Maximizing Furnace Efficiency with Heat Exchanger Design: A Comprehensive Guide

[Ricardo Ferreira]

Hello guys,

I have a problem with my work!

I have a furnance that have a "heat exchanger" to heat the air that will be used on the burners!
(You can work every value close to 1 atm pressure on the entire exercice)

The furnance is 1100~900 ºC

The fan sends in a tube 7500m^3 air per hour! (arround 30ºC)
The tube have
-400 mm Outside Diameter
-395 mm Inside Diameter

So the velocity is something like, Q= Qm*p=7500*1,164=8730 Kg/h

Im going calculate the time the tube need to asborb heat to get 300+ ºC
Im using that diameters, but i want to know how to calculate to change some values and see if i can get better performance.

That this tube create 42 ramifications:
-50 mm Inside Diameter
-60 mm Outside Diameter
-3300 mm Length
So, Q=v(m/s)*A(m^2) = 7500/3600 / (0,05*42)^2* 3,14/4 = 0,601 m/s

Example: With that temperature, we are getting 250ºC+. So the air is getting

Q=m*cp*(Tf-Ti) = 8730*1,033*(270-30)= 2 164 341 KJ/h = 601 KW / 42 tubes = 14,3 KW per tube
Tm = (270+30)/2=150ºC
Re=(ρ*v*D∅)/μ = (0,83 * 0,601 * 0.05)/ 0,000024 = 1039,23 Re, So its Laminar
Now I am Lost, i want to know What i can do to improve, number of tubes, etc...
The best temperature its 400 ºC but the tube is expensive, 13 000 euros the 42 tubes.

Tell me if i need to give you more variables

Regards,
Ricardo Ferreira

 

[Ricardo Ferreira]

Ahhh the material of the "heat exchanger" is Silicon Carbide,

k= 20 W/mK

 

[anorlunda]

I did not follow all your calculations. But you should know about the usual mode of calculation, LMTD. https://en.wikipedia.org/wiki/Logarithmic_mean_temperature_difference

But to your question, there are limited ways to improve a heat exchanger. Increased area of the exchange surface, decreased thermal resistance in the materials, increase the residence time. Regardless of your calculations, increased area meaning more tubes or longer tubes may be your only option.

You also don't mention if your heat exchanger is parallel flow or counter flow. Counter flow is the most common.

 

[Chestermiller]

I don't see any consideration of the heat transfer coefficients in your calculations to determine whether the total heat transfer area is adequate. As @anorlunda seems to be indicating, please tell us more about the detailed design of the heat exchange, like tube layout and external geometry (and properties of the furnace gas).

 

[Ricardo Ferreira]

Thanks for the reply,
The problem that's not a "normal" Heat exchanger.

Its just 42 tubes, that pass in the furnance to pre heat the air.
the tubes are perpendicular.

The problem using the LMTD how i know the correct heat transfer coeficiente(U)? If i don't know the exact heat convection transfer coeficiente h?
i just want to know if passing a tube with 60OD and 50ID with 3300mm lenght, how much energy will get? and what temperature will have on the 3,3meters.
If i don't have the mass flow of the furnance.

I don't have anything more.

Thanks.

 

[Ricardo Ferreira]

Thanks for the reply,
The problem that's not a "normal" Heat exchanger.

Its just 42 tubes, that pass in the furnance to pre heat the air.
the tubes are perpendicular.

The problem using the LMTD how i know the correct heat transfer coeficiente(U)? If i don't know the exact heat convection transfer coeficiente h?
i just want to know if passing a tube with 60OD and 50ID with 3300mm lenght, how much energy will get? and what temperature will have on the 3,3meters.
If i don't have the mass flow of the furnance.

I don't have anything more.

Thanks.

 

[Ricardo Ferreira]

I don't see any consideration of the heat transfer coefficients in your calculations to determine whether the total heat transfer area is adequate. As @anorlunda seems to be indicating, please tell us more about the detailed design of the heat exchange, like tube layout and external geometry (and properties of the furnace gas).

I put everything i have there, more than that just the draw, on solidworks... If helps!
Thanks for the reply!

 

[Chestermiller]

Thanks for the reply,
The problem that's not a "normal" Heat exchanger.

