This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Solution 3 Given: Thermal conductivity and dimensions of bars of two materials Sketch: Two thermal circuits (one series and one parallel) [8 points] Find: Effective thermal conductivity in two directions Modes of Heat Transfer: Conduction Assumptions: Steady-state, constant properties [2 points] Basic Equations: [3 points] q = T R tot R cond = L kA q cond =- kA dT dx Simplified Basic Equations: [4 points for k eff equation] q = k eff A T x = T R tot k eff = x R tot A Solution: (a) across width, thermal resistance in series [4 points] R tot = n L k A A + L k B A L = W 2 n A = tD R tot = W 2 tD 1 k A + 1 k B = W 2 tD k A + k B k A k B k eff = W R tot tD k eff = 2 k A k B k A + k B (b) through thickness, thermal resistance in parallel [4 points] 1 R tot = n k A A L + k B A L = nA L ( k A + k B ) L = t 1 A = WD 2 n 1 R tot = WD 2 t ( k A + k B ) k eff = t R tot WD k eff = 1 2 ( k A + k B ) Computing Equations: k eff,a = 2 k A k B k A + k B k eff,b = 1 2 ( k A + k B ) 2 Solution 4...
View Full Document
- Spring '09
- Heat Transfer