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surface temperature with given conditions

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Mechanical
Oct 12, 2010
4
Good morning,
I have the following problem and I would like your advice:

I have an pipe (DN 100) insulated (50 mm, Rockwool) + Cladding (aluminum). Then, I have inside of pipe an temperature of 200 °C, the ambient temperature (20 °C) and the air speed (1 m/s). How can I calculated the surface temperature of the cladding (aluminum)?

thx for help and watching
 
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You have to equal the heat transferred by conduction throughout the pipe wall, insulation and cladding from the inner to the outer surface (assume Tinner = 200 °C and Touter is your unknown) to the heat dissipated by convection and radiation from the outer surface.

Iterate until the two amounts of heat (Qcond and Qdissipated) converge (equilibrium reached).

For the convection heat transfer coefficient you can assume 8 W/(m2*°C)
 
This a homework problem?

Patricia Lougheed

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assuming insulation Conductivity of 0.0387 (kcal/h-m-C)
the surf temp will be 28.3 (C)
U=1.1 (kcal/h-m2-C)
 
thx for help. I have an small program from Rockwool (company that make insulation), putted inside all the value, gives me the temperature of cladding (conform the above conditions)in the end of the temperature is 39.3°C. I try to remake the rockwool formula to added in an excel sheet.
So, I have the equation of the Heat transfer, Total Coefficient of conductivity (Convection+Radiation)

Q r = ? ( e1 T14 - e2T24 ) A1 (radiation)

A 1 = Area of Radiating surface (m2)
T 1 = Temperature or radiating body (K)
T 2 Temperature or Suroundings (K)
? = Stefan Boltzman constant = 5,673 x 10-8 W m-2 K-4
e 1 = Emissivity of Radiating surface
e 2 = Emissivity of Surroundings

Qco = (k.A /x). (t 1-t 2) (conductivity)

t 1 = inside(hot)temperature,( K )
t 2 = outside(cooler)temperature,( K )
k = Thermal conductivity (W/mK)
x = thickness of insulation (m)

Q.total = Qco + Q r ? (this is correct)

Q.total = pi*(t1-t2)/((1/2landa*ln(d.ext/d.int)+1/alfa*d.ext)
t 1 = inside(hot)temperature,( K )
t 2 = outside(cooler)temperature,( K )
alfa = coefficient .radiation + coefficient .convection (W/m^2K)
landa = (W/mK)

but "alfa = coefficient .radiation + coefficient .convection" is related to temperature of the surface (to be find)
I am a bit lost on this equation (no experience in the field), I do not understand were the speed of wind can be inserted, then to find the surface temperature of the cladding
 
blueflag,

Have you seen my previous post (12 Oct 10 7:04)?

Qconduction = Qconvection + Qradiation

For Qconduction, you can consider the pipe wall and the thermal insulation as two thermal resistances in series.

Assume a first trial temperature for the external temperature to calculate the convection heat transfer coefficient (here the speed of wind enters into play).

You can use this calculator for forced convection


Then solve the above equation with outer surface temperature as unknown and iterate until the result and the trial temperature converges.
 
thx, ione

I saw your first post, also, thx for the link (really nice one). so please correct me, if I made mistakes:

1. this is the base to calculate the Heat Flow:?

Q=2*pi*k*L*(t.inside-t.ext)/ln(r.ext/r.int)
were:
r.ext = exterior radius of cladding (m)
r.int = exterior radius of pipe (m)
L = length (m)
k = thermal conductivity

in this formula t.ext = t.surface.ext? or t.ext is t.ambiant air temperature

or the equation of heat flow has to be based on coefficient .radiation + coefficient .convection.. there I saw that the wind is involve but for the moment I am very confused. and I do not know what it is good for me.
 
Text is the outer cladding surface temperature.

T1 is the inner surface pipe temperature (assume it equal to 200 °C)

For a unit length of pipe (L = 1 m)

Qcond = (T1 – Text)*2PI*[Kpipe/ln(R1/Rint) + Kins/ln(R2/R1) + Kcladding/ln(Rext/R2)]

Rint = pipe inner radius
R1 = piper outer radius = insulation inner radius
R2 = insulation outer radius = cladding inner radius
Rext = cladding outer radius
 
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