A furnace wall is to be designed to transmit a maximum heat flux of 200 Btu/hr f

Question

A furnace wall is to be designed to transmit a maximum heat flux of 200 Btu/hr ft2 of wall area. The inside and outside wall temperatures are to be 2000 and 300 degrees fahrenheit respectively. Determine the most economical arrangement of bricks measuring 9 by 4.5 by 3 in. if they are made from two materials, one with a k of .44 Btu/h ft F and a maximum usuable temperature of 1500F and another with a k of .94 Btu/h ft F and a maximum usuable temperature of 2200 F. Bricks made of each material cost the same amount and may be laid in any manner.

My understanding is that the question wants the length of the wall to be determined.
I know that the resistivity equation is q=(T1-T4 /R) and that,in this case, R= (L1/k1A) + (L2/k1A)

Am I on the right track with the use of these equations? What I don't understand is how to determine the length.

Explanation / Answer

Given:

q = 200 Btu/hr.ft2

Ti = 2000 0F

To = 300 0F

k1 = 0.44 Btu/h.ft.0F T1allowable = 1500 0F

k2 = 0.94  Btu/h.ft.0F T2allowable = 2200 0F

Brick dimensions = 9 x 4.5 x 3 in = 0.75 x 0.375 x0.25 ft

Solution:

q = T/R =(2000 - 300)/R

R = 1700/200 = 8.5 0F/(Btu/hr.ft2)

Resistance of the wall should be R =  8.5 0F/(Btu/hr.ft2)

The wall with a low thermal conductivity will have a greater thermal resistance and so a greater temperature gradient. Hence to make a wall which satisfies the given temperature conditions and also involves the least number of bricks we have to use a brick of lower thermal conductivity, but the lower thermal conductivity brick cannot withstand a temperature of 2000 0F and so cannot be used.Hence the higher thermal conductivity brick or the higher heat resistant brick shouls be used till the temperature falls to 1500 0F.

Resistance of wall 2, R = T/q = (2000 - 1500)/200 = 2.5 0F/(Btu/hr.ft2)

Thickness of the wall 2 , L2 = k2xR = 0.94 x 2.5 = 2.35 ft

This could be done efficiently by placing 10 bricks along the thickness of the wall and the extra lenghth contributes to some resistance.

Resistance of wall 2, R = L2/k2 = 2.5/0.94 = 2.66 0F/(Btu/hr.ft2)

The remaining resistance could be produced by brick 1

R = 8.5 - 2.66 = 5.84  0F/(Btu/hr.ft2) = L1/k1 = L1/(0.44)

L1 = 5.84 x 0.44 = 2.57 ft

This length could be filled by 10 bricks of the first type.

Thus the wall consists of 10 bricks of type 2 followed by 10 bricks of type 1 with each brick placed with 0.25 ft along the direction of heat flow.

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