature actual ?

Please, should I face with humid air, how should I consider the relative humidity ?

many thanks in advance

 

[ione]

If you want to take into account relative humidity then use:

Flow standard = (Pactual/Pstandard)*Flow actual*(Tstandard/Tactual)

Flow standard = Flow actual *(Tstandard/Tactual)*(Pactual/Pstandard)*[Patm-(RHactual*PVstandard)]/[Pstandard-(RHstandard*PVactual)]

Where:

Pactual = actual pressure
Pstandard = standard pressure
Patm =atmospheric pressure (it depends on elevation too)
T Standard = absolute temperature standard
T actual = absolute temperature actual
RHactual = actual relative humidity
RHstandard = standard relative humidity
PVstandard = saturated vapour pressure of water at standard temperature
PVactual = saturated vapour pressure of water at actual temperature

 

[ione]

http://www.engineeringtoolbox.com/scfm-acfm-icfm-d_1012.html

Regards

 

[AO1958]

Dear Mr. ione,

I'd like to thank you for your kind help.

I have just a last step.
That is:
shall I convert the normalized volumetric flow to a mass flow, how to handle it ?

Is the basic equation to consider

p/rho = nRT

where

p = pressure
rho = density
n = number of moles
R = ideal gas constant
T = absolute temperature

therefore the therm

p/(rho*T ) = const ?

therefore to add the following term to your last formula

Flow standard = Flow actual *(Tstandard/Tactual)*(Pactual/Pstandard)*[Patm-(RHactual*PVstandard)]/[Pstandard-(RHstandard*PVactual)]

Mass Flow = Flow Standard * rho standard = Flow actual * rho actual ?

where

rho standard = density @ standard conditions
rho actual = density @ actual conditions


All in a all the mass flow conserves



Sorry, very much to learn.

thanks in advance

 

[ione]

What is behind the conversion between standard and actual conditions is the fact the moles of air remains the same and so does the mass flow rate.
Honestly I cannot understand why you’ve to go back to mass flow rate once you can pass from actual to standard conditions volumetric flow rates. This should allow you to use your fan performance curve.

 

[AO1958]

Hello, many thanks

there are two reasons.

1) please remember that I have got three heat exchangers.
This means that the volumetric flow I find at the inlet of the first heat exchanger is not the same at the inlet of the second and both are different to the volumetric flow at the inlet of the third one.

This is the reason why I supposed it was much more simple to determine the mass flow with a first calculation and then to keep the same mass flow for the calculation of all the three heat exchangers.

2) I have been a dispute concerning the fact that I had to consider mass flow of dry air and not mass flow.

This puzzles me, since I suppose that the mass flow is always the mass flow.

Please remember my target is to get from fan curve - given volumetric air flow and pressure - the mass flow.
In my opinion this would simplify a lot all the three selections.

Many thanks

 

[ione]

I see your arrangement foresees three heat exchangers in series so outlet conditions of one heat exchanger are the inlet conditions of the following heat exchanger and so on. For each heat exchanger you have anyway to know the required flow rate at actual conditions. You can then convert to standard conditions and access the fan performance curve.

 

[speco]

AO1958,

Are you actually trying to cool water coming from the furnace or hot gas? In the original posting, you said you were trying to cool water. This make the whole subject of the air flow a big red herring.

However, if you are trying to cool furnace gas using water, then you need to know the volume flow at point where the fan is.

Regards,

Speco

 

[AO1958]

Hello,

many thanks to everybody for your kind patience, and apologizes for not being clear.

Mr Ione,
please, the point is that I have the resistance of three heat exchangers but a sole fan, with the curve with conditions given at standard conditions.
Therefore I think I can't determine the air flow at standard conditions of the second heat exchanger, unless considering the bank without the first heat exchanger. This is the reason why I am trying to consider a mass flow instead than a volumetric flow.

Ex: I can consider an heat exchanger thick as the sum of the thickness of the three heat exchangers.
This will determine the air flow as standard conditions and would allow me to calculate the first heat exchanger.
But for the second ?
This air flow once passed through the heat exchangers cools water, therefore is hotter, and if I consider a volumetric flow, this is no longer equal to the volumetric flow at the inlet of the first heat exchanger.

Therefore I am stuck in trying to understand which air flow to consider to dimension the second heat exchanger - and third, as consequence -


Mr. Speco.
I am trying to cool water.
But the target is to dimension the heat exchanger.
To dimension the heat exchanger, I need to know inlet temperature and air flow (volumetric or massic).

Therefore I'd need to understand which is this air flow at the inlet of the second, and third heat exchanger when the fan curve is known
What you say is correct, I need to know the volume air flow at the point where the fan is - therefore I need to convert from standard conditions to actual conditions with the formula kindly given by Mr. Ione.

But the point is: which flow to consider at each heat exchanger inlet ?
Shall I consider the mass flow at the conditions of fan position and then keep it for the whole heat exchanger bank ?
But wouldn't be enough to consider the volume flow got from fan curve multiplying it by standard air density ?

