- #1

- 26

- 0

How long will it take for a surface to drop to say 800°C

should i use the heat transfer eq: Q=KA[itex]\Delta[/itex]T to find the power in joules per second, then transpose to find seconds somehow???

Any help will be appreciated

Adam.

- Thread starter 1988ajk
- Start date

- #1

- 26

- 0

How long will it take for a surface to drop to say 800°C

should i use the heat transfer eq: Q=KA[itex]\Delta[/itex]T to find the power in joules per second, then transpose to find seconds somehow???

Any help will be appreciated

Adam.

- #2

- 3

- 0

[itex]-\dot{E}_{out} = \dot{E}_{st}[/itex]

Then:

[itex]-hA_{s}(T - T_{\infty}) = \rho Vc\dfrac{dT}{dt}[/itex]

With a bunch of exciting math, we can arrive at:

[itex]\dfrac{\rho Vc}{hA_{s}} \ln\dfrac{T - T_{\infty}}{T_{i} - T_{\infty}} = t[/itex]

Where:

[itex]

\rho = \text{material density} \\

V = \text{volume} \\

c = \text{specific heat capacity} \\

h = \text{convective heat transfer coefficient} \\

A_{s} = \text{surface area exposed to the air} \\

T = \text{final temperature} \\

T_{i} = \text{initial temperature} \\

T_{\infty} = \text{temperature of the air} \\

t = \text{time taken to cool from } T_{i} \text{ to } T \text{ in seconds}\\

[/itex]

At this point, consult some references to determine the specific heat capacity and the convective heat transfer coefficient, and then it's plug and chug. Watch out for units!

According to my reference here:

[itex]

c_{steel} = \text{440 } \dfrac{\text{J}}{\text{kg}\cdot{\text{K}}} \\

h_{air} = \text{20} \dfrac{\text{W}}{\text{m}^{2}\cdot{\text{K}}}

[/itex]

********************

Keep in mind this is an approximation because we are assuming the gradient in the block has no effect on the rate of cooling (although most likely good enough for your needs). A measure of the validity of the approximation can be found through calculating the Biot number.

[itex]

Bi = \dfrac{hL_{c}}{k}\\

\text{Where:} \\

L_{c} = \dfrac{V}{A_{s}} = \text{characteristic length} \\

k = \text{thermal conductivity of the material} \\

[/itex]

For a good approximation, Bi should be much less than 1.

Hope that helps!

- #3

- 1,198

- 5

If you cannot do this you can approximate the effect of radiation and lump it into the convection coefficient. It's crude but better than ignoring the major heat loss.

- #4

- 366

- 16

Last time I was there, I remember red hot ingots emerging from where they were made and them sitting (still red) before they were moved to where they would be worked.

- #5

- 94

- 1

- #6

- 1,198

- 5

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 8

- Views
- 1K

- Replies
- 4

- Views
- 5K

- Replies
- 13

- Views
- 13K

- Replies
- 4

- Views
- 705

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 9K

- Last Post

- Replies
- 6

- Views
- 6K

- Replies
- 3

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 1K