Question in (Conduction ) A steel plate 2 cm thick

In summary, the conversation discusses the calculation of heat transfer through a steel plate with a thickness of 2 cm and temperature difference of 500 C. The thermal conductivity of stainless steel at 300 C is also mentioned. The formula used to calculate heat transfer is provided and the final answer is 477.5 w/m^2. However, there may be an error in the problem statement as the answer should be per unit area and may need further clarification.
  • #1
manal950
177
0
Hi

A steel plate 2 cm thick is maintained at a temperature of 550 C at one face and 50 C on the other .
The thermal conductivity of stainless stress at 300 C is 19.1 W/mK .

Compute the heat transferred through the material per unit length ?

Answer : -

we have q = - k . A . (dt/dx)

now
q = - k . (l.h) . (dt/dx)

q / (l.h) = -k( (dt/dx) )
= 477.5 w/m^2
 
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  • #2
There is something wrong with the problem statement. It should be "per unit area". Also, it looks like the answer is off by a factor of 1000.
 
  • #3
Pls explains your variables in the context of the problem.
 

1. What is the purpose of the steel plate in conduction?

The steel plate serves as a medium for heat conduction, allowing heat to flow through it from one side to the other.

2. How thick is the steel plate in this scenario?

The steel plate is 2 cm thick.

3. Does the thickness of the steel plate affect the rate of conduction?

Yes, the thicker the steel plate, the slower the rate of conduction. Thicker materials have more resistance to heat flow.

4. How does the composition of the steel plate impact conduction?

The composition of the steel plate can affect its thermal conductivity, which is the measure of its ability to conduct heat. Different types of steel may have different thermal conductivities, which can impact the rate of conduction.

5. Can the steel plate insulate against heat transfer?

No, the steel plate is not an insulator. It is a conductor and allows heat to flow through it. Insulators, on the other hand, prevent or slow down the transfer of heat.

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