Anonymous
Anonymous asked in 科學及數學其他 - 科學 · 6 years ago

# conduction and convection

The inner and the outer surfaces of the brick wall of a furnace, whose thickness is 23cm, are kept at 820oC and 170oC respectively. Thermal conductivity of the brick is 0.865W/mK.

a) Determine the steady-state one-dimensional heat flux through the brick wall.

Attempt: 0.865*(820-179)/(23/100)=2445W/^2, which is correct.

b) An insulating wall (k=0.26W/mK) of 23cm thick is added to the outer surface of the brick wall of the furnace. At the external surface of the insulating wall, heat is losing to the ambient fluid at 20oC by convection (by assuming radiation to be negligible) with a heat transfer coefficient of 11.9W/m^2K. Temperature at the inner surface of the brick wall remains at 820oC. Determine by assuming one-dimensional steady-state condition:

i) Reduction in heat loss from the furnace. (ANS: 72.6%)

ii) Temperature at the interface of the two walls. (ANS: 648oC)

iii) Temperature at the external surface of the insulating wall. (ANS: 74oC)

Why it seems I have to find ii) first in order to find i)?

And I don't know how to find ii)

Is there some less complicated way to solve i)?

I would really appreciate it if you could help...

Rating

(b)(i) Original conductance of wall = 0.865/0.23 w/m^2.K = 3.761 w/m^2.K

Resistance of composite wall

= (0.23/0.865 + 0.23/0.26) K.m^2/w = 1.151 K.m^2/w

Conductance of composite wall = 1/1.151 w/m^2.K = 0.8692 w/m^2.K

Reduction in heat flow = [(3.761 - 0.8692)/3.761] x 100% = 77%

(ii)&(iii) Let Ti and Tf be the temperatures at the interface and external surface of the walls.

Hence, rate of heat flow per unit area Q is given by,

Q = (0.865/0.23).(820 - Ti) = (0.26/0.23).(Ti - Tf) = 11.9.(Tf - 20)

solve for Ti and Tf

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