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Surface Conductance of Mild Steel?

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Xyimon

Electrical
Joined
May 20, 2014
Messages
12
Location
GB
Hi all,

I am having issues trying to work out a value needed to formulate the Biot number which is needed to work out what temperature a bath of water would have to be kept at to cool a specific amount of mild steel in an allotted period of time. (Quite a mouthful, I know)

So far I have worked out the following:

K = 43w/(m.k)
Cp = 620 J/(Kg.K)
SpGr = 7.86ps OR 7860Kg/m3
Cylinder volume = 0.00216 m2
Cylinder Surface = 0.177 m2

From this I am close to working out the Biot number, but to do so I need the unit surface conductance of my specific size of mild steel, which is worked out by the Thermal transmittance, Φ = A × U × (T1 - T2)

To do this I need to calculate the U-Value, which involves working out the Lambda Value which is where I get lost...

Any help would be greatly appreciated.

Regards,
 
What's the formula for the Biot number? You did not provide complete information.


Tunalover
 
Bi = h*Lc/k

Where:
h expressed in [W/m2/K] or equivalent units is heat transfer coefficient between the fluid (water in your case) and the body (steel bar in your case). *
Lc expressed in [m] or equivalent units is characteristic length of the body (generally Volume/Transverse area).
k expressed in [W/m/K] or equivalent units is thermal conductivity of the body.

I am struggling to work out h.
 
Please refrain from multiple postings. This is related to your other post, and should have been asked there.

TTFN
faq731-376
7ofakss

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