Consider the oven of Problem 1.54. The walls of the oven consist of L = 30-mm-th

Question

Consider the oven of Problem 1.54. The walls of the oven consist of L = 30-mm-thick layers of insulation characterized by kins = 0.03W/m.K that are sandwiched between two thin layers of sheet metal. The exterior surface of the oven is exposed to air at 23 degreeree C with hext = 2 W/m2.K. The interior oven air temperature is 180 degreeree C. Neglecting radiation heat transfer, determine the steady-state heat flux through the oven walls when the convection mode is disabled and the free convection coefficient at the inner oven surface is hb = 3 W/m2.K. Determine the heat flux through the oven walls when the convection mode is activated in which case the forced convection coefficient at the inner oven surface is hf0 = 27 W/m2 . K Does operation of the oven in its convection mode result in significantly increased heat losses from the oven to the kitchen? Would your conclusion change if radiation were included in your analysis?

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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