Heat transfer coefficient

Why?

A heat transfer is from the plate to some other material, which one ? under which conditions?

If you define this then it is possible to determine the value you want to have.

The coefficient is independent of the reason you want to have it only on the conditions of contact and conditions.

The convection heat transfer process is:

Q = h*A*(T2-T1)

where

Q = heat flow (heat / time); positive Q means heat flows from T2 to T1

h = heat transfer coefficient ( [ heat ] / [time area degree] )

A = area

T2, T1 = hot and cold temperatures (degree)

Given the equation, the heat transfer coefficient may be determined if the other values are known.

However, TIG welding may be more of a heat generation process than a convection process. "The TIG (Tungsten Inert Gas) welding process (also known as gas tungsten arc welding, GTAW, or HELIARC, a trade name of Linde) generates heat from an electric arc maintained between a non consumable tungsten electrode and the part being welded." Thus treating TIG as a direct heat source may be more appropriate than modeling as convection. If you know the power consumption of the TIG welder, perhaps the heat can be estimated as some fraction of the power consumption, i.e. electrical to heat conversion is usually very efficient, on the order of 95%.

Your equation is correct but the guy asked for the "h" value it seems that he knew how to use it but did not know the value. I presume - I may be of course wrong- that he wanted to estimate the cooling time of the plate after the welding which of course will increase the temperature. The cooling will depend on several factors the main being the convection conditions: free ? position of plate horizontal or vertical? forced with an air jet on one surface? a fan ? and so on....

Dear friend

Is it about heat loss to atmospheric conditions/surrounding? If so, following equations can be used.

The convective heat transfer coefficient for a flat plate is given by:

h = 1.957[(Ts-Ta)^(1/4)]SQRT(1+2.857V)

where

h = film coefficient. W/m2K

Ts = surface temperature, K

Ta = ambient temperature, K

V = wind speed, m/s

The heat loss is then given by

Q = h A (Ts - Ta)

If the temperature of the surface is significant, you need to add heat of radiation which is given by:

Q = 0.548 e A [ (Ts/55.55)^4 - (Ta/55.55)^4]

where Q = radiative heat loss, W

e = emissivity ( = 1 for black body, = 0.3 for ss)

A = surface area, m2

Hope the above is helpful