Many thanks for your help

Regards

 

[speco]

AO1958,

If your goal is to cool the water using air, why do you want to put the exchangers in series with respect to the air? The normal way to do this would be to run them all in parallel. Otherwise, you are cooling water in the second and third exchangers with hotter and then even hotter air.

Am I totally missing something here?

Thanks,

Speco (

http://www.stoneprocess.com)

 

[AO1958]

Mr. Speco

the goal is to cool water but with this boundary conditions

1 - the water is coming from 3 different circuits.
Moreover in one circuit flows pure water, the second water with 10% Ethylene Glycol, the third water with 20% EG.
This is the reason why three separate heat exchangers are needed.

2 - the space for utilizing this cooler is limited, so that the solution of the three heat exchangers has been chosen by designer.

 

[rmw]

Knowing the conditions of each fluid would help those of us who are reading along trying to help draw some conclusions.

Are all 3 fluids the same temperature, or do they vary? If the latter how much they vary could have a big impact on how you arrange your coolers.

rmw

 

[speco]

AO1958,

So, we we really have here is an air-cooled heat exchanger problem. A very unusual one, since bundles are very seldom stacked on top of each other.

If you start with these two assumptions:

1. The plot area (or tube bundle face area) is very limited. And.
2. The three tube bundles must be stacked.

I would start with this approach:

The first thing is to use up all the available face area, because you are going to need it. I will make the wild assumption that you are using finned tubes for this application. A "normal" air-side mass velocity for such an exchanger would be in the range of 5000 to 7500 (very high) pounds mass per sq. ft. of free area per hour. If you are really going to stack these bundles, you need to keep the total number of rows to a minimum so that the total static pressure for the fan is not too high for an available fan. A "normal" static pressure would be in the range of .5" H20, but some axial flow fans can handle higher one, up to about 1.0" of static.

In a "normal" finned tube bundle, the free (unobstructed) area is about half the total face area, by the way.

Put the bundle with the coldest process temperatures on the inlet side, and the hottest on on the outlet side with respect to the air.

If you assume that you are using a forced-draft fan arrangement, you only need to know the altitude and temperature to convert from mass flow to ACFM of air. It's a simple matter of calculating the density, and converting the flows.

The cooling air mass flow will be the same through all three bundles. The outlet air temperature from the first bundle will be the inlet temperature to the second one, and so forth.

After going through the exercise of figuring out the sizes of the three tube bundles, you might really want to reconsider this approach. Designs are not done in a vacuum. There are also cost considerations. That's why stacked tube bundles are generally not used in air-cooled applications.

Regards,

Speco (

http://www.stoneprocess.com)

 

[AO1958]

Hello thank you very much.

apologizes for the delay of my answer.

To Mr rmw

A basic composed heat exchanger - a modular solution has been considered - considered has following design data

first heat exchanger
water inlet 140 F
water flow 7300 gph approx

second heat exchanger
EG 20% inlet 158 F
EG 20% flow 5200 gph

third heat exchanger
EG 30% inlet 194 F
EG 30% flow 9800 gph

But mr.rmw I think this is outside of the nature of the problem that is : how to move from curve data to real air flow.

To Mr. Speco

Many thanks

quoting
<<
If you assume that you are using a forced-draft fan arrangement, you only need to know the altitude and temperature to convert from mass flow to ACFM of air.  It's a simple matter of calculating the density, and converting the flows.
>>

Please, shall I use the formula suggested by mr. ione ?

Just a final clarification: the original project - and the request to put these heat exchangers in this way, is from final customer


Thanks for your help !

 

[speco]

AO1958,

As usual, Ione is correct. The one piece of missing information is how to get the ambient pressure at altitudes above sea level. Here's a formula for the altitude correction factor:

Alt CF = e^(-.000037 * Alt)

Where Alt CF is the density correction factor,

Alt is in feet above sea level,

and e is base of natural logarithms.

It's a useful formula if you work with fans.

Regards,

Speco

 

[ione]

AO1958,

You have definitely received valid input from speco and I suggest you to reconsider this arrangement.The “customer wants it that way” is really not an engineering argument, even if it is quite strong argument. Be that as it may the arrangement remains really odd. With such an arrangement the increasing air temperature at the outlet of a bundle will worsen the performance of the following bundle, as the driving force of the heat exchange process will be progressively reduced (this is the reason why the coldest process temperature has to be on the inlet side, and the hottest on the outlet side with respect to the air).


Speco,

Are in the altitude correction factor more zeros than required?

I have this one

Alt CF = e^(-.00037 * Alt)

But I practically never use the formula above as I’ve to deal with sea level installations, and so I could be mistaken.

 

[speco]

Ione,

I have it in my programs as .000037. I rechecked it with the Moore Fans data. They show it as 3.7*10^(-5). I would go with four zeros.

Regards,

Speco

 

[ione]

Thanks Speco