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Temperature distribution in slab with variable thermal conductivity. Chapter 2 This holds good in the case k = ko (1 – βT) also. 4 × 10–4 T) where T is in °C and k is in W/mK. 6 m. The pipe surface is at 300°C and the outside insulation temperature is 60°C. Determine the heat flow for a length of 5 m. Also find the mid layer temperature. 32. The data are shown in Fig. Ex. 8. Quarter section is shown due to symmetry. 42 m (a) (b) Fig. Ex. 8. Temperature variation in hollow cylinder with variable thermal conductivity.
1. (b) Elemental volume in cylindrical coordinates. Fig. 1. (c) Elemental volume in spherical coordinates. In spherical coordinates (r, Φ, θ), Fig. 1(c), we get FG H FG H IJ K ∂T ∂ ∂T 1 ∂ 1 k . + 2 kr 2 2 ∂ 2 ∂r ∂Φ r r r sin θ ∂Φ + With k constant eqn. 5 reduces to FG H FG H IJ K FG IJ H K IJ K ∂ ∂T ∂T k sin θ + q = ρc ∂θ ∂τ r sin θ ∂θ 1 . 2 F GH IJ K 1 ∂ 1 ∂ 2T 2 ∂T . 5) . 3. The complete solutions to the general model is rather complex. Some of the simplified models for which solutions are attempted are listed below: 1.
The complete solutions to the general model is rather complex. Some of the simplified models for which solutions are attempted are listed below: 1. One dimensional steady flow (x or r directions) with constant or variable properties, without heat generation. 2. Same as above but with heat generation 3. Two dimensional steady flow (with constant properties, without heat generation) and 4. One dimensional unsteady state without heat generation. The simplified expressions in these cases in the various coordinate systems are Cartesian LM N OP Q ∂ ∂T ∂ 2T = 0 and =0 k ∂x ∂x ∂x 2 ∂ 2T ∂x + 2 q =0 k ∂ 2T ∂x = 2 1 ∂T .