3 replies to this topic

[white_flag]

Hello

I have the following situation and I do not know if I am doing correct:

so ..I like to find out temperature of outside pipe with:
diameter: DN 300
given temperature of fluid (inside of the pipe): 200°C
ambiental temperature: 5 °C
wind speed: 5 m/s
thickness of pipe: 5 mm

so:
I have Q.condution(3) = Q.convection(1) + Q.radiation(2)

(1)
Q.convection: h*pi*D*L*(T.outside.pipe-T.ambient)

to find h:
h=Nu*k/D - Heat transfer coefficient (watt/m°C)
Nu=0,193*Re^0,618*Pr^(1/3) - Nusselt Number
Re=u.m*D/v - Reynold Number
Pr=(v/a) - Prantl Number
v - Air Kinematic Viscosity (m²/s)
k - Air Conductivity (watt/m°C)
a - Air Thermal Diffusivity (m²/s)
D - diameter
T.ambient - °C
T.outside.pipe << to find out
L = lenght (m)

(2)
Q.radiation = α*( e1 T.outside.pipe^4 - e2 T.ambient^4 )*A (radiation)

A = Area of Radiating surface (m2)
T.outside.pipe = Temperature or radiating body (K) << to find out
T.ambient = Temperature or Suroundings (K)
α = Stefan Boltzman constant = 5,673 x 10-8 W m-2 K-4
e 1 = Emissivity of Radiating surface
e 2 = Emissivity of Surroundings

(3)
Q.conduction = 2*pi*(T.inside.pipe-T.outside.pipe)/[ln(r2/r1)/k.pipe]

r2 - radius outside pipe (m)
r1 - radius inside pipe (m)
T.inside.pipe (°C)
T.outside.pipe (°C) --- to find out
k.pipe = thermal conductivity of pipe (watt/m°K)

from
Q.condution(3) = Q.convection(1) + Q.radiation(2)
equilibru will be reached and I will find out T.outside.pipe??

then I like to know if Q.condution = heatloss
(2*pi*r.pipe*k.pipe*(T.inside.pipe-T.ambient))?

thank you

[ElSid]

Chris has a great BLOG article here http://www.cheresour...with-insulation
Also, free insulation software at http://www.insulation.org
Previously posted insulation article at http://www.raeng.org...2_SteamPipe.pdf

Edited by ElSid, 15 January 2013 - 01:04 PM.

[DB Shah]

Referred post may be useful -
http://www.cheresour...ipe/#entry66568


Quote from the post
"
Dear process85

My experience which relates to your case-
We have ~ 500 mts of 4" uninsulated CO2 line. CO2 at source is 20 Barg & 205°C & at ambient conditions of 30°C, guess CO2 temperature at exit, It was 34°C, APPROACH OF JUST 4°C, Both temps have indication on DCS, I physically touched the line at exit plant battery limits.

Considering the line as a single tube exchanger I backcalculated the Ud as 8 kcal/hr-m2C (In Aspen you have an option to plug ambient temp & U value for pipes to match outlet temp).
We have another uninsulated CO2 line of 12" 600 mtrs where CO2 at -5°C is sent to another plant. First 10 mts of the line is SS and rest is CS. Here also I back calculated the Ud & it was in the range of 6~7 kcal/hrm2C.

I think you can safely take the values of ~8 kcal/hrm2C to cal your heat loss and exit temp."

[gregga]

The procedure you propose will work fine and is easily adaptable to excel spreadsheet (I'll let you write your own)...hint, equate the conduction to the convection and radiation (as you've shown) and use goal seek in excel to solve for the surface temperature.  You can expand the analysis to include multiple layers including the weather barrier and pipe wall using the discussion from Chris

 

or.... the easiest way is to use the following free program which is widely used in the U.S. by major companies

 

http://www.pipeinsul...vailablecontent