A square chimney has an outside dimension of 60 cm x 60 cm. The wall thickness o
ID: 1853898 • Letter: A
Question
A square chimney has an outside dimension of 60 cm x 60 cm. The wall thickness of the chimney is L=15 cm and has a thermal conductivity of k =0.85 W/m/K. The height of the chimney is H = 1 m. The inner wall temperature of the chimney is Thot=350 degree and outer wall temperature is Tcold =25 degree . Using the finite difference method, write an equation for the steady-state temperature of each node shown in the figure. Solve your system of equations, and present a table of the node temperatures. Calculate the heat loss from the chimney.Explanation / Answer
The heat conductivity by conduction though a medium is..
(Just the formula for heat transfer though conduction above.)
The next part of the problem is asking what would the thickness of an insulation with a thermal conductivity of k=0.08 to make the heat transfer through the composite wall reduced by a factor of 2.
Conduction will take place if there exist a temperature gradient in a solid (or stationary fluid) medium.
Energy is transferred from more energetic to less energetic molecules when neighboring molecules collide. Conductive heat flow occur in direction of the decreasing temperature since higher temperature are associated with higher molecular energy.
Fourier's Law express conductive heat transfer as
q = k A dT / s (1)
where
A = heat transfer area (m2, ft2)
k = thermal conductivity of the material (W/m.K or W/m oC, Btu/(hr oF ft2/ft))
dT = temperature difference across the material (K or oC, oF)
s = material thickness (m, ft)
A plane wall constructed of solid iron with thermal conductivity 70 W/moC, thickness 50 mm and with surface area 1 m by 1 m,temperature 150 oC on one side and 80 oC on the other.
Conductive heat transfer can be calculated as:
q = 70 (W/moC) 1 (m) 1 (m) (150 (oC) - 80 (oC)) / 0.05 (m)
= 98,000 (W)
= 98 (kW)
Heat conducted through several walls in good thermal contact can be expressed as
q = (T1 - Tn) / ((s1/k1A) + (s2/k2A) + ... + (sn/knA)) (2)
where
T1 = temperature inside surface (K or oC, oF)
Tn = temperature outside surface (K or oC, oF)
A furnace wall of 1 m2 consist of a 1.2 cm thick stainless steel inner layer covered with a 5 cm this outside insulation layer of asbestos board insulation. The inside surface temperature of the steel is 800 K and the outside surface temperature of the asbestos is 350 K. The thermal conductivity for stainless steel is 19 W/m.K and for asbestos board 0.7 W/m.K.
The conductive heat transport through the wall can be calculated as
q =(800 (K) - 350 (K)) / ((0.012 (m) / 19 (W/mK) 1 (m2)) + (0.05 (m) / 0.7 (W/m.K) 1 (m2)))
= 6245 (W)
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