Its just 42 tubes, that pass in the furnance to pre heat the air.
the tubes are perpendicular.

The problem using the LMTD how i know the correct heat transfer coeficiente(U)? If i don't know the exact heat convection transfer coeficiente h?
i just want to know if passing a tube with 60OD and 50ID with 3300mm lenght, how much energy will get? and what temperature will have on the 3,3meters.
If i don't have the mass flow of the furnance.

I don't have anything more.

Thanks.

Have you tried to determine the heat transfer coefficient inside the tubes, and the heat transfer resistance through the tube walls?

 

[Ricardo Ferreira]

Have you tried to determine the heat transfer coefficient inside the tubes, and the heat transfer resistance through the tube walls?

Im trying doing that, so i got that Reynolds arround 1000, that's mean laminar flow, now i don't know how to do it... I know i need Nussel number to calculate hint and hext, and have many formulas and i don't know how to use it yet... So i put there to know if have any simple way to do it :)

 

[anorlunda]

http://www.enggcyclopedia.com/2011/10/calculation-heat-transfer-coefficient/

That page goes over how to calculate the heat transfer coefficient from the dimensions, operational data, and from the thermal conductivity of steel. See if it fits your case.

 

[Chestermiller]

Im trying doing that, so i got that Reynolds arround 1000, that's mean laminar flow, now i don't know how to do it... I know i need Nussel number to calculate hint and hext, and have many formulas and i don't know how to use it yet... So i put there to know if have any simple way to do it :)

The Nussult number solution for laminar flow in a tube as a function of Re, Pr, and L/D is in every heat transfer book.

 

[Ricardo Ferreira]

The Nussult number solution for laminar flow in a tube as a function of Re, Pr, and L/D is in every heat transfer book.

Hello Chestermiller,
I was looking on some books and i found that,
Circular Tubes (Laminar Flow):
Nu=(h*L)/kf = 3,66 to Constant Surface Temperature
Nu=(h*L)/kf = 4,36 to Constant Surface Heat Flux
But then on other book i found:
Nu=1,86*(RePr)^0,33 * (de/L)^0,33 * (μ/μw)^0,14 if i use that one, Nu will be 0,28... Thats low, right? (Re = 881,781, Pr=0,6974, de=0,06m, L= 3,3m)
What you think about that?

Nu=h*L/kf = 0,28 = h * 0,05/0,027, h=0,1512 W/m^2ºC , that's impossible... Where I am doing wrong, because that formula is used to Reynolds < 2000.

If you can explain me that i think i can finish that Project!
Thanks for beeing so nice!

 

[Ricardo Ferreira]

http://www.enggcyclopedia.com/2011/10/calculation-heat-transfer-coefficient/

That page goes over how to calculate the heat transfer coefficient from the dimensions, operational data, and from the thermal conductivity of steel. See if it fits your case.

Thanks for the help, i just got stuck on the Nusselt Number, but I am starting handle that!
That link still helps me, anyway!
Regards!

 

[Chestermiller]

Hello Chestermiller,
I was looking on some books and i found that,
Circular Tubes (Laminar Flow):
Nu=(h*L)/kf = 3,66 to Constant Surface Temperature
Nu=(h*L)/kf = 4,36 to Constant Surface Heat Flux
But then on other book i found:
Nu=1,86*(RePr)^0,33 * (de/L)^0,33 * (μ/μw)^0,14 if i use that one, Nu will be 0,28... Thats low, right? (Re = 881,781, Pr=0,6974, de=0,06m, L= 3,3m)
What you think about that?

The first pair of equations apply in the limit of very large L/D. The 3rd equation applies over the region of small L/D. So you calculate Nu from the 1st and 3rd equations (assuming the tube wall is closer to being constant temperature than constant heat flux...this is conservative), and choose the value that is higher. Or, to be even more conservative, you just use the first equation.

 

[Ricardo Ferreira]

The first pair of equations apply in the limit of very large L/D. The 3rd equation applies over the region of small L/D. So you calculate Nu from the 1st and 3rd equations (assuming the tube wall is closer to being constant temperature than constant heat flux...this is conservative), and choose the value that is higher. Or, to be even more conservative, you just use the first equation.

You are really good, thanks for the help!
Hope you help me more in the future! :)
Regards