[Bibinou]

Hello,

in order to prepare my transfer exam, I decided to prepare by attempting this exercise. However, I encountered problem in resolving it.

One is required to construct a furnace with with a refractory (heat resistant) material . This material remains solid at high temperature but it is not necessay a good heat insulating material. The heat conductivity of the refractory material is 1.5 W m^-1K^-1. The temperature inside the furnace can reach a maximum of 1400K and the temperature of the outer surface area must remain constant at 300K. The estimated heat heat loss at the outer surface area is 5000W/m².

a) What is the thickness of the outer wall?

b) We have at disposal refractory bricks of 15cm thickness. In order to have an minimum thickness of the outer surface, one will use only one partition of refractory bricks and we will complete this 15cm thickness with a insulating material of  0.3 W.m^-1K^-1. A good thermal insulator is not ncessary a refractory material. The contact temperature between the refractory material and the insultaing material should be such that T_contact < T_fusion of the insulating material.

- Determine T_contact?
- Deduce the range of T_fusion
- Determine the thickness of the insulating material
- Deduce the total thickness (refractory material and insulating material) of the outer surface for a heat loss 5000 W/m².

Thank you very much for your help

EDITED: For grammatical and spelling error

[Donaldson Tan]

Ref: http://www.chemicalforums.com/index.php?topic=7208

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For the benefit of the doubt, I shall answer some part of your question.

1. I guess T_fusion of a substance refers to its melting point

2. Assume negligble heat transfer by radiation to simplify my calculations (although the furnance combustion chamber is capable of reaching 1400K)

3. Assume linear temperature profile within the each layer.

Part (a) requires you to calculate the thickness of the single-layered wall made of of refractory material.The internal temperature of the wall is 1400K and the external wall temperature is 300K. Since the temperature profile of wall is assumed linear, using Fourier's Law for single-dimension heat diffusion, can you work out the thickness?

Part (b) is just a multi-layered wall problem. It is conceptually similar to part (a)

[Donaldson Tan]

To solve (a),

According to Fourier's Law, q = -k.(dT/dX) where
k is conductivity
q is heat flux

dT/dX = ?T/?X for linear profile where
?X is the thickness of wall
?T is the temperature difference between the external surface and internal surface.

[Donaldson Tan]

For (b), the furnace wall consist of 2 layers.

The internal layer is made of refractory brick and the external layer is made of the insulating material. The outer surface temperature of the insulating material is 300K and the internal surface area of the refractory material is 1400K.

T_contact refers to the temperature of the interface between the refractory material and the insulating material. The refractory material is 15cm thick. The thickness of the insulating material is such that the heat loss remains at 5000W/m². At steady state, the heat loss of refractory layer equals to the heat loss of the insulating layer.

Hence, q_R = q_I = 5000W/m² where
q is heat flux
R is refractory layer
I is insulating layer

[Donaldson Tan]

Bibinou, have you derrived the required thickness already?

1. -k_R(?T/?X )_R = 5000
2. -k_I(?T/?X )_I = 5000

Solving Equation 1 gives you the intermediate temperature.

Solving Equation 2 gives you the thickness of the insulating layer.

You must solve Equation 1 before Equation 2.