The composite wall of an oven consists of three materials as shown. k1=20 W/m.K,

Question

The composite wall of an oven consists of three materials as shown. k1=20 W/m.K, k3=50 W/m.K,
L1=0.3, L2=0.15 m and L3=0.030 m.

Under steady-state conditions :
T?,i=900 oC ( air temperature inside the oven)
hi=20 W/m2.K (convective heat transfer coefficient on the oven inner wall)
T4=700 oC (the oven wall temperature)
T?,o=20 oC (outside air temperature)
ho=15 W/m2.K (convective heat transfer coefficient on the outer wall)

a)Draw a thermal circuit for the composite wall. b) Calculate k2. c) Calculate the temperaures T1, T2 and
T3. d) Calculate the overall heat transfer coefficient, Ui (or Uo are they equal).

Explanation / Answer

It may help you a lot Solution: The hot face (LHS) is at T = 370 ºC and the cold face (RHS) is at 66 ºC. The first layer on the left is material A with kA = 150 W/m-ºC and LA = 2.5 cm, second layer (middle) is a layer with materials B and D in parallel with kB = 30 W/m-ºC and kD = 70 W/m-ºC and LB = LD = 7.5 cm and AB = AD = 0.5*AA, and the right-most layer is material C with kC = 50 W/m-ºC and AC = AA = 0.1 m2. Calculating the individual resistances according to R = L/kA gives the following: RA = LA/kAAA = (0.025 m)/{(150 W/m-ºC)*(0.1 m2)} = 1.667 x 10-3 ºC/W RB = LB/kBAB = (0.075)/{(30)*(0.1/2)} = 0.05 ºC/W RD = LD/kDAD = (0.075)/{(70 W/m-ºC)*(0.1/2)} = 2.14286 x 10-2 ºC/W For the parallel resistors: Req = RBRD/(RB + RD) = (0.05)*(2.14286 x 10-2)/(0.5 + 2.14286 x 10-2) = 2.0548 x 10-3 ºC/W RC = LC/kCAC = (0.05)/{(50)*(0.1)} = 0.01 ºC/W Overall heat flow is given by q = ?T/?R = (370 – 66)/(1.667 x 10-3 + 2.0548 x 10-3 + 0.01) = 22155 W (A = 0.1 m2) gives q/A = 221550 W/m2.

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