How to calculate Nusselt number

  • #1

Main Question or Discussion Point

Hello I need to calculate the Nusselt number for liquid sodium in a pipeline. Length of the pipeline is 3 meter. Diameter of the pipeline is 0.3 meter. Heat flux through outer surface of the pipeline is equal zero. Temperature on the right side and left side of the pipeline is 573 K and 773 K, respectively. Solution of the problem is free convective heat transfer.

Nu=[tex]\lambda_{eff}[/tex]/[tex]\lambda[/tex]

[tex]\lambda[/tex] is heat flux coefficient for liquid sodium at 673 K (average temperature)
[tex]\lambda_{eff}[/tex] is effective heat flux coefficient including effect of free convective heat transfer. [tex]\lambda_{eff}[/tex]=Q[tex]\delta[/tex]/S[tex]\Delta[/tex]T

Q is heat flux through surface
[tex]\delta[/tex] is the length
S is cross sectional area of the pipe
[tex]\Delta[/tex]T is temperature difference (200 K)

Which the surface have be used for calculation of Q? (the right side, the left side, the outer surface) How to calculate Nusselt number?

TIA
 

Answers and Replies

  • #2
8
0
Nu=(h*Lc)/k
k is the thermal conductivity
Lc is the length
and i can say that the nusselt number is the ratio of convection and conduction ==>[q(conv)=h*DT]/[q(cond)=(k*DT)/L
 
  • #3
minger
Science Advisor
1,495
1
The simplest equation is the Colburn equation
[tex] Nu_D = 0.023 {Re}^{4/5}_D {Pr}^{1/3} [/tex]

The Dittus-Boelter equation is slightly different
[tex] Nu_D = 0.023 {Re}^{4/5}_D {Pr}^{n} [/tex]
Where n is 0.4 for heating and 0.3 for cooling. These equations are valid for 0.7<Pr<160; Re>10,000; L/D > 10.

Another more complex correlation is attributed to Petukhov
[tex] {Nu}_D = \frac{ (f/8){Re}_D {Pr}}{ 1.07 + 12.7(f/8)^{1/2}({Pr}^{2/3} - 1)}[/tex]
Where f is the friction factor. This correlation is good for 0.5<Pr<2000 and 10^4<Re<5x10^6. For smaller Reynolds numbers
[tex] {Nu}_D = \frac{ (f/8)({Re}_D-1000) {Pr}}{ 1.00 + 12.7(f/8)^{1/2}({Pr}^{2/3} - 1)}[/tex]
This is valid for the same Prandlt number, but with Reynolds down to 3000.

For smooth pipes, one should use the following for the friction factor
[tex] f = (0.790 \ln{Re}_D - 1.64)^{-2}[/tex]

For fully developed laminar flows in pipes with a circular cross section
[tex]{Nu}_D = 4.36[/tex] for uniform heating and
[tex]{Nu}_D = 3.66[/tex] for uniform Temperature imposed

All of this was taken from "Introduction to Heat Transfer" by Incropera and DeWitt, 4th edition.
 
  • #4
You may want to restate your problem a little bit.
First of, there is no heat transfer through the outer wall. Does this means the pipe is insulated? Is the pipe heated surounded by insulation? Furthermore, what do you mean with free convection here? is there boyuancy driven flow? All equation given by Minger are for FORCED convection. What do you mean with left side and right side? Entrance and exit of the pipe?

If you know the entrance and exit temperature and flow conditions (Reynolds number) you can simply calculate the average Nu.
 
  • #5
minger
Science Advisor
1,495
1
Oh, guess maybe I should have read the post a little better as well. Hoping that he means free convection outside of the pipe.
 
  • #6
You may want to restate your problem a little bit.
First of, there is no heat transfer through the outer wall. Does this means the pipe is insulated? Is the pipe heated surounded by insulation? Furthermore, what do you mean with free convection here? is there boyuancy driven flow? All equation given by Minger are for FORCED convection. What do you mean with left side and right side? Entrance and exit of the pipe?

If you know the entrance and exit temperature and flow conditions (Reynolds number) you can simply calculate the average Nu.
The pipe is insulated. There is boyuancy driven flow. On the left side and the right side only heat flux is set (temperature). Problem is solved into the pipe.
 
  • #7
1
0
The pipe is insulated. There is boyuancy driven flow. On the left side and the right side only heat flux is set (temperature). Problem is solved into the pipe.


Hello Guys, I have a very simple situation but I am not an expert so I need a bit of help. I got a simple Aluminium plate with square dimension which I am heating in a room. How will I calculate the Nusselt number for that. I mean I know the simple solution is to use Nu=hL/k but somehow I think it is not that simple, and anybody can give me an idea or reference to find the value of h at room temperature in ambient air?
thanks in advance
 
  • #8
1
0
hello all,

how can i calculate Nu number at 1 wall of a room as a function of Rayleigh number floor and ceiling, if i accept the other walls insulated?

Thanks for attention,

nasilbir
 

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