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posted by  Ry@n on 7/29/2008 5:27:12 AM  |  status: Closed  

Urgent!!!Tks for helping!

Course Textbook Chapter Problem
Calculus Based Physics Fundamentals of Physics Extended (8th) by Resnick, Halliday, Walker 18 61P
Question Details:
 Figure below shows (in cross section) a wall consisting of four layers, with thermal conductivities k1 = 0.060 W/m·K, k3 = 0.040 W/m·K, and k4 = 0.12 W/m·K (k2 is not known). The layer thicknesses are L1 = 1.5 cm, L3 = 2.8 cm, and L4 = 3.5 cm (L2 is not known). The known temperatures are T1 = 30°C, T12 = 25°C, and T4 = -10°C. Energy transfer through the wall is steady. What is interface temperature T34? 
 

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posted by Steve (retired) on 7/29/2008 7:41:29 AM  |  status: Live
Asker's Rating: Lifesaver   
Ry@n's comment:
"Tks a lots~~~"
Response Details:
from the theory we have that the rate Pcond at which the energy is conducted through a slab
   whose faces are maintained at temperatures TH and TC is  
   Pcond = (Q / T)
             = k A [(T - T4) / L]
   if we divide the equation by A on both sides we get
   Pcond / A = k A [(T - T4) / A L]
   so that we get the rate of heat conduction per unit area as
   Pcond / A = k1 [(T - T12) / L1]
                   = k4 [(T34 - T4) / L4]
   TH = 30oC 
   T1 = 25oC 
   TC = - 10oC
   solve for T34

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