Estimation of heat flux from burner

[NovaStark]

Hi all,

I am using API 530 to calculate the maximum tube temperature of a reformer tube.

However, I am unable to find a correct value for the heat flux from the burner (Average radiant heat flux density for outside surface). I believe I should use the heat flux from a burner to get this value, however, the data sheet for the burner only gives the heat output from the burner as approximately 6.7 mmBTU/hr.

Is there any way to estimate this heat flux? Otherwise, the maximum temperature I get is 1103 degC which is not really an ideal temperature to have the tube at.

I have attached the calculation sheet I made (mathcad pdf) and the section from API 530 that I am using.

 

[davefitz]

NovaStark,

There are several ways to obtain the heat flux onto the tube. The following might be progressively sophisticated ways:

a)contact another vendor of similar equipment that has a reliable history

b)use field data from a past facility that was properly insturmented to provide heat flux vs burner heat input for a similar fuel, geometry , and heat duty

c) use the "russian normative method( author A. Blokh, or see S. Kakac's chapter 8.2 of "boilers evaporators and condensers")to predict furnace overall duty vs fuel heat input and flame adiabatic temperature

d) use a numerical model of the radiation heat transfer of the furnace, as as per Woodrow Fiveland's 3D CFD "discrete ordinate solution", but obtaiing accurate optical properties of participaitng media ( compex index of refraction) is tough.

"Nobody expects the Spanish Inquisition! "

 

[IRstuff]

A picture would help immensely.

Additionally, you have two terms that are greater than 1, F[sub]L[/sub] and F[sub]circ[/sub], for which there may be plausible reasons. But, in your equation, they imply that you can get double the heater of the burners into the whatever it is that it goes into.

TTFN
faq731-376

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[NovaStark]

A picture would help immensely.

I don't have a diagram at the moment, but I can describe the basic layout of it. The unit consists of 5 rows of tubes (each row has 56 tubes and 5 risers where outlet gas flows into an effluent transfer line). Between each row of tubes are 16 burners (so there are 6 rows of burners to give a total of 96 burners each with an average heat output of 6.7 mmBTU/hr)

Additionally, you have two terms that are greater than 1, FL and Fcirc, for which there may be plausible reasons. But, in your equation, they imply that you can get double the heater of the burners into the whatever it is that it goes into.

I used what was described from API 530:

The circumferential variation factor, Fcir, is given as a function of tube spacing and coil geometry in Figure B.1. The
factor given by this figure is the ratio of the maximum local heat flux density at the fully exposed face of a tube to
the average heat flux density around the tube. This figure was developed from considerations of radiant heat
transfer only. As mentioned above, influences such as conduction around the tube and flue gas convection act to
reduce this factor. Since these influences are not included in this calculation, the calculated value will be somewhat
higher than the actual maximum heat flux density.


The longitudinal variation factor, FL, is not easy to quantify. Values between 1,0 and 1,5 are most often used. In a
firebox that has a very uniform distribution of heat flux density, a value of 1,0 can be appropriate. Values greater
than 1,5 can be appropriate in a firebox that has an extremely uneven distribution of heat flux density (for example,
a long or a tall, narrow firebox with burners in one end only).

For Fcirc, the tube spacing is 9.5 inches and I was not too sure which curve to use from the code, so I just averaged a figure from the sample calculation to see what kind of temperature output I'd get. For FL, I just used a value near to 1 since the heat distribution should be uniform due to the presence of the 96 burners.


As to daveftiz's suggestions, I don't think I can get the data to make those plots or if it even existed.

 

[lappygamer]

The heat flux information is usually found in the heater data sheet not the burner data sheet. If you have it, that would be the best place to start.

As an estimate, take the absorbed duty in that section of the heater divided by the external surface area of the tubes. That will give you an average radiant flux. Then apply Figure B.1. For the heater you are describing, Curve 4 is applicable.

Good luck!

 

[NovaStark]

The heat flux information is usually found in the heater data sheet not the burner data sheet. If you have it, that would be the best place to start.

As an estimate, take the absorbed duty in that section of the heater divided by the external surface area of the tubes. That will give you an average radiant flux. Then apply Figure B.1. For the heater you are describing, Curve 4 is applicable.

Good luck!

I contacted the burner manufacturer and at that point in time they never really did tests to determine the heat flux but only the heat emitted. The designer of the reformer in question said that they designed it based on a temperature of 1675F. However, the only way for me to get that value using API 530 is if I reduce the heat flux value as given in the pdf file I attached.

According to the data I have, the duty is 323 mmBTU/hr and each tube will have a surface area of 57.18 ft^2. So the heat flux would be approximated to 20175.45 BTU/hrft^2 = 63602.72 W/m^2. Using that value, I would get the temperature to be well over the 1675F. This lies my problem and also the Re number is < 2300 so I had to use a different correlation to find K for laminar flow.

 

[IRstuff]

Again, not having a picture makes it difficult to determine anything. One issue I have is that you seem to be using the entire heat output of the burner in the calculations. But, I can't even tell what you are doing there, since you appear to be doing a calculation, but simply entering a value. Transferring all the heat is not at all plausible, as that would imply that the exhaust stack's effluent would be at room temperature, which we know is never the case.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[NovaStark]

Again, not having a picture makes it difficult to determine anything.

I don't have a diagram showing the entire thing in one view so I tried to describe it as below:

I don't have a diagram at the moment, but I can describe the basic layout of it. The unit consists of 5 rows of tubes (each row has 56 tubes and 5 risers where outlet gas flows into an effluent transfer line). Between each row of tubes are 16 down-fired burners (so there are 6 rows of burners to give a total of 96 burners each with an average heat output of 6.7 mmBTU/hr)

One issue I have is that you seem to be using the entire heat output of the burner in the calculations. But, I can't even tell what you are doing there, since you appear to be doing a calculation, but simply entering a value. Transferring all the heat is not at all plausible, as that would imply that the exhaust stack's effluent would be at room temperature, which we know is never the case

The burner data sheet gives the following:

Heat release per burner = 6.77 mmBTU/hr
Firebox Bridgewall Temperature = 1890F
Minimum Burner Tile Service Temp = 3000F

If I use Flux = k(T[sub]2[/sub][sup]4[/sup]-T[sup]1[/sup][sup]4[/sup]) where k = stefan boltzman's constant (T2 and T1 are given above) then the heat flux comes out to over 600,000 W/m^2.

I understand your point about utilizing all of the heat emitted from the burner, I didn't even notice that before. As for the calculation, I am just following the format of the code as they just stated the value and did no calculation to show it. I was just messing around with the numbers to see what heat flux I would need to obtain a value around 1675F.

I didn't input a calculation for the heat flux as the values I was getting were too